{ "Tag": [ "inequalities" ], "Problem": "How do you prove that $ \\frac{a^2}{b}\\plus{}\\frac{b^2}{c}\\plus{}\\frac{c^2}{a}\\ge a\\plus{}b\\plus{}c$ for positive reals $ a, b, c$ with Cauchy-Schwarz?\r\n\r\nThanks.", "Solution_1": "[hide=\"Useful lemma\"][b]Lemma[/b] (Titu Andreescu). $ \\sum \\frac {x_i^2}{y_i}\\geq\\frac {\\left(\\sum x_i\\right)^2}{\\sum y_i}$.\n\n[i]Proof.[/i] This is a direct consequence of Cauchy-Schwarz. Multiply both sides by $ \\sum y_i$ to get $ \\left(\\sum y_i\\right)\\left(\\sum\\frac {x_i^2}{y_i}\\right)\\stackrel{\\text{\\tiny{C\\minus{}S}}}{\\geq} \\left(\\sum\\frac {x_i\\sqrt {y_i}}{\\sqrt {y_i}}\\right)^2 \\equal{} \\left(\\sum x_i\\right)^2$. $ \\Box$[/hide]\r\n\r\nApplying that, $ \\text{LHS}\\geq\\frac {(a \\plus{} b \\plus{} c)^2}{a \\plus{} b \\plus{} c} \\equal{} a \\plus{} b \\plus{} c$.", "Solution_2": "Thank you :D", "Solution_3": "[hide=\"Just a little-How to do it\"]\n\nYou want Cauchy-Schwarz\n\n1) Get rid of denominators, so the second multiple has to be like (xa+yb+zc), so the a,b,c in the denominators would disappear\n\n2) You have already squares in numerators, so you can put x=y=z=1\n\nNow we have $ (a\\plus{}b\\plus{}c)(\\frac{c^2}a\\plus{}\\frac{a^2}b\\plus{}\\frac{b^2}c)\\ge(a\\plus{}b\\plus{}c)^2$, from which the inequality follows...\n\n[/hide]", "Solution_4": "not hsb really is it?" } { "Tag": [ "integration", "algebra", "polynomial", "real analysis", "real analysis unsolved" ], "Problem": "Prove that equation $\\int_{0}^{a}e^{-x}(1+\\frac{x}{1!}+\\frac{x^{2}}{2!}+...+\\frac{x^{100}}{100!})dx=50$ has root in $(50,100)$.", "Solution_1": "${k=100}$\r\n${\\sum_{n=0}^{k}\\frac{x^{n}}{n!}=\\frac{e^{x}\\Gamma (k+1,x)}{k!}}$\r\n\r\nSimplifying your equation\r\n\r\n${\\int_{0}^{a}\\frac{\\Gamma (k+1,x)}{k!}\\, dx=\\frac{a \\Gamma (k+1,a)+\\Gamma (k+2,0)-\\Gamma (k+2,a)}{k!}}$\r\n\r\n\r\nthis helps?!\r\n\r\ni think no. h ehe", "Solution_2": "$f(a)=\\int_{0}^{a}e^{-x}(1+\\frac{x}{1!}+\\frac{x^{2}}{2!}+...+\\frac{x^{100}}{100!})dx$ is continuous, $f(50)<50$, and $f(51)>50$ so by the intermediate theorem there exists a root in the range $]50,51[$.", "Solution_3": "[quote=\"puuhikki\"] $f(50)<50$, and $f(51)>50$ [/quote]\r\n\r\nWhy? :maybe:", "Solution_4": "$f(50) < 50$ because the polynomial in that expression is bounded above by $e^{x}$ when $x \\geq 0$. The other one, I don't know :-)", "Solution_5": "[quote=\"N.T.TUAN\"]\nWhy? :maybe:[/quote]\r\nWrite $f$ as an enough accurate Taylor polynomial and compute its value. Then approximate its error term.", "Solution_6": "[quote=\"JBL\"]$f(50) < 50$ because the polynomial in that expression is bounded above by $e^{x}$ when $x \\geq 0$. [/quote]\r\n\r\nThat also is all what I know on this problem :D" } { "Tag": [], "Problem": "What is the sum of the two least, distinct positive integers each of which has exactly eight positive factors?", "Solution_1": "The two are\r\n$ 2 * 3 * 5 \\equal{} 30$\r\nand\r\n$ 2^3 * 3 \\equal{} 24$\r\nThe sum is $ 30 \\plus{} 24 \\equal{} \\boxed{54}$." } { "Tag": [], "Problem": "This might be off-topic for this forum but from what I've read here there seems to be a few Exeter students. Just wanted to gauge the amount of Exeter people there are.", "Solution_1": "There are Exeterians, dont u worry. What's so special about them necessarily? :-P", "Solution_2": "I'm not from Exeter,the other phillips school.\r\nthe better phillips school. \r\nGO ANDOVER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!", "Solution_3": "Ha - I wish I went to Exeter.", "Solution_4": "[quote]Exeterians...[/quote]\r\n\r\nAhem ... Exonians?", "Solution_5": "[quote=\"tetrahedr0n\"][quote]Exeterians...[/quote]\n\nAhem ... Exonians?[/quote]\r\n\r\nLOL :lol:", "Solution_6": "[quote=\"tetrahedr0n\"][quote]Exeterians...[/quote]\n\nAhem ... Exonians?[/quote]\r\n\r\n-.- whatever", "Solution_7": "[quote=\"MithsApprentice\"][quote=\"tetrahedr0n\"][quote]Exeterians...[/quote]\n\nAhem ... Exonians?[/quote]\n\n-.- whatever[/quote]\r\n\r\nhehe, exeterians. I like that better than exonians. :D", "Solution_8": "[quote=\"guangweicao\"]I'm not from Exeter,the other phillips school.\nthe better phillips school. \nGO ANDOVER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/quote]\r\n\r\nI also go to Philip's school...", "Solution_9": "All of you are from exeter?\r\nAm I the only one from Andover?\r\n[quote]filletwho said: I also go to Philip's school...[/quote]\r\n\r\nAnd who is filletwho?\r\ndo i know u", "Solution_10": "No, my name is philip, thus I go to Philip's school", "Solution_11": "I think mathfanatic goes to Exeter.", "Solution_12": "[quote]\"Exonians\"?????[/quote]\r\n\r\nAll along I thought you guys were called Exies\r\n :D" } { "Tag": [ "function", "integration", "calculus", "calculus computations" ], "Problem": "Solve the differential equation $y'=x+y$.\r\n\r\nThe obvious solution is $y=Ce^x-x-1$. I'm just interested in the method in solving it.", "Solution_1": "Let me show you how to solve $y' + f(x)y = g(x)$. Find a function $F(x)$ which fulfills $F'(x) = f(x)$. I.e, it is a primitive to $f$. Now multiply the equation by $e^{F(x)}$ (this can be done without loss of information, since $e^x \\ne 0$. Now, check that your equation now indeed is equivalent to $(ye^{F(x)})' = e^{F(x)}g(x)$. Now find a primitive to $e^{F(x)}g(x)$ to find all solutions \r\n\\[ y = e^{-F(x)}\\int e^{F(x)}g(x)dx \\]\r\n\r\nTry it on your equation, where $f(x) = -1$ and $g(x) = x$.", "Solution_2": "we can also write $y' - y =x$ and charasteristic equation $\\lambda - 1 =0$ then general solution of homogeneous equation is $y_0 = C_1 e^{1x}$, and find partial solution for nonhomogeneous equation in form $y_1(x)=Ax+B$." } { "Tag": [ "function", "number theory unsolved", "number theory" ], "Problem": "Im not sure if this is the right forum for this or not, but anyways:\r\n\r\nLet $ f$ be a function such that $ f(n)$ is an integer whenever n is an integer. Prove that there is some integer $ n$ so that $ f(f(n))\\not \\equal{}n\\plus{}3$.", "Solution_1": "Let's say $ f(f(n)) \\equal{} n \\plus{} 3$ for every integer $ n$. Then $ f(n) \\plus{} 3 \\equal{} f(f(f(n))) \\equal{} f(n\\plus{}3)$. From this, we can see that $ f$ preserves residue classes modulo $ 3$. Combined with $ f(f(n)) \\equal{} n \\plus{} 3$, we see that $ f(0), f(1), f(2)$ are respectively $ 0, 1, 2$ modulo $ 3.$\r\n\r\nFrom this, we see that $ f(3n) \\equal{} 3(n \\plus{} a)$ for some constant $ a$. Now $ f(f(3n)) \\equal{} 3n \\plus{} 3 \\equal{} 3(n\\plus{}1)$ but $ f(f(3n)) \\equal{} f(3(n\\plus{}a)) \\equal{} 3(n \\plus{} 2a)$. So $ 3(n \\plus{} 2a) \\equal{} 3(n\\plus{}1)$ for all integers $ n$. This means $ 2a \\equal{} 1$ for some integer $ a$, impossible.", "Solution_2": "Let exist f, suth that $ f(f(n))\\equiv n\\plus{}3$. Then $ f(n\\plus{}3)\\equal{}f(f(f(n)))\\equal{}f(n)\\plus{}3$. Therefore $ f(n)\\equal{}n\\plus{}a_n$, were $ a_n\\in \\{a_0,a_1,a_2\\}$ and $ a_n\\equal{}a_m$ if $ n\\equal{}m\\mod 3$. $ n\\plus{}3\\equal{}f(n\\plus{}a_n)\\equal{}n\\plus{}a_n\\plus{}a_{n\\plus{}a_n}$, therefore $ a_n\\plus{}a_{n\\plus{}a_n}\\equal{}3$.\r\nIf $ 3|a_n$, then $ a_n\\plus{}a_n\\equal{}3$ - contradition. Therefore $ a_n\\not \\equal{}0\\mod 3$ and exist only two means $ a, 3\\minus{}a, 3\\not |a$ and one of them meet twice in $ \\{a_0,a_1,a_2\\}$ Let $ g(n)\\equiv f(n)\\mod 3$, then we get $ g: \\{0,1,2\\}\\to \\{1,2\\}$, suth that $ g(x)\\plus{}g(x\\plus{}g(x))\\equal{}3$.\r\nIt give contradition." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Anybody has the answer key to 2007 and 2008 national mathcounts? Thank you.", "Solution_1": "Post this is the MATHCOUNTS resources topic." } { "Tag": [ "abstract algebra", "vector", "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "Let $L$ be a Banach space. $M\\subset L$ - closed linear subspace, and $x_0\\notin M$. Does exist a point $y\\in M$ such that $\\rho(x_0,y)=\\rho (x_0,M)$?", "Solution_1": "No, I don't think such a $y$ always exists. \r\n\r\nFirst notice that if we restrict our attention to the subspace spanned by $M$ and $x_0$, this is a closed linear subspace of $L$, so it's also a Banach space. We can thus assume that within $L,\\ M$ is actually a hyperplane: the kernel of a bounded linear functional. From now on, this is the setting we're working in.\r\n\r\nNow, given a hyperplane $M$ of the Banach space $L$, there is essentially a unique continuous linear functional $f$ whose kernel is $M$. This is because every element of $L$ is of the form $ax_0+y$ for $a\\in K$ (the scalar field, which we take to be either $\\mathbb R$ or $\\mathbb C$ for now), and given another functional $\\tilde f$ with kernel $M$ such that $\\tilde f(x_0)=\\lambda f(x_0),\\ \\lambda\\in K$, we have $\\tilde f(ax_0+y)=\\tilde f(ax_0)=a\\tilde f(x_0)=\\lambda af(x_0)=\\lambda f(ax_0)=\\lambda f(ax_0+y)$. In other words, $\\tilde f=\\lambda f$. Now take $K=\\mathbb R$. Then a functional $\\tilde f$ with the same kernel $M$ as $f$ (which we assume has norm $1$) and norm $1$ is either $f$ or $-f$. This means that the conditions $\\ker f=M,\\ \\|f\\|=1,\\ f(x_0)>0$ determine $f$ uniquely.\r\n\r\nA consequence of the above: if $f$ is a continuous linear functional on a real Banach space $L$ with norm $1$, kernel $M$, and $x_0$ such that $f(x_0)>0$, then $f(x_0)=d=\\rho(x_0,M)$ (I'll denote this distance by $d$ from now on). To see this, notice that it's easy to construct, using the Hahn-Banach Theorem on the extension of functionals, a functional $\\tilde f$ such that $\\tilde f(y)=0,\\ \\forall y\\in M,\\ \\|\\tilde f\\|=1$, and $\\tilde f(x_0)=d$. The observations made in the previous paragraph imply that $\\tilde f=f$, and we're done.\r\n\r\nNow let $L=\\ell^1$, the real Banach space of summable real sequences, and let the functional $f$ be defined by $f(x)=\\sum_{n\\ge 1}\\left(1-\\frac 1n\\right)x^n,\\ \\forall x=(x^1,x^2,\\ldots)\\in\\ell^1$. Clearly, $\\|f\\|=\\sup_{n\\ge 1}\\left(1-\\frac 1n\\right)=1$. Let $M=\\ker f$, and take $x_0\\in\\ell^1$ such that $f(x_0)>0$. By the above, $f(x_0)=d$. Now assume we can find $y\\in M$ such that $\\|x_0-y\\|=d$. We then have $f(x_0-y)=f(x_0)=d=\\|x_0-y\\|$. We have thus found a vector $t=x_0-y$ such that $f(t)=\\|t\\|$. It's easy to see from the expression of $f$ though that there can be no such vectors; in other words, the norm of $f$ is bever reached: we always have $\\|f(x)\\|<\\|x\\|$ if $x\\ne 0$. We have a contradiction.\r\n\r\nI hope it's Ok. I'd hate to have written so much in vain :).", "Solution_2": "Thanks grobber! I think it's OK :)" } { "Tag": [ "induction" ], "Problem": "Compute $ \\sum_{k \\equal{} 1}^n \\frac1{k(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)}$", "Solution_1": "$ \\frac{1}{k(k\\plus{}1)(k\\plus{}2)}\\minus{}\\frac{1}{(k\\plus{}1)(k\\plus{}2)(k\\plus{}3)}\\equal{}...$ :)", "Solution_2": "Precisely... :D", "Solution_3": "What result did you get the calculation of my hint?", "Solution_4": "$ \\sum_{k \\equal{} 1}^{n}\\frac {1}{k(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)}$\r\n$ \\equal{} \\frac {1}{3}[\\sum_{k \\equal{} 1}^{n}\\frac {1}{k(k \\plus{} 1)(k \\plus{} 2)} \\minus{} \\sum_{k \\equal{} 1}^{n}\\frac {1}{(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)}]$\r\n$ \\equal{} \\frac {1}{3}(\\frac {1}{6} \\minus{} \\frac {1}{(n\\plus{} 1)(n \\plus{} 2)(n \\plus{} 3)} )$\r\n$ \\equal{} \\frac {1}{18} \\minus{} \\frac {1}{3(n \\plus{} 1)(n \\plus{} 2)(n \\plus{} 3)}$\r\n\r\nP.S. Corrected, Kunny, using your hint.", "Solution_5": "Resume summation:\r\n\r\n$ \\equal{} \\frac {1}{6}(\\frac {2}{n \\plus{} 2} \\minus{} \\frac {1}{n \\plus{} 3})$\r\n\r\n$ \\equal{} \\frac{1}{6}[\\frac{n \\plus{} 4}{(n \\plus{} 2)(n \\plus{} 3)}]$", "Solution_6": "[quote=\"stephencheng\"]$ \\sum_{k \\equal{} 1}^{n}\\frac {1}{k(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)}$\n$ \\equal{} \\frac {1}{2}(\\sum_{k \\equal{} 1}^{n}\\frac {1}{k(k \\plus{} 3)} \\minus{} \\sum_{k \\equal{} 1}^{n}\\frac {1}{(k \\plus{} 1)(k \\plus{} 2)})$\n$ \\equal{} \\frac {1}{2}[\\frac {1}{3}(\\sum_{k \\equal{} 1}^{n}(\\frac {1}{k} \\minus{} \\frac {1}{k \\plus{} 3}) \\minus{} \\sum_{k \\equal{} 1}^{n}(\\frac {1}{k \\plus{} 1} \\minus{} \\frac {1}{k \\plus{} 2})]$\n$ \\equal{} \\frac {1}{2}[\\frac {1}{3}(\\frac {1}{1} \\plus{} \\frac {1}{2} \\minus{} \\frac {1}{n \\plus{} 2} \\minus{} \\frac {1}{n \\plus{} 3}) \\minus{} (\\frac {1}{2} \\minus{} \\frac {1}{n \\plus{} 2})]$\n$ \\equal{} \\frac {1}{6}(\\frac {2}{n \\plus{} 2} \\minus{} \\frac {1}{n \\plus{} 3})$[/quote]\r\n\r\nThat's wrong in the 4th line.\r\n\r\nWhy do you use Telescoping Method that I gave it as hint? :wink:\r\n\r\nand you can assure whether your answer is correct or not by plugging $ n \\equal{} 1$ into the result.", "Solution_7": "[quote=\"stephencheng\"]$ \\sum_{k \\equal{} 1}^{n}\\frac {1}{k(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)}$\n$ \\equal{} \\frac {1}{3}[\\sum_{k \\equal{} 1}^{n}\\frac {1}{k(k \\plus{} 1)(k \\plus{} 2)} \\minus{} \\sum_{k \\equal{} 1}^{n}\\frac {1}{(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)}]$\n$ \\equal{} \\frac {1}{3}(\\frac {1}{6} \\minus{} \\frac {1}{(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)} )$\n$ \\equal{} \\frac {1}{18} \\minus{} \\frac {1}{3(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)}$\n\nP.S. Corrected, Kunny, using your hint.[/quote]\r\n\r\nIncorrect :( , Take it easy! :lol:", "Solution_8": "hello, prove by induction that $ \\sum_{k\\equal{}1}^n\\frac{1}{k(k\\plus{}1)(k\\plus{}2)(k\\plus{}3)}\\equal{}\\frac{1}{18}\\minus{}\\frac{1}{3(n\\plus{}1)(n\\plus{}2)(n\\plus{}3)}$.\r\nSonnhard.", "Solution_9": "[quote=\"Dr Sonnhard Graubner\"]hello, prove by induction that $ \\sum_{k \\equal{} 1}^n\\frac {1}{k(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)} \\equal{} \\frac {1}{18} \\minus{} \\frac {1}{3(n \\plus{} 1)(n \\plus{} 2)(n \\plus{} 3)}$.\nSonnhard.[/quote]\r\nWorning :D\r\nBecause we don't yet calculate an answer." } { "Tag": [ "geometry", "3D geometry", "prism", "conics", "ellipse", "congruent triangles", "AMC" ], "Problem": "A cylindrical tank with radius $ 4$ feet and height $ 9$ feet is lying on its side. The tank is filled with water to a depth of $ 2$ feet. What is the volume of the water, in cubic feet?\r\n\r\n$ \\textbf{(A)}\\ 24\\pi \\minus{} 36 \\sqrt {2} \\qquad \\textbf{(B)}\\ 24\\pi \\minus{} 24 \\sqrt {3} \\qquad \\textbf{(C)}\\ 36\\pi \\minus{} 36 \\sqrt {3} \\qquad \\textbf{(D)}\\ 36\\pi \\minus{} 24 \\sqrt {2} \\\\ \\textbf{(E)}\\ 48\\pi \\minus{} 36 \\sqrt {3}$", "Solution_1": "[hide]\nThe area of the cylinder's base = $ 16\\pi$\n\nand the area of that covered by the 2 ft of water = $ \\frac{16\\pi}{3} \\minus{} 4\\sqrt{3}$\n\nand $ 9 (\\frac{16\\pi}{3} \\minus{} 4\\sqrt{3}) \\equal{} \\boxed{48\\pi \\minus{} 36\\sqrt{3} \\Rightarrow E}$\n[/hide]", "Solution_2": "[quote=\"undefined117\"][hide]\nThe area of the cylinder's base = $ 16\\pi$\n\nand the area of that covered by the 2 ft of water = $ \\frac {16\\pi}{3} \\minus{} 4\\sqrt {3}$\n\nand $ 9 (\\frac {16\\pi}{3} \\minus{} 4\\sqrt {3}) \\equal{} \\boxed{48\\pi \\minus{} 36\\sqrt {3} \\Rightarrow E}$\n[/hide][/quote]\r\n\r\nCould you explain what you mean by the \"area covered by the 2 feet of water\"?\r\nPretty much everything that you said? :huh: \r\nThanks", "Solution_3": "[hide]Gosh I really need to learn asymptote..\n\nSince the cylinder is on its side, the water inside will form a prism with height 9 and the area of the base being however much of the original cylinder's base is covered by the water level (see image).\n\nThat area can be represented as being the area of sector AB minus the area of triangle ABC.\n\nNote that triangle ABC, when the two congruent halves are combined, is an equilateral triangle with side length 4.\n\n$ \\frac{120^\\circ}{360^\\circ}\\pi4^2 \\minus{} 4\\sqrt{3} \\equal{} \\frac{16}{3}\\pi \\minus{} 4\\sqrt{3}$ is the area of the prism's base.\n\nmultiplying that by 9 will give us the volume, so\n\n$ 9(\\frac{16}{3}\\pi \\minus{} 4\\sqrt{3}) \\equal{} \\boxed{48\\pi \\minus{} 36\\sqrt{3}} \\Rightarrow E$[/hide]", "Solution_4": "Just notice that it has to be a 120 degree sector.\r\nThen 1/3*4^2pi=16/3pi\r\ndont worry about the area of the triangle...\r\nbecause 16pi/3*9=48pi, so only E is a logical choice (saves like 30 secs if you really needed the time at the end)", "Solution_5": "Sorry I think I'm missing something easy, but could you tell me how you figured out AB = 4 ?\r\nOr how you know that it is 120 degrees?", "Solution_6": "AB does not equal 4, but $ AC \\equal{} BC \\equal{} 4$ because they are the radii (-us?), as was given in the problem.\r\n\r\nAB (the water level) and the radius that is represented by the red line are perpendicular bisectors of each other.\r\n\r\nIt can be seen that the two congruent triangles formed have hypotenuse length 4, and a shorter leg of length 2. Because of the right angle adjacent to the side length 2, it can be seen that the triangles are 30-60-90 triangles (for more info, go [url=http://mathworld.wolfram.com/30-60-90Triangle.html]here[/url]).\r\n\r\n$ 60^\\circ \\times 2 \\equal{} 120^\\circ$", "Solution_7": "All right Thanks!\r\nThat helps!\r\n :D", "Solution_8": "In the end, how did you multiply by 9? Like, what type of volume formula did you use since this isn't a cylinder or half of one?", "Solution_9": "You multiply by 9 because the height is 9.", "Solution_10": "I know, but this isn't a cylinder and you didn't compute the volume of one of 9 things, you computed the area.", "Solution_11": "The volume we're trying to find is a sliced off part of a cylinder. \nWe multiply by nine just like we multiply by the height in a prism. (You find the area of the base, and then multiply by height.)\nIn other words, we first found the area of a slice of the \"prism\" one foot high, then multiplied it by nine.", "Solution_12": "Sorry, I'm confused now. You found the area of the slice, not the volume. And the water can't be half of a cylinder either, since it has an ellipse bases. Please help, thanks guys!", "Solution_13": "You currently have the area, correct? When you multiply it be $9$, you get the volume.", "Solution_14": "I agree completely, but my question is, what type of a solid is this? If it's not half of a cylinder, what is it?", "Solution_15": "[quote=\"AruKasera\"]I agree completely, but my question is, what type of a solid is this? If it's not half of a cylinder, what is it?[/quote]\n\nIt doesn't matter what it is (I don't think it has a name), we can calculate its area using Cavalieri's principle." } { "Tag": [], "Problem": "what happens when sym-tri nitro toluene is treated with $ \\ce{(NH4)2S}$", "Solution_1": "The group in the para posn wrt to CH3 will get selectively reduced :D \r\n[hide]\nThis may be bcos the CH3 grp might be under steric strain due to the other 2 NO2 grps. So they mght not get reduced :D \n\n[/hide]", "Solution_2": "r u sure about ur answer....i am not sure of ur reasoning", "Solution_3": "im very sure of my product . See the NO2 grp adjacent to the CH3 grp is sterically hindered and hence the reducing agent cant reduce the NO2 grp easily (as it cant easily approach) so the NO2 grp which is free and not sterically hindered is easily reduced :D . Think i have to make finetunings to my level of clarity :D", "Solution_4": "yes this seems better :)", "Solution_5": "There is no reasoning, because this is not a usefull reaction: several products can be formed, and this reaction can only be applied to about three compounds. Other synthetic routes are better." } { "Tag": [ "function", "LaTeX", "search", "real analysis", "real analysis unsolved" ], "Problem": "Letting (x) denote the fractional part of the real number x, consider the function f(x)=sum of (nx)/n^2 (x real).\r\nFind all discontinuities of f, and show that they form a countable dense set. Show that f is nevertheless Riemann-integrable on every bounded interval.", "Solution_1": "Learn to use latex:\r\n\r\n$f(x) = \\sum_{n = 1}^\\infty \\frac{\\{nx\\}}{n^{2}}$, where I have replaced your $(x)$ with $\\{x\\}$.", "Solution_2": "The first part was discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=discontinuous&t=113150]here[/url]. The second part follows from the fact that\r\n[quote=\"Kent Merryfield\"]The function $f$ is discontinuous at every rational number and continuous at every irrational number. [/quote]\r\n\r\nNext time, you should mention the source. It's easier to search. Or, better yet, you could search topics on your own :D" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "find lim : \r\n$ \\lim_{n \\minus{} > \\infty } \\frac {1.3.5.7...(2n \\minus{} 1)}{2.4.6.8...2n}$\r\nsori, i edited it", "Solution_1": "Sorry that this post doesn't answer your question but some tips to improve your TeX. The command \"\\cdot\" can be used for multiplication instead of \".\", and to create an arrow, just type $ \\rightarrow$.", "Solution_2": "Here is my solution (has other solution simple and I will post it later.\r\nLet $ a_n$ is this sequence \r\n${ \\frac {1}{a_n} = \\prod_{k = 0}^{n - 1}(1 + \\frac {1}{2k + 1}} > \\sum_{k = 0}^{n-1}\\frac {1}{2k + 1}$\r\nBut $ \\lim_{k=0}^{+\\infty} \\frac {1}{2k + 1} = + \\infty$\r\nSo $ \\lim_{n\\to\\infty}a_n = 0$", "Solution_3": "[quote=\"Hong Quy\"]find lim : \n$ \\lim_{n \\minus{} > \\infty } \\frac {1.3.5.7...(2n \\minus{} 1)}{2.4.6.8...2n}$\nsori, i edited it[/quote]\r\nWhat if we wrote :\r\n$ \\lim_{n \\minus{} > \\infty } \\frac {1.3.5.7...(2n \\minus{} 1)\\sqrt{n}}{2.4.6.8...2n}$ ? :) \r\nanswer will be $ \\frac{1}{\\sqrt{\\pi}}$", "Solution_4": "[quote=\"TTsphn\"]Here is my solution (has other solution simple and I will post it later.\nLet $ a_n$ is this sequence \n${ \\frac {1}{a_n} = \\prod_{k = 0}^{n - 1}(1 + \\frac {1}{2k + 1}} > \\sum_{k = 0}^{n - 1}\\frac {1}{2k + 1}$\nBut $ \\lim_{k = 0}^{ + \\infty} \\frac {1}{2k + 1} = + \\infty$\nSo $ \\lim_{n\\to\\infty}a_n = 0$[/quote]\r\nBut $ \\lim_{k = 0}^{ + \\infty} \\frac {1}{2k + 1} = + \\infty$ \r\nwho can solve this problem: $ \\lim_{k = 0}^{ + \\infty} \\frac {1}{2k + 1} = + \\infty$ \r\n ? with a solution's beatyful", "Solution_5": "$ a_n\\equal{}\\sum_{i\\equal{}0}^n\\frac{1}{2i\\plus{}1}$\r\nThen \r\n$ a_{2n}\\minus{}a_n\\geq \\frac{1}{4}$\r\n$ a_{2^kn}\\minus{}a_n\\geq \\frac{2^{k\\minus{}1}}{4}$ \r\n$ a_n$ increasing .\r\nSo $ a_n\\to\\infty$" } { "Tag": [ "search", "ARML", "MATHCOUNTS", "AMC 12 A" ], "Problem": "Is there an actual stateside contest for Wisconsin?\r\nWhen i used to live in Ohio, there was like a OCTM...", "Solution_1": "The UW runs a talent search ([url]http://math.wisc.edu/~talent[/url]), and some schools participate in a statewide Math League (ask Mustafa, my school doesn't do that). The WI chapter of the MAA also administers a statewide test, which was in December.\r\n\r\nOther than that, I think all our competitions are either regional or national.", "Solution_2": "Really? Our school takes MAA...um...I don't exactly know when but it's later", "Solution_3": "yea yea, i took the maa like last month. i got a perfect onit :) but they don't acutally give like a statewide ranking on it or anything like that right...\r\n\r\nhey, anybody here went to last year's arml?", "Solution_4": "I didn't because my school doesn't do AMC even though I have tried...\r\nFor further ARML discussion you can go [url=http://www.artofproblemsolving.com/Forum/index.php?f=348]here[/url]", "Solution_5": "You should seriously find some way to do the AMC tests, they are very cool.", "Solution_6": "it's not as if I haven't tried...anyway Michael said that if in no way no how I could take it at my school, he would let me take it at Magellan, but I don't know if that would exactly work out :D Anyway, no laughing matter. I GOTTA TAKE THE AMC!!!!!!", "Solution_7": "All you have to do is sign up and then drive over to the right place for the right day and take the test. It would work.", "Solution_8": "right place? say where?", "Solution_9": "I just found a nearby high school by asking around so I am taking both A and B on the test dates.", "Solution_10": "[quote=\"kstan013\"]right place? say where?[/quote]\r\n\r\nMagellan if you're going to do it there. Just wherever the test it being administered.", "Solution_11": "I DO live 2 hours from Appleton :wink:", "Solution_12": "Sleep in a hotel that night :D . Or find somewhere closer. You made the drive plenty of times for Mathcounts Practice...", "Solution_13": "[quote=\"mustafa\"]Sleep in a hotel that night :D . Or find somewhere closer. You made the drive plenty of times for Mathcounts Practice...[/quote]on weekends :wink:", "Solution_14": "Yeah, yeah, whatever. This is far more important than school :D .", "Solution_15": ":yup: tell me bout it", "Solution_16": "Magellan does it? My MC teammates went to a high school to do it!", "Solution_17": "Michael did say \"his school\" and since he has no real affiliation with a school (e.g., a teacher), the high school and middle school are both \"his school\" from my understanding", "Solution_18": "*cough* The math teacher at Magellan didn't say anything about the AMC *cough*", "Solution_19": "Does anybody know Wisconsin can allow competitors to protest?", "Solution_20": "[quote=\"ghjk\"]Does anybody know Wisconsin can allow competitors to protest?[/quote]\r\n\r\nWhat do you want to protest?", "Solution_21": "I want to protest my problem 1! Since I compared my solution to Wisonsin's solution and I see it's just different from the last part which is not a big cause to me. I just get 1 point for that problem 1 :( . Do you know how to protest, mustafa?", "Solution_22": "I [b]go to Magellan[/b], they don't offer the AMC. We go to Appleton West High School to take it, which is like a minute drive from Magellan middle school. They only offer AMC 12 A though. This information might not be accurate but its what I've been thinking this whole time.", "Solution_23": "[quote=\"Stargirl240\"]*cough* The math teacher at Magellan didn't say anything about the AMC *cough*[/quote]\nMr. Blom's an idiot.", "Solution_24": "WMTS is op" } { "Tag": [ "modular arithmetic" ], "Problem": "[b]Problema 2096:[/b]\r\nDeputatii unui Parlament au format 2008 comisii, fiecare comisie are cel mult 10 membrii. Oricare 11 comisii exista un parlamentar care face parte din fiecare dintre ele. Demonstrati ca exista un parlamentar care face parte din fiecare comisie. ([i]Petrov[/i])\r\n\r\n[b]Problema 2097:[/b]\r\nGasiti toate numerele prime $ p$ de forma $ a^2\\plus{}b^2\\plus{}c^2$, care au proprietatea ca: $ p|a^4\\plus{}b^4\\plus{}c^4.$ ([i]Senderov[/i])", "Solution_1": "$ p \\equal{} 2$ \u015fi $ p \\equal{} 3$", "Solution_2": "2096. Este un rezultat de folclor c\\u a, fiind dat\\u a o familie (finit\\u a sau infinit\\u a) $ \\mathcal{F} \\equal{} (X_i)_{i \\in I}$ de mul\\c timi cu $ | X_i | \\leq n$, astfel \\^ inc\\^ at intersec\\c tia a oricare $ n\\plus{}1$ dintre ele este nevid\\u a, atunci intersec\\c tia tuturor este nevid\\u a. Fie $ A \\neq B$ dou\\u a dintre mul\\c timi (dac\\u a toate sunt egale, rezultatul este trivial); atunci $ 0 < | A \\cap B | < n$, deci $ A \\cap B \\equal{} \\{x_1, \\ldots, x_k \\}$, cu $ 1 \\leq k \\leq n\\minus{}1$. Presupun\\^ and rezultatul fals, pentru fiecare element $ x_i$ va exista (m\\u acar) o mul\\c time $ X_i$ astfel ca $ x_i \\not \\in X_i$ (mul\\c timile $ X_i$ nu sunt neap\\u arat distincte). Dar atunci, cele cel mult $ n\\plus{}1$ mul\\c timi $ A, B, X_1, \\ldots, X_k$ vor avea intersec\\c tia nevid\\u a; fie $ x$ un element comun, vom avea evident $ x \\neq x_i$ pentru orice $ i$, dar \\c si $ x \\in A \\cap B$, absurd.\r\n\r\n2097. Avem $ p \\equal{} a^2\\plus{}b^2\\plus{}c^2 \\ | \\ a^4\\plus{}b^4\\plus{}c^4 \\equal{} a^4\\plus{}(b^2\\plus{}c^2)^2\\minus{}2b^2c^2 \\equal{} a^4\\plus{}(p\\minus{}a^2)^2\\minus{}2b^2c^2 \\equal{}$ $ p^2 \\minus{} 2pa^2 \\plus{} 2(a^4\\minus{}b^2c^2)$, deci $ p \\ | \\ 2(a^4\\minus{}b^2c^2) \\equal{} 2(a^2\\plus{}|bc|)(a^2\\minus{}|bc|)$. Atunci $ p\\equal{}2$, pentru dou\\u a dintre variabile egale cu $ \\pm 1$ \\c si a treia $ 0$; sau $ p \\ | \\ a^2\\plus{}|bc| \\leq a^2\\plus{}2|bc| \\leq a^2\\plus{}b^2\\plus{}c^2 \\equal{} p$, f\\u ar\\u a solu\\c tii; sau $ p \\ | \\ |a^2\\minus{}|bc\\parallel{} \\leq a^2 \\plus{} |bc| \\leq a^2\\plus{}b^2\\plus{}c^2 \\equal{} p$, conduc\\^ and la $ a^2 \\equal{} |bc|$, \\c si similar, $ b^2 \\equal{} |ca|$, $ c^2 \\equal{} |ab|$, cu singura solu\\c tie $ p \\equal{} 3$, pentru toate cele trei variabile egale cu $ \\pm 1$.\r\n(Ideea vine mai u\\c sor dac\\u a lucr\\u am \\^ in $ \\mathbb{Z}_p$, unde pentru $ c\\neq 0$ se ob\\c tine $ (ac^{\\minus{}1})^2 \\plus{} (bc^{\\minus{}1})^2 \\equiv \\minus{}1 \\pmod{p}$, $ (ac^{\\minus{}1})^4 \\plus{} (bc^{\\minus{}1})^4 \\equiv \\minus{}1 \\pmod{p}$, apoi $ (ac^{\\minus{}1})^4 \\plus{} (bc^{\\minus{}1})^4 \\plus{} 2(abc^{\\minus{}2})^2 \\equiv 1 \\pmod{p}$, etc.)\r\n\r\nmoldovan, rezultatul t\\u au este cel bun, dar solu\\c tia este gre\\c sit\\u a, vezi factorizarea ...", "Solution_3": ":maybe: am lasat doar raspunsurile - mi-am dat seama ce nu este bine" } { "Tag": [ "floor function", "real analysis", "real analysis unsolved" ], "Problem": "What is the measure of the number $ 1 1$.\r\n\r\nHardy, G. H. and Littlewood, J. E. \"Some Problems of Diophantine Approximation.\" [i]Acta Mathematica[/i] 37, 193-239, 1914.", "Solution_2": "Have you this article?", "Solution_3": "Actually I got the citation wrong, the page numbers should be 155-191 (193-239 is the second part of the same article, but it does not pertain to your question).\r\n\r\nActa Mathematica is a fairly popular journal. Judging by your location, I'm assuming you're at Babe\u015f-Bolyai; I would think a research institution of that caliber would almost certainly have it available in their library. Even if the library doesn't have it, it likely has some kind of sharing agreements with other universities and publishes, so they should be able to get it for you.\r\n\r\nFailing all this, you can also purchase a (rather expensive) [url=http://www.springerlink.com/content/r6h60873n592p442/?p=999a6b285673498d8af9d39f14dfe59c&pi=5]copy from Springer[/url], though this particular article would cost you about $ \\$$32.00 plus shipping.", "Solution_4": "OK,but you know the solution?" } { "Tag": [ "analytic geometry", "trigonometry", "geometry solved", "geometry" ], "Problem": "Assume South is +-0\r\nAssume North is +-180\r\nAssume East is -90\r\nAssume West is +90\r\nKnown variables:\r\nCurrent X, Y and Heading\r\nTarget X, Y\r\n\r\nI need a formula that looks at your current heading, then your desired target, and calculates the shortest distance so I can instruct \r\nto turn left or right to get there.\r\n\r\n\r\nAny help would be great!", "Solution_1": "Your coordinate system is oriented so that the positive x-axis (0o) is in the south direction and the positive y-axis (90o) is in the west direction. Assume that heading means the oriented angle from -180o to 180o between the south direction (the positive x-axis) and the direction of travel.\r\n\r\nLet $x_C, y_C$ be the current position and $\\alpha_C$ the current heading. In your coordinate system, this line has an equation\r\n\r\n$y - y_C = \\tan{\\alpha_C} \\cdot (x - x_C)$\r\n\r\nThe line from the current position to the destination $x_D, y_D$ has an equation\r\n\r\n$y - y_C = \\frac{y_D - y_C}{x_D - x_C} \\cdot (x - x_C) = \\tan{\\alpha_D} \\cdot (x - x_C)$\r\n\r\nwhere the destination heading $\\alpha_D$ is\r\n\r\n$\\alpha_D = \\arctan{\\frac{y_D - y_C}{x_D - x_C}} + (x_D - x_C < 0) \\cdot \\frac{y_D - y_C}{|y_D - y_C|} \\cdot 180^o$\r\n\r\nTo calculate the change of heading $\\alpha_D - \\alpha_C$,\r\n\r\n$\\tan{(\\alpha_D - \\alpha_C)} = \\frac{\\tan{\\alpha_D} - \\tan{\\alpha_C}}{1 + \\tan{\\alpha_D} \\cdot \\tan{\\alpha_C}} = \\frac{\\frac{y_D - y_C}{x_D - x_C} - \\tan{\\alpha_C}}{1 + \\frac{y_D - y_C}{x_D - x_C} \\cdot \\tan{\\alpha_C}} =$\r\n\r\n$= \\frac{y_D - y_C - \\tan{\\alpha_C} \\cdot (x_D - x_C)}{x_D - x_C + \\tan{\\alpha_C} \\cdot (y_D - y_C)}$\r\n\r\n$\\alpha_D - \\alpha_C = \\arctan{\\left(\\frac{y_D - y_C - \\tan{\\alpha_C} \\cdot (x_D - x_C)}{x_D - x_C + \\tan{\\alpha_C} \\cdot (y_D - y_C)}\\right)}\\ +$\r\n\r\n$+ (x_D - x_C < 0) \\cdot \\frac{y_D - y_C}{|y_D - y_C|} \\cdot 180^o$" } { "Tag": [ "inequalities", "trigonometry", "geometry", "circumcircle", "function", "geometry solved" ], "Problem": "A+B+C=180 and A >= B >= C\r\nProve that a(C-B) + b(A-C) + c(B-A) >= 0", "Solution_1": "You didn't tell us what a, b, c are, but it's almost sure that a, b, c are the sidelengths of the triangle with the angles A, B, C, so that your question is:\r\n\r\n[i]Let A, B, C be the angles and a, b, c the corresponding sidelengths of a triangle. Assume that $A\\geq B\\geq C$. Prove that $a\\left(C-B\\right)+b\\left(A-C\\right)+c\\left(B-A\\right)\\geq 0$.[/i]\r\n\r\nIndeed, after the Extended Sine Law, we have a = 2R sin A, b = 2R sin B and c = 2R sin C, where R is the circumradius of our triangle. In other words, a = f(A), b = f(B) and c = f(C), where the function f(x) is defined by f(x) = 2R sin x. Now, since the function sin x is concave on the interval $\\left[0;\\;\\pi\\right]$, and 2R is positive, the function f(x) = 2R sin x is also concave on the interval $\\left[0;\\;\\pi\\right]$. This interval is, of course, the interval where the angles A, B, C lie. Thus, our problem will immediately follow from the following more general problem:\r\n\r\n[i][b]General problem.[/b] Let A, B, C be three arbitrary real numbers in an interval I, and let f(x) be a concave function on I. Assume that $A\\geq B\\geq C$. Then, $f\\left(A\\right)\\left(C-B\\right)+f\\left(B\\right)\\left(A-C\\right)+f\\left(C\\right)\\left(B-A\\right)\\geq 0$.[/i]\r\n\r\nThe solution of this general problem is easy: Since $A\\geq B\\geq C$, we have $B-C\\geq 0$, $A-C\\geq 0$ and $A-B\\geq 0$, so that $\\frac{B-C}{A-C}\\geq 0$ and $\\frac{A-B}{A-C}\\geq 0$. Also, clearly, $\\frac{B-C}{A-C}+\\frac{A-B}{A-C}=\\frac{A-C}{A-C}=1$. Thus, since the function f(x) is concave on the interval I, we have by the definition of a concave function\r\n\r\n$\\frac{B-C}{A-C}f\\left(A\\right)+\\frac{A-B}{A-C}f\\left(C\\right)\\leq f\\left(\\frac{B-C}{A-C}A+\\frac{A-B}{A-C}C\\right)$;\r\n\r\nsince $\\frac{B-C}{A-C}A+\\frac{A-B}{A-C}C=\\frac{\\left(B-C\\right)A+\\left(A-B\\right)C}{A-C}=\\frac{B\\left(A-C\\right)}{A-C}=B$, this rewrites as\r\n\r\n$\\frac{B-C}{A-C}f\\left(A\\right)+\\frac{A-B}{A-C}f\\left(C\\right)\\leq f\\left(B\\right)$.\r\n\r\nMultiplication by A - C (remembering that $A-C\\geq 0$) yields\r\n\r\n$\\left(B-C\\right)f\\left(A\\right)+\\left(A-B\\right)f\\left(C\\right)\\leq \\left(A-C\\right)f\\left(B\\right)$.\r\n\r\nNow, subtracting $\\left(B-C\\right)f\\left(A\\right)+\\left(A-B\\right)f\\left(C\\right)$, we get\r\n\r\n$0\\leq \\left(A-C\\right)f\\left(B\\right)-\\left(\\left(B-C\\right)f\\left(A\\right)+\\left(A-B\\right)f\\left(C\\right)\\right)$\r\n$= \\left(A-C\\right)f\\left(B\\right)+\\left(C-B\\right)f\\left(A\\right)+\\left(B-A\\right)f\\left(C\\right)$\r\n$=f\\left(A\\right)\\left(C-B\\right)+f\\left(B\\right)\\left(A-C\\right)+f\\left(C\\right)\\left(B-A\\right)$,\r\n\r\nand the general problem is solved. $\\blacksquare$\r\n\r\n darij", "Solution_2": "The initial inequality is very easy [tex] a(C-B)+b(A-C)+c(B-A)\\geq c(C-B)+c(A-C)+c(B-A)=0 [/tex]\r\n\r\n[b]EDIT:[/b] I am sorry, this is wrong." } { "Tag": [], "Problem": "Evaluate \r\n\r\n$ \\sum_{k\\equal{}1}^{n}\\sum_{r\\equal{}0}^k r C_{r}^{n}$", "Solution_1": "$ \\sum_{k\\equal{}1}^{n} \\sum_{r\\equal{}1}^{k} rC_n^r \\equal{}$ $ \\sum_{k\\equal{}1}^{n} \\sum_{r\\equal{}1}^{k} nC_{n\\minus{}1}^{r\\minus{}1} \\equal{}$ $ n\\sum_{k\\equal{}0}^{n\\minus{}1} (n\\minus{}k)C_{n\\minus{}1}^k \\equal{}$ $ n\\sum_{k\\equal{}0}^{n\\minus{}1} ((n\\minus{}1)C_{n\\minus{}2}^k \\plus{} C_{n\\minus{}1}^k) \\equal{}$ $ n\\left ( (n\\minus{}1)2^{n\\minus{}2} \\plus{} 2^{n\\minus{}1} \\right )$." } { "Tag": [], "Problem": "For integers $ a, b$, define $ a*b\\equal{}a^{b} \\plus{} b^{a}$. If $ 2*x\\equal{} 100$, find the value of $ x$.", "Solution_1": "$ 2 * x \\equal{} x^2 \\plus{} 2^x \\equal{} 100$\r\n\r\nSince the LHS increases very rapidly, we can just guess and check for *positive, even integers.\r\n\r\n$ x \\equal{} 6 \\implies 6^2 \\plus{} 2^6 \\equal{} 36 \\plus{} 64 \\equal{} 100$\r\n\r\nSo $ \\boxed{x \\equal{} 6}$.\r\n\r\n\r\n*Note: x must be positive, because if it's not, LHS is not an integer.\r\nx must be even because if not $ x^2$ is odd so LHS is odd." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $x_{1},\\,x_{2},\\,x_{3},\\,x_{4},\\,x_{5}$ be the real number ssuch that: $\\sum_{i=1}^{5}x_{i}=1.$ Prove that:\r\n\\[x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+x_{3}x_{4}x_{5}+x_{4}x_{5}x_{1}+x_{5}x_{1}x_{2}\\le\\frac{1}{25}\\]", "Solution_1": "[quote=\"nobody1\"]Let $x_{1},\\,x_{2},\\,x_{3},\\,x_{4},\\,x_{5}$ be the real number ssuch that: $\\sum_{i=1}^{5}x_{i}=1.$ Prove that:\n\\[x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+x_{3}x_{4}x_{5}+x_{4}x_{5}x_{1}+x_{5}x_{1}x_{2}\\le\\frac{1}{25}\\]\n[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=112980" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Let A \\in Mn(C) such that there exist k \\geq 2, integer \r\nA^(k+1)+A^k=0. Let B=I+A\r\nProve that C=I+AB+(AB)^2+...+(AB)^(k-1) is invertible\r\n---------------------------------------------------------------\r\nI think we must prove that (AB) is nilpotent of order k\r\nC^(-1)=I-AB\r\nCan someone write a proof that (AB)^k=0 \r\nThanks", "Solution_1": "Isn't it obvious?\r\n\r\nWe know that A k *(A+I_n) = 0_n and we want to prove that [A(A+I_n)] k =0_n, but [A(A+I_n)] k = A k *(A+I_n)*(A+I_n) k-1 = 0_n*(A+I_n) k-1 = 0_n." } { "Tag": [], "Problem": "Janice earns a monthly salary of $ \\$600$ plus a $ 2\\%$ commission on all her sales. How many dollars worth of sales must Janice make in a month to earn a total of $ \\$1400$ that month?", "Solution_1": "That means Janice wants to earn $ 1400 \\minus{} 600 \\equal{} 800$ dollars commission. Thus, she must make $ 800 \\div 0.02 \\equal{} 40,000$ dollars worth of sales.", "Solution_2": "In an equation this would look like\n\nLet $x$=the amount of money she earns in sales\n\n$600+0.02(x)=1400$\n\n$x=\\boxed{40000}$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "We have that $x,y,z,t$ are 4 real nimbers in the interval $[-1,1]$.\r\nShow that $t^2(xy+yz+zx)+2t(x+y+z)+3\\ge 0$. When does equality hold?", "Solution_1": "The inequality is the same as $(tx+1)(ty+1)+(ty+1)(tz+1)+(tz+1)(tx+1)\\geq 0$, with equality for $\\alpha=\\beta=-t\\in\\{\\pm 1\\}$, where $\\alpha,\\beta$ are two of $x,y,z$.", "Solution_2": "[quote=\"gnosis\"]The inequality is the same as $(tx+1)(ty+1)+(ty+1)(tz+1)+(tz+1)(tx+1)\\geq 0$, with equality for $\\alpha=\\beta=-t\\in\\{\\pm 1\\}$, where $\\alpha,\\beta$ are two of $x,y,z$.[/quote]\r\n\r\n\r\nGreat!\r\nIn the original solution we make the substitution $u=1/t$ and then we go on in standart way." } { "Tag": [ "inequalities", "quadratics" ], "Problem": "Find the integers (x,y) if $2(x^{3}-xy+y^{3})=3(x^{2}+y^{2})$.", "Solution_1": "Could you say what's the origin of this problem? 'cause I found it non-trivial, although quite easy (it took me only 5-10 minutes), but still remains the fact that until this year I wouldn't b able to attack it... \r\n\r\n[hide=\"Anyway...\"]\n$2(x^{3}-xy+y^{3})=3(x^{2}+y^{2})-\\to 2x^{3}+2y^{3}-3x^{2}-3y^{2}-2xy=0$\n$(x^{2}+y^{2})(2x-3)=2xy$. Since the RHS is even, then $x^{2}+y^{2}|2xy$. But by the QM-GM inequality, $x^{2}+y^{2}\\geq 2xy$.\nThen it must be the case that $x^{2}+y^{2}=2xy$, which is $(x-y)^{2}=0$ and finally $x=y$.\nSubstituting, $2x^{2}(2x-3)=2x^{2}$ , $2x=4$ , $x=y=2$.\nThat's all folks... (at least, it should be!) :wink: [/hide]", "Solution_2": "this is a romanian shortlist problem for nineth grade since 2002.", "Solution_3": "[quote=\"Ponnamperuma\"]Could you say what's the origin of this problem? 'cause I found it non-trivial, although quite easy (it took me only 5-10 minutes), but still remains the fact that until this year I wouldn't b able to attack it... \n\n[hide=\"Anyway...\"]\n$2(x^{3}-xy+y^{3})=3(x^{2}+y^{2})-\\to 2x^{3}+2y^{3}-3x^{2}-3y^{2}-2xy=0$\n$(x^{2}+y^{2})(2x-3)=2xy$. Since the RHS is even, then $x^{2}+y^{2}|2xy$. But by the QM-GM inequality, $x^{2}+y^{2}\\geq 2xy$.\nThen it must be the case that $x^{2}+y^{2}=2xy$, which is $(x-y)^{2}=0$ and finally $x=y$.\nSubstituting, $2x^{2}(2x-3)=2x^{2}$ , $2x=4$ , $x=y=2$.\nThat's all folks... (at least, it should be!) :wink: [/hide][/quote]\r\n\r\nnice solution. shouldnt it be am-gm instead of qm-gm?", "Solution_4": "[quote=\"jli\"][quote=\"Ponnamperuma\"]Could you say what's the origin of this problem? 'cause I found it non-trivial, although quite easy (it took me only 5-10 minutes), but still remains the fact that until this year I wouldn't b able to attack it... \n\n[hide=\"Anyway...\"]\n$2(x^{3}-xy+y^{3})=3(x^{2}+y^{2})-\\to 2x^{3}+2y^{3}-3x^{2}-3y^{2}-2xy=0$\n$(x^{2}+y^{2})(2x-3)=2xy$. Since the RHS is even, then $x^{2}+y^{2}|2xy$. But by the QM-GM inequality, $x^{2}+y^{2}\\geq 2xy$.\nThen it must be the case that $x^{2}+y^{2}=2xy$, which is $(x-y)^{2}=0$ and finally $x=y$.\nSubstituting, $2x^{2}(2x-3)=2x^{2}$ , $2x=4$ , $x=y=2$.\nThat's all folks... (at least, it should be!) :wink: [/hide][/quote]\n\nnice solution. shouldnt it be am-gm instead of qm-gm?[/quote]\r\n\r\nWell, the qm-gm is the quadratic mean inequality, I think, so I think it is supposed to be there.\r\n\r\nBut nice solution.\r\n\r\nAnd I hate to ruin your day, but $(0,0)$ is also a solution.", "Solution_5": "Quadratic Mean and Root Mean Square are the same thing, correct?", "Solution_6": "[quote=\"mathgeniuse^ln(x)\"] \n\nWell, the qm-gm is the quadratic mean inequality, I think, so I think it is supposed to be there.\n\n [/quote]\r\n\r\n$x^{2}+y^{2}\\geq 2xy$ follows from both am-gm or qm-gm :wink:", "Solution_7": "[quote=\"Ponnamperuma\"]Since the RHS is even, then $x^{2}+y^{2}|2xy$. But by the QM-GM inequality, $x^{2}+y^{2}\\geq 2xy$.\nThen it must be the case that $x^{2}+y^{2}=2xy$[/quote]\r\n\r\nI think you have a problem here, because if $2xy$ is negative, we can't say anything. Like $2^{2}+(-2)^{2}|(2 \\cdot 2 \\cdot-2)$.\r\n\r\nEDIT: But you can say $|x^{2}+y^{2}| \\ge |2xy|$ so either $x^{2}+y^{2}= 2xy$ or $x^{2}+y^{2}=-2xy$." } { "Tag": [ "vector" ], "Problem": "Post erased by moderator.", "Solution_1": "This is pure insanity. How you managed to obtain the last question to STEP II 2009 of the pure maths section BEFORE people in the UK had even gone to school for the eam (which started at 9 am) begs belief and it is rather appalling that you think what you have done here is acceptable." } { "Tag": [ "algorithm", "Additive Number Theory" ], "Problem": "Prove that any positive integer can be represented as an aggregate of different powers of $3$, the terms in the aggregate being combined by the signs $+$ and $-$ appropriately chosen.", "Solution_1": "Is it correct?\r\n\r\nLet $ A\\equal{}(a_0, a_1,\\cdots,a_k)$ the digits of $ n$ in base $ 3$ $ (n\\equal{}a_0\\plus{}3a_1 \\plus{}\\cdots \\plus{}3^k a_k)$ and $ a_i \\in \\{0,1,2\\}$ \r\nLet $ g(A)\\equal{}$ the amount of digits $ 2$ in $ A$\r\n$ 1)$ If $ a_k\\equal{}2$ then redefine $ A\\equal{}(a_0,a_1,\\cdots,a_{k\\minus{}1},\\minus{}1,1)$ notice that $ g(A)$ decrease in one unit from the previous one.\r\n$ 2)$ If $ g(A)\\equal{}0$ then we are done.\r\n$ 3)$ Otherwise let $ a_j\\equal{}$ the first element from the right of $ A$ equal to $ 2$ then \r\nfrom $ 2$ $ 3^i\\equal{}3^{i\\plus{}1}\\minus{}3^{i}$\r\nredefine $ A\\equal{}(a_0,a_1,\\cdots,a_{j\\minus{}1},\\minus{}1,a_{j\\plus{}1}\\equal{}a_{j\\plus{}1}\\plus{}1,a_{j\\plus{}2},\\cdots,a_k)$ \r\nnotice that the new $ a_{j\\plus{}1} \\in \\{0,1,2\\}$ thus $ g(A)$ is nonincreasing\r\nRepeating the steps $ 2$ or $ 3$ and because the array $ A$ is finite, $ g(A)$ for same step reachs the rightmost element of $ A$ and then decrease by one.\r\nAt the end we get to $ g(A)\\equal{}0$ and the result follows.", "Solution_2": "[hide=\"Seems too simple\"]Consider the algorithm of taking every $ 2$ in the base $ 3$ expansion of the number, replacing it with $ \\minus{}1$, and adding $ 1$ to the digit to the left, using standard rules for carrying in base $ 3$. Repeat this as much as needed until no $ 2$s are left. It's pretty obvious the algorithm works; if $ x$ is a digit for $ \\minus{}1$ in base $ 3$, we can see that $ 2 \\equal{} 1x$ in base $ 3$, justifying our procedure. The algorithm terminates because for there to be an infinite number of $ 2$s to replace, there has to be an infinite number of $ 1$s or $ 2$s to start out with, which is absurd. If there aren't infinite $ 2$s, we can only get more by adding $ 1$ (through the algorithm) to the $ 1$s. In the end, we are left with a number with $ 1,0,\\minus{}1$ for digits in base $ 3$. Then expanding it solves the problem.[/hide]", "Solution_3": "In fact, that algorithm works for any base $ n$ and numbers $ a_1, a_2, ..., a_n$ that cover all residues $ \\mod n$. It just may happen that you get an infinite representation (we get $ n$-adic numbers, thus for example $ 1\\plus{}n\\plus{}n^2\\plus{}n^3\\plus{}... \\equal{}1\\minus{}n$).", "Solution_4": "Another way is that for a given positive integer $ n$, the decompositions with the form $ a_0 \\plus{} \\cdots a_n3^n$, where $ a_i \\in \\{\\minus{}1,0,1\\}$ are pairwise distincts.\r\nMoreover any number with that form belongs to $ E_n \\equal{} \\{ \\minus{}\\frac {3^{n\\plus{}1}\\minus{}1}{2}, \\cdots ,\\frac {3^{n\\plus{}1}\\minus{}1}{2}\\}$ and this set contains exactly $ 3^{n\\plus{}1}$ integers. In another hand, there are exactly $ 3^{n\\plus{}1}$ decompositions (three choices for each $ a_i$), thus each integer in $ E_n$ can be written as desired.\r\nSince, this is true for any $ n$ we get the desired conclusion.\r\n\r\nPierre." } { "Tag": [ "geometry", "3D geometry", "perimeter" ], "Problem": "Do these problems fast. If you get two other people to do with you, that's even better as well. :) \r\n\r\nR1 - 1. The product of five consecutive positive integers is divisible by both 13 and 31. If that product is as small as possible, compute the smallest of the five integers.\r\n\r\nR1 - 2. Let $ T \\equal{} TNYWR$ and let $ K \\equal{} \\frac{T\\minus{}1}{15}$.\r\nCompute the value of $ (1 \\minus{} i)^{2K}$.\r\n\r\nR1 - 3. Three vertices of a cube are connected to form a triangle. Compute the maximum possible perimeter of this triangle if the edge of the cube is $ T$.", "Solution_1": "[hide=\"1\"]61,62,63,64,65\n\nso 61[/hide]\n\n[hide=\"2\"](1-i)^8=16.[/hide]\n\n[hide=\"3\"]equilater triangle: 48 sqrt2[/hide]" } { "Tag": [ "function" ], "Problem": "Determine $ a,b\\in \\mathbb{R}$ such that the function $ f: [0,2]\\rightarrow [\\minus{}1,3]\\ ,\\ f(x)\\equal{}ax\\plus{}b$ is bijective.", "Solution_1": "If $ f(x) \\equal{} ax \\plus{} b$ is bijective, then all the real numbers in the interval $ [0,2]$ must map to all the real numbers in the interval $ [\\minus{}1,3]$. Note that the length of the first interval is $ 2$ and that the length of the second interval is $ 4$. Therefore, we let $ a \\equal{} 2$ so we have $ [0,2] \\mapsto [0,4]$. Now we let $ b \\equal{} \\minus{}1$ so that we have $ [0,4]\\mapsto [\\minus{}1,3]$. Thus, $ f(x) \\equal{} \\boxed{2x \\minus{} 1}$." } { "Tag": [ "number theory", "Eulers function", "system of equations", "arithmetic sequence", "IMO Shortlist", "imo 1991", "Laurentiu Panaitopol" ], "Problem": "Let $ \\,n > 6\\,$ be an integer and $ \\,a_{1},a_{2},\\cdots ,a_{k}\\,$ be all the natural numbers less than $ n$ and relatively prime to $ n$. If\r\n\\[ a_{2} \\minus{} a_{1} \\equal{} a_{3} \\minus{} a_{2} \\equal{} \\cdots \\equal{} a_{k} \\minus{} a_{k \\minus{} 1} > 0,\r\n\\]\r\nprove that $ \\,n\\,$ must be either a prime number or a power of $ \\,2$.", "Solution_1": "I'll use Bertrand's Postulate: for $u\\ge 2$ is a positive integer, there is a prime in the interval $(u,2u)$. \r\n\r\nClearly, $a_2$ must be equal to the smallest prime $p$ which does not divide $n$. If $p=2$, then $n$ is a prime since the common difference $a_{i+1}-a_i$ is equal to $1$, i.e. all positive integers less than $n$ are coprime to $n$. If, on the other hand, $p=3$, we find $n$ to be a power of $2$: the positive integers less than $n$ and coprime to it are precisely the odd ones. We may thus assume that $p\\ge 5$. Furthermore, since $n>6$, the positive integers less than $n$ and coprime to it cannot be $a_1,a_2$ alone, i.e. $k=\\varphi(n)\\ge 3$.\r\n\r\nBy Bertrand's Postulate, the largest prime less than $p$ is strictly larger than $\\frac{p-1}2$, so it cannot divide $p-1$. We will denote this prime by $q$. We know that $q|n$. It's easy to check that for $p\\le 31$ there is a prime $r$ strictly between $a_2=p,\\ a_3=2p-1$. $rq|n$, so, in particular, $pq32$, the two largest primes $r_12.\r\n 1. If p=3, then 1, 3, ..., n-1 is our progression :arrow: n is a power of 2.\r\n 2. If p>3, then n is divisible by 3. we have n-1=a[k]=1+(p-1)(k-1) :arrow: p-1 is divisor of n-2. :idea: Let q be a random prime divisor of p-1. q | p-1 | n-2, but q|n, because q

3, i is even.\r\n But a[3]=2p-1=2^(i+1) + 1 is divisible by 3, but n is divisible by 3 too :arrow: gcd(n, a[3])>1. This contradiction ends our proof. :lol:", "Solution_3": "The sequence is $ a_i \\equal{} 1 \\plus{} (p \\minus{} 1)(i \\minus{} 1), 1 \\le 1 \\le \\phi(n), a_{\\phi(n)} \\equal{} 1 \\plus{} (p \\minus{} 1)(\\phi(n) \\minus{} 1) \\equal{} n \\minus{} 1$, where $ p$ is the smallest prime not dividing $ n$.\r\nIf $ p \\equal{} 2$, then $ \\phi(n) \\equal{} n \\minus{} 1$ so $ n$ is a prime.\r\nIf $ p \\equal{} 3$, then $ n$ is coprime to all odd primes $ \\le n$, so $ n$ is a power of $ 2$.\r\nOtherwise, $ p \\ge 5$. It implies that $ 6 | n$ and that if you take 4 (=5-1) consecutive integers ($ \\le n$) then at most one of them is coprime to $ n$. Let $ t \\equal{} \\frac {n}{6}$.\r\nConsider $ 3t \\minus{} 1, 3t, 3t \\plus{} 1$. At most one of $ 3t \\minus{} 1$ and $ 3t \\plus{} 1$ should be coprime to $ n$. But they have the same gcd with $ n$, so $ 3t \\pm 1$ are not coprime with $ n$, which implies that $ t$ is odd, because $ (3t \\minus{} 1, n) \\equal{} (3t \\minus{} 1,6t) \\equal{} (3t \\minus{} 1,2)$.\r\n\r\nIt implies that $ (3t \\minus{} 2,6t) \\equal{} (3t \\minus{} 4,6t) \\equal{} 1$:2 number with distance 2 that are coprime to $ n$, a contradiction.\r\n(note: we need $ n > 6$ for $ 3t \\minus{} 4 \\ge 1$, because $ n \\equal{} 6$ also satisfies the requirements)", "Solution_4": "here is my solution of this problem :) \r\n\r\nLet, $ a_2 \\minus{} a_1 \\equal{} a_3 \\minus{} a_2 \\equal{} ... \\equal{} a_k \\minus{} a_{k \\plus{} 1} \\equal{} s$ and $ a_k > a_{k \\minus{} 1} > ... > a_2 > a_1 \\equal{} 1$\r\n\r\n[u]Case:1[/u]: $ s > 2$\r\n\r\nif $ a_2$ is not a prime number it means that there exist a divisor $ d$ of $ a_2$ such that $ 1 < d < a_2$. Hence $ \\gcd(d,n) > 1\\Rightarrow\\gcd(a_{2},n) > 1$ which is a contradiction !!\r\n\r\nNow let's suppose that $ a_i$ is not a prime for some $ i$. Hence there exist a prime divisor $ l$ of $ a_i$ such that $ 1 < l < a_i$ if $ l\\neq a_j$ $ \\forall j \\le i$ then $ \\gcd(l,n) > 1 \\Rightarrow \\gcd(a_i,n) > 1$ which is a contradiction !! and so there exist $ j > 1$ such that $ \\forall i$ we have $ a_i \\equal{} a_j^c$ and $ j < i$\r\nBy the last argument we have $ a_i \\equal{} a_j^c \\equal{} a_{j_2}^{c_2} \\equal{} a_{j_3}^{c_3}... \\equal{} a_2^d$\r\nUsin our condition we get $ a_2^d \\minus{} a_2^{d'} \\equal{} a_2 \\minus{} 1$ which is not true !!\r\nwe conclude that $ \\forall i , a_i$ is a prime number . if $ a_2^2 < n$.Since $ a_i$ is a prime number it means that $ 2a_2 \\minus{} 1$ is prime and $ 3a_3 \\minus{} 2$ is prime and $ 4a_4 \\minus{} 3$ is prime...$ (a_2 \\plus{} 1)a_2 \\minus{} a_2$ is prime $ \\Leftrightarrow$ $ a_2^2$ is prime which is not true. Hence $ a_2 > \\sqrt {n}$and it means that $ \\forall k < \\sqrt {n}$ we have $ \\gcd(n,k) > 1$which implies that $ \\gcd(k,n \\minus{} k) > 1$,$ \\forall k < \\sqrt {n}$ . On the other hand there exist a $ l$ such that $ n \\minus{} k \\equal{} a_l$ and so $ \\gcd(n,a_l) > 1$ which is not true !!\r\nSo the hypothesis of the case 1 is not true !! :D \r\nLet's go now to the case $ s \\equal{} 2$\r\nwe have \r\n$ \\gcd(n,1) \\equal{} 1 \\Rightarrow \\gcd(n,3) \\equal{} 1 \\Rightarrow \\gcd(n,5) \\equal{} 1 ...\\Rightarrow \\gcd(n,2k \\plus{} 1) \\equal{} 1$ and so the power of two are a solution.\r\nWhen $ s \\equal{} 1$ we get easily the fact that n is prime\r\n\r\nand we have done :) !!", "Solution_5": "if n is odd then the problem is trivial. if not then n must be square free because \r\nif n=(2^k)((p_1)^ a_1)...((p_t)^a_t) (we suppose that n#2^m) then this two number are relatively prime to n:\r\n2(p_1)...(p_t) + 1 and 2(p_1)...(p_t) - 1 \r\nif this two number are less than n then the problem woud be trivial. so n must be on this form\r\nn=2(p_1)...(p_t) but u can easily see that in this case (n/2)+2 and (n/2)-2 are relatively prime to n. the rest is trivial(u can find that 3|n but (9,n)=1!!!!)", "Solution_6": "[quote=\"davicoelhoamorim\"]a_k>a_(k-1)>...>a_2>a_1 => k=phi(n), a_k=n-1 and a_1=1\n\nbecause a_2-a_1=a_3-a_2=...=a_k-a_(k-1)=r, we have that a_i=a_(i-1) + r => a_k=a_1 + (k-1).r => n-1=1 + (phi(n)-1).r =>\n\n[n-2]/[phi(n)-1]=r => phi(n)-1|n-2\n\nif phi(n)-1=n-2 => phi(n)=n-1 => n is a prime\n\nif no, 2[phi(n)-1]= n/2>=phi(n) => n=2^k[/quote]\n$a_k=n-1<=>n $be prime???", "Solution_7": "No. For any $n\\geq 2$ we have $\\gcd(n-1,n)=1$, so $a_k = n-1$. It is only when, as said, $\\phi(n) = n-1$, that the conclusion $n$ prime was reached.", "Solution_8": "Nice problem. Really liked it. Sorry that I don\u00b4t know LaTex yet!\nOk, my solution is really simple:\n\nFirst, solve the equalities two - by - two. Ater that, you get a general formula, from now on called G(n):\n2a_i = a_i+1 + a_i-1\n\nHence:\n1) 2a_2 = a_3 + a_1\n2) 2a_3 = a_4 + a_2\n .\n .\n .\nk-2) 2a_k-1 = a_k + a _k-2\n\nNow, by definition, we know that 1 is relatively prime to every natural number > 1.\nTherefore, a_i = 1 , i element of euler\u00b4s function of n (sorry, don\u00b4t know how to use latex).\n\nThen, since 2a_2 = a_3 + a_1, a_3 + a_1 = 0 (mod 2).\nNow, since a_1 = 1, which is uneven, a_3 must be uneven.\n\nThen, since 2a_4 = a_5 + a_3, a_5 + a_3 = 0 (mod 2). Since we already established a_3 is uneven, so is a_5.\n\nThis process goes on, and we arrive to the conclusion that both terms on the right hand side of the equals are uneven, for the equalities on uneven places, starting with 2a_2 = a_3 + a_1.\n\nThen, we see two cases: 1) a_2 Uneven. 2) a_2 Even.\n\n[u]Case 1[/u]: a_2 Uneven.\nUsing the same analysis as before, since 2a_3 = a_4 + a_2, a_4 + a_2 =0 (mod 4). Now, since a_2 is uneven, a_4 must also be uneven.\n\nThen, using the same analysis in the remaining equalities we see that both terms to the right hand side of the equals will be uneven in the equalities on even places (starting in 2a_3 = a_4 + a_2).\n Earlier, we already established that both terms to the right hand side of the equals on uneven places were both uneven, independent of the parity of a_2.\nTherefore, all terms are uneven.\nTherefore, a_i is uneven, for i = 1,2,...,k.\n\nNow, that means that n is not relatively prime to even numbers, so n must be even.\n\nNow, lets assume n has at least one uneven factor.\n\nThat means that at least one of the uneven integers from 1 to n won\u00b4t be relatively prime to n.\n\nThen, assume that p and p+2 are both relatively prime to n. \nSince both are uneven, that means that a_i = p , a_i+1 = p + 2.\n\nNow, assume m-2, m and m+4 are relatively prime to n, but m+2 is not relatively prime to n.\nThat means a_i-1 = m-2 , a_i = m , a_i+1 = m + 4\nBy G(n):, we get:\n 2m = m + 4 + m - 2\n Therefore 2m = 2m + 2, so 0 = 2, which is obviously false, so this is a condradiction.\nTherefore, n doesn\u00b4t have uneven factors, so [b]n = 2^t[/b].\n\n[u]Case 2[/u]: a_2 Even.\nSince 2a_3 = a_4 + a_2, a_4 + a_2 = 0 (mod 2). Since a_2 is even, a_4 is also even.\n\nUsing the same analysis, we see that both terms to the right side of the equal sign will be even, for the equalities on even places.\nWe already established that both terms to the right hand side of the equals on uneven spaces were both uneven, independent of the parity of a_2.\n\nSince this terms are even, that means n is relatively prime to some even numbers, so n is uneven.\n\nTherefore, n is relatively prime to 2.\n\nSince a_1 = 1, that means a_2=2.\nTherefore, solving in G(n), a_3= 3.\nReplacing in the next equality we get a_4 = 4.\nIn general, we get a_i = i , for i = 1,2,3,...,k.\n\nTherefore, n is relatively prime to 1,2,3,...,k.\n\n[b]That means n is prime[/b], thus finishing the demonstration.\n\nCould somebody please check my solution? Thanks!!! :)", "Solution_9": "I found a really simple solution.\nLet $d=a_{i+1}-a_i$. Then we have $a_i=1_1+d(i-1)=1+d(i-1)$. Since $\\gcd(n-1,n)=1$, so $n-1$ is the largest positive integer smaller than $n$ and co-prime to $n$. Thus, $a_k=n-1$ and $1+d(k-1)=n-1$ which shows $d(k-1)=n-2$ and so $d$ divides $n-2$. We make two cases.\n[b]Case 1:[/b] $n$ is odd. Then $\\gcd(n-2,n)=1$. So, every divisor of $n-2$ is co-prime to $n$ too, so is $d$. Thus, there is a $j$ such that $d=a_j=1+d(j-1)$ which implies $d|1$. Therefore, in this case, $d=1$ and we easily find that all numbers less than $n$ are co-prime to $n$, so $n$ must be a prime.\n[b]Case 2:[/b] $n$ is even. But then $\\gcd(n-2,n)=2$ and also $d=1$ is ruled out. So $d>1$. Say, $n=2l$. Then, $d$divides $2(l-1)$. If $d$ is odd, $d|l$ and $d|l-1$ forcing $d|1$, contradiction! So, $d=2s$ with $s|l-1$. If $l=2r+1$ for some $r$, then $\\gcd(r,n)=1$. So, again $r=1+d(j-1$ for a $j$. If $s$ is odd, then $s|r$ and thus, $s|1$. Similarly, we find that actually $d|2$ must occur. We finally have, $d=2$. But then all odd numbers less than $n$ are co-prime to $n$. So, $n$ does not have any odd factor i.e. $n=2^k$ for some $k\\in\\mathbb N$.\n\nAnd learning $\\text{\\LaTeX}{}$ is really easy. You just need to learn some symbols and put your formulas between two dollar signs. For more, you can see here:\nhttp://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About", "Solution_10": "Here is another solution. If $n$ is odd, then $a_{2}=2$, and the conclusion follows. ($n$ has no common divisor with all integers less than it). Now assume $n$ is even.\n$i)$ $a_{2}=3$ , in this case n is pairwisely coprime with all odd integers less than itself, so that it should be a power of $2$.\n\n$ii)$ $a_{2} > 3$. Observe that $3|n$. In this case, let $p$ be the smallest prime number not dividing $n$. We claim that $p \\equiv1\\; (mod\\; 3)$. Because if $p \\equiv2\\; (mod\\; 3)$, then $a_{3}=2p-1$ will be divisible by $3$, contradicting with the fact that $3|n$. Now, we have an arithmetic sequence with common difference $p-1$. So, $(k-1)(p-1)=n-2$. But if we check both sides of this equation in $mod 3$, we see that $3|(k-1)(p-1)$; however, $3 \\nmid n-2$, contradiction.\n\nHence the problem is done.", "Solution_11": "[quote=\"grupyorum\"][quote=\"AFdrolf\"][quote=\"grupyorum\"]Here is another solution. If $n$ is odd, then $a_{2}=2$, and the conclusion follows. ($n$ has no common divisor with all integers less than it). Now assume $n$ is even.\n$i)$ $a_{2}=3$ , in this case n is pairwisely coprime with all odd integers less than itself, so that it should be a power of $2$.\n[/quote]\n\nThe fact that a_2 = 3 doesn\u00b4t imply that it is comprime with all odd integers less than itself... Take 20 for example:\na_2=3, but it is not coprime with 5[/quote]\n\nActually, if $a_{2}=3$, then $a_{3}=5$, $a_{4}=7$, .. $a_{k}=n-1$ (remember $n$ is even). So, n is pairwisely coprime with all odd integers less than itself ;) \n\nBy the way, $20$ does not satisfy the conditions of the problem, so that it's wrong ;)[/quote]\n\nOh, sorry about that, I thought you meant that if ANY integer n is coprime with 3, n was coprime with all pairwise integers less than itslef.. I actually used a similar approach as you in my solution. Well thought solution that of yours!", "Solution_12": "Let $p$ be the smallest prime number such that $p$ doesn't divide $n$.It is trivial that $a_1=1$ , $a_2=p$ , $a_k=n-1$ so all the equalities are equal to $p-1$.Now for cases $p=2$ and $p=3$ we trivialy obtain that prime $n$ and every $n=2^k$ are solutions.Now we consider the case $p>3$.By summing the equalities we get $(p-1)(k-1)=n-2$. Now because $p>3$ we have $3|n$ so $3$ doesn't divide $n-2$ so we obtain $p \\equiv 2 $ modulo $3$.That means $p-1 \\equiv 1$ which immediately implies $3|a_3$ which is a contradiction so we are finished.", "Solution_13": "If $n$ is odd then $a_1 = 1, a_2 = 2 \\rightarrow a_k = k$ for all $k$, so if $n$ is not prime then $n = pq$ for some $p, q > 1$, which means $p$ should not be the value of $a_i$; but it is, contradiction.\n\nSuppose $n$ is even. If $n$ is not divisible by $3$ but has an odd factor greater than 1, then $a_1 = 1, a_2 = 3 \\rightarrow a_k = 2k - 1$ and we are done in a similar manner to above. Thus assume $n$ is divisible by $3$. Because $n > 6$, it is clear that $\\phi(n) > 2$ (by considering the prime factorization of the largest odd factor of $n$), so there exists $p \\neq 1, n-1$ relatively prime to $n$. Thus $a_2 \\neq n-1$. Because if $p$ is relatively prime to $n$ then $n - p$ is also, we have $a_2 < \\frac{n}{2}$. Now, because $1$ and $n-1$ are values of some $a_k$, the common difference, $p-1$, must divide $(n-1) - 1 = n-2$, so since $3 | n$, we have $3$ relatively prime to $n-2$ and thus $p-1$. Therefore, $a_1, a_2, a_3$ form a complete set of residues modulo $3$, which means one of these numbers is not relatively prime to $3$ and thus $n$, contradiction. Thus $n$ is a power of $2$ if $n$ is even, completing the proof.", "Solution_14": "I'll use Bonse's Inequality: $\\[p_{m+1}^2 < p_1p_2\\cdots p_m\\]$, for all $m \\ge 4$, $p_1 = 2$. \\\\\n\nLet $p$ denote the smallest prime which does not divide $n$. \\\\\n\nCase 1) $p = 2$. This implies $n$ is a prime.\nCase 2) $p = 3$. This implies $n = 2^k$ for some $k \\in \\mathbb{N}$. \\\\\nCase 3) $p = 5$. Since $n > 6$, we have that $n > 9$. Therefore $a_3 = 9$, but $\\gcd (n, 9) > 1$.\nCase 4) $p \\ge 7$. Write $n = \\[p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_k^{\\alpha_k}\\]$. We have $(p - 2)^2 < p_1p_2\\cdots p_k \\le n = \\[p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_k^{\\alpha_k}\\]$. Hence $(p-2)^2 = a_i$, a contradiction since $\\gcd(p-2, n) > 1$. \n\n\n ", "Solution_15": "I think I have carefully avoided advanced NT, so here is my solution:\n\nClearly, $a_1 = 1$. Now let $p$ be the smallest prime number which is not a divisor of $n$. Hence, $a_2 = p$, and the common difference $d = p-1$. Observe that since $a_k = n-1$, we have that \\[ d \\mid a_k - a_1 \\], or \\[ p - 1 \\mid a_k - a_1 = n-2 \\]. This means that all prime factors of $p-1$ are prime factors of $n-2$. However, due to the minimality of $p$, all primes factors of $p-1$ are prime factors of $n$. By the Euclidean algorithm, the only prime factors of $p-1$ are $2$, so $p = 2^{2^k} + 1$ or $p = 2$, by the theory of Fermat primes. We will now do casework on $p$.\n\nCase 1: $p = 2$\nThis case is relatively easy. $a_2 = 2$, so all numbers less than $n$ must be relatively prime to $n$. However, if $n = pq$, where $p > 1$ and $q > 1$, observe that $p$ is not relatively prime to $n$, so $n$ must be prime for this case to work.\n\nCase 2: $k = 0$ and $p = 3$.\nThis case is somewhat tricky. Observe that $n$ is even. Since $a_2 = 3$, we have that all numbers less than $n$ which are odd are relatively prime to $n$. However, if $n$ has a odd divisor $p > 1$, clearly $p$ is not relatively prime to $n$, so $n$ must be a power of two.\n\nCase 3: $k \\ge 1$.\nNote that $3$ must be a divisor of $n$, so $a_k$ cannot be a multiple of $3$ for any $k$. I will contradict this statement. First, I claim that $k = \\phi(n) > 2$ for all integers $n > 6$. $\\phi{n} = 1$ is trivial, so let me show that $\\phi(n) = 2$ has only solutions less than or equal to $6$. Since $\\phi$ is a multiplicative function, observe that $n$ cannot be a multiple of a prime greater than $3$, because for such primes $p$ we have that $\\phi(p^k) = (p-1)p^{k-1} \\ge p-1 > 2$, which is contradiction. Hence, $n$ is just composed of factors of $2$ and $3$. If it divides both, and is of the form $2^m 3^n$ for $m$ and $n$ positive, we get that we need $2^{m} 3^{n-1} = 2$, so $n = 1$ and $m = 1$ in this case. If it is of the form $n = 2^k$, we get $k = 2$. And if it is of the form $3^k$, we get $k = 1$. $3$, $4$, and $6$ are all less than or equal to $6$, so $\\phi(n) > 2$. Now we're almost done. Since $k > 2$, the number $a_3 = 2p-1$ must be in this sequence. However, $a_3 = 2p-1$ is of the form $2^{2^n+1} + 1$, which is a multiple of $3$, and we have contradiction, so we are done.", "Solution_16": "Here is my solution.\n\nClearly that $a_1=1$. Let $a_{i+1}-a_i = d \\; (1 \\le i \\le \\varphi(n)-1)$ then $$a_{i}=(i-1)d+1 \\; \\; (1 \\le i \\le \\varphi(n)). \\qquad (1)$$\nWe can see that if $n$ is a prime then $n$ satisfies the condition. We consider the case when $n$ is a composite number. Let $p$ be the smallest prime divisor of $n$ then $p \\le \\sqrt{n}$. Note that $\\varphi (n) > \\sqrt{n} \\ge p$ for all prime $n>6$ since $p_i^{k_i}-p_i^{k_i-1} \\ge p_i^{k_i/2}$ for all $p \\ge 3$. Therefore, if $p \\nmid d$ then there will exists $1 \\le i \\le \\varphi(n)$ such that $(i-1)d \\equiv -1 \\pmod{p}$ or $p \\mid a_i$, a contradiction since $\\gcd (a_i,n)=1$. Thus, $p \\mid d$. We consider two cases:\n\nIf $d \\ge 2p$ then $p+1 \\ne a_i \\; (1 \\le i \\le \\varphi(n))$. Hence, $\\gcd (p+1,n)>1$. This means there exists a prime $q>p \\ge 2$ such that $q \\mid p+1$. Since $q>p$ so we get $q=p+1$ or $p=2,q=3$, a contradiction since $n>6$. Thus, $d <2p$ or $d=p$.\n\nIf $p=d=2$ then all odd numbers $a_i \\le n$ are relatively prime to $n$. This can only happen when $n=2^x \\; (x \\in \\mathbb{N})$.\n\nIf $p=d >3$ then $p+3 \\ne a_i \\; \\forall 1 \\le i \\le \\varphi(n)$ since $a_i \\equiv 1 \\pmod{p} \\; \\forall 1 \\le i \\le \\varphi(n)$. Thus, $\\gcd (p+3,n)>1$ or there exist a prime $q>p>3$ such that $q \\mid p+3$. Note that $2 \\mid p+3$ so $q \\le \\tfrac{p+3}{2} 6\\,$ be an integer and $ \\,a_{1},a_{2},\\cdots ,a_{k}\\,$ $(k\\geq5)$ be all the natural numbers less than $ n$ and relatively prime to $ n$ . If $k$ is even ,and $a_{2} \\minus{} a_{1} \\equal{} a_{4} \\minus{} a_{3} \\equal{}a_{6} \\minus{} a_{5} \\cdots \\equal{} a_{k} \\minus{} a_{k \\minus{} 1} ,$ find all such positive integers $n$.", "Solution_19": "[quote=sqing][url=http://www.artofproblemsolving.com/community/c6h1343797p7308398]China Mathematical Olympiad 2017 Q5[/url][/quote]\n\nHi sqing, what is your reason behind linking this topic to Q5 China MO 2017 ? I don't see any connection between these two problems. ", "Solution_20": "[quote=Rewritten Problem Statement]Let $n\\in\\mathbb Z_{\\geq 2}$ be an integer and $ \\,a_{1},a_{2},\\cdots ,a_{k}\\,$ be all the natural numbers less than $ n$ and relatively prime to $ n$. If\n\\[ a_{2} \\minus{} 1 \\equal{} a_{3} \\minus{} a_{2} \\equal{} \\cdots \\equal{} \\left(n-1\\right) \\minus{} a_{\\varphi(n) \\minus{} 1}=\\Delta> 0,\n\\]\nprove that $ \\,n\\,$ must be either a prime number or a power of $ \\,2$.[/quote]\nIt is clear that we have $\\Delta\\mid n-2$. Hence $\\gcd(n,\\Delta)\\in\\{1,2\\}$. Let $n=2^{\\alpha}p^{\\beta}\\prod q_i^{\\gamma_i}$ where $p$ is the lowest odd prime divisor of $n$. If $p$ does not exist then we necessarily have $n=2^{\\alpha}$, a case that is easily checked to give $\\Delta=1$. If $n\\not \\in\\left\\{p,2p\\right\\}$ then we have $\\varphi(n)\\geq p$, meaning that our arithmetic sequence contains a multiple of $p$, a clear contradiction, since $p\\nmid \\Delta$. The case $n=p$ is also easily checked, giving $\\Delta=2$. For $n=2p$, if $p\\geq 5$ we have $\\Delta=2$ by considering $a_1=1,a_2=3$, but then $p$ is on the arithmetic sequence which is ludicrous. Note that $n=6$ also works. We conclude that the solutions are $\\boxed{n\\in\\left\\{p,6,2^{\\alpha}\\right\\}}$", "Solution_21": "[quote=shinichiman][quote=sqing][url=http://www.artofproblemsolving.com/community/c6h1343797p7308398]China Mathematical Olympiad 2017 Q5[/url][/quote]\n\nHi sqing, what is your reason behind linking this topic to Q5 China MO 2017 ? I don't see any connection between these two problems.[/quote]\nI judged wrong .\nSorry .", "Solution_22": "Let $p$ be the smallest prime that does not divide $n$. Let $p \\ge 5$ (the other cases were treated in the other solutions). $a_{1}=1$, $a_{2}=p$, so $a_{k}-a_{k-1}=p-1$. We easily get by induction that $a_{i}=(i-1)p - i + 2$, so $a_{p-2}=p(p-3) - p+ 4=(p-2)^2$, so $(n,p-2)=1$. This contradicts the minimality of $p$.", "Solution_23": "[quote=pavel kozlov]There is simple solution without using Bertrand's Postulate...\n\nWe have a[1]=1, a[2]=p, where p is the least prime, which is not a divisor of n; a[k]=n-1, r=p-1 is the difference of the progression a[i].\n If n is odd, then a[2]=2 :arrow: 1, 2, ..., n-1 is our progression :arrow: n is a prime.\n If i is even, then p>2.\n 1. If p=3, then 1, 3, ..., n-1 is our progression :arrow: n is a power of 2.\n 2. If p>3, then n is divisible by 3. we have n-1=a[k]=1+(p-1)(k-1) :arrow: p-1 is divisor of n-2. :idea: Let q be a random prime divisor of p-1. q | p-1 | n-2, but q|n, because q

3, i is even.\n But a[3]=2p-1=2^(i+1) + 1 is divisible by 3, but n is divisible by 3 too :arrow: gcd(n, a[3])>1. This contradiction ends our proof. :lol:[/quote]\n\nThis is exactly what I did. ", "Solution_24": "Let the smallest prime not dividing $n$ be $p$. Then the sequence has to be $1, p, 2p-1, 3p-2, 4p-3, \\dots, n-1$.\n\nSince the common difference in this arithmetic sequence is $p-1$, we have that $p-1$ must divide $(n-1)-1=n-2$. But by the assumption of $p$ being the smallest prime, we also have that $p-1$ divides $n$. Therefore, $p-1$ divides $n-(n-2)=2$.\n\nSo $p$ is either $2$ or $3$. If $p$ is $2$, then the sequence is $1, 2, 3, \\dots. n-1$, so $n$ is prime. If $p$ is $3$, then the sequence is $1,3,5,\\dots n-1$ so $n$ is relatively prime to all odd numbers less than it and must be a power of $2$.", "Solution_25": "I solved the problem with the condition that $n\\in \\mathbb{N}.$\n\nObviously $n=1$ is a solution$,$ as well as $n=2, 3, 4,$ so assume that $n\\geq 5.$ Let the common difference be $d.$ The condition implies $$a_{\\varphi(n)}=a_1+d(\\varphi(n)-1)\\iff n-2=(\\varphi(n)-1)d\\hspace{1cm}(*)$$ as it's easy to notice that $a_{\\varphi(n)}=n-1.$\n\nIf $d=1,$ we get that $n$ is prime$,$ which is easily seen to work$.$\nIf $d=2,$ we get that $n$ is a power of $2,$ as $$2=\\frac{n}{\\varphi(n)}=\\prod_{p\\mid n}\\bigg(\\frac{p}{p-1}\\bigg)\\geq 2$$ with equality only when $\\text{rad}(n)=2.$ It is again not that hard to verify that all such $n$ work$.$\nSuppose $d\\geq 3$ and notice that $\\varphi(n)$ is even for $n>2,$ so if $d$ is odd$,$ by equation $(*)$ we obtain $n$ must also be odd$,$ but then $\\gcd(2, n)=1,$ meaning $a_2=2$ and $d=1,$ contradiction$.$ Therefore $d\\geq 4$ and $n$ is even$.$\nLet $p$ be the smallest prime not dividing $n.$ Now all the numbers $2, \\cdots , p-1$ are not relatively prime to $n$ and so $a_2=p \\Rightarrow d=p-1.$\n\nSuppose that $r$ is a prime factor of $p-1.$ Because $p$ is minimal$,$ $r$ must divide $n.$ By equation $(*)$ we see that $r\\mid 2,$ therefore $p-1=2^k$ for some $k\\geq 2$ (recall that $4\\leq d=p-1$)$.$ But since $p$ is a prime$,$ $k$ cannot have any odd factor$,$ from which we get $k=2^t$ and $p=1+2^{2^t}.$\n\nFinally$,$ suppose $\\varphi(n)\\geq 3$ so that $a_3=2p-1$ exists$.$ Note that $p\\equiv 2\\pmod 3 ,$ meaning $a_3\\equiv 0\\pmod 3$ and as $p>3,$ we see that $3$ also divides $n,$ a clear contradiction$.$ This means that $\\varphi(n)\\leq 2$ and the only possibility is $n=6,$ which is also easily verified$.$", "Solution_26": "Firstly $n \\text{ odd} \\iff n \\text{ prime }$:\n$2 \\nmid n \\implies a_2=2 \\implies a_i \\text{ are } 1, 2, 3, \\ldots \\implies n \\text{ prime}$\nSo, let $n$ even:\nLet $p$ be the smallest prime with $p \\nmid n$\nCase 1: $p=3$\n$\\implies a_i \\text{ are } 1, 3, 5, \\ldots \\implies n \\text{ is a power of 2 }$\nCase 2: $p>3 \\implies 3 \\mid n$\n$\\implies$ none of $m \\in \\{ 2, 3, \\ldots, p-1 \\}$ can be an $a_i$ else: $\\gcd(m, n) = 1 \\implies$\nfor any prime $q \\mid m \\text{ we must have } q \\nmid n \\text{ but } q \\leq m 6$. (The only case where it is violated is where $p_1=3$ so $n = 6$). Therefore the common difference of the AP at maximum is $2$, so it would cover all odd primes in the factorisation, a contradiction. The only case where it does not, is when there is no odd prime in the prime factorisation ! This means that $n$ is a power of $2$ which is easily seen to work $\\blacksquare$\n", "Solution_28": "Does this work?\n\nThe condition says that the positive integers less than and relatively prime to $n$ form an arithmetic sequence.\n\nCase 1: $n$ is odd. Then $1$ and $2$ are relatively prime to $n$, so the common difference of the arithmetic sequence is $1$. But since $n-1$ is in the sequence, this means all of the positive integers less than $n$ are relatively prime to $n$, implying $n$ is prime.\n\nCase 2: $4 | n$. Then $\\frac{n}{2}-1$ and $\\frac{n}{2}+1$ are relatively prime to $n$, so the common difference of the arithmetic sequence is $2$. But since $n-1$ is in the sequence, this means all the odd positive integers less than $n$ are relatively prime to $n$, so since $n$ is even $n$ must be a power of $2$.\n\nCase 3: $2 || n$. Then $\\frac{n}{2}-2$ and $\\frac{n}{2}+2$ are relatively prime to $n$, so the common difference of the arithmetic sequence is $4$. Since the sequence contains $1$ and $n-1$, it contains all the positive integers less $n$ which are $1$ mod $4$. In particular, $3 | n$. But if $n = 9$, $7$ is not in the sequence but relatively prime to $n$, a contradiction. If $n > 9$, $9$ is in the sequence but not relatively prime, also a contradiction.", "Solution_29": "[hide=solution(MONT)][quote=IMO 1991#2]Let $ \\,n > 6\\,$ be an integer and $ \\,a_{1},a_{2},\\cdots ,a_{k}\\,$ be all the natural numbers less than $ n$ and relatively prime to $ n$. If\n\\[ a_{2} \\minus{} a_{1} \\equal{} a_{3} \\minus{} a_{2} \\equal{} \\cdots \\equal{} a_{k} \\minus{} a_{k \\minus{} 1} > 0,\n\\]\nprove that $ \\,n\\,$ must be either a prime number or a power of $ \\,2$.[/quote]\nClearly $a_1=1,a_k=n-1$.\nNow the question asks for an AP of coprime integers.\nSay the common difference be $d$.\n$1+(k-1)d=n-1 \\implies d=\\frac{n-2}{k-1}$.\nNow say $p$ be a prime factor of $n$ and assume $pk,k=p-1$ or $n$ is a prime.[/hide]", "Solution_30": "My soln not seen above solutions yet when n is a prime it is trivial, now let n be prime power then it's easy to see n is power of 2, So now let n be non prime power composite number , then let it,s smallest divisor greater than 1 be p , then 1,2...p-1 are co prime to n so common difference would be 1,we can easily check p =2 gives the previous soln \n[color=#f00]Claim 1:[/color] : No such n is there \nProof : $2-1...= a_p-(p-1) $\nBut this mean a_p is p and is coprime to n CONTRADICTION!!!\nHence the claim and result.", "Solution_31": "It is easy to check that primes and powers of $2$ satisfy the condition. Also, the condition readily gives $a_16$, $n-1=a_k\\neq a_2=2$, and to keep the condition satisfied we must have $\\gcd(n,k)=1$ for all $1\\le k1$ coprime to $n$ smaller then $p$ then $q$ would have a prime factor $r$ smaller than $p$, which must in turn be also relatively prime to $n$, which is non-sense. Hence, in order for the condition to be fulfilled, we must have \n\t$$a_1=1,a_2=p,a_3=2p-1,a_4=3p-2,\\ldots,a_k=n-1$$\n\tHence, $p-1|(n-1)-1$. But clearly $p-1|n$, so $p-1|2$, i.e. $p=2$ or $p=3$. In these cases, we get that $n$ must be prime or a power of $2$, respectively, so we are done.", "Solution_32": "Really good problem!\nWe do casework on $a_2$. If $a_2=2,3$, we can easily obtain $n$ is a prime or a power of two, respectively. Now let $a_2=p>3$, so that we obtain $a_3=2p-1$ (the difference is apparently $p-1$).\nNote that by summing, $(k-1)(p-1)=n-2$. Now we finish by modulo $3$: if $p \\equiv 2 \\pmod 3$, then $3|2p-1$, but $3|n$, absurd; otherwise $3|n-2$, but we have again $3|n$ due to the fact that $a_2>3$, contradiction.", "Solution_33": "Note that there are $\\phi(n)$ terms, with the first term being $1$ and the last term being $n-1$, so the common difference is $$d=\\frac{n-2}{\\phi(n)-1}.$$ Now, consider any prime $p$ dividing $n$.\\\\\n\nIf $p>2$, then $d$ is relatively prime to $p$ since $p$ does not divide $n-2$ if it divides $n$. However, if $d$ is relatively prime to $p$, there exists some $0\\leq s\\leq p-1$ for which $$1+ds\\equiv 0\\pmod{p}.$$ However, since we need the entire arithmetic progression to be relatively prime to $n$ (and thus to $p$), we must have $$\\phi(n)-1\\leq p-2\\rightarrow \\phi(n)\\leq p-1.$$ Since $\\phi(p)$ is already $p-1$, and $\\phi$ is multiplicative, we also have $\\phi(n)\\geq p-1$, with equality at $n=p$ and $n=2p.$ We can rule out $n=2p$ since then $$d=\\frac{2p-2}{p-2}=2+\\frac{2}{p-2},$$ which is only an integer for $p=3$ but we require $n>6$, so this case is resolved.\\\\\n\nOtherwise, there is no prime $p>2$ dividing $n$, so $n$ is a power of 2 and we are done.", "Solution_34": "$a_2-a_1=a_3-a_2=...a_k-a_{k-1}>0 \\implies a_13$. Now take the least prime $p_i$, from the canonical factorization of $n$, such that $p_i$ and $p_{i+1}$ has at least one prime between them. The least of these 'in-between' primes $=p_c$. (If theres no prime between any $p_i$ and $p_{i+1}$ for $1 \\leq i \\leq k-1$ then the same argument we give below holds because then there would be no primebetween $p_k$ and $p_c$.)Thus the prime factors of $p_c-1$ will also be the factor of $n$. Now if any prime besides $2$ divides $p_c-1$ we get a contradiction in $(i) \\implies p_c-1=2^z$. Now as $z$ has to be odd since otherwise $3|2^z-1=p_c$. But then$3|a_3=2^{z+1}-1$ which is a contradiction since $(3,n)=3$. So we are done.\nThe existence of $a_3$ is assured by the fact that for all $n>6, k \\geq4$ by the formula for eulers totient function.\n", "Solution_35": "Forgot that I did this problem a year ago already so here is another solution (which I like but which is more complicated):\n\nIf $n$ is odd then $a_{i+1}-a_i=1$, so in fact $a_i=i$ as $a_1=1$. Thus $n$ is relatively prime to all numbers smaller than $n$, so $n$ is prime. Otherwise $n$ is even. If now assume that $n$ has at least one odd prime divisor, and let $q$ be the smallest. Also, let the smallest odd prime not dividing $n$ be $p$, so that $a_1=1, a_2=p, a_3=2p-1, a_4=3p-2,\\ldots$ so that in general, $a_i=(i-1)p-(i-2)\\iff a_i=1+(i-1)(p-1)$. \n\t\n\tThe idea is now to seek a contradiction by showing that some $a_i$ is $0\\mod q$ at least once. Consider $i-1\\in I=\\{0,1,\\ldots,q-1\\}$. We now notice that in total, there are $\\phi(n)$ numbers $a_k$ relatively prime to $n$, but as $\\phi(n)\\ge\\sqrt{n}\\ge q$ for $n>6$ the, index set $I$ delivers $a_i$ that are all relatively prime to $n$. However note that as $\\gcd(p-1,q)=1$, the set $(p-1)I+1$ is still a complete residue system modulo $q$, so there exists one $i\\in I$ such that $a_i\\equiv 0\\mod q$, a contradiction." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let a,b,c>0 such that a+b+c=1 Prove that\r\n$ 1\\leq\\sum\\frac{(a\\plus{}b)b}{c\\plus{}a}$", "Solution_1": "by AM_GM is very easy . Good lucky!", "Solution_2": "[quote=\"tranquoc\"]Let $ a,b,c>0$ such that $ a\\plus{}b\\plus{}c\\equal{}1$ Prove that\n$ 1\\leq\\sum\\frac {(a \\plus{} b)b}{c \\plus{} a}$[/quote]\r\n$ 1\\leq\\sum_{cyc}\\frac {(a \\plus{} b)b}{c \\plus{} a}\\Leftrightarrow\\sum_{cyc}(a^4\\minus{}a^2b^2\\plus{}a^3c\\minus{}a^2bc)\\geq0,$ which obviously true.", "Solution_3": "but I By AM_GM ---> OK easy", "Solution_4": "this solution isn't nice.", "Solution_5": "$ \\sum\\frac{b(a\\plus{}b)}{c\\plus{}a} \\geq \\sum a$\r\n<-> $ \\sum\\frac{a(a\\plus{}b\\plus{}c)\\plus{}b(a\\plus{}b\\plus{}c\\minus{}a\\minus{}c)}{c\\plus{}a} \\geq 2\\sum a$\r\n<->$ \\sum\\frac{a\\plus{}b}{a\\plus{}c} \\geq 3$\r\ndone :rotfl:" } { "Tag": [], "Problem": "propozitia matematica \"80 este mai mare sau egal cu 70\" este adevarata?", "Solution_1": "Da, este adevarata", "Solution_2": ":D $ 80 \\ge 70$ este adevarat din cate stiu eu" } { "Tag": [ "projective geometry", "geometry proposed", "geometry" ], "Problem": "Let ABC be a triangle. The A-excircle (Ia) touches BC at D. AD meets the excircle (Ia) at D and K. E,F are on AK such that CF=CD, BE=BD. Show that F lies on the polar of E with respect to the excircle (Ia).", "Solution_1": "It is enough to prove that the point $ F$ is the harmonic conjugate of $ E,$ with respect to the points $ D,\\ K.$\r\n\r\nFrom the similar isosceles triangles $ \\bigtriangleup BDE$ and $ \\bigtriangleup CDF,$ we have that $ \\frac {ED}{FD} \\equal{} \\frac {BE}{CF}$ $ ,(1)$\r\n\r\nLet be the point $ L\\equiv BE\\cap CK$ and based on the below [b][size=100]Lemma[/size][/b], we conclude that $ BE \\equal{} EL$ $ ,(2)$\r\n\r\nSo, because of $ (2)$ and $ CF\\parallel EL,$ we have that $ \\frac {EL}{CF} \\equal{} \\frac {EK}{FK}$ $ \\Longrightarrow$ $ \\frac {BE}{CF} \\equal{} \\frac {EK}{FK}$ $ ,(3)$\r\n\r\nFrom $ (1),$ $ (2)$ $ \\Longrightarrow$ $ \\frac {ED}{FD} \\equal{} \\frac {EK}{FK}$ $ \\Longrightarrow$ $ \\frac {ED}{EK} \\equal{} \\frac {FD}{FK}$ $ ,(4)$\r\n\r\nHence, from $ (4)$ we conclude that the point $ F,$ as the harmonic conjugate of $ E,$ with respect to the points $ D,\\ K,$ lies on the polar of $ E$ with respect to the $ A$-excircle $ (I_{a})$ of $ \\bigtriangleup ABC$ and the proof is completed.\r\n\r\n[b][size=100][color=DarkBlue]LEMMA. \u2013 A triangle $ \\bigtriangleup ABC$ is given and let $ D$ be, the tangency point of the $ A$-excircle $ (I_{a}),$ to its side-segments $ BC.$The line segment $ AD$ intersects $ (I_{a})$ at one point so be it $ K$ and let be the point $ L\\equiv BE\\cap CK,$ where $ E$ is the point on $ AD,$ such that $ BE \\equal{} BD.$ Prove that $ BE \\equal{} EL.$[/color][/size][/b]\r\n\r\nKostas Vittas.\r\n\r\nPS. I will post here later the proof of the above [b][size=100]Lemma[/size][/b], I have in mind.", "Solution_2": "[quote=\"vittasko\"]\n[b][size=100][color=DarkBlue]LEMMA. \u2013 A triangle $ \\bigtriangleup ABC$ is given and let $ D$ be, the tangency point of the $ A$-excircle $ (I_{a}),$ to its side-segments $ BC.$The line segment $ AD$ intersects $ (I_{a})$ at one point so be it $ K$ and let be the point $ L\\equiv BE\\cap CK,$ where $ E$ is the point on $ AD,$ such that $ BE \\equal{} BD.$ Prove that $ BE \\equal{} EL.$[/color][/size][/b][/quote]\r\n[b][size=100]PROOF.[/size][/b] - Let $ M,\\ N$ be, the tangency points of $ (I_{a})$ to the sidelines $ AB,\\ AC$ respectively and we denote the point $ Q\\equiv BC\\cap MN.$\r\n\r\nBased on the [b][size=100]Ceva theorem[/size][/b], we conclude that $ P\\equiv AD\\cap BN\\cap CM$ and so, fron the complete quadrilateral $ BCNMAQ,$\r\nwe have that the points $ B,\\ D,\\ C,\\ Q,$ are in harmonic conjugation.\r\n\r\nHence, the pencil $ K.BDCQ,$ is also in harmonic conjugation and so, it is enough to prove that $ BL\\parallel KQ.$\r\n\r\n$ \\bullet$ The line segment $ MN,$ is the polar of $ Q$ with respect to $ (I_{a})$ and because it passes through the point $ Q,$\r\nwe conclude that the polar of $ Q$ wrt $ (I_{a}),$ passes through the point $ A.$\r\n\r\nSo, the line segment $ AD,$ is the polar of $ Q$ with respect to $ (I_{a}),$ because it connects the point $ A$ with the point $ D,$\r\nas the tangency point of $ (I_{a})$ with the line through the point $ Q.$\r\n\r\n\r\nHence, the line segment $ QK,$ tangents to $ (I_{a})$ at point $ K$ and then, we have $ I_{a}K\\perp KQ$ and it is enough to prove that $ I_{a}K\\perp BL.$\r\n\r\n$ \\bullet$ Let be the point $ T\\equiv BL\\cap I_{a}K$ and it is easy to prove that the quadrilateral $ BFKT,$ where $ F$ is the orthogonal projection of $ B$ on $ AD,$\r\n\r\nis cyclic, because of $ \\angle DKT \\equal{} \\angle KDI_{a} \\equal{} 90^{o} \\minus{} \\angle KDC \\equal{} 90^{o} \\minus{} \\angle FDB \\equal{} \\angle FBD \\equal{} \\angle FBE$ $ ,(1)$\r\n\r\nSo, we conclude that $ \\angle BTK \\equal{} \\angle BFK \\equal{} 90^{o}$ $ \\Longrightarrow$ $ BL\\perp I_{a}K$ $ ,(2)$\r\n\r\nFrom $ (2),$ we conclude that $ BL\\parallel KQ$ $ ,(3)$\r\n\r\nFrom $ (3)$ and because of the pencil $ K.BDCQ$ is in harmonic conjugation, we conclude that $ BE \\equal{} EL$ and the proof is completed.\r\n\r\nKostas Vittas.", "Solution_3": "The A-excircle $ (I_a)$ touches $ AB, AC$ at $ U, V.$ $ (B)$ is a circle with center $ B$ and radius $ BD \\equal{} BE.$ $ (C)$ is a circle with radius $ CD \\equal{} CF.$ $ (B), (C)$ are externally tangent at $ D,$ $ I_aD$ is their single common internal tangent. Points $ A, D, E$ are collinear, their polars $ UV, BC, e$ WRT $ (I_a)$ are concurrent at $ P.$ By Menelaus theorem for the $ \\triangle ABC$ cut by the line $ PUV,$ $ \\frac {\\overline{PB}}{\\overline{PC}} \\equal{} \\frac {\\overline{UB}}{\\overline{UA}} \\cdot \\frac {\\overline{VA}}{\\overline{VC}} \\equal{} \\frac {\\overline{UB}}{\\overline{VC}} \\equal{} \\minus{}\\frac {\\overline{DB}}{\\overline{DC}}$ $ \\Longrightarrow$ $ P$ is the external similarity center of the circles $ (B), (C).$ Line $ I_aE$ cuts $ (B)$ again at $ X.$ Power of $ I_a$ to $ (B)$ is $ I_aD^2 \\equal{} I_aE \\cdot I_aX$ $ \\Longrightarrow$ $ e \\equiv PX \\perp I_aX$ is the polar of $ E$ WRT $ (I_a),$ cutting $ (B)$ again at a point $ Y$ diametrically opposite point to $ E.$ From similarity of the isosceles $ \\triangle DBE \\sim \\triangle DCF,$ $ BY \\equiv EB \\parallel CF$ $ \\Longrightarrow$ $ FY$ goes through the external similarity center $ P$ of $ (B), (C),$ which means $ F \\in PY \\equiv PX \\equiv e.$" } { "Tag": [ "calculus", "integration" ], "Problem": "A 7.00 kg object starting from rest falls through a viscous medium and experiences a resistive force = -b , where is the velocity of the object. The object reaches one half its terminal speed in 4.94 s. \r\n\r\nHow far has the object traveled in the first 4.94 s of motion?", "Solution_1": "[quote=\"kl2836\"]...a resistive force = -b , where is the velocity of the object.[/quote]\r\nI think you're missing something. From Newton's second law, $ \\sum\\mathbf{F}\\equal{}m\\mathbf{a}$. It looks like the two forces are $ \\mathbf{g}$ and the viscous force, which at the moment, I can't quantify.", "Solution_2": "if acceleration is time varying, and i want to find the distance s= (1/2)at^2, \r\n\r\nhow do i write the integral form of s", "Solution_3": "it travelled about 97.74 meters", "Solution_4": "The resistive force seems to be = -bv.\r\nIf so, castigioni's answer is OK, but to came to it you can't put $ \\;s \\equal{} (1/2)at^2\\;$ since (you alone said it) [b]acceleration is time varying[/b].\r\n\r\nDo you manage differential equation?", "Solution_5": "yes, you need a differential equation\r\n\r\n[hide]\nso you make the equation $ ma \\equal{} mg \\minus{} bv$\nset $ a \\equal{} \\frac {dv}{dt}$\n\n$ m\\frac {dv}{dt} \\equal{} mg \\minus{} bv$\nsolve...\n\nget to $ m*\\frac {dv}{mg \\minus{} bv} \\equal{} dt$\nintegrate $ \\minus{} \\frac {1}{b}ln(mg \\minus{} bv)*m \\equal{} t$\nthen, solve and get $ v \\equal{} \\frac {m(g \\minus{} e^{\\frac { \\minus{} bt}{m}})}{b}$\nand then, you can integrate that from 0 to 4.94\n\n[/hide]\r\n\r\nwondering if there's any other way?\r\n\r\nnote: $ \\int{(s)}dt \\equal{} v$\r\n\r\nif you want me to, i can go into a little more detail", "Solution_6": "There may be another way, but I can't think of one at the moment. The presence of the drag force $ bv(t)$ sets this up as a nice equation to use integrating factors on." } { "Tag": [ "calculus", "integration", "logarithms", "complex analysis", "complex analysis unsolved" ], "Problem": "Hi, could anyone help me on these integrals please?\r\n\r\n1. Verify the following:\r\n$ \\int_{0}^{\\infty}\\frac{\\log{x}}{(1\\plus{}x^2)^2}dx\\equal{}\\minus{}\\frac{\\pi}{4}$\r\n\r\n2. Compute\r\n$ I\\equal{}P.V. \\int_{\\minus{}\\infty}^{\\infty}\\frac{xe^{i{\\lambda}x}}{x^2\\minus{}b^2}dx$\r\nwhere $ \\lambda>0,b>0$.\r\n\r\nThanks in advance,", "Solution_1": "For the first one, note that the numerator is a multi-valued function. So first, specify a branch cut. The one running along the positive real axis is a good choice (why?). Call the integral we want $ I$. When we cross the branch, we get $ \\minus{}2\\pi i I$ (why?). The poles occur at $ x\\equal{}\\pm i$, each with multiplicity 2. So we need the residues. After that,\r\n$ \\oint_{\\Gamma}f(z)\\ dz\\equal{}(1\\minus{}2\\pi i)I\\equal{}2\\pi i(\\text{Res}(i)\\plus{}\\text{Res}(\\minus{}i))\\Rightarrow I\\equal{}\\frac{2\\pi i(\\text{Res}(i)\\plus{}\\text{Res}(\\minus{}i))}{1\\minus{}2\\pi i}$.\r\nNow you need to calculate the residues, which I'll leave to you as an exercise.", "Solution_2": "Thanks JRav", "Solution_3": "No problem. Now the second problem seems a bit simpler. First note there are two poles: $ x\\equal{}\\pm b$ so we know where we need to take residues. Construct a contour that runs along the real axis, with indentations around the two poles, since they both lie on the real axis. Close the contour with an upper-semicircle. Now we constructed our contour so that there are no poles inside so\r\n$ \\oint_{\\Gamma}f(z)\\ dz\\equal{}0\\equal{}I\\plus{}\\int_{S_{\\delta_{1}^{\\minus{}}}}f(z)\\ dz\\plus{}\\int_{S_{\\delta_{2}^{\\minus{}}}}f(z)\\ dz\\plus{}\\int_{C_{R}^{\\plus{}}}f(z)\\ dz$\r\nThe last integral is the upper-semicircle, which goes to 0 by Jordan's Lemma. The second and third integral are given by $ \\minus{}\\pi i\\text{Res}(f,\\minus{}b)$ and $ \\minus{}\\pi i\\text{Res}(f,b)$, respectively. Now just take the residues and you're done." } { "Tag": [ "email", "LaTeX", "Olimpiada de matematicas" ], "Problem": "Como ya sabran hoy se cumple la segunda semana de venir celebrando el [b]Reto de la Semana[/b].\r\n\r\nEste conciste en una lista de problemas cuyas reglas son:\r\n- No colgar las soluciones y/o comentarios en los respectivos topics durante la semana. Solo comentarlos luego de que se conlcuya el tiempo.\r\n \r\n- Enviar sus soluciones a Email : jibarotimbalero@hotmail.com (manuel) , pascual2k1@hotmail.com (pascual)\r\n\r\n- Duracion: 1 semana\r\n \r\n- Luego de la semana, se publicaran las soluciones mas sobresalientes.\r\n \r\nAsi que la dinamica es simple: Colgamos problemas, los resuelven, nos mandan las brillantes soluciones en cualquier formato (preferiblemente Latex, pero no es requisito), Las revisamos, Colgamos las mas lindas con elk respectivo nombre del autor y finalmente las discutimos.\r\n\r\nPara que esto sea grande deben participar todos y enviar soluciones todos! Piensen que les sirve como un entrenamiento ya que nosotros hacemos las veces de jurado!\r\n\r\nSuerte y espero participen.\r\n\r\nFinalmente una encuesta. Espero sugerencias.\r\n\r\n\r\n[b]que tal le parece la iniciativa? Alguna sugerencia? Aproveche y opine aqui:[/b]", "Solution_1": "hola!! :) Bueno a mi parecer esto que realizan esta muy bien pero hasta ahora no entiendo en que semana se esta...adem\u00e1s el tipo de problemas que proponen en algunos casos son complicados( para mi son un poco dificiles \u00a1\u00a1\u00a1\u00a1pero no imposibles!!! ;) ), no se si puedan insertar sus soluciones oficiales (es decir de los organizadores)eso es mi parecer...\r\nHasta otra oportunidad :P", "Solution_2": "lo que pasa es que estamos como en un receso durante vacaciones y mientras se acababa el a\u00f1o escolar, ya vamos a iniciar otra tempoorada de problemas esta semana. Acerca de las soluciones la idea es que los usuarios compartan las suyas, claro esta que si hay una que no han podido resolver entonces nos toca a los moderadores.", "Solution_3": "hola!!! yo tambien participare en el reto de la semana\r\n\r\n por ahora no tengo ningun pregunta :D ... \r\n pero me parece muy buena la iniciativa\r\n\r\n\r\n fabiola-peru :ninja:" } { "Tag": [ "geometry", "analytic geometry", "trigonometry", "search" ], "Problem": "In triangle ABC, BD is a median. CF intersects BD at E so that BE=ED. Point F is on AB. THen if BF=5, what is BA?", "Solution_1": "I dont think theres enough information.", "Solution_2": "Does anyone know how to do this?\r\n\r\nTHis was an old ASHME problem and the solution was a little sketchy..but taht was the problem, with all the information that I gave..", "Solution_3": "Note that if you stretch or skew the plane in any direction, if (generically) A is the midpoint of BC then A will always be the midpoint of BC. That won't change. There are certain constructions that will not change under stretch and skew transformations.\r\n\r\nNotice that through just stretching and skewing, one can transform any triangle into any other triangle. Since the construction presented in this problem does not change with stretching and skewing, it applies to all triangles.\r\n\r\nYou can therefore pick whichever triangle is most convenient for you.\r\n\r\nA good pick would be a 45-45-90 right triangle with the right angle at B or an equilateral triangle.", "Solution_4": "But I dont see how taht helps..I swear this problem is mistyped or something in my book", "Solution_5": "Using that method\r\n\r\n[hide]Ok, so let's assume for a moment that the triangle is a 45-45-90 triangle formed by $A(4k,0), \\; B(0,0), \\; C(0,4k)$\n\nSince $BD$ is a median, $D\\equiv (2k,2k)$, and the line $\\overline{BD}$ is the line that connects $(0,0)$ and $(2k,2k)$, namely $y=x$\n\n$E$ is the midpoint of $\\overline{BD}$, so $E \\equiv (k,k)$\n\nSince $\\overline{CF}\\equiv \\overline{CE}$, $\\overline{CF}$ is the line passing through $(0,4k)$ and $(k,k)$. This line is clearly $y=4k-3x$, and intercepts $\\overline{AB}$ when $y=0 \\implies 3x=4k \\implies x=\\frac{4}{3}k$. Thus, $F \\equiv \\left( \\frac{4}{3}k , 0 \\right)$ Hence, $BF=\\frac{4}{3}k$ and $BA=4k=3\\frac{4}{3}k = 3 BF = 3(5)=15$[/hide]\r\n\r\nA more natural solution would involve mass points and/or using area ratios. You may consider looking for a good resource on mass points (also called barycentric coordinates) - I'm not the right person to ask for a clear explanation on that. Although there are plenty of other people on this forum that can.", "Solution_6": "[hide=\"solution, there is probably an easier one\"]Using a 45-45-90 triangle with $\\angle B$ as the vertex as TZF suggested, we must first note that since the triangle is isosceles the median from the vertex is the altitude and thus $\\overline{CD}\\perp\\overline{AC}$. Now label $ED$ as $x$. Obviously, $EB=x$ as well. Since the median from the right angle is half the side it meets in a right triangle, $DC=2x$. Because $BDC$ is also a 45-45-90, $BC=2\\sqrt{2}x$. Now, obviously, $\\tan{DCE}=\\frac{1}{2}$, $\\tan{BCF}=\\tan{BCE}=\\frac{5}{2\\sqrt{2}x}$. To combine these, we have that \\begin{eqnarray*}\\tan{BCD}&=&1\\\\ &=&\\tan{(DCE+BCE)}\\\\ &=&\\frac{\\tan{DCE}+\\tan{BCE}}{1-\\tan{DCE}\\tan{BCE}}\\end{eqnarray*}\nI don't feel like typing out all the simplification here to find $x$, so you can do that yourself and then multiply by $2\\sqrt{2}$. I may have messed it up, but the answer I got was $\\boxed{15}$.[/hide]", "Solution_7": "The problem has a unique solution...\r\n\r\nLets use mass points. [use the aops search if you don't know what that is] Let $w_{A}$ denote the weight at $A$. WLOG $w_{a}=1$, since $AM=MC$, $w_{C}=1$ and $w_{M}=1+1=2$. Since $BE=EM$, $w_{B}=2$. Then $w_{A}*AF=w_{B}*BF\\implies AF=10$, so $AB=5+10=\\boxed{15}$", "Solution_8": "[hide=\"umm yeaaa ok\"]\nSince $BE=EC,$ $[BEC]=[EDC]$ and since $AD=DC,$ $[AED]=[EDC]$\n\n$\\frac{BF}{AF}=\\frac{[BEC]}{[AEC]}=\\frac{[BEC]}{[AED]+[EDC]}=\\frac{[BEC]}{2[BEC]}=\\frac{1}{2}$\n\nso\n$AF=10$ and $BA=15$\n[/hide]", "Solution_9": "[hide]Yay mass points!\nPut a mass of $1$ on $A$ and $C$, so there's a mass $2$ on $D$ and thus $B$, so $2*BF=AF \\implies AF=10 \\implies AB=\\boxed{15}$[/hide]", "Solution_10": "[quote=\"Altheman\"]Lets use mass points. [/quote]\r\n\r\nDoes anyone know a good introductory text for mass points? (I know what they are. :P )", "Solution_11": "[quote=\"cincodemayo5590\"][quote=\"Altheman\"]Lets use mass points. [/quote]\n\nDoes anyone know a good introductory text for mass points? (I know what they are. :P )[/quote]\r\n\r\nAn introductory text?\r\n\r\nI don't really think it warrants an entire book, it's just a clever trick derived from Menelaus...\r\n\r\nHere's an explanation, though: [url]http://mathcircle.berkeley.edu/mpgeo/mpgeo.html[/url]" } { "Tag": [ "function", "geometry", "3D geometry", "sphere", "analytic geometry", "integration", "real analysis" ], "Problem": "$f: \\mathbb{R} \\to \\mathbb{R}$ a $C^2$ function .\r\nSuppose there exists a solution $u$ to the equation $(E)$, where $u$ is a function fom the unit ball to $\\mathbb{R}$:\r\n \r\n$(E):$\r\n1/$\\Delta{u}+f(u)=0$ on the open unit ball\r\n2/ $u(x)=0$ on the unit sphere\r\n3/ $u(x) \\geq 0$ everywhere on the unit ball\r\n\r\nDoes it follows that $u(r)$ is decreasing (decreases on every radius) ?", "Solution_1": "Do you mean $f(x)$, a $C^2$ function on the unit ball, or $f(u)$, a $C^2$ function on $\\mathbb R$?", "Solution_2": "sorry, I've edited ...", "Solution_3": "The answer is YES. The problem seems to assume that the solution $u$ depends only on $|x| = r$, i.e. $u(x) = U(r)$. That's correct but nontrivial (afaik).\r\n\r\nIn polar coordinates the problem becomes $U''(r) + \\frac{n-1}{r}U'(r) + f(U(r)) = 0$, with $n$ = space dimension. Also, $U'(0) = 0$ and $U(1) = 0$. It's clear that the solution is in $C^4$. We must show that $U'(r) \\le 0$ for $0 < r < 1$.\r\n\r\nSuppose not, then there exist $0 \\le a < c < b \\le 1$ such that $U(a) = U(b) < U(c)$. We may assume that $U'(a) = 0$, i.e. $U$ has a local minimum at $a$.\r\n\r\nMultiply the equation with $U'(r)$ and integrate from $r=a$ to $r=b$. Then\r\n$\\frac{1}{2}\\left((U'(b))^2 - (U'(a))^2 \\right) + \\int_a^b \\frac{n-1}{r}(U'(r))^2dr + F(U(b))-F(U(a)) = 0$\r\nwhere $F'=f$. That is,\r\n$\\frac{1}{2}(U'(b))^2 + \\int_a^b \\frac{n-1}{r}(U'(r))^2 dr = 0$\r\nimplying $U' = 0$ on $[a,b]$. This contradicts the assumption that $U$ is non-constant on this interval and proves the assertion. :)" } { "Tag": [], "Problem": "Linarite is a mineral containing copper and lead. Its chemical formula is $ PbCuSO_{4}(OH)_{2}$. This mineral is soluble in both nitric acid and sulfuric acids, but its non reactive in HCL. A student was given a sample and asked to dissolve it to produce a solution of lead, copper and sulfate ions.Which acid should the student use? and why?", "Solution_1": "[hide=\"Opinion\"]Perhaps nitric acid should be more appropriate, since by using sulphuric acid the sulphate concentration could be high enough to precipitate some of the Pb2+ ions.[/hide]" } { "Tag": [ "calculus", "integration", "function", "real analysis", "real analysis unsolved" ], "Problem": "Here is a nice integral. It can be calculated exactly.\r\n\r\nLet $x>0$. Calculate\r\n\r\n$\\int_{0}^{\\infty}\\frac{t^{x}}{(e^{t}+1)^{2}}dt=$", "Solution_1": "Let $x>0$ and set\r\n\r\n\\[J: =\\int_{0}^{\\infty}\\frac{t^{x}}{(e^{t}+1)^{2}}dt\\]\r\n\r\nRecall that \r\n\r\n\\[\\sum_{k\\geq 0}(-1)^{k}x^{k}=\\frac{1}{1+x}\\Rightarrow-\\sum_{k\\geq 1}(-1)^{k}kx^{k-1}=\\frac{1}{(1+x)^{2}}\\]\r\n\r\nso that we have that\r\n\r\n\\[\\frac{1}{(e^{t}+1)^{2}}=\\frac{e^{-2t}}{(1+e^{-t})^{2}}=-e^{-2t}\\sum_{k\\geq 1}(-1)^{k}ke^{-2tk+2t}=-\\sum_{k\\geq 1}(-1)^{k}ke^{-2tk}\\]\r\n\r\nand hence that\r\n\r\n\\[J=-\\int_{0}^{\\infty}t^{x}\\sum_{k\\geq 1}(-1)^{k}ke^{-2tk}dt=-\\sum_{k\\geq 1}(-1)^{k}k\\int_{0}^{\\infty}t^{x}e^{-2tk}dt\\]\r\n\r\nnow substitute $u=2kt\\Rightarrow du=2kdt$ so that\r\n\r\n\\[J=-\\sum_{k\\geq 1}(-1)^{k}k\\int_{0}^{\\infty}\\left(\\frac{u}{2k}\\right)^{x}e^{-u}\\frac{du}{2k}=-\\frac{1}{2^{x+1}}\\sum_{k\\geq 1}\\frac{(-1)^{k}}{k^{x}}\\int_{0}^{\\infty}u^{x}e^{-u}du\\]\r\n\\[=-\\frac{\\Gamma (x+1)}{2^{x+1}}\\sum_{k\\geq 1}\\frac{(-1)^{k}}{k^{x}}= \\frac{\\Gamma (x+1)}{2^{x+1}}(1-2^{1-x})\\sum_{k\\geq 1}\\frac{1}{k^{x}}\\]\r\n\\[=\\frac{1-2^{1-x}}{2^{x+1}}\\Gamma (x+1)\\zeta(x)\\]", "Solution_2": "Very nice benorin. There is another approach of this integral involving Mellin's transform.\r\n\r\nLet $f(t)=\\frac{1}{e^{t}+1}$, for $t>0$.\r\n\r\nThen $f'(t)=-f(t)+\\frac{1}{(e^{t}+1)^{2}}$.\r\n\r\nNow take the Mellin transform of the two functions and we get that:\r\n\r\n$M[f,\\alpha]=g(\\alpha)\\xi(\\alpha)$ and \r\n\r\n$M[f', \\alpha]=-g(\\alpha)\\xi(\\alpha)+M[\\frac{1}{(e^{t}+1)^{2}},\\alpha]$\r\nwhere $g(x)=\\Gamma(x)(1-2^{1-x})$.\r\n\r\n\r\nThe two Mellin transforms are related by the formula:\r\n\r\n$M[f',\\alpha]=-(\\alpha-1)M[f,\\alpha-1]$, provided that $t^{\\alpha-1}f(t)$ vanishes at 0 and infinity.\r\n\r\nIt follows that:\r\n\r\n$M[\\frac{1}{(e^{t}+1)^{2}},\\alpha]=g(\\alpha)\\xi(\\alpha)-(\\alpha-1)g(\\alpha-1)\\xi(\\alpha-1)$.\r\n\r\nReplacing $\\alpha$ by $x+1$ we recover your result.", "Solution_3": "Note (for 1st post):\n\\[\\sum_{k\\geq 1}\\frac{(-1)^{k}}{k^{x}}+\\sum_{k\\geq 1}\\frac{1}{k^{x}} =\\sum_{k\\geq 1}\\frac{2}{(2k)^{x}}\\Rightarrow \\sum_{k\\geq 1}\\frac{(-1)^{k}}{k^{x}}= \\left( 2^{1-x}-1\\right)\\sum_{k\\geq 1}\\frac{1}{k^{x}}\\]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Hello !!\r\n\r\nlets $ x,y,z >0$ and $ x\\plus{}y\\plus{}z\\equal{}1$ \r\n\r\nprove that : \r\n\r\n$ \\frac{12}{xyz} \\plus{} (\\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z})^3 \\geq (\\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z})^2 \\plus{}\\frac{4}{xyz}(\\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z} )$", "Solution_1": "[quote=\"*210*\"]Hello !!\n\nlets $ x,y,z > 0$ and $ x \\plus{} y \\plus{} z \\equal{} 1$ \n\nprove that : \n\n$ \\frac {12}{xyz} \\plus{} (\\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z})^3 \\geq (\\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z})^2 \\plus{} \\frac {4}{xyz}(\\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} )$[/quote]\r\nWell,that's my inequality that I posted before in a moroccan math forum.\r\n \r\nHere is my generalization:\r\n\r\nIf $ x,y,z,k > 0$ and $ x \\plus{} y \\plus{} z \\equal{} k$ then:\r\n\r\n$ \\frac {9k \\plus{} 3k^2}{xyz} \\plus{} k.(\\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z})^3 \\geq (\\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z})^2 \\plus{} \\frac {4k^2}{xyz}.(\\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} )$\r\n\r\n$ *$ The case $ k \\equal{} 1$ gives the inequality you've posted." } { "Tag": [ "function", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Find all functions $ \\ f: R^\\plus{}\\rightarrow\\ R^\\plus{}$ such that\r\n\r\n$ f(x \\plus{} \\frac{1}{y}) \\plus{} f(y \\plus{} \\frac{1}{x}) \\equal{} f(x \\plus{} y \\plus{} \\frac{1}{x} \\plus{} \\frac{1}{y})$ for all $ x,y > 0$", "Solution_1": "[hide]$ f(x)\\equal{}0$ is the obvious solution. The equation satisfies Cauchy's Functional Equation, so its continous solution is $ f(x)\\equal{}xf(1)$[/hide]", "Solution_2": "[quote=\"basted\"][hide]$ f(x) \\equal{} 0$ is the obvious solution. The equation satisfies Cauchy's Functional Equation, so its continous solution is $ f(x) \\equal{} xf(1)$[/hide][/quote]\r\n\r\nIt not satisfies Cauchy's Functional Equation because the equation not holds for every positive real numbers a ,b\r\nthat $ f(a\\plus{}b)\\equal{}f(a)\\plus{}f(b)$", "Solution_3": "For any $ a,b>0$ with $ ab\\ge4$, we can solve \r\n$ x\\plus{}\\frac1y\\equal{}a$ and $ y\\plus{}\\frac1x\\equal{}b$.\r\nHence, we have\r\n$ f(a\\plus{}b)\\equal{}f(a)\\plus{}f(b)$ when $ ab\\ge4,a>0,b>0$.\r\nNow we assume $ x\\ge2$. Let $ n\\ge m>0$ be integers. Then we have\r\n$ f(nx)\\equal{}nf(x)\\equal{}mf(\\frac{nx}m)$.\r\nIn other words, $ f(rx)\\equal{}rf(x)$ whenever $ r\\ge1$ is rational and $ x\\ge2$.\r\nNow for any $ q>r$, we can find integer $ n$ such that $ n(q\\minus{}r)>1$. \r\nThen we have for $ x\\ge 2$,\r\n$ nf(qx)\\minus{}nf(rx)\\equal{}f(nqx)\\minus{}f(nrx)\\equal{}f(nqx\\minus{}nrx)>0$, or $ f(qx)>f(rx)$.\r\nUsing $ r_n$ increasing converging to $ q$ and another $ r_m$ decreasing converging to $ q$, monotonicity will gives us $ f(qx)\\equal{}qf(x)$ for any $ q\\ge1$ and $ x\\ge2$.\r\nHence, we have $ f(x)\\equal{}\\frac x2f(2)$ for $ x\\ge2$.\r\n\r\nFor any $ x<2$, we have\r\n$ f(x)\\equal{}f(x\\plus{}\\frac4x)\\minus{}f(\\frac4x)\\equal{}\\frac{f(2)}2(x\\plus{}\\frac4x\\minus{}\\frac4x)\\equal{}\\frac x2f(2)$.\r\nSo we prove $ f(x)$ is linear. Note we do not need continuity condition.\r\nQ.E.D." } { "Tag": [ "floor function", "function" ], "Problem": "What is $f^{-1}{(x)}$ given $f(x)=x+\\lfloor{x} \\rfloor$?", "Solution_1": "[hide]When x is an integer, this is trivial. Inverse is y=x/2\nWhen x is a positive non-integer, we have x+a=y where a is x's integer part as the function. The inverse is x-a=y.\nWhen x is a negative non-integer, we have x+a-1=y where a is x's integer part as the funtion. The inverse is x+1-a=y.\nNow, when x is an integer, basementfuntion(x)=x\nWhen x is a positive non-integer, basementfunction(x)=int. part of x.\nWhen x is a negative non-integer, basementfunction(x)=int. part of x-1.\nTherefore, when y=x+basementfunction(x) is the function, x-basementfunction(y)=y is the inverse.[/hide]", "Solution_2": "But how can you write $f^{-1}{(x)}$ as some function of x explicitly?", "Solution_3": "[hide]For all y, y=x-basementfunction(x)/2.\nThis works because whatever that's added to the ys in the original equation, there was at most one non-integer and one integer. This causes the non-integer part of x to be the same as y's corresponding part. All I had to do was to isolate that part and remove the excess. If we set y as a(Int. Part)+b(Not Int. Part) and basementfunction(y) as a for positives and integers in the original function, we add these to equal x. We get 2a+b. To get y back, we need to get a+b with only functions involving x, which is 2a+b. Since basementfunction(x) is 2a and x is 2a+b, I can synthesize y=a+b if I divide by two. But the basementfunction for negatives works not as the int. part but as the int. part-1. In other words, when y is negative non-integer, a+b=y and basement function(y)=a-1. Adding these up, we get 2a+b-1 as x but the basement function of this is 2a-2, so dividing by two works here also, yielding 2a+b-1-(a-1) or a+b=y.[/hide]" } { "Tag": [], "Problem": "there are 3 Roses the white the red and the black rose\r\neach is a symbol for a wizard clan\r\neach has 2 or 3 wizards and the rest a rest warriors\r\nit works like this each rose will secretly discuss and decide to try to kill a person on one of the other roses by use of a tournament. one rose challenges another and the remaining rose does a vote on who they want to win and pm me then i or someone i tell to post the event will tell who won and who died\r\nsignups are now. o and wizards are more powerful and can only be killed in a war of the roses or a special challenge.\r\n[b]\nPM ME IF YOU WANT TO JOIN I WILL GIVE MORE RULES AND WHICH ROSES PEOPLE ARE IN SOON.[/b]\r\n\r\nBATTLES:\r\n\r\ntournament challenge:\r\none rose puts a warrior up and challenges another rose then challenged rose places a warrior or wizard for battle. if it is a warrior remaining rose puts it to a vote to decide the winner, but if it is a wizard i will discuss with the wizard(s) in the remaining rose to decide.\r\n\r\nspecial challenge: two warriors from a rose challenge a person from another. if that person is a wizard the wizard dies but if he isnt the team losesa warrior.\r\n\r\nWar of the roses:\r\none wizard goes out and declares war with another rose its kinda like a wizards duel except each wizard brings 2 warriors and there are 3 battles which the remaining rose and i decide the outcome of each one. the rose that wins 2 of them win the \"war\" and the mage and warrior survive but the other wizard does not, but the other warriors do. Can only be declared if team declaring have only 1 wizard.\r\n\r\nmore roses will be added as necessary to make the game intresting. i will announce the time when we have enough roses to create alliances of dragons", "Solution_1": "I especially like the lack of clear sentences. Also, explain later? No thanks.", "Solution_2": "ok ive showed whats happening if you want to join you can i might add sorcerers and enchanters to the lot. sorcerers make talismans that enhance winning chance and enchanters enchant warriors so they switch to enchanters rose.", "Solution_3": "Mathblitz, TALK IN COMPLETE SENTENCES.\r\n\r\nAnyway, I join :P", "Solution_4": "[quote=\"mathblitz\"]ok ive showed whats happening if you want to join you can i might add sorcerers and enchanters to the lot. sorcerers make talismans that enhance winning chance and enchanters enchant warriors so they switch to enchanters rose.[/quote]\r\n\r\nyeah... eh... no\r\n\r\nJust use the first rules.\r\n/in", "Solution_5": "[quote=\"mathblitz\"]ok ive showed whats happening if you want to join you can i might add sorcerers and enchanters to the lot. sorcerers make talismans that enhance winning chance and enchanters enchant warriors so they switch to enchanters rose.[/quote]\r\n\r\nyeah... eh... no\r\n\r\nJust use the first rules.\r\n/in", "Solution_6": "[quote=\"Dojo\"][quote=\"mathblitz\"]ok ive showed whats happening if you want to join you can i might add sorcerers and enchanters to the lot. sorcerers make talismans that enhance winning chance and enchanters enchant warriors so they switch to enchanters rose.[/quote]\n\nyeah... eh... no\n\nJust use the first rules.\n/in[/quote]\r\nagree with dojo why did dojo make two posts", "Solution_7": "It's called a glitch.", "Solution_8": "fine no more ill edit agian", "Solution_9": "[quote=\"eggylv999\"][quote=\"Dojo\"][quote=\"mathblitz\"]ok ive showed whats happening if you want to join you can i might add sorcerers and enchanters to the lot. sorcerers make talismans that enhance winning chance and enchanters enchant warriors so they switch to enchanters rose.[/quote]\n\nyeah... eh... no\n\nJust use the first rules.\n/in[/quote]\nagree with dojo why did dojo make two posts[/quote]\r\n\r\nIt's when you click submit, and the internet lags, and you click submit again to try to make it go faster.", "Solution_10": "Mathblitz, could you post everything you posted in the first post in COMPLETE SENTECES?\r\n\r\nThank you. That is all.", "Solution_11": "Proper capitilization would be nice too. And punctuation. no one wants to read rules when they are written like this if they are like this you cant tell where they end and begin so it becomes confusing.", "Solution_12": "YEA SURE. There is one problem though. i can only edit once for some reason and im trying to contact Klebian about this", "Solution_13": "Make a new post...", "Solution_14": "The deadline for editing posts is no more than 48 hours. (It may be less, but I know it's definitely not more than that.)", "Solution_15": "Lol, I would like to join????", "Solution_16": "my games tend to nt get past the sign up stages because not enough people join.\r\nDo not revive this thread. I am working on a better game for less people." } { "Tag": [], "Problem": "Have the entry forms been processed yet? Because I faxed mine in like two weeks ago, and approval hasn't showed up yet. Or do we actually need to fax it at the same time in which we e-mail the solutions?", "Solution_1": "Well it said to mail the registration forms with the solutions but I'm not sure what to do with emailed solutions...", "Solution_2": "It says that mine has been received; I faxed mine in the same day I emailed my solutions in, though no confirmation of the solutions yet. Not fussed, though.", "Solution_3": "[quote=\"13375P34K43V312\"]Have the entry forms been processed yet? Because I faxed mine in like two weeks ago, and approval hasn't showed up yet.[/quote]\r\n\r\nMaybe you should resend your entry form. I mailed mine weeks before sending in my solutions, and they processed it about five days later." } { "Tag": [ "quadratics", "algebra" ], "Problem": "The quadratic equation $ x^2 \\plus{} mx \\plus{} n \\equal{} 0$ has roots that are twice those of $ x^2 \\plus{} px \\plus{} m \\equal{} 0$, and none of $ m,n,$ and $ p$ is zero. What is the value of $ n/p$?\r\n\r\n$ \\textbf{(A)}\\ 1\\qquad\r\n\\textbf{(B)}\\ 2\\qquad\r\n\\textbf{(C)}\\ 4\\qquad\r\n\\textbf{(D)}\\ 8\\qquad\r\n\\textbf{(E)}\\ 16$", "Solution_1": "[hide=\"Answer\"]$m=2(a+b)$, $n=4ab$, $p=a+b$, $m=2(a+b)=ab$\n$n=4ab=4m=8(a+b)$, $\\frac{n}{p}=\\frac{8(a+b)}{a+b}=8\\Rightarrow \\boxed{D}$[/hide]", "Solution_2": "[hide=\"Attempt\"]Roots of first equation be $2r,2s$, so $2r+2s=-m,4rs=n$. From second equation, roots are $r,s$, so $r+s=-p,rs=m$. Now,\n\\[ \\frac{n}{p}=\\frac{4rs}{-(r+s)}=\\frac{8rs}{-2(r+s)}=\\frac{8m}{m}=8 \\]\nMy answer is [b]D[/b][/hide]", "Solution_3": "Quite a fun problem. My solution is what I would do during a contest, although it does make a foolish assumption at the start...\n\n[hide = \"Solution\"]\nNote from the start that I assume the roots are integers :blush: Let the roots of the first equation be $2a$ and $2b$. Then the roots of the first equation are $a$ and $b$. Vieta's on both equations gives $ab=m, 4ab=n, a+b=-p, 2a+2b=-m$. Seeing a $-m$ and $m$, we add those two equations to get $2a+2b+ab=0$. SFFT on this gives $(a+2)(b+2)=4$. Note that $a+2$ and $b+2$ cannot be equal to $2$ since then that would make them zero, and the question wouldn't work out. Instead let $a+2=4$ and $b+2=1$. Then we get $a=2$ and $b=-1$. Since $n/p=4ab/(a+b)$ and $a=$2 and $b=-1$, we plug in the values to get a final answer of $8$, or [b]D[/b].", "Solution_4": "[hide]Let the roots of $x^2+px+m$ be $r_1,r_2$. Then $r_1+r_2=-p$, so $2r_1+2r_2=-m=-2p$. Similarly, $r_1r_2=m\\Rightarrow 4r_1r_2=n=4m$. So $\\frac np=\\frac{4m}{m/2}=\\boxed{8}$.[/hide]" } { "Tag": [ "calculus", "geometry", "MIT", "college", "AMC", "USA(J)MO", "USAMO" ], "Problem": "I'm not sure if this really belongs here since it's about choosing my high school AP class, but I'd like to know:\r\n\r\nI've finished taking Biology and Chemistry Honors, and I'm planning to double up on science for my sophomore year. I'll need to take Physics Honors and another AP science class, but I'm not sure if I should take Bio or Chem.\r\n\r\nHow relevant is math to biology? To chemistry? I'm guessing it's probably [i]very[/i] relevant to physics. I'm still not really sure what I'll do for a living when I grow up, but I'd like to do something that I might have an easier time with, due to my current math abilities. Of course, I'd continue studying that 'something' through college.\r\n\r\n\r\nOn a slightly different note, what do you guys usually take after having taken AP Calculus BC? I still want a more thorough understanding of the subject, so I'm going to sign up for the AoPS Calculus class for 10th grade, but.. Multivar? Linear Algebra?\r\n\r\nI'd really appreciate your input. :)\r\n\r\n\r\nEDIT: I forgot to ask, how loaded should my schedule be? I haven't really put in much community service yet, and my mom worries that if I don't push ahead now, I may not have enough time for it in my junior/senior years.. which would look pretty bad on an essay.\r\n\r\nFor example, about how many \"empty\" periods should I have in my junior/senior years? That is, subjects that won't really hurt to slack off on, such as a college math course since I've already finished the high school curriculum. My tentative schedule for my sophomore year:\r\n\r\n1. AoPS Calculus\r\n2. Physics H\r\n3. Eng 10 H\r\n4. American Gov\r\n5. French III\r\n6. Spanish I\r\n7. AP Bio\r\nExtracurriculars: tennis, violin (these will become *official* once I start attending public school again next January or so)\r\n\r\nIt'll be living hell, but how much would it pay off?", "Solution_1": "[hide=\"As for AP Biology\"]\nLittle math concepts are needed. The most difficult would probably be the chi-squared analysis but that is easily taught. The majority of the course is focused on connecting individual concepts to one another and analyizing the big picture.\n[/hide]\n\n[hide=\"As for AP Courses\"]\nI'm only taking three AP courses next year as a sophmore, while I have two friends that are taking four. I believe you should gradually build up APs, with junior year being the most difficult, and senior year being the most relaxing. \n\nColleges barely check your senior year grades, so there is no sense in potentially destroying them by taking full APs. That being said, you also don't want to slack off and take all easy courses. I'd recommend one or two AP courses or three at the most. Junior year is definately the most loaded though, as most high schoolers start APs in most subject areas. For example, in junior year, I can take AP english, AP US History, AP Calc AB, AP Spanish, etc. That's already 4 APs, not counting electives and science. \n\nIn determining which AP courses to take, I wouldn't recommend taking every AP course you're offered. Come junior year, there is potential to rack up a full schedule of APs - which is not a good idea. If you're weak in a subject area, definately debate whether that AP course is good for you or not. \n\nEDIT - After completion of an AP course - it is a good idea to take the SAT II subject test in that area that year. \n[/hide] \n\n[hide=\"As for Community Service\"]\nI finished my required community service hours in 7th grade - I suggest you complete it ASAP. There will be little time during the school year (it might even affect your grades) and you'll need your summers to prep for next year and relax. That being said, go above and beyond in your community service hours. We have a required amount of 75 and if you complete over 100 hours, then you get a special recommendation (which the colleges might like). \n[/hide]", "Solution_2": "Building off of what Henry said, do not just take APs because they are offered; take them only if you are interested in the subject area or if you are up to the challenge. How can your application be unique is you are jamming it full of \"generic\" courses just for the sake of it?\r\n\r\nAbout community service, I have been doing community service every summer since the summer of seventh grade - :oops: except this one. Colleges usually have you write about one \"activity\" you enjoyed or got the most out of, so make sure you volunteer where you will enjoy it; make the possible list for that question as long as possible. :wink: \r\n\r\nThe big thing is to do things you enjoy and write it all over you application.", "Solution_3": "Wow you really know this stuff; I need to do my research. :oops:\r\n\r\n@7h3: That's a lot of AP courses.. :D but yeah I'll take your advice on taking more AP courses in my junior year and relaxing a bit in my senior year.. also about taking the SAT subject tests.\r\n\r\n@AstroPhys: hehe yep I'm being careful about taking too many AP courses.. especially about continuing to study the subject through a college course - \r\n\r\nexactly how hard are college courses compared to high school ones? For example, is one college course equivalent to maybe taking two or three high school courses at once? If I wanted, I [i]could[/i] take several college courses (math/French/sciences..) in my senior year, though it's probably not a good idea.\r\n\r\n\r\nI googled community service a bit, and it seems more flexible than I thought. I always wanted to have the privilege of helping teach a school math team (not a circle just yet :roll:) - would that count towards my community service hours? There's also this site that lets you \"track your service hours.\" It seems the record-keeping of hours (thus meeting the requirements) isn't that strict, either..", "Solution_4": "I don't agree with the, junior year should be the busiest year. It just happens to turn out like that for a lot of people, but don't try to take more classes just because you feel like you need them to show up on your transcript. High school is only a step in your education, and you should take the classes because you feel that they are useful. Don't pressure yourself to take more classes one year to flatter the colleges. Just do what you feel like YOU can gain the most out of.\r\nI'm going to take BC calculus (probably) and AP physics B as a sophmore (regular everything else), then AP physics C and Chemistry AP (if I don't get to skip this year, which I probably won't) as a junior, along with college math. Now I could cram in AP bio, but that would be suicide. I need time for other things, so I decided to put that off till senior year.", "Solution_5": "The answers to most of these questions really depend on where you want to go to college. Some people want to go to a state school and graduate in 2 years, others want to go to MIT and graduate in 5, or go where I'm going and get a masters in 4, and so on.\r\n\r\nAfter AP Calculus BC, you can take either multivar or linear algebra. Most people take multivar. I took it senior year and got bored and dropped out. It's exactly like regular calculus, except with more variables.\r\n\r\nEveryone I know that's been good at math has done well in biology and chemistry. But neither AP course has anything to do with any math over like Algebra I. I found biology easier, since in chemistry, you have to memorize things like the colors of solutions and stuff.\r\n\r\nI took 5 AP classes junior year, and taught myself calc BC and APES on my own, and it wasn't busy. I think everyone overestimates the rigors of high school.\r\n\r\nBut yeah, as I said, it depends where you want to end up. I wanted to go to MIT or CalTech, but I knew if I went to like Oklahoma or some other state school, I'd stilll end up doing just as well in life. So I took it easy and did what I wanted junior and senior years. I did no community service and wrote something like \"I prefer to help myself before I help others, thus I work 40 hours a week instead of volunteering\" on my applications. It's what made me happy, so meh.", "Solution_6": "Your question on what math you should study next seems to me more interesting and important than questions about AP courses. My advice is to move away from calculus and study some topic that you find interesting, e.g., modern algebra. Once you've completed calculus, math learning becomes more non-linear than ever. There is no specific sequence of topics you should study; math branches out in many different directions. I like modern algebra (also called abstract algebra) personally.\r\n\r\nAnyway, if you don't learn something really well in high school, you'll probably learn it well in college, so don't fret too much about it. To be honest, the courses you take in high school will probably have almost no impact on your career as far as qualification and ability go. Of course, you could find something really wonderful that inspires you to take a certain path.\r\n\r\nOh, also, you might want to consider olympiad math, if you haven't done much of that yet. I notice you're only going into tenth grade. There are really only one or two new ideas in calculus. The rest of the course is investigation of specialized applications of these ideas; these can be viewed as exercises in mathematical versatility as much as anything else. Olympiad math can be a good place to develop mathematical vesatility.\r\n\r\nHm, and in general, writing essays emphasizing your defects is probably not a good idea.", "Solution_7": "True that deciding a certain year to be the busiest one is not the best idea, but getting things out of the way in my junior year does seem to have its upsides.. it'd make more time for SAT tests, college apps, and such (or is that common sense? :|)\r\n\r\nOsud also PMed me and recommended computational biology as something that integrated math and biology. I think I could pursue my programming interests as well, so it's very appealing. I'll look at non-usual (I don't want to say unusual, but..) courses to take after AP Calc as Boy Soprano suggested. The prerequisites for abstract algebra seem to include a lot of contest math, too. :)\r\n\r\nbut ehheh yes I should look at olympiad math a lot more (last year's USAMO 0 still hurts :()\r\n\r\n@iin77: individuality is always good, but I prefer to be more.. orthodox ;)\r\n\r\nI guess I should just take whatever course seems interesting, and hope to get a better idea of what I'm really interested in after I taste all the subject areas. Thanks for all your info and ideas!", "Solution_8": "Actually, one of the reasons I recommended abstract algebra is that it has [i]no[/i] prerequisites other than a little set theory and some mathematical maturity.", "Solution_9": "hehe that's even better! :lol: Thanks.", "Solution_10": "I disagree with the advice of ever taking an easy-looking course load, even in senior year. You should be challenging yourself every year. \r\n\r\nEvidently some states have required community service hours, but that varies from state to state. There is no such requirement here, for instance. \r\n\r\nStudying what YOU like is always a good idea, because then you will be motivated to do good work. This year, my son is really enjoying a creative writing class at our state flagship university (taken as his first semester \"eleventh grade\" English by dual enrollment). He puts in extra reading for that course because he is interested in the subject. That would be the kind of thing to look for when choosing courses. \r\n\r\nAlas, it is very difficult to find a high school course, or even a dual enrollment college course, in science that presupposes much math background. That's because most American students have lousy math backgrounds and so the science courses are designed with few prerequisites beyond beginning algebra, as mentioned in a reply above.", "Solution_11": "[quote=\"tokenadult\"]I disagree with the advice of ever taking an easy-looking course load, even in senior year. You should be challenging yourself every year. [/quote]\r\n\r\nI agree; you definately shouldn't take all honors classes senior year, and I myself would probably be taking 4 to 5 APs (I still have to complete a standard year of gym) and I consider that a pretty relaxed year. Even though it's 4 to 5 APs, in the context of my school, they are the easiest classes to get an A with (out of the APs). \r\n\r\nAs for studying what for what [i]I[/i] like, is it okay if I still haven't made up my mind about what I want to be? And when should I definately have a choice on this?", "Solution_12": "I didn't feel like starting a new thread about this, so I'm just going to ask it here.\r\n\r\nDo A-'s count the same to colleges? I know that most/many schools have 90-100 worth 4 or 5 gpa, 80-89 3 or 4, etc....last year I had about half A-'s and that made my gpa go down by .2/.3 or so...going from what would be a perfect gpa to a good bit less...if a college sees all a's and a-'s on your transcript, do they sort of disregard the gpa part of it?", "Solution_13": "I mean, if your HS gives A-'s it certainly is not going to help your transcript to have them instead of A's. I'm sure they won't be too critical of A-'s, though." } { "Tag": [], "Problem": "Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike.\r\nShe bikes at a uniform rate of 12 miles per hour. Brian leaves\r\nMarquette at 9:00 AM heading for Escanaba on his bike. He bikes at a\r\nuniform rate of 16 miles per hour. They both bike on the same\r\n62-mile route between Escanaba and Marquette. At what time in the\r\nmorning do they meet? (Express your answer in standard hh:mm time form.)", "Solution_1": "Cassie leaves Escanaba at 8:30, she can walk for 30 minutes while Marquette heads to Escanaba at 9:00.\r\n\r\nso she rode 6 miles.\r\n\r\nand that makes remaining length 62-6=56 miles.\r\n\r\n56 miles are remaining, and it's 9:00.\r\n\r\n56/(12+16)=56/28=2.\r\n\r\nso it takes 2 hours them to meet.\r\n\r\n9:00 + 2:00 = 11:00\r\n\r\nanswer : 11:00" } { "Tag": [ "function", "geometry", "incenter", "induction", "algebra", "polynomial", "Vieta" ], "Problem": "So, now it's over. Overall it was on the easy side; in my efforts to simulate a real AMC 12 I seemed to have dumbed it down just a tad too much :P. Plus having number 17, which would probably have one of the lowest averages, be free made it easier. Oh well. In any case, here we go. There are probably plenty of people who can help you with solutions if you need them. Every question was solved multiple times. For reference, the test can be found [url=http://wangsblog.com/jeffrey/mockAMC.pdf]here[/url].\r\n\r\n[hide=\"Winner\"] CornOnTheCobbe - 150 [/hide]\n[hide=\"Second\"] EFuzzy - 145.5 [/hide]\n[hide=\"Third\"] beta - 144 [/hide]\n[hide=\"Honorable Mention\"] SplashD - 138\nmysmartmouth - 138[/hide]\n\nStatistics:\n\n[hide=\"Average Scores\"]1. 5.785714286\n2. 6\n3. 5.571428571\n4. 5.571428571\n5. 5.785714286\n6. 5.410714286\n7. 5.357142857\n8. 5.196428571\n9. 5.571428571\n10. 4.928571429\n11. 5.357142857\n12. 3\n13. 5.625\n14. 4.821428571\n15. 5.25\n16. 4.071428571\n17. 6\n18. 4.392857143\n19. 3.803571429\n20. 5.089285714\n21. 3.321428571\n22. 3.696428571\n23. 2.892857143\n24. 4.392857143\n25. 2.089285714\nTOTAL: 118.9821429[/hide]\n\n[hide=\"Hardest Questions\"]25, 23, 12, 21, 22[/hide]\n\n[hide=\"Participants\"]28[/hide]\n\n[hide=\"Qualifiers (100+)\"]23[/hide]\n\n[hide=\"Answers\"] 1. D\n2. E\n3. A\n4. E\n5. C\n6. C\n7. D\n8. E\n9. A \n10. B\n11. E\n12. C\n13. D\n14. A\n15. C\n16. C\n17. - \n18. B\n19. B\n20. B\n21. C\n22. C\n23. B\n24. A\n25. C[/hide]", "Solution_1": "EH!\r\n\r\nmisread 23 completely\r\n\r\nwho knows how to do 22? I couldn't figur ethat out", "Solution_2": "Everyone got number 2 :woah:", "Solution_3": "oh man, out of the three recent mock amcs, this is the one i didn't do, and unfortunately this one had the best questions :(\r\n\r\nAlthough the scores seem much higher than the other two ;)", "Solution_4": "Here's to careless mistakes! \r\n\r\n...SplashD, aren't you in 8th grade? :read: \r\n\r\n[quote=\"Elemennop\"]EH!\n\nmisread 23 completely\n\nwho knows how to do 22? I couldn't figur ethat out[/quote]\r\n\r\n22 I proved (to myself) for every case the maximum number, then found a case of 9. It was nowhere near rigorous though.\r\n\r\nMi me gusta el problemo numero vienticinco. Es muy dificil para ver, pero no es dificil para hacer.", "Solution_5": "yea i found a case of 9 but i couldn't even begin to figure out how to show you dont' need more htan 9", "Solution_6": "Anyone have solutions to 12,16,24,25 :fool:", "Solution_7": "[quote=\"mysmartmouth\"]Here's to careless mistakes! \n\n...SplashD, aren't you in 8th grade? :read: \n\n[quote=\"Elemennop\"]EH!\n\nmisread 23 completely\n\nwho knows how to do 22? I couldn't figur ethat out[/quote]\n\n22 I proved (to myself) for every case the maximum number, then found a case of 9. It was nowhere near rigorous though.\n\nMi me gusta el problemo numero vienticinco. Es muy dificil para ver, pero no es dificil para hacer.[/quote]\r\nYes, splash D is in 8th grade. I'm getting really, really, really scared.\r\n[hide=\"24\"]I don't think this is the most elegant way, but I expanded heron's formula, then used Newton's sums. [/hide]\n[hide=\"I got 12 wrong, this was what my idea was though\"] Ok, after n people enter the number is $2^{n}-1$ so for 10 this is $2047$. 10 people leave and we have 2037 people. After that the number of people rises too quickly too get something smaller than this[/hide]\r\nCan someone point out what was wrong in my number 12?\r\nHah, for #19 I graphed it and saw that B was to only true thing for the value I saw.", "Solution_8": "Wow... three completely idiotic mistakes. I guess I dont' read English or something.", "Solution_9": "[quote=\"bpms\"]\n[hide=\"24\"]I don't think this is the most elegant way, but I expanded heron's formula, then used Newton's sums. [/hide]\n[hide=\"I got 12 wrong, this was what my idea was though\"] Ok, after n people enter the number is $2^{n}-1$ so for 10 this is $2047$. 10 people leave and we have 2037 people. After that the number of people rises too quickly too get something smaller than this[/hide]\nCan someone point out what was wrong in my number 12?\n[/quote]\r\n$2^{10}=1024\\ne2048$\r\nFor 24, you always have the calculator :wink: \r\n(find the roots with solve function, use heron's for the area. The decimal expansion comes out awesomely simple)\r\n\r\n5 stupid mistakes for me.....FIVE grrr....I really need to be careful and watch out for mistakes......", "Solution_10": "[hide]\n\n\n\nfor 12, it's for people entering\n\n\n\nfor 16, let , then we have and , ie. is an arithmetic sequence. Thus since it's arithmetic we have , and it's first non positive value (since it's monitonically decreasing) is when .\n\n\n\nfor 24, , so by Heron's we want . Suppose , then we have , so our answer is .\n\n\n\nfor 25, it's just really BRUTAL diophantine equations, something like and then anotehr value I think (3/2?)...\n\n\n\n[/hide]", "Solution_11": "Could someone post a solution for 21?", "Solution_12": "12. is just looking binary...we get\r\n\r\n2047=11,111,111,111 , so once they leave, that is minus 11, so 2036\r\n\r\n16. a_n^2007-a_n^2007=-2007...an explict form is easy now\r\n\r\n24. herons can be written as A^2=s* prod (s-a)=4* f(4)\r\n[assuming the lengths satisfy the triangle inequality...]", "Solution_13": "21\n\n\n\n[hide]AP and CP are angle bisectors of triangle ADC, thus P is the incenter of triangle ADC, and consequently DP is the angle bisector of angle ADC. so we have \n2nd move: $a_{2}=\\frac12$ in the negative direction <----\n3rd move: $a_{3}=\\frac14$ in the positive direction ---->\nand so on\n\nWhere does it end up? We add the $a_{n}$ with the respective signs:\n\\[\\sum_{n=1}^{\\infty}(-1)^{n+1}a_{n}=\\sum_{n=1}^{\\infty}(-2)\\left(-\\frac12\\right)^{n}=(-2)\\frac{\\left(-\\frac12\\right)}{1-\\left(-\\frac12\\right)}=\\frac23 \\]\n[/hide]\n[hide=\"17\"]Solve the inequalities carefully noting the [b]domain[/b] of each logarithm:\n$1+\\log\\log|x|\\leq0\\Rightarrow \\log\\log|x|\\leq-1\\Rightarrow 0<\\log|x|\\leq10^{-1}\\Rightarrow 1<|x|\\leq10^{\\frac1{10}}$\nSimilarly, $1<|y|\\leq10^{\\frac1{10}}$\nIf you put these two inequalities together, they are four squares, one in each quadrant, where each square has side $10^{\\frac1{10}}-1$ and area $(10^{\\frac1{10}}-1)^{2}$\nAnswer: $\\boxed{4(10^{\\frac1{10}}-1)^{2}}$\n[/hide]\n[hide=\"19\"]For this one, solve the equation into the form:\n\\[\\frac{\\sin x}{\\cos x}=\\frac{\\sqrt3-1}{\\sqrt3+1}\\]\nFrom this form, since $(\\sqrt3-1)^{2}+(\\sqrt3+1)^{2}=(2\\sqrt2)^{2}$\n$\\sin x=\\frac{\\sqrt3-1}{2\\sqrt2}$ and $\\cos x=\\frac{\\sqrt3+1}{2\\sqrt2}$\nYou can also do this by simply drawing a right triangle with one angle marked 'x' and since you have the ratio of the opposite leg to the adjacent leg, you can find the hypotenuse and then find sin and cos from the definition :).\nSo, which one of the choices is true now that you know the sin and cos value? :D\n[/hide]\nat last, the final one\n[hide=\"23\"]This one you have to read the problem carefully. It says exactly two vertices of an isosceles right triangle are on $C$. If the triangle were inside of $C$, all 3 vertices must be on $C$ because it is a right triangle and whose hypotenuse has same length as the diameter of $C$. So the triangle cannot be wholly inside of $C$. Your drawing should be something like this: draw chord $XY$ of length $\\sqrt2$ inside of $C$, this way you have exactly 2 vertices on $C$. Draw the other leg $XZ$ of length $\\sqrt2$ directing away from the center of $C$ and perpendicular at $X$. You will see that the triangle is partially outside of $C$, vertex $Z$ is outside of $C$. $R$ is the union of all of these triangles, which is the annulus with outer radius $OZ$ and inner radius $OM$, where $M$ is the midpoint of $XY$. See if you can do the rest :D \n[/hide]", "Solution_23": "Could someone do #13 out.. I'm extremely rusty in my geometry.\r\n\r\nThanks.", "Solution_24": "For 13, since the length of the diameter $AB$ is $8$, you know the radius of the circle is $4$, and $AD, DC, CE, EB$ are all equal to $2$. So the triangle with vertices $D$ and $E$ has a base of length $4$, height length $4$, and area $8$. For the triangle with vertices $A, D$, draw a radius from $C$ to the vertex of the triangle on the semicircle. Then draw the perpendicular from the vertex on the semicircle to $AD$. Since it's an isosceles triangle, the perpendicular will hit the midpoint of $AD$. So you have a right triangle with the hypotenuse as the radius of the semicircle, and one of the legs of length $3$, so the other leg, the height of the triangle, is $\\sqrt{7}$. So the area of that triangle is $\\sqrt{7}$. The triangle with $E$ and $B$ as vertices is the same, so the total area is $8+2\\sqrt{7}$.", "Solution_25": "Awesome. You rock! Knew I had to draw something in, shoulda figured that oh well, heh.", "Solution_26": "I agree with the test writer that the solution set for \r\n\r\n 2007/n1 + 2007/n2 = 1/2 \r\n\r\ncontains 45 ordered pairs, obtainable by counting the positive factors of 4014^2. I do not agree that 23 more solutions exist corresponding to some of the negative factors of 4014^2. I checked them all with Mathematica, and none of those meet the n1,n2 > 2007 criterion. I get 3 additional ordered pairs by solving\r\n\r\n 2007/n1 + 2007/n2 = 3/2\r\n\r\nand 40 more by solving\r\n\r\n 2007/n1 + (1-2007/n2) = 3/2 \r\n\r\nfor a total of 45 + 3 + 40 = 88.\r\n\r\nAt one point I think I convinced myself that no other equations would produce solutions that meet the n1,n2 > 2007 criterion.", "Solution_27": "Interesting claim by svonlaven...\r\nWhat I've found out is that the solution I wrote months ago do indeed lack analysis and is not complete.\r\n\r\nAfter a careful consideration of the problem, this is what I have found out (I hope this is correct):\r\n\r\nSkipping the formalities, let $ \\alpha \\equal{} \\frac {2007\\pi}{n_1}$ and $ \\beta \\equal{} \\frac {2007\\pi}{n_2}$\r\n\r\nwe have the equation $ \\sin^2\\alpha \\plus{} \\sin^2\\beta \\equal{} 1$\r\n\r\nThen, $ \\sin\\beta \\equal{} \\pm\\cos\\alpha$\r\n\r\nAll possible solutions can be characterized into\r\n\\[ \\pm\\cos\\alpha \\equal{} ( \\minus{} 1)^{\\frac {k \\minus{} 1}{2}}\\sin\\left(\\frac {k\\pi}2\\pm\\alpha\\right)\r\n\\]\r\nfor all odd integers $ k$. Note that we can assume the positive value for sine since\r\n\\[ ( \\minus{} 1)^{\\frac {k \\minus{} 1}{2}}\\sin\\left(\\frac {k\\pi}2\\pm\\alpha\\right) \\equal{} ( \\minus{} 1)^{\\frac {k \\plus{} 1}{2}}\\sin\\left(\\frac {|k|\\pi}2\\mp\\alpha\\right)\r\n\\]\r\nIn any case, we choose the positive value without losing any of the solutions. Therefore,\r\n\r\n$ \\cos\\alpha \\equal{} \\sin\\left(\\frac {k\\pi}2\\pm\\alpha\\right)$ for odd integers $ k$.\r\n\r\nThen, it follows that $ \\beta \\equal{} \\frac {k\\pi}2\\pm\\alpha$\r\n\r\nHaving settled with this, we want to choose the permissible values of $ k$ under the restrictions of $ n_1,n_2 > 2007$\r\n\r\nSuppose $ \\beta \\plus{} \\alpha \\equal{} \\frac {k\\pi}2$ or $ \\frac {2007}{n_2} \\plus{} \\frac {2007}{n_1} \\equal{} \\frac {k}2$. Then, $ 0 < \\frac {2007}{n_2} \\plus{} \\frac {2007}{n_1} < 2$; so in this case we have $ k \\equal{} 1,3$.\r\n\r\nNow suppose $ \\beta \\minus{} \\alpha \\equal{} \\frac {k\\pi}2$ or $ \\frac {2007}{n_2} \\minus{} \\frac {2007}{n_1} \\equal{} \\frac {k}2$. Then, $ \\minus{} 1 < \\frac {2007}{n_2} \\minus{} \\frac {2007}{n_1} < 1$; so in this case we have $ k \\equal{} \\minus{} 1,1$.\r\n\r\n[b]CASE 1[/b]\r\n\r\nPreviously, we solved $ \\frac {2007}{n_2} \\plus{} \\frac {2007}{n_1} \\equal{} \\frac {k}2$ for $ k \\equal{} 1$, in which there are 45 ordered pairs:\r\n\r\n$ (n_2 \\minus{} 2\\cdot3^2\\cdot223)(n_1 \\minus{} 2\\cdot3^2\\cdot223) \\equal{} 2^2\\cdot3^4\\cdot223^2$\r\n\r\nIf we try to check for all the divisors $ n_1 \\minus{} 2\\cdot3^2\\cdot223$ satisfying $ n_1,n_2 > 2007$, all divisors are feasible, therefore there are $ 3\\cdot5\\cdot3 \\equal{} 45$ ordered pairs.\r\n\r\n[b]CASE 2[/b]\r\n\r\nFor $ k \\equal{} 3$, we have: $ \\frac {2007}{n_2} \\plus{} \\frac {2007}{n_1} \\equal{} \\frac32\\Rightarrow \\frac {3\\cdot223}{n_2} \\plus{} \\frac {3\\cdot223}{n_1} \\equal{} \\frac12$,\r\n\r\n$ (n_2 \\minus{} 2\\cdot3\\cdot223)(n_1 \\minus{} 2\\cdot3\\cdot223) \\equal{} 2^2\\cdot3^2\\cdot223^2$\r\n\r\nHere we must have, with $ n_1 \\minus{} 2\\cdot3\\cdot223$ being a divisor of $ 2^2\\cdot3^2\\cdot223^2$,\r\n\r\n$ n_1 > 2007 \\equal{} 3^2\\cdot223\\Rightarrow n_1 \\minus{} 2\\cdot3\\cdot223 > 3^2\\cdot223 \\minus{} 2\\cdot3\\cdot223 \\equal{} 3\\cdot223$\r\nand $ n_2 \\minus{} 2\\cdot3\\cdot223 \\equal{} \\frac {2^2\\cdot3^2\\cdot223^2}{n_1 \\minus{} 2\\cdot3\\cdot223} > 3^2\\cdot223 \\minus{} 2\\cdot3\\cdot223\\Rightarrow n_1 \\minus{} 2\\cdot3\\cdot223 < 2^2\\cdot3\\cdot223$\r\n\r\nPut together, $ 3\\cdot223 < n_1 \\minus{} 2\\cdot3\\cdot223 < 2^2\\cdot3\\cdot223$. So there are three divisors for $ n_1 \\minus{} 2\\cdot3\\cdot223$, ie $ 2^2\\cdot223,2\\cdot3\\cdot223,3^2\\cdot223$.\r\n\r\nTherefore, we have 3 ordered pairs in this case.\r\n\r\n[b]CASE 3[/b]\r\n\r\nNow for the last case $ k \\equal{} 1$, $ \\frac {2007}{n_2} \\minus{} \\frac {2007}{n_1} \\equal{} \\frac12$,\r\n\r\nThis case also covers the case when $ k \\equal{} \\minus{} 1$ because $ \\frac {2007}{n_2} \\minus{} \\frac {2007}{n_1} \\equal{} \\frac { \\minus{} 1}2\\Rightarrow \\frac {2007}{n_1} \\minus{} \\frac {2007}{n_2} \\equal{} \\frac12$.\r\n\r\n$ (n_2 \\minus{} 2\\cdot3^2\\cdot223)(n_1 \\plus{} 2\\cdot3^2\\cdot223) \\equal{} \\minus{} 2^2\\cdot3^4\\cdot223^2$\r\n\r\nSince $ n_1 > 2007$, $ n_1 \\plus{} 2\\cdot3^2\\cdot223 > 3^3\\cdot223$.\r\n\r\nNow, let $ d \\equal{} n_1 \\plus{} 2\\cdot3^2\\cdot223$.\r\n\r\n$ n_2 \\minus{} 2\\cdot3^2\\cdot223 \\equal{} \\frac { \\minus{} 2^2\\cdot3^4\\cdot223^2}{d} > 3^2\\cdot223 \\minus{} 2\\cdot3^2\\cdot223\\Rightarrow d > 2^2\\cdot3^2\\cdot223$.\r\n\r\nThus, the possible values of $ d$ are $ 2\\cdot3^3\\cdot223,3^4\\cdot223,2^2\\cdot3^3\\cdot223,2\\cdot3^4\\cdot223,2^2\\cdot3^4\\cdot223$ and those of the form $ 2^a\\cdot3^b\\cdot223^2$ for integers $ a,b$, $ 0\\leq a\\leq2$ and $ 0\\leq b\\leq4$. So the number of ordered pairs in this case is $ 5 \\plus{} 3\\cdot5 \\equal{} 20$ ordered pairs.\r\n\r\nLikewise, for the case when $ k \\equal{} \\minus{} 1$, we have another $ 20$ ordered pairs.\r\n\r\nThere are no instances in this case where $ n_1\\equal{}n_2$ :)\r\n\r\n[b]Answer[/b]\r\nTherefore, there are a total of $ 45 \\plus{} 3 \\plus{} 20 \\plus{} 20 \\equal{} 88$ ordered pairs.", "Solution_28": "Thanks for grinding through all that. I think this problem turned out to be a little hard for the AMC 12. However, I'm glad you included it. It was a good lesson on how to organize a problem.", "Solution_29": "is there an answer sheet for the mock amc 12?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ be non-negative numbers.\r\n(a) Prove that\r\n$(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab) \\geq (ab+bc+ca)^{3}$;\r\n\r\n(b) Prove that\r\n$(2a^{2}+bc)(2b^{2}+ca)(2c^{2}+ab) \\geq (ab+bc+ca)^{3}$;\r\n\r\n(c) Find the larger $p$ such that\r\n$(a^{2}+pbc)(b^{2}+pca)(c^{2}+pab) \\geq (1+p)^{3}$\r\nfor $ab+bc+ca=3$;\r\n\r\n(d) Find the larger $p$ such that\r\n$(pa^{2}+bc)(pb^{2}+ca)(pc^{2}+ab) \\geq (p+1)^{3}$\r\nfor $ab+bc+ca=3$.", "Solution_1": "[quote=\"Vasc\"]Let $a,b,c$ be non-negative numbers.\n(a) Prove that\n$(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab) \\geq (ab+bc+ca)^{3}$;\n[/quote]\r\n\r\n$\\Rightarrow \\sum{a^{3}b^{3}}+4\\sum{a^{4}bc}+3a^{2}b^{2}c^{2}\\geq 3 \\sum{a^{3}b^{2}c}+3\\sum{a^{3}bc^{2}}$, true by schur ($\\sum{a^{3}b^{3}}+3a^{2}b^{2}c^{2}\\geq \\sum{a^{3}b^{2}c}+\\sum{a^{3}bc^{2}}$) and muirhead ($2(4,1,1)\\succ2(3,2,1)$)", "Solution_2": "[quote=\"Vasc\"]Let $a,b,c$ be non-negative numbers.\n(b) Prove that\n$(2a^{2}+bc)(2b^{2}+ca)(2c^{2}+ab) \\geq (ab+bc+ca)^{3}$;\n[/quote]\r\n\r\n$\\Rightarrow 2\\sum{a^{3}b^{3}}+2\\sum{a^{4}bc}+(\\sum{a^{3}b^{3}+3a^{2}b^{2}c^{2}) \\geq 3\\sum{a^{3}b^{2}c}+3\\sum{a^{3}bc^{2}}}$, true by schur and muirhead for $(3,3,0)\\succ(3,2,1)$ and $(4,1,1)\\succ(3,2,1)$" } { "Tag": [ "inequalities", "search", "function", "inequalities theorems" ], "Problem": "Could someone give a proof of Karamata's inequality? I tried AoPS wiki and Wikipedia but it's not there.", "Solution_1": "Do you know how to prove Muirhead using repeated applications of AM-GM? It's virtually the exact same proof except using Jensen over and over again.", "Solution_2": "try downloading the pdf of Hungkhtn [url=http://www.mathlinks.ro/portal.php?t=196913]here[/url]", "Solution_3": "[quote=\"MellowMelon\"]Do you know how to prove Muirhead using repeated applications of AM-GM? It's virtually the exact same proof except using Jensen over and over again.[/quote]\nNo, I don't. Could you show me please?\n\n[quote=\"Albanian Eagle\"]try downloading the pdf of Hungkhtn [url=http://www.mathlinks.ro/portal.php?t=196913]here[/url][/quote]\r\nI'm not sure where to download the pdf in that link. :?:", "Solution_4": "I know of several proofs:\r\n\r\n1) Karamata's inequality is Theorem 9 in my [url=http://www.cip.ifi.lmu.de/~grinberg/Popoviciu.pdf]Generalizations of Popoviciu's inequality[/url], and is proven there up to the proof of a lemma which is proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=19097 post #11 as Lemma 1 ( http://www.mathlinks.ro/Forum/viewtopic.php?p=140145#140145 ). But this proof is long and far not the best.\r\n\r\n2) As Albanian Eagle mentioned, Hungkhtn's article contains a proof, but only for the case of a twice differentiable function.\r\n\r\n3) A very nice and short proof was given by [url=http://www.imocompendium.com/tekstkut/ineq_im.pdf]Ivan Matic[/url] and generalized by BG Yoda at http://www.mathlinks.ro/Forum/viewtopic.php?t=18701 . However, just as the proof in point 1), it has some theoretical disadvantages when it comes to generalizing... I wish I had the time to discuss this now :)\r\n\r\nMore proofs can probably be found by the search function, but for some I am not sure of their correctness. A point rather far on my todo list is to write up the combinatorial proof of Karamata, as there seem to be few places where it is properly explained, but unfortunately this has a reason...\r\n\r\n darij" } { "Tag": [ "graph theory", "combinatorics unsolved", "combinatorics" ], "Problem": "let $G$ be a graph without loop.show that there is abipartitespanning subgraph of $G$ like $H$ such that $d_H(V) \\geq \\dfrac{1}{2}d_G(v),v \\in V_G$", "Solution_1": "is it $d_H(v)$ or $d_H(V)$? if it is the second one, what does it mean? does the ineq has to hold for ever vertex of $H$?", "Solution_2": "[quote=\"manuel\"]is it $d_H(v)$ or $d_H(V)$? if it is the second one, what does it mean? does the ineq has to hold for ever vertex of $H$?[/quote]\r\nno different it means degree of vertices like $v$.", "Solution_3": "this should do the trick:\r\n\r\nbasically we need to create a bipartite graph from the given graph by dropping edges. For any $S\\subset V(G)$ define\r\n$b(S): =\\frac{e(S,\\overline{S})}{e(S,S)+e(\\overline{S},\\overline{S})}$ to be the $\\textit{bipartiteness}$ of the set $S$, where by $e(X,Y)$ we denote the number of edges between the sets $X,Y$. Consider a set $S$ such that $b(S)$ is maximum.\r\ni claim that for such a set, for any vertex $v\\in S,d_S(v)\\leq \\frac{d(v)}{2}$ where $d_X(v)$ denotes the degree of $v$ in the set $X$.\r\n suppose not. Then consider $T=S\\setminus\\{v\\}$. Note that $e(T,\\overline{T})=\\sum_{x\\in T,y\\in \\overline{T}} 1_{xy\\in E(G)},e(T,T)=\\sum_{x,y\\in T} 1_{xy\\in E(G)}$ and likewise for $e(\\overline{T},\\overline{T})$.\r\nNow $ b(T)=\\frac{\\sum_{x\\in T,y\\in \\overline{T}} 1_{xy\\in E(G)}}{\\sum_{x,y\\in T} 1_{xy\\in E(G)}+\\sum_{x,y\\in \\overline{T}} 1_{xy\\in E(G)}}=\\frac{e(S,\\overline{S})+(d_T(v)-d_{\\overline{S}}(v))}{e(S,S)+e(\\overline{S},\\overline{S})-(d_T(v)-d_{\\overline{S}}(v))}$.\r\nIn the last expression ,the term in brackets (present both in the numerator and denominator) is positive(by assumption) and hence it follows that $b(S)0$, so I'm posting a new solution here:\r\n\r\nFrom $f(f(x))=x+1$ we can easily infer that $f(t)=t+n$ for some $t$ and some positive integer $n$. Then $f(t+n)=t+1$. We now use induction to prove that $f(m)=m+n$ for all $m>t$: suppose this is true when $m=k$. Then $f(k)=k+n$, $f(k+n)=k+1$, and $f(k+1)=k+n+1$ follows. Setting $m=t+n$ yields a contradiction." } { "Tag": [ "topology", "function", "advanced fields", "advanced fields unsolved" ], "Problem": "Suppose that $A, B \\subset [0, 1]$ are measurable with Hausdorff dimensions $a$ and $b$ respectively. \r\n\r\nWe know that $A \\times B \\subset [0, 1]^{2}$ has Hausdorff dimension at least $a+b$. But then if we project that set onto the x axis or the y axis, we get smaller (in dimension) sets (if $0 < a, b < 1$, say). So, even though the dimension is decreased (or =1, if $a+b > 1$) for almost every projection, the two 'obvious' ones are bad. \r\n\r\nSo let's say I want a set $E \\in [0, 1]^{2}$ such that the projection of $E$ onto the x axis is $A$ and onto the y axis is $B$. In a way $A\\times B$ is the most obvious choice, but the above observation suggests that that choice is kinda weird. Specifically, $A \\times B$ is the largest possible set $E$ with the desired property, and it seems like there ought to be smaller ones that would behave better when we did stuff to them. \r\n\r\nSo here are the things I want to prove (or did myself but am giving to you for your enjoyment) [difficulty varies]:\r\n\r\n1) $A\\times B$ is the largest possible set with the property mentioned above. \r\n2) There always exists a set $E$ with the desired property and also with $\\text{dim}_\\mathcal{H}(E) = \\text{max}(a, b)$. \r\n3) There always exist sets $E$ with the desired properties and with dimension $s$, $\\text{max}(a, b) < s < \\text{dim}_\\mathcal{H}(A\\times B)$.\r\n4) What can we say about the lower Minkowski dimensions of the sets in 2 and 3, if the lower Minkowski dimensions of $A$ and $B$ are $\\alpha$ and $\\beta$?", "Solution_1": "1. $E \\subset \\pi_{1}(E)\\times\\pi_{2}(E)$.. anything more to be said? :lol: :blush: \r\n2. pick $\\overline{a}\\in A, \\overline{b}\\in B$, and choose $E = A\\times\\{\\overline{b}\\}\\cup \\{\\overline{a}\\}\\times B$. $\\dim_\\mathcal{H}E = \\max\\{a,b\\}$.\r\n\r\njust to shoot out the two obvious ones :wink:\r\n\r\n3. this could follow from a proposition like: for every set of hausdorff measure $d>0$ the image through the function hausdorff dimension of its measurable subsets is $[0,d]$.. i guess it's true, but i have no idea if it's a well-known fact or not.", "Solution_2": "More generally, is it true that any metric space $X$ contains a subset of Hausdorff dimension $s$ for any $00$).I thought it was equal $(A^{-1})^{n}$.", "Solution_1": "can someone says if its right??", "Solution_2": "Yes, that is the meaning, which means that this is only defined for invertible matrices.", "Solution_3": "thank you :)" } { "Tag": [], "Problem": "There are two colors Jessica can use to color this 2 by 2 grid. If non-overlapping regions that share a side cannot be the same color, how many possible ways are there for Jessica to color the grid?\n\n[asy]size(101);\ndraw(unitsquare);\ndraw((0,.5)--(1,.5));\ndraw((.5,0)--(.5,1));\nlabel(\"$A$\",(0,1),NW); label(\"$B$\",(1,1),NE); label(\"$C$\",(1,0),SE); label(\"$D$\",(0,0),SW);[/asy]", "Solution_1": "Say the colors are blue and red.\r\n\r\nThe only possible ways to color this grid are:\r\n\r\nB R\r\nR B\r\n\r\nor\r\n\r\nR B\r\nB R\r\n\r\nthus, the answer is 2." } { "Tag": [ "email", "inequalities", "pigeonhole principle", "HMMT", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Hello, everyone! We are now announcing some of the basic information about our contest, such as the date and format. \r\n\r\nThis information is all available [url=https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition]here[/url]. \r\n\r\nAll updates will be posted on that page as well. We'll have a schedule available soon (possibly today). If you're interested, you're welcome to post here, PM me, or email me (my email address is on my profile), in addition to using the contact information on our Math Club page. \r\n\r\nFeel free to ask any questions! We'll answer as quickly as we are able.", "Solution_1": "Please note that the date of the contest is now Dec 16! We had to change the date due to conflicts. Hopefully the new date will be more convenient for everyone. \r\n\r\nAlso note that the schedule will soon change to make the overall length of the contest shorter. Hopefully that will be more convenient for people (especially those who have to travel) as well. I'll post again to bring it to attention when that happens.", "Solution_2": "wait, this thing is serious? I thought Peter was just joking...", "Solution_3": "Haha, I don't think we'd have gotten our own forum if this were just a joke!", "Solution_4": "Please note that our date is now set to Saturday, December 16th, and not Saturday, November 11th. Feel free to send questions or concerns to Nathan or any of the rest of us.", "Solution_5": "What kind of stuff is on the Advanced Topics test?", "Solution_6": "I can't tell you exactly what's on there now that we've started writing the questions. However, I can tell you that we initially wanted the advanced topics test to include popular competition topics such as inequalities, sequences and series, combinatorics, and problems using methods like the pigeonhole principle and the invariance principle. So basically intended for people with a significant amount of contest experience.", "Solution_7": "For those wishing to practice, our tests will be similar in format to Rice and HMMT tests (both have their old tests online), and reasonably similar in content and difficulty as well. If you want to take the Number Theory test, just any general studying of number theory is probably good. If you can do most USAMO number theory problems, you will not find our test too difficult. :P", "Solution_8": "Nathan, how many people do you know who can do most USAMO number theory problems??\r\n\r\nDon't be intimidated by Nathan...he has no concept of how smart the average math competitor is... ;)", "Solution_9": "Haha I can't do USAMO NT problems. :ninja:", "Solution_10": "I have one question. For the tiebreakers you say that it's for the highest scoring people on the Individual. But is it for each subject? So say the highest score for algebra is 7 and two people got that, and the highest score for geometry is 8, and three people got that. Would all five people go to the tiebreaker round? Also, in the tiebreaker round, are the tiebreaker questions the same subject? For instance, if the two people mentioned above with 7s on algebra go to tiebreaker round, do they both get algebra tiebreaker questions?", "Solution_11": "For the first question: yes, we'll break ties for each individual test. \r\n\r\nFor the second: that would make sense. I think that's what we intended, but don't be totally shocked if I'm wrong, since the format of the tiebreaker is WindingFunction's idea, and I don't remember exactly what she said.", "Solution_12": "Ok, so tiebreaker format, unless there are a large number of complaints:\r\n\r\nEveryone writes the same tiebreaker, regardless of which individual test you took. The tiebreaker consists of six questions, two from each of three topics - Algebra and Geometry (combined), Number Theory, and Advanced Topics. (Sorry, calculus-lovers, but we can't penalize people who don't know calculus.) At the beginning of the tiebreaker, everyone gets one question from each of the three topics. When you're ready, you may hand in a FULL SOLUTION with the answer boxed. If your answer is correct, you will receive the next (harder) question in the same topic. When thirty minutes is up, or when someone finishes all six questions (unlikely, but maybe some of you geniuses out there will surprise us), we will grade the solutions that you have handed in at that point, and you will get a Tiebreaker Score out of 42 (7 marks per question).\r\n\r\nPS - Nathan, this is not just a random concoction of mine...I ran most of it by all of you, and I ran the final version by Aaron. However, it is subject to change if there's enough dislike. :P", "Solution_13": "[quote=\"WindingFunction\"]PS - Nathan, this is not just a random concoction of mine...I ran most of it by all of you, and I ran the final version by Aaron. [/quote]\r\nI know. But I have no memory capacity and I'm incompetent. :D", "Solution_14": "[quote=\"WindingFunction\"]Sorry, calculus-lovers, but we can't penalize people who don't know calculus.[/quote]\r\n\r\nYes you can!\r\n\r\nalso why are algebra and geometry combined they are most decidedly not the same thing", "Solution_15": "Let's just push it back indefinitely, why don't we?\r\n\r\nSeriously, there's always going to be something that's going to come up. And while we hope as many of you as possible can make it, we really have looked at a lot of possible dates and chosen December 16th as the best one. So we don't really want to change it now. We hope you understand, and we hope to see you there anyways.", "Solution_16": "I have exams the subsequent week. I'll be there. :lol:", "Solution_17": "[quote=\"Pakman2012\"]I have exams the subsequent week. I'll be there. :lol:[/quote]\r\n\r\nGlad to hear it. :)", "Solution_18": "[quote=\"WindingFunction\"][quote=\"JSteinhardt\"]Out of curiosity, how does one write 6 proofs in 30 minutes? Unless the problems are trivial and the solutions are < 5 lines.\n\nThat wouldn't be much of a tiebreaker though...\n\nI realize that people aren't expected to solve all 6 problems, but shouldn't it at least be possible to do so?\n\nAlso if it stops as soon as somewhat submits all 6 tiebreaker questions, what is to stop me from finishing the first one as quickly as possible, then submitting the next 5 as blank pages immediately after?[/quote]\n\nWell, it's mainly meant so that you can \"prioritize\" and solve the questions in the topics that are most suited to you. Also, you can't submit the other 5 problems after just solving one problem, because you only receive problems 4, 5, and 6 by handing in the correct answer to an earlier problem in the same subject area. So in order for your master plan to work, you would have to finish the first THREE problems as quickly as possible, which may take you the whole 30 minutes anyways.[/quote]\r\n\r\nSo are you suggesting that the optimal strategy would be to just solve the first three as fast as possible? Or will the 4,5,6 questions not be that much harder than the 1,2,3 questions such that it might be better to not try to do 1,2,3 first?", "Solution_19": "[quote=\"mathdool\"]So are you suggesting that the optimal strategy would be to just solve the first three as fast as possible? Or will the 4,5,6 questions not be that much harder than the 1,2,3 questions such that it might be better to not try to do 1,2,3 first?[/quote]\r\nI think that would depend on your strengths. Personally, I can do a \"very hard\" inequality problem much more easily than a \"very easy\" geometry problem, so...", "Solution_20": "This question wasn't 'exactly' asked before but on the website it only says \"2. The top students from each Individual Test will be asked to participate in the Tiebreaker\", but it never specifies how many, and from some posts on the first page of this thread, it seems as if only the top single person from each Individual Test gets to do the tiebreaker, is that true?", "Solution_21": "[quote=\"SplashD\"]This question wasn't 'exactly' asked before but on the website it only says \"2. The top students from each Individual Test will be asked to participate in the Tiebreaker\", but it never specifies how many, and from some posts on the first page of this thread, it seems as if only the top single person from each Individual Test gets to do the tiebreaker, is that true?[/quote]\r\nSorry for the lack of clarity, but actually we're just going to have people who tie for one of the top 3 spots in their individual test participate in the tiebreaker.", "Solution_22": "Can we make it like a november tournament next year?", "Solution_23": "[quote=\"Pakman2012\"]Can we make it like a november tournament next year?[/quote]\r\nYes. In fact, we've already selected a date and it is in November. We'll post details once everything's set (which should be soon).", "Solution_24": "Aww... All of my Fall contests are in November, and I liked it right before break... I hope I can still come.", "Solution_25": "Hey, is it possible to post the relay competition as its own file (not just as part of the solutions)? Thanks. I was looking through the solutions and realized I missed a part of the competition :-). I want to go back and do it the honest way.", "Solution_26": "Is there any news on next year's contest? In particular, I would like to know whether you can make sure that PUMAC won't conflict with the Duke Math Meet. If they conflict, you may very well lose NC/SC/GA/TJ and others.", "Solution_27": "[quote=\"jb05\"]Is there any news on next year's contest? In particular, I would like to know whether you can make sure that PUMAC won't conflict with the Duke Math Meet. If they conflict, you may very well lose NC/SC/GA/TJ and others.[/quote]\r\nAs far as I can tell from their website, the Duke people haven't scheduled their contest (or haven't announced the date). We were planning to do ours Nov 10, I think, but don't consider that an official announcement, since my own computer died and I am just producing this from memory. We'll have an official announcement of the 2007 contest soon and then we'll open the registration. But before that can happen, I need to fix my computer and get reliable internet access!", "Solution_28": "November 10 is correct.", "Solution_29": "We need to archive some of this stuff out. It's pretty distracting." } { "Tag": [], "Problem": "Prove that the difference between any two odd perfect squares is divisible by 8.", "Solution_1": "The square of every odd number is $\\equiv1(mod8)$ (this is a good exercise)\r\nand the conclusion results." } { "Tag": [ "summer program", "MathPath" ], "Problem": "Please post - thanks!", "Solution_1": "Considering that the camp is over, I don't think it would be bad to post the questions in interm topics that you have trouble with.", "Solution_2": "In fact, if anyone here has ANY answers or solutions from ANY prior Mathpath qualifying test - any year - could you please share either on site, or if easier, pm me! Thanks" } { "Tag": [], "Problem": "Let a and b be integers and n a positive integer. Prove that $\\frac{b^{n-1}a(a+b)(a+2b)...(a+(n-1)b)}{n!}$ is an integer.", "Solution_1": "anyone wish to try or should i post a hint?", "Solution_2": "hint please", "Solution_3": "Link please ;)\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=21347", "Solution_4": "you've done it again ZetaX, thanks. :) (Sarcasm is not necesary)\r\njhcreinhardt, if you still need help, feel free to pm me." } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "Hi,\r\nCan you help me solve this ODE with fourier series?\r\n\r\nu'' - c^2 u + x = 0 ; u (0) = u (l) = 0", "Solution_1": "This doesn't seem like you'd need Fourier series to me. Suppose that your solution has the form $U(x)=Ax+B$. Then $c^{2}(Ax+B)=-x \\Rightarrow Ac^{2}=-1$ and $B=0$. Now from standard theory use the auxilary equation $r^{2}-c^{2}=0 \\Rightarrow (r-c)(r+c)=0\\Rightarrow r=\\pm c$ so $u(x)=C_{1}e^{cx}+C_{2}e^{-cx}-\\frac{x}{c^{2}}$ from the initial conditions we get $u(0)=C_{1}+C_{2}=0$ and $u(1)=C_{1}e^{c}+C_{2}e^{-c}-1=0$ so $C_{1}=\\frac{1}{e^{c}-e^{-c}}$ and $C_{2}=-C_{1}$ so $u(x)=\\frac{e^{cx}}{e^{c}-e^{-c}}-\\frac{e^{cx}}{e^{c}-e^{-c}}-\\frac{x}{c^{2}}$.\r\n\r\n(My guess is that even this is more complicated than necessary and that there is a better trig way (or maybe I'm missing or miss interpreting something that would need Fourier series to be used.) When I use Maple to check this answer it doesn't seem to simplify correctly, so the answer isn't correct, but I feel like since it is a second-order linear, method of undetermined coefficients should work." } { "Tag": [ "geometry open", "geometry" ], "Problem": "Is there something like this:\r\n\r\nPoint $M$ in the triangle $ABC$ for which $d_1\\cdot MA+d_2\\cdot MB+d_3\\cdot MC$ is minimum is......\r\n\r\n$d_1,d_2,d_3>0$.", "Solution_1": "[quote=\"dule_00\"]Is there something like this:\n\nPoint $M$ in the triangle $ABC$ for which $d_1\\cdot MA+d_2\\cdot MB+d_3\\cdot MC$ is minimum is......\n\n$d_1,d_2,d_3>0$.[/quote]\r\n\r\nI know a proof in \"the mathematical gazette\" page 316,317 of Nguyen Minh Ha.\r\nIf $d_1,d_2,d_3$ is three side of a triangle with angles $A_1,A_2,A_3$ then min of sum when $M$ is $N$ that $\\angle BNC=\\Pi-A_1, \\angle CNA=\\Pi-A_2$ \r\nIn other case it is quite intricate..." } { "Tag": [], "Problem": "I think I saw Christmas trees in sale at my town (ugh, earlier every year) ... so time for a Christmas problem!\r\n\r\nEvery Christmas Day, Jimmy is presented, by his very rich true love, with every gift given in the 12 Days of Christmas song. He enjoys them for one day, then decides to give them away, one gift per day. Supposing leap year never happens, on what date does Jimmy give away his last gift every year?\r\n\r\n(Yes, it's just an adding problem.)", "Solution_1": "Well...\r\n\r\n[hide=\"Hint\"]\nOne the first day he gets 1, and he gets that every day. So thats 12*1 presents\nOne the second day he gets 2, and gets that 11 times, so thats 11*2\nOne the third day he gets 3.... 10*3\n...\n[/hide]", "Solution_2": "[hide=\"Answer\"]He receives a total of $ \\sum_{i\\equal{}1}^{12}\\sum_{j\\equal{}1}^{i} j$ or $ 364$ gifts. So he gives away the last one on December 24, only to get another load of gifts the next day. Poor Jimmy.[/hide]" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "$ A$ hermitian \r\n$ B$ is positive semidefinite \r\n$ f(t) \\equal{} Tr(e^{A \\minus{} tB})$\r\n\r\nProve that for any $ k\\in N$ and any $ t\\in [0, \\plus{} \\infty[$\r\n\r\n$ ( \\minus{} 1)^{k \\plus{} 1}f^{(k)}(t)\\geq 0$", "Solution_1": "[quote=\"Moubinool\"]$ A$ hermitian \n$ B$ is positive semidefinite \n$ f(t) \\equal{} e^{A \\plus{} tB}$\n\nProve that for any $ k\\in N$ and any $ t\\in [0, \\plus{} \\infty[$\n\n$ ( \\minus{} 1)^{k \\plus{} 1}f^{(k)}(t)\\geq 0$[/quote]\r\nMoubinool, I think you missed a trace somewhere, or you need to define what \"$ \\geq 0$\" means.\r\nIf you wanted to asked that about the Bessis-Moussa-Villani conjecture, I doubt that it can be solved on this forum in its entirety, and I refer to http://www.msri.org/people/members/chillar/files/BMVtraceconjcounterex.pdf" } { "Tag": [], "Problem": "Simplify: $ \\frac{x^{10}\\plus{}x^{10}\\plus{}x^{10}}{x^{10}}$ if $ x\\neq0$", "Solution_1": "we can just plug $ 1$ into each of the $ x$'s\r\n\r\nso we have $ \\frac {1^{10} \\plus{} 1^{10} \\plus{} 1^{10}} {1^{10}} \\equal{} \\frac {3} {1} \\equal{} 3$", "Solution_2": "$ x^{10} \\plus{} x^{10} \\plus{} x^{10} \\equal{} 3\\times x^{10}$\r\n\r\n$ \\frac {3\\times x^{10}}{x^{10}} \\equal{} 3$\r\n\r\nanswer : 3", "Solution_3": "You could imagine that $ x^{10} \\equal{} n$.\r\n\r\nThen you would have $ \\frac{n \\plus{} n \\plus{} n}{n}$, or $ \\frac{3n}{n}$. That's how I like to do these problems." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $x_{1},...,x_{2n}$ be a permutation of the numbers $1,2,...,2n$ .Find by proof the minimum and the maximum of \r\n$A=\\left|a_{1}-a_{2}\\right|+\\left|a_{2}-a_{3}\\right|+....+\\left|a_{2n-1}-a_{2n}\\right|$", "Solution_1": "Maybe not hard:\r\n$A\\leq n+1+\\dots+2n-(1+2+\\dots+n)=n^{2}$\r\n$A\\geq n$", "Solution_2": "it's same with [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=495837#p495837]Balkan MO 1994 #3[/url]" } { "Tag": [ "ratio" ], "Problem": "Question 3 (45 minutes)\r\nData: E0\r\n(Ag+/Ag) = 0.800 V log Kstab ([Ag(CN)2]\u2013) = 21.1\r\nE0\r\n(SCE) = 0.285 V F = 96485 C mol\u20131\r\npKsp (AgCN) = 15.8 R = 8.315 J K\u20131 mol\u20131\r\npKsp (AgCl) = 9.75 = 8.206 x 10\u20132 L atm J K\u20131 mol\u20131\r\nAn electrochemical cell is prepared by placing a silver electrode in 25.00 mL of a neutral\r\nsolution containing potassium chloride and potassium cyanide, then joining a standard\r\ncalomel electrode with a potassium nitrate salt bridge. The KCl/KCN solution is then\r\npotentiometrically titrated with a standard 0.1000 M silver nitrate solution at 25 \u00b0C. The\r\npotentiometric curve obtained (cell potential against burette reading) is shown below.\r\n\r\nThroughout this problem, you may assume that protonation of the cyanide ion is negligible.\r\n(a) The endpoints of the reactions taking place during the titration are marked with A, B and C.\r\nWrite a balanced ionic equation for each reaction.\r\n(b) What volume of the titrant is required to reach point B?\r\n(c) Calculate the concentrations of KCl and of KCN in the sample solution.\r\n(d) Calculate the emf readings at points A and C in volts.\r\n(e) What is the molar ratio Cl\u2013/CN\u2013 in the solution at point C?\r\n(f) What is the molar ratio Cl\u2013/CN\u2013 in the precipitate at point C?\r\n\r\nI don't know what the reactions are", "Solution_1": "anybody have any idea\r\n\r\nwhich reactions endpoint happens first" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "for all $ x,y,z\\succ 0$ such that : $ x \\plus{} y \\plus{} z \\equal{} 1$\r\nShow that :\r\n\r\n$ (1 \\plus{} \\frac {1}{x})(1 \\plus{} \\frac {1}{y})(1 \\plus{} \\frac {1}{y})\\geq\\,64$", "Solution_1": "let [b]x+y+z=p[/b] and [b]xy+yz+zx=q[/b] and [b]xyz=r[/b]\r\n\r\n now we have [b]p=1[/b] and we must prove that [b]2+q>= 63r[/b] and by AM-GM we have [b]q>=9r[/b] so we must prove that [b]1>=27r[/b] that is AM-GM :wink:", "Solution_2": "because $ x\\plus{}y\\plus{}z \\equal{} 1$\r\nLet $ x\\equal{}\\frac{a}{a\\plus{}b\\plus{}c} , y\\equal{}\\frac{b}{a\\plus{}b\\plus{}c} , z\\equal{}\\frac{c}{a\\plus{}b\\plus{}c}$\r\n$ \\therefore LHS \\equal{} \\frac{(2a\\plus{}b\\plus{}c)(a\\plus{}2b\\plus{}c)(a\\plus{}b\\plus{}2c)}{abc} \\geq 64$ (By AM-GM)", "Solution_3": "A.M.-G.M. gives $ \\frac {x \\plus{} y \\plus{} z}{3}\\geq \\sqrt [3]{xyz}\\Longleftrightarrow \\frac {1}{3}\\geq \\sqrt [3]{xyz}\\ (x,\\ y,\\ z > 0)$, yielding $ \\frac {1}{xyz}\\geq 27$.\r\n\r\n$ \\therefore \\left(1 \\plus{} \\frac {1}{x}\\right)\\left(1 \\plus{} \\frac {1}{y}\\right)\\left(1 \\plus{} \\frac {1}{z}\\right)$\r\n\r\n$ \\equal{} 1 \\plus{} \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\plus{} \\frac {1}{xy} \\plus{} \\frac {1}{yz} \\plus{} \\frac {1}{zx} \\plus{} \\frac {1}{xyz}$\r\n\r\n$ \\geq 1 \\plus{} 3\\sqrt [3]{\\frac {1}{xyz}} \\plus{} 3\\sqrt [3]{\\frac {1}{xy}\\cdot \\frac {1}{yz}\\cdot \\frac {1}{zx}} \\plus{} \\frac {1}{xyz}$\r\n\r\n$ \\equal{} 1 \\plus{} 3\\sqrt [3]{27} \\plus{} 3\\sqrt [3]{27^2} \\plus{} 27$\r\n\r\n$ \\equal{} 1 \\plus{} 9 \\plus{} 27 \\plus{} 27$\r\n\r\n$ \\equal{} 64$. Q.E.D.", "Solution_4": "Multiply both side by $ xyz$ to get\r\n\r\n$ \\left (1 \\plus{} x \\right) \\left(1 \\plus{} y \\right) \\left(1 \\plus{} z \\right) \\geq 64xyz$\r\n\r\n$ \\Longleftrightarrow \\left( \\frac {x \\plus{} x \\plus{} y \\plus{} z}{4} \\right) \\left( \\frac {x \\plus{} y \\plus{} y \\plus{} z}{4} \\right) \\left( \\frac {x \\plus{} y \\plus{} z \\plus{} z}{4} \\right) \\geq xyz$\r\n\r\n But \r\n\r\n$ \\left( \\frac {x \\plus{} x \\plus{} y \\plus{} z}{4} \\right) \\left( \\frac {x \\plus{} y \\plus{} y \\plus{} z}{4} \\right) \\left( \\frac {x \\plus{} y \\plus{} z \\plus{} z}{4} \\right)$$ \\geq \\sqrt [4]{x^2yz} \\sqrt [4]{xy^2z} \\sqrt [4]{xyz^2} \\equal{} \\sqrt [4]{x^4y^4z^4} \\equal{} xyz$", "Solution_5": "nice sollutions :blush:" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "$A$ - An infinite collection of equilateral hexagons with length of side 1.\r\n$B$ - An infinite collection of equilateral hexagons with length of side 2.\r\nCan we face the plane using at least one hexagon from each collection ?", "Solution_1": "just wondering:\r\n\r\nBy \"equilateral\", you meant regular? like all angles and sides are equal?\r\n\r\nAnd also, your last sentence is rather confusing. You mean we have to tile the plane with only A and B, and use at least one of each?", "Solution_2": "[quote=\"al.M.V.\"]just wondering:\n\nBy \"equilateral\", you meant regular? like all angles and sides are equal?\n\nAnd also, your last sentence is rather confusing. You mean we have to tile the plane with only A and B, and use at least one of each?[/quote]\r\nYes, I meant that, sorry for my English. :blush:", "Solution_3": "No. Firstly notice that each hexagon has an angle of 120 degrees at its vertice. At some point, two hexagons of different sizes will touch. Then, have several cases:\r\n1. They touch at just 1 point. Then, both angles that have vertices at that pt will be less than 120 degrees, and it is impossible to tile that area with regular hexagons.\r\n2. They touch at a line segment. Then, because they have different sizes, at the one of the endpoints of that line segment we have a line through it and a 120 degree angle. Hence the third angle is 60 degress. But this is less than 120 and hence cannot be tiled\r\nThat's all the cases and we are done." } { "Tag": [], "Problem": "Let $a_{1}, \\cdots, a_{44}$ be natural numbers such that \\[00$ darim:\n\n$ \\frac{a_{1}}{a_{2}+a_{3}}+\\frac{a_{2}}{a_{3}+a_{4}}+...+\\frac{a_{n}}{a_{1}+a_{2}}>\\frac{n}{4}$\n\n$ \\boxed{\\text{problem 2}}$(Moldova 2005):$ x_{1},...,x_{n}>0$ sabet konid:\n\n$ \\sum_{i=1}^{n}\\frac{x_{i}}{x_{i+1}+x_{i+2}}>(\\sqrt{2}-1)n$[/b]\r\n\r\n[u]@N.T.TUAN:please dont answer these two problems,and let others to think on them... :) [/u]", "Solution_4": "jofte mesaalhat ghalatan.tooye ghaziyeye shapiro faghat adadhaye mosbat made nazar hastan.", "Solution_5": "jofte masaleham dorostano hich iradiam nadaran,rahe halle jalebiam daran,pas age fek mikoni halatan mesale naghz bezano khodet emtehanesh kon... :wink: \r\ntooye soorate shapiro ham neveshti ke bayad mosbat bashan,dge dobare vase chi gofti? :huh:", "Solution_6": "[quote=\"BaBaK Ghalebi\"]jofte masaleham dorostano hich iradiam nadaran,rahe halle jalebiam daran,pas age fek mikoni halatan mesale naghz bezano khodet emtehanesh kon... :wink: \ntooye soorate shapiro ham neveshti ke bayad mosbat bashan,dge dobare vase chi gofti? :huh:[/quote]\r\nsefr($ 0$) mosbate?\r\nhala fahmidi?\r\nmanzooram MESAAL bood na MASALE!!!\r\ngerefti?", "Solution_7": "$ 1+5\\epsilon,6\\epsilon,1+4\\epsilon,5\\epsilon,1+3\\epsilon,4\\epsilon,1+2\\epsilon,3\\epsilon,1+\\epsilon,2\\epsilon,1+2\\epsilon,3\\epsilon,1+\\epsilon,2\\epsilon,1+4\\epsilon,3\\epsilon,1+5\\epsilon,4\\epsilon,1+6\\epsilon,5\\epsilon$\r\n\r\nfek konam in yeki dge doros bashe... :wink:", "Solution_8": "[url]https://artofproblemsolving.com/community/c6h1331833p7184392[/url]\n[url]https://artofproblemsolving.com/community/c6h589p1818[/url]\n[url]https://olympiads.mccme.ru/lktg/2010/5/5-1en.pdf[/url]\n[url]https://artofproblemsolving.com/community/c6h3131209p28366381[/url]\n[url]https://www.doc88.com/p-8922218882672.html[/url]" } { "Tag": [], "Problem": "one anagram for art of problem solving! is girls from a bloop vent!\r\n\r\ntry to find some more", "Solution_1": "There, let's anagram! Arrange mathletes.", "Solution_2": "Hate Melts?", "Solution_3": "Lets keep the anagrams to here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35980", "Solution_4": "[quote=\"DarkKnight\"]Lets keep the anagrams to here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35980[/quote]\r\n\r\nI'd say that's a different topic.", "Solution_5": "[quote=\"maniacman384\"]Hate Melts?[/quote]\r\n\r\nActually what I wrote originally was an anagram, but that's ok :P." } { "Tag": [], "Problem": "Propose structures for compounds A - C:\r\n\r\n1) Cathecol + $ \\ce{ClCH2COCl}/\\ce{POCl3}$ -----> A\r\n\r\n2) A + $ \\ce{CH3NH2}$ -----> B\r\n\r\n3) B + $ \\ce{H2}/\\ce{Pd}$ -----> C", "Solution_1": "long time since i posted coz i was outta town and no time/access to the internet.\r\nneways- the product is this i think--> http://en.wikipedia.org/wiki/Adrenaline", "Solution_2": "Yes, that's the correct answer. How about a mechanistic explanation?", "Solution_3": "well ok.\r\nin the first step- OH attacks carbonyl group, foll by fries rearrangement to para position.\r\nsecond is again nucleophilic substitution of the remaining Cl.\r\nthrid is reduction of carbonyl group to get a racemic mixture.", "Solution_4": "The first step is not correct. What is the purpose of $ \\ce{POCl3}$?", "Solution_5": "$ \\ce{POCl3}$ helps in the fromation of acylium ion and also helps in fries rearrangement. :maybe:" } { "Tag": [ "logarithms", "function", "real analysis", "real analysis unsolved" ], "Problem": "$ u\\equal{}x^{y^z} \\Rightarrow \\frac{\\partial u}{\\partial z}\\equal{}?$\r\n$ lnu\\equal{}ln(x^{y^z})\\equal{}y^{z} \\cdot lnx \\Rightarrow \\frac{du}{u}\\equal{} (y^{z} \\cdot lny \\cdot lnx)dz$\r\n$ \\frac{du}{dz}\\equal{}u \\cdot (y^{z} \\cdot lny \\cdot lnx) \\equal{} x^{y^z} \\cdot (lnx \\cdot lny \\cdot y^{z})$\r\n$ \\Rightarrow \\frac{\\partial u}{\\partial z}\\equal{} x^{y^z} \\cdot (lnx \\cdot lny \\cdot y^{z}) $\r\nIs this correct ?", "Solution_1": "I got the same thing when I tried it. As an alternative approach, write $ a^b \\equal{} e^{b\\ln a}$. In this case, $ x^{y^z} \\equal{} e^{y^z\\ln x} \\equal{} e^{e^{z\\ln y}\\ln x},$ which allows you to simply apply the chain rule and basic facts about the exponential function.", "Solution_2": "let u=x^f(z),f(z)=y^z.u is a compoud function depend on z .then use the chain rule" } { "Tag": [], "Problem": "Hi everyone\r\nis there any one who has this book: \r\nFunctional Equations and How to Solve Them by Christopher G. Small by springer verlag \r\nIf you have please attach here. \r\nthanks \r\nBye.", "Solution_1": "Maybe that I (or who) have but we can upload here because it is copyright! :)", "Solution_2": "[quote=\"N.T.TUAN\"]Maybe that I (or who) have but we can upload here because it is copyright! :)[/quote]\r\n\r\n\"p\u00e1c\" Tu\u00e2n \u01a1i,b\u1ea1n \u00fd h\u1ecfi cu\u1ed1n j\u00ec dz\u1eady?? :?:", "Solution_3": "Please don't post Vietnamese on different forum , except Vietnam forum :mad:" } { "Tag": [ "inequalities", "inequalities unsolved", "109" ], "Problem": "Given $ a, b, c \\geq\\ 0$. Prove that:\r\n$ \\frac {a^3 \\plus{} b^3}{(a \\plus{} b)^3} \\plus{} \\frac {b^3 \\plus{} c^3}{(b \\plus{} c)^3} \\plus{} \\frac {c^3 \\plus{} a^3}{(c \\plus{} a)^3} \\plus{} \\frac {27}{4}.\\frac{abc}{a^3 \\plus{} b^3 \\plus{} c^3} \\leq\\ 3$", "Solution_1": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. Prove that:\n$ \\frac {a^3 \\plus{} b^3}{(a \\plus{} b)^3} \\plus{} \\frac {b^3 \\plus{} c^3}{(b \\plus{} c)^3} \\plus{} \\frac {c^3 \\plus{} a^3}{(c \\plus{} a)^3} \\plus{} \\frac {27}{4}.\\frac {abc}{a^3 \\plus{} b^3 \\plus{} c^3} \\leq\\ 3$[/quote]\r\n$ \\sum_{cyc}\\frac {a^3 \\plus{} b^3}{(a \\plus{} b)^3} \\plus{} \\frac {27abc}{4(a^3 \\plus{} b^3 \\plus{} c^3)} \\leq\\ 3\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\left(1 \\minus{} \\frac {a^2 \\minus{} ab \\plus{} b^2}{(a \\plus{} b)^2}\\right)\\geq\\frac {27abc}{4(a^3 \\plus{} b^3 \\plus{} c^3)}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac {ab}{(a \\plus{} b)^2}\\geq\\frac {9abc}{4(a^3 \\plus{} b^3 \\plus{} c^3)}.$\r\nBut by AM-GM we obtain: $ \\sum_{cyc}\\frac {ab}{(a \\plus{} b)^2}\\geq3\\sqrt [3]{\\frac {a^2b^2c^2}{(a \\plus{} b)^2(a \\plus{} c)^2(b \\plus{} c)^2}}.$\r\nId est, it remains to prove that $ 64(a^3 \\plus{} b^3 \\plus{} c^3)^3\\geq27(a \\plus{} b)^2(a \\plus{} c)^2(b \\plus{} c)^2abc,$ which is obvious.\r\nThe following stronger inequality is true too. See here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=221924", "Solution_2": "Thank you, arqady. My proof is same to you.\r\nThe stronger is nice and also easy:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=221924\r\n :)", "Solution_3": "The following more general statement holds: \r\n\r\nIf $ a, b, c$ be real numbers such that $ 7bc\\plus{}a^2\\geq0,7ca\\plus{}b^2\\geq0,7ab\\plus{}c^2\\geq0$ and $ (b\\plus{}c)(c\\plus{}a)(a\\plus{}b)\\neq0$, then\r\n\r\n$ \\frac {b^3 \\plus{} c^3}{(b \\plus{} c)^3} \\plus{} \\frac {c^3 \\plus{} a^3}{(c \\plus{} a)^3}\\plus{}\\frac {a^3 \\plus{} b^3}{(a \\plus{} b)^3} \\plus{}\\frac {27abc}{4(a^3 \\plus{} b^3 \\plus{} c^3)} \\leq\\ 3.$" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Let $ A,B,C$ be three distinct points on a circle whose radius is 1. Let $ G,H$ be the centroid and the orthocenter of triangle $ ABC$, respectively. Let $ F$ be the midpoint of $ GH$. Evaluate $ AF^2\\plus{}BF^2\\plus{}CF^2$.", "Solution_1": "[color=darkblue]By median theorem in triangles $ AGH,BGH$ and $ CGH$ we have: $ \\left\\| \\begin{array}{cc} \nAF^2\\equal{}\\frac{1}{4}\\big(2(AG^2\\plus{}AH^2)\\minus{}GH^2\\big)\\\\\n\\\\BF^2\\equal{}\\frac{1}{4}\\big(2(BG^2\\plus{}BH^2)\\minus{}GH^2\\big)\\\\\n\\\\CF^2\\equal{}\\frac{1}{4}\\big(2(CG^2\\plus{}CH^2)\\minus{}GH^2\\big)\n \\end{array}\\right\\|$\n So $ \\sum AF^2\\equal{}\\frac{1}{4}(2\\sum AG^2\\plus{} 2\\sum AH^2\\minus{}3GH^2)$.\n\n By [b]Leibniz[/b] for point $ H$ we have $ \\sum AH^2\\equal{} \\sum AG^2 \\plus{}3GH^2$.\n \nThen $ \\sum AF^2\\equal{}\\sum AG^2\\plus{}\\frac{3}{4}GH^2$.\n\n Finally $ \\sum AG^2\\equal{}\\frac{1}{3}\\sum a^2$ and $ GH^2\\equal{}\\frac{4}{9}(9R^2\\minus{}\\sum a^2)$ and\n\\[ \\sum AF^2\\equal{}\\frac{1}{3}\\sum a^2\\plus{}\\frac{1}{3}(9R^2\\minus{}\\sum a^2)\\equal{}3R^2\\equal{}3 \\]\n[/color]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c\\in\\mathbb{R}^{+}$, such that $a^{2}+b^{2}+c^{2}=3abc$. Prove that \\[\\frac{a}{b^{2}c^{2}}+\\frac{b}{a^{2}c^{2}}+\\frac{c}{a^{2}b^{2}}\\ge\\frac{9}{a+b+c}\\] Good Luck!", "Solution_1": "$\\sum_{\\text{cyclic}}a \\cdot \\sum_{\\text{cyclic}}\\frac{a}{b^{2}c^{2}}= \\sum_{\\text{cyclic}}a \\cdot \\sum_{\\text{cyclic}}\\frac{9a^{3}}{9a^{2}b^{2}c^{2}}= \\sum_{\\text{cyclic}}a \\cdot \\sum_{\\text{cyclic}}\\frac{9a^{3}}{(a^{2}+b^{2}+c^{2})^{2}}$ $=$ $9 \\frac{\\sum_{\\text{cyclic}}a^{3}\\cdot \\sum_{\\text{cyclic}}a}{\\left( \\sum_{\\text{cyclic}}a^{2}\\right)^{2}}\\ge 9$ by Cauchy-Schwarz.", "Solution_2": "The condition gives us a way to increase or decrease the degree of the terms on each side by $1$. Each term on the LHS has degree $-3$ whereas each the RHS has degree $-1$.\r\n\r\nSo we want to use identities like $bc=\\frac{a^{2}+b^{2}+c^{2}}{3a}$. \r\n\r\nApplying this to each of the terms on the LHS twice gives \\[\\sum \\frac{a}{b^{2}c^{2}}=\\sum \\frac{a}{\\frac{(a^{2}+b^{2}+c^{2})^{2}}{9a^{2}}}= \\sum \\frac{9a^{3}}{(a^{2}+b^{2}+c^{2})^{2}}\\] and the resulting homogeneous inequality, \\[\\sum \\frac{a^{3}}{(a^{2}+b^{2}+c^{2})^{2}}\\geq \\frac{1}{a+b+c}\\] is easily handled.", "Solution_3": "[quote=\"MysticTerminator\"]$\\sum_{\\text{cyclic}}a \\cdot \\sum_{\\text{cyclic}}\\frac{a}{b^{2}c^{2}}= \\sum_{\\text{cyclic}}a \\cdot \\sum_{\\text{cyclic}}\\frac{9a^{3}}{9a^{2}b^{2}c^{2}}= \\sum_{\\text{cyclic}}a \\cdot \\sum_{\\text{cyclic}}\\frac{9a^{3}}{(a^{2}+b^{2}+c^{2})^{2}}$ $=$ $9 \\frac{\\sum_{\\text{cyclic}}a^{3}\\cdot \\sum_{\\text{cyclic}}a}{\\left( \\sum_{\\text{cyclic}}a^{2}\\right)^{2}}\\ge 9$[/quote]\r\n\r\nThis is (only a little bit) simpler: \\[\\sum_{\\text{cyclic}}a \\cdot \\sum_{\\text{cyclic}}\\frac{a}{b^{2}c^{2}}\\geq \\left(\\sum_{\\text{cyclic}}\\frac{a}{bc}\\right)^{2}= \\left(\\frac{a^{2}+b^{2}+c^{2}}{abc}\\right)^{2}= 3^{2}= 9.\\]", "Solution_4": "The [i]equivalent[/i] inequality,\r\n$a,b,c>0,\\ a+b+c=1:$ \\[\\sum_{cyc}\\sqrt{a}\\sum_{cyc}a\\sqrt{a}\\ge 1 !`\\]" } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "If Zn denotes the ring od integers modulo n, for which the pairs of rings (Zm,Zn) given below does there exist a ring homomorphism \u03a6:Zm->Zn such that \u03a6(1)=1?\r\nA(Z4,Z6) B(Z6,Z4) C(Z5,Z6) D(Z6,Z5) E(Z5,Z4) F(Z3,Z6) G(Z6,Z3) H(Z3,Z4)", "Solution_1": "Note that necessarily $ m \\equal{} m\\cdot\\Phi(1) \\equal{}\\Phi(m) \\equal{}\\Phi(0) \\equal{} 0$ in $ \\mathbb Z/n\\mathbb Z$. owk", "Solution_2": "Thank you." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Consider the 3D system\r\ndx/dt = a(y-x)\r\ndy/dt = (c-a)x- xz+cy\r\ndz/dt = xy-bz\r\nwhere a>0, b>0, c0>c.\r\n\r\nunsolved part:\r\nhow can i find the conditions for stability for the two other fixed points??\r\n\r\nan extra part:\r\nis there any ellipsoidal or spherical trapping region for the system?" } { "Tag": [ "geometry", "circumcircle", "incenter", "inequalities" ], "Problem": "Let S,I,O be the circumcenter, incenter and orthocenter of a \\[ \\Delta ABC\r\n\\]\r\nprove\r\n\\[ SI \\ge IO\\sqrt 2 \r\n\\]", "Solution_1": "Wrong. See\r\nhttp://www.mathlinks.ro/viewtopic.php?t=42335\r\nhttp://www.mathlinks.ro/viewtopic.php?t=54012\r\nhttp://www.mathlinks.ro/viewtopic.php?t=38700\r\nhttp://www.mathlinks.ro/viewtopic.php?t=2950\r\nfor similar problems that happen to be right.\r\n\r\n darij", "Solution_2": "hello, by a well-known formual is $ IO\\equal{}\\sqrt{R^2\\minus{}2rR}\\equal{}\\sqrt{\\left(\\frac{abc}{4A}\\right)^2\\minus{}\\frac{abc}{a\\plus{}b\\plus{}c}}$ and \r\n$ IO\\equal{}\\sqrt{2r^2\\plus{}4R^2\\minus{}S_\\omega}\\equal{}\\sqrt{\\frac{2A^2}{s^2}\\plus{}4\\frac{(abc)^2}{16A^2}\\minus{}\\frac{1}{2}(a^2\\plus{}b^2\\plus{}c^2)}$\r\nSetting $ a\\equal{}1,b\\equal{}2,c\\equal{}2$ we have $ \\frac{2}{15}\\sqrt{15}\\minus{}\\frac{1}{6}\\sqrt{23}\\sqrt{3}\\sqrt{2}\\geq0$. This is not true like Darij pointed out.\r\nSonnhard.", "Solution_3": "I know the problem it should be \\[ SO \\ge IO\\sqrt 2\\]", "Solution_4": "hello, now, this make sense, we have $ SO=\\sqrt{9\\left(\\frac{abc}{4A}\\right)^2-(a^2+b^2+c^2)}$ and $ IO=\\sqrt{2\\left(\\frac{A}{s}\\right)^2+4\\left(\\frac{abc}{4A}\\right)^2-\\frac{1}{2}(a^2+b^2+c^2)}$. So we have th show\r\n$ SO^2-2IO^2\\geq0$. This is equivalent with $ \\frac{abc}{4A}\\geq\\2\\frac{A}{s} \\Leftrightarrow abc\\geq 8(s-a)(s-b)(s-c)$. The last inequality is true because we have by AM-GM $ (x+y)(x+z)(y+z)\\geq8xyz$ with $ a=y+z,b=x+z,c=x+y$.\r\nSonnhard." } { "Tag": [ "probability", "expected value" ], "Problem": "I can't figure out how to do this problem. Can someone tell me if it is possible or not and how to do it?\r\n\r\nSay you started at 1 on a number line, with each turn you moving either 1 unit to the left or 1 unit to the right, with equal chances of both. The goal is to get to 0. What are the odds that you never make it?\r\n\r\nI got $ 1 \\minus{} \\sum_{k \\equal{} 1}^{\\infty}{\\frac {(2k)!}{(2^kk!)^2}}$.\r\n\r\nCan someone tell me if this is right or wrong, and how would you simplify this?", "Solution_1": "Let the probability that you never make it be $ p$, such that the probability that you eventually make it is $ 1\\minus{}p$. You need to go one step to the left, and you're expected to be in the same place at the end--this means $ E\\equal{}0\\equal{}p\\cdot\\text{\"}\\infty\\text{\"}\\plus{}0(1\\minus{}p)\\implies p\\equal{}0$. This should also make sense intuitively...", "Solution_2": "If you move 1 step to the right on the first turn, you are also expected to be in the same place in the end. And if you happen to get to #3, you are also expected to be there \"in the end\"... I don't understand.", "Solution_3": "I think I might have been unclear/messed up.\r\n\r\nBy the definition of expected value, $ E\\equal{}\\frac12\\cdot1\\plus{}\\frac12\\cdot\\minus{}1\\equal{}0$, where $ E$ is the expected number of places to the right that you will move. This means that you are always expected to stay in the same place. Let the probability that you don't ever get to $ 0$ be $ p$. Since you have to eventually either get to $ 0$ or not get to $ 0$, let the probability that you get to $ 0$ be $ 1\\minus{}p$. Again, by the definition of expected value, $ F\\equal{}\\minus{}1\\equal{}p\\cdot\\infty\\plus{}(1\\minus{}p)\\cdot\\minus{}1$ (it is $ \\minus{}1$ because you must move a net value of $ 1$ to the left to get to $ 0$). This equation gives us $ \\minus{}1\\equal{}p\\cdot\\infty\\minus{}1\\plus{}p\\implies p\\equal{}0$.\r\n\r\nI hope this makes more sense...\r\n\r\nIf it doesn't, notice that it's impossible to not get to $ 0$ eventually.", "Solution_4": "Oh thanks, I understand now. :)" } { "Tag": [ "inequalities", "parameterization", "function", "inequalities unsolved" ], "Problem": "I have a couple of nice inequalities that need help to prove:\r\n\r\n1.\r\n\r\n (a + b) (a + c) (b + c) (a + d) (b + d) (c + d) >= 4 a b c d (a + b + c + d)^2\r\n\r\nfor all positive a, b, c, d\r\n\r\n\r\n2. \r\n\r\n3 (a/b + b/a + b/c + c/b + a/c + c/a) + (1 + a) (1 + b) (1 + c) (c/b + c/a) (b/a + b/c) (a/b + a/c) \r\n>= 6 abc + 6 + 9(ab + bc + ac + a + b + c) + 3(ab/c + bc/a + ac/b)\r\n\r\nwith a, b, c are positive reals\r\n\r\nIf it helps, I know the equality occurs when a=b=c=2 (although I'm not sure if it's the only one).\r\n\r\nAny hints are appreciated...", "Solution_1": "For the first one i think $ SMV$ might be useful .", "Solution_2": "for the first one I have another solution,but it's based on the assumption $ a\\geq b\\geq c\\geq d$.\r\ncan we make this assumption here? :maybe:", "Solution_3": "[quote=\"BaBaK Ghalebi\"]for the first one I have another solution,but it's based on the assumption $ a\\geq b\\geq c\\geq d$.\ncan we make this assumption here? :maybe:[/quote]\r\n\r\nYes we can :) and can we see your solution ?", "Solution_4": "I think so since when you interchange any two of the variables, say $ a$ and $ b$, \r\n$ (a\\plus{}b)(a\\plus{}c)(a\\plus{}d)(b\\plus{}c)(b\\plus{}d)(c\\plus{}d) \\geq 4abcd(a\\plus{}b\\plus{}c\\plus{}d)^2$ => $ (b\\plus{}a)(b\\plus{}c)(b\\plus{}d)(a\\plus{}c)(a\\plus{}d)(c\\plus{}d) \\geq 4bacd(b\\plus{}a\\plus{}c\\plus{}d)^2$ which are equivalent, so the assumption is valid. So can you put your solution here? \r\n\r\n(P.S. i have a rather ugly solution so can you post yours? looking forward to it)", "Solution_5": "Because in both inequalities, the parameters have equal roles so we definitely can make assumption: a>= b>= c>=d.\r\n\r\nAny solutions (even socalled ugly ones) are good for me, so can you guys post it for further discussion?\r\n\r\nCheers", "Solution_6": "[quote=\"dinoamo\"]\n\n1.\n\n $ (a \\plus{} b) (a \\plus{} c) (b \\plus{} c) (a \\plus{} d) (b \\plus{} d) (c \\plus{} d) >\\equal{} 4 a b c d (a \\plus{} b \\plus{} c \\plus{} d)^2$\n\nfor all positive $ a, b, c, d$\n\n\n[/quote]\r\n$ (a\\plus{}b)(b\\plus{}c)(c\\plus{}d)(d\\plus{}a)\\geq(abc\\plus{}abd\\plus{}acd\\plus{}bcd)(a\\plus{}b\\plus{}c\\plus{}d)\\Leftrightarrow(ac\\minus{}bd)^2\\geq0$\r\nand $ (abc\\plus{}abd\\plus{}acd\\plus{}bcd)(a\\plus{}c)(b\\plus{}d)\\geq4abcd(a\\plus{}b\\plus{}c\\plus{}d)\\Leftrightarrow$\r\n$ \\Leftrightarrow(a\\minus{}c)^2(b\\plus{}d)bd\\plus{}(b\\minus{}d)^2(a\\plus{}c)ac\\geq0.$ Done! :)", "Solution_7": "nice one arqady!!\r\nThanks for that...\r\n\r\nAnyone gave a crack on the second inequality? \r\n\r\nI managed to prove it by showing that (2,2,2) is the only positive solution of the equality and by showing that this point is the local minimum of the function LHS-RLS (consider it to be a function of variable a). Therefore the function LHS - RHS is always positive given a, b, c are positive. However, this solution is not rigorous because I needed to use Mathematica to computationally show that (2,2,2) is the only positive point giving equality. And also it involves calculating the Hessian matrix. So it's can be called an...ugly solution :)\r\n\r\nWould be nice if someone can come up with a simpler solution!!!", "Solution_8": "See also here:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=547802" } { "Tag": [], "Problem": "The mass of 1L of a gas at 750 torr and 27\u00baC is 1.50g. Find the molar mass of the gas, assuming ideal behaviour.", "Solution_1": "Density of an ideal gas is given by-\r\n\r\n$ \\rho\\equal{}\\frac{PM}{RT}$\r\n\r\n$ \\frac{m}{V}\\equal{}\\frac{PM}{RT}$\r\n\r\n$ M\\equal{}\\frac{mRT}{PV}$\r\n\r\nSubstituting values,\r\n\r\n$ M\\equal{}\\frac{1.5 * 0.0821 * 300}{ \\frac{750}{760} * 1}$\r\n\r\n$ \\boxed{M\\equal{}37.4376 g/mol}$" } { "Tag": [ "inequalities" ], "Problem": "Given \\[ a, b, c >\\equal{}0; a \\plus{}b \\plus{}c \\equal{}2\\]\r\n\r\nProve: \\[ (a^2b^2 \\plus{} a^2c^2 \\plus{} b^2c^2)(a^2\\plus{}b^2\\plus{}c^2)^2 <\\equal{} 4\\]\r\nHave A Great Day!", "Solution_1": "[quote=\"wonderful\"]Given\n\\[ a, b, c > \\equal{} 0; a \\plus{} b \\plus{} c \\equal{} 2\n\\]\nProve:\n\\[ (a^2b^2 \\plus{} a^2c^2 \\plus{} b^2c^2)(a^2 \\plus{} b^2 \\plus{} c^2)^2 < \\equal{} 4\n\\]\nHave A Great Day![/quote]\r\n[hide]We need to prove that $ (a\\plus{}b\\plus{}c)^8\\geq64(a^2\\plus{}b^2\\plus{}c^2)^2(a^2b^2\\plus{}a^2c^2\\plus{}b^2c^2).$\nBut $ (a\\plus{}b\\plus{}c)^8\\geq\\left(2\\sqrt{2(a^2\\plus{}b^2\\plus{}c^2)(ab\\plus{}ac\\plus{}bc)}\\right)^4\\equal{}$\n$ \\equal{}64(a^2\\plus{}b^2\\plus{}c^2)^2(ab\\plus{}ac\\plus{}bc)^2\\geq64(a^2\\plus{}b^2\\plus{}c^2)^2(a^2b^2\\plus{}a^2c^2\\plus{}b^2c^2).$\n[/hide]", "Solution_2": "Well-done Arqady! \r\n\r\nThanks and Have A Great Day!" } { "Tag": [ "algebra", "polynomial" ], "Problem": "We know that if $f(x)=(g(x))^2+(h(x))^2$, where $g(x)$ and $h(x)$ are real polynomial (polynomial with real coefficients), then $f(x)\\geq 0$ for all real number $x$. \r\nProve that the converse of the statement above is also true, that is, if $f(x)$ is a real polynomial with $f(x)\\geq 0$ for all real number $x$, we can always find real polynomials $g(x)$ and $h(x)$ such that $f(x)=(g(x))^2+(h(x))^2$", "Solution_1": "$f$ has no real zeros, so all of them are complex and pairwise conjugated ($z'$ means the conjugate of $z$):\r\n\r\n$f(x)=\\prod (x-a_i)(x-a_i') = \\prod (x-a_i) \\prod (x-a_i') = u(x) v(x)'$ with $u(x)= \\prod (x-a_i) , v(x)= \\prod (x-a_i')$ and some $a_i$\r\n\r\nNow the coefficients of $u$ and $v$ are conjugated, so when you write $u(x)=g(x)+i\\cdot h(x)$ with real polynomials $g,h$, you get $v(x)=g(x)-i\\cdot h(x)$.\r\n\r\nSo $f(x)=u(x)v(x)=(g(x)+i\\cdot h(x))(g(x)-i\\cdot h(x))=g(x)^2+h(x)^2$", "Solution_2": "[quote=\"ZetaX\"]$f$ has no real zeros, so all of them are complex and pairwise conjugated ($z'$ means the conjugate of $z$):\r\n\r\n$f(x)=\\prod (x-a_i)(x-a_i') = \\prod (x-a_i) \\prod (x-a_i') = u(x) v(x)'$ with $u(x)= \\prod (x-a_i) , v(x)= \\prod (x-a_i')$ and some $a_i$\r\n\r\nNow the coefficients of $u$ and $v$ are conjugated, so when you write $u(x)=g(x)+i\\cdot h(x)$ with real polynomials $g,h$, you get $v(x)=g(x)-i\\cdot h(x)$.\r\n\r\nSo $f(x)=u(x)v(x)=(g(x)+i\\cdot h(x))(g(x)-i\\cdot h(x))=g(x)^2+h(x)^2$[/quote]\r\n\r\nI think that f may have real roots wich must be double root.When we move as factor the expression related to those roots we can write the reamining as product of second order polynomial that have no real root(which is related to what you did in your proof) so the discriminat each is less than zero and each of those second order polynomial can be put as sum of two squares.\r\nTHE RAMAINING OF THE PROOF RESULT FROM THE FACT THAT \r\n the produt of the sum ot two squares can be written as the sum of two square(consider the complex form)", "Solution_3": "sorry, I read only $f(x)>0$ :blush: \r\n\r\nBut at the thing with 'product of two sums of two squares is a sum of two squares', I have problems understanding why this should give the result...\r\nThe real reason I think is a theorem by Artin I asked for in the suberior algebra section\r\n(for that who understand: the algebraic closure of a field has degree one, two or infinity (over the field), where at degree two it's 'similar' to the relation of the complex to the real numbers, because the only thing you have to adjunct is $\\sqrt {-1}$)", "Solution_4": "[quote=\"ZetaX\"]sorry, I read only $f(x)>0$ :blush: \r\n\r\nBut at the thing with 'product of two sums of two squares is a sum of two squares', I have problems understanding why this should give the result...\r\nThe real reason I think is a theorem by Artin I asked for in the suberior algebra section\r\n(for that who understand: the algebraic closure of a field has degree one, two or infinity (over the field), where at degree two it's 'similar' to the relation of the complex to the real numbers, because the only thing you have to adjunct is $\\sqrt {-1}$)[/quote]\r\n\r\n(a+ib)(c+id)=(ac-bd)+i(ad+bc)\r\nfrom here since the norm in both side is the same we get:\r\n(a^2 +b^2)(c^2 + d^2)=(ac-bd)^2 +(ad+bc)^2\r\nwe apply this equality step by step to the product of the polynomial of degree two written in the canonical form as sum of two squares." } { "Tag": [ "geometry", "circumcircle", "Euler", "power of a point", "radical axis", "geometry proposed" ], "Problem": "Let $ k$ be a positive real number. Triangle ABC is acute and not isosceles, O is its circumcenter and AD,BE,CF are the internal bisectors. On the rays AD,BE,CF, respectively, let points L,M,N such that $ \\frac {AL}{AD} \\equal{} \\frac {BM}{BE} \\equal{} \\frac {CN}{CF} \\equal{} k$. Denote $ (O_1),(O_2),(O_3)$ be respectively the circle through L and touches OA at A, the circle through M and touches OB at B, the circle through N and touches OC at C. \r\n1) Prove that when $ k \\equal{} \\frac{1}{2}$, three circles $ (O_1),(O_2),(O_3)$ have exactly two common points, the centroid G of triangle ABC lies on that common chord of these circles. \r\n2) Find all values of k such that three circles $ (O_1),(O_2),(O_3)$ have exactly two common points", "Solution_1": "I only proof that three circles have a common $ X$. $ X$ is the intersection of $ AP, BQ, CW$, with $ P, Q, W$ respect are the intersection of the incicle (of $ \\Delta ABC$) with $ BC, CA, AB$.", "Solution_2": "@windrock: In my diagram, your comment maybe wrong.\r\n@mr.danh: You posted this problem in Geom Proposed Problems, and no solution after about 4 months, so please post your proof(if have). Thanks.", "Solution_3": "Yes, I'll post my proof. Wait...", "Solution_4": "And I think Can we use CANADA MO 2003 to solve this problem? \r\nCANADA MO 2003 is a strong lemma to proof group circles have two common point when we dertermined one of them", "Solution_5": "OK, maybe there are many solution for this, and here is mine\r\n[hide=\"Solution\"]Let $ d_A, d_B, d_C$ be the tangent of (O) at A,B,C resp. Denote $ A'\\equal{} d_B\\cap d_c, B' \\equal{} d_C\\cap d_A, C' \\equal{} d_A\\cap d_B$. \n1) When $ k \\equal{} \\frac {1}{2}$, we will prove that $ (O_1),(O_2),(O_3)$ have two common points and the Euler line of triangle ABC passes through these points.\nLet $ K \\equal{} d_A\\cap BC$. Easy to see $ KD \\equal{} KA$, thus $ O_1$ is the midpoint of AK and AK is the diameter of cirlce $ (O_1)$, so $ (O_1)$ passes through the foot of the perpendicular from A to BC. Similar properties for $ (O_2), (O_3)$ force the orthocenter H of triangle ABC has the same power to three circles $ (O_1),(O_2),(O_3)$. Obviously, the point O has the same power to that cirlces since the lengths of tangents from O to them are equal. This implies that OH is the common radical axis of $ (O_1),(O_2),(O_3)$. Because of acute triangle ABC, H lies on the inside of that triangle, so OH cuts all of three circles. Hence result.\n2) Assume that three cirlces $ (O_1),(O_2),(O_3)$ have exactly two common points.\nThen, the centers $ O_1,O_2,O_3$ are collinear\nLet $ q \\equal{} 1 \\minus{} k, B'C' \\equal{} a, C'A' \\equal{} b, A'B' \\equal{} c,p \\equal{} \\frac {a \\plus{} b \\plus{} c}{2}$. \nBy some algebraic transformations, we have that\n$ \\frac {O_1B'}{O_1C'} \\equal{} \\frac {(p \\minus{} b)(a \\minus{} 2q(p \\minus{} c))}{(p \\minus{} c)(a \\minus{} 2q(p \\minus{} b))}$\n$ \\frac {O_2C'}{O_1A'} \\equal{} \\frac {(p \\minus{} c)(b \\minus{} 2q(p \\minus{} a))}{(p \\minus{} a)(b \\minus{} 2q(p \\minus{} c))}$\n$ \\frac {O_3A'}{O_3B'} \\equal{} \\frac {(p \\minus{} a)(c \\minus{} 2q(p \\minus{} b))}{(p \\minus{} b)(a \\minus{} 2q(p \\minus{} a))}$ \nApply the Menelaus\u2019s theorem to the line $ (O_1O_2O_3)$ with triangle A'B'C', we get $ q \\equal{} 0$ or $ q \\equal{} \\frac {1}{2}$. Hence $ k \\equal{} 1$ or $ k \\equal{} \\frac {1}{2}$\n+) $ k \\equal{} \\frac {1}{2}$, solved by part 1)\n+) $ k \\equal{} 1$\nThree points $ O_1,O_2,O_3$ form a line, say $ m$.\nThe segments AD,BE,CF meets at a point, so if $ (O_1)$ passes through A,D; $ (O_2)$ passes through B,E; $ (O_3)$ passes through C,F; then any pair of these cirlces have common point, and they have the same radical axis (the perpendicular line from O to $ m$), hence result.\nAnswer: $ k \\equal{} 1$ or $ k \\equal{} \\frac {1}{2}$\n[/hide]\n[hide=\"To windrock\"]The Euler line does not pass through the Gergonne-point, I think your proof made a mistake :maybe: ?[/hide]", "Solution_6": "Oh sorry, yes, I wrong. Here is true proof (part a of problem)\r\nAssume that the targent at $ A$ of $ (O)$ cut $ BC$ at $ X$. Similar, we have $ Y, Z.$\r\nAnd call the intersection of targent at $ B$ cut targent at $ C$( with $ (O)$ at $ P$. Similar, we have $ Q, R.$\r\nWe know that $ AP, BQ, CR$ are concurent at a point, we call it is $ T$ \r\nEasy to see that the circle with diameter $ XA$ is circle $ (O_{1})$ and it pass through $ T$.\r\nOh, we can proof that $ O_{1}, O_{2},O_{3}$ are conlinear\r\nThree circles, with 3 centers are conlinear, they have one common point($ T$). Hence they have two common point.\r\n(I think it\u2019s true!)\r\nPart 2), who can help me?", "Solution_7": "Here is my solution for this problem \n[b]Solution[/b] \na) Let $A_1$, $B_1$, $C_1$ be orthogonal projections of $A$, $B$, $C$ on $BC$, $CA$, $AB$, respectively; $H$ be orthocenter of $\\triangle ABC$ \nWe have: $(LA; LO_1) \\equiv (AO_1; AL) \\equiv \\dfrac{\\pi}{2} - (AD; AO) \\equiv \\dfrac{\\pi}{2} - (AA_1; AD)$ (mod $\\pi$) \nThen: $LO_1$ $\\perp$ $AA_1$ \nBut: $L$ is midpoint of $AD$ so: $LO_1$ is perpendicular bisector of $AA_1$ \nHence: $A_1$ $\\in$ $(O_1)$ \nSimilarly: $B_1$ $\\in$ $(O_2)$, $C_1$ $\\in$ $(O_3)$ \nWe have: $P_{H / (O_1)} = \\overline{HA} . \\overline{HA_1}$, $P_{H / (O_2)} = \\overline{HB} . \\overline{HB_1}$, $P_{H / (O_3)} = \\overline{HC} . \\overline{HC_1}$ and $\\overline{HA} . \\overline{HA_1} = \\overline{HB} . \\overline{HB_1} = \\overline{HC} . \\overline{HC_1} < 0$ \nThen: $P_{H / (O_1)} = P_{H / (O_2)} = P_{H / (O_3)} < 0$ \nCombine with: $P_{O / (O_1)} = P_{O / (O_2)} = P_{O / (O_3)} = R^2 > 0$ (With $R$ is radius of $(ABC)$), we have: $(O_1)$, $(O_2)$, $(O_3)$ have 2 common points", "Solution_8": "Here is my rest part of the solution for this problem \n[b]Solution[/b] \nb) We need a lemma: \n[b]Lemma[/b]: Given $\\triangle ABC$ and a arbitrary point $S$. Let $D$ $\\equiv$ $AP$ $\\cap$ $BC$, $E$ $\\equiv$ $BP$ $\\cap$ $CA$, $F$ $\\equiv$ $CP$ $\\cap$ $AB$, $G$ $\\equiv$ $EF$ $\\cap$ $BC$, $H$ $\\equiv$ $FD$ $\\cap$ $CA$, $I$ $\\equiv$ $ED$ $\\cap$ $AB$. Let $A_k$ $\\in$ $BC$, $B_k$ $\\in$ $CA$, $C_k$ $\\in$ $AB$ which satisfy: $\\dfrac{\\overline{DA_k}}{\\overline{DG}} = \\dfrac{\\overline{EB_k}}{\\overline{EH}} = \\dfrac{\\overline{FC_k}}{\\overline{FI}} = k$. Prove that: $A_k$, $B_k$, $C_k$ are collinear $\\Leftrightarrow$ $k = 1$ or $k = \\dfrac{1}{2}$ \n[b]Proof[/b]: +) We have: $\\dfrac{\\overline{GB}}{\\overline{GC}} . \\dfrac{\\overline{HC}}{\\overline{HA}} . \\dfrac{\\overline{IA}}{\\overline{IB}} = - \\dfrac{\\overline{DB}}{\\overline{DC}} . \\dfrac{\\overline{EC}}{\\overline{EA}} . \\dfrac{\\overline{FA}}{\\overline{FB}} = 1$ \nThen: $G$, $H$, $I$ are collinear \nLet $M$, $N$, $P$ be midpoint of $AG$, $BH$, $CI$ so: $M$, $N$, $P$ lie on Newton line of completed quadrilateral $ABMN.CP$ \nHence: $k = 1$, $k = \\dfrac{1}{2}$ satisfy the problem \n+) Suppose there exist $k \\ne 1$, $k \\ne \\dfrac{1}{2}$ satisfy the problem \nIt's easy to prove that: $\\dfrac{\\overline{A_kM}}{\\overline{A_kG}} = \\dfrac{\\overline{B_kN}}{\\overline{B_kH}} = \\dfrac{\\overline{C_kP}}{\\overline{C_kI}} = \\dfrac{1 - 2k}{2 - 2k}$ \nThen: $\\overline{M, N, P}$ $\\parallel$ $\\overline{A_k, B_k, C_k}$ $\\parallel$ $\\overline{G, H, I}$ or $\\dfrac{\\overline{HG}}{\\overline{HI}} = \\dfrac{\\overline{NM}}{\\overline{NP}}$ \nBut: $\\dfrac{\\overline{DM}}{\\overline{DG}} = \\dfrac{\\overline{EN}}{\\overline{EH}} = \\dfrac{\\overline{FP}}{\\overline{FI}} = \\dfrac{1}{2}$ and $D$, $E$, $F$ aren't collinear so: $\\dfrac{\\overline{HG}}{\\overline{HI}} = \\dfrac{\\overline{NM}}{\\overline{NP}}$ can not happen \nCombine with: $\\overrightarrow{NM} = \\dfrac{1}{2} (\\overrightarrow{HG} + \\overrightarrow{ED})$, $\\overrightarrow{NP} = \\dfrac{1}{2} (\\overrightarrow{HI} + \\overrightarrow{EF})$ and $D$, $E$, $F$ aren't collinear then: $\\overline{M, N, P}$ $\\parallel$ $\\overline{A_k, B_k, C_k}$ $\\parallel$ $\\overline{G, H, I}$ can not happen \nHence: There don't exist $k \\ne 1$, $k \\ne \\dfrac{1}{2}$ satisfy the problem \nTherefore: $A_k$, $B_k$, $C_k$ are collinear $\\Leftrightarrow$ $k = 1$ or $k = \\dfrac{1}{2}$ \n \n\n ", "Solution_9": "[b]Back to the main problem[/b]: From the lemma, we have: $O_1$, $O_2$, $O_3$ are collinear $\\Leftrightarrow$ $k = 1$, $k = \\dfrac{1}{2}$ \nWith: $k = \\dfrac{1}{2}$ then from part a, we have: $(O_1)$, $(O_2)$, $(O_3)$ have 2 common points \nWith: $k = 1$ so: $(O_1)$, $(O_2)$, $(O_3)$ are 3 Apollonius circles of $\\triangle ABC$ then: $(O_1)$, $(O_2)$, $(O_3)$ have 2 common points \nHence: $(O_1)$, $(O_2)$, $(O_3)$ have 2 common points $\\Leftrightarrow$ $k = 1$ or $k = \\dfrac{1}{2}$" } { "Tag": [ "function", "trigonometry" ], "Problem": "whenever i see a sigma, i get scared out of my wits and don't even try the problem! i don't know why, i've been working with them for over a year, what can i do to prevent this?", "Solution_1": "I can totally relate to that! The first thing you need to do is to \"break the barrier.\" Even though you don't want to solve a particular problem, force yourself to do it. But don't just force yourself to do it, also make sure you understand it.\r\n\r\nFirst, you should replace every instance of the $ \\sum$ symbol with the corresponding sum, for example if a problem had\r\n\\[ \\sum_{i \\equal{} 0}^2 i^2\r\n\\]\r\nGo ahead and replace that with\r\n\\[ 0^2 \\plus{} 1^2 \\plus{} 2^2\r\n\\]\r\nIn my experience, the expanded version provides more insight into the problem most of the time. (And also helps you not to get turned off!)\r\nEventually, you'll become more comfortable with these types of problems.\r\n\r\nThe second thing after you break the barrier is to practice realizing the the small characters are actually the most important parts of the sum. Another major reason I get turned off by sigma symbols is because I don't visually distinguish stuff like\r\n\\[ \\sum_{i \\equal{} 0}^6 i^2, \\sum_{i \\equal{} 1}^9 i^2, \\sum_{i \\equal{} 3}^\\infty i^2\r\n\\]\r\nPractice distinguishing them.", "Solution_2": "thank you very much yongyi! your explanation was very thorough and helpful :)", "Solution_3": "I'm going to go ahead and second Yongyi's post. I endorse this way of looking at the Sigma notation, as it helps you break through and really get a grasp on what the underlying concept is.\r\n\r\nPerhaps ONE example, however, where using the sigma notation is more useful is when doing a proof with generating functions (index shifting is VERY clear here) or perhaps a Taylor Series problem.", "Solution_4": "One of the things I don't particularly like about sigma notation is that it is a bit visually awkward, perhaps in the same way that [b]yongyi[/b] finds it. There's this huge sigma, and the bounds are these tiny numbers tucked away on the top and on the bottom.\r\n\r\nIn principle, however, it is very nice -- [i]add together all the terms in this form between said bounds[/i].\r\n\r\nIn your own work, feel free to distort summation notation to suit your needs. Make things bigger or smaller. Put things where you want them. If you are dealing with sums of a particular form and all thats changing are the bounds, feel free to write something like $ S(a,b)$ where $ a,b$ are the bounds. Get rid of all the clutter and ugliness that using sigma notation can lead to. In short, make your own, convenient way of expressing to yourself the same thing:[i] \"add together all terms in this form between said bounds\"[/i].\r\n\r\nThis is idea is not just limited to summations. Trigonometric problems can get pretty ugly too. I like to do things like writing $ s$ for $ \\sin x$ and $ c$ for $ \\cos x$ and $ s_2$ for $ \\sin 2x$, etc. It makes everything look more condensed, and therefore simpler and more manageable!" } { "Tag": [ "LaTeX", "inequalities", "AMC", "AIME", "triangle inequality" ], "Problem": "$ f(x,y)=\\sqrt{x^2+y^2}+\\sqrt{x^2+y^2-2x+1}+\\sqrt{x^2+y^2-2y+1}+\\sqrt{x^2+y^2-6x-8y+25}$\r\nFind the minimum value of $ f(x,y)$ if $ x,y \\in \\mathbb{R}$\r\n\r\n[hide=\"Dangit\"]How do I make this look better in $ \\text{\\LaTeX}$ ? The spacing is stupid.[/hide]", "Solution_1": "You have one single expression that is too long, so it is rendering strangely.\r\n\r\nOne way to get around it is to split it up into multiple sections:\r\n\r\n$ f(x,y)=\\sqrt{x^{2}+y^{2}} +\\sqrt{x^{2}+y^{2}-2x+1}$ $ +\\sqrt{x^{2}+y^{2}-2y+1} +\\sqrt{x^{2}+y^{2}-6x-8y+25}$\r\n\r\nAnother way that helps make it more readable is to use the eqnarray* environment.\r\n\r\n\\begin{eqnarray*} f(x,y) & = & \\sqrt{x^{2}+y^{2}} \\\\ & & + \\sqrt{x^{2}+y^{2}-2x+1} \\\\ & & + \\sqrt{x^{2}+y^{2}-2y+1} \\\\ & & + \\sqrt{x^{2}+y^{2}-6x-8y+25} \\end{eqnarray*}", "Solution_2": "i think i am wrong but ... \r\nwrite it as ,\r\n$ f(x,y) = \\sqrt{x^2+y^2}+\\sqrt{(1-x)^2+y^2} + \\sqrt{(1-y)^2+x^2}+\\sqrt{(3-x)^2+(4-y)^2}$ and by Cauchy,\r\n$ \\sqrt{2}\\cdot f(x,y) \\ge (x+y+1-x+y+1-y+x+3-x+4-y) = 9$ \r\nso.. the min is - $ \\frac{9}{\\sqrt{2}}$\r\nam i right :?: \r\nif i am then please give me the justification of whether while i applied cauchy writing $ a^2$ as $ (-a)^2$ is correct so to get - $ (b-a)^2$.. \r\nnever mind if it is correct then i think that .. equality case holds for - $ \\frac{-a}{1} = \\frac{b}{1}$.. but a is a positive real..", "Solution_3": "I believe the minimum occurs when $ x = 3/7$ and $ y = 4/7$. By the [url=http://www.artofproblemsolving.com/Wiki/index.php/Minkowski_Inequality]Minkowski inequality[/url],*\r\n\\begin{align*} & \\sqrt {x^2 + y^2} + \\sqrt {(1 - x)^2 + y^2} + \\sqrt {x^2 + (1 - y)^2} + \\sqrt {(3 - x)^2 + (4 - y)^2} \\\\\r\n= & \\left(\\sqrt {x^2 + y^2} + \\sqrt {(3 - x)^2 + (4 - y)^2}\\right) + \\left(\\sqrt {(1 - x)^2 + y^2} + \\sqrt {x^2 + (1 - y)^2}\\right) \\\\\r\n\\ge & \\sqrt {(x + (3 - x))^2 + (y + (4 - y))^2} + \\sqrt {((1 - x) + x)^2 + (y + (1 - y))^2} \\\\\r\n= & 5 + \\sqrt {2}, \\end{align*}\r\nwith equality when\r\n\\[ \\frac {x}{y} = \\frac {3 - x}{4 - y} = \\frac {x + (3 - x)}{y + (4 - y)} = \\frac34\\]\r\nand\r\n\\[ \\frac {1 - x}{y} = \\frac {x}{1 - y} = \\frac {(1 - x) + x}{y + (1 - y)} = 1,\\]\r\nor equivalently $ x = 3/7,y = 4/7$.\r\n\r\n*We can also interpret the second to third step as sums of the hypotenuses of right triangles with legs $ x,y$ and $ 3 - x,4 - y$ for the first parentheses and $ 1 - x,y$ and $ x,1 - y$ for the second one. For each pair, place the two triangles next to each other so that their hypotenuses are going in the same direction and the left and right corners of different triangles touch (in either way). Then the sum of the hypotenuses is at least the hypotenuse of the triangle with the sums of the legs, by the Triangle Inequality. For example, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=432791#432791]the AIME problem[/url] from the above link.", "Solution_4": "[quote=\"math154\"]I believe the minimum occurs when $ x = 3/7$ and $ y = 4/7$. By the [url=http://www.artofproblemsolving.com/Wiki/index.php/Minkowski_Inequality]Minkowski inequality[/url],*\n\\begin{align*} & \\sqrt {x^2 + y^2} + \\sqrt {(1 - x)^2 + y^2} + \\sqrt {x^2 + (1 - y)^2} + \\sqrt {(3 - x)^2 + (4 - y)^2} \\\\\n= & \\left(\\sqrt {x^2 + y^2} + \\sqrt {(3 - x)^2 + (4 - y)^2}\\right) + \\left(\\sqrt {(1 - x)^2 + y^2} + \\sqrt {x^2 + (1 - y)^2}\\right) \\\\\n\\ge & \\sqrt {(x + (3 - x))^2 + (y + (4 - y))^2} + \\sqrt {((1 - x) + x)^2 + (y + (1 - y))^2} \\\\\n= & 5 + \\sqrt {2}, \\end{align*}\nwith equality when\n\\[ \\frac {x}{y} = \\frac {3 - x}{4 - y} = \\frac {x + (3 - x)}{y + (4 - y)} = \\frac34\\]\nand\n\\[ \\frac {1 - x}{y} = \\frac {x}{1 - y} = \\frac {(1 - x) + x}{y + (1 - y)} = 1,\\]\nor equivalently $ x = 3/7,y = 4/7$.\n\n*We can also interpret the second to third step as sums of the hypotenuses of right triangles with legs $ x,y$ and $ 3 - x,4 - y$ for the first parentheses and $ 1 - x,y$ and $ x,1 - y$ for the second one. For each pair, place the two triangles next to each other so that their hypotenuses are going in the same direction and the left and right corners of different triangles touch (in either way). Then the sum of the hypotenuses is at least the hypotenuse of the triangle with the sums of the legs, by the Triangle Inequality. For example, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=432791#432791]the AIME problem[/url] from the above link.[/quote]\r\nwhy am i gettin a diff. answer ? am i wrong?", "Solution_5": "No, you're not wrong, it's just that your bound $ 9/\\sqrt2\\approx6.36$ is not actually achievable (note that $ 5\\plus{}\\sqrt2\\approx6.41$).", "Solution_6": "Another way to do it is to realize that this question is really asking where the minimum value is from the sum of the distances from the points (0,0), (1,0), (0,1), (3,4)\r\nI haven't gotten it rigorous yet, but the idea is that the shortest distance between (0,0) and (3,4) is a line with length 5, and thus, the shortest sum of distances from a point on that line to the other two points will be at (1/2,1/2), which will lead to a distance of $ \\sqrt2$, hence the answer 5+$ \\sqrt2$", "Solution_7": "[b]@zapi2007[/b]: Bang on target! That's the easiest way to do this problem. The sum of distances from the vertices of a quadrilateral is minimum at the intersection of the diagonals. [hide=\"Hint\"]Triangle inequality[/hide]" } { "Tag": [ "calculus", "analytic geometry", "graphing lines", "slope", "derivative" ], "Problem": "I was wondering what are some applications of Calculus in the real world? I have done many calculus word problems, but are they actually relevent to the real world? What is calculus used for?", "Solution_1": "Optimization was the first thing I could think of", "Solution_2": "I think it would be hard to overstate the significance of calculus to understanding the physical world. Everything (and I really mean everything) is a solution to a differential equation, some of which are simple and most of which are not. Engineers and physicists are constantly required to use calculus (or, I suppose, to use computer programs in which the calculus part is disguised). Any problem concerning aero- or hydrodynamics is a calculus problem, for instance.", "Solution_3": "Economics is the first thing that comes to mind, besides what is already mentioned. Someone has to find the slope to those supply/demand curves. Of course, there are more applications than finding slopes in economics.", "Solution_4": "A couple things are:\r\n\r\n1) Your daily weather forecasts. (I believe, fractional derivatives?)\r\n\r\n2) architecture\r\n\r\n3) aeronautics\r\n\r\n4) ... and much much much more......\r\n\r\nI'm doing a calculus project on how Nasa landed Apollo on the moon (For class), It's rather interesting. :)", "Solution_5": "[quote=\"4everwise\"]1) Your daily weather forecasts. (I believe, fractional derivatives?)[/quote]\r\n\r\nMore generally, any computer simulation of any real-world situation will in some way rely upon the ideas of calculus.\r\n\r\nEconomics is an excellent example, CheesyPi -- lots and lots of economics is mathematical in nature." } { "Tag": [ "trigonometry" ], "Problem": "Prove that tan(x)+2tan(2x)+4tan(4x)+8cot(8x)=cot(x)", "Solution_1": "Make use of the identity\r\n$ \\tan \\theta \\equal{} \\cot \\theta \\minus{} 2cot 2 \\theta$\r\n(tell me if you want me to prove this :) )\r\n\r\nNow, \r\n$ LHS \\equal{} \\cot x\\minus{} 2cot 2 x \\plus{} 2(\\cot 2x\\minus{} 2cot 4 x) \\plus{} 4(\\cot 4x\\minus{} 2cot 8 x) \\plus{} 8 \\cot 8x$\r\n$ \\equal{} \\cot x \\equal{} RHS$", "Solution_2": "Proof :lol: :D \r\n[hide]\n$ \\cot \\theta \\minus{} \\tan \\theta \\equal{} \\frac{\\cos \\theta}{\\sin \\theta} \\minus{} \\frac{\\sin \\theta}{\\cos \\theta} \\equal{} \\frac{\\cos^2 \\theta \\minus{} \\sin^2 \\theta}{\\sin \\theta . \\cos \\theta} \\equal{} \\frac{2\\cos 2\\theta}{\\sin 2\\theta} \\equal{} 2\\cot 2\\theta$[/hide]", "Solution_3": "hey ,nobody told u to prove that :D :rotfl: :P" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "prouve that for evry integr n http://www.servimg.com/image_preview.php?i=97&u=10033511", "Solution_1": "It is on the forum, right [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148948]here[/url]. (In fact, it appears that you have taken the image from that post and exported it, to then link back to it. Why? I don't have any idea.)" } { "Tag": [ "MATHCOUNTS", "logarithms" ], "Problem": "$x^{2}\\cdot e^{x}+x \\cdot e^{x}-e^{x}=0$\r\n\r\n$e^{x}-12e^{-x}-1=0$\r\n\r\nThe half-life of a certain element is 1337 years. Assume exponential decay. I have a 30 mg sample. When will the sample be 5 mg? How much will there be in 50 years?\r\n \r\n:wink:", "Solution_1": "This should be moved, MATHCOUNTS never has exponential growth and decay, or the irrational [i]e[/i].", "Solution_2": "I'm moving this topic to Getting Started, which should be the right place for it. If not, a Getting Started mod can move it to a different forum.", "Solution_3": "The first question doesn't require e at all. :D \r\n\r\nThe second one a little bit of e... \r\n\r\n[hide=\"third one\"]\nThe first part calls for the equation \n\n$30(\\frac{1}{2}^{(x/1337)})=5$\n\nDividing by 30 and taking logarithms: \n\n$(\\frac{x}{1337})(\\log{\\frac{1}{2}})=\\log\\frac{1}{6}$. \n\n$x=3456$ years. \n\nThe second part is a simple computation: \n\n$30(\\frac{1}{2}^{(50/1337)})=29.23$ mg[/hide]", "Solution_4": "[quote=\"surge\"]$x^{2}\\cdot e^{x}+x \\cdot e^{x}-e^{x}=0$\n$e^{x}-12e^{-x}-1=0$[/quote]\r\n[hide] $x^{2}\\cdot e^{x}+x \\cdot e^{x}-e^{x}=0$.\n$e^{x}(x^{2}+x-1)=0$.\n$e^{x}=0$ (No solution) or $x^{2}+x-1=0$.\n$x=\\boxed{\\frac{-1\\pm\\sqrt5}{2}}$.\n\n$e^{x}-12e^{-x}-1=0$. Let $k=e^{x}$.\n$k-\\frac{12}{k}-1=0$.\n$k^{2}-k-12=0$.\n$(k-4)(k+3)=0$.\n$k=-3$ or $k=4$.\n$e^{x}=-3$ (No Solution) or $e^{x}=4$.\n$x=\\boxed{\\ln4}$. [/hide]", "Solution_5": "[quote=\"236factorial\"]The first question doesn't require e at all. :D \n\nThe second one a little bit of e... \n\n[hide=\"third one\"]\nThe first part calls for the equation \n\n$30(\\frac{1}{2}^{(x/1337)})=5$\n\nDividing by 30 and taking logarithms: \n\n$(\\frac{x}{1337})(\\log{\\frac{1}{2}})=\\log\\frac{1}{6}$. \n\n$x=3456$ years. \n\nThe second part is a simple computation: \n\n$30(\\frac{1}{2}^{(50/1337)})=29.23$ mg[/hide][/quote]\r\nWow, I did it a different way that took so much longer.", "Solution_6": "[quote=\"surge\"][quote=\"236factorial\"][hide=\"third one\"]\nThe first part calls for the equation \n\n$30(\\frac{1}{2}^{(x/1337)})=5$\n\nDividing by 30 and taking logarithms: \n\n$(\\frac{x}{1337})(\\log{\\frac{1}{2}})=\\log\\frac{1}{6}$. \n\n$x=3456$ years. \n\nThe second part is a simple computation: \n\n$30(\\frac{1}{2}^{(50/1337)})=29.23$ mg[/hide][/quote]\nWow, I did it a different way that took so much longer.[/quote]\r\n\r\nDid you use e instead (it's possible)?", "Solution_7": "Yeah, I used $e$." } { "Tag": [ "function", "integration", "trigonometry", "real analysis", "calculus", "topology", "real analysis unsolved" ], "Problem": "Let $f: (0,\\infty) \\rightarrow \\mathbb{R}$ be differentiable, and assume that $f(x)+f'(x) \\rightarrow 0$ when $x \\rightarrow \\infty$. Show that $f(x) \\rightarrow 0$ as $x \\rightarrow \\infty$.", "Solution_1": "We discussed a similar problem recently, with $f+f'+f''$. \r\n\r\nLet $f(x)=e^{-x}g(x)$, then $f(x)+f'(x)=e^{-x}g'(x)$. Thus, for any $\\epsilon>0$ there is $M$ such that $|g'(x)|M$. By Mean Value theorem $|g(x)-g(x-1)|\\le e^{x}\\epsilon$ for $x>M+1$. Iterating this estimate, obtain $|g(x)-g(x-n)|\\le \\frac{e^{x}\\epsilon}{1-e^{-1}}$ for $x>M+n$. Choose $n$ so that $x-n\\in [M,M+1]$. Since $g$ is bounded on $[M,M+1]$, it follows that $\\limsup_{x\\to\\infty}e^{-x}|g(x)|\\le \\frac{\\epsilon}{1-e^{-1}}$. \r\n\r\nThere ought to be a better proof. :|", "Solution_2": "The solution in the book used $g(x) = f(x)+f'(x) \\Rightarrow f(x) = e^{-x}\\int_{a}^{x}e^{t}g(t)dt+Ce^{-x}$ (by using an integrating factor). Then we can say $|g(x)| < \\epsilon$ for all $x > a$, so\r\n\r\n$|f(x)| \\le \\epsilon (1-e^{a-x})+|Ce^{-x}| < 2\\epsilon$\r\n\r\nfor sufficiently large $x$.", "Solution_3": "How do you know that $\\int_{a}^{x}e^{t}g(t)dt$ is defined? Here is a classical example: $f(x)=x^{2}\\sin\\frac{1}{x^{3}}$ is a differentiable function but $f'(x)=2x\\sin\\frac{1}{x^{3}}-\\frac{3}{x^{2}}\\sin\\frac{1}{x^{3}}$ fails to be integrable in either Riemann or Lebesgue sense.\r\n\r\nThe problem goes away if one uses the [url=http://en.wikipedia.org/wiki/Henstock-Kurzweil_integral]Henstock-Kurzweil integral[/url], which can recover any differentiable function from its derivative. But I don't know if your book introduces this integral.", "Solution_4": "I see. That's a good point. Larson does not seem to address that problem, nor does it say anything about integrals other than the Riemann integral.", "Solution_5": "By generalizing the above argument based on repeated application of Mean Value Theorem, you can prove the following\r\n\r\n[b]Lemma[/b]. Suppose that $f$ is differentiable on an open interval containing $[a,b]$. Let $g$ be a continuous function on $[a,b]$ such that $|f'(x)|\\le g(x)$ for all $x\\in [a,b]$. Then $|f(b)-f(a)|\\le\\int_{a}^{b}g(x)dx$. \r\n\r\nThis may be helpful in other situations when you want to estimate a differentiable function from its derivative.\r\n\r\nThis also brings us to a topic of current research: if $f$ is defined on an abstract metric space, a function $g$ is called an \"upper gradient\" of $f$ provided that $|f(\\gamma(b))-f(\\gamma(a))|\\le\\int_{a}^{b}g(\\gamma(t))dt$ for any rectifiable curve $\\gamma\\colon[a,b]\\to X$ parametrized by arclength. The point here is that although we can't differentiate in a metric space, we can define something like the magnitude of the gradient, and for some purposes (like the Sobolev embedding) the magnitude is enough. Recommended reading: \"Lectures on Analysis on Metric Spaces\" by Heinonen." } { "Tag": [ "logarithms", "trigonometry", "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "Find the order and degree ( if defined ) for the below two differential equations .\r\n\r\n1) $ \\frac{d^{2} y}{d x^{2} } \\equal{} x \\ln(\\frac{dy}{dx} )$\r\n\r\n2 ) $ \\frac{d^{2} y}{d x^{2} } \\equal{} \\sin (\\frac{dy}{dx})$\r\n\r\nplease help me and if the answers are not defined please tell me a reason for those.", "Solution_1": "hello \r\n[hide=\"1\"]\n$ \\frac{d^{2}y}{d x^{2}}\\equal{} x\\ln(\\frac{dy}{dx})$\norder=$ 2$\nthis can't be written as polynomial in all differential coefficient , the degrees is not defined. [/hide]", "Solution_2": "[hide=\"2\"]\n$ \\frac{d^{2}y}{d x^{2}}\\equal{}\\sin (\\frac{dy}{dx})$\nagain order=$ 2$\ndegree can't be defined .Same reason. \n[/hide]\n[hide=\"NOTE\"]\nRefer to NCERT .I believe they have explained it in details. [/hide]\r\nthank u", "Solution_3": "Thanks Kabi, even i thought abt the same thing just wasn't sure.\r\n\r\nthank u very much :)" } { "Tag": [ "calculus", "integration", "greatest common divisor", "number theory" ], "Problem": "The ordered pair of positive integers $ (x,y)\\equal{}(1000,1)$ is one of the many solutions to the equation\r\n\\[ 2x\\plus{}8y\\equal{}2008.\r\n\\]\r\nAmong the $ 81$ ordered pairs of positive digits $ (a,b)$, how many of the equations\r\n\\[ ax\\plus{}by\\equal{}2008\r\n\\]\r\nhave solutions $ (x,y)$ which are ordered pairs of positive integers?", "Solution_1": "The equation comes out to be:\r\n[size=18][/size][color=violet][/color][b] y^2+ x^2= 2008\r\n\r\nThen you slove it like a circle equaton. :D [/b]", "Solution_2": "Does someone have a more complete solution? I'm still stuck on this problem.", "Solution_3": "I'm sorry; I don't understand. Are you talking about the intersection of ax+by=2008 and 2x+8y=2008 or are you just talking about an integral solution to ax+by=2008 (given a and b are positive digits)?", "Solution_4": "Can someone please provide a full solution? I haven't been able to solve the problem yet. Thanks.", "Solution_5": "I believe each of a and b are numbers from 1 to 9 inclusive and we are asked to find how many of the 81 possible (a,b) have solutions.\r\n\r\n[hide=\"hint that I am not too sure of\"] The greatest common factor of a and b must divide 2008.\n\nFor example, (7,7) does not work.[/hide]\n\n[hide=\"solution?\"] We will count the number of (a,b) that do not satisfy that hint.\n\n$ 2008\\equal{}251\\cdot2^8$. We clearly do not need to worry about 251. \n\nNow, we can use 3, 5, 7, 9 as the greatest common factors.\n\nIf a=3, then b can be 3, 6 or 9. \nIf a=5, then b must be 5.\nIf a=7, then b must be 7.\nIf a=9, then b must be 3, 6, or 9.\n\nI count 8 that do not work, perhaps 73?[/hide]\r\n\r\nI do not believe my solution is right.", "Solution_6": "Let $ a \\equal{} gcd(a,b) \\cdot p, b \\equal{} gcd(a,b) \\cdot q$\r\n\r\nIt's easy to see that $ gcd(a,b)|2008$ must be satisfied.\r\n\r\nThen the question becomes we can find positive integer $ x$ such that $ ax \\le 2008 \\minus{} b.......(1)$ and $ px \\equal{} \\frac {2008}{gcd(a,b)}(mod q).........(2)$. As $ p,q$ coprime , we can find $ 0 < x \\le q$ satisfies (2), and obviously the smallest solution $ x_1 \\le q \\le 9 < 2008 \\minus{} 9 \\le 2008 \\minus{} b$. So $ gcd(a,b)|2008$ is a sufficient and necessary condition.\r\n\r\nPossible values of $ gcd(a,b) \\equal{} 1,2,3,4$ for $ a \\neq b$ , among them only $ 3 \\nmid 2008$\r\n\r\nSo only $ (3,6),(3,9),(6,9),(6,3),(9,3),(9,6)$ and $ (3,3),(5,5),(6,6),(7,7),(9,9)$ give no such solution. The answer $ \\equal{} 81 \\minus{} 11 \\equal{} 70$", "Solution_7": "What about (a,b)=(5,5), (7,7)?\r\n\r\nIf $ 5a\\plus{}5b\\equal{}2008$, then 5 divides the left side but not the right, so there are no solutions.\r\nSame for $ 7a\\plus{}7b\\equal{}2008$.", "Solution_8": "The correct answer [hide=\"is\"]$ 70$[/hide], but I can't follow stephencheng's solution. Could someone please explain why he did what he did? (It looks like a solution with a lot of number theory, and I'm quite weak on my number theory.) Thanks!", "Solution_9": "Ugh I missed (6, 6), (6, 3), and (6, 9).\r\n\r\nstephencheng's solution is complicated. I believe mine is easier to understand, but it's somewhat less rigorous. Basically, you know gcf(a,b) must be 3, 5, 6, 7, or 9. 5, 6, 7, and 9 are easy to get rid of; only (5, 5), (6, 6), (7, 7), (9, 9) do not work.\r\n\r\nThe 3 requires some more work, but not too bad." } { "Tag": [], "Problem": "Wow, that's amazing that you did the sprint in 15 minutes. Did you get the rest of the 28 correct?\r\n\r\n-interesting_move", "Solution_1": "[quote=\"Rep123max\"]This isnt really a question but it seems most people come to this forum so i thought i would ask it here.\n\ndo any of you guys do UIL (i dont know if its only in Texas). Number Sense, Calculator, Math, and Science. Ive done it for 2 years and I feel it helps and hurts on these contests. It helps because in number sense you look for quick shortcuts which really is valuable in some of these tests (AMC MATHCOUNTS....[/quote]\r\n\r\nUIL (University Interscholastic League) is based in Austin at U. Texas - to my knowledge you need to be a resident of Texas to officially participate. I did it during my 2 years of living in Texas, and definitely agree that the number sense problems helped immensely with other contests (especially the old AHSME - now AMC 12).", "Solution_2": "Yea rep123max, UIL of course exists in Texas. As far as I know though, in most major cities its only offered for High school (but correct me if im wrong). The State Round of MC was extremely unfair this year. Problems were way too easy and (if you will recall our state rankings) most of the top 20 were stuck within 2 pts of each other. I was devestated when I saw the problems b/c my specialty is making stupid errors on easy problems and getting harder problems correct that nobody else would know how to do." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "USAMTS", "search" ], "Problem": "What are some distinguished proof contests?(other than USAMO and IMO)", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/resources.php[/url]\r\n\r\nthese are the ones linked under contests at the top of each aops page.", "Solution_2": "USAMTS, TST (Team Selection Test), and various international olympiads (for example, the USAMO would count, as well as CMO (Canadian Math Olympiad) and others).", "Solution_3": "Specifics, please? I have checked the link, hurdler, and I will see which ones are proof contests, but any others?", "Solution_4": "There is a separate category of \"mail-in\" proof contests. USAMTS falls into this category, as do some more local things, like the Wisconsin and Pomona College \"Talent Search\" contests." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Here s a very simple geometric interpretation of our Nesbitt Inequality that I happened to notice:\r\nIn a triangle $ ABC$ if all letters have their usual meaning: then,\r\nprove that:\r\n$ \\sum_{i \\equal{} 1}^{3} \\frac {h_{i} \\plus{} r}{h_{i} \\minus{} r} \\ge 6$\r\n\r\nHere s yet another related problem:\r\nFor positive reals $ x,y,z$ prove that\r\n$ \\sum \\frac {2}{x \\plus{} y} \\ge \\frac{9}{x\\plus{}y\\plus{}z}$", "Solution_1": "[quote=\"Euclidean Geometer\"]\nFor positive reals $ x,y,z$ prove that\n$ \\sum \\frac {2}{x \\plus{} y} \\ge 9$[/quote]\r\nTry $ x\\equal{}y\\equal{}z\\equal{}1.$ :wink:", "Solution_2": "OOPS!!! Sorry\r\nThanks Arqady, I mistyped it. The correct one is:\r\n\r\n$ \\sum \\frac {2}{x \\plus{} y} \\ge \\frac {9}{x \\plus{} y \\plus{} z}$\r\n\r\nEditted in the original post as well. :)", "Solution_3": "[quote=\"Euclidean Geometer\"]Here s a very simple geometric interpretation of our Nesbitt Inequality that I happened to notice:\nIn a triangle $ ABC$ if all letters have their usual meaning: then,\nprove that:\n$ \\sum_{i \\equal{} 1}^{3} \\frac {h_{i} \\plus{} r}{h_{i} \\minus{} r} \\ge 6$[/quote]$ \\frac{h_a\\plus{}r}{h_a\\minus{}r}\\plus{}\\frac{h_b\\plus{}r}{h_b\\minus{}r}\\plus{}\\frac{h_c\\plus{}r}{h_c\\minus{}r}\\geq 6$\r\n$ \\Longleftrightarrow \\frac{2A\\plus{}ar}{2A\\minus{}ar}\\plus{}\\frac{2A\\plus{}br}{2A\\minus{}br}\\plus{}\\frac{2A\\plus{}cr}{2A\\minus{}cr}\\geq 6$\r\n$ \\Longleftrightarrow \\frac{r(a\\plus{}b\\plus{}c)\\plus{}ar}{r(a\\plus{}b\\plus{}c)\\minus{}ar}\\plus{}\\frac{r(a\\plus{}b\\plus{}c)\\plus{}br}{r(a\\plus{}b\\plus{}c)\\minus{}br}\\plus{}\\frac{r(a\\plus{}b\\plus{}c)\\plus{}cr}{r(a\\plus{}b\\plus{}c)\\minus{}cr}\\geq 6$\r\n$ \\Longleftrightarrow \\frac{(b\\plus{}a)\\plus{}(c\\plus{}a)}{b\\plus{}c}\\plus{}\\frac{(a\\plus{}b)\\plus{}(c\\plus{}b)}{a\\plus{}c}\\plus{}\\frac{(a\\plus{}c)\\plus{}(b\\plus{}c)}{a\\plus{}b}\\geq 6$ true by AM-GM." } { "Tag": [ "MATHCOUNTS" ], "Problem": "IP adresses are composed of 4 8-digit base 2 numbers. How many total IP addresses are there?\r\n\r\nExtra credit to anyone who can find a pattern for (2n-1)^m (some number of 1s to the something power; example: 111^11)", "Solution_1": "Technically, Tare, you're wrong :-p Several IPs are reserved for networks seperate from the internet - all starting in 192.168 for example. And I don't believe that 0.0.0.0 is a valid IP. Anyway, interesting problem for the kiddies.\r\n\r\nedit: fixed the typo", "Solution_2": "I thought it was 192.168, not 198.168.", "Solution_3": "Ignoring all restrictions:[hide] There are 2 possible values (0 & 1) for each digit). There are 32 digits, so 2^32, =4,294,967,296. It's a big Internet out there.[/hide]", "Solution_4": "Yup.\r\n\r\ndoo-whoop, I'm trying to introduce some people into the magnificent world of base 2...", "Solution_5": "[quote=\"Ravi B\"]I thought it was 192.168, not 198.168.[/quote]\r\nEh, yeah. Typo." } { "Tag": [ "function", "combinatorics unsolved", "combinatorics" ], "Problem": "In a round robin chess tournament each player plays every other player exactly once. The winner of each game gets $ 1$ point and the loser gets $ 0$ points. If the game is tied, each player gets $ 0.5$ points. Given a positive integer $ m$, a tournament is said to have property $ P(m)$ if the following holds: among every set $ S$ of $ m$ players, there is one player who won all her games against the other $ m\\minus{}1$ players in $ S$ and one player who lost all her games against the other $ m \\minus{} 1$ players in $ S$. For a given integer $ m \\ge 4$, determine the minimum value of $ n$ (as a function of $ m$) such that the following holds: in every $ n$-player round robin chess tournament with property $ P(m)$, the final scores of the $ n$ players are all distinct.", "Solution_1": "Answer: $2m-4$\n\nConsider an arbitrary tournament with property $P(m)$ with $n\\ge2m-3$ players. We call a set $S$ of $m$ players \"good\" if there is a player who won all her games against the other $m-1$ players in $S$ and a player who lost all her games against the other $m-1$ players in $S$. Otherwise, call it \"bad\".\n\n(i) There is at least one tie.\nAssume $P$ tied with $Q$. If there are $m-2$ players who lost to $P$, then these players with $P$ and $Q$ form a bad set. So there are at most $m-3$ players who lost to $Q$. Similarly, there are at most $m-3$ players who won are tied with $P$. Thus there are at most $m-3+m-3+2=2m-4$ players, a contradiction.\n\n(ii) There are three players $P,Q,R$ such that $P$ won against $Q$, $Q$ won against $R$, and $R$ won against $P$ (call this a cycle).\nSimilar as case (i), we find that at most $m-4$ players lost against $P$ and at most $m-4$ players won against $P$. Hence there are at most $m-4+m-4+3=2m-5$ players, a contradiction.\n\n(iii) There is no tie and no cycle.\nThere is a player $P$ with the largest number of wins. If $Q$ won against $P$, then $Q$ beat all players who lost against $Q$, so $Q$ has more wins, a contradiction. So $P$ won against all other players. Among players other than $P$, there is a player who won against all other players. Continuing this, we can label the players as $P_1,P_2,\\ldots,P_n$ such that $P_i$ won against $P_j$ for all $i>j$. Clearly their final scores are ditinct.\n\nSo for $n\\ge2m-3$, the result holds. Now we give a construction to show that $2m-4$ is not enough: $P_1,P_2,\\ldots,P_{2m-4}$ are such that $P_i$ won against $P_j$ for all $i>j$, except that $P_{m-1}$ tied with $P_{m-2}$." } { "Tag": [], "Problem": "A polyhedron (a solid with polygonal faces, not a regular one) has an equal number of quadrilaterals and pentagons as its faces. Exactly three faces meet at each vertex. How many faces does this polyhedron have?", "Solution_1": "wrong forum", "Solution_2": "[hide]Let the polyhedron have $ X$ quadrilaterals and $ X$ pentagons. \nThere are $ 2X$ faces in the polyhedron. \nSince each edge is shared by $ 2$ faces, there are $ \\frac {1}{2}(4X + 5X) = \\frac {9}{2}X$ edges. \nSince each vertex is shared by $ 3$ faces, there are $ \\frac {1}{3}(4X + 5X) = \\3X$ edges. \nBy Euler's identity, $ V - E + F = 2 \\Rightarrow 3X - \\frac {9}{2}X + 2X = 2 \\Rightarrow X = 4$. So, the polyhedron has $ 2X = 8$ faces.[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "Check this: local Lipshitz condition (I know I misspelled Lipschitz but the punishment I got doesn't match the crime :P)", "Solution_1": "Nothing to do. Try not to miss-spell :D", "Solution_2": "I first saw this about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=143623]six months ago[/url].\r\nThat was a poster that didn't know the correct spelling rather than a typo." } { "Tag": [ "symmetry", "search" ], "Problem": "Please help me in solving the following question.\r\n\r\nWhen all diagonals are drawn in a regular octagon, the interior is divided into how many regions?\r\n\r\nThanks in advance!", "Solution_1": "Trying what athunder said, we get \r\n\r\nSquare: $ 4$ regions\r\nPentagon: $ 11$ regions\r\nHexagon: $ 20$ regions.\r\n\r\nAnd, then I cam up with this formula: $ r\\equal{}s^2\\minus{}2s\\minus{}4$, which works. Now plugging in $ 8$, we get $ r\\equal{}8^2\\minus{}2(8)\\minus{}4\\equal{}64\\minus{}16\\minus{}4\\equal{}64\\minus{}20\\equal{}44$, which is hopefully correct..... No way to prove though..... :(", "Solution_2": "This is not right. The answer for octagon is 80. This problem is from 1986 National Team 5. The tricky part is some of the lines intersect so it not possible to generalize easily.\r\n\r\nAlso it is different for a polygon with odd sides and a polygon with even sides.", "Solution_3": "[url=http://mathworld.wolfram.com/RegularPolygonDivisionbyDiagonals.html]This[/url] might be useful.\r\n\r\nFunny that that's not on Victor's formula sheet...", "Solution_4": "Any easier way to do it?\r\nCan anyone produce a direct calculation?", "Solution_5": "Draw a large diagram and count up?\r\n\r\nYou can use symmetry for this problem - it is discussed more [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=210200562&t=202511]here[/url]." } { "Tag": [ "probability", "ratio", "expected value" ], "Problem": "The Detroit Tigers and the Chicago White Sox play each other in the first round of the playoffs, a series which ends when one team gets a total of 3 wins. The probability that each team wins is fixed across all games -- that is, the probability the Tigers win game 2 is the same as the probability that they win every other game -- and there are no ties. After 3 games, Detroit is found to be leading the series 2 wins to 1, which is exactly what one would expect given theirs respective probabilities of winning. What is the probability that Detroit goes on to win the series?\r\n\r\n\r\nNow, in the World Series, Detroit plays the Chicago Cubs. If it is found that after 4 games, Detroit leads 3-1 and this is exactly what one would expect given their relative chances of winning each game, what is the probability that Detroit goes on to win the series?\r\n\r\nWhy doesn't the same analysis work if instead this were a seven-game series and Detroit lead 3-2 after 5 games?", "Solution_1": "Joel, maybe these questions will activate this strand.\n\nClarification of wording and intent?\n\n\n\nfirst, by the phrase \"Detroit is found to be leading the series 2 wins to 1, which is exactly what one would expect given theirs respective probabilities of winning\" I inferred that you meant that their probability of winning each game is 2/3? Is that your intent?\n\n\n\nsecond, I don't understand your cryptic comment in the third question. To confirm whether I'm on the right track, would you confirm if [hide]8/9[/hide] is correct for #1 and 'by same method' I get [hide]21/25[/hide] for #3. \n\n\n\nBy the way, hint: [hide]use a probablity tree diagram[/hide]", "Solution_2": "My intent was to say that given the probability that Detroit would win each game, after three games Detroit could have been expected to have 2 wins. It is your job to figure out what the probability that Detroit won each game would be.\n\n\n\nOh, and I left something out of the second one -- the World Series is a \"best of 7\" series, so the first team to win 4 games wins.\n\n\n\n\n\nYour answer to number 1 is [hide]correct[/hide], and your answer to number 3 is [hide]not.[/hide]", "Solution_3": "The same analysis doesn't work for the third one because:[hide]\n\nFor the first two you can see that the other team has to win all of the games so you can do \n\n1-(1/3) :^2: =8/9\n\nBecause the probibility the other team wins is 1/3 and the other team has to win 2 games consecutivly to win the series\n\nand for the second one you do \n\n1-(1/4) :^3: =63/64 \n\nBecause the probibility the other team wins is 1/4 and the other team has to win 3 games consecutivly to win the series\n\nThe same analysis doesn't work for the third one because the other team doesn't have to all of the rest of the games. \n\n[/hide]\n\nMy answer for the third one is[hide]477/625[/hide]", "Solution_4": "Good thing you didn't say the Chicago CUBS, or I would have had to say that the probability is 0.", "Solution_5": "JMI wrote:[hide]\nThe same analysis doesn't work for the third one because the other team doesn't have to all of the rest of the games. \n[/hide]\nMy answer for the third one is[hide]477/625[/hide]\n\n\n\nThey actually do have to win all games. if they lose a single game, the Detroit has 4 wins, and thus wins the series. My answer is [hide]1-(2/5) :^2: =1-4/25=21/25[/hide] So I'd say the same reasoning does work.", "Solution_6": "I've seen correct answers for 1 and 2. I'm interested in where the more complicated-looking answer to number 3 came from, but I think it's wrong. The \"obvious\" answer for number 3, [hide]21/25[/hide] is wrong, though.", "Solution_7": "Oh I just realized I completely messed up number 3 but I can't find the right way to do it either.", "Solution_8": "JBL wrote:which is exactly what one would expect given theirs respective probabilities of winning\n\nJBL wrote:The \"obvious\" answer for number 3, [hide]21/25[/hide] is wrong, though.\n\n\n\nIt seems to me you have given the classic definition of expected value. The probability of Detroit winning a game follows directly as p=3/5.\n\n\n\nYour previous responses indicate that's incorrect. So you must be supplying an expected value for some conditional probability.\n\n\n\nI looked at the problem another way. Let's assume your phrase \"what one would expect\" applies only to the series that are still in play. The series may have ended 4-0, 4-1, 0-4, or 1-4. The only series that are still in play after game 5 are those that now stand 3-2 or 2-3.\n\n\n\nConsider the binomial expansion of [tex](p+q)^5[/tex], where p is the probability that Detroit wins and [tex]q=(1-p)[/tex]. The terms representing a series that's 3-2 and 2-3 after 5 games are given by: [tex]{{5}\\choose {3}}p^3q^2[/tex] and [tex]{{5}\\choose {2}}p^2q^3[/tex]. Solving these two terms for p and q, such that the ratio of the terms is 3:2 also results in p=3/5, q=2/5.\n\n\n\nYou have stumped me.", "Solution_9": "[quote=\"JBL\"]Now, in the World Series, Detroit plays the Chicago Cubs. If it is found that after 4 games, Detroit leads 3-1 and this is exactly what one would expect given their relative chances of winning each game, what is the probability that Detroit goes on to win the series?\n\nWhy doesn't the same analysis work if instead this were a seven-game series and Detroit lead 3-2 after 5 games?[/quote]\r\n\r\nIs the 3-2 in the third question exactly what one would expect given their relative chances of winning each game?", "Solution_10": "Okay:\r\n\r\nDetroit has some winning percentage against the Cubs, p. After 5 games [b][i]in a best of 7 series[/i][/b] with this chance of winning, it is expected that Detroit would have won 3 games. I will tell you that actually solving for Detroit's chances of winning the series is hard; however, to explain why it isn't 21/25 is less hard. I've given a hint in the content of this message. Just in case you're wondering, the fault lies here:\r\n\r\n[quote=\"rcv\"]The probability of Detroit winning a game follows directly as p=3/5. [/quote]\r\n\r\nI am using the traditional meaning of expected value that all of you are working with, though. You need to figure out why it doesn't work out to 3/5.", "Solution_11": "But I thought E(X+Y) = E(X) + E(Y) with a slightly warped sense of \"+\", so wouldn't we have E(5 games) = 5*E(1 game), and so 3 = 5*E(1 game), and E(1 game) = 3/5?", "Solution_12": "That would be true, except that you are ignoring a very important part of the problem. Let's look at the definition of (finite) expected value: the weighted average of all possible outcomes. Now, the formula you are given is correct if we simply had the Cubs and Tigers play 5 games. However, they aren't just playing 5 games -- they are playing a best of seven series. This means that our available outcomes are limited -- in particular, Detroit is incapable of winning 5 games. Thus, the expected value, were Detroit to win with probability 3/5, would actually be [i]smaller[/i] than 3 games in 5. Detroit must have a chance of winning greater than three fifths in order to be expected to win three fifths of its games.\r\n\r\nIn fact, it is relatively easy to calculate that with a winning chance of 3/5, Detroit can expect to win 2.90688 of its first 5 games in the series." } { "Tag": [], "Problem": "Let $D$ represent a repeating decimal. If $P$ denotes the $r$ figures of $D$ which do not repeat themselves, and $Q$ denotes the $s$ figures which do repeat themselves, then the incorrect expression is:\r\n\\[ \\text{(A)}\\ D = .PQQQ\\cdots \\qquad \\text{(B)}\\ 10^rD = P.QQQ\\cdots \\]\r\n\\[ \\text{(C)}\\ 10^{r+s}D = PQ.QQQ\\cdots \\qquad \\text{(D)}\\ 10^r(10^s-1)D = Q(P-1) \\]\r\n\\[ \\text{(E)}\\ 10^r \\cdot 10^{2s}D = PQQ.QQQ\\cdots \\]", "Solution_1": "i'm going towards $D$, :huuh:", "Solution_2": "That was my gut.. but not in the mood to try and work it.", "Solution_3": "Same here. D looks awkward to me.", "Solution_4": "A, B, C and E are correct. C-B gives the correct left side of D but the incorrect right side - it should be P(Q-1).", "Solution_5": "[hide]d... :? :huh: [/hide]", "Solution_6": "D...because it doesn't look like a decimal :|", "Solution_7": "D, definitely. I checked all the others and they worked. Checked D and it didn't work, so I'm going with everybody else on this one. :)", "Solution_8": "It has to be D because I have no idea how they got that since there is nothing that repeats. Plus, everyone else is saying D because it looks really wierd and different from the other choices, so I pick D. :?" } { "Tag": [ "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "If $a 0$ , and\r\n[color=green] [b] 2)[/b] [/color] for every $f \\in C[a,b]$ there is $\\theta \\in [a,b]$ such that $\\int\\limits_{a}^{b}f(x)w(x)\\; dx = f(\\theta) \\int\\limits_{a}^{b} w(x)\\; dx ,$\r\n [b] prove that $w \\ge 0$ on $[a,b]$ . [/b]", "Solution_1": "We admit that statement of problem is not true, so $\\exists x\\in [a,b]$ such that $w(x)<0.$\r\nNow $w\\in C[a,b]$ so $w^{-1}(-\\infty;0)$ is open in $[a,b]$, and $U=\\{y\\in [a,b] | w(y)<0\\}=\\bigcup_{k} I_k=\\bigcup_{k} (\\alpha_k;\\beta_k)$ where $\\alpha_{k+1}\\ge\\beta_{k}$ . And $x\\in U$\r\nThan $\\exists l$ such that $x\\in I_l$. (We assume that $\\alpha_l,\\beta_l \\in (a,b)$. If this statement not true the solution is almost the same. I think it's clear)\r\n(In other way $\\alpha_k=\\inf_{y\\in[a,b]} w(y)<0$ and $\\beta_k=\\sup_{y\\in [a,b]} w(y)<0$.)\r\nThan $w(\\alpha_l)=w(\\beta_l)=0,$ and $w(y)<0$ when $y\\in (\\alpha_l;\\beta_l)$.\r\nWe define $f(y)$ as $w(y)$ when $y\\in [\\alpha_l,\\beta_l]$ and $f(y)=0$ in other case. Than $f(y)\\in C[a,b]$. (Even if $\\alpha_k=a$ or $\\beta_k=b$)Thus $f(y)\\le 0$. And since $f(x)<0$ we have \r\n\\[ 0<\\int\\limits_{a}^{b}f(t)f(t)\\; dt =\\int\\limits_{a}^{b}f(t)w(t)\\; dt = f(\\theta) \\int\\limits_{a}^{b} w(t)\\; dt \\le 0 \\] because $\\int\\limits_{a}^{b} w(x)\\; dx \\ge 0$ and $f(\\theta)\\le 0 \\forall \\theta.$\r\nContradiction... And $w(y)\\ge 0.$ :)", "Solution_2": "Assume that there exists $x_0$ such that $w(x_0)<0$. Take $f(x)=|\\min(0,w(x))|$ and $g(x)=f(x)w(x)=-\\min(0,w(x))^2<0$. Is easy to prove that $f \\in C[a,b]$. \r\n$\\int_a^b f(x)w(x)dx=\\int_a^b g(x)dx<0$.\r\nOn the other hand we have that \r\n$f(\\theta)\\int_a^bw(x)dx>0$, since $f(x)>0,\\forall x\\in [a,b]$. Contradiction. So $w$ can take only nonegative values.", "Solution_3": "xirti\r\n$f(\\theta) \\int\\limits_{a}^{b} w(x)\\; dx \\ge 0$ and $f\\ge 0$ but you solution correct in any way :)", "Solution_4": "I also didn't prove that if g continuous , $g(x)\\ge 0$ and $g(x_0)>0$ we have $\\int_0^1g(x)dx >0$. \r\nSince g is continuous someone can find a neibourghood $[c,d]$ of $x_0$ such that $g(x)>\\epsilon>0,\\forall x \\in [c,d]$.\r\n$\\int_0^1g(x)dx=\\int_0^cg(x)dx+\\int_c^dg(x)dx+\\int_d^1g(x)dx\\ge\\int_c^dg(x)dx\\ge\\epsilon (d-c)>0$" } { "Tag": [ "calculus", "integration" ], "Problem": "Show that $ i^{n}$ takes on one of four values $ \\pm1$ or \\pm{i} if n is an integral", "Solution_1": "[hide]\n$ n\\equal{}4k\\Rightarrow i^{4k}\\equal{}(i^{4})^{k}\\equal{}1$\n$ n\\equal{}4k\\plus{}1\\Rightarrow i^{4k\\plus{}1}\\equal{}i\\cdot i^{4k}\\equal{}i$\n$ n\\equal{}4k\\plus{}2\\Rightarrow i^{4k\\plus{}2}\\equal{}i^{2}\\cdot i^{4k}\\equal{}i^{2}\\equal{}\\minus{}1$\n$ n\\equal{}4k\\plus{}3\\Rightarrow i^{4k\\plus{}3}\\equal{}i^{3}\\cdot i^{4k}\\equal{}i^{3}\\equal{}\\minus{}i$\n[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "why does sinx equal square root of (1-(cosx)^2)? isn't it +-square root of (1-(cosx)^2)?", "Solution_1": "hello, yes, it is $ \\sin(x)\\equal{}\\pm\\sqrt{1\\minus{}\\cos(x)^2}$ because $ \\sqrt{x^2}\\equal{}|x|$.\r\nSonnhard.", "Solution_2": "Are you asking this question because that substitution was used in a trigonometric proof? If so, can you show us the problem?\r\n\r\nOtherwise, in general, it is as Sonnhard said.", "Solution_3": "NOTE: $ \\sin{x} \\equal{} \\color{red}{\\text{one of}\\color{black}\\pm\\sqrt {1 \\minus{} \\cos^2{x}}}$it depends from interval of argument", "Solution_4": "This is covered in AoPS volume 2 chapter 2. You can refer to it there.", "Solution_5": "[quote=\"batteredbutnotdefeated\"]This is covered in AoPS volume 2 chapter 2. You can refer to it there.[/quote]\r\n\r\nThat's best for review. I haven't done any trig, can't wait till the AoPS trig comes out. What do I use before then?", "Solution_6": "Any Precalc book should work, if you want basics.", "Solution_7": "I haven't used it, but I've heard that it was good for trig:\r\n\r\nPlane Trigonometry by S. L. Loney", "Solution_8": "Actually I think all you need for AMC 12 trig is right triangle trig and some identities" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let a,b,c are positive numbers. Prove that\r\n$ 27(a\\plus{}b)^2(b\\plus{}c)^2(c\\plus{}a)^2\\ge 64abc(a\\plus{}b\\plus{}c)^3$\r\nIt have a nice solution. :D", "Solution_1": "[quote=\"thegod277\"]Let a,b,c are positive numbers. Prove that\n$ 27(a \\plus{} b)^2(b \\plus{} c)^2(c \\plus{} a)^2\\ge 64abc(a \\plus{} b \\plus{} c)^3$\nIt have a nice solution. :D[/quote]\r\n\\[ \\left( {a \\plus{} b} \\right)\\left( {b \\plus{} c} \\right)\\left( {c \\plus{} a} \\right) \\ge \\frac{8}{9}\\left( {a \\plus{} b \\plus{} c} \\right)\\left( {ab \\plus{} bc \\plus{} ca} \\right)\r\n\\]\r\n :|", "Solution_2": "oh yes :) . Why don't I think it early? :maybe: >So, can you prove that this inequality\r\n$ (a^2\\plus{}ab\\plus{}b^2)(b^2\\plus{}bc\\plus{}c^2)(c^2\\plus{}ca\\plus{}a^2)\\ge abc(a\\plus{}b\\plus{}c)^3$", "Solution_3": "[quote=\"thegod277\"]oh yes :) . Why don't I think it early? :maybe: >So, can you prove that this inequality\n$ (a^2 \\plus{} ab \\plus{} b^2)(b^2 \\plus{} bc \\plus{} c^2)(c^2 \\plus{} ca \\plus{} a^2)\\ge abc(a \\plus{} b \\plus{} c)^3$[/quote]\r\nIt is Holder. :wink:", "Solution_4": "can you describe how you used holder here...i am not used to use holder.please help.........", "Solution_5": "[quote=\"ishfaq420haque\"]can you describe how you used holder here...i am not used to use holder.please help.........[/quote]\r\n\\[ \\left( {a^2 \\plus{} ab \\plus{} b^2 } \\right)\\left( {bc \\plus{} c^2 \\plus{} b^2 } \\right)\\left( {a^2 \\plus{} c^2 \\plus{} ac} \\right) \\ge \\left( {\\sqrt[3]{{a^4 bc}} \\plus{} \\sqrt[3]{{abc^4 }} \\plus{} \\sqrt[3]{{ab^4 c}}} \\right)^3 \\equal{} abc\\left( {a \\plus{} b \\plus{} c} \\right)^3 \r\n\\]" } { "Tag": [ "number theory", "relatively prime", "prime numbers", "algebra unsolved", "algebra" ], "Problem": "Find all polunomials $P$ with integer coefficients which satisfay the property that, for any relatively prime integers $a,b$ the sequence $\\left\\{ P(an+b)\\right\\}$ for $n\\geq 1$ contains an infite number of terms , any two of which are relatively prime.", "Solution_1": "Could anyone post a solution ?Harazi could you give a hint or a solution ?", "Solution_2": "It is nesesary and sufficiently, that $gsd(a_0,a_1,\\dots,a_n)=1,$ were $P(x)=a_0+a_1x+\\dots+a_nx^n.$", "Solution_3": "Can you write a proof for this fact. Maybe is easy but I cannnot figure it out. thank you!", "Solution_4": "Can you write a proof for this fact. Maybe is easy but I cannnot figure it out. I can only see that it is necesary. thank you!", "Solution_5": "I was wrong. Nesessary and sufficiently conditions are\r\n1)$m_a=gcd(a_0,a_1,\\dots ,a_n)=1,$\r\n2)$m_b=(a,P(b))=1,$\r\n3)for all primes $p\\not|a$ (p-1 $A^2=\\frac{P}{2}(\\frac{P}{2}-\\frac{P}{3})^3$\n\nSo we need to prove that the value \n($\\frac{P}{6})^3$ is maximized for all P, with nothing added to or taken from any of the sides.\n\nNow consider all cases where P is not equilateral. Let the two longer sides (isoscoles works in this case too) be $\\frac{P}{3}+a$ and $\\frac{P}{3}+b$, and then the short side is $\\frac{P}{3}-(a+b)$. \n\nApplying Herons again, the expansion turns out to be $\\frac{P^3}{6^3}+\\frac{Pab}{6}-\\frac{Pa^2}{6}-\\frac{Pb^2}{6}-ba^2-ab^2$. \n\nTo show that this is always smaller than the equlateral case, we need only show that all the negative terms (remember, a and b are 0 or positive) have an absolute value greater than $\\frac{Pab}{6}$. \n\nTo do this we see that $\\frac{Pab}{6}-\\frac{P(a^2+b^2)}{6}$ must always be less than zero, because either a or b must be greater than the other (or equal), so $a^2$ or $b^2$ must be greater than ab (or they are equal and $a^2+b^2=2ab$).\n\nTherefore, the case where the sides all equal $\\frac{P}{3}$ gives the maximum area for any perimeter P.[/hide]" } { "Tag": [ "induction", "real analysis", "real analysis solved" ], "Problem": "Consider the real sequence (a_n)n>=1 defined by a1=a and a_(n+1)=a_n/n for all n>=1.\r\nProve that there is a unique a>0 such that lim(n-->oo) ((a_n)^(1/2^n))=1.\r\n\r\ncheers! :D :D", "Solution_1": "Are you sure? Please modify your post so that it would be clear if an is a_n or a*n. Or if you use a_nn for a_(n*n) please make it clear.\r\n\r\nIf the text is the one I understand, then for every a>0 that limit is 1.", "Solution_2": "Prove by induction a_n=a/n!\r\nlim(n->00)((a_n)^(1/2^n))=1 <=>\r\nlim (1/2^n) ln(a_n)=0 <=>lim (lna/(2^n)-(lnn!)/2^n)=-lim lnn!/2^n=(Stolz-Cesaro) =-lim ln(n+1)/2^n=0. So you can see that if a>0, then it doesn't count in the limit, so forget about uniqueness." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $n \\geq 3$ be an integer. Consider positive real numbers $a_1,..., a_n$ such that their product is 1. Show that: $\\sum_{i=1}^n \\frac{a_i+3}{(a_i+1)^2} \\geq 3$", "Solution_1": "I have no idea how this problem arrived in UK, since I sent it to our team leader this year for the national olympiad and TST. Quite interesting if both people had exactly the same inequality in mind. Anyway, it's not that difficult, but I would really like to see a smart solution, without IPt, SUV, GRT and so on...", "Solution_2": "It can be done with Ipt, but what does mean SUV and GRT ?", "Solution_3": "[quote=\"harazi\"]I have no idea how this problem arrived in UK, since I sent it to our team leader this year for the national olympiad and TST. Quite interesting if both people had exactly the same inequality in mind. Anyway, it's not that difficult, but I would really like to see a smart solution, without IPt, SUV, GRT and so on...[/quote]\r\n\r\noh, it was in the balkan shortlists for this year.", "Solution_4": "Please tell me what Ipt , SUV , GRT are !!", "Solution_5": "For future, do not ask also, what are \"iran1996\", or \"manlio239.17.8b\" inequalities unless you want to seem an ignoramus. Every educated mathlinker must know such basic things.\r\n\r\nThere is a plenty of three-letters abbreviations for some very useful theorems devised by forum members, using which one may easily solve almost any inequality. \r\n\r\nI guess these are various facts about minimazing a sum $\\sum f(x_i)$ for a fixed $\\sum x_i$, when $f$ is not convex or concave, but has not too much inflection points. But I think so just because it is a standard way to solve such inequalities.", "Solution_6": "I figured SUV and GRT were jokes :D", "Solution_7": "[quote=\"harazi\"]...Anyway, it's not that difficult, but I would really like to see a smart solution, without IPt, SUV, GRT and so on...[/quote]\r\n\r\nI found it difficult...", "Solution_8": "As nobody posted a solution, let me post mine.\r\nWe can break the initial inequality into two parts:\r\n1. $\\sum_{i=1}^{n} \\frac{1}{a_i+1} \\ge 1$ and \r\n2. $\\sum_{i=1}^{n} \\frac{1}{a_i+1}^2 \\ge 1$.\r\nProof of 1.\r\nNotice that \\[\r\n\\begin{array}{l}\r\n \\frac{1}{{x + 1}} + \\frac{1}{{y + 1}} \\ge \\frac{1}{{xy + 1}} \\\\ \r\n \\Leftrightarrow (x + y + 2)(xy + 1) \\ge xy + x + y + 1 \\\\ \r\n \\Leftrightarrow xy(x + y) + 2xy + x + y + 2 \\ge xy + x + y + 1 \\\\ \r\n \\Leftrightarrow xy(x + y) + xy + 1 \\ge 0 \\\\ \r\n \\end{array}\r\n\\]\r\nwhich is true.\r\nSo, $\\sum_{i=1}^{n} \\frac{1}{a_i+1} \\ge \\frac{1}{a_1+1} + \\frac{1}{a_2 \\cdots a_n +1} = 1$.\r\nProof of 2.\r\nNotice that\r\n\\[\r\n\\begin{array}{l}\r\n \\frac{1}{{\\left( {x + 1} \\right)^2 }} + \\frac{1}{{\\left( {y + 1} \\right)^2 }} \\ge \\frac{1}{{\\left( {xy + 1} \\right)^2 }} \\\\ \r\n \\Leftrightarrow \\left( {x^2 + y^2 + 2x + 2y + 2} \\right)\\left( {x^2 y^2 + 2xy + 1} \\right) \\ge \\\\ \r\n x^2 y^2 + 2x^2 y + 2xy^2 + x^2 + y^2 + 4xy + 2x + 2y + 1 \\\\ \r\n \\Leftrightarrow \\left( {x^2 + y^2 + 2x + 2y + 1} \\right)x^2 y^2 + 2xy(x^2 + y^2 + x + y) + 1 \\ge 0 \\\\ \r\n \\end{array}\r\n\\]\r\nSo,$\\sum_{i=1}^{n} \\frac{1}{(a_i+1)^2} \\ge \\frac{1}{(a_1+1)^2} + \\frac{1}{(a_2+1)^2}+\\frac{1}{(a_3+1)^2} + \\frac{1}{(a_4 \\cdots a_n)^2 +1}$.\r\nIt remains to prove if $xyzt=1$, then $\\sum \\frac{1}{(x+1)^2} \\ge 1$.\r\nWe can then use $\\frac{1}{(x+1)^2}+\\frac{1}{(y+1)^2} \\ge \\frac{1}{xy+1}$ to finish it. \r\n\r\nThen, we have proved the inequality for $n \\ge 4$ by adding (1)+2(2).\r\n\r\nFor n=3, denote $f(x)=\\frac{x+3}{(x+1)^2}$. Troublesome checking yields that $f(x)+f(y) \\ge 2 f(\\sqrt{xy})$ for $xy \\ge 1$. WLOG assume $xy \\ge 1$, then we have to prove $2f(\\sqrt{xy})+f(z) \\ge 3$ or $2f(\\frac{1}{a})+f(a^2) \\ge 3$ for $a \\le 1$\r\nThis is equivalent to prove $\\frac{2(a^2+3a)}{(a+1)^2}+\\frac{a^2+3}{(a^2+1)^2} \\ge 3$.\r\nBy cauchy, $L.H.S. \\ge \\frac{a^2+3a}{a^2+1}+\\frac{a^2+3}{(a^2+1)^2}$\r\nExpanding of $\\frac{a^2+3a}{a^2+1}+\\frac{a^2+3}{(a^2+1)^2} \\ge 3$ yields $3a^3+3a \\ge 2a^4+4a^2$ which is true by $a \\le 1$ and $a^3+a \\ge 2a^2$.\r\nQED.", "Solution_9": "I think we should only prove $n=3$\r\nand $n=3$ implies $n\\geq3$", "Solution_10": "[quote=\"siuhochung\"]\nFor n=3, denote $f(x)=\\frac{x+3}{(x+1)^2}$. Troublesome checking yields that $f(x)+f(y) \\ge 2 f(\\sqrt{xy})$ for $xy \\ge 1$. WLOG assume $xy \\ge 1$, then we have to prove $2f(\\sqrt{xy})+f(z) \\ge 3$ or $2f(\\frac{1}{a})+f(a^2) \\ge 3$ for $a \\le 1$\nThis is equivalent to prove $\\frac{2(a^2+3a)}{(a+1)^2}+\\frac{a^2+3}{(a^2+1)^2} \\ge 3$.\nBy cauchy, $L.H.S. \\ge \\frac{a^2+3a}{a^2+1}+\\frac{a^2+3}{(a^2+1)^2}$\nExpanding of $\\frac{a^2+3a}{a^2+1}+\\frac{a^2+3}{(a^2+1)^2} \\ge 3$ yields $3a^3+3a \\ge 2a^4+4a^2$ which is true by $a \\le 1$ and $a^3+a \\ge 2a^2$.\nQED.[/quote]\r\n\r\nFor n=3 you can put $a_1=1/x,a_2=1/y,a_3=1/z$ and use Cauchy.\r\nI remember the similar method was given on 5'th mathlinks contest.($abc\\ge 8... $)", "Solution_11": "[quote=\"hardsoul\"]For n=3 you can put $a_1=1/x,a_2=1/y,a_3=1/z$ and use Cauchy.\nI remember the similar method was given on 5'th mathlinks contest.($abc\\ge 8... $)[/quote]\r\n\r\nCould you make that more explicit? I can't get Cauchy to work.", "Solution_12": "[quote=\"jhaussmann5\"][quote=\"hardsoul\"]For n=3 you can put $a_1=1/x,a_2=1/y,a_3=1/z$ and use Cauchy.\nI remember the similar method was given on 5'th mathlinks contest.($abc\\ge 8... $)[/quote]\n\nCould you make that more explicit? I can't get Cauchy to work.[/quote]\r\nWell,unfortunately I have lost the solution paper,and couldn't remember my solution by Cauchy(it seems it was wrong)\r\nHere's another solution:\r\n\r\n1)We can reduce the inequality to the case n=3 by repeadately using the inequality $\\displaystyle\\frac{x+3}{(x+1)^2}+\\frac{y+3}{(y+1)^2}\\ge \\frac{xy+3}{(xy+1)^2}$ for $x\\ge1,y\\leq 1$\r\n2)Put $a_1=1/a,a_2=1/b,a_3=1/c$ to get $\\displaystyle\\sum\\frac{a+3a^2}{(a+1)^2}\\ge 3$ that is equivalent to \r\n$\\displaystyle\\sum\\frac{2a^2-a-1}{(a+1)^2}\\ge 0\\Leftrightarrow\\displaystyle\\sum\\frac{2a^2}{(a+1)^2}\\ge \\displaystyle\\sum\\frac{1}{a+1}$ Now use the inequality $\\sum_{cyc}\\left(\\displaystyle\\frac{a^2}{(a+1)^2}+\\displaystyle\\frac{b^2}{(b+1)^2}\\right)\\ge \\displaystyle\\sum_{cyc}\\frac{1}{c+1}$", "Solution_13": "[quote=\"zhaobin\"]I think we should only prove $n=3$\nand $n=3$ implies $n\\geq3$[/quote]\r\n\r\nWhy is this? Can you show?", "Solution_14": "[quote=\"indybar\"][quote=\"zhaobin\"]I think we should only prove $n=3$\nand $n=3$ implies $n\\geq3$[/quote]\n\nWhy is this? Can you show?[/quote]\r\n\r\n Well, I think zhaobin said is right. Because \r\n$ \\sum_{i=1}^4 \\frac{a_i+3}{(a_i+1)^2} \\geq\\sum_{i=1}^3 \\frac{a_i+3}{(a_i+1)^2} $\r\nSo if $n=3$ is true. Obviously$ n=4$ is true too.\r\nIs there anyone use Jensen to solve it? :?", "Solution_15": "[quote=\"Gibbenergy\"]So if $n=3$ is true. Obviously$n=4$ is true too.[/quote]\r\n\r\nWhy \"obviously\"?", "Solution_16": "It should be:\r\nif $abc \\le 1$ then:\r\n$\\sum \\frac{a+3}{(a+1)^2} \\ge 3$\r\n\r\nFor n number, we suppose that $a=a_1 \\le b=a_2 \\le c=a_3 \\le a_4 \\le ... \\le a_n$ then $abc \\le 1$\r\n\r\nApply that inequality we have done.\r\n(Note that $abc \\le 1$, not $abc=1$)", "Solution_17": "hungkhtn do you have a solution? :?" } { "Tag": [ "abstract algebra", "number theory", "prime numbers", "group theory", "number theory unsolved" ], "Problem": "Find all positive integers $n>1$ for which there exists a unique integer $a$ with $0 < a\\leq n!$ such that $a^n+1$ is divisible by $n!$.", "Solution_1": "The answer is all prime numbers $n$.\r\n\r\nHere is my not very beautiful but I hope correct solution:\r\n\r\n$n=2$ is a solution, for $n\\ge 4$ even, $a^{n}$ is a square and $3\\mid n!$ but $3\\nmid a^{n}+1$, so $n$ must be odd if $n\\ge 3$.\r\n\r\nLet $n!=p_{1}^{k_{1}}\\ldots p_{r}^{k_{r}}$ with $p_{i}$ distinct primes. By Chinese Remainder Theorem, $a^{n}+1\\equiv 0\\mod n!$ has unique solution in $a$ iff $a^{n}+1\\equiv 0 \\mod p_{i}^{k_{i}}$ has a unique solution for all $i=1\\ldots r$.\r\n\r\nWe shall prove first, that $a^{n}+1\\equiv 0 \\mod 2^{k}$ has a unique solution.\r\nWe have $a^{n}+1=(a+1)(\\sum_{i=0}^{n-1}(-1)^{i}a^{i})\\equiv 0 \\mod 2^{k}$, but $\\sum_{i=0}^{n-1}(-1)^{i}a^{i}$ is odd, so $a+1\\equiv 0\\mod 2^{k}$.\r\nThus, we do not need to worry about the oddest prime $2$ anymore.\r\n\r\n\r\nFirst, let $n\\ge 3$ be a prime. Clearly, $v_{n}(n!)=1$.\r\nBy FLT, $a^{n}\\equiv a \\equiv-1\\mod n$, so there is a unique solution modulo $n$.\r\nNow let $p\\geq 3$ be a prime dividing $n!$ and let $k=v_{p}(n!)$ and $p\\neq n$.\r\nFurthermore let $g$ be a primitive root $\\mod p^{k}$. \r\nThen $-1\\equiv g^{\\frac{\\phi(p^{k})}{2}}\\mod p^{k}$ and $a\\equiv g^{x}\\mod p^{k}$ for some $x\\in \\{1,2,\\ldots,\\phi(p^{k})\\}$.\r\nSo $a^{n}\\equiv-1\\mod p^{k}\\Leftrightarrow g^{nx}\\equiv g^{\\frac{\\phi(p^{k})}{2}}\\mod p^{k}\\Leftrightarrow nx\\equiv \\frac{\\phi(p^{k})}2 \\mod \\phi(p^{k})$.\r\nHence there is an unique solution $a\\mod p^{k}$ iff there is a unique solution $x\\mod \\phi(p^{k})$. But $nx\\equiv \\phi(p^{k})/2 \\mod \\phi(p^{k})$ has a unique solution in $x$ since $gcd(n,phi(p^{k}))=gcd(n,(p-1)p^{k-1})=1$.\r\nThus there is a unique $a\\mod n!$ for $n\\ge 3$ prime.\r\n\r\n\r\nNow let $n$ be composite and let $p\\ge 3$ be a prime dividing $n$ and different from $n$ (which exists since we assumed that $n$ is odd).\r\nClearly, $n\\ge 2p$ so $v_{p}(n!)\\ge 2$.\r\nNow if $a$ would be unique $\\mod n!$ then it would be unique $\\mod p^{k}$ where $k=v_{p}(n!)\\ge 2$. \r\nLet $g$ be a primitive root $\\mod p^{k}$. Thus again, $-1\\equiv g^{\\frac{\\phi(p^{k})}2}\\mod p^{k}$ and $a\\equiv g^{x}\\mod p^{k}$ for some $x\\in\\{1,\\ldots,\\phi(p^{k})\\}$. \r\nNow $a^{n}\\equiv-1 \\mod p^{k}\\Leftrightarrow nx\\equiv \\phi(p^{k})/2 \\mod \\phi(p^{k})$, so by the assumption that $a$ is unique $\\mod p^{k}$, there is a unique $x\\mod \\phi(p^{k})$, i.e. that $nx\\equiv \\phi(p^{k})/2 \\mod \\phi(p^{k}) \\Leftrightarrow nx\\equiv \\frac{(p-1)p^{k-1}}2 \\mod (p-1)p^{k-1}$ has a unique solution in $x$.\r\nBut this is a contradiction since $p\\mid n$ and $p\\mid (p-1)p^{k-1}$.", "Solution_2": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=85073 also.\r\n\r\n@Yimin: I agree to your signature :)", "Solution_3": "Ghm, in all solutions to this problem that I have seen the part where one shows that prime $n$ works is handled somewhat meticulously. Correct me if I am wrong but it follows immediately from the fact that mapping $x\\rightarrow x^{a}$ is a bijection for the set $A=\\{x|(x, b)=1\\}\\subset \\mathbb{Z}_{b}$ if $(a, \\phi(b))=1$ (which is the case for the problem since $(p, \\phi(p!))=1$).", "Solution_4": "The converse holds also (it is only bijective iff they are coprime), giving the whole statement. Thats the way I proved it in our TST, with proving this equivalence\r\nI think this was the source of the problem also.", "Solution_5": "Can you please prove the converse?", "Solution_6": "Take prime power $p^{r}\\parallel b$ such that $d=(a, \\phi(p^{r})) \\geq 2$. Write $a=kd$ and let $g$ be the primitive root $\\mod{p^{r}}$. We have:\r\n$(g^{\\frac{\\phi(p^{r})}{d}})^{d}\\equiv 1 \\mod{p^{r}}$.\r\nBecause $g$ is the primitive root $g^{\\frac{\\phi(p^{r})}{d}}$ is not $1$.\r\nAlso, $g^{\\frac{\\phi(p^{r})}{d}}\\equiv x^{k}$ for some $x$ because $(k, \\phi(p^{r}))=1$. Therefore: $(x^{k})^{d}= x^{a}\\equiv 1$. Now, write $x \\equiv \\frac{s}{t}$ and we have $s^{a}\\equiv t^{a}$. Since $x \\not \\equiv 1$ $k \\not \\equiv t$ and this mapping isn't bijective for some highest prime power dividing $b$. so it isn't a bijective mapping $\\mod{b}$.\r\nActually, the case $b=2^{k}$ should be treated seperately.", "Solution_7": "More general claim:\r\n\r\nLet $G$ be any finite group. Then $x \\mapsto x^{k}$ is bijective iff $\\gcd(k,|G|)=1$.\r\n\r\nProof:\r\nIf they are coprime, we find by Bezout's theorem integers $a,b$ with $ak-b|G|=1$. By Lagrange we have $x^{|G|}=e$ for all $x \\in G$, thus $x^{ak}= x^{1+b|G|}= x$.\r\nThus the map $x \\mapsto x^{a}$ is an inverse mapping, thus both mappings are bijective.\r\nConverse:\r\nWhen they are not coprime, let $p$ be a common prime divisor of $|G|$ and $k$. On various ways we can get that $x \\mapsto x^{p}$ is already not injective, so also not $x \\mapsto x^{k}$. The most appealing one (since it is near to be equivalent) is using Cauchy's theorem (see http://www.mathlinks.ro/Forum/viewtopic.php?t=27036 , post #6 for a proof) which gives that there is an $x$ with order $p$, so an $x \\neq e$ with $x^{p}=e$, showing that it is not injective.\r\n\r\n\r\nSince my proof for the residue case was more like the one of TomciO (for the original problem one could avoid using primitive roots, but well...), I will not post it.", "Solution_8": "[quote=\"venkata\"]Find all positive integers $n>1$ for which there exists a unique integer $a$ with $0 < a\\leq n!$ such that $a^{n}+1$ is divisible by $n!$.[/quote]\r\nIf n is odd then $n!|(n!-1)^{n}+1 \\ \\ (a=n!-1)$.", "Solution_9": "@Rust: this post was completely useless.\r\nRead the previous posts and especially the links given there.\r\nAnd note that this helps as good as nothing ;)", "Solution_10": "[quote=\"ZetaX\"]The converse holds also (it is only bijective iff they are coprime), giving the whole statement. Thats the way I proved it in our TST, with proving this equivalence\nI think this was the source of the problem also.[/quote]\r\n\r\nIt seems as though I'm too dumb to understand your solution :maybe: \r\n\r\nHow do you prove the problem with the converse? I don't really understand how you deduce from $f$ being not bijective that $-1$ is hit more than once? How do you prove that e.g. the case that only $f(-1)=-1$ and the rest is not bijective (e.g. that there are different $x,y$ so that $f(x)=f(y)\\neq-1$ or whatever) cannot occur?\r\n\r\nEdit: Here $f$ is of course the map $x\\mapsto x^{k}$ you were talking about", "Solution_11": "Sorry for not mentioning/writting that last step:\r\n\r\n\r\nThe general one for groups again:\r\nWhen $f: G \\to H$ is a homomorphism of groups (so $f(ab)=f(a)f(b)$) but not injective, then $f(x)=a$ has never exactly one solution $x$ for given $a \\in H$.\r\n\r\nProof:\r\nAt first, note that $f(t^{-1})=f(t)^{-1}$ for all $t \\in G$ (simple to see when not clear from beginning).\r\nThere are two different $x_{1},x_{2}\\in G$ with $f(x_{1})=f(x_{2})$. Thus setting $z: =x_{1}x_{2}^{-1}\\neq e_{G}$ we get $f(z)=f(x_{1}x_{2}^{-1})=f(x_{1})f(x_{2}^{-1})=e_{H}$, the neutral element (this is one direction of the well known property $f$ injective $\\iff$ $\\ker(f)=\\{e_{H}\\}$).\r\nNow let that $a$ be given. When there is $x$ with $f(x)=a$ (when there is none, we are done) we find for $y: =xz \\neq x$ that $f(y)=f(xz)=f(x)f(z)=a e = a$.\r\nSo there are at least two different solutions, done.\r\n\r\n\r\nNow the mapping $x \\mapsto x^{k}$ is a homomorphism $G \\to G$ at least when $G$ is abelian, which is the case here.\r\n\r\n\r\n\r\nAll in all, this seems to be pure group theory :lol:", "Solution_12": "Or note that $n$ must be odd (on the contrary $-a\\neq a$ satisfies the congruence as well) and deduce that for a congruence $a^{n}\\equiv-1\\mod{n!}$ to have unique solution the congruence $(-a)^{n}\\equiv 1\\mod{n!}$ must have unique solution. But if the mapping is not bijective then there exist $b\\neq c\\colon \\ b^{n}\\equiv c^{n}\\mod{n!}$. But then $\\frac{b}{c}\\neq 1$ and $(\\frac{b}{c})^{n}\\equiv 1$. Hence $-1$ is not unique and we are done." } { "Tag": [ "geometry", "trigonometry", "geometry proposed" ], "Problem": "Consider C(O,R) a circle and C'(O',R)another circle sunh that O' is on C.Consider A and B the two intersection points between the two circles.Let C be on the arc AB and on the other part of AB again on the AB arc a point D.\r\na)Find the maximum area of ADC\r\nb)Find the maximum area of ABCD\r\n Luca V. Iliesiu", "Solution_1": "For a given $D \\in \\mathcal C'$, the triangle $\\triangle ACD$ has maximum area, when C is the tangency point of is the circle $\\mathcal C$ with the tangent $t \\parallel AD.$ Then $CO \\perp t,\\ CO \\perp AD$ is the C-altitude line. Let F be the foot of this altitude. Denote the directed angle $\\angle O'AD = \\phi$ (can be positive or negative). Since the circles $\\mathcal C' \\cong \\mathcal C$ are congruent and $O' \\in \\mathcal C,$ the triangle $\\triangle AOO'$ is equilateral and $\\angle OAO' = 60^\\circ.$ Then $\\angle OAF \\equiv \\angle OAD = \\angle OAO' + \\angle O'AD = 60^\\circ + \\phi,$ $OF = OA \\sin \\widehat{OAF} = R \\sin (60^\\circ + \\phi)$ and $CF = OC + OF = R[1 + \\sin(60^\\circ + \\phi)].$ In addition, $AD = 2R \\cos \\phi.$ Area of the triangle $\\triangle ACD$ is then\r\n\r\n$S = |\\triangle ADC| = \\frac{CF \\cdot AD}{2} = R^2 [1 + \\sin(60^\\circ + \\phi)] \\cos \\phi$\r\n\r\nFor the maximum area, $\\frac{\\text dS}{\\text d\\phi} = 0.$\r\n\r\n$\\cos(60^\\circ + \\phi) \\cos \\phi - [1 + \\sin(60^\\circ + \\phi)] \\sin \\phi] = 0$\r\n\r\n$\\sin \\phi = \\cos(60^\\circ + 2\\phi) = \\sin(90^\\circ - (60^\\circ + 2\\phi)] = \\sin (30^\\circ - 2 \\phi)$\r\n\r\n$\\phi = 30^\\circ - 2 \\phi,\\ \\ \\ \\phi = 10^\\circ$\r\n\r\nStarting from an arbitrary point $C \\in \\mathcal C,$ we can similarly find $\\angle OAC = 10^\\circ.$ Hence, the triangle $\\triangle ADC$ is isosceles with $CD \\parallel OO'$ and $\\angle CAD = \\angle CAO + \\angle OAO' + \\angle O'AD = 80^\\circ.$\r\n\r\nThe 2nd problem is trivial, even if the circles $\\mathcal C, \\mathcal C'$ are not congruent or $O'$ is not on $\\mathcal C.$ The area of the quadrilateral ABCD is $|ABCD| = \\frac 1 2 AB \\cdot CD \\sin\\widehat{(AB, CD)}.$ AB is fixed and both CD and $\\sin\\widehat{(AB, CD)}$ reach their maximum values when $C \\equiv OO' \\cap \\mathcal C,\\ D \\equiv OO' \\cap \\mathcal C',$ where C is outside of $\\mathcal C'$ and D is outside $\\mathcal C.$", "Solution_2": "[color=darkblue]Sorry, Yetti, your proof isn't clearly a bit for me. I will try to prove more clearly what you proved above. Please, see you if I made a mistake somewhere ![/color]\r\n\r\n[color=darkred]Define $x\\ .s.s.\\ y$ if and only if $x=y$ or $xy>0$. I will suppose that: the point $D$ belongs to the little arc $\\frown\\above AB\\end$ of the circle $C'(O',R')$; the point $C$ belongs to the little arc $\\frown\\above AB\\end$ of the circle $C(O,R)$ so that $CO\\perp AD$. Denote $m(\\widehat {O'AD})=x$ and the projection $F$ of the point $C$ on the line $AD$. We observe that: $30^{\\circ}\\le x\\le 90^{\\circ}$; $AD=2R'\\cos x$ and $CF=R[1+\\sin (x-60^{\\circ}]$. Therefore, $[ACD]=AD\\cdot CF=$ $R\\cdot R'f(x)$, where\n$\\boxed {f(x)=\\cos x[1+\\sin (x-60^{\\circ})]\\mathrm {\\ ,where\\ }x\\in \\left[30^{\\circ},90^{\\circ}\\right]}$.\n$f'(x)=-\\sin x+\\cos \\left(2x-60^{\\circ}\\right)=\\cos \\left(2x-60^{\\circ}\\right)-\\cos \\left(90^{\\circ}-x\\right)\\Longrightarrow$\n$f'(x)\\ .s.s.\\ \\left[\\left(90^{\\circ}-x\\right)-\\left(2x-60^{\\circ}\\right)\\right]=3\\left(50^{\\circ}-x\\right)$, i.e. $\\boxed {f'(x)\\ .s.s.\\ \\left(50^{\\circ}-x\\right)}\\ .$\nWe came to the conclusion that for any $x\\in \\left[30^{\\circ},90^{\\circ}\\right]$\nwe have $f(x)\\le f\\left(50^{\\circ}\\right)$ a.s.o.[/color]", "Solution_3": "nice solution...well mine was similar when I made the problem...Just using trigonometry", "Solution_4": "[quote=\"Virgil Nicula\"][color=darkblue]Sorry, Yetti, your proof isn't clearly a bit for me. I will try to prove more clearly what you proved above. Please, see you if I made a mistake somewhere ![/color]...[/quote]\r\nMaybe you did not notice that the circles $\\mathcal C \\cong \\mathcal C'$ are congruent, there is no $R' \\neq R.$ The minor arcs AB of both circles $\\mathcal C, \\mathcal C'$ are to be ignored - if either C or D or both are on the minor arcs, then the line CD meets the corresponding circle at C' or D' and areas of the triangles $\\triangle AC'D, \\triangle ACD', \\triangle AC'D'$ are larger than the area of $\\triangle ACD.$ If D on the major arc of $\\mathcal C'$ is arbitrary, with an angle $\\angle O'AD = \\phi$, then the base AD is fixed and to maximize the altitude, it must be $CO \\perp AD.$ Both AD and the C-altitude $CO \\equiv CF$ can be calculated in terms of $\\phi = \\angle O'AD.$ It does not matter how $\\phi$ is directed, if we direct it the wrong way, it will come out negative. Acording to my direction,\r\n\r\n$AD = 2R \\cos \\phi,\\ \\ \\ CF = R[1 + \\sin(60^\\circ + \\phi)], \\ \\ \\ \\phi \\in (-30^\\circ, +90^\\circ).$\r\n\r\nAccording to your direction, $x = -\\phi$ and\r\n\r\n$AD = 2R \\cos x,\\ \\ \\ CF = R[1 + \\sin(60^\\circ - x)],\\ \\ \\ x \\in (-90^\\circ, +30^\\circ).$\r\n\r\nHowever, if $R' \\neq R,$ the side AD and the altitude CF would be\r\n\r\n$AD = 2R' \\cos x,\\ \\ \\ CF = CO + OF = R + R' \\sin(60^\\circ - x)$\r\n\r\n$\\frac{CF \\cdot AD}{2} = RR' \\cos x \\left[1 + \\frac{R'}{R}\\ \\sin(60^\\circ - x)\\right] \\neq RR' f(x),$ where $f(x) = \\cos x\\ [1 + \\sin(60^\\circ - x)]$\r\n\r\nand it would not be easy to solve $f'(x) = 0$ for x, except numerically. Assuming the problem condition $\\mathcal C \\cong \\mathcal C',$ then maximizing $f(x) = f(-\\phi)$ with respect to the angles $\\phi$ or $x = -\\phi$ gives $\\sin (30^\\circ - 2 \\phi) - \\sin \\phi = 0$ or $\\sin(30^\\circ + 2x) + \\sin x = 0.$ Instead of concluding $30^\\circ - 2 \\phi = \\phi$ or $30^\\circ + 2x = -x$ leading to $\\phi = +10^\\circ,\\ x = -10^\\circ,$ we could rigorously write\r\n\r\n$\\sin (30^\\circ - 2 \\phi) - \\sin \\phi = 2 \\cos \\frac{30^\\circ - \\phi}{2} \\sin \\frac{30^\\circ - 3 \\phi}{2} = 0$\r\n\r\n$\\sin (30^\\circ + 2x) + \\sin x = 2 \\sin \\frac{30^\\circ + 3x}{2} \\cos \\frac{30^\\circ + x}{2} = 0$\r\n\r\n$x = -\\phi \\in (-90^\\circ, +30^\\circ)\\ \\Longrightarrow$\r\n\r\n$\\frac{30^\\circ + x}{2} = \\frac{30^\\circ - \\phi}{2} \\in (-30^\\circ, +30^\\circ)\\ \\ \\&\\ \\ \\frac{30^\\circ + 3x}{2} = \\frac{30^\\circ - 3 \\phi}{2} \\in (-120^\\circ, +60^\\circ)$\r\n\r\nThus $\\cos \\frac{30^\\circ + x}{2} = \\cos \\frac{30^\\circ - \\phi}{2} \\neq 0$ and sine has the only root 0 in $(-120^\\circ, +60^\\circ),$ hence $\\frac{30^\\circ + x}{2} = \\frac{30^\\circ - \\phi}{2} = 0,$ $x = -10^\\circ, \\phi = +10^\\circ.$ The maximum triangle area is then\r\n\r\n$|\\triangle ADC|_{\\text{max}} = R^2 \\cos 10^\\circ (1 + \\sin 70^\\circ) = R^2 \\cos 10^\\circ (1 + \\cos 20^\\circ) =$\r\n\r\n$= 2R^2 \\cos^3 10 = \\frac 1 2 (2R \\cos 10^\\circ)^2 \\sin 80^\\circ$\r\n\r\nHaving calculated the maximum triangle area, we can stop right here.", "Solution_5": "O.K. Thanks. I understood wrong the initial problem,." } { "Tag": [], "Problem": "$\\sum_{x=1}^{200}\\frac{x^{4}+4x^{3}+10x^{2}+12x+4}{(x^{2})(x+1)^{2}(x+2)^{2}}= \\frac{a}{b}$. Find $\\frac{a}{b}-\\left(\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+...+\\frac{1}{102}\\right)$.", "Solution_1": "you need a two in the partial fraction decomp or something...\r\n\r\ncause i get 1/(x+2)^2+1/(x)^2-1/(x+2)^2\r\n\r\nand not enough stuff cancels...", "Solution_2": "Sorry, see post below.", "Solution_3": "It's 3/2?", "Solution_4": "Wow, I posted this at 3 AM and it shows. I rewrote the problem. Altheman, you were on the right track. Sorry!" } { "Tag": [ "inequalities", "rearrangement inequality", "algebra solved", "algebra" ], "Problem": "Prove that\r\n\r\nx^(a+1)/y^a + y^(a+1)/z^a + z^(a+1)/x^a \\geq x+y+z\r\n\r\nwhere x,y,z,a are all >0.\r\n\r\nIt is very simple by rearrangement inequality so I have found a solution without using rearrangement inequality.\r\n\r\nI think the beauty of inequality resides in the possibility to find different solutions.", "Solution_1": "Indeed, another solution is the following:\r\n Let us suppose that x+y+z=1. Then using the convexity of f(t)=t^(a+1) we find that y*f(x/y)+z*f(y/z)+x*f(z/x)>=f(y*x/y+z*y/z+x*z/x)=f(1)=1, which is excatly the desired inequality." } { "Tag": [ "percent" ], "Problem": "So I understand how to do stoichiometric calculations with chemical equations containing strong acids/bases, but how do I do neutralization/tritration calculations when a weak acid or base or both take part in the reaction? I don't know how to take into account for the incomplete dissociation of the weak acids/bases into the calculation. Do I just multiply the result by the percent of the weak acid/base that dissociates?", "Solution_1": "Here is a starting point:\r\nWhen a weak acid is titrated by a strong base, or vice versa, the reaction between the two is assumed to proceed to completion, due to the constant stress placed on the system (this is in accord with the Le Chatelier principle). Therefore, when doing calculations with titration, start with the stoichiometric calculations and then move on to calculating the equilibrium concentrations.", "Solution_2": "This is an equilibrium problem. Don't talk about \"percent of dissociation\" - that won't get you anywhere. You need to know the pKa for each of the species involved.", "Solution_3": "You will almost certainly never see a problem involving titration both a weak acid/weak base. I have never worked one myself, but I think you would need to do a lot of algebra. One of the reason these problems never come up is that in practice, they are never used. You always have a choice of which titrant to use, so why use a weak acid/base when you could make things more simple by using a strong acid/base?\r\n\r\nFor how to actually perform these calculations, you may find the following site useful:\r\n\r\n[url]http://www.chembuddy.com/?left=pH-calculation&right=toc[/url]" } { "Tag": [ "symmetry", "geometry", "geometric transformation", "reflection", "similar triangles", "angle bisector" ], "Problem": "Let AM be a median of triangle ABC, where M is the midpoint of BC. Point B, C are reflected over AM to B', C', respectively. AC' intersects BC at D. B'C' intersects AC at E. AE=6, EC=12, BD=10, find AB.", "Solution_1": "What does \"reflected over AM\" mean??", "Solution_2": "$\\begin{array}{ccc}\r\n&A&\\\\C&--------&D\\\\&B&\\end{array}$\r\n\r\nA is B, reflected over CD", "Solution_3": "anyone?? :maybe:", "Solution_4": "I've got no clue. Sorry. So far I have:\n\n\n\n[hide]Triangle ACC' is isocelis so AC'=AC=AE+EC=18[/hide]\n\n\n\nI don't think it helps much, though.", "Solution_5": "I'm having a terrible time drawing this monstrosity. =D Anyone wanna help me out?", "Solution_6": "Solution: [hide]First refer to diagram 1:\n\n\n\n1. Prove that BNM, B'NM, CQM, C'QM are all congruent.\n\n\n\n2. Then the shape NB'CQ must be rectangular. Since AE = 1/2 EC, we must have AM = 1/2 B'C by similar triangles. However NM = 1/2 B'C, so A and N are the same point.\n\n\n\nNow refer to diagram 2:\n\n\n\n3. Then by symmetry, BD = B'E. AE = 6, EC = 12, B'E = 10, by similar triangles EM = 5. Let AM = x, B'C = 2x, AB = y. Then\n\n\n\ny2 + x2 = 152\n\ny2 + 4x2 = 182\n\n\n\nThen x2 = 33, y2 = 192, y = 8:rt3:.[/hide]", "Solution_7": "Nice!! my solution:\n\n\n\n[hide]By definition of reflection, , so AD=AE=6, angle DAM=angle DAE. So AM is the angle bisector of triangle DAC. \n\n\n\nBy angle bisector theorem,\n\nAC/AD=18/6=MC/DM=3, MC=3MD. \n\n\n\nSince M is the midpoint of BC. BD+MD=MC, 10+MD=3MD, MD=5, MC=15, BC=30, CD=20\n\n\n\nApply Stewart's Theorem to ABC, we get\n\nAB^2(CD)+AC^2(BD)=AD^2(BC)+(BD)(CD)(BC)\n\n\n\nAB^2(20)+(18^2)(10)=6^2(30)+(10)(20)(30)\n\n AB^2=192, AB=8 :sqrt: 3 [/hide]" } { "Tag": [ "calculus", "derivative", "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "In Evans' book \"PDE\", page 29, thm7: \r\n$ u$ harmonic in $ U$, \r\nprove: $ |D^{\\alpha}u(x_0)|<\\equal{}\\frac{C_k}{r^{n\\plus{}k}}\\parallel{}u\\parallel{}_{L_1(B(x_0,r))}$ for each ball $ B(x_0,r) \\subset U$.\r\n\r\nIn the proof: case $ k\\equal{}0$ already proved, for case $ k\\equal{}1$, \r\nlet $ x \\in \\partial B(x_0,r/2)$, then $ x \\in B(x,r/2) \\subset B(x_0,r)$, so use case $ k\\equal{}0$, we have:\r\n$ u(x)<\\equal{}\\frac{C_0}{r^{n}}\\parallel{}u\\parallel{}_{L_1(B(x_0,r))}$.\r\n\r\nI think it should be $ u(x)<\\equal{}\\frac{C_0}{r^{n}}\\parallel{}u\\parallel{}_{L_1(B(x,r/2))}$ by using case $ k\\equal{}0$ ( take \r\n$ x_0\\equal{}x$). Did I miss something? \r\n\r\n\r\nMaybe $ \\parallel{}u\\parallel{}_{L_1(B(x,r/2))}<\\equal{}\\parallel{}u\\parallel{}_{L_1(B(x_0,r))}$ because $ B(x,r/2) \\subset B(x_0,r)$? but why?", "Solution_1": "[quote=\"kalazealot\"]\nMaybe $ \\parallel{}u\\parallel{}_{L_1(B(x,r/2))} < \\equal{} \\parallel{}u\\parallel{}_{L_1(B(x_0,r))}$ because $ B(x,r/2) \\subset B(x_0,r)$? but why?[/quote]\r\nBecause the integral of a non-negative function $ |u|$ over a larger set is not smaller ;)", "Solution_2": "[quote=\"fedja\"][quote=\"kalazealot\"]\nMaybe $ \\parallel{}u\\parallel{}_{L_1(B(x,r/2))} < \\equal{} \\parallel{}u\\parallel{}_{L_1(B(x_0,r))}$ because $ B(x,r/2) \\subset B(x_0,r)$? but why?[/quote]\nBecause the integral of a non-negative function $ |u|$ over a larger set is not smaller ;)[/quote]\r\n\r\n I see. :blush:" } { "Tag": [ "USAMTS" ], "Problem": "Are we allowed to use mathematica while solving a problem?", "Solution_1": "Yes.\r\n[size=0]asfosrd3oj;ed[/size]", "Solution_2": "From the USAMTS Rules page ([url]http://www.usamts.org/About/U_AbRules.php[/url]):\r\n[quote=\"USAMTS Rules\"]Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose. Therefore, if a key step of a participant's solution uses these tools, the participant must demonstrate that the tool completed the step correctly.[/quote]" } { "Tag": [], "Problem": "(a) Find all five-digit numbers which are reversed on multiplying by $ 4$.\r\n\r\n(b) Find all five-digit numbers which are reversed on multiplying by $ 9$.\r\n\r\n(c) Find all five-digit numbers which are reversed on multiplying by $ 8$.", "Solution_1": "[hide=\"a.\"] We can set bounds as $ 20000\\le x < 25000$ (since our end result is also $ 5$ digits and no multiple of $ 4$ ends in $ 1$). We can then set further bounds at $ 21000\\lex\\le21999$ and $ 23000\\lex\\le23999$ (since $ 2$, $ 22$, and $ 42$ are not divisible by $ 4$). We can next eliminate the $ 23abc$ case by noting that when this is multiplied by $ 4$ the ten thousands digit will be $ 9$ ($ a$, $ b$, and $ c$ are digits). But any number that ends in $ 9$ when multiplied by $ 4$ will end in $ 6$. Thus we look at the $ 21ab8$ cases. By systematically checking (we can limit checking by noting that $ \\frac {85000}{4} = 21250$, $ \\frac {86000}{4} = 21500$, and $ \\frac {87000}{4} = 21750$), we find that the only is $ \\boxed{21978}$, since $ 21978\\times{4} = 87912$.[/hide]", "Solution_2": "[hide=\"c.\"] I don't think there are any. We have that $ \\frac{100000}{8}=12500$, so any number which reverses on multiplying by $ 8$ would need to be in the bounds $ 10000\\lex\\le12500$. But any multiple of $ 8$ must be even, and that cannot be accomplished with a $ 1$ as the last digit. So there are no numbers. [/hide]", "Solution_3": "[hide=\"b.\"] By using a similar process to the other problems, we can set bounds on $ 9$. These bounds are $ 10000\\le\\11111$. Additionally, our number must end in $ 9$ (since when it is multiplied by $ 9$ the number will begin with a $ 9$) , and since it is a multiple of $ 9$, the number must sum up to $ 9$. Running through the cases, we find that the only number that works is $ \\boxed{10989}$, since $ 10989\\times{9}=98901$.[/hide]" } { "Tag": [], "Problem": "pLease heLp me..\r\n\r\n\r\nh0w t0 use the trig0n0metric tabLe with aLL the degrees, minutes, and sec0nds..\r\n\r\nt0gether with tan, csc, cos, sin..\r\n\r\n\r\npLease heLp me..\r\n\r\n\r\nurgent pLease..\r\n\r\nhuhuhu", "Solution_1": "...?\r\nA minute is $\\frac{1}{60}$ of a degree, and a second is $\\frac{1}{60}$ of a minute. There are 360 degrees in a circle.\r\nBut I'm not quite clear on what you're asking about.", "Solution_2": "in the t0pic..\r\n\r\n\r\ns0Lving angLes inv0Lving 0bLique triangLes..\r\n\r\n\r\ninstead 0f using a caLcuLat0r..\r\n\r\n\r\ny0u can use the said tabLe..\r\n\r\n\r\nbut h0w can i use the tabLe?..\r\n\r\npLease repLy..", "Solution_3": "And, as you can see, it distracts the rest of us from your actual problem. :D\r\n\r\n[jli edit]: ok...we dont need 4 ppl telling him that its annoying. we only need 3...jk. so i deleted all the other ones." } { "Tag": [ "analytic geometry", "geometry", "graphing lines", "slope" ], "Problem": "If a Cartesian coordinate grid were placed on top of the map of Math City, Algebra Lane would go through the points $ (\\minus{}1,3)$ and $ (1,\\minus{}1)$. Geometry Boulevard runs perpendicular to Algebra Lane and passes through the points $ (2,2)$ and $ (\\minus{}2,y)$. What is the value of $ y$?", "Solution_1": "the slope of the algebra lane is $ \\frac{\\minus{}1\\minus{}3}{1\\minus{}(\\minus{}1)}$, which is $ \\minus{}2$.\r\n\r\nand geometry boulevard's slope is 1/2 (perpendicular to -2)\r\n\r\n$ \\frac{y\\minus{}2}{\\minus{}2\\minus{}2}\\equal{}1/2$\r\n\r\n-4=2y-4.\r\ny=0\r\n\r\nanswer : 0" } { "Tag": [ "probability", "Alcumus", "Support" ], "Problem": "twenty five points are arranged in a 5x5 grid. If two distinct points are chosen at random from th egrid, what is the probability that the segment connecting these two points will not pass through any other points", "Solution_1": "Please post problems in the Alcumus Problem Discussion forum.", "Solution_2": "Are you sure it is a probability problem? Re-read the problem.\r\n (I did this problem already)", "Solution_3": "hello Mathlearner2012. The question says what is the probability....?\r\nSo It is is it not? Anyway can you help me?", "Solution_4": "[quote=\"PowerOfPi\"][b]Please post problems in the Alcumus Problem Discussion forum.[/b][/quote]\r\n\r\nThis forum is for issues with the program, not problems/solutions", "Solution_5": "To tell you the truth, you are partly right. This problem is a \"basic probability as counting(no combinations)\" problem." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "[color=blue]\nEvaluate the following integrals:\n\n1. $ \\int \\frac {1}{\\left(a\\plus{}b\\cos x \\right)^{2}} dx$ where $ a \\ne b$\n\n2. $ \\int \\frac {\\left(\\cos x \\plus{}2\\sin x \\plus{}3\\right)}{\\left(4\\cos x \\plus{}5\\sin x \\plus{}6\\right)^{2}} dx$ \n[/color]", "Solution_1": "$ \\tan \\frac{x}{2}\\equal{}t$", "Solution_2": "[quote=\"kunny\"]$ \\tan \\frac {x}{2} \\equal{} t$[/quote]\r\nit will work but very-very lengthy solution. Try something else or some elegant approach", "Solution_3": "hello, i think the tan half angle substitution works here.\r\nSonnhard.", "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]hello, i think the tan half angle substitution works here.\nSonnhard.[/quote]\r\n[color=blue]\nI do agree with you but i want to say that don't put it at the outset of the problem. First try to rewrite it to reduce the length of solution considerably then you should go for it otherwise what you people are saying is theoretically easy practically not advisable also very complex and lengthy solution. Please write your answers at least if not the solution it self. \n[/color]", "Solution_5": "[url=http://www.mathlinks.ro/viewtopic.php?t=302445]Generalization[/url]", "Solution_6": "A slight correction thanks kunny,\r\n[color=blue]\nEvaluate the following integrals:\n\n1. $ \\int \\frac {1}{\\left(a \\plus{} b\\cos x \\right)^{2}} dx$ where $ |a| \\ne |b|$\n\n2. $ \\int \\frac {\\left(\\cos x \\plus{} 2\\sin x \\plus{} 3\\right)}{\\left(4\\cos x \\plus{} 5\\sin x \\plus{} 6\\right)^{2}} dx$ \n[/color]" } { "Tag": [ "probability" ], "Problem": "I have 5 red plates and 4 blue plates. If I randomly select two plates to serve dinner on, what is the probability that they're both the same color?", "Solution_1": "Case 1: Both are red.\r\n\r\nThere is a 5/9 times 4/8 = 5/18 of this occurring.\r\n\r\nCase 2: Both are blue.\r\n\r\nThere is a 4/9 times 3/8 = 1/6.\r\n\r\nAdding these two fractions together gives 4/9.", "Solution_2": "Alternatively, the probability that the two plates are not the same color is $ 2\\left(\\frac59\\cdot\\frac48\\right)\\equal{}\\frac59$, so the probability that they are is $ 1\\minus{}\\frac59\\equal{}\\boxed{\\frac49}$." } { "Tag": [ "algebra", "polynomial", "quadratics", "AMC" ], "Problem": "ever since I saw this problem on the AMC 12, I have never understood it, so would someone please expain to be the logic behind the problem?\r\n\r\nA polynomial of degree 4 w/ leading coefficient 1 and integer coefficients has 2 real roots, both of which are integers. Which of the following can also be a zero of the polynomial?\r\n\r\na. (1+i :sqrt: 11)/2\r\n\r\nB. (1+i)/2\r\n\r\nC. 0.5 + i\r\n\r\nD. 1 + i/2\r\n\r\nE. (1+i :sqrt: 13)/2", "Solution_1": "[quote=\"keta\"]ever since I saw this problem on the AMC 12, I have never understood it, so would someone please expain to be the logic behind the problem?\n\nA polynomial of degree 4 w/ leading coefficient 1 and integer coefficients has 2 real roots, both of which are integers. Which of the following can also be a zero of the polynomial?\n\na. (1+i :sqrt: 11)/2\n\nB. (1+i)/2\n\nC. 0.5 + i\n\nD. 1 + i/2\n\nE. (1+i :sqrt: 13)/2[/quote]this is what i think:\r\nlet the two integer roots be a and b. the equation is now P(X)=Q(N)(x-a)(x-b) where Q(N) is a quadratic with integer coefficients. so whichever the answer is, (lets say it is m+a) then (m+a)*(m-a) is an integer(follows from quadratic).\r\n\r\nDISCLAIMER: i may be completely wrong", "Solution_2": "yeah...except it's the first one (a) since the product is 3.", "Solution_3": "[quote=\"polymorphic\"]yeah...except it's the first one (a) since the product is 3.[/quote]\r\nWhat?", "Solution_4": "The product of (a) and its complex conjugate is 3.", "Solution_5": "cool, im in!" } { "Tag": [ "ratio", "geometry", "geometry unsolved" ], "Problem": "For every triangle $ ABC$, let $ D,E,F$ be a point located on segment $ BC,CA,AB$, respectively. Let $ P$ be the intersection of $ AD$ and $ EF$. Prove that:\r\n\\[ \\frac{AB}{AF}\\times DC\\plus{}\\frac{AC}{AE}\\times DB\\equal{}\\frac{AD}{AP}\\times BC\\]", "Solution_1": "[quote=\"wangsacl\"]For every triangle $ ABC$, let $ D,E,F$ be a point located on segment $ BC,CA,AB$, respectively. Let $ P$ be the intersection of $ AD$ and $ EF$. Prove that:\n\\[ \\frac {AB}{AF}\\times DC \\plus{} \\frac {AC}{AE}\\times DB \\equal{} \\frac {AD}{AP}\\times BC\\]\n[/quote]\r\nIt is nice but easy!.\r\nIn the case $ AB2) rectangular region which consists of n unit squares.\r\nWe start by putting three coins on the last three squares on the right.\r\nEach player then takes turns to move any one of the coins to its left hand side.\r\nThe coins cannot overlap.\r\nThe first one who cannot move a coin loses.\r\n\r\n[b]Remark[/b]:\r\n [ ] [ ] [ ] [ ] [ ] [ ] [ ] [C] [C] [C] -> [ ] [C] [ ] [ ] [ ] [ ] [ ] [C] [ ] [C]\r\nThis is a legal move.\r\n\r\n[b]For example[/b]:\r\nWhen n = 4: [ ] [C] [C] [C]\r\nIf the first player move the right most coin to the left, i.e. [C] [C] [C] [ ], then the first player win.\r\n\r\nWhen n = 5: [ ] [ ] [C] [C] [C]\r\nIf the first player move the left most coin to the left, either [ ] [C] [ ] [C] [C] or [C] [ ] [ ] [C] [C].\r\nThen no matter how the second player move any one of the two coins on the right. The first player then move the remaining coin and win the game.\r\n\r\n[b]Question[/b]: Does any one of the players has a winning strategy for all possible values of n? If so, what is it?\r\n\r\nPersonally I agree that one of the players has a winning strategy, by the Zermelo's Theorem, this game cannot end in a draw, result follows.\r\nBut then I cannot think of a winning strategy for all n. Any comment ?\r\n\r\n=====\r\n\r\nGAME 2)\r\n[b]Rule[/b]:\r\nThis is the mis\u00e8re tic-tac-toe, i.e. the same as ordinary tic-tac-toe but the first player who completes a row, column, or diagonal with that player's marks loses.\r\n\r\n[b]Question[/b]: Want to show each of the two players has a non-losing strategy.\r\n\r\nA non-losing strategy for the first player is simple.\r\nJust put the mark at the centre first.\r\nThen no matter how the second player put his or her mark, the first player just need to put the mark opposite to that of the second player.\r\n\r\nHowever, I cannot think of a non-losing strategy for the second player, or equivalently, to show the first player has no winning strategy.\r\n\r\n=====\r\n\r\nGAME 3)\r\n[b]Rule[/b]:\r\nThis is the mis\u00e8re SIM, i.e. the same as ordinary SIM but the first player who completes a triangle with that player's color wins.\r\n\r\n[b]Question[/b]: Want to show one of the two players has a winning strategy and find the winning strategy.\r\n\r\nI am not sure if my argument works.\r\nAs in the proof that the ordinary SIM cannot end in a draw, so the mis\u00e8re SIM cannot end in a draw too.\r\nBy the Zermelo's Theorem, one of the two players has a winning strategy.\r\n\r\nNow, as in the ordinary SIM, the second player has a winning strategy, that is to prevent himself or herself to complete a triangle earlier than the first player.\r\nFor the mis\u00e8re SIM, the first player can then adopt the same idea to prevent the second player to complete a triangle earlier than himself or herself.", "Solution_1": "For Game Two, the second player can first pick a corner as close as possible to player one's first move (assume it's the upper left), and then for his second move there will be an empty spot a knight's move away from his first spot (assume it's the middle of the right side). Then he tries to make one of the marked moves (marked with an M):\r\n\r\n[code](O) (M) ( )\n(*) ( ) (O)\n(M) (M) ( )[/code]\r\n\r\nIf he can get one of those squares, he is guaranteed to have a nonlosing fourth move. If the first player has already taken the three marked spots, then the player moves in the *'d square, guaranteeing a win :)\r\n\r\nBrute force FTW", "Solution_2": "[quote=\"zeb\"]For Game Two, the second player can first pick a corner as close as possible to player one's first move (assume it's the upper left), and then for his second move there will be an empty spot a knight's move away from his first spot (assume it's the middle of the right side). Then he tries to make one of the marked moves (marked with an M):\n\n[code](O) (M) ( )\n(*) ( ) (O)\n(M) (M) ( )[/code]\n\nIf he can get one of those squares, he is guaranteed to have a nonlosing fourth move. If the first player has already taken the three marked spots, then the player moves in the *'d square, guaranteeing a win :)\n\nBrute force FTW[/quote]\r\nOh, really :P what I did is to contruct a game tree, and certainly not complete yet :( \r\nWhat do you think about GAME one? :maybe: I wonder if there are different strategy for different value of n.", "Solution_3": "[quote=\"jerryjerrychan\"]\nOh, really :P what I did is to contruct a game tree, and certainly not complete yet :( \nWhat do you think about GAME one? :maybe: I wonder if there are different strategy for different value of n.[/quote]\r\nBy now I've solved GAME one.\r\nFor even n, the first player move the right most coin to the left most square.\r\nIn the remaining part of the game, the first palyer just need to keep the two remaining coins are next to each other, and the number of empty squares between the left most coin and the middle coin(after moving) is even.\r\nFor odd n, nearly the same, only difference is to move the left most coin to the left most square at the begining instead of moving the right most coin." } { "Tag": [], "Problem": "Given that $ m$ and $ n$ are digits, what is the sum of the values for $ m$ and $ n$ which yield the greatest six-digit number $ 5m5,62n$ that is divisible by 44?", "Solution_1": "Set n as the largest it can be divisble by 4, or 8. Then use the divisibility rule to find that m = 9. 9+8=17", "Solution_2": "What divisibility rule? Explain it, please (because I don't know it, and whoever asked this probably doesn't either).", "Solution_3": "Divisibility rule for 11:\r\n\r\nSay you have number $ abcdefghijkl$\r\n\r\nThen for that number to be divisible by 11, $ a\\minus{}b\\plus{}c\\minus{}d\\plus{}e\\minus{}f\\plus{}g\\minus{}h\\plus{}i\\minus{}j\\plus{}k\\minus{}l\\equal{}\\text{multiple of 11}$\r\n\r\nYou just have to alternately add and subtraact digits. This has to do with the fact that 10 is -1mod11", "Solution_4": "You should probably use the divisibility rule for 4 as well: it seems most straightforward to do it in these steps:\r\n0: To be divisible by 44, of course you need to be divisible by 4 and 11 at the same time.\r\n\r\n1: Divisibility test for Four: Four divides the number abc,def,...,...,xyz iff 4 divides the number yz. This is so because 4 is certain to divide abc,def,...,x00, which is a multiple of 100. Based on this step, n is either 0, 4 or 8.\r\n\r\n2: Divisibility test for Eleven, as described above. That says n-2+6-5+m-5=n+m-6 is divisible by eleven. \r\n\r\n3. Since n+m is at most 18, n+m-6 is at most 12. So n+m-6= 0 or 11. Then n+m= 6 or 17. To make 6, you'd use n=0 m=6 or n=4 m=2, but n=8 doesn't work. To make 17, you can notice that 0 and 4 for n won't work! That leaves you with n=8 m=9. So, that's going to turn out to be the solution, to get the largest.\r\n\r\nThis problem could be a lot trickier if the pair (8,9) weren't already so high :p In a similar problem you could have m=8 n=9 as a solution and m=9 n=0 as a solution... in that case the second solution is larger!\r\n\r\nPS: You probably know the divisibility test for Three: If the sum of all the digits in the number is divisible by three, the original number is also.\r\nThere is also a divisibility test for 9: If the sum of digits is divisible by 9, so is the original number!\r\nPPS: Those are the most common ones... simliar rules are NOT true for all numbers." } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "For a,b,c,d positive real numbers prove that\r\n\r\n(a^4+b^4+c^4+d^4) /4abcd + 4(abcd)^(1/4)/(a+b+c+d) \\geq 2", "Solution_1": "Using the inequalities between means of order 4 and 1, and AM/GM, we have :\r\n(a^4+b^4+c^4+d^4)/4 \\geq ((a+b+c+d)/4)\r\n 4 \r\nand ((a+b+c+d)/4) 3 \\geq (abcd) 3/4 .\r\nThus :\r\n(a^4+b^4+c^4+d^4) /4abcd + 4(abcd)^(1/4)/(a+b+c+d) \\geq ((a+b+c+d)/4)/(abcd) 1/4 + 4(abcd)^(1/4)/(a+b+c+d) \\geq 2 since x+1/x \\geq 2 for each positive x.\r\n\r\nNote that equality occurs iff a=b=c=d.\r\n\r\nPierre." } { "Tag": [ "topology", "real analysis", "real analysis unsolved" ], "Problem": "$X$ is a metric space and $E\\subset X$.\r\n\r\nThe following statements are equivlant:\r\n\r\n(i) $E$ is sequentially compact\r\n(ii) $E$ is limit point compact\r\n(iii) $E$ is compact(here the compact is defined by finite-covering)\r\n(iv) $E$ is totally bounded and complete.\r\n\r\n(ii)=>(i) and (iii)=>(ii) are not hard.\r\n\r\nHow to deduce (iv) from (i) and (iv) => (iii)?\r\n\r\nand why we need *totally* bounded in (iv)? What will be wrong if it is only bounded, not totally?", "Solution_1": "[quote]and why we need *totally* bounded in (iv)? What will be wrong if it is only bounded, not totally?[/quote]\r\nThe problem is that \"only\" bounded isn't a topological property, it's a property of a particular metric.\r\n\r\nTake this example. Let the set be $\\mathbb{R}$ and define the metric by $d(x,y)=\\frac{|x-y|}{1+|x-y|}$. Then we have a complete metric space that is bounded. (All distances are less than 1; the whole thing is contained in a ball of radius 1 centered anywhere you please.) It is quite clearly not a compact set.\r\n\r\nIn fact, the obvious mapping is a homeomorphism between this metric space and $\\mathbb{R}$ with the usual metric. This is another way of saying that \"compact\" is a topological property but \"bounded\" isn't.", "Solution_2": "I have a question regarding all this:\r\n\r\nWhat does sequentially compact really mean? I've found two definitions: on Planetmath it says that a set is sequentially compact if all sequences have a convergent subsequence. However, I knew that it meant that all sequences have a subsequence converging to a point in the set. Which one is correct? (I tend to think the second is correct).\r\n\r\nI think in a metric space both compacity and sequential compacity are equivalent to the fact that the set is closed and bounded $(*)$, and from this we can prove (i)$\\iff$(iii). It's easy to see how (i) implies the completeness, and (iii) implies the total boundedness, so (i)$\\Rightarrow$(iv). However, I'm not sure about equivalence $(*)$.", "Solution_3": "[quote=\"grobber\"]I have a question regarding all this:\n\nWhat does sequentially compact really mean? I've found two definitions: on Planetmath it says that a set is sequentially compact if all sequences have a convergent subsequence. However, I knew that it meant that all sequences have a subsequence converging to a point in the set. Which one is correct? (I tend to think the second is correct).\n[/quote]\n\nYes, it does mean the second one.\n\n[quote=\"grobber\"]\nI think in a metric space both compacity and sequential compacity are equivalent to the fact that the set is closed and bounded $(*)$, and from this we can prove (i)$\\iff$(iii). It's easy to see how (i) implies the completeness, and (iii) implies the total boundedness, so (i)$\\Rightarrow$(iv). However, I'm not sure about equivalence $(*)$.[/quote]\r\n\r\nIn Euclidean spaces, a set is compact if and only if it is both closed and bounded. However, it is not true for metric spaces in general...\r\n\r\nI can proof (i)=>(iii) by assuming Linderlofness :blush: But I think that (i)=>Linderlofness..., more precisely, (i)=>seperable=>Linderlofness... I currently have no idea about the proof...", "Solution_4": "[quote=\"liyi\"][quote=\"grobber\"]\nI think in a metric space both compacity and sequential compacity are equivalent to the fact that the set is closed and bounded $(*)$, and from this we can prove (i)$\\iff$(iii). It's easy to see how (i) implies the completeness, and (iii) implies the total boundedness, so (i)$\\Rightarrow$(iv). However, I'm not sure about equivalence $(*)$.[/quote]\n\nIn Euclidean spaces, a set is compact if and only if it is both closed and bounded. However, it is not true for metric spaces in general...\n[/quote]\r\n\r\nI really need to be enlightened regarding these issues :).\r\n\r\nIs it at least true that if a set is closed and bounded then it's compact (or sequentially compact)? (in a metric space, of course)", "Solution_5": "[quote=\"grobber\"]Is it at least true that if a set is closed and bounded then it's compact (or sequentially compact)? (in a metric space, of course)[/quote]\r\n\r\nNo..As Kerry Merryfield says, \"boundness\" tells you little. Suppose you have a metric $d$ and we define a new metric $d' = \\min (d, 1)$. Compactness (even completeness) is same for $(X,d)$ and $(X,d')$. Consider the metric space $(R,d)$ where $d(x,y)= \\min (|x-y|,1)$. $R$ is closed and bounded but not compact.\r\n\r\nI worked out the proof later. Seperable=>Linderlofness is not difficult, and I here write the outline of the proof of \"seq compact => separable\"\r\n\r\nSuppose $E$ is uncountable. Pick any $x_1$. Define $d_1 = \\sup\\{d(x,x_1) | x\\in E\\}$, choose $x_2$ such that $d(x_1,x_2) > \\frac{d_1}{2}$.\r\n\r\nChosen $x_1,\\dots,x_k$, define $d_k = \\sup\\{\\min\\limits_{i\\leq k} d(x,x_i) | x\\in E\\}$, choose $x_{k+1}$ such that $\\min\\limits_{i\\leq k} d(x_{k+1},x_i) > \\frac{d_k}{2}$.\r\n\r\nWe get a sequence $x_1,x_2,\\dots,x_n$. we need to prove that it is dense in $X$. If not, there exists $y$ which has a neighbour $O(y,\\delta)$ such that $O(y,\\delta) \\cap \\{x_n\\} = \\emptyset$. thus we have $delta < d_n, \\forall n$ and \r\n$d(x_m, x_n) > \\frac{d_{m-1}}{2} > \\frac{\\delta}{2}$\r\nFrom Cauchy's Convergence Principle, we know that $\\{x_n\\}$ doesn't have a convergent subsequence, which contradicts with sequentially compactness.", "Solution_6": "I see. Thanks for clearing that up.\r\n\r\nWe still need (iv)$\\Rightarrow$(iii), right? \r\n\r\nWe can prove it by using (i)$\\Rightarrow$(iii). Take any infinite sequence of $E$. One of the balls of a finite covering of $E$ with bals of radius $1$ (one such covering exists because of the total boundedness) contains infinitely many terms of the sequence. pick one, and then cover this ball with finitely many balls of radius $\\frac 12$. One of these balls contains infinitely many terms of the sequence, so we can pick one again, and cover the ball with smaller balls, and so on. You can see that we have defined an infinite Cauchy sequence, which must converge to a point in $E$ (because of the completeness), so $E$ is sequentially compact (this is (i)).\r\n\r\nSeparability (and thus Lindelofness) is even easier to prove here: cover the set with finitely many balls of radius: first $1$, then $\\frac 12$, then $\\frac 13$,etc. Take a point in each of these balls. We have found a dense and countable subset of $E$.\r\n\r\nI think (i)$\\Rightarrow$(iii) isn't hard to show using Lindelofness, so we're sort of done.\r\n\r\nIs this correct? I'm a topology newbie :)." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Find all possible sets of 4 consecutive integers such that the sum of the cubes of the smallest three is the cube of the 4th.", "Solution_1": "Let :alpha: be the smallest integer.\r\nThen :alpha: :^3: + ( :alpha: +1) :^3: + ( :alpha: +2) :^3: = ( :alpha: +3) :^3:\r\nSimply solve it and we get one real root :alpha: = 3 and two complex roots which should be rejected. :(" } { "Tag": [ "probability", "summer program", "Mathcamp" ], "Problem": "In poker, a straight flush--five consecutive cards of the same suit--beats four of a kind. In each suit, a straight flush can start with an ace, a deuce (a two), or any other card up to a 10 (the ace may rank hight or low), making ten pssibilities in all. Since there are four suits, there are four times ten, or fory, different hands that are straight flushes. There are only thirteen different four-of-a-kind hands. So, if there are thirteen four-of-a-kind hands and forty straight flushes, why does a straight flush beat four of a kind?\r\n\r\n\r\nPosted by MrMcGeek", "Solution_1": "[hide]\n\nFour of a kind is more probable because the fifth card in your hand doesn't matter, while it does for a straight flush.[/hide]", "Solution_2": "you fail to account for fifth card in the four of a kind. you have 4 suits * 13 possible sets of four cards * 48 other cards to complete the hand. so a total of 2496 four-of-a kind.", "Solution_3": "re-read that, you made one mistake, but your idea is right\r\n\r\n\r\nPosted by MrMcGeek", "Solution_4": "Tom Davis about poker.\r\n\r\nhttp://www.geometer.org/mathcircles/poker.pdf", "Solution_5": "[quote=\"MrMcGeek\"]re-read that, you made one mistake, but your idea is right\n\n\nPosted by MrMcGeek[/quote]\r\n\r\nIn particular, there is only one way to choose four cards from a set of four cards.", "Solution_6": "This problem is for those of you who know the poker game called Texas Hold'em (If you don't know the game, don't bother to learn it until you are 18 ).\n\n\n\nProblem If you are holding a pair, what is the probability of flopping a set (three of a kind) or better?\n\n\n\n[hide]Somebody is actually talking about having a Texas Hold'em poker tournament at Mathcamp, he must be watching the travel channel too much. [/hide]" } { "Tag": [ "symmetry", "geometry", "parallelogram", "geometry solved" ], "Problem": "ABCDE is a REGULAR pentagon \r\nSHOW THIS \r\n(AB/AC)+(AC/AB)=3", "Solution_1": "I think $\\frac{AB}{AC}+\\frac{AC}{AB}=\\sqrt{5}$ :?", "Solution_2": "Indeed, what is correct is the following assertion: If ABCDE is a regular pentagon, then $\\frac{AB}{AC}+\\frac{AC}{AB}=\\sqrt{5}$.\r\n\r\nHere is the simplest proof of this: Let the diagonals AC and BD of the pentagon ABCDE intersect at T. By the symmetry properties of the regular pentagon ABCDE, we have AD || BC, and thus Thales yields $\\frac{AT}{AD}=\\frac{CT}{CB}$.\r\n\r\nBut the symmetries of the regular pentagon ABCDE imply DE || CA and EA || DB; in other words, DE || TA and EA || TD. Hence, EATD is a parallelogram, so that AT = DE. Thus, CT = AC - AT = AC - DE. Thus, the equation $\\frac{AT}{AD}=\\frac{CT}{CB}$ becomes $\\frac{DE}{AD}=\\frac{AC-DE}{CB}$. Since ABCDE is a regular pentagon, DE = AB, AD = AC and CB = AB; thus, this transforms into $\\frac{AB}{AC}=\\frac{AC-AB}{AB}$. Equivalently, $\\frac{AB}{AC}=\\frac{AC}{AB}-1$. In other words, $\\frac{AB}{AC}-\\frac{AC}{AB}=-1$. Hence,\r\n\r\n$\\left(\\frac{AB}{AC}+\\frac{AC}{AB}\\right)^2=\\left(\\frac{AB}{AC}-\\frac{AC}{AB}\\right)^2+4\\cdot\\frac{AB}{AC}\\cdot\\frac{AC}{AB}=\\left(-1\\right)^2+4=5$,\r\n\r\nso that $\\frac{AB}{AC}+\\frac{AC}{AB}=\\sqrt{5}$, and the proof is complete.\r\n\r\n Darij", "Solution_3": "I am agree with kunny.\r\nplease chek the problem :!: :!:" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "let a,b,c is real number\r\nfind a,b,c so that (a*b*c) 2 \\geq ( a+b+c) ^3 \r\na 2+b2 +c 2 = (a+b)(b+c)(c+a)\r\n :) :(", "Solution_1": "Find all solutions or what?" } { "Tag": [ "induction", "number theory", "prime numbers", "number theory unsolved" ], "Problem": "we have 2n - 1 natural numbers we have to proove that we can choose n of them , such that their sum is a multiple of n .", "Solution_1": "We can use induction.\r\n\r\n1) Prove the result for all prime numbers n.\r\n2) If n is not prime, say n = pq, then prove it using the induction hypothesis for p, q.", "Solution_2": "Then prove the proble for primes by Chevale theorem" } { "Tag": [ "topology" ], "Problem": "Show that any basic open set about a point on the \"top edge,\" that is, a point of form $ (a, 1)$, where $ a<1$, must intersect the \"bottom edge.\"\r\n\r\n[b]Background:[/b]\r\n\r\n[b]Definition-[/b] The [b]lexicographic square[/b] is the set $ X\\equal{}[0,1] \\times [0,1]$ with the dictionary, or lexicographic, order. That is $ (a, b)<(c, d)$ if and only if either $ ab\\plus{}1$ it mean $ p|a^{p\\plus{}1}\\minus{}b$ or $ p|a^2\\minus{}b$ contradition.", "Solution_2": "See here:\r\nhttp://www.mathlinks.ro/viewtopic.php?p=466215#466215\r\nhttp://www.mathlinks.ro/resources.php?c=1&cid=17&year=2005&p=466215" } { "Tag": [ "function", "calculus", "derivative", "limit", "algebra unsolved", "algebra" ], "Problem": "Given is the function $f(x)=-x+\\sqrt{(x+a)(x+b)}$, where $a$, $b$ are distinct given positive real numbers. Prove that for all real numbers $s\\in (0,1)$ there exist only one positive real number $\\alpha$ such that \\[ f(\\alpha)=\\sqrt [s]{\\frac{a^s+b^s}{2}} . \\]", "Solution_1": "-f(x) is increase. ${f(0)=\\sqrt{ab},f(\\infty}=0.$ Therefore it is true if s<0, and it is wrong for s>0. For example a=1,b=4, f(x)<2, but ${\\sqrt^s{\\frac{a^s+b^s}{2}}}>\\sqrt{ab}$, when s>0.", "Solution_2": "[quote=\"Rust\"]-f(x) is increase. ${f(0)=\\sqrt{ab},f(\\infty})=0.$ Therefore it is true if s<0, and it is wrong for s>0. For example a=1,b=4, f(x)<2, but ${\\sqrt^s{\\frac{a^s+b^s}{2}}}>\\sqrt{ab}$, when s>0.[/quote]I don't agree with you ,Rust.I think ${f(\\infty})=(a+b)/2$but not ${f(\\infty}=0$\r\nfor $f(x)=(a+b)/2-\\frac{((a-b)^2)/4}{\\sqrt{(x+a)(x+b)}}$\r\nso this conclusion is right.", "Solution_3": "We claim that $f(x)$ is a strictly increasing function on the interval $[0,\\infty)$. This can be shown by taking the derivative and doing a little algebra. We also know that $f(0) = \\sqrt{ab}$.\r\n\r\nNow we evaluate $\\lim_{x \\rightarrow \\infty} (-x+\\sqrt{(x+a)(x+b)}) = \\lim_{x \\rightarrow \\infty} \\left( \\frac{(a+b)x+ab}{\\sqrt{(x+a)(x+b)}+x}\\right) = \\frac{a+b}{2}$.\r\n\r\nSince $\\sqrt{ab} < \\sqrt[s]{\\frac{a^s+b^s}{2}} < \\frac{a+b}{2}$ (by power mean, strict because $a,b$ distinct) and $f(x)$ is strictly increasing, there must be exactly one positive real $\\alpha$ such that $f(\\alpha) = \\sqrt[s]{\\frac{a^s+b^s}{2}}$, as desired.\r\n\r\nOf course the more algebraic (and less calculus) way to do it is outlined in the above post, but this is what I came up with.", "Solution_4": "with a little calculate it can be obtained $\\alpha=\\frac{\\sqrt[s]{(\\frac{a^s+b^s}{2})^2}-ab}{(a+b)-2\\sqrt[s]{\\frac{a^s+b^s}{2}}}$\nand obviously this fraction is positive and $\\alpha$ is unique." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that\r\n\\[ k^4+(k+1)^4+(k+2)^4=n^2+(n+1)^2 \\]\r\nhas no integer solutions.", "Solution_1": "Take a look here my friend. I think this will help you\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=56146", "Solution_2": "Thank a lot,friend.", "Solution_3": "Another way by using mod\r\n\r\nTaking mod 9 both side and after some analytic , you found that\r\n\r\nLHS $\\equiv 2,8 (\\mod 9)$ \r\n\r\nRHS $\\equiv 1,4,5,7 (\\mod 9)$\r\n\r\nso obviously no solution here" } { "Tag": [ "logarithms", "calculus", "integration", "limit", "geometric sequence", "real analysis", "real analysis solved" ], "Problem": "series for Ln(x) centered at 1\r\nsum { (-1)^k (x-1)^k } / k! n=1 to infinity\r\n\r\nnow this converges for 0$\\infty$ i am guessing $\\int_0^1{x^n}dx$ bounds the above integral but why ?( i dont have a concrete explanation x^n integral is bigger and goes to zero as n goes to $\\infty$ so the above integral goes to zero as well??)\r\nthanks,", "Solution_6": "nevermind... :)" } { "Tag": [ "geometry" ], "Problem": "[img]http://foto.terpe.lt/inkelti/20081118/apskritimai.JPG[/img]\r\n\r\nProve that AC=DB", "Solution_1": "AD=BC (it's semiperimeter of the triangle CDX where X is the intersection point of the inner tangents) hence result.", "Solution_2": "Why it's semiperimeter? :maybe:", "Solution_3": "Let E,F be tangency points of inner tangents to T (assum E is the higher one). Lep P(ABX)=CD+DX+CX=CD+CX+DE+XE=CB+CF=2CB and analogously P(ABX)=2AD. Is it clear now?", "Solution_4": "Thank you very much, look's pretty easy :) \r\n\r\nI'm really bad at geometry :oops:" } { "Tag": [ "percent", "Support", "function" ], "Problem": "Hello,\r\nonce again I would like to ask quite a serious question.\r\nRather than discussing policies of nations, or whether democracy is the best system, I would like to ask which democratic system has which flaws. Judging from the huge amount of nations represented here, that should be quite interesting. Let me explain :\r\n\r\nIn Belgium we elect a federal government (normally every four years). Parliament seats are divided among the several sections of Belgium, and each section divides them according to the vote results in that section.\r\nHowever the actual government (ministers and prime ministers) come from the coalition : a set of parties whose sum of votes is more than fifty percent.\r\nWe also have king but his purpose is to provide comic relief and sign laws, everyone is offended when the slighest thing like an opinion comes out of him and clever schemes are thought of when he refuses to sign something.\r\n\r\n\r\nThe main reason why I want to compare is that in Belgium we have the Vlaams Belang (Flemish Value/Importance). The party is accused of racism, extreme right, and wants to split Belgium( it however is not the only party with that aim). Couple of years back they were still Flemish Block, but that party got convicted for racism so they immediately formed a new party to escape that conviction. \r\nThe party becomes bigger and bigger, and last election it turned out be one of the biggest (30%). However all other parties have an agreement, the 'cordon sanitaire' to keep them out of the 50% coalition, by never going in a government with them.\r\nMany consider this undemocratic, to keep 30% voters out of politics, and to ignore the largest party. On the other hand, 30% is still not 50%, so the majority did not vote on them. Isn't this just the basic right of a group opposing something, to still be divided into factions among themselves?\r\n\r\n\r\nIn USA, parliament is, as I understand it, not chosen simultaneously with the president? \r\nA thing that is different with the president there is that he in chosen by having the highest number of votes? This creates a situation where two parties always will rule in my eyes, as voting for something like green party immediately means voting for the other end.\r\n\r\nIn France and Finland for instance, when no presidential candidate gets more than fifty percent, a second round is organised. Last time in France, this caused many leftists to vote for right wing Chirac , because he was up against Le pen (far left).\r\n\r\n\r\nSo share your ideas. Have I said something utterly incorrect or something you disagree with? Please explain.", "Solution_1": "Let me just give you a primer on the American system: the American federal government has three elected portions, the Senate and the House of Representatives (which together make up Congress) and the Presidency. (We don't call our legislature a Parliament.) Each of the 435 Representatives is elected by the voters of an individual districts apportioned by population and serve 2 year terms. Each of the 100 Senators is elected from a particular state (2 per state) and serve 6 year terms. The president is elected through a confusing system: on election day, everyone goes and votes. They aren't voting directly for the president, however: they are voting for Electors who have agreed to vote for one of the candidates for president. In all but two small states, this happens on a \"winner-takes-all\" basis, so (for example) in New York, any candidate winning more than 50% of the vote gets all of New York's 31 electoral votes. (Each state has a number of electoral votes equal to its number of Senators and Representatives combined; thus, roughly according to population but with a bonus tilt towards small states.)\r\n\r\nSo, there is your brief primer on American government. There will be a quiz on this material next week ;-)", "Solution_2": "thanks for the explanation\r\nIndeed parliament you don't use that word\r\n\r\nI wouldn't really say I said anything wrong, but on the other hand I hadn't said that much about USA system. \r\nBut I am correct that all the ministres (in USA you call them secretaries?) are from one party, the one the president is from, and thus now republican? Well in belgium that is so not true. \r\n\r\nThat was my point. You can't go a bit more to the left than usual as it is synonymous with going much more to the right. On the other hand you can imagine a system of ministers from three parties has its flaws as well :(", "Solution_3": "Another interpretation of the American system: it is inefficient by design. The writers of the constitution wanted to restrict the government's power, and that was accomplished by splitting that power among several branches and bodies, which would rarely all be controlled by the same faction. When that wasn't enough, the Bill of Rights was added to list some things the government would never be allowed to do.\r\n\r\nThe other key feature of the American system: Nearly all legislative seats are based on districts, winner takes all. This makes it extremely difficult for a small party to win anything, so everything is split between the two major parties. A regional party could break through this system, but there are no regional parties in current politics.", "Solution_4": "I see. So jmerry what exactly do you mean? Could you elaborate a bit about the regional party (maybe i don't understand your words exactly)\r\n\r\nAnd, although perhaps Belgian politics in particular aren't that influential outside our tiny country, is it undemocratic what we do right now? Personally I dont' think so. I keep my opinion that one party should be surprised that the majority opposing, still has some more things to worry about constantly.", "Solution_5": "By \"regional party\" I mean a party that is very strong in one part of the country and has essentially no support elsewhere. A party of this type is small by national standards, but still has influence because it wins seats in its home region. The current American system has minor national parties instead that have small amounts of support everywhere, and never win anything.", "Solution_6": "Something that I thought about deeply as I used to work as an interpreter for official visitors to the United States is that political parties in the United States are much less ideological than in most other countries. For example, the Republican Party and the Democratic Party have EXCHANGED positions on many issues compared to their positions a century ago. One reason the two-party de-facto system in the United States is so stable is that whenever a third party discovers an appealing position on issues not supported by a major party, one major party or the other will change its position to gain support of the voters who have that position. There is a fairly strong third party, called the Independence Party, in my state, but it has very few positions on issues that cannot be adopted by either the Republican Party or Democratic-Farmer-Labor Party of my state. The biggest influence small parties have on United States politics is nudging the major parties into new ideological positions (at least until the next nudge comes along). That makes the whole overall national non-system very responsive to deeply felt popular concerns, with some time lag. \r\n\r\nAFTER EDIT: The edit was only to correct a typing error that looked like a spelling error.", "Solution_7": "In 1995 (I think), President Clinton, (belonging to the Democratic Party) proposed a budget that the House of Representatives (which had a majority from the Republican Party) refused to pass. The impasse over the budget escalated to a temporary shutdown of many of the functions of the federal government. And yet Clinton continued as President for six more years, winning re-election in 1996. All the while the House of Representatives continued to have a Republican majority.\r\n\r\nThis sequence of events would be unimaginable in a parliamentary system. If a prime minister (a head of government) were ever to fail to get the parliament to pass such a basic measure as the yearly budget for the whole government, then his administration would collapse. It would have to collapse, and there would have to be new elections.\r\n\r\nThe system of separate election of legislatures and executives also exists at the state level in the U.S.A. Ask tokenadult if he can explain (at something less then book length) the political career of Jesse Ventura in a way someone from another country could understand. (I wouldn't even attempt such an explanation.)", "Solution_8": "The short answer about Jesse Ventura (a former professional wrestler who had been mayor of a suburban city) becoming a third-party governor of Minnesota for one term goes like this: \r\n\r\nVentura ran on a position saying that it was time to change and have an ordinary member of the public (although he is anything but ordinary) in office rather than a professional politician. Both of his opponents had been in politics for a long time. Many voters (I asked them) voted for him as a protest vote, \"to stir things up,\" as they said. His previous experience as a mayor helped make him a credible candidate, and his campaign director put some GREAT ads on TV near the end of his campaign. His election victory was a stunning surprise. The other big surprise is that voter turnout was up for the election, and the Democratic-Farmer-Labor candidate got many fewer votes than expected. Ventura filled his state cabinet with mavericks from both major parties, and was moderately successful in attaining his state platform in a divided state legislature. \r\n\r\nAFTER EDIT: Again, the edit was only to correct typing errors.", "Solution_9": "I had never heard of Jesse Ventura. I found an interesting quote of him though :\r\n\r\n[quote]\"I believe patriotism comes from the heart. Patriotism is voluntary. It is a feeling of loyalty and allegiance that is the result of knowledge and belief. A patriot shows their patriotism through their actions, by their choice [such as voting, attending community meetings and speaking out when needed]. No law will make a citizen a patriot.\" [/quote]\r\n\r\nI think that is true. I , as a Belgian, find it a bit weird and totalitarian that kids needs to pledge allegiance and stuff.\r\n\r\nIs he unique in his situation?\r\n\r\n\r\nSlightly off topic : Tokenadult, you seem to be regarded by many as a person who knows a lot. Are you professional historian/mathematician / professor or something, or several of them? ;)", "Solution_10": "[quote=\"fredbel6\"]Tokenadult, . . . . Are you professional historian/mathematician / professor or something, or several of them? ;)[/quote]\r\n\r\nMy undergraduate degree was in Chinese language, so my first professional training was in Sinology. I have lived in Chinese-speaking countries for six years of my life (1982-1985 and 1998-2001). I also have a law degree although I am not currently a lawyer in active practice. Anything I know about mathematics is through self-education, recreational reading I do to keep up with my son's interest in math. Like many Americans, I take advantage of lavish public library systems and a free press to keep up my knowledge of the world after graduating from school." } { "Tag": [ "number theory", "greatest common divisor", "modular arithmetic" ], "Problem": "I don't know if they go in this forum, if not\r\nPlease move it to the correct place\r\n\r\n\r\n\r\n1. Prove that if $gcd (a,m) = 1$, then the congruence $ax \\equiv b\\pmod m$ has the solution \r\n$x \\equiv a^{\\phi(m) -1 } b \\pmod m$\r\n\r\n\r\n\r\n\r\n\r\n2. The definition of a primitive root can be extended to composite numbers. Say $w$ is a primitive root modulo $n$ if $\\phi (n)$ is the smallest power of $w$ which is congruent to 1 modulo $n$. \r\n\r\na.) Find any primitive roots of 10.\r\nb.) Show that 12 has no primitive roots\r\n\r\n\r\n\r\n\r\n\r\n3. Prove that any multiple of a perfect number, other than the perfect number itself, is abundant.\r\n\r\nfyi: Abundant numbers are those whose integer factors excluding the number itself, have a sum larger than the value of the number. For example take 20. The integer factors are 1, 2, 4, 5, 10 and 20 if you add all the factors (except 20) you get 22 which is larger than 20 therefore it is abundant.", "Solution_1": "Should I repost this in another place?\r\n\r\nI feel that it may be a little difficult for Intermediate....\r\n\r\nIf anyone could help me out, I'd appreciate it", "Solution_2": "I'll move it up to Pre Olympiad.", "Solution_3": "[hide=\"Solution to 1\"]\n\\[\nax \\equiv b \\pmod m \\iff a^{\\phi (m)} x \\equiv x \\equiv a^{\\phi m - 1} b \\pmod m\n\\]\nby Fermat's little theorem (Or is it Euler's theorem? I can't remember which...).\n[/hide]\n[hide=\"Solution to 3\"]\nLet $n = p_1^{e_1} p_2^{e_2} \\cdots p_n^{e_n} \\cdots$ be our perfect number where $p_1, p_2, \\cdots$ are all the primes and $e_1, e_2,\\cdots$ are infinitely many integer exponents. Because the sum of $n$'s divisors is $n$, then we can say that \n\\[\nn = \\left(\\frac{p_1^{e_1+1}-1}{p_1-1}\\right) \\cdots\n\\]\nSuppose we are multiplying $n$ by a prime; WLOG let it be $p_1$. So we get\n\n\\begin{eqnarray*}\nnp_1&=&p_1^{e_1 + 1} p_2^{e_2} p_3^{e_3} \\cdots\\\\\n &=& \\left(\\frac{p_1^{e_1+2}-p_1}{p_1-1}\\right) \\cdots \\\\\n&<& \\left(\\frac{p_1^{e_1+2}-1}{p_1-1}\\right)\\left(\\frac{p_2^{e_2+1}-1}{p_2-1}\\right) \\cdots\\\\\n&=& \\mbox{the sum of the divisors of }np_1\n\\end{eqnarray*}\nSince $np_1$ is less than the sum of its divisors, it is abundant. This can easily be extended for general $nm$ with $m=p_1^{f_1} p_2^{f_2}\\cdots$ with integers $f_i$ :P\n[/hide]", "Solution_4": "Thanks for the help on those two\r\n\r\nI really appreciate it", "Solution_5": "[quote=\"ankur87\"]2. The definition of a primitive root can be extended to composite numbers. Say $w$ is a primitive root modulo $n$ if $\\phi (n)$ is the smallest power of $w$ which is congruent to 1 modulo $n$. \n\na.) Find any primitive roots of 10.\nb.) Show that 12 has no primitive roots\n[/quote]\r\n$\\phi (10)=\\phi (12)=4$\r\nFor a.) $w^4\\equiv 1 \\mod10$ and $w^3,w^2,w \\neq 1 \\mod10$. Only odds will work, so, $\\pm 1, \\pm 3$ work with $\\pm 1$ being eliminated because $(\\pm 1)^2 \\equiv 1 \\mod10$. So, 3 and 7 work.\r\nFor b.) $w^4\\equiv 1 \\mod12$ and $w^3,w^2,w \\neq 1 \\mod12$. Again only odds will work, so, $\\pm 1, \\pm 3, \\pm 5$ may work. $\\pm 1, \\pm 5$ are eliminated because $(\\pm 1)^2 \\equiv 1 \\mod12$ and $(\\pm 5)^2 \\equiv 1 \\mod12$. $\\pm 3$ is eliminated because $gcd(3,12)=3$ so any power of 3 will differ from 12 by a multiple of 3. So none can work.\r\n\r\nIn general:\r\n$w^{\\phi (n)} \\equiv 1 \\mod n$ tells us that $w^{\\phi (n)}-nk = 1$. If $gcd(w,n)=m$, $m \\mid w^{\\phi (n)} - nk$, therefore $w^{\\phi (n)} \\equiv 1 \\mod n$ can only happen if $m=1$.", "Solution_6": "Thanks for the answer to that. I was going over it and kept on calculating things out...\r\n\r\nAppreciate it." } { "Tag": [ "linear algebra", "matrix", "trigonometry", "inequalities", "vector", "linear algebra unsolved" ], "Problem": "[color=green] [b] Let $A$ be the real matrix $A=\\left(\\begin{array}{cccc} 0&a&b&c\\\\ a&0&d&e\\\\ b&d&0&f\\\\ c&e&f&0\\\\ \\end{array}\\right)\\; .$ where $a^2+b^2+c^2+d^2+e^2+f^2 = 6$. \nConsider that $r_1 \\le r_2 \\le r_3 \\le r_4$ are eigenvalues of $A$. It is known that $r_4= 3.$\n1) It's true that all eigenvalues are integer numbers ?\n2) If $A$ is non-singular, which is the inverse matrix $A^{-1}$ ?\n3) If $f$ $: {\\mathbb R} \\to {\\mathbb R}$ find the matrix $f(A)$ (holonomic calculus). For instance consider cases : \n[/b] [/color]\r\n\\[ \\begin{array}{c} f_1(x) = x^{n} \\; \\; , \\; ( n\\in {\\mathbb N}^{*}) \\; \\; ,\\;\\; f_2 (x) =\\sin{x} \\\\ f_3(x)=\\arctan{x} \\; \\; ,\\; \\; f_4(x)=\\left(1+x^2\\right)^{-1}\\; . \\end{array} \\]", "Solution_1": "Here's the answer to part (1):\r\n\r\n$\\text{Tr}(A^2)=2(a^2+b^2+c^2+d^2+e^2+f^2)=12=r_1^2+r_2^2+r_3^2+r_4^2.$\r\n\r\nSo, with $r_4=3,$ we have\r\n\r\n$r_1+r_2+r_3=-3$\r\n\r\n$r_1^2+r_2^2+r_3^2=3$\r\n\r\nApply Cauchy's inequality to the vectors $(-1,-1,-1)$ and $(r_1,r_2,r_3).$ We get that\r\n\r\n$(-r_1-r_2-r_3)^2\\le3(r_1^2+r_2^2+r_3^2).$\r\n\r\n$3^2\\le3\\cdot3$\r\n\r\nSince this is an equality, we must satisfy the equality condition of Cauchy's inequality. Thus $r_1=r_2=r_3=-1$\r\n\r\nSo yes, all of the eigenvalues must be integers.\r\n\r\nSince none of the eigenvalues are zero, then $A$ is invertible.\r\n\r\nAll that remains is to compute $A^{-1}.$", "Solution_2": "You posted this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=15169]in 2004[/url], and it was solved then.", "Solution_3": "[quote=\"jmerry\"]You posted this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=15169]last year[/url], and it was solved then.[/quote] OOps ! Indeed ! I make some non-important changes in the original posted text. Many thanks,Alex" } { "Tag": [ "function", "symmetry", "trigonometry" ], "Problem": "If $ g(x)\\equal{}$ |$ f(x)$| and $ p$ be the period of $ f$.Find the period of $ g$.", "Solution_1": "The answer isn't uniquely determined by the period of $ f$ alone. If $ f$ has glide symmetry, as in the case of the sine or cosine, the period is cut in half, but it could also be unchanged or be cut even further (by $ n$ for any positive integer $ n$).", "Solution_2": "Also, $ g(x)$ may not have period because $ g(x)$ can be $ f(x)$ as well as $ \\minus{}f(x)$ for every $ x$ .", "Solution_3": "Certainly it's possible for $ f$ to be periodic but for $ g$ to be the constant function; is that what you're saying?" } { "Tag": [ "calculus", "derivative", "search", "number theory", "relatively prime" ], "Problem": "1)Let a,b element of Z+\r\n a+b=12 then calculate (1/a)+(4/b) maximum value.(please don't use derivative)\r\n\r\n2)Show that i) p and q relatively prime iff p+q and p.q relatively prime\r\n ii)p and q relatively prime iff p-q and p.q relatively prime", "Solution_1": "Hi mathmaniack,\n\n\n\nMy attempt for (2) :\n\n[hide]\n\n(i) Let p and q are relatively prime. Now, assume d is a common prime dividing both p+q and pq. Since d is prime and it divides pq, so it either divides p or q. If it divides p (say), the as d | (p+q), so d | q too, and hence p and q aren't coprime.\n\n\n\nConversely, let p+q and pq be coprime. Assume d is a common divisor of p and q. Then \n\nd| (p+q) and d|(pq), and hence p+q and pq aren't coprime.\n\n\n\n(ii) Exactly similar to (i)\n\n[/hide]\n\n****************************************************\n\nSambit", "Solution_2": "To (1): I think you search for the minimum value because for a->0 you see there is no upper bound.", "Solution_3": "If my suggestion with the minimum value is true then here is the most elegant solution I found: 1/a+4/b=((a+b)/a+(4a+4b)/b)/12=(5+b/a+4a/b)/12. With AM-GM you have (5+b/a+4a/b)/12=>(5+2sqrt(4))/12=3/4. Equality holds for b/a=4a/b so for b=8,a=4." } { "Tag": [ "vector", "induction", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $\\mathbb{R}^n$ be the Euclidean space. Prove that for any vectors $a_1, a_2, ... ,a_{n+2}$ there exist $i\\ne j \\text{ such that } (a_i,a_j)\\geq 0$.\r\nTo view all the problems visit [url]http://www.mathlinks.ro/Forum/topic-81099.html[/url]", "Solution_1": "Similar questions have been discussed before. For instance, the same holds for $2n+1$ instead of $n+2$ and $>$ instead of $\\ge$.", "Solution_2": "To be specific, http://www.mathlinks.ro/Forum/viewtopic.php?t=39329 .", "Solution_3": "Provided this is false, I can take a set of such vectors {ai}. Since (a1,aj)<0 any j>1, for all vectors v of the subspace spanned by {aj} j>1, which we denote by V1, we have (a1,v)<0. But if dimV1=n, this would be impossible since a1 is contained in V1 itself. So dimV1<=n-1. But the number of non-zero vectors we have for V1 is n+1, so the induction concludes the problem.", "Solution_4": "I've proved for n=3:\r\nLet be to the contrary that $\\forall i\\ne j \\Rightarrow (a_i,a_j)< 0$. \r\nIf all the vectors are colleniar than there exist 2 vectors that will have same direction wich contradicts to our assumption.\r\nSo there exist 2 non-colleniar vectors a and b. Let the $\\alpha$ be the plane such that vectors a and b belongs to it.\r\nFrom our assumtion all the vectors are not ortogonal i.e $(a_i,a_j)\\ne 0$. \r\nSo the projection to this plane of the rest 3 vectors is non zero vector and there exist a pair of vectors c and d such that they look to the same half-space.\r\nLet $c=: pra_{\\alpha} c + \\gamma \\text{ and } d=: pra_{\\alpha} d + \\delta$ such that $(\\gamma,a)=(\\delta,a)=0$ and $(\\gamma,\\delta )>0$ since they are colleniar.\r\nThen from assumtion we have that $0>(c,d)=(\\gamma,\\delta ) + (pra_{\\alpha} c, pra_{\\alpha} d )>(\\gamma,\\delta )$ so we have that there are 4 vectors in plane such that the angle between any 2 of them is >90 which s contradiction.\r\n\r\nI am a beginner in a linear and euclidean space. I just wanted to ask whether my solution could be transformed to the general case?" } { "Tag": [ "function", "calculus", "integration", "trigonometry", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "Hallo to everybody, I have some question that I can't an answer, if some of you can answer me I'll be very pleased of this\r\n1) Let be f:[a,b]->R a derivable and bounded variation function on [a,b], is it true that f', the derivate of f is bounded on [a,b]?\r\n2)Let be f:[a,b]->[c,d] an invertible and bounded variation function on [a,b], is it true that f^-1:[c,d]->[a,b] have to be a bounded variation function on [c,d]?\r\n3)Let be f:[a,b]->[c,d] an invertible and Riemann integrable function on [a,b], is it true that f^-1:[c,d]->[a,b] have to be an integrable function on [c,d]?\r\nThanks to everybody bye", "Solution_1": "1. False.\r\n\r\nConsider $ f(x) \\equal{} x^{3/2}\\sin\\left(\\frac1x\\right)$ on $ [0,1].$ Note that $ f'(0) \\equal{} 0$ and other than that, $ f(x) \\equal{} \\frac32x^{1/2}\\sin\\left(\\frac1x\\right) \\plus{} x^{ \\minus{} 1/2}\\cos\\left(\\frac1x\\right).$ This derivative is unbounded near zero, but since $ |f'(x)|\\le Cx^{ \\minus{} 1/2}$ is absolutely integrable, we do have that $ f$ is of bounded variation.\r\n\r\n2. True for continuous functions, since an invertible continuous function has to be monotone, the inverse of a monotone function is monotone, and monotone functions on closed bounded intervals are of bounded variation. \r\n\r\nAs to whether you can cause any trouble with discontinuous invertible functions - I haven't worked through all of that just yet.", "Solution_2": "Yes, you can cause trouble with discontinuous functions. Let $ [a,b]\\equal{}[\\minus{}1,1]$ and $ [c,d]\\equal{}[0,1]$. Define $ x_0\\equal{}x_1\\equal{}0, x_n\\equal{}(\\minus{}1)^n(1\\minus{}\\frac1n)$ for $ n\\ge 2$ and $ y_n\\equal{}\\frac1n$ for $ n\\ge 1$. Map $ (x_k,x_{k\\plus{}2}]$ (or $ [x_{k\\plus{}2},x_k)$; just make sure it's open on the $ x_k$ side and closed on the other) to $ [y_{k\\plus{}2},y_{k\\plus{}1})$ linearly for each $ k>0$. For the first intervals $ k\\equal{}0$, do the same but omit the points $ x_0$ and $ y_1$; we've already accounted for the image of $ 0$. For the two remaining points, map $ \\minus{}1$ to $ 0$ and $ 1$ to $ 1$.\r\nThe graph looks a bit like a kid's drawing of a Christmas tree.\r\n\r\nThis has bounded variation because it's increasing on $ (\\minus{}1,0)$ and decreasing on $ (0,1)$. The total variation is $ 3$. The inverse doesn't have bounded variation, because it jumps back and forth between near $ \\minus{}1$ and near $ 1$ infinitely often." } { "Tag": [], "Problem": "Can you find the next two terms in each of the following sequences?\r\n\r\nTHESE ARE PRETTY CHALLENGING\r\n\r\nA)\r\n2, 12, 1112, 3112, 132112, ..., ...\r\n\r\nB) 1, 1, 1, 2, 3, 7, 23, ..., ...\r\n\r\nC) 1, 1, 1, 2, 3, 5, 11, ..., ...\r\n\r\n$ \\text{HINT:}$ patterns B and C are very similar", "Solution_1": "Hint: Say part (A) aloud.", "Solution_2": "(A)2, 12, 1112, 3112, 132112, 1113122112, 311311222112", "Solution_3": "I thought I had C (sum of two previous numbers), but then the 11 ended everytning.", "Solution_4": "C) 1, 1, 1, 2, 3, 5, 11, 26, 81\r\n\r\nEach term is the sum of the previous term and the product of the two terms before it.", "Solution_5": "I just got B:\r\n\r\nB) 1, 1, 1, 2, 3, 7, 23, 167, 3862\r\n\r\nEach term is the sum of the product of the previous 2 terms and the product of the 2 terms before that. \r\n\r\nFor example:\r\n a, b, c, d, e\r\n\r\ne=ab +cd" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "compute detA\r\n$ A\\equal{}\\begin{pmatrix} a_1b_1 & a_1b_2 & a_1b_3 & ... & a_1b_n\\\\ a_1b_2 & a_2b_2 & a_2b_3 &... &a_2b_n\\\\ a_1b_3 & a_2b_3 & a_3b_3 & ... &a_3b_n\\\\ ...\\\\a_1b_n&a_2b_n&a_3b_n&...&a_nb_n\\end{pmatrix}$", "Solution_1": "Let the value of the determinant be $ D$. We make the following observations:\r\n1) If $ a_1b_n \\equal{} 0$ then $ A$ has an all-zero row and so $ D \\equal{} 0$.\r\n2) Rows 1 and 2 are almost multiples of each other. If $ a_1b_2 \\equal{} a_2b_1$ then actually they are linearly dependent.\r\n3) Similarly, rows 2 and 3 are almost multiples of each other, and if $ a_2b_3 \\equal{} a_3b_2$ then actually they are linearly dependent.\r\n4) Et cetera.\r\nSince $ D$ is a polynomial in the $ a_i$ and $ b_i$, we have that actually $ D$ must be divisible by $ a_1b_n$ and by $ a_ib_{i \\plus{} 1} \\minus{} a_{i \\plus{} 1}b_i$ for $ i \\equal{} 1, \\ldots, n \\minus{} 1$. Thus, $ D \\equal{} F \\cdot a_1b_n \\cdot \\prod_{i \\equal{} 1}^{n \\minus{} 1} a_ib_{i \\plus{} 1} \\minus{} a_{i \\plus{} 1}b_i$, where $ F$ is some other polynomial in the $ a_i$ and $ b_i$. Comparing degrees on the left and right, we have that $ 2n \\equal{} 2n \\plus{} \\deg F$ and thus $ F$ is just a constant. Finally, the coefficient of $ \\prod_{i \\equal{} 1}^n a_ib_i$ in $ D$ is $ 1$, while (if I haven't made any mistake) the coefficient of $ \\prod_{i \\equal{} 1}^n a_ib_i$ in $ a_1b_n \\cdot \\prod_{i \\equal{} 1}^{n \\minus{} 1} a_ib_{i \\plus{} 1} \\minus{} a_{i \\plus{} 1}b_i$ is $ (\\minus{}1)^{n \\minus{} 1}$. It follows that $ F \\equal{} (\\minus{}1)^{n \\minus{} 1}$ and so the requested determinant is\r\n\\[ (\\minus{}1)^{n \\minus{} 1} \\cdot a_1b_n \\cdot \\prod_{i \\equal{} 1}^{n \\minus{} 1} a_ib_{i \\plus{} 1} \\minus{} a_{i \\plus{} 1}b_i \\equal{} a_1b_n \\cdot \\prod_{i \\equal{} 1}^{n \\minus{} 1} a_{i \\plus{} 1}b_i \\minus{} a_ib_{i \\plus{} 1}.\\]\r\n\r\n(By the way, how did I actually figure this out? Well, I guessed that it would be \"nice\" and so wrote out the cases $ n \\equal{} 1, 2, 3$ by hand. From here, it wasn't hard to guess a pattern, and then looking at the matrix it's pretty easy to see why the pattern holds.)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "$ \\sum_{k} {2n \\plus{} 1 \\choose 2p \\plus{} 2k \\plus{} 1} {p \\plus{} k \\choose k}$", "Solution_1": "The answer ([i]sans preuve[/i]) is $ 4^{n \\minus{} p} \\binom{2n \\minus{} p}{p}$." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Prove that every Linear Continuum is normal.", "Solution_1": "this is a nice proof:\r\n\r\nhttp://ocw.mit.edu/NR/rdonlyres/Mathematics/18-901Fall-2004/51A1BA10-27FB-4B44-ACD5-225EED9A15BD/0/notes_e.pdf" } { "Tag": [ "logarithms" ], "Problem": "Solve the following equalities.\r\n\r\n(1) $\\log_{10}(2x-1)+\\log_{10}(x-9)=2$\r\n\r\n(2) $2^{x}=3^{x-1}$\r\n\r\n(3) $4^{x+1}-2^{x}-2^{-x}+4^{-x+1}=60$\r\n\r\n(4) $x^{1+\\log_{10}x}=100$", "Solution_1": "[hide=\"1\"]\n$\\log_{10}{(2x-1)(x-9)}= \\log_{10}100$\n$2x^{2}-19x-91 = 0$\n${x}={{-7}\\over{2}}$ or $x=13$\nClearly ${-7}\\over 2$ is incorrect, so $x=13$\n[/hide]\n\n[hide=\"2\"]\n${x}.{\\log 2}={(x-1)}{\\log 3}$\n$x(\\log 2-\\log 3) =-\\log 3$\n$x ={{-\\log 3}\\over{\\log 2-\\log 3}}$\n$x = 2.7095$\n[/hide]\n\n[hide=\"4\"]\nTake $\\log_{10}$ of both sides and we get:\n$(1+\\log_{10}x)(\\log_{10}x) = 2$\n${(log_{10}x)}^{2}+\\log_{10}x = 2$\nLet $k=\\log_{10}k$\n$k^{2}+k-2=0$\n$(k-1)(k+2)=0$\n$k=1,-2$\n$\\log_{10}{x}= 1,-2$\n$x = 10,{1 \\over 100}$\n[/hide]", "Solution_2": "[hide=\"3\"]$4^{x+1}-2^{x}-2^{-x}+4^{-x+1}=60$\n$=4\\cdot 2^{2x}-2^{x}-\\frac{1}{2^{x}}+\\frac{4}{2^{2x}}= 60$\n\nLet $u=2^{x}+\\frac{1}{2^{x}}$, $4u^{2}-8 = 4\\cdot 2^{4x}+\\frac{4}{2^{2x}}$ \n\n$4u^{2}-8-u=60$\n$4u^{2}-u-68=0$\n$(4u-17)(u+4)=0$\n$u>0$ $u=\\frac{17}{4}$\n$2^{x}+2^{-x}=\\frac{17}{4}$\n$4(2^{2x})-17(2^{x})+4=0$\n$(4(2^{x})-1)(2^{x}-4) = 0$\n$2^{x}= \\frac{1}{4}$ OR $2^{x}=4$\n$x=-2$ OR $x=2$\n$x = \\pm2$[/hide]\r\n\r\nThanks for pointing that out, rjt.", "Solution_3": "[hide=\"3\"]$4^{x+1}-2^{x}-2^{-x}+4^{-x+1}=60$\n$4\\cdot 2^{2x}-2^{x}-\\frac{1}{2^{x}}+\\frac{4}{2^{2x}}=60$\n$4(2^{2x}+\\frac{1}{2^{2x}})-(2^{x}+\\frac{1}{2^{x}})=60$\n\nLet $y=2^{x}+\\frac{1}{2^{x}}$.\n\n$4(y^{2}-2)-y=60$\n$4y^{2}-y-68=0$\n$(y+4)(4y-17)=0$\n\nSince $y>0$, we have $y=\\frac{17}{4}$.\n\n$2^{x}+\\frac{1}{2^{x}}=\\frac{17}{4}$\n$4\\cdot 2^{2x}+4=17\\cdot 2^{x}$\n$(4\\cdot 2^{x}-1)(2^{x}-4)=0$\n$2^{x}=\\frac{1}{4},4$\n$x=\\pm 2$[/hide]", "Solution_4": "[hide=\"vishalarul\"]\nLine 2 should be:\n$=4\\cdot 2^{2x}...$, although this doesn't affect your answer as a similar mistake a few lines down cancels it out. Your main mistake is:\n\n$4u^{2}-8-u=60$\nIt's $-u$. Just a typo really :P\n\nYour answer works perfectly with the minus. \n[/hide]", "Solution_5": "1)\r\n[hide]This simplifies out to\n\n$\\log_{10}{2x^{2}-19x+9}=2$\n\nSo this means that\n\n$2x^{2}-19x-91=0$\n\nSo, factoring we get\n\n$(2x+7)(x-13)=0$\n\nSo $x=-\\frac{7}{2}$ or $x=13$\n\nBut x>0, so $x=13$\n\n$13$[/hide]" } { "Tag": [ "greatest common divisor" ], "Problem": "The general formula for the num. people in a group is $\\text {total num of people } = \\text { num in A } + \\text { num in B } + \\text { num in neither } - \\text { num in both }$.\r\n\r\nWould the formula for 3 groups be\r\n\r\n$\\text {total} = A + B + C + \\text { neither } - (A,B) - (B,C) - (C,A) + (A,B,C)$\r\n\r\nwhere $(x,y)$ means in both $x$ and $y$?", "Solution_1": "almost. You have to add (A, B, C) because if you look at someone in all three,\r\n\r\nthey get added three times at the beginning, then subracted 3 times because they are in each set of 2. This means they aren't being counted! so you muct add (A,B,C) to make it complete.", "Solution_2": "What about 4 groups?", "Solution_3": "$A+B+C+D+neither-(A,B)-(A,C)-(A,D)-(B,C)-(B,D)-(C,D)+(A,B,C)+(A,B,D)+(A,C,D)+(B,C,D)-(A,B,C,D)$", "Solution_4": "I just realised something.... This formula is the exact same thing that got moved to the getting started forum but instead of GCF, we're using groups with given intersections.\r\n\r\nhere's the general formula (but this could easily be moved)\r\n\r\nlet $\\Omega()$ be the intersection of the groups in the input\r\n\r\nwith x groups:\r\n\r\n[b]total people $=$ neither $+\\sum_{i=1}^{x} (-1)^{i+1}($ sum of all $\\Omega(\\ldots)$ with $i$ groups in the input $)$ [/b]\r\n\r\nsorry that I had to make up a name for the function...don't know how to do the intersection symbol.\r\n\r\nAnyway, this means taht for 4 groups it would be:\r\n\r\n[b]neither + A + B + C + D - (A,B) -(B,C)-(C,D)-(D,A)-(A,C)-(B,D)+(A,B,C)+(B,C,D)+(C,D,A)+(D,A,B)-(A,B,C,D)[/b]", "Solution_5": "[quote=\"Hamster1800\"]I just realised something.... This formula is the exact same thing that got moved to the getting started forum but instead of GCF, we're using groups with given intersections.\n\nhere's the general formula (but this could easily be moved)\n\nlet $\\Omega()$ be the intersection of the groups in the input\n\nwith x groups:\n\n[b]total people $=$ neither $+\\sum_{i=1}^{x} (-1)^{i+1}($ sum of all $\\Omega(\\ldots)$ with $i$ groups in the input $)$ [/b]\n\n\nsorry that I had to make up a name for the function...don't know how to do the intersection symbol.\n\nAnyway, this means taht for 4 groups it would be:\n\n[b]neither + A + B + C + D - (A,B) -(B,C)-(C,D)-(D,A)-(A,C)-(B,D)+(A,B,C)+(B,C,D)+(C,D,A)+(D,A,B)-(A,B,C,D)[/b][/quote]\r\nYour sum there is kinda funny. It just looks like there are more words than math symbols", "Solution_6": "yes it is more words than math symbols. Is that a problem? The (few) math symbols shorten it a lot. I never said (in this one) that is was mostly math symbols. Do sums have to have only math symbols? I don't think so!\r\n\r\nor we could if you would prefer, define Q as what's in teh sum other than $(-1)^{i+1}$ so we have $\\text{neither } + \\sum_{i=1}^{x} (-1)^{i+1}(Q)$", "Solution_7": "Well, it works, and shortens up the notation. But when your actually adding it up you'll have to expand it and sum it up like the old way, cause you can't just plug that into your calculator using the sum(seq(...)), which is very useful", "Solution_8": "well you can.... :P :P \r\n\r\nbut you need programming with for loops. It would ask you how many groups, how many are in each group, how many are in each intersection of 2, how many....up to how many are in all groups and how many are in no groups.\r\n\r\nI'm not going to write up a whole program but if you feel the need, go ahead.", "Solution_9": "That would be more difficult than actually doing the problem :D", "Solution_10": "[quote=\"nat mc\"]That would be more difficult than actually doing the problem :D[/quote]\r\n\r\nyeah, but you write the program once, and you'll never have to do the problem again", "Solution_11": "Well lets say its more than n times more difficult than doing the problme, if you do it n times. In other words, ITS REALLY HARD", "Solution_12": "No, it couldn't be that. That would mean that it gets harder the more problems you do :D instead of being [i]harder the more subsets there are[/i], which makes a lot more sense.", "Solution_13": "actually it doesn't seem that hard.... (30 element lists, anyone?)\r\n\r\nbasic layout:\r\n\r\nyou have a list:\r\n\r\n{0,0,0,0,0,0,0}\r\n\r\nThis has 7 elements, so the groups are groups 1, 2, 3, 4, 5, 6, 7.\r\n\r\nFirst, It asks you how many are in no groups, and adds that to 0.\r\n\r\nThen it cycles through each time changing the last element until it is greater than 7 (the number of elements) Once that happens:\r\n\r\n{0,0,0,0,0,0,8} -> {0,0,0,0,0,1,1}\r\n\r\ncontinue like this until it becomes:\r\n\r\n{7,7,7,7,7,7,7}\r\n\r\nThen you'll have done it all. The total number of cycles needed is $n^n$ where $n$ is the number of groups. You figure out how to tell whether to add or subtract it. It's very easy but let's see you think about it.", "Solution_14": "Go ahead and make a program then. :P", "Solution_15": "lol, yeah just try it" } { "Tag": [ "geometry", "calculus", "function", "integration", "derivative", "algebra", "polynomial" ], "Problem": "I have a few questions about finding the area under very, very basic curves. I have yet to take calculus but my teacher has taught me enough to solve most calculus questions that appear in competitions around here (using the helpful functions the TI calculators have). Please correct any errors I might make while writing what I \"know\", also.\r\n\r\nI know the power rule for finding integrals and derivatives of basic polynomials, so I can find the area under a curve of, say, $\\int_{1}^{2}{x^2}$. I find the integral of $x^2$ which is $\\frac{1}{3}x^3$. Then I plug in 2 for x, solve, and subtract from that the value when x is 1. (again, correct me if that is wrong)\r\n\r\nWhat I don't know how to do is find the area between curves. Probably the most basic I can think of is the area between $f(x)$ and $g(x)$ where\r\n$f(x) = x^2$\r\n$g(x) = -x^2 + 8$\r\nThe intersection points will be $(-2, 4)$ and $(2,4)$, that doesn't require calculus to figure out. I don't know how to find the area between them thouogh... $\\int_{-2}^{2}{f(g(x))}$? Perhaps $\\int_{-2}^{2}{g(x)} - \\int_{-2}^{2}{f(x)}$. Actually I think that last one makes sense. It would be the area under the whole thing, minus the area under the smaller one. But does that always work? Hm.", "Solution_1": "Yes, you're right. $\\int_a^b g(x) dx-\\int_a^b f(x) dx$\r\nJust a technicality, remember to put $dx$ or whatever variable you are integrating with respect to at the end of the integral. This is because each strip under the curve has a width $dx$, a height $f(x)$, and an area $dA$, so $dA=f(x)dx\\Rightarrow A=\\int_a^bf(x)dx$.\r\nHere's an exercise, prove the power rule.", "Solution_2": "Okay, thanks. So for any two functions $f(x), g(x)$ the area between them can be \r\n$| \\int_a^b{f(x)dx} - \\int_a^b{g(x)dx} |$ where $a$ and $b$ are the x-values of the intersection points. Well what if they don't intersect? Then there are no values for $a$ and $b$. I assume the area would be either $0$ or $\\infty$\r\n\r\nProve the power rule? I don't even know what the power rule IS, I just know that's what it's called when you manipulate coefficients and exponents to integrate/derivate polynomials. I don't know what TO prove. It's like saying \"prove $log_2{8}$\". Okay, well I know that's equal to 3. Maybe I should look up what the power rule actually is. Then I'd have a better chance of understanding the chain rule...", "Solution_3": "In order to find the area between curves in a general sense, you should graph the curves and use the geometric principles that led you to $\\int_a^bg(x)dx-\\int_a^bf(x)dx$.\r\n\r\nYou said in your first post that you know the power rule for finding derivitives and integrals, if you know what a derivitive represents, you should be able to prove (basically derive) it.\r\n\r\nHint:\r\n[hide]$f'(x)=\\lim_{h\\to 0} \\frac{f(x+h)-f(x)}{h}$.[/hide]\r\n\r\nFor 2-dimentional calculus, the chain rule is $\\frac{dy}{dx}=\\frac{du}{dx} \\cdot \\frac{dy}{du}$, it basically allows you to use an intermediate function $u(x)$ in order to take the derivative.\r\n\r\nHere's an example: $y=\\sqrt{x^3+2}$\r\n$u(x)=x^3+2$\r\n$y=\\sqrt{u}$.\r\n$\\frac{du}{dx}=3x^2$.\r\n$\\frac{dy}{du}=\\frac{1}{2\\sqrt{u}}=\\frac{1}{2\\sqrt{x^2+2}}$\r\n$\\frac{dy}{dx}=\\frac{du}{dx} \\cdot \\frac{dy}{du}=\\frac{3x^2}{2\\sqrt{x^3+2}}$", "Solution_4": "When I said I knew the power rule, I meant I knew that the power rule is what you're using when finding the derivative of a simple polynomial. I never saw that limit function you showed. See, the academic team math tests our state makes goes into pretty deep calculus for the last, oh, 1/10 of the tests, and they brought my score from a great score to a decent one because the seniors had all taken calculus and I'm just a young'un who hasn't. So our coach (also the calculus teacher) taught me how to use all these special functions the TI-83s have to solve these questions. The premise is to plug in the equation, the calculator finds the derivative or integral based on the x you give it, then I plug in all the multiple choice answers there are with the x I used, and the one with the same answer is the right one. I asked what dx and dy etc meant, and was told to just forget about them.\r\n\r\nSince I don't know much about dy or dx, my understanding of the chain rule suffers. Your example has made me understand it the most. I see that, in your example, $\\frac{du}{dx}$ is the derivative of $du$. Those are terms I can understand. But what are you doing with $\\frac{dy}{du}$? I see a correlation with the $y = \\sqrt{u}$ that you put earlier, but I can't draw any conclusion from just that problem.", "Solution_5": "$\\frac{dy}{du}$ is the derivative of $y$ with respect to $u$ as opposed to $x$.\r\nThe idea behind the limit I showed you is this, take two points on a function $a$ and $b$ let the cooridinates of $a$ be $(x,f(x))$, and let the coordinates of $b$ be $(x+h, f(x+h))$. Now, the slope of the line connecting these two points is referred to as the secant slope and is $\\frac{f(x+h)-f(x)}{x-(x+h)}=\\frac{f(x+h)-f(x)}{h}$. So what if we wanted to find the slope of the line tangent to the curve at a certain point? (Also know as the instantaneous rate of change of the function), then $a$ and $b$ would have to become one point, meaning $h\\to 0$, therefore $f'(x)=\\lim_{h\\to 0}\\frac{f(x+h)-f(x)}{h}$, and this is what a derivative is (the slope of the tangent line at a given point).\r\nFor the power rule, plug in $f(x)=x^n$ and use the binomial expansion, you will see that all terms cancel except $nx^{n-1}$.\r\n\r\nHere is another example that might help you with the chain rule.\r\n$y=(3x^5+16)^{38}$ (I doubt that you would want to expand that).\r\n$u=3x^5+16$.\r\n$y=u^{38}$.\r\n$\\frac{du}{dx}=15x^4$\r\n$\\frac{dy}{du}=38u^{37}$\r\n$\\frac{dy}{dx}=\\frac{du}{dx} \\cdot \\frac{dy}{du} = 15x^4(38u^{37})=15x^4(38(3x^5+16)^{37})=570x^4(3x^5+16)^{37}$\r\n\r\nHere's a simple exercise to help you understand the concept of a derivative a little better.\r\nGraph the function $f(x)=x^2$, with a ruler mark tangent lines at unit intervals of $x$ ($x=1,2,3,4,5,...$).\r\nIf you look at the slopes of all of these tangent lines, you will see that they are $2x$.", "Solution_6": "wow, thanks, the chain rule became a lot clearer with your first line. The same with the purpose of the \"d\"s. That last example was easier than the first, mainly because I don't know how to find the derivative of $\\sqrt x$. If I did, I might have made the connection that $\\frac{dy}{du}$ is the derivative of dy with respect to u. Now, what happens if there are two terms that I can take the derivative of together? For example\r\n$y=sin x cos x$\r\nI know the derivative of sin x is cos x, and of cos x is -sin x (correct me there if I'm wrong), but how do I derivate when they're multiplied?\r\nNow, for this particular equation, I could simplify to one term:\r\n$y=\\frac 1 2 (sin(2x))$\r\n$y'=cos(2x)$\r\nI've picked up on multiplying the 2 out from inside a trig function when derivating, although I'm not positive that is correct.\r\nSo, I solved that by simplifying and thus was a bad example, but how would I solve that if I didn't know I could simplify that? Since the power rule is no use here, how could I use the chain rule (or some other rule I don't know)", "Solution_7": "Your answer is correct, but there is a product rule and a quotient rule for derivatives.\r\nIf $y=u(x)v(x)$, then $y'=u'(x)v(x)+v'(x)u(x)$.\r\nIf $y=\\frac{u(x)}{v(x)}$, then $y'=\\frac{u'(x)v(x)-v'(x)u(x)}{v(x)^2}$.\r\nHere's your example:\r\n$y=\\sin(x)\\cos(x)$ $u(x)=\\sin(x)$ and $v(x)=\\cos(x)$\r\n$y'(x)=cos(x)^2-\\sin(x)^2=\\cos(2x)$.\r\n\r\nFor the derivative of $\\sqrt{x}$, just use the power rule with $n=\\frac{1}{2}$.", "Solution_8": "You know... I get it now. thanks so much. Do you know where I can find lists of derivatives/integrals of trig functions or logs etc? That would be helpful. Then I'll find some problems to do, or make some up and have the calculator check.\r\nAgain, thanks.", "Solution_9": "Try this site http://www.math2.org/." } { "Tag": [ "ratio" ], "Problem": "(this is a classic problem, modified to be easier...)\r\n\r\nThree men and a monkey are shipwrecked on an island. On the first day, they all gather mangoes from the mango trees on the island. That evening, they put their mangoes in a pile and go to sleep. \r\n\r\nDuring the night, the first man wakes up. Since he is worried that someone will steal his share, he splits the mangoes up into three equal piles with one left over, which he gives to the monkey. He hides one pile and puts the other two piles back together. \r\n\r\nLater that night, the second man wakes up, splits the remaining mangoes into three equal piles with one left over, and gives the leftover mango to the monkey. The second man hides his pile and puts the remainder back together. \r\n\r\nLater that night, the third man wakes up and does the same thing; again, there is one mango left over which is given to the monkey. \r\n\r\nThe next morning, the men all wake up, split the remaining mangoes into three piles with one left over, and give the leftover mango to the monkey. \r\n\r\nWhat is the minimum number of mangoes that were initially in the pile?", "Solution_1": "Isn't the original problem with five men and six iterations? It isn't really significantly harder since you'll still do exactly the same process. (At least, that was how I solved it.) The answer suggests something very elegant to me, but I never figured out a neat combinatorial explanation.", "Solution_2": "[hide]Say we start off with a mangoes. Next round we have 2(a-1)/3 mangoes. Let's call this b. After second time, we have 2(b-1)/3 mangoes, let's call this c. After round 3 we get 2(c-1)/3, let's call this d, then after round four, each of the men will get (d-1)/3 mangoes, which we'll call e. Substituting back in, we get c=(4(a-1)-6)/9. Substituting that in, we get d=(8(a-1)-30)/27. After our final round, we substitute again to get (8(a-1)-57)/81=e. Now we just have a good ole fashioned Diophantine. Solving for a we get (81e+65)/8. We have to find the smallest integer value of e for which 8|(81e+65). Just to overcomplicate things, let's use mods. This statement is equivelent to 81e+65 :equiv: 0 (mod 8), which leads to e+1 :equiv: 0 (mod 8). Clearly e=7. Plugging that into the original equation, we get 79 mongoes. Yay. [/hide]", "Solution_3": "Does anyone know if there's a good reason for why it's 79=[hide]34-3+1[/hide]?", "Solution_4": "ComplexZeta wrote:Does anyone know if there's a good reason for why it's 79=[hide]34-3+1[/hide]?\n\n\n\nThis would be true of the more complicated cases, too: that is, for five men and a monkey, there are [hide]5^4 - 5 + 1[/hide], for seven men and a monkey, there are [hide]7^4 - 7 + 1[/hide], etc.\n\n\n\nThis is explained in Gardner's Colossal Book of Mathematics. From what I understand, that's because -2 is a special number when you have three men, -3 is a special number when you have four men, etc.\n\n\n\nSuppose it were possible to have negative mangoes. Then if we start with -2 mangoes, when the first man wakes up, there are enough to make 3 piles of -1, with one (positive) mango left over (which is given to the monkey). The man takes one pile, leaving -2. The second and third men repeat this, and the next morning, split -3 more mangoes. So this would work with -2. It should also work with any number of mangoes 3^4 more than -2, i.e. 79, since 3^4 more initial mangoes will not change the modular residues after 1, 2, or 3 \"operations\". (For that matter, it would also work with 3^4 + 79 mangoes.)\n\n\n\nWe can extend this argument to other cases, too. Just take the number of men, subtract one, tack on a negative sign, and add (number of men)^4.", "Solution_5": "I don't think that's right. I think in general it's [hide]nn+1-n+1[/hide]. Maybe I'll work it out at some time if I don't have enough to do. (Too bad that never happens.)", "Solution_6": "ComplexZeta wrote:I don't think that's right. I think in general it's [hide]nn+1-n+1[/hide]. Maybe I'll work it out at some time if I don't have enough to do. (Too bad that never happens.)\n\n\n\nYou're right. I was stuck in 3-mode.", "Solution_7": "[hide]\n\nYou can prove that nn+1-n+1 is the answer by induction.\n\n\n\nClaim: If we start with nn+1-n+1 mangoes, \n\nafter k iterations of the process there will be nn+1-knk-n+1. \n\n\n\nAfter n iterations there will be n(n-1)n-n+1, so each man gets (n-1)n-1 mangoes.\n\n\n\nTo show that this is minimal, say we start out with nn+1-n+1+M mangoes. We see that nn+1 divides M. So we are done.\n\n[/hide]\n\n\n\n--Dan", "Solution_8": "It occurred to me that this problem is a lot like AIMEB #6: \r\n\r\n6. Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio 3:2:1, what is the least possible total for the number of bananas?", "Solution_9": "That was my first reaction too." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "If the followings are given:\r\n\r\n$ 02 + Bt + C\n\nf(t+1) - f(t) = (2t+1)A + B = 2t\n\n\n\nBecause it holds for all t: 2A = 2 and A+B = 0. Then A = 1, B = -1. Given that f(4) = 6, we can find C = -6. Our polynomial is f(t) = t2 - t - 6.[/hide]", "Solution_2": "Quote:1. Circle O has center (0,0) and radius 1, while circle Q has center (4,3) and radius 2. From point P secants are drawn to the circles, intersecting O at A and B, and Q at C and D. If P moves so that (PA)(PB) = (PC)(PD), then determine the intersection of the locus of P with the x axis. \n\n\n\n[hide]\n\nI got x=11/4\n\nFirst, using those power of a point theorems with a secant and a tangent, the tangent to the respective circle squared is equal to (PA)(PB) or (PC)(PD).\n\n\n\nSo um, P = (x,0), and the distance from P to the circle is y and then drawing in tangents and using a right triangle with Circle O,\n\n1+y^2=x^2\n\nand then with a triangle for Circle Q,\n\n2^2+y^2 = (4-x)^2+3^2\n\n\n\nso then -3= x^2-x^2+8x-16-9\n\n22=8x\n\nx=11/4\n\n\n\n[/hide]\n\nMind disclosing where you got these problems from? Because if this was an ARML problem I dont think i would have come close to getting that in the allowed time.", "Solution_3": "Thanks for the solutions. This problem was on the team round of the MAML state meet a few years back. I'd say you could have spent 5-10 minutes on each depending on how good your team is :P", "Solution_4": "For #2, does a,b,c have to be integers?", "Solution_5": "[quote=\"beta\"]For #2, does a,b,c have to be integers?[/quote]\r\n\r\nThe problem doesn't state that they have to be integers, but when you solve, you do get integral solutions.", "Solution_6": "tim, are you going to that tomorrow?", "Solution_7": "[quote=\"cats...\"]tim, are you going to that tomorrow?[/quote]\r\n\r\nYeah, I'm going to the state meet. I'm on the Boston Latin team... are you going?", "Solution_8": "Boston Latin is going to New englands right?\r\n\r\nWas anyone else there today?" } { "Tag": [], "Problem": "cut cut cut cut cut cut cut cut cut cut \r\nif you got lower than a 100 in a latin test(think 90% range)", "Solution_1": "i know!!! cut cut cut bleeeaaaat cut bleeatt\r\nbllleeeeeddd\r\nfirst under 100\r\ncut cut cut cut cut cut cutcut", "Solution_2": "i am not really emo just feeling like being emo for those who are :ninja:", "Solution_3": "I am emo. cut cut cut cut cut cut cut. I'm probably the first emo goat you know! cut cut cut cut cut cut\r\np.s. This is not spam!!!!! :spam: :spam: :spam:", "Solution_4": "How do you get out of the Olympiad section?" } { "Tag": [ "geometry", "\\/closed" ], "Problem": "It occured to me how hard it actually was to find old problems in the archives on MathLinks due to the poor choice of topic subjects. I mean, is it so hard to pick an appropriate topic which isn't that ambigous and which actually resembles something about the problem itself, not only; \"difficult one\", \"really simple\"; \"try this one\".\r\n\r\nThanks for letting me whine.", "Solution_1": "Good idea Mindspa, I will rename some threads in the Geometry forum.\r\n\r\n dg", "Solution_2": "yes I must admit that the subjects fields are not always original :D:D" } { "Tag": [ "calculus", "integration", "geometry", "derivative", "function", "rectangle", "limit" ], "Problem": "How do you prove that a definite integral yields the area under a curve? I never understood...", "Solution_1": "See [url=http://en.wikipedia.org/wiki/Riemann_sum]Riemann sum[/url].", "Solution_2": "But how did they show that definite integration is the same as Riemann sums?", "Solution_3": "(The first time you learn about) definite integration (, it) is defined in terms of Riemann sums.\r\n\r\nI think the question you actually want to ask is: \"why does indefinite integration give an antiderivative?\" In other words, what you seem to be looking for is a proof of the [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus]fundamental theorem of calculus[/url]. Correct me if I'm wrong, though.", "Solution_4": "Yes, but I want a proof of the second part of the Fundamental Theorem of Calculus, not the first part. And how do u use Riemann sums to explain definite integration? Integration should be like anti-differentiation...why does it have any relation to area?\r\n\r\nBasically, I don't see why integrals give you the area under a curve...what's the relation between the two?", "Solution_5": "Proofs are given in the article. What you mean is \"I don't see why antiderivatives give you the area under a curve.\"\r\n\r\nIf I can attempt to give an intuitive explanation: differentiation takes a function $ f(x)$ to another function $ f'(x)$ that describes the rate of change of that function. Integration is the process by which we find how much a function has changed by adding up its rates of change $ f'(x)$ over intervals that become smaller. In terms of the graph of $ f'(x)$, this describes the sum of the areas of rectangles that approximate the area under the curve.", "Solution_6": "Perhaps something you're a bit more familiar with might help. Just for a moment, lets think physics: velocity is a rate, yes? One form of the particular rate is $ v\\equal{}\\frac{d}{t}$. So obviously $ vt\\equal{}d$. Now lets extend this to a graph by allowing $ v\\equal{}f(t)$. So the y-axis is velocity and the x-axis is time. Furthermore, for our purposes assume $ v\\equal{}c$ where $ c$ is some constant greater than zero. Assume whatever this velocity is describing is traveling at 4 meters per second for three seconds.\r\n\r\nLooks like the area under our velocity line from t=0 to t=3. In this case, it just so happened to be a rectangle. Very simple. But if you recall, a reimann sum is just a whole lot of rectangles right? The reason we developed limit summation and then antidifferentiation is because not all curves are so simple. It's easy to general the area of regions bounded by functions of degree two or less, but beyond that is troublesome.\r\n\r\nNow to extend a wee bit further. $ f^{\\prime}(x)$ is the rate of change in $ y\\equal{}f(x)$. Exactly as $ v$ is to $ d$. Using notation similar to kinematics,\r\n\\[ y_b\\equal{}y_a\\plus{}y^{\\prime}\\Delta x\\] Let $ \\lim_{\\Delta x \\to 0} \\Delta x \\equal{}dx$, \\[ y_b\\equal{}y_a\\plus{}y^{\\prime}dx\\] Now apply the fact that limit summation of these tiny dx values over an interval results in a finite value; Using the definition of a reimann sum we know this is a definite integral from a to b: \\[ y_b\\minus{}y_a\\equal{}\\int^b_a y^{\\prime}dx\\].", "Solution_7": "I think I kind of get it now.\r\n\r\nThis isn't exactly rigorous, but we know that $ \\frac {dy}{dx} \\equal{} f'(x)$, so $ dy \\equal{} f'(x) \\cdot dx$. The RHS is kind of like the formula for a rectangle, with a very very very small width, and height of $ f'(x)$. Therefore, if we add up all the $ dy$'s from $ a$ to $ b$, we get the sum of the area from $ a$ to $ b$. And the change in height from $ a$ to $ b$ is just simply $ f(b) \\minus{} f(a)$.\r\n\r\nIs that a way of looking at it?\r\n\r\nThanks for everyone's explanation. It was hard for me to visualize the concept, and most people's explanations were helpful, but failed to address why the Riemann sum can just be directly translated to a definite integral.\r\n\r\nBut igiul, your explanation really helped me to understand it. Thanks.", "Solution_8": "Yes sir that is. Keep in mind that though it may be explaining the change in $ y$ the whole reason the area gets involved is because if we graphed $ y$ in $ y$ we couldn't see a thing: that's one dimension. So we introduced a second axis along which the change in y is shown. Area happened.\r\n\r\nI don't know whether you know this or not (and if you don't some people might disaprove (of me)). You won't be able to do much with it yet, but I figured it was pretty associated with the subject at hand.\r\n[hide]Just for kicks, I'm gonna show you something interesting. You now know that the change if $ f$ is very much governed by the behaviour of $ f^{\\prime}$... which is governed by the behaviour of $ f^{\\prime \\prime}$. And so on.\n\nSO lets think differentials for a sec. As you just stated $ dy\\equal{}f^{\\prime}(c)\\Delta x\\Rightarrow f(x)\\approx f(c)\\plus{}f^{\\prime}(c)\\Delta x$. And as you probably already know, $ \\Delta x\\equal{}x\\minus{}c$, so\n\n$ f(x)\\approx f(c)\\plus{}f^{\\prime}(c)(x\\minus{}c)$ and $ f(x)\\minus{}f(c)\\approx \\int^x_c f^{\\prime}(c)dx$\n($ f^{\\prime}(c)$ is a constant.)\nIt's important you realize that this is an integral because we use this system to better approximate the first derivative using a differential of the second etc.\n\nRather, $ f(x)\\equal{}\\sum^{\\infty}_{n\\equal{}0} \\int^n_{\\delta} f^{[n]}(c)dx^n$\n\nI know that was a big leap, but hold on. That represents the nth integral (you integrate the term repetitively n times) of the nth derivative evaluated at c. the little $ \\delta$ represents the initial bounds of integration, b=x, c=a. But what is this doing? It's closing in on the error in the first approximation, $ f(x)\\approx f(c)$. Another way of thinking of it is 'using one differential to approximate the one before it; then you integrate that (indefinitely) the necessary amount of times so that it affects the first approximation $ f(c)$'. If possible (I say this because of the first term where n=0), you're integrating $ n\\minus{}1$ times indefinitely after one definite integral from c to x. It actually has a pattern so that we don't have to ugly up the general term of summation with an integrand.\n\n$ f(x)\\approx f(c)\\plus{}\\frac{f^{\\prime}(c)(x\\minus{}c)}{1}\\plus{}\\frac{f^{\\prime \\prime}(c)(x\\minus{}c)^2}{1\\cdot 2}\\plus{}\\frac{f^{\\prime \\prime \\prime}(c)(x\\minus{}c)^3}{1\\cdot 2\\cdot 3} \\plus{}\\ldots \\plus{}\\frac{f^{[n]}(c)(x\\minus{}c)^n}{n!}$\nAnd so the sum is\n$ f(x)\\equal{}\\sum^{\\infty}_{n\\equal{}0} \\frac{f^{[n]}(c)}{n!}(x\\minus{}c)^n$\nThis is what's known as a taylor series centered at $ c$. They're very useful, but can also be quite problematic due to their covergent or divergent natures. You won't see them for another semester (If my assumption is correct. Sorry if I'm wrong) but they're great fun.\n\nTry approximating the roots of $ x\\minus{}\\cos{x}\\equal{}0$ with a second degree taylor series for cosine.[/hide]", "Solution_9": "[quote=\"igiul\"]Perhaps something you're a bit more familiar with might help. Just for a moment, lets think physics: velocity is a rate, yes? One form of the particular rate is $ v \\equal{} \\frac {d}{t}$. So obviously $ vt \\equal{} d$. Now lets extend this to a graph by allowing $ v \\equal{} f(t)$. So the y-axis is velocity and the x-axis is time. Furthermore, for our purposes assume $ v \\equal{} c$ where $ c$ is some constant greater than zero. Assume whatever this velocity is describing is traveling at 4 meters per second for three seconds.\n\nLooks like the area under our velocity line from t=0 to t=3. In this case, it just so happened to be a rectangle. Very simple. But if you recall, a reimann sum is just a whole lot of rectangles right? The reason we developed limit summation and then antidifferentiation is because not all curves are so simple. It's easy to general the area of regions bounded by functions of degree two or less, but beyond that is troublesome.\n\nNow to extend a wee bit further. $ f^{\\prime}(x)$ is the rate of change in $ y \\equal{} f(x)$. Exactly as $ v$ is to $ d$. Using notation similar to kinematics,\n\\[ y_b \\equal{} y_a \\plus{} y^{\\prime}\\Delta x\n\\]\nLet $ \\lim_{\\Delta x \\to 0} \\Delta x \\equal{} dx$,\n\\[ y_b \\equal{} y_a \\plus{} y^{\\prime}dx\n\\]\nNow apply the fact that limit summation of these tiny dx values over an interval results in a finite value; Using the definition of a reimann sum we know this is a definite integral from a to b:\n\\[ y_b \\minus{} y_a \\equal{} \\int^b_a y^{\\prime}dx\n\\]\n.[/quote]\r\n\r\nWhat do you mean the definition of a Riemann sum?\r\nSorry, but I don't understand how your explanation shows why Riemann sums is equivalent to anti-differentation, which leads to area. It seems that all you did in your algebra is definite what a definite integral is.", "Solution_10": "I basically want a way to visualize why the second part of the Fundamental Theorem of Calculus is true:\r\n\r\n[quote]Let $ f$ be a continuous real-valued function defined on a closed interval $ [a, b]$. Let $ F$ be an antiderivative of $ f$, that is one of the infinitely many functions such that, for all $ x$ in $ [a, b]$,\n\\[ f(x) \\equal{} F'(x).\n\\]\nThen\n\\[ \\int^b_a f(x)\\,dx \\equal{} F(b) \\minus{} F(a).\n\\]\n[/quote]", "Solution_11": "Try looking at the [url=http://en.wikipedia.org/wiki/Euler_method]Euler method[/url]. The intuitive idea is that to find the total change over an interval we add up the simultaneous changes over subdivisions of the interval.", "Solution_12": "Even though I'm not sure if this is what I'm looking for, could you explain to me what the $ f(t, y(t))$ means?\r\n\r\n[url]http://en.wikipedia.org/wiki/Numerical_ordinary_differential_equations#The_Euler_method[/url]\r\n[quote]Starting with the differential equation (1), we replace the derivative $ y'$ by the finite difference approximation\n\\[ y'(t) \\approx \\frac{y(t\\plus{}h) \\minus{} y(t)}{h},\\]\nwhich yields the following formula\n\\[ y(t\\plus{}h) \\approx y(t) \\plus{} hf(t,y(t)).\\][/quote]", "Solution_13": "The assumption of the Euler method is that we are trying to approximate the solution to a differential equation $ \\frac{dy}{dx} \\equal{} f(x, y)$. In the special case that we are trying to understand a differentiable function $ \\frac{dy}{dx} \\equal{} f'(x)$, we merely have $ f(t, y(t)) \\equal{} f'(t)$.", "Solution_14": "I still don't really see how this shows why Riemann sums imply anti-differentiation.\r\n\r\nOn Wolfram, [url]http://mathworld.wolfram.com/IndefiniteIntegral.html[/url], after displaying the definition of the definite integral, it says that you can use indefinite integrals to evaluate it. Then it says that while it is taught early on in elementary calculus, the result is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral.\r\n\r\nHow did they find that deep result?\r\n\r\nHow did they find out that definite integrals can be evaluated in terms of indefinite integrals?", "Solution_15": "Historically (to the best of my knowledge), the fundamental theorem was developed intuitively, driven by physical applications (which most of the examples here do a good job of conveying). As far as a rigorous proof goes, I'll post one if you'd like, but to really understand it takes a fair bit more work than what a typical single variable calculus course would give you, an intuitive notion of it should work fine for now.\r\nI think I had the same issue as you when I first learned the Fundamental Theorem, and that's because I mistook the definition of a definite integral with the fundamental theorem itself. The definition of what it means for a function to be (in any way) integrable is in [b]no way[/b] connected to anti-derivatives, in fact, you can calculate some definite integrals without the fundamental theorem at all! \r\nWhen you ask how anti-derivatives are connected with area, an even more fundamental question to ask first is, what is a rigorous definition of \"area\"? (I think that this is actually a much deeper question than I intend it to be, as I'm not versed in Measure Theory). \r\nFrom what I know, I'd define area through the definite integral itself, so the anti-derivative's connection to area is explicitly expressed in the fundamental theorem.", "Solution_16": "But I don't understand why you can evaluate Riemann sums (definite integrals) using indefinite integrals. Like I know that $ \\displaystyle\\int^b_a f(x)\\,dx$ is DEFINED as the area trapped between $ f(x)$ and the $ x$-axis, but I don't why you can evaluate this expression the same way as you would with a indefinite integral.", "Solution_17": "[hide=\"If you would like...\"]\nLet $ f$ be integrable on $ [a,b]$. Suppose that $ F$ is differentiable on $ [a,b]$ with $ F' \\equal{} f$. Let $ \\epsilon > 0$ be arbitrary. Choose $ \\delta > 0$ so that for every partition $ X$ of $ [a,b]$ with the mesh of $ X$ less than $ \\delta$, we have $ |\\displaystyle\\int^b_a f(x)\\,dx \\minus{} \\displaystyle\\sum_{i \\equal{} 1}^{n}f(t_i)\\Delta_i x| < \\epsilon$ for any choice of sample points $ t_i \\in [x_{i \\minus{} 1},x_i]$. (Where $ \\Delta_i x \\equal{} (x_i \\minus{} x_{i \\minus{} 1})$, i.e. the inequality holds for every Riemann sum of $ f$ on $ X$).\nChoose $ t_i \\in [x_{i \\minus{} 1},x_i]$ as in the Mean Value Theorem so that $ F'(t_i) \\equal{} \\frac {F(x_i) \\minus{} F(x_{i \\minus{} 1})}{x_i \\minus{} x_{i \\minus{} 1}}$, that is $ f(t_i)\\Delta_i x \\equal{} F(x_i) \\minus{} F(x_{i \\minus{} 1})$. Then $ |\\displaystyle\\int^b_a f(x)\\,dx \\minus{} \\displaystyle\\sum_{i \\equal{} 1}^{n}f(t_i)\\Delta_i x| < \\epsilon$, and $ \\displaystyle\\sum_{i \\equal{} 1}^{n}f(t_i)\\Delta_i x \\equal{} \\displaystyle\\sum_{i \\equal{} 1}^{n}(F(x_i) \\minus{} F(x_{i \\minus{} 1}))$ which telescopes to $ F(b) \\minus{} F(a)$.\nSo $ |\\displaystyle\\int^b_a f(x)\\,dx \\minus{} (F(b) \\minus{} F(a))| < \\epsilon$. Since $ \\epsilon$ was arbitrary, $ |\\displaystyle\\int^b_a f(x)\\,dx \\minus{} (F(b) \\minus{} F(a))| \\equal{} 0$.\n\n(not my own work, pulled more or less off my course notes)\n[/hide]\r\n\r\nThat being said, I don't think this gives any better insight (in fact, probably in this case worst insight) than the intuitive arguments mentioned above.", "Solution_18": "Why is it so difficult to Google such results?\r\n\r\nI kind of understand your proof, but where's the motivation for it? What's the motivation between the integral and Riemann sums?", "Solution_19": "[quote=\"123456789\"]How do you prove that a definite integral yields the area under a curve? I never understood...[/quote]\r\n\r\nA Riemann sum is defined as:\r\n\r\n\t \\[ \\lim_{\\Delta x\\to 0}\\sum_{i \\equal{} 1}^{\\infty} f(x_i)\\Delta x\r\n\\]\r\n\r\nIn this sum $ \\displaystyle{f(x_i)}$ is the height of the curve at $ \\displaystyle{x_i}$ , an arbitrary point within a given interval that is $ \\displaystyle{\\Delta x}$ wide. Considering any particular rectangular strip, $ \\displaystyle{f(x_i)\\Delta x}$ , is the area of that strip by definition. Summing the areas of all such strips consequently yields the total area under the curve.\r\n\r\nFrom the perspective of the definite integral, the antiderivative $ \\displaystyle{F(a)}$ represents the area under the curve $ \\displaystyle{f(x)}$ to the left of $ \\displaystyle{a}$ ; likewise, $ \\displaystyle{F(b)}$ represents the area under the curve $ \\displaystyle{f(x)}$ to the left of $ \\displaystyle{b}$ . Hence, if $ \\displaystyle{a\\leq b}$ then $ \\displaystyle{F(b) \\minus{} F(a)}$ represents the area under the curve $ \\displaystyle{f(x)}$ between $ \\displaystyle{a}$ and $ \\displaystyle{b}$ .", "Solution_20": "[quote=\"Dr. No\"]From the perspective of the definite integral, the antiderivative $ \\displaystyle{F(a)}$ represents the area under the curve $ \\displaystyle{f(x)}$ to the left of $ \\displaystyle{a}$[/quote]You write that the antiderivative represents the area, but why? I'm trying to get past just the definition.", "Solution_21": "The motivation behind the proof is that you're picking a sample point so that the area of a strip under $ f$ is equal to the change of $ F$ (the anti-derivative) over the width of the strip (which is why we employ the mean value theorem: an understanding of the mean value theorem is key to understanding our motivation). When you add these up all the strips in the interval it is equal to the change of the anti-derivative over the entire interval", "Solution_22": "Thanks, I see the proof, but what was the original non-rigorous intuition that got the mathematicians to see the link?", "Solution_23": "[quote=\"123456789\"]Thanks, I see the proof, but what was the original non-rigorous intuition that got the mathematicians to see the link?[/quote]\r\n\r\nWhile Newton was developing the calculus he kept tabulated pairs of derivatives and their corresponding anti-derivatives in his notes. Leibniz, while developing the calculus independently of Newton, created the concept of what we know now as the Riemann sum, which he called [i]sum omnium[/i], the sum over all. It is from Leibniz that we got the notation $ \\displaystyle{\\int}$ to represent the integral (the integral symbol is a script \"S\" ). Leibniz was able to show that the Riemann sum of a given function is equal to its anti-derivative.", "Solution_24": "http://www.google.com/search?hl=en&q=history+of+the+integral&btnG=Google+Search", "Solution_25": "http://www.math.wpi.edu/IQP/BVCalcHist/calc1.html\r\n\r\nIn that website, it says that Wallis \"noticed\" the pattern (Power Rule). So there was no reasoning involved? Just got lucky with a pattern?\r\n\r\nAm I asking dumb questions? I feel like none of the answers are what I'm looking for, or am I just not understanding?", "Solution_26": "Although I'm not sure this is necessarily the case, I think you'd be surprised how often noticing a pattern can lead someone on the right track, quoting from Zeitz: \"It is a well-kept secret that much high level mathematical research is the result of low-tech plug and chug methods. The great Carl Gauss [...] painstakingly computed the number of integer solutions to $ x^2 \\plus{} y^2 \\le 90 000$.\"", "Solution_27": "[quote=\"123456789\"]http://www.math.wpi.edu/IQP/BVCalcHist/calc1.html\n\nIn that website, it says that Wallis \"noticed\" the pattern (Power Rule). So there was no reasoning involved? Just got lucky with a pattern?[/quote]\nHave you read the the entire thing? There's rather a lot of reasoning involved in the approximation using Cavalieri's principle.\n\n[quote=\"123456789\"]Am I asking dumb questions? I feel like none of the answers are what I'm looking for, or am I just not understanding?[/quote]\r\nI'm still not sure what you're looking for. We've given proofs and we've given intuitions. Does the concept still not make sense to you? Let me go through the steps.\r\n\r\n- Differentiation takes a function and returns its rate of change.\r\n- Definite integration takes a function and returns the area under it on some interval.\r\n- Indefinite integration takes a rate of change and returns the function it corresponds to.\r\n- The area under a function that represents a rate of change is a sum of rates of change (Riemann sum), which is a definite integral. On the other hand, because it calculates the difference between two values of a function (to which the rate of change corresponds), it is also a difference between two indefinite integrals.\r\n\r\nIt seems like the fourth step is what you're hung up on. Again, do you need more in the way of proof or more in the way of intuition? Do you understand why Riemann sums give definite integrals on one hand and indefinite integrals on the other?" } { "Tag": [ "function", "probability", "trigonometry" ], "Problem": "Consider a function $f(x)$ that has zeroes $4$ and $9$. Given that Patrick randomly selects a number from the set $\\{-10,-9,-8, ... 8, 9, 10\\}$, what is the probability that Patrick chooses a zero of $f(x^{2})$? Express your answer as a common fraction.", "Solution_1": "[quote=\"ragnarok23\"]Consider a function $f(x)$ that has zeroes $4$ and $9$. Given that Patrick randomly selects a number from the set $\\{-10,-9,-8, ... 8, 9, 10\\}$, what is the probability that Patrick chooses a zero of $f(x^{2})$? Express your answer as a common fraction.[/quote]\r\n[hide]Note that $f(x^{2})$ has the same zeros as $f(x)$. So there are 2 choices out of 21 numbers, giving you $\\frac{2}{21}$[/hide]", "Solution_2": "[quote=\"bpms\"]Note that $f(x^{2})$ has the same zeros as $f(x)$.[/quote]\r\n\r\nDo think this over.", "Solution_3": "f(x^2) has zeros -2,2,-3,3\r\n\r\nanswer is 4/21", "Solution_4": "But the $f(x)$ can be factored as $(a-4)(a-9)$. So squaring this is $(a-4)(a-9)(a-4)(a-9)$. Unless I am misinterpreting something...", "Solution_5": "[quote=\"bpms\"]But the $f(x)$ can be factored as $(a-4)(a-9)$. So squaring this is $(a-4)(a-9)(a-4)(a-9)$. Unless I am misinterpreting something...[/quote]\r\n\r\nYou are - you think of $f^{2}(x)$, i.e. $f(x)\\cdot f(x)$, instead of $f(x^{2})=f(x\\cdot x)$", "Solution_6": "[hide]if $f(x)$ can be factored as $(x-4)(x-9)$ then $f(x^{2})$ can be factored as $(x^{2}-4)(x^{2}-9)=(x+2)(x-2)(x+3)(x-3)$ so the above answer is correct[/hide]", "Solution_7": "[quote=\"Farenhajt\"][quote=\"bpms\"]But the $f(x)$ can be factored as $(a-4)(a-9)$. So squaring this is $(a-4)(a-9)(a-4)(a-9)$. Unless I am misinterpreting something...[/quote]\n\nYou are - you think of $f^{2}(x)$, i.e. $f(x)\\cdot f(x)$, instead of $f(x^{2})=f(x\\cdot x)$[/quote]\r\nYou are right.\r\nLet me point this out, $f^{2}(x)$ is actually $f(f(x))$, unlike the trig thing $\\sin^{2}x=(\\sin(x))^{2}$", "Solution_8": "The zeroes, for the $f(x^{2})$ function have been given to you.\r\nThere are four zeroes.\r\n\r\nIn the set of integers from $[-10,10]$\r\nThere are 21 elements.\r\n\r\nFour of these are the zeroes.\r\n\r\nTherefore, the probability is $P = \\frac{4}{21}$", "Solution_9": "[quote=\"bpms\"]\nLet me point this out, $f^{2}(x)$ is actually $f(f(x))$, unlike the trig thing $\\sin^{2}x=(\\sin(x))^{2}$[/quote]\r\n\r\nThat depends on the context - if there's no confusion, $f^{2}(x)$ is used freely for $f(x)\\cdot f(x)$, and if there CAN be a confusion, some use $f^{2}(x)$ for square and $f^{[2]}(x)$ for composition, or $f(x)^{2}$ for square and $f^{2}(x)$ for composition. But whatever notation you use, make sure it's clear what you mean." } { "Tag": [ "search", "combinatorics proposed", "combinatorics" ], "Problem": "The top and bottom edges of a chessboard are identified together, as are the left and right edges, yielding a torus. Find the maximum number of knights which can be placed so that no two attack each other.", "Solution_1": "Posted:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88102[/url]\r\n\r\n(Found by doing a search on \"knights\" and \"torus\".)" } { "Tag": [ "trigonometry", "inequalities", "inequalities solved" ], "Problem": "Prove that\r\n$\\frac{cosA}{1-cosA}+\\frac{cosA}{1-cosA}+\\frac{cosA}{1-cosA}\\geq 3$\r\nSorry if it has been posted before.", "Solution_1": "I think there must be some typo in it. Otherwise you could just take [tex]A=\\frac{\\pi}{2}[/tex] :?", "Solution_2": "Use AM-GM for 3 real numbers and following well-known ineq:\r\n[tex]cosAcosBcosC\\geq\\ (1-cosA)(1-cosB)(1-cosC)[/tex]", "Solution_3": "[quote=\"keira_khtn\"]Use AM-GM for 3 real numbers and following well-known ineq:\n[tex]cosAcosBcosC\\geq\\ (1-cosA)(1-cosB)(1-cosC)[/tex][/quote]\r\n\r\nThe sign is reserver!", "Solution_4": "Sorry for my stupid mistake,Hungkhtn.", "Solution_5": "It is enough to prove that:\r\n$\\sum\\left(\\frac{2\\cos\\alpha-1}{1-\\cos\\alpha}\\right)\\geq 0$\r\nor\r\n$\\sum\\frac{1}{\\sin\\alpha^2}\\geq 12$\r\nThe last ineqaulity is true by Jensen" } { "Tag": [ "calculus", "integration", "pigeonhole principle", "binomial coefficients", "Pascal\\u0027s Triangle", "combinatorics proposed", "combinatorics" ], "Problem": "How many ways are there to choose $ k$ numbers from $ n$ numbers $ 1,2,\\cdots , n$ such that no two consecutive numbers is chosen?", "Solution_1": "[quote=\"Nguyen Van Linh\"]How many ways are there to choose $ k$ numbers from $ n$ numbers $ 1,2,\\cdots , n$ such that no two consecutive numbers is chosen?[/quote]\r\n[color=blue]\nLet the numbers be arranged in a line as $ \\ 1,\\ 2,\\ 3,\\cdots ,\\ n$\nLet there be $ \\ x_1$ and $ \\ x_{k \\plus{} 1}$ numbers left respectively before the first and after the last selected number $ \\ \\implies x_1,\\ x_{k \\plus{} 1}\\ge 0$ and $ \\ y_{i \\plus{} 1}$ numbers be left between $ \\ i^{th}$ and $ \\ (i \\plus{} 1)^{th}$ selected numbers for $ \\ i \\equal{} 1,\\ 2,\\ 3,\\cdots,k \\minus{} 1 \\implies y_{i \\plus{} 1} > 0\\ \\forall \\ i \\equal{} 1,\\ 2,\\ 3,\\cdots,k \\minus{} 1.$\n\n$ \\implies x_1 \\plus{} y_2 \\plus{} y_3 \\plus{} \\cdots \\plus{} y_{k} \\plus{} x_{k \\plus{} 1} \\equal{} n \\minus{} k$\n\nLet $ \\ x_1 \\equal{} y_1 \\minus{} 1$ and $ \\ x_{k \\plus{} 1} \\equal{} y_{k \\plus{} 1} \\minus{} 1\\implies y_1,\\ y_{k \\plus{} 1} > 0$\n\n$ \\implies y_1 \\plus{} y_2 \\plus{} y_3 \\plus{} \\cdots \\plus{} y_{k} \\plus{} y_{k \\plus{} 1} \\equal{} n \\minus{} k \\plus{} 2$ ---(i)\n\nHence required number of ways = number of positive integral solutions of (i)$ \\ \\equal{} \\binom {n \\minus{} k \\plus{} 1}{k}$\n[/color]", "Solution_2": "Denote by $ F_{n.k}$ this number. It is obvious that $ F_{n,0} \\equal{} 1$ (there is exactly one way to choose no element!), $ F_{n,1} \\equal{} n$, and $ F_{n,k} \\equal{} 0$ for $ k > (n \\plus{} 1)/2$ (from pigeonhole principle). Now, looking at $ F_{n \\plus{} 1,k}$, either the $ k$ elements are chosen from among the first $ n$ (therefore in $ F_{n,k}$ ways), or the element $ n \\plus{} 1$ is chosen, so $ n$ is not, hence the remaining $ k \\minus{} 1$ are chosen from among the first $ n \\minus{} 1$ (in $ F_{n \\minus{} 1,k \\minus{} 1}$ ways). This leads to the recurrence relation $ F_{n \\plus{} 1,k} \\equal{} F_{n,k} \\plus{} F_{n \\minus{} 1,k \\minus{} 1}$. But this is exactly the recurrence relation for some binomial coefficients; if one builds the array of its values, one will notice it is in fact Pascal's triangle, read diagonally. Therefore $ F_{n,k} \\equal{} \\genfrac {(} {)} {0pt} {} {n \\minus{} k \\plus{} 1} {k}$ (with the convention this value is zero for $ k>n\\minus{}k\\plus{}1$, i.e. $ k>(n\\plus{}1)/2$).", "Solution_3": "[quote=\"mavropnevma\"]Denote by $ F_{n.k}$ this number. It is obvious that $ F_{n,0} \\equal{} 1$ (there is exactly one way to choose no element!), $ F_{n,1} \\equal{} n$, and $ F_{n,k} \\equal{} 0$ for $ k > (n \\plus{} 1)/2$ (from pigeonhole principle). Now, looking at $ F_{n \\plus{} 1,k}$, either the $ k$ elements are chosen from among the first $ n$ (therefore in $ F_{n,k}$ ways), or the element $ n \\plus{} 1$ is chosen, so $ n$ is not, hence the remaining $ k \\minus{} 1$ are chosen from among the first $ n \\minus{} 1$ (in $ F_{n \\minus{} 1,k \\minus{} 1}$ ways). This leads to the recurrence relation $ F_{n \\plus{} 1,k} \\equal{} F_{n,k} \\plus{} F_{n \\minus{} 1,k \\minus{} 1}$. But this is exactly the recurrence relation for some binomial coefficients; if one builds the array of its values, one will notice it is in fact Pascal's triangle, read diagonally. Therefore $ F_{n,k} \\equal{} \\genfrac {(} {)} {0pt} {} {n \\minus{} k \\plus{} 1} {k}$ (with the convention this value is zero for $ k > n \\minus{} k \\plus{} 1$, i.e. $ k > (n \\plus{} 1)/2$).[/quote]\r\n\r\n[color=blue]\nImpressive solution!! good application of recurrence!\n\nI propose an alternate using bijection principle: \n\nConsider $ \\ k$ identical dominoes and $ \\ n \\minus{} 2k \\plus{} 1$ identical squares. arrange all of them in a line.\n\nNumber of ways = $ \\frac {(n \\minus{} 2k \\plus{} 1 \\plus{} k)!}{k!(n \\minus{} 2k \\plus{} 1)!} \\equal{} \\frac {(n \\minus{} k \\plus{} 1)!}{k!(n \\minus{} 2k \\plus{} 1)!} \\equal{} \\binom {n \\minus{} k \\plus{} 1}{k}$\n\nNow after arranging all in a line, write $ \\ 1,\\ 2,\\ 3,\\cdots,\\ n$ on them starting from left most position.\nNote: Rightmost position will be blank as there are $ \\ n\\plus{}1$ positions and we are writing $ \\ n$ numbers and on a domino we will write two numbers but if there is a domino at the end of the arrangement then its right place will be blank.\nFinally pick left number of every domino and we are done with the problem.\n\n$ \\boxed{\\boxed{\\text{Hence desired answer} \\equal{} \\binom {n \\minus{} k \\plus{} 1}{k}}}$\n[/color]", "Solution_4": "[quote=\"Nguyen Van Linh\"]How many ways are there to choose $ k$ numbers from $ n$ numbers $ 1,2,\\cdots , n$ such that no two consecutive numbers is chosen?[/quote]\r\nLet the numbers be $ x_1,x_2,..,x_k$, wlog $ x_1 < x_2 < \\dots < x_k$ and consider $ (y_1,y_2,..,y_k) \\equal{} (x_1 \\minus{} 1,x_2 \\minus{} 2,...,x_k \\minus{} k)$. Obviously $ 0 \\le y_1 < y_2 < \\dots < y_k \\le n \\minus{} k$. And there is a unique relationship between any $ y_1,..,y_k$ and $ x_1,..,x_k$. So we should find the number of ways to choose $ k$ different numbers from the set $ \\{0,1,2,...,n \\minus{} k\\}$. This set has $ n \\minus{} k \\plus{} 1$ elements so the answer is $ \\binom{n \\minus{} k \\plus{} 1}{k}$.", "Solution_5": "[quote=\"Mathias_DK\"][quote=\"Nguyen Van Linh\"]How many ways are there to choose $ k$ numbers from $ n$ numbers $ 1,2,\\cdots , n$ such that no two consecutive numbers is chosen?[/quote]\nLet the numbers be $ x_1,x_2,..,x_k$, wlog $ x_1 < x_2 < \\dots < x_k$ and consider $ (y_1,y_2,..,y_k) \\equal{} (x_1 \\minus{} 1,x_2 \\minus{} 2,...,x_k \\minus{} k)$. Obviously $ 0 \\le y_1 < y_2 < .. y_k \\le n \\minus{} k$. And there is a unique relationship between any $ y_1,..,y_k$ and $ x_1,..,x_k$. So we should find the number of ways to choose $ k$ different numbers from the set $ \\{0,1,2,...,n \\minus{} k\\}$. This set has $ n \\minus{} k \\plus{} 1$ elements so the answer is $ \\binom{n \\minus{} k \\plus{} 1}{k}$.[/quote]\r\n\r\n[color=blue]\nIt is another application of bijection principle\nExcellent!!\n\nsome typo error please fix it\n[/color]", "Solution_6": "What if the numbers are written on a circle, so $ n$ and $ 1$ are considered consecutive?\r\n\r\n[hide]Use the notations from my previous post. Denote by $ G_{n.k}$ this number. It is obvious that $ G_{n,0} \\equal{} 1$ (there is exactly one way to choose no element!), $ G_{n,1} \\equal{} n$, and $ G_{n,k} \\equal{} 0$ for $ k > n/2$ (from pigeonhole principle). Now, for $ k>0$, looking at $ G_{n,k}$, either the $ k$ elements do not contain $ n$ (therefore $ F_{n\\minus{}1,k}$ ways), or the element $ n$ is chosen, so $ 1$ and $ n\\minus{}1$ are not, hence the remaining $ k\\minus{}1$ are chosen from among the other elements $ 2,3,\\ldots,n\\minus{}2$ (in $ F_{n\\minus{}3,k\\minus{}1}$ ways). This leads to the relation $ G_{n,k} \\equal{} F_{n\\minus{}1,k} \\plus{} F_{n\\minus{}3,k\\minus{}1} \\equal{} \\genfrac {(} {)} {0pt} {} {n\\minus{}k} {k} \\plus{} \\genfrac {(} {)} {0pt} {} {n\\minus{}k\\minus{}1} {k\\minus{}1} \\equal{} \\frac {n} {k} \\genfrac {(} {)} {0pt} {} {n\\minus{}k\\minus{}1} {k\\minus{}1}$.[/hide]" } { "Tag": [ "geometry", "trapezoid", "circumcircle", "perpendicular bisector", "geometry unsolved" ], "Problem": "Let ABCD be a trapezoid such that AB IS parallel to CD and AB+CD=AD. Let P be the point on AD such that AP=AB and PD=CD. Prove that:\r\n\r\na) angleBPC=90\r\nb) Let Q be the midpoint of BC and R the intersection of AD with the circle through B, A and Q. Then B,P,R and C are concyclic.", "Solution_1": "a) is simply angle chasing.\r\nangle BPC\r\n=180 - angle BPA -angle CPD\r\n=(360 - 2 angle BPA - 2 angle CPD)/2\r\n=(360 - angle BPA - angle PBA - angle CPD - angle PCD)/2\r\n=(angle BAP + angle CDP)/2\r\n=180/2\r\n90\r\nNot sure about b) though.", "Solution_2": "The trapezoid angles $\\angle DAB + \\angle CDA = 180^o$ add up to a straight angle. The triangles $\\triangle ABP, \\triangle CDP$ are both isosceles with $AP = AB, DP = DC$. Hence,\r\n\r\n$\\angle APB + \\angle DPC = 90^o - \\frac{\\angle DAB}{2} + 90^o - \\frac{\\angle CDA}\r\n{2} = 180^2 - \\frac{180^o}{2} = 90^o$\r\n\r\n$\\angle BPC = 180^o - (\\angle APB + \\angle DPC) = 90^o$\r\n\r\nThe perpendicular bisector of the segment $BP$ bisects the angle $\\angle PAB \\equiv \\angle RAB \\equiv \\angle DAB$. Since it is perpendicular to the leg $BP$ of the right angle triangle $\\triangle BPC$, it is parallel to the other leg $CP$. In addition, it cuts the leg $BP$ is half, hence it cuts the hypotenuse $BC$ also in half, i.e., at its midpoint $Q$. Since the angles $\\angle BAQ = \\angle RAQ = \\frac{\\angle DAB}{2}$ are equal, the chords $BQ = RQ$ of the circumcircle $(O)$ of the triangle $\\triangle ABQ$ are also equal. Hence, the triangle $\\triangle BRC$ has the median $RQ$ equal to half the opposite side $BC$, which implies that the angle $\\angle BRC = 90^o$ is right. But then the circumcircle $(Q)$ of the right angle triangle $\\triangle BPC$ is also the circumcircle of the right angle triangle $\\triangle BRC$, i.e., the quadrilateral $BPRC$ is cyclic.", "Solution_3": "yetti that is the oficial solution :D any others??? :lol:" } { "Tag": [ "geometry", "parallelogram", "geometry proposed" ], "Problem": "Dear everyone.\r\n\r\nPlease think of the following.\r\n\r\nA circle with center O is internally tangent to two circles inside it, with centers O1 and O2, at points S and T respectively. Suppose the two circles inside intersect at points M, N with N closer to ST.\r\nShow that S, N, T are collinear if and only if SO1/OO1 = OO2/TO2.\r\n\r\nThank you for reading. :)", "Solution_1": "$ O,O_1,S$ and $ O,O_2,T$ are collinear and triangles $ OST,O_1SN,O_2NT$ are isosceles whose legs are the radii of $ (O),(O_1),(O_2),$ respectively. If $ S,N,T$ are collinear, from $ \\triangle O_1SN \\sim \\triangle O_2NT,$ due to $ \\angle OST \\equal{} \\angle OTS$ and $ \\angle O_1NS \\equal{} \\angle O_2NT,$ we obtain $ \\frac {_{NO_1}}{^{TO_2}} \\equal{} \\frac {_{SO_1}}{^{NO_2}}.$ But it's easy to see that $ OO_1NO_2$ is a parallelogram $ \\Longrightarrow$ $ NO_1 \\equal{} OO_2 $ and $NO_2 \\equal{} OO_1$ $ \\Longrightarrow$ $ \\frac {_{OO_2}}{^{TO_2}} \\equal{} \\frac {_{SO_1}}{^{OO_1}}.$\n\nAssume that $ \\frac {_{OO_2}}{^{TO_2}} \\equal{} \\frac {_{SO_1}}{^{OO_1}},$ which implies $ \\frac {_{OO_2 \\plus{} TO_2}}{^{TO_2}} \\equal{} \\frac {_{SO_1 \\plus{} OO_1}}{^{OO_1}}$ \n\n$\\Longrightarrow$ $ TO_2 \\equal{} OO_1 \\equal{} NO_2$ and similarly we'll have $ OO_2 \\equal{} SO_1 \\equal{} NO_1$ $ \\Longrightarrow$ $ OO_1NO_2$ is a parallelogram." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Quite intreseting problem :\r\n\r\nHere by Construction : i mean with just using $RULER$.\r\n\r\n\r\n[color=blue]Problem[/color]A circle ,a point and a line ,are given in the plane.\r\nconstruct a perpendicular line from the point to the line.", "Solution_1": "nice problem Arman today I solve it in the Physic class :D \r\nNice physic class :mad:" } { "Tag": [], "Problem": "How many feet are in 3.5 bronessts, where a bronesst is 4/5 of a third of a half of a mile? :D", "Solution_1": "[hide]4/5*1/3=.26666666 of a mile. 3.5 whatever that is times .266666666...=.9333333...\n*5280=4928 ft.[/hide]", "Solution_2": "close, aznmonkey, there's one part you missed.", "Solution_3": "[quote=\"236factorial\"]How many feet are in 3.5 bronessts, which is 4/5 of a third of a half of a mile? :D[/quote]\r\n\r\n[hide]No brainer. :rotfl: \n\nAnyways, 4/5*1/3*1/2 = 4/30 = 2/15\n\n3 1/2*2/15 = 7/2*2/15 = 14/30 = 7/15 of a mile\n\n5280/15 = 352 ft.\n\n352*7= 2464 ft.[/hide]", "Solution_4": "[quote=\"236factorial\"]How many feet are in 3.5 bronessts, which is 4/5 of a third of a half of a mile? :D[/quote]\r\n[hide]3.5* 4/5(5280*1/3*1/2)=2464 feet?[/hide]", "Solution_5": "correct, aznmonkey1992 just missed the half part :P", "Solution_6": "State stuff clearer next time:\r\n\r\n[quote=\"236Factorial and anirudh\"]Is 4/5 of a third of a half of a mile 1 bronner or 3 broonners. :bruce: [/quote]", "Solution_7": "[quote=\"anirudh\"]State stuff clearer next time:\n\n[quote=\"236Factorial and anirudh\"]Is 4/5 of a third of a half of a mile 1 bronner or 3 broonners. :bruce: [/quote][/quote]\r\n\r\nnow you can solve it since I fixed it ;)", "Solution_8": "[quote=\"236factorial\"]How many feet are in 3.5 bronessts, where a bronesst is 4/5 of a third of a half of a mile? :D[/quote]\r\n\r\n1bronner: 4/5*1/3*1/2*5280=2/15*5280=704\r\n\r\n704*3.5=2464, unless my calculator made a mistake", "Solution_9": "[hide]\n$\\frac{4}{5} * \\frac{1}{3} * \\frac{1}{2} = \\frac{2}{15}$\n$\\frac{2}{15}$ of $5820 = 704$\nThe answer is $704 * 3.5 = \\boxed{2464}$ [/hide]", "Solution_10": "your answer does not show up on my computer", "Solution_11": "[hide][quote=\"236factorial\"]How many feet are in 3.5 bronessts, where a bronesst is 4/5 of a third of a half of a mile? :D[/quote]\n\n$\\frac45\\times\\frac13\\times\\frac12=\\frac{2}{15}$\n\n$\\frac{2}{15}\\cdot3.5= \\frac{7}{15}$\n\n$\\frac{7}{15}\\cdot5280=2464$\n\n2464 feet[/hide]", "Solution_12": "[hide]2464[/hide]", "Solution_13": "[hide]2464 feet[/hide] did you make this up yourself?", "Solution_14": "[quote=\"jhollenbeck\"] did you make this up yourself?[/quote]\r\n\r\nYep, as with most of the other problems I post :)" } { "Tag": [ "trigonometry", "limit", "algebra unsolved", "algebra" ], "Problem": "let sequence $ u_k\\equal{}\\frac{\\cos1\\minus{}\\cos k.\\cos (k\\plus{}1)}{\\cos k.\\cos (k\\plus{}1)}$.Defined $ a_k$,$ b_k$ are the number of postive number and negative number of sequence $ u_1,u_2,...,u_n$.\r\n(i) Prove that $ a_k 0$ . find minimal value of expression : \r\n$\\frac{1}{x^2}+ \\frac{1}{y^2}+ \\frac{1}{z^2} + 8x^2 +108y^2 +16z^2$\r\n where $xy+yz+zx=1$", "Solution_1": "[quote=\"tranthanhnam\"]let $x,y,z >0$ . prove that : \n$\\frac{1}{x^2}+ \\frac{1}{y^2}+ \\frac{1}{z^2} + 8x^2 +108y^2 +16z^2$\n where $xy+yz+zx=1$[/quote]\r\n\r\nWhat exactly to prove ? :huh:", "Solution_2": "Not [b]prove that [/b]but [b]find minimal value of expression [/b]" } { "Tag": [], "Problem": "What are the limitations on value k when\r\nx^4 + 4 x^3 + 4 x^2 + k\r\nhas 4 real roots", "Solution_1": "[hide]For f(x) = x^4 + 4x^3 + 4x^2 + k to have 4 real roots, two turning points must be either above or below the x-axis and one must be below or above the x-axis respectively. \n\nNow, d(x^4 + 4x^3 + 4x^2 + k)/dx = 4x^3 + 12x^2 + 8x. Turning points occur when f'(x) = 0. 4x^3 + 12x^32 + 8x = 0 ===> x(x+1)(x+2) = 0. .'. x = 0, -1, -2. \n\nCase 1:\n\nf(-1) > 0, which means that f(-2) and f(0) < 0\n\nThe condition for this situation is k > -1 and k < 0, so -1 < k < 0. \n\nCase 2: \n\nf(-1) < 0, which means that f(-2) and f(0) > 0\n\nThe condition for this situation is k < -1 and k > 0, which is a paradox. Therefore, the only condition for four real roots is -1 < k < 0.[/hide]", "Solution_2": "[hide=\"Solution\"]This is equivalent to $ ((x\\plus{}1)^2\\minus{}1)^2\\equal{}\\minus{}k$, so $ k$ must be nonpositive. Let $ \\minus{}k\\equal{}m^2$; then each of the four real solutions must satisfy $ (x\\plus{}1)^2\\equal{}1\\plus{}m$ or $ (x\\plus{}1)^2\\equal{}1\\minus{}m$. But $ 1\\plus{}k\\equal{}1\\minus{}m^2\\equal{}(1\\minus{}m)(1\\plus{}m)\\equal{}(x\\plus{}1)^4\\ge0$, so $ k\\ge\\minus{}1$ and our answer is $ \\minus{}1\\le k\\le0$.[/hide]" } { "Tag": [ "floor function", "function", "algebra" ], "Problem": "7. The symbolism $ \\lfloor x\\rfloor$ denotes the largest integer not exceeding x. For example, $ \\lfloor3\\rfloor = 3$, and $ \\lfloor\\frac{9}{2}\\rfloor = 4$. Compute $ \\lfloor\\sqrt{1}\\rfloor + \\lfloor\\sqrt{2}\\rfloor + \\lfloor\\sqrt{3}\\rfloor +\\ldots+\\lfloor\\sqrt{16}\\rfloor$.\r\n\r\n$ (A) 35 \\qquad (B) 38 \\qquad (C) 40 \\qquad (D) 42 \\qquad (E) 136$", "Solution_1": "This one is a lot easier if you know the trick\r\n[hide]The floor function makes this problem nice. We know that everything between $ \\sqrt{1}$ and $ \\sqrt{3}$ is bigger than one but less than two, so it \"floors\" to 1. Similarly, the numbers between $ \\sqrt{4}$ and $ \\sqrt{8}$ simplify to 2. We get 1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+4=3+10+21+4=[b]38, [size=150]B[/size][/b][/hide]" } { "Tag": [ "geometry", "trapezoid" ], "Problem": "The bases of a trapezoid have lengths $10$ and $20$, and the lengths of its other two sides are $6$ and $8$. Find the area of the trapezoid.", "Solution_1": "k you can form a $6-8-10$ triangle clearly..if you draw it you'll see what i mean\r\n\r\nthen the height of the triangle therefore trapezoid is\r\n\r\n$h=\\frac{48}{10}=4.8$\r\n\r\nSo then the Area is\r\n\r\n$A=\\frac{(10+20)(4.8)}{2}=\\boxed{72}$" } { "Tag": [ "Harvard", "college", "MIT" ], "Problem": "I went to mu alpha theta competition and competed in algebra II division. i feel that even if i finished the entire Algebra II book, i can't do all of the problems. What do i need to learn to get a perfect score in MAO?", "Solution_1": "The purpose behind our online school here at AoPS is to help students reach a higher level of problem solving. We specifically offer an Intermediate Algebra class that is great for students who are in or have completed Algebra II.\r\n\r\nAlso, have you seen the Art of Problem Solving books before? I would suggest making sure that you know all of the topics in the first volume which covers most problem solving material tested on the AMC. After that, the second volume will take you even farther.", "Solution_2": "Gee he's done all of those.. He's read the entire book and done research on things that are Pre-cal and so on.. He's asking how is it possible to find some hard Algebra II practice problems and things like that.. He always misses first place by a few points... that's why", "Solution_3": "If you're just missing first place by a couple points in MAO stuff, it's time to start working on harder problems. \r\n\r\nCheck out the Harvard-MIT materials, for example.\r\n\r\nDon't worry about taking that last step in MAO from second to first. Aim at learning more challenging material and that last step will take care of itself.", "Solution_4": "Yeah but he's GOTTA win at teams so I can leech off him so I can make my momma proud.", "Solution_5": "David, stop acting like an immature man....................... :D", "Solution_6": "Hmm...Rui, I feel your pain.\r\n\r\nThe same thing is happening to me right now, except I'm off by like 3 mistakes at every competition.", "Solution_7": "MAO has nothing to do w/ learning hard math. just do a million old tests, they always repeat questions anyway.\r\nit works for me and gaku :lol:", "Solution_8": "yeah MAO reuses material all the time. also if you look at the old test you can see what kinds of things they ask at each point in the year." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that \r\n\r\n \\sum (m_a)(h_a) \\leq s 2 \r\n\r\n\r\nwhere h_a,h_b,h_c and m_a,m_b,m_c are heights and medians of a triangle of semiperimeter s.", "Solution_1": "it seems to me that you are missing something from the statement manlio, please edit your post to make it correct :)", "Solution_2": "Does anyone have a solution for this one? Dzeta gave me this problem and I couldn't solve it except for the case when the triangle is acute. For obtuse triangles I cannot prove it. And it's so nice. It seems that I'm unable to prove any inequality related to medians of a triangle. :(", "Solution_3": "Now solved in http://www.mathlinks.ro/Forum/viewtopic.php?t=21462 post #16. Argh, that was a piece of work...\r\n\r\n Darij" } { "Tag": [ "algorithm", "number theory", "prime numbers", "number theory open" ], "Problem": "I apologize if this is an inappropriate place for this question. Is there a way to accurately test for prime numbers that are so large they exceed the limit of the programming langauge? I am running a PERL program to find primes and I beleive PERL has a 16 digit limit. Is there a way mathematically (or by other means) to work with numbers that are more than 16 digits?", "Solution_1": "You need a package (data types, operations) that can handle large integers.The default types use a fixed amount of memory storage, and thus have an upper limit. Very large integers can be represented with structures that use more memory as the numbers get larger.\r\nThe details of that are a computer science question- I'm sure there are standard packages, and I don't know the details of the structures used.\r\nThe simplest version of this I can think of would do calculations in base $N$, where $N$ is the size of a standard unsigned integer.\r\n\r\nOnce you can handle large integers, you can use standard algorithms to test for primality.", "Solution_2": "Create an array representing the digits of the large number, and code the division algorithm yourself.", "Solution_3": "Thank you both - I got it running." } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO", "induction", "geometry" ], "Problem": "Given $x+y+z=1$ and $x,y,z>0$ Prove\r\n\\[ \\frac{x}{\\sqrt{1-x}}+\\frac{y}{\\sqrt{1-y}}+\\frac{z}{\\sqrt{1-z}} \\geq \\sqrt{\\frac{3}{2}} \\]", "Solution_1": "By Holder, we have\r\n$\\left(\\sum_{\\text{cyc}} \\frac{1}{\\sqrt{1-x}} \\right)^2 \\left( \\sum_{\\text{cyc}} (1-x) \\right) \\geq 3^3 \\Longrightarrow \\sum_{\\text{cyc}} \\frac{1}{\\sqrt{1-x}} \\geq \\frac{3\\sqrt{6}}{2}$\r\nBy Chebyshev,\r\n$\\sum_{\\text{cyc}} \\frac{x}{\\sqrt{1-x}} \\geq \\frac 13 \\sum_{\\text{cyc}} x \\sum_{\\text{cyc}} \\frac{1}{\\sqrt{1-x}}$\r\nCombining these yields the result :) \r\nJensen probably works too...\r\n\r\nMarathon? Ok, here's a prob:\r\n\r\nLet $x_1, x_2, \\dots , x_n$ be positive reals with product equal to $1$. Prove that\r\n\r\n\\[ \\frac{1}{1+2x_1} + \\frac{1}{1+2x_2} + \\cdots + \\frac{1}{1+2x_n} \\geq \\frac n3 \\]", "Solution_2": "I didn't use that highpowered of a result but now time to solve yours :)", "Solution_3": "Heres the solution to the problem\r\n\r\n[hide]\nFor ThAzN1's problem, we first apply cauchy\n\n \\[ \\frac{1}{1+2x_1} + \\frac{1}{1+2x_2} + \\cdots + \\frac{1}{1+2x_n} \\geq \\frac{(1 + 1 + ... + 1)^2}{(1 + 1 + .... + 1) + 2(x_1 + x_2 + ... + x_n)} \\vskip .2in = \\displaystyle\\frac{n^2}{n + 2(x_1 + x_2 + ... + x_n)} \\]\n\nnow by am-gm, \n\n\\[ \\frac{x_1 + x_2 + ... + x_n}{n} \\geq (x_1x_2x_3...x_n)^{\\displaystyle\\frac{1}{n}} = 1 \\Rightarrow x_1 + x_2 + ... + x_n \\geq n \\]\n\nCombining this with the result above, we get\n\n\\[ \\frac{n^2}{n + 2(x_1 + x_2 + ... + x_n)} \\geq \\frac{n^2}{n + 2(n)} = \\frac{n}{3} \\]\n\nand we are done.\n[/hide]\r\n\r\nFor the next problem... how about\r\n\r\nLet a,b,c be real numbers greater than or equal to zero. Prove that \r\n\r\n\\[ \\sqrt{3(a+b+c)} \\geq \\sqrt{a} + \\sqrt{b} + \\sqrt{c} \\]", "Solution_4": "[hide=\"solution\"]First square both sides and simplify:\n\n$a + b + c \\geq \\sqrt{a}\\sqrt{b} + \\sqrt{a}\\sqrt{c}+ \\sqrt{b}\\sqrt{c}.$\n\nThis statement is true by applying the re-arrangement inequality to $\\sqrt{a}, \\sqrt{b}, \\sqrt{c}$ and assuming WLOG that $a \\geq b \\geq c$.[/hide]\r\n\r\nA new problem: Let $a,b,c > 0$ and $abc = 1$. Prove that\r\n\r\n$1 + \\frac{3}{a + b + c} \\geq \\frac{6}{ab + ac + bc}$", "Solution_5": "[quote=\"DPopov\"]Given $x+y+z=1$ and $x,y,z>0$ Prove\n\\[ \\frac{x}{\\sqrt{1-x}}+\\frac{y}{\\sqrt{1-y}}+\\frac{z}{\\sqrt{1-z}} \\geq \\sqrt{\\frac{3}{2}} \\][/quote]\r\n[hide=\"my solution\"]Jensen's inequality also works for this, btw. Since if we let\n\n$f(x) = \\frac{x}{\\sqrt{1-x}} \\Rightarrow f''(x) = \\frac{1}{(1-x)^{3/2}} + \\frac{3x}{4(1-x)^{5/2}} > 0$\n\nfor $x > 0$, we have by Jensen:\n\n$\\frac{x}{\\sqrt{1-x}}+\\frac{y}{\\sqrt{1-y}}+\\frac{z}{\\sqrt{1-z}} \\geq 3 f\\left(\\frac{x + y + z}{3}\\right) = 3 f\\left(\\frac{1}{3}\\right) = \\frac{3}{\\sqrt{6}} = \\sqrt{\\frac{3}{2}}$[/hide]", "Solution_6": "[quote=\"The Original Pi Guy\"]Heres the solution to the problem\n\n[hide]\nFor ThAzN1's problem, we first apply cauchy\n\n \\[ \\frac{1}{1+2x_1} + \\frac{1}{1+2x_2} + \\cdots + \\frac{1}{1+2x_n} \\geq \\frac{(1 + 1 + ... + 1)^2}{(1 + 1 + .... + 1) + 2(x_1 + x_2 + ... + x_n)} \\vskip .2in = \\displaystyle\\frac{n^2}{n + 2(x_1 + x_2 + ... + x_n)} \\]\n\nnow by am-gm, \n\n\\[ \\frac{x_1 + x_2 + ... + x_n}{n} \\geq (x_1x_2x_3...x_n)^{\\displaystyle\\frac{1}{n}} = 1 \\Rightarrow x_1 + x_2 + ... + x_n \\geq n \\]\n\nCombining this with the result above, we get\n\n\\[ \\frac{n^2}{n + 2(x_1 + x_2 + ... + x_n)} \\geq \\frac{n^2}{n + 2(n)} = \\frac{n}{3} \\]\n\nand we are done.\n[/hide][/quote]\r\n\r\nThat's actually not correct. You applied AM-GM, but when you took the reciprocal you didn't flip the inequality. In fact\r\n\r\n$\\frac{n^2}{n+2(x_1+\\cdots+x_n)} \\le \\frac{n}{3}$.", "Solution_7": "oh... :blush: \r\n\r\num... evidently cauchy isn't strong enough...\r\n\r\nbut you're definitely right\r\n\r\n :wallbash:", "Solution_8": "My solution to the first problem basically reduced it to Nesbitt's from which you can procede in w/e fashion you would like.", "Solution_9": "The Original Pi Guy wrote:\n\n\n\nEven quicker with Cauchy's.\n\n\n\n[hide]\n\n\n\n\n\n\n\nTake square root and you're done.\n\n\n\n[/hide]", "Solution_10": "[quote=\"ThAzN1\"]\n\nMarathon? Ok, here's a prob:\n\nLet $x_1, x_2, \\dots , x_n$ be positive reals with product equal to $1$. Prove that\n\n\\[ \\frac{1}{1+2x_1} + \\frac{1}{1+2x_2} + \\cdots + \\frac{1}{1+2x_n} \\geq \\frac n3 \\][/quote]\r\n\r\nShouldn't $\\sum x_i=n$? This way the result follows immediately from a direct application of Jensen's.\r\n\r\n*Perhaps you can multiply both sides by the product in some fashion so that the inequality is homogenized.*", "Solution_11": "No, $\\sum x_i\\ge n$.", "Solution_12": "Wow...woops, yeah, AM-GM *ahem* Well in that case it again follows immediately from Jensen:\r\n\r\n[hide=\"Solution Perhaps\"]\nLet $f(x)=\\frac{1}{ax+b}$. With quick computation we see that $f''(x)>0$. It follows from Jensen's that\n\\[{ \\sum_{i=1}^n f(x_i) \\geq n f (\\frac{x_1+...+x_n}{n}}) \\]\nBy AM-GM $\\sum x_i \\geq n$. Hence ${\\frac{x_1+...+x_n}{n}} \\geq 1$. Using this fact we see that \\[{ n f (\\frac{x_1+...+x_n}{n}} ) \\leq \\frac{n}{a+b} \\] In this case $a=2, b=1$ so we have\n\\[{ \\sum_{i=1}^n f(x_i) \\geq n f (\\frac{x_1+...+x_n}{n}} ) \\]\n\\[ \\sum_{i=1}^n f(x_i) \\geq \\frac{n}{a+b} \\]\n\\[ \\sum_{i=1}^n f(x_i) \\geq \\frac{n}{3} \\]\n\nas desired.\n\n*Equality holds when $\\sum x_i=n$*\n\nThis was a bit rushed so I might have some errors ;)\n\n[/hide]", "Solution_13": "[quote=\"DPopov\"][hide]Wow...woops, yeah, AM-GM *ahem* Well in that case it again follows immediately from Jensen:\n\nLet $f(x)=\\frac{1}{ax+b}$. With quick computation we see that $f''(x)>0$. It follows from Jensen's that\n\\[{ \\sum_{i=1}^n f(x_i) \\geq n f (\\frac{x_1+...+x_n}{n}}) \\]\nBy AM-GM $\\sum x_i \\geq n$. Hence ${\\frac{x_1+...+x_n}{n}} \\geq 1$. Using this fact we see that \\[{ n f (\\frac{x_1+...+x_n}{n}} ) \\leq \\frac{n}{a+b} \\] In this case $a=2, b=1$ so we have\n\\[{ \\sum_{i=1}^n f(x_i) \\geq n f (\\frac{x_1+...+x_n}{n}} ) \\]\n\\[ \\sum_{i=1}^n f(x_i) \\geq \\frac{n}{a+b} \\]\n\\[ \\sum_{i=1}^n f(x_i) \\geq \\frac{n}{3} \\]\n\nas desired.[/hide][/quote]\r\nNope - that doesn't work. Your second-to-last step doesn't follow from your given inequalities (you have $\\sum f(x_i) \\geq n \\ f(...) \\leq n / 3$, which doesn't tell you anything). Jensen isn't powerful enough for this inequality.", "Solution_14": "Yeah woops, I just realized but I think this is the correct approach? ;)", "Solution_15": "[quote=\"probability1.01\"]Huh? They are the same problem![/quote]\r\n\r\n#14 and #15?", "Solution_16": "Amazing solution on #12.\n\n\n\nFor #15 I did some work on it (note all products are cyclic)\n\n\n\n[hide]\n\n\n\nFrom the given condition we can substitute and so on such that . Now we have \n\n\n\nThe only problem is that plugging in a couple values into the end equation showed me that it wasn't true, so I'm at a loss right now. The beginning is fine, I believe.\n\n\n\n[/hide]", "Solution_17": "[quote=\"Elemennop\"]Amazing solution on #12.\n\nFor #15 I did some work on it (note all products are cyclic)...\n\n[/quote]\r\n\r\nYour last inequality is true for acute triangle. I actually created #15 from that. But there is a simple algebraic solution for #15 without using any substitution.", "Solution_18": "Shouldn't it be true for all triangles, though? Or maybe that's a stipulation of the substitution.", "Solution_19": "[quote=\"Elemennop\"]Shouldn't it be true for all triangles, though? Or maybe that's a stipulation of the substitution.[/quote]\r\n\r\nIf the triangle is not acute, your 2nd last step doesn't imply the last step because you were dividing a negative number. Anyway, since $x,y,z>0$, the triangle has to be acute with your substitution.", "Solution_20": "Aha, I see now.\r\n\r\nWoot, I actually did an inequality correct.\r\n\r\nedit> bah, I can't finish it up. I don't know how to prove that the final inequality is true for acute triangles.", "Solution_21": "Yeah $x,y,z>0$ basically implies acute-ness\r\n\r\n#14 and #15 (they're equivalent) I'll just show #15 here:\r\n[hide]Note that $x^2 = (\\sqrt{x^2+1}-1)(\\sqrt{x^2+1}+1)$ etc so the inequality becomes\n\n$\\prod (\\sqrt{x^2+1}-1) \\geq 1$ (which in fact is the secant inequality above) Note that\n\n$\\sqrt{yz}\\sqrt{x^2+1} = \\sqrt{x(x+y+z) + yz} = \\sqrt{(x+y)(x+z)} \\geq x+\\sqrt{yz}$ (by Cauchy)\n\n$\\Rightarrow \\sqrt{x^2+1} -1 \\geq \\frac{x}{\\sqrt{yz}}$ and yeah...[/hide]", "Solution_22": "Anyone care elaborating how #14 and 15 are equivalent?", "Solution_23": "So basically just normalize $xy+yz+zx = 1$ in #14 then replace everything with their reciprocals and voila", "Solution_24": "Ok, I see it now.\r\n\r\nIs there any specific guide to normalizing? Like for this one, we basically show that if (x,y,z) is a solution than (tx,ty,tz) is a solution as well, then we can normalize...but can you normalize any symmetric sum? ie. could we also have normalized x+y+z=1 or xyz=1 ??", "Solution_25": "I've moved this topic into the Olympiad Inequalities forum.\r\n\r\nPlease only start topics in the AMC forum that are [i]directly relevant[/i] to the AMC contests. The AMC forum has way too many topics as is without having general mathematical discussions in there as well.\r\n\r\nBesides, the AMC forum is going to be locked tomorrow for the Alternate AIME, but there's no reason that this topic needs to be locked as well!", "Solution_26": "So no one new problem?", "Solution_27": "[quote=\"jin thynj\"]So no one new problem?[/quote]\r\nFor you, jin thynj :) \r\nLet $ a,$ $ b$ and $ c$ are positive numbers. Prove that\r\n\\[ \\frac {a^3 \\plus{} b^3 \\plus{} c^3}{ab \\plus{} ac \\plus{} bc} \\plus{} \\frac {4abc}{a^2 \\plus{} b^2 \\plus{} c^2}\\geq\\frac {7}{9}(a \\plus{} b \\plus{} c)\\]", "Solution_28": "I have done that we can assume $ a \\plus{} b \\plus{} c \\equal{} 1$, and considering that:\r\n$ \\sum{a^3} \\equal{} (\\sum{a})(\\sum{a^2} \\minus{} \\sum{ab}) \\plus{} 3abc \\equal{} (1 \\minus{} 3\\sum{ab}) \\plus{} 3abc$, \r\nthe statement is equivalent to:\r\n\r\n\"let a,b,c>0 with sum 1 then $ \\frac {1 \\plus{} 3abc}{\\sum{ab}} \\plus{} \\frac {4abc}{1 \\minus{} 2\\sum{ab}} \\ge \\frac {34}{9}$.\"\r\n\r\nBut nothing else (I have tried with fixing abc and derivate all respect to sum(ab) with regating to bound, but it would be not olympic and ugly..), any idea?", "Solution_29": "[quote=\"ThAzN1\"]By Holder, we have\n$\\left(\\sum_{\\text{cyc}} \\frac{1}{\\sqrt{1-x}} \\right)^2 \\left( \\sum_{\\text{cyc}} (1-x) \\right) \\geq 3^3 \\Longrightarrow \\sum_{\\text{cyc}} \\frac{1}{\\sqrt{1-x}} \\geq \\frac{3\\sqrt{6}}{2}$\nBy Chebyshev,\n$\\sum_{\\text{cyc}} \\frac{x}{\\sqrt{1-x}} \\geq \\frac 13 \\sum_{\\text{cyc}} x \\sum_{\\text{cyc}} \\frac{1}{\\sqrt{1-x}}$\nCombining these yields the result :) \nJensen probably works too...\n\nMarathon? Ok, here's a prob:\n\nLet $x_1, x_2, \\dots , x_n$ be positive reals with product equal to $1$. Prove that\n\n\\[ \\frac{1}{1+2x_1} + \\frac{1}{1+2x_2} + \\cdots + \\frac{1}{1+2x_n} \\geq \\frac n3 \\][/quote] \nJensen works here" } { "Tag": [ "LaTeX" ], "Problem": "If a * b is defined as \u201chalf the average of a and b\u201d, determine the positive difference between: (3 * 5) * 6 and 3 * (5 * 6). Express answer as a fraction in simplest form or as an exact decimal.\r\n\r\nif a * b = \u00bd (a+ b)/2 then a * b = a + b /4\r\n\r\nso 3 * 5 = 3 + 5 /4 = 2 and 2 * 6 = 2+6/4 =2\r\n\r\n5 * 6 = 5+6/4 = 11/4 3 * 11/4 = (3 + 11/4)/4 = 23/16\r\n\r\n\r\npositive difference between = 2 - 23/16 = 9/16 = 0.5625 = 0.57", "Solution_1": "I've checked, and I do not see any problem (though be warned, my arithmetic is awful), except that I don't think you were supposed to round the decimal in the last step (it says \"exact decimal\").\r\n\r\nIncidentally, you might want to learn LaTeX if you're going to post any more math expressions. The main thing being that it looks a bit nicer, and sometimes clearer.\r\n\r\n$ a * b \\equal{} \\frac{1}{2}\\cdot\\frac{a \\plus{} b}{2}$" } { "Tag": [], "Problem": "\u0394\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bf \u03c0\u03af\u03bd\u03b1\u03ba\u03b1\u03c2 \u0391=$ \\left( \\begin{array}{c c c} 2 & 1 & 0 \\\\\r\n3 & 0 & 0 \\\\\r\n0 & 0 & 1 \\end{array} \\right)$\r\n\r\n\u03b1) \u03bd\u03b1 \u03b2\u03b5\u03b8\u03b5\u03af \u03b7 \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03ae \u03bc\u03bf\u03c1\u03c6\u03ae Jordan\r\n\u03b2) \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03bf \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03c1\u03ad\u03c8\u03b9\u03bc\u03bf\u03c2 \u03c0\u03af\u03bd\u03b1\u03ba\u03b1\u03c2 P \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c2 \u03ce\u03c3\u03c4\u03b5 $ P^ - ^1AP$ \u03bd\u03b1 \r\n \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03bc\u03bf\u03c1\u03c6\u03ae Jordan \u03c4\u03bf\u03c5 \u0391\r\n\r\n*\u0392\u03b1\u03c3\u03b9\u03ba\u03ac \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03b4\u03b5\u03bd \u03b8\u03c5\u03bc\u03ac\u03bc\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03bd \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03c1\u03ad\u03c8\u03b9\u03bc\u03bf \u03c0\u03af\u03bd\u03b1\u03ba\u03b1 P, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c0\u03bf\u03b9\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b3\u03b5\u03bd\u03b9\u03ba\u03ae \u03bc\u03b5\u03b8\u03bf\u03b4\u03bf\u03bb\u03bf\u03b3\u03af\u03b1 \u03c0\u03bf\u03c5 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03bf\u03cd\u03bc\u03b5 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03c4\u03bf\u03bd \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03bf\u03cd\u03bd?\r\n\r\n\u03a0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03ce \u03b2\u03ac\u03bb\u03c4\u03b5 \u03c4\u03b1 \u03c3\u03c4\u03bf \u03ba\u03b1\u03c4\u03ac\u03bb\u03bb\u03b7\u03bb\u03bf \u03bc\u03ad\u03c1\u03bf\u03c2", "Solution_1": "[quote=\"spinos\"]\u0394\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bf \u03c0\u03af\u03bd\u03b1\u03ba\u03b1\u03c2 \u0391=$ \\left( \\begin{array}{c c c} 2 & 1 & 0 \\\\\n3 & 0 & 0 \\\\\n0 & 0 & 1 \\end{array} \\right)$\n\n\u03b1) \u03bd\u03b1 \u03b2\u03b5\u03b8\u03b5\u03af \u03b7 \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03ae \u03bc\u03bf\u03c1\u03c6\u03ae Jordan\n\u03b2) \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03bf \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03c1\u03ad\u03c8\u03b9\u03bc\u03bf\u03c2 \u03c0\u03af\u03bd\u03b1\u03ba\u03b1\u03c2 P \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c2 \u03ce\u03c3\u03c4\u03b5 $ P^ - ^1AP$ \u03bd\u03b1 \n \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03bc\u03bf\u03c1\u03c6\u03ae Jordan \u03c4\u03bf\u03c5 \u0391\n\n*\u0392\u03b1\u03c3\u03b9\u03ba\u03ac \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03b4\u03b5\u03bd \u03b8\u03c5\u03bc\u03ac\u03bc\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03bd \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03c1\u03ad\u03c8\u03b9\u03bc\u03bf \u03c0\u03af\u03bd\u03b1\u03ba\u03b1 P, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c0\u03bf\u03b9\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b3\u03b5\u03bd\u03b9\u03ba\u03ae \u03bc\u03b5\u03b8\u03bf\u03b4\u03bf\u03bb\u03bf\u03b3\u03af\u03b1 \u03c0\u03bf\u03c5 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03bf\u03cd\u03bc\u03b5 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03c4\u03bf\u03bd \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03bf\u03cd\u03bd?\n\n\u03a0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03ce \u03b2\u03ac\u03bb\u03c4\u03b5 \u03c4\u03b1 \u03c3\u03c4\u03bf \u03ba\u03b1\u03c4\u03ac\u03bb\u03bb\u03b7\u03bb\u03bf \u03bc\u03ad\u03c1\u03bf\u03c2[/quote]\r\nhttp://en.wikipedia.org/wiki/Jordan_normal_form#Example" } { "Tag": [ "logarithms", "number theory", "greatest common divisor" ], "Problem": "How many distinct four-tuples $ (a,b,c,d)$ of rational numbers are there with\r\n\r\n$ a \\log_{10} 2 \\plus{} b \\log_{10} 3 \\plus{} c \\log_{10} 5 \\plus{} d \\log_{10} 7 \\equal{} 2005$?\r\n\r\n$ \\textbf{(A)}\\ 0\\qquad\r\n\\textbf{(B)}\\ 1\\qquad\r\n\\textbf{(C)}\\ 17\\qquad\r\n\\textbf{(D)}\\ 2004\\qquad\r\n\\textbf{(E)}\\ \\text{infinitely many}$", "Solution_1": "[hide=\"Answer\"]$\\log_{10}2^a+\\log_{10}3^b+\\log_{10}5^c+\\log_{10}7^d=\\log_{10}2^a3^b5^c7^d=2005$\n$10^{2005}=2^a3^b5^c7^d\\Rightarrow a,c=2005$, $b,d=0$, so we have one four-tuple. $\\boxed{B}$[/hide]", "Solution_2": "I'm not exactly sure as to how you got the answer... i only get the part until you solve the entire thing (right before the last step).\n\nUm, I got the first couple of steps until you actually solved the equation. How do you know that there [hide]is only one solution[/hide] and not infinitely many? :?", "Solution_3": "I'm assuming you understood at least up until $\\log_{10}2^a3^b5^c7^d=2005$. From there, by one of the basic properties of logarithms, we have $10^{2005}=2^a3^b5^c7^d$. $10^{2005}=2^{2005}5^{2005}$, so we only have one solution: $(a,b,c,d)=(2005,0,2005,0)$. :)", "Solution_4": "No, I understand all the stuff about logarithms, but I don't understand why there can be only one solution. :?", "Solution_5": "well, b, d have to equal 0 because 3, 7 are not factors of 10 to any power. Now with a and c, try to find another pair of numbers which satisfies that...\r\n\r\nif you think about it in your head, 2 * 5 = 10, right?\r\n\r\nso to get 2005 10's you need 2005 2's and 2005 5's, and since 2 and 5 dont have any common factors, there is only one solution, which is 2^2005 * 5^2005", "Solution_6": "Oh! Thanks so much Cheesy :lol:", "Solution_7": "[hide]We can rewrite this as $\\log_{10}(a^2b^3c^5d^7)=2005$. \n\n$a^2b^3c^5d^7=10^{2005}$. Since this is a power of 10, we want the 3 and the 7 to stay out of this, so b and d must be 0. If a and c are both 2005, this works. \n\ntHus there is one solution set. [/hide]", "Solution_8": "Does \"four-tuples $(a,b,c,d)$ of rational numbers\" mean a, b, c, and d have to be integers?", "Solution_9": "No, rational means they must be able to be written in fraction form (i.e. no sqare roots or transcendentals).", "Solution_10": "uhhh 236factorial, I think you meant $ log_{10} {2^a3^b5^c7^d} \\equal{} 2005$. :D", "Solution_11": "What would be the rigorous proof to show that b,d must be 0? That is, how can we show that there is no way for $3^{b}5^{d}$ to work in this problem for any [b]rational[/b] values of b and d?\n\nIn advance, sorry for reviving this thread.", "Solution_12": "[quote=\"Justong\"]What would be the rigorous proof to show that b,d must be 0? That is, how can we show that there is no way for $3^{b}5^{d}$ to work in this problem for any [b]rational[/b] values of b and d?\n\nIn advance, sorry for reviving this thread.[/quote]\n\nJust consider the GCD of the denominators of all the fractions, let it be $m$. Then, we have $2^a3^b5^c7^d=10^{2005m}$ for $a,b,c,d,m$ integers, and the result follows." } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ n$ be a positive integer, and let $ A_1$, $ A_2$, ..., $ A_{n\\plus{}1}$ be non-empty subsets of $ \\left\\{1,2,...,n\\right\\}$.\r\nProve that: we can choose two non-intersecting groups of $ A_i$'s such that the union of the sets in one group is equal to the union of the sets in another group.", "Solution_1": "Is it really hard?\r\nwell, let U be the set of all subsets of M={A1,...,An+1} not including M and empty set.\r\nthere are 2^(n+1) -2>n such subsets, and therefore there two subsets with the same union. remove the common elements and you will end up with two non-empty and disjoint groups of Ai with same union.", "Solution_2": "i don't think that's quite what he means by the question :?\r\nanyway what u consider a \"union\" (sorry for my bad english) ?", "Solution_3": "I consider the union of two sets A and B two be the set of all elements that belong either to A or B or both.", "Solution_4": "[quote=\"ali\"]remove the common elements and you will end up with two non-empty and disjoint groups of Ai with same union.[/quote]\r\n\r\n\r\nNo.\r\n\r\nFor example, Take A1={1,2},A2={1,2,3},A3={2,3}\r\nThen we have A1 \\cup A2=A2 \\cup A3, but A1!=A3.", "Solution_5": "What is CWMO?", "Solution_6": "China West Mathematical Olympiad, sort of a district olympiad, before the actuall All Chinese Mathematical Olympiad.", "Solution_7": "I have a solution.\r\nCall degree of an element the number of sets it belongs to.\r\nFirstly we should analyse the case all elements of A have degree at least 2.\r\n(1)I claim there are two groups of subsets with union the set A.\r\nIf a set has just one element, we can easy delete it and use induction( as there is another set containing it).\r\nSo suppose every set has at least two elements.\r\n(2)I claim there is a set A[k] such that at least |A[k]|-1 elements have degree at least 3.\r\nSuppose not. Then every subset A[i] correspond at least 2 elements with degree 2 contained in it.\r\nSumming and dividing by 2( each element belongs to just 2 subsets) we get at least n+1 elemnts, contradiction. The claim is proven.\r\nNow let this subset be A[n+1]. Take x belonging to A[1] with smallest degree.\r\nThen there is another set containing x, let it be A[n]. If we delete x and A[n+1]\r\nwe respect the induction hypothesis so there are two groups with union A/{x}.\r\nIf we add x, we have three cases:\r\n*neither of groups contains x. Then we add A[n],A[n+1] to them\r\n**one of groups contains x. Then we add A[n+1] to the another\r\n***Both groups contain x. Then we are done.\r\nThis claim is also proved.\r\nNow pass to the general case. Let Z be the set of all elements with degree at least 2.\r\n(3)I claim that at least |Z|+1 of A[i] are subsets of Z.\r\nIf not, then to at least n+1-|Z| subset we put in correspondence an element with degree 1 contained in it. Summing we get at least n+1-|Z| elements with degree 1 and |Z| elements with degree 2 so at least n+1 elements contradiction.\r\nThe claim is proved.\r\nNow we can apply (1) to set Z and make the conclusion.", "Solution_8": "That's a nice solution, Andrew! I also found it in a book. I don't really understand iura's solution, but statement (1) seems wrong to me. For example, you could take n=4 and A_1={1,3},A_2={1,2},A_3={2,3},A_4={4},A_5={4}. You can't divide these sets into 2 groups both having the union [4] ([4]:={1,2,3,4}). Maybe I missunderstood the whole thing in iura's solution...", "Solution_9": "This problem was again discussed in this forum.\r\n\"Set of subsets\" . I think there my solution is written more explicitely.\r\nNote that I supposed any 2 subsets are distinct and any subset has at least 2 elements( or else delete it an use induction). So your example doesn't work according to this criteria.", "Solution_10": "I'm sorry, iura, but I still don't really understand. As far as I understand, your conditions are: every element has degree >=2, #A_i>=2 and A_i!=A_j (I mean different by !=). Now you could take n=6 and A_1={1,2}, A_2={1,3}, A_3={2,3}, A_4={4,5}, A_5={4,6}, A_6={5,6}, A_7={4,5,6}. If I still don't understand, maybe I'll ask you at the IMO :D ...", "Solution_11": "Yes, I finally must assume you are right.\r\nBut I have abother idea: suppose the set is s.t. the union of any k subsets has at least k elements( if not, use induction). Can't we prove now there are 2 groups of sets with union A. I mean we can delete one element as in my previous proofs and to hold the condition?\r\nPlease anwer me Dzeta, I see you are the most interested in this problem. Thank you.", "Solution_12": "That\u2019s a very nice idea. I really think your claim is true, but I still think there\u2019s a small gap in your induction argument. I think the problem is the case when there are no weak elements. It is possible that by removing an arbitrary element of A_1, not to keep that property of the sets. For example, take n=4, A_1=A_2=A_3={1,2,3,4} and A_4=A_5={1,2}. By removing 1 and A_1, you get the sets A_4\u2019={2} and A_5\u2019={2}, which have union {2}." } { "Tag": [ "complex analysis", "function", "advanced fields", "advanced fields unsolved" ], "Problem": "Consider a sequence of distinct numbers $a_{k}\\in \\mathbb{C}$ with $|a_{k}| < 1$. Let $e_{k}= \\{1, a_{k}, a_{k}^{2}, a_{k}^{3}, ...\\}$. Let $S$ be the closure of the span of the $e_{k}$. Is $S$ all of $l^{2}$? \r\n\r\nObserve: $\\exists f \\in S^\\bot$. If $f=\\{c_{0}, c_{1}, c_{2}, ...\\}$, then we have $F(x) = \\sum_{n}c_{n}x^{n}$ and $F(a_{k}) = 0$ for all $k$. [b]IF[/b] the $a_{k}$ have an accumulation point in the open ball of radius 1, then $F$ is a holomorphic function on the ball with an accumulation point of zeros, so it must be 0. Then $f \\equiv 0$, and $S = l^{2}$ as desired. \r\n\r\nThe question: do we still have the result if $a_{k}\\to 1$ (and has no other accumulation point)?", "Solution_1": "The question is equivalent to the description of possible zeroes of a function in the Hardy class $H^{2}$ in the unit disk. The answer is well-known: the condition $\\sum_{k}(1-|a_{k}|)<+\\infty$ is necessary and sufficient for existence of a non-trivial function with zeroes $a_{k}$.", "Solution_2": "In other words, the $a_{k}$ have to go to 1 \"fast enough.\" I can see why that might be the condition, but could you please give me a hint about how to prove it? (Or a reference; I wouldn't be surprised if it's already been discussed on the forum. I don't have Halmos' book, though.)", "Solution_3": "The proof with all details is a bit too long to post, so I'll just indicate the main ideas.\r\nTo construct such a function, just consider the infinite Blaschke product $B(z)=\\prod_{k}\\frac{|a_{k}|}{a_{k}}\\frac{a_{k}-z}{1-\\bar a_{k}z}$. The condition $\\sum_{k}(1-|a_{k}|)<\\infty$ ensures that it converges to some non-zero function bounded in the unit disk. \r\nTo show that a function $f\\in H^{2}$ with zeroes $a_{k}$ satisfying $\\sum_{k}(1-|a_{k}|)=\\infty$ must be identically $0$ is a bit harder. Basically, you have to apply some version of maximum principle to the function $f(z)/B_{n}(z)$ where $B_{n}(z)$ are the finite Blaschke products corresponding to the first $n$ zeroes $a_{k}$. Note that $|B_{n}(z)|=1$ on the unit circumference and $B_{n}(z)\\to 0$ as $n\\to\\infty$ for all $|z|<1$.\r\nAs to the references, just take any book on Hardy spaces. You'll find all the details there.", "Solution_4": "Thanks. The girl who asked the question understood that. I guess I'll have to wait till I take a course on complex analysis (I'd teach myself, but I already have too long a list of things I'd like to teach myself...)." } { "Tag": [], "Problem": "I consider myself okay at math. I can easily get an A in school (i am currently taking Alg II). I just really want to get GOOD at math. I want to be able to do well in competitions at maybe one day take the USAMO. Whats the best way to get good at math? I practice everyday, but there's just so much stuff to learn, and I dont know where to start. Should I get a tutor to teach me or just buy a textbook and work through the problems myself, or should I just practie old tests from competitions?", "Solution_1": "You should probably start by getting AoPS I and II and possibly the introduction series books. Also, try to solve problems that are posted on the forum, post questions that you need help on, and look at past contests.", "Solution_2": "I've got both AoPS books. I'm going to start the second one in the summer.", "Solution_3": "Practice tests are useful because they help you identify subjects you are not yet familiar with so that you know what to study. The alternative is using a textbook, presumably AoPS as you have mentioned, because you get a pretty comprehensive overview of high school mathematics.", "Solution_4": "wut grade are you taking alg 2? b/c if it's middle school, give yourself a pat on the back. not many ppl can do that!( whoops! I forgot that this is AOPS!)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "All the squares of a drawn into squares board with $(n + 1) (n - 1)$ squares are painted with $3$ colours\r\nsuch that, for each two different columns and each two different rows, the four squares\r\nof their intersections are not painted with the same colour. Find the maximum value\r\nof $n$.", "Solution_1": "nothing about the number of diferent colors? ;)", "Solution_2": "[quote=\"M4RI0\"]All the squares of a drawn into squares board with $(n + 1) (n - 1)$ squares are painted\nsuch that, for each two different columns and each two different rows, the four squares\nof their intersections are not painted with the same colour. Find the maximum value\nof $n$.[/quote]\r\nHow many colours in your problem?", "Solution_3": "Sorry, it's now corrected. :D" } { "Tag": [ "inequalities", "function" ], "Problem": "Let $ x,y,z\\ge0, x+y+z=2$, prove that $ \\frac{1}{1+x^{2}}+\\frac{1}{1+y^{2}}+\\frac{1}{1+z^{2}}\\ge2$", "Solution_1": "[quote=\"zkw\"]Let $ x,y,z\\ge0, x+y+z=2$, prove that $ \\frac{1}{1+x^{2}}+\\frac{1}{1+y^{2}}+\\frac{1}{1+z^{2}}\\ge2$[/quote]\r\nI wish to show the Zhaobin's method. :wink: \r\nLet $ x\\leq y\\leq z$ and $ f(x)=\\frac{1}{1+x^{2}}.$ Hence, $ f''(x)=\\frac{2(3x^{2}-1)}{(x^{2}+1)^{3}}.$\r\nId est, $ f$ is convex function on $ [\\frac{1}{\\sqrt3},2]$ and $ f$ is concave function on $ [0,\\frac{1}{\\sqrt3}].$\r\nCase 1. $ \\frac{1}{\\sqrt3}\\leq x.$\r\nHence, $ \\sum_{cyc}f(x)\\geq3f\\left(\\frac{x+y+z}{3}\\right)=3f\\left(\\frac{2}{3}\\right)>2.$\r\nCase 2. $ x\\leq\\frac{1}{\\sqrt3}\\leq y.$\r\nHence, $ \\sum_{cyc}f(x)\\geq f(x)+2f\\left(\\frac{y+z}{2}\\right)=f(x)+2f\\left(\\frac{2-x}{2}\\right)=\\frac{1}{1+x^{2}}+\\frac{8}{x^{2}-4x+8}.$\r\nThus, it remain to prove that $ \\frac{1}{1+x^{2}}+\\frac{8}{x^{2}-4x+8}\\geq2,$ which true for $ x\\in[0,\\frac{1}{\\sqrt3}].$\r\nCase 3. $ x\\leq y\\leq\\frac{1}{\\sqrt3}$ and $ x+y\\geq\\frac{1}{\\sqrt3}.$\r\nHence, $ f(x)+f(y)\\geq f\\left(x+y-\\frac{1}{\\sqrt3}\\right)+f\\left(\\frac{1}{\\sqrt3}\\right)=f\\left(2-z-\\frac{1}{\\sqrt3}\\right)+f\\left(\\frac{1}{\\sqrt3}\\right).$\r\nHence, it remains to prove that $ f\\left(2-z-\\frac{1}{\\sqrt3}\\right)+f\\left(\\frac{1}{\\sqrt3}\\right)+f(z)\\geq2,$\r\n where $ z\\in [2-\\frac{2}{\\sqrt3},2-\\frac{1}{\\sqrt3}],$ which easy to check.\r\nCase 4. $ x\\leq y\\leq\\frac{1}{\\sqrt3}$ and $ x+y\\leq\\frac{1}{\\sqrt3}.$\r\nHence, $ f(x)+f(y)\\geq f(0)+f(x+y)=1+f(2-z).$\r\nThus, it remains to prove that $ f(2-z)+f(z)\\geq1,$ where $ z\\in[2-\\frac{1}{\\sqrt3},2],$ which triviality.", "Solution_2": "More proof. \r\nAfter expanding and homogenization we obtain that \r\n$ \\frac{1}{1+x^{2}}+\\frac{1}{1+y^{2}}+\\frac{1}{1+z^{2}}\\ge2\\Leftrightarrow$\r\n$ \\Leftrightarrow (x+y+z)^{6}-16(x+y+z)^{2}(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})-128x^{2}y^{2}z^{2}\\geq0,$\r\nwhich true because $ (x+y+z)^{4}\\geq16(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+11xyz(x+y+z).$", "Solution_3": "Well.. $ \\frac{1}{1+x^{2}}=1-\\frac{x^{2}}{1+x^{2}}\\geq 1-\\frac{x}{2}$ :)\r\nSum it up and you`re done :wink:", "Solution_4": ":rotfl: \r\nAre you sure? When does equality hold?", "Solution_5": "What is not clear in my proof?\r\nEquality holds for (1,1,0) and permutations.", "Solution_6": "[quote=\"andyciup\"]Well.. $ \\frac{1}{1+x^{2}}=1-\\frac{x^{2}}{1+x^{2}}\\geq 1-\\frac{x}{2}$ :)\nSum it up and you`re done :wink:[/quote]\r\nThis is easy and right...thank you!", "Solution_7": "See also here the Drij's post:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=82172" } { "Tag": [ "function", "calculus", "integration", "trigonometry", "Euler", "calculus computations" ], "Problem": "For each $ f : [0, \\pi] \\to \\mathbb{R}$ satisfying $ f(0) \\equal{} f(\\pi) \\equal{} 0$, define the integral\r\n\r\n$ I(f) \\equal{} \\int_{0}^{\\pi} f'(t)^2 \\plus{} 2\\sin(t)f''(t) \\,dt$\r\n\r\nFind a function $ f$ that minimizes this integral.", "Solution_1": "$ f'(t)^2 \\plus{}f(t)f''(t) \\equal{} (f(t)f'(t))'$\r\n\r\n$ \\int_0^\\pi (f(t)f'(t))'dt \\equal{} f(\\pi)f'(\\pi)\\minus{}f(0)f'(0)\\equal{}0$\r\n\r\nso letting $ f(t)\\equal{}2\\sin t$ seems logical?", "Solution_2": "From your computation it follows that letting $ f(t) \\equal{} 2\\sin t$ gives $ I(f) \\equal{} 0$, but this could be achieved with $ f\\equiv 0$ anyway.\r\n\r\nLet $ y \\equal{} f'$ and $ L(t,y,y') \\equal{} y^2 \\plus{} 2y' \\sin t$, so that we're looking to minimize $ J(y) \\equal{} \\int_0^{\\pi}L(t,y,y')dt$. The relevant [url=http://en.wikipedia.org/wiki/Euler-Lagrange_equation]Euler-Lagrange equation[/url] is, formally, $ \\frac {\\partial L}{\\partial y} \\equal{} \\frac {d}{dt}\\frac {\\partial L}{\\partial y'} \\Rightarrow y \\equal{} \\cos t$ which yields a stationary value for $ J$, and hence our solution is $ f \\equal{} \\sin t$ for which $ I(f) \\equal{} \\minus{} \\frac {\\pi}{2}$.", "Solution_3": "Hm, can you explain more thoroughly how you used the $ \\frac {\\partial L}{\\partial y} \\equal{} \\frac {d}{dt}\\frac {\\partial L}{\\partial y'} \\Rightarrow y \\equal{} \\cos t$ part? I'm not too familiar with the Euler-Lagrange equations." } { "Tag": [], "Problem": "The common application for undergraduate applications has a brief section for academic honors and gives National Merit and Cum Laude Society as two examples. What are other examples of academic honors high school students should shoot for?", "Solution_1": "Pretty much anything that you can arguably describe as an academic honor. That's the place to put your valedictorian status, AMC scores, scores on other academic competitions, and so forth.", "Solution_2": "Oh. \r\n\r\nWould it be inappropriate to ask about cum laude society in this thread? I checked the website and got really confused about what it is and such...", "Solution_3": "its just a high honor organization for hi skool, similar to phi beta kappa in college. all you have to do to get nominated for it is to be in the top 20% of your class , which is pretty easy. Its basically a national honor society thats more selective and serious. At least thats what i heard." } { "Tag": [], "Problem": "What is 10$ \\%$ of 20$ \\%$ of 30$ \\%$ of 10,000 ?", "Solution_1": "This is equivalent to $ ((10000\\cdot .3)\\cdot .2)\\cdot .1\\equal{}(3000\\cdot .2)\\cdot .1\\equal{}600\\cdot .1\\equal{}60$", "Solution_2": "Or you could just do:\r\n\r\n$ (0.1 \\times 0.2 \\times 0.3) \\times 10000 \\equal{} \\boxed{60}$" } { "Tag": [], "Problem": "Does anyone know if it is possible to get a copy of our solutions from 2004-2005? Are they filed somewhere, or have they all been discarded? I would be highly grateful for any information on this (in private message or public post). Thanks.", "Solution_1": "They're posted.... I believe.", "Solution_2": "I'm sorry for not being clear. I was referring to my own solutions, which I sent by fax and have since misplaced." } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that if $R$ is a finite ring, then every prime ideal of $R$ is a maximal ideal.", "Solution_1": "sorry, found it's been on the forum before, so disregard it." } { "Tag": [ "geometry", "parallelogram", "trigonometry", "similar triangles", "geometry proposed" ], "Problem": "[color=darkred]Let be given a parallelogram $ABCD.$ A point $M$ inside its so that $\\angle MAB=\\angle MCB$\nProve that: $\\angle MBC=\\angle MDC$ [/color]", "Solution_1": "i)we easily get that, by sines law, $\\frac{sin(MBA)}{sin(MBC)}= \\frac{sin(MDA)}{sin(MDC)}$!\r\n\r\nii)$\\angle{MBA}+\\angle{MBC}= \\angle{MDA}+\\angle{MDC}$\r\n\r\nwith this, you can kill your problem.", "Solution_2": "Yeah, with trigo. we can kill all things easily :mad:", "Solution_3": "[quote=\"April\"][color=darkred]Let be given a parallelogram $ABCD.$ A point $M$ inside its so that $\\angle MAB=\\angle MCB$\nProve that: $\\angle MBC=\\angle MDC$ [/color][/quote]\r\n[color=indigo][b]Proof. (See the figure)[/b] $(EF\\parallel AB,\\,GH\\parallel BC)$\nWe have: \n$\\angle AGM = \\angle CFM\\Longrightarrow\\triangle AGM\\sim\\triangle CFM\\Longrightarrow\\frac{AG}{CF}=\\frac{MG}{MF}$ $\\Longrightarrow\\frac{DH}{MH}=\\frac{BF}{MF}\\quad (1)$ \nOn th otherhand, we have: $\\angle DHM = \\angle MGB = \\angle BFM\\quad (2)$ \nFrom $(1)$ and $(2)$ we have: $\\triangle DHM\\sim\\triangle BFM$ \nHence: $\\angle MDH = \\angle MBF$ i.e. $\\angle MDC = \\angle MBC$ [/color]", "Solution_4": "I had the same proof (but you posted first!! :ninja: )\r\n\r\nAngry\u00bf?\r\n\r\nWhy should I?\r\n\r\nSend me a PM please, 'cause I don't know what you're talking about.", "Solution_5": "first solution\r\n$AM$ meets $CD$ at $K$, $CM$ meets $AB$ at $N$\r\nso triangle $ADK$ and $NBC$ are similar. Thus $KD/BC=AK/CN$\r\nBut we also have $CM/CN=KM/KA$ so triangle $DMK$ and $BMC$ are similar\r\nThus $\\angle MDC=\\angle MBC$ $q.e.d$\r\n\r\nsecond solution\r\n$AM$ meets $BC$ at $E$, $CM$ meets $AB$ at $N$\r\nwe have triangles $BNF$ and $DAC$ are similar\r\ntriangles $AMC$ and $NMF$ are similar\r\nso $CM/MF=CA/NF=CD/FB$\r\nbut $\\angle MCD=\\angle MAD=\\angle MFB$ so triangle $CMD$ and $FMB$ are similar\r\nThus $\\angle MDC=\\angle MBC$ $q.e.d$" } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra solved" ], "Problem": "Let A,B\\in M_N(C) s.t. 5A-2B=AB.\r\na)Prove that (A-B)(A+B)=A^2-B^2.\r\nb)Prove that for any n\\geq 2 there exists an infinity of matrix A,B\\in M_n(Z) s.t. 5A-2B=AB.", "Solution_1": "For a) AB-5A+2B=0 (1) can be factor as \r\n(A+2I)(B-5I) = -10.I \r\nA+2I, B-5I are invertible, they commute \r\n(B-5I)(A+2I) = - 10.I \r\nBA +2B -5A =0 (2) \r\n\r\n(1) & (2) => AB=BA\r\n\r\n(A+B)(A-B)=A^2-AB+BA-B^2= A^2 - B^2", "Solution_2": ":D\r\ni found a little generalisation of this one moubi...it sounds like this:\r\nLet A,B \\in M_p{C} satisfying the relation : aA+bB=AB, a,b \\in C*. Prove that there exists an infinity of matrix A,B and a,b \\in C with the property that aA+bB=AB.\r\n\r\nfor the first one the proof is like yours moubi... for the second one(quite easy:)) ).\r\nif aA+bB=AB then\r\nwe have that (A-bI_2)(aI_2-B)=-abI_2=>A-bI_2 is invertible\r\nwe choose the matrix B=a(A-bI_2)^(-1).A which veryfies the hypothesis . so there exists an infinity of matrices with the property that aA+bB=AB. \r\n:D:D", "Solution_3": "[quote=\"Kuba\"]\nb)Prove that for any n\\geq 2 there exists an infinity of matrix A,B\\in M_n(Z) s.t. 5A-2B=AB.[/quote]\r\n\r\nLet \r\nN=\r\n(0 0 ...... 0 1)\r\n(0 0 ...... 0 0)\r\n......................\r\n......................\r\n(0 0 ...... 0 0) \r\n\r\nN is nilpotent matrix with indice 2 that's mean N^2 = 0 \r\n\r\nA = - 4I_n + 2tN \r\nB = 10I_n + 5tN where t any integer \r\n\r\n5A - 2B = (-20.I_n +10tN) + (-20.I_n -10tN) = - 40.I_n \r\nAB = -40.I_n -20tN + 20t.N + 10t.N^2 = - 40t.I_n \r\n\r\n\r\nWe have found an infinity matrices A,B in M_n(Z) such that 5A-2B=AB\r\n\r\nA=\r\n( -4 0 0 ..........0 0 2t)\r\n( 0 -4 0 ..........0 0 0)\r\n..................................\r\n( 0 0 0 ...........0 0 -4)\r\n\r\nB=\r\n( 10 0 0 ..........0 0 2t)\r\n( 0 10 0 ..........0 0 0)\r\n..................................\r\n( 0 0 0 ...........0 0 10) with t \\in Z \r\n\r\n\r\nThis problem should be in College Playground -> Superior Algebra", "Solution_4": "[quote=\"Moubinool\"]\nThis problem should be in College Playground -> Superior Algebra[/quote]\r\n\r\n\r\ndone that! :D" } { "Tag": [ "MIT", "college", "blogs", "Support", "Princeton", "Stanford", "calculus" ], "Problem": "Thanks to all who came and stayed through the nearly 3 hours of tonight's MIT Math Jam. We'll post a link to the chat transcript here when it goes online.\r\n\r\nI'm happy to take more questions here or elaborate on the answers I provided. Some were incomplete and at least one was incorrect (thanks to the student who called my attention to this): I recommended the NASA SHARP summer program, but two weeks ago NASA announced it is cutting funding to the program, retiring it. I am disappointed to say the least. \r\n\r\nI'm also happy to take questions on my MIT Admissions web log: [url]http://matt.mitblogs.com[/url]\r\n\r\nThanks again!", "Solution_1": "Thank you Matt for participating in the Math Jam. I will be telling a lot of friends about the transcript after the transcript is posted. \r\n\r\nYour blog already mentions that you will turn to the issue of homeschooled students, and as a leader of a homeschooling support group I'm sure we'd like to hear all about it. I recall a remark by Marilee Jones, quoted in some place that can be found by Google, that some homeschooled applicants are \"extraordinary\" while others \"are not even close.\" I think it would be interesting to explore how much pursuit of personal passions results in adequate preparation for MIT and what kind of preparation might be plainly insufficient. \r\n\r\nIn general, I admire your school's commitment, from the admissions director and all of the admissions staff, to open the windows on the process and make MIT admissions less of a \"black box.\" Keep up the good work.", "Solution_2": "As promised, the link to this week's Math Jam transcript:\r\n\r\n[url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=114[/url]", "Solution_3": "Thanks. I missed a large part of the Math Jam because I was at my MIT interview.", "Solution_4": "[quote=\"frankthetank\"]Thanks. I missed a large part of the Math Jam because I was at my MIT interview.[/quote]\r\n\r\nHow ironic is that?", "Solution_5": "Matt, \r\nI just sent my early application, and i'm looking foward to go to MIT (with a lot of luck of course)\r\nI have some questions:\r\nI have some friends in other universities like Princeton and Stanford and they can take the classes they want since their freshmen year (like there are not many required classes). I've seen and heard that in MIT its quite different. I want to study math and I wanted to know if i could take 2 math classes every year, and if i can take any math class I want even in the freshmen year. I am really interested in combinatorics and I've seen some great courses MIT offers for undergradautes so I was wondering if I can take at least one of them the first year and the required math class (i gues it is calculus or something like that...though i am taking calculus this year...)\r\nI think MIT has a research group in combinatorics for undergrads., can I be part of this group even in my freshmen year? if not, when?\r\n\r\nthanks for answering them!", "Solution_6": "I'm not sure of the answers to your questions, manuel, but I'll note that few (very few) universities give undergraduates as many research opportunities as MIT does...", "Solution_7": "That's the main reason why MIT is my first choice! ;)", "Solution_8": "[quote=\"manuel\"]\nI want to study math and I wanted to know if i could take 2 math classes every year, and if i can take any math class I want even in the freshmen year. [/quote]\r\n\r\nThe requirements for a math major are clearly spelled out on the math dept website. They differ somewhat depending on which type of math major you are going for, and you should go to the website for an \"official\" answer to your question. http://math.mit.edu/undergraduate/degree-options.html \r\n\r\nA general math major (Option I) only has 3 required courses (single and multivar calc, plus diff eq). You can test out of any of those that you feel you have already mastered, and then you take at least 8 more \"substantially different\" math courses of your choice. \r\n\r\nAs long as you take your required 8 semesters of humanities, there doesn't seem to be anything keeping you from taking all math courses with the rest of your time. That would potentially be 3 math courses per semester as a freshman if you wanted, and no limit after freshman year. It also seems very easy to take advanced courses, including grad courses, without proving that you have the prereq.", "Solution_9": "thanks...the math programs look great", "Solution_10": "Hi manuel,\r\nrrusczyk and texas137 covered nicely for me while I was away from the forum. The short answers to your questions are: \r\nThere's little preventing you from taking 2+ math classes each term at MIT, with the exception of the need for sleep and sanity. There is no \"per unit\" cost at MIT so exceptional students may take as many classes as they wish, though I generally do not recommend more than 5 each term. In the first year, there is a credit limit (~4.5 classes/term), but in theory that may include more than one math class, including advanced classes.\r\nAnd generally, research (or 'UROP,' for the Undergraduate Research Opportunities Program) is open to all students, including first term freshmen. I don't know whether specific groups will necessarily have openings, though most students can find a number of groups that meet their needs.\r\nGood luck with your EA application!\r\n\r\nAnd hi tokenadult, \r\nThanks for the comment. I do want to write more about homeschooled applicants, though I feel I would write a better post about them after early action, when we'll all be back in the application-evaluating mindset. How's that sound? \r\n\r\nIt's starting to get quite busy here in Cambridge. It's an exciting time of year...", "Solution_11": "[quote=\"MattMcGannMIT\"]I do want to write more about homeschooled applicants, though I feel I would write a better post about them after early action, when we'll all be back in the application-evaluating mindset. How's that sound?[/quote]\r\n\r\nIt's perfectly understandable that this is a busy time of year for you. When you post your collected thoughts about homeschooled applicants on your blog, I'll be sure to let lots of parents know.", "Solution_12": "Quick question: is it worth mentioning a 5 on last year's AIME, or in MIT's pool would that be detrimental?\r\n\r\nThanks,\r\nsran", "Solution_13": "i think that would be a plus. just because mit has many applicants who have scored almost perfectly on the aime and are usamo qualifiers, the majority of the applicant pool is not like that. the aime is a very difficult exam. just being able to qualifiy for it is impressive. a 5 out of 15 on the aime is certainly helpful.", "Solution_14": "Here is the statistics from the last AIME (2005), which showed about 2200 contestants got 5 or more. About 680 of them among the 11th grade contestants (current seniors who are supposedly applying for colleges), got 5 or more. \r\n\r\n[url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2005-aa/05aimestats.html]AIME stats[/url]", "Solution_15": "sran, \r\nA 5 on the AIME is definitely worth mentioning. It's a nice score, good for you.\r\n\r\nInterestingly, I've already seen some mentions of AoPS on this year's applications...", "Solution_16": "Matt, I will be visiting MIT in NOV 19-24. I would like to talk to somebody of the math departmenet because i am comparing the math depts. of different universities. Do you think I should meet someone from the admission office too? Also, I am interested in playing tennis in college so it would be nice if I could meet the coach or someone from the team. So it would be nice if you could tell me what to do during my visit and who to contact for the math dept. info..\r\n\r\nThanks!\r\n\r\nPS. I didn't mention AOPS in my app... :( if you read the app. of a guy named Manuel from PR interested in maths, remember he is a member of AoPS :D", "Solution_17": "[quote=\"manuel\"]Matt, I will be visiting MIT in NOV 19-24. I would like to talk to somebody of the math departmenet because i am comparing the math depts. of different universities. Do you think I should meet someone from the admission office too? Also, I am interested in playing tennis in college so it would be nice if I could meet the coach or someone from the team. So it would be nice if you could tell me what to do during my visit and who to contact for the math dept. info..[/quote]Although I believe that Matt reads this board fairly frequently, I would recommend that you contact him directly (at MIT) if you have a specific request such as this.", "Solution_18": "Hi everyone! I'm Maria, and am currently a frosh at MIT... I didnt' go on AoPS for a while cause things have been pretty busy, but the term is winding down and i shoudl have more free time till the end of January or so. I really like it here- the people are great and the professors are quite interested in the education of the studnets. i haven't gotten a UROP, but know some first term frosh who have... if you have any questions feel free to PM me (or post on this thread...)\r\n\r\nto manuel, the core requirement is 8 terms of humanities, 2 terms of calc (single and multi variable), 2 terms of physics (maechanics and E&M), a term of chem and a term of bio. you can place out of calc I with BC, mechanics with physics C, bio with AP bio. you can also place out of the other sciences by taking a placement test, though you have to take all classes if you dont' place out. most people take them as soon as they get here, but you dont' have to...", "Solution_19": "thanks mariag!\r\n\r\ntwo weeks ago i was hanging out in mit! i liked it a lot!!! def, my first choice..\r\ni stayed in simmons hall and in burton third..\r\nalso, it was cool to visit wellesley too........... :D", "Solution_20": "2-3 more days for early desicions!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :blush:" } { "Tag": [], "Problem": "mikham inja ba ejazeye hameye bozorgan basaate esteghra va lane kabootari ro pahn konam ta hame lezzat bebearim", "Solution_1": "avvalin soaal:\r\nhaddeaghal chand nemoone az shekle \"tromino\" ra bayad dar morabbae $8*8$ khanei ja dad ta digar natavan hatta yeki az in shekl ha ra bedoone in ke rooye digaran gharar girad dar an mostaghar kard?(shoravi)", "Solution_2": "$n$ pich o mohre az nazare zaheri shabih be ham hastand,dade shode ast.midanim ke har pich tanha be yek mohre mikhorad(ba an ham andaze ast) va hich do pichi ham andaze nistand.\r\namale \"azmoon\" yani bardashtane yek pich va yek mohre va emtehan kardane anha.ba in kar tashkhis midahim ke pich az mohre bozorgtar ast ya mohre az pich bozorg tar ast ya ham andaze hastand.mikhahim ba anjame tedaadi amale \"azmoon\" koochek tarin pich va koochek tarin mohre(ke mosallaman be ham mikhorand) ra peyda konim.tavajjoh konid ke nemitavan do mohre ba do pich ra mostaghiman ba ham moghayese kard.\r\na)neshan dahid ke baraye $n=2$ masale ra dar badtarin halat mitavan ba do azmoon hal kard.\r\nb)raveshi erae dahid ta betavan masale ra dar haalate kolli ba $2n-2$ azmoon hal kard.(injash esteghra dare!)\r\nbedrood", "Solution_3": "tromino ine ke ye khane bechasbooni be ye tarafe domino (na dar emtedadesh) va shabih be L shavad albatte ba se khane.", "Solution_4": "dar yek mehmaani n nafar hozoor darand.saabet konid $2$ nafar vjood daran be tori ke betavan az miyan $n-2$ nafare baghi mande haddeaghal $[\\frac n{2}]-1$ nafar joda kard ke harkodam yahar do in afraad ra beshnasand va hich kodam ra nashnasand.dar nazar dashte bashid ke ashnayi rabetei do tarafe ast.\r\n($[x]\\leq{x}<[x]+1$ yani joze sahihe $x$ ast)", "Solution_5": "hala ham ye soaale bi rabt be topic va dar mabhase \"dirty mathematics\":\r\nma yek noshaabe darim ke mikhahim oon ro be daste $10$ nafar beyne hamoon $10$ nafar taghsim konim.hame ham ba ham sare taghsim kardan ekhtelaaf darand (masalan taghsim bandie hich kas ro baraye khodeshoon ghabool nadaran.)hala shoma rah kari eraae konid ke betoonim nooshabe ro taghsim konim.albatte deghat konid ke eddeaye har kas bayad manteghi bashad va masalan kasi nemitoone bege ke man hameye nooshabe ro mikham.", "Solution_6": "inam jalebe albatte be topic kheili rabt nadare:\r\ndo nafar bazie zir ra ba zrafi ke shaamele n keshmesh ast anjaam midahand:\r\nhar bazikon dar nobate khod mitavanad $1,3$ ya $5$ keshmesh bar daarad.bazi koni ke aakharin keshmesh ra bardaarad baazande ast.aya bazikone avval esterategie bord darad?", "Solution_7": "heyfe ke jaee bahs esteghra beshe va in masale kheily ghashang az esteghra matrah nashe az tot\r\nthere is a large pile of cards. on each card one of numbers 1,2,3,...,n is written. it is know that thesum of all numbers of all the cards is equal to k.n! for some integer k. prove that it is possible to arrange cards into k stacks so that the sum of numbers written on teh cards in each stack is equal to n! .", "Solution_8": "agha ma mokhlese hameye combinatorics baazaye roozegarim.manam age ye hafte tarkibiyat nakhoonam degh mikonam.\r\nye soale dige:\r\nfarz konid $x_1$ va $x_2$ rishe haye moaadele ye $x^{2}+px-1=0$ ke $p$ adadi fard ast bashand.gharar dahid $y_n={x_1}^n+{x_2}^n$ .saabet konid ke $y_n$ va $y_{n+1}$ be ham avval hastand.", "Solution_9": "in ham badak nist:\r\n$51$ hashareye riz daroone yek morabba be zale $1$ gharar darand.neshan dahid dar har lahze hadde aghal $3$ hashare vojood darand ke mitavan anha ra daroone dayerei be shoaae $\\frac 1{7}$ ja dad.", "Solution_10": "saabet konid ke tamaame adad be shekle $12008$,$120308$,$1203308,...$ bar $19$ bakhsh pazir hastan.\r\nva hamin tor saabet konid ke tamaame adaade be shekle $1007$,$10017$,$100117,...$ bar $53$ bakhsh pazir hastand.", "Solution_11": "say konid chand leme fibonacci ro ke dar ketaabe sterategi hast sabet karde va sepas oona ro ba ye jozi taghirat dar in donbale ham be vojood biyarin va ba esteghra saabet konid\r\n${ g_n }$ ra in tori tarif mikonim:\r\n$g_{n}+g_{n-1}+g_{n-2}=g_{n+1}, n\\geq2$ va $g_0=g_1=0$ va $g_2=1$\r\nshayad donbale jalebi bashe.", "Solution_12": "[quote=\"mehdi\"]avvalin soaal:\nhaddeaghal chand nemoone az shekle \"tromino\" ra bayad dar morabbae $8*8$ khanei ja dad ta digar natavan hatta yeki az in shekl ha ra bedoone in ke rooye digaran gharar girad dar an mostaghar kard?(shoravi)[/quote]\r\n\r\nage manzoorte \"domino\" e, \r\nhamoon mostatil e 2*1 e. ;)", "Solution_13": "na manzooresh triminoe ke age az yek morabae $2*2$yek khane ra hazf konim hasel mishavad. ;)", "Solution_14": "[quote=\"mohamadhosein\"]na manzooresh triminoe ke age az yek morabae $2*2$yek khane ra hazf konim hasel mishavad. ;)[/quote]\r\n\r\nBe tore kolli be har majmoye hamband az $n$ moraba $n$-mino ggoyand. Masalan be yek mostatile $3\\times1$ ham trimino gooyand. ;) ;)", "Solution_15": "sabet dar beine har $2n-1$ addade tabiei .daghighan $n$ adda vojod darad ke majomeshan bar $n$ bakhshpazir ast.\r\n\r\n$\\mathrm{Paul\\ Erdos}$", "Solution_16": "in ham ye soale toooooop.\r\n\r\nbe ezaye har addae tabei manande $k$,yek $n$ vojod darad ke be ezaye har efraz az addade $1,2,\\dots,n$ be $k$ daste ,\r\ndastei daraye se ozve$x,y,z$ bashad toori ke $x+y=z$ :D", "Solution_17": "[quote=\"BaBaK Ghalebi\"]yek dayere be shoae $1$ darim\nshakhsi ba gam be toole $\\alpha$ ke adadi gong ast dore an ghadam mizanad dar yek ghesmat az in dayere godali be toole $\\epsilon>0$ vojood darad,neshan dahid $\\epsilon$ har ghadr ham ke koochak bashad in shakhs dir ya zood dar godal mioftad :?:[/quote]\r\nebteda ghaziei ra esbat mikonim:\r\n[b]ghazie:[/b]\r\nfarz konid $\\alpha<0$ adadi gong bashad hamchenin $a,b \\in [0,1]$,$an$ hokme masale sabet mishavad\r\n ;) :D", "Solution_21": "[quote=\"BaBaK Ghalebi\"]khanehaye yek safhe moshabake $7*7$ (ke az $49$ khane tashkil shode ast) ra ba do rang rang kardeim\nneshan dahid haddeaghal $21$ mostatil ba $4$ goosheye yek rang va azlae movazie azlae safheye moshabak dar in safhe vojood darad :?:[/quote]\r\nyek joft az khanehaye yek satr ra joft $khobb$ minamim har gah az yek rang bashand\r\nfarz konid dar yek satr $k$ khane sefid va $7-k$ khane siah bashad\r\ndar in soorat dar in satr tedad jofthaye khoob barabar ast ba:\r\n${{\\frac{k(k-1)}{2}}+\\frac{(7-k)(6-k)}{2}}=k^2-7k+21$\r\nkamtarin meghdare ebarate fogh $9$ ast va be ezaye $k=2,k=4$ be dast miaiad\r\nbanabarin dar har satr haddeaghal $9$ jofte $khoob$ darim\r\npas dar kolle safhe haddeaghal $63$ jofte khoob darim\r\ndo jofte $khoob$ ra $aali$ minamim har gah dar sotoonhaye yeksan bashand\r\nbe vozooh do jofte $aali$ yek mostatile moredenazar masale ra moshakhas mikonand\r\ntedad halathai ke baraye yek jofte $khoob$ az lahaze mogheiat sotooni vojood darad barabare $\\left(\\begin{array}{c}{1cr}7\\2\\end{array}\\right)=21$ ast\r\nbanabarin bishtar az $42=2*21$ zoje $khoob$e gheire $aali$ nadarim pas haddeaghal $21$ jofte $aali$ vojood darad(lazem be zekr ast ke adade $21$ daghigh mibvashad)", "Solution_22": "[quote=\"BaBaK Ghalebi\"]aadade tabii $a_1\\leq a_2\\leq...\\leq a_n\\leq 2n$ tori hastand ke $k.m.m$ har do ta az anha az $2n$ bozorgtar ast\nneshan dahid $a_1>[\\frac{2n}{3}]$ :?:[/quote]\r\n[b]$Lemma:$$n+1$ adade az $\\{1,2,...,2n\\}$ entekhab shode ast.az miane aadade entekhab shode yeki bar digari bakhshpazir ast.\nborhan:farz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$ bashand anha ra be shekle $a_i=2^{k_i}b_i$ minivisim ke $b_i$ adadi fard bashad dar in soorat $b_1,b_2,...,b_n_+_1$ $n+1$ adade far mibashand ke beine $1,2n$ gharar darand chon dar in baze $n$ adade fard bishtar nadarim pas tebghe asle hojreha do ta az $b_i$ ha manande $b_p,b_q$ barabarand banabarin az do adade $a_p,a_q$ yaki bar digari bakhshpazir ast[/b]\r\nhal be soraghe masale miravim ba borhane kholf farz konid $a_1\\leq [\\frac{2n}{3}]$ dar in soorat $3a_1\\leq 2n$ majmooe $\\{2a_1,3a_1,a_2,...,a_n\\}$ ke az $n+1$ adade sahihe koochektar ya mosavie $2n$ tashkil shode ast hichkodam bar digari bakhshpazir nist,in matlab ba $Lemma$ bala tanaghoz darad\r\npas be tanaghoz residim\r\npas $a_1>[\\frac{2n}{3}$\r\n ;) :D", "Solution_23": "[quote=\"amir2\"]sabet dar beine har $2n-1$ addade tabiei .daghighan $n$ adda vojod darad ke majomeshan bar $n$ bakhshpazir ast.\n\n$\\mathrm{Paul\\ Erdos}$[/quote]\r\nfarz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$ bashand anha ra be shkle $a_i=2^{k_i}b_i$ minivisim ke $b_i$ adadi far bashad\r\ndar in soorat $b-1,b_2,...,b_n_+_1$ $n+1$ adade fard mibashand ke beine $1,n$ gharar darand va chon dar in baze $[\\frac{n}{2}]$ ta adade fard vojood darad pas tebghe asle hojreha haddeaghal do ta az $b_i$ha manande $b_r,b_s$ barabarand banabarin az do adade $a_r,a_s$ yeki bar digari bakhshpazir ast\r\ndoroste??\r\n ;) :D", "Solution_24": "[quote=\"BaBaK Ghalebi\"]noghate safhe ra be vasileye do rang rang kardeim neshan dahid yek mostatil ba roose tak rang mitavan peida kard :?:[/quote]\r\n$7$ noghte hamkhat dar nazar migirim\r\ntebghe asle hojreha haddeaghal $4$ ta az in noghta az yek rang masalan ghermez mibashand in noghat ra $R_1,R_2,R_3,R_4$ minamim\r\nin noghat ra rooye do khate movazie khate avalie $tasvir$ mikonim $tasvir$e in noghat rooye khate aval ra $S_1,...,S_4$ va rooye khate dovom $T_1,...,T_4$ minamim agar do noghte az $S_i$ ha ghermez bashand masale hal ast va agar do noghte az $T_i$ ha ghermez bashand baz ham masale hal ast\r\npas har kodam az do khate $T,S$ haddeaghal $3$ noghte abi darand\r\ndar natije yek mostatil ba rooose abi be dast miaiad va baz ham masale hal ast\r\n ;) :D", "Solution_25": "[quote=\"BaBaK Ghalebi\"][quote=\"amir2\"]sabet dar beine har $2n-1$ addade tabiei .daghighan $n$ adda vojod darad ke majomeshan bar $n$ bakhshpazir ast.\n\n$\\mathrm{Paul\\ Erdos}$[/quote]\nfarz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$ bashand anha ra be shkle $a_i=2^{k_i}b_i$ minivisim ke $b_i$ adadi far bashad\ndar in soorat $b-1,b_2,...,b_n_+_1$ $n+1$ adade fard mibashand ke beine $1,n$ gharar darand va chon dar in baze $[\\frac{n}{2}]$ ta adade fard vojood darad pas tebghe asle hojreha haddeaghal do ta az $b_i$ha manande $b_r,b_s$ barabarand banabarin az do adade $a_r,a_s$ yeki bar digari bakhshpazir ast\ndoroste??\n ;) :D[/quote]\n :o :o \n\n[quote]farz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$[/quote]\nkodoom $n+1$ addade mafrooz.\n\naz inke $2n-1$ adad darim che estefadei kardi?\n\n[quote] dar in soorat $b-1,b_2,...,b_n_+_1$ $n+1$ adade fard mibashand ke beine $1,n$ gharar darand [/quote]\r\nChera? beine har $2n-1$ addad ,pas in addad mitoonand kheili bozorg bashand.", "Solution_26": "[quote=\"amir2\"][quote=\"BaBaK Ghalebi\"][quote=\"amir2\"]sabet dar beine har $2n-1$ addade tabiei .daghighan $n$ adda vojod darad ke majomeshan bar $n$ bakhshpazir ast.\n\n$\\mathrm{Paul\\ Erdos}$[/quote]\nfarz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$ bashand anha ra be shkle $a_i=2^{k_i}b_i$ minivisim ke $b_i$ adadi far bashad\ndar in soorat $b-1,b_2,...,b_n_+_1$ $n+1$ adade fard mibashand ke beine $1,n$ gharar darand va chon dar in baze $[\\frac{n}{2}]$ ta adade fard vojood darad pas tebghe asle hojreha haddeaghal do ta az $b_i$ha manande $b_r,b_s$ barabarand banabarin az do adade $a_r,a_s$ yeki bar digari bakhshpazir ast\ndoroste??\n ;) :D[/quote]\n :o :o \n\n[quote]farz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$[/quote]\nkodoom $n+1$ addade mafrooz.\n\naz inke $2n-1$ adad darim che estefadei kardi?\n\n[quote] dar in soorat $b-1,b_2,...,b_n_+_1$ $n+1$ adade fard mibashand ke beine $1,n$ gharar darand [/quote]\nChera? beine har $2n-1$ addad ,pas in addad mitoonand kheili bozorg bashand.[/quote]\r\noh\r\nsharmande man masalaro eshtebahi gereftam :wallbash_red: :stretcher:", "Solution_27": "plz!!!!!!!!!!!!!!!!!!!!!!!!\r\nage momkene hale in soval ru baram bezarid \r\nman montazer hal haye shoma hastam ;) \r\n $14.$ The number $1$ or $-1$ is written in each cell of a $2000 * 2000$ array. The sum of all the numbers in the array is non-negative. Show that there are $1000$ rows and $1000$ columns such that the sum of the numbers at their intersections is at least $1000$??????? :?:", "Solution_28": "Rahnamiyi baraye masaleye zibaye Erdos. Kafi ast barye aadade avval sabet konid. tamame $(x_{i_1}+\\dots+x_{i_{p}})^{p-1}$ ha ra begirid va ba ham jam konid. Intori hokme ghavitari sabet mishe va sabet mishe teedae zirmajmeehaye $n$ ozvi az $\\{1,\\dots,n\\}$ mesle $\\{i_1,\\dots,i_n\\}$ ke $p|x_{i_1}+\\dots+x_{i_{n}}$ be peimaneye $p$ barabar ba $1$ ast.", "Solution_29": "besiar jaleb va ziba bood :lol:" } { "Tag": [ "trigonometry" ], "Problem": "Find the smallest positive angle $\\theta$ in degrees satisfying\r\n\r\n\\[\\sin^2(2004\\theta) +\\cos^2(2005\\theta)=1.\\]", "Solution_1": "$\\sin^2(2004\\theta)=\\sin^2(2005\\theta)$\r\n$\\sin(2004\\theta)=\\sin(2005\\theta)$ since $\\sin\\theta$ is positive for $0<\\theta<180^\\circ$.\r\n\r\n$\\sin(2004\\theta)=\\sin(2005\\theta)\\rightarrow180^\\circ-2004\\theta=2005\\theta$ since we want the smallest value of $\\theta$.\r\n$\\theta=\\frac{180}{4009}^\\circ$" } { "Tag": [ "calculus", "geometry", "geometric transformation", "algorithm", "linear algebra", "superior algebra", "superior algebra solved" ], "Problem": "Evaluate:\r\n\r\n[img]http://www.stud.uni-karlsruhe.de/~uzaxx/math/det.gif[/img]", "Solution_1": "((n-1)!)n\r\nWho can prove it?", "Solution_2": "I agree with Myth.I hope that someone will bring up a proof for this one :)", "Solution_3": "I have a proof based on identity\r\n(x+k)k=\\sum {i=0..k-1} (-1)k+1-iC(k,i)(x+i)k + k!.\r\n\r\nUsing this identity we can reduce initial determinant to standart Wan der Warden one.\r\n\r\nEnjoy!", "Solution_4": "Van der Waerden", "Solution_5": "Myth,can you show us the introductory part of your proof?Thank you.", "Solution_6": "What do you mean?\r\n\r\nUnfortunately, I have no time to write detailed proof.", "Solution_7": "I will wait until you're not so busy :)", "Solution_8": "Take this idea:\r\nA=(a_ij)1 \\leq i,j \\leq n, where\r\na_ij=(a_i+a_j) n-1 . If a_i=i, we have our problem.\r\n\r\nLets compute detA.\r\nTake Pi(X)=detAi(X), where A_i(X) is obtained by replacing a_i by X.\r\nP_i has degree n-1 in X.\r\n\r\nNotice that P_i(a_j) is 0 if j<>i (because the determinant has 2 equal lines) and det(A) if a_j=a_i. This means, by Bezout, that (X-a_1)*...*(X-a_(i-1))(X-a_(i+1))*...*(X-a_n)|P_i(X).\r\nYou continue from here, I'm lazy and I should study for a Differential Equations Exam.", "Solution_9": "and I should sleep for the same exam :D:D God have mercy on us, please! ...", "Solution_10": "quite by accident, i found that this problem is included in a book of linear algebra problem set.\r\n\r\nRussians have some famous problem set books, the most famous one in China is a math analysis problem set (i don't know how to spell its author in English)", "Solution_11": "Which book Liyi?", "Solution_12": "i have the translation..and i don't know its original name...", "Solution_13": "chinese version?\r\nMay I knwo the name?\r\nSee if I can get it.", "Solution_14": "[quote=\"Valentin Vornicu\"]and I should sleep for the same exam :D:D God have mercy on us, please! ...[/quote]\r\nwell :( he didn't have too much mercy. we passed but not with full-score, neither me, nor amfulger, nor pliviu. Fortunately we will get a second chance in autumn ...", "Solution_15": "[quote=\"Valentin Vornicu\"]well :( he didn't have too much mercy. we passed but not with full-score, neither me, nor amfulger, nor pliviu. Fortunately we will get a second chance in autumn ...[/quote]\r\n\r\ndo all of you require full-score?", "Solution_16": "[quote=\"siuhochung\"]chinese version?\nMay I knwo the name?\nSee if I can get it.[/quote]\r\n\r\nthey are old ones... i don't know if they are out of print now\r\n\r\nthe linear algebra problem set:\r\n\u7ebf\u6027\u4ee3\u6570\u4e60\u9898\u96c6 by \u0418. B.\u666e\u7f57\u65af\u5e93\u5217\u67ef\u592b\r\n\r\nthe calculus problem set:\r\n\u6570\u5b66\u5206\u6790\u4e60\u9898\u96c6 by \u0411. \u041f. \u5409\u7c73\u591a\u7ef4\u5947\r\n\r\ni don't know how to translate the Russian names....�ion op_start() { windowop_start() { window", "Solution_17": "[quote=\"liyi\"]do all of you require full-score?[/quote] we like to belive that we are capable of perfect scores at least in a usual math college exam :D unfortunately it wasn't the case. we under-prepared for the test and now we sufer the consequences.", "Solution_18": "If you guys,dont achieve perfect scores,i bet that most of the other students failed the whole exam :D .Siochung,give me that translation on MSN,thx.", "Solution_19": "[quote=\"PiDeltaPhi\"]If you guys,dont achieve perfect scores,i bet that most of the other students failed the whole exam :D .Siochung,give me that translation on MSN,thx.[/quote]\r\n\r\ni dunno how to translate russian names too...", "Solution_20": "[quote=\"PiDeltaPhi\"]If you guys,dont achieve perfect scores,i bet that most of the other students failed the whole exam :D .Siochung,give me that translation on MSN,thx.[/quote]\r\nyeah unfortunately many papers were graded 2/10. The passing note is 5/10. We got 8/10 :(", "Solution_21": "Were the exercises that hard?Or did the prof consider that your given solution was not flawless?", "Solution_22": "The exam was not all problems. It had a theoretic part also. That was a problem for us if we didn't study. That was not all. Me and Valentin got 8/10 and Pliviu got 9. 2 of our colleagues got 10 and another colleague 9, so there is room for improovement.\r\nThe problems were easy but I have managed to err in each of them. Valentin made the exercises correctly but he made less contact with the theoretic subjects.\r\nOur results were bettered by our active presence in seminaries, so the real results are smaller. I don't know about the others, but I am ashamed :(. I have ignored the course over the entire semester and have comed, int he end, to realize that it is quite hard. There is a big difference between the courses and the seminaries.", "Solution_23": "To put it simple I can solve almost any ODE, but I couldn't prove correctly why I can apply the algorithms I use :D funny isn't it?", "Solution_24": "The author of the Calculus problem set is B. P. Demidovich (\u0411.\u041f.\u0414\u0435\u043c\u0438\u0434\u043e\u0432\u0438\u0447). I can't find the author of the linear algebra problem set on Google :?", "Solution_25": "Demidovich's analysis problem set is very famous among all the math dept students. Everyone knows it -- but few of them do the problems. :D \r\nIn Western world, his book \"5,000 problems of mathematical analysis\" is famous, but this books seems to be designed for engineering courses, not math specialty. The book which is popular in China contains less than 5000 problems. :)", "Solution_26": "[quote=\"amfulger\"]Take this idea:\nA=(a_ij)1 \\leq i,j \\leq n, where\na_ij=(a_i+a_j) n-1 . If a_i=i, we have our problem.\nLets compute detA.\nTake Pi(X)=detAi(X), where A_i(X) is obtained by replacing a_i by X.\nP_i has degree n-1 in X.\nNotice that P_i(a_j) is 0 if j<>i (because the determinant has 2 equal lines) and det(A) if a_j=a_i. This means, by Bezout, that (X-a_1)*...*(X-a_(i-1))(X-a_(i+1))*...*(X-a_n)|P_i(X).\n[/quote]\r\n\r\nI must correct the idea. a_ij=a_i+b_j. I must do this to have P_i(X) of degree n-1. I don't actually know how to finish the proof.\r\nIt must be smth like P_i(X)=c*(X-a_1)*...*(X-a_(i-1))(X-a_(i+1))*...*(X-a_n).", "Solution_27": "Was it ODE using Lie groups or something? Because it's pretty strange that you learn about ODE's \"again\"? So I assume it has to be some other way, like Lie groups.", "Solution_28": "The solution to the determinant:\r\n\r\n[url]http://www.blacksun.de.tc/math/determinant.pdf[/url]\r\n\r\nEnjoy!", "Solution_29": "Here is a texed web version of the determinant:\r\n\r\n[url]http://www.sosmath.com/matrix/determ3/determ3.html[/url]" } { "Tag": [ "probability" ], "Problem": "Jack is walking around in Luna Park, and notices an alleyway called 'Infinite Ice Cream'. Jack notes that the 'Infinite Ice-Cream' appears to possess an infinitely large number of people selling ice-cream. Upon walking outside any particular shop, Jack feels a huge compulsion to purchase an ice-cream. For every shop that Jack visits, he is 37% less likely to purchase an ice-cream then at the previous shop. After purchasing an ice-cream, Jack leaves Luna Park. What is the probability of Jack purchasing an ice-cream at the second shop he visits?\r\n\r\nThanks guys !", "Solution_1": "This depends on how much the probability is that he will buy an ice cream :icecream: at the first store and how the probability goes down (linear, exponential, etc.)", "Solution_2": "ohhh sorry i left out that accidently lol i have edited my post, it is 37%", "Solution_3": "hmm quite tricky" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "I need a sanity check, please.\r\n\r\nI'm testing this for uniform convergence (using Weierstrass M-Test):\r\n\r\n$\\sum_{n=1}^{\\infty} \\frac{n}{n+1} x^n$\r\n\r\nIt looks like it would converge for $0 \\epsilon > 0$ \r\nprove that there are infinitely many integers $n$ for wich $\\cos n \\geq 1 - \\epsilon$", "Solution_1": "A well known theorem ( :D ) says that every $x \\in [0;1]$ is 'cluster point' of the series $\\{an\\}$, where a is irrational. Since $\\pi$ is irrational, you find by this that $0$ is cluster point of the series $n \\mod \\pi$, which gives immediatly the result.", "Solution_2": "Yes the density of {$a_n$} follows from Weyl's criterion.", "Solution_3": "and $cos(n^2) > 1-\\eps$ ?", "Solution_4": "This is Weyl theorem. It works for $ cos(P(n))$ with $P$ having at least one coefficient incomensurable with $\\pi$, I think." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": "Let $ \\mathbb{K}$ be a field, $ \\Phi : \\mathrm{GL}_n(\\mathbb{K}) \\to \\mathbb{K}^*$ a morphism.\r\nSuppose that for $ M\\equal{}(m_{i,j})$, $ \\Phi(M) \\in \\mathbb{K}(m_{1,1,}, \\ldots, m_{n,n})$.\r\nFind $ \\Phi$.\r\n\r\nP.S.: It might (should) be in this forum already, but I couldn't find it.", "Solution_1": "The transvections $ T_{ij}=I+E{ij}$, where, if $ i\\not=j$, $ E_{ij}$ is the elementary matrix with the $ {i,j}$ element is $ 1$ and the other $ 0$, span $ SL_n(\\mathbb{K})$.\r\n$ T_{ij}T_{jk}T_{ij}^{-1}T_{jk}^{-1}=T_{ik}$ if $ i\\not=k$. Then $ \\phi(T_{ik})=1$ and $ \\phi=1$ on $ SL_n(\\mathbb{K})$.\r\nNow $ GL_n(\\mathbb{K})$ is the semi direct product of $ SL_n(\\mathbb{K})$ and ${ \\mathbb{K}}^*$ with the product $ (n,k)(n',k')=(n\\bar{k}n'\\bar{k}^{-1},kk')$ where $ \\bar{k}=diag(k,1,\\cdots,1)$. $ \\phi((n^{-1},1)(n,k))=\\phi(1,k)=\\phi(n^{-1},1)\\phi(n,k)=\\phi(n,k)$ and $ \\phi(n,k)=\\phi(\\bar{k})\\in{\\mathbb{K}(k)}$.\r\nWe assume that $ \\mathbb{K}(k)$ is the field of fractions.\r\nThere exists a morphism $ \\psi: \\mathbb{K}^*\\rightarrow{\\mathbb{K}}^*$ s.t. $ \\phi(\\bar{k})=\\psi(k)$. Here $ \\psi$ is a rational function. Then there exists $ \\alpha\\in{\\mathbb{Z}}$ s.t. $ \\psi(k)=k^{\\alpha}$.\r\nConclusion: $ \\phi(A)=(det(A))^\\alpha$.", "Solution_2": "Proof of $ \\psi(k)\\equal{}k^\\alpha$.\r\nCase 1: $ K$ is an infinite field:\r\nThere exists $ \\alpha \\in\\mathbb{Z}$ s.t. $ \\phi(k)\\equal{}k^\\alpha{P(k)/Q(k)}$ where $ P,Q$ are polynomials s.t. $ P(1)\\equal{}Q(1)\\equal{}1,P(0)\\not\\equal{}0,Q(0)\\not\\equal{}0$ and $ gcd(P,Q)\\equal{}1$.\r\n$ \\phi(kk')\\equal{}\\phi(k)\\phi(k')$ gives $ P(kk')Q(k)Q(k')\\equal{}P(k)P(k')Q(kk')$ for all $ k,k'\\in\\mathbb{K}^*$ i.e. for an infinity of values of $ k,k'$. Then this relation is also satisfied when $ k\\equal{}0$: $ P(0)Q(0)Q(k')\\equal{}P(0)P(k')Q(0)$; thus $ P(k')\\equal{}Q(k')$ for all $ k'\\not\\equal{}0$ and $ \\psi(k)\\equal{}k^\\alpha$.\r\nCase 2: $ K$ is a finite field:\r\nThen the supplementary hypothesis upon $ \\phi$ is useless. $ \\mathbb{K}^*\\equal{}$ (cyclic group). Let $ \\psi(a)\\equal{}a^\\alpha$; then $ \\psi(a^u)\\equal{}\\psi(a)^u\\equal{}(a^u)^\\alpha$ and $ \\psi(k)\\equal{}k^\\alpha$." } { "Tag": [ "geometry", "incenter", "analytic geometry" ], "Problem": "Hey I was working on some problems which I don't have the answers and I would like to check if I got them right (which I don't think so, they seem strange)\r\n\r\n[hide=\"Problem 1\"]In triangle $ ABC$ with $ AB\\equal{}c,AC\\equal{}b,BC\\equal{}a$ , find the distance between the baricenter and side $ \\overline{BC}$\n\nAnswer I found: $ d\\equal{}\\frac{\\sqrt{4a^2c^2\\minus{}b^2\\plus{}a^2\\plus{}c^2}}{4a}$[/hide]\n\n[hide=\"Problem 2\"]Find the relation between cevians $ x$ and $ y$ of a right triangle with hypothenuse equal to $ 9$ knowing that the feet of the cevians divide the hypothenuse in three equal parts.\n\nI found $ 4y^2\\minus{}x^2\\equal{}135$[/hide]\n\n[hide=\"Problem 3\"]In a right isosceles triangle with leg $ a$ find the distance between incenter $ I$ and baricenter $ G$\n\nI found $ IG\\equal{}a\\left(\\frac{6\\minus{}5\\sqrt{2}}{3}\\right)$[/hide]", "Solution_1": "[quote=\"triplebig\"]Hey I was working on some problems which I don't have the answers and I would like to check if I got them right (which I don't think so, they seem strange)\n\n[hide=\"Problem 1\"]In triangle $ ABC$ with $ AB \\equal{} c,AC \\equal{} b,BC \\equal{} a$ , find the distance between the baricenter and side $ \\overline{BC}$\n\nAnswer I found: $ d \\equal{} \\frac {\\sqrt {4a^2c^2 \\minus{} b^2 \\plus{} a^2 \\plus{} c^2}}{4a}$[/hide]\n\n[hide=\"Problem 2\"]Find the relation between cevians $ x$ and $ y$ of a right triangle with hypothenuse equal to $ 9$ knowing that the feet of the cevians divide the hypothenuse in three equal parts.\n\nI found $ 4y^2 \\minus{} x^2 \\equal{} 135$[/hide]\n\n[hide=\"Problem 3\"]In a right isosceles triangle with leg $ a$ find the distance between incenter $ I$ and baricenter $ G$\n\nI found $ IG \\equal{} a\\left(\\frac {6 \\minus{} 5\\sqrt {2}}{3}\\right)$[/hide][/quote]\r\n\r\n i think that for the problem 1 coordinate geometry will do the trick.", "Solution_2": "Is it wrong? And the other ones?" } { "Tag": [ "integration", "calculus", "algebra", "function", "domain", "probability and stats" ], "Problem": "I'm not sure what is going on in this class so I would appreciate it if someone would explain how to do this to me...\r\n\r\n1. Let the random variables X and Y have the joint pmf\r\na) p(x,y) = 1/3, (x,y) = (0,0), (1,1), (2,2), zero elsewhere\r\nb) p(x,y) = 1/3, (x,y) = (0,2), (1,1), (2,0), zero elsewhere\r\nc) p(x,y) = 1/3, (x,y) = (0,0), (1,1), (2,0), zero elsewhere\r\n\r\n2. Let f(x,y) = 2, 00$.\r\n\r\nIf you want to know why it all should be done this way, ask your teacher first. If you still have some trouble after that, post your questions here :)." } { "Tag": [ "trigonometry", "algebra", "polynomial", "function", "geometry", "LaTeX", "cyclic quadrilateral" ], "Problem": "Can someone explain to me the following quote and derviations of the \"18n trig family\", very slowly and in simple words? :wink: \r\n[quote=\"Mildorf\"]The quickest way I know to get into the \"18n trig family\" is to use Ptolemy on a regular pentagon. You come up with $\\cos{108^{\\circ}}= \\frac{1-\\sqrt{5}}{4}$. Then the complete set $\\frac{\\pm 1+\\sqrt{5}}{4}, \\sqrt{\\frac{5\\pm\\sqrt{5}}{8}}$ of sines and cosines follows easily. If you don't memorize exact values, at least try to become familiar with some derivation.[/quote]\n\nAlso, what is the solution to the following related problem?\n[quote] Using the roots of $x^{4}+x^{3}+x^{2}+x+1$ find $\\sin 18^\\circ$ and $\\cos 18^\\circ.$[/quote]", "Solution_1": "Maybe this will help: http://www.math.fau.edu/yiu/EuclideanGeometryNotes.pdf\r\n\r\nLook at page $11$", "Solution_2": "Thank you! Do you have a solution for that polynomial problem?\r\nAlso, how do you show (ii) in the the third exercise?", "Solution_3": "$x^{4}+x^{3}+x^{2}+x+1=0$, $x \\neq 0$ so:\r\n\r\n$x^{2}+x+1+\\frac{1}{x}+\\frac{1}{x^{2}}=0$\r\n\r\n$(x+\\frac{1}{x})^{2}+(x+\\frac{1}{x})-1=0$\r\n\r\n$x+\\frac{1}{x}=\\frac{-1+\\sqrt{5}}{2}$ or $\\frac{-1-\\sqrt{5}}{2}$, i think you can continue :P", "Solution_4": "In the exercises, I can't find the segment lengths, but I know how to continue from there to find the trig values, I think. Could someone explain this?\r\n\r\n(page 11 of the packet Mario linked to)", "Solution_5": "How do I find all of the 18n values or each of the three main functions? Or are there only a few that can be found easily? Because Mildorf implies that there are more...", "Solution_6": "[color=darkblue]Let $ABC$ be an isosceles triangle with $AB=AC=1$ and $A=36^{\\circ}\\ .$ Denote the point $M\\in (AC)$ so that $BM=BC=x\\ .$ Thus, $AM=BM=BC=x$ and $CM=1-x\\ .$ From the similarity $\\triangle ABC\\sim\\triangle BCM$ obtain $\\frac{1}{x}=\\frac{x}{1-x}\\ ,$ i.e. $x^{2}+x-1=0$ $\\Longrightarrow$ $x=\\frac{-1+\\sqrt 5}{2}\\ .$ Denote the projection $N$ of the point $M$ to the line $AB$ (the middlepoint of the side $[AB]$). Then $\\cos 36^{\\circ}=\\frac{AN}{AM}=$ $\\frac{1}{2x}=\\frac{1}{-1+\\sqrt 5}=$ $\\frac{1+\\sqrt 5}{4}\\ ,$ i.e. $\\boxed{\\ \\cos 36^{\\circ}=\\frac{1+\\sqrt 5}{4}\\ }\\ .$\n\n[b]Remark 1.[/b] $\\|\\begin{array}{c}2\\sin^{2}18^{\\circ}=1-\\cos 36^{\\circ}\\Longrightarrow\\sin 18^{\\circ}=\\frac{-1+\\sqrt 5}{4}\\\\\\\\ 2\\cos^{2}18^{\\cic}=1+\\cos 36^{\\circ}\\Longrightarrow \\cos 18^{\\circ}=\\frac{\\sqrt{10+2\\sqrt 5}}{4}}\\end{array}$\n\n[b]Remark 2.[/b] Otherwise, $x=36^{\\circ}\\ .$ $\\Longrightarrow$ $\\|\\begin{array}{c}3x+2x=\\pi\\\\\\ \\sin 3x=\\sin 2x\\end{array}$ $\\Longrightarrow$ $\\sin x(3-4\\sin^{2}x)=2\\sin x\\cos x$ $\\Longrightarrow$ $3-4\\sin^{2}x=2\\cos x$ $\\Longrightarrow$ $4\\cos^{2}x-2\\cos x-1=0$ $\\Longrightarrow$ $\\cos x=\\frac{1+\\sqrt 5}{4}\\ .$[/color]", "Solution_7": "Thank you! How do you find the cosine and sine values for 18, 54, 72, and 108 geometrically? I'm on a computer without a pdf reader, but doesn't the packet have a derivation of the 108 or 18 ones? If so, could a person expliciting type that up on the wiki so I could see it?\r\n\r\nEDIT: I did not mean through the regular method (addition formula), I ment geometrically, though now I see the other derivation is nothing special and I rescind my question.", "Solution_8": "[color=darkblue][b][u]A geometrical proof of the relation[/u][/b] $\\boxed{\\ \\cos 2x=2\\cos^{2}x-1\\ }\\ .$ Let $ABC$ be an $A$-isosceles triangle and denote $m(\\widehat{BAC})=2x<90^{\\circ}\\ .$ Define the middlepoint $D$ of the side $[BC]$ and the projection $E$ of the point $B$ on the line $AC\\ .$ Observe that $\\cos 2x=\\frac{AE}{AB}$ , $\\cos x=\\frac{AD}{AB}$ , the quadrilateral $ABDE$ is cyclically , $BD=DE$ and $BE=2\\cdot DE\\cdot \\cos x\\ .$ Apply the [u]Ptolemeu's theorem[/u] in the cyclic quadrilateral $ABDE\\ : \\ AB\\cdot DE+BD\\cdot AE=BE\\cdot AD$ $\\Longleftrightarrow$ $DE+BD\\cdot\\cos 2x=BE\\cdot \\cos x$ $\\Longleftrightarrow$ $DE+DE\\cdot \\cos 2x=2\\cdot DE\\cdot\\cos^{2}x$ $\\Longleftrightarrow$ $2\\cos^{2}x=1+\\cos 2x\\ .$\n\n[b][u]A geometrical proof of the relation[/u][/b] $\\boxed{\\ \\sin (x+y)=\\sin x\\cos y+\\sin y\\cos x\\ }\\ ,$ where $\\{x,y\\}\\subset (0,90^{\\circ})\\ .$ Let $ABCD$ be a quadrilateral inscribed in the circle with the diameter $[AC]$ and $m(\\widehat{BAC})=x$ , $m(\\widehat{DAC})=y\\ .$ Prove easily that $\\|\\begin{array}{c}AB=AC\\cdot\\cos x\\ ,\\ BC=AC\\cdot\\sin x\\\\\\ AD=AC\\cdot\\cos y\\ ,\\ DC=AC\\cdot\\sin y\\\\\\ BD=AC\\cdot\\sin (x+y)\\end{array}\\|\\ .$ Apply the [u]Ptolemeu's theorem[/u] for this quadrilateral : $AB\\cdot CD+AD\\cdot BC=AC\\cdot BD$ $\\Longleftrightarrow$ $AC^{2}\\cdot\\cos x\\cdot\\sin y+AC^{2}\\cdot \\cos y\\cdot\\sin x=AC^{2}\\cdot\\sin (x+y)$ $\\Longleftrightarrow$ $\\sin (x+y)=\\sin x\\cdot\\cos y+\\sin y\\cdot\\cos x\\ .$\n\n[b]Remark 1.[/b] $18^{\\circ}=\\frac{1}{2}\\cdot 36^{\\circ}\\ ;\\ 54^{\\circ}=18^{\\circ}+36^{\\circ}\\ ;\\ 72^{\\circ}=2\\cdot 36^{\\circ}=18^{\\circ}+54^{\\circ}$ a.s.o.\n\n[b]Remark 2.[/b] $\\boxed{\\ \\|\\begin{array}{c}x>0\\ ,\\ y>0\\\\\\ x+y=180^{\\circ}\\end{array}\\Longrightarrow\\|\\begin{array}{c}\\sin x=\\sin y\\\\\\ \\cos x=-\\cos y\\end{array}\\ }$[/color]", "Solution_9": "[quote=\"Virgil Nicula\"][color=darkblue][b][u]A geometrical proof of the relation[/u][/b] $\\boxed{\\ \\cos 2x=2\\cos^{2}x-1\\ }\\ .$ Let $ABC$ be an $A$-isosceles triangle and denote $m(\\widehat{BAC})=2x<90^{\\circ}\\ .$ Define the middlepoint $D$ of the side $[BC]$ and the projection $E$ of the point $B$ on the line $AC\\ .$ Observe that $\\cos 2x=\\frac{AE}{AB}$ , $\\cos x=\\frac{AD}{AB}$ , the quadrilateral $ABDE$ is cyclically , $BD=DE$ and $BE=2\\cdot DE\\cdot \\cos x\\ .$ Apply the [u]Ptolemeu's theorem[/u] in the cyclic quadrilateral $ABDE\\ : \\ AB\\cdot DE+BD\\cdot AE=BE\\cdot AD$ $\\Longleftrightarrow$ $DE+BD\\cdot\\cos 2x=BE\\cdot \\cos x$ $\\Longleftrightarrow$ $DE+DE\\cdot \\cos 2x=2\\cdot DE\\cdot\\cos^{2}x$ $\\Longleftrightarrow$ $2\\cos^{2}x=1+\\cos 2x\\ .$\n\n[b][u]A geometrical proof of the relation[/u][/b] $\\boxed{\\ \\sin (x+y)=\\sin x\\cos y+\\sin y\\cos x\\ }\\ ,$ where $\\{x,y\\}\\subset (0,90^{\\circ})\\ .$ Let $ABCD$ be a quadrilateral inscribed in the circle with the diameter $[AC]$ and $m(\\widehat{BAC})=x$ , $m(\\widehat{DAC})=y\\ .$ Prove easily that $\\|\\begin{array}{c}AB=AC\\cdot\\cos x\\ ,\\ BC=AC\\cdot\\sin x\\\\\\ AD=AC\\cdot\\cos y\\ ,\\ DC=AC\\cdot\\sin y\\\\\\ BD=AC\\cdot\\sin (x+y)\\end{array}\\|\\ .$ Apply the [u]Ptolemeu's theorem[/u] for this quadrilateral : $AB\\cdot CD+AD\\cdot BC=AC\\cdot BD$ $\\Longleftrightarrow$ $AC^{2}\\cdot\\cos x\\cdot\\sin y+AC^{2}\\cdot \\cos y\\cdot\\sin x=AC^{2}\\cdot\\sin (x+y)$ $\\Longleftrightarrow$ $\\sin (x+y)=\\sin x\\cdot\\cos y+\\sin y\\cdot\\cos x\\ .$\n\n[b]Remark 1.[/b] $18^{\\circ}=\\frac{1}{2}\\cdot 36^{\\circ}\\ ;\\ 54^{\\circ}=18^{\\circ}+36^{\\circ}\\ ;\\ 72^{\\circ}=2\\cdot 36^{\\circ}=18^{\\circ}+54^{\\circ}$ a.s.o.\n\n[b]Remark 2.[/b] $\\boxed{\\ \\|\\begin{array}{c}x>0\\ ,\\ y>0\\\\\\ x+y=180^{\\circ}\\end{array}\\Longrightarrow\\|\\begin{array}{c}\\sin x=\\sin y\\\\\\ \\cos x=-\\cos y\\end{array}\\ }$[/color][/quote]\r\n\r\nI should have remembered that those can add...\r\n\r\nFunny though, I remembered it for the sin 15 derivations in the packet to get sin 30, 45, 105, etc. Their method with the polygon is a good way to memorize those quickly for a contest if you know the addition formula.\r\n\r\nAnyway, I find the demonstration in AoPS 2 of the addition formula to be much clearer and elementary, also the first relation can be easily derived from the second given sinx squared plus cos x squared equals one.\r\n\r\nLet ABC be a general triangle. Turn it so BC is the base and draw a perpendicular to BC at H. Set AH=1 and HAB=a, HAC=b. Area ABC=X.\r\n\r\nX=1/2*AC*AB*sinBAC=1/2*1/cosb*1/cosa*sinBAC.\r\nX=Area ABH plus Area AHC=1/2*AH*BH+1/2*AH*CH=1/2(tan a + tan b)\r\nSince sinBAC=sin(a+b), sin(a+b)/cosa*cosb=sina/cosa+sinb/cosb and multiply by cosacosb to get the desired relation. Quick and easy to memorize.", "Solution_10": "Quick unrelated question: How do you put something in a box as Virigil has done using LaTex?", "Solution_11": "[quote=\"mna851\"]Quick unrelated question: How do you put something in a box as Virigil has done using LaTex?[/quote]\r\n\r\nYou can find out yourself by hovering your cursor over his LaTeX or quoting his text (you don't have to post it, just click quote and look at it, of course) :D", "Solution_12": "How do we get BC=AC*sin(x+y) is the cyclic quadrilateral?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find a positive integer $n$ with five non-zero different digits, which satisfies to be equal to the sum of all the three-digit numbers that can be formed using the digits of $n$.", "Solution_1": "If the digits are all different, then the sum of all possible three-digit numbers is\r\n\r\n$n = 10000a+1000b+100c+10d+e = 12 \\cdot 111(a+b+c+d+e)$\r\n\r\nNote that $\\bmod 9$ we conclude that $9 | a+b+c+d+e$. Let $a+b+c+d+e = 9d$; then\r\n\r\n$n = 11988d$\r\n\r\nWhen $d = 3$ we have $n = \\boxed{ 35964}$, with digit sum $27 = 9d$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "MATHCOUNTS", "Alcumus", "geometry" ], "Problem": "Hello!!!\r\n\r\nI am your host, Dojo, in Survivor.\r\n\r\nSign up now and I will put you into 3 (or 4) different tribes.\r\nEach tribe will be given a set of questions. I will grade the questions and/or proofs. \r\nThe tribe with the least score will have to vote off one unproductive player.\r\n\r\nYour goal is to be the last person standing and to have your team win.\r\n\r\nRules:\r\n\r\n1. You must not use any resources other than each other.\r\n2. You may talk to other team mates though (not restricted to) PM, forum, FTW.\r\n3. You must select only [b]one[/b] team member to submit your answers.\r\n4. You must turn in your answers by the due date.\r\n5. You may not help team members even if you have been voted off.\r\n\r\nRemember, have fun!!!\r\n\r\nI need 12 or 16 people, depending on turnout by a certain date.", "Solution_1": "Do we PM you to join, or just post on this topic?\r\n\r\nEither way, 007math joins Survivor...AoPS Style!", "Solution_2": "I phailed usamo so i'll do this.", "Solution_3": "join. ", "Solution_4": "I'm in.", "Solution_5": "JOIN JOIN JOIN!\r\n\r\nWait, you put us into tribes? Use random.org then.\r\n\r\nBTW, because this is the MC forum, I don't think there should be proofs :wink:", "Solution_6": "I'm in.\r\n\r\nThe message is too small. Please make the message longer before submitting.", "Solution_7": "Join. It would be good if we could pick our own tribes.", "Solution_8": "@AIME_is_hard: No I don't want any \"stacked teams\".\r\n\r\n@AIME15: It's okay if there are no proofs on MATHCOUNTS. This forum is Middle School Tournaments not MATHCOUNTS. :wink: \r\n\r\n[u][i][b]Sign ups!!![/b][/i][/u]\r\n1. 007math\r\n2. iphailedusamo\r\n3. BOGTRO\r\n4. isabella2296\r\n5. AIME15\r\n6. high_flyer10\r\n7. AIME_is_hard\r\n\r\nWe need 5 more.\r\n\r\nI have finished writing the tests, and need to type them out.\r\n\r\nThe test will consist of 25 normal answer response questions: No explanation will be needed.\r\nThere will be one proof question. This is worth 5 points.\r\n\r\nIn the case of a tie, the team with more of the challenge questions (last five) correct will win.", "Solution_9": "Dojo, it's a subforum of the MathCounts forum. Look at some other tourneys; mods have given warnings about level.", "Solution_10": "All the problems should be solvable. If I can do them, you probably can too. :P", "Solution_11": "Just don't want to get this cancelled...\r\n\r\nAs long as all the problems can be solved using mc techniques we'll be fine.", "Solution_12": "All righty got all the questions all nicely typed up in LaTeX.\r\n\r\nJust get more peeps!!!", "Solution_13": "i'll join \r\n\r\ni passed u up in alcumus points btw dojo :P", "Solution_14": "[u][i][b]Sign ups!!![/b][/i][/u]\r\n1. 007math\r\n2. iphailedusamo\r\n3. BOGTRO\r\n4. isabella2296\r\n5. AIME15\r\n6. high_flyer10\r\n7. AIME_is_hard\r\n8. theprodigy\r\n\r\nNot definite on how many more people, possibly 1 or 4, we will see by tomorrow.", "Solution_15": "Oh I forgot to say I won.\r\n\r\n[b] BOGTRO vs. nikeballa96:\n\nBOGTRO wins! 3 games to 2 [/b]", "Solution_16": "[quote=\"BOGTRO\"]Oh I forgot to say I won.\n\n[b] BOGTRO vs. nikeballa96:\n\nBOGTRO wins! 3 games to 2 [/b][/quote]\r\n\r\nHm...doesn't nikebella have to state this is true or false now?", "Solution_17": "W/E, but i don't htink any1 was watching", "Solution_18": "What? No I won. Ask nike.", "Solution_19": "nike... where are you? Thi needs to get clarified.", "Solution_20": "Okay get it done.\r\n\r\n[b]nikeballa96[/b] please choose a challanger.\r\n\r\nBTW: You will have to play against everyone else.", "Solution_21": "YO NIKE!!!\r\n\r\nPick your challanger.", "Solution_22": "The FTW idea for Round 2 was a very bad idea. I think you should just make a test...by the speed this is going, tis season will be dead...", "Solution_23": "Okay fine. Will post questions soon.", "Solution_24": "In genetics, a Punnett Square is used determine the probabilites of a certain trait in an organism.\r\n\r\nThere are two basic types of inheritance that can happen within punnett squares. \r\na) Mendelian \r\nb) Incomplete Dominance\r\n\r\nMendelian Inheritance is where if there are two different traits, one dominates over the other.\r\nMendel's famous pea plant experiment showed just that.\r\nFor example, there were two purebred plants. One with green pods and one with yellow pods.\r\nThose two plants were then crossbred.\r\n\r\n(Y=yellow, y=green)\r\nThe yellow parent is represented by YY while the green is represented by yy.\r\n\r\n[asy]draw((0,0)--(0,2));\ndraw((1,0)--(1,2));\ndraw((2,0)--(2,2));\ndraw((0,0)--(2,0));\ndraw((0,1)--(2,1));\ndraw((0,2)--(2,2));\nlabel(\"$Y$\",(0.5,2),N);\nlabel(\"$Y$\",(1.5,2),N);\nlabel(\"$y$\",(0,0.5),W);\nlabel(\"$y$\",(0,1.5),W);[/asy]\r\n\r\nWhen the plants were mixed the resulting genotypes (or genetic types) were Yy, Yy, Yy, and Yy.\r\n\r\n[asy]draw((0,0)--(0,2));\ndraw((1,0)--(1,2));\ndraw((2,0)--(2,2));\ndraw((0,0)--(2,0));\ndraw((0,1)--(2,1));\ndraw((0,2)--(2,2));\nlabel(\"$Y$\",(0.5,2),N);\nlabel(\"$Y$\",(1.5,2),N);\nlabel(\"$y$\",(0,0.5),W);\nlabel(\"$y$\",(0,1.5),W);\nlabel(\"$Yy$\",(0.5,0.5),N);\nlabel(\"$Yy$\",(0.5,1.5),N);\nlabel(\"$Yy$\",(1.5,0.5),N);\nlabel(\"$Yy$\",(1.5,1.5),N);[/asy]\r\n\r\nAll the plants now had the yellow and the green trait. However, the yellow trait dominated the green trait resulting in all yellow pods. (Which is why it is represented by a capital.)\r\n\r\nb) \r\nLater when scientists discovered Mendel's work, they tried to put it to work. Using flowers that were red and white, they expected either red or white flowers:\r\n\r\n[asy]draw((0,0)--(0,2));\ndraw((1,0)--(1,2));\ndraw((2,0)--(2,2));\ndraw((0,0)--(2,0));\ndraw((0,1)--(2,1));\ndraw((0,2)--(2,2));\nlabel(\"$R$\",(0.5,2),N);\nlabel(\"$R$\",(1.5,2),N);\nlabel(\"$W$\",(0,0.5),W);\nlabel(\"$W$\",(0,1.5),W);[/asy]\r\n\r\nHowever, the got:\r\n\r\n[asy]draw((0,0)--(0,2));\ndraw((1,0)--(1,2));\ndraw((2,0)--(2,2));\ndraw((0,0)--(2,0));\ndraw((0,1)--(2,1));\ndraw((0,2)--(2,2));\nlabel(\"$R$\",(0.5,2),N);\nlabel(\"$R$\",(1.5,2),N);\nlabel(\"$W$\",(0,0.5),W);\nlabel(\"$W$\",(0,1.5),W);\nlabel(\"$RW$\",(0.5,0.5),N);\nlabel(\"$RW$\",(0.5,1.5),N);\nlabel(\"$RW$\",(1.5,0.5),N);\nlabel(\"$RW$\",(1.5,1.5),N);[/asy]\r\n\r\nWhich resulted in all pink flowers. (Which is why there were R and W.)\r\n\r\nNow for the questions:\r\n\r\n1. Assume that there are only two colors in pea plants, green and yellow.\r\n\r\na) If we have a purebred yellow plant (YY) crossbred with a mixed yellow plant (Yy), what are the resulting genotypes and phenotypes, with probabilities (phenotypes are the final result you can see, such as yellow)?\r\n\r\n(List all possible types, even if they don't show up).\r\n\r\nb) If we have a mixed yellow (Yy) and another mixed yellow plant, what are the resulting genotypes and phenotypes, with probabilities?\r\n\r\n(List all possible types, even if they don't show up).\r\n\r\nc) If you have two green pods, what is the probability that the offspring will be yellow? Green?\r\n\r\n2. \r\na) If we have a red flower and a white flower that are crossbred, what are the resulting genotypes and phenotypes, with probabilities?\r\n\r\n(List all possible types, even if they don't show up).\r\n\r\nb) Using the offspring from above, we crossbred 2 of them. What are the resulting genotypes and phenotypes, with probabilities?\r\n\r\n(List all possible types, even if they don't show up).\r\n\r\n3.\r\nBlood types are a little different then the above. There are three different traits (or alelles).\r\nA, B and O. O is the only recessive (or non-dominant) trait and when A and B are put together, they result in incomplete dominance.\r\n\r\na) If you have a parent that is mixed A and a parent that is full A, what are the resulting genotypes and phenotypes, with probabilities? \r\n\r\n(List all possible types, even if they don't show up).\r\n\r\nb) If you have a parent that is mixed B and a parent that is AB, what are the resulting genotypes and phenotypes, with probabilities? \r\n\r\n(List all possible types, even if they don't show up).\r\n\r\nc) If you have a parent that is mixed B and a parent that is mixedl A, what are the resulting genotypes and phenotypes, with probabilities? \r\n\r\n(List all possible types, even if they don't show up).\r\n\r\n4)\r\nLet there be a planet with flowers that have 3 parents. The traits carried by them are T which is T and short, which is t. If three parents are, respectively, TTT, TTT and TTt, what are the resulting genotypes and phenotypes, with probabilities? \r\n\r\n\r\nTable:\r\n\r\nTTT: Tall\r\nTTt: Medium Tall\r\nTtT: Medium Tall\r\ntTT: Medium Tall\r\nTtt: Medium Short\r\nttT: Medium Short\r\nttt: Short\r\n\r\n(List all possible types, even if they don't show up).", "Solution_25": "As always, pm me the answers.\r\nThese are due Jan. 14th.", "Solution_26": "lol..did you make these or are they homework....(you probably made them,i think)", "Solution_27": "[quote=\"Dojo\"]In genetics, a Punnett Square is used determine the probabilites of a certain trait in an organism.\n\nThere are two basic types of inheritance that can happen within punnett squares. \na) Mendelian \nb) Incomplete Dominance\n\nMendelian Inheritance is where if there are two different traits, one dominates over the other.\nMendel's famous pea plant experiment showed just that.\nFor example, there were two purebred plants. One with green pods and one with yellow pods.\nThose two plants were then crossbred.\n\n(Y=yellow, y=green)\nThe yellow parent is represented by YY while the green is represented by yy.\n\n[asy]draw((0,0)--(0,2));\ndraw((1,0)--(1,2));\ndraw((2,0)--(2,2));\ndraw((0,0)--(2,0));\ndraw((0,1)--(2,1));\ndraw((0,2)--(2,2));\nlabel(\"$Y$\",(0.5,2),N);\nlabel(\"$Y$\",(1.5,2),N);\nlabel(\"$y$\",(0,0.5),W);\nlabel(\"$y$\",(0,1.5),W);[/asy]\n\nWhen the plants were mixed the resulting genotypes (or genetic types) were Yy, Yy, Yy, and Yy.\n\n[asy]draw((0,0)--(0,2));\ndraw((1,0)--(1,2));\ndraw((2,0)--(2,2));\ndraw((0,0)--(2,0));\ndraw((0,1)--(2,1));\ndraw((0,2)--(2,2));\nlabel(\"$Y$\",(0.5,2),N);\nlabel(\"$Y$\",(1.5,2),N);\nlabel(\"$y$\",(0,0.5),W);\nlabel(\"$y$\",(0,1.5),W);\nlabel(\"$Yy$\",(0.5,0.5),N);\nlabel(\"$Yy$\",(0.5,1.5),N);\nlabel(\"$Yy$\",(1.5,0.5),N);\nlabel(\"$Yy$\",(1.5,1.5),N);[/asy]\n\nAll the plants now had the yellow and the green trait. However, the yellow trait dominated the green trait resulting in all yellow pods. (Which is why it is represented by a capital.)\n\nb) \nLater when scientists discovered Mendel's work, they tried to put it to work. Using flowers that were red and white, they expected either red or white flowers:\n\n[asy]draw((0,0)--(0,2));\ndraw((1,0)--(1,2));\ndraw((2,0)--(2,2));\ndraw((0,0)--(2,0));\ndraw((0,1)--(2,1));\ndraw((0,2)--(2,2));\nlabel(\"$R$\",(0.5,2),N);\nlabel(\"$R$\",(1.5,2),N);\nlabel(\"$W$\",(0,0.5),W);\nlabel(\"$W$\",(0,1.5),W);[/asy]\n\nHowever, the got:\n\n[asy]draw((0,0)--(0,2));\ndraw((1,0)--(1,2));\ndraw((2,0)--(2,2));\ndraw((0,0)--(2,0));\ndraw((0,1)--(2,1));\ndraw((0,2)--(2,2));\nlabel(\"$R$\",(0.5,2),N);\nlabel(\"$R$\",(1.5,2),N);\nlabel(\"$W$\",(0,0.5),W);\nlabel(\"$W$\",(0,1.5),W);\nlabel(\"$RW$\",(0.5,0.5),N);\nlabel(\"$RW$\",(0.5,1.5),N);\nlabel(\"$RW$\",(1.5,0.5),N);\nlabel(\"$RW$\",(1.5,1.5),N);[/asy]\n\nWhich resulted in all pink flowers. (Which is why there were R and W.)\n\nNow for the questions:\n\n1. Assume that there are only two colors in pea plants, green and yellow.\n\na) If we have a purebred yellow plant (YY) crossbred with a mixed yellow plant (Yy), what are the resulting genotypes and phenotypes, with probabilities (phenotypes are the final result you can see, such as yellow)?\n\n(List all possible types, even if they don't show up).\n\nb) If we have a mixed yellow (Yy) and another mixed yellow plant, what are the resulting genotypes and phenotypes, with probabilities?\n\n(List all possible types, even if they don't show up).\n\nc) If you have two green pods, what is the probability that the offspring will be yellow? Green?\n\n2. \na) If we have a red flower and a white flower that are crossbred, what are the resulting genotypes and phenotypes, with probabilities?\n\n(List all possible types, even if they don't show up).\n\nb) Using the offspring from above, we crossbred 2 of them. What are the resulting genotypes and phenotypes, with probabilities?\n\n(List all possible types, even if they don't show up).\n\n3.\nBlood types are a little different then the above. There are three different traits (or alelles).\nA, B and O. O is the only recessive (or non-dominant) trait and when A and B are put together, they result in incomplete dominance.\n\na) If you have a parent that is mixed A and a parent that is full A, what are the resulting genotypes and phenotypes, with probabilities? \n\n(List all possible types, even if they don't show up).\n\nb) If you have a parent that is mixed B and a parent that is AB, what are the resulting genotypes and phenotypes, with probabilities? \n\n(List all possible types, even if they don't show up).\n\nc) If you have a parent that is mixed B and a parent that is mixedl A, what are the resulting genotypes and phenotypes, with probabilities? \n\n(List all possible types, even if they don't show up).\n\n4)\nLet there be a planet with flowers that have 3 parents. The traits carried by them are T which is T and short, which is t. If three parents are, respectively, TTT, TTT and TTt, what are the resulting genotypes and phenotypes, with probabilities? \n\n\nTable:\n\nTTT: Tall\nTTt: Medium Tall\nTtT: Medium Tall\ntTT: Medium Tall\nTtt: Medium Short\nttT: Medium Short\nttt: Short\n\n(List all possible types, even if they don't show up).[/quote]\r\n\r\nFor me, this is 6th grade HW...", "Solution_28": "We did this is $ 7^{\\text{th}}$ grade.", "Solution_29": "We did this recently and I added a twist on to it.\r\n(See question 4)" } { "Tag": [], "Problem": "While waiting for the bishop to arrive at St Stephen's Anglican Church to lead a service, every person present greeted every other person with a hand shake. Arriving late, the bishop shook hands with only some of the people as he made his way in. Altogether 1933 handshakes had taken place. How many people shook hands with the bishop?", "Solution_1": "[hide=\"Solution\"]If every person shook hands with every other person, then $ \\binom{n}{2}$ handshakes took place. We are looking for a value of n which satisfies both:\n\n$ \\binom{n}{2} < 1933$\n\n$ \\binom{n}{2} \\plus{} n > 1933$\n\nThe only value of n which suits is 62.\n\nWe now solve $ \\binom{n}{2} \\plus{} x \\equal{} 1933$\n\n$ \\binom{62}{2} \\plus{} x \\equal{} 1933$\n\n$ 1891 \\plus{} x \\equal{} 1933$\n\n$ x \\equal{} 42$\n\nThus 42 people shook hands with the bishop.[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $a,b,c>0$ and $abc=1$ prove that \r\n\r\n$\\frac{1}{\\sqrt{b+\\frac{1}{a}+\\frac{1}{2}}}+\\frac{1}{\\sqrt{c+\\frac{1}{b}+\\frac{1}{2}}}+\\frac{1}{\\sqrt{a+\\frac{1}{c}+\\frac{1}{2}}}\\geq\\sqrt{2}$", "Solution_1": "[quote=\"silouan\"]If $a,b,c>0$ and $abc=1$ prove that \n$\\frac{1}{\\sqrt{b+\\frac{1}{a}+\\frac{1}{2}}}+\\frac{1}{\\sqrt{c+\\frac{1}{b}+\\frac{1}{2}}}+\\frac{1}{\\sqrt{a+\\frac{1}{c}+\\frac{1}{2}}}\\geq\\sqrt{2}$[/quote]\r\nWhat if $a=b=c=1$? ;)", "Solution_2": "For $a=b=c=1$ the ineq becomes $\\frac{3}{\\sqrt{5}}\\geq 1$ which is true.\r\n\r\nI think the ineq is true. Also it is from Serbia 2004 MO", "Solution_3": "[quote=\"silouan\"]For $a=b=c=1$ the ineq becomes $\\frac{3}{\\sqrt{5}}\\geq 1$ which is true.\n[/quote]\r\nOK. Sorry.", "Solution_4": "[quote=\"silouan\"]If $a,b,c>0$ and $abc=1$ prove that \n\n$\\frac{1}{\\sqrt{b+\\frac{1}{a}+\\frac{1}{2}}}+\\frac{1}{\\sqrt{c+\\frac{1}{b}+\\frac{1}{2}}}+\\frac{1}{\\sqrt{a+\\frac{1}{c}+\\frac{1}{2}}}\\geq\\sqrt{2}$[/quote]\r\n\r\nIsn;t this a classical one?? \r\n\r\nWe use $\\sqrt{2x+1}0$. Thus, $\\frac {XB}{XC}=\\frac cb\\cdot \\frac {s-b}{s-c}\\cdot \\frac{b-r_b}{c-r_c},\\ a.s.o.$\r\n\r\nBut I consider that the [b]Grobber's idea [/b]and the complete [b]Darij's solution [/b] of the initiated problem are the [u][b]BEST[/b][/u] !." } { "Tag": [ "integration", "geometry", "3D geometry", "sphere", "calculus", "trigonometry", "function" ], "Problem": "Let F be $ F \\equal{} ( x^2 z^2 ) i \\plus{} (sin xyz) j \\plus{} (e^x z) k.$$ Find \\int\\int \\nabla \\times F \\cdot n dS$\r\n\r\nwhere the region E is above the cone $ z^2 \\equal{} x^2 \\plus{} y^2$ and inside the sphere centered at (0,0,1) and with radius 1. (so it is $ x^2 \\plus{} y^2 \\plus{} (z\\minus{}1)^2 \\equal{} 1).$. I know that they intersect at z = 1, z = 0. So the boundary is at z = 1. \r\n\r\n===\r\nSo I know that I can apply Stokes Theorem here or evaluate the surface integral directly. \r\n\r\nIf I try Stokes Theorem, I apply it to the boundary $ x^2 \\plus{} y^2 \\plus{} (z\\minus{}1)^2 \\equal{} 1$, which can be parametrized by $ r(\\theta) \\equal{} cos(\\theta) i \\plus{} sin(\\theta) j \\plus{} k, r'(\\theta) \\equal{} \\minus{}sin(\\theta) i \\plus{} cos(\\theta) j \\plus{} 0 k$\r\n\r\nBut when I try $ F(r(\\theta)) \\cdot r'(\\theta)$, I get $ \\int_0^{2 \\pi} \\minus{}sin(\\theta) cos(\\theta)^2 \\plus{} cos(\\theta) sin( cos(\\theta) sin(\\theta)).$. Which is pretty much impossible to evaluate, since I never heard of integrating a sine function within a sine function\r\n\r\nas for trying to evaluate the surface integral directly... I get the cross product which is...\r\n\r\n$ \\nabla \\times F \\equal{} (\\minus{}x y cos(xyz))i \\plus{} (2x^2 z \\minus{} e^x z)j \\plus{} (y z cos(xyz) k).$\r\n\r\nSo I try to evaluate the surface integral through the northern hemisphere. Of $ x^2 \\plus{} y^2 \\plus{} (z\\minus{}1)^2 \\equal{} 1)$, I get...\r\n\r\n$ \\frac{\\partial g}{\\partial x} \\equal{} 2x, \\frac{\\partial g}{\\partial y} \\equal{} 2y$\r\n\r\nSo...\r\n\r\n$ P \\frac{\\partial g}{\\partial x}\\minus{} Q \\frac{\\partial g}{\\partial y} \\plus{} R$\r\n$ \\equal{} 2x xy cos(xyz) \\minus{} 2y 2x^2 z \\plus{} 2y e^y z \\plus{} yz cos(xyz)$... which is like totally impossible to evaluate with a surface integral.\r\n\r\nSo what am I doing wrong?", "Solution_1": "$ \\iint {\\nabla \\times \\overrightarrow F .ndS}$ mean? there is some misprints, isnt it?", "Solution_2": "You mean that $ \\iint {\\frac{\\partial }\r\n{{\\partial x}}\\left( {x^2 z^2 } \\right)dydz \\plus{} \\frac{\\partial }\r\n{{\\partial y}}\\left( {\\sin xyz} \\right)dzdx \\plus{} \\frac{\\partial }\r\n{{\\partial y}}\\left( {e^x z} \\right)dzdx}$?", "Solution_3": "No. The notation refers to the [url=http://en.wikipedia.org/wiki/Curl]curl[/url]." } { "Tag": [], "Problem": "Phenol + ${ \\ce{NaNO2}/\\ce{HCl}/\\ce{H2O}}$ -----> ?", "Solution_1": "para nitroso phenol? :maybe:", "Solution_2": "Plus the ortho isomer. Very good." } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Let $p,q,r$ are primes number. Find all $(p,q,r)$ triples make $p^{q}+p^{r}$ is a square.", "Solution_1": "$p^{q}+p^{r}= x^{2}$. wlog $q \\leq r$. Then $p^{q}(1+p^{r-q}) = x^{2}$. Since the two factors are coprime and $p$ is a prime, then $q = 2$ and it rewrites as $p^{2}(1+p^{r-2}) = x^{2}$. So $(1+p^{r-2}) = y^{2}$ and $p^{r-2}= (y-1)(y+1)$. If $p$ is odd it's clear the only solutions are $(3, 2, 3)$ and $(3, 3, 2)$.\r\nIf $p = 2$ then $(2, 2, 5)$ and $(2, 5, 2)$ are the only solutions.", "Solution_2": "1)$r=q$ then we have $2p^{q}=x^{2}\\longrightarrow x=2k \\longrightarrow p^{q}=2k^{2}\\longrightarrow p=2 \\longrightarrow 2^{q}=2k^{2}\\longrightarrow 2^{q-1}=k^{2}$ $k=2^{\\frac{q-1}{2}}$ , in this case we have $(p,q,r)=(2,q,q)$ where $q$ is odd prime number .\r\n2)WLOG$q>r$ then we have $p^{r}(p^{q-r}+1)=x^{2}$ but $(p^{r}, p^{q-r}+1)=1 \\longrightarrow p^{r}=x^{2}$ and $p^{q-r}+1=y^{2}\\longrightarrow r=2$ . we get $p^{q-2}+1=y^{2}\\longrightarrow p^{q-2}=(y-1)(y+1)$, now if $p\\neq2$ then $(y-1,y+1)=1 \\longrightarrow y-1=1 , y+1=p^{q-2}\\longrightarrow p=q=3$ which gives 2 solution $(p,q,r)=(3,3,2)$ or $(p,q,r)=(3,2,3)$ if $p=2$ then $y=2k+1 \\longrightarrow 2^{q-2}=(2k)(2k+2)$. if$q>4$then we have $2^{q-4}=k(k+1)$ but one of $k$ or $k+1$ is odd $\\longrightarrow k=1$ which gives $(p,q,r)=(2,5,2)$ or $(p,q,r)=(2,2,5)$\r\nif $q=3$ then $4k(k+1)=2$ which gives contradiction :)", "Solution_3": "[quote=\"mayhem\"]Let $p,q,r$ are primes number. Find all $(p,q,r)$ triples make $p^{q}+p^{r}$ is a square.[/quote]\r\n\r\nyou can use [url=http://www.mathlinks.ro/viewtopic.php?search_id=1855663038&t=145692]this[/url]. also, it can be showed using the same idea that if $p^{a}+p^{b}$ is a perfect square and $a\\ge b$, then either 1.$p=2,a=b=2k+1$ or 2.$p=2,a=2k,b=2k-3$ or 3.$p=3,a=2k+1,b=2k$. so your problem has soluitons $(2,q,q),(2,2,5),(3,3,2)$ for an odd prime $q$." } { "Tag": [ "AMC", "AIME", "USAMTS", "USA(J)MO", "USAMO" ], "Problem": "I was just wondering... not that I'll do that, but the USAMTS website says \"high enough\". Not very specific :wink:", "Solution_1": "It varies from year to year. This past year, it was 68/75.", "Solution_2": "What would your AMC score be if it becomes necessary during USAMO qualification?", "Solution_3": "[quote=\"Ignite168\"]What would your AMC score be if it becomes necessary during USAMO qualification?[/quote]\r\n\r\nI think it's treated as a score of 100 on the AMC 12.", "Solution_4": "[quote=\"Adunakhor\"][quote=\"Ignite168\"]What would your AMC score be if it becomes necessary during USAMO qualification?[/quote]\n\nI think it's treated as a score of 100 on the AMC 12.[/quote]\r\n\r\nLook [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105637]here[/url].\r\n\r\nAlthough Richard Rusczyk doesn't know exactly what the scores become, they actually do base your index on your USAMTS score.", "Solution_5": "Oh ok, thanks for point that out :)" } { "Tag": [ "Alcumus", "Support" ], "Problem": "can u restart alcumus \r\n\r\nlike make a new account\r\n\r\nthanks!!! :lol: :rotfl:", "Solution_1": "to get a new account, you'd have to make a new account on AoPS, which is forbidden\r\n\r\nanyway, you should use Alcumus to help you learn math\r\nif you want to restart, the most likely reason i can think of is that you want to get higher on the scoreboard\r\nAlcumus should not be used in a competitive way\r\njust keep doing problems, learn from them, and get better at math", "Solution_2": "no its not because of the competition \r\nits because all my problems are chapters ahead of what i am doing", "Solution_3": "no i dont want to restart because of the competition\r\n\r\nbut i want to start again because the problems were suited for the lessons", "Solution_4": "You can 3.5 years later.", "Solution_5": "If you haven't already, post your request [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=497&t=429784]here[/url].", "Solution_6": "asf check the date.", "Solution_7": "I don't think they reset stuff anymore because I made a request and my account still hasn't been resetted.", "Solution_8": "It can take a long time because Paul could get many requests a day, and yours could take a couple months.", "Solution_9": "[quote=\"tmathman\"]It can take a long time because Paul could get many requests a day, and yours could take a couple months.[/quote]\n\nUnless Paul or whoever else manages this stuff decides that your account does not need to be reset.", "Solution_10": "[quote=\"paulhryu\"]You are allowed to reset your student progress on Alcumus on a case-by-case basis. To avoid abuse of this feature, I have made it such that resetting your progress not only resets your subject and topic scores, it will also [b]erase your quests, secret honors, settings, and levels[/b]. Really, for all purposes, it will be like you never existed on Alcumus before.\n\n[b]This process is completely irreversible. Please do not take this lightly.[/b]\n\nIf you would like your student progress reset, please [url=http://aops.com/Alcumus/Reset][b]click here[/b][/url] (don't worry, there's a confirm dialog) and then choose the \"Yes, delete my progress!\" button.\n\nI'll only allow one reset per year. Zero exceptions.[/quote]", "Solution_11": "Actually, you could because if you click the \"click here,\" it will show you to a page that you can restart." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "An integer $ n>101$ is divisible by $ 101$. Suppose that every divisor $ d$ of $ n$ with $ 1d$?", "Solution_2": "Nope.\r\nSuppose d is the maximum divisor of n. All divisors of n (besides n) satisfy: d1<=n/2.\r\nThen if d<=n/2 we have d=n-d1>=n/2 (n is the largest divisor of n besides d). \r\nAnd that means d=n/2.\r\nn is divisible by 2. There's no contradiction here.", "Solution_3": "[quote=\"Fuaran\"]Nope.\nSuppose d is the maximum divisor of n. All divisors of n (besides n) satisfy: d1<=n/2.\nThen if d<=n/2 we have d=n-d1>=n/2 (n is the largest divisor of n besides d). \nAnd that means d=n/2.\nn is divisible by 2. There's no contradiction here.[/quote]\r\nBut then $ d=\\frac{n}{2}$ is $ d_{i}-d_{j}$ for some $ i,j$ by condition of the problem,which shows that $ d_{i}>\\frac{n}{2}$ ?", "Solution_4": "yes, and that $ d_{i}$ is exactly $ n$.\r\nevery divisor SMALLER than $ n$ is a difference of SOME(including $ n$) 2 divisors of $ n$. :wink:\r\nso $ \\frac{n}{2}=n-\\frac{n}{2}$ and its good\r\nedit:\r\ni think i've found a solution\r\n[hide]\nwe know $ n=101k$ for some $ k$ and we know $ k>1$ from the condition so\n$ k=a-b$ where $ a,b$ are divisors of $ n$. assume $ 101\\not|a$ then from $ a|n=101k$ and since $ 101$ is prime we have $ a|k$ which isn't true...so $ a=101a_{0}$ and now we get that $ 101a_{0}|n=101k$ so $ a_{0}|k$ so $ k=la_{0}$ and from that it's easy to get that $ b=ma_{0}$ so $ k=(101-m)a_{0}$\n$ n=101(101-m)a_{0}$. since $ b$ is a divisor of $ n$ we have that $ m|101(101-m)$ for some $ m$ such that $ 1\\leq m\\leq 100$ but again since $ 101$ is prime it can only happen when $ m=1$\nso this works if $ 101$ is replaced with prime $ p$ and $ 100$ with $ p-1$\ni didn't find an example that such a number exists yet...[/hide]", "Solution_5": "[quote=\"goc\"]... i didn't find an example that such a number exists yet...[/quote]\r\n\r\n$ n=10100$\r\n\r\n$ 2=4-2$\r\n$ 4=5-1$\r\n$ 5=10-5$\r\n$ 10=20-10$\r\n$ 20=25-5$\r\n$ 25=50-25$\r\n$ 50=100-50$\r\n$ 100=101-1$\r\n$ 101=202-101$\r\n$ 202=404-202$\r\n$ 404=505-101$\r\n$ 505=1010-505$\r\n$ 1010=2020-1010$\r\n$ 2020=2525-505$\r\n$ 2525=5050-2525$\r\n$ 5050=10100-5050$", "Solution_6": "[quote=\"goc\"]yes, and that $ d_{i}$ is exactly $ n$.\nevery divisor SMALLER than $ n$ is a difference of SOME(including $ n$) 2 divisors of $ n$. :wink:\nso $ \\frac{n}{2}=n-\\frac{n}{2}$ and its good\nedit:\ni think i've found a solution\n[hide]\nwe know $ n=101k$ for some $ k$ and we know $ k>1$ from the condition so\n$ k=a-b$ where $ a,b$ are divisors of $ n$. assume $ 101\\not|a$ then from $ a|n=101k$ and since $ 101$ is prime we have $ a|k$ which isn't true...so $ a=101a_{0}$ and now we get that $ 101a_{0}|n=101k$ so $ a_{0}|k$ so $ k=la_{0}$ and from that it's easy to get that $ b=ma_{0}$ so $ k=(101-m)a_{0}$\n$ n=101(101-m)a_{0}$. since $ b$ is a divisor of $ n$ we have that $ m|101(101-m)$ for some $ m$ such that $ 1\\leq m\\leq 100$ but again since $ 101$ is prime it can only happen when $ m=1$\nso this works if $ 101$ is replaced with prime $ p$ and $ 100$ with $ p-1$\ni didn't find an example that such a number exists yet...[/hide][/quote]\r\nTo [u]goc[/u]: Good job\r\nTo [u]pco[/u]: Very convincing", "Solution_7": "ok,seems that i'm just a lazy bastard for not checking the trivial case... :blush: \r\nthanks pco..." } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Something related, Let H be any group with an automorphism f such that the only fixed point of \"f\" is \"e\" and such that ff(x)=x for all x in H. Prove that H is abelian and f(x)x=e for all x in H.", "Solution_1": "I think $H$ should be finite, right? :? If this is indeed the case, we could do this:\r\n\r\nIf $x_1^{-1}f(x_1)=x_2^{-1}f(x_2)$, then $f(x_1x_2^{-1})=x_1x_2^{-1}$, meaning that $x_1=x_2$. This shows that $x\\mapsto x^{-1}f(x)$ is injective, and thus surjective, since $H$ is finite. Now take any $t\\in H$ and some $x$ s.t. $x^{-1}f(x)=t$. We have $tf(t)=x^{-1}f(x)[f(x)]^{-1}(f\\circ f)(x)=x^{-1}(f\\circ f)(x)=1$, which means that for all $t\\in H$ we have $f(t)=t^{-1}$. The fact that $H$ is abelian follows from this." } { "Tag": [], "Problem": "This problem was found in a romanian book, (Mircea Fianu, Olimpiadele bucurestene de Matematica, page 36) but the solution is so long,,,, Do you think there is a shorther one?\r\nLet $a,b,c>0$ such that $a|b+c-a|+b|c+a-b|+c|a+b-c|=a^2+b^2+c^2.$ Prove that $a=b=c.$\r\nThank you!", "Solution_1": "Nickolas , \r\n\r\n I took $a , b , c$ as sides of a triangle( so absolute values have no role) , put $a+b-c = x , b+c-a = y , c+a-b = z$ ect , and the proof was easy. I have not tried for the other cases. I had not time. I ' ll try it in the evening.\r\n\r\n [u]Babis[/u]" } { "Tag": [ "MATHCOUNTS", "ARML" ], "Problem": "In the upcoming year, im going to be teaching the mathcounts team at my old middle school, but being that it wasnt offered when I was there, I dont know what Im supposed to teach. \r\nIf someone could please post the main topics and give an example of 1 or 2 of the more popular types of questions asked for each topic, I would be very appreciative. Also, a general overview of the competitions would be very helpful as well.", "Solution_1": "Use the coaching kit at http://mathcounts.org/Program/program.html", "Solution_2": "I think that coahcng kit is the most useless idea ever invented. Its just schedules. Anyways, if anyne can answer my original question, I would be greatly appreciative.", "Solution_3": "I'm still learning what jelyman is asking about myself, so I would appreciate other, more experienced coaches jumping in here with additional suggestions.", "Solution_4": "Well I can hardly be considered \"an experienced coach\". My MC experience just consists of 3 years of competition, but I'll be teaching a MC class (not on AoPS) this year (and ppl are actually paying to be in the class...what a surprise) so I'll make a comment\r\n\r\n[quote=\"jelyman\"]I think that coahcng kit is the most useless idea ever invented. Its just schedules. Anyways, if anyne can answer my original question, I would be greatly appreciative.[/quote]\r\n\r\njelyman, in my personal opinion the coaching kit is useless unless u've got no idea what ur doing and neither does anyone else u know...\r\n\r\nHOWEVER, it does include an EXCELLENT toolbox that covers just about every formula that will be covered in the MC competitions...if u can make ur students understand those then they'll have everything they need. after that, just practice...and apply those formulas...doing problems is the best way. MC resources can be purchased at http://www.mathcounts.org or they're available online in some places...", "Solution_5": "Ooh, I liked the coaching kit I got from my MathCounts coach last year. So many interesting formulas :) .\r\n\r\nLet me give a tip: hand out snacks. My MathCounts coach does that, and most people like snacks.", "Solution_6": "[quote=\"Treething\"]\nLet me give a tip: hand out snacks. My MathCounts coach does that, and most people like snacks.[/quote]\r\n\r\nSounds nuts to say, but this is very important when you're bringing new people in. Don't overlook it.\r\n\r\nTry to make it social - try some team games in addition to you just standing up there and teaching. If you stand up there and lecture all the time, your team will get small in a hurry.\r\n\r\nAs for teaching - start from the basics. Don't assume anything is too easy. If you want to teach them mods, do base numbers first. If you want to teach them combinations and permutations, do 'using multiplication to count' first. And so on.", "Solution_7": "[quote]Let me give a tip: hand out snacks. My MathCounts coach does that, and most people like snacks.[/quote]\r\n\r\nI coach ARML, not Mathcounts. But yes, by all means provide snacks. Once you're up and established, maybe you can get some parents to help you with that. After all, there's a pretty widely followed tradition along those lines for youth sports teams.", "Solution_8": "[quote=\"Kent Merryfield\"] Once you're up and established, maybe you can get some parents to help you with that. After all, there's a pretty widely followed tradition along those lines for youth sports teams.[/quote]\r\n\r\nThis is how we handle it at the San Diego Math Circle.", "Solution_9": "I m in high school so we don't do mathcounts. My teacher usually has food at the meetings like pizza and hamburgers and we still don't have all that high of turnouts. It s probably because math is \"nerdy\" and some of the kids on the team are...well very nerdy to say the least. I usually never make a practice until halfway into the second quarter of school since I have cross country until then. So I can safely say that I am not scaring away the people :D ( I think I am one of the least likely persons on my team to scare ppl away but its still nice to have proof). I think we ll have to start recruiting people for our jv team (our varsity is pretty good this year). I think I might start talking to my teacher about ways we can expand the team. \r\n\r\nWell the point of all of this was that food alone wont necessarily make your team larger. It never hurts though!", "Solution_10": "[quote]If someone could please post the main topics and give an example of 1 or 2 of the more popular types of questions asked for each topic, I would be very appreciative. Also, a general overview of the competitions would be very helpful as well.[/quote]\r\n\r\nHmm... I don't know how much I can help you with that but if you want to look some not-textlike questions, meaning some creative questions, scan through the SC post I did long time ago. There is first one which many people competed (winner was theone853) and second one which very few people did. It is good test with questions though. And if you still want more, look at the my competition which I'll post the questions for the round.\r\nThat's official competition I made and for details, read the another thread.\r\n\r\n :cool: Plus, don't worry too much. Pretty sure you'll do fine.", "Solution_11": "Also, try hiring helpers (or getting volunteers) to help out with the advanced kids...\r\n\r\nOh, if you get food, give some to me. ;)", "Solution_12": "Food is key. I remember at my second Mathcounts practice, I found a huge stash of starbursts. As soon as that news spread around, we had 4 new members the next week. :D \r\n\r\nDo some practice tests to see what level each person is at. You'll ned to know which 4 are best to take on your team.", "Solution_13": "my math teacher was the coach of the california mathcounts team (won first place 3 years in a row) ;) .\r\n\r\nshe doesnt do much. i think she is kind of lucky because our school has many intelligent students, but all she does is give warm-ups and workouts and previous mathcounts contests", "Solution_14": "My MathCounts coach writes out problems that people don't understand...", "Solution_15": "What's the point? Try not to bump meaninglessly. ", "Solution_16": "Sorry since this is off-topic - but you both have identical avatars! :wow:", "Solution_17": "[quote=falcon22]Sorry since this is off-topic - but you both have identical avatars! :wow:[/quote]\n\n[img]https://i.kym-cdn.com/entries/icons/facebook/000/007/666/_57c8a1a431a592af806925e57258202f.jpg[/img]", "Solution_18": "@above that's creepy if i die tonight I'm suing you", "Solution_19": "[quote name=\"falcon22\" url=\"/community/p17297959\"]\nSorry since this is off-topic - but you both have identical avatars! :wow:\n[/quote]\n\nThe avatar was copied, not a coincidence.", "Solution_20": "My gosh MSM is gonna get locked at this rate. Please don't let that happen. ", "Solution_21": "[quote=falcon22]Sorry since this is off-topic - but you both have identical avatars! :wow:[/quote]\n\nyeah they both copied a certain 3Blue1Brown thumbnail", "Solution_22": "[quote=HumanCalculator9][quote=falcon22]Sorry since this is off-topic - but you both have identical avatars! :wow:[/quote]\n\nyeah they both copied a certain 3Blue1Brown thumbnail[/quote]\n\nI see. \n\nAnyway, stop posting in this thread - I've reported the post, it should be handled soon! :)" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "If $ \\displaystyle m$ is a positive interger, show that\r\n\r\n$ \\displaystyle \\int_0^{\\frac{\\pi}{2}} \\cos ^{m}x \\sin^{m}x \\,dx\\: =\\: 2^{-m}\\int_0^{\\frac{\\pi}{2}} \\cos^{m} x \\,dx$", "Solution_1": "[hide]$ \\int_0^{\\frac {\\pi}{2}} \\cos ^{m}x \\sin^{m}x \\,dx \\equal{} 2^{ \\minus{} m}\\int_0^{\\frac {\\pi}{2}} \\sin^{m}(2x) \\,dx \\equal{} 2^{ \\minus{} m \\minus{} 1}\\int_0^{\\pi} \\sin^{m}t \\,dt \\equal{} 2^{ \\minus{} m}\\int_0^{\\pi/2} \\sin^{m}t \\,dt \\equal{} 2^{ \\minus{} m}\\int_0^{\\pi/2} \\sin^{m}(\\pi/2 \\minus{} s) \\,ds \\equal{} 2^{ \\minus{} m}\\int_0^{\\pi/2} \\cos^{m}s \\,ds$[/hide]", "Solution_2": "Awesome Thanks :D" } { "Tag": [], "Problem": "Emily used a standard metric ruler to measure the length of a piece of yarn she was using for a project. The yarn was as long as 24 of the smallest divisions on the ruler. How many centimeters long was the piece of yarn?", "Solution_1": "[hide]$2.4$ i counted them out on the ruler[/hide]", "Solution_2": "[hide]24 mm = 2.4 cm.[/hide]", "Solution_3": "[hide]the smallest measure on a metric ruler is millimeters. so 24mm=2.4cm.[/hide]", "Solution_4": "Well, you could also think that if you count 24 dividers, the length is 23. There are two dividers around 1 mm, three around 2mm, so I guess you could also say 2.3mm", "Solution_5": "oh yeah, nice job natmc", "Solution_6": "The smallest mark on a metric ruler is a millimeter. On an \"inch ruler\" it's 1/8 of an inch.\r\n\r\nSo the yarn is 24 mm, which is 2.4 cm. :)" } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "absolute value", "superior algebra", "superior algebra unsolved" ], "Problem": "Let P(z) be a polynomial with complex coefficients and complex roots. Is it true that the coefficient in P(z) with largest absolute value always is greater (that is, has greater absolute value) than the absolute value of the roots of P(z)?", "Solution_1": "Try $ \\frac{x}{2}\\minus{}1$.", "Solution_2": "Sorry ZetaX, I messed one thing up. I meant to write integral coefficients (and complex roots). :)" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "If f and g are continuously differentiable functions for all real numbers, which of the following definite integrals is equal to f(g(4))-f(g(2))?\r\n\r\n\r\n${ \\int_{2}^{4}{f'(g(x))}}\\ dx$\r\n${ \\int_{2}^{4}{f(g(x))f'(x)}}\\ dx$\r\n${ \\int_{2}^{4}{f(g(x))g'(x)}}\\ dx$\r\n${ \\int_{2}^{4}{f'(g(x))g'(x)}}\\ dx$\r\n${ \\int_{2}^{4}{f(g'(x))g'(x)}}\\ dx$", "Solution_1": "use the chain rule; what is $ [f(g(x)]'$?", "Solution_2": "i d k . . i d k . .", "Solution_3": "you don't know what $ \\frac {d}{dx} f(g(x))$ is? it is very hard to do integral calculus if you don't know essentially the most important result in differential calculus.\r\n\r\nin particular if you think hard enough you probably do know what the answer is, and i am not inclined to answer this sort of multiple choice question for you with no apparent effort on your part.", "Solution_4": "Calculus BC? November? Either your teacher doesn't review material at all or you haven't been paying attention in class. In either case,\r\n\r\n$ [f(g(x))]' \\equal{} f'(g(x)) g'(x)$\r\n\r\nUsing the Fundamental Theorem of Calculus, can you solve the problem now?", "Solution_5": "do you just add the\r\n\r\n$ \\int_{2}^{4}$\r\n\r\nto get\r\n\r\n${ \\int_{2}^{4}{f'(g(x))g'(x)}}\\ dx$", "Solution_6": "Basically, yes..." } { "Tag": [ "function", "linear algebra" ], "Problem": "Let $ V$ be a subspace of the complex inner product space $ E$, let $ T: V \\rightarrow E$ be linear, and define $ Q(x) = $ for $ x \\in V$. Show that\r\n\r\n(i) If $ Q(x) = 0$ everywhere, then $ T(x) = 0$ for all $ x$.\r\n\r\n(ii) If $ Q(x)$ is always real, then $ T$ is hermitian.", "Solution_1": "(ii) Well if $ Q(x)$ is real for all $ x$ then\r\n\r\n$ Q(x)= = = $ \r\n\r\nfor all $ x$ thus $ T(x)=T^{*}(x)$\r\n\r\n\r\n(i) If $ Q(x)=0$ everywhere then choose a nonzero $ x$ such that $ Q(x)=0$. Now since $ Q(x)=0$ we have that \r\n\r\n$ =0$ for all $ x$. Also $ = x^{*}Q(x)=\\left(x^{*}\\right)^{2}T(x)$\r\n\r\nsince $ Q(x)=x^{*}T(x)$. now since we assumed that $ x\\neq 0$ we also have that $ x^{*}\\neq 0$ thus $ T(x)=0$ for all $ x$." } { "Tag": [ "geometry", "trigonometry", "geometry unsolved" ], "Problem": "In a triangle ABC, D is a point on side BC such that DB = 14, DA = 13 and DC = 4. Knowing that the circle circumscribed to the triangle ADB is equal to the radius of the circle circumscribed to the triangle ADC, calculate the area of triangle ABC.\r\n\r\nPortuguese question:\r\n\r\nEm um tri\u00e2ngulo ABC, seja D um ponto sobre o lado BC tal que DB = 14, DA = 13 e DC = 4. Sabendo que o c\u00edrculo circunscrito ao tri\u00e2ngulo ADB tem raio igual ao do c\u00edrculo circunscrito ao tri\u00e2ngulo ADC, calcule a \u00e1rea do tri\u00e2ngulo ABC.", "Solution_1": "Applying the formulae for radius we concluded that b=c=15 and the area is 36*15^.5", "Solution_2": "since for any $ \\triangle ABC$ we have $ R\\equal{}\\frac{abc}{4L}\\equal{}\\frac{abc}{4.\\frac12ab\\sin X}\\equal{}\\frac{c}{2\\sin X}$ , then\r\n\r\n$ \\frac{c}{2\\sin X}\\equal{}\\frac{b}{2\\sin (180\\minus{}x)}\\equal{}\\frac{b}{2 \\sin X}$ , so , $ b\\equal{}c\\equal{}k$ , ($ X\\equal{}\\angle{ADB}$)\r\n\r\nthen by stewart , we get $ 4k^2\\plus{}14k^2\\minus{}14.4.18\\equal{}13^2.18\\rightarrow k^2\\equal{}225\\rightarrow k\\equal{} 15\\equal{}b\\equal{}c$\r\n\r\nso by Heron we get $ L\\triangle \\equal{} \\sqrt{24.9.9.6}\\equal{}108$", "Solution_3": "Well ... the area is $ 8\\sqrt{3}\\minus{}6$\r\nHow to do it ?", "Solution_4": "[quote=\"luiseduardo\"]Well ... the area is $ 8\\sqrt {3} \\minus{} 6$\nHow to do it ?[/quote]\r\nHow come ?\r\nOr maybe there are some mistakes in my solution huh ?", "Solution_5": "No, I think your solution that's correct, but the question wrong :o \r\nI copy two questions differents in one ... :blush: \r\nI'm sorry, but thanks for your help." } { "Tag": [ "floor function", "limit" ], "Problem": "$lim$ $[x+0.6]$=?\r\n$x \\to 2$", "Solution_1": "javab: $2$\r\nhanooz montazere rahe hal haye ghashange shma hastam. :D", "Solution_2": "age as $[$ $]$ manzooret $\\lfloor$ $\\rfloor$ hast,\r\nsoel kheyli rahateh ;)", "Solution_3": "[quote=\"amirhtlusa\"]age as $[$ $]$ manzooret $\\lfloor$ $\\rfloor$ hast,\nsoel kheyli rahateh ;)[/quote]\r\ndaghighan manzooram hamoone va khob man ham neveshtam ke kheili rahate vali doost daram ba rahe hal haye motafavet va ghashang hal beshe. :D", "Solution_4": "$\\\\\\lim_{x\\rightarrow 2}\\lfloor x+0.6\\rfloor\\\\ =\\lfloor x+0.6\\rfloor |_{x=2}\\\\ =\\lfloor 2.6\\rfloor=2$\r\n\r\nEDIT: Mitooni $\\delta -\\epsilon$ estefade koni.\r\n\r\nMasoud Zargar" } { "Tag": [ "geometry", "ratio", "inequalities" ], "Problem": "Now that the test is over, I'd like to ask anyone who took it, or who didn't to show me how they proved this problem. I did it really quickly, but I think I may be wrong.\r\n----------------------------------------------------------------\r\nBAMO (Bay Area Math Olympiad) 2007\r\nQuestion 5 of 5\r\n\r\nTwo sequences of positive integers $x_{1}, x_{2}, x_{n}...$ and $y_{1}, y_{2}, y_{n}...$ are given such that $\\frac{y_{n+1}}{x_{n+1}}> \\frac{y_{n}}{x_{n}}$\r\nfor each $n \\geq 1$. Prove that there are infinitely many values of $n$ such that $y_{n}> \\sqrt{n}$.\r\n----------------------------------------------------------------\r\n[hide=\"Possible Solution\"]\nLet $y_{n}= n$, and $x_{n}= n+1$\n\nOur sequence then looks like\nY: 1, 2, 3, 4, 5, 6\nX: 2, 3, 4, 5, 6, 7\n\nThe ratio of terms is like this:\n$\\frac{1}{2}< \\frac{2}{3}< \\frac{3}{4}< \\frac{4}{5}< \\frac{n}{n+1}$\nThis satisfies the conditions, because $lim_{n \\to \\infty}\\frac{n}{n+1}$ converges to one.\n\nThus, we want all $y_{n}> \\sqrt{n}$\nBecause $y_{n}= n$, we get $n > \\sqrt{n}$, and because N is positive, we can multiply both sides by it without changing direction or multipying by zero.\nThus, we get $n^{2}> n$, which is true for all real numbers.\nThe cardinality of the set of all real numbers is infinite.\nTherefore, there are infinite solutions.\nQED.\n[/hide]\r\nIf you read my proof, can you comment on either what I did wrong, or what I could do better? I'm looking for some constructive criticism, as this was my first ever Olympiad.", "Solution_1": "[hide]The problem with your solution is that you are assuming that each sequence is a certain type of sequence. Another set of sequences could be $x_{n}=F_{n}$ where $F_{n}$ is the nth Fibonacci number and $y_{n}=3^{n}$. I'm sorry, but I don't have a proof.\n[/hide]\r\n\r\n[b]Edit: I think the directors of BAMO would appreciate if we wait till at least tommorow to discuss the exam.[/b]", "Solution_2": "Don't you just have to show that if you can find two sequences with the first property (ratios one) that there are an infinite amount of solutions to the second?\r\n\r\ni.e. does it matter what the series are, as long as they match the properties?", "Solution_3": "[quote=\"DiscreetFourierTransform\"]does it matter what the series are, as long as they match the properties?[/quote]\r\n\r\nOf course it does. You need to show that [b]any possible sequences that satisfy the first condition[/b] also satisfy the second.\r\n\r\nEdit: You seem to have written the problem incorrectly. Should it be $y_{n}> \\sqrt{n}$?", "Solution_4": "Oh snap. Missed that one for sure then.", "Solution_5": "yeah i just got home from taking the test; wait till at least tomorrow", "Solution_6": ":blush: Yes it should. Changed.", "Solution_7": "[hide=\"it is the next day now\"]\nWe are given $\\frac{x_{n}}{y_{n}}>\\frac{x_{n+1}}{y_{n+1}}$. (It is better to have the $y$'s in the denominators; you'll see why soon.) After clearing denominators,\n\\[x_{n}y_{n+1}-x_{n+1}y_{n}>0\\Rightarrow x_{n}y_{n+1}-x_{n+1}y_{n}\\geq 1\\]\nWe now have the sharpened inequality\n\\[\\frac{x_{n}}{y_{n}}\\geq \\frac{x_{n+1}}{y_{n+1}}+\\frac{1}{y_{n}y_{n+1}}\\]\nWe now prove infinitely many $n$ with $y_{n}>\\sqrt{n}$ by contradiction. Suppose that for some $N$, $y_{n}\\leq \\sqrt{n}$ for all $n\\geq N$. Then for any $n\\geq N$, we get\n\\[\\frac{x_{n}}{y_{n}}\\geq \\frac{x_{n+1}}{y_{n+1}}+\\frac{1}{\\sqrt{n(n+1)}}\\]\n\\[>\\frac{x_{n+1}}{y_{n+1}}+\\frac{1}{n+1}\\]\nSo we end up with\n\\[\\frac{x_{N}}{y_{N}}>\\sum_{k=N+1}^\\infty \\frac{1}{k}\\]\n :o THE HARMONIC SERIES DIVERGES!!! CONTRADICTION!!! YAY!!!\n[/hide]" } { "Tag": [ "geometry", "perimeter", "combinatorics unsolved", "combinatorics" ], "Problem": "Nine 1*1 cells of a 10*10 squre are infected. In one time unit, the cells with at least two infected neighbors(having a common side)become infected.Can the infection spread to the whole squre?", "Solution_1": "It seems that if the 9 cells are in one row, the answer is obviously \"no\".", "Solution_2": "It's been discussed before here: the idea is that whenever a new cell is infected, the perimeter of the infected region decreases (or stays the same), so since in the end you want the whole board to be infected, thus having a perimeter of $4\\cdot 10$ for the infected region, you cannot achieve this starting with $9$ squares, which have a perimeter $\\le 4\\cdot 9$." } { "Tag": [ "geometry", "rectangle", "perimeter", "LaTeX", "calculus", "derivative", "calculus computations" ], "Problem": "I don't quite understand what I am doing wrong.. but the answer I am getting is not correct.\r\nAnd this is my last problem and its due online at 11pm tonight =/...\r\nI get this problem right and I get a 100% on the assignment!\r\nI went to my math lab at my school and this is the answer/ work I got. \r\nAny help is appreciated! \r\n----------\r\n\r\nThe problem asks:\r\n[b]A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 25 feet?[/b]\r\n\r\n\r\nSo this is how I approached it: we need the areas of the rectangle and the semi circle... so we get:\r\n\r\nA = bh + pi(b/2)^2 = bh + (pi/4)(b^2)\r\n\r\nfor simplicity, b = x (easier on the eyes :))\r\n\r\nA= xh + (pi/4)(x^2)\r\n\r\nso I got that h (the height) is = (25-(1 + pi)x)/2\r\n\r\nthis means that A(x) = (25x - (1+pi)x^2)/2 + (pi/4)(x^2) \r\n\r\nSo from here I differentiated the equation and got...\r\n[25x - (1 + pi) x^2]/2 + (pi/4)(x^2) = \r\n(25x)/2 - (1/2)(x^2) - (pi/2)(x^2) + (pi/4)(x^2) =\r\ni solve for x...\r\n\r\n25/2 - x - pix + (pi/2)x\r\n\r\nand then i attempt to solve for x and get...\r\nx= (-25/2)/(-1 - pi + pi/2) \r\n\r\nand i get that x is equal to 4.862306620604284\r\n\r\nI then plug that into my area equation and that should be my answer for the problem...\r\nwhich is:\r\n[b]42.210429215162904[/b]\r\n\r\nAny help? Suggestions? \r\nThanks once again.\r\nSorry if my work is hard to understand.. hopefully I made no typing errors.", "Solution_1": "1. Learn LaTeX. There is a very helpful FAQ in the LaTeX forum.\r\n\r\n2. Right in your first step, you used the formula for the area of a circle, not a semicircle.", "Solution_2": "I know it is later, but here is a way to solve the problem using first derivatives and maximums\r\n\r\nWe have two equations\r\n$ P$ is perimeter. $ A$ is area. $ l$ and $ w$ are length and width respectively.\r\n\r\nIf the semicircle's diameter is the width of the rectangle, then we can say that the radius is half the width ($ r=\\frac{w}{2}$).\r\n\r\nThe area of the semicircle is $ \\frac{1}{2}\\pi r^{2}$\r\nArea of Semicircle(plugging in for $ r$) is $ \\frac{\\pi w^{2}}{8}$\r\n\r\n$ P=2l+w+\\frac{\\pi w^{2}}{2}= 25$\r\nSolving for $ ;$ we get\r\n$ l={\\frac{25-\\frac{w \\pi}{2}-w}{2}= \\frac{25}{2}-\\frac{w \\pi}{4}-\\frac{w}{2}}$\r\n\r\nNext, the area of the window\r\n$ A = wl+\\frac{\\pi w^{2}}{8}$\r\n\r\nPlug in for $ l$\r\n\r\n$ A = w({\\frac{25}{2}-\\frac{w \\pi}{4}-\\frac{w}{2})+\\frac{\\pi w^{2}}{8}}$\r\n\r\n$ A = \\frac{(-\\pi-4)w^{2}+100w}{8}$\r\n\r\n\r\nWe want to maximize this, so take the first derivative and set it equal to zero\r\n\r\n$ \\frac{dA}{dw}= \\frac{25}{2}-\\frac{(\\pi+4)w}{4}= 0$\r\n\r\nSolve for $ w$\r\n$ w = \\frac{50}{\\pi+4}$\r\n\r\nPlug into our equation for $ l$\r\n$ l = \\frac{25}{\\pi+4}$\r\n\r\nNow to find the area, plug $ l$ and $ w$ back into our original area equation\r\n\r\n$ A = 43.74 ft^{2}$\r\n\r\nAnd to check to make sure it is right, plug $ l$ and $ w$ back into the perimeter equation to make sure it equals 25 ft, which it does.", "Solution_3": "i was fortunately able to get it before the deadline! yay =)\r\n\r\nbut i do like your method of solving it ankur... my way seems more confusing.\r\n\r\nthanks!", "Solution_4": "[quote=\"JBL\"]1. Learn LaTeX. There is a very helpful FAQ in the LaTeX forum.\n\n2. Right in your first step, you used the formula for the area of a circle, not a semicircle.[/quote]\r\n\r\n\r\n ill be sure to look it up.\r\nthanks for the help." } { "Tag": [ "FTW", "LaTeX" ], "Problem": "I am running a tourney, as my graduation from newbdom. The first five contestants are lucky, as I already have the first five places down:\r\n\r\n1. biblioman\r\n2. \r\n3. \r\n4. \r\n5. \r\n\r\nThis is for anyone, but [i]post your rating[/i]! Else I don't know who to match you against. I want suggestions on the minimum number of contestants, but it starts the day after I have the agreed number.\r\n\r\nWARNING: I am semi-active, so IF I do not post for 7 days, I personally will pass on the duties of running the tournament to somebody else, so PM me if the deadline has passed, and I will choose the person. who will be the person who PMed me. Feel free to pass the title down as needed.\r\n\r\nI just realized I haven't said the format. Here goes:\r\n\r\n[b]FORMAT[/b]:\r\n\r\n2 and only 2 people against each other, with lowest rating against 2nd lowest, 3rd against 4th, et cetera up to 2nd highest vs. highest. Winner moves on. 3 games per deciding match, which means that you play 2 games. If one person (A) has beaten the other (B) both times, then A moves on. Else, there is one more match. NO COUNTDOWNS. PERIOD. 10 questions, 45 seconds each, Responses style.\r\n\r\nMore moderators would be appreciated.\r\n\r\nMy rating is 1340-1360 depending on the day.\r\n\r\nThanks,\r\nPLEASE SIGN UP!,\r\nbiblioman", "Solution_1": "Join, even though I stink at FTW. :P \r\n\r\nRating: 1160-1180 (depends when you catch me.)\r\nFTW test rating: 1231\r\n\r\nYou should totally have the tourney happen in FTW test. :P", "Solution_2": "I sign up, rating:1050-1100 Depends on the day.\r\n1. biblioman \r\n2. westiepaw\r\n3. monkeygirl13\r\n4. \r\n5.", "Solution_3": "Hi westiepaw and monkeygirl,\r\n\r\nThanks! Spread the word, because I would like to have 8 or 16 people.", "Solution_4": "I will join. 1416 rating currently.", "Solution_5": "Thanks! Again, if you could spread the word, I would appreciate it. My current goal is 8 sign-ups.\r\n\r\n1. biblioman \r\n2. westiepaw \r\n3. monkeygirl13 \r\n4. ksun48\r\n\r\nMatchups:\r\n\r\nMonkeygirl13 vs. westiepaw\r\nbiblioman (me!) vs. ksun48", "Solution_6": "What time is our match???", "Solution_7": "I'll join.\r\n\r\n1. biblioman\r\n2. westiepaw\r\n3. monkeygirl13\r\n4. ksun48 \r\n5. $ \\text{\\LaTeX}$\r\n\r\nCurrent rating is 1570, high is 1663.", "Solution_8": "I'll join if you change scoring type.\r\nI don't like responses :wink: \r\nMy current rating is 1345\r\nTest rating is 1291\r\nhigh for normal is 1444...hmmmmmmm i seem to have lost a lot recently :(", "Solution_9": "Okay, ranking or time?\r\n\r\n1. biblioman \r\n2. westiepaw \r\n3. monkeygirl13 \r\n4. ksun48 \r\n5. Chompy\r\n\r\nThanks!", "Solution_10": "ranking rocks", "Solution_11": "Did you guys kick me out? :(\r\n\r\nIf not: Time pwns rating.", "Solution_12": "Oh, sorry Latex:\r\nLets retry this:\r\n\r\n1. biblioman \r\n2. westiepaw \r\n3. monkeygirl13 \r\n4. ksun48\r\n5. $ LaTeX$\r\n6. Chompy\r\n\r\nSorry, I could not get your name to look right, LaTeX.\r\n\r\nAll right, take votes:\r\n\r\nksun48 and biblioman vote Ranking\r\n$ LaTeX$ votes Time.\r\n\r\nAs for the time we start, it will be established the day I get 8 people. It will most likely be the day after that, in the evening, about 7:00-8:00 PM.", "Solution_13": "I'll join iff [that's not a typo] you reduce the time to 25 seconds or less. Either ranking or time is fine with me.", "Solution_14": "Tentatively... what do you guys think? I vote that we remain at 45 seconds, but that is just me.\r\n\r\n1. biblioman \r\n2. westiepaw \r\n3. monkeygirl13 \r\n4. ksun48 \r\n5. LaTeX\r\n6. Chompy\r\n7. (depending on opinions) ernie", "Solution_15": "I beat monkeygirl today.\r\n\r\nNew rating: 1170", "Solution_16": "$ LaTeX$ 2-0 slightly beats ksun48\r\n\r\nBFTR1M3KvL \r\n\r\n[size=75][color=red]i'm not exactly sure what the above means, but i'm pretty sure you don't have to post it 10+ times. \nplease refrain from doing that in the future\n~vallon[/color][/size]", "Solution_17": "$ \\text{\\LaTeX}$ defeats ksun48 in 2 games. 15-5, 10-7.", "Solution_18": "All right. I PMed ernie and lotsofmath. If he doesn't get a reply within 2 days, we move on to Round 2.\r\n\r\nI will be away this weekend, so we will start Round 2 today or next Monday.\r\n\r\nEDIT: If he doesn't hear from him within 2 hours, I will send out the PMs so that we aren't held up.", "Solution_19": "Wait until next Monday, its not entirely their fault if they can't find a suitable time. Of course, if lotsofmath is intentionally avoiding a game (with I doubt), then we start Round 2 today.", "Solution_20": "I PM'd ernie yesterday, asking for a suitable time. He hasn't responded yet.", "Solution_21": "[quote=\"lotsofmath\"]I PM'd ernie yesterday, asking for a suitable time. He hasn't responded yet.[/quote]\r\n\r\nAnd I shall respond here. How about tomorrow?", "Solution_22": "All right. Round two starts on Monday. We are patiently waiting. Yes, I stole that from you, ernie. :P", "Solution_23": "I forfeit. lotsofmath wins.", "Solution_24": "All right. I will send the PMs out. And thank you, ernie.\r\n\r\nEDIT: The PMs will go out Monday.", "Solution_25": "It's Tuesday...", "Solution_26": "I PMed the Round 2 contestants with their matches. Also, we are NOT placing everybody.", "Solution_27": "Just wondering, is this dead? I haven't seen lotsofmath or biblioman on in a while.....", "Solution_28": "No, it is not.\r\n\r\nNOTE TO PARTICIPANTS IN ROUND 2: PM ME YOUR RESULTS!!\r\n\r\n-biblioman", "Solution_29": "lotsofmath wins 2-0, 14-8, 12-9..... :(" } { "Tag": [ "floor function" ], "Problem": "Fie numarul natural $A$. Notam cu $s$ suma cifrelor lui $A$ cu $S$ suma cifrelor lui $5\\cdot A$ si fie $T=(2\\cdot S-s): 9$.\r\nSa se arate ca $T$ este egal cu numarul de cifre impare din scrierea lui $A$.", "Solution_1": "fie A format cu cifre din $\\{0 , 1\\}$.\r\natunci $5\\cdot A$ e clar de determinat si cu siguranta\r\n$S=5\\cdot s$ => $T=(10s-s): 9=s$ adica numarul de 1\r\ndin A care este si numarul de cifre impare.\r\nsa presupunem ca mai adaugam un 2 la multimea de apartententa\r\na cifrelor.(unde $k$ numarul cifrei nou adaugate la multimea\r\nde aparatenenta a cifrelor)\r\nacel 2 va influenta pe \r\n$s'=s+k*2$\r\n$S'=S+k$(deoarece 2*5=10 => 1+0=1 => adaugam cate un\r\n1 pentru fiecare 2)\r\ndeci $T'=(2S'-s'): 9=( 2S+2k-s-2k): 9=(2S-s): 9=T$\r\nsa presupunem acum ca adaugam un 3 la multimea de cifre.\r\n$s'=s+k*3$\r\n$S'=S+k*(1+5)$ (de la 15 care apare din 5*3)\r\natunci $T'=(2S'-s'): 9=(2S+k*12-s-k*3): 9=(2S-s+k*9): 9=T+k$\r\npentru 7\r\n$s'=s+k*7$\r\n$S'=S+k*8$(5*7=35 => 3+5=8)\r\navem $T'=(2S'-s'): 9=(2S+k*16-s-k*7)=T+k$\r\npentru 5\r\n$s'=s+k*5$\r\n$S'=S+k*7$(5*5=25 => 2+5=7)\r\navem $T'=(2S'-s'): 9=(2S+k*14-s-k*5)=T+k$\r\nlafel si pentru 9 si pentru celelalte numere pare...\r\nde ex pentru $8$ avem $s'=s+k*8$\r\nsi $S'=s+k*4$\r\ncazurile care au ramas se fac lafel\r\n(initial am vrut sa fac o inductie dar nu prea iese...deoarece\r\nsuma_cifre(5*cifra) - cifra nu se imparte la 9,decat daca se considera\r\nfiecare caz in parte)", "Solution_2": "Cam multe cazuri...\r\nExista o solutie mult mai eleganta. :)", "Solution_3": "7 cazuri nu-s asa multe... sau sunt ?", "Solution_4": "prezentati va rog si solutia dvs", "Solution_5": "Eu zic sa mai asteptam putin. Problema mi-a placut mult si nu cred ca merita sa-i dau solutia asa repede.", "Solution_6": "Fie $f : \\mathbb N \\to \\mathbb N, \\, f(n) = \\sum_{k=0}^{m}a_{k}$, unde $n = \\sum_{k=0}^{m}a_{k}10^{k}, \\, a_{k}\\in \\overline{0,9}, \\, \\forall k \\in \\overline{0,m}$.\r\nFolosind aceasta notatie, $T_{A}= \\frac{2 \\cdot f(5A)-f(A)}9$.\r\n\r\nFie $A = 10p+q$, unde $p,q \\in \\mathbb N, \\, q \\in \\overline{0,9}$.\r\n\\begin{eqnarray*}2 \\cdot f(5A)-f(A) &=& 2 \\cdot f(50p+5q)-f(10p+q) \\\\ \\ &=& 2 \\cdot f(50p)+2 \\cdot f(5q)-f(10p)-f(q) .\\end{eqnarray*}\r\nAvem\r\n\\begin{eqnarray*}2 \\cdot f(5q)-f(q) &=& 2 \\cdot f \\left( 10 \\cdot \\left\\lfloor \\frac{q}2 \\right\\rfloor+10 \\cdot \\left\\{ \\frac{q}2 \\right\\}\\right)-f(q) \\\\ \\ &=& 2 \\cdot \\left\\lfloor \\frac{q}2 \\right\\rfloor+20 \\cdot \\left\\{ \\frac{q}2 \\right\\}-q \\\\ \\ &=& \\begin{cases}0, \\ \\ \\ \\ \\ \\ q \\ \\textrm{par}\\\\ 9, \\ \\ \\ \\ \\ \\ q \\ \\textrm{impar}\\end{cases}\\end{eqnarray*}\r\nAsadar, $T_{A}= T_{p}$ daca $q$ este par si $T_{A}= T_{p}+1$ daca $q$ este impar.\r\n\r\nAcuma problema poate fi rezolvat\u0103 prin induc\u0163ie.", "Solution_7": "Fie $A=\\overline{a_{1}a_{2}...a_{n}}.$ \r\nPutem scrie $a_{p}=2\\cdot k_{p}+r_{p}$, $k_{p}\\in \\overline{0,4}$, $r_{p}\\in \\overline{0,1}$, $\\forall{p\\in\\overline{1,n}}$.\r\nAtunci\r\n\\[s=a_{1}+a_{2}+...+a_{n}=2(k_{1}+k_{2}+...+k_{n})+(r_{1}+r_{2}+...+r_{n}), \\]\r\n\r\n\\[S=(k_{1}+k_{2}+...+k_{n})+5(r_{1}+r_{2}+...+r_{n}) \\]\r\nsi\r\n\\[T=(2S-s): 9=r_{1}+r_{2}+...+r_{n}, \\]\r\nadica ceea ce trebuia demonstrat." } { "Tag": [ "function", "algebra", "domain", "analytic geometry", "graphing lines", "slope" ], "Problem": "$ 2.)$ Find $ x \\in \\mathbb{R}$ :\r\n\\[ \\sqrt[3]{x\\plus{}7} \\minus{} \\sqrt[3]{2x\\minus{}1} \\equal{} \\sqrt[3]{x}\\] .", "Solution_1": "This was a problem that wasn't getting much attention [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180938]here[/url].\r\n\r\nI quickly noticed that $ \\boxed{x\\equal{}1}$ is a solution, but I'm still curious as to how we can manipulate or work with this more elegantly (and not brute-forcing it to such an extent of cubing it again and again until we get something workable and then check the possible solutions :wacko: ).", "Solution_2": "[hide=\"Completion of the above\"][b]Sign of x:[/b] If $ x < 0$, then since $ x \\plus{} 7 > x$, we must have that $ 2x \\minus{} 1$ is positive. However, the latter cannot be true for $ x < 0$. Since, clearly, $ x \\neq 0$, we must have that $ x > 0$.\n\n[b]Term-wise rates of change:[/b] Now, note that the rate of change of $ (x \\plus{} 7)^{\\frac {1}{3}}$ is less than that of $ x^{\\frac {1}{3}}$ because the cuberoot function is concave-down over the positives (imagine a cuberoot graph - it gets less steep).\n\nToo, since the cuberoot function is everywhere increasing, $ (2x \\minus{} 1)^{\\frac {1}{3}}$ has a positive rate of increase as well. \n\n[b]Overall rate of change:[/b] Thus, the combined rate of increase of $ \\sqrt [3]{x} \\plus{} \\sqrt [3]{2x \\minus{} 1}$ is greater than that of $ \\sqrt [3]{x \\plus{} 7}$, and thus $ f(x) \\equal{} \\sqrt [3]{x} \\plus{} \\sqrt [3]{2x \\minus{} 1} \\minus{} \\sqrt [3]{x \\plus{} 7}$ is strictly increasing.\n\n[b]Conclusion:[/b] Since $ f(x)$ is strictly increasing over the valid domain, there can be at most one solution. We notice that $ \\boxed{x \\equal{} 1}$ is said solution, and we are done.[/hide]", "Solution_3": "There is a strategy at this kind of equations . I'm not quite sure that it works here .\r\n\r\nFirst we note like this : $ \\sqrt[3]{x\\plus{}7}\\equal{}a$ $ \\sqrt[3]{2x\\minus{}1} \\equal{}b$ and $ \\sqrt[3]{x}\\equal{}c$ .\r\nWe have that $ a \\minus{} b \\equal{} c$ , $ a^3 \\equal{} c^3 \\plus{} 7$ , and $ b^3 \\equal{} 2c^3 \\minus{}1$ . \r\nWe have to solve this sistem of 3 equations now .", "Solution_4": "[quote=\"The Zuton Force\"][hide]Now, note that the rate of change of $ (x \\plus{} 7)^{\\frac {1}{3}}$ is less than that of $ (x \\plus{} 7)^{\\frac {1}{3}}$ because the cuberoot function is concave-down over the positives (imagine a cuberoot graph - it gets less steep).[/hide][/quote]\r\n\r\n :huh: Is that a typo? Anyway, I don't get your logic...", "Solution_5": "Fixed, the rate of change of $ \\sqrt[3]{x\\plus{}7}$ is strictly less than the rate of change of $ \\sqrt[3]{x}$.\r\n\r\nHere's the same logic, heuristically (and possibly somewhat inaccurately worded).\r\n\r\nThe solution(s)s to that equation is/are the root(s) of $ f(x) \\equal{} \\sqrt[3]{x} \\minus{} \\sqrt[3]{x\\plus{}7} \\plus{} \\sqrt[3]{2x\\minus{}1}$\r\n\r\n(1) Consider the graph of $ \\sqrt[3]{x}$ over the positive numbers. As $ x$ increases, the steepness decreases.\r\n\r\n(2) Thus, for any $ x$, $ \\sqrt[3]{x}$ has greater slope than $ \\sqrt[3]{x\\plus{}7}$.\r\n\r\n(3) Thus, the slope of the difference $ \\sqrt[3]{x} \\minus{} \\sqrt[3]{x\\plus{}7}$ wrt x is positive.\r\n\r\n(4) Also, $ \\sqrt[3]{2x\\minus{}1}$ has positive slope. Thus, their sum, $ \\sqrt[3]{x} \\minus{} \\sqrt[3]{x\\plus{}7}\\plus{}\\sqrt[3]{2x\\minus{}1}\\equal{}f(x)$, has positive slope. Since $ f(x)$ is strictly increasing, it can only cross the $ x$ axis once.\r\n\r\n(5) Thus, there is only one solution.", "Solution_6": "Nice logic. :) \r\nThough, is there a way to obtain $ x\\equal{}1$ other then using brute force or the system of equations.", "Solution_7": "[hide=\"Another approach\"]\nWe have $ \\sqrt[3]{x\\plus{}7}\\equal{}\\sqrt[3]{x}\\plus{}\\sqrt[3]{2x\\minus{}1} \\equal{} 2 \\frac{\\sqrt[3]{x}\\plus{}\\sqrt[3]{2x\\minus{}1}}{2} \\leq 2\\sqrt[3]{\\frac{3x\\minus{}1}{2}}$\n\nCubing gives $ \\frac{x\\plus{}7}{8} \\leq \\frac{3x\\minus{}1}{2} \\implies x \\geq 1$\n\nCan we make the other side (ie, $ x \\leq 1$) in a similar way?\n[/hide]" } { "Tag": [ "limit", "logarithms", "calculus", "calculus computations" ], "Problem": "$ \\lim_{n\\to \\infty} \\left(\\frac{\\sqrt[n]{a}\\plus{}\\sqrt[n]{b}}{2} \\right)^n$", "Solution_1": "The full result is this:\r\n\r\nFix $ a,b>0$ with $ a\\ne b.$ Define $ M(p)\\equal{}\\left(\\frac{a^p\\plus{}b^p}2\\right)^{\\frac1p}$ for $ p\\ne0$ and define $ p(0)\\equal{}\\sqrt{ab}.$ Then $ M(p)$ is continuous on $ \\mathbb{R}$ (which notably includes the statement that $ \\lim_{p\\to0}M(p)\\equal{}M(0),$ which is what this problem asks for). Also: $ M(p)$ is strictly increasing, $ \\lim_{p\\to\\infty}M(p)\\equal{}\\max(a,b),$ and $ \\lim_{p\\to\\minus{}\\infty}M(p)\\equal{}\\min(a,b).$\r\n\r\nThis is a pretty famous collection of results. After all, $ M(0)0} \\ln{M(p)} \\equal{} \\lim_{p\\minus{}>0}\\frac{f(p)\\minus{}f(0)}{p} \\equal{} f'(0)$ by definition of the derivative. We have $ f'(p) \\equal{} \\frac {2}{a^p\\plus{}b^p}\\frac{a^p\\ln{a} \\plus{}b^p\\ln{b}}{2}$, and so $ f'(0)\\equal{}(\\ln{a} \\plus{}\\ln{b})/2$ as desired." } { "Tag": [ "trigonometry", "inequalities", "geometry solved", "geometry" ], "Problem": "Prove that in any triangle we have:\r\n\r\n$\\sum\\left(a^2\\left(b+c-a\\right)\\right) \\leq 3abc$\r\n\r\nthank you!!", "Solution_1": "[b]here is a solution:[/b]\r\n\r\n$3abc-a^2b-a^2c+a^3-b^2c-ab^2+b^3-ac^2-bc^2+c^3\\geq 0$ we get that\r\n$\\sum\\left(a^3-ab^2-ac^2+abc\\right)\\geq 0$ which gives us : $\\sum \\left(a\\left(a^2-b^2-c^2+bc\\right)\\right)\\geq 0$ which is equivalent to : $\\sum\\left(a\\left(bc-2bc\\cdot\\cos A\\right)\\right) \\geq 0$ $\\Longleftrightarrow$ $abc\\left[3-2\\left(\\sum\\left(\\cos A\\right)\\right)\\right] \\geq 0$ which is true.\r\nwe have equality if a=b=c.\r\n\r\n[b]Solution 2:[/b]\r\n\r\nthinking more over this one I came with another solution:\r\n\r\n$\\sum\\left(a\\left(ab+ac-a^2-bc\\right)\\right)\\leq 0$ gives us $\\sum a\\left(b-a\\right)\\left(a-c\\right)\\leq 0$ we multiply it now by 2 and arrange it like so:\r\n\r\n$\\sum\\left(\\left(b-a\\right)\\left[a\\left(a-c\\right)+b\\left(c-b\\right)\\right]\\right) \\leq 0$ which gives:\r\n\r\n$\\sum\\left(\\left(b-a\\right)^2\\left(c-a-b\\right)\\right) \\leq 0$ which is true cause cn>0 and a,b,c>0 .Prove that\r\n$ \\frac{a^{m}}{b^{n}}\\plus{}\\frac{b^{m}}{c^{n}}\\plus{}\\frac{c^{m}}{a^{n}}\\geq a^{m\\minus{}n}\\plus{}b^{m\\minus{}n}\\plus{}c^{m\\minus{}n}$", "Solution_1": "It is difficult?I don't think so :lol: \r\n\r\nHint:\r\n$ (m\\minus{}n)\\frac{a^{m}}{b^{n}}\\plus{}nb^{m\\minus{}n}\\ge ma^{m\\minus{}n}$\r\nby\r\n$ xa\\plus{}yb\\ge a^{x}b^{y}$\r\nfor $ a,x,b,y>0,x\\plus{}y\\equal{}1$", "Solution_2": "[quote=\"zhaobin\"]It is difficult?I don't think so :lol: \n\nHint:\n$ (m\\minus{}n)\\frac{a^{m}}{b^{n}}\\plus{}nb^{m\\minus{}n}\\ge ma^{m\\minus{}n}$\nby\n$ xa\\plus{}yb\\ge a^{x}b^{y}$\nfor $ a,x,b,y > 0,x\\plus{}y \\equal{} 1$[/quote]\r\nthank u 4 your solution but... can you make it more detail ? I'm a newbie", "Solution_3": "I think zhaobin mean to sum these three ineq:\r\n$ (m\\minus{}n)\\frac{a^{m}}{b^{n}}\\plus{}nb^{m\\minus{}n}\\ge ma^{m\\minus{}n}$\r\n$ (m\\minus{}n)\\frac{b^{m}}{c^{n}}\\plus{}nc^{m\\minus{}n}\\ge mb^{m\\minus{}n}$\r\n$ (m\\minus{}n)\\frac{c^{m}}{a^{n}}\\plus{}na^{m\\minus{}n}\\ge mc^{m\\minus{}n}$", "Solution_4": "[quote]\n$ xa\\plus{}yb\\ge a^{x}b^{y}$\nfor $ a,x,b,y > 0,x\\plus{}y \\equal{} 1$[/quote]\r\nCan you prove this ?", "Solution_5": "[quote=\"imathsvn\"][quote]\n$ xa\\plus{}yb\\ge a^{x}b^{y}$\nfor $ a,x,b,y > 0,x\\plus{}y \\equal{} 1$[/quote]\nCan you prove this ?[/quote]\r\nWeighted AM-GM inequality!\r\n[i]Case 1:[/i] x,y arw rational numbers\r\nSet $ x\\equal{}\\frac{m}{m\\plus{}n},y\\equal{}\\frac{n}{m\\plus{}n}, m,n\\in N$\r\nBy AM-GM inequality, we have:\r\n$ ma\\plus{}nb\\ge (m\\plus{}n)a^{\\frac{m}{m\\plus{}n}}b^{\\frac{n}{m\\plus{}n}}$\r\nIt follows that: $ ax\\plus{}by\\ge a^{x}b^{y}$\r\n[i]Case 2:[/i] x,y arw real numbers, there exist two sequences rational numbers $ (r_{n})_{n\\ge 0}$ and $ (s_{n})_{n\\ge 0}, r_{n}\\rightarrow x, s_{n}\\leftarrow y, r_{n}\\plus{}s_{n}\\equal{}1$.\r\nSo:\r\n$ ar_{n}\\plus{}bs_{n}\\ge a^{r_{n}}b^{s_{n}}\\Leftrightarrow ar_{n}\\plus{}b(1\\minus{}r_{n})\\ge a^{r_{n}}b^{1\\minus{}r_{n}}$\r\nTaking the limit when $ n\\rightarrow\\propto$, we have $ xa\\plus{}yb\\ge a^{x}b^{y}$", "Solution_6": "I thinks it is very easy if you use Rearrangement Inequality for two sequences:\r\n$ a^{m}\\geq b^{m}\\geq c^{m}$\r\n$ \\frac{1}{a^{n}}\\leq\\frac{1}{b^{n}}\\leq\\frac{1}{c^{n}}$", "Solution_7": "Choosing a positive rational sequence a1,a2,.. such that Lim an (n->oo) =x and letting bi=1-ai, we get limbn (n->oo) =y .Okay ,The problem 's really upset\r\nThank u 4 yr response !" } { "Tag": [], "Problem": "Let us take the three integers $ 1$, $ 2$ and $ 3$.\r\n\r\nThey can be arranged in $ 6$ ways\r\n$ 123$\r\n$ 132$\r\n$ 213$\r\n$ 231$\r\n$ 312$\r\n$ 321$\r\n\r\nFor each permutation, if it is represented as ($ a$, $ b$, $ c$), compute :\r\n$ S\\equal{} (a\\minus{}b)^2 \\plus{} (b\\minus{}c)^2$. So we get the following:\r\n\r\n$ 123$ $ \\Rightarrow$ $ 2$\r\n$ 132$ $ \\Rightarrow$ $ 5$\r\n$ 213$ $ \\Rightarrow$ $ 5$\r\n$ 231$ $ \\Rightarrow$ $ 5$\r\n$ 312$ $ \\Rightarrow$ $ 5$\r\n$ 321$ $ \\Rightarrow$ $ 2$\r\nThe average value of $ S$ over the $ 6$ permutations here is $ \\frac{24}{6}$ = $ 4$.\r\n\r\n\r\nCan you compute the average value of $ S$, for the general case where you take the integers $ 1$,$ 2$,$ 3$,....$ n$ for any value of $ n$?\r\n\r\nThe general formulate for $ S$ being: $ (a_1\\minus{}a_2)^2 \\plus{} (a_2\\minus{}a_3)^2 \\plus{} (a_3\\minus{}a_4)^2 \\plus{} .... \\plus{} (a_{n\\minus{}1} \\minus{} a_n)^2$", "Solution_1": "Are you talking about the average (when using the formula for S) of [b]all arrangements[/b] of 1,2,3,.....n?\r\nIf so, how would you find all the arrangements because there were 6 arrangements in your original but there are infinitely many in this set of numbers (1,2,3...n)?", "Solution_2": "There are $ n!$. The key here is linearity of expectation; show that it suffices to compute the average value of $ (a_1 \\minus{} a_2)^2$ and multiply by $ n \\minus{} 1$." } { "Tag": [ "summer program", "Mathcamp", "Ross Mathematics Program", "PROMYS", "email", "MathPath", "number theory" ], "Problem": "Hi, I thought I should open a thread that directly asks all of you to compare the [url=http://www.math.ohio-state.edu/ross/introduction.html]Ross Program[/url] (and its sister program the [url=http://www.claymath.org/programs/outreach/PROMYS/]PROMYS[/url] program) to [url=http://www.mathcamp.org]MathCamp[/url]. \r\n\r\nThe thing that I find puzzling, for planning purposes, is that the minimum age for MathCamp is lower than it is for PROMYS, and thus my son is going to have one summer when MathCamp is the only logical choice for a summer math program for him (unless lightning strikes and he qualifies for MOP that year, which I'm not planning on). But I seem to recall threads here suggesting that after one year of MathCamp, especially a year at MathCamp taking its number theory courses, there isn't a lot left to learn at Ross/PROMYS. \r\n\r\nI'm not trying to stir up a fight here, as I have heard MANY good things about the Ross/PROMYS programs, and I have a high regard for the alumni of those programs who post here, but who has made the head-to-head comparison? After a person has been to one year of MathCamp, there is plenty more to go back for for at least two more years (according to what I have read on this site). \r\n\r\nI'm the kind of guy who could readily be persuaded to send my children to a different program the following year just for the sake of variety, but after a first year of MathCamp, what would be the best different program to go to?", "Solution_1": "I don't know about PROMYS, but the Ross program does allow age exceptions on both ends, so a 13 year old can go as long as he or she is qualified.", "Solution_2": "[quote=\"zscool\"]I don't know about PROMYS, but the Ross program does allow age exceptions on both ends, so a 13 year old can go as long as he or she is qualified.[/quote]\r\n\r\nThanks for letting me know that. I particularly had PROMYS in mind, because I think I read once (on an email list where Alison's mom posts) that PROMYS has a hard younger age limit because of Boston University dorm rules. If that doesn't apply to Ross, and I can thus disregard the age rules I see on the Ross Web site, that opens up another possibility for that summer. It happens that A+MATH has a summer birthday that falls just after the program start dates for most summer programs. So when a program says, \"Must be age [x] by [date of program start],\" he always has to wait an \"extra\" year by strict application of the program rules. For example, this year he will be just a few weeks too young for MathPath, which is our top plan for next year. Next year, he will be just too young for the strict interpretation of the rules for MathCamp, which would be our plan in 2006, but if he is legal for Ross in 2006 I might rather have him try Ross. (We know a Ross alumnus from the 1970s locally, and I am pretty impressed with what he knows about math.) \r\n\r\nRules, rules. I'm a big guy for \"rules are rules,\" and not trying to test the limits, but one always has to take into account exceptional situations. I guess I'll just have him fill out an application form and do an application quiz and see what he decides after he gets a response. It may be that my son is stretching the lower age limit in one of his summer programs this year, but the program didn't seem to have any hard limit on its Web site, and they let him in. \r\n\r\nThanks for the tip. Just one more thing to think deeply about.", "Solution_3": "Hi! I'm one of the directors of Mathcamp.\r\n\r\nI'm always a little surprised that there isn't more crossover between Mathcamp and Ross/PROMYS: though there are some exceptions, people who come to Mathcamp once tend to keep coming back, and the same for Ross or PROMYS. Part of it, no doubt, is wanting to see old friends again, but I think that experiencing the variety of different programs is an admirable thing to do. There's enough going on at Mathcamp that you can come for a summer, stay away from number theory classes, and still be challenged by Ross's offering; conversely, there's still plenty to do at Mathcamp if one has previously been to Ross/PROMYS. I think there are few enough people who are alums of both programs that there's no consensus on a preferred order for trying the programs, but I would be very interested if there's someone who feels otherwise!" } { "Tag": [ "search", "\\/closed" ], "Problem": "When a topic is moved, how can we find out where it goes?", "Solution_1": "You can find the topic again with a search. If it's your topic that was moved, take the hint about where to post next time.", "Solution_2": "Try this: look up your own profile. Click on \"Find all posts by (your username)\". Since the one you're wondering about should be one of your more recent posts, it should be near the top.", "Solution_3": "But what if it's a topic that I haven't posted in?", "Solution_4": "[quote=\"Zakary\"]When a topic is moved, how can we find out where it goes?[/quote]\r\nDoesn't the topic remain in the forum it was originally posted in, but with a blue arrow thingy instead of the new post indicator?", "Solution_5": "Moderators have an option of whether or not to insert the \"blue arrow thingy.\" In practice, it's used only small portion of the time - too many topics get moved to make it the default. And the topic itself doesn't remain - the \"blue arrow thingy\" is just a pointer, not the topic itself.\r\n\r\nTo answer Zachary: you can go through the profile of a user that you remember having posted in the topic and do what I suggested above, or you can use the search feature to search for some word or words you are sure were used in the topic. You can also pretend you're a moderator - if you were going to move this, what would be the most likely destination?\r\n\r\nAlternatively: is there some reason why the topic might have been deleted altogether? One possible reason for that would be if it contains information about some contest that is still in progress.", "Solution_6": "[quote=\"Zakary\"]When a topic is moved, how can we find out where it goes?[/quote]\r\n\r\nWait when a topic is moved there is an arrow and also the link...click the link and look at what forum it is in? unless im missing something? :maybe:", "Solution_7": "D3m0n, go read the post directly above yours.", "Solution_8": "Ok. Thanks. It looks like I'll have to go searching for the topic. I was just wondering that if, in case the topic was moved and I couldn't find the topic in the original forum, there was some way of locating it. Hopefully I have a good memory then. Thank you!!!\r\n\r\nEDIT: Oh, and I didn't immediately see where the search button at the top was was. >_> :rotfl: <_< But I see it now.", "Solution_9": "Can't you just click the topic title, and you will be redirected to the forum that it was moved to? It worked for me.", "Solution_10": "Why don't people read posts in the threads they respond to? 1=2, go read Kent Merryfields post in this thread." } { "Tag": [ "inequalities", "geometric inequality" ], "Problem": "Let $ ABC$ be an acute trianle.Prove that:\r\n$ \\frac{h_a}{l_a}\\plus{}\\frac{h_b}{l_b}\\plus{}\\frac{h_c}{l_c} \\ge \\frac{21}{10}\\plus{}\\frac{9r}{5R}$", "Solution_1": "This one is not hard if we take into account the following [url=http://artofproblemsolving.com/community/c1756h1019905_geometrical_inequality_with_sum_of_h_al_a]super sharp inequality[/url]:\n\n\\[\\boxed{\\frac {h_a}{l_a} + \\frac {h_b}{l_b} + \\frac {h_c}{l_c} \\ge 1 + \\sqrt{\\frac {s^2 + r^2 + 2Rr}{2R^2}}}\\, .\\]\n\nAfter a couple of computations it remains to show that in any [b]acute-angled[/b] triangle the following inequality holds:\n\n\\[50\\Big(s^2 + r^2 + 2Rr\\Big) \\ge \\Big(11R + 18r\\Big)^2.\\]\n\nBut the latter can be proven without too much effort via [url=http://artofproblemsolving.com/community/c6h342626][b]Walker[/b]'s inequality[/url], i.e. $\\boxed{s^2\\ge 2R^2 + 8Rr + 3r^2}$ as well as the known inequality $\\boxed{s\\ge 2R + r}$ which holds in acute or right-angled triangles. Specifically, we can show that:\n\n\\[\n\\begin{array}{rcr}\n\\displaystyle 50\\Big[\\Big(2R^2 + 8Rr + 3r^2\\Big) + r^2 + 2Rr\\Big] \\ge \\Big(11R + 18r\\Big)^2 & \\text{holds for} & 2 \\le \\displaystyle \\frac Rr \\lessapprox 2.95238 \\phantom{.} \\\\\\\\\n\\displaystyle 50\\Big[\\Big(2R + r\\Big)^2 + r^2 + 2Rr\\Big] \\ge \\Big(11R + 18r\\Big)^2 & \\text{holds for} & \\displaystyle \\frac Rr \\gtrapprox 2.39773.\n\\end{array}\n\\]", "Solution_2": "[quote=Mateescu Constantin]This one is not hard if we take into account the following [url=http://artofproblemsolving.com/community/c1756h1019905_geometrical_inequality_with_sum_of_h_al_a]super sharp inequality[/url]:\n\n\\[\\boxed{\\frac {h_a}{l_a} + \\frac {h_b}{l_b} + \\frac {h_c}{l_c} \\ge 1 + \\sqrt{\\frac {s^2 + r^2 + 2Rr}{2R^2}}}\\, .\\]\n\nAfter a couple of computations it remains to show that in any [b]acute-angled[/b] triangle the following inequality holds:\n\n\\[50\\Big(s^2 + r^2 + 2Rr\\Big) \\ge \\Big(11R + 18r\\Big)^2.\\]\n\nBut the latter can be proven without too much effort via [url=http://artofproblemsolving.com/community/c6h342626][b]Walker[/b]'s inequality[/url], i.e. $\\boxed{s^2\\ge 2R^2 + 8Rr + 3r^2}$ as well as the known inequality $\\boxed{s\\ge 2R + r}$ which holds in acute or right-angled triangles. Specifically, we can show that:\n\n\\[\n\\begin{array}{rcr}\n\\displaystyle 50\\Big[\\Big(2R^2 + 8Rr + 3r^2\\Big) + r^2 + 2Rr\\Big] \\ge \\Big(11R + 18r\\Big)^2 & \\text{holds for} & 2 \\le \\displaystyle \\frac Rr \\lessapprox 2.95238 \\phantom{.} \\\\\\\\\n\\displaystyle 50\\Big[\\Big(2R + r\\Big)^2 + r^2 + 2Rr\\Big] \\ge \\Big(11R + 18r\\Big)^2 & \\text{holds for} & \\displaystyle \\frac Rr \\gtrapprox 2.39773.\n\\end{array}\n\\][/quote]\n\nBut\n\n\\[\\boxed{\\frac {h_a}{l_a} + \\frac {h_b}{l_b} + \\frac {h_c}{l_c} \\ge 1 + \\sqrt{\\frac {s^2 + r^2 + 2Rr}{2R^2}}}\\, .\\]\n\nis yang xue zhi and yinghua yan inequality(In 1999.11)\n\nsee \"Inequality Research\" P594", "Solution_3": "Mateescu Constantin:\n\nsay you e_mainl,and me:\nwshr987@163.com,\n\nand I post you \"inequality Research\" .pdf", "Solution_4": "[quote=Mateescu Constantin]This one is not hard if we take into account the following [url=http://artofproblemsolving.com/community/c1756h1019905_geometrical_inequality_with_sum_of_h_al_a]super sharp inequality[/url]:\n\nAfter a couple of computations it remains to show that in any [b]acute-angled[/b] triangle the following inequality holds:\n\n\\[50\\Big(s^2 + r^2 + 2Rr\\Big) \\ge \\Big(11R + 18r\\Big)^2.\\]\n[/quote]\n\n\\[f=50\\Big(s^2 + r^2 + 2Rr\\Big)-\\Big(11R + 18r\\Big)^2.\\]\n\nBut\n\n${\\it f}\\,{s}^{2}= \\left( R-2\\,r \\right) \\left( s-2\\,R-r \\right) \n \\left( 1350\\,{r}^{2}+270\\,sr \\right) +164\\, \\left( R-2\\,r \\right) \n \\left( {s}^{2}-16\\,Rr+5\\,{r}^{2} \\right) r+ \\left( R-2\\,r \\right) \n \\left( {s}^{2}-2\\,{R}^{2}-8\\,Rr-3\\,{r}^{2} \\right) \\left( 158\\,R+270\n\\,r \\right) \n$\n$+\\left( s-2\\,R-r \\right) \\left( {s}^{2}-16\\,Rr+5\\,{r}^{2} \\right) \n \\left( {\\frac {833}{2}}\\,R+{\\frac {1913}{4}}\\,r+{\\frac {833}{4}}\\,s\n \\right) + \\left( s-2\\,R-r \\right) \\left( {s}^{2}-2\\,{R}^{2}-8\\,Rr-3\n\\,{r}^{2} \\right) \\left( {\\frac {1867}{2}}\\,R+{\\frac {787}{4}}\\,r+{\n\\frac {1867}{4}}\\,s \\right) $\n$+{\\frac {791}{4}}\\, \\left( {s}^{2}-16\\,Rr+5\\,{r}^{2} \\right) \\left( 4\n\\,{R}^{2}+4\\,Rr+3\\,{r}^{2}-{s}^{2} \\right) +{\\frac {1709}{4}}\\,\n \\left( 4\\,{R}^{2}+4\\,Rr+3\\,{r}^{2}-{s}^{2} \\right) \\left( {s}^{2}-2\n\\,{R}^{2}-8\\,Rr-3\\,{r}^{2} \\right)\\geq 0\n$\n" } { "Tag": [ "calculus", "derivative", "real analysis", "integration", "linear algebra" ], "Problem": "What is a good calculus book for self study for a junior entering senior year? I understand limits and derivatives pretty well, they're kinda easy, and want to continue so that next year i can learn calc 2.", "Solution_1": "It depends how advanced you want to go. Stewart's book is considered one of the standards, but I've never used it. If you want a really hard-core, rigorous treatment of calculus, you have to go with Spivak's book. I've heard Apostol is good but I don't know anything about it.", "Solution_2": "Also, there is Intro to Calculus and Analysis by Richard Courant and Fritz John. \r\nPerhaps even ... baby rudin.", "Solution_3": "Which one is \"baby\" Rudin?", "Solution_4": "Oh, right. It's Principles of Mathematical Analysis by Walter Rudin.", "Solution_5": "[quote=\"Cyrazeno\"]Also, there is Intro to Calculus and Analysis by Richard Courant and Fritz John. \nPerhaps even ... baby rudin.[/quote]\r\n\r\nThat's really bad advice. Rudin's book is by no means meant for an \"introduction to Calculus\".\r\n\r\nMy advice is to try George Thomas' [u]Calculus[/u] (my favorite intro to calculus book) or James Stewart's [u]Calculus[/u]. If you know everything in these books already, then perhaps try Michael Spivak's [u]Calculus[/u] or Tom Apostol's [u]Calculus, Vol. 1: One-Variable Calculus with an Introduction to Linear Algebra[/u]. Rudin should be saved until you've mastered all of the topics of introductory calculus, and perhaps even (computational) multivariable calculus.", "Solution_6": "Whooaa I just looked at Spivak and it is WAY more interesting then all the multivariable computations I've been at recently.\r\n\r\nBut yeah don't do it if you have no idea what integration is...", "Solution_7": "[quote]Whooaa I just looked at Spivak and it is WAY more interesting then all the multivariable computations I've been at recently. \n[/quote]\r\nSpivak has a really rigorous book but if you sit down with it, you can get a lot out of it. Some of the problems are really, really interesting." } { "Tag": [ "combinatorics open", "combinatorics" ], "Problem": "hmm, just a random thought, can't we just treat every country on a map, through some topological transformation, as a square (or any fixed shape) of a different size? This way, the map (or plane) is composed of whole bunch of squares of different side lengths. :?", "Solution_1": "Just curious.. How does that help in simplifying the problem? :?", "Solution_2": "so then we can start to coordinatize the plane and play with equations? (i don't know, otherwise i'd have published my results already :P ) i'm just wondering if the people who actually have an idea what's going on can tell me whether or not this is helpful. :lol:", "Solution_3": "I see.. I agree with your square argument..\r\n\r\nIntuitively, it seems right, and probably is, like many math problems. I just don't see where to go from it, like you said, I would be working on my paper right now if I knew that. :D", "Solution_4": "okay, what's wrong with this:\r\n\r\nTake an arbitrary square ABCD in the plane. Let O be its centre. Coordinatize the plane with O as the origin, x and y-axes perpendicular to BC and AB, respectively. Label the direction of positive y-axis N, nega y-axis S, posi x-axis E and nega x-axis W. Now we only consider the squares that touch square ABCD, and are to the N or to the E of square ABCD. Let D{N} and D{E} denote these two sets of squares. Temporarily suppose that there is no other square in the plane except square ABCD, and D{N}. Then we can color D{N} alternatively with only two colors s.t. any two neighbouring squares are of a different color. Similarly, if only square ABCD, and D{E} were in the plane, we could color D{E} with only two colors as well. Now consider the existence of all of square ABCD, D{E} and D{N}. Let D{N_E} denote the most eastern square in D{N}, similarly, D{E_N} denote the most northern square in D{E}. Then we have two cases:\r\n\r\nCase 1: D{N_E} is the same as D{E_N}. In this case, it's just a square that shares exactly point B with square ABCD. Hence it requires a different color. And D{N} and D{E} each requires another different color, so we need 4 colors in total.\r\n\r\nCase 2: D{N_E} touches D{E_N}. This way we only need to give each of them a distinct color from each other and from ABCD. then D{N} and D{E} can all be colored alternatively using the same colors. So we need only 3 colors in total.\r\n\r\nThis shows that for any arbitrary square, we can use at most 4 colors to color all the squares touching it in two neighbouring directions. But now we add in the other squares inductively, and using the same argument, we can ensure using still at most 4 colors in total. Right??? :P :? :D", "Solution_5": "I don't find the square statement intuitively right. How will you for example do the situation of croatia-bosniaherzegovina-jugoslavia? one country surrounded by only 2 other countries that touch eachother at both sides?", "Solution_6": "hmm, i see, but by \"squaring\" the map, we can only end up with MORE countries, with the extra countries assigned arbitrary colors. So in that regard, suppose that there exists a map could not be colored with 4 colors, then \"squaring\" it in the plane must also not be able to be colored with 4 colors...isn't it? So it's like a generalized case....:?\r\n\r\np.s. So my argument above is correct IF we can indeed convert a map of countries into squares in a plane???!!! :D :P :D", "Solution_7": "I don't get it... can you give me a practical explanation of how you would do do for the thing I listed? I don't think adding countries is a generalisation! Because if 2 adjacent countries cannot have the same color, you are avoiding this by putting in another country? :?", "Solution_8": "yeah i see the problem, an even bigger problem is what if a country is COMPLETELY within another? Then we can't very well take it out.... :( damn... :(", "Solution_9": "[quote=\"al.M.V.\"]yeah i see the problem, an even bigger problem is what if a country is COMPLETELY within another? Then we can't very well take it out.... :( -... :([/quote] That is not a problem since that is completely trivial, it doesn't influe the amount of colors ;) but indeed, I'm afraid your lemma is false." } { "Tag": [ "calculus", "derivative", "algebra", "system of equations" ], "Problem": "Solve in $ \\mathbb{R}$ the following system of equations:\r\n$ \\begin{array}{l}x^{3}\\plus{}y^{3}\\plus{}z^{3}\\equal{} 0\\\\ x^{5}\\plus{}y^{5}\\plus{}z^{5}\\equal{} 0\\\\ x^{2007}\\plus{}y^{2007}\\plus{}z^{2007}\\equal{} 0.\\\\ \\end{array}$", "Solution_1": "I don't know how to find the solutions you actually have to think on to figure out, but couldn't x, y, and z just equal 0?", "Solution_2": "If x,y,z is solution, then -x,-y,-z is solution. So wlog assumume that two of them (WLOG x and y) are non-negative and z is negative. Let z=-a and now x,y and a are non-negative, \r\n$ x^{3}\\plus{}y^{3}\\equal{}a^{3}$\r\n$ x^{5}\\plus{}y^{5}\\equal{}a^{5}$\r\n$ t\\equal{}x^{3}$,$ r\\equal{}y^{3}$\r\n$ t^{5/3}\\plus{}r^{5/3}\\equal{}(r\\plus{}t)^{5/3}$\r\nOne possibility is that t=0, the other is\r\nr/t=k\r\n$ 1\\plus{}k^{5/3}\\equal{}(k\\plus{}1)^{5/3}$\r\n$ 1\\plus{}k^{5/3}\\minus{}(k\\plus{}1)^{5/3}\\equal{}0$\r\nLHS is 0 for k=0 and it's derivative is\r\n$ 5/3.k^{2/3}\\minus{}5/3(k\\plus{}1)^{2/3}<0$, so its decreasing and cannot be 0 for k>0.\r\nIn both cases one of r, t is 0, so we know either x,y,or z is 0. The solutions are\r\n[0,t,-t], [t,0,-t], [t,-t,0], $ t\\in\\mathbb{R}$" } { "Tag": [ "email", "\\/closed" ], "Problem": "I know this is a stupid question...\r\n\r\nI'm thinking of taking a class soon, but I don't know how to get on the classes.\r\n\r\nI mean, how do you go to class?\r\n\r\nThanks.", "Solution_1": "It's ok. Everybody has to learn sometime.I didn't now for sometime too.\r\nClick 'classroom' on the left hand side. If you are logging in during your class time it should say that the classroom is loading. This takes some time so click on embedded classroom. It has no pop-ups, that is the only difference.\r\nHope this was helpful.", "Solution_2": "You can also go to My Classes and there's a link to the class, especially if there is a math jam running", "Solution_3": "When you enroll, you'll receive an email with some instructions about how to get to the class homepage. The homepage has a Course Information document that explains many details about the courses, including how to reach the classroom.", "Solution_4": "And if some reason the both the embedded classroom and the regular classroom does not work, you can always view a transcript of the class while it is going on. :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove if a,b,c>0 then\r\n$ 2(\\sum\\frac {a \\plus{} b}{c}) > \\equal{} \\sum\\frac {a^2}{bc}\\plus{}9$\r\nThis inequality is true sure and it is very easy :P", "Solution_1": "[quote=\"tranquoc\"]Prove if a,b,c>0 then\n$ 2(\\sum\\frac {a \\plus{} b}{c}) > \\equal{} \\sum\\frac {a^2}{bc} \\plus{} 9$\nThis inequality is true sure and it is very easy :P[/quote]\r\nAre you sure it is true? Try $ a\\equal{}b\\equal{}1,c\\equal{}3$. :)" } { "Tag": [ "geometry", "calculus", "search", "superior algebra", "superior algebra theorems" ], "Problem": "Hello, I need to develop a project on the close link b/w algebra and geometry\r\nIf you can provide some simple algebraic identities proved with geometry(preferably Euclidean)\r\nPlease avoid use of calculus beyond the elementary level", "Solution_1": "you can try to get your hand on this book: [url]http://www.amazon.com/exec/obidos/tg/detail/-/0883857006/002-6477257-6168037?v=glance[/url]\r\nor you can search for some proofs without words via google" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that there exist infinitely many positive integers $ k$ for which equation $ n^m\\minus{}m^n\\equal{}k$ has unique solution in positive numbers.", "Solution_1": "[quote=\"rogue\"]Prove that there exist infinitely many positive integers $ k$ for which equation $ n^m \\minus{} m^n \\equal{} k$ has unique solution in positive numbers.[/quote]\r\n$ \\forall k>0$: $ (n,m)\\equal{}(k\\plus{}1,1)$ is solution\r\nthen then equation has unique solution \r\n$ \\iff$ $ \\forall m>1$ we have $ k\\neq n^{m}\\minus{}m^n$\r\n$ \\iff$ $ k\\in E\\equal{}\\mathbb{N}^*\\minus{}\\{n^m\\minus{}m^n: \\ m>1\\}$\r\n\r\nso we must pove that $ E$ can't be finite \r\ni continue after :wink:", "Solution_2": "We will prove that if $ k \\equal{} 8m \\plus{} 3$ or $ k \\equal{} 8m \\plus{} 5$ then the equation has unique solution in positive integers.\r\nwe have $ n^{m} \\minus{} m^{n} \\equal{} k$ . now we have 2 case :\r\n1)$ m$ is odd and $ n$ is even . now if $ m\\geq 3$ then $ n^{m}\\equiv 0(mod8)$ and $ m^{n}\\equiv 1(mod8)$ . so $ k\\equiv 7 (mod8)$ . which is contradiction .\r\nif $ m \\equal{} 1$ then $ n \\equal{} k \\plus{} 1$ \r\n2)$ m$ is even and $ n$ is odd . now $ n\\geq 3$ because $ n^{m} > 1$ \r\n$ m^{n}\\equiv 0(mod8)$ , $ n^{m}\\equiv 1 (mod8)$ so $ k\\equiv 1(mod8).$ which is contradiction :) \r\nso the equation has unique solution $ (m,n) \\equal{} (1,k \\plus{} 1)$ in positive integers ." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello, \r\n\r\n\r\nLet G be a finite group and a,b two elements of order p prime, with a \r\nnot in . Show that G contains at least p^2-1 elements of order p. \r\n\r\n\r\nI am stuck in the case where all the p-Sylow subgroups of G are cyclic \r\n(there are at least p+1 such p-Sylows, but they may overlap, which \r\nprevents me from concluding). \r\n\r\n\r\nThanks for your help, \r\n\r\n\r\n-- \r\nJulien Santini", "Solution_1": "A very useful theorem, which I proved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22490]here[/url], says that if $m$ divides the order of $G$, the number of elements of $G$ with order dividing $m$ is divisible by $m$.\r\n\r\nFor this problem, we note that the number of elements of order $p$ must be $\\equiv-1\\mod p$ and must also be divisible by $p-1$. This means that the number of such elements is of the form $(p-1)(kp+1)$ for some nonnegative integer $k$, and the smallest such value greater than $p-1$ is $p^2-1$.\r\n\r\nActually, the special case of prime $m$ is quite easy:\r\nConsider the set $S$ of all ordered $m$-tuples $(a_1,a_2,\\dots,a_m)$ of elements of $G$ with product $a_1a_2\\cdots a_m=e$. Let $n$ be the number of elements of $G$. Clearly, $S$ has $n^{m-1}$ members, which is a multiple of $m$ since $m|n$. The shift operator $T$ acts by $T(a_1,a_2,\\dots,a_{m-1},a_m)=(a_2,a_3,\\dots,a_m,a_1)$.\r\nDecompose $S$ into orbits under $T$. All have size 1 or $m$, and the orbits of size 1 are $m$-tuples $(a,a,\\dots,a)$ with $a^m=e$. Simple counting verifies that the number of such elements is divisible by $p$, and we are done.\r\n\r\nThe Sylow theorems are a waste of time on this problem.", "Solution_2": "\"The Sylow theorems are a waste of time on this problem.\"\r\n\r\nThe problem is directly connected to a generalizaton of one of the Sylow theorems: \"whenever p^r divides the cardinal of a group G, where p is of course prime, then the number of sungroups of G with cardinal p^r is 1 mod p\" -which makes the problem trivial-." } { "Tag": [ "geometry", "parallelogram", "similar triangles" ], "Problem": "ABCD is a parallelogram. Points Q, R, S, and T are the midpoints of AB, BC, CD, and DA, respectively. If ST=8 and BD=14, then AC+QT=?", "Solution_1": "[hide]Use similar triangles (i know, i hate them too)\n$\\frac{ST}{AC}= \\frac{1}{2}= \\frac{QT}{BD}$\n\n$AC =16, QT =7$\n$AC+QT=23$[/hide]\r\n\r\n[size=75][color=darkred]Don't forget to hide your answers/solutions.\n-nebula42[/color][/size]" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Let a.b >0. Show that\r\n1/2 (a+b) 2 +1/4(a+b) \\geq a \\sqrt b+b \\sqrt a", "Solution_1": "The left-hand side equals a2/2 + b2/2 + ab + a/4 + b/4. Now note that \r\na2/2 + ab/2 + a/8 + b/8 \\geq 4(a2/2 ab/2 a/8 b/8)1/4 = a\\sqrtb by AM-GM.\r\nSimilarly \r\nb2/2 + ab/2 + a/8 + b/8 \\geq 4(b2/2 ab/2 a/8 b/8)1/4 = b\\sqrta by AM-GM. \r\nAdding these inequalities gives the result.", "Solution_2": "Here my solution :\r\n 1/2 (a+b) 2 + 1/4(a+b) \\geq a \\sqrt b + b \\sqrt a\r\n<=> 2(a-b) 2 + b(2 \\sqrt a - 1) 2 + a(2 \\sqrt b - 1) 2 \\geq 0 (obvious)" } { "Tag": [ "linear algebra", "matrix", "logarithms", "inequalities", "function", "inequalities proposed" ], "Problem": "Prove that:$\\frac{a^{3}}{b}+\\frac{b^{3}}{c}+\\frac{c^{3}}{a}\\geq a\\sqrt{ac}+b\\sqrt{ba}+c\\sqrt{bc}$\r\nwhere $a;b;c$ are positive real numbers\r\nCan you prove it by many solution?", "Solution_1": "Let $\\vec{x}= \\begin{bmatrix}\\ln a\\\\ \\ln b\\\\ \\ln c\\end{bmatrix}$, $P= \\begin{bmatrix}3 &-1 & 0\\\\ 0 & 3 &-1\\\\-1 & 0 & 3\\end{bmatrix}$ and $Q= \\begin{bmatrix}\\frac{3}{2}& 0 & \\frac{1}{2}\\\\ \\frac{1}{2}& \\frac{3}{2}& 0\\\\ 0 & \\frac{1}{2}& \\frac{3}{2}\\end{bmatrix}$. Then the inequality rewrites as\r\n$\\sum_{i}\\exp\\left(\\left[P\\cdot\\vec{x}\\right]_{i}\\right)\\geqslant \\sum_{i}\\exp\\left(\\left[Q\\cdot\\vec{x}\\right]_{i}\\right)$. \r\nSince the matrix $P$ is nonsingular, we can make a substitution $\\vec{y}= P\\cdot\\vec{x}$. Then $\\vec{x}= P^{-1}\\cdot\\vec{y}$ and the inequality becomes\r\n$\\sum_{i}\\exp\\left(\\left[\\vec{y}\\right]_{i}\\right)\\geqslant \\sum_{i}\\exp\\left(\\left[Q\\cdot P^{-1}\\cdot\\vec{y}\\right]_{i}\\right)$. \r\nA direct calculation shows that the matrix $Q\\cdot P^{-1}$ is doubly - stochastic (the entries of the first row are $\\frac{15}{26}, \\frac{5}{26}, \\frac{3}{13}$; the second and third row are its cyclical shifts), so $\\vec{y}\\succ Q\\cdot P^{-1}\\cdot\\vec{y}$ and by the Karamata inequality for the convex function $f(x)= \\exp\\left(x\\right)$ we are done.", "Solution_2": "Who have other solution?", "Solution_3": "since$(3.-1.0)$ majorize$(1.\\frac{1}{2}.\\frac{1}{2})$ and . ur inequlitie is equivalent to $\\sum_{sym}a^{3}b^{-1}\\geq \\sum_{sym}ab^{\\frac{1}{2}}c^{\\frac{1}{2}}$ so ur inequaliti holds .", "Solution_4": "[quote=\"arkhammedos\"]since$\\(3.-1.0)$ majorize$(1.\\frac{1}{2}.\\frac{1}{2}$ and . ur inequlitie is equivalent to $\\sum_{sym}a^{3}b^{-1}\\geq ab^{\\frac{1}{2}}c^{\\frac{1}{2}}$ so ur inequaliti holds .[/quote]\r\nBut the sum at the left hand side is not symmetric.", "Solution_5": "ah ur right sorry :blush:", "Solution_6": "help me!\r\nI need about five or six solutions!", "Solution_7": "Easy I think.\r\n\r\nBy Cauchy,\r\n\r\n\\[LHS.(ab+bc+ca) \\ge (a^{2}+b^{2}+c^{2})^{2}.\\]\r\n\r\nMoreover, we have certainly,\r\n\r\n\\[2RHS \\le a^{2}+b^{2}+c^{2}+ab+bc+ca.\\]\r\n\r\nSorry but I don't think this one is a very nice and good problem.", "Solution_8": "I know this solution on \"Math for children magazine\"\r\nDo you have other solution?" } { "Tag": [ "LaTeX", "integration", "calculus", "trigonometry", "logarithms", "limit", "modular arithmetic" ], "Problem": "Hey, I just want to create a thread for testing tex.\r\n\r\n[tex]\\large i=i_{0}[e^{\\frac{\\alpha_{A}n\\Im}{RT}\\eta} - e^{\\frac{\\alpha_{C}n\\Im}{RT}\\eta}][/tex][/tex]", "Solution_1": "[quote=\"fanzha\"]Hey, I just want to create a thread for testing tex.\n\n[tex]\\large i=i_{0}[e^{\\frac{\\alpha_{A}n\\Im}{RT}\\eta} - e^{\\frac{\\alpha_{C}n\\Im}{RT}\\eta}][/tex][/tex][/quote]\r\n\r\nuse \\displaystyle{.....} around your tex, it will make it look much better:\r\n\r\n[tex]\\displaystyle{i=i_{0}[e^{\\frac{\\alpha_{A}n\\Im}{RT}\\eta} - e^{\\frac{\\alpha_{C}n\\Im}{RT}\\eta}]}[/tex]", "Solution_2": "hmm... maybe that particular sequnce of characters, it won't :o.", "Solution_3": "Fractions in exponents tend not to work very well.", "Solution_4": "I also noticed that regular latex formatting tags like \\large or \\small don't work. Is there a reason or is the same effect reproducible some other way?", "Solution_5": "You can use \\large and \\small in an mbox.\r\nThus normal size is [tex]\\displaystyle \\int_{0}^{1}\\dfrac{x^{4}\\left( 1-x\\right) ^{4}}{1+x^{2}}dx=\\dfrac{22}{7}-\\pi[/tex]\r\nTiny size is\r\n[tex]\\mbox {\\tiny \\displaystyle \\int_{0}^{1}\\dfrac{x^{4}\\left( 1-x\\right) ^{4}}{1+x^{2}}dx=\\dfrac{22}{7}-\\pi}[/tex]\r\nand Large size is\r\n[tex]\\mbox {\\Large \\displaystyle \\int_{0}^{1}\\dfrac{x^{4}\\left( 1-x\\right) ^{4}}{1+x^{2}}dx=\\dfrac{22}{7}-\\pi}[/tex]", "Solution_6": "[quote=\"Spoon\"]use \\displaystyle{.....} around your tex, it will make it look much better:\n\n[tex]\\displaystyle{i=i_{0}[e^{\\frac{\\alpha_{A}n\\Im}{RT}\\eta} - e^{\\frac{\\alpha_{C}n\\Im}{RT}\\eta}]}[/tex][/quote]\r\n\r\nHere's one way to make the exponents clear - using \\mbox, as mentioned in the previous post. Try the various sizes \\large, \\Large, \\LARGE, \\huge, \\Huge\r\nThis is what \\LARGE looks like\r\n[tex]\\mbox{\\LARGE \\displaystyle{i=i_{0}[e^{\\frac{\\alpha_{A}n\\Im}{RT}\\eta} - e^{\\frac{\\alpha_{C}n\\Im}{RT}\\eta}]}}[/tex]\r\n\r\nThe full range of sizes is:\r\n\\tiny, \\scriptsize, \\footnotesize, \\small, \\normalsize, \\large, \\Large, \\LARGE, \\huge, \\Huge", "Solution_7": "This is my first time using Latex...heh\r\n\r\n[tex](3-25)^2(x+3)=5[/tex]\r\n\r\nBTW...how do you use integrals and other symbols?", "Solution_8": "[tex]\\mbox {\\Large \\displaystyle \\int_{1}^{\\sqrt[3]3}}z^{2}dz\\ast\\cos{\\dfrac{3\\pi}{9}}=\\ln{\\sqrt[3]e}[/tex]\r\n\r\nDid that work?", "Solution_9": "Here are a bunch of the more common symbols:\r\n\r\n[tex]\\displaystyle\\sum_{n=0}^\\infty a_n \\prod_{k=1}^n b_k[/tex]\r\n\r\n[tex]\\displaystyle \\lim_{t\\to \\infty} \\int_{-t}^t e^{-x^2}\\; dx=\\sqrt{\\pi}[/tex]\r\n\r\n[tex]\\displaystyle 19\\equiv 3 \\pmod 4[/tex]\r\n\r\n[tex]\\displaystyle n\\in\\mathbb{Z}\\subset\\mathbb{Q}[/tex]", "Solution_10": "testing \\times vs. \\cdot\r\n\r\n[tex]\\mbox{\\Large a\\cdot b ... a\\times b}[/tex]\r\n\r\n\r\n*EDIT* - DUH! lol", "Solution_11": "\\ldots instead of three periods is the preferred way to enter an ellipsis.\r\n\r\n[tex]\\ldots \\mathrm{~v.s.~} ...[/tex]", "Solution_12": "Hm. I usually use \\cdots (at least for multiplication), as in [tex]n!=n\\cdot(n-1)\\cdots 2\\cdot 1[/tex].", "Solution_13": "how does \\ work, which parts does the \\ modify?", "Solution_14": "[tex]\\Huge\\displaystyle\\ 2^2+2_2[/tex]", "Solution_15": "\\ introduces a macro, but some things don't take \\s. The best examples of things that don't take \\s are ^ (superscript) and _ (subscript). More or less everything else that you don't want to see in plain text will take a \\, such as \\sum for [tex]\\sum[/tex], \\log for [tex]\\log[/tex] (note that this is not the same as log without the \\, which would look like [tex]log[/tex]), \\sqrt{\\pi} for [tex]\\sqrt{\\pi}[/tex], etc.", "Solution_16": "Also, you may have noticed that a { or a } won't appear in your text normally. If you want them, you have to do \\{ and \\} (and I think also \\[ and \\], although I don't remember for sure).", "Solution_17": "You don't need to do \\[ and \\] to get [ and ]. I think those do something funny with spacing, but I have never used them.", "Solution_18": "The characteristic polynomial $ f(\\lambda)$ of the\r\n$ 3\\times 3$ matrix\r\n\\[ \\begin{pmatrix}a & b & c\\\\ d & e & f\\\\ g & h & i\\end{pmatrix}\\] \r\nis given by the equation\r\n\\[ f(\\lambda) \\equal{}\\begin{vmatrix}\\lambda\\minus{}a &\\minus{}b &\\minus{}c\\\\ \\minus{}d &\\lambda\\minus{}e &\\minus{}f\\\\ \\minus{}g &\\minus{}h &\\lambda\\minus{}i\\end{vmatrix}.\\]", "Solution_19": "[hide=\"Unparseable or potentially dangerous latex formula.Error: 3\"]$ \\frac{1}{4\\pi}\\oint_{\\Sigma}\\frac{1}{r}\\frac{\\partial U}{\\partial n}ds$[/hide]", "Solution_20": "There is now a [url=http://www.artofproblemsolving.com/Forum/index.php?f=224]Test Forum[/url] for this sort of thing, rather than reviving an ancient thread." } { "Tag": [ "blogs", "\\/closed" ], "Problem": "Well since Vista has been launched, and it seems that it includes a nicely done nav. panel with embedded RSS feed reader (as IE 7 and other modern browsers do) I thought we might use this opportunity to launch the AoPS - MathLinks RSS 2.0 Feeds.\r\n\r\nFor AoPS we have [code]http://www.artofproblemsolving. com/Forum/feed.php[/code] (which feeds topics from the HighSchool Forums) and for MathLinks we have [code]http://www.mathlinks. ro/Forum/feed.php[/code] (which feeds topics from the Olympiad forums).\r\n\r\nMore feeds will be available soon aswell.", "Solution_1": "That sounds good!\r\nUmmm...what is an RSS feed?", "Solution_2": "Google for RSS (I'm sure you'll find tons of info about it). To use it the simplest way would be to input those links above in the code tags (without the space between them) in IE 7 and subscribe to the feed.\r\n\r\nHere's how it looks on my computer \r\n[img]7080[/img]", "Solution_3": "It says somethng about not proper authorization", "Solution_4": "How about blog RSS feeds???", "Solution_5": "[quote=\"solafidefarms\"]How about blog RSS feeds???[/quote]Will come soon.", "Solution_6": "Is it something like google reader?\r\nLike I don't have to go to the site to read it?\r\n(In fact, when I used to have google desktop, and had visited a friends blog, I used to get the blog entries on it.)", "Solution_7": "You can use Google to read the feed (although if you do that you might as well go on the forum as you do now :D). It's recommended for Windows Vista (and/or) Internet Explorer 7 users.", "Solution_8": "Isn't the feed just the same thing as viewing the regular web page?\r\nhow do i add feeds on IE7?", "Solution_9": "[b]Instructions for Firefox 2.0+\n[/b]\r\n\r\nGo to this link:\r\n[url]http://www.artofproblemsolving.com/Forum/feed.php[/url]\r\nor\r\n[url]http://www.mathlinks.ro/Forum/feed.php[/url]\r\n\r\nSelect \"Live Bookmarks\" in the drop down box.\r\n\r\nHit \"Subscribe Now\".\r\n\r\nYou should now see an orange square which has a dot and 2 quarter circles on the top part of your browser, under the address bar. Click it, and a drop down box of recent post titles will appear. You can click on one of those titles, and it will take you to the topic.\r\n\r\n[b]Instructions for IE 7\n[/b]\r\n\r\nGo to this link:\r\n[url]http://www.artofproblemsolving.com/Forum/feed.php[/url]\r\nor\r\n[url]http://www.mathlinks.ro/Forum/feed.php[/url]\r\n\r\nHit \"Subscribe to this feed\".\r\n\r\nNow, press your favorites icon and tab over to Feeds. Then click the Art of Problem solving one. It takes you to a page where you can preview all of the posts and if you click on their titles you can go to those topics.", "Solution_10": "You might want to add them to your Gmail web clips :D", "Solution_11": "I didn't want to post the links because both links will point to the same site (the one that the user is using to access the forum) :)", "Solution_12": "[quote=\"#H34N1\"]It says somethng about not proper authorization[/quote]\r\n\r\nIt's giving us Sorry_auth_view_album when we click on the picture link you posted.", "Solution_13": "That's what its saying.", "Solution_14": "When I click on the firefox link, it takes me to a bunch of code with the following error message at the top: \"This XML file does not appear to have any style information associated with it. The document tree is shown below.\"", "Solution_15": "[quote=\"#H34N1\"]That's what its saying.[/quote]\r\n\r\nI know. I was trying to get Valentin's attention...", "Solution_16": "Indeed it is a problem. I may be a gallery mod, but I can't even see Valentin's pictures.", "Solution_17": "I dont' understand, are you having trouble with the image posted or with the RSS feed?\r\n\r\nEdit: the picture was indeed problematic, as I put it my personal album pictures :D Fixed now, you should be able to see it.\r\n\r\nRSS Feeds work properly for me at this point on both Vista and XP and on both IE 7 and Firefox 2.0", "Solution_18": "I am having no problems with the RSS feeds, I only had problems with the picture. :lol:", "Solution_19": "What about RSS feeds for comments?\r\n\r\nWhy need RSS feeds when http://www.artofproblemsolving.com/Forum/marquee.php?ls=4&width=800&height=600", "Solution_20": "[quote=\"#H34N1\"]What about RSS feeds for comments?\n\nWhy need RSS feeds when http://www.artofproblemsolving.com/Forum/marquee.php?ls=4&width=800&height=600[/quote]\r\n\r\nUmm...RSS is like a newspaper. Multiple news stories from multiple sources are delivered to one place, so you can read them all at once in one place. You can also use RSS to pop up a message on your computer desktop every time a new thread is posted, if you are so inclined.", "Solution_21": "How can you do that?", "Solution_22": "There's a bunch of downloadable RSS readers that do that. I know the old version of Pluck used to do it, but it got annoying, so I uninstalled it.", "Solution_23": "Awesome :) . It's good.", "Solution_24": "[quote=\"#H34N1\"]What about RSS feeds for comments?\n\nWhy need RSS feeds when http://www.artofproblemsolving.com/Forum/marquee.php?ls=4&width=800&height=600[/quote]Feeds are a totally different thing than that marquee. That marquee is a list of the latest topic links.\r\n\r\nThe comments you will see by visiting the site :D", "Solution_25": "Wrong topic? :huh:", "Solution_26": "Very nice ! :)", "Solution_27": "[quote=\"Valentin Vornicu\"]\nFor AoPS we have [code]http://www.artofproblemsolving. com/Forum/feed.php[/code] (which feeds topics from the HighSchool Forums) and for MathLinks we have [code]http://www.mathlinks. ro/Forum/feed.php[/code] (which feeds topics from the Olympiad forums).\n\nMore feeds will be available soon aswell.[/quote]\r\n\r\nAre there more feeds yet? It would be awesome to be able to choose feeds on a forum-by-forum basis. \r\n\r\nThanks for all your hard work!", "Solution_28": "We will do it in the near future.", "Solution_29": "Are there different feeds now?\r\n\r\nI recently tried to access the links in the first post and they no longer work :(.", "Solution_30": "What? They do work. http://www.artofproblemsolving.com/Forum/feed.php and http://www.mathlinks.ro/Forum/feed.php", "Solution_31": "Is it possible to get the feed of a particular forum?\r\n\r\neg: Combinatorics forum?", "Solution_32": "Nope, not yet.", "Solution_33": "Sorry I was wondering, but is it possible to get a feed of the middle school and particular forums?\r\n\r\nThanks in advance" } { "Tag": [ "geometry", "3D geometry", "pyramid" ], "Problem": "can someone please explain breifly what are the scaling principles? I was looking an old uil calc test.\r\n\r\nthanks", "Solution_1": "it's a proportion...\r\nif you are scaling a 3-d object, then you will have to cube it\r\n2-d, then square it\r\n\r\nex:\r\na turkey (a 3-d object) that is 11 feet long feeds 10 people, how many people can a 15 feet turkey feed??\r\n\r\nit would be:\r\nlet x be the answer\r\n11^3/10 = 15^3/x\r\nas you can see, the proportion of the turkey must be cubed because it is in a 1-d format.\r\n\r\n\r\nnext for 2d:\r\na 7 feet tall pyramid has a 15 sqr feet base, then what would be the base of a 9 feet tall pyramid.\r\n\r\nans:\r\n7/(sqrt of 15) = 9/(sqrt of x)\r\nthis is because the base is already cubed, so you're looking for the length of the base in proportion with the height...." } { "Tag": [], "Problem": "Se da un triunghi $\\ ABC$ in care notam $\\ G$ centrul de greutate si $\\ L$ punctul lui Lemoine, $\\ AM,BN$ mediane si $\\ R,S$ doua puncte in interiorul triunghiului $\\ ABC$(nu pe laturi).Daca exista relatia \\[(a^{2}-b^{2})(S(CAR)-S(ABR))(S(ABC)-S(BSC))+(b^{2}-c^{2})S(BCR)(S(BSC)-S(ABS))+(c^{2}-a^{2})S(CAS)(S(CAR)-S(ABR))=0\\] demonstrati ca dreptele $\\ LG,MR,SN$ sunt concurente.\r\n\r\nPrin $\\ S(XYZ)$ intelegem aria triunghiului $\\ XYZ$.", "Solution_1": "Impartim tot monstrul ala la $S_{ABC}$, dupa care folosim [b]coordonate baricentrice[/b] (Daca in punctele $A,B,C$ se pun greutatile $S_{BTC}, S_{CTA}, S_{ATB}$, atunci centrul de greutate al sistemului e chiar $T$.). Urmeaza niste calcule plictisitoare, dupa care (cel putin teoretic) obtinem concluzia.\r\n\r\nAcuma n-am timp sa fac calculele. Poate diseara sau maine.", "Solution_2": "Da, asta era si ideea, Tiberiu.Asa am compus-o, folosind coordonatele baricentrice.Poate ceva mai greut era sa stii ca $\\ L\\left(\\frac{a^{2}}{a^{2}+b^{2}+c^{2}},\\frac{b^{2}}{a^{2}+b^{2}+c^{2}},\\frac{c^{2}}{a^{2}+b^{2}+c^{2}}\\right)$", "Solution_3": "Se poate si mai general: Daca $P$ are coordonate $\\left( x, y, z \\right)$ (nu neaparat normalizate), atunci izogonalul sau (nu sunt sigur cum ii zice exact in romana) are coordonate $\\left( \\frac{a^{2}}{x}, \\frac{b^{2}}{y}, \\frac{c^{2}}{z}\\right)$.", "Solution_4": "Da, punctul ala se numeste izogonal.Mai exista si puncte izotomice, dar oricum.." } { "Tag": [ "calculus", "integration", "derivative", "calculus computations" ], "Problem": "$ \\int\\frac {a}{cos^2ax}$\r\n\r\nanswer:tanx+C\r\n\r\nI got the answer but how do I apply the double angle formula to get the same result.\r\nIf it cannot be integrated then it what cases can I apply the double angle formula to integrate\r\n\r\n\r\n$ cos2ax \\equal{} 2cos^2ax \\minus{} 1$\r\n\r\n$ acos^2ax \\equal{} \\frac {cos2ax \\plus{} 1}{2}$\r\n\r\n$ \\int\\frac {a}{cos^2ax} \\equal{} 2a\\int\\frac {1}{cos2ax \\plus{} 1}$\r\n\r\nhow to proceed from here to get $ tanax \\plus{} c$\r\n\r\n--------------------------------------------------------------------------\r\n\r\nfor $ \\int{cos^2x}$\r\n\r\ncan I say =$ \\minus{} \\frac {cos^3x}{sinx}$ and proceed\r\n\r\nBut if I use double angle formula\r\n\r\n$ \\int{cos^2x}$=$ \\int\\frac {cos2x \\plus{} 1}{2}$\r\n\r\ndoes not equal the above expression! wacko wacko!", "Solution_1": "[quote=\"Bos1234\"]how to proceed from here to get tanax + c[/quote]\n\nMake the substitution tan(2ax/2) = tan(ax) = t.\n\n[quote=\"Bos1234\"]can I say = $ \\frac {\\cos^3x}{\\sin x}$ and proceed[/quote]\r\n\r\nWhere did that cos^3x/sinx came from?", "Solution_2": "I don't think you need to use the double angle formula for the first one. It will only make things more complicated. Set $ u\\equal{}ax\\equal{}>du\\equal{}adx$ so you have to integrate $ 1/cos^2 u$ with respect to u. But $ 1/cos^2 u\\equal{}sec^2u\\equal{}d/du(tanu)$ \r\nSo the Integral that you want is $ tanu\\plus{}c\\equal{}tan(ax)\\plus{}c$ where c is a constant.\r\nFor the second one you cannot do what you said. If you take the derivative of $ \\minus{}cos^3x/sinx$ it is not equal to $ cos^2x$\r\nFor the second one you have to use the double angle formula. You can do it from here right?", "Solution_3": "[quote=\"Carcul\"][quote=\"Bos1234\"]how to proceed from here to get tanax + c[/quote]\n\n[quote=\"Bos1234\"]can I say = $ \\frac {\\cos^3x}{\\sin x}$ and proceed[/quote]\n\nWhere did that cos^3x/sinx came from?[/quote]\r\n\r\nsry I misunderstood the theory :blush: \r\n\r\nok thanks I can do it from here" } { "Tag": [ "email", "blogs", "AMC 10", "AMC" ], "Problem": "Welcome to the Mock AMC 12 9/09! [hide=\"85 participants?\"]\n\nbargello\nrd5493\nisabella2296\nPoincare\nbokagadha\nPowerOfPi\nprofessordad\nIhatepie\nnarik\nmodularmarc101\nthemorninghlighttt\nAIME15USAMO\nTachyonPulse\nzapi2007\nbasketball9\nAwesomeToad (assuming you're gonna take it)\nComplex_Ninja\npascal12: preference Friday\nathunder\ncamath08, time assuming\nThunder365\nmaybach\nCyanRedYellow\nveezbo: preference Friday\nTriple-AAA\nlimac\nrts2007\nAlphaBetaTheta\nNuncChaos\nmodx07\nlightning\nAIME15\nmathemonster\nJongao\nbbgun34\nCountdownKing\nSnowEverywhere\nMrBob\nvarunrocks\nwormABC\nmathrat, time assuming\ncrazypianist1116\nPaiev\nyaofan\nSorcerorofDM (and his 9 friends)\nStokes93\nMewto55555\nshentang\nFantasyLover\nmathemagician1729\nbulutcocuk\nBrut3Forc3's math club\nalkjash\nzhou_qiang\npumpkinpi\nblackbelt14253\nSephiroth\nwinternights1\nandersonw\ndnkywin\ncognos599\nzserf\nwestiepaw\nJust_Beginner\ndgreenb801\nBOGTRO\niYOA\nn0tn0t\nDavid2143\npytheagle\ninfinite_sigma\nProtestanT\n15!!!\nnohomo332\ngogators\nEstaticPotter\nBacteria\n[/hide]\n[hide=\"Official Friday participants\"]\nDavid2143 \npascal12 \nveezbo \ncountdownking\nthemorninglighttt\nSnoweverywhere\nBacteria \nprofessordad\n[/hide]\r\n\r\nYou may still take it even if you have not signed up. Please take the Saturday time if possible. However, we will be lenient and let you submit before the beginning of Sunday. It would be nice if you followed the time limit though. :wink: :wink: \r\n\r\n[b]OFFICIAL TEST DATES:[/b]\r\n(Alternate) Friday: 6:00 P.M.-7:15 P.M. / 7:30 P.M. EDT (3:00-4:15/4:30 PDT)\r\nSaturday 9/19 4:00 P.M.-5:15 P.M. / 5:30 P.M. EDT (4:00-5:15/5:30 PDT)\r\n\r\n[b]You will get 75 minutes to officially take this exam. [/b] However, you may turn it in late (Saturday night) and we'll still grade it, but please put how much time it took you to complete the mock AMC 12 for statistics purposes. \r\n\r\n[b]How you should take the test:[/b]\r\nDownload the file. Print it out (hopefully double-sided); close your computer (come back on if you have questions, there may be updates); shut yourself from everyone (take some food/drinks if you need to), and time yourself! Once you're done, get back online and get ready to submit.\r\n\r\n[b]How to submit:[/b]\r\n\r\nPut your answers in blocks of five. Blank answers get a -. For instance, if the answer to every question except for #25 (you left that one blank) was J, you'd submit your answers like this: \r\n\r\nJJJJJ JJJJJ JJJJJ JJJJJ JJJJ- \r\n\r\nAfterwards, complete the survey. There are five questions. We will give you extra time to complete it. (You should take only five extra minutes at most for the survey). Remember, your input will benefit future mock AMCs!\r\n\r\nFinally, PM Caelestor and gfour84 (faster grading), or email us both at caelestor@gmail.com (alternate: caelestor@aol.com) and king-of-cards@cox.net.\r\n\r\n[b]Remember, failure to follow any of these instructions may delay everyone's results![/b]", "Solution_1": "Hello. I looked at some of your blog questions and some of them are really a lot harder than any AMC 12 I have ever seen! Will this be 100% a killer AMC 12?", "Solution_2": "shoot i'm a friday participant?\r\n\r\nerr well... i know for sure i won't be available tonight.. maybe saturday, but i doubt it. :/ stupid father.\r\nsorry~", "Solution_3": "To all ppl signed up today: please check either your PM or email (the one you used to sign up on aops). Have fun! :wink: If you have issues, please contact either me @ caelestor@gmail.com or king-of-cards@cox.net.\r\n\r\nGrr, I hate typos. Note in #15, $ \\frac {\\sqrt {b}}2$ should be $ \\sqrt {b}$. That will be edited in Form B. \r\n\r\nFor the rest of you looking, please do not discuss the test material until Sunday 5:00 A.M. EDT. Feel free to prattle about its difficulty. :rotfl:", "Solution_4": "I may have missed it...but where do we download the test from?", "Solution_5": "[quote=\"limac\"]I may have missed it...but where do we download the test from?[/quote]\r\n\r\nI don't recall you specifying the Friday time. Nevertheless, your email does not exist apparently.\r\nTake it tomorrow.", "Solution_6": "I am a retarded buffoon :mad: :wallbash_red: \r\n\r\nSorry, it completely slipped my mind. If however you are changing the turn-in date (and hopefully keeping the test up?) until Sunday, then I'll still be able to take it tomorrow.", "Solution_7": "[quote=\"veezbo\"]I am a retarded buffoon :mad: :wallbash_red: \n\nSorry, it completely slipped my mind. If however you are changing the turn-in date (and hopefully keeping the test up?) until Sunday, then I'll still be able to take it tomorrow.[/quote]\r\n#\r\nI emailed you the test. Just take it and give yourself a reasonable amount of time (preferably the official 75 minutes).", "Solution_8": "Caelestor, oh, I meant like where can I download the test from, but I want to take it on Saturday (today).", "Solution_9": "Hello. Sorry, I didn't take the test yesterday even though I was one of the Friday participants. Is it alright if I do it today and submit it by tonight?", "Solution_10": "Are you sending it to us via pm? Or posting it here?", "Solution_11": "Email or PM, I think.\r\n\r\nEDIT: [b]Is it going to start in half an hour?[/b] Email or PM?", "Solution_12": "READDDDY?\r\n\r\nThere really is nothing more to say . . . Everyone ENJOY, and good luck (the typos from yesterday have been corrected)", "Solution_13": "Any questions?", "Solution_14": "would a number like 121 be modular for number 18?", "Solution_15": "Well, why would it not be (thats a yes).", "Solution_16": "and also, in the first question, is there meant to be a minus sign in front of the 2009?", "Solution_17": "not that I am aware.", "Solution_18": "Does the polygon in 8 have to be convex?", "Solution_19": "If it does not say, then why would it have to be?", "Solution_20": "Hello everybody,\r\n\r\nThe test was apparently very difficult. Thus, we will try to make it easier in the revision.\r\n\r\nNote that we will be less rigid and take submissions until tomorrow afternoon.", "Solution_21": "so I can solve the problems I didn't have time to do today?", "Solution_22": "The test is really hard... :lol: . When we submit our answers, how do we type blanks? Should we just use Xs or something like that?", "Solution_23": "[quote=\"Just_Beginner\"]The test is really hard... :lol: . When we submit our answers, how do we type blanks? Should we just use Xs or something like that?[/quote]\r\n\r\nBlanks get a -.", "Solution_24": "Oh oops! I just read the directions again. :blush:", "Solution_25": "If you get over 90 on this test, be proud of yourself (unless you cheated :P ).\r\n\r\nWe may offer another \"form\" of this test later, probably an easier version of the current one.", "Solution_26": "Will there be a solutions thread up soon for discussing answers?", "Solution_27": "In passing through the instructions (I am not trying to take this\r\nmock exam or the actual AMC test), you have inconsistent grading.\r\n\r\n\"A correct answer is worth 6 points, an unanswered question is \r\nworth 1.5 points, and an incorrect answer is worth no points.\"\r\n\r\n1) Any unanswered question should be marked as \"incorrectly \r\nanswered\" because it did not get the correct mark by it.\r\n\r\n2) With that said, all unanswered questions should get no points\r\nas well because they are incorrect answers.\r\n\r\n3) a) Look at this disparity in an example: Student A gets 4 problems\r\ncorrect, 8 problems incorrect, and leaves 13 problems unanswered.\r\n6(4) + 8(1.5) + 13(0) = 36 . . . (Score)\r\n\r\nb) Student B turns in an exam that has no answered questions at all.\r\n25(1.5) = 37.5 . . . (Score)\r\n\r\nStudent B's score is higher than Student A's, but Student B shows\r\nno apparent knowledge.", "Solution_28": "Is it possible for you guys to post the answer key since the submitting deadline is over already? I just realized today that there was going to be a Mock AMC 12 -_-.", "Solution_29": "Arrange your tan: Err, this is the exact same way the AMC is graded. Your example is wrong, student A receives 43.5 points. If person A were to receive 36 points with 4 correct answers, he would need 13 wrong and 8 blank, and this doesn't really show any knowledge either. This grading isn't inconsistent, the same concept is used on tons of tests. It discourages guessing (since guessing has a lower expected value). \r\n\r\nRegarding the actual test, I totally failed. Probably shouldn't have taken it at 1 am Friday night. :maybe:", "Solution_30": "gfour84 is doing something in excel. He likes to do statistics. Expect the full results tomorrow, maybe? (You should've received a grade).\r\n\r\nAnd for statistics reasons, try Form B of this test. Hopefully, this is more like a regular mock AMC 12 of yore (note this is still harder than a regular AMC 12). Hey, we all could use some practice.\r\n\r\nMeanwhile, we will open up a tentative results page.", "Solution_31": "When do we have to get the Form B answers turned in? I know that I probably can't get it in until the weekend unless I get really lucky and my teachers decide to go easy on the homework. :P", "Solution_32": "No need to turn in. Just compare your answers once the answer key is completed, and post a response saying how much you improved. Since about 80-90% of the questions are the same, there's no need to time yourself. \r\n\r\nDon't worry, the Mock AMC 10-09 will resemble the actual test easier. Hey, we've got about 10-15 questions unused from the previous one." } { "Tag": [ "geometry", "trapezoid", "parallelogram", "geometry unsolved" ], "Problem": "[color=darkblue]Let $ABCD$ be a trapezoid $(AB\\parallel CD)$ for which exists an interior point $P$ so that\n\n$m(\\widehat{ABP})=2\\cdot m(\\widehat{ADP})$ and $m(\\widehat{DCP})=2\\cdot m(\\widehat{DAP})\\ .$\n\nProve that $\\boxed{\\ m(\\widehat{PAD})-m(\\widehat{PDA})=\\frac{A-B}{2}\\ }$ and in this case\n\n$m(\\widehat{BAP})+m(\\widehat{ADP})=\\frac{A+B}{2}$, $m(\\widehat{CDP})+m(\\widehat{DAP})=180^{\\circ}-\\frac{A+B}{2}\\ .$\n\n[b]Particular case.[/b] If the quadrilateral $ABCD$ is a parallelogram, then $\\frac{A+B}{2}=90^{\\circ}$ and\n\n$m(\\widehat{BAP})+m(\\widehat{ADP})=m(\\widehat{CDP})+m(\\widehat{DAP})=90^{\\circ}$, i.e. $AB=BP=PC=CD\\ .$\n\n[b]Remark.[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=106618 [/color]", "Solution_1": "I've got a solution but I don't have time to put it" } { "Tag": [ "geometry", "similar triangles", "Pythagorean Theorem" ], "Problem": "In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB$, $BC$, and $CD$ are diameters of circle $O$, $N$, and $P$, respectively. Circles $O$, $N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has length\n\n[asy]\nsize(250);\ndefaultpen(fontsize(10));\npair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1];\ndraw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7));\ndot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P);\nlabel(\"$A$\", A, W);\nlabel(\"$B$\", B, SE);\nlabel(\"$C$\", C, NE);\nlabel(\"$D$\", D, dir(0));\nlabel(\"$P$\", P, S);\nlabel(\"$N$\", N, S);\nlabel(\"$O$\", O, S);\nlabel(\"$E$\", E, dir(120));\nlabel(\"$F$\", F, NE);\nlabel(\"$G$\", G, dir(100));[/asy]\n\n$\\textbf {(A) } 20 \\qquad \\textbf {(B) } 15\\sqrt{2} \\qquad \\textbf {(C) } 24 \\qquad \\textbf{(D) } 25 \\qquad \\textbf {(E) } \\text{none of these}$", "Solution_1": "[hide=\"Solution\"]Draw a perpendicular from point N to EF, and let them meet at $M$. Thus, $MF=\\frac 12 EF$. Also notice that $\\triangle NMF$ is right and $NF=15$, as a radius of circle $N$.\n\nNow, we apply similar triangles to see that $\\frac{MN}{GP}=\\frac{AN}{AP}$ and $\\frac{MN}{15}=\\frac{45}{75}$. Solving this, $MN=9$. \n\nNow, we apply the Pythagorean theorem of triangle $NMF$ to get that $MF=12$, and $EF=24$. [/hide]", "Solution_2": "[hide]Yeah dude, just use similar triangles you know!\n\n$GP=15$\n$AP=75$\n$AG=15\\sqrt{26}$ (not really important, unless you wanna brute force)\n\nDrop a perpendicular from point $S$ on $EF$ to point $N$. Similar triangles.\n\n$\\frac{AN}{AP}=\\frac{NS}{PG}$\n\n$\\frac{45}{75}=\\frac{NS}{15}$\n\n$NS=9$\n\nHere was the tricky part of the problem, realizing that $NSF$ was right. At least to me\n\n$9^2+x^2=15^2$\n$x=12$\n\nSince $N$ was the midpoint of the diameter, they're both $ES$ and $SF$ are equal\n$12+12=\\boxed{24}$\n$\\boxed{C}$[/hide]", "Solution_3": "Is the picture available, because the problem means nothing without the diagram in the resource section? :(", "Solution_4": "[geogebra]b5bdb0e94ffadadf1db9e9a8b821ba77327ac27f[/geogebra] \r\n\r\nHere's a rough drawing of it." } { "Tag": [ "function", "linear algebra", "matrix", "advanced fields", "advanced fields unsolved" ], "Problem": "Show that if f is a diffeomorphism M -> N and g is C^1 close enough to f (and is\r\nsmooth), then g is a diffeomorphism.\r\n\r\nI am not even sure what \"C^1 close enough\" means.\r\n\r\nThanks.", "Solution_1": "It means close enough in the $ C^1$ norm, that is both $ |f\\minus{}g|$ and $ |f'\\minus{}g'|$ are bounded by a constant that you can take as small as you wish. \r\nThe tools to use: (1) inverse function theorem; (2) the set of invertible matrices is open (under any kind of matrix norm).", "Solution_2": "I wasn't quite able to prove it...Could you please give me more details?\r\nthanks" } { "Tag": [ "inequalities", "number theory", "relatively prime", "algebra unsolved", "algebra" ], "Problem": "let s dinote the set of rational numbers in the interval(0,1),dertermine with proof if there exists a subset T of S such that erery element in s can be uniquely witten a the sum of finitely many distinct elements in T", "Solution_1": "The answer, I think, is no. Suppose that $ T$ exists.\r\n\r\n[b]Lemma.[/b] If $ a$ and $ b$ are elements of $ T$ for which $ a>b$, then $ a\\ge 2b$.\r\n\r\n[i]Proof.[/i] Suppose otherwise. Then $ a\\minus{}b < a < b$. There exists a set $ R \\subset T$ such that $ \\sum_{r\\in R}r\\equal{}b\\minus{}a$; clearly, $ a,b\\notin R$. Then\r\n\\[ a \\equal{} b\\plus{}(a\\minus{}b) \\equal{} b \\plus{} \\sum_{r\\in R}r, \\]\r\ncontradicting the hypothesis that $ a$ has a [i]unique[/i] representation as a finite sum of elements of $ T$. $ \\blacksquare$\r\n\r\nIt follows from the lemma that every nonempty subset of $ T$ has a maximal element. In particular, if $ \\alpha$ is the supremum of a nonempty subset $ R$ of $ T$, then we can pick $ a\\in R$ larger than $ \\alpha/2$, and for any $ b\\in T$ for which $ b>a$, we have the inequality $ b\\ge 2a > \\alpha$, so $ b\\notin T$; hence $ a$ is the maximum of $ T$.\r\n\r\nWe can form a sequence $ a_0, a_1, \\dotsc$ of elements of $ T$ recursively, so that\r\n\\[ a_k \\equal{} \\max T\\setminus \\{a_j\\}_{0\\le j 0(+)? (Right limit) \r\n\r\nWhat is the limit of f (x) for x => 0(-)? (Left limit) \r\n\r\nWhat is the limit of f (x) for x => 0? (Limit at x=0) \r\n\r\nThanks\r\nleonsotelo", "Solution_1": "$ f(x)\\equal{}0^{x}$? That's just $ f(x)\\equal{}0$, or the $ x$-axis.", "Solution_2": "[quote=\"JRav\"]$ f(x) \\equal{} 0^{x}$? That's just $ f(x) \\equal{} 0$, or the $ x$-axis.[/quote]\r\n\r\nNo, 0^x doesn't exist for nonpositive numbers, so the limit from the right side is 0, from the left side is undefined, so the limit at x=0 is also undefined.", "Solution_3": "The limit of x^x as x tends to zero (from the right) is 1. In other words, if we want the x^x function to be right continuous at 0, we should define it to be 1. \r\n\r\nleon-sotelo@hotmail.com" } { "Tag": [ "algorithm", "Support", "calculus", "integration" ], "Problem": "I am a 4th year High school student and am thinking about participating in the Informatics Olympiad(the discussion regarding learning languages ec. is in the \"Reality Check?\" post in this forum). I Believe the most important thing(besides knowing th elanguages of course) is the Algorithms. There are a number of books out there and though most people think\"Introuction to Algorithms\" by Cormen is the authoritative book, I don't think I have the time to read 1200 pages. I have considered the \"Art of Computer Programming\" series(the BIBLE) and have got the first part. Which other bokks on Algorithms would you recommend? Please suggest books that are not tooo long ( Probably < 600- 700 pages) and are introductory while good for the IOI. Are the books recommended at the official site ( [url=http://olympiads.win.tue.nl]olympiads.win.tue.nl[/url] Good enough? Which are the other books that might be used for the Olympiad???", "Solution_1": "I think there are more forums of the same character as this one, and I have already posted my opinions in those.\r\n\r\nDon't start reading books, start coding, do lots of PRACTICE.\r\n\r\nAnd the way of thinking that \"I don't have time to read a 1200 page book\" seems a little odd to me. If you don't have time to read a book, you don't have time to practice. And then, you probably don't have enough enthusiasm into programming. You either like it and devote your time to it, or you don't.\r\n\r\nThere is no recipe to learn to get gold at IOI overnight. Sorry.", "Solution_2": "Although you are write that practice is THE KEY, I am a total beginner and as you might know, I cant writing codes starightaway!! That is why I am asking for books on algorithms. Also, which programming language would you most recommnd? Where can I get it and its learning documentation? Is Java good? What's so special in Java? ALso, can I get Visual C++???", "Solution_3": "In that case, do not start reading books on algorithms. Try handling the basics (I mean - really basics) of the language first from any kind of a book about a language or web documentation (google: c++ tutorial), and then start coding, from the simplest algos. USACO is good for this.\r\n\r\nI started with Pascal, this year switched to C++. Maybe Pascal is a little easier to learn for a beginner but their essence is almost the same.", "Solution_4": "Is Java good? Does it have wide applicabiliy??? I have a book on C++ and a web tutorial . Should I continue with it? Pascal is a bit less well known. Is it because it had some shortcoming. Also, which is the best compiler for C++. When should I start studying algorithms? I was thinking if reading \"knuth\" alongside learning languages!!!", "Solution_5": "Java is a really useful language, if I was starting to learn a language I'd definitely pick Java. They say it's easier to learn, easier to use, it is supposed to be secure and reliable and platform independent and multimedial and all kinds of things that you could use if you were writing everyday real-world applications... however, it's main disadvantage at algorithm competitions is that it runs much slower. Fortunately, the people that run these competitions are aware of it and they usually set time limits for Java submits higher (5x the normal time limit or so...). Not all the competitions support Java submissions though.\r\nAs for Pascal, it is not good for anything except for the fact that there is Delphi, a (kind of) free development enviroment with which you can create windows GUI applications easily.\r\nC/C++ is considered [i]the[/i] language for algorithm competitions. (At least by me :).) This would be my choice of language to learn if you're planning to compete at IOI-like competitions. As I've already written at a different forum, a very good compiler + development enviroment is Dev-C++ which is completely free. That is for Windows. Under Linux, C/C++ compilers are usually integral part of the operating system.", "Solution_6": "C sucks, learn C++.", "Solution_7": "Does Knuths \"Art of Computer Programming: Volume 1(Fundamental Algorithms) \" cover all algorithm related issues that might appear in competitions or besides olympiads, does the book cover all you need to know[hide]( by 'all' I obviously don't mean what it says-- just 'all' means covering matter useful for atleast 80% of the population)[/hide] about algorithms ,etc??? Many people recommended the \"Algorithm Design Manual\" by Skiena. Which obe would you prefer. ALso, does \"Practice of Programming\" by Kernighan include algorithms only?? What is its use in IOI? I got \"Programming Pearls\" -- is it good???" } { "Tag": [ "induction", "ceiling function", "algebra proposed", "algebra" ], "Problem": "The real sequence is defined as follows: $ x_0 \\equal{} 1, x_1 \\equal{} 1 \\plus{} k$, where $ k$ is a positive real, $ x_{2n\\plus{}1} \\minus{} x_{2n} \\equal{} x_{2n} \\minus{} x_{2n\\minus{}1}$, and $ \\frac{x_{2n}}{x_{2n\\minus{}1}} \\equal{}\\frac{x_{2n\\minus{}1}}{x_{2n\\minus{}2}}$. Show that $ x_n > 1994$ for all sufficiently large $ n$.", "Solution_1": "But can a sequence be a gp as well as an ap???? only a,a,a,.... is possible as far as i know.....", "Solution_2": "doesn't such a definition generate them\r\nP.S hell_ever i have sent u a Private Message please check it at the top right hand side of this page", "Solution_3": "It is simple enough to prove by induction the formula\r\n\\[ a_{2n}\\equal{}(1\\plus{}nk)^2,a_{2n\\plus{}1}\\equal{}(1\\plus{}nk)(1\\plus{}n(k\\plus{}1))\\]\r\nThis gives $ a_M>1994$ for $ M>2 \\left\\lceil \\frac{\\sqrt{1994}\\minus{}1}{k} \\right\\rceil$." } { "Tag": [ "calculus", "integration", "trigonometry", "limit", "derivative", "calculus computations" ], "Problem": "compute $\\int_0^{+\\infty} \\frac{sin(tx) dx}{x(x^2+1)}$", "Solution_1": "We assume that $t>0$ here.\r\nLet $F(t) = \\int_0^{+\\infty} \\frac{\\sin(t x) dx}{x(x^2+1)}$.\r\nIt is clear that $F(t)$ exists for all $t>0$.\r\nand\r\n$\\int_0^{+\\infty} \\frac{\\partial}{\\partial t}\\left(\\frac{\\sin(t x)}{x(x^2+1)}\\right) dx = \\int_0^{+\\infty} \\frac{\\cos tx}{x^2+1} dx$\r\nconverges uniformly for all $t>0$. (from Weierstrass's Test).\r\nHence\r\n$F'(t) = \\int_0^{+\\infty} \\frac{\\cos tx}{x^2+1} dx$.\r\nNow let us consider $F''(t)$.\r\n$\\int_0^{+\\infty} \\frac{\\partial}{\\partial t}\\frac{\\cos tx}{x^2+1} dx = -\\int_0^{+\\infty} \\frac{x\\sin tx}{x^2+1} dx$ which uniformly converges for all $t\\in [a,A]$ (from Dirichlet's Test).\r\nThus\r\n$F''(t) = -\\int_0^{+\\infty} \\frac{x\\sin tx}{x^2+1} dx$.\r\nNow we see that\r\n$F''(t)-F(t) = -\\int_0^\\infty \\frac{\\sin{tx}}{x}dx = -\\frac{\\pi}{2}$.\r\nSolve the differential equation,\r\n$F(t) = \\frac{\\pi}{2} + C_1 e^t + C_2 e^{-t}$.\r\nFrom $F(0) = 0$, we have $C_1 + C_2 = -\\frac{\\pi}{2}$. \r\nFrom $\\lim_{t\\to\\infty} F(t) = 0$ (using Riemann's Theorem) , we know that $C_1 = 0$, and then $C_2 = -\\frac{\\pi}{2}$.\r\nAnd $F(t) = \\frac{\\pi}{2}(1-e^{-t})$.", "Solution_2": "The Residue Theorem!\r\n\r\nConsider an indented (around the origin) semicircular contour in the upper half plane. We know that the contribution along the upper part goes to zero by the Jordan Lemma. \r\n\r\nAs the indented portion shrinks to the origin, the residue at $0$ is $1$ (independent of $t$), so\r\n\r\n$\\int^{\\infty}_{-\\infty} \\frac {e^{iz}dz}{z(z^2+1)}=\\pi i+2\\pi i\\text{Res}_{z=i}f(z)=\\pi i-\\pi ie^{-t}$\r\n\r\nUse the evenness of the integral to get $\\frac {\\pi}{2}(1-e^{-t})$\r\n\r\nJust as a note, I used\r\n\r\n$\\text{Res}_{z=z_0} \\frac {e^{itz}}{z(z^2+1)}=\\frac {e^{itz}}{3z^2+1}$ to calculate the residues.", "Solution_3": "I was trying not to use complex analysis :oops:", "Solution_4": "of course that is the idea Liyi. Nevertheless, though it is not hard to prove, I can't understand how you justify the computation of $F\\\"(t)$. (what is this dirichlet test ?).", "Solution_5": "Exchanging differentiation and integration signs need some conditions. In the frame of mathematical analysis (Lebesgue's integral is not introduced), the condition is strong. We have the following theorem.\r\n\r\n[u]Theorem[/u]. Suppose $f(x,y)$ and $f'_y(x,y)$ are continuous on $[a,\\infty)\\times[c,d]$ and $\\int_a^\\infty f(x,y)dx$ converges for all $y\\in[c,d]$; $\\int_a^\\infty f'_y(x,y)dx$ uniformly converges for all $y\\in[c,d]$. Then we have that $I(y) = \\int_a^\\infty f(x,y)dx$ is differentiable on $[c,d]$ and \\[I'(y) = \\int_a^\\infty f'_y(x,y)dx\\].\r\n\r\n[u]Dirichlet's Test[/u]. If $f(x,y)$ and $g(x,y)$ satisfies\r\n(1) $\\int_a^A f(x,y) dx$ is uniformly bounded;\r\n(2) $g(x,y)$ is monotone about $x$ (for a fixed $y$);\r\n(3) $g(x,y)$ uniformly tends to 0 about $y$ as $x$ tends to $\\infty$\r\nthen $\\int_a^\\infty f(x,y)g(x,y)dx$ uniformly converges for all $y\\in [c,d]$.", "Solution_6": "[quote=\"liyi\"]... $\\int_a^\\infty f'_y(x,y)dx$ uniformly converges for all $y\\in[c,d]$. ...[/quote]\r\n\r\nI don't understand what you mean here.", "Solution_7": "Okay.\r\n$\\int_a^\\infty f(x,y)dx$ uniformly converges for all $y\\in S$ means that\r\n\r\nfor $\\epsilon > 0$, there exists $A>a$ such that $\\left|\\int_A^\\infty f(x,y)dx\\right| < \\epsilon$ for all $y\\in S$.", "Solution_8": "[img]https://sites.google.com/site/atzahrani/example-artofproblemsolving30.png[/img]" } { "Tag": [ "function", "quadratics", "number theory solved", "number theory" ], "Problem": "Let $g(n)$ be define as follows: $g(1)=0,g(2)=1$ and\r\n$g(n+2)=g(n)+g(n+1)+1$, ($n\\geq 1$).\r\nProve that if $n>5$ is a prime, then $n$ divides $g(n)(g(n)+1)$.", "Solution_1": "Some problems like this one have been posted beforer on the forum. As usual, I'm stuck half-way :). In my notation $g(n)=a_n$ (I'm more comfortable with that :)).\r\n\r\nBy computing the generating function of the sequence (pretty boring stuff; I don't want to do it here), we can find $a_n=\\frac {(-1)^n}{\\sqrt 5}(r_+^{n+1}-r_-^{n+1})$, where $r_+=\\frac {-1+\\sqrt 5} 2$ and $r_-=\\frac {-1-\\sqrt 5} 2$. \r\n\r\nWe have two cases: if (a) $(\\frac 5p)=1$ then we can regard $r_{\\pm}$ as numbers in $\\mathbb Z_p$, and we easily reach the conclusion that $p|a_p$. \r\n\r\nOn the other hand, if (b) $(\\frac 5p)=-1$ then we have to look at the $r$'s in the field (it's now a field) $\\mathbb Z_p[\\sqrt 5]$. We should show that $p|a_p+1$, but in order to do this we must show that in $\\mathbb Z_p[\\sqrt 5]$ we have $r^{2(p+1)}=1$, where $r$ is any of $r_{\\pm}$. I don't know how to do this because, as I've mentioned, $\\mathbb Z_p[\\sqrt 5]$ is a field so it has $p^2-1$ invertible elements, and $2(p+1)$ is much smaller than $p^2-1$.\r\n\r\nAnother approach might work.", "Solution_2": "There I go again missing obvious things :):\r\n\r\nIn the case (b) we just lift $r_+$ to the power $p+1$ if we get $\\pm 1$ we're Ok. We also use the fact that $\\binom {p+1}k=1 (\\mod p)$ for $k=0,1,p,p+1$ and $0 (\\mod p)$ otherwise. We get, after all the computations, $\\frac{1-\\sqrt 5-5^{\\frac {p-1}2}\\sqrt 5+5^{\\frac {p+1}2}}4$. Because $5$ isn't a quadratic residue of $p$, we have $5^{\\frac {p-1}2}=-1 (\\mod p)$, so that thing is equal to $-1 (\\mod p)$, so $r^{p+1}=-1\\Rightarrow r^{2(p+1)}=1$, Q.E.D. \r\n\r\nWe don't need any freaky fields or rings or anything like that :).", "Solution_3": "I think you may find smth interesting here: \r\n[url]http://www.mathlinks.ro/viewtopic.php?t=1428[/url]\r\n\r\nFor grobber: I think you are not doing well working with $a_n$ as linear combination of (n+1)th powers of $r^+$ and $r^-$. nth powers are easier to work with because you know that $(a+b)^p=a^p+b^p$ when working mod p. Of course you can adapt, but nth powers are just nicer." } { "Tag": [ "algebra", "function", "domain", "algorithm", "polynomial", "Ring Theory", "superior algebra" ], "Problem": "Let $K$ a comutative field. Find all prime ideals from $\\mathbb{K} [X].$. ;)", "Solution_1": "Yeah- $K[X]$ is a principal ideal domain by the divison algorithm, so the prime ideals are precisely the ideals generated by irreducible polynomials.", "Solution_2": "can u be more explicit?\r\n ;)", "Solution_3": "$\\mathbb{K}[x]$ is an euclidean domain (by the degree of the polynomials), so also a principal ideal one. Now if $\\mathbf{o}$ is an ideal of $\\mathbb{K}[x]$, there is an $a \\in \\mathbb{K}[x]$ with $\\mathbf{o} = a \\mathbb{K}[x]$.\r\nIf $\\mathbf{o}$ is prime, we get that $a$ is prime $=$ irreducible.\r\nNow if $xy=az \\in a\\mathbb{K}[x]$, you get that $a$ divides $x$ or $y$, so the inverse also holds." } { "Tag": [ "LaTeX" ], "Problem": "Is there any way to make text smaller than \\tiny?", "Solution_1": "It depends on the font you are using. If you are using a standard font then the [url=http://tug.ctan.org/pkg/fix-cm]fix-cm package[/url] will scale them to any size.\r\n[code]\\usepackage{fix-cm} \n...\n{\\tiny This is tiny}\\\\\n\\fontsize{3}{5pt}\\selectfont This is size 3[/code]" } { "Tag": [], "Problem": "$S$ is a subset of $\\left\\{1,2,3,...,2n\\right\\}$ with $n+1$ elements. Show that we can always choose $a,b \\in S$ so that $a | b$.", "Solution_1": "This has been posted before. We write each number in the form $2^{a}b$ where $b$ is odd and $a$ is a nonnegative integer. Then we have every even number in the set is a power of 2 times some odd number in the set. Then, we have only n odd numbers, so we have $n$ psosible values for $b$. Then two numbers written this way have the same $b$, as desired.", "Solution_2": "Indeed. I heared this question from one of the Dutch team members on IMC. \r\n\r\n(Did I give you this problem in fact, Jan? I only remember giving it to Arne and Christophe?)", "Solution_3": "An absolute Erd\u00f6s classic. I've known this problem since... 4 years or something.", "Solution_4": "We dealed with this problem on the training weekend in Beersel :)", "Solution_5": "And there will always be two such that $gcd(a,b)=1$", "Solution_6": "That holds because there will always be two consecutive integers." } { "Tag": [ "geometry", "circumcircle", "Euler", "incenter", "geometry unsolved" ], "Problem": "$\\triangle ABC$ is scalene with circumcircle $(O)$ and incircle $(I).$ Let $X,Y,Z$ be the midpoints of the arcs $BC,CA,AB$ of $(O).$ Show that the inverse lines of the circles with diameters $IX,$ $IY,$ $IZ,$ under the inversion with respect to $(I),$ cut the corresponding sidelines of the medial triangle of $\\triangle ABC$ at three collinear points. Furthermore, show that this line also passes through the pole of the Euler line of $\\triangle ABC$ WRT $(I).$", "Solution_1": "After massive calculations, I found that all three points satisfy the equation\r\n\\[ XI^2 \\minus{} XG^2 \\equal{} \\frac{16Rr \\plus{} r^2 \\minus{} s^2}{9},\\]\r\nwhere $ I$ is the incenter and $ G$ is the centroid. Thus, all three points lie on a line that is perpendicular to $ GI$. I don't think the pole of the Euler line lies on this line. It would be nice to see a synthetic solution.", "Solution_2": "Thanks for your interest dear nsato, the problem is indeed correct.\n\n[url=http://www.xtec.cat/~qcastell/ttw/ttweng/resultats/r276.html]Lemma 1[/url]. If $ M_c, M_b$ are midpoints of $ AB$ and $ AC,$ then the pole of $ M_bM_c$ WRT $(I)$ is the orthocenter $ H_1$ of $\\triangle IBC.$\n\n[url=http://www.xtec.cat/~qcastell/ttw/ttweng/definicions/d_Schiffler_p.html]Lemma 2[/url]. Euler lines of $\\triangle ABC,\\triangle IBC,\\triangle ICA,\\triangle IAB$ concur at Shiffler's point of triangle $\\triangle ABC.$\n\nThe inverse line $p_a$ of the circle with diameter $ IX,$ under inversion WRT $(I),$ is the polar of $X$ WRT $(I).$ The polar of $ H_1$ WRT $(I)$ is the A-midline $M_bM_c$ of $\\triangle ABC.$ Therefore, $J_A \\equiv M_bM_c \\cap p_a$ is the pole of $XH_1,$ the Euler line of $\\triangle IBC.$ Similarly, $J_B,J_C$ are the poles of the Euler lines of $\\triangle ICA,$ $\\triangle IAB$ $\\Longrightarrow$ $J_A,J_B,J_C$ are collinear on the polar of the Schiffler's point of $\\triangle ABC$ WRT $(I)$ $\\Longrightarrow$ $J_AJ_BJ_C$ passes through the pole of the Euler line WRT $(I),$ as desired.", "Solution_3": "First, I misread the question. But I still got three collinear points. :| I'll be posting it as another problem.\r\n\r\nSecond, thanks for your very nice problem and solution." } { "Tag": [], "Problem": "\u0633\u0644\u0627\u0645 \u0628\u0647 \u0647\u0645\u0647 \u0645\u0646 \u0633\u0627\u0646\u0627\u0632\u0645 . \u0645\u0646 4 \u0631\u0648\u0632\u0647 \u062f\u0627\u0631\u0645 \u0631\u0648 \u0627\u06cc\u0646 \u0633\u0648\u0627\u0644 \u0641\u06a9\u0631 \u0645\u06cc \u06a9\u0646\u0645 \u06a9\u0633\u06cc \u0627\u0633\u062a \u06a9\u0645\u06a9\u0645 \u06a9\u0646\u0647 \u061f\r\nR ----->R+\r\n\r\nF(x^2+y^2)=F(x^2-y^2)+F(2xy)\r\n\r\n So F(x) = ? .....\r\n \r\n\r\n\u0644\u0637\u0641\u0627 \u06a9\u0645\u06a9\u0645 \u06a9\u0646\u06cc\u062f ...", "Solution_1": "age gharar bedim $ x\\equal{}0$ natije mishe $ f(y^2)\\equal{}f(\\minus{}y^2)$ yeni dar vaghe baraye har $ x\\in\\mathbb{R}$ darim $ f(x)\\equal{}f(\\minus{}x)$ khob pas kafie $ f$e aadade mosbat ro hesab konim:\r\n\r\ngharar bedid $ x^2\\minus{}y^2\\equal{}a,2xy\\equal{}b$ khob hala deghat konid ke darim:\r\n\r\n$ a^2\\plus{}b^2\\equal{}x^4\\plus{}y^4\\minus{}2x^2y^2\\plus{}4x^2y^2\\equal{}x^4\\plus{}y^4\\plus{}2x^2y^2\\equal{}(x^2\\plus{}y^2)^2$\r\n\r\npas dar vaghe:\r\n\r\n$ x^2\\plus{}y^2\\equal{}\\sqrt{a^2\\plus{}b^2}$\r\n\r\npas ba tavajoh be farze masale khahim dasht:\r\n\r\n$ f(\\sqrt{a^2\\plus{}b^2})\\equal{}f(a)\\plus{}f(b)$\r\n\r\nya be ebarati baraye har $ a,b\\in\\mathbb{R}$ darim:\r\n\r\n$ f(\\sqrt{a\\plus{}b})\\equal{}f(\\sqrt{a})\\plus{}f(\\sqrt{b})$\r\n\r\nkhob hala gharar bedid $ g(x)\\equal{}f(\\sqrt{x})$ khob pas avvalan $ g: \\mathbb{R}^\\plus{}\\to\\mathbb{R}^\\plus{}$ hamchenin $ g(a\\plus{}b)\\equal{}g(a)\\plus{}g(b)$ khob hala inam moadele kooshye,ke ye moadele maroofe,dar har soorat gharar bedid $ b\\equal{}\\epsilon\\to 0^\\plus{}$ pas darim:\r\n\r\n$ g(a\\plus{}\\epsilon)\\equal{}g(a)\\plus{}g(\\epsilon)>g(a)$\r\n\r\npas dar vaghe tabamoon soudie,khob pas javabamoon hast $ g(a)\\equal{}ca$ yeni $ f(x)\\equal{}cx^2$... :wink:", "Solution_2": "\u0622\u06cc\u0627 \u06a9\u0633\u06cc \u0631\u0627\u0647 \u062d\u0644 \u062f\u06cc\u06af\u0631\u06cc \u062f\u0627\u0631\u062f \u061f", "Solution_3": "[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cauchy&t=51585[/url]" } { "Tag": [ "calculus", "integration", "logarithms", "trigonometry", "special factorizations", "calculus computations" ], "Problem": "Find the following integrals:\r\n\r\n1) Find the indefinite integral $ \\int \\frac{3^{\\tanh x}}{\\cosh ^{2}x}dx$.\r\n\r\n2) Find the indefinite integral $ \\int \\frac{dx}{x\\sqrt{1\\plus{}(\\ln x)^{2}}}$.\r\n\r\n3) Find the indefinite integral $ \\int e^{2x}\\sin ^{2}xdx$.\r\n\r\n4) Find the indefinite integral $ \\int \\frac{x\\plus{}1}{x^{2}\\plus{}x\\plus{}1}dx$.", "Solution_1": "Hi.\r\n\r\nFor 3 I would write $ \\sin^{2} x$ in terms of $ \\cos 2x$ and use integration by parts.\r\n\r\nFor 4. I would split the integrand into $ \\frac{1}{2}\\frac{2x\\plus{}1}{x^{2}\\plus{}x\\plus{}1} \\plus{}\\frac{1}{2}\\frac{1}{x^{2} \\plus{} x \\plus{} 1}$. The first one is a log and the second, after completing the square is an $ \\tan^{\\minus{}1}$, I think.\r\n\r\nSorry, gotta go.", "Solution_2": "For 2 we first substitute lnx=u so we have\r\n$ \\begin{array}{l}\r\n I \\equal{} \\int {\\frac{{du}}{{\\sqrt {1 \\plus{} {u^2}} }}} \\mathop \\equal{} \\limits_{du \\equal{} \\frac{1}{{{{\\cos }^2}t}}dt}^{u \\equal{} \\tan t} \\int {\\frac{{\\cos t}}{{{{\\cos }^2}t}}dt \\equal{} } \\int {\\frac{{{{\\left( {\\sin t} \\right)}^\\prime }}}{{\\left( {1 \\minus{} \\sin t} \\right)\\left( {1 \\plus{} \\sin t} \\right)}}dt \\equal{} } \\\\ \r\n \\equal{} \\frac{1}{2}\\left[ {\\ln \\left( {1 \\plus{} \\sin t} \\right) \\minus{} \\ln \\left( {1 \\minus{} \\sin t} \\right)} \\right] \\plus{} c \\equal{} \\\\ \r\n \\equal{} \\frac{1}{2}\\ln \\left( {\\frac{{1 \\plus{} \\sin t}}{{1 \\minus{} \\sin t}}} \\right) \\plus{} c \\equal{} \\\\ \r\n \\equal{} \\frac{1}{2}\\ln \\left( {\\frac{{1 \\plus{} \\sin (arc\\tan u)}}{{1 \\minus{} \\sin (arc\\tan u)}}} \\right) \\plus{} c \\equal{} \\\\ \r\n \\equal{} \\frac{1}{2}\\ln \\left( {\\frac{{1 \\plus{} \\sin (arc\\tan (\\ln x))}}{{1 \\minus{} \\sin (arc\\tan (\\ln x))}}} \\right) \\plus{} c \\\\ \r\n \\end{array}$", "Solution_3": "$ \\frac{d}{dx}(\\tanh x)\\equal{}\\frac{1}{\\cosh ^2x}$." } { "Tag": [], "Problem": "[i]Source[/i]:[b][u]Spanish Math Olympiad[/u][/b]\r\n\r\nShow that $\\sqrt{x} + \\sqrt{y} + \\sqrt{xy} = \\sqrt{x} + \\sqrt{(y + xy + 2y\\sqrt{x} )}$.", "Solution_1": "When you square and multiple $ \\sqrt{y} and\\ \\sqrt {xy} $ , I belive, the signs of x and y come into picture. Rest is ok. Isn't it?", "Solution_2": "I think probably it has got more to it but it is form an olympiad, that is weird!", "Solution_3": "what to do you think", "Solution_4": "I think it is very obvious that it is equivalent :roll: \r\n\r\n[hide] $\\sqrt{(\\sqrt{y} + \\sqrt{xy})^2} = \\sqrt{y+xy+2y\\sqrt{x}}$ [/hide]", "Solution_5": "Hi,\r\nIs this really an exercise from Olympiad tests??? :o \r\nIt's very easy from the start we have [tex]\\displaystyle \\sqrt{x} + \\sqrt{y} + \\sqrt{xy} = \\sqrt{x} + \\sqrt{(y + xy + 2y\\sqrt{x} )}[/tex] we can simplify by [tex] \\sqrt{x} [/tex] the rest is more difficult :D \r\nSo how can this be an Olympiad test :P", "Solution_6": "A catch is that when we have $ 2 \\times \\sqrt {y} \\times \\sqrt {xy} $ , we can write it as $2 \\sqrt{ x y^2 } $ if and only if at least one of $ y $ & $ xy $ is positive. Maybe thats what they are asking for", "Solution_7": "No,\r\nWhen [tex] \\sqrt{x}[/tex] and [tex] \\sqrt{y} [/tex] are defined so [tex] x \\ge 0 [/tex] and [tex] y \\ge 0 [/tex]", "Solution_8": "If you can write $x$ in such way that $\\sqrt{x}$ is real , then definitely we can say $x$ is positive . :P", "Solution_9": "look here\r\n[url]http://www.kalva.demon.co.uk/spain/ome90.html[/url]", "Solution_10": "Ah yes, Maybe the goal is to show the other equality It's more difficult. Did you show it?", "Solution_11": "of course , my point is valid in $ \\mathbb{C} $ nos. Also, the 2nd equality is by no means difficult." } { "Tag": [ "probability" ], "Problem": "What is the probability that a randomly-selected two-digit\nnumber is a multiple of 14? Express your answer as a common\nfraction.", "Solution_1": "There are 90 two digit numbers. Out of these, 14*1 thru 14*7, or 7 numbers, are multiples of 14. This is $ \\boxed{7/90}$", "Solution_2": "[hide]1.there are 90 2 digit #'s from 1-100 '\n\n2.there are 14*1-14*7 multiple of 14 from 1-99 and thats 7 \n\n3.so we get our answer as 7/90[/hide]\n\n :spam: :spam: :spam: :spam:" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "The vertices of triangle BAC are lattice points (they have integer coordinates). There are no other lattice points on the boundary of the triangle, but there is exactly one lattice point inside the triangle. Prove that it must be the centroid of ABC.", "Solution_1": "See here: [url]http://www.mathlinks.ro/viewtopic.php?t=150968[/url]." } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "Prove that for any integers $ n \\ge 2$ there is a set $ A_n$ of $ n$ distinct positive integers such that for any two distinct elements $ i,j \\in A_n, |i\\minus{}j|$ divides $ i^2\\plus{}j^2.$", "Solution_1": "[quote=\"moldovan\"]Prove that for any integers $ n \\ge 2$ there is a set $ A_n$ of $ n$ distinct positive integers such that for any two distinct elements $ i,j \\in A_n, |i \\minus{} j|$ divides $ i^2 \\plus{} j^2.$[/quote]\r\nThis problem has been posted before. You should notice some ideas : \r\n Let A be a set of numbers , we defind $ A \\plus{} k \\equal{} \\{x \\plus{} k|x\\in A\\}$ . \r\n1) If $ A$ be a good set then $ A \\plus{} d$ also is a good set where $ d \\equal{} \\prod_{i < j}(j \\minus{} i)$ . \r\n2)If A be a good set ,then $ A \\plus{} \\{0\\}$ is a good set .\r\nInduction on N will do the rest .", "Solution_2": "Take $ A_n \\equal{} \\{ n!,2*n!,...,n*n!\\}$." } { "Tag": [ "geometry", "perimeter", "inradius", "circumcircle", "HCSSiM", "incenter", "area of a triangle" ], "Problem": "The circle inscribed in a triangle with perimeter 17 cm. has a radius of 2 cm. What is the area of the circle? What is the area of the triangle?", "Solution_1": "[hide]The area of the circle will be 4pi. The area of the triangle will be 17 because using the formula rs. r=inradius,s=semi perimter[/hide]", "Solution_2": "[quote=\"mecrazywong\"][quote=\"Altheman\"]for the other question...\n\nnote that $p=a+b+c$, $A=\\frac{1}{2}rp$, $R=\\frac{abc}{4A}=\\frac{abc}{2rp}$, \n\nby am-gm, $\\frac{a+b+c}{3}\\ge\\sqrt[3]{abc}$\n\n$\\frac{p^{3}}{27}\\ge 2rpR$\n\n$R\\le \\frac{p^{2}}{54r}$ with equality when $a=b=c$\n\np.s. this is nice![/quote]\nSorry to spoil the fun, you have only found a upper bound for the circumradius of $\\triangle ABC$.[/quote]\n\nAnd... that's what I had asked for in my follow-up question, if you hadn't read it in the post above.\n\n[quote=\"I\"][i]Another question:[/i] With the same perimeter and inradius, what's the maximum radius of the circumcircle?[/quote]", "Solution_3": "[quote=\"Shivar\"]The circle inscribed in a triangle with perimeter 17 cm. has a radius of 2 cm. What is the area of the circle? What is the area of the triangle?[/quote]\r\n\r\n[hide]\n\nThe area of the circle is trivial. The radius is given, so $\\pi \\cdot r^{2}= 4\\pi$.\n\nI'm still thinking about the triangle, though.\n\n[/hide]", "Solution_4": "[quote=\"erik-the-red\"][quote=\"Shivar\"]The circle inscribed in a triangle with perimeter 17 cm. has a radius of 2 cm. What is the area of the circle? What is the area of the triangle?[/quote]\n\n[hide]\n\nThe area of the circle is trivial. The radius is given, so $\\pi \\cdot r^{2}= 4\\pi$.\n\nI'm still thinking about the triangle, though.\n\n[/hide][/quote]\r\n\r\nUse the fact that the area of a triangle is $A=rs$, where $A$ is the area, $r$ is the inscribed circle's radius, and $s$ is the semiperimeter.", "Solution_5": "for the other question...\r\n\r\nnote that $p=a+b+c$, $A=\\frac{1}{2}rp$, $R=\\frac{abc}{4A}=\\frac{abc}{2rp}$, \r\n\r\nby am-gm, $\\frac{a+b+c}{3}\\ge\\sqrt[3]{abc}$\r\n\r\n$\\frac{p^{3}}{27}\\ge 2rpR$\r\n\r\n$R\\le \\frac{p^{2}}{54r}$ with equality when $a=b=c$\r\n\r\np.s. this is nice!", "Solution_6": "[hide=\"Question\"]\nI take it semiperimeter is half the perimeter?\n\nIf so, then $A = rs = (2) \\cdot \\frac{17}{2}= 17$?\n[/hide]", "Solution_7": "[quote=\"erik-the-red\"][hide=\"Question\"]\nI take it semiperimeter is half the perimeter?\n\nIf so, then $A = rs = (2) \\cdot \\frac{17}{2}= 17$?\n[/hide][/quote]\r\n\r\nyes...", "Solution_8": "Haha the alternate solution to the problem is to realize that this is a HCSSiM problem that has a numerical answer, so clearly the answer must be 17 heh.", "Solution_9": "erik-the-red, this formula (A=rs) comes from drawing the line from the incenter of the triangle to the three verticies. This creates 3 triangles, all of whose altitude is r, the radius of the incircle. The bases of these three triangles add up to the perimeter. multiplying r*p/2 to get the area of all the triangles (you can try assigning other variables to the other sides and adding them up individually but it will eventually come down to this) gives the area of the entire triangle. \r\n\r\nIt doesn't make sense to me to write it as A=rs just because s=p/2. what does the semiperimeter have anything to do with the area of this triangle from the way this formula is derived?", "Solution_10": "[quote=\"Altheman\"]for the other question...\n\nnote that $p=a+b+c$, $A=\\frac{1}{2}rp$, $R=\\frac{abc}{4A}=\\frac{abc}{2rp}$, \n\nby am-gm, $\\frac{a+b+c}{3}\\ge\\sqrt[3]{abc}$\n\n$\\frac{p^{3}}{27}\\ge 2rpR$\n\n$R\\le \\frac{p^{2}}{54r}$ with equality when $a=b=c$\n\np.s. this is nice![/quote]\r\nSorry to spoil the fun, you have only found a upper bound for the circumradius of $\\triangle ABC$.", "Solution_11": "Wow, I'm sorry about my above post ^. Somehow I ended up editing my original post, instead of making a new one to respond to [b]mecrazywong[/b]; but I can't edit the post anymore because of the forum restrictions. :ninja:", "Solution_12": "[quote=\"cincodemayo5590\"]Wow, I'm sorry about my above post ^. Somehow I ended up editing my original post, instead of making a new one to respond to [b]mecrazywong[/b]; but I can't edit the post anymore because of the forum restrictions. :ninja:[/quote]\r\n\r\n :ninja:", "Solution_13": "[quote=\"Altheman\"][quote=\"erik-the-red\"][hide=\"Question\"]\nI take it semiperimeter is half the perimeter?\n\nIf so, then $A = rs = (2) \\cdot \\frac{17}{2}= 17$?\n[/hide][/quote]\n\nyes...[/quote]\r\n\r\nJust making sure." } { "Tag": [ "calculus", "integration", "search", "algorithm", "linear algebra", "matrix", "function" ], "Problem": "I would appreciate help with this problem and also an explanation of what a line integral is:\r\n\r\nEvaluate explicitly each of the line integrals\r\n\\[ \\int (x \\textup{ d}x \\plus{} y \\textup{ d}y \\plus{} z \\textup{ d}z), \r\n\\int (y \\textup{ d}x \\plus{} x \\textup{ d}y \\plus{} \\textup{ d}z),\r\n\\int (y \\textup{ d}x \\minus{} x \\textup{ d}y \\plus{} e^{x\\plus{}y} \\textup{ d}z),\\] along (i) the straight line path joining the origin to $ x \\equal{} y \\equal{} z \\equal{} 1$, and (ii) the parabolic path given parametrically by $ x \\equal{} t, y \\equal{} t, z \\equal{} t^2$ with $ 0 \\le t \\le 1$.\r\n\r\nFor which of these integrals do the two paths give the same results, and why?", "Solution_1": "[quote=\"aidan\"]I would appreciate help with this problem and also an explanation of what a line integral is[/quote] It seems to me that you have these written down in the wrong order. The appropriate thing to do here is to read the section on line integrals in any standard multivariable calculus text and/or a few of the websites (like wikipedia) that pop up when you do a search in your favorite search engine for the words \"line integral.\" A few minutes of reading will be enough to uncover an algorithm that will allow you to evaluate the integrals in question; a somewhat more thorough effort will introduce you to the idea of a conservative field, which is what the \"why\" part is asking about.", "Solution_2": "Wikipedia tells me the line integral along a piecewise smooth curve $ C$ is defined as\r\n\\[ \\int_C f \\textup{ d}s\\]\r\nwhere $ \\textup{ d}s$ is the length of an infinitestimal segment of the curve and $ f$ is a scalar field. But in the integrals given, e.g. $ \\int (x\\textup{ d}x \\plus{} y\\textup{ d}y \\plus{} z\\textup{ d}z)$ how do I deal with the differentials dx, dy, dz or convert this to one where there is only one differential, say dt?\r\n\r\nEdit: Wait, do you write them as\r\n\\[ \\int \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\cdot \\begin{pmatrix} \\textup{ d}x \\\\ \\textup{ d}y \\\\ \\textup{ d}z \\end{pmatrix}, \\, \\int \\begin{pmatrix} y \\\\ x \\\\ 1 \\end{pmatrix} \\cdot \\begin{pmatrix} \\textup{ d}x \\\\ \\textup{ d}y \\\\ \\textup{ d}z \\end{pmatrix}, \\, \\int \\begin{pmatrix} y \\\\ \\minus{}x \\\\ e^{x\\plus{}y} \\end{pmatrix} \\cdot \\begin{pmatrix} \\textup{ d}x \\\\ \\textup{ d}y \\\\ \\textup{ d}z \\end{pmatrix}?\\]\r\nIs it correct to say that $ \\textup{ d}s \\equal{} \\left|\\begin{pmatrix} \\textup{ d}x \\\\ \\textup{ d}y \\\\ \\textup{ d}z \\end{pmatrix}\\right|$?", "Solution_3": "Keep reading :). That's the line integral of a scalar function -- you have here the line integral of a vector field, which is the next step down the page. What you wrote in your edit is correct, but we don't need to have $ ds$ in this context. The next thing you need to do is to translate into the parametrization of the curves over which you're integrating.", "Solution_4": "Ok, since\r\n\\[ \\int_{C}\\mathbf{F}(\\mathbf{r}) \\cdot \\textup{ d}\\mathbf{r} \\equal{} \\int_a^b \\mathbf{F}(\\mathbf{r}(t)) \\cdot \\mathbf{r'} \\textup{ d}t\\]we can consider things in the light of $ x \\equal{} y \\equal{} z \\equal{} t$ for the first integral. Then $ \\textup{ d}x \\equal{} \\textup{ d}y \\equal{} \\textup{ d}z \\equal{} \\textup{ d}t$. So the first integral simply becomes $ \\int_0^1 3t \\textup{ d}t$?\r\n\r\nI have another question at this point: say you have a vector field $ \\mathbf{r}$ of a curve dependent on a parameter $ t$. Later on in most courses they will want to parametrize the field in terms of the arc length $ s$ and consider the function as $ \\mathbf{r}(s(t))$. What is the purpose of this?", "Solution_5": "[quote=\"aidan\"]So the first integral simply becomes $ \\int_0^1 3t \\textup{ d}t$?[/quote] Yes. (And similarly with the others.)\n\n[quote=\"aidan\"]I have another question at this point: say you have a vector field $ \\mathbf{r}$ of a curve dependent on a parameter $ t$. Later on in most courses they will want to parametrize the field in terms of the arc length $ s$ and consider the function as $ \\mathbf{r}(s(t))$. What is the purpose of this?[/quote] In the notation of the first part of your post, the vector field is $ \\textbf{F}(\\textbf{r})$ and the curve is parametrized by $ \\textbf{r} \\equal{} \\textbf{r}(t)$. Arc-length (or constant-speed) parametrization has a number of nice features. Notably, the tangent vector $ \\textbf{r}'(t)$ is always a unit vector and so the second derivative $ \\textbf{r}''(t)$ is a normal vector to the curve. This allows a canonical coordinate frame at every point of the curve.", "Solution_6": "I see.. that's nice indeed. And earlier there was something about conservative forces - that means the integral depends only on the endpoints and not on the path taken?", "Solution_7": "Yes -- some vector fields have this property and others do not, and there is a nice characterization of those fields that do have this property.", "Solution_8": "Alrighty, thanks! :)" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all naturals $ x,y,z$, which satisfy the equation $ 1\\plus{}{2^{x}}{3^{y}}\\equal{}z^{2}$", "Solution_1": "[quote=\"aleksandar\"]Find all naturals $ x,y,z$, which satisfy the equation $ 1 \\plus{} {2^{x}}{3^{y}} \\equal{} z^{2}$[/quote]\r\n$ 2^x3^y\\equal{}(z\\minus{}1)(z\\plus{}1)$\r\n\r\n1) $ x\\equal{}0$ $ \\implies$ $ 3^y\\equal{}(z\\minus{}1)(z\\plus{}1)$\r\nBoth $ z\\minus{}1$ and $ z\\plus{}1$ cant be multiple of $ 3$, so one is $ 1$ and the other is $ 3^y$ $ \\implies$ $ 1\\plus{}2^03^1\\equal{}2^2$\r\n\r\n2) $ x>0$ $ \\implies$ $ z$ even $ \\implies$ $ x\\geq 2$\r\nBoth $ z\\minus{}1$ and $ z\\plus{}1$ cant be multiple of $ 3$\r\nBoth $ z\\minus{}1$ and $ z\\plus{}1$ cant be multiple of $ 4$\r\nSo 4 cases :\r\n\r\n2.1) $ z\\minus{}1\\equal{}3^y2^{x\\minus{}1}$ and $ z\\plus{}1\\equal{}2$ impossible\r\n\r\n2.2) $ z\\minus{}1\\equal{}3^y2$ and $ z\\plus{}1\\equal{}2^{x\\minus{}1}$, so $ ^2\\cdot 3^y\\plus{}2\\equal{}2^{x\\minus{}1}$ so $ 3^y\\plus{}1\\equal{}2^{x\\minus{}2}$\r\nIf $ y\\equal{}0$, we have the solution $ 1\\plus{}2^33^0\\equal{}3^2$\r\nIf $ y>0$, the equation $ 3^y\\plus{}1\\equal{}2^{x\\minus{}2}$ mod 3 implies $ x\\minus{}2\\equal{}2v$ even and then $ 3^y\\equal{}(2^v\\minus{}1)(2^v\\plus{}1)$\r\nOnly one of the two factors may be divisible by 3 and so $ 2^v\\minus{}1\\equal{}1$ and the solution $ 1\\plus{}2^43^1\\equal{}7^2$\r\n\r\n2.3) $ z\\minus{}1\\equal{}2^{x\\minus{}1}$ and $ z\\plus{}1\\equal{}2\\cdot 3^y$, so $ ^2\\cdot 3^y\\minus{}2\\equal{}2^{x\\minus{}1}$ so $ 3^y\\minus{}1\\equal{}2^{x\\minus{}2}$\r\nIf $ x\\equal{}2$, this gives no solution\r\nIf $ x\\equal{}3$, this gives the solution $ 1\\plus{}2^33^1\\equal{}5^2$\r\nIf $ x\\geq 4$, the equation $ 3^y\\minus{}1\\equal{}2^{x\\minus{}2}$ mod 4 implies $ y\\equal{}2v$ odd and so $ (3^v\\minus{}1)(3^v\\plus{}1)\\equal{}2^{x\\minus{}2}$\r\nOnly one of the two factors may be divisible by $ 4$ and so the other is $ 2$, so $ 3^v\\minus{}1\\equal{}2$ and the solution $ 1\\plus{}2^53^2\\equal{}17^2$\r\n\r\n2.4) $ z\\minus{}1\\equal{}2$ and $ z\\plus{}1\\equal{}2^{x\\minus{}1}3^y$ and the solution $ 1\\plus{}2^33^0$ already found.\r\n\r\nSo the five solutions :\r\n$ 1\\plus{}2^03^1\\equal{}2^2$\r\n$ 1\\plus{}2^33^0\\equal{}3^2$\r\n$ 1\\plus{}2^43^1\\equal{}7^2$\r\n$ 1\\plus{}2^33^1\\equal{}5^2$\r\n$ 1\\plus{}2^53^2\\equal{}17^2$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be positive real numbers and $ k\\ge 0$. Prove that \r\n\\[ \\frac{a}{\\sqrt{a+kb}}+\\frac{b}{\\sqrt{b+kc}}+\\frac{c}{\\sqrt{c+ka}}\\le \\sqrt{\\frac{3(a+b+c)}{k+1}}\\]", "Solution_1": "In our topic, VASC proposed the general problem for my inequality\r\n\r\n[quote=\"Vasc\"][quote=\"nayel\"]I think I just proved that the following inequality holds for $ k,a,b,c>0$ and $ a+b+c=1$\n\\[ \\frac{a}{\\sqrt{a+kb}}+\\frac{b}{\\sqrt{b+kc}}+\\frac{c}{\\sqrt{c+ka}}\\le \\sqrt{\\frac{3}{k+1}}\\]\n[/quote]\nActually, the following inequality holds for $ k,a,b,c>0$ :\n\\[ \\frac{a}{\\sqrt{a+kb}}+\\frac{b}{\\sqrt{b+kc}}+\\frac{c}{\\sqrt{c+ka}}<\\sqrt{\\frac{(k+1)(a+b+c)}{k}}\\]\n[/quote]\r\n\r\nAnd Nayel, also in this topic, you show a solution to it (actually, the solution is totally the same as the particular case). But at the moment, I think your problem is wrong, It is wrong just for $ k=2$. I think you can check it easily." } { "Tag": [ "MATHCOUNTS", "inverse proportion" ], "Problem": "1. I can paint the floor alone in 2 hours. My dad can paint it alone in 5 hours. How long will it take us to paint the floor together?\r\n\r\n2. In four hours, Adam, Ben, and Chris can paint the whole house. If Adam works alone, he can paint the house in 28 hours. If Ben works alone, he can paint it in 7 hours. How long does it take Chris to paint the house alone?\r\n\r\n3. Amber and Brittney can make a bucket of paint in 3 hours. Amber and Chelsea can make a bucket of paint in 4 hours. Brittney and Chelsea can make a bucket in 6 hours. How long will it take all 3 girls to make a bucket of paint if they work together?\r\n\r\n4. Pump A can fill a pool in 6 hours. Pump B can fill the pool in 4 hours. Pump C can drain the pool in 5 hours. Pump A is opened and begins filling an empty pool. 2 hours later, pump C is opened and starts draining. An hour after that, pump B is opened. How many more hours will it take to fill up the pool?\r\n\r\n5. 4 people can make 10 pounds of flour in 3 days. How many days will it take 3 people to make 4 pounds of flour?\r\n\r\nThat's it for now. These are the types of questions (work/rate/time) that you should never miss on a Mathcounts test.", "Solution_1": "[hide=\"hint for 1-4\"]\nHow much of the house/bucket/pool is made/filled every hour?\n[/hide]\n\n[hide=\"hint for 5\"]\nThe people and flour are directly proportional\nThe people and days are in inverse proportion\nThe flour and days are directly proportional.\n[/hide]", "Solution_2": "I may be wrong on some...\r\n\r\n[hide=\"Answer\"]1.$ \\frac{10}7$ hours\n\n2.$ 14$ hours\n\n3.$ 40$ minutes ??\n\n4.$ \\frac37$hours\n\n5.$ \\frac85$days??[/hide]", "Solution_3": "Check your work on 3 and 4", "Solution_4": "Fixed..I think..\r\n\r\n[hide]3.$ \\frac{8}{3}$hours\n\n4.$ \\frac{18}{37}$hours[/hide]", "Solution_5": "I don't think I got those as well. Someone correct me if I'm wrong. I just made these questions out of the blue.", "Solution_6": "[quote=\"Chinaboy\"]I don't think I got those as well. Someone correct me if I'm wrong. I just made these questions out of the blue.[/quote]\r\nYou can't be wrong....\r\n\r\nI got a new answer for 4..(miscomprehension of a problem..and I still might be wrong)\r\n\r\n[hide]$ \\frac{13}{42}$hour[/hide]", "Solution_7": "hmm, nope. check your work carefully.", "Solution_8": "[quote=\"Chinaboy\"]\n3. Amber and Brittney can make a bucket of paint in 3 hours. Amber and Chelsea can make a bucket of paint in 4 hours. Brittney and Chelsea can make a bucket in 6 hours. How long will it take all 3 girls to make a bucket of paint if they work together?\n\n[/quote]\r\n\r\n[hide=\"What I did on #3\"]let $ A$=Amber, $ B$=Brittney, $ C$=Chelsea.\n\n$ A\\plus{}B\\equal{}\\frac13$\n\n$ A\\plus{}C\\equal{}\\frac14$\n\n$ B\\plus{}C\\equal{}\\frac16$\n\nAdding those up, we get $ 2A\\plus{}2B\\plus{}2C\\equal{}\\frac{4\\plus{}3\\plus{}2}{12}\\equal{}\\frac9{12}\\equal{}\\frac34$\n\nDivide by 2 to get $ A\\plus{}B\\plus{}C\\equal{}\\frac{3}{8}$\n\nWe want $ (A\\plus{}B\\plus{}C)x\\equal{}1$, so $ x\\equal{}\\boxed{\\frac83}$[/hide]", "Solution_9": "good on 3 DK.", "Solution_10": "[quote=\"Chinaboy\"]4. Pump A can fill a pool in 6 hours. Pump B can fill the pool in 4 hours. Pump C can drain the pool in 5 hours. Pump A is opened and begins filling an empty pool. 2 hours later, pump C is opened and starts draining. An hour after that, pump B is opened. How many more hours will it take to fill up the pool? [/quote]\r\n\r\nhmm... would this be it?\r\n\r\n[hide]$ \\frac16(3)\\minus{}\\frac15\\plus{}(\\frac16\\plus{}\\frac14\\minus{}\\frac15)x\\equal{}1$\n\n$ \\frac12\\minus{}\\frac15\\plus{}(\\frac{10\\plus{}15\\minus{}12}{60})x\\equal{}1$\n\n$ \\frac3{10}\\plus{}(\\frac{13}{60})x\\equal{}1$\n\n$ (\\frac{13}{60})x\\equal{}\\frac7{10}$\n\n$ x\\equal{}\\frac7{10}\\times \\frac{60}{13}\\equal{} \\boxed{\\frac{42}{13}}$[/hide]", "Solution_11": "Nicejob , DK!" } { "Tag": [ "conics", "parabola", "geometry", "perpendicular bisector", "geometry proposed" ], "Problem": "Hi, everyone.\r\n\r\nProve that triangles subtended at the focus of a parabola by any two tangents are similar.\r\n\r\n[b]with synthetic proofs (no analytic, no trigonometric)[/b]\r\n\r\nThanks in advance. :blush:", "Solution_1": "Let $ F$ be the parabola focus and $ d$ its directrix. Parabola main axis $ m \\perp d$ through $ F$ cuts the directrix at $ D$ and the perpendicular bisector $ v$ of $ FD$ is the parabola vertex tangent. Let arbitrary parabola tangents $ s, t$ at $ S, T$ intersect at $ X.$ Let $ K, L \\in d$ be feet of perpendiculars from $ S, T$; $ SK \\equal{} SF, TL \\equal{} TF.$ Parabola tangents $ s, t$ bisect the angles $ \\angle FSK, \\angle FTL$ $ \\Longrightarrow$ they cut the segments $ FK, FL$ at their midpoints $ M, N \\in v.$ Since $ FM \\perp SMX$ and $ FN \\perp TNX,$ quadrilateral $ FMXN$ is cyclic. Then $ \\angle FXT \\equal{} \\angle FXN \\equal{} \\angle FMN \\equal{} \\angle MSK \\equal{} \\angle FSM \\equal{} \\angle FSX$ and similarly, $ \\angle FXS \\equal{} \\angle FTX$ $ \\Longrightarrow$ $ \\triangle FSX \\sim \\triangle FXT$ are similar." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Solve this equation:\r\n$x\\in \\mathbb{R}$\r\n$3^x+4^x+5^x=6^x$ :)", "Solution_1": "Let $f(x)=(3/6)^x+(4/6)^x+(5/6)^x$. We have $f^{'}(x)<0$. Therefore equation had unique solution x=3.", "Solution_2": "why the solution is $3$ when we have $f^{'}(x)<0$? :)\r\ncan you explain more?", "Solution_3": "That means that the function is strictly decreasing, thus is injective." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "For all positive integer $ n\\geq 2$, prove that product of all prime numbers less or equal than $ n$ is smaller than $ 4^{n}$.", "Solution_1": "It equavalent to $ \\sum_{p\\le n} ln(p) p > \\frac{n}{2}$ then $ \\prod_{\\frac{n}{2} < p < n} p < {n \\choose \\frac{n}{2}}$\r\nBy applying Stirling's formula you can see $ {n \\choose \\frac{n}{2}} < 4^{\\frac{n}{2}}$\r\nSo then we have $ \\prod_{p} p \\equal{} (\\prod_{\\frac{n}{2} < p < n} p)(\\prod_{\\frac{n}{4} < p < \\frac{n}{2}} p) (\\ldots) < 4^{\\frac{n}{2}}4^{\\frac{n}{4}} \\ldots \\equal{} 4^{n}$" } { "Tag": [ "AMC", "AIME" ], "Problem": "I just had a dream that I didn't qualify for the AIME last year and was really sad and decided to go on AoPS and talk about how I didn't make it but then AoPS was acing really weird and I didn't know how to go to any of the forums besides Fun and Games. It was really weird.", "Solution_1": "There is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159549]thread[/url] for this you know.", "Solution_2": "omg", "Solution_3": "Okay, so tell me again why you have to tell us about it???", "Solution_4": "wow what a nightmare", "Solution_5": "OMG, i didnt get all As :ewpu: Call the police!" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "-the group has order 143. \r\n\r\n\r\n\r\n-is Z24 isomorphic to S4?\r\n\r\n\r\n\r\nnot sure of the source.", "Solution_1": "1) Note 143 = 11*13. There is a theorem about this.\r\n\r\n2) Note Z24 is abelian whereas S4 is not.", "Solution_2": "sorry, but for 1), could you point me @ the theorem?", "Solution_3": "new to group theory, but i'm looking at these sylow theorems and trying to figure it out...feeling stupid. anymore tips?", "Solution_4": "well apply sylow theorems. in the case of a group G of order 143, we have two normal sylow subgroups order 11 and 13, which must be cyclic (prime order!), and G is the direct product of these, hence cyclic (11 and 13 are coprime)." } { "Tag": [], "Problem": "Find the value(s) of $ x$ which make the series converge.\r\n\r\n$ \\displaystyle\\sum_{n\\equal{}0}^{\\infty}\\frac{(\\minus{}1)^n(2x\\minus{}3)^n}{4^n}$", "Solution_1": "$ \\sum_{n \\equal{} 0}^{\\infty} r^{n}$ converges $ \\Longleftrightarrow |r| < 1$.", "Solution_2": "Thanks, I think I got it now!" } { "Tag": [ "conics", "parabola", "ellipse", "hyperbola" ], "Problem": "What conic does 25X^2 + Y^2 - 100X - 2Y + 76 = 0 represent?\r\n\r\nA: PARABOLA\r\nB: CIRCLE\r\nC: ELLIPSE\r\nD: HYPERBOLA\r\nE: NOT ENOUGH INFORMATION", "Solution_1": "Conic in Alg 1... Either it is something else or TX educations is declining we didn't do this stuff until midterm Alg 2...", "Solution_2": "It's an ellipse. Here's why:\r\n\r\nThere are both X^2 and Y^2 terms, thus it is not a parabola (a parabola only has 1 variable squared)\r\nThen look at the signs of the X^2 and Y^2 terms - they are both positive, thus you know it's not a hyperbola (in a hyperbola, they are of different signs)\r\nLastly look at the coefficients of the X^2 and Y^2 terms - they are not the same, thus you know it's not a circle.\r\n\r\nThat leaves an ellipse.", "Solution_3": "[quote]a parabola only has 1 variable squared) [/quote]\r\n\r\nNot necessary : for example x^2+y^2+2xy=16 is a parabola.\r\n(You look at b^2-4*a*c value assuming the given eq is ( a*x^2 +b*x*y + c*y^2 + .other_.terms =0) amd see if it is positive ( Hyparaobla), zero (parabola) or negative (ellipse).. .. circle is special case when a=c and b=0 etc.. \r\n\r\nprob. the first thing one learns when one does conic section .. :maybe:\r\nEdited later to remove typos", "Solution_4": "My bad. sorry", "Solution_5": "[quote=\"marucho8d7\"]My bad. sorry[/quote]\r\n\r\nNever mind. I won't give the answer to this problem, though.", "Solution_6": "[quote=\"Yrael\"]It's an ellipse. Here's why:\n\nThere are both X^2 and Y^2 terms, thus it is not a parabola (a parabola only has 1 variable squared)\nThen look at the signs of the X^2 and Y^2 terms - they are both positive, thus you know it's not a hyperbola (in a hyperbola, they are of different signs)\nLastly look at the coefficients of the X^2 and Y^2 terms - they are not the same, thus you know it's not a circle.\n\nThat leaves an ellipse.[/quote]\n\n[quote=\"Zratonx\"]Conic in Alg 1... Either it is something else or TX educations is declining we didn't do this stuff until midterm Alg 2...[/quote]\r\n\r\nThe way Yreal said it, I could prove with what I learned in midterm Geometry.", "Solution_7": "[quote=\"Gyan\"][quote]a parabola only has 1 variable squared) [/quote]\n\nNot necessary : for example x^2+y^2+2xy=16 is a parabola.\n(You look at b^2-4*a*c value assuming the given eq is ( a*x^2 -b*x*y + c*x^2 + .other_.terms =0) amd see if it is positive ( Hyparaobla), zero (parabola) or negative (ellipse).. .. circle is special case when a=c and b=0 etc.. \n\nprob. the first thing one learns when one does conic section .. :maybe:[/quote]\r\n\r\nyour method seems to be good \r\n25X^2 + Y^2 - 100X - 2Y + 76 = 0\r\nyou said [i]the given eq is ( a*x^2 -b*x*y + c*x^2 + .other_.terms =0)[/i]\r\nI think it should be a*x^2 -b*x*y + c*[b]y[/b]^2 + .other_.terms =0\r\nand a=25, b=0, c=2, b^2-4*a*c=-4*25*2=-200, an ellipse.", "Solution_8": "P. - Thanks, typo is corrected in the original post.", "Solution_9": "[quote=\"marucho8d7\"]What conic does 25X^2 + Y^2 - 100X - 2Y + 76 = 0 represent?\n\nA: PARABOLA\nB: CIRCLE\nC: ELLIPSE\nD: HYPERBOLA\nE: NOT ENOUGH INFORMATION[/quote]\r\n\r\n25x2 + y2 - 100x - 2y + 76 = 0\r\n25[x2 - 4x] + y2 - 2y + 76 = 0\r\n25[(x - 2)2 - 4] + (y - 1)2 - 1 + 76 = 0\r\n25(x - 2)2 - 100 + (y - 1)2 - 1 + 76 = 0\r\n25(x - 2)2 + (y - 1)2 = 52\r\n\r\nBecause the coefficient of the x brackets is different to the y bracket the shape will be an ellipse, if they are equal it would be a circle." } { "Tag": [ "limit", "calculus", "algebra unsolved", "algebra" ], "Problem": "Prove that the equation $ x^4\\minus{}x^3\\minus{}1\\equal{}0$ has exactly two roots. If $ a$ and $ b$ denote their sum and product respectively, prove that $ b<\\minus{}\\frac{11}{10}$ and $ a> \\frac{6}{11}$.", "Solution_1": "Can someone please post a solution for this problem?\r\nI can't seem to link the bounds for $ a,b,c$ with the condition for the occurence of a pair of complex conjugates as roots.", "Solution_2": "Let $ x^4\\minus{}x^3\\minus{}1\\equal{}(x^2\\minus{}ax\\plus{}b)(x^2\\minus{}(1\\minus{}a)x\\minus{}\\frac 1b)$.\r\nThen $ a(1\\minus{}a)\\plus{}b\\minus{}\\frac 1b \\equal{}0$ and $ \\frac ab\\equal{}b(1\\minus{}a)$ or \r\n$ b^2\\equal{}\\frac{a}{1\\minus{}a}, \\frac 1b \\minus{}b\\equal{}a(1\\minus{}a)$. Therefore $ 00$, then $ \\frac{\\sqrt{67}\\minus{}1}{8}\\le b<1$. It give $ 4b>a^2$ and $ x^2\\minus{}ax\\plus{}b$ had not solution. Therefore $ b<0$ and $ \\frac{\\minus{}1\\minus{}\\sqrt{67}}{8}\\le b<\\minus{}1$. It give $ (1\\minus{}a)^2\\plus{}\\frac{4}{b}<0$ - only 2 real solution.\r\nTherefore $ 1<\\frac{a}{1\\minus{}a}\\le \\frac{34\\plus{}\\sqrt{67}}{68\\plus{}\\sqrt{67}}<\\frac 59$. Therefore $ \\frac{20}{81}<\\frac 1b \\minus{}b\\le \\frac 14$.\r\nIt give $ b<\\minus{}\\frac{10\\plus{}\\sqrt{6661}}{81}\\equal{}\\minus{}1.131048761...<\\minus{}\\frac 98<\\minus{}\\frac{11}{10}$ and $ a\\equal{}\\frac{b^2}{1\\plus{}b^2}>\\frac{1}{2}\\plus{}\\frac{5}{\\sqrt{6661}}$ or $ a>0.561263286..>\\frac 59 >\\frac{6}{11}$", "Solution_3": "[quote=\"Agr_94_Math\"]Prove that the equation $ x^4 \\minus{} x^3 \\minus{} 1 \\equal{} 0$ has exactly two roots. If $ a$ and $ b$ denote their sum and product respectively, prove that $ b < \\minus{} \\frac {11}{10}$ and $ a > \\frac {6}{11}$.[/quote]\r\n\r\nLet $ P(x)\\equal{}x^4\\minus{}x^3\\minus{}1$ \r\n\r\n1) two real roots\r\n$ P'(x)\\equal{}4x^3\\minus{}3x^2\\equal{}4x^2(x\\minus{}\\frac 34)$ and so $ P(x)$ is decreasing on $ (\\minus{}\\infty,\\frac 34)$ and increasing on $ (\\frac 34,\\plus{}\\infty)$\r\nSince $ \\lim_{x\\to\\minus{}\\infty}P(x)\\equal{}\\plus{}\\infty$ and $ \\lim_{x\\to\\plus{}\\infty}P(x)\\equal{}\\plus{}\\infty$ and $ P(\\frac 34)\\equal{}\\minus{}\\frac{283}{256}<0$, we have exacty two real roots $ x_1<\\frac 34\\frac{91}{66}$\r\n\r\n$ P(\\minus{}\\frac 45)\\equal{}\\minus{}\\frac{49}{5^4}<0$ $ \\implies$ $ x_1<\\minus{}\\frac 45$\r\n\r\n$ P(\\minus{}\\frac {55}{66})\\equal{}\\frac{9559}{2376^2}>0$ $ \\implies$ $ x_1>\\minus{}\\frac {55}{66}$\r\n\r\nSo $ a\\equal{}x_1\\plus{}x_2>\\frac{91}{66}\\minus{}\\frac {55}{66}\\equal{}\\frac 6{11}$\r\n\r\nAnd $ b\\equal{}x_1x_2<\\minus{}\\frac{91}{66}\\frac 45<\\minus{}\\frac {11}{10}$\r\n\r\nQ.E.D.", "Solution_4": "[quote=\"pco\"]\n2) bounds\n$ P(\\frac {91}{66}) \\equal{} \\minus{} \\frac {135461}{66^4} < 0$ $ \\implies$ $ x_2 > \\frac {91}{66}$\n\n$ P( \\minus{} \\frac 45) \\equal{} \\minus{} \\frac {49}{5^4} < 0$ $ \\implies$ $ x_1 < \\minus{} \\frac 45$\n\n$ P( \\minus{} \\frac {55}{66}) \\equal{} \\frac {9559}{2376^2} > 0$ $ \\implies$ $ x_1 > \\minus{} \\frac {55}{66}$\n\n[/quote]\r\nHow these numbers came to your mind dear [b]pco[/b]???\r\n\r\nDimitris", "Solution_5": "[quote=\"Dimitris X\"]How these numbers came to your mind dear [b]pco[/b]???[/quote]\r\n\r\n :blush: I computed the exact roots thru calculus and then looked for approx values not too complicated to use" } { "Tag": [ "geometry", "conics", "parabola", "ellipse" ], "Problem": "Find the focus for y=3x^2+6x+4\r\n\r\nhow do u find the focus...? is there an equation to use..", "Solution_1": "In geometry, the foci (singular focus) are a pair of special points used in describing conic sections. The four types of conic sections are the circle, parabola, ellipse, and hyperbola.\r\n$Source: Wikipedia$", "Solution_2": "that equation is equivalent to\r\n$y=3(x+3)^{2}-5$\r\nso the vertex is at (-3, -5)\r\nremember the equation\r\n$y=4px$ where $p$ is the focus?\r\ntranslating the vertex back to the origin, we find the focus\r\n$y=3x^{2}$ is the base parabola, so $3x^{2}=4px$\r\nso $p=\\frac{3x}{4}$\r\n...and I can't remember what to do from here. sorry - I didn't take my first term of precal when they covered this stuff.", "Solution_3": "thanks... but can anyone finish showing me..?", "Solution_4": "[hide=\"probably wrong...\"]$y=3x^{2}+6x+4=3(x+1)^{2}+1$. The general form for a parabola is $y-k=\\frac{1}{4a}(x-h)^{2}$(?), where (h,k) is the vertex, and a is the distance from vertex to focus.\n$(-1,-1)$ is the vertex.\n$a=1/12$. So the focus is at $(-1,-11/12)$(?).[/hide]", "Solution_5": "The vertex is actually $(-1,1)$, so the focus is at [hide]$(-1,\\frac{13}{12})$.[/hide]", "Solution_6": "yea the answer is (-1,13/12) but how did you get that could you do it from squaring it.. i keep trying it but cant get that 13/12 focus is (h,p+k) right..?", "Solution_7": "The focus is $(h,k+p)$, and $(h,k)$ is the vertex, which is $(-1,1)$. Since $\\frac{1}{4p}=3$ you get $p=\\frac{1}{12}$. \r\n\r\nThus, the focus is at $(-1,1+\\frac{1}{12})$ or $\\boxed{(-1,\\frac{13}{12})}$.", "Solution_8": "wait how u get 1/4p=3 ? doesnt that equal to 12 and not 1/12??", "Solution_9": "$\\frac{1}{4p}=3$ comes from the the factored equation $y-1=3(x+1)^{2}$. Compare this with $y-k=\\frac{1}{4p}(x-h)^{2}$." } { "Tag": [ "geometry", "3D geometry", "geometric transformation", "rotation" ], "Problem": "Each of the faces of a cube is colored by a different color. How many of the coloring are distinct?", "Solution_1": "Here's another Burnside's Lemma approach.\r\n[hide]\nLet $ G$ be a group of all rotations on a cube. Now $ |G|=24$ because there are 6 ways to place the face down and another 4 different positions after that. \n\nLet $ X$ be the set of all possible positions of the cube, not counting rotations. So $ |X|=6!=720$. We say two arrangements in $ x_{1},x_{2}\\in X$ are equivalent if there is some rotations that brings $ x_{1}$ to $ x_{2}$, i.e. $ x_{1},x_{2}$ are in the same orbit under $ G$ acting by rotations on $ X$.\n\nNow if $ g=e$, i.e. the identity element then $ X_{g}= X$ and so $ |X_{g}|=720$. If $ g\\not = e$ then $ |X_{g}|=0$ because all other positions are changed.\n\nAs explained above we are counting the total number of [b]orbits[/b] and by Burnside's Lemma we have:\n$ \\frac{1}{|G|}\\cdot \\sum_{g\\in G}|X_{g}| = \\frac{1}{24}\\cdot \\left(720+0+...+0 \\right) = 30$.\n[/hide]", "Solution_2": "While the above solution is certainly more educative and easier to generalize with, below is the most basic solution to this particular problem:\r\n\r\n[hide]Without loss of generality, let the face with color 1 be the top face.\n\nThere are 5 remaining possibilities for the bottom face's color.\nThere are 4! ways to color the four sides.\nWith the top face locked, there are 4 rotational positions.\n\n$ N=\\frac{5 \\times 4!}{4}=30$[/hide]", "Solution_3": "yep, choose a specific color for the upper face of the cube, say A. Then we have five choices for colouring the lower face of the cube, say with color B. Rotate the cube so that some color C is facing us. Now the remaining sides are fixed with respect to these three. We can distribute the three colors in 3 \u00d72 \u00d71 = 6 ways, giving 5 \u00d76 = 30 possibilities." } { "Tag": [ "algebra", "polynomial", "induction", "geometry", "calculus" ], "Problem": "[Possibly posted before.]\r\n\r\nSolve\r\n\r\n\\begin{eqnarray*}1-{x\\over 1}+\\frac{x(x-1)}{1\\cdot 2}-\\frac{x(x-1)(x-2)}{1\\cdot 2\\cdot 3}+\\dots+(-1)^{n}\\frac{x(x-1)(x-2)\\dots(x-n+1)}{n!}=0\\end{eqnarray*}", "Solution_1": "I believe you mean \\begin{eqnarray*}1-{x\\over 1}+\\frac{x(x-1)}{1\\cdot 2}-\\frac{x(x-1)(x-2)}{1\\cdot 2\\cdot 3}+\\dots+(-1)^{n}\\frac{x(x-1)(x-2)\\dots(x-n+1)}{n!}=0\\end{eqnarray*}\r\n\r\n[i]Solution:[/i]\r\n\r\nWe will prove by induction that \\begin{eqnarray*}1-{x\\over 1}+\\frac{x(x-1)}{1\\cdot 2}-\\frac{x(x-1)(x-2)}{1\\cdot 2\\cdot 3}+\\dots+(-1)^{n}\\frac{x(x-1)(x-2)\\dots(x-n+1)}{n!}=(-1)^{n}\\frac{(x-1)(x-2)\\dots(x-n)}{n!}\\end{eqnarray*}\r\nFor $n=1$, we have $1-x=-(x-1)$ which is true. Assume that it is true for all $n$. By adding $(-1)^{n+1}\\frac{x(x-1)(x-2)\\dots(x-n)}{(n+1)!}$ to both sides of the equation, we see that after simplification our conjecture is proven by the principle of mathematical induction. As a result, we are to solve the for $x$ in \\[(-1)^{n}\\frac{(x-1)(x-2)\\dots(x-n)}{n!}=0\\] Clearly, we have that $x=1,2,\\dots,n$.\r\n\r\nMasoud Zargar", "Solution_2": "Yup, there was a typo in the statement which is now edited.", "Solution_3": "What is induction. I'm dumb.", "Solution_4": "im dumber :lol: \r\n\r\n\r\nI am only taking geometry, and am still working hard for tournaments lol. \r\n\r\nYou already are probably taking pre cal or calculus or even higher", "Solution_5": "[quote=\"neelnal\"]What is induction. [/quote]\r\n\r\nWe have a statement that we want to prove is true for all positive integers $n$. This usually takes the form of showing that some sequence has some form; i.e. $s_{n}= f(n)$.\r\n\r\nFirst, we verify that it is true for $n = 1$, that is, $s_{1}= f(1)$. \r\n\r\nThen we show that\r\n\r\n[b]If[/b] the statement is true for $n = k$,\r\n\r\n[b]Then[/b] the statement must also be true for $n = k+1$.\r\n\r\nBecause it is true for $n = 1$, it is true for $n = 2$, which means it is true for $n = 3$, ... which means it is true for all positive integers.\r\n\r\n[hide=\"A more formal response\"] Call $S \\subseteq \\mathbb{N}$ an [b]inductive set[/b] if $a \\in S \\implies a+1 \\in S$.\n\nThe Principle of Induction states that any inductive set that contains $1$ is $\\mathbb{N}$. [/hide]\r\n\r\nEdit: A more direct solution to the problem is not hard, either. We suppose $x = 1, 2, 3, ... n$. We see that after the $n^{th}$ term the numerator is zero, so for each $x$ value we are really dealing with the summation\r\n\r\n${x \\choose 0}-{x \\choose 1}+{x \\choose 2}-+...$\r\n\r\nWhich is, of course, $P(1)$, where\r\n\r\n$P(t) = (t-1)^{x}$\r\n\r\nOf course, this is identically equal to $0$.\r\n\r\nMoreover, the original polynomial has degree $n$, so we have found all the roots." } { "Tag": [ "function", "induction" ], "Problem": "1) f is an injective function N -> N such that:\r\n\r\nf(f(n)) $\\leq$ (n+f(n))/2. \r\n\r\nDetermine all possibilities for f.\r\n\r\n2) Find all integers n such that :sqrt: ((4n-2)/(n+5)) is rational.\r\n\r\nI get the answers to be [color=white] (1) f(n) = n and (2) n = 13 only [/color].\r\n\r\nWhat do people think?", "Solution_1": "I got the same answer for 2.\r\n\r\n$\\frac{4n-2}{n+5}=\\frac{a^2}{b^2}$ with (a,b)=1.\r\n$n=\\frac{2b^2+5a^2}{4b^2-a^2}$\r\nWe see from $2*(2b^2+5a^2)-(4b^2-a^2)=11a^2$ that $d=(2b^2+5a^2, 4b^2-a^2)|(11a^2)$. However, because $(a,b)=1$, $2b^2+5a^2$ cannot be divisible by $a^2$. So $d|11$, which means that d=1 or 11. We then set $4b^2-a^2=1$, to which a simple mod 4ing reveals that there there are no solutions, and $4b^2-a^2=11$, whose only positive solution is $a=5, b=3$. This implies that $n=13$.", "Solution_2": "[quote]1) f is an injective function N -> N such that:\nf(f(n)) $\\leq$ (n+f(n))/2. [/quote]\r\n\r\n. if f(n) < n then f(f(n)) < n and by induction 0 <= f^k(n) < n which is impossible since the f^k(n) are all different. So f(n) >= n for all n.\r\n\r\n. if f(n) > f(n) then f(n) < n, we have just seen it's not possible. So f(n) = f(n) for all n and f(n)=n since f is injective.\r\n\r\nSo f(n)=n is the only solution.\r\n\r\nHope it's correct." } { "Tag": [ "function", "algebra", "domain", "calculus", "derivative", "analytic geometry", "graphing lines" ], "Problem": "It is true that Q(x)=x^5+x^3+x is a one to one function whose domain and range are all numbers.\r\n\r\nSuppose that R is the function inverse to Q, There is no simpe algebraic way to compute values of R. compute R(3), R'(3), and R''(3).", "Solution_1": "Well the first derivative is\r\n5x^4+3x^2+1\r\nand you can see that setting x^2 to.. X, the new equation 5X^2 + 3X + 1 has no real roots when equated to 0.\r\nThat is, its determinate (is that the right word? ) b^2 - 4ac = 9 - 20 < 0.\r\nTherefore the function has no critical points so though it could increase or decrease (figure out where if possible)\r\nanytime f(x) = f(y), x=y. That is, the graph never slopes \"back down\" to a 'y' already visited.\r\n\r\nSo yes it is 1:1.\r\n\r\nAlso, f(-inf) = -inf\r\nf(+inf) = +inf\r\nTherefore domain and range are both the Real set.\r\nBy observation F^1(3) = 1. Because F(1) = 3.\r\nI've given F`up top, so try it's inverse question if possible. Though you might need a restriction on the domain/range." } { "Tag": [], "Problem": "Assume that penguins live with a density of 1,000 penguins per square mile and can run at an average speed of 7 miles per hour on land and swim at 20 miles per hour. Also assume that a polar bear has a territory of 10 square miles, can run at 25 miles per hour and swim at 10 miles per hour. \r\n\r\nHow many penguins will an average polar bear eat in any given month, remembering that a polar bear could, as a maximum, only eat one penguin per hour and 7% of the land is next to the sea.", "Solution_1": "you only need to post the topic once and please be more discriptive with your title", "Solution_2": "[quote=\"jli\"]you only need to post the topic once and please be more discriptive with your title[/quote]\r\n\r\nYou're a mod; just change the title. (title changed from ??? to Penguins and Polar Bears by me)", "Solution_3": "[quote=\"chess64\"][quote=\"jli\"]you only need to post the topic once and please be more discriptive with your title[/quote]\n\nYou're a mod; just change the title. (title changed from ??? to Penguins and Polar Bears by me)[/quote]\r\n\r\nSorry about that..." } { "Tag": [ "combinatorial geometry", "point set", "combinatorics", "Triangle", "IMO Shortlist" ], "Problem": "Determine whether or nor there exist two disjoint infinite sets $ A$ and $ B$ of points in the plane satisfying the following conditions:\r\n\r\na.) No three points in $ A \\cup B$ are collinear, and the distance between any two points in $ A \\cup B$ is at least 1.\r\n\r\nb.) There is a point of $ A$ in any triangle whose vertices are in $ B,$ and there is a point of $ B$ in any triangle whose vertices are in $ A.$", "Solution_1": "[hide]This is not possible. \n\nFirst lemma: There exist four points in $ A$ that are the vertices of a quadrilateral, convex or concave, which contains a fifth point in $ A$ inside it. That is, there exists some quadrilateral (whose vertices are in $ A$) that has another point in $ A$ inside it.\nProof: Choose any five points in $ A$. Either these five satisfy the lemma, or the five points are the vertices of a convex pentagon. If the five points form a convex pentagon $ VWXYZ$, then each of the three disjoint triangles $ VWX$, $ VXY$, and $ VYZ$ must contain one point in $ B$. These three points in $ B$ form a triangle in which there must be a point in $ A$. But this new point in $ A$ is inside $ VWXYZ$! So we will be able to find four vertices of the pentagon which form a quadrilateral that contain this new point. Thus the lemma is proved.\n\nSecond lemma: If a quadrilateral $ VWXY$ contains a point $ P$ in its interior as in the first lemma (All five points are in set $ A$), then there exists another point in set $ A$ that lies in the convex hull of the interior of the quadrilateral.\nProof: If $ VWXY$ is concave with $ \\angle WXY > \\pi$, then the three disjoint triangles $ VWX$, $ VXY$, $ WXY$ each contain a point in $ B$. These three points in turn determine a triangle in which a new point in $ A$ lies. (If this point happens to be $ P$, then we just consider triangle $ PWX$ instead of $ VWX$, or triangle $ PXY$ instead of $ VXY$, depending on where $ P$ lies.)\nIf $ VWXY$ is convex, then the three triangles $ VWP$, $ WXP$, $ XYP$ determine three points in $ B$ which in turn determine a new point in $ A$ in the interior.\n\nAfter applying the second lemma to some five points, we will be left with a quadrilateral with two points in its interior. We can now choose one of the points such that the remaining points satisfy lemma 1. The new quadrilateral will then be no larger than the first. wee can repeatedly aply lemma 2 and keep getting smaller quadrilaterals. But all of this must stop eventually, because there is a minimum distance between any two points! So this is impossible. [/hide]", "Solution_2": "[hide]\nLemma 1: Given any convex set of $5$ points in $A$, there is another point inside the points that is in $A$.\nProof: Triangulate the set, we have enough points to form a triangle with points in $B$, thus there must be a another point in $A$. $\\boxed{}$.\n\nTake $25$ points that belong to $A$. If we take the convex hull, by Lemma 1 there are at least five points inside this convex hull. Assume that are $a$ points in this convex hull that belong to $A$ (including the points on the convex hull) and let $b$ be the number of points in the convex hull that are in $B$. (Both of these numbers are finite since there is at most one point in any disc of diameter one.)\n\nLemma 2: Given a set of $d$ points with $c$ points in the convex hull (not on the convex hull), there exists a triangulation of the $d$ points with $d-2+c$ triangles.\nProof: If $c=0$, we do a usual triangulation.\nIf $c=1$, draw a line from each point in the convex hull to the point.\nWe can prove it for $c\\ge 2$ by adding another point $P$ inside this triangulation and drawing lines from that $P$ to the verticies of the triangle that it is in. $\\boxed{}$ \n\nSince there are at least $5$ points in $A$ in the convex hull (not counting the points on the convex hull), there are at least $(a-2)+5$ triangles in the triangulation of these $a$ points in $A$. Thus, $b\\ge a+3$. If we triangulate the $b$ points in $B$, there are at least $b-2$ triangles. Thus, $a\\ge b-2$. In conclusion, $b\\ge a+3\\ge b+1$, contradiction.\n\nedit: So no such sets of points exists.[/hide]", "Solution_3": "@archimedes1,\nI am sorry if this is dumb question, but is it possible for the quadrilateral to decrease every time but decrease \"asymptotically,\" so it decreases but stays above side length 1? Thanks!", "Solution_4": "MathPanda1, I see what you are saying. If you look at each of these \"concentric\" quadrilaterals individually, then it is not obvious that any of them have side length less than 1. Instead, note that archimedes1's lemma 2 proves that any given quadrilateral $WXYZ$ with finite area must have an infinite amount of distinct points inside of it. From here it is easy to see that there must be two points with distance less than 1. :)", "Solution_5": "Ah, I see, that is really nice, thank you so much for all your help!" } { "Tag": [ "geometry", "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "This problem requires the knowledge of multivariable calculus. If it is too advanced for this part of the forum feel free to move it.\r\n\r\nProve that the average area of the two dimensional orthogonal projections (i.e. shadows) of a convex three dimensional body is $1/4$ its surface area.", "Solution_1": "It looks like my poor little topic has gotten lost in this forum. :( Maybe someone can move it to Unanswered and Proposed Problems.", "Solution_2": "OK, it's been moved.", "Solution_3": "Read some books on Integral Geometry. This is the 3D case of the general theorem." } { "Tag": [ "probability" ], "Problem": "Here are two bets that are offered in craps:\r\nBig Six: A six will be rolled before a seven.\r\nBig Eight: An eight will be rolled before a seven.\r\nBoth bets pay out at 1:1. This means that you will either win 1 for every 1 that you bet (plus you keep the original bet), or you will lose that bet.\r\n\r\nOne dollar is bet on Big Six and one dollar is bet on Big Eight.\r\nCompute the probability that:\r\n1. Both bets will lose.\r\n2. Only one of the two bets will win.\r\n3. Both bets will win.", "Solution_1": "Are you rolling two dice and taking the sum, or what?", "Solution_2": "*punches self in face*\r\nYes. You are rolling two fair six-sided dice and taking their sum. Sorry, I forgot to include that." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c>0$. Prove that\r\n\r\n$ (a\\plus{}b\\plus{}c)^5\\geq81(a\\plus{}b\\minus{}c)(a\\minus{}b\\plus{}c)(\\minus{}a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)$\r\n\r\n------------------\r\n :blush: Panagiote LIGOURAS", "Solution_1": "$ (a\\plus{}b\\plus{}c)^2 \\ge 3(ab\\plus{}bc\\plus{}ac)$, so we must prove that $ (a\\plus{}b\\plus{}c)^3\\ge 27(a\\plus{}b\\minus{}c)(a\\minus{}b\\plus{}c)(\\minus{}a\\plus{}b\\plus{}c)$, \r\nlet $ a\\equal{}x\\plus{}y,b\\equal{}y\\plus{}z,c\\equal{}x\\plus{}z$, so our inequality becomes $ (2(x\\plus{}y\\plus{}z))^3\\ge 27.8xyz$, and this is true by AM-GM", "Solution_2": "[quote=\"Ligouras\"]Let $ a,b,c > 0$. Prove that\n\n$ (a \\plus{} b \\plus{} c)^5\\geq81(a \\plus{} b \\minus{} c)(a \\minus{} b \\plus{} c)( \\minus{} a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)$\n\n------------------\n :blush: Panagiote LIGOURAS[/quote]\r\n\r\nyour inequality is equivalent as: if $ a,b,c$ are the sides of triangle\r\n\r\n$ ((a\\plus{}b\\minus{}c)\\plus{}(a\\minus{}b\\plus{}c)\\plus{}(\\minus{}a\\plus{}b\\plus{}c))^3(a\\plus{}b\\plus{}c)^2\\geq3^3(a \\plus{} b \\minus{} c)(a \\minus{} b \\plus{} c)( \\minus{} a \\plus{} b \\plus{} c)3(ab \\plus{} bc \\plus{} ca)$\r\n\r\nwhich is obviously true !! :oops:" } { "Tag": [ "symmetry", "geometry", "3D geometry", "tetrahedron", "octahedron", "icosahedron", "dodecahedron" ], "Problem": "Prove that\r\n\r\n\r\n[b]The group of rotational symmetries of a tetrahedron is \nA(4).\n\n\n\nThe full group of symmetries of a tetrahedron is \nS(4).\n\nThe group of rotational symmetries of a cube or of an octahedron\nis S(4).\n\nThe group of rotational symmetries of an icosahedron or of a dodecahedron is A(5).[/b]\r\n\r\n\r\nbased on:\r\n\r\n\r\nA regular solid is a 3-\r\ndimensional polyhedron in which each face is a regular polygon. Any two\r\nfaces as well as any two vertices can be matched by an isometry (a rigid\r\nmotion) of the 3-dimensional space. It is convenient to use a symbol \r\n{p, q}\r\n\r\n\r\nfor a regular solid whose faces are regular \r\np-gons with q of them situated\r\naround each vertex. We have proved that there are FIVE regular solids:\r\n\r\n\r\n\u2022 \r\nTETRAHEDRON {3, 3},\r\n\r\n\r\n\u2022 \r\nCUBE {4, 3},\r\n\r\n\r\n\u2022 \r\nOCTAHEDRON {3, 4},\r\n\r\n\r\n\u2022 \r\nDODECAHEDRON {5, 3},\r\n\r\n\r\n\u2022 \r\nICOSAHEDRON {3, 5}.\r\n\r\nAny regular solid may be inscribed in a sphere, and then any symmetry of\r\nany regular solid will leave the center of the sphere fixed and will transform\r\nthe surface of the sphere onto itself. We call a rotational symmetry of a\r\nregular solid any rotation of the sphere (with respect to an axis passing\r\nthrough its center) mapping the regular solid into inself. A symmetry of a\r\nregular solid is either a rotational symmetry or a reflection with respect to\r\na plane (also passing through the center of the sphere), mapping the regular\r\nsolid onto itself.", "Solution_1": "Group theory goes [url=http://www.artofproblemsolving.com/Forum/index.php?f=8]here[/url]." } { "Tag": [], "Problem": "Simplyfy\r\n$ \\sqrt[3]{a\\plus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}\\plus{}\\sqrt[3]{a\\minus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}$", "Solution_1": "Maybe you mean $ 1\\minus{}a$ instead of $ a\\minus{}1$. Then the answer is $ 2$. :ninja:", "Solution_2": "Actually, it is correct as is. \r\n[hide=\"hint\"]If $ X \\equal{} p\\plus{}q$, then: \n$ X^3 \\equal{} p^3\\plus{}3p^2q\\plus{}3pq^2\\plus{}q^3$\n$ X^3 \\equal{} p^3\\plus{}q^3\\plus{}3pq(p\\plus{}q)$ \n$ X^3 \\equal{} p^3\\plus{}q^3\\plus{}3pqX$\n[/hide]", "Solution_3": "@panos lo: i think it's the same, or maybe i'm mistaken\r\n\r\ni've found it, \r\n$ \\sqrt[3]{a\\plus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}\\plus{}\\sqrt[3]{a\\minus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}\\equal{}x$\r\n\r\nWe know that if $ a\\plus{}b\\plus{}c\\equal{}0$ we have $ a^3\\plus{}b^3\\plus{}c^3\\equal{}3abc$\r\nLet, $ \\sqrt[3]{a\\plus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}\\equal{}a,\\sqrt[3]{a\\minus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}\\equal{}b,x\\equal{}c$\r\nThen $ c\\minus{}a\\minus{}b\\equal{}0$, so $ c^3\\minus{}a^3\\minus{}b^3\\equal{}3abc$\r\n$ x^3\\minus{}2a\\equal{}3x.\\sqrt[3]{a\\plus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}.\\sqrt[3]{a\\minus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}$\r\nSimplifying this, we'll get $ x^3\\minus{}2a\\equal{}4x\\minus{}ax$\r\n$ x^3\\minus{}4x\\minus{}a(2\\minus{}x)\\equal{}0$\r\n$ x(x\\plus{}2)(x\\minus{}2)\\plus{}a(x\\minus{}2)\\equal{}0$\r\n$ (x\\minus{}2)(x^2\\plus{}2x\\plus{}a)\\equal{}0$\r\n We get $ x\\equal{}2$ or $ x^2\\plus{}2x\\plus{}a\\equal{}0$, but this doesn't satisfy the equation since $ a\\le 1$ which is contradict with $ a\\ge 1$, so $ \\sqrt[3]{a\\plus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}\\plus{}\\sqrt[3]{a\\minus{}\\frac{a\\plus{}8}{3}\\sqrt{\\frac{a\\minus{}1}{3}}}\\equal{}2$" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Does anyone happen to know whether the following problem was solved or not (and if yes, could you please indicate the solution):\r\n For a fixed n>1 find all functions f defined on the positive integers with values in the same set such that f(a^n+b^n)=f^n(a)+f^n(b). For n=2 we find a very nice French problem (I think it was given in 1994). I'm really interested in this problem, but I'm not smart enough to solve it even when n=3. Any help?", "Solution_1": "Maybe it is a time to move it to Open Questions." } { "Tag": [ "\\/closed" ], "Problem": "I came upon this message when I tried to reply to TripleM's post under the now non-existant (which I didn't cause) thread \"nominations\":\r\nSorry, but only can reply to posts in this forum.\r\n\r\nI'm pretty positive this was because just when I tried to reply to the post someone deleted the entire thread, is that the case?", "Solution_1": "yeah, stop it", "Solution_2": "That's what happens when someone deletes a post while you're writing a reply -- it happened to me when I clicked on one of the posts that was deleted between the time I entered the forum and the time I clicked on the message." } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra solved" ], "Problem": "Show that a finite abelian group is not cyclic iff it contains a subgroup\r\n isomorphic to $\\mathbb{Z}_{p}\\times\\mathbb{Z}_{p}$ for some prime $p$.", "Solution_1": "Let the group be $G$.\r\n\r\nIf $G$ is cyclic, so are it's subgroups. So if such a subgroup of type $(\\mathbb{Z}/p \\mathbb{Z})^{2}$ exists, it has to be cyclic. But every element has at most order $p$, never giving all the $p^{2}$ elements.\r\nThus if $G$ is cyclic, such subgroups will not occure.\r\n\r\nNow write $G$ as product of finite cyclic groups by the structure theorem. By the chinese remainder theorem, we know $\\mathbb{Z}/a \\mathbb{Z}\\times \\mathbb{Z}/b \\mathbb{Z}\\cong \\mathbb{Z}/(ab) \\mathbb{Z}$ for coprime $a,b$. Thus we merge factors $\\mathbb{Z}/a \\mathbb{Z}$ and $\\mathbb{Z}/b \\mathbb{Z}$ untill it's not possible anymore, meaning that all factors $\\mathbb{Z}/a \\mathbb{Z}$ and $\\mathbb{Z}/b \\mathbb{Z}$ are such that $a,b$ have a common prime divisor $p$.\r\nIf there is only one factor left, $G$ is cyclic, so assume there is more than one factor. Then choose any two factors, let them be $\\mathbb{Z}/a \\mathbb{Z}$ and $\\mathbb{Z}/b \\mathbb{Z}$, and take prime $p$ such that $p|a,b$. Then $\\mathbb{Z}/a \\mathbb{Z}\\times\\mathbb{Z}/b \\mathbb{Z}$ is a subgroup of $G$ and since $\\mathbb{Z}/p \\mathbb{Z}$ is a subgroup of $\\mathbb{Z}/a \\mathbb{Z}$ and $\\mathbb{Z}/b \\mathbb{Z}$, we get that $(\\mathbb{Z}/p \\mathbb{Z})^{2}$ is a subgroup of $G$." } { "Tag": [ "inequalities" ], "Problem": "If $ a,b$are posivite Real numbers then prove the inequality : \r\n $ \\frac {a^2}{b^2} \\plus{} 1 > 2ab \\minus{} \\frac {b^2}{b^2 \\plus{} 1}$", "Solution_1": "[quote=\"enndb0x\"]If $ a,b$are posivite Real numbers then prove the inequality : \n $ \\frac {a^2}{b^2} \\plus{} 1 > 2ab \\minus{} \\frac {b^2}{b^2 \\plus{} 1}$[/quote]\r\n\r\nyour inequality is symply absurd.suppose it is true.then make $ a\\equal{}b\\equal{}1000$ .then we\u00b4ll have that $ 2>2*{1000}^2\\minus{}\\frac{1000^2}{1000^2\\plus{}1}$.absurd.", "Solution_2": "[quote=\"v235711\"][quote=\"enndb0x\"]If $ a,b$are posivite Real numbers then prove the inequality : \n $ \\frac {a^2}{b^2} \\plus{} 1 > 2ab \\minus{} \\frac {b^2}{b^2 \\plus{} 1}$[/quote]\n\nyour inequality is symply absurd.suppose it is true.then make $ a \\equal{} b \\equal{} 1000$ .then we\u00b4ll have that $ 2 > 2*{1000}^2 \\minus{} \\frac {1000^2}{1000^2 \\plus{} 1}$.absurd.[/quote]\r\n\r\nyou've shown that it's true for one value of $ a$ and $ b$. now how do we prove that it's true for ALL values of $ a$ and $ b$?", "Solution_3": "[quote=\"nikeballa96\"][quote=\"v235711\"][quote=\"enndb0x\"]If $ a,b$are posivite Real numbers then prove the inequality : \n $ \\frac {a^2}{b^2} \\plus{} 1 > 2ab \\minus{} \\frac {b^2}{b^2 \\plus{} 1}$[/quote]\n\nyour inequality is symply absurd.suppose it is true.then make $ a \\equal{} b \\equal{} 1000$ .then we\u00b4ll have that $ 2 > 2*{1000}^2 \\minus{} \\frac {1000^2}{1000^2 \\plus{} 1}$.absurd.[/quote]\n\nyou've shown that it's true for one value of $ a$ and $ b$. now how do we prove that it's true for ALL values of $ a$ and $ b$?[/quote]\r\n\r\ni did not show it is true. i simply shown a counterexample.in other words, this inequality does not work for all posiible $ a$ and $ b$. this inequality is false.", "Solution_4": "Yes,sorry my Bad.I can see it now :(" } { "Tag": [ "trigonometry", "geometry proposed", "geometry" ], "Problem": "In a triangle $ABC$ with $BC=a$ , $AC=b$ , $AB=c$ we have $\\angle{B}=4\\angle{A}$ .\r\n\r\nProve that \r\n$ab^{2}c^{3}=(b^{2}-a^{2}-ac)((a^{2}-b^{2})^{2}-a^{2}c^{2})$", "Solution_1": "[hide=\"i think\"]We have $\\frac{B}{2}=2A\\Longrightarrow \\cos{\\frac{B}{2}}=\\cos{2A}\\Longrightarrow \\cos^{2}{\\frac{B}{2}}=\\cos^{2}{2A}\\Longrightarrow \\frac{1+\\cos{B}}{2}=(2\\cos^{2}{A}-1)^{2}$. Now use cosin law.[/hide]", "Solution_2": "N.T.TUAN could you complete your solution ? I tried to follow your hint but...\r\nI also found that the relation is equivalent to to prove that\r\n\r\n$a^{4}-2a^{2}b^{2}+b^{4}+a^{3}c-ab^{2}c-a^{2}c^{2}-ac^{3}=0$ \r\n\r\nCould anyone help in solving this ?", "Solution_3": "If your problem is true then my hint will works, I am sure. But it can long. Please try again! :wink:", "Solution_4": "$ab^{2}c^{3}=(b^{2}-a^{2}-ac)((a^{2}-b^{2})^{2}-a^{2}c^{2})=(b^{2}-a^{2}-ac)^{2}(b^{2}-a^{2}+ac)$\r\nLet $\\angle A=x \\Longrightarrow B=4x$.From law of cosine $b^{2}-a^{2}=c^{2}-2a.c.cos{4x}=c(2R.sin{5x}-2.2R.sin{x}cos{4x})=2R.csin{3x}$\r\nThen $ab^{2}c^{3}=c^{3}(2Rsin{3x}-a)^{2}(2Rsin{3x}+a)$\r\n$\\Longrightarrow sin{x}sin{4x}^{2}=(sin{3x}-sin{x})^{2}(sin{3x}+sin{x})$\r\n$sin{x}sin{4x}^{2}=4sin{x}^{2}cos{2x}^{2}.2.sin{2x}cos{x}$\r\n\r\nThat's all :)", "Solution_5": "It is known that when in $\\triangle ABC \\angle B=2\\angle A$ $b^{2}-a^{2}=ac$.\r\nLet now $\\angle B=3\\angle A$. Let $M \\in AC$ is such that $\\angle CBM=\\angle A$. As $\\angle ABM=2\\angle A$ $AM^{2}-BM^{2}=BM.c$. As $\\triangle ACB \\sim \\triangle BCM$ we obtain $BM=\\frac{ac}{b}$, $CM=\\frac{a^{2}}{b}$ and $AM=b-CM=\\frac{b^{2}-a^{2}}{b}$. Now sustitute in $AM^{2}-BM^{2}=BM.c$ to get\r\n\r\n$(b^{2}-a^{2})(b-a)=ac^{2}$ - true when $\\angle B= 3\\angle A$.\r\n\r\nWhen $\\angle B= 4\\angle A$ take $K$ on $AC$ such that $\\angle CBK=\\angle A$. Now $(AK^{2}-BK^{2})(AK-BK)=BK.c^{2}$. Replace $BK=\\frac{ac}{b}$ and $AK=\\frac{b^{2}-a^{2}}{b}$ to obtain the desired equality.\r\nIt is evident that this process can be continued the same way to obtain more and more compicated formulas when $\\angle B=5\\angle A$ etc." } { "Tag": [ "probability and stats" ], "Problem": "can anyone show me the relation between poisson process and semi", "Solution_1": "Semi what?", "Solution_2": "semi-martingale ?", "Solution_3": "semi-question :lol:" } { "Tag": [ "geometry", "AMC", "AIME", "trigonometry", "\\/closed" ], "Problem": "I was wondering if this course was being thought about. I was looking at the Winter schedule, and aside from Olympiad classes and Problem Series, nothing else was being offered. I don't think I'm that able at geometry to the point where I could do Olympiad Geometry, but I do feel I can do Introduction to Geometry post test with ease. I would be left behind in Olympiad Geometry but bored in Introduction to Geometry.\r\n\r\nI think more than a few people feel the same way.\r\n\r\nJust curious.", "Solution_1": "isnt the post test to intro geometry almost identical to the pretest to olympiad geometry?", "Solution_2": "There is definitely a level in between Intro Geometry and Olympiad and unfortunately we do not yet have a course for that level. Students interested in this level of geometry might consider taking the AIME Problem Series which covers a lot of hard AIME geometry which is around the level we might call Intermediate Geometry. It's not as good as a whole course, but it's part of a good comprehensive class.\r\n\r\nThere are 3 days spent on geometry and one spent on trigonometry and related algebraic geometry topics during the problem series.", "Solution_3": "[quote=\"Pakman2012\"]I was wondering if this course was being thought about. I was looking at the Winter schedule, and aside from Olympiad classes and Problem Series, nothing else was being offered. I don't think I'm that able at geometry to the point where I could do Olympiad Geometry, but I do feel I can do Introduction to Geometry post test with ease. I would be left behind in Olympiad Geometry but bored in Introduction to Geometry.\n\nI think more than a few people feel the same way.\n\nJust curious.[/quote]\r\n\r\nUnfortunately, it will be a little while before we have such a course - I'll be developing it concurrently with the Intermediate Geometry textbook I'll be writing, which I hope to begin this winter.", "Solution_4": "Just a suggestion as far as scheduling is concerned. The Introduction to Geometry Course should be offered this spring/summer, then Intermediate Geometry in the summer/fall (i.e. Now), and then Olympiad Geometry in the winter as planned. \r\n\r\nI just wish I could use it to prepare for this year's AIME rather than next lol.\r\nSelf studying just isn't as fun" } { "Tag": [ "geometry", "Columbia", "probability", "HCSSiM", "Ross Mathematics Program", "number theory" ], "Problem": "Does anyone know of any math (math- not science) programs in the NYC metro area? I am very interested in Math, am a junior, and am a girl. I applied for the columbia summer program \"A Survey for Modern Mathematics\" but that is only a month (july), and I would like to do more- during the schoolyear and August, perhaps.\r\n\r\nAny ideas?\r\nthanks for any help :)", "Solution_1": "Have you heard of NYCIML?", "Solution_2": "I just googled it, and it seems like a math competition in NYC that is mainly run through high schools...\r\n\r\nI think that I'm looking more for a program than a math competition, mostly because this is my situation:\r\nI had always been somewhat proficient in math compared to others in my grade- I was a year ahead. Then, last year I became really interested in math, so I skipped to extended honors, which I am taking this year, and which is my best subject. I'll be taking BC calc next year.\r\nSo, I am way below par compared to those who participate in math competitions. I'm not sure that would be the best option.\r\n\r\nI don't live in NYC (and don't go to a NYC school either,) but I just figured that the programs I was looking for would most likely exist there.\r\nAny other suggestions?", "Solution_3": "The City College of New York offers a free five-week summer program in math, but I believe it is open only to students who live in New York City. Here is a link:\r\n[url]http://www.collegenow.cuny.edu/students/summer/ccny.html[/url].", "Solution_4": "Unfortunately, that program runs through July, which is when I hope to go to the columbia program. I'll keep it in mind if I can't go, though.\r\n\r\nAny other suggestions? ;)", "Solution_5": "Where do you live (and go to school)? Many programs are run by public institutions (cities, states, or their universities) and thus only accept students who live in the particular region (examples: the CCNY program and the New Jersey Governor's School for Science). Also, many programs are afterschool, which you might not be able to reach if you lived to far away. (I also gave you an answer on the NYC forum, although it doesn't contain much that people haven't said here already.)", "Solution_6": "Ask your guidance counselor about college classes after school. I will be taking a Probability and Statistics course at Queens College starting in about 3 weeks. Last term, I did Number Theory at Columbia. CUNY also offers College Now, college courses that can be taken for credit. Most of the College Now classes aren't too difficult.\r\nMost summer programs are only a month long. :( I did CCNY three years ago and had a great time. I went to HCSSiM two years ago, which was 6 weeks long. I wanted to go to the Ross summer program at U.Chicago last year, but it started too soon, and I needed to take Regents and didn't want to take the August ones.\r\nDo some volunteer work or travel or independent study in August." } { "Tag": [ "puzzles" ], "Problem": "Find the shape and position of pentomino, that according to the same rule, should fit into the empty square. Please, send the solution and rule (logic behind the solution).\r\n\r\n[img]http://img86.imageshack.us/img86/7025/pentominopuzzle4rb.jpg[/img]", "Solution_1": "I know what the piece is \r\nI just don't know how it's supposed to go in.", "Solution_2": "oh\r\n\r\n___________________\r\nl___l___l___l___l___l--1\r\nl___l___l___l___l___l--2\r\nl___l___l___l___l___l--3\r\nl___l___l___l___l___l--4\r\nl___l___l___l___l___l--5\r\n--A---B--C---D--E--\r\n\r\nIt's a w-shaped piece in C3, C4, D4, D5, and E5\r\n\r\ni think that's right", "Solution_3": "[quote=\"sapphyre571\"]\nIt's a w-shaped piece in C3, C4, D4, D5, and E5\n[/quote]\r\nHow did you figure it out? :?", "Solution_4": "my mom is a pre-k teacher so she has all these cool brain-developing things for kids\r\nI play with them when I'm bored\r\nthe w is missing\r\ni just looked for a pattern as to is placement\r\n[hide=\"placement solution\"]\neach vertical row is set up that if the three 5-by-5 squares (with pentominos included) is overlapped,\nthen none of the pentominos overlap.\nUsing this as a basis,\nthe only possible placement for a w that satisfies this condition\nis the place I specified earlier.[/hide]", "Solution_5": "Why the W and not the cross? No one has argued for the uniqueness of this solution which seems like a fundamental part of solving this problem.", "Solution_6": "[quote=\"JBL\"]Why the W and not the cross? No one has argued for the uniqueness of this solution which seems like a fundamental part of solving this problem.[/quote]\r\n\r\nI have the official solution, and it is unique. In other words, there is a simple logic that eliminates all other possibilities and leaves only sapphyre571's solution which is correct. As you say, although sapphyre571's logic seems pretty, unfortunately it leaves multiple solutions. Maybe someone can find rule that will lead to only one solution? :)", "Solution_7": "I guess I kinda cheated - the bottom-right squre looks more darkened so I knew that it would have to cover that square. The cross cannot do that. Working on complete solution..." } { "Tag": [ "AMC" ], "Problem": "Where can I find solutions to the IMO Team Selection Tests?", "Solution_1": "http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=35" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "perimeter", "inequalities", "function", "geometry unsolved" ], "Problem": "Given a tetrahedron $ABCD$, $E$, $F$, $G$, are on the respectively on the segments $AB$, $AC$ and $AD$. Prove that:\r\n\r\ni) area $EFG \\leq$ max{area $ABC$,area $ABD$,area $ACD$,area $BCD$}.\r\nii) The same as above replacing \"area\" for \"perimeter\".", "Solution_1": "Fix $E,F$, and move $G$ on $AD$. If we regard the area/perimeter of $EFG$ as a function of $x=AG$, then it's easy to see that this function is convex, so it reaches its maximum in one of the endpoints of the segment $AD$. We can thus move $G$ to a vertex of the tetrahedron and increase the area/perimeter. We repeat the procedure until all three vertices $E,F,G$ are placed in some vertices of the tetrahedron, and the conclusion follows." } { "Tag": [ "geometry", "circumcircle", "incenter", "trigonometry", "geometric transformation", "reflection", "parallelogram" ], "Problem": "The exinscribed circle of a triangle $ABC$ corresponding to its vertex $A$ touches the sidelines $AB$ and $AC$ in the points $M$ and $P$, respectively, and touches its side $BC$ in the point $N$. Show that if the midpoint of the segment $MP$ lies on the circumcircle of triangle $ABC$, then the points $O$, $N$, $I$ are collinear, where $I$ is the incenter and $O$ is the circumcenter of triangle $ABC$.", "Solution_1": "this should be easy with trilinears :P", "Solution_2": "Maybe, but while trying to solve it I also tried not to spoil it with trilinears or trigonometry :).", "Solution_3": "Cuzzez! I have been thinking upon this problem for two hours, and the only solution I found is a kind of trig bash. However, since nobody has posted any synthetic solution, here is mine:\r\n\r\nWe know that if $I_{a}$, $I_{b}$, $I_{c}$ are the excenters of triangle ABC, then\r\n\r\n- the lines $AI_{a}$, $BI_{b}$, $CI_{c}$ are the three altitudes of the triangle $I_{a}I_{b}I_{c}$;\r\n- the triangle ABC is the orthic triangle of the triangle $I_{a}I_{b}I_{c}$;\r\n- the incenter I of the triangle ABC is the orthocenter of the triangle $I_{a}I_{b}I_{c}$;\r\n- the circumcircle of the triangle ABC is the nine-point circle of triangle $I_{a}I_{b}I_{c}$.\r\n\r\nWe also know that\r\n\r\n- the circumcenter V of the triangle $I_{a}I_{b}I_{c}$ lies on the line $I_{a}N$ (and the two similar lines through $I_{b}$ and $I_{c}$);\r\n- this circumcenter V is the reflection of the incenter I of triangle ABC in the circumcenter O of triangle ABC;\r\n- the circumradius of triangle $I_{a}I_{b}I_{c}$ equals $I_{a}V = 2R$, where R is the circumradius of triangle ABC.\r\n\r\n(Note that the circumcenter V of triangle $I_{a}I_{b}I_{c}$ is called the [i]Bevan point[/i] of triangle ABC.)\r\n\r\nSince the points M and P are the points of tangency of the A-excircle of triangle ABC with the sidelines AB and AC, these points M and P must be symmetric to each other with respect to the angle bisector of the angle CAB. Hence, the midpoint of the segment MP must lie on the angle bisector of the angle CAB. But this angle bisector is the line $AI_{a}$, and this line is the $I_{a}$-altitude of triangle $I_{a}I_{b}I_{c}$. So we see that the midpoint of the segment MP lies on the $I_{a}$-altitude of triangle $I_{a}I_{b}I_{c}$. On the other hand, by our hypothesis, the midpoint of the segment MP lies on the circumcircle of triangle ABC, i. e. on the nine-point circle of triangle $I_{a}I_{b}I_{c}$. Hence, the midpoint of the segment MP is the point of intersection of the $I_{a}$-altitude of triangle $I_{a}I_{b}I_{c}$ with the nine-point circle of this triangle. But, by the definition of a nine-point circle, the $I_{a}$-altitude of triangle $I_{a}I_{b}I_{c}$ intersects the nine-point circle of this triangle at two points: at the foot A of the $I_{a}$-altitude, and at the midpoint of the segment joining the vertex $I_{a}$ of the triangle $I_{a}I_{b}I_{c}$ with the orthocenter I of this triangle. Since the midpoint of the segment MP is distinct from A, we can therefore conclude that the midpoint of the segment MP is the midpoint of the segment joining the vertex $I_{a}$ of the triangle $I_{a}I_{b}I_{c}$ with the orthocenter I of this triangle. In other words, the midpoint of the segment MP is the midpoint of the segment $I_{a}I$.\r\n\r\nTherefore, the quadrilateral $M I_{a}P I$ is a parallelogram, and $MI\\parallel I_{a}P$. But $I_{a}P \\perp CA$; hence, $MI \\perp CA$, and thus the triangle AYM is right-angled, where Y is the point where the incircle of triangle ABC touches the side CA. Hence, $AY=AM\\cdot \\cos \\measuredangle YAM=AM\\cdot \\cos A$. Now, since the points M and P are symmetric to each other with respect to the angle bisector of the angle CAB, we have AM = AP, so that $AY=AP\\cdot \\cos A$, and\r\n\r\n$\\cos A=\\frac{AY}{AP}$.\r\n\r\nBut Thales shows\r\n\r\n$\\frac{AY}{AP}=\\frac{r}{r_{a}}$,\r\n\r\nwhere r, $r_{a}$, $r_{b}$, $r_{c}$ are the inradius and the exradii of triangle ABC. Thus,\r\n\r\n$\\cos A=\\frac{r}{r_{a}}$.\r\n\r\nNow, what do we want to prove? We want to show that the points O, N, I are collinear. This will obviously hold if we show that N = V (since we know that the points O, V, I are collinear, for V is the reflection of I in O). The point V lies on the line $I_{a}N$; hence, in order to obtain N = V, it is enough to show that the segments $I_{a}N$ and $I_{a}V$ are equal. But $I_{a}N = r_{a}$ and $I_{a}V = 2R$. Hence, we must show that $r_{a}= 2R$.\r\n\r\nHere the battle begins. Denoting x = cos A, y = cos B, z = cos C, we know the formulas\r\n\r\n$r=R\\left( x+y+z-1\\right)$;\r\n$r_{a}=R\\left(-x+y+z+1\\right)$.\r\n\r\nAlso, we have given\r\n\r\n$x=\\cos A=\\frac{r}{r_{a}}$,\r\n\r\nso that $r=xr_{a}$. Thus, we can substitute the above formulas to get\r\n\r\nR (x + y + z - 1) = x R (- x + y + z + 1).\r\n\r\nAfter some algebra, this simplifies to\r\n\r\n(x - 1) (y + z - x - 1) = 0.\r\n\r\nNow, since $x-1\\neq 0$ (the cosine of a triangle's angle is never 1), this becomes y + z - x - 1 = 0, so that y + z = x + 1, and\r\n\r\n$r_{a}=R\\left(-x+y+z+1\\right) =R\\left(-x+x+1+1\\right) =2R$,\r\n\r\nqed..\r\n\r\nThe converse of the problem also holds and can be proved just by inverting the observations above.\r\n\r\n[b]EDIT:[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=105101 for a discussion of this problem.\r\n\r\n Darij", "Solution_4": "[hide=\"Only for the fans and not for the professionals.\"][color=darkred]The $A$- exincircle $C_{a}=C(I_{a},r_{a})$ of a triangle $ABC$ touches the sidelines $AB$, $BC$ and $AC$ in the points $R$, $N$ and $P$ respectively.\nThe its incircle $C(I,r)$ touches the lines $BC$, $CA$, $AB$ in the points $D$, $E$, $F$ respectively. Denote : the circumcircle $w=C(O,R)$\nof the triangle $ABC$ ; the middlepoint $M$ of the side $[BC]$ ; the orthocenter $H$ ; the middlepoint $L$ of the segment $PR$.\nProve that $H\\in IM$ $\\Longleftrightarrow$ $N\\in OI$ $\\Longleftrightarrow$ $\\cos A=\\frac{r}{2R}$ $\\Longleftrightarrow$ $r_{a}=2R$ $\\Longleftrightarrow$ $L\\in w$ $\\Longleftrightarrow$ $H\\in EF\\ .$[/color]\n\n[color=darkblue][b]Proof.[/b] Suppose w.l.o.g. that $b>c$. Then,\n\n$1.\\blacktriangleright\\ DB=NC=p-b\\ ;\\ MD=MN=\\frac{b-c}{2}\\ ;\\ ID=r\\ ;\\ OM=R\\cos A$.\nTherefore, $\\boxed{\\ N\\in OI\\ }$ $\\Longleftrightarrow$ $ID=2\\cdot OM$ $\\Longleftrightarrow$ $\\boxed{\\ \\cos A=\\frac{r}{2R}\\ }\\ .$\n\n$2.\\blacktriangleright$ If $l_{a}$ is the length of the bisector of the angle $\\widehat{BAC}$, then $\\boxed{\\ L\\in w\\ }\\Longleftrightarrow AL\\cdot l_{a}=bc\\Longleftrightarrow$\n$p\\cos\\frac{A}{2}\\cdot \\frac{2bc\\cos\\frac{A}{2}}{b+c}=bc$ $\\Longleftrightarrow$ $p(1+\\cos A)=b+c$ $\\Longleftrightarrow$ $\\cos A=\\frac{p-a}{p}$ $\\Longleftrightarrow$ $\\boxed{\\ \\cos A=\\frac{r}{r_{a}}\\ }\\ .$\n\n$3.\\blacktriangleright$ Denote the intersections $S\\in AH\\cap IM$, $V\\in AI\\cap BC$ and the projection $U$ of the point $A$ on the line $BC$. \nApply the Menelaus' theorem to the transversal $\\overline{MIS}$ for the triangle $AUV$ : $\\frac{MV}{MU}\\cdot \\frac{SU}{SA}\\cdot \\frac{IA}{IV}=1$ $\\Longleftrightarrow$\n$\\frac{\\frac{a(b-c)}{2(b+c)}}{\\frac{b^{2}-c^{2}}{2a}}\\cdot \\frac{SU}{SA}\\cdot\\frac{b+c}{a}=1$ $\\Longleftrightarrow$ $\\frac{SA}{a}=\\frac{SU}{b+c}=\\frac{h_{a}}{2p}$ $\\Longleftrightarrow$ $SA=\\frac{ah_{a}}{2p}=\\frac{2pr}{2p}$ $\\Longleftrightarrow$ $SA=r\\ .$\nTherefore, $\\boxed{\\ H\\in IM\\ }$ $\\Longleftrightarrow$ $AH=AS$ $\\Longleftrightarrow$ $2R\\cos A=r$ $\\Longleftrightarrow$ $\\boxed{\\ \\cos A=\\frac{r}{2R}\\ }\\ .$\n\n$4.\\blacktriangleright\\ \\boxed{\\ r_{a}=2R\\ }$ $\\Longleftrightarrow$ $(p-b)(p-c)=2Rr$ $\\Longleftrightarrow$ $p_{w}(D)=p_{w}(I)$ $\\Longleftrightarrow$ $OI=OD$ $\\Longleftrightarrow$\n$2\\cdot OM=ID$ $\\Longleftrightarrow$ $\\boxed{\\ \\cos A=\\frac{r}{2R}\\ }$, where $p_{w}(X)$ is the power of the point $X$ w.r.t. the circle $w\\ .$\n\n$5.\\blacktriangleright\\ H\\in EF$ $\\Longleftrightarrow$ $\\frac{2R\\cos A}{\\cos\\frac{A}{2}}=\\frac{p-a}{\\sin \\left (B+\\frac{A}{2}\\right)}$ $\\Longleftrightarrow$ $2R\\cos A\\sin\\left(B+\\frac{A}{2}\\right)=(p-a)\\cos\\frac{A}{2}$ $\\Longleftrightarrow$ $2R\\cos A\\sin\\left(B+\\frac{A}{2}\\right)\\cdot 2\\cos \\frac{A}{2}=(p-a)\\cos\\frac{A}{2}\\cdot 2\\cos\\frac{A}{2}$ $\\Longleftrightarrow$ $2R\\cos A(\\sin C+\\sin B)=(p-a)(1+\\cos A)$ \n$\\Longleftrightarrow$ $(b+c)\\cos A=(p-a)(1+\\cos A)$ $\\Longleftrightarrow$ $\\cos A=\\frac{p-a}{p}$ $\\Longleftrightarrow$ $\\cos A=\\frac{r}{r_{a}}\\ .$\n\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=5397[/color][/hide]", "Solution_5": "Let $X$ be midpoint of $MP$ ,WE know $I_{a}X.I_{a}A=OI_{a}^{2}-R^{2}=2Rr_{a}=r_{a}^{2}\\rightarrow r_{a}=2R$\r\nlet $D,D'$ be reflection of $I,O$ on the $BC$ we have $OX\\perp BC$ and $D'D=D'N$ so cause $OX=R\\ ,\\ I_{a}N=2R$ and $X$ is midpoint of $II_{a}$ hence $O,I,N$ are collinear.", "Solution_6": "Inversion in the excircle $(I_{a})$ carries the vertices A, B, C into the midpoints X, Y, Z of the excircle chords MP, MN, PN and the circumcircle (O) of $\\triangle ABC$ into the 9-point circle of $\\triangle MNP.$ If $X \\in (O),$ the circumcircle is carried into itself, hence it is the 9-point circle of $\\triangle MNP$ (and $r_{a}= 2R)$. But (O) is also the 9-point circle of the triangle $\\triangle I_{b}II_{c},$ always centrally similar to the triangle $\\triangle MNP$ (with parallel sides). Thus in case $X \\in (O),$ these 2 triangles are congruent. The external bisector $I_{b}I_{c}\\perp AI_{a}\\equiv AX$ of $\\angle A$ meets (O) again at the midpoint Q of $I_{b}I_{c}$ and XQ is a diameter of (O), hence the midpoint O of XQ is the similarity center of $\\triangle MNP \\cong \\triangle I_{b}II_{c},$ so that I, N, O are collinear and O is the midpoint of IN.", "Solution_7": "[quote=\"Virgil Nicula\"]\n\n$2.\\blacktriangleright$ If $l_{a}$ is the length of the bisector of the angle $\\widehat{BAC}$, then $\\boxed{\\ L\\in w\\ }\\Longleftrightarrow AL\\cdot l_{a}=bc\\Longleftrightarrow$\n$p\\cos\\frac{A}{2}\\cdot \\frac{2bc\\cos\\frac{A}{2}}{b+c}=bc$ [/quote]\r\n\r\nWhy is this?", "Solution_8": "$2.\\blacktriangleright$ If $l_{a}$ is the length of the bisector of the angle $\\widehat{BAC}$, then is well-known that $l_{a}=\\frac{2bc}{b+c}\\cdot\\cos \\frac{A}{2}$. Denote $S\\in AI\\cap BC$. Then $\\boxed{\\ L\\in w\\ }$ $\\Longleftrightarrow$ $\\left\\{\\begin{array}{c}\\triangle ABL\\sim \\triangle ASC\\\\\\ LA=RA\\cdot \\cos \\frac{A}{2}\\\\\\ RA=p\\end{array}\\right\\|$ $\\Longleftrightarrow$ $\\left\\{\\begin{array}{c}AL\\cdot l_{a}=bc\\\\\\ LA=p\\cdot \\cos \\frac{A}{2}\\end{array}\\right\\|$ $\\Longleftrightarrow$\r\n$p\\cos\\frac{A}{2}\\cdot \\frac{2bc\\cos\\frac{A}{2}}{b+c}=bc$." } { "Tag": [ "calculus", "integration", "inequalities", "function", "derivative", "probability", "expected value" ], "Problem": "Let $f : \\left[ 0, 1 \\right] \\to \\mathbb R$ be a continuous function and $g : \\left[ 0, 1 \\right] \\to \\left( 0, \\infty \\right)$.\r\nProve that if $f$ is increasing, then\r\n\\[\\int_{0}^{t}f(x) g(x) \\, dx \\cdot \\int_{0}^{1}g(x) \\, dx \\leq \\int_{0}^{t}g(x) \\, dx \\cdot \\int_{0}^{1}f(x) g(x) \\, dx .\\]", "Solution_1": "Let $f: [0,1]\\to \\mathbb{R}$ and $g: [0,1]\\to \\mathbb{R}^{+}$ be continuous functions. Suppose $f$ is an increasing function.\r\n\r\nNow define $h$ on $[0,1]$ as\r\n\r\n$h(t) = \\int_{0}^{t}g(x) \\, dx \\cdot \\int_{0}^{1}f(x)g(x) \\, dx-\\int_{0}^{1}g(x) \\, dx \\cdot \\int_{0}^{t}f(x)g(x) \\, dx$.\r\n\r\nThen $h$ is differentiable and the derivative is given by\r\n\r\n$h'(t) = g(t) \\int_{0}^{1}f(x)g(x) \\, dx-f(t)g(t) \\int_{0}^{1}g(x) \\, dx = g(t) \\int_{0}^{1}(f(x)-f(t))g(x) \\, dx$.\r\n\r\nSince $\\frac{h'(t)}{g(t)}= \\int_{0}^{1}(f(x)-f(t))g(x) \\, dx$ is a (continuous) decreasing function with $\\frac{h'(0)}{g(0)}> 0$ and $\\frac{h'(0)}{g(0)}< 0$, there is only one point $c \\in (0, 1)$ satisfying $h'(c) = 0$. Moreover,\r\n\r\n(1) $h$ is increasing on $[0, c]$, so $0 = h(0) \\leq h(t)$ for $t \\in [0, c]$\r\n(2) $h$ is decreasing on $[c, 1]$, so $h(t) \\geq h(1) = 0$ for $t \\in [c, 1]$\r\n\r\nIn any case, $h(t) \\geq 0$.", "Solution_2": "I was kind of blind with this problem :) I started in the contest with the idea of sos440 but I could not finish...My solution is a little different. I took H the antiderivate of fg, G the antiderivate of g and the inequality was equivalent with:\r\n$\\frac{H(t)-H(0)}{G(t)-G(0)}\\leq \\frac{H(1)-H(0)}{G(1)-G(0)}$ ( I assumed t different of 0 and 1)\r\nEven not I was not able to find a solution using the dervative so applying the trick from the convex function I got that the inequality is equivalent with:\r\n$\\frac{H(t)-H(0)}{G(t)-G(0)}\\leq \\frac{H(1)-H(t)}{G(1)-G(t)}$\r\nNow just apply cauchy for the 2 expresions...the first is f(a) with a in (0,t) and the second f(b) with b in (t,1). The inequality is thus proved :)", "Solution_3": "[quote=\"perfect_radio\"]Let $f : \\left[ 0, 1 \\right] \\to \\mathbb R$ be a continuous function and $g : \\left[ 0, 1 \\right] \\to \\left( 0, \\infty \\right)$.\nProve that if $f$ is increasing, then\n\\[\\int_{0}^{t}f(x) g(x) \\, dx \\cdot \\int_{0}^{1}g(x) \\, dx \\leq \\int_{0}^{t}g(x) \\, dx \\cdot \\int_{0}^{1}f(x) g(x) \\, dx . \\]\n[/quote]\r\n\r\nI assume g is integrable, though you didn't say so.. :p\r\n\r\nUse the additivity of interval theorem to reduce the given inequality to:\r\n\\[\\int_{0}^{t}f(x) g(x) \\, dx \\cdot \\int_{t}^{1}g(x) \\, dx \\leq \\int_{0}^{t}g(x) \\, dx \\cdot \\int_{t}^{1}f(x) g(x) \\, dx . \\]\r\nfrom here convert\r\n\\[\\int_{0}^{t}f(x) g(x) \\, dx \\]\r\nto a Riemann Stieltjes integral with f as the integrand. Note that the integrator is of bounded variation since g is everywhere positive. Do the same for\r\n\\[\\int_{t}^{1}f(x) g(x) \\, dx \\]\r\nto have f as the integrator. (to be definite, the integrator in both cases is the integral of g over (0, x))\r\n\r\nThen the mean value theorem for Riemann Stieltjes integrals tells you that you can write:\r\n\\[f(c) \\int_{0}^{t}g(x) \\, dx \\cdot \\int_{t}^{1}g(x) \\, dx \\leq \\int_{0}^{t}g(x) \\, dx \\cdot f(d) \\int_{t}^{1}g(x) \\, dx . \\]\r\n(where I have converted the Rieman Stieltjes integral back to a Riemann integral for clarity).\r\n\r\nBut (if you know the proof of the mean value theorem for Riemann Stieltjes integrals) you will notice that 0 <= c <= t and t <= d <= 1. Since f is increasing, f(c) <= f(d) and the inequality is proven.\r\n\r\n\r\nOf course if you really wanted to prove it you had better work backwards from what I've done. Heh. Hurrah for the first problem on this forum that I even remotely understsand! This is a great problem for anyone learning the Riemann integral. It certainly took me a while to figure it out. The main trick was to split up the intervals of integration on both sides. That seems to be the only way to justify that c < d.\r\n\r\nP.S. I can't believe I just solved my first ever Olympiad question. There must be some error here. In the meanwhile, I shall sleep well. :)", "Solution_4": "$g$ is continuous (but can be chosen to be only Riemann integrable, I think) and $t$ is in $[0,1]$.\r\nFor some reason, I forgot to write these conditions :blush: \r\n\r\nAnyway, your solution is fine. Actually, on a close inspection, you'll discover that it's equivalent to ciprian's solution.\r\n\r\nBTW, here's my version:\r\nLet $h(t) = \\left( \\int_{0}^{t}f(x) g(x) \\, dx \\right) \\left\\slash \\left( \\int_{0}^{t}g(x) \\, dx \\right) \\right.$. The conclusion is equivalent to proving that $h(t) \\leq h(1)$.\r\nWe can prove more, $h$ is actually increasing. Now compute $h^\\prime(t)$:\r\n\\[\\left( f(t) g(t) \\int_{0}^{t}g(x) \\, dx-g(t) \\int_{0}^{t}f(x) g(x) \\, dx \\right) \\left\\slash \\left( \\int_{0}^{t}g(x) \\, dx \\right)^{2}\\right. . \\]\r\n$g$ is positive (and integrable), so the Mean Value Theorem works here.\r\nNow just use the fact that $f$ is increasing to get that $h^\\prime \\geq 0$.", "Solution_5": "[quote=\"perfect_radio\"]BTW, here's my version:\nLet $h(t) = \\left( \\int_{0}^{t}f(x) g(x) \\, dx \\right) \\left\\slash \\left( \\int_{0}^{t}g(x) \\, dx \\right) \\right.$[/quote]\r\n\r\nThis seems to be by far the most natural way to approach the problem to me.", "Solution_6": "Here's another approach. Let $h : [0, 1] \\to \\mathbb{R}$ be the function defined by\r\n\\[h(x) = \\begin{cases}1 & \\textrm{if }x \\le t; \\\\ 0 & \\textrm{otherwise}. \\end{cases}\\]\r\nNote that $h$ is a decreasing, piecewise-continuous function. Then the inequality can be rewritten as\r\n\\[\\int_{0}^{1}g(x) f(x)h(x) \\, dx \\cdot \\int_{0}^{1}g(x) \\, dx \\le \\int_{0}^{1}g(x) f(x) \\, dx \\cdot \\int_{0}^{1}g(x) h(x) \\, dx. \\]\r\nYou might already recognize this inequality as Chebyshev's inequality. If not, note that because $f$ is increasing and $h$ is decreasing, we have\r\n\\[g(x) g(y) (f(x)-f(y)) (h(x)-h(y)) \\le 0 \\]\r\nfor all $x$ and $y$ in $[0, 1]$. Now integrate both sides over all $x$ and $y$ in $[0, 1]$, and you will get the desired inequality.\r\n\r\nOr you might like to view the inequality via probability. By scaling, we may assume that\r\n\\[\\int_{0}^{1}g(x) \\, dx = 1. \\]\r\nIn other words, $g$ is a probability density function. Let $X$ be a random variable with density function $g$. Then our inequality becomes\r\n\\[E(f(X)h(X)) \\le E(f(X)) E(h(X)) \\, . \\] \r\nAgain, this is Chebyshev's inequality. For a proof from scratch, start with the inequality\r\n\\[(f(X)-f(Y))(h(X)-h(Y)) \\le 0, \\]\r\nwhere $Y$ is an independent copy of $X$. Now take the expected value of both sides." } { "Tag": [ "function", "summer program", "Mathcamp", "set theory" ], "Problem": "Prove, using only mathematics (no altering of the equation), that 1=1. Basically,\n\nProve: 1=1.\n\nRead my signature if you find this hard.:rotfl:", "Solution_1": "Huh? How do you prove an axiom?", "Solution_2": "many axioms are based on logic and therefore proven in teh human mind. Think about logical things in mathematics such as set theory (simple ones).", "Solution_3": "[quote=\"boxedexe\"]many axioms are based on logic and therefore proven in teh human mind.[/quote]What are you talking about?\r\n\r\nCouldn't there be an axiom like $(\\forall{a})(a=a)$?", "Solution_4": "http://mathworld.wolfram.com/PeanosAxioms.html", "Solution_5": "[quote=\"Singular\"]http://mathworld.wolfram.com/PeanosAxioms.html[/quote]Then you could say... since the successor of 1 is 2, and the successor of 1 is 2, then $1=1$", "Solution_6": "[quote=\"eryaman\"][quote=\"Singular\"]http://mathworld.wolfram.com/PeanosAxioms.html[/quote]Then you could say... since the successor of 1 is 2, and the successor of 1 is 2, then $1=1$[/quote]\n\nYou meant to say, since the successor of 1 is 2 and the successor of 2 is 1, $1=1$, right?", "Solution_7": "How do you know 2=2?\r\nAnd the successor of 2 isn't 1...", "Solution_8": "equality is transitive, so as long as $1$ is well defined, we have $1=1$.", "Solution_9": "[quote=\"boxedexe\"][quote=\"eryaman\"][quote=\"Singular\"]http://mathworld.wolfram.com/PeanosAxioms.html[/quote]Then you could say... since the successor of 1 is 2, and the successor of 1 is 2, then $1=1$[/quote]\n\nYou meant to say, since the successor of 1 is 2 and the successor of 2 is 1, $1=1$, right?\n\nMasoud Zargar[/quote]\r\n\r\nNah, he meant exactly what he said. Remember, if 2 is the successor of 1, then 1 is not the successor, but the predecessor of 2.\r\nAnd in any case, this is a postulate upon which everything is built, so you'd need some other form of math to prove it. I remember an interesting conversation with my friends about a universe in which the reflexive property does not hold...But it's like Godel said: in any system of mathematics, there will be something which cannot be proven by something else in the system.", "Solution_10": "This is a beautiful solution, which is very, yet logical and correct. There are other proofs using set theory.", "Solution_11": "Proof by Contradiction\r\n\r\nAssume this is not true, that $1\\ne1$ or 1 is \"NOT 1\" \r\n\r\nThen every time i have 1 cup of coffee \r\nim having NOT 1 cup of coffee which means \r\nim having NOT NOT 1 cup of coffee... \r\nwhich is a beatiful paradox, because NOT NOT, is a double negative and so 1=1, and thus proven. (but i guess the double negative is itself an axiom? - or at least derived from one?)\r\n\r\nI wonder if this logic works?", "Solution_12": "[quote=\"zeus_three\"]Proof by Contradiction\n\nAssume this is not true, that $1\\ne1$ or 1 is \"NOT 1\" \n\nThen every time i have 1 cup of coffee \nim having NOT 1 cup of coffee which means \nim having NOT NOT 1 cup of coffee... \nwhich is a beatiful paradox, because NOT NOT, is a double negative and so 1=1, and thus proven. (but i guess the double negative is itself an axiom? - or at least derived from one?)\n\nI wonder if this logic works?[/quote]This should be submitted to a site about comically fallacious proofs.", "Solution_13": ":P", "Solution_14": "Okay well set theoretically (if I still remember from 8th grade):\r\n\r\n1 is defined as set of the null set.\r\n\r\nNow 1=1 becuase we can define an isomorphism that maps every element to another element in 1, namely the identity function!", "Solution_15": "[quote=\"AntonioMainenti\"][quote=\"zeus_three\"]Proof by Contradiction\n\nAssume this is not true, that $1\\ne1$ or 1 is \"NOT 1\" \n\nThen every time i have 1 cup of coffee \nim having NOT 1 cup of coffee which means \nim having NOT NOT 1 cup of coffee... \nwhich is a beatiful paradox, because NOT NOT, is a double negative and so 1=1, and thus proven. (but i guess the double negative is itself an axiom? - or at least derived from one?)\n\nI wonder if this logic works?[/quote]This should be submitted to a site about comically fallacious proofs.[/quote]\r\n\r\n :rotfl: :rotfl: , Zeus_three, I think my teacher would love to have you in his philosophy class :lol: .\r\n\r\nBy the way, about the proof $1=1$, I don't remember actually :lol: , but I once heard John Conway talk about that [b]beautiful proof[/b] in mathcamp this year. I wish I had remembered :( .", "Solution_16": "I thought i could chip in my dumb thoughts...\r\n0=0\r\n0+1=0+1\r\n1=1 :| \r\n\r\nI know my sign says think twice before you speak :D" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "calculus", "function", "derivative" ], "Problem": "This is from some AHSME....\r\n\r\nIf $x^2+y^2=14x+6y+6$,what is the largest possible value of $3x+4y$?", "Solution_1": "[hide=\"Answer\"]$x^2+y^2=14x+6y+6\\Rightarrow x^2-14x+y^2-6y=6 \\Rightarrow (x-7)^2+(y-3)^2=8^2$\nTherefore, we have a circle with center at $(7,3)$ and $r=8$. Okay... I know there's another way to solve it, but it's not coming to mind right now, so I'm going to take the derivative. $f'(x)=2x-14+2y\\frac{dy}{dx}-6\\frac{dy}{dx}=0\\Rightarrow \\frac{dy}{dx}=\\frac{14-2x}{2y-6}$. Therefore, our maximum is at $x=7$ and our answer is $3(7)+4(11)=\\boxed{65}$.[/hide]\r\n\r\nHope I did that right...", "Solution_2": "[hide]i dont know if this is the exact approach, but consider the line $3x+4y=n$, we want to maximize $n$ clearly, the line will be tangent to the circle (and be above), \nthis line has a slope of $\\frac{-3}{4}$, and if this line is tangent to the circle, then the line through the center to the point of tangency will be perpendicular to that line, and the slope will $\\frac{4}{3}$, and since it passes through the center, the equation will be $y-3=\\frac{4}{3}(x-7)$\nso if we take the intersection of that line, and the circle, we will have 2 point of intersection, and if we take our maximized point, the answer follows\n(note, you do not actually have to go through the algebra to solve those 2 equations, you can just draw the x and y components of that line) but anyway, we get our point to be $\\left(\\frac{59}{5},\\frac{47}{5}\\right)$ and\n$3x+4y=\\boxed{73}$[/hide]", "Solution_3": "AHSME problems never require calculus. Your solution is wrong anyway. You totally disregarded that we were trying to maximize 3x+4y...it's nowhere in your solution. Altheman's solution is the intended solution (but please use the Hide tags next time...I put them in for you).", "Solution_4": "sorry about that, i will remember to hide my solutions now", "Solution_5": "I know that AHSME problems never require Calculus, but there's nothing wrong about using it when applicable.\r\n\r\nAs for my answer... I know we're trying to maximize $3x+4y$, and I did use that at the end with the values of $x$ and $y$ that I got, but my mistake was that the maximum of the circle equation is not the maximum of $3x+4y$...", "Solution_6": "Using Cauchy Schwarz always makes me happy\r\n\r\n[hide]$8^2\\cdot 5^2=((x-7)^2+(y-3)^2)(3^2+4^2)\\geq (3(x-7)+4(y-3))^2$\n\nThis is the same as $40\\geq 3x-21+4y-12\\implies 73\\geq 3x+4y$[/hide]", "Solution_7": "[quote=\"JesusFreak197\"]I know that AHSME problems never require Calculus, but there's nothing wrong about using it when applicable.\n\nAs for my answer... I know we're trying to maximize $3x+4y$, and I did use that at the end with the values of $x$ and $y$ that I got, but my mistake was that the maximum of the circle equation is not the maximum of $3x+4y$...[/quote]\r\n\r\nThe main problem was that you used calculus blindly. It should have been obvious that the two values you got would be the 'top' and 'bottom' of the circle, which you could have found by inspection anyways.", "Solution_8": "[quote=\"blahblahblah\"]Using Cauchy Schwarz always makes me happy\n\n[hide]$8^2\\cdot 5^2=((x-7)^2+(y-3)^2)(3^2+4^2)\\geq (3(x-7)+4(y-3))^2$\n\nThis is the same as $40\\geq 3x-21+4y-12\\implies 73\\geq 3x+4y$[/hide][/quote]\r\n\r\nThat's pretty brilliant!", "Solution_9": "[quote=\"JesusFreak197\"]I know that AHSME problems never require Calculus, but there's nothing wrong about using it when applicable.\n\nAs for my answer... I know we're trying to maximize $3x+4y$, and I did use that at the end with the values of $x$ and $y$ that I got, but my mistake was that the maximum of the circle equation is not the maximum of $3x+4y$...[/quote]\r\nYes, you want to maximize $3x+4y$ not the other one eg $(x-7)^2+(y-3)^2-8^2$ (which BTW you know is always zero!). \r\n\r\nOk mods close your eyes and don't look at the hidden part. :) (Did not want to dabble in calculus, just commenting on what was wrong in the method) \r\n\r\n[hide]So what you have to do is find y in terms of x from the second equation, substitute it so you $3x+4y$ becomes a function of $x$ and now do your thing with calculus to maximize it.\n\nThis does become a little messy, an easy method, if you have done Lagrange Multiplier Method would be to write:\n$z=3x+4y - \\lambda ((x-7)^2+(y-3)^2-8^2)$ Now take partial derivative wrt $x,y, \\lambda$ equate each to zero and find $x,y,\\lambda$ substitute and you are done. [/hide]\r\n\r\nAbove part is calculus, I would delete it in a day or so after JesusFreak197 had a look at it.. hope that is useful.", "Solution_10": "[quote=\"blahblahblah\"]Using Cauchy Schwarz always makes me happy\n\n[hide]$8^2\\cdot 5^2=((x-7)^2+(y-3)^2)(3^2+4^2)\\geq (3(x-7)+4(y-3))^2$\n\nThis is the same as $40\\geq 3x-21+4y-12\\implies 73\\geq 3x+4y$[/hide][/quote]\r\n\r\nI dind't understand...Why start with $8^2*5^2$?...and what about the equation of the question,where is that coming in use in your solution blahblahblah?", "Solution_11": "The 'equation of the question' is just $(x-7)^2+(y-3)^2=8^2$. I know that if I use Cauchy-Schwarz appropriately, I can extract a $3x+4y$ somehow. That is where the $3^2+4^2=5^2$ comes in.", "Solution_12": "Hmm... I've learned Cauchy-Schwarz, but apparently I don't really have much practice with it, so I'm not sure what you did. The basic formula is $\\displaystyle\\left(\\sum_{k=1}^na_kb_k\\right)^2\\le \\left(\\sum_{k=1}^na_k^2\\right)\\left(\\sum_{k=1}^nb_k^2\\right)$; what numbers did you put in where?\r\n\r\nGyan, thanks for the clarification on what I did wrong. I had put $y$ in terms of $x$ and $m$ at one point, but I saw that I had one variable and one unknown, which didn't quite seem congenial to a derivative, so I didn't use that. :oops:", "Solution_13": "[quote=\"JesusFreak197\"]Hmm... I've learned Cauchy-Schwarz, but apparently I don't really have much practice with it, so I'm not sure what you did. The basic formula is $\\displaystyle\\left(\\sum_{k=1}^na_kb_k\\right)^2\\le \\left(\\sum_{k=1}^na_k^2\\right)\\left(\\sum_{k=1}^nb_k^2\\right)$; what numbers did you put in where?\n\nGyan, thanks for the clarification on what I did wrong. I had put $y$ in terms of $x$ and $m$ at one point, but I saw that I had one variable and one unknown, which didn't quite seem congenial to a derivative, so I didn't use that. :oops:[/quote]\r\n\r\nHe used $a_1 = x-7, a_2 = y-3$ and $b_1 = 3, b_2 = 4$ to get $((x-7)^2+(y-3)^2)(3^2+4^2) \\ge (3(x-7)+4(y-3))^2$.", "Solution_14": "Ah, very nice solution there. :)" } { "Tag": [ "probability" ], "Problem": "Two distinct numbers are selected simultaneously and at\nrandom from the set $ \\{1, 2, 3, 4, 5\\}$. What is the probability that\nthe smaller one divides the larger one? Express your answer as a\ncommon fraction.", "Solution_1": "I think the only way to do this is if you actually do it out.\r\n\r\n1/2, no\r\n1/3, no\r\n1/4, no\r\n1/5, no\r\n2/1, yes\r\n2/3, no\r\n2/4, no\r\n2/5, no\r\n3/1, yes\r\n3/2, no\r\n3/4, no\r\n3/5, no\r\n4/1, yes\r\n4/2, yes\r\n4/3, no\r\n4/5, no\r\n5/1, yes\r\n5/2, no\r\n5/3, no\r\n5/4, no\r\n\r\nWe have 5 yes-es out of 20 possible, so the answer is $ \\boxed{\\frac{1}{4}}$.", "Solution_2": "A little trick to go faster:\r\n\r\nRather than list them all out like ernie did, which is very time-consuming, you can start with 5 and work your way down.\r\n\r\nSince 5 is prime, there is only one integer less than 5 that divides - 1.\r\n\r\nFor the number 4, there are two integers - 2 and 1.\r\n\r\nFor 3, we have only one integer - 1.\r\n\r\nFor 2, we have one integer - 1.\r\n\r\nFor 1, we have none.\r\n\r\nCounting these up, we have 5 ways out of a possible $ 5 \\times 4 \\equal{} 20$ ways, so $ \\boxed{\\frac{1}{4}}$.", "Solution_3": "I did it by thinking the smaller divides INTO the larger...\r\n\r\nSo 1 divides into everything else, thats 4.\r\n2 divides into only 4.\r\n(Note that choosing 2 then 4 has the same result as when you choose 4 and then 2)\r\n3 and 5 are prime, and 4 is too big, so they don't divide into anything there...\r\n\r\nAnd there are 5x4=20 possibilities, so $ \\frac{5}{20}\\equal{} \\boxed{1/4}$.", "Solution_4": "it says the answer is 1/2....", "Solution_5": "There are $ 5$ pairs which satisfy the condition ($ (1,2)$, $ (1,3)$, $ (1,4)$, $ (1,5)$, $ (2,4)$), and the number of ways we can pick two integers is $ \\binom{5}{2}\\equal{}10$, so $ \\frac{5}{10}\\equal{}\\boxed{\\frac12}$.", "Solution_6": "Yeah, two years ago, I made the mistake using permutations. Stupid me. :P", "Solution_7": "The only way is if the denominator is 1 or 2 which there are 5 ways out of 5choose2=10 so 1/2" } { "Tag": [ "trigonometry" ], "Problem": "If $\\sin{\\beta}=2\\sin{(2\\alpha+\\beta)}$, then prove:\r\n\\[\\tan{(\\alpha+\\beta)}+3\\tan{\\alpha}=0\\]", "Solution_1": "I wrote this down. I'll try to verify it doing the day.", "Solution_2": "[quote=\"shobber\"]If $\\sin{\\beta}=2\\sin{(2\\alpha+\\beta)}$, then prove:\n\\[\\tan{(\\alpha+\\beta)}+3\\tan{\\alpha}=0\\][/quote]\r\n\r\nFirst I Prove $\\tan A + \\tan B = \\frac{{\\sin (A + B)}}{{\\cos A\\cos B}}$\r\n\r\nby writing $TanA$ as $\\frac{{\\sin A}}{{\\cos A}}$ and $TanB$ as $\\frac{{\\sin B}}{{\\cos B}}$\r\n\r\nGiven $\\frac{{\\sin (2\\alpha + \\beta )}}{{\\sin \\beta }} = \\frac{1}{2}$\r\n\r\nwriting it as \r\n\r\n$\\frac{{\\frac{{\\sin [(\\alpha + \\beta ) + \\alpha ]}}{{\\cos (\\alpha + \\beta )\\cos \\alpha }}}}{{\\frac{{\\sin [(\\alpha + \\beta ) - \\alpha ] }}{{\\cos (\\alpha + \\beta )\\cos \\alpha }}}} = \\frac{1}{2}$\r\n\r\n$ \\Rightarrow \\frac{{\\tan (\\alpha + \\beta ) + \\tan \\alpha }}{{\\tan (\\alpha + \\beta ) - \\tan \\alpha }} = \\frac{1}{2}$\r\n\r\n$ \\Rightarrow \\frac{{\\tan (\\alpha + \\beta ) + \\tan \\alpha }}{{\\tan (\\alpha + \\beta ) - \\tan \\alpha }} = \\frac{{3 - 1}}{{3 + 1}}$\r\napplying componendo-dividendo, we get\r\n\r\n\r\n$\\tan{(\\alpha+\\beta)}+3\\tan{\\alpha}=0$" } { "Tag": [ "calculus", "integration", "limit", "function", "geometry", "rectangle", "derivative" ], "Problem": "Please excuse me if this the wrong board to post this in.\r\n\r\nI am currently taking AP Calc AB in High School, and the book introduces terrible equations for the limit definition of the integral. They are:\r\n$ \\lim_{\\parallel{}P\\parallel{}\\rightarrow0}\\sum_{k \\equal{} 1}^{n}f(c_{k})\\Delta x_{k}$\r\nand\r\n$ \\lim_{n\\rightarrow\\infty}\\sum_{k \\equal{} 1}^{n}f(c_{k})\\Delta x$\r\nalso, I don't know the code the little triangle delta, but that's what those $ \\text{delta}$s are. (EDIT: fixed)\r\n\r\nAs such, I have decided to develop my own, however it has issues and doesn't evaluate the integrals properly. It looks like it should work, but it doesn't, incorrectly evaluating $ \\int_{0}^{2}[3x^{2}]dx$ as 16. It is, in its two variations:\r\n\r\n$ \\lim_{i\\rightarrow\\infty}\\sum_{n \\equal{} a}^{b\\cdot i}\\frac {f\\left(a \\plus{} \\frac {n \\minus{} a}{i}\\right)(b \\minus{} a)}{i}$\r\nor\r\n$ \\lim_{i\\rightarrow\\infty}\\left(\\frac {b \\minus{} a}{i}\\sum_{n \\equal{} a}^{b\\cdot i}f\\left(a \\plus{} \\frac {n \\minus{} a}{i}\\right)\\right)$\r\n\r\nThe idea is for these to equal $ \\int_{a}^{b}[f(x)]dx$, $ a$ and $ b$ have the same definitions, $ n$ is the sum variable, and $ i$ is how many times the interval $ [a,b]$ is divided.\r\n\r\nHow it should work is it takes the initial function value ($ f(a)$) and multiplies it against the width ($ \\lim_{i\\rightarrow\\infty}\\frac {b \\minus{} a}{i}$), and then takes $ \\lim_{i\\rightarrow\\infty}f\\left(a \\plus{} \\frac {1}{i}\\right)$ multiplies that against the width, it takes all these values and adds them together. The second functions in the same way, except that is multiplies the constant width afterwards.\r\n\r\nI thought I'd put it up so that you could completely annihilate it, and tell me what's wrong. Thanks in advance.", "Solution_1": "The definition of the integral given (the real one) is rigorous and was proven so that is how you should define it. Basically, this definition selects some test point in an infinitesimal square (or rectangle) in the region, finds its area, and then distributes it throughout the entire region. I'm not sure you can define the integral to be in terms of the way you defined it. If you (or no one) can prove it, then it is wrong. Anyway, the definition itself is not particularly important in AP. And the code is $ \\Delta$ is \\Delta.", "Solution_2": "[quote=\"JRav\"]The definition of the integral given (the real one) is rigorous and was proven so that is how you should define it. Basically, this definition selects some test point in an infinitesimal square (or rectangle) in the region, finds its area, and then distributes it throughout the entire region. I'm not sure you can define the integral to be in terms of the way you defined it. If you (or no one) can prove it, then it is wrong. Anyway, the definition itself is not particularly important in AP. And the code is $ \\Delta$ is \\Delta.[/quote]\r\n\r\nThanks, JRav. It's not that I think that the definitions given by the book are [i]wrong[/i] per say, just that they're aren't functional (least ways, I don't get them). For instance, they don't limit the function to just $ [a,b]$ in their definitions, and good luck getting a calculator to process them without some extreme coding.\r\n\r\nThe derivative limit definition works splendidly in the calculator, these do not. I guess what I want is something evaluable.\r\n\r\nThe way I defined the integral is obviously incorrect as it produces incorrect values. I'd like someone to help me figure how to make it work. Or just explain the book's definitions, I'm not picky.", "Solution_3": "It looks like you don't understand the notation- your sums are nonsense. The standard definition of $ \\sum$ notation includes that the indices must be integers, and yours aren't.\r\n\r\nNow, if you want to fix your ideas- it looks like you're trying to restrict to the case of an evenly spaced partition. A way to express this correctly:\r\n$ \\lim_{N\\to\\infty}\\sum_{j \\equal{} 1}^N\\frac {b \\minus{} a}{N}f\\left(a \\plus{} j\\frac {b \\minus{} a}{N}\\right)$\r\n\r\nThat works- we don't actually need unevenly spaced partitions. In practice, it's better to make the definition broad- those other partitions have their uses. Two examples:\r\n\r\nUse a geometrically spaced partition to calculate $ \\int_a^b x^r\\,dx$ for some real $ r\\neq \\minus{} 1$ and fixed $ a,b > 0$. What happens if $ r \\equal{} \\minus{} 1?$\r\nGeometric sums are a lot easier to handle than the stuff that comes up with uniform partitions.\r\n\r\nShow that $ \\int_0^1 f(x)\\,dx$ exists and is zero, where $ f(x) \\equal{} \\begin{cases}\\frac1q & x \\equal{} \\frac pq\\text{ in lowest terms} \\\\\r\n0 & \\text{otherwise}\\end{cases}$.", "Solution_4": "There's also the upper and lower bounds approach you could consider.\r\n\r\nI surmise you're looking for something you can develop for your TI-whatever number. I'm sure there are programs like that.\r\n\r\nIf you are interested in learning integration your best bet is to get an analysis book and start from scratch.\r\n\r\nOh, and you also want to check that your definition shows that continuous functions are Riemann integrable.\r\n\r\nHonestly though AP doesn't require you to spend your time on the definition of Riemann integral.", "Solution_5": "Maybe it's because I never focused on the definition in AP, but once I learned the rigorous definition of double and triple integrals, I found it a lot easier to understand the single variable case.", "Solution_6": "Thanks very much, guys. I'll stop wasting my time trying to develop this. I would have gotten back sooner, but my e-mail wasn't (and isn't) receiving e-mails, so I had no idea that these replies were posted. Sorry.\r\n\r\nOn a tangent, during my time away, I realized that the $ b\\minus{}a$ in my formula is incorrect.\r\n\r\nThanks, again." } { "Tag": [ "AMC", "AIME", "calculus", "college", "articles", "geometry", "function" ], "Problem": "Do you ever get frustrated because your math teacher won't or can't answer your questions in class? Do your classmates groan when you raise your hand because they want to get through class as quickly as possible whereas you actually want to understand what's being taught?", "Solution_1": "if that is the case, you gotta find another class, buddy...", "Solution_2": "What if it's the only Calc BC section in your school?", "Solution_3": "I ask on AoPS if I have any questions....which does happen often\r\n\r\n\r\nMy math teacher likes to rush through sections and assign homework as fast as possible. My classmates are helpful, but I'm the type that doesn't pay much attention in class, and I like to learn it at home. I feel I know it better if I can teach it to myself through a book.\r\n\r\nI used to ask my math teacher questions about different types of math. I was greatly disappointed when he couldn't teach me basic mod functions. With the AP Test tomorrow, class is just a formality for me....I'm only going because I want to graduate high school....\r\n\r\n\r\nIf you do have any questions, I suggest that you bring them to AoPS...I have yet to encounter a question that someone here can't answer (barring milennium problems and what not.)\r\n\r\n\r\n_peace", "Solution_4": "At the beginning of the semester I would ask my AlgII teacher how different formulas were derived, or things like that, but he said that it didn't matter as long as I knew how to use the formula. So I basically stopped asking questions like that in my class. I do, however, enjoy pointing out when he has made a mistake on a problem that he's doing in front of the class. :lol:", "Solution_5": "[quote=\"ankur87\"]If you do have any questions, I suggest that you bring them to AoPS...I have yet to encounter a question that someone here can't answer (barring milennium problems and what not.)\n\n\n_peace[/quote]\r\n\r\nGive us a little more time ;)", "Solution_6": "you can PM me any questions.... not about BC, because i am not in yet. But anything else is okay", "Solution_7": "Welcome to American classroom math :P. You should ask people here or try to figure things out yourself (it is sometimes within reach!). Or ask your math teacher very nicely and cross your fingers and toes... :?", "Solution_8": "You ask questions in math class??? :P\r\n\r\nI never really bothered to after like middle school, because I figured out that anything I don't understand (not that it happens often ;)), someone else will not either and he or she will ask and I'll get my explanation. Then again, I usually don't come across this very much, so maybe I shouldn't be talking.", "Solution_9": "A great skill to acquire is being able to look stuff up and learn it on your own. Becoming an independent learner goes a long way in furthering your education. On the other hand, do NOT be afraid to ask questions when you don't get something. You won't (or at least shouldn't) lose any pride in the process.", "Solution_10": "Part of the problem is that so few (i.e. almost none) of the better math students want to go on to become teachers when they grow up. :(\r\n\r\nThe other problem is that too many American students know that they can go on and have comfortable middle-class lives in this country (if not become President of the US) by simply going through the motions of education without really becoming intellectually engaged in the process.", "Solution_11": "[quote=\"joml88\"]A great skill to acquire is being able to look stuff up and learn it on your own. Becoming an independent learner goes a long way in furthering your education. On the other hand, do NOT be afraid to ask questions when you don't get something. You won't (or at least shouldn't) lose any pride in the process.[/quote]\r\n\r\nTHAT helps a lot when you want to get better at AIME or USAMO. Alot of independent studying and studying other solutions... and other stuff helps.", "Solution_12": "In my BC calculus class, we rush in order to get through the material. We occasionally do a proof, so it's not a \"memorize and regurgitate\" class. I knew a good portion of what we did cover, so I didn't have any questions really. Most of my comments were just pointing out faster or significantly different ways of solving problems, for the benefit of the class, really.\r\nI save theory and proof questions for after class because I know no one else is interested in that kind of stuff and because I feel that if we went through all of my questions, the curriculum would never be finished. So far, I've gotten relatively thorough answers with this approach and haven't annoyed any peers. I also learned that the point of the class is to get through the AP exam. Although I do not like that, I've accepted that it's what a majority of the students in the class care about and it's their loss if they're not interested in the theory.\r\nI find independent study helpful as well, and AOPS has been great in terms of providing challenging questions as well as complete answers.\r\n\r\nOn the other hand, if you don't understand how to do the problem after following an example and making an honest attempt, you should ask for help. It does you no good to leave class confused. If your teacher can't/won't explain these types of questions, s/he's not much of a teacher, so get someone else. (Ignore your class when they groan.)", "Solution_13": "Agreed. Asking questions after class can do wonders. I am very fortunate this year and have an excellent math teacher who will make every effort to answer my questions. If he doesn't know, he'll look it up. \r\n\r\nI've been less fortunate in my other classes, however, where asking questions in class can elicit groans and rolled eyes (yes, from the teacher) and sometimes even openly hostile reactions. Haha...ok, so it doesn't happen that much, but it does happen! So like joml88 says, it's a good thing it doesn't hurt my pride to ask questions. \r\n\r\nSo if you're stuck in one of the less fortunate circumstances, here are some of the things I've done:\r\n\r\n-Give the teacher a chance and instead of asking questions while they're teaching, ask them afterwards. (i.e. First accept blindly and try to follow along as you jot down questions; later, once you've picked through the lesson and know what you can understand/accept, ask questions about what is left to be desired.)\r\n\r\n-Select question-asking representatives. (No joke.) lol. My friends have been great in this respect. If I can interest them in a question enough that they want to know the answer, they'll be willing to ask the question for me. :) This diversifies (to some extent) the question-askers in class! (My friends also enjoy seeing me squirm when the teacher glances over at the true source of the question.)\r\n\r\n-Self-study. As other posters have indicated, self-study is an EXCELLENT way to learn material. I cannot emphasize this more. I have taken up self-study in a certain unnamed and absolutely frustrating HL nightmare (I mean...class), and since then, it's become much more enjoyable. I've learned a lot, and because I also have to teach the material to my peers, I can now say that I really understand the material. :)\r\n\r\n-And if all else fails, go look for help. Teachers, for the most part, WANT you to succeed. Hopefully that's why they went into the profession. If your classroom teacher doesn't want to help for whatever reason, go ask another one. Besides, if you ask a teacher that doesn't teach you, you're in no danger of playing teacher's pet, and it's a complement of that teacher's teaching ability. \r\n\r\nOh, and of course, I'm sure AoPS would be happy to answer any/most/all of your math questions!", "Solution_14": "Well, I haven't really been in a normal classroom setting since first semester of last year, but back in the day, I had the exact same problem. Well, sometimes they would groan, and sometimes they would know that I was about to pwn my math teacher, because she was entirely incompetent.", "Solution_15": "yea... same here. in the MIddle school, i used to pwn my teacher. To keep her superiority, she would just try to shut my mouth... which didnt work. Then i would get in trouble... which was.... really ridiculous. School teachers just try to impress the students, although half of them are already smarter then them. ;)", "Solution_16": "[quote=\"xxreddevilzxx\"]School teachers just try to impress the students, although half of them are already smarter then them. ;)[/quote]\r\n\r\nPerhaps... (Sorry if this is getting slightly off-topic, but I do feel quite compelled to respond to xxreddevilzxx's comment.)\r\n\r\nI'm a pretty hard person to impress (ok...so a lot of you on this site regularly impress me, but still!), but I've also had the good fortune of having excellent teachers. \r\n\r\nI've heard many fantastic things about my math teacher: he's so smart, he's so good at math, he's so this...he's so that. Since finally attending his class, I find that it's not his intellect that impresses me (he is quite good), but the way he carries himself, treats others, and approaches life. I have seen so many students literally flunk his class (i.e. D or below), but NONE of them have ever hated him OR his class. How many teachers can pull that off? Not very many. What's more, he has a way of always being a student's biggest cheerleader. He fished me out of some obscure classroom when I was in 8th grade and has been watching over me since. His ability to inspire and give courage to those who lack it is absolutely phenomenal. \r\n\r\nI guess the thing is that in life, you'll probably come across many brilliant people. And initially, it may be their brilliance that impresses you. But eventually, for most people you meet, the qualities that will sustain the \"impress\" are personal qualities.\r\n\r\nOk...so, how off-topic was THAT?", "Solution_17": "Thanks for all the responses. Actually this is more my friend's situation than my own. I was wondering how common this problem is. She started the year asking her calc teacher for proofs and derivations but soon gave up. She says no one in her class asks questions. Maybe the teacher doesn't know the proofs. More likely he has so much material to cover, he can't stop to answer questions. I will pass along your advice to self-study and make use of the vast knowledge available on AoPS.\r\n\r\nIt just seems too bad that our schools, which should be encouraging bright, curious students, are instead turning out passive robots. If students can't ask questions in the classroom, wouldn't we be better off videotaping the best math teachers in the country and showing their lectures to students? It would be more effective and much cheaper than the current system.\r\n\r\nI think a lot of teachers don't like to be asked questions because they're afraid they won't know the answers. One of the other forum threads mentioned Professor Hung-Hsi Wu of UC Berkeley, who has written some excellent articles on math education. He says that many elementary school math teachers are terrified of fractions!!! That's not surprising. I remember a grade school teacher who didn't understand the commutative law of multiplication: \"To find the area of a triangle, you MUST multiply 1/2 * base * height. You CANNOT multiply base * height * 1/2.\" \r\n\r\nMel, your strategy of enlisting friends to ask questions for you is hilarious. :lol:", "Solution_18": "our Alg2 teacher can't answer a lot of my classmates' questions", "Solution_19": "Just for the record, I'd like to point out that it is very difficult to answer questions that you haven't prepared for. Often, the really interesting questions are about proofs or derivations that a teacher may not have seen in a long, long time. A teacher is a store of knowledge, but (eep! I'm going to slip into the first person) we don't always have all of our knowledge readily available because we may not have thought about the question that's being asked when preparing the lesson. So, not all of your in-class questions will be answered on the spot. But, that doesn't mean they shouldn't be asked -- coming up to a teacher after class is an excellent way to have questions answered that go beyond the scope of the lesson. Beyond that, there're always other kinds of resources -- textbooks, other individuals than your teacher, and of course the internet.", "Solution_20": "Teachers should feel comfortable saying, \"I don't know but I'll look that up for you later\", especially if they're teaching a difficult subject. We shouldn't expect teachers to be all-knowing but we should expect them to encourage and respect their students. Mel mentioned her experience with teachers who roll their eyes when students ask questions. Such teachers don't belong in the classroom.", "Solution_21": "i really disliked my BC teacher last term, his response to my questions were usually \"i dont' know, what do you think\" and if other peopel asked something, he occassionally said Maria, why is that true... hated him!!!\r\nthis term, my teacher was much nicer and he actually cares about the theroy stuff. if he felt the class got too bored he would stop with the proof and just talk to me about it after class. i did alot of learning by example and talking to people though, that was alright\r\nalso, at least in my school, a few very bright math people have come back to teach (2 just this year!) so don't give up on smart math people :)", "Solution_22": "I had a teacher similar to you'res last term...everytime I asked him a question that was not in his curriculum,he ran to another teacher to get an answer...The problem is that I don't see many teachers giving an honest effort anymore...But that's only half the equation...It seems that many students have lost interest in mathematics...I truly believe that a year or two after I graduate, there will no real interest in mathematics from students in my school...apart from my year, you can only find a handful of people who show interest...\r\n Stan", "Solution_23": "[quote=\"pendulum\"]...She started the year asking her calc teacher for proofs and derivations but soon gave up...[/quote]\r\n\r\nYou have to understand that a high school calculus class is not the right place place for proving theorms. Its about building students intuition for the concepts of calculus and giving them some dexterity with their applications. A few of the proofs of even elementary things in calculus (like the Intermediate Value Theorem for instance) are fairly deep, and would take way too long to develop in a normal AP Calculus class. So a teacher has to pick and choose which theorems to prove and which theorems to \"explain.\"\r\n\r\nI tell my students right off the bat that my aim is not to teach them how to [i]prove[/i] every theorem of calculus. My aim is to build their intuition for the theorems and concepts, and to give them plausible arguments for why they are true. So I might explain the Intermediate Value Theorem and its consequencesm and then tell them that its proof hinges on the not-quite-as-obvious-as-you-might-think fact that the Real Numbers have no \"gaps\" in them -- but not go into the nitty gritty details of its proof. [This year I actually did go into a discussion about how to prove that the rational numbers have \"gaps\" in them, to which I got the usual question, \"Is this going to be on the AB Exam?\" sigh].\r\n\r\nStudents will learn about the proofs of the theorems of calculus when they take an advanced calculus class or a real analysis class in college. For students being exposed to calculus for the first time though, its not really the best place to expose them to too many proofs.\r\n\r\nI'm sure most high school calculus teachers have at some point seen the proofs of elementary calculus theorems, they just may not be able to reproduce them on the spot. For example, if someone asked me for a proof of the Chain Rule, I probably couldn't do it right there on the spot. And even if I could, it would depend on if it were instructive for me to do so for the whole class.\r\n\r\nThat being said, on the [i]rare[/i] occasion when I do get a student who asks me such a question -- I tell them where they can find an answer to it or to come back and see me later. Sometimes that's just the best I can do at the moment.", "Solution_24": "It sounds like you don't already know what the teacher is teaching. If you are on AoPS, I would think that this would mean that the class is relatively advanced, so I don't see why the other kids would have the mentality that you mentioned. If you actually do already know what the teacher is teaching, then why are you in that class (unless it is the highest class that your school offers)?", "Solution_25": "[quote=\"gauss202\"]You have to understand that a high school calculus class is not the right place place for proving theorms. Its about building students intuition for the concepts of calculus and giving them some dexterity with their applications. A few of the proofs of even elementary things in calculus (like the Intermediate Value Theorem for instance) are fairly deep, and would take way too long to develop in a normal AP Calculus class. So a teacher has to pick and choose which theorems to prove and which theorems to \"explain.\"\n\nFor example, if someone asked me for a proof of the Chain Rule, I probably couldn't do it right there on the spot. And even if I could, it would depend on if it were instructive for me to do so for the whole class.[/quote]\r\n\r\nAgreed. High school calculus for the most part isn't about rigorously proving all the theorems. (Try [i]Calculus: An Intuitive and Physical Approach[/i] by Kline.)\r\n\r\nHowever, even though we don't do rigorous proofs in my math class, the teacher still shows the development of the theorems and rules to make them more intuitive and to show students how/why they should accept them. It's true that most students don't care about why something is true, but for me, if I am to understand something, I have to be able to accept it. And acceptance comes from being able to understand its development/proof. It's not always possible, but one more proof understood is more mathematical insight gained.", "Solution_26": "There's a lot to be said for how subtle calculus really is, and how crucially it depends on countless theorems which we would think are 'obvious' but are actually far harder to prove.\r\n\r\nIn a thread a few days ago, I asked someone to prove that for every continuous function $f$ on an interval $[a,b]$, that there exists $c\\in [a,b]$ such that $f(c)=\\frac {1}{b-a}\\int^b_a f(t)dt$\r\n\r\nThis is a relatively simple problem. Continuity implies integrability. Continuous functions attain their maximum $M$ and minimum $m$ on compact sets (such as $[a,b]$). We have $m(b-a)\\leq \\int^b_a f(t)dt\\leq M(b-a)$ by a simple theorem, which gives us the result by the intermediate value theorem.\r\n\r\nBut this is not so easy - this hinges on four theorems, none of which are really suitable for a calculus class. And this is really an easy proof. If you ask me to prove, say, L'Hopital, I'd be using the mean value theorem until every student in the class had nodded off to sleep.", "Solution_27": "gauss202, thanks for the teacher's perspective. I think this is not so much a calculus class problem as a general passive learning problem across all subject areas. Because of large classes and a desire to get through the curriculum, there is little time for questions. Maybe questions and discussions should be an integral part of the curriculum. Curiosity and active questioning should be promoted. At a minimum, teachers should encourage self-learning by pulling aside bright students and saying, \"I'm glad you want to learn more about this topic. Here's a list of excellent books written by experts in the field. You will probably find the answers to your questions in these books.\" I don't think that happens often enough.", "Solution_28": "well, my teacher always ignores my geometry question. it seems that i shall never get the answer from her!!", "Solution_29": "Anyways, I think that the important thing to get out of a first calculus class is not knowledge of proofs (the existence of, say, the Riemann integral is not exactly a useful thing in the immediate future for most students) but rather a good geometrical interpretation of why things work\r\n\r\nThey should be able to understand roughly why the chain rule takes the form it does, they should be able to understand why the derivative of the integral gives back the original function, they should be able to understand why the arclength formula takes that form, etc. If you understand why this stuff works, then the proofs just become a matter of speaking in the necessary epsilon language later on.", "Solution_30": "[quote=\"blahblahblah\"]Anyways, I think that the important thing to get out of a first calculus class is not knowledge of proofs (the existence of, say, the Riemann integral is not exactly a useful thing in the immediate future for most students) but rather a good geometrical interpretation of why things work\n\nThey should be able to understand roughly why the chain rule takes the form it does, they should be able to understand why the derivative of the integral gives back the original function, they should be able to understand why the arclength formula takes that form, etc. If you understand why this stuff works, then the proofs just become a matter of speaking in the necessary epsilon language later on.[/quote]\r\n\r\nAhhh...well said! That's basically what I was trying to say. Now...how is it that you managed to put it to better words? :P", "Solution_31": "Hopefully its because I've had a year more to think about it than you. :)", "Solution_32": "At my school, the teacher isn't as smart as I am and it's difficult to get questions answered." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "hola amiguitos saludo su conocimiento adquiridos y compartidos a sus projimos\r\n\r\n\r\nque solucion le darian a estos 2 problemita para calcular dicho \u00e1ngulo", "Solution_1": "Al primero ya lo propusieron ac\u00e1 en la comunidad espa\u00f1ola, aunque sin respuesta, pero sigamos la discusi\u00f3n all\u00e1.", "Solution_2": "x, 2x ,7x lo resolvi usando ley de senos y el problema se complica al resolver la ecuacion trigonometrica", "Solution_3": "chekalo eso me lo dio un profesor y todavia no me sale", "Solution_4": "http://www.mathlinks.ro/Forum/viewtopic.php?t=107386", "Solution_5": "El problema de x,2x,7x es un problema de construcci\u00f3n(no me gusta ese tipo de problemas), asi que si lo intentan por trigo no les va a salir o se van a demorar mucho. Aqui les presento esta soluci\u00f3n, un poco forzada pero v\u00e1lida.\r\n\r\nSea H un punto sobre AC tal que DHC=2x, entonces ADH=x y AH=HD=DC. Sea Q el sim\u00e9trico de H respecto a la recta AD, entonces AQ=QD=BD. Ademas QAD=QDA=x=DAC, entonces QD es paralelo a AC, BDQ=BCA=2x. Hemos construido una nueva figura, el caudril\u00e1tero no convexo (en Q) ABDQ, con BD=DQ=AQ y QAD=QDA=x, QDB=2x y ABD=7x. Sea M el punto medio del tri\u00e1ngulo is\u00f3sceles QDB y sea P el pie de la perpendicular a AB trazada desde Q y sea y=BAM. Pueden ocurrir tres casos:\r\n\r\n2PQBQ entonces PBQ=7x-(90-x)>30, entonces x>15. Ademas PQ>MQ y AQ=QD entonces y>x. Sumando los angulos del tri\u00e1ngulo ABD, tenemos\r\n(7x)+(3x)+(y+x)=11x+y>12x>180, contradicci\u00f3n.\r\n\r\nPor lo tanto debe ocurrir que 2PQ=BQ, es decir 7x-(90-x)=30, esto es x=15.", "Solution_6": "Una solucion trigonometrica para la primera de x , 2x , 7x :\r\n\r\nAplicando ley de senos : sen(10x)/sen(7x)=sen(x)/sen(2x)\r\n 2sen(10x)cos(x)=sen(7x)\r\n sen(11x)+sen(9x)=sen(7x)\r\n sen(11x)-sen(7x)=-sen(9x)\r\n 2sen(2x)cos(9x)=-sen(9x)\r\n 2sen(2x)=-tg(9x)\r\n\r\npero como la suma de x , 2x , 7x es menor que 180 entonces :\r\n x < 18 y analizando las graficas de :\r\n\r\n2sen(2x) = -tg(9x) se observa que l primer termino es positivo \r\nasi que 9x>90 entonces x > 10 ; de esta manera analizando las graficas de : 2sen(2x) y -tg(9x) en el intervalo <10,18> nos damos cuenta que la primera es continua estrictamente creciente y la segunda es continua estrictamente decreciente \r\nasi que a lo mas existira una solucion . pero vemos que al tanteo x=15 cumple luego x=15 es la unica solucion .", "Solution_7": "[quote=\"DarthTenebrus\"]Una solucion trigonometrica para la primera de x , 2x , 7x :\n\nAplicando ley de senos : sen(10x)/sen(7x)=sen(x)/sen(2x)\n 2sen(10x)cos(x)=sen(7x)\n sen(11x)+sen(9x)=sen(7x)\n sen(11x)-sen(7x)=-sen(9x)\n 2sen(2x)cos(9x)=-sen(9x)\n 2sen(2x)=-tg(9x)\n\npero como la suma de x , 2x , 7x es menor que 180 entonces :\n x < 18 y analizando las graficas de :\n\n2sen(2x) = -tg(9x) se observa que l primer termino es positivo \nasi que 9x>90 entonces x > 10 ; de esta manera analizando las graficas de : 2sen(2x) y -tg(9x) en el intervalo <10,18> nos damos cuenta que la primera es continua estrictamente creciente y la segunda es continua estrictamente decreciente \nasi que a lo mas existira una solucion . pero vemos que al tanteo x=15 cumple luego x=15 es la unica solucion .[/quote]\r\n\r\n :rotfl: jejeje Edson si ubieras aceptado mi solucion no la ubieras recibido dos veces jejeje" } { "Tag": [ "linear algebra", "matrix", "vector", "linear algebra solved" ], "Problem": "It is known that all columns of $A$ are pairwise orthogonal. Prove that $|\\det A|$ equals to product of lengths of vector-columns.\n\n[i](6 points)[/i]", "Solution_1": "$A^TA$ is a diagonal matrix, the $i$-th diagonal element is $\\|\\alpha_i\\|^2$, where $\\alpha_1,\\dots,\\alpha_n$ are the column vectors of $A$.\r\nThus\r\n$(\\det A)^2 = \\det (A^T A) = \\prod \\|\\alpha_i\\|^2$", "Solution_2": ":first: \r\n\r\nI think it was the easiest problem on the contest. Try other problems! I believe in you!" } { "Tag": [ "Support" ], "Problem": "The US Open, a tennis Grand Slam event, begins August 31.\r\n\r\nPlease vote and discuss!\r\n\r\nSorry the poll is only for men's singles.", "Solution_1": "Until someone takes out Federer at the Open, I'm sticking with him. Some people will point to Nadal's victory in Australia this year but Nadal has a history of falling in Flushing.\r\n\r\nI would love to see Roddick win, but I think Federer will take his sixth in a row.", "Solution_2": "I'm hoping for Roddick, Murray or Nadal. Basically, anyone but Federer :D \r\nI feel like it's someone else's turn. With Federer winning the slam on \"Nadal's Turf\" maybe it's time for Nadal to reciprocate? And Roddick and Murray both had really good seasons, and really want to win. \r\n\r\nWe'll have to see what happens!", "Solution_3": "Federer is the biggest favorite, and in my opinion, he will win. Although I hate him.", "Solution_4": "@Bugi:\r\nWhy do you hate Federer?", "Solution_5": "Federer FTW.\r\n :)", "Solution_6": "[b]Federer[/b]\r\n\r\nWho voted other??", "Solution_7": "nobody votes for Nadal? I noticed quite a lot of people do not believe the last year favourite can win the title, but I do. Still I support Federer.", "Solution_8": "[b] GO NADAL!!!!!!!!!!![/b] Federer is overrated.", "Solution_9": "i think federer will win.", "Solution_10": "i hope Andy Roddick wins Go America\r\nAlso does anyone know a american channel coverring the us open.", "Solution_11": "ESPN2.", "Solution_12": "[quote=\"yaofan\"][b] GO NADAL!!!!!!!!!!![/b] Federer is overrated.[/quote]\r\n\r\nWait what? How can you be overrated if you have the most grandslams, most consecutive weeks at #1, most consecutive wimbledons, most consecutive US Opens, and have won a grandslam at every turf, something that only 5 others in the history of tennis were able to accomplish?", "Solution_13": "[quote=\"yaofan\"][b] GO NADAL!!!!!!!!!!![/b] Federer is overrated.[/quote]\r\n\r\nHow is Federer overated???\r\n15 grand slams, most wimbledons, most #1, etc. etc. ETC.\r\n\r\nneed i say more??", "Solution_14": "OK. Federer is not overrated. And I think he's going to win. Unfortunately.", "Solution_15": "Federer more or less lost the match. I give del Potro credit for getting his act together after the first set, but Federer really went downhill. In the first set, del Potro's forehand was flat and erratic, and Federer was putting every shot (save for his serves) exactly where he wanted. In the later sets, especially the fourth and fifth, you could sense that Federer was tired and agitated. I noted several points in which Federer just gave up. Granted, del Potro's forehands became more like Federer's in those sets, but Federer just lost it-just look at how many unforced errors he had.", "Solution_16": "[quote=\"JRav\"]Federer more or less lost the match. I give del Potro credit for getting his act together after the first set, but Federer really went downhill. In the first set, del Potro's forehand was flat and erratic, and Federer was putting every shot (save for his serves) exactly where he wanted. In the later sets, especially the fourth and fifth, you could sense that Federer was tired and agitated. I noted several points in which Federer just gave up. Granted, del Potro's forehands became more like Federer's in those sets, but Federer just lost it-just look at how many unforced errors he had.[/quote]\r\n\r\nI don't know if it is more Del Potro Pwn or Federer fail :?: . Federer did make lot of errors but look at what Del Potro did against Nadal.\r\n\r\nAlthough Nadal was not doing well either.", "Solution_17": "Again, I give del Potro credit for forcing the unforced errors :P , but on most other days, Federer is getting to those shots and hitting winners. Also, while del Potro's first serve percentage was terrible, Federer's was even worse, which is pretty rare.", "Solution_18": "Yea, I know.\r\n\r\n\r\nI don't know what happened to him today, underestimated?", "Solution_19": "Great tournament! Great thread too!\r\n\r\nCongrats Juan Martin!", "Solution_20": "Yea,\r\n\r\nCBS got like 7 mil from that in PROFIT hosting that final (30 sec commercials are like 1 mil dollars).\r\n\r\n\r\nWow.", "Solution_21": "Urgghh I hate to see Federer lose but yeah, Juan Martin played a great game.", "Solution_22": "In Argentina they having a party right now (not really).", "Solution_23": "[quote=\"polkadotman1\"]In Argentina they having a party right now (not really).[/quote]\r\n\r\nAh my dad said Argentina sent lots of people to the USOPen\r\n\r\nI could tell...", "Solution_24": "lol yea (not really).\r\n\r\n\r\n\r\naudience screaming: roger, roger, roger, roger...etc.", "Solution_25": "Unfortunately Federer is already 28 so we'll be saying bye in about 5-6 years\r\n\r\nOf course DP is only 20, will pwn for years", "Solution_26": "AwesomeToad, you may be a bit optimistic. It's very unlikely in today's game that a player can perform at such a consistently high level into their 30's. I say he's got less time than that: 3-4 years.", "Solution_27": "Fine. But I am a naturally optimistic person. [hide=\"This is a little off topic, but...\"]Didn't Djokovic strip to his underwear once because he was bored? Sorry :blush: [/hide]", "Solution_28": "idk.\r\n\r\nBut i do know that nadal has a tendency to pick wedgies.\r\n\r\n\r\nand Djokovic imitates them.", "Solution_29": "How old is Djokovic?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Given $ n$ consecutive integers starting from $ 1$, is it possible to arrange them in such a way such that the sum of any two consecutive integers is a perfect square for all $ n$(repetition is not allowed)? \r\nIf not, then for what values of $ n$ is it possible ? If it is true for some number, then how many such combinations exist?", "Solution_1": "are you sure that you mean sum of 'any' 2 'consecutive' numbers is a square?\r\nif so then the squares shall be seperated by a number, leading to no such possiblity except for 1.\r\nif not so, then could you explain it a bit clearer?", "Solution_2": "consider:\r\n$ 16,9,7,2,14,11,5,4,12,13,3,6,10,15,1,8,17$\r\n\r\nIt contains numbers from $ 1$-$ 17$. Sum of any two consecutive numbers is a square." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Considere un cuadrillatero $ABCD$ circunscrito a una circunferencia $\\Gamma$ de centro $O$. Sean $P$ y $Q$ los puntos de contacto de $\\Gamma$ con $AB$ y $BC$ ,respectivamente. Si $X$ es el punto de interseccion de $PQ$ y $AC$, pruebe que:\r\na) $XO$ es ortogonal a $BD$.\r\nb) Los puntos donde las tangentes a $\\Gamma$ que pasan por $X$ interceptan a $\\Gamma$ estan sobre $BD$.\r\n\r\nObs: Si $R$ y $S$ son los puntos de contacto de $\\Gamma$ con $CD$ y $DA$ , \"creo que\" $P$, $B$, $Q$, $R$, $D$ y $S$ estan sobre una elipse.", "Solution_1": "[hide]Sugerencia: Usar polos y polares[/hide]", "Solution_2": "Bueno, aqui mi solucion: \r\nPirmero llamemos T a la interseccion de PR, QS, AC y BD(es conocido que concurren). \r\nUtilizando polos y polares (o algun menelao por alla), observamos que PQ, AC y RS concurren en X. Analogamente observamos que PS, QR y BD concurren en un punto Y. \r\na) Bueno, ahora tenemos un cuadrilatero PQRS ciclico. Ahora usaremos el siguiente teoremita que dice:\r\n\"Sea PQRS un cuadilatero ciclico de centro O, donde PQ y RS se intersectan en X, PS y QR se intersectan en Y, PR y QS se intersectan en T. Entonces O es ortocentro del triangulo XTY\" Vemos que aplicando este teorema al problema ya demostramos que XO es perpendicular a BD.\r\n\r\nb)Ahora, usaremos el sig. teorema:\r\n\"Sea PQRS un cuadrilatero ciclico, donde PQ y RS se intersectan en X, PS y QR se intersectan en Y, PR y QS se intersectan en T. Trazamos la recta YT y a los puntos donde corte a la circunferencia circuscrita de PQRS les llamamos J y K. Entonces XJ y XK son las tangentes desde X a la circunferencia\" \r\nVemos que aplicando esto al problema obtenemos un resultado rapido.", "Solution_3": "Aqui pongo un link donde se puede encontrar la demostracion del teorema usado para el inciso a). \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=108432\r\nJeje", "Solution_4": "Definamos $ M \\equiv AD \\cap BC,$ $ N \\equiv AB \\cap DC,$ $ X \\equiv PQ \\cap SR,$ $Y \\equiv PS \\cap QR.$ Por teorema de Newton se tiene que $ SQ,PR,AC,DB$ concurren en un punto, ll\u00e1mese $ E$ y como $ SQ$ y $ PR$ son las polares de $ M$ y $ N$ respecto a $ (O),$ se sigue que $ MN$ es la polar de $ E$ respecto a $ (O).$ En el cuadril\u00e1tero c\u00edclico completo $ PQRS$ la diagonal $ XY$ es polar de su circunc\u00edrculo $ (O)$ respecto a $ E \\equiv PR \\cap SQ,$ as\u00ed pues $ M,N,X,Y$ estan alineados. Las rectas $ PQ$ y $ SR$ se cortan en el polo $ X$ de la recta $ DB,$ por consiguiente $ OX \\perp DB$ y los puntos de contacto de las tangentes desde $ X$ a $ (O)$ est\u00e1n en $ DB.$\n\n[quote=\"jonfv\"]Obs: Si $ R$ y $ S$ son los puntos de contacto de $ \\Gamma$ con $ CD$ y $ DA,$ \"creo que\" $ P,$ $ B,$ $ Q,$ $ R,$ $ D$ y $ S$ estan sobre una elipse.[/quote]\nEsta conjetura es cierta y facil de demostrar. Los puntos $ P,B,Q,R,D,S$ estan en una misma c\u00f3nica ya que los lados opuestos del hex\u00e1gono $ PBQRDS$ se cortan en los tres puntos alinedos $ M,N,Y.$ Por el teorema de Pascal $ PBQRDS$ est\u00e1 inscrito en una c\u00f3nica en la cual $ MN$ es una recta de Pascal." } { "Tag": [ "geometry", "3D geometry", "inequalities", "inequalities proposed" ], "Problem": "(1) Let $a,b,c$ be positive reals such that $ab+bc+ca=1$. Prove that\r\n\r\n\\[ \\sqrt[3]{\\displaystyle \\frac{a^2}{5+12bc}} + \\sqrt[3]{\\frac{b^2}{5+12ca}} + \\sqrt[3]{\\frac{c^2}{5+12ab}} \\geq 1 . \\]\r\n\r\n(2) Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that\r\n\r\n\\[1 \\leq \\sqrt[3]{\\displaystyle \\frac{a^2}{1+18bc}} + \\sqrt[3]{\\frac{b^2}{1+18ca}} + \\sqrt[3]{\\frac{c^2}{1+18ab}} < \\sqrt[3]{3}.\\]", "Solution_1": "Can you give a hint for the first one?", "Solution_2": "Hint:\r\n\r\n[hide]Holder[/hide]", "Solution_3": "I spent 3 hours trying to solve it :( Can anybody post the solution? Sorry for my bad English :blush:", "Solution_4": "I solved second one's left ineq. \r\n\r\nLet $f(x)= \\frac {1}{x^{\\frac{1}{3}}}$. It is self-evident that $f(x)$ is convex.\r\n\r\nthen $\\sqrt[3]{\\frac{a^2}{1+18bc}}= a\\sqrt[3]{\\frac{1}{a+18abc}}$\r\n\r\nWe know $a+b+c=1$ and $f(x)$ is convex. \r\n\r\nUsing Jensen, We get $\\sqrt[3]{\\frac{a^2}{1+18bc}} + \\sqrt[3]{\\frac{b^2}{1+18ca}} + \\sqrt[3]{\\frac{c^2}{1+18ab}} \\geq \\sqrt[3]{\\frac{1}{\\sum_{cy}(a^2 + 18a^2bc)}} $\r\n\r\nSo, It is enough to prove that $\\sqrt[3]{ \\frac{1}{\\sum_{cy}(a^2 + 18a^2bc)}} \\geq 1$\r\n\r\n$\\Leftrightarrow 1\\geq \\sum_{cy}(a^2 + 18a^2bc)= \\sum_{cy} a^2 + 18abc$\r\n\r\n$\\Leftrightarrow 2\\sum_{cy} ab \\geq 18abc$\r\n\r\n$\\Leftrightarrow \\sum_{cy} ab \\geq 9 abc$\r\n\r\nusing AD-GD, We get $\\sum_{cy}ab \\geq 3\\sqrt[3] {a^2b^2c^2} $ and $ 1=a+b+c\\geq 3\\sqrt[3]{abc}$\r\n\r\n$Q.E.D.$", "Solution_5": "Okay, here are my solutions:\r\n\r\n(1) [hide]We rewrite the inequality as\n\n\\[\\sum_{\\text{cyc}} \\frac{a}{\\sqrt[3]{5a^2b+17abc+5a^2c}} \\geq 1.\\]\n\nFrom Holder,\n\n\\[\\left(\\sum_{\\text{cyc}} \\frac{a}{\\sqrt[3]{5a^2b+17abc+5a^2c}}\\right)^{3/4} \\left(\\sum_{\\text{cyc}}a(5a^2b+17abc+5a^2c) \\right)^{1/4} \\geq a+b+c,\\]\n\n\\[\\sum_{\\text{cyc}} \\frac{a}{\\sqrt[3]{5a^2b+17abc+5a^2c}} \\geq \\sqrt[3]{\\frac{(a+b+c)^4}{\\sum_{\\text{cyc}}a(5a^2b+17abc+5a^2c)}},\\]\n\nso we need\n\n\\[(a+b+c)^4 \\geq \\sum_{\\text{cyc}}a(5a^2b+17abc+5a^2c)\\]\n\\[\\iff \\sum_{\\text{cyc}} a^4+4a^3b+4ab^3+12a^2bc+6a^2b^2 \\geq \\sum_{\\text{cyc}}5a^3b+5ab^3+17a^2bc\\]\n\\[\\iff \\sum_{\\text{cyc}} a^4-a^3b-ab^3-5a^2bc+6a^2b^2 \\geq 0,\\]\n\nbut this follows from\n\n$\\sum_{\\text{cyc}} 6a^2b^2-6a^2bc \\geq 0 \\iff \\sum_{\\text{cyc}} a^2(b-c)^2 \\geq 0,$\n$\\sum_{\\text{cyc}} a^4+a^2bc \\geq \\sum_{\\text{cyc}} a^3b+ab^3$ (Schur, r=2)\n\nsince adding gives the result.[/hide]\n\n(2) [hide]The LHS is similar, i.e. after Holder, it reduces to\n\n\\[(a+b+c)^4 \\geq (a^2+b^2+c^2)(a+b+c)^2+18abc(a+b+c),\\]\n\nbut this follow from\n\n$(a+b+c)^4 = (a^2+b^2+c^2)(a+b+c)^2+2(ab+bc+ca)(a+b+c)^2$\n$\\geq (a^2+b^2+c^2)(a+b+c)^2+6(ab+bc+ca)^2$\n$\\geq (a^2+b^2+c^2)(a+b+c)^2+18abc(a+b+c).$\n\nNow, for the RHS, we apply Holder once again (in the other direction)\n\n\\[\\sum_{\\text{cyc}} \\sqrt[3]{\\displaystyle \\frac{a^2}{1+18bc}} \\leq \\left(\\sum_{\\text{cyc}} \\frac{1}{1+18bc}\\right)^{1/3} (a+b+c)^{2/3} < (1+1+1)^{1/3} = \\sqrt[3]{3}.\\][/hide]\r\n\r\nWell, now that I look at it, it would probably be nicer with the condition $a,b,c \\geq 0$ and the $<$ replaced with a $\\leq$ (for RHS of 2). But anyways, did you like these problems? :lol:", "Solution_6": "[quote=\"Chang Woo-JIn\"]Using Jensen, We get $\\sqrt[3]{\\frac{a^2}{1+18bc}} + \\sqrt[3]{\\frac{b^2}{1+18ca}} + \\sqrt[3]{\\frac{c^2}{1+18ab}} \\geq \\sqrt[3]{\\frac{1}{\\sum_{cy}(a^2 + 18a^2bc)}} $[/quote]\r\n\r\nI don't understand your idea..", "Solution_7": "He WLOG set $\\sum a^2=1$; the rest is just the definition of Jensen's inequality.", "Solution_8": "I just used $\\displaystyle \\sqrt[3]{\\frac{a^2}{1+18bc}}= a\\sqrt[3]{\\frac{1}{a+18abc}}$ and $a+b+c=1$\r\n\r\nHere, $a, b, c$ are weights.", "Solution_9": "Yes ,You're right!\r\nIt's very interesting!" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c>0$ such that $ a\\plus{}b\\plus{}c\\equal{}1$. Prove that\r\n\r\n$ \\frac{1\\minus{}a}{2a\\plus{}b}\\plus{}\\frac{1\\minus{}b}{2b\\plus{}c}\\plus{}\\frac{1\\minus{}c}{2c\\plus{}a}\\geq2$", "Solution_1": "[quote=\"Ligouras\"]Let $ a,b,c > 0$ such that $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that\n\n$ \\frac {1 \\minus{} a}{2a \\plus{} b} \\plus{} \\frac {1 \\minus{} b}{2b \\plus{} c} \\plus{} \\frac {1 \\minus{} c}{2c \\plus{} a}\\geq2$[/quote]\r\n\r\n$ \\sum \\frac {1 \\minus{} a}{2a \\plus{} b}\\geq2$\r\n<=>\r\n$ \\sum\\frac {(b \\plus{} c)^2}{(2a \\plus{} b)(b \\plus{} c)}\\geq2$\r\nby Chausy\r\n$ \\sum \\frac {(b \\plus{} c)^2}{(2a \\plus{} b)(b \\plus{} c)}$\r\n$ \\ge \\frac {(2\\sum a)^2}{\\sum (2a \\plus{} b)(b \\plus{} c)}$\r\n$ \\equal{} \\frac {(2\\sum a)^2}{\\sum a^2 \\plus{} 5\\sum ab}$\r\n$ \\equal{} \\frac {(2\\sum a)^2}{(\\sum a)^2 \\plus{} 3\\sum ab}$\r\n$ \\ge \\frac {(2\\sum a)^2}{(\\sum a)^2 \\plus{} (\\sum a)^2}$\r\n$ \\equal{} 2$" } { "Tag": [ "abstract algebra", "geometry", "geometric transformation", "reflection", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $\\displaystyle G$ be a group with $\\displaystyle 10$ elements such that there are $\\displaystyle a,b \\in G \\setminus \\left\\{ e \\right\\}$, distinct, which fulfill $\\displaystyle a^2=b^2=e$.\r\n\r\nIs $\\displaystyle G$ abelian?", "Solution_1": "There are just two groups with ten elements: the cyclic one and the dihedral group. Both fulfill your property (and the last one is not abelian: take $a,b$ as any reflection here).", "Solution_2": "There's only one element of order two in the cyclic group. That makes the group dihedral.", "Solution_3": ":blush: ", "Solution_4": "[quote=\"ZetaX\"]There are just two groups with ten elements: the cyclic one and the dihedral group.[/quote]\r\n\r\nIs that true for all $2p$, where $p \\geq 3$ is a prime (the only groups of order $2p$ are the cyclic one and the dihedral group)? I'm asking this because I looked at a list of small groups, and I noted this fact.\r\n\r\nIf it's true, can you give some hints to prove it? Thanks.", "Solution_5": "i think you can find it somewhere on the forum..\r\nanyway, the question can be poses with a prime $q$ instead of $2$: there are at most two groups of order $pq$ for any pair of primes $p\\le q$ (not necessarily different): if $p=q$, there are the cyclic one and the product of the two cyclic ones, if $p|q-1$ there are a semidirect product of cyclic groups, and the cyclic one, otherwise there's just the cyclic one...\r\n\r\nas for the proof.. just a short sketch.\r\nsuppose $p$. then, $ab$ and $ba$ is $H$'s element.(by \u2172) \r\n$ab=a^{-1}b^{-1}=(ba)^{-1}$. since $ab$ is $H$'s element, \r\n$ab$ is $c$ or $c^{2}$ or $c^{3}$ or $c^{4}$. and \r\n$ba$ is $c^{4}$ or $c^{3}$ or $c^{2}$ or $c$. \r\ntherefore $ab\\neq ba$.\r\n\r\nnot good reply.. :blush:", "Solution_7": "The official solution is very short: If $G$ is abelian, then $\\left\\{ e,a,b,ab \\right\\}$ is a subgroup. By Lagrange's Theorem, $4 | 10$, contradiction.\r\n\r\nBut the other problem (with $pq$) is much more interesting. I'll look into it later.", "Solution_8": "[quote=\"perfect_radio\"]\nBut the other problem with $pq$) is much more interesting. I'll look into it later.[/quote]\r\nThere was a topic about it, about a year ago (at that time I didn't understand it as well as I do now) :\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=semi&t=45051[/url]" } { "Tag": [ "MATHCOUNTS" ], "Problem": "How hard do you think MOEMS is? I gave it a 1.", "Solution_1": "Very very easy. Of course, I'm not the slow and careful type, so I missed about 4 this year. But I beat in Mathcounts someone who got only 1 wrong, so...", "Solution_2": "1. Simply 1.", "Solution_3": "nvr did it before...but the problems sound easy enough to be a 1.", "Solution_4": "I think such a poll on AoPS will be severely biased by the math levels of the \"regulars\" who are mostly older and certainly more experienced relative to the MOEMS target age groups. Remember that mathematical maturity is only beginning to show itself during grades 4-8.\r\n\r\nI think using any standard at which MATHCOUNTS is a 3 (where only a few students per year score 90% of the problems) is examining something other than the appropriate level of MOEMS.\r\n\r\nMOEMS introduces students to problem solving and I think that is on the level that it should be examined.", "Solution_5": "Well yes, I did MOEMS and it was an amazing entry to problem solving. It really is the best competition for elementary schoolers. I'm not sure how good it is for real competitive middle-schoolers, but it is still great for the casual participater.", "Solution_6": "i gave it a 1. It is for grades 4-8.", "Solution_7": "Moems is easier than mathcounts. For one thing, there is no speed element - if you know how to solve the problems there is more than enough time. Doing the elementary olympiads in 5th grade and the middle school olympiads in 6th grade would be a good warm-up for mathcounts in 7th grade. I have seen students follow this pattern, with considerable success.", "Solution_8": "Can you give me a link where i can find the former tests of MOEM?", "Solution_9": "http://www.moems.org/sample.htm\r\nthat's where you can find sample problems", "Solution_10": "I don't think it's too easy, for example, try this:\r\n\r\nLet $a_1a_2a_3...a_n$ be a natural number, where $a_1$, $a_2$...are digits. What must this number be, so that $12\\cdot2a_1a_2a_3...a_n1 = 21\\cdot1a_1a_2a_3...a_n2$?\r\n\r\nYeah yeah, it's not [i]that[/i] hard, but it is on mathcounts level.", "Solution_11": "It depends on whether you're talking about school, chapter, state, or national mathcounts. That's why a gave it a 2 instead of a 1.", "Solution_12": "I gave it a two also. Looks like chess64 and I are the lonely 2's (as of 6/5 at 1:53 GMT - 4 hrs.)\r\n\r\nSome of the problems can get nasty. I believe a problem on the marathon I gave out was solved wrong, but was let loose by me anyway.", "Solution_13": "I also gave it a 2.", "Solution_14": "I used to have trouble with MOEMS, but after math counts it seems pretty easy. All the top scorers from our school did MC.", "Solution_15": "[quote=\"chess64\"]It depends on whether you're talking about school, chapter, state, or national mathcounts. That's why a gave it a 2 instead of a 1.[/quote]\r\n\r\nLol, think ur in the wrong forum. This is math olympiads." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "To this problem I do not like tha solution I have.I'd like to see something more elegant.\r\nThanks!", "Solution_1": "Dear Stergiu and Mathlinkers,\r\nI don't know your solution.\r\nMy quikly approach : consider the Miquel's point and circles leading to this point od the triangle MAB and the transversal NCD, then a simple angle chasing and we are done...\r\nSincerely\r\nJea-Louis", "Solution_2": "Because ABCD are concyclic,you can find a point G , that A,G,N,D A,G,M,B are concyclic\r\nG,D,C,M G,B,C,N are concyclic => \u25b3BGC\u223d\u25b3ADG => AG/BG=AD/BC=AD/AE \u2220AGB=\u2220CGD=\u2220CMD=\u2220EAD => \u25b3AGB\u223d\u25b3DAE => AD/AG=DE/AB=DE/CE \u2220DEC=\u2220DEA+\u2220CEA=\u2220GBA+\u2220ABC=\u2220GBC=\u2220GAD => \u25b3AGD\u223d\u25b3CED \r\nBut we have \u25b3AGD\u223d\u25b3CNF => \u25b3CNF\u223d\u25b3CED => E,D,N,F are concyclic", "Solution_3": "Thanks to both of you :) !\r\n\r\n I'm leaving in the morning for vacation for a month, so ... have a nice summer ! See you again in September with your nice solutions!\r\n\r\n Babis" } { "Tag": [ "geometry", "AMC", "AIME", "USA(J)MO", "USAMO", "algebra", "function" ], "Problem": "Just wanted to talk about the upcoming UGA tournament. who is coming? I am. Also here is a good site:[url]http://www.math.uga.edu/mathmeet/[/url]\r\nI went on that site and i saw many previous years' tests, but they weren't divided into jv and varsity like most others. this is the first time i am going to UGA, so are there two different tests each for jv and varsity, or is it one test?", "Solution_1": "It's all one test.\r\n\r\nI'm coming! Everybody, come, especially because Vestavia is coming!\r\n\r\nBy the way, the registration deadline is tomorrow.", "Solution_2": "i'll be there as well.", "Solution_3": "I come, hope I make as good or better than last year, down with Vestavia :P ;). I may actually have a team, plus there will (as far as I know) be another homeschool team there :), so it should be a lot of fun. :) I wonder, how many people think Harrison will win? or will Harrison even come?", "Solution_4": "i should be there", "Solution_5": "ha! i'm not letting harrison win", "Solution_6": "I'll be there as an individual, not as a team due to Wheeler policy. \r\n\r\nBut, I'll still be there, that is the important thing.", "Solution_7": "Yes I am coming too with a team...thx for answering the question. has everyone registered? i have", "Solution_8": "I think we're registered.", "Solution_9": "i wanna come. is there a geometry section :rotfl: \r\n\r\nespecially if Vestavia is coming, jk, who cares if vestavia comes or not.", "Solution_10": "No. As a matter of fact, most of UGA is not geometry. In fact, last year had more geometry than usual.", "Solution_11": "question: is uga more amc 10 question difficulty, amc 12 difficulty, or aime difficulty? (or USAMO??? i doubt that HEAVILY :D )", "Solution_12": "ok scratch that, i dont think i'm going. our math team hasnt met since cobb. i think its roughly amc 12 difficulty, maybe a tiny bit easier", "Solution_13": "We're down to two seniors. :(", "Solution_14": "Unfortunately, almost all local regional competitions don't even approach olympiad difficulty.\r\n\r\nYeah, I'm having some trouble getting people from our school to show up as well. It's unfortunate. But I will still be there!", "Solution_15": "why even bother with that miles", "Solution_16": "[quote=\"Boy Soprano II\"][quote=\"Cicatriz\"]Conjecture: each Georgia person is weirder than the last.[/quote]\nWell, I assume that by \"the last\" you meant Billy, but weirdness does not assume a local minimum at Billy unless there exists an open interval containing Billy consisting entirely of irrational ARML members, which is false, since rational people are dense. :P (No offense intended, Billy\u2014I think possibly $\\omega(\\mbox{Billy}) \\leq \\omega(\\mbox{Miles})$, where $\\omega$ is the weirdness function :wink: )[/quote] \r\n\r\nI don't think there are many irrational or rational ARML members since most of them are complex. \r\n\r\nBut I'm sure that rational people are dense. \r\n\r\nBTW: Is Howard a moderator? It doesn't list him as a moderator, but he might be a special moderator.", "Solution_17": "^^^boy soprano^^ after that, you guys can never accuse me of wasting time posting stuff. almost all of the stuff i post i type as im thinking it, so i rarely spend more than one minute per post. i dont even wanna imagine how much time you spent on that when you could have been sleeping", "Solution_18": "Proof by induction:\r\n \r\nAssume the nth Georgian is weirder than the n-1th. The nth Georgian spreads his or her weirdness to the n+1th Georgian by HIV. This weirdness adds to an inbred genetic weirdness possessed by each Georgian.\r\n\r\nThus each Georgia person is weirder than the last.", "Solution_19": "haha nice proof. you should enter the national proof competition \r\n\r\nu wud win 1st place, with 2nd elemmnop", "Solution_20": "[quote=\"captcha000\"]Excuse you, Billy. I actually do read most of what gets posted in this forum at some point or another. Most of the stuff being posted here is quite useless anyway and I never bothered reading much of it. I don't mind \"fun\" stuff being posted, but if you guys are going to throw around insults and stuff do that on private messages or stuff, not on a public forum for the UGA tournament.\n\nI'll keep this up for a while, but please think before you post.\n\nThanks[/quote]\n\nI didn't say you didn't read everything ;) and if I implied that I am sorry for the misunderstanding... [quote]Somehow I think that Howard will delete this whole thread. On the other hand, having some superficial though firsthand knowledge of Howard, he isn't :D.[/quote]\n\nOK :maybe: :huh: so I'm a contender for the Weirdest Georgian award. Meh, I wonder, do they give awards at ARML for the weirdest ARMLer? :D[/quote]", "Solution_21": "Hmm... Thinking before posting would be a good idea, but then again, sometimes thinking too much gets you wound around in circles...\r\n\r\nLet's divide the worth of Miles's post by the time he took, and compare that to the worth of certain other people's posts (admittedly mine included) and divide by the time taken... \r\n\r\nOkay, okay, that may be interpreted as an insult of some sort. If so, I apologize--just pointing out a technicality. \r\n\r\nI pulled myself back together, but sorry if I bothered anyone. And... maybe some of you won't be so quick to assume things, if you realize that sometimes something that sounds absurd is the sincere belief of the person posting it. Forgive me for having difficulty distinguishing between jokes and seriousness--I tend to chat with people who must be unlike what you're used to, and there are so many gradations of meaning, so many different sorts of intentions, and so many different standards of normal. I can say something that is a joke, a nod to a memory, an expression of anger and uncertainty... And I'm sometimes poor at controlling myself and making sure not to say unclear things. \r\n\r\nI may apologize excessively, too--it doesn't make things any better, I know, but it's a good alternative to the rage and irritability that I gave myself over to last year. \r\n\r\nOkay, okay, my real sin is probably talking too much about myself and thinking that my posts have impact, even to annoy y'all, but I don't have much else to say. Except that math and mania make a mouthwatering mix. \r\n\r\nCantor would have understood. Or maybe it was not understanding that got him in so much trouble...", "Solution_22": "[quote=\"Xantos C. Guin\"]I don't think there are many irrational or rational ARML members since most of them are complex.[/quote]\nTrue, but many of them have irrational real parts, and the set of complexes with rational rectangular coefficients is dense over the complex plane.\n\n[quote=\"Xantos C. Guin\"]BTW: Is Howard a moderator? It doesn't list him as a moderator, but he might be a special moderator.[/quote]\nHe is a (the) moderator of the Georgia memberlist, therefore he is a moderator of this forum.\n\n[quote=\"Cicatriz\"]Proof by induction:\n\nAssume the nth Georgian is weirder than the n-1th. The nth Georgian spreads his or her weirdness to the n+1th Georgian by HIV. This weirdness adds to an inbred genetic weirdness possessed by each Georgian.\n\nThus each Georgia person is weirder than the last.[/quote]\nYou have two sizable holes in your \"proof\" which I won't even bother to point out, and many smaller ones. For instance, you assume that each Georgian has positive weirdness, whereas many are weird in a very negative sort of way. Plus, your knowledge of genetics is clearly cursory at best. Everybody knows HIV is a somatic bacterium. Plus, you are not even proving that each Georgian is weirder than the last; you're merely using it as your inductive step for some other obscure intent which you are clearly not sharing with us because you do not support the dissemination of knowledge and would clearly prefer a return to the Dark Ages wherein everybody but you would be so ignorant that you wouldn't even have to bother with learning new things to maintain the intellectual superiority over the masses necessary to maintain your power.\n\n[quote=\"solafidefarms\"]so I'm a contender for the Weirdest Georgian award.[/quote]\nSorry, it was only because you were the last GA person who posted before Cicatriz's post, and that particular misinterpretation was a crucial element of my argument. :wink: \n\n[quote=\"jhcreinhardt\"]i dont even wanna imagine how much time you spent on that when you could have been sleeping[/quote]\r\nDid I mention that I spent two months learning topology, hyperbolic geometry, and set theory just so I could make that post? I even learned two new notations. That was the high point of my life, and now you have just ruined it for me. On the other hand, you have opened my eyes to the wonderful value of sleep. In fact, I think that from now on, I will spend my days sleeping. Forget this math stuff! What is the point of living and being awake at the same time? In dreams, you can make everything how you want.\r\n\r\nAnyhow, I would like to iterate (or reiterate\u2014it's all recursive, anyhow) that if I have posted something offensive, it was not meant to be taken seriously, but only as a rebuttal that was intended (however dismally it might have failed) to be mildly entertaining, or at the very least interesting.", "Solution_23": "[quote=\"Boy Soprano II\"]What is the point of living and being awake at the same time? In dreams, you can make everything how you want.[/quote]\r\n\r\nAll kidding aside, is there anything wrong with that? Life is alot better asleep than awake.", "Solution_24": "sleep GOOD", "Solution_25": "But can it really be said that you MAKE things the way you want in dreams? It's all rather subconscious and confused... \r\n\r\nToo bad ARML members are only partially real... \r\n\r\nAnd I don't believe it's possible to measure weirdness accurately when you consider whether it is positive or negative--someone possessed of great weirdness with both positive and negative (and are these the only directional choices?) impact could then be considered to have only a low level of weirdness, which probably would not be useful in applications. I propose we assign weirdness vectors for each behavior exhibited by a person and define weirdness as the mean of the magnitudes of all weirdness vectors.\r\n\r\nAnd, whatever the biology of weirdness... Wouldn't the trend toward being deader (corpses have 0 weirdness, right?) that eventually results from HIV counteract the weirdness being spread?", "Solution_26": "dude wtf are you talking about\r\n\r\nno one cares", "Solution_27": "everything everyone is saying is \r\n\r\n\r\nedible poop :)", "Solution_28": "Isn't that redundant?", "Solution_29": "okay\r\n\r\nguys\r\n\r\nwtf?\r\n\r\nnote how the topic of this thread is: \"stuff for uga tournament\" and not \"talk about random crap which is so pointless that it is retarded\"\r\n\r\nhoward, please lock this thread..." } { "Tag": [ "USAMTS" ], "Problem": "so if we're registered already, when should we send in the paper works for year 18???", "Solution_1": "[quote=\"pkerichang\"]so if we're registered already, when should we send in the paper works for year 18???[/quote]\r\n\r\nWhenever you like, but we don't have the paperwork up on the site yet.\r\n\r\nMost students send them in with the first round of solutions.", "Solution_2": "And when would the first round start?" } { "Tag": [ "vector", "inequalities", "triangle inequality", "inequalities unsolved" ], "Problem": "vectors a,b,c,d satisfy a+b+c+d=0\r\n\r\nprove |a|+|b|+|c|+|d| is not less than |a+d|+|b+d|+|c+d|", "Solution_1": "Adding up the vectors $ \\vec{AB}\\equal{}\\vec{a},\\vec{BC}\\equal{}\\vec{b},\\vec{CA}\\equal{}\\vec{c},\\vec{DA}\\equal{}\\vec{d}$, we get a closed polygon ABCD. It is easily to prove.\r\n\r\nIf $ ACBD$ is convex quadrilateral. $ AB cut CD\\equal{}E$ .\r\n We can easily see that at least one of the six possible arrangements yields such a polygon.\r\n Adding up \r\n$ \\vec{AB}\\equal{}\\vec{a},\\vec{CD}\\equal{}\\vec{b},\\vec{BC}\\equal{}\\vec{c},\\vec{DA}\\equal{}\\vec{d}$\r\n$ |\\vec{AE}|\\plus{}|\\vec{CE}|\\geq |\\vec{AC}|$,\r\n$ |\\vec{BE}|\\plus{}|\\vec{DE}|\\geq |\\vec{BD}|$, we get \r\n $ |\\vec{AB}|\\plus{}|\\vec{CD}|\\geq |\\vec{AC}|\\plus{}|\\vec{BD}|$\r\nor $ |\\vec{a}|\\plus{}|\\vec{b}|\\geq |\\vec{b}\\plus{}\\vec{d}|\\plus{}|\\vec{a}\\plus{}\\vec{d}|$\r\nThe triangle inequality yields \r\n$ |\\vec{c}|\\plus{}|\\vec{d}|\\geq |\\vec{c}\\plus{}\\vec{d}|$\r\nAdding up the last two inequalites, we get\r\n $ |\\vec{a}|\\plus{}|\\vec{b}|\\plus{}|\\vec{c}|\\plus{}|\\vec{d}|\\geq |\\vec{b}\\plus{}\\vec{d}|\\plus{}|\\vec{a}\\plus{}\\vec{d}|\\plus{}|\\vec{c}\\plus{}\\vec{d}|$" } { "Tag": [ "inequalities", "geometry", "rectangle", "analytic geometry", "induction", "triangle inequality", "inequalities unsolved" ], "Problem": "Prove that if $ a_1,...,a_k$ e $ b_1,...,b_k$ are reals we have the inequality\r\n\\[ {\\sqrt{{({\\sum_{k\\equal{}1}^n a_k^2})}^2\\plus{}{({\\sum_{k\\equal{}1}^n b_k^2})}^2} \\leq \\sum_{k\\equal{}1}^n \\sqrt{(a_k^2\\plus{}b_k^2)}}\\]", "Solution_1": "[quote=\"Damien Hannigan\"]$ {\\sqrt {{({\\sum_{k = 1}^n a_k^2})}^2 + {({\\sum_{k = 1}^n b_k^2})}^2} \\leq \\sum_{k = 1}^n \\sqrt {(a_k^2 + b_k^2)}}$[/quote]\r\n\r\nAre you sure that you have post the right question?", "Solution_2": "Sorry,\r\n\r\n$ {\\sqrt {{({\\sum_{k = 1}^n a_k})}^2 + {({\\sum_{k = 1}^n b_k})}^2} \\leq \\sum_{k = 1}^n \\sqrt {(a_k^2 + b_k^2)}}$", "Solution_3": "[quote=\"Damien Hannigan\"]\nHow to prove\n$ {\\sqrt {{({\\sum_{k = 1}^n a_k})}^2 + {({\\sum_{k = 1}^n b_k})}^2} \\leq \\sum_{k = 1}^n \\sqrt {(a_k^2 + b_k^2)}}$[/quote]\r\n\r\nActually I also didn't know how to proof it.But I used it as a special case of Minkowski's inequality. Anyway i googled and got some good links about your question.\r\n[url=http://en.wikipedia.org/wiki/Minkowski_inequality]Wikipedia[/url]\r\n[url=http://planetmath.org/encyclopedia/ProofOfMinkowskiInequality.html]PlanetMath[/url]\r\nBoth of these proofs uses Holder's Inequality and some very advanced math (to be honest I haven't understood the proofs :blush: )\r\n[url=http://planetmath.org/encyclopedia/HolderInequality2.html]Proof of holder's inequality[/url]", "Solution_4": "to solve the ineq,we can assume some rectangles which length of the side is $ a_1,b_1$......\r\nI just write an example about when $ n\\equal{}2$,and for $ n\\ge 3$,the method is almost same.\r\nWe consider:\r\n$ A(0,0),B(a_1,0),C(a_1,b_1),D(0,b_1),E(a_1\\plus{}a_2,b_1),F(a_1\\plus{}a_2,b_1\\plus{}b_2),G(a_1,b_1\\plus{}b_2)$\r\n$ LHS\\equal{}|AC|\\plus{}|CF|$\r\n$ RHS\\equal{}|AF|$\r\nof course,$ LHS \\ge RHS$", "Solution_5": "[quote=\"Damien Hannigan\"]Sorry,\n\n$ {\\sqrt {{({\\sum_{k = 1}^n a_k})}^2 + {({\\sum_{k = 1}^n b_k})}^2} \\leq \\sum_{k = 1}^n \\sqrt {(a_k^2 + b_k^2)}}$[/quote]\r\nConsider Cartesian coordinate system.\r\nLet $ O,X_1,X_2,\\dots,X_n$ be the points with coordinates $ (0,0),(a_1,b_1),(a_1+a_2,b_1+b_2),\\dots,(a_1+a_2+\\dots+a_n,b_1+b_2+\\dots+b_n)$,respectively.\r\nObviously $ {\\sqrt {{({\\sum_{k = 1}^n a_k})}^2 + {({\\sum_{k = 1}^n b_k})}^2}=OX_n\\leq OX_1+X_1X_2+\\dots+X_{n-1}X_n=\\sum_{k = 1}^n \\sqrt {(a_k^2 + b_k^2)}}$", "Solution_6": "Suppose that $ z_n\\equal{}a_n\\plus{}i\\cdot b_n$ for $ n\\equal{}1,2,3,...,k$.\r\n\r\nThen the inequality is just \\[ |z_1\\plus{}z_2\\plus{}...\\plus{}z_k|\\le |z_1|\\plus{}|z_2|\\plus{}...\\plus{}|z_n|\\] This is obvious by the triangle inequality (or induction or whatever).\r\n\r\nTo prove the general form, we'd like to have \\[ \\sqrt[p]{(a\\plus{}c)^p\\plus{}(b\\plus{}d)^p}\\le \\sqrt[p]{a^p\\plus{}b^p}\\plus{}\\sqrt[p]{c^p\\plus{}d^p}\\] Now...I'm not sure what to do exactly...I'm sure we can do it with calculus...or holder's. Perhaps there is a nice proof..." } { "Tag": [ "geometry", "rectangle", "function", "conics", "parabola", "quadratics" ], "Problem": "A rectangular pen which uses the wall of a house as one side, is enclosed by 1000 m of fencing.\r\n\r\na. What is w in terms of L?\r\n\r\nb. Express the area in terms of L only?\r\n\r\nc. Use the eq to find the area of the pen for x=50,100,200,300 and 400m\r\n\r\nd. Describe the area of the pen as the length w increases?\r\n\r\ne. Estimate the maximum are of the pen and the dimensions od the rectangle that would give one maximum possible area?", "Solution_1": "It is a hard problem for pre-algebra or beginning algebra. At higher levels, you may say \u201cbeen there, done that.\u201d\r\n\r\na) It would be best to take advantage of the \u201cfree side\u201d by making the longer side of the pen (the length, $ L$) along the house. \r\nThere will be a length of fence $ (L)$ parallel to the side of the house connected to the house by two shorter sides of length $ W$ (the width of the pen).\r\nThe total fence length is $ 2W \\plus{} L \\equal{} 1000$\r\nYou can solve for $ W$ to get $ W$ as a function of $ L$.\r\n[hide]$ W \\equal{} 500 \\minus{} L/2$[/hide]\n\nb) The area is the product of the length $ L$, and the width $ W$ $ ($which you have as a function of $ L)$. That will give you the area as a function of $ L$.\n[hide]$ A \\equal{} 500L \\minus{} L^2$[/hide]\n\nc) The $ x$ in this part of the problem is meant to be the same as $ W$. We could have called the length of the side along the house $ x$ and the length of the other side $ y$ instead of $ L$ and $ W$, and the area as a function of $ x$ would be the same function. [hide]$ A(x) \\equal{} 500L \\minus{} x^2$[/hide]\r\nThe only difference is that when we called it $ L$ we meant for it to be longer than the other side. Calling it $ x$ makes it more general.\r\n\r\nd) Part c) must have shown you that as you increased the length of the side along the house $ (x)$ from 50 to 400 the area of the pen increased.\r\n\r\ne) IF you had studied parabolas (or quadratic functions) you would know that the area as a function of the length along the side of the house, $ A(x)$ or $ A(L)$, has a maximum, and would know how to find where that maximum occurs. Not knowing that, you can still estimate the location of the maximum by calculating $ A(x)$ for various values of $ x$. Since you did it for x=50,100,200,300 and 400m in part c), try x=500, 600m now." } { "Tag": [ "geometry", "trigonometry", "trig identities", "Law of Sines" ], "Problem": "in triangle ABC, side AC is of length 1, angle A is 30 degrees, and angle B is 15 degrees. what is the length of side B?", "Solution_1": "[hide=\"My sorution\"]\nhmm side B...AB or BC? well we'll just have to do both i guess. :maybe: \n\nSo Law of Sines gives us \\[\\frac{\\text{sin}15}{1}=\\frac{\\text{sin}30}{BC}=\\frac{\\text{sin}135}{AB}\\]\nAnd so we gots\n\\[BC=\\frac{\\text{sin}30}{\\text{sin}15}\\]\nand\n\\[AB=\\frac{\\text{sin}135}{\\text{sin}15}\\]\n\n[/hide]", "Solution_2": "side B normally refers to the side opposite angle B a.k.a segment AC\r\nbut sides should be in lower case, not in capitals", "Solution_3": "Does the problem statement have a mistake? Or does it just give the answer away..", "Solution_4": "[hide=\"My Solution\"]Use an image:\n\n[img]http://i96.photobucket.com/albums/l167/adasarathy/triangle_marked.gif[/img]\n\nThe Law of Sines states: $\\frac{a}{\\sin A}=\\frac{b}{\\sin B}=\\frac{c}{\\sin C}$.\n\nBy using it, $\\frac{1}{\\sin 15^{\\circ}}=\\frac{BC}{\\sin 30^{\\circ}}=\\frac{AB}{\\sin 135^{\\circ}}$\n\nTherefore, $ BC=\\frac{\\sin 30^{\\circ}}{\\sin 15^{\\circ}}=\\frac{\\frac{1}{2}}{\\sqrt{\\frac{1-\\cos 30^{\\circ}}{2}}}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{6}-\\sqrt{2}}{4}}\\implies \\frac{\\sqrt{6}+\\sqrt{2}}{2}$\n\nAnd, $AB=\\frac{\\sin 135^{\\circ}}{\\sin 15^{\\circ}}\\implies\\sqrt{3}+1$.[/hide]\n\n[hide=\"@NeverOddOrEven\"]You inverted the Law of Sines, even though you still got the right answer. :rotfl: [/hide]\n\n[hide=\"@sapphyre571\"]In that case, the answer is already given. :wink: [/hide]", "Solution_5": "[quote=\"anirudh\"][hide=\"@NeverOddOrEven\"]You inverted the Law of Sines, even though you still got the right answer. :rotfl: [/hide]\n[/quote]\r\nI do that frequently myself. Keep in mind that $a=b$ means $\\frac{1}{a}=\\frac{1}{b}$. It only makes a difference in the general form...", "Solution_6": "It's obviously 1 since AC=b.\r\n\r\nI'll find all three sides.\r\n\r\n[hide=\"solution\"]By the Law of Sines(which means $\\frac{a}{\\sin{A}}=\\frac{b}{\\sin{B}}=\\frac{c}{\\sin{C}}$),\n\n$2a=\\frac{\\sqrt{6}+\\sqrt{2}}{2}=\\frac{c\\sqrt{2}}{2}$\n\na happens to be $\\frac{\\sqrt{6}+\\sqrt{2}}{4}$\n\nand c happens to be $1+\\sqrt{3}$[/hide]\r\n\r\n@anirudh: how do you get those images????\r\n\r\n@NeverOddOrEven: Contests usually want the sides in simplest radical form, and side b is always the side opposite of angle B in problems that don't give a diagram.", "Solution_7": "[quote=\"nutz_for2.718281828\"]@anirudh: how do you get those images????[/quote]\r\n\r\nGeometers Sketchpad. You can download a demo to try it out [url=http://www.chartwellyorke.com/sketchpad/gspdemos.html]here[/url]. It is very useful in calculating those stuff, and if I input data, it'll do the math for me. :D \r\n\r\nAnd, I use [url=http://www.webtree.ca/newlife/printkey_info.htm]Printkey 2000[/url] and capture the image in sketchpad. That saves it as an image, which I upload to [url=http://photobucket.com]Photobucket[/url] and post it in Image tags here. :)" } { "Tag": [ "analytic geometry", "conics", "parabola", "quadratics", "graphing lines", "slope", "algebra" ], "Problem": "A circle of radius $4\\sqrt{5}$ has its center on the y-axis and is tangent to the parabola $y= \\frac{1}{8}x^2$. Point $(p,q)$ is their common point in the first quadrant. Find $p+q$", "Solution_1": "The equation of the circle can be written as $x^2+(y-b)^2=(4\\sqrt{5})^2$. Then, since the parabola $y=\\frac{1}{8}x^2$ is tangent to the circle, we can substitute $x^2=8y$ into the equation of the circle (which leads to quadratic equation of $y$) and then use the fact that the discriminant ($D$) of the quadratic is 0.\r\n\r\nOr maybe there's an easier way... In my method, I have to find the equation of the circle first. Is there a way that you can directly find $p$ and $q$?", "Solution_2": "I think your method is good :) \r\n\r\nMaybe another method is to see that the tangent line has slope $\\frac{p}{4}$" } { "Tag": [ "probability", "absolute value" ], "Problem": "1. In the expansion:\r\n$ (3x^8\\minus{}2x^6\\plus{}x^5\\plus{}2x^4\\minus{}x^2\\plus{}1)^5\\equal{}a_0\\plus{}a_1x\\plus{}a_2x^2...a_{40}x^{40}$,\r\nthe sum $ a_0 \\plus{}a_2\\plus{}a_4...a_{38}\\plus{}a_{40}\\equal{}$\r\n\r\n2. Suppose that we label each of 8 cards with a +1 on one side and -1 on the other. The cards are randomly dealt onto a table, so that either side of each card is equally likely to be facing up. Let $ x$ be the absolute value of the sum of the numbers that are facing up. The most likely value of $ x$ is\r\n\r\n3. Find the product $ cos(20^\\circ)cos(40^\\circ)cos(80^\\circ)$\r\n\r\n4. A number $ m\\equal{}111\\dots 11$ consists entirely of 1's in its decimal representation, and is divisible by 13. Which number need not divide $ m$? 3, 7, 11, 17, 37", "Solution_1": "[hide=\"4\"]\nIt is well know that $ 1001 \\equal{} 7\\cdot11\\cdot13$ and $ 111 \\equal{} 3\\cdot37$\n\nMultiplying, we have $ 111111 \\equal{} 3\\cdot7\\cdot11\\cdot13\\cdot37$ \n\nThus, 17 doesn't have to divide.[/hide]", "Solution_2": "Mewto, I think you mean Problem 4?\r\n\r\nAnd number 3 is just a specific case of a more general formula.", "Solution_3": "[hide=\"1\"]\n$ f(x)\\equal{}(3x^8\\minus{}2x^6\\plus{}x^5\\plus{}2x^4\\minus{}x^2\\plus{}1)^5\\equal{}a_0\\plus{}a_1x\\plus{}a_2x^2...a_{40}x^{40} \\\\\n f(1)\\equal{}4^5\\equal{}a_0\\plus{}a_1\\plus{}a_2\\plus{}a_3\\plus{}\\cdots \\plus{}a_{40} \\\\\nf(\\minus{}1)\\equal{}2^5\\equal{}a_0\\minus{}a_1\\plus{}a_2\\minus{}a_3\\plus{}....\\cdots \\plus{}a_{40} \\\\\nf(1)\\plus{}f(\\minus{}1)\\equal{}1056\\equal{}2(a_0\\plus{}a_2\\plus{}a_4\\plus{} \\cdots \\plus{}a_{38}\\plus{}a_{40}) \\\\\na_0\\plus{}a_2\\plus{}a_4\\plus{} \\cdots \\plus{}a_{38}\\plus{}a_{40}\\equal{}\\boxed{528}$\n[/hide]", "Solution_4": "[hide=\"2\"]\nIf I understood well, expected value?\nIf so, 4a.[/hide]", "Solution_5": "For number 3, let $ cos 20 \\equal{} x$\r\n$ cos (3 \\cdot 20) \\equal{} 4x^3 \\minus{} 3x \\equal{} \\frac {1}{2}$\r\nWe seek $ x(2x^2 \\minus{} 1)(2(2x^2 \\minus{} 1)^2 \\minus{} 1)$\r\n$ (2x^3 \\minus{} x)(8x^4 \\minus{} 8x^2 \\plus{} 1)$\r\n$ (\\frac {4x^3 \\minus{} 3x}{2} \\plus{} \\frac {x}{2})(2x(4x^3 \\minus{} 3x) \\minus{} 2x^2 \\plus{} 1)$\r\n$ (\\frac {x}{2} \\plus{} \\frac {1}{4})(x \\minus{} 2x^2 \\plus{} 1)$\r\n$ \\minus{} x^3 \\plus{} \\frac {3x}{4} \\plus{} \\frac {1}{4}$\r\n$ 1/8$", "Solution_6": "For $ \\#3$ there are $ 2^{8}$ possible trials. Let's consider five cases:\r\n$ 1.$ The sum is $ + 2$. \r\nHence, we must have that there are 5 $ + 1's$ and 3 $ - 1's$. This can happen in $ {8 \\choose 5} = 56$ ways. Thus, the probability of this happening is $ \\frac {56}{2^{8}}$. \r\n$ 2.$ The sum is $ + 4$.\r\nHence, we must have that there are 6 $ + 1's$ and 2 $ - 1's$. This can happen in $ {8 \\choose 6} = 28$ ways. The probability of this happening is thus $ \\frac {28}{2^{8}}$.\r\n$ 3.$ The sum is $ + 6$.\r\nHence, we must have that there are 7 $ + 1's$ and 1 $ - 1$. This can happen in $ {8 \\choose 7} = 8$ ways. The probability of this happening is $ \\frac {8}{2^{8}}$.\r\n$ 4.$ The sum is $ + 8$\r\nHence, we must have that there are 8 $ + 1's$ and 0 $ - 1's$. This can happen in $ {8 \\choose 8} = 1$ way. The probability of this happening is $ \\frac {1}{2^{8}}$.\r\n$ 5.$ The sum is $ + 0$\r\nHence, we must have that there are 4$ + 1's$ and 4 $ - 1's$. This can happen in $ {8 \\choose 4} = 70$ ways. The probability of this happening is $ \\frac {70}{2^{8}}$.\r\n\r\nWe do not even have to list when the the totals are negative, since the probability of those happening will be the same as when the totals are positive, but have the same magnitude (i.e. a total of $ - 4$ and $ + 4$ will have the same value). Thus, since we're looking for the absolute values of $ x$, $ -2$ and $ +2$ will have the same value, so we can just multiply each value by 2, except the sum of $ 0$. Hence, the answer is, as the greatest probability, $ \\boxed { + 2}$.\r\n\r\n\r\nEDIT: just reworded the last paragraph, it was hard to understand.", "Solution_7": "An easier way for 3:\r\n\r\nLet $ \\cos(20^\\circ)cos(40^\\circ)cos(80^\\circ) \\equal{} P$\r\n\r\n$ \\sin(20^\\circ)P \\equal{} \\sin(20^\\circ)cos(20^\\circ)cos(40^\\circ)cos(80^\\circ)$\r\n\r\n$ \\equal{} \\frac{1}{2} \\sin(40^\\circ)cos(40^\\circ)cos(80^\\circ) \\equal{}\\frac{1}{4} \\sin(80^\\circ)cos(80^\\circ)$\r\n\r\n$ \\equal{}\\frac{1}{8} \\sin(160^\\circ) \\equal{} \\frac{1}{8} \\sin(20^\\circ)$\r\n\r\nSo\r\n$ P\\equal{}\\frac{1}{8}$" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory", "Zsigmondy" ], "Problem": "Solve the equation\n\\[ 3^x \\minus{} 5^y \\equal{} z^2.\\]\nin positive integers.\n\n[i]Greece[/i]", "Solution_1": "[quote=\"Ahiles\"]Solve the equation\n\n$ 3^x \\plus{} 5^y \\equal{} z^2$\n\nin positive integers.[/quote]\r\nWe denote that a perfect square must be 3k or 3k+1, which k is a interger.\r\nSo $ 5^y$ must be 3k+1, then y divise by 2.\r\nWe recall $ y\\equal{}2a$, a is a postisive interger.\r\n$ 3^x \\plus{} 5^(2a) \\equal{} z^2<\\equal{}> 3^a \\equal{}(z\\minus{}5^a)(z\\plus{}5^a)<\\equal{}>3^m\\equal{}z\\minus{}5^a,3^n\\equal{}z\\plus{}5^a$\r\nThen $ 3^n\\minus{}3^m\\equal{}2.5^a$ so 3 is divise by 2 or 5 that wrong if m,n is positive interger.\r\nSo m must be zero, n=a.\r\nWe have:$ 3^m\\equal{}2.5^a\\plus{}1$\r\nWe consider m with modulo 4. Then m is 4e+1 which e is apositive interger.\r\nBut like that so $ 3^m\\minus{}1$ is divise by 4, but right hand is not. So there is no pair (x,y,z) satify for this problem.", "Solution_2": "Considering modulo 3, $ 5^y\\equiv1\\pmod3$, so $ y$ is even. Let $ y\\equal{}2a$.\r\n\r\nSo $ 3^x \\equal{} z^2 \\minus{} 5^{2a} \\equal{}(z\\plus{}5^a)(z\\minus{}5^a)$\r\n\r\nThus $ z\\plus{}5^a\\equal{}3^b,z\\minus{}5^a\\equal{}3^c$, where $ b\\plus{}c\\equal{}x$ and $ b,c\\in\\mathbb{N}_0$.\r\n\r\nWe have $ 2\\cdot 5^a \\equal{} 3^b\\minus{}3^c$. LHS is not divisible by 3, so $ c\\equal{}0$, $ b\\equal{}x$.\r\n\r\n$ z\\minus{}5^a\\equal{}1\\implies z\\equal{}5^a\\plus{}1$\r\n\r\n$ 5^a\\plus{}1\\plus{}5^a\\equal{}3^b$\r\n\r\nSo $ 3^b \\equal{} 2\\cdot 5^a\\plus{}1$\r\n\r\nConsider modulo 5, $ b$ is even. Consider modulo 4, $ b$ is odd. Contradiction, so there are no solutions.", "Solution_3": "I'm sorry. The question was incorrect.\r\nNow everything is correct.", "Solution_4": "Looking mod 4, x is even. Let $ x\\equal{}2a$\r\n$ 9^a\\minus{}5^y\\equal{}z^2$\r\n$ (3^a\\minus{}z)(3^a\\plus{}z)\\equal{}5^y$\r\nSo let $ 3^a\\minus{}z\\equal{}5^b$, $ 3^a\\plus{}z\\equal{}5^c$\r\nThen $ 3^a\\equal{}\\frac{5^b\\plus{}5^c}2$, which is divisible by 5 unless $ b\\equal{}0$ (since $ b\\leq c$)\r\nSo $ 2\\cdot3^a\\equal{}5^c\\plus{}1$\r\nSince the only primes dividing the LHS are 2 and 3, this has no solutions for $ c>1$ by Zsigmondy's theorem. So the only solution is $ x\\equal{}2,y\\equal{}1,z\\equal{}2$", "Solution_5": "[quote=\"caffeineboy\"]... by Zsigmondy's theorem.[/quote]\r\n\r\nWow...\r\n\r\nA sledgehammer is not required here. :wink: \r\nJust look modulo $ 9$ to see that if $ a>1$ then $ c$ is divisible by $ 3$. Then deduce that $ 1\\plus{}5^c$ is divisible by $ 7$ which is not the case of the $ LHS$.\r\n\r\nPierre.", "Solution_6": "alternatively, you might note that $ z \\equal{} 3^a \\minus{} 5$, and plug this into the main equation, we get:\r\n\r\n$ \\minus{}5^{y}\\equal{}\\minus{}10\\cdot 3^{a} \\plus{} 25$, so $ y\\equal{}1$ and $ a\\equal{}1$.", "Solution_7": "Like [b]pbornsztein[/b] I look modulo 9. Then again I get c is divisible by 3. Let c=3m. Then $ 2.3^{a}\\equal{}\\left(5^{m}\\plus{}1\\right)\\left(5^{2m}\\plus{}5^{m}\\plus{}1\\right)$ so $ 5^{m}\\plus{}1\\equal{}2.3^{h}$ and using the method of the infinite descent we get $ c\\equal{}0$. Contradiction! So $ a\\equal{}1$ and $ c\\equal{}1$.", "Solution_8": "[quote=\"SAPOSTO\"]Like [b]pbornsztein[/b] I look modulo 9. Then again I get c is divisible by 3. Let c=3m. Then $ 2.3^{a} \\equal{} \\left(5^{m} \\plus{} 1\\right)\\left(5^{2m} \\plus{} 5^{m} \\plus{} 1\\right)$ so $ 5^{m} \\plus{} 1 \\equal{} 2.3^{h}$ and using the method of the infinite descent we get $ c \\equal{} 0$. Contradiction! So $ a \\equal{} 1$ and $ c \\equal{} 1$.[/quote]\r\n\r\nAnother way to finish it is by noticing that \r\n\\[ 5^m\\plus{}1\\equal{}2\\cdot 3^{\\alpha} \\textrm{ and } 5^{2m}\\plus{}5^m\\plus{}1\\equal{}3^{\\beta} \\Longrightarrow 5^m\\equal{}4\\cdot3^{2\\alpha}\\minus{}3^{\\beta},\\]\r\nwhich leads to a contradiction.", "Solution_9": "We can also find out that $ c$ is divisible by 3 in following way:\r\n\r\nFrom $ 2.3^{a}\\equal{}5^{c}\\plus{}1$ we conclude that $ c$ is odd (modulo 3) so $ c\\equal{}2m\\plus{}1$ and we get:\r\n\r\n$ 2.3^{a}\\equal{}(5\\plus{}1)(5^{2m}\\minus{}5^{2m\\minus{}1}\\plus{}...\\plus{}5^{2}\\minus{}5\\plus{}1)$ \r\n\r\nFrom last one it is obvious that (for $ a>1$ , case $ a\\equal{}1$ is easy) we get that 3 divides $ 5^{2m}\\minus{}5^{2m\\minus{}1}\\plus{}...\\plus{}5^{2}\\minus{}5\\plus{}1$ and since each of these $ 2m\\plus{}1$ summands is congruent 1 modulo 3 we get $ 2m\\plus{}1$ is congruent 0 modulo 3 being equaivalent to $ c\\equal{}3d$.", "Solution_10": "there are 100 solutions for this question and I got only 4 points in BMO. I shuold kill myself!", "Solution_11": "I think it is a very easy problem for BMO and it can be a good problem for JBMO :lol:", "Solution_12": "Hello dear mathlinkers.\n\nCould someone PLEASE answer the following question.\n\nApparently,the choice of modulo 9 is not random. So,what should lead us to consider this equation modulo 9,and why not modulo any other integer?\n\nI am preparing for a team selection test in Greece and, since I do not live in a city,I dont have the resources or any teacher to explain these things to me.\n\n\nSo I would strongly appreciate your help.......\n\n\nWaiting for a response,\nNick", "Solution_13": "[quote=\"nickthegreek\"]Hello dear mathlinkers.\n\nCould someone PLEASE answer the following question.\n\nApparently,the choice of modulo 9 is not random. So,what should lead us to consider this equation modulo 9,and why not modulo any other integer?\n\nI am preparing for a team selection test in Greece and, since I do not live in a city,I dont have the resources or any teacher to explain these things to me.\n\n\nSo I would strongly appreciate your help.......\n\n\nWaiting for a response,\nNick[/quote]\nBut only $\\mod 9$ does not work,others work too,that depends on the way you have proceeded.Specially here $3^x$ is divisible by $9$ for $x>1$,so it was taken for providing informations.", "Solution_14": "[quote]Just look modulo $ 9$ to see that if $ a>1$ then $ c$ is divisible by $ 3$. Then deduce that $ 1\\plus{}5^c$ is divisible by $ 7$ which is not the case of the $ LHS$[/quote]\n\nI'm confused.. :maybe: Can you explain why??? I don't see... :!:", "Solution_15": "@fighter\nHey,brother! :w00t:\nCould you teach me solution by Pythagorean triple?", "Solution_16": "This thread is $7$ years old.But I want to solve problem by using [b]Zsigmondy's theorem[/b]. :bruce: So I revive this thread. :D\n\nThanks,\nTakeya.O", "Solution_17": "$$3^x-5^y=z^2$$\nfirst we use $\\pmod 4$ and get $x=2m$ and $y$ is odd.\nso we can write $(3^m-z)(3^m+z)=5^y(1)$\nlet $d=\\gcd(3^m-z,3^m+z)=\\gcd(3^m-z,2z)$ so $d|2z$ but from $(1)$ we get $d$ is in the form $5^k$. Note that $5\\nmid z$ because if $5|z$ then $5|3^x$ which is impossible so we can conclude that $d=1$\n\nso $3^m-z=1,3^m+z=5^y$ and $2\\cdot 3^m=5^y+1$ by zygmondy theorem we cannot find $y>1$ because if for all $y>1$ RHS will get e new primitive prime so we conclude $y=1$ and $m=1\\implies z=2$. Only $(2,1,2)$ works", "Solution_18": "hey brother don't be rood please because I didn't read any post just solved it sorry", "Solution_19": "@fighter\n\nHey,brother.I want to know Pythagorean triple solution. :D\n\nCould you teach me?", "Solution_20": "[quote=AMN300]Very standard problem\n[hide=Solution]\nTaking$\\pmod{4}$, $(-1)^x-1=z^2 \\implies x=2a$ for positive integer $a$.\nBy difference of squares, \n\\[ (3^a-z)(3^a+z)=5^y \\]\nSuppose $3^a-z=5^m, 3^a+z=5^n$ where $0 \\le m < n \\le y$. Adding the equations, $2 \\cdot 3^a=5^m(5^{n-m}+1)$. This means $m=0$, so \n\\[ 2 \\cdot 3^a=5^y+1\\]\nWe can finish through LTE now but here is a more elementary approach. Suppose $a \\ge 2$, so$\\pmod{9}$ reveals $y=6w-3$ since the order of $5\\pmod{9}$ is $6$. By sum of cubes, $2 \\cdot 3^a = (5^{2w-1}+1)(5^{4w-2}-5^{2w-1}+1)$. $5^{2w-1}\\equiv -1\\pmod{3}$, so the right hand side isn't divisible by $3$, contradiction. \nThis means that $a=1$. From here, we recover the solution \n\\[ \\boxed{(x, y, z)=(2, 1, 2)} \\]\n[/hide][/quote]\n\nnot entirely sure your contradiction is a contradiction?\n[hide=My solution]\nFirst, we take $\\pmod 4$. This gives us $x$ is even; if $x$ is odd, it follows that $3^x \\equiv 3 \\pmod 4$, and as $5^y \\equiv 1 \\pmod 4$, then the LHS would be $2 \\pmod 4$; $2$ is not a quadratic residue $\\pmod 4$, contradiction.\n\nNow consider the equation $\\pmod 3$. Clearly the LHS is equivalent to $-2^y \\pmod 3$; from this we get $y$ is odd, as if $y$ is even, $-2^y \\equiv 2 \\pmod 3$, and $2$ is not a quadratic residue $\\pmod 3$.\n\nLet $x=2k$. Rearranging the equation gives us $3^{2k} - z^2 = 5^y$, and as $3^{2k}$ is a square we can apply difference of squares. The equation becomes $$(3^k - z)(3^k+z) = 5^y.$$\nIt follows that both $3^k-z$ and $3^k+z$ are multiples of $5$. \n\nFrom this it follows that $(x,y,z) = (2,1,2)$ is the only solution; basically what we want is for $3^k$ to be the average of two different powers of $5$ where the exponent is nonzero. If it is, then it follows that the last digit of $3^k$ is $5$, which is clearly impossible. Therefore $3^k - z = 1$, from which it follows that\n$$2 \\cdot 3^{k} = 5^y + 1.$$\n$k=1$ yields the solution $(x,y,z) = (2,1,2)$; now assume $k \\ge 2$. Taking the equation $\\pmod 9$ yields $0 \\equiv 5^y + 1 \\pmod 9$, from which it follows that $y$ is of the form $6m+3$ as $5^6 \\equiv 1 \\pmod 9$ and $5^3 \\equiv -1 \\pmod 9$. \n\nSubstituting in $y = 6m+3$ gives us $2\\cdot 3^k = 125^{2m+1} + 1$. However, taking this equation $\\pmod 7$ gives us $2\\cdot 3^k \\equiv 0 \\pmod 7$, contradiction. Therefore the only solution is $(x,y,z) = (2,1,2)$.\n", "Solution_21": "The only solution is $\\boxed{(x,y,z)=(2,1,2)}$. Taking the given equation modulo $4$, we learn that\n\\[(-1)^x-1\\equiv 0,1\\pmod{4},\\]\nso the only possibility is that $x$ and $z$ are even. We actually don't need that $z$ is even, but we will et $x=2b$. Thus, we have\n\\[3^{2b}-z^2=5^y,\\]\nso\n\\[(3^b-z)(3^b+z)=5^y.\\]\nLet the first factor be $5^p$, the second $5^q$. We see that $5^p+5^q=2\\cdot 3^b$, which means that we must have $p=0$. Thus, $3^b-z=1$ and $3^b+z=5^y$, so\n\\[2\\cdot 3^b=5^y+1.\\]\nFor the left side to be divisible by $3$, we need $y=2c+1$ to be odd. Then, we see that $3\\nmid 5^y-1$, so\n\\[\\nu_3(5^y+1)=\\nu_3(25^y-1)=1+\\nu_3(y)\\]\nby LTE. Thus, $b=1+\\nu_3(y)$, so\n\\[6\\cdot 3^{\\nu_3(y)}=5^y+1.\\]\nBut note that the LHS is at most $6y$, so we have $5^y+1\\le 6y$, or $y\\le 1$. When $y=1$, we get that $b=1$, so $x=2$. This means that $z^2=9-5=4$, so $z=2$, so the solution must be the one that we claimed. It is easy to see that it works, so we're done. $\\blacksquare$", "Solution_22": "[hide=Solution][quote=Balkan MO 2009 P1]Solve the equation\n\\[ 3^x \\minus{} 5^y \\equal{} z^2.\\]\nin positive integers.\n[/quote]\n[b][color=#000]Solution:[/color][/b] Clearly, $z$ is even. Taking $\\pmod{4}$, we get, $x$ must be even too. Let $z=2k$ & $x=2x'$\n$$5^y=(3^{x'}-2k)(3^{x'}+2k) \\implies \\begin{cases} 3^{x'}-2k=5^a \\\\ 3^{x'}+2k=5^b \\end{cases}$$\nSo, $\\implies$ $2\\cdot 3^{x'}=5^a(1+5^{b-a})$ Since, $\\text{ LHS }$ is not divisible by $5$ $\\implies$ $a=0$. So,\n$$1+5^y=2 \\cdot 3^{x'}$$\nIf $x'=1$ $\\implies$ $(x,y,z)=(2,1,2)$. When, $x' \\geq 2$, Taking, $\\pmod{9}$ $\\implies$ $y=6y'+3$. Now, Taking $\\pmod{7}$ $\\implies$ $5^y+1$ is $0 \\pmod{7}$, but $7$ does not divide $\\text{ RHS }$, a contradiction! So, we have $\\boxed{(x,y,z)=(2,1,2)}$ $\\qquad \\blacksquare$\n[/hide]", "Solution_23": "I claim that the only solution is $(x,y,z) = (2,1,2)$.\nFirst, if we take $\\mod 4$, since $z^{2} \\equiv 0,1\\mod 4$ and $5^{y}\\equiv 1\\mod4$, we must have $x$ as even, or else $3^{x} - 5^{y}\\equiv 2\\mod4$ which can't be equal to $z^{2}$. Therefore, we denote $x = 2x_{1}$.\nRearranging the terms, we get \n$$3^{2x_{1}} - z^{2} = 5^{y} \\Rightarrow (3^{x_{1}} - z)(3^{x_{1}} + z) = 5^{y}$$\nSince the only factors of $5^{y}$ is powers of $5$, we must have $3^{x_{1}} - z$ and $3^{x_{1}} + z$ as powers of $5$. I claim that $3^{x_{1}} - z = 1$. If $3^{x_{1}} - z$ is a power of $5$ greater than $1$, then $z\\equiv 3^{x_{1}}\\mod 5$. However, this implies that $3^{x_{1}} + z \\equiv 2\\cdot 3^{x_{1}}\\mod 5$, so $3^{x_{1}} \\equiv 0\\mod5$, which is clearly impossible.\nThus, we have $$3^{x_{1}} - z = 1 \\Rightarrow z = 3^{x_{1}} - 1 \\Rightarrow 3^{x_{1}}+z = 2\\cdot 3^{x_{1}} -1 = 5^{k}$$ where $k$ is a positive integer. I claim that for all $x_{1} \\geq 2$, there are no solutions. This is because then $2\\cdot 3^{x_{1}} \\equiv 0 \\mod 9$, thus $5^{k} \\equiv 8 \\mod 9$, which implies $k$ is a multiple of $3$. Rearranging, we get $2\\cdot 3^{x_{1}} = 5^{k} + 1$. However, then because $k$ is a multiple of $3$, we have $5^{k} + 1 \\equiv 0 \\mod 7$. Thus $2\\cdot 3^{x_{1}}$ must be divisible by $7$, which is clearly impossible. Thus, the only solution is when $x_{1} = 1$, so our only solutions for $(x,y,z)$ is $(2,1,2)$.", "Solution_24": "Not hard but beautiful problem!\nLet us consider $mod 4$. We get that $x$ is even and let's denote it as $x = 2k$. Then $5^y = (3^k - z)(3^k + z)$. Then it follows that $3^k - z = 5^a$ $(1)$ and $3^k + z = 5^b$ $(2)$ where $a+b=y$ and $a1$ by taking $mod 9$ we get that $b$ is divisible by $3$ and let $b=3l$. Now we get that $5^l + 1 = 2*3^c$ and that $5^2l + 5^l + 1 = 3^d$, then it follows that $5^2l = 3^d - 2*3^c$ which gives us contradiction when watching $mod 3$ so $k=1$. Then plugging back we get that $y=1$, $x=2k=2$ and $z=2$. With this we are done! ", "Solution_25": "Take $\\pmod{4}$ we get that $x = \\text{even}$\n\n$$\\implies x = 2m$$\n\nSubstituting back this in our original equation we get that\n$$3^{2m} - 5^y = z^2$$\n$$\\implies (3^m + z)(3^m - z) = 5^y$$\n$$\\implies 3^m + z = 5^a \\quad \\text{and} \\quad 3^m - z = 5^b, \\quad a+b=y, \\quad a > b$$\n$$\\implies 3^m = 5^b \\left(\\frac{1 + 5^{a-b}}{2}\\right) = 5^b \\cdot q$$\n$$\\implies b = 0 \\quad \\text{because power of 3 can't contain 5 in it}$$\nPutting $b=0$ we get that\n$$3^m = \\frac{5^y + 1}{2}$$\nNote that taking $\\pmod{3}$ in out original equation we get that $y = \\text{odd}$\n$$\\implies m = v_3\\left(\\frac{5^y + 1}{2}\\right) = v_3(6) + v_3(y) - v_3(2)$$\n$$\\implies m = 1 + v_3(y)$$\n$$\\implies y = 3^{m-1} \\cdot k \\quad \\text{(k is co-prime to 3)}$$\n$$\\implies y \\ge 3^{m-1}$$\nPlugging it back in we get\n$$2\\cdot 3^m = 5^{k \\cdot 3^{m-1}} + 1 \\ge 5^{3^{m-1}} + 1$$\nBut notice that $RHS > LHS$ for $m-1 > 0$ (Rigorous proof : Induction)\n$$\\implies m-1 = 0 \\implies m = 1$$\n$$\\implies 3^1 - z = 1 \\implies z = 2$$\n$$\\implies y = 1$$\nHence the solutions are $\\boxed{(x,y,z) = (2,1,2)}$", "Solution_26": "I have done this task with mod:\n3\"x=5\"y+z\"2\nEuler's theorem:\n3*2/3=2\nz\"2=1(mod3)\n5\"y must be:\n5\"y=2(mod3)\nbecause z\"2+5\"y must divided by 3.\nif y is 2k+1 then 5\"y=2(mod3)\nbecause of that y is single integer.\nlet k=0\nthen x=2 y=1 z=2\nproblem solved\nand its the only solve we can have", "Solution_27": "[quote=Ahiles]Solve the equation\n\\[ 3^x \\minus{} 5^y \\equal{} z^2.\\]\nin positive integers.\n\n[i]Greece[/i][/quote]\n$3^x-5^y=z^2$\n$(-1)^x-1\\equiv z^2 \\pmod{4}$\n$\\Rightarrow x\\equiv 0 \\pmod{2} \\Rightarrow x=2a$\n$\\Rightarrow (3^a-z)(3^a+z)=5^y$\n$\\Rightarrow 3^a-z=5^m , 3^a+z=5^n$\n$\\Rightarrow 3^a=\\frac{5^m+5^n}{2}$\n$\\Rightarrow m=0$\n$\\Rightarrow z=3^a-1, z=5^n-3^a$\n$\\Rightarrow 2\\times 3^a=5^n+1$\nIf $n>1:$\n$5^1+1=2\\times 3$\nBy Zsigmondy\u00b4s Theorem:\n$\\exists$ prime $p\\neq 2,3/ p|5^n+1 (\\Rightarrow \\Leftarrow)$\n$\\Rightarrow n\\le 1$\nIf $n=1:$\n$\\Rightarrow a=1 \\Rightarrow (x,y,z)=(2,1,2)$ is solution\nIf $n=0:$\n$\\Rightarrow a=0 \\Rightarrow x=0 (\\Rightarrow \\Leftarrow)$\n$\\Rightarrow (x,y,z)=(2,1,2)$ is a unique solution $_\\blacksquare$", "Solution_28": "[b][color=#6AA84F]Claim:[/color][/b]The only solution is $\\boxed{(x,y,z)=(2,1,2)}$\n[i]Proof:[/i]\nChecking$\\pmod 3$ gives us $z^2=3^x-5^y\\equiv 1,2\\pmod 3$ furthermore from the fact that $z^2\\equiv 0,1\\pmod 3$ we get that $5^y\\equiv 1\\pmod 3\\Longrightarrow 2\\nmid y$, $y$ is odd.\nFurthermore, by taking $\\pmod 4$, we get that $z^2=3^x-5^y\\equiv (-1)^x-1\\pmod 4$ furthermore $z^2\\equiv 0,1\\pmod 4$ implies that $x$ is even. ($x=2k$)\nNow taking $\\pmod 8$, we get that $z^2=3^{2k}-5^y\\equiv 4\\pmod 8$ which implies that $z^2\\equiv 4\\pmod 8\\Longrightarrow z$ is even.\n\nAfter these observations, the equation can be rewritten as: $\\left(3^k\\right)^2-z^2=5^y\\Longleftrightarrow (3^k-z)(3^k+z)=5^{a+b}\\text{ where } a+b=y\\Longrightarrow 3^k-z=5^a\\text{ and } 3^k+z=5^b$, now notice that if we sum the equations we get $2\\cdot 3^k=5^a+5^b\\Longrightarrow 3^k=\\frac{5^a+5^b}{2}$ however notice that for $RHS\\in \\mathbb{Z^+}$, $a$ must be equal to $0$. Thus $a=0\\text{ and } b=y$\nThus the equation transforms into $2\\cdot3^k=5^y-(-1)^y$ so now we can use $LTE$ to finish up our problem.\n$\\nu_3(2\\cdot3^k)=\\nu_3(5^y-(-1))\\Longrightarrow k=\\nu_3(y)+1\\Longrightarrow y\\ge3^{k-1}$, furthermore by plugging in our result into our original equation we get $6\\cdot 3^k-5^{3^{k-1}}-1\\ge0$ now let $f(k)=6\\cdot3^k-5^{3^{k-1}}-1$\n[i]Claim:[/i] The previously stated inequality in integers if and only if, $k=1$\n[i]Proof:[/i]\n$f'(k)=6\\cdot3^k\\ln {3}-3^{k-1}\\cdot 5^{3^{k-1}}\\ln {3}\\ln {5}<0, \\forall k\\ge2$, thus $k=1$ $\\square$\n\nThus $k=1\\Longrightarrow x=2$, furthermore form the equation $3^k=z+1$ we get that $z=2$, thus by plugging into our original equation we get $9-4=5^y\\Longrightarrow y=1$\nSo the only solutions are the ones previously stated in the claim$\\blacksquare$.", "Solution_29": "Taking $\\pmod4$ we see $x$ is a multiple of $2$. Let $x=2x_1.$ From $\\pmod3$ we also see that $y\\equiv1\\pmod 2.$ Say $y=2y_1+1.$ We have $$(3^{x_1})^{2}-z^{2}=(3^{x_1}-z)(3^{x_1}+z)=5^{2y_1+1}.$$ If $3^{x_1}\\neq z+1$ we have $2(3^{x_1})\\equiv0\\pmod 5$ a contradiction. Thus $3^{x_1}=z+1$ so, $$2(3^{x_1})=5^{2y_1+1}+1.$$\nIf $x_1\\geq2$ we have $5^{2y_1+1}\\equiv -1\\pmod 9$ and from Euler's Theorem $y_1\\equiv 1 \\pmod 6.$\nSay $y_1=6y_2+1.$ We have $2(3^{x_1})=5^{12y_2+3}+1.$ From sum of cubes we have $$3^{x_1-1}=5^{8y_2+2}-5^{4y_2+1}+1.$$ If $x_1\\geq 3$ we have $5^{8y_2+2}-5^{4y_2+1}+1\\equiv 0\\pmod 9.$ Since $4y_2+1\\equiv\\{1,3,5\\}\\pmod 6$ we have $5^{8y_2+2}-5^{4y_2+1}+1\\equiv3\\pmod 9$ a contradiction. Thus $$x_1\\leq2.$$ For $x_1=2$ we have $81-5^{y}=64$ no solution. For $x_1=1$ we have $9-5^{y}=4$ and $\\boxed{(x,y,z)=(2,1,2)}$ as the only solution." } { "Tag": [], "Problem": "Translate each the sentence given into a matematical equation. Be sure to identify the meaning of all symbols.\r\n\r\nHere it is:\r\n\r\nThe total variable cost of manufacturing x dishwashers is 150 dollars per dishwasher times the number of dishwashers manufactured.", "Solution_1": "[quote=\"Interval\"]Translate each the sentence given into a matematical equation. Be sure to identify the meaning of all symbols.\n\nHere it is:\n\nThe total variable cost of manufacturing x dishwashers is 150 dollars per dishwasher times the number of dishwashers manufactured.[/quote]\r\nWouldn't the equation just be $150x$?", "Solution_2": "If you say it is, okay.", "Solution_3": "is it just me, or does it seem like we are doing intervals algebra 1 homework for him?", "Solution_4": "I am not in algebra 1 or 2. \r\n\r\nI graduated high school over 20 years ago.\r\n\r\nI am just preparing for a math state exam.\r\n\r\nThe questions come from that textbook.\r\n\r\nIf the questions are bothering you, then skip my post." } { "Tag": [], "Problem": "Hey, I just recently heard of the UNSCO from this forum and other people. I was wondering, how can one participate in this competition and what materials and resources do you recommend to practice for it? I appreciate any comments. Thanks in advance.", "Solution_1": "i am going to be the first person in the history of my school to participate in it for next year.\r\nall you have you to do is to talk to your chem teacher, then your chem teacher has to e-mail someone (the coordinator) and say that there are some interested people. then you'll take the local examination. if you pass you go to nationals. then you can qualify for summer camp\r\nfor more info go to\r\nhttp://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education%5Cstudent%5Colympiad.html" } { "Tag": [ "calculus", "integration", "LaTeX", "trigonometry", "calculus computations" ], "Problem": "Can't seem to figure this one out, would appreciate some help.\r\n\r\nEvaluate $ \\int \\frac{1}{x^4 \\sqrt{a^2\\minus{}x^2}} \\, dx$", "Solution_1": "Try doing this substitution....\r\n\r\n$ {x\\equal{}acosy}$...", "Solution_2": "@Valeri\r\n$ \\text{\\LaTeX}$ tip use [code]$\\cos x$[/code] instead of [code]$ cos x$[/code] to get the more nice looking $ \\cos x$ :wink:", "Solution_3": "I've tried that which gets it down to $ \\int \\frac {1}{\\cos^4y}dy$ or $ \\int \\sec^4y dy$, but I'm not sure the best way to attack that either :/\r\n\r\nI also received this problem before learning anything about trig substitutions so I'm wondering if I am supposed to take a different approach...", "Solution_4": "THen pcz.... use integration by parts.... split ur integral as $ {sec^2x}$*$ {sec^2x}$....\r\n\r\nthis is split like this because $ {sec^2x}$ is an integrable function...", "Solution_5": "Okay, got it from there. Thanks for the help :)", "Solution_6": "[quote=\"pcz\"]Can't seem to figure this one out, would appreciate some help.\n\nEvaluate $ \\int \\frac {1}{x^4 \\sqrt {a^2 \\minus{} x^2}} \\, dx$[/quote]\r\nSubstitute $ x\\equal{}\\frac1u,$ then you'll get an easy integral.", "Solution_7": "[quote=\"pcz\"]I've tried that which gets it down to $ \\int \\frac {1}{\\cos^4y}dy$ or $ \\int \\sec^4y dy$, but I'm not sure the best way to attack that either :/\n\nI also received this problem before learning anything about trig substitutions so I'm wondering if I am supposed to take a different approach...[/quote]\r\n\r\n\r\nYou can also do $ \\int \\sec^2 y [1\\plus{} \\tan^2 y] dy$ which you can do pretty easily since $ (\\tan\\theta)' \\equal{} \\sec^2 \\theta$", "Solution_8": "pcz the method mentioned by chickendude will be better than mine....." } { "Tag": [ "function", "inequalities theorems", "inequalities" ], "Problem": "Does there exist a generalization of Karamata?\r\ni do not mean the generalization with two function or...,i am seeking one that puts some extra conditions on sequences,which yields the enequlity,that is some conditions weaker than majorizing.", "Solution_1": "See here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=322728[/url]", "Solution_2": "There is also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23644[/url]" } { "Tag": [ "induction", "inequalities", "college contests" ], "Problem": "f(x) is twice differentiable and it's defined on (-\\infty,\\infty). It satisfies f(0)=f'(0)=0 and |f''(x)|\\leq C|f(x)f'(x)| for any x\\in [b]R[/b], where C is a constant. Prove that f(x)=0.\r\n\r\n[i]year 1990[/i]", "Solution_1": "Furthermore, this problem can be generalized as follows:\r\n\r\nf:[b]R[/b]->[b]R[/b] is [i]n[/i]-times(n is an positive integer) differentiable. It satisfies f(0)=f'(0)=...=f(n-1)(0)=0 and there exists positive constant c and j\\in {0,...,n-1} s.t. |f n(x)|\\leq c|f(j)(x)| for any x\\in [b]R[/b]. Prove that f(x)=0.", "Solution_2": "I don't know it the following is good or not, but at least I try:\r\n Take any a>0. We show that for any x in [0,a] f(x)=0. I believe we can treat similarly the case x<0. \r\n Let M_1 the maximum value of |f(x)*f'(x)| when x is in [0,a] and M-2 the maximum value of |f'(x)| on [0,a]. For arbitrary x in [0,a] integrate the relation in the problem between a and x. We find that |f'(x)|<=cM_1*x for all x. Then integrate the same relation and use the previous result to find that for all x in [0,a] we have |f'(x)|<=c^2*m_1*M_2*x^2/2. Integrate again and use the previous result. It follows that for all x in [0,a] we have |f'(x)|<=c^3*M_1*(M_2)^2/3!*x^3. By induction, we find that |f'(x)|<=c^k*M_1*(M_2)^k-1*x^k/k! for all k and all x in [0,a]. For fixed x, take k very large. It follows that f'(x)=0 for all x in [0,a] and thus f is constant of [0,a]. So f(x)=f(0)=0 for x in [0,a]. Since a is arbitrary, f is zero for positive values of x. The same for the negative ones.", "Solution_3": "I study your proof Harazi it's good, it seems f'(0)=0 is useless.\r\nWhat is your proof Liyi ? Do you really need f'(0) = 0 ?", "Solution_4": "suppose f(x) \\neq 0, for a x \\geq 0. WLOG f(x) \\neq 0 on [0,\\varepsilon]. Hence f'(x)\\neq 0 [0,\\varepsilon] (if not, we can make an translation).\r\n\r\nTake x_1>0 s.t. 1/x_1 > C, f is continous, so we can suppose that\r\nmax_[0,x_1] |f(x)| = M_0 \\in (0,1).\r\n\r\nLet M_1 = max_[0,x_1] |f'(x)|, M_1 > 0.\r\n\r\nTake x_2 \\in [0,x_1] such that |f'(x_2)| = M_1. From Lagrange's mean value theorem, we know there is a \\xi \\in (0,x_2) such that\r\n\r\n|f''(\\xi)| = |f'(x_2) - f'(0)| / |x_2 - 0| = |f'(x_2)/x_2|\r\n\r\nthus\r\n\r\nCM_1 \\geq C|f'(\\xi)| > C|f(\\xi)f'(x_i)| \\geq |f''(\\xi)| = |f'(x_2)/x_2| > CM_1\r\n\r\na contradiction....\r\n\r\nhence for all x\\geq 0, f(x) = 0.\r\n\r\nin the similar way we get proof that for x\\leq 0, f(x) = 0", "Solution_5": "the codes r not showing up, i have NO idea wat u guys r saying, lol...", "Solution_6": "I believe some people here forgot teh dollar signs.", "Solution_7": "[quote=\"boxedexe\"]I believe some people here forgot teh dollar signs.[/quote]\r\n\r\nYeah, really....", "Solution_8": "First,we choose an interval $[a,b]$ containing $x=0$,then it is obvious that$|f''(x)|\\leq C'|f'(x)|$,where $C'$ is a constant on $[a,b]$.So according to Gronwall's lemma,we have $f'(x)=0$ on $[a,b]$,that is, $f(x)$ is a constant.And then $f(x)=0$.Cover the real axis with such successive intervals and the desired result obtained.", "Solution_9": "I also dont know if the following is good or not but I requires the condition f'(0)=0!\r\nTake abitrarily a and b>a.For all x in [a,b] choose M as max{$ \\frac{f(x)*f'(x)}{x}*\\frac {(b \\plus{} a)}{(b \\minus{} a)}$}.So,we get:\r\n\r\n$ |f''(x)|\\leq M*x*\\frac {(b \\minus{} a)}{(b \\plus{} a)}$.Now let's intergrate between a and b both two sides of the inequality we get:\r\n\r\n$ |f'(a) \\minus{} f'(b)|\\leq M*(b \\minus{} a)^2/2$ or $ |f'(a) \\minus{} f'(b)/(a \\minus{} b)|\\leq M*(b \\minus{} a)/2.$\r\nLet b tend to a and by the defination of derivation we gain:$ f''(a) \\equal{} 0$.Hence $ f'(a) \\equal{} f'(0) \\equal{} 0$ and $ f(a) \\equal{} f(0) \\equal{} 0$.", "Solution_10": "I still doubt on what I wrote in the above message.Any check???" } { "Tag": [ "geometry", "function", "rectangle", "quadratics", "parallelogram", "real analysis", "real analysis solved" ], "Problem": "You have a linear function $f(x)=-\\frac{2}{3}x+4$ where $0\\leq x\\leq 6$ and $0\\leq y\\leq 4$. What is the maximum area of a rectangle that has one side on the line of the function?\r\n\r\nI know how to optimize this, I am just having trouble finding the equation for area of such a rectangle in terms of the function.\r\n\r\nI have included an image of what I am talking about. Thanks a lot.", "Solution_1": "Here's a slightly offbeat approach. The region that is the rectangle is the result of starting with a big triangle and deleting from it three nonoverlapping triangles. All of the triangles are similar, and similar to a right triangle with legs 2 and 3 and hypotenuse $\\sqrt{13}.$\r\n\r\nThe large triangle has area $12.$\r\n\r\nSuch a triangle with longer leg $u$ has area $\\frac13\\,u^2.$\r\n\r\nSuch a triangle with hypotenuse $u$ has legs $\\frac{2u}{\\sqrt{13}}$ and $\\frac{3u}{\\sqrt{13}}$ and thus area $\\frac3{13}\\,u^2.$\r\n\r\nNow suppose the point at which the rectangle intesects the $x$ axis is $(z,0).$ The three triangles we subtract from the big triangles have a longer leg of $z$, a hypotenuse of $6-z$ and a hypotenuse of $4-\\frac{2z}3$.\r\n\r\n[oops - original post was in error. Everything from here down, deleted, fixed in next post.]", "Solution_2": "Thus, the area of the rectangle is\r\n\r\n$12-\\frac13\\,z^2-\\frac3{13}\\,(6-z)^2-\\frac3{13}\\,\\left(4-\\frac{2z}3\\right)^2= 4z - \\frac{2}{3}\\,z^2$\r\n\r\nThis quadratic function can be maximized by precalculus means. The maximum occurs when $z=3$ and the maximum area is $6,$ which is half of the area of the triangle.\r\n\r\n---\r\n\r\nThat was clumsier than it had to be. There has to be a nicer solution.", "Solution_3": "Here's an argument. I'll won't do the drawing myself - you'll have to make your own.\r\n\r\nWe have a right triangle $ABC$ with right angle at $C.$ We would like to fit a rectangle of maximal area with one side along $AB.$ Let this rectangle be $DEFG$ with $D$ in $AC,$ $E$ in $BC$ and $FG$ a subsegment of $AB.$\r\n\r\nShear the rectangle into a parallelogram by fixing segment $DE$ and sliding $FG$ along $AB$ until it becomes $F'A$. We now have parallelogram $DEF'A$ in which side $EF'$ is parallel to $AC.$ Since this was a parallel shear, the area of $DEF'A$ is the same as the area of $DEFG.$\r\n\r\nNow shear this again, fixing segment $EF'$ and sliding $AD$ along $AC$ until it becomes $HC.$ We now have a rectangle $CEF'H$ which has the same area as the original rectangle $DEFG.$\r\n\r\nSo our original problem is equivalent to finding the rectangle of largest area with one corner at $C$ that fits inside the triangle. This is a much easier problem than the original problem, and leads readily to the condition that point $F'$ must be the midpoint of the hypotenuse $AB$ and that the area of the rectangle $CEF'H$ (or of $DEFG$) must be half of the area of triangle $ABC.$" } { "Tag": [ "LaTeX", "Asymptote" ], "Problem": "very diffucult question :( \r\n[img]http://img217.imageshack.us/img217/2678/adszur9.png[/img]", "Solution_1": "Oh! Problem is nice. I will solution it.", "Solution_2": ":what?: i didnt solve...i think not nice problem :(", "Solution_3": "This is picture of my solution!", "Solution_4": "Thank you for your help..Can you show me your solution way?\r\nThanks and best regards!", "Solution_5": "$ \\bullet$ Let $ P$ is midpoint of $ EC$, let $ AB \\equal{} EC \\equal{} a$ and $ BE \\equal{} b$ $ \\Longrightarrow BC \\equal{} a \\plus{} b$.\r\n\r\n$ \\bullet$ In $ \\triangle ABC$ we have $ AC^2\\equal{}AB^2\\plus{}BC^2\\minus{}2AB.BC.cos120^0\\equal{}a^2\\plus{}(a\\plus{}b)^2\\plus{}a(a\\plus{}b)$\r\n\r\n$ \\bullet$ We have $ DP // AE \\Longrightarrow EF//DP \\Longrightarrow \\frac{BF}{BD}\\equal{}\\frac{EF}{DP} \\Longleftrightarrow \\frac{x}{BD}\\equal{}\\frac{4}{DP}$.\r\n\r\n$ \\bullet$ $ BD^2\\equal{} \\frac{AB^2\\plus{}BC^2}{2} \\minus{} \\frac {AC^2}{4} \\equal{} \\frac{a^2\\plus{}(a\\plus{}b)^2}{2} \\minus{} \\frac {a^2\\plus{}(a\\plus{}b)^2\\plus{}a(a\\plus{}b)}{4}\\equal{}$ $ \\frac {a^2\\plus{}b^2\\plus{}ab}{4}$ \r\n$ \\Longrightarrow BD \\equal{} \\frac {1}{2} \\sqrt {a^2\\plus{}b^2\\plus{}ab}$ , (1).\r\n\r\n$ \\bullet$ But $ DP \\equal{} \\frac {1}{2}AE \\equal{} \\frac {1}{2} \\sqrt {AB^2\\plus{}BE^2\\minus{}2AB.BE.cos120^0}\\equal{}\\frac {1}{2}\\sqrt {a^2\\plus{}b^2\\plus{}ab}$ , (2).\r\n\r\n$ \\bullet$ Since (1) and (2) $ \\Longrightarrow BD \\equal{} DP$ $ \\Longrightarrow x \\equal{} 4$.\r\n\r\nOk! :rotfl:", "Solution_6": "Is it nice? Wish you good luck.", "Solution_7": "Thank you very much :)", "Solution_8": "[hide=\"Geometrical solution\"]\nExtend BD to X such that DX=BD.\nJoin X to E. Clearly triangle CEX is equilateral.\nSo /_ XEB=120\n\nNow, triangles ABE and BEX are congurent\nHence /_ XFE=/_AEB\n=>FB=FE [/hide]\n\nI am new to this forum, can u tell me how to add pictures and other symbols to posts.[/hide]", "Solution_9": "[quote=\"guru_bhai\"][I am new to this forum, can u tell me how to add pictures and other symbols to posts.[/hide][/quote]\r\n\r\nYou can learn about [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]LaTeX[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]Asymptote[/url] from the AoPS Wiki.\r\n\r\nLaTeX can be enclosed in dollar signs, while Asymptote needs to be enclosed in [ a s m ] tags (no spaces).", "Solution_10": "Why can't I see OP?", "Solution_11": "[quote=\"fishythefish\"]Why can't I see OP?[/quote]\r\n\r\nProbably your browser. Use Firefox.", "Solution_12": "I'm on Firefox.", "Solution_13": "Dear Guru_bhai can you show me your solution by picture ?\r\nThanks and best regards!i added by imageshack :lol:", "Solution_14": "You can use sketpad to draw picture and post it.", "Solution_15": "Oh! I have other solution for problem. I will post it soon.", "Solution_16": "Really? :o \r\nI wonder your another way solution...", "Solution_17": "[asy]size(300);\nimport olympiad;\npair A,B,C,D,E,F,G,H;\nA=(0,0);\nB=10*dir(120);\nC=(15,0);\nG=(-8,0);\nD=midpoint(B--C);\nE=(5,0);\nF=intersectionpoint(B--E,A--D);\ndraw(A--G,blue,MidArrow);\ndraw(A--C--B--A,blue);\ndraw(A--D,blue);\ndraw(B--E,blue);\nfill(anglemark(B,A,G,40),green);\ndot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(G,red);dot(E,red);dot(F,red);\nlabel(\"$G$\",G,S);\nlabel(\"$A$\",A,S);\nlabel(\"$B$\",B,W);\nlabel(\"$C$\",C,S);label(\"$E$\",E,S);\nlabel(\"$D$\",D,NE);label(\"$F$\",F,N);\nlabel(\"$x$\",A--F,NW);label(\"$4$\",F--E,NE);\nlabel(\"$60^\\circ$\",anglemark(B,A,G),4*NW);\nadd(pathticks(A--B,2,0.5,0,25,red));\nadd(pathticks(E--C,2,0.5,0,25,red));\nadd(pathticks(D--B,2,.5,7,25,red));add(pathticks(D--C,2,0.5,7,25,red));[/asy]\r\n\r\n\r\n[b] Hepl me to draw $ DH$ such that $ AD\\equal{}DH$[/b]\r\n\r\n[code][asy]\nsize(300);\nimport olympiad;\npair A,B,C,D,E,F,G,H;\nA=(0,0);\nB=10*dir(120);\nC=(15,0);\nG=(-8,0);\nD=midpoint(B--C);\nE=(5,0);\nF=intersectionpoint(B--E,A--D);\ndraw(A--G,blue,MidArrow);\ndraw(A--C--B--A,blue);\ndraw(A--D,blue);\ndraw(B--E,blue);\nfill(anglemark(B,A,G,40),green);\ndot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(G,red);dot(E,red);dot(F,red);\nlabel(\"$G$\",G,S);\nlabel(\"$A$\",A,S);\nlabel(\"$B$\",B,W);\nlabel(\"$C$\",C,S);label(\"$E$\",E,S);\nlabel(\"$D$\",D,NE);label(\"$F$\",F,N);\nlabel(\"$x$\",A--F,NW);label(\"$4$\",F--E,NE);\nlabel(\"$60^\\circ$\",anglemark(B,A,G),4*NW);\nadd(pathticks(A--B,2,0.5,0,25,red));\nadd(pathticks(E--C,2,0.5,0,25,red));\nadd(pathticks(D--B,2,.5,7,25,red));add(pathticks(D--C,2,0.5,7,25,red));\n[/asy][/code]", "Solution_18": ":) Am i right?", "Solution_19": "[hide=\"Complete solution :\"] \n\nConstruction :\n\nExtend $ AD$ to $ H$ such that $ AD=DH$.\nJoin $ E$ and $ C$ to $ H$.\n\n\n[asy]size(300);\n import olympiad;\n pair A,B,C,D,E,F,G,H;\n A=(0,0);\n B=10*dir(120);\n C=(15,0);\n G=(-8,0);\n D=midpoint(B--C);\n E=(5,0);\n F=intersectionpoint(B--E,A--D);\n draw(A--G,blue,MidArrow);\n draw(A--C--B--A,blue);\n draw(A--D,blue);\n draw(B--E,blue);\n fill(anglemark(B,A,G,40),green);\n dot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(G,red);dot(E,red);dot(F,red);\n label(\"$G$\",G,S);\n label(\"$A$\",A,S);\n label(\"$B$\",B,W);\n label(\"$C$\",C,S);label(\"$E$\",E,S);\n label(\"$D$\",D,NE);label(\"$F$\",F,N);\n label(\"$x$\",A--F,NW);label(\"$4$\",F--E,NE);\n label(\"$60^\\circ$\",anglemark(B,A,G),4*NW);\n add(pathticks(A--B,2,0.5,0,25,red));\n add(pathticks(E--C,2,0.5,0,25,red));\n \n H=2*D-A;\n draw(E--H--D,dashed+blue);\n draw(H--C,dashed+blue);\n dot(H,red);\n add(pathticks(E--H,2,.5,0,25,red));add(pathticks(H--C,2,0.5,0,25,red));\n \n label(\"$H$\",H,NE);\nfill(anglemark(E,H,C,40),green);fill(anglemark(H,C,E,40),green);fill(anglemark(C,E,H,40),green);\nlabel(\"$60^\\circ$\",anglemark(E,H,C),4*S);label(\"$60^\\circ$\",anglemark(H,C,E),4*NW);label(\"$60^\\circ$\",anglemark(C,E,H),4*NE);[/asy]\n\n\nClearly,\n$ \\triangle ABD \\cong \\triangle HCD$\n$ \\Rightarrow \\angle BAD = \\angle CHD$\n$ \\Rightarrow AB \\parallel CH$ \n$ \\Rightarrow \\angle HCE = 60^\\circ$\n\nNow in $ \\triangle CEH, EC=HC \\Rightarrow \\triangle CEH$ is equilateral.\n$ \\Rightarrow \\angle HEA = 120^\\circ$\n\n \nNow clearly $ \\triangle ABE \\cong \\triangle AEH$\n$ \\Rightarrow \\angle FAE = \\angle FEA$\n\nSo, In $ \\triangle FAE, FA=FE.$\n\nHence Proved[/hide]", "Solution_20": "[quote=\"guru_bhai\"][asy]size(300);\nimport olympiad;\npair A,B,C,D,E,F,G,H;\nA=(0,0);\nB=10*dir(120);\nC=(15,0);\nG=(-8,0);\nD=midpoint(B--C);\nE=(5,0);\nF=intersectionpoint(B--E,A--D);\ndraw(A--G,blue,MidArrow);\ndraw(A--C--B--A,blue);\ndraw(A--D,blue);\ndraw(B--E,blue);\nfill(anglemark(B,A,G,40),green);\ndot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(G,red);dot(E,red);dot(F,red);\nlabel(\"$G$\",G,S);\nlabel(\"$A$\",A,S);\nlabel(\"$B$\",B,W);\nlabel(\"$C$\",C,S);label(\"$E$\",E,S);\nlabel(\"$D$\",D,NE);label(\"$F$\",F,N);\nlabel(\"$x$\",A--F,NW);label(\"$4$\",F--E,NE);\nlabel(\"$60^\\circ$\",anglemark(B,A,G),4*NW);\nadd(pathticks(A--B,2,0.5,0,25,red));\nadd(pathticks(E--C,2,0.5,0,25,red));\nadd(pathticks(D--B,2,.5,7,25,red));add(pathticks(D--C,2,0.5,7,25,red));[/asy]\n\n\n[b] Hepl me to draw $ DH$ such that $ AD\\equal{}DH$[/b]\n\n[code][asy]\nsize(300);\nimport olympiad;\npair A,B,C,D,E,F,G,H;\nA=(0,0);\nB=10*dir(120);\nC=(15,0);\nG=(-8,0);\nD=midpoint(B--C);\nE=(5,0);\nF=intersectionpoint(B--E,A--D);\ndraw(A--G,blue,MidArrow);\ndraw(A--C--B--A,blue);\ndraw(A--D,blue);\ndraw(B--E,blue);\nfill(anglemark(B,A,G,40),green);\ndot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(G,red);dot(E,red);dot(F,red);\nlabel(\"$G$\",G,S);\nlabel(\"$A$\",A,S);\nlabel(\"$B$\",B,W);\nlabel(\"$C$\",C,S);label(\"$E$\",E,S);\nlabel(\"$D$\",D,NE);label(\"$F$\",F,N);\nlabel(\"$x$\",A--F,NW);label(\"$4$\",F--E,NE);\nlabel(\"$60^\\circ$\",anglemark(B,A,G),4*NW);\nadd(pathticks(A--B,2,0.5,0,25,red));\nadd(pathticks(E--C,2,0.5,0,25,red));\nadd(pathticks(D--B,2,.5,7,25,red));add(pathticks(D--C,2,0.5,7,25,red));\n[/asy][/code][/quote]\n\nadd\n[code]\nH=2*D-A\n[/code]", "Solution_21": "[asy]\n\npair A, B, C, D, EE;\n\nEE=(0,0);\nA=(-1,0);\nB=(0,-1.5);\nD=(0,.8);\nC=(1,0);\n\ndraw(A--B--C--D--A--C);\ndraw(D--B);\nlabel("$A$",A,W);\nlabel("$B$",B,S);\nlabel("$C$",C,E);\nlabel("$D$",D,N);\nlabel("$E$",EE,SE);\n\n//pathticks(path, how many ticks, where on segment (0=start, 1=end), ?, size of ticks)\nadd(pathticks(A--B, 1, .5, 6, 2));\nadd(pathticks(B--C, 1, .5, 6, 2));\n\nadd(pathticks(A--D, 2, .5, 1, 2));\nadd(pathticks(C--D, 2, .5, 1, 2));\n\n[/asy]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": ":maybe: i think it was on poland math olympiad.\r\n\r\n\r\nDetermine all possible functions $f: Q\\to Q$\r\nwhich satisfies\r\n\r\n$f(x^{2}+y) =xf(x)+f(y)$ \r\n\r\n$x,y are rational numbers$ :", "Solution_1": "Letting $x=1$, $y=0$ we get $f(1)=f(1)+f(0)\\rightarrow f(0)=0$. Putting $y=0$ gives then $f(x^{2})=xf(x)$, so $f(x^{2}+y)=f(x^{2})+f(y)$ and, in particular $f(2x^{2})=2f(x^{2})$.\r\nNow, let $x=z^{2}$ and $y=2z^{2}+1$; then \r\n\r\n$(z^{2}+1)[f(z^{2})+f(1)]=(z^{2}+1)f(z^{2}+1)=$\r\n$=f((z^{2}+1)^{2})=f(z^{4}+2z^{2}+1)=z^{2}f(z^{2})+f(2z^{2}+1)=$\r\n$=z^{2}f(z^{2})+f(1)+f(2z^{2})=z^{2}f(z^{2})+2f(z^{2})+f(1)$\r\n\r\nComparing the first and final expressions we get:\r\n\r\n$f(1)z^{2}=f(z^{2})=zf(z)\\rightarrow f(z)=f(1)z \\;\\;\\forall z\\in \\mathbb Q$\r\n\r\nThe function works." } { "Tag": [ "geometry", "parallelogram", "perpendicular bisector", "geometry proposed" ], "Problem": "Let $ ABCD$ be a convex quadrilateral, such that\r\n\r\n$ \\angle CBD\\equal{}2\\cdot\\angle ADB, \\angle ABD\\equal{}2\\cdot\\angle CDB$ and $ AB\\equal{}CB$.\r\n\r\nProve that quadrilateral $ ABCD$ is a kite.", "Solution_1": "[color=darkblue]Let $ P \\in AD$, $ Q \\in CD$, such that $ BP$ and $ BQ$ bisects angles $ ABD$ and $ DBC$ respectively. Then $ \\angle PBD=\\angle CDB \\Rightarrow PB \\parallel CD$, and, similliarly $ AD %Error. \"parralel\" is a bad command.\nBQ$. So $ DPBQ$ is a paralelogram and $ BD$ bisects $ PQ$. Also\n\n$ \\dfrac{DQ}{QC}=\\dfrac{BD}{BC}=\\dfrac{BD}{AB}=\\dfrac{DP}{PA}$\n\nSo $ PQ\\parallel AC$ and $ BD$ bisects $ AC$. Thus, $ BD$ is perpendicular bisector of $ AC$. The conclusion follows. [/color]", "Solution_2": "[hide=\"Solution in short notes\"]\nFirst, note $ E$ and $ F$ be the intersections of $ BD$ with the circle $ (B,AB)$, which contains $ C$, such that $ E$ is on segment $ BD$. Looking central and inscribed angles, we get easily $ AFCD$ is a parallelogram, so its diagonals are bisecting each other, and we get $ BD$ is a bisector of $ AC$. \nLet $ \\{S\\}\\equal{}AC\\cap BD$. Then $ \\triangle ASB\\cong\\triangle CSB$, so $ BD\\bot AC$. [/hide]", "Solution_3": "$AB=b$ $BD=k$ accoring $ \\triangle ABD$ and $ \\triangle BCD$\n$\\frac{k}{sin(\\alpha+2\\lambda)}=\\frac {b}{sin\\alpha}$\n$\\frac{k}{sin(\\lambda+2\\alpha)}=\\frac {b}{sin\\lambda}$\ndivede by side-by side\n$cos(\\alpha+\\lambda)-cos(\\lambda+3\\alpha)=cos(\\alpha+\\lambda)-cos(\\alpha+3\\lambda)$ then $\\boxed{\\alpha=\\lambda}$\nNOTE: $\\angle{ADB}=\\alpha$ $\\angle{ABD}=2\\lambda$ $\\angle{BDC}=\\lambda$ $\\angle{DBC}=2\\alpha$ \n" } { "Tag": [ "complex analysis", "function", "complex analysis unsolved" ], "Problem": "Suppose that $ f$ in holomorphic inside and on $ \\alpha(0;1)$ , with taylor expansion $ \\sum_{n>\\equal{}0}c_{n}z^{n}$. give that $ f$ has $ m$ zeros inside $ \\alpha(0;1)$, prove that\r\n\r\n\\[ min\\{|f(z)|: |z|\\equal{}1\\}<\\equal{}|c_{0}|\\plus{}........\\plus{}|c_{m}|\\] :maybe:", "Solution_1": "Assume the opposite and apply the Rouche theorem to the pair $ f(z)$, $ f(z)\\minus{}\\sum_{k\\equal{}1}^m c_k z^k$. Note that the second function has a zero at the origin of multiplicity $ m\\plus{}1$." } { "Tag": [ "LaTeX", "inequalities unsolved", "inequalities" ], "Problem": "Let x, y, z be real numbers in the interval (0; 1). Prove that\r\n\\frac{1}{x(1-y)}+\\frac{1}{y(1-Z)}+\\frac{1}{Z(1-x)}\\geq \\frac{3}{xyz + (1-x)(1-y)(1-z)}", "Solution_1": "[quote=\"lnnvnu\"]Let $ x, y, z$ be real numbers in the interval $ (0; 1).$ Prove that\n$ \\frac{1}{x(1\\minus{}y)}\\plus{}\\frac{1}{y(1\\minus{}z)}\\plus{}\\frac{1}{z(1\\minus{}x)}\\geq \\frac{3}{xyz \\plus{} (1\\minus{}x)(1\\minus{}y)(1\\minus{}z)}$[/quote]\r\n\r\nUse the dollar sign to write in LATEX dear friand.....", "Solution_2": "It's equivalent to\r\n$ \\displaystyle (xyz\\plus{}(1\\minus{}x)(1\\minus{}y)(1\\minus{}z))\\sum_{cyc}\\frac{1}{x(1\\minus{}y)}\\geq3$\r\n$ \\displaystyle\\Leftrightarrow \\sum_{cyc}\\frac{yz}{1\\minus{}y}\\plus{}\\sum_{cyc}\\frac{(1\\minus{}x)(1\\minus{}z)}{x}\\geq3$\r\n$ \\displaystyle\\Leftrightarrow \\sum_{cyc}\\frac{(1\\minus{}(1\\minus{}y))z}{1\\minus{}y}\\plus{}\\sum_{cyc}\\frac{(1\\minus{}x)(1\\minus{}z)}{x}\\geq3$\r\n$ \\displaystyle\\Leftrightarrow \\sum_{cyc}\\left(\\frac{z}{1\\minus{}y}\\minus{}z\\right)\\plus{}\\sum_{cyc}\\left(\\frac{1\\minus{}z}{x}\\minus{}(1\\minus{}z)\\right)\\geq3$\r\n$ \\displaystyle\\Leftrightarrow \\sum_{cyc}\\frac{z}{1\\minus{}y}\\plus{}\\sum_{cyc}\\frac{1\\minus{}z}{x}\\geq6$ which is obviously true by AM-GM." } { "Tag": [ "algebra", "polynomial", "function", "complex analysis" ], "Problem": "Hi all,\r\n\r\nI have a homework question I could use some help with. I don't want anyone to give the full answer so I won't type out the whole question but I will give the gist of it and maybe some kind soul out there can let me know what theorem or approach would be helpful in solving the problem.\r\n\r\nAnywho, the question asks to find the # of zeros in the closed, right half complex plan (x >=0) of a degree 4 polynomial. I have only ever solved problems like these asking for # of zeros in a disk, and so I usually just apply Rouche's theorem. I can't think of a way to generalize my approach to this problem though. Any thoughts?\r\n\r\nThanks for your time.", "Solution_1": "Try the [url=http://en.wikipedia.org/wiki/Routh-Hurwitz_theorem]Routh-Hurwitz theorem[/url] - it's based on the same idea (the argument principle) as Rouche's theorem.", "Solution_2": "Another possibility is to create some continuous family $ P_t$, $ t\\in[0,1]$ of polynomials such that none of them has any roots on the imaginary axis, $ P_1$ is your polynomial, and $ P_0$ is a polynomial for which the answer is obvious. For instance, if your polynomial is something like $ z^4\\plus{}10z^3\\plus{}5z\\plus{}1$, you can take $ P_t(z)\\equal{}z^4\\plus{}1\\plus{}t(10z^3\\plus{}5z)$ (note that $ \\Re P_t(z)>0$ on the imaginary axis for all $ t$). The numbers of roots of $ P_0$ and $ P_1$ in the right half-plane are the same. This doesn't work always, of course, but it is a useful trick nevertheless. Note that it is even better than Rouche's theorem: if you can represent your polynomial as $ Q\\plus{}S$ with $ |Q|>|S|$ everywhere on the imaginary axis, you can always take $ P_t\\equal{}Q\\plus{}tS$ but in the example I gave above $ |z^4\\plus{}1|$ does not dominate $ |10z^3\\plus{}5z|$.", "Solution_3": "Thanks for the responses. Fedja, what theorem is that, do you know?", "Solution_4": "Essentially it is the same argument principle. (By the way, I forgot to mention one little but important detail: the degree of all polynomials $ P_t$ should be the same. In practice you can always do it by fixing the leading coefficient). The idea is that if you have a closed contour and a continuous family $ f_t$, $ t\\in[0,1]$ of nice analyic functions none of which vanishes on the contour, then the increment of the argument over the contour should be continuous on one hand and can take only a discrete set of values on the other hand, so it must stay constant, which implies that the number of zeroes inside the contour also does not change. Of course, the right half plane isn't bounded by a closed contour, but you can take a long interval on the imaginary axis, complement it by a semicircle, let the radius of the semicircle be large enough, etc." } { "Tag": [ "function", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Find a formula f(x) where x is the number of posts, and f(x) is the number of your stars. I doubt if it is a linear function.", "Solution_1": "In what way is this related to the American Mathematics Competitions?", "Solution_2": "Well...\r\n\r\n0-19: 0 stars\r\n20-49: 1/2 star\r\n50-99: 1 star\r\n100-249: 2 stars\r\n250-499: 5/2 stars\r\n500-999: 3 stars\r\n1000-2499: 4 stars\r\n2500+: 5 stars\r\n\r\nIt's kind of weird, now that you look at it. Someone else want to come up with a formula? ;)", "Solution_3": "I need 6 to 2 stars!!!!", "Solution_4": "The way I see it, it $f$ will either have to be piecewise defined, or use the floor function. :wacko:", "Solution_5": "i don't think it is linear...\r\ncan't see any \"mappings\" from it...", "Solution_6": "[quote=\"dts\"]The way I see it, it $f$ will either have to be piecewise defined, or use the floor function. :wacko:[/quote]\r\n\r\nThe floor function is piecewise defined.", "Solution_7": "Formula? well, henryyangrui is bored of his prepration for USAMO.", "Solution_8": "What exactly does this have to do with AMC? Was this an old USAMO problem? :P", "Solution_9": "You guys are also forgetting that administrators, regardless of the number of posts, get 6 stars automatically." } { "Tag": [ "geometry", "geometric transformation", "rotation" ], "Problem": "Link is coloring a triforce, which consists of four equilateral triangles and is depicted below. He has three colors to use: gold, black, and green. So that it remains recognizable, he doesn't want to color any two triangles the same color if they share a side. How many different ways can he color the triforce?\n[asy]size(50);\ndraw((0,0)--(2,0)--(1,1.732)--cycle);\ndraw((1,0)--(1.5,.866)--(.5,.866)--cycle);[/asy]", "Solution_1": "I didn't realize you were supposed to count rotations.", "Solution_2": "It never says that rotations should not be distinct.", "Solution_3": "the answer is 24. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=357525&p=1950491&hilit=Link+is+coloring+a+triforce#p1950491]here[/url]", "Solution_4": "It's always black in the middle and golden in the three others.", "Solution_5": "[hide]There are $3$ ways to color the middle triangle. Since the other triangles cannot be the same color as the middle triangle, there are $2$ ways to choose the color for each of the $3$ triangles (excluding the middle one). The total ways to color is $3\\times 2^3=\\boxed{24}$ ways.[/hide]", "Solution_6": "This is definetly not level 16 lol." } { "Tag": [ "geometry", "rhombus", "geometric transformation", "reflection", "circumcircle", "angle bisector", "geometry unsolved" ], "Problem": "Hello\r\n\r\nI don't know if this has already been posted, but this problem is good, and I need help.\r\n\r\nLet ABC be a triangle with measure of angle ABC 80 degrees. Let BD be the angle bisector of angle ABC with D on AC. If AD=DB+BC, determine angle A, using only a geometric argument.\r\n\r\nThanks", "Solution_1": "Let $ A_0BC_0D$ be a rhombus with $ \\angle A_0BC_0 \\equal{} 80^\\circ.$ Let a line $ b$ through $ D$ cut the rays $ BA_0, BC_0$ at $ A, C,$ respectively, forming a $ \\triangle ABC$ with bisector $ BD$ of the angle $ \\angle ABC \\equal{} 80^\\circ.$ The angle $ \\angle ADB$ must be in $ (40^\\circ, 140^\\circ),$ otherwise $ b$ does not intersect both rays. We demand $ DA \\equal{} DB \\plus{} BC.$ Let $ P$ be foot of the perpendicular from $ D$ to $ BA_0$ and $ A_1 \\in BA_0$ reflection of $ B$ in $ DP$ so that $ DA_1 \\equal{} DB.$ Because of our demand $ DA > DB \\equal{} DA_1,$ it must be $ BA > BA_1$ and this further limits the angle $ \\angle ADB > \\angle A_1DB \\equal{} 100^\\circ,$ or to $ (100^\\circ, 140^\\circ).$ Since $ \\angle BAD < \\angle BA_1D < BPD \\equal{} 90^\\circ,$ $ DA$ is strictly increasing and $ BC$ strictly decreasing with increasing $ BA.$ Thus $ DA \\equal{} DB \\plus{} BC$ can hold only for one position of $ A$ and a unique angle $ \\angle ADB.$\r\n\r\nLet $ \\triangle ADE$ be isosceles triangle with $ DA \\equal{} DE$ and $ \\angle ADE \\equal{} 120^\\circ,$ let $ M$ be the midpoint of $ AE,$ $ DM \\perp AE,$ and let $ N$ be a point on the segment $ DM,$ such that $ \\angle NAE \\equal{} \\angle NEA \\equal{} 10^\\circ.$ Let $ AN$ cut $ DE$ at $ B.$ Since\r\n\r\n$ \\angle BNE \\equal{} \\angle NAE \\plus{} \\angle NEA \\equal{} 20^\\circ \\equal{}$\r\n$ \\equal{} 30^\\circ \\minus{} 10^\\circ \\equal{} \\angle DEA \\minus{} \\angle NEA \\equal{} \\angle DEN \\equal{} \\angle BEN,$\r\n\r\nthe $ \\triangle BEN$ is isosceles with $ BE \\equal{} BN.$ Let $ C$ be reflection of $ N$ in $ DE,$ so that $ BD$ bisects the angle $ \\angle ABC,$ $ B$ is circumcenter of the isosceles $ \\triangle ENC,$ and $ DA \\equal{} DE \\equal{} DB \\plus{} BE \\equal{} DB \\plus{} BC.$ Due to $ \\angle ADE \\equal{} 120^\\circ,$ the line $ DA$ is a reflection of the line $ DN$ in the line $ DE,$ therefore $ C \\in DA$ and $ \\angle ABC \\equal{} 2 \\angle NBD \\equal{} 4 \\angle NED \\equal{} 80^\\circ.$ As for the remaining angle(s) of the $ \\triangle ABC,$ $ \\angle CAB \\equal{} \\angle DAN \\equal{} 20^\\circ.$" } { "Tag": [ "videos", "Alcumus", "pigeonhole principle", "rate problems", "Support" ], "Problem": "What happens when you run out of Karma?\r\n(no, i have not run out, i'm just curious.)", "Solution_1": "You need to go back and rate solutions and such to build karma back up. Otherwise you won't have access to more problems. So far, no one has ever run out of karma.", "Solution_2": "I have once :D \r\nbecause I never rated problems :D", "Solution_3": "Oh wow! I guess I should have said, \"as far as I know, no one has run out of karma.\" :) Did you stop getting the problems?", "Solution_4": "If you miss all your questions, even on your second chance and first, and rate the solution, you can still run out of karma. You'll lose 5 points for every problem. Then when you run out, you can't rate any solutions, you can't watch a video and rate that, or anything else. So it is possible to run out of karma (if you are not good at Alcumus).", "Solution_5": "You get more karma for rating solutions then getting problems.", "Solution_6": "[quote]\nAnswering a problem\tSpend 5\nRate problem/solution\tReceive 5\nSubmit bug report\tSpend 10\nBug report determined valid\tReceive 25\nWatch video\tSpend 5\nRate video\tReceive 5\nComment on video\tReceive 5\n\n[/quote]\r\n\r\nSO if you answer a question twice, you lose 10, but gain 5 for rating, so a loss of 5.", "Solution_7": "wait, [b]lets say [/b]i ran out of karma, and have 0 karma. if i rate a problem, i must first answer a question, and that takes karma, which i dont have any.\r\n\r\nif i watch a video, and rate it, i spend karma for wattching the video.\r\n\r\nhow do i earn the karma then??", "Solution_8": "So...\r\n\r\nyou can gain up to 15 karma per problem (10 because spending 5)\r\nsince you can rate 3 different questions?\r\n\r\nor would you just have to rate all the Questions to get your 5 karma back?", "Solution_9": "wait, how do u earn 10 (or 15?) karma per problem?\r\n\r\ni dont see it right now......\r\n\r\nand wat do u mean by \"since you can rate 3 different questions? \r\n\r\nor would you just have to rate all the Questions to get your 5 karma back?\"?", "Solution_10": "I think in your situation you have to click on the subject (the line with the bar next to it) so you can go back and rate problems and gain karma to get more problems.", "Solution_11": "oh......\r\n\r\nim gonna earn xtra karma that way!", "Solution_12": "For each problem you can rate three things: The difficulty of problem, the quality of solution, and something else.", "Solution_13": "The \"something else\" is how much you learned from the solution.", "Solution_14": "Unfortunatley you won't be able to do anything\r\ndid you actually run out of karma!!! :D", "Solution_15": "If you ever run out of karma, you can go to the report and rate unrated problems (click the category and then the magnifying glass).\r\nThat way, you won't run out of karma (I think).", "Solution_16": "I have run out of Karma before!! I used to be on the leaderboard in Karma I think... But then I ran out. I had to go back to my report, and then select a problem and then rate it... Thanks for the little graph there 007math! :lol:", "Solution_17": "What is karma? is 100 a lot? how do you rate problems?", "Solution_18": "Karma is the currency of Alcumus. You need to spend a little bit of karma to answer a question or watch an Alcumus video. \r\n\r\n100 karma is very little. If you only have 100 karma, that means you have really been spending a lot of karma and not earning any back. If you continue with 100 karma, you could only afford 20 more questions or Alcumus video views (since they cost 5 karma each). \r\n\r\nTo rate problems, go to the bottom of the solution screen; there, there are three questions you can choose a rating on. For each one of those you answer, you receive 5 karma. Thus, if you answer all three questions for each solution, you receive 15 karma. That means that if you rate all three questions for each solution and answer each question only once, you will end up with a net profit of 10 karma. This is the main way karma is gained. For more information, consult the instructions.", "Solution_19": "Oh ok yea i found out. So you have to rate the questions when you see the answer.", "Solution_20": "I only have a karma of 500", "Solution_21": "ok cool. try [url]http://www.artofproblemsolving.com/Alcumus/Instructions.php#karmapoints[/url]", "Solution_22": "Haha, I only have 700\r\nI think I got negative karma once...somehow", "Solution_23": "I don't like Karma. It's a good idea, but I don't like it because I usually forget to rate....", "Solution_24": "Please understand the great quality of Alcumus in that it is free; the administrators of AoPS created Alcumus as a service brought to you free of charge; thus, don't you think you should at least do them a favor by rating the questions so as to help them improve the site?", "Solution_25": "once you start rating problems, and keep it up, you will start rating problems naturally.", "Solution_26": "Yeah, is the difficulty of the questions affected by how you rate them?", "Solution_27": "Why bother using Karma as \"currency\" in AOPS? Most people will probably just do problems regardless of right or wrong, and just rate the problems....people will try to get Karma instead of trying to get the questions right. Why do you have Karma anyways? You really need to depend on a type of \"currency\" to get kids to study??? The point of Alcumus is to do questions and learn....while some people thin it is a goal to get the most Karma!!\r\n\r\n :ninja: :ninja: :ninja: :ninja: :ninja: :ninja: :ninja: :ninja: :ninja: :ninja:\r\n\r\nBTW, what do you use Karma on anyway? do you spend it on like a certificate or something? Because if you just use it to watch videos, or something, people won't want to watch videos anymore....they want to get RICH!\r\n\r\n\r\n\r\n\r\n ______________________________________________________________________________________________\r\n\r\nThe Math Kid", "Solution_28": "Let's see...I'll take it apart piece by piece for you. \r\n\r\n[quote=\"faush101\"]The point of Alcumus is to do questions and learn....while some people thin it is a goal to get the most Karma!![/quote]\n\nThose that think that way aren't going to be very successful, and most AoPSers would know that learning is the point. \n\n[quote=\"faush101\"]people will try to get Karma instead of trying to get the questions right[/quote]\n\nAbsolutely not true. Ask most any experienced AoPSer that frequently does Alcumus. It is a greater pleasure to get questions right than to get more Karma. \n\n[quote=\"faush101\"]Why do you have Karma anyways? [/quote]\n\nNote that Alcumus is free. The makers of Alcumus are just asking for a little in return from you in the form of feedback on the quality of the questions. That way, they know which questions are preferred by users, and will make more of those, which makes users happy, and also attracts more users, so that users would regard AoPS as a better learning tool (which it really is) and buy more AoPS material. It's a win-win situation.\n\n[quote=\"faush101\"]people won't want to watch videos anymore....they want to get RICH! [/quote]\r\n\r\nApparently, you do not know the \"people\" of AoPS. Please refrain from assumptions when you do not know what is actually going on. Not everybody is like you, and just because you think a certain way doesn't mean everyone else thinks along the same lines.", "Solution_29": "click the laughing dude!\r\n[hide=\" :rotfl: \"]then you cannot answer the questions, to get Kamara, you haz to rate the answers![/hide]", "Solution_30": "i have run out of Kamara, but that is cuz i didnt know about kamara, LOLZ!", "Solution_31": "We have a temporary maintenance right now, but we're able to watch videos, so do we still spend karma when we watch them?", "Solution_32": "I Dont think so because when you watch it youre not allowed to rate the video so you would not get any karma for that.", "Solution_33": "You can still post comments though, but I don't think you get awarded Karma for that. And think of this as your free opportunity to watch as many videos as you like without having to worry about karma at all! :)\r\n\r\nAlso, how come there are no videos on the Pigeonhole Principle in Alcumus?", "Solution_34": "[quote=\"maybach\"]Also, how come there are no videos on the Pigeonhole Principle in Alcumus?[/quote]\r\n\r\nSame reason with having no Pigeonhole Principle Alcumus problem, they haven't made any yet. :)", "Solution_35": "I think they might be making it right now", "Solution_36": "[quote=\"nsun48\"]I think they might be making it right now[/quote]\n\nProbably not, because\n\n[quote=\"DPatrick\"]We have nothing to add to what is already stated.[/quote]\n\n[quote=\"rrusczyk\"]We're thinking 1-2 more weeks. We will still have all your work from before, but will be using a different model to determine where you are. That will be the main initial change, with several others coming later in the fall (adding algebra, more counting problems in [b]existing topics[/b], and giving students more control over what topics they see).[/quote]\r\n\r\nSo we don't get to have additional topics.", "Solution_37": "But when Alcumus was actually up, under my report it said \r\n Answered Correctly Out of\r\nBasic Pigeonhole NA NA\r\n \r\nSo was there once Pigeonhole, and then they removed it or what?", "Solution_38": "Oh... :huh: Then maybe it's a glitch. I'm sure it will come back. I also had a time when there were Hockey Stick Identity videos available and soon disappeared, but after a while it came back.", "Solution_39": "[quote=\"maybach\"]But when Alcumus was actually up, under my report it said \n Answered Correctly Out of\nBasic Pigeonhole NA NA\n \nSo was there once Pigeonhole, and then they removed it or what?[/quote]\r\n\r\nNo, they inputted the category before they made the questions, and they still haven't made the questions yet.", "Solution_40": "i have a lot of karma!!!!! :starwars: :o", "Solution_41": "[quote=\"scfliu\"]i have a lot of karma!!!!! :starwars: :o[/quote]\r\n\r\nPlease do not revive very old topics, especially for irrelavent spam like this." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "incenter", "angle bisector", "geometry proposed" ], "Problem": "[color=darkred]Given $\\triangle ABC$, where $\\measuredangle\\ B=60^\\circ$. Angle bisector of $\\measuredangle\\ BAC$ meets $\\overline{BC}$ in point $D$. Point $K$ lies on $\\overline{AC}$ such that $\\measuredangle\\ ADK=30^\\circ$. Find $\\measuredangle\\ OKC$ where $O$ is the circumcenter of $\\triangle ADC$[/color]", "Solution_1": "Nice problem!\r\n\r\nDenote $\\angle{\\frac{A}{2}}=t$, and $\\angle{OKC}=x$, therefore after a simple angle chasing we obtain: \r\n$\\angle{OCK}=t-30, \\angle{CDK}=\\angle{CKD}=30+t$\r\nand also $\\angle{OKD}=30+t+x, \\angle{ODK}=2t-60$. \r\n\r\nNow:\r\n1) From the sin law in $\\triangle{OCK}$ we obtain: $\\frac{OC}{\\sin{x}}=\\frac{KC}{\\sin{(x+t-30)}}$.\r\n\r\n2) From the sin law in $\\triangle{OCD}$ we obtain: $\\frac{OC}{\\sin{(90-t)}}=\\frac{CD}{\\sin{2t}}$.\r\n\r\nBy dividing them and from $CD=CK$, we obtain:\r\n\r\n$\\sin{2t}\\cdot \\sin{x}= \\cos{t}\\cdot \\sin{(x+t-30)}$\r\n\r\n$\\iff \\sin{2t}\\cdot \\sin{x}= \\cos{t}\\cdot \\sin{(t+[x-30])}$\r\n\r\n$\\iff \\sin{2t}\\cdot \\sin{x}= \\cos{t}\\sin{t}\\cos{(x-30)}+\\cos^{2}{t}\\sin{(x-30)}$\r\n\r\n$\\iff \\sin{2t}\\cdot (\\sin{x}-\\frac{1}{2}\\cdot \\cos{(x-30)})=\\cos^{2}{t}\\cdot \\sin{(x-30)}$\r\n\r\n$\\iff \\sin{2t}\\cdot (\\sin{x}-\\sin{30}\\cdot \\cos{(x-30)}) =\\cos^{2}{t}\\cdot \\sin{(x-30)}$\r\n\r\n$\\iff 2\\sin{t}\\cdot (\\frac{\\sin{x}-\\sin{(60-x)}}{2}) = \\cos{t}\\cdot \\sin{(x-30)}$\r\n\r\n$\\iff 2\\sin{t}\\cdot \\sin{(x-30)}\\cdot \\cos{30}= \\cos{t}\\cdot \\sin{(x-30)}$.\r\n\r\nFrom here we have 2 cases: \r\n\r\n1) $\\sin{(x-30)}=0$, therefore $x=30^\\circ \\iff \\angle{OKC}=30^\\circ$.\r\n\r\n2) $\\tan{t}=\\frac{1}{\\sqrt{3}}\\iff A=60^\\circ$, therefore $ABC$ is equilateral, and then it remains us to check again, but this time the problem is trivial.\r\n\r\nTherefore the problem is solved.", "Solution_2": "[quote=\"\u30b3\u30ea\u30cd\u30a2\u30eb\"][color=darkred]Given $\\triangle ABC$, where $\\measuredangle\\ B=60^\\circ$. Angle bisector of $\\measuredangle\\ BAC$ meets $\\overline{BC}$ in point $D$. Point $K$ lies on $\\overline{AC}$ such that $\\measuredangle\\ ADK=30^\\circ$.\nFind $\\measuredangle\\ OKC$ where $O$ is the circumcenter of $\\triangle ADC$[/color][/quote]\r\n[color=darkblue][b][u]Proof.[/u][/b] Suppose that $A\\ne 60$. $A=2k\\ (k\\ne 30)\\ , m(\\widehat{OKC})=x\\Longrightarrow$ $\\left\\{\\begin{array}{c}m(\\widehat{CDK})=30+k\\ ,\\ m(\\widehat{OCK})=30-k\\\\\\ C=120-2k\\ ,\\ m(\\widehat{ADC})=60+k\\\\\\ b=2\\cdot OC\\cdot\\sin (60+k)\\end{array}\\right\\|$ $\\Longrightarrow$ $\\frac{KA}{KC}=$ $\\frac{DA}{DC}\\cdot\\frac{\\sin\\widehat{KDA}}{\\sin \\widehat{KDC}}=$ $\\frac{\\sin (120-2k)}{\\sin k}\\cdot \\frac{\\sin 30}{\\sin (30+k)}=$ ${\\frac{\\sin (60+2k)}{2\\sin k\\sin (30+k)}}=$ ${\\frac{\\cos (30+k)}{\\sin k}}$ $\\Longrightarrow$ $\\frac{KA}{\\sin (60-k)}=\\frac{KC}{\\sin k}=\\frac{b}{\\cos (30-k)}$ $\\Longrightarrow$ $\\boxed{KC=\\frac{b\\sin k}{\\cos (30-k)}}$. Therefore, $\\frac{\\sin (x+30-k)}{\\sin x}=$ $\\frac{\\sin \\widehat{COK}}{\\sin \\widehat{OKC}}=$ $\\frac{KC}{OC}=$ $\\frac{KC}{b}\\cdot 2\\sin (60+k)=$ $\\frac{2\\sin k\\cos (30-k)}{\\cos (30-k)}=2\\sin k$ $\\Longrightarrow$ $\\sin (x+30-k)=2\\sin k\\sin x$ $\\implies$ $\\sin (x+30-k)=\\cos (k-x)-\\cos (k+x)$ $\\implies$ $\\cos (k+x)=\\cos (x-k)-\\cos (k-x+60)$ $\\implies$ $\\cos (k+x)=2\\sin 30\\sin (k-x+30)$ $\\implies$ $\\cos (k+x)=\\cos (60+x-k)$ $\\implies$ $\\emptyset$. In conclusion, the triangle $ABC$ is equilateral !?[/color]", "Solution_3": "Yeah, it's a really nice problem.\r\n\r\nDoes anybody has a synthetic proof?", "Solution_4": "Indeed, it is a very nice problem, but pretty computational, i hope to see a synthetic solution, but i doubt, that it will be a very short and nice one :).\r\n\r\n@Virgil Nicula: i think you misstyped something somewhere at the beggining, but your method, similar to mine, is very straightforward.", "Solution_5": "[quote=\"pohoatza\"]... $\\angle{OCK}=t-30$ ...[/quote]\r\nPohoatza, it is wrong. Correctly, $m(\\widehat{OCK})=30-t$ !!!!!", "Solution_6": "[quote=\"pohoatza\"]... $\\angle{OCK}=t-30$ ...[/quote]\r\nPohoatza, it is wrong. In the my opinion, $m(\\widehat{OCK})=30-t$ is correctly !!!!!", "Solution_7": "Let M be the midpoint of CA, R' circumradius of $\\triangle ADC$ and usual notation for $\\triangle ABC.$ $\\angle ADC = \\frac{\\angle A}{2}+60^\\circ,$\r\n\r\n$|OM| = \\left|R' \\cos \\left(\\frac{A}{2}+60^\\circ\\right)\\right| = \\left|\\frac{b}{2 \\tan \\left(\\frac{A}{2}+60^\\circ\\right)}\\right| = \\frac{b}{2}\\left|\\frac{1-\\sqrt{3}\\tan \\frac{A}{2}}{\\sqrt{3}+\\tan \\frac{A}{2}}\\right|$\r\n\r\nSubstituting\r\n \r\n$\\tan \\frac{A}{2}= \\frac{r}{s-a}= \\frac{\\triangle}{s(s-a)}= \\frac{ca \\sqrt{3}}{(b+c)^{2}-a^{2}}$\r\n\r\n$|OM| = \\frac{b}{2 \\sqrt{3}}\\cdot \\left|\\frac{(b+c)^{2}-a^{2}-3ca}{(b+c)^{2}-a^{2}+ca}\\right| = \\frac{b}{2 \\sqrt{3}}\\left|1-\\frac{4ca}{(b+c)^{2}-a^{2}+ca}\\right|$\r\n\r\nSubstituting $ca = c^{2}+a^{2}-b^{2}$ in the denominator yields\r\n\r\n$|OM| = \\frac{b}{2 \\sqrt{3}}\\left|1-\\frac{2a}{b+c}\\right|$\r\n\r\n$\\angle CKD = \\frac{A}{2}+30^\\circ = \\angle CDK,$ $\\triangle CKD$ is isosceles with $CK = CD = \\frac{ab}{b+c},$\r\n\r\n$|KM| = \\left|\\frac{b}{2}-\\frac{ab}{b+c}\\right| = \\frac{b}{2}\\left|1-\\frac{2a}{b+c}\\right|$\r\n\r\n$\\tan{\\widehat{OKM}}= \\frac{|OM|}{|KM|}= \\frac{1}{\\sqrt{3}},\\ \\ \\ \\angle{OKM}= 30^\\circ.$\r\n\r\nIf c > b, then b > a, b + c > 2a, $\\frac{b}{2}-\\frac{ab}{b+c}> 0,$ and C, K, M, A follow on CA in this order. In addition, $\\angle A < 60^\\circ,$ $\\angle ADC < 90^\\circ$ and O is inside $\\triangle ADC.$ Then $\\angle OKC = 180^\\circ-\\angle OKM = 150^\\circ.$ On the other hand, if c < b, then b < a, ..., and C, M, K, A follow on CA in this order. In addition, $\\angle A > 60^\\circ,$ $\\angle ADC > 90^\\circ$ and O is outside $\\triangle ADC.$ Then $\\angle OKC = \\angle OKM = 30^\\circ.$ If c = b, then b = a, ..., $O \\equiv M \\equiv K$ are identical and $\\angle OKC$ undefined.", "Solution_8": "My solution:\r\nLet $I$ be the incenter of $ABC$ and $P$ be the point outside triangle $ABC$ s.t. $ACP$ is equilateral. We have $AKD \\sim AIB$ so $AKI \\sim ADI$ thus $\\angle AIK=60$. $\\angle AIC=120$ so points $A$, $I$, $C$, $P$ lie on a circle so $I$, $K$, $P$ are colinear. By angle chasing $IP$ is perpendicular to $OC$ and $CK$ is perpendicular to $OP$. Thus $K$ is orthocenter of $COP$ so $\\angle OKC=180-\\angle OPC=150$." } { "Tag": [ "geometry", "geometric transformation", "rotation", "3D geometry", "calculus", "sphere", "reflection" ], "Problem": "Find the area of the solid created by rotating Triangle ABC with A(0.0) B(9,0) C(9,40) first about the line x=9 and then the y-axis.", "Solution_1": "[quote=\"unimpossible\"]Find the area of the solid created by rotating Triangle ABC with A(0.0) B(9,0) C(9,40) first about the line x=9 and then the y-axis.[/quote]\r\n\r\nuhh i know how to do from just x=9.\r\n\r\nI dont what a rotation of a cone looks like (calculus?)", "Solution_2": "Correct me if I'm wrong, but rotating the cone around the x-axis would produce two cones stacked on top of each other, I think.\r\n\r\nEdit: Yes, it would, since the sphere formed by rotating the base is completely enveloped by the shape formed by rotating the rest of the cone.", "Solution_3": "[quote=\"uldivad9\"]Correct me if I'm wrong, but rotating the cone around the x-axis would produce two cones stacked on top of each other, I think.\n\nEdit: Yes, it would, since the sphere formed by rotating the base is completely enveloped by the shape formed by rotating the rest of the cone.[/quote]\r\n\r\nNo it woundt..thats if you directly reflect it across the x-axis\r\n\r\nA sphere would work because everything is equidistant from a point, but thats the not the case with a sphere.\r\n\r\nTry rotating it only 90 degrees around hte x-axis. Where you stopped, you would get another cone 90 degrees away from it, so if it were rotated 360 degrees, you would get like inifnity many cones.", "Solution_4": "if you rotate the figure about the line x=9, you get two cones that meet at the vertex." } { "Tag": [ "\\/closed" ], "Problem": "I seem to have made a mistake in my Online Classes enrollment...I thought I was signed up for the unevaluated enrollment, but I recently received an e-mail asking for my Midterm challenges...does this mean that even if you do not have the evaluated enrollment, you should still send in your Midterm challenge? Or did I make a mistake when I signed up for the courses?", "Solution_1": "Our records show that you do indeed have an evaluated enrollment in both the Counting and the Algebra classes." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Si a y b son enteros positivos que cumplen:\r\n$(\\sqrt{b+a^{3}}+29)a^{3}=b$\r\nDeterminar todas las soluciones $(a,b)$ de la ecuacion.", "Solution_1": "Este problema vino en el examen selectivo IBERO 1998 en PERU como problema # 4 .", "Solution_2": "[quote=\"DarthTenebrus\"]Este problema vino en el examen selectivo IBERO 1998 en PERU como problema # 4 .[/quote]\r\nsi, postea , mi pregunta es que si tiene o no solucion, ya q ami me salio de un paso :blush: y para comprovar que esta bien o no. gracias, espero soluciones. :D", "Solution_3": "hay 4 pares ordenados que resuelven dicha ecuacion", "Solution_4": "bueno ami mi me salio que la unica solucion es b=35, k=6, a=1, (a menos que me hayan faltado reemplazar otros valores admisibles) ;en el transcurso de la semana pondre la solucion :D", "Solution_5": "La primera vez que lo resolvi me salieron las 4 soluciones pero lo que no hice muy bien debido a mi inexperiencia de aquel entonces fue probar que no habian mas soluciones osea no \r\nsabia resolver muy bien las ecuaciones diofanticas \r\n\r\nEspero ver las 4 soluciones .", "Solution_6": "Si a y b son enteros positivos que cumplen: \r\n$(\\sqrt{a^{3}+b}+29)a^{3}=b$\r\nDeterminar las soluciones $(a;b)$ de la ecuacion.\r\n$Solucion$\r\nComo a y b son enteros positivos -> $b=ka^{3}$(k entero positivo) de aqui reemplazandolo en la ecuacion:$\\sqrt{a^{3}(k+1)}=k-29$ ....($k>29$), luego elevando al cuadrado a ambos miembros:$a^{3}(k+1)=k^{2}-2.29+29^{2}$->$0=k^{2}-(a^{3}+2.29)+(29^{2}-a^{3})$ de donde: $k=\\frac{a^{3}+58 \u00b1\\sqrt{a^{6}+120a^{3}}}{2}$ Ahora $k>29$ ->$\\frac{a^{3}+58 \u00b1\\sqrt{a^{6}+120a^{3}}}{2}>29$ ....$a^{3}\u00b1\\sqrt{a^{6}+120a^{3}}>0$ $...(f)$; como $a^{3}<\\sqrt{a^{6}+120a^{3}}$ entonces $(f)$ tan solo se cumple cuando el signo es positivo de donde: $k=\\frac{a^{3}+58+\\sqrt{a^{6}+120a^{3}}}{2}\\[: 147b3ff598....(i)$\r\nComo a y k son enteros ->$a^{6}+120a^{3}=n^{2}$...$a^{4}\\frac{(a^{3}+120)}{a}=n^{2}$->$\\frac{a^{3}+120}{a}=m^{2}$\r\nde donde: $120=a(m-a)(m+a)$, ahora sea $(m+a)=X$, $(m-a)=Y$ (X e Y enteros positivos)->$\\frac{X-Y}{2}=a$\r\nentonces: $120=\\frac{X-Y}{2}XY\\]: 147b3ff598...(ii)$; como $(X-Y)$(X-Y)<\\sqrt{120}$->$\\frac{X-Y}{2}<6$; ademas $\\frac{X-Y}{2}|120$-> $\\frac{X-Y}{2}=a$ puede tomar los valor: 1, 2, 3, 4 y 5\r\nEn $(ii)$::: con a=1, se tiene que X=12 y Y=10 (admite el valor de 1); Con a=2, X=10 y Y=6(admite 2); con a=3, X=10 y Y=4(admite 3); con a=4, no hay soluciones de X e Y en enteros(no admite); con a=5, X=12 y y=2\r\nAhora reemplazando los valores de a en $(i)$:\r\n$a=1$->$k=35$ de ahi $b=35$\r\n$a=2$->$k=49$ de ahi $b=49.8$\r\n$a=3$->$k=74$ de ahi $b=74.27$ \r\n$a=5$->$k=179$ de ahi $b=125.179$ \r\nEntonces solo hay 4 soluciones.\r\nCreo que ha sido algo extenso :P \r\n$Am\u00e9rico Arone$", "Solution_7": "Si a y b son enteros positivos que cumplen: \r\n$(\\sqrt{a^{3}+b}+29)a^{3}=b$\r\nte falto para $a=4$. aca la explicacion:D\r\ndando $b=k.a^{3}$, y $k=t^{2}-1$, donde $t$ pertenece a los enteros.\r\nreemplasando.\r\n$(\\sqrt{a^{3}(t^{2})}+29)a^{3}=a^{3}.(t^{2}-1)$\r\n$\\Rightarrow , a=4, 8t+29=t^{2}-1$\r\n$t^{2}-8t-30=0$, su discriminante, es igual a $64+120=184$, el cual no es un cuadrado perfecto, entonces, $t$, no pertenece a los enteros.\r\npor lo tanto, para $a=4$, no hay solucion.", "Solution_8": "[quote=\"aev5peru\"]Si a y b son enteros positivos que cumplen: \n$(\\sqrt{a^{3}+b}+29)a^{3}=b$\nte falto para $a=4$. aca la explicacion:D\ndando $b=k.a^{3}$, y $k=t^{2}-1$, donde $t$ pertenece a los enteros.\nreemplasando.\n$(\\sqrt{a^{3}(t^{2})}+29)a^{3}=a^{3}.(t^{2}-1)$\n$\\Rightarrow , a=4, 8t+29=t^{2}-1$\n$t^{2}-8t-30=0$, su discriminante, es igual a $64+120=184$, el cual no es un cuadrado perfecto, entonces, $t$, no pertenece a los enteros.\npor lo tanto, para $a=4$, no hay solucion.[/quote]\r\nCuando a=4 se tiene:\r\n$120=4XY$ $X-Y=2.4$ y $XY=30$, no hay solucion.\r\nAm\u00e9rico Arone :wink:", "Solution_9": "Veo que tiene un serio problema de teoria de numeros :\r\nporque para empezar no se puede aceptar que k es entero y ustedes lo afirman como si fuese obvio . Espero que sepan remendar ese error .", "Solution_10": "[quote=\"DarthTenebrus\"]Veo que tiene un serio problema de teoria de numeros :\nporque para empezar no se puede aceptar que k es entero y ustedes lo afirman como si fuese obvio . Espero que sepan remendar ese error .[/quote]\r\nhola DarthTenebrus ...me interesa mucho que postees los errores cometidos en la solucino, xfavor y tambien como poder corregirlo para soluciones posteriores...\r\nGracias :roll:" } { "Tag": [ "geometry", "parallelogram", "geometry proposed" ], "Problem": "#Let, $ ABCD$ be a cyclic quadrilateral. diangonal intersect at $ O$. Prove that centroids of $ \\triangle AOB$, $ \\triangle BOC$, $ \\triangle COD$, $ \\triangle AOD$ forms a parallelogram.\r\n\r\n#A quadrilateral $ ABCD$. the centroids of $ \\triangle ABD$, $ \\triangle ABC$, $ \\triangle BCD$, $ \\triangle ACD$ forms a quadrilateral $ A'B'C'D'$. prove: $ [A'B'C'D'] \\equal{} \\frac {[ABCD]}{9}$\r\next. for a cyclic quadrialteral $ A'B'C'D'$ is cyclic.", "Solution_1": "I am going to assume that you are taking about ''Centroids''\r\n\r\nto the first part. \r\n\r\nIt is not necessary that ABCD has to be cyclic. See that this quadrilateral is centrally similar to the parallelogram of the midpoint of the sides of ABCD.Center of similarity is the diagonal intersection O and the coefficent is 2/3." } { "Tag": [], "Problem": "Sa se determine functia $f : \\mathbb{N}\\rightarrow \\mathbb{N}$ astfel incat:\r\n\r\n$f(1)=1$\r\n$f(3)=3$\r\n$f(2n)=f(n)$\r\n$f(4n+1)=2f(2n+1)-f(n)$\r\n$f(4n+3)=3f(2n+1)-2f(n)$", "Solution_1": "scriind $n$ si $f(n)$ in baza 2 pt cateva valori ale lui $n$, se observa ca $f(n)$ inverseaza cifrele lui $n$: de exemplu, $f(110_{2}) = 011_{2}$. demonstram prin inductie. pt fiecare nr binar $x$, notam cu $x'$ nr-ul obtinut prin inversarea cifrelor lui $x$. pt. numere binare de o singura cifra sau de doua cifre, ipoteza se verifica usor. fie acuma $n$ un numar binar de cel putin 3 cifre, si sa presupunem ca pt orice $k$ de mai putine cifre ca $n$, $f(k) = k'$. daca $n = x0$ (un sir binar $x$ urmat de 0), atunci $f(n) = f(x0) = f(2\\cdot x) = f(x) = x' = n'$. daca $n = x01$, atunci $f(n)=f(x01)=f(4\\cdot x+1)=2f(x1)-f(x)=$$2\\cdot 1x'-x'=1x'+1x'-x' = 10x'=n'$. si daca $n = x11$, atunci $f(n) = f(x11) = f(4\\cdot x+3) = 3f(2\\cdot x+1)-2f(x) = 3f(x1)-2f(x) =$ $3\\cdot 1x'-2\\cdot x' = 1x'+1x'+1x'-x'-x' = 1x'+2\\cdot 10...0$ (unde sunt atatea 0'uri cat sunt cifre in $x'$) $= 1x'+10...0$ (unde sunt atatea 0'uri cat sunt cifre in $x'$, plus 1) $= 11x' = n'$" } { "Tag": [ "calculus", "inequalities", "analytic geometry", "graphing lines", "slope", "continued fraction", "real analysis" ], "Problem": "[size=125]I don't understand this part in my advanced calculus textbook:\n\nFor S = {m/n | 0 <= m/n and m\u00b2 < 2n\u00b2}\nit is almost obvious that the smallest upper bound for this set has to be \\sqrt {2} and hence outside [b]Q[/b], for if m/n is an element of S, then\nm/n < 2(m+n)/(m+2n) and {2(m+n)/(m+2n)}\u00b2 < 2...\n\nMy problem is: how do we derive the inequalities m/n < 2(m+n)/(m+2n) and {2(m+n)/(m+2n)}\u00b2 < 2? :huh: [/size]", "Solution_1": "[b]Lemma:[/b] Let $ a, b, c, d$ be positive integers. If $ \\frac {a}{b} < \\frac {c}{d}$, then $ \\frac {a}{b} < \\frac {a \\plus{} c}{b \\plus{} d} < \\frac {c}{d}$.\r\n\r\nProof: consider the slopes of the lines from the origin to the lattice points $ (b, a), (d, c), (b \\plus{} d, a \\plus{} c)$.\r\n\r\nThen $ \\frac {2m}{2n} < \\frac {2m \\plus{} 2n}{2n \\plus{} m} < \\frac {2n}{m}$. I do not have a similarly elegant argument for the other inequality (of this type), but I believe both inequalities are in some sense a consequence of the [url=http://en.wikipedia.org/wiki/Continued_fraction#Periodic_continued_fractions]continued fraction[/url] expansion of $ \\sqrt {2}$." } { "Tag": [], "Problem": "Determine a formula for \r\n\r\n$ 1^{2} \\minus{} 2^{2} \\plus{} 3^{2} \\minus{} 4^{2} \\plus{} ... \\plus{} ( \\minus{} 1)^{n \\plus{} 1} n^{2}$\r\n\r\nfor every possible integer $ n$.\r\n\r\n[hide=\"answer\"]\n$ (\\frac{n^{2} \\plus{} n}{2}) (\\minus{}1)^{n\\plus{}1}$\n[/hide]", "Solution_1": "If n is even then,\r\n1^2+3^2+...+n^2-(2^2+4^2+...+(n-1)^2)\r\n=(n(n+1)(2n+1))/6-4.2(((n-1)/2)((n+1)/2)n)/6)\r\n=(2n^3+3n^2+n)/6-(n^3-n)/3\r\n=3n^2-n/6.\r\nSimilarly we can get the results for odd n and the conclusion follows." } { "Tag": [ "trigonometry", "geometry", "similar triangles", "geometry proposed" ], "Problem": "Let $ABC$ be a triangle with $b\\neq c$. Points $D$ is the midpoint of $BC$ and let $E$ be the foot of angle $A$ bisector. In the exterior of the triangle we construct the similar triangles $AMB$ and $ANC$ . Prove:\r\na) $MN\\bot AD \\Longleftrightarrow MA \\bot AB$\r\nb) $MN\\bot AE \\Longleftrightarrow M,A,N$ are colinear.", "Solution_1": "Sorry, what is MN?", "Solution_2": "[img]8136[/img]", "Solution_3": "[quote=\"felixmann\"][color=darkred]Let $ABC$ be a triangle with $b\\neq c$. Denote the middlepoint $D$ of the side $[BC]$. In the exterior of the triangle $ABC$ construct the points $M$ and $N$ so that $\\widehat{BAM}\\equiv\\widehat{CAN}$ and $\\frac{AM}{c}=\\frac{AN}{b}$. Prove that $MN\\bot AD \\Longleftrightarrow AM \\bot AB$.[/color][/quote]\r\n[color=darkblue][u][b]Proof.[/b][/u] Denote $\\{\\begin{array}{c}m(\\widehat{BAM})=m(\\widehat{CAN})=x\\\\\\ m(\\widehat{DAB})=u\\ ,\\ m(\\widehat{DAC})=v\\end{array}$. Observe that\n\n$DB=DC\\Longleftrightarrow$ $\\frac{\\sin u}{\\sin v}=\\frac{b}{c}$ $\\Longleftrightarrow$ $\\frac{AM}{AN}=\\frac{\\sin v}{\\sin u}$ $\\Longleftrightarrow$ $\\frac{\\sin\\widehat{ANM}}{\\sin\\widehat{AMN}}=\\frac{\\sin v}{\\sin u}$.\n\nThus, $MN\\perp AD$ $\\Longleftrightarrow$ $\\{\\begin{array}{c}m(\\widehat{AMN})=90^{\\circ}-(x+u)\\\\\\ m(\\widehat{ANM})=90^{\\circ}-(x+v)\\end{array}\\|$ $\\Longleftrightarrow$\n\n$\\frac{\\cos (x+v)}{\\cos (x+u)}=\\frac{\\sin v}{\\sin u}$ $\\Longleftrightarrow$ $\\sin u\\cdot\\cos (x+v)=\\sin v\\cdot\\cos (x+u)$ $\\Longleftrightarrow$\n\n$\\sin (x+v-u)=\\sin (x+u-v)\\ \\ (u\\ne v)$. But $b\\ne c$ $\\Longleftrightarrow$ $u\\ne v$.\n\nTherefore, $(x+v-u)+(x+u-v)=\\pi$ $\\Longleftrightarrow$ $x=\\frac{\\pi}{2}$, i.e. $AM\\perp AB$.[/color]" } { "Tag": [ "probability", "MATHCOUNTS" ], "Problem": "A bowl contains only red marbles, blue marbles, and green marbles. The probablilty of selecting a red marble from the bowl is $\\frac{3}{13}$. The probablilty of selecting a blue marble is $\\frac{2}{5}$. There are fewer than 100 marbles in the bowl. What is the probability of selecting, at random and without replacement, a green marble and then a red marble from the bowl on the first two selections?\r\nEdit: sorry. :blush: I thought I was in the mathcounts forum. Can a mod move this? Or it this okay where it is?", "Solution_1": "[hide]\nObviously there are an integer number of marbles. The number of marbles must be a multiple of 13. It also must be a multiple of 5.\nSo the number of marbles must be a multiple of 65.\nSince our answer is under 100, $\\boxed{65}$[/hide]", "Solution_2": "[hide]There must be 65 marbles, because 65 is the only positive multiple of lcm(13,5) less than 100. $\\frac{3}{13}$ red marbles becomes $\\frac{15}{65}$, and $\\frac{2}{5}$ blue marbles becomes $\\frac{26}{65}$. 24 marbles remain, so there are $\\frac{24 \\cdot 15}{65 \\cdot 64}$, which simplifies to our answer, $\\boxed{\\frac{9}{104}}$[/hide]", "Solution_3": "[quote=\"diophantient\"][hide]\nObviously there are an integer number of marbles. The number of marbles must be a multiple of 13. It also must be a multiple of 5.\nSo the number of marbles must be a multiple of 65.\nSince our answer is under 100, $\\boxed{65}$[/hide][/quote]\r\nThe question wasn't for how many marbles there were. It asked for the probablity of pulling a green and then a red without replacement." } { "Tag": [], "Problem": "For example, does 3 to the power of -2 = -9 or positive 9?\r\n\r\nI've heard an explanation on negative exponents that I'm not sure is true. If it's an even negative exponent, then you'll get a positive answer, and if it's an odd negative exponent, you'll get a negative answer. Is that true?", "Solution_1": "Neither.\r\n$ 3^{\\minus{}2} \\equal{} \\frac{1}{3^2} \\equal{} \\frac{1}{9}$, with nothing to do with positive/negative. \r\nBasically, when you have a negative exponent, it's the reciprocal of the positive exponent of the same number.", "Solution_2": "$ x^{\\minus{}y}\\equal{}\\frac{1}{x^y}$\r\n$ \\therefore$ $ 3^{\\minus{}2}\\equal{}\\frac{1}{3^2}\\equal{}\\frac{1}{9}$", "Solution_3": "[quote]it's the reciprocal of the positive exponent of the same number.[/quote]\r\nSo it's the reciprocal of a positive 2? What do you mean by reciprocal?\r\n\r\n3 to the power of negative 2 turns into 1/9th?", "Solution_4": "[quote=\"leroyjenkens\"]3 to the power of negative 2 turns into 1/9th?[/quote]\r\n\r\nYes because if $ 3^2 \\cdot 3^{\\minus{}2} \\equal{} 3^{2 \\plus{} (\\minus{}2)} \\equal{} 1$ (by the rules of exponents), then $ 3^{\\minus{}2}$ must be equal to $ 1/9$.", "Solution_5": "[quote=\"gauss202\"][quote=\"leroyjenkens\"]3 to the power of negative 2 turns into 1/9th?[/quote]\n\nYes because if $ 3^2 \\cdot 3^{ \\minus{} 2} \\equal{} 3^{2 \\plus{} ( \\minus{} 2)} \\equal{} 1$ (by the rules of exponents), then $ 3^{ \\minus{} 2}$ must be equal to $ 1/9$.[/quote]\r\nOk thanks. One more question. Is -5^2= -25 or 25? I'm getting two different answers from different places. Is a negative base always negative regardless of the exponent?", "Solution_6": "By writing $ (\\minus{}5)^2$, you're taking the number -5, and squaring it. \r\n\r\nBy simply writing $ \\minus{}5^2$, you are implying that you are taking the negative of the square of 25, so the answer would be $ \\minus{}25$.\r\n\r\nMoved to Classroom Math.", "Solution_7": "The reciprocal means the number that, when multiplied by the original, equals 1. The easiest way to obtain the reciprocal is putting it as the denominator, and having one as the numerator.\r\n\r\ne.g:\r\n\r\nThe reciprocal of $ a$ equals $ \\frac{1}{a}$\r\n\r\nAlso, memorize the formulas given above; this will help avoid further confusion.", "Solution_8": "[quote=\"7h3.D3m0n.117\"]By writing $ ( \\minus{} 5)^2$, you're taking the number -5, and squaring it. \n\nBy simply writing $ \\minus{} 5^2$, you are implying that you are taking the negative of the square of 25, so the answer would be $ \\minus{} 25$.\n\nMoved to Classroom Math.[/quote]\nSo if it's written as -5^2 or whatever the exponent is, it's going to be a negative number as the answer. But if it's written as (-5)^2 or any even number will give you a positive answer and any odd number will give you a negative answer? \n[quote]The reciprocal means the number that, when multiplied by the original, equals 1. [/quote]\nMultiplied by the original what?\n[quote]The easiest way to obtain the reciprocal is putting it as the denominator, and having one as the numerator. [/quote]\r\nAnd I do that with every number that has a negative exponent?\r\n\r\n\r\nOh and I have one more exponent question. Say for example I have the equation (5 + 5)^2. I'm supposed to add the 5's together to make (10)^2, instead of making it (5^2 + 5^2), right? Doesn't that contradict the order of operation?", "Solution_9": "Sorry: The reciprocal of a number is what you multiply that number by to get 1. It's easier to think of it as $ \\frac{1}{a}$\r\n\r\nThe order of operations is not contradicted because it goes:\r\n\r\nParentheses\r\n\r\nExponents\r\n\r\nMultiplication/Division\r\n\r\nAddition/Subtraction\r\n\r\n\r\nExponents do not distribute.", "Solution_10": "[quote=\"fishythefish\"]Sorry: The reciprocal of a number is what you multiply that number by to get 1. It's easier to think of it as $ \\frac {1}{a}$[/quote]\nedit: Oh nevermind, I'm dumb. I edited cause I figured this out in my sleep. lol. I woke up thinking \"oh crap I hope he didn't respond yet\"\n\n[quote]The order of operations is not contradicted because it goes:\n\nParentheses\n\nExponents\n\nMultiplication/Division\n\nAddition/Subtraction\n\n\nExponents do not distribute.[/quote]\r\nOh yeah. I got mixed up because you can simplify the expressions.\r\nThanks.\r\n\r\nActually I have another question. This is in regards to the parentheses around the negative base.\r\n\r\nFor example -2^2(3). The answer to that will be -12. But if I had put it in parentheses, like this (-2)^2(3). It would be a positive 12 instead of a negative. So putting parentheses can change the answer, so putting parentheses needs to be done with caution. But that's not what I was taught. I wasn't taught that the parentheses around a single number made a difference like that.\r\nSo when do I know when I can put the parentheses around the number or not?\r\n\r\nI'll show you what I'm referring to.\r\nOn this page http://www.purplemath.com/modules/evaluate.htm the top question is a^2b a= -2, b= 3\r\nHe puts the answer like this (-2)^2(3). If I had done it, I wouldn't have put the 2 in parentheses and I would've got the wrong answer. How do I know when and when not to use parentheses in a situation like that?\r\nThanks.", "Solution_11": "someone please respond\r\nthanks", "Solution_12": "You put parentheses around numbers that are operated apon, so if you are substituting numbers into an equation, you should always use them.", "Solution_13": "[quote=\"jackdillon\"]You put parentheses around numbers that are operated apon, so if you are substituting numbers into an equation, you should always use them.[/quote]\r\nThanks for the response.\r\n\r\nSo I should always put parentheses around numbers that I substitute into an equation? Because as you can see from my example, the parentheses make the difference between having the right answer and the wrong answer.\r\nWhat I don't understand, really, is why I need to put parentheses around the first number.", "Solution_14": "parentheses are extremely important. parentheses are arguably as important as the order of operations. here are a few examples of what can happen if you dont pay attnetion to your parentheses...\r\n\r\n5-(3-2)=5-1=4 \r\n(5-3)-2=2-2=0;4 doesnt equal 0.\r\n\r\nalso, about the (5+5)^2 problem, theres this cool formula about a general (a+b)^2.\r\n\r\n$ (a\\plus{}b)^2\\equal{}a^2\\plus{}2ab\\plus{}b^2$\r\n\r\nso (3+4)^2=7^2=49obviously. but using the formula....let a=3 and b=4\r\n\r\n(3+4)^2=(3)^2+2(3)(4)+(4)^2=9+24+16=49.\r\nbut in general, unless you've had some exposure to algebra, try to follor order of operations instead of using shortcuts like this lol.\r\nyou could use this trick to evaluate 1001^2. 1001^2=(1000+1)^2=1000^2+2(1000)(1)+1^2=1,000,000+2,000+1=1002001 NEATO!!", "Solution_15": "Basically, the reason you put parentheses around the -2 was because of the confusion of whether the \"-\" applied to the whole quantity, or just the \"2\". If it was part of an expression, without the parentheses it would have been subtraction.\r\n\r\nBTW: (-2)^2(3)=(-2)^6, not (-2)(6).\r\n\r\n(-2)^2(3)=(-2)^6=64 :lol:", "Solution_16": "A note on the original question:\r\n\r\nThe motivation for defining negative exponents this way is so that a familiar law of exponents holds: in particular, we want $ a^{x \\plus{} y} \\equal{} a^x a^y$ even if one (or both) of $ x, y$ are negative. In order for this to hold, it is necessary that\r\n\r\n$ a^{x \\minus{} x} \\equal{} a^x a^{\\minus{}x} \\equal{} a^0 \\equal{} 1 \\implies$\r\n$ a^{\\minus{}x} \\equal{} \\frac{1}{a^x}$.", "Solution_17": "Thanks for the responses. I think Mathking misunderstood my problem. My problem is I don't know when I'm supposed to put parentheses around negative numbers. Like if I'm given the problem -2^2+8 for example. The original problem was A^2+B, but I filled in the blanks where the variables were. If I fill it in and put (-2)^2 for the beginning part instead of -2^2 for the beginning part, it changes the answer. (-2)^2 gives you a different answer than -2^2.\r\n\r\n[quote]#16Basically, the reason you put parentheses around the -2 was because of the confusion of whether the \"-\" applied to the whole quantity, or just the \"2\". If it was part of an expression, without the parentheses it would have been subtraction. \n[/quote]\nWhy would the minus sign in -2^2(3) apply to the whole thing and not just the 2 in front?\n[quote]BTW: (-2)^2(3)=(-2)^6, not (-2)(6). \n\n(-2)^2(3)=(-2)^6=64[/quote]\r\nThe 3 is a big 3, it's not part of the exponent.", "Solution_18": "If you want the 3 to be a \"big 3\", then consider starting to use LaTeX. Go to the guide in the Wiki for help.\r\n\r\nI'll leave the explaining about the negative sign to someone else. I'm having a little trouble making it this basic.", "Solution_19": "[quote=\"fishythefish\"]If you want the 3 to be a \"big 3\", then consider starting to use LaTeX. Go to the guide in the Wiki for help.\n\nI'll leave the explaining about the negative sign to someone else. I'm having a little trouble making it this basic.[/quote]\r\nI can't see why. I'm breaking it down by asking simple questions about a step I don't understand in the process. Each step has a reason for why it is how it is and if I don't understand that step, how else can I begin to understand it unless I ask?", "Solution_20": "[quote=\"leroyjenkens\"]My problem is I don't know when I'm supposed to put parentheses around negative numbers. [/quote]\r\n\r\nWhen in doubt, [b]always.[/b]", "Solution_21": "[quote=\"fishythefish\"]Exponents do not distribute.[/quote]\r\nHere's question #1 on a the multiple choice section of a high school contest that I'm responsible for:\r\n\r\n$ (a^{-2}+b^{-2})^{-1}$ equals to \r\n\r\n\\[ \\mathbf{(A)} \\ a^2b^2 \\quad \\& \r\n\\mathbf{(B)} \\ a^2+b^2 \\quad \\& \r\n\\mathbf{(C)} \\ \\frac{1}{a^2+b^2} \\quad \\& \r\n\\mathbf{(D)} \\ \\frac{a^2+b^2}{a^2b^2} \\quad \\& \r\n\\mathbf{(E)} \\ \\frac{a^2b^2}{a^2+b^2} \\]\r\nLet's just say that this wasn't even the easiest question on the test - there were people who got it wrong.", "Solution_22": "(E) is an right answer .It seems to be very easy ...", "Solution_23": "[quote=\"Kent Merryfield\"][quote=\"fishythefish\"]Exponents do not distribute.[/quote]\nHere's question #1 on a the multiple choice section of a high school contest that I'm responsible for:\n\n$ (a^{ - 2} + b^{ - 2})^{ - 1}$ equals to\n\\[ \\mathbf{(A)} \\ a^2b^2 \\quad \\& \\mathbf{(B)} \\ a^2 + b^2 \\quad \\& \\mathbf{(C)} \\ \\frac {1}{a^2 + b^2} \\quad \\& \\mathbf{(D)} \\ \\frac {a^2 + b^2}{a^2b^2} \\quad \\& \\mathbf{(E)} \\ \\frac {a^2b^2}{a^2 + b^2}\n\\]\nLet's just say that this wasn't even the easiest question on the test - there were people who got it wrong.[/quote]\nIt's not C?\n[quote]When in doubt, always.[/quote]\r\nSo always put the parentheses around the negative numbers. I don't need to do that for non-negatives though, right?", "Solution_24": "[quote=\"long14893\"](E) is an right answer .It seems to be very easy ...[/quote]\r\nI didn't say that it was supposed to be hard, but people did miss it. Yes, (E) is the correct answer. The most commonly given incorrect answer was (B).", "Solution_25": "In case some people are still having trouble with it, I'll break it down.\r\n\r\n[hide=\"Clickness\"]\n\nYou have $ (a^{\\minus{}2} \\plus{} b^{\\minus{}2})^{\\minus{}1}$.\n\nFirst, according to the Order of Operations, we do the operation inside of the parentheses. \n\nAt first, you may think to simplify the problem as $ \\frac{1}{a^2\\plus{}b^2}$. \n\nHowever, this is incorrect, as you add the terms together and they cannot be grouped like that. You do them individually and then use a common denominator to group them together. \n\n[u]Therefore, the process then goes like this:[/u]\nFirst, you simplify each exponent. \n$ \\frac{1}{a^2} \\plus{} \\frac{1}{b^2}$\n\nThe common denominator is fairly easy to find. You simply multiply both of the denominators to get $ a^2b^2$\n\nMultiply to get the common denominators on both. It now looks like this:\n$ \\frac{b^2}{a^2b^2} \\plus{} \\frac{a^2}{a^2b^2}$\n\nNow we add:\n$ \\frac{a^2\\plus{}b^2}{a^2b^2}$\n\nWe're not done yet. Remember that $ \\minus{}1$ exponent? Now we flip what we currently have for the final answer of: $ \\boxed {\\frac{a^2b^2}{a^2\\plus{}b^2}}$ or $ \\boxed{(E)}$ [/hide]\r\n\r\nFairly simply once you know what you're doing. Just remember the basics. Small mistakes could easily get you an answer of $ \\boxed{(B)}$ or something else. It never hurts to double-check your work. :) Hope that helped.", "Solution_26": "@leroy: Just remember that subtracting is the same as adding a negative. Now see if you can figure out why this would cause confusion if there were no parentheses.\r\n\r\nTake this as an example:\r\n\r\n$ 3\\minus{}(\\minus{}2)^2$\r\n\r\nNow, work that out. The answer is -1, right? That is because the parentheses say \"treat this as a whole quantity\". In this case, the parentheses are telling you to square the -2, sign and all. That's what parentheses are for. They tell you to perform all operations on the quantity as a whole.\r\n\r\nAnother way of expressing this would be:\r\n\r\n$ X\\equal{}\\minus{}2 \\rightarrow$\r\n$ 3\\minus{}X^2$\r\n\r\nCompare this to:\r\n\r\n$ 3\\minus{}\\minus{}2^2$\r\n\r\nIt's the same thing without the parentheses.\r\nBecause the parentheses are gone, this is not interpreted as \"3 minus -2 squared\", it's interpreted as 3 minus the negative of $ 2^2$. Now work out the new answer. It's completely different. (Remember, subtracting a negative means adding the positive. e.g. 2--2=2+2=4)\r\nBasically, due to the order of operations, anytime you need to perform any operations on parentheses, you need to perform that operation on the quantity inside.", "Solution_27": "Thanks for the response, fishy.\r\nMy problem is that with your example, the -2 is right next to the minus sign on the right side (which means the two minus signs are right next to each other), so I was told always to put parentheses around negative numbers to keep negative signs seperate from other signs. Like in your example of 3 - (-2). But for instance if the 2 comes first in the equation, ie -2 - 3, is it still necessary to put the parentheses?\r\nWhich brings me to my next point. If that 2 has an exponent, parentheses around that -2 would cause the answer to be different than if it didn't have parentheses.\r\n\r\nFor a short question and answer, should I just put parenthses around all negative numbers? Because I'm wondering if there would ever be an instance in which the negative number isn't supposed to be in parentheses.\r\n\r\nHope all this makes sense. Thanks.", "Solution_28": "Well, generally, when you have 3--2. (3 minus a negative two), you don't necessarily [i]need[/i] parentheses. However, some schools tell kids to put parentheses around the negative 2, just for clarity. You don't need them. That eliminates one problem right there. You see, when you put the negative two in front, it eliminates all this confusion. :)\r\n\r\nIf you really still need the parentheses, just think them or something else in stead. Learn to diifferentiate. Since you don't need them, it would be an added burden of constantly adding them. Break that bad habit. :lol:", "Solution_29": "for your case, 3 - -2 does not have to be changed to 3 - (-2), it just makes it easier to visualize that way. If it was the negative of 2^2, just do\r\n\r\n3 - (-2^2)", "Solution_30": "Yeah, a lot of people get confused with the negtive signs, but just remember, parenthesis around a number means that whole term squared or cubed, etc.. \r\ni.e. (-2)^2 means negative 2 squared as opposed to -2^2 which means the negative of two squared aka -4", "Solution_31": "[quote]Well, generally, when you have 3--2. (3 minus a negative two), you don't necessarily need parentheses. However, some schools tell kids to put parentheses around the negative 2, just for clarity. You don't need them. That eliminates one problem right there. You see, when you put the negative two in front, it eliminates all this confusion.[/quote]\nThis is what I was confused about because...\n\n(-2)^2= 4\nBut\n-2^2= -4\nThe parentheses are the difference between the right and wrong answer. See how one results in negative 4 and the other in positive 4. So my original question was just asking how do I know when to put the parentheses and when not to? If the answer is always, then I'm cool with that.\n[quote]i.e. (-2)^2 means negative 2 squared as opposed to -2^2 which means the negative of two squared aka -4[/quote]\r\nExactly.\r\nSo if I'm substituting variables, like this A = 2\r\nAnd the question is -A^a, when I substitute the 2 for the A, I need to put parentheses around that negative 2, right?\r\nOr what if the question wants me to get a negative 4? Which means I'm not supposed to put parentheses? That's where I'm confused. How do I know which answer the question is looking for?", "Solution_32": "Review PEMDAS if you're not sure.\r\n\r\n[url]http://en.wikipedia.org/wiki/Order_of_operations#Examples_from_arithmetic[/url]\r\n\r\n[quote]5. Evaluate negation on the same level as subtraction, starting from the left[/quote]\r\n\r\nThe question has been answered. Be confident in the answer!", "Solution_33": "[quote=\"leroyjenkens\"]\n[quote]i.e. (-2)^2 means negative 2 squared as opposed to -2^2 which means the negative of two squared aka -4[/quote]\nExactly.\nSo if I'm substituting variables, like this A = 2\nAnd the question is -A^a, when I substitute the 2 for the A, I need to put parentheses around that negative 2, right?\nOr what if the question wants me to get a negative 4? Which means I'm not supposed to put parentheses? That's where I'm confused. How do I know which answer the question is looking for?[/quote]\r\n\r\nbasically assume as if the whole number is squared when you insert\r\n\r\nso with your question. You insert 2 into the equation.\r\n\r\nSo [b]only[/b] the two is squared, since that is what A is.\r\n\r\nso -2^2=-4\r\n\r\nTry some others out.\r\n\r\n-b^2 b=3\r\n-b^2 b=-2\r\n-b^2 b=-5\r\n-b^2 b=4", "Solution_34": "[quote]basically assume as if the whole number is squared when you insert \n\nso with your question. You insert 2 into the equation. \n\nSo only the two is squared, since that is what A is. \n\nso -2^2=-4 \n\nTry some others out. \n\n-b^2 b=3 \n-b^2 b=-2 \n-b^2 b=-5 \n-b^2 b=4[/quote]\r\nOh, duh. Thanks.\r\n\r\nSo had it been A= -2, the answer would be 4 instead of -4.\r\n\r\nFor the ones you gave me\r\n-b^2 b=3 The answer is -9 because the negative stays there while you insert the positive 3 and square it, so the answer becomes negative.\r\n\r\n-b^2 b=-2 The answer I believe is -4. I insert the -2, square it, that becomes a positive 4, then the negative sign that was there before makes it become a negative 4.\r\n\r\n-b^2 b=-5 Same story as above, but the answer is -25.\r\n\r\n-b^2 b=4 As I did in the first one, the 4 is positive which I insert, but the negative stays there and makes the answer become a negative number, which is -16.\r\n\r\nSo are my answers right and is my reasoning correct?\r\n\r\nThanks.", "Solution_35": "yep, it's all right.\r\n\r\njust a bit more to make sure.\r\n\r\n-b^3 b=7\r\nb^2 b=5\r\n-b^3 b=-6\r\n(-b)^2 b=4\r\n-b^2 b=2\r\n(-b)^3 b=-5\r\n(-b)^3 b=3\r\nb^3 b=6\r\n(-b)^2 b=-6", "Solution_36": "[quote=\"Ihatepie\"]yep, it's all right.\n\njust a bit more to make sure.\n\n-b^3 b=7\nb^2 b=5\n-b^3 b=-6\n(-b)^2 b=4\n-b^2 b=2\n(-b)^3 b=-5\n(-b)^3 b=3\nb^3 b=6\n(-b)^2 b=-6[/quote]\r\n\r\nAlright let's see here.\r\n\r\n1) -343\r\n2) 25\r\n3) 216\r\n4) 16\r\n5) -4\r\n6) 125\r\n7) -27\r\n8) 216\r\n9) 36 <- Not so sure about this one. I don't know if I square the -6 and then add the - to it after I get my answer of positive 36, or does the -6 automatically become a positive 6 once I put it in the quotations with that other minus sign?", "Solution_37": "[quote=\"leroyjenkens\"]I don't know if I square the -6 and then add the - to it after I get my answer of positive 36, or does the -6 automatically become a positive 6 once I put it in the quotations with that other minus sign?[/quote]\r\n\r\nWhere are the parentheses?", "Solution_38": "well you are right with 1-8.\r\n\r\nI'll let you figure out if you are right for number 9 yourself. (read the post above.)\r\n\r\nRemember the order of operations.", "Solution_39": "First I do what's in the parentheses, which is make the negative of a negative a positive and then I do the exponents?", "Solution_40": "yep, so you change -6 to 6 by taking it's negative, and you square it.", "Solution_41": "[quote=\"Ihatepie\"]yep, so you change -6 to 6 by taking it's negative, and you square it.[/quote]\r\n\r\nCool, thanks for all your help." } { "Tag": [ "ratio" ], "Problem": "The students in Mrs. Reed's English class are reading the same\n760-page novel. Three friends, Alice, Bob and Chandra, are in the\nclass. Alice reads a page in 20 seconds, Bob reads a page in 45\nseconds and Chandra reads a page in 30 seconds. \\\\ \\\\\nChandra and Bob, who each have a copy of the book, decide that they\ncan save time by `team reading' the novel. In this scheme, Chandra\nwill read from page 1 to a certain page and Bob will read from the\nnext page through page 760, finishing the book. When they are\nthrough they will tell each other about the part they read. What is\nthe last page that Chandra should read so that she and Bob spend the\nsame amount of time reading the novel?", "Solution_1": "Since the problem only mentions Bob and Chandra, we can kick Alice out of the reading club for now. Their reading ratios are 2:3, so Chandra will read 3/2 times as many pages as Bob, so Chandra should read $ \\frac{3}{5}\\cdot 760\\equal{}456$ pages. Pages 456 happens to be the 456th page in the book, so Chandra will stop on page 456. Now, we can let Alice back in.", "Solution_2": "3/5*760=$ \\boxed{456}$", "Solution_3": "Quit spamming. This was already answered. Thank you.", "Solution_4": "[quote] Chandra will read 3/2 times as many pages as Bob[/quote]\r\n\r\nhow do you know that", "Solution_5": "Their reading ratios are $ 2$ to $ 3$, so she reads $ \\frac{3}{2}$ as much as he does.", "Solution_6": "Thanks!!!!!", "Solution_7": "No Problem. :)" } { "Tag": [ "function", "number theory proposed", "number theory" ], "Problem": "Let $n$ be a positive integer which ends with exactly $k$ zeros but $n>10^k$ holds.\r\nFor an integer $n$, only $k=k(n)\\geq 2$ is known. At least how many ways (as function of $k=k(n)$ ) can $n$ be written as the difference of two perfect squares?", "Solution_1": "The answer is (k^2+k-2)/2." } { "Tag": [ "trigonometry", "algebra", "binomial theorem", "combinatorics unsolved", "combinatorics" ], "Problem": "Compute the sum: $ S_n\\equal{}\\displaystyle\\sum_{k\\equal{}0}^{n}q^k \\binom {n}{k} \\cos (a\\plus{}kx),$ where $ a,q,x$ are real numbers.", "Solution_1": "I would approach this as follows.\r\n\r\n$ S_n \\equal{} cos(a) \\sum_{k \\equal{} 0}^{n} q^k {n\\choose k} cos(kx) \\minus{} sin(a) \\sum_{k \\equal{} 0}^{n} q^k {n\\choose k} sin(kx)$\r\n\r\nFurther, one can evaluate using binomial theorem with $ sin (a) \\equal{} \\frac {e^{ika} \\minus{} e^{ \\minus{} kia}}{2i}$ and $ cos(a) \\equal{} \\frac {e^{ika} \\plus{} e^{ \\minus{} kia}}{2}$\r\n\r\nDo we have a combinatorial interpretation ?" } { "Tag": [ "AoPS Books" ], "Problem": "i am thinking about getting one of the AoPS books (ex, vol 1), which one do your reccomend?\r\n\r\nand whats the difference between the 11$one and the 30$ one?, (it says solutions text) does that mean that one has solutions and one doesnt?", "Solution_1": "The solution manual and the text are two different books. The solution manual contains full solutions (not just answers) to all the questions in the text. Almost everyone who buys the text for their own individual work buys the solutions as well." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Hi\r\nI should have 3-4 way of solving this problem.\r\nI appriciate everyonr who help me\r\nthank you.\r\n\r\n\r\n(sorry for bad english)", "Solution_1": "What is the problem?\r\nAnyway, it is known that any ordinary (ruler and compass) construction can be performed with compass only (Mascheroni's theorem).", "Solution_2": "I know. \r\nbut i'm looking for diffrent ways of solving the problem not the mascheroni's way. \r\ni'm really need help!!!thank you :) \r\nthanks.", "Solution_3": "[quote=\"shinytick\"]I know. \nbut i'm looking for diffrent ways of solving the problem [/quote]\r\n\r\nAny problem ? Or do you have some specific problem that you're referring to ?", "Solution_4": "khoobi :!: :!: :?: :!: :?: \r\nina migan masalehat koo :?: :?: \r\nwhat and where is your problem :?: :?: :?: :?: :?: :mad:", "Solution_5": "the problem is just this:\r\nHow can we carry out the constructions with a compass only:\r\nthis means:\r\n1-finding the intersection of two lines\r\n2-foinding the intersection of a line and circle\r\n\r\nand i know the macheroni's way for solving the problem.i;m looking for new and diffrent ways of solving this problem.\r\n\r\n\r\n\r\n\r\nkheyli mamnoon man motavajjeh shodam ina chi migan ina nefimahman man chi migam.", "Solution_6": ":rotfl: :rotfl: :rotfl: Sorry but let me just Speak persain:\r\nkheli bahal bod.Shoma masalaro matrah nakardi chon fekr kardid az mozoye masale mifahman :?: :?: \r\nBe nazaret ye zare ajib nist ke begi :\r\n\r\n[quote]Hi \nI should have 3-4 way of solving this problem. \nI appriciate everyonr who help me \nthank you. \n[/quote] bad nagi masale chiye ? ;) ;) ;) \r\nvali rast migi , moshkele in hast , khob nemifahman. :D\r\n\r\nPS. aga in alamataro ( :rotfl: ) jedi nagir , shokhiye :) :) your sincerely", "Solution_7": "Help???!!!! :? :o" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "We choose 5 points $A_1, A_2, \\ldots, A_5$ on a circumference of radius 1 and centre $O.$ $P$ is a point inside the circle. Denote $Q_i$ as the intersection of $A_iA_{i+2}$ and $A_{i+1}P$, where $A_7 = A_2$ and $A_6 = A_1.$ Let $OQ_i = d_i, i = 1,2, \\ldots, 5.$ Find the product $\\prod^5_{i=1} A_iQ_i$ in terms of $d_i.$", "Solution_1": "We have $ \\frac {\\overline{A_iQ_i}}{\\overline{Q_iA_{i + 2}}} = \\frac {\\bigtriangleup A_iPA_{i + 1}}{\\bigtriangleup A_{i + 2}PA_{i + 1}},i = 1,2,3,4,5$\r\n\r\nMutiply all these five equations and we can get $ \\prod_{i = 1}^{5}\\frac {\\overline{A_iQ_i}}{\\overline{Q_iA_{i + 2}}} = 1$ ..........(i)\r\n\r\nHowever, suppose $ \\overrightarrow{A_iO}$ meets the circle at $ {A}'_i$, then the circle power of $ Q_i = \\overline{A_{i + 1}Q_i}\\cdot\\overline{Q_i{A}'_{i + 1}} = (1 + d_{i + 1})(1 - d_{i + 1}) = \\overline{A_iQ_i}\\cdot\\overline{Q_iA_{i + 2}}$ \r\n\r\nThat is ${ \\overline{A_iQ_i}\\cdot\\overline{Q_iA_{i + 2}} = (1 - d_i^2}),i = 1,2,3,4,5$\r\nMutiply all these five equations and we can get ${ \\prod_{i = 1}^{5}\\overline{A_iQ_i}\\cdot\\overline{Q_iA_{i + 2}} = (1 - d_i^2})$ .........(ii)\r\n\r\n$ \\sqrt {(i)\\times (ii)}\\Rightarrow \\prod_{i = 1}^{5}{\\overline{A_iQ_i}} = \\prod_{i = 1}^{5}\\sqrt{(1 - d_i^2)}$ \r\n\r\nQED#" } { "Tag": [], "Problem": "Can someone help me come up with the right formula to calculate the number of subscriptions that we'd be getting paid for on a given month k given the following assumptions:\r\n\r\n Here are my assumptions:\r\n\r\n 1. Assume 250 users sign up to use a subscription-based service during\r\n the 1st month and pay for their subscriptions.\r\n 2. Monthly subscription is automatically renewed unless the user cancels.\r\n 3. Assume that 20% of them (50) will drop out, i.e. their membership will not be renewed during the second month, so only 200 previous subscribers stay on.\r\n\r\n 4. Assume that 12% more subscribers start using our service during the\r\n second month as compared to the first month\u2013 which means that we will\r\n have 280 (250 * 1.12) new subscriptions during that second month.\r\n 5. Add to that the 200 users that stayed on from the previous month (see\r\n #3) and get 480.\r\n 6. The same pattern continues repeating itself, where the 20% of the previous month's new subscribers drop out during a given month and 12% more sign up each month than have signed up the previous month.\r\n\r\n What is the formula to calculate the number of subscriptions that we will be getting paid on a given month k?", "Solution_1": "[hide=\"Solution\"]\n\nEach month, there will be 80% of the subscribers from the last month; there will also be 112% new subscribers, so the number of subscribers is 192% of the old number of subscribtions. Then the number of subscribers after $k$ months is $250\\times (1.92)^{k}$.[/hide]", "Solution_2": "That formula can't be the right one because it seems to grow too fast.", "Solution_3": "No, I think it works.\r\nYou can check the numbers for the first few months if you want.", "Solution_4": "If you plug in 10, the answer is 170,197.1548\r\nIt's highly unlikely that a business can go from 250 subscriptions to 170,197 in the span of 10 months.", "Solution_5": "Perhaps it's your assumptions that are wrong, then. Why did you multiply by $1.12$, and then add on the users from the previous month again?", "Solution_6": "It intuitively seems to make sense to me that if there's automatic subscription renewal, then 80% of the previous month's subscribers will stay on during the present month.\r\nThe only question then is at what rate the amount of new subscribers grow. It seems logical that a business (through advertising, etc.) can attract 12% more people than it attracted the previous month.\r\nWhat do you think?", "Solution_7": "The problem is customer base. If I own a store and I have 12% subscriber growth then I get (say) 100 customers this month, 112 the next, 125 the third month, etc. This means perhaps 20% customer loss rate and 32% customer gain rate. However, in this example, we have a 20% loss and 112% gain rate. In other words, If I have X subscribers this month I will gain more than X new people over the month.\r\n\r\nNew problem: using the store data from this post, at what month will less then 1% of the current customers be among the originals?", "Solution_8": "[b]Bill The Maniac:[/b] I believe the 112% growth refers to the number of [i]new[/i] subscribers, not the number of [i]total[/i] subscribers. \r\n\r\nThe assumptions notwithstanding, therefore, samath's approach is quite off, but the whole statement is quite confusing.\r\n\r\n[quote=\"mathman25\"]the 20% of the previous month's [i]new[/i] subscribers drop out during a given month and [i]12% more[/i] sign up each month than have signed up the previous month.[/quote]\r\n\r\nSo, let me see if I get this straight. In a given month, some number of new subscribers join, which is 12% more than the number of new subscribers that joined last month (so $1.12$ times as many new subscribers). But then 20% of the subscribers from [i]only the previous month[/i] (and [i]not[/i] any of the other months) drop out?\r\n\r\nThe way it's worded here, if a subscriber does not drop out within a month, then they are assumed to stay forever. Was that your intention?", "Solution_9": "My solution matches exactly with the problem :P", "Solution_10": "We all know that. However, the information given in the problem is a little odd, giving a 92% growth rate. My growth rate converges to 12% in my alternate interpretation.", "Solution_11": "The 112% growth definitely refers to the number of new subscribers, not the number of total subscribers.\r\n\r\nI'm trying to say the following: the amount of people paying for month k subscriptions are from two categories:\r\n1) 80% of the people who paid last month (k-1)\r\n2) The new subscribers. The amount of [i]new[/i] subscribers in month k are determined by the amount of [i]new[/i] subscribers in month (k-1) * 1.12\r\n\r\nSo if month 1 has 100 subscribers. Then 20 of these drop out by month 2. Also, 112 [i]new [/i]subscribers sign up in month 2. So the total paying subscribers in month 2 is 112+80=192. The 80% of 192 stay on for month 3 = 153.6 (I'm not rounding). The amount of [i]new[/i] subscribers in month 3 is 112 * 1.12 = 125.44. The total paying subscribers in month 3 is 125.44+153.6=279.04. etc.", "Solution_12": "In that case, it's my formula that should work, not 1.92^i.", "Solution_13": "Bill the Maniac:\r\nWhat was your formula?", "Solution_14": "[quote=\"mathman25\"]1) 80% of the people who paid last month (k-1)[/quote]\r\n\r\nOkay, that's what I thought. And that isn't what you said before.\r\n\r\n[hide=\"Solution\"] We have the recursion\n\n$s_{k+1}= 0.8s_{k}+250(1.12)^{k}$\n\nWhere $s_{1}= 250$. Clearly $s_{k}$ is in the form $A \\cdot 0.8^{k}+B \\cdot 1.12^{k}$. Substituing and simplfying,\n\n$1.12B = 0.8B+250 \\Leftrightarrow$\n$B = 781.25$\n\nThen $s_{1}= 0.8A+1.12B = 250 \\Leftrightarrow$\n$A =-781.25$\n\nSo we have\n\n$s_{k}= \\boxed{-781.25 \\cdot 0.8^{k}+781.25 \\cdot 1.12^{k}}$[/hide]\r\nThe number of subscribers in month $10$ is then $\\approx 2343$, much more reasonable.", "Solution_15": "I must have posted it somewhere else. *looks for it* *can't find it*\r\n\r\nT_i=N_0(1.12^(i-1))(1-K^i)/(1-K) where\r\n\r\nT_i=ith total number of customers\r\nN_0=250\r\nK=.8/1.12", "Solution_16": "Good, our answers are equivalent.", "Solution_17": "Thank you for your help!" } { "Tag": [ "geometry", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Find the area of the minor segment cut off from $ r \\equal{} a(1 \\plus{} cos \\theta)$ by the half line $ \\theta \\equal{}\\pi/2$.\r\n\r\nI did it like,\r\n $ \\frac {1}{2}\\int r^2 d \\theta \\equal{} \\frac {a^2}{2}\\{\\frac {3}{2} \\theta \\plus{} 2 sin \\theta \\plus{} \\frac {sin 2 \\theta}{4}\\}$\r\n\r\nBut I can't get the limits of this, means $ \\theta$ where to where.", "Solution_1": "By \"minor segment\" I'm assuming they mean the bump that occurs from $ \\frac{\\pi}{2}$ to $ \\pi$. (Try sketching it, that is the only thing that makes sense to me)." } { "Tag": [ "articles" ], "Problem": "I need an advice. I need a program to make any kind of animations ( i saw some on Art of problem solving).\r\nI'm writing an scientific article and i need it.", "Solution_1": "Macromedia Flash?", "Solution_2": "Indeed, all of our [url=http://www.artofproblemsolving.com/Gallery/AoPS_G_Gallery.php]animations[/url] are written using [url=http://www.macromedia.com/software/flash/flashpro/?promoid=BINT]Macromedia Flash[/url].", "Solution_3": "Thank you very much :)" } { "Tag": [ "USAMTS", "email" ], "Problem": "I sent my solutions to USAMTS and got it postmarked at 8:00 AM Oct. 13, 2009\r\n\r\nI see a check on the \"My USAMTS\" page, so my solutions were received.\r\n\r\nHowever, I am a little worried. No problems have been graded yet.\r\n\r\nJust wondering, usually, how long does the grading take?", "Solution_1": "nobody has been graded yet. scores aren't released until NOV.", "Solution_2": "The scores should be available in the second week of November.", "Solution_3": "Are they all uploaded at the same time? Or do you post the scores as you grade them?", "Solution_4": "They will all be posted at the same time. Not just scores, but commentary as well.", "Solution_5": "I sent in my solution by mail before the day it is due, but I haven't even received the check mark yet... is there something wrong?", "Solution_6": "[quote=\"mathwizarddude\"]I sent in my solution by mail before the day it is due, but I haven't even received the check mark yet... is there something wrong?[/quote]Please email with usamts@usamts.org with the details, and include your USAMTS ID#.\r\n\r\nTo our knowledge, all of the round 1 solutions that we have received have been logged in, but we did get a small handful with no ID# on them.", "Solution_7": "Hmm, I'm pretty sure I have my ID number on every page of the solution... so could there be another reason for that? My USAMTS ID number is 11825, so have the staff received the solution with this ID number? Thanks.", "Solution_8": "We have received it. Although it was mailed on time, we only received it this weekend.", "Solution_9": "Update: due to the unexpectedly large number of participants this year (up over 20% from last year), the grading is taking longer than expected. We are aiming for the week of Nov 16-20 for the Round 1 scores and comments to be available.\r\n\r\nPlease keep in mind that one of the unique features of the USAMTS is that you get written comments to your solutions, not just numeric scores, and this takes time." } { "Tag": [ "probability", "geometry", "3D geometry" ], "Problem": "You are blindfolded, and there are $6$ cubes in front of you. Of these $6$ cubes, $3$ are red, $2$ are blue, and $1$ is yellow. If you pick four cubes at random, what is $m^n$ if $\\frac{m}{n}$ is the probability that you will obtain $2$ red cubes and $2$ blue cubes?\r\n\r\n\r\n[hide=\"my solution\"]$\\frac{m}{n} = \\frac{5*3*2*2!}{30} = \\frac{1}{6}$\n$m^n = 1$[/hide]", "Solution_1": "[hide=\"Err...\"]\nThere are $\\frac{4!}{2!2!} = 6$ different ways to pick $2$ red and $2$ blue, and the odds of any one of these configurations is $\\frac{ 3 \\times 2 \\times 2 \\times 1}{6 \\times 5 \\times 4 \\times 3} = \\frac{1}{30}$, so the probability is $6 \\left( \\frac{1}{30} \\right) = \\frac{1}{5}$, so $m^n = 1^5 = \\boxed{1}$...?\n\nThis seems like a very bad problem because $m^n = 1$ whenever $m = 1$ so it is very possible to get the wrong value of $n$ and still get the problem right.\n\nAlso, it seems too easy for Intermediate.\n[/hide]\r\n\r\nEdit: Your solution seems wrong. It seems like the $5$ term is for picking either red or blue, but as there aren't an equal number of reds and blues this way of counting will result in an incorrect answer later on.\r\n\r\nEdit: Furthermore, $\\frac{5 \\times 3 \\times 2 \\times 2!}{30} = 2$... :huh: however, if you used $\\frac{5 \\times 3 \\times 2 \\times 2!}{6 \\times 5 \\times 4 \\times 3}$ then you would get $\\frac{1}{6}$, which still disagrees with what we have gotten...", "Solution_2": "[hide=\"hmm\"]$\\frac{m}{n}$ is also the probability that, in picking two cubes at random, you get 1 red 1 yellow. So label the cubes as $R_1$, $R_2$, $R_3$, $B_1$, $B_2$, $Y$. Then there are $6*5=30$ ways to pick two cubes, order mattering, and the number of ways to pick the yellow and 1 red is 6: $(Y, R_1), (Y, R_2), (Y, R_3), (R_1, Y), (R_2, Y), (R_3, Y)$. Thus, \\[ \\frac{m}{n}=\\frac{6}{30}=\\frac{1}{5}\\implies m^n=\\boxed{1} \\][/hide]\r\nOur final answers were the same, but something went wrong. How did you get your answer?", "Solution_3": "Doesn't $\\frac{4!}{2!2!}$ treat R1R2B1B2 and R2R1B2B1 as the same permutation? Because the sample space treats it as two seperate permutations.\r\n\r\nThis is the part I am confused on.\r\n\r\nEDIT: Nevermind, I get can visualize this now." } { "Tag": [ "HMMT", "algebra", "polynomial", "analytic geometry", "function", "calculus", "derivative" ], "Problem": "The polynomial $3x^5 - 250x^3 + 735x$ is interesting because it has the maximum possible relative extrema and points of inflection at integer lattice points out of any quintic. Find the the sum of the x coordinates of said points.\r\n\r\n[hide=\"this is what confuses me\"]The relative extrema will occur when $15x^4 - 750x^2 + 735 = 0$, which is at $x = 1, 7, -1, -7$. The points of inflection will have x-coordinates that satisfy $60x^3 - 1500x = 0$ which are $x = 0, 5, -5$. Yet the answer is $75$.[/hide]", "Solution_1": "Tha looks like it might be a sum of squares. Are you sure you have the right statement?", "Solution_2": "yes, this is the exact problem.", "Solution_3": "Since the function $f=3X^5-250X^3+735X$ is odd, it's derivative is even(simetric about x-axis), and it's second derivative is odd again(simetric about origin). So the x-coordinates of their roots is $0$ even without knowing their value. Probably the problem is wrong." } { "Tag": [ "geometry", "rotation", "trigonometry" ], "Problem": "Two strips of width 1 overlap at an angle of $\\alpha$ as shown. The area of the overlap (shown shaded) is\n\n[asy]\npair a = (0,0),b= (6,0),c=(0,1),d=(6,1);\ntransform t = rotate(-45,(3,.5));\npair e = t*a,f=t*b,g=t*c,h=t*d;\npair i = intersectionpoint(a--b,e--f),j=intersectionpoint(a--b,g--h),k=intersectionpoint(c--d,e--f),l=intersectionpoint(c--d,g--h);\ndraw(a--b^^c--d^^e--f^^g--h);\nfilldraw(i--j--l--k--cycle,blue);\nlabel(\"$\\alpha$\",i+(-.5,.2));\n//commented out labeling because it doesn't look right.\n//path lbl1 = (a+(.5,.2))--(c+(.5,-.2));\n//draw(lbl1);\n//label(\"$1$\",lbl1);[/asy]\n\n$\\text{(A)} \\ \\sin \\alpha \\qquad \\text{(B)} \\ \\frac{1}{\\sin \\alpha} \\qquad \\text{(C)} \\ \\frac{1}{1 - \\cos \\alpha} \\qquad \\text{(D)} \\ \\frac{1}{\\sin^2 \\alpha} \\qquad \\text{(E)} \\ \\frac{1}{(1 - \\cos \\alpha)^2}$", "Solution_1": "$h= \\frac{1}{\\sin{\\alpha}}$, thus $B$.", "Solution_2": "Please hide your answer. :dry:\r\n\r\nAnyway, it would be nice if you gave some explanation.\r\n\r\n[hide=\"Answer\"]I think that the figure formed is a rhombus... so, $\\sin \\alpha=\\frac{h}{x}$, where $x$ is the side length of the rhombus. However, we know that $h=1$, and $A=xh=x$. From the previous equation, we have $x=\\frac{h}{\\sin \\alpha}=\\frac{1}{\\sin \\alpha}\\Rightarrow \\boxed{B}$.[/hide]" } { "Tag": [ "function", "linear algebra", "linear algebra unsolved" ], "Problem": "Hi,\r\nI have an interesting problem on my hands and as it is of general interest - I suppose - I hoped you guys could give me some pointers.\r\n\r\nHere is a particular example that would help me explain it. Let's say that in a company you decide to test and assess your employees, by giving them each a mark based on single skills. The skills are structured in a tree, and each node in the tree can get a grade, like in the following image:\r\n\r\nhttp://www.interact-studio.com/alex/skill-tree.jpg\r\n\r\n(sorry for the poor originality while choosing the example :) )\r\n\r\nNow, when it comes to grading the employees, I use the following grades (this is a must for me):\r\n0 - unknown\r\n1 - poor, the lowest grade\r\n2 - basics\r\n3 - advanced\r\n4 - expert, the best grade\r\n\r\nAssume that after grading two employees, Tom and Jerry, their skills and grade look like this (please note that 0 means that they have not been graded for that skill). Just look at the red numbers, I'll explain next what the blue ones are\r\n\r\nhttp://www.interact-studio.com/alex/tom-jerry.jpg\r\n\r\nThanks for your patience, as now comes the true problem. I hope I made myself clear until now. As you may see, Tom (the first guy) has 0 (unknown) at Windows XP, as no-one has graded him on that particular skill. However, by using the children nodes of Windows XP (SP1 and SP2) for which he has a 3 and a 2, we could approximate that for Windows XP he has a skill of 2.5 (the blue number :) ). By generalising, for each unknown node we could use some sort of an average (I am now using the Arithmetic mean) based on its children's grades.\r\n\r\nWe can go all the way up to the root (please look at the image again), and after some simple computation, we can see that Tom gets an average of 2,16, while Jerry gets an average of 2,33. To some this may seem right, but this is not the case. Just compare the known grades from Tom to those of Jerry, which indicate that Tom is better than Jerry and really this is the case. It's just that Tom didn't get graded on the other skills Jerry has been.\r\n\r\nSo, my question is what kind of mean do you suggest I would use ? (as the arithmetic one doesn't seem to do the job). Please suggest anything which comes to mind, as I am open to interesting ideas.\r\n\r\nThanks for your patience and I hope I will wake up some grey cells :),\r\nAlex", "Solution_1": "Well I think there is some problem in your rating system... let me try to explain:\r\n\r\nSuppose you have a function that does the mean of a set of numbers, let's call it $f$. For example, the arithmetic mean is \r\n$f(a_{1}, \\ldots, a_{n}) = \\frac{a_{1}+\\cdots+a_{n}}{n}$\r\nand the geometric mean is \r\n$f(a_{1}, \\ldots, a_{n}) = \\sqrt[n]{a_{1}a_{2}\\cdots a_{n}}$.\r\nBut generally, you want a mean to satisfy the condition\r\n$\\text{min}(a_{1}, \\ldots, a_{n}) \\le f(a_{1}, \\ldots, a_{n}) \\le \\text{max}(a_{1}, \\ldots, a_{n})$\r\n\r\nAccording to your tree, Jerry has a 3 for Windows XP even though he has a 2 for both of the children of that node. But no mean will give $f(2, 2) = 3$. Actually any mean should give $f(2, 2) = 2$. So if your rating system allows for a parent node to be larger than all of its children (or smaller than all of its children), then you need to account for that in your estimation of unknown skills. \r\n\r\nIt's hard to tell what kind of function you want -- it should be based on your rating system. Think about what kind of factors go into the ratings for parent nodes. Do you give a bonus to Windows XP because Jerry is familiar with both service packs? Then you might want to include the number of non-1 children of a node in your estimation function for that node. \r\n\r\nI hope this helps. \r\n\r\nPS. Another thing I just noticed is that when you do Tom's mean for the Windows node, you counted 0 as 0. Shouldn't you count it as 1? Why should \"unknown\" be worse than \"unfamiliar with the software\"? Then, Tom gets a 2.5, and is still better than Jerry even though Jerry somehow has a 3 for Windows XP even though his skill with each SP is 2. And if Tom turns out to know Windows 2000 as well, you got a great deal for hiring him." } { "Tag": [], "Problem": "Diamonds increase in value by 10\\% each year. If you inherit your grandmother's diamond pendant which is worth $ \\$36,000.00$ in 1994, what will its value be in 1998?", "Solution_1": "Hello\r\nnew prize =initial prize$ {\\frac{11}{10}}^4\r\n \\equal{}$52707.6\r\nthank u." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $x,y,z>0$ such that $xyz=1$ . Prove that \r\n$\\frac{x^{2}}{x^{2}+x+1}+\\frac{y^{2}}{y^{2}+y+1}+\\frac{z^{2}}{z^{2}+z+1}\\geq 1$", "Solution_1": "[quote=\"silouan\"]Let $x,y,z>0$ such that $xyz=1$ . Prove that \n$\\frac{x^{2}}{x^{2}+x+1}+\\frac{y^{2}}{y^{2}+y+1}+\\frac{z^{2}}{z^{2}+z+1}\\geq 1$[/quote]\r\nVery nice ineq\r\nLet $x=\\frac{a^{2}}{bc}$\r\n$\\Rightarrow\\sum\\frac{x^{2}}{x^{2}+x+1}\\geq 1 \\Leftrightarrow\\sum\\frac{a^{4}}{a^{4}+a^{2}bc+b^{2}c^{2}}\\geq 1$\r\nWe have :$\\sum\\frac{a^{4}}{a^{4}+a^{2}bc+b^{2}c^{2}}\\geq \\frac{(a^{2}+b^{2}+c^{2})^{2}}{\\sum (a^{4}+a^{2}bc+b^{2}c^{2})}$\r\nSo we need to prove $\\frac{(a^{2}+b^{2}+c^{2})^{2}}{\\sum(a^{4}+a^{2}bc+b^{2}c^{2})}\\geq 1\\Leftrightarrow \\sum b^{2}c^{2}\\geq \\sum a^{2}bc$ ( It's obviously true)\r\n\r\n\r\nLet $x=\\frac{a}{b}$ then we also have a very nice ineq:\r\n$\\sum\\frac{a^{2}}{a^{2}+ab+b^{2}}\\ge 1$", "Solution_2": "Another way(a bit shorter)[But unfortunately it is wrong :( ]:\r\nUsing AM-GM,we get that $x\\leq \\frac{1+x^{2}}{2}$ so \\[\\frac{x^{2}}{x^{2}+x+1}\\geq\\frac{x^{2}}{x^{2}+1}\\cdot\\frac{2}{3}\\] and the other two cyclic results.Now we simply set $x^{2}=\\frac{a}{b}$ etc and we get that the ineq is equivalent to the famous \\[\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a}\\geq\\frac{3}{2}\\] The end :)", "Solution_3": "[quote=\"andyciup\"] \\[\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a}\\geq\\frac{3}{2}\\] [/quote]\r\nI'm sorry but it is wrong :( (maybe you mistoke it with Nesbitt). Set $a=1$, $b=2$, $c=4$. Then you get $\\mathrm{LHS}=\\frac{22}{15}<\\frac32$.", "Solution_4": "OOh..sorry bout that..it seemed so nice,I didn`t check it :blush:" } { "Tag": [ "puzzles" ], "Problem": "2. What numbers come next in the series, 3-6,7-14? \r\n a) 11-30,15-64 b) 15-30,31-62 \r\n c) 21-42,100-200 d) 12-24,28-56 \r\n e) none of these", "Solution_1": "i believe the answer is b", "Solution_2": "The question isn't specific enough, so I think it is safe to assume that the number of terms in each case doubles each time, and uses the next set of terms. Thus, the answer is $ \\fbox{B}$, which has $ 8\\times2\\equal{}16$ terms in the first group and $ 16\\times2\\equal{}32$ terms in the second group." } { "Tag": [ "modular arithmetic" ], "Problem": "1.) Show by modulo division that $ a^3 \\equiv a (mod 6)$.\r\n2.) Show by modulo division that $ a^2 \\equiv 1 (mod 8)$.", "Solution_1": "[hide]For the first, $ a^3 \\equiv a \\pmod{6} \\iff a^3 \\minus{} a \\equiv 0 \\pmod{6} \\iff a(a\\plus{}1)(a\\minus{}1) \\equiv 0 \\pmod {6}$\n\nClearly, at least one of the three is $ \\equiv 0 \\pmod{2}$, and one of the three is $ \\equiv 0 \\pmod{3}$.\n\n\nFor the second, I assume you mean, prove that $ a \\equiv 1 \\pmod{2} \\implies a^2 \\equiv 1 \\pmod{8}$:\n\n$ a: \\equal{}2b\\plus{}1 \\implies a^2 \\equal{} 4b^2\\plus{}4b\\plus{}1 \\equal{} 4b(b\\plus{}1)\\plus{}1$\n\nSince either $ 2|b$ or $ 2|b\\plus{}1$, $ 8| 4b(b\\plus{}1)$, and $ a^2 \\equiv 0\\plus{}1 \\equiv 1 \\pmod{8}$[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ x,y,z>0$ such that $ (2\\minus{}x)(2\\minus{}y)(2\\minus{}z)\\equal{}xyz$. Prove that:\r\n\r\n$ x\\plus{}y\\plus{}z\\le 3$", "Solution_1": "[quote=\"Pain rinnegan\"]Let $ x,y,z > 0$ such that $ (2 \\minus{} x)(2 \\minus{} y)(2 \\minus{} z) \\equal{} xyz$. Prove that:\n\n$ x \\plus{} y \\plus{} z\\le 3$[/quote]\r\n$ x\\equal{}y\\rightarrow2,$ $ z\\rightarrow0.$ :wink:", "Solution_2": "[quote=\"arqady\"][quote=\"Pain rinnegan\"]Let $ x,y,z > 0$ such that $ (2 \\minus{} x)(2 \\minus{} y)(2 \\minus{} z) \\equal{} xyz$. Prove that:\n\n$ x \\plus{} y \\plus{} z\\le 3$[/quote]\n$ x \\equal{} y\\rightarrow2,$ $ z\\rightarrow0.$ :wink:[/quote]\r\n\r\nYes, sorry i was careless. I meant the following one:\r\n\r\nLet $ x,y,z > 0$ such that $ (2 \\minus{} x^2)(2 \\minus{} y^2)(2 \\minus{} z^2) \\equal{} x^2y^2z^2$. Prove that:\r\n\r\n$ x \\plus{} y \\plus{} z\\le 3$\r\n\r\nThis one should be true." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "$ f: R\\longrightarrow R$ $ f(xf(y))\\plus{}f(yf(x))\\equal{}2xy$ find all functions. :!:", "Solution_1": "I don't think this problem is easy at all... :maybe:\r\n[hide]first of all prove that $ f(1) \\equal{} 1\\textrm{ or} \\minus{} 1$,then prove that $ f$ is injective,then prove that $ f(\\frac 1x) \\equal{} \\frac 1{f(x)}$ from which it follows that $ f(x) \\equal{} x$ or $ f(x) \\equal{} \\minus{} x$ for every $ x\\in\\mathbb{R}$...[/hide]", "Solution_2": "yes you're right. maybe isnt easy . sorry! .. .. .. .. :(" } { "Tag": [ "induction", "combinatorics proposed", "combinatorics" ], "Problem": "Given n different positive numbers a_1, a_2, ... , a_n.\r\nWe construct all the possible sums (from 1 to n terms).\r\nProve that among those sums there are at least n(n+1)/2 different ones.", "Solution_1": "This one is really easy. Just use induction on $n$.\r\nThe result clearly holds for $n=1$.\r\nNow assume it holds for any set of $n$ positive real numbers.\r\nLert's give a set of $n+1$ positive real numbers. Wlog, we may assume that $a_1 < \\cdots < a_{n+1}$.\r\nThen, let $s_i$, where $i=1, \\cdots, \\frac {n(n+1)} 2$ be the distinct sums from the set $\\{a_1, \\cdots , a_n \\}$ (by the induction hypothesis).\r\n. Then, the sums $a_1, a_2, \\cdots, a_{n+1}, a_{n+1}+s_1, a_{n+1} + s_2, \\cdots, a_{n+1} + s_{\\frac {n(n+1)} 2}$ are pairwise distinct and we are done.\r\n\r\nPierre;" } { "Tag": [ "algebra", "system of equations" ], "Problem": "Source: 360 PMC\r\nProve that for any integer $n$ the number \r\n$5^{5^{n+1}}+5^{5^n}+1$ is not prime.\r\n\r\nOk..I let $5^{5^n}=x$ and got\r\n$x^5+x+1$..and tried to factor it out, but couldn't...\r\nAnd on the solution it just showed that \r\n$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$. Can anyone plz show me how to factor like that? Thanks,", "Solution_1": "well you know that\r\n$x^5+x+1=(x^2+ax+1)(x^3+bx^2+cx+1)$ expand this out and solve the system of equations and you get that\r\n$a=1$, $b=0$, and $c=-1$.\r\n\r\nim not sure what you get if you try, but go ahead and give it a shot.\r\n$x^5+x+1 = (x^2+ax-1)(x^3+bx^2+cx-1)$", "Solution_2": "you see that $x^2+x+1$ is a factor of the LHS by plugging in 3rd roots of unity. the motivation to do that is that the powers of each term on the LHS are all distinct mod 3.", "Solution_3": "[quote=\"kimby_102\"]Source: 360 PMC\nProve that for any integer $n$ the number \n$5^{5^{n+1}}+5^{5^n}+1$ is not prime.\n\nOk..I let $5^{5^n}=x$ and got\n$x^5+x+1$..and tried to factor it out, but couldn't...\nAnd on the solution it just showed that \n$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$. Can anyone plz show me how to factor like that? Thanks,[/quote]\r\nbut what about $x=1$ it doesn't gives natural $n$ indeed but... :?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $a,b,c,d$ are positive numbers, then\r\n(1) $(a+b+c+d)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}) \\geq 12+\\sqrt{(a^2+b^2+c^2+d^2)(\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}+\\frac{1}{d^2})}$;\r\n(2) $(a+b+c+d)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}) -14 \\geq \\sqrt{(a^2+b^2+c^2+d^2)(\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}+\\frac{1}{d^2})-12}$.", "Solution_1": "Your inequalities can give me some inspiration. I like the first, and from this one we can write down more. The second is one of them. Vasc, have you got any generalizations for such?" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "For a,b,c,d real positive numbers prove that\r\n\r\n(a^4+b^4+c^4+d^4)/(abcd) + (abcd)^(1/4)/(a+b+c+d) \\geq 1", "Solution_1": "Did I miss something or the same argument as in m14 gives a lower bound equals to 2, which is better than desired...\r\n\r\nPierre.", "Solution_2": "indeed, it's done in the same way! \r\n\r\ncheers! :D :D" } { "Tag": [ "search", "email", "inequalities unsolved", "inequalities" ], "Problem": "Prove that $\\forall$ $x_{1} , x_{2} ,....., x_{n} \\in R$ ,we have:\r\n$\\frac{x_{1}}{1+x_{1}^2} +\\frac{x_{2}}{1+x_{1}^2+x_{2}^2}+.....+ \\frac{x_{n}}{1+x_{1}^2+x_{2}^2+....+x_{n}^2} \\leq \\sqrt{n}$", "Solution_1": "It's really IMO short list 2000. I guess it's quite well-known.", "Solution_2": "Anyone can post the solution.I don't know this problem", "Solution_3": "no one help me????", "Solution_4": "[quote=\"James Potter\"]Prove that $\\forall$ $x_{1} , x_{2} ,....., x_{n} \\in R$ ,we have:\n$\\frac{x_{1}}{1+x_{1}^2} +\\frac{x_{2}}{1+x_{1}^2+x_{2}^2}+.....+ \\frac{x_{n}}{1+x_{1}^2+x_{2}^2+....+x_{n}^2} \\leq \\sqrt{n}$[/quote]\r\n\r\n[b]1.[/b] Stop spamming.\r\n[b]2.[/b] Don't post from two accounts.\r\n[b]3.[/b] Use the \"search\" button. A search for *\\frac{x_n}{1+x_1^2*, for instance, will give you this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=48859 . And there you can find some links to solutions of the problem.\r\n\r\n Darij", "Solution_5": "[b]1)[/b]I cannot find any solution from your three links\r\n[b]2)[/b]Buihaiha_2000 is not me.", "Solution_6": "[b]1.[/b] The second link leads to a solution. (The first one leads to a thread where I hope a solution will be posted one day.)\r\n\r\n[b]2.[/b] This is a bit strange, as your email addresses differ by 1 letter.\r\n\r\n Darij" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find the remainder of the division $ \\displaystyle\\frac {69!( 60! \\plus{} 1 )}{71}$\r\n\r\nGreetings :D", "Solution_1": "[hide]$ 71$ is prime, so $ 70! \\equiv 70$ modulo $ 71$. $ 69! \\equiv 1$ modulo $ 71$ as a result, so we need only consider $ (60!\\plus{}1)$ modulo $ 71$.\n\n$ 70! \\equiv (\\minus{}1)(\\minus{}2)(\\minus{}3)(\\minus{}4)(\\minus{}5)(\\minus{}6)(\\minus{}7)(\\minus{}8)(\\minus{}9)(\\minus{}10)60!$ modulo $ 71$, so if we can compute $ (\\minus{}1)(\\minus{}2)(\\minus{}3)(\\minus{}4)(\\minus{}5)(\\minus{}6)(\\minus{}7)(\\minus{}8)(\\minus{}9)(\\minus{}10)$ modulo $ 71$, since $ 70! \\equiv \\minus{}1$ modulo $ 71$, we can compute $ 60!$ modulo $ 71$.\n\n$ (\\minus{}1)(\\minus{}2)(\\minus{}3)(\\minus{}4)(\\minus{}5)(\\minus{}6)(\\minus{}7)(\\minus{}8)(\\minus{}9)(\\minus{}10) \\equiv (1)(2)(3)(4)(5)(6)(7)(8)(9)(10) \\equiv 61$ modulo $ 71$. Therefore, $ 60! \\equiv 64$ modulo $ 71$. Therefore, $ 60! \\plus{} 1 \\equiv 65$ modulo $ 71$, and the desired answer is $ 65$.[/hide]" } { "Tag": [ "geometry", "geometric transformation", "rotation", "analytic geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "Given a n*n grid filled with the numbers from 1 to n^2, in the obvious order with the exception of 1 and 2 that are swapped, determine whether a sequence of cyclic rotations of rows or columns can result in a fully ordered grid.\r\n\r\nMore precisely, \"obvious order\" means that the cell at (x, y), with coordinates in the [0, n) interval , contains y * n + x + 1 and \"cyclic rotation of a row\" means that for a given y, for each x (x, y) is moved to (x + k mod n, y) while cyclic rotations of a column are similarly defined.", "Solution_1": "If $n$ is odd, I think the permutations generated by cyclic rotations of rows/columns are even, so we will never be able to do that. However, when $n$ is even, it might be possible in the general case, I'm not sure. Think about $n=2$: it's clearly possible to interchange $1$ and $2$." } { "Tag": [ "\\/closed" ], "Problem": "Hi, I'm a new member. I'm curious, how does one get stars and/or the name of a cool mathematical formula under their member name.", "Solution_1": "That all depends on the number of posts you have." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ x_1,x_2,...,x_n$ be real numbers such that $ x_1>1,x_2>2,...,x_n>n$. Find the minimum value of \r\n\r\n$ \\frac{(x_1\\plus{}x_2\\plus{}...\\plus{}x_n)^2}{\\sqrt{x_1^2\\minus{}1^2}\\plus{}\\sqrt{x_2^2\\minus{}2^2}\\plus{}...\\plus{}\\sqrt{x_n^2\\minus{}n^2}}$", "Solution_1": "Here $ \\sum$ denotes $ \\sum_{i\\equal{}1}^n$. Let $ y_i>0$ such that $ y_i^2\\equal{}x_i^2\\minus{}i^2$, $ i\\equal{}1,2,\\dots,n$. Then we are looking for the minimum value of \r\n\\[ Y\\equal{}\\frac{\\left(\\sum\\sqrt{y_i^2\\plus{}i^2}\\right)^2}{y_1\\plus{}y_2\\plus{}\\dots\\plus{}y_n}\\]\r\nUsing $ \\sqrt{y_i^2\\plus{}i^2}\\ge\\frac{1}{\\sqrt 2}(y_i\\plus{}i)$ we deduce that \r\n\\[ Y\\ge\\frac{\\left(\\sum y_i\\plus{}\\sum i\\right)^2}{2\\sum y_i}\\ge 2\\sum i\\equal{}n(n\\plus{}1)\\]\r\nwhere the last inequality follows from $ (a\\plus{}b)^2\\ge 4ab$. Equality holds iff $ y_i\\equal{}i$ for all $ i$. That is, when $ x_i\\equal{}i\\sqrt 2$, $ i\\equal{}1,2,\\dots,n$." } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "limit", "complex numbers", "real analysis" ], "Problem": "OK, here goes....... show :D\r\n\r\n$ \\int_{0}^{\\infty}\\;\\;\\frac{4\\cdot\\tan^{\\minus{}1}\\,(4\\,x)}{e^{2\\pi\\,x}\\minus{}1}\\;\\;\\textbf dx\\;\\;\\equal{}\\;\\;\\boxed{\\ln\\left(\\Gamma^{2}\\left(\\frac{1}{4}\\right)\\right)\\minus{}2\\ln 2\\minus{}\\ln\\pi\\plus{}\\frac{1}{2}}$", "Solution_1": "For all complex numbers $ z\\neq 0$ with $ \\text{Re}(z)\\ge 0$ define\r\n\r\n$ \\text{Ei}(z) \\equal{} \\gamma \\plus{} \\ln(z) \\plus{} \\sum_{k \\equal{} 1}^\\infty \\frac {z^k}{k\\, k!} \\quad , \\quad \\text{Ci}(z) \\equal{} \\gamma \\plus{} \\ln(z) \\plus{} \\sum_{k \\equal{} 1}^\\infty ( \\minus{} 1)^k$$ \\frac {z^{2k}}{2k\\, (2k)!} \\quad , \\quad \\text{Si}(z) \\equal{} \\sum_{k \\equal{} 0}^\\infty ( \\minus{} 1)^k \\frac {z^{2k \\plus{} 1}}{(2k \\plus{} 1) \\, (2k \\plus{} 1)!}$ .\r\n\r\nBecause of $ \\ln(iz) \\equal{} \\ln(i) \\plus{} \\ln(z) \\equal{} i\\frac {\\pi}{2} \\plus{} \\ln(z) \\qquad \\boxed{\\text{Ei}(iz) \\equal{} \\text{Ci}(z) \\plus{} i\\left(\\text{Si}(z) \\plus{} \\frac {\\pi}{2}\\right)}$\r\n\r\nFor $ a,s > 0$ define $ f(a,s) \\equal{} \\int_0^\\infty \\frac {e^{ \\minus{} sx}}{x \\minus{} ia} \\, dx$ and substitute $ x \\equal{} \\minus{} \\frac {1}{s} t$\r\n\r\n$ \\minus{} \\int_{ \\minus{} \\infty}^0 \\frac {e^t}{t \\plus{} ias} \\, dt \\equal{} \\minus{} \\int_{ \\minus{} \\infty \\plus{} ias}^{ias} \\frac {e^{t \\minus{} ias}}{t} \\,dt \\equal{} \\minus{} e^{ \\minus{} ias}\\, \\Big(\\text{Ei}(ias) \\minus{} i\\pi\\Big) \\equal{}$ $ \\minus{} \\Big(\\cos(as) \\minus{} i\\, \\sin(as)\\Big)\\, \\Big(\\text{Ci}(as) \\plus{} i\\left(\\text{Si}(as) \\minus{} \\frac {\\pi}{2}\\right)\\Big)$\r\n\r\nThus $ \\int_0^\\infty \\frac {a\\, e^{ \\minus{} sx}}{x^2 \\plus{} a^2} \\, dx \\equal{} \\frac {f(a,s) \\minus{} \\overline{f(a,s)}}{2i} \\equal{} \\text{Im}\\big(f(a,s)\\big) \\equal{}$ $ \\sin(as)\\, \\text{Ci}(as) \\minus{} \\cos(as)\\,\\Big(\\text{Si}(as) \\minus{} \\frac {\\pi}{2}\\Big)$\r\n\r\nwhereat $ \\int_0^\\infty \\frac {a\\, e^{ \\minus{} sx}}{x^2 \\plus{} a^2} \\, dx$ is the same as $ \\int_0^\\infty s \\, e^{ \\minus{} sx} \\, \\arctan\\left(\\frac {x}{a}\\right)\\, dx$ after integration by parts.\r\n\r\nTherefore $ \\boxed{\\int_0^\\infty e^{ \\minus{} sx} \\, \\arctan\\left(\\frac {x}{a}\\right)\\, dx \\equal{} \\frac {1}{s} \\left[\\sin(as)\\, \\text{Ci}(as) \\minus{} \\cos(as) \\left(\\text{Si}(as) \\minus{} \\frac {\\pi}{2}\\right)\\right] }$\r\n\r\nFor $ n\\in \\Bbb{N} \\quad I_n: \\equal{} \\int_0^\\infty \\frac {\\arctan\\left(\\frac {x}{n}\\right)}{e^{2\\pi x} \\minus{} 1} \\, dx \\equal{} \\sum_{k \\equal{} 1}^\\infty \\int_0^\\infty e^{ \\minus{} 2\\pi k x} \\, \\arctan\\left(\\frac {x}{n}\\right) \\, dx \\equal{} \\sum_{k \\equal{} 1}^\\infty \\frac {1}{2\\pi k} \\left[0 \\minus{} 1\\cdot \\left(\\text{Si}(2\\pi kn) \\minus{} \\frac {\\pi}{2} \\right)\\right]$\r\n\r\n$ \\equal{} \\sum_{k \\equal{} 1}^\\infty \\frac {1}{2\\pi k} \\int_{2\\pi kn}^\\infty \\sin(t) \\, \\frac {dt}{t}$ . Replace $ t$ by $ kx \\quad I_n \\equal{} \\frac {1}{2\\pi} \\sum_{k \\equal{} 1}^\\infty \\int_{2\\pi n}^\\infty \\frac {\\sin(kx)}{k} \\, \\frac {dx}{x}$ and split $ \\int_{2\\pi n}^\\infty$ up to $ \\sum_{j \\equal{} n}^\\infty \\int_{2\\pi j}^{2\\pi (j \\plus{} 1)}$ \r\n\r\nto get $ I_n \\equal{} \\frac {1}{2\\pi} \\sum_{j \\equal{} n}^\\infty \\sum_{k \\equal{} 1}^\\infty \\int_{2\\pi j}^{2\\pi (j \\plus{} 1)} \\frac {\\sin(kx)}{k} \\, \\frac {dx}{x}$ then use the formula $ \\sum_{k \\equal{} 1}^\\infty \\frac {\\sin(kx)}{k} \\equal{} \\frac {(2j \\plus{} 1)\\pi \\minus{} x}{2}$ for $ 2\\pi j < x < 2\\pi (j \\plus{} 1)$ .\r\n\r\n$ 2 I_n \\equal{} \\frac {1}{\\pi} \\sum_{j \\equal{} n}^\\infty \\int_{2\\pi j}^{2\\pi (j \\plus{} 1)} \\frac {(2j \\plus{} 1)\\pi \\minus{} x}{2} \\, \\frac {dx}{x} \\equal{} \\lim_{m\\to\\infty} \\sum_{j \\equal{} n}^{m \\minus{} 1} \\left[\\left(j \\plus{} \\frac12\\right)\\, \\ln\\left(\\frac {j \\plus{} 1}{j}\\right) \\minus{} 1\\right]$ , whereas\r\n\r\n$ \\sum_{j \\equal{} n}^{m \\minus{} 1} j \\, \\ln\\left(\\frac {j \\plus{} 1}{j}\\right) \\equal{} \\ln \\prod_{j \\equal{} n}^{m \\minus{} 1} \\left(\\frac {j \\plus{} 1}{j}\\right)^j \\equal{} \\ln\\left(\\frac {m^m}{m!}\\right) \\minus{} \\ln\\left(\\frac {n^n}{n!}\\right) \\;\\, , \\;\\, \\sum_{j \\equal{} n}^{m \\minus{} 1} \\ln\\left(\\frac {j \\plus{} 1}{j}\\right)$ $ \\equal{} \\ln(m) \\minus{} \\ln(n)$ and $ \\sum_{j \\equal{} n}^{m \\minus{} 1} 1 \\equal{} m \\minus{} n$ .\r\n\r\nSo $ 2I_n \\equal{} \\lim_{m\\to\\infty} \\left(\\ln\\left(\\frac {m^m}{m!}\\right) \\plus{} \\frac12 \\ln(m) \\minus{} m\\right) \\minus{} \\left(\\ln\\left(\\frac {n^n}{n!}\\right) \\plus{} \\frac12 \\ln(n) \\minus{} n\\right)$ , whereas $ \\lim_{m\\to\\infty} (...) \\equal{} \\minus{} \\ln \\sqrt {2\\pi}$ after Stirling's formula.\r\n\r\n$ \\Rightarrow \\frac12 \\ln(2\\pi) \\plus{} 2I_n \\equal{} \\ln(n!) \\minus{} n\\ln(n) \\minus{} \\frac12 \\ln(n) \\plus{} n \\; \\Rightarrow \\;$ $ \\ln(\\Gamma(n)\\cdot n) \\minus{} \\frac12 \\ln(n) \\equal{} n\\ln(n) \\minus{} n \\plus{} \\frac12 \\ln(2\\pi) \\plus{} 2I_n$ ,\r\n\r\nwhich shows at least [url=http://mathworld.wolfram.com/BinetsLogGammaFormulas.html]Binet's Log Gamma Formula[/url] $ \\ln\\Gamma(z) \\equal{} z\\, \\ln(z) \\minus{} \\frac12 \\ln(z) \\minus{} z \\plus{} \\frac12 \\ln(2\\pi) \\plus{} 2I_z$ for $ z \\equal{} \\frac {1}{n}$.\r\n\r\nShould it be true for all $ z$ with $ \\text{Re}(z) > 0$ then write $ 4I_z \\equal{} \\ln\\Big(\\Gamma^2(z)\\Big) \\minus{} 2z\\ln(z) \\plus{} \\ln(z) \\plus{} 2z \\minus{} \\ln(2\\pi)$ and set $ z \\equal{} \\frac14$ ." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "number theory", "relatively prime" ], "Problem": "Below is an example of a square grid of dots with 3 dots along each side. In a similar square grid of dots with 8 dots along each side, how many distinct lines pass through the dot at the bottom left corner and at least one other dot of the grid?\n[asy]dot((0,0)); dot((0,1)); dot((0,2));\ndot((1,0)); dot((1,1)); dot((1,2));\ndot((2,0)); dot((2,1)); dot((2,2));[/asy]", "Solution_1": "The slopes of the lines are\r\ninfinity\r\nzero\r\n1/1 2/1 3/1 4/1 5/1 6/1 7/1\r\n1/2 2/2 3/2 4/2 5/2 6/2 7/2\r\n1/3 2/3 3/3 4/3 5/3 6/3 7/3\r\n1/4 2/4 3/4 4/4 5/4 6/4 7/4\r\n1/5 2/5 3/5 4/5 5/5 6/5 7/5\r\n1/6 2/6 3/6 4/6 5/6 6/6 7/6\r\n1/7 2/7 3/7 4/7 5/7 6/7 7/7\r\nHow do we count the unique fractions on the top? Well there are:\r\n3+2(totient of 2-7)=1+2(1 +2 +2 +4 +2 +6) = 3+2(17)=[b]37[/b]\r\n\r\nNot too enlightening.", "Solution_2": "I shall explain what he did.\r\n\r\n[hide=\"Solution\"]Two lines with the given properties are different if and only if they have different slopes. Therefore we are counting the number of different slopes we can get. Now the slope of a line is the rise over the run, so if the line passes through the coordinate (n,m), then it has a slope of m/n. The coordinate that the line passes through has the following two restrictions placed on the coordinates:\n\n1. mn=0 (at least one of m or n is zero)\nor\n2. $ 1\\leq m\\leq 7$ and $ 1\\leq n\\leq 7$.\n\nIf 1 holds, then the slope of the line is either 0 or undefined, and there are two lines that have this property.\n\nIf 2 holds, then we will have to count distinct slopes. Now we only have to count the pairs (m,n) such that m and n are relatively prime, because the slopes to all of these points are each different, and we will have counted them all (because if GCD(m,n) is greater than 1, then we can cancel out something from the numerator and denominator of m/n to get that some point (y,x) such that (0,0), (y,x), and (n,m) are collinear, and GCD(y,x)=1, and we would have already counted this line.\n\nNow the n such that m=1 and GCD(m,n)=1 are 1, 2, 3, 4, 5, 6, and 7. This gives 7 lines.\nThe n such that m=2 and GCD(m,n)=1 are 1, 3, 5, and 7. This gives 4 more lines.\nIf n=3, then m is either 1, 2, 4, 5, or 7. This gives 5 lines.\nIf n=4, then m is either 1, 3, 5, or 7. This gives 4 more lines.\nIf n=5, then m is either 1, 2, 3, 4, 6, or 7. This gives 6 more lines.\nIf n=6, then m is either 1, 5, or 7. This gives 3 more lines\nIf n=7, then m is either 1, 2, 3, 4, 5, or 6, and we hae 6 more lines.\nWe therefore have 7+4+5+4+6+3+6=35 lines. Add the two lines we found in case 1 to get [b]37[/b] total lines.[/hide]", "Solution_3": "The slopes of these lines are in the form x/y where x and y are integers from 1-7. Using this, we 49 different lines, but we have to subtract ones we counted twice. The slope 1 was counted 7 times (1/1, 2/2 ...) so we subtract 6 and get 43. The slope 2 was counted 3 times (2/1, 4/2, 6/3) and 1/2 is the same, because it is just the inverse. So, we subtract 4 to get 39. Lastly, 3 and 1/3 were counted twice (3/1, 6/2), so we subtract 2 to get [b]37[/b]." } { "Tag": [], "Problem": "If 3+ x + x^2 + x^3 + x^4 + x^5 ... = 3.5\r\nThen x=?\r\n\r\n1/3\r\n1/4\r\n1/7\r\n2/7\r\n2/3\r\n\r\nThanks a lot.", "Solution_1": "[hide]\nFirst think what is the formula for an infinite geometric series? :) \n[/hide]", "Solution_2": "[hide=\"Hint\"]$ \\sum_{i\\equal{}0}^\\infty x^i \\equal{} \\frac{1}{1\\minus{}x}$[/hide]" } { "Tag": [], "Problem": "For those addicted to neverending stories- the same rules apply as in Never Ending Story- try to keep all content at at a PG level.\r\n\r\nWord limit- 5-25\r\n\r\n\r\n--------------------------------------------------------------------------------------------------------------------\r\n\r\nOne sunny day, hero-wannabe Randy started out on his quest to...", "Solution_1": "find the biggest porkchop in the land of Udoria. He goes to", "Solution_2": "why do we need two of these? :?", "Solution_3": "Because the other one's really pointless, and this one will be soon too.\r\n\r\nDoes that make any sense? I didn't think so either..", "Solution_4": "A moderator should put this Neverending Story as a place where posts don't count, so people can't spam to earn posts.\r\n\r\nThe story begins:\r\n\r\nIt all started with a dark and windy night in the land of America, Samuel Gates was looking out the back window at the clouds racing above...\r\n\r\n(Sorry, 27 words...:blush:)", "Solution_5": "[quote=\"hwenterprise\"]A moderator should put this Neverending Story as a place where posts don't count, so people can't spam to earn posts.\n\nThe story begins:\n\nIt all started with a dark and windy night in the land of America, Samuel Gates was looking out the back window at the clouds racing above...\n\n(Sorry, 27 words...:blush:)[/quote]\r\n\r\nand suddenly lightning struck a tree that was right next to his house.", "Solution_6": "c'mon guys, one never ending story is bad enough, now two? :huh: This is not good..... \r\n\r\n\r\n.. which shut out all the power in the city...", "Solution_7": "The city was plummeted into darkness, and then, the terrible rath of the storm knocked down a pillar supporting the main water system...", "Solution_8": "...Samuel Gates watched as the storm became fiercer and fiercer, the sky darker and darker, the rain heavier and heavier, to the point when...", "Solution_9": "the water all flowed into the city called Nwe Olreans and...", "Solution_10": ".....cardwacko9999 lost his AoPs account in the violent tides of the admin, for creating an additional pointless topic.......", "Solution_11": "except the admin relented because he was cool and he realized spammers are basically really fun people at heart", "Solution_12": "No guys. Just no. One of these things is plenty. -.-\r\n\r\nLocked." } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": ":) Hey People!!! :) \r\n\r\nThis is a problem I don't have any idea of solving it:\r\n\r\nLet p(x) be the polynomial x^3 + 14*x^2 - 2*x + 1. \r\nLet p(n)(x) denote p(p(n-1)(x)), p(1)(x)=p(x)\r\nShow that there is an integer N such that p(N)(x) - x is divisible by 101 for all integers x. \r\n\r\n(PS.: Do exist infinitely many N?)", "Solution_1": "Already on this site. Go to :\r\nhttp://www.mathlinks.ro/viewtopic.php?p=14148&highlight=#14148\r\n.........." } { "Tag": [ "function" ], "Problem": "Is there an easy way to use generating functions to find the number of positive integer solutions to $ 3x\\plus{}4y\\plus{}5z\\equal{}100$?", "Solution_1": "How about this ... Let\r\n$ g_3(u) \\equal{} u^3 \\plus{} u^6 \\plus{} u^9 \\plus{} \\dots$, exponents multiples of $ 3$, and similarly\r\n$ g_4(u) \\equal{} u^4 \\plus{} u^8 \\plus{} u^{12} \\plus{} \\dots$ ,\r\n$ g_5(u) \\equal{} u^5 \\plus{} u^{10} \\plus{} u^{15} \\plus{} \\dots$ .\r\nNow in the product $ g_3(u) g_4(u) g_5(u)$ isn't the coefficient of $ u^a$ equal to the number of solutions of $ 3x\\plus{}4y\\plus{}5z\\equal{}a$ ?", "Solution_2": "Yea, I got stuck on finding that coefficient without casework.", "Solution_3": "[hide=\"solution\"]\nThe generating function would be \\[ (x^3\\plus{}x^6\\plus{}x^9\\plus{}\\dots)(x^4\\plus{}x^\\plus{}x^{12}\\plus{}\\dots)(x^5\\plus{}x^{10}\\plus{}x^{15}\\plus{}\\dots)\\] and the coefficient we want is the coefficient of $ x^{100}.$\n\nFactor $ x^3$ out of the first, $ x^4$ out of the second, and $ x^5$ out of the third. It would be \\[ x^{12}(1\\plus{}x\\plus{}x^2\\plus{}x^3\\plus{}\\dots)^3\\equal{}x^{12}(1\\plus{}3x\\plus{}6x^2\\plus{}10x^3\\plus{}\\dots).\\] We just need to find the coefficient of $ x^{88}$ in the second term. In general, the coefficient of $ x^{k}$ in the power series \\[ 1\\plus{}3x\\plus{}6x^2\\plus{}10x^3\\plus{}\\dots\\] is $ \\frac{(k\\plus{}1)(k\\plus{}2)}{2}.$ Just substitute $ k\\equal{}88$ to get \\[ \\frac{89\\cdot90}{2}x^{100}\\equal{}\\boxed{4005}x^{100}.\\] Therefore, there are $ 4005$ solutions to the equation \\[ 3x\\plus{}4y\\plus{}5z\\equal{}100.\\][/hide]", "Solution_4": "When you factor $ x^3$ out, it should be $ (1\\plus{}x^3\\plus{}x^6\\plus{}...)$.", "Solution_5": "Ah oops...:oops:", "Solution_6": "Thanks! That solution was pretty good. :D", "Solution_7": "Sure, except that it's also wrong. The correct generating function is $ \\frac {x^{12}}{(1 \\minus{} x^3)(1 \\minus{} x^4)(1 \\minus{} x^5)}$, and computing the coefficient of $ x^{100}$ from this generating function doesn't significantly simplify the computation; it's easier to divide into cases, for example based on the value of $ z$. Computing an exact formula from this generating function is possible but highly tedious and won't help much unless you're either interested in asymptotics or in exact values for very large $ n$.\r\n\r\nIf you're interested, the asymptotic is $ \\frac{(n\\minus{}12)^2}{2 \\cdot 3 \\cdot 4 \\cdot 5}$, so there should be approximately $ 64$ solutions. The problem is that the terms of lower order are periodic in a nontrivial way.", "Solution_8": "How did you get the asymptotic from the generating function? :)", "Solution_9": "The slick way is to compute the volume of the region $ 4y \\plus{} 5z \\le 88$ and divide by $ 3$. The more general way is to look at the smallest pole on the positive real axis, which is $ x \\equal{} 1$, and look at what happens to the function \"locally\" there. It's not hard to see that near $ x \\equal{} 1$,\r\n\r\n$ \\frac{1}{(1 \\minus{} x)^3(1 \\plus{} x \\plus{} x^2)(1 \\plus{} x \\plus{} x^2 \\plus{} x^3)(1 \\plus{} x \\plus{} x^2 \\plus{} x^3 \\plus{} x^4)} \\approx \\frac{1}{(1 \\minus{} x)^3 3 \\cdot 4 \\cdot 5}$\r\n\r\nand the Taylor series expansion of $ \\frac{1}{(1 \\minus{} x)^3}$ is precisely $ {n\\plus{}2 \\choose 2}$ by the generalized binomial theorem. Usually one only takes the leading term because periodicity in the other terms makes those terms less useful, but in this case the denominator has no other double roots so the linear term is also trustworthy, which I should've included.", "Solution_10": "The method of looking at the nearest singularity ... there is a basic explanation in the book \"generatingfunctionology\" by Wilf" } { "Tag": [ "geometry", "trapezoid", "circumcircle", "trigonometry", "cyclic quadrilateral", "trig identities", "Law of Sines" ], "Problem": "(Maybe the problem is a bit easy, but it is nice. :) )\r\n\r\nLet $ A_1A_2B_2B_1$ be a cyclic quadrilateral (circumcircle $ \\mathcal{O}$) for which $ A_1A_2//B_1B_2$. Let the tangent to $ \\mathcal{O}$ at $ A_1$ intersect $ B_1B_2$ at $ Y$ and let the tangent to $ \\mathcal{O}$ at $ B_1$ intersect $ A_1A_2$ at $ X$. Let $ M_1$ and $ M_2$ be the midpoints of $ A_1X$ and $ A_2X$, respectively. Let $ N_1$ and $ N_2$ be the midpoints of $ B_1Y$ and $ B_2Y$, respectively. Prove that $ M_1M_2$, $ N_1N_2$, and $ XY$ are concurrent.", "Solution_1": "[quote=\"Altheman\"](Maybe the problem is a bit easy, but it is nice. :) )\n\nLet $ A_1A_2B_2B_1$ be a cyclic quadrilateral (circumcircle $ \\mathcal{O}$) for which $ A_1A_2//B_1B_2$. Let the tangent to $ \\mathcal{O}$ at $ A_1$ intersect $ B_1B_2$ at $ Y$ and let the tangent to $ \\mathcal{O}$ at $ B_1$ intersect $ A_1A_2$ at $ X$. Let $ M_1$ and $ M_2$ be the midpoints of $ A_1X$ and $ A_2X$, respectively. Let $ N_1$ and $ N_2$ be the midpoints of $ B_1Y$ and $ B_2Y$, respectively. Prove that $ M_1M_2$, $ N_1N_2$, and $ XY$ are concurrent.[/quote]\r\nSorry, but isn't the line $ M_1M_2$ equal to $ A_2X$ and the line $ N_2N_1$ equal to $ B_2Y$ and thus $ XY$, $ M_1M_2$, and $ N_1N_2$ are not concurrent? Is the problem $ M_1N_1$, $ XY$, and $ M_2N_2$? Or did I misinterpret the problem?", "Solution_2": "Assuming we want to prove $ M_1N_1$, $ M_2, N_2$ and $ XY$ are concurrent, it suffices to show that $ \\frac {M_1X}{N_1Y} = \\frac {M_2X}{N_2Y}$. Rearranging things a bit, it suffices to show $ \\frac {A_1X}{A_2X} = \\frac {B_1Y}{B_2Y}$. Because $ B_1X$ is the geometric mean of $ A_1X$ and $ A_2X$ and $ A_1Y$ is the geometric mean of$ B_1Y$ and $ B_2Y$, it suffices to show that $ \\frac {A_2X}{B_1X} = \\frac {B_2Y}{A_1Y}$. This follows by the law of Sines, because $ m\\angle{B_2A_1Y}=m\\angle{A_2B_1X}$ and $ \\sin{\\angle{B_1A_2X}}=\\sin{\\angle{A_1B_2Y}}$.", "Solution_3": "Alternatively, we can show that $ \\frac {YB_1}{XA_1} \\equal{} \\frac {B_1B_2}{A_1A_2}$ as follows. \r\n$ \\angle YA_1B_1 \\equal{} \\angle A_1A_2B_1$, $ \\angle A_1B_1Y \\equal{} \\angle B_1A_1A_2$, $ \\angle XB_1A_1 \\equal{} \\angle B_1B_2A_1$, and $ \\angle XA_1B_1 \\equal{} \\angle A_1B_1B_2$. Now, \r\n\\[ \\triangle YB_1A_1\\sim \\triangle B_2A_2A_1\\implies \\frac {YB_1}{B_2A_2} \\equal{} \\frac {B_1A_1}{A_2A_1}\r\n\\]\r\n\\[ YB_1\\times A_2A_1 \\equal{} B_2A_2\\times B_1A_1 \\equal{} (A_1B_1)^2\r\n\\]\r\nAlso, we notice that\r\n\\[ \\triangle XA_1B_1\\sim \\triangle A_1B_1B_2\\implies \\frac {XA_1}{A_1B_1} \\equal{} \\frac {A_1B_1}{B_1B_2}\\implies XA_1\\times B_1B_2 \\equal{} (A_1B_1)^2\r\n\\]\r\nThus, $ XA_1\\times B_1B_2 \\equal{} YB_1\\times A_1A_2\\implies \\frac {YB_1}{XA_1} \\equal{} \\frac {B_1B_2}{A_1A_2}$." } { "Tag": [ "ratio", "LaTeX", "geometric series", "absolute value" ], "Problem": "1. A set of consecutive positive integers beginning with 1 is written. Then one number is deleted. The average of the remaining number is 35 7/17. What number was deleted.\r\n\r\n2.S1 is the sum of an infinite geometric series with first term a and common ratio r.\r\nS2 is the sum of an infinite geometric series with first term a^2 and common ratio r^2. This is true for all Sn and absolute value of r is less than 1 and absolute value of a is greater than 1. then determine the sum of the infinite series\r\n1/S1+1/S2+1/S3+1/S4+...", "Solution_1": "1. Since the average is 35 5/17 we know that highest number would be in the 70 ish place. We know that the number of numbers is 69 since it is close to 70 and 69-1 is divisible by 17. Then we find out that the missing numbers has to be 15 since (69*70/2)-(35 5/17*68)=15.", "Solution_2": "[hide=\"2\"]\nS1=$ \\frac{a}{1-r}$\n\nS2=$ \\frac{a^2}{1-r^2}$\n\netc\n\nso we are looking for\n\n$ S=\\frac{1-r}{a}+\\frac{1-r^2}{a^2}....=\\left(\\frac{1}{a}+\\frac{1}{a^2}+...\\right) - \\left(\\frac{r}{a}+\\frac{r^2}{a^2}+...\\right)$\n\nso we have two different infinite geometric series.\n\n$ S=\\frac{\\frac{1}{a}}{1-\\frac{1}{a}}-\\frac{\\frac{r}{a}}{1-\\frac{r}{a}}=\\frac{1}{a-1}-\\frac{r}{a-r}=\\frac{a-r-r(a-1)}{(a-1)(a-r)=\\frac{a-ar}{(a-1)(a-r)}$[/hide]", "Solution_3": "[quote=\"Mewto55555\"][hide=\"2\"]\nS1=$ \\frac {a}{1 - r}$\n\nS2=$ \\frac {a^2}{1 - r^2}$\n\netc\n\nso we are looking for\n\n$ S = \\frac {1 - r}{a} + \\frac {1 - r^2}{a^2}.... = \\left(\\frac {1}{a} + \\frac {1}{a^2} + ...\\right) - \\left(\\frac {r}{a} + \\frac {r^2}{a^2} + ...\\right)$\n\nso we have two different infinite geometric series.\n\n$ S = \\frac {\\frac {1}{a}}{1 - \\frac {1}{a}} - \\frac {\\frac {r}{a}}{1 - \\frac {r}{a}} = \\frac {1}{a - 1} - \\frac {r}{a - r} = \\frac {a - r - r(a - 1)}{(a - 1)(a - r)} = \\frac {a - ar}{(a - 1)(a - r)}$[/hide][/quote]\r\n\r\nFixed your $ \\text{\\LaTeX}$; the preview button's there for a reason ;)" } { "Tag": [ "logarithms", "inequalities", "inequalities proposed" ], "Problem": "Let $ a, b, c$ be real numbers with $ abc\\equal{}1$. \r\n\r\nLet $ p(x)\\equal{}2x^3\\plus{}3x^2\\minus{}6x\\plus{}1$. \r\n\r\nProve that $ \\sum \\frac{p(a)}{a^2} \\ge 0$.", "Solution_1": "[hide=\"solution\"] The sum is $ \\frac {p(a)}{a^2} \\plus{} \\frac {p(b)}{b^2} \\plus{} a^2b^2p\\left( \\frac {1}{ab}\\right)$. Expanding everything out, after some tedious algebra you find it can be factored as\n\\[ \\frac {(a \\plus{} b \\minus{} 3ab \\plus{} a^2b^2)^2}{a^2b^2}\n\\]\nwhich is non-negative. [/hide]", "Solution_2": "[quote=\"djuro\"]Let $ a, b, c$ be real numbers with $ abc \\equal{} 1$. \n\nLet $ p(x) \\equal{} 2x^3 \\plus{} 3x^2 \\minus{} 6x \\plus{} 1$. \n\nProve that $ \\sum \\frac {p(a)}{a^2} \\ge 0$.[/quote]\r\n$ \\sum_{cyc}\\frac {p(a)}{a^2}\\geq0\\Leftrightarrow\\sum_{cyc}\\left(\\frac {p(a)}{a^2} \\minus{} 6\\ln a\\right)\\geq0,$ which is true because\r\n$ \\left(\\frac {p(a)}{a^2} \\minus{} 6\\ln a\\right)' \\equal{} \\frac {2(a \\minus{} 1)^3}{a^3}.$", "Solution_3": "I have a little nicer solution. Keep trying! :wink: \r\n\r\nWhat do you think is this inequality sharp or not?", "Solution_4": "[quote=\"arqady\"]\n$ \\sum_{cyc}\\frac {p(a)}{a^2}\\geq0\\Leftrightarrow\\sum_{cyc}\\left(\\frac {p(a)}{a^2} \\minus{} 6\\ln a\\right)\\geq0$[/quote]\r\n\r\nThat'd be a slick proof for $ a,b,c > 0$, but they need not be positive, so the right hand sum is complex.", "Solution_5": "[quote=\"JoeBlow\"][quote=\"arqady\"]\n$ \\sum_{cyc}\\frac {p(a)}{a^2}\\geq0\\Leftrightarrow\\sum_{cyc}\\left(\\frac {p(a)}{a^2} \\minus{} 6\\ln a\\right)\\geq0$[/quote]\n\nThat'd be a slick proof for $ a,b,c > 0$, but they need not be positive, so the right hand sum is complex.[/quote]\nOh yes, you are right. Thank you!\nI have read \"abc=1, cyclic sum p(a)/a^2 is positive\" : \" $ a,b,c$ positive. :( \n\n\n[quote=\"djuro\"]Let $ a, b, c$ be real numbers with $ abc \\equal{} 1$. \n\nLet $ p(x) \\equal{} 2x^3 \\plus{} 3x^2 \\minus{} 6x \\plus{} 1$. \n\nProve that $ \\sum \\frac {p(a)}{a^2} \\ge 0$.[/quote]\r\nIt's equivalent to $ (ab \\plus{} ac \\plus{} bc \\minus{} 3)^2\\geq0.$", "Solution_6": "[quote=\"arqady\"]It's equivalent to $ (ab \\plus{} ac \\plus{} bc \\minus{} 3)^2\\geq0.$[/quote]\r\n\r\nYeah, that's exactly the way I have created it! :D" } { "Tag": [ "ratio", "geometry" ], "Problem": "Let XYZT be a rectangle. Let's take A the midpoint of XY and B the midpoint of XT. Let C be the intersection point of TA with ZB. Find AX/XY,BX/XT,CB/CZ, CT/CA.", "Solution_1": "[hide=\"Solution\"]For convenience, I will solve the problem stated as follows:\n[b]Let $ ABCD$ be a rectangle. Denote by $ M$ and $ N$ the midpoints of $ AB$ and $ AD$, respectively. The lines $ DM$ and $ CN$ meet in $ P$. Find the ratios $ \\frac{AM}{AB}$, $ \\frac{AN}{AD}$, $ \\frac{PN}{PC}$, $ \\frac{PD}{PM}$. [/b]\nIt is trivial that $ \\frac{AM}{AB}\\equal{}\\frac{AN}{AD}\\equal{}\\frac{1}{2}$.\nTo compute the last two ratios, I will use areas. We get $ \\frac{PN}{PC}\\equal{}\\frac{[DNM]}{[DCM]}\\equal{}\\frac{[DAM]}{2\\cdot [DCB]}\\equal{}\\frac{[DAB]}{4\\cdot [DCB]}\\equal{}\\frac{1}{4}$. At last, $ \\frac{PD}{PM}\\equal{}$ $ \\frac{[NDC]}{[NMC]}\\equal{}$ $ \\frac{[ADC]}{2\\cdot ([ABCD]\\minus{}[DNC]\\minus{}[NAM]\\minus{}[MBC])}\\equal{}$ $ \\frac{[ADC]}{4\\cdot [ABC]\\minus{}[DAC]\\minus{}[DAB]/2\\minus{}[ABC]}\\equal{}\\frac{2}{3}$.\n[b]Remark[/b]. The result is the same when $ ABCD$ is a parallelogram. However, this is an easy example - areas can be successfully used on problems which are hardly attackable by other elementary means.\n[/hide]" } { "Tag": [ "group theory", "abstract algebra", "topology", "invariant", "linear algebra", "matrix", "absolute value" ], "Problem": "Here's a nice one:\r\n There exists a neighbourhood $V$ of $I_n$ in $GL_n(C)$ that does not contain any non-trivial (with more than one element) subgroup of $GL_n(C)$.", "Solution_1": "We may assume that we're working with the norm $\\|M\\|=\\sqrt{\\sum_{i,j=1}^n|a_{ij}|^2}$, because it induces the usual topology on $\\mathcal M_n(\\mathbb C)$. \r\n\r\nSince our norm is invariant to unitary similarities, if we assume that our neighbourhood contains a non-trivial subgroup which thus has at least one non-trivial element $A$, then it also contains $UAU^*$ for any unitary matrix $U$. We can choose $U$ s.t. $UAU^*$ is upper triangular, so we may assume WLOG that $A$ is upper triangular, for simplicity. If we choose the neighbourhood small enough s.t. the eigenvalues (which must clearly have absolute value $1$) all have positive real part (i.e. the distance between such an eigenvalue and $1$ is $<\\sqrt 2$), we're done, because given any $|z|=1,z\\ne 1$, there is an integer $n$ s.t. the real part of $z^n$ is negative. \r\n\r\nI hope it's correct.", "Solution_2": "I don't know what I was thinking :?. All I have shown here is that we can find a small enough neighbourhood $\\mathcal V$ around $I_n$ s.t. if $A$ is an invertible matrix s.t. $A^n\\in\\mathcal V,\\ \\forall n\\in\\mathbb Z$, then the eigenvalues of $A$ are all equal to $1$. This doesn't mean that the matrix itself is $I_n$. :?\r\n\r\nI'll give it more thought.\r\n\r\nEdit: Actually, that's easy to take care of, since we can now assume we're working with upper-triangular matrices $A$ with $1$ on the main diagonal. By looking at the entries $1,2$ in $A^n$, we can show that $a_{1,2}=0$. Then we go on to $a_{1,3}$, and so on. We will find all the entries of $A$ which are not on the main diagonal to be $0$.", "Solution_3": "i can't understand why if $A$ is in the group then $UAU*$ is in the group :? \r\nwhat am I missing ?", "Solution_4": "[quote=\"alekk\"]if $A$ is in the group then $UAU*$ is in the group :? [/quote]\r\n\r\nThat is not what I said. This is what I meant:\r\n\r\nAssume you have a ball $\\mathcal B$ of a certain radius $r$ (under the mentioned norm) around $I_n$ which contains a group $$, where $A\\ne I_n$. Then $U\\mathcal B U^*$ is the same neighbourhood around $I_n$ (this transformation preserves the norms of the matrices) and it contains $$. From here we continue as above. \r\n\r\nI've already received two PM's with questions about this problem, so you people are beginning to make me doubt it. Now, is it correct or not? :?", "Solution_5": "Grobber, probably people wouldn't doubt about it if you wrote it a little bit clearer. ;)", "Solution_6": "Is at least the proof that there is a small enough neighbourhood of $I$ s.t. any matrices $A$ with the property that the group $$ belongs to this neighbourhood must have all eigenvalues equal to $1$ clear now?", "Solution_7": "Yes, grobber, but this was quite clear. :P ;)", "Solution_8": "Ok, then, let's take it step by step: is the part where I said that we could assume $A$ to be upper-triangular clear and correct as well? :)", "Solution_9": "Grobber, I don't need to understand like a baby, you know? I just said that the fact some peoples which are not quite idiot did not understand means that you weren't clear." } { "Tag": [ "function", "inequalities" ], "Problem": "first of all question says $ 4^x \\plus{} 3^x \\equal{} 2^x \\plus{} 5^x$\r\nLet f(x) = $ 4^x \\plus{} 3^x \\minus{} 2^x \\minus{} 5^x$\r\nf(1) = 0 so clearly 1 is a soln\r\nf(0) = 0 so clearly 0 is a soln\r\nf(2) < 0 and $ 5^x > 4^x \\plus{} 3^x$ for x > 2\r\nimplies no soln for x > 2\r\nalso for x < -2 clearly there is no soln as $ 4^x \\plus{} 3^x < 2^x$\r\nLemme think after this :)", "Solution_1": "This is just an application of Mean Value theorem isnt it?\r\nOne can use it for the function f(t) = t^{x} to the t in intervals [2,3] and [4,5] by writin the problemas 3 (to the x) - 2(to the x) = 5 (to the x)- 4 (to the x).indeed one must have c1 in (2,3) and c2 in(4,5) so that xc1(to the x-1) =3 (to the x) - 2(to the x) and similarly for the other thing. Now since c1 and c2 are in disjoint intervals, c1 and c2 are different forcing x=1 as the only solution apart from the trivial x=0. :D", "Solution_2": "perfect da this was the solution...though it looks easy the method is so ingenious :) :lol: \r\nok ur suppose dto post it as a different thread...i will move it :)", "Solution_3": "Write it as $ 5^x \\minus{} 4^x \\equal{} 3^x \\minus{} 2^x$\r\n\r\nThe function $ x^a \\minus{} (x \\minus{} 1)^a$ is strictly increasing when $ x \\geq2$ if a is not 0 or 1. So, if $ 5^x \\minus{} 4^x \\equal{} 3^x \\minus{} 2^x$, then, f'(x)=0 for all x. i.e. a=0 or a=1.", "Solution_4": "One more method occurred to me later\r\n\r\nAgain write as $ 5^x$ - $ 4^x$ = $ 3^x$ - $ 2^x$\r\n\r\nNow consider the function f(x) = $ x^a$\r\n\r\nf'(x) = $ ax^{a\\minus{}1}$\r\n\r\nf\"(x) = $ a(a\\minus{}1)x^{a\\minus{}1}$\r\n\r\nNow, f\"(x) is analogous to acceleration, it is rate at which f'(x) is increasing and here f\"(x) is non-zero when a is not 0 or 1.\r\n\r\nBut here rate of increase is zero. Hence f\"(x) = 0 i.e. a=0 or a=1", "Solution_5": "i don't get ur last method can u explain that a bit?", "Solution_6": "I meant this\r\n\r\nf'(x) > 0 tells us $ 5^x > 4^x$ and $ 3^x > 2^x$\r\n\r\nBut if f\"(x)>0 then $ \\frac {5^x \\minus{} 4^x} {5 \\minus{} 4} > \\frac {3^x \\minus{} 2^x} {3 \\minus{} 2}$\r\n\r\nAnd the opposite inequality if f\"(x)<0\r\n\r\nBut here $ 5^x \\minus{} 4^x \\equal{} 3^x \\minus{} 2^x$\r\n\r\nThis means f\"(x) = 0" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Problem1\r\nLet ABC be a triangle and construct squares ABED,BCGF,ACHI externally\r\non the sides of ABC.Show that the points D,E,F,G,H,I are concyclic \r\nif and only if ABC is equilateral or isoceles right.", "Solution_1": "Try complex numbers.. I think.. :? \r\n\r\nAnd I also think that vecten's theorem should be of some help...I think..\r\n\r\n:D", "Solution_2": "The center of this circle must be O, because the middle perpendiculars to DE and AB are coinside.\r\n\r\nAB=2R sin C\r\n\r\nOE^2=R^2+EB^2-2*R*EB cos EBO = R^2+4R^2 sin^2 C - 4R^2 sinC cos C=R^2(3+2cos(2C)+2sin(2C)).\r\n\r\nSo this six points concyclic iff cos 2C+sin2C=cos 2A+sin2A=cos 2B+sin2B\r\nbut this can be only in one case (very easy)." } { "Tag": [ "algebra", "polynomial", "search", "algebra unsolved" ], "Problem": "$ P(x),Q(x),R(x),S(x)$ are polynomial such that\r\n$ P(x^5)\\plus{}Q(x^5)\\plus{}x^2R(x^5)\\equal{}(x^4\\plus{}x^3\\plus{}x^2\\plus{}x\\plus{}1)S(x)$\r\nProve that $ x\\minus{}1|P(x)$.", "Solution_1": "I think you have a slight typo.\r\n\r\nIf I'm right, then http://www.mathlinks.ro/Forum/viewtopic.php?search_id=509040721&t=89664" } { "Tag": [], "Problem": "1 -1\\2+ 1\\ 3- 1 \\4 + ... +1\\2003 - 1\\2004 =\r\n1\\1003 + 1\\1004 +1\\1005 + ..... +1\\2003 +1\\2004", "Solution_1": "[quote=\"ashrafmod\"]1 -1\\2+ 1\\ 3- 1 \\4 + ... +1\\2003 - 1\\2004 =\n1\\1003 + 1\\1004 +1\\1005 + ..... +1\\2003 +1\\2004[/quote]\r\n$LHS=(1+\\frac{1}{2}+\\frac{1}{3}+...+\\frac{1}{2004})-2(\\frac{1}{2}+\\frac{1}{4}+...+\\frac{1}{2004})=1+\\frac{1}{2}+\\frac{1}{3}+...+\\frac{1}{2004}-1-\\frac{1}{2}-\\frac{1}{3}-...-\\frac{1}{1002}=RHS$" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "[b] (1) find the sum of 1/1+1/2+1/3+....1/n[/b]", "Solution_1": "It is Harmonic series. As I know, There is no answer of closed form.", "Solution_2": "if you can give a simple forme to this sum ..then am gonna give you Fields Medal :rotfl:", "Solution_3": "it diverges." } { "Tag": [ "geometry", "trigonometry" ], "Problem": "Given the side length of a regular n-gon, find the area.", "Solution_1": "is it to something to do with a sequence?", "Solution_2": "I'm pretty sure there's some trig way...\r\n\r\nEDIT: http://mathworld.wolfram.com/RegularPolygon.html\r\n\r\nEDIT 2: This problem is also in the intermediate forum, so this topic is locked." } { "Tag": [ "trigonometry" ], "Problem": "Find the exact value of $ \\frac {\\frac {2 \\sin (140) \\sec (280)}{\\sec (220)} \\plus{} \\frac {\\sec (340)}{\\csc (20)}}{\\frac {\\cot (200) \\minus{} \\tan(280)}{\\cot (200)}}$\r\n\r\nIt's a lot less cluttered now.", "Solution_1": "Let $ \\theta \\equal{} \\pi/9$ (radians). Then\r\n\r\n$ \\frac{2\\frac{\\sin 140 \\, \\sec 280}{\\sec 220} \\plus{} \\frac{\\sec 340}{\\csc 20}}{\\frac{\\cot 200 \\minus{} \\tan 280}{\\cot 200}} \\equal{} \\frac{2\\frac{\\sin 7\\theta \\, \\sec 14\\theta}{\\sec 11\\theta} \\plus{} \\frac{\\sec 17\\theta}{\\csc \\theta}}{\\frac{\\cot 10\\theta \\minus{} \\tan 14\\theta}{\\cot 10\\theta}} \\equal{} \\frac{2\\frac{\\sin 2\\theta \\, \\sec (\\minus{}4\\theta)}{\\minus{}\\sec 2\\theta} \\plus{} \\frac{\\sec \\theta}{\\csc \\theta}}{1 \\minus{} \\tan (\\minus{}4\\theta) \\, \\tan \\theta}$\r\n\r\n$ \\equal{} \\frac{\\minus{}2 \\sin 2\\theta \\, \\sec 4\\theta \\, \\cos 2\\theta \\plus{} \\frac{\\sin \\theta}{\\cos \\theta}}{1 \\plus{} \\tan 4\\theta \\, \\tan \\theta} \\equal{} \\frac{\\minus{}\\frac{\\sin 4\\theta}{\\cos 4\\theta} \\plus{} \\tan \\theta}{1 \\plus{} \\tan \\theta \\, \\tan 4\\theta}$\r\n\r\n$ \\equal{} \\frac{\\tan\\theta \\minus{} \\tan 4\\theta}{1 \\plus{} \\tan\\theta \\, \\tan 4\\theta} \\equal{} \\tan(\\theta \\minus{} 4\\theta) \\equal{} \\minus{}\\tan 3\\theta$\r\n\r\n$ \\equal{} \\minus{}\\tan \\frac{\\pi}{3} \\equal{} \\minus{}\\sqrt{3}$." } { "Tag": [ "trigonometry", "logarithms", "integration", "calculus", "calculus computations" ], "Problem": "$\\textbf{show}\\;:$ :D \r\n\r\n$\\sum_{n=0}^{\\infty}\\;\\left[ \\left( \\frac{3}{3n+1}\\;+\\;\\frac{1}{3n+2}\\right)\\cdot \\frac{1}{27^{n}}\\right] \\;=\\;\\boxed{\\sqrt 3 \\cdot \\tan^{-1}\\left( \\frac{\\sqrt 3}{7}\\right) }$", "Solution_1": "i think nobody can show that. because there is a error on it. the rigth formulae is\r\n\r\n${\\sum_{n=0}^{\\infty }\\frac{\\frac{1}{3 n+2}+\\frac{3}{3 n+1}}{27^{n}}=3 \\log \\left(\\frac{13}{4}\\right)}$\r\n\r\njust put n=0 and see that the sumation is greater than 1", "Solution_2": "$S = 3\\sum_{n=0}^{\\infty}\\frac{1}{3^{3n}(3n+1)}+\\sum_{n=0}^{\\infty}\\frac{1}{3^{3n}(3n+2)}$.\r\n\r\nNow:\r\n\r\n$\\sum_{n=0}^{\\infty}x^{3n}= \\frac{1}{1-x^{3}}\\Rightarrow \\sum_{n=0}^{\\infty}\\frac{x^{3n+1}}{3n+1}= \\int \\frac{1}{1-x^{3}}\\,dx$\r\n$= \\frac{1}{3}\\ln\\left(\\frac{\\sqrt{x^{2}+x+1}}{1-x}\\right)+\\frac{\\sqrt{3}}{3}\\arctan\\left(\\frac{2x+1}{\\sqrt{3}}\\right)$,\r\n\r\nand so\r\n\r\n$\\sum_{n=0}^{\\infty}\\frac{x^{3n}}{3n+1}= \\frac{1}{3x}\\ln\\left(\\frac{\\sqrt{x^{2}+x+1}}{1-x}\\right)+\\frac{\\sqrt{3}}{3x}\\arctan\\left(\\frac{2x+1}{\\sqrt{3}}\\right)$.\r\n\r\nSimilarly:\r\n\r\n$\\sum_{n=0}^{\\infty}x^{3n+1}= \\frac{x}{1-x^{3}}\\Rightarrow \\sum_{n=0}^{\\infty}\\frac{x^{3n+2}}{3n+2}= \\int \\frac{x}{1-x^{3}}\\,dx$\r\n$= \\frac{1}{3}\\ln\\left(\\frac{\\sqrt{x^{2}+x+1}}{1-x}\\right)-\\frac{\\sqrt{3}}{3}\\arctan\\left(\\frac{2x+1}{\\sqrt{3}}\\right)$\r\n\r\nand so\r\n\r\n$\\sum_{n=0}^{\\infty}\\frac{x^{3n}}{3n+2}= \\frac{1}{3x^{2}}\\ln\\left(\\frac{\\sqrt{x^{2}+x+1}}{1-x}\\right)-\\frac{\\sqrt{3}}{3x^{2}}\\arctan\\left(\\frac{2x+1}{\\sqrt{3}}\\right)$.\r\n\r\nTo evaluate S we take $x = \\frac{1}{3}$:\r\n\r\n$S = 6\\ln\\left(\\frac{\\sqrt{\\frac{1}{9}+\\frac{1}{3}+1}}{2/3}\\right) = 3\\ln\\left(\\frac{13}{4}\\right)$." } { "Tag": [], "Problem": "Let $a$ and $b$ be positive integers such that $\\frac{2}{3} < \\frac{a}{b} < \\frac{5}{7}.$.Find the value of $a + b$ when $b$ has the minimum value....\r\n :? A thorough explanation would be appreciated....", "Solution_1": "2/3 is about 0.66\r\n5/7 is about 0.71\r\n\r\nx/3 doesn't work obviously (5/7 < 3/3)\r\nx/4 doesn't work (0.5, 0.75)\r\nx/5 doesn't work (0.6, 0.8)\r\nx/6 doesn't work (0.66, 0.83)\r\nx/7 doesn't work obviously (4/7 < 2/3)\r\nx/8 doesn't work (0.625, 0.75)\r\nx/9 doesn't work (0.66, 0.77)\r\n\r\n7/10" } { "Tag": [ "vector", "linear algebra", "matrix" ], "Problem": "I know this is a VERY silly question but I don't really see how to do this right now:\r\n\r\nI want to show that an inner product (defined on a real vector space) is a non degenerate bilinear form, so basically I need to show that the matrix $A$ of the bilinear form is invertible (where $\\langle v, w \\rangle = x^TAy$, I guess this is standard notation). But how? I know that $x^TAx = 0$ implies $x = 0$ for all $x$, but then I'm stuck.", "Solution_1": "If A is a diagonalyzing matrice then you can write $y^tDy=0$ where D is the diagonal matrice with eigenvalues onto the diagonal. Write? If $d_i=0$ for some i then we cam take a y with y_i not 0 then we show that there exist an y not identically zero, which contradicts that y=0. So d_i isn\u00b4t zero for all i and thus A is invertible. Is this write? \r\n And if A isn\u00b4t a diagonayizing matrice, how to procede?", "Solution_2": "You can write A in the Jordan form...", "Solution_3": "if it were not invertible, there would be a vector $0 \\neq x \\in \\ker A$, contradiction...", "Solution_4": "Right, how could I not see that? Thanks.", "Solution_5": "I don't think you even have to use matrices\r\n\r\ndefinition of an inner product f :\r\n\r\n$f(v,w)=f(w,v)$\r\nf is linear in its first component (or the second, that is convention)\r\n$f(v,v)\\geq 0$ with equality only if $v=0$\r\n\r\ndefinition of a nondegenerate (symmetric) bilinear form\r\nthere is no nontrivial vector v such that for all w in the space $f(v,w)=0$\r\n\r\nWell if f is an innerproduct, and v is a vector, let us suppose $f(v,w)=0$ for all w in the space\r\nthen also it holds for $w=v$ thus $f(v,v)=0$ \r\nthe last part of the definition of inner product implies that $v=0$, thus f is a nondegenerate symmetric bilinear form\r\n\r\ndoes this answer your question?", "Solution_6": "Yes, but the point is that we got that exercise in class before we learnt the theorem that there is no nontrivial vector v such that for all w in the space f(v,w)=0 iff f is non-degenerate. So I was looking for a way to do it without using the theorem. Anyways, I was really tired last night :D", "Solution_7": "Strange, then what is the definition of nondegenerate? Because in our class that was the definition?", "Solution_8": "Our definition: \r\n\r\nLet $\\langle \\cdot, \\cdot \\rangle$ be a bilinear form on a finite dimensional vector space $V$ with matrix $A$ (with respect to some basis of $V$). The rank of $\\langle \\cdot, \\cdot \\rangle$ is defined as the rank of $A$. Then $\\langle \\cdot, \\cdot \\rangle$ is non degenerate if and only if the rank of $\\langle \\cdot, \\cdot \\rangle$ is equal to $\\dim V$.", "Solution_9": "Aah i see\r\nwell tastes differ of course, but I'd go for the other way around, which is more geometrical\r\nA little subtly is now that one might wonder if the rank is not dependent upon the basis, and that requires proof\r\n(not difficult however, as the matrices are congruent regarding a different basis)", "Solution_10": "Yes, that was another exercise we got in class: verify that the rank of a bilinear form is independent of the choice of a basis. By the way, I don't like the matrix approach either :(" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "khane haye yek matrix $ n \\times n$ ra ba addadhaye namanfi por mikonim be shartike jame adade har satr va sotun $ 1$ shavad sabet konid mitavan $ n$ khane az matrix entekhab kard ke mosbat bashand va hich kodam ham satr va ham sotun nabashand.", "Solution_1": "ye ghazie ghavitar has,avval ye tarif,badam ghazie:\r\n\r\n[b]Tarif:[/b] matrise $ P$ ra yek matrise tartibi namim,hargah derayehaye an $ 0,1$ bashand va dar har satro sotoone an daghighan yek $ 1$ be kar rafe bashad.\r\n\r\n[b]Ghazie:[/b] [i]farz konid $ Q$ matrisi $ n\\times n$ bashad ke derayehaye an aadadi haghighi va namanfi bashand,hamchenin jame derayehaye har satr va har sotoon barabare $ \\ell$ bashad.sabet konid chenin matrisi ra mitavan be soorate $ Q \\equal{} c_1P_1 \\plus{} c_2P_2 \\plus{} \\ldots \\plus{} c_mP_m$ nevesht ke dar an $ c_1,c_2,\\ldots ,c_m$ aadadi haghighio namanfiand ke majmooeshan barabare $ \\ell$ ast va $ P_1,P_2,\\ldots ,P_m$ matrishai $ n\\times n$ va tartibi hastand.[/i]", "Solution_2": "masale kheily rahat ba phillip hall dar miad.\r\n\r\ndar morede ghazie ke BaBaK Ghalebi goft ,esbatesh kheily rahat ba hamin masale dar miad.mishe ro tedad arraye haye nasefre matrix esteghra zad. tebghe masale yek matrix be shekle $ aB$ ke $ B$ tartibie va $ a\\leq 1$ vojode darad(a adad minimum dar n addade entekhab shodast) .agar matrixe avaliaro $ A$ begirim un mughe $ A\\minus{}aB$ tedae arraye nasefre kamtari darad(albate $ A\\minus{}aB$ sharte masala ro darad)\r\nbra ghazie esbate digari ham darin?", "Solution_3": "oon ghaziei ke goftam ye tamrine maroofe ghazie Halle,hamoontori ke khodetam gofti..." } { "Tag": [ "ARML" ], "Problem": "If anybody has information and/or application material for the 2006 ARML team, could you please post it here? Thanks.", "Solution_1": "Here it is:" } { "Tag": [ "geometry", "geometric transformation", "reflection", "projective geometry", "geometry unsolved" ], "Problem": "A line meets the lines containing sides $ BC,CA,AB$ of a triangle $ ABC$ at $ A_1,B_1,C_1,$ respectively. Points $ A_2,B_2,C_2$ are symmetric to $ A_1,B_1,C_1$ with respect to the midpoints of $ BC,CA,AB,$ respectively. Prove that $ A_2,B_2,$ and $ C_2$ are collinear.", "Solution_1": "If $ {A_1}'$ is the harmonic conjugate of $ A_1$ WRT $ BC,$ then the harmonic conjugate $ {A_2}'$ of its reflection $ A_2$ about the midpoint $ M_a$ is clearly the reflection of $ {A_1}'$ about $ M_a.$ By similar reasoning, $ A{A_2}',B{B_2}',C{C_2}'$ concur at the isotomic conjugate $ U'$ of the tripole $ U$ of the transversal $\\overline{A_1B_1C_1}.$ Then, $ A_2,B_2,C_2$ are collinear on the trilinear polar of $ U'.$", "Solution_2": "Let $ M,N,P$ respectively be the midpoint of $ BC,CA,AB$. Since $ A_1,B_1,C_1$ are collinear, we obtain:\r\n\r\n$ \\frac {A_1B}{A_1C}.\\frac {B_1C}{B_1A}.\\frac {C_1A}{C_1B} \\equal{} 1$\r\n\r\nSince $ A_2,B_2,C_2$ respectively are the symmentric points of $ A_1,B_1,C_1$ wrt $ M,N,P$, therefore, we obtain:\r\n\r\n$ \\frac {A_2B}{A_2C}.\\frac {B_2C}{B_2A}.\\frac {C_2A}{C_2B} \\equal{} 1$, which follows that $ A_2,B_2,C_2$ are collinear. $ \\square$" } { "Tag": [], "Problem": "Let a and b be integers not divisible by the prime number p. If a^p = b^p (mod p), prove that a^p = b^p (mod p^2). \r\nPlease help me!", "Solution_1": "Have you posted it in the right forum? Anyway use Fermats little theorem to show that $a-b$ is congruent to zero mod $p$ then use the fact that $a^{p}-b^{p}=(a-b)(a^{p-1}+\\dots+b^{p-1})$ to finish your proof.", "Solution_2": "[hide=\"Solution\"] Let $c \\equiv \\frac{a}{b}$. We are given $c^{p}\\equiv c \\equiv 1 \\bmod p$. Let $c = 1+pk$. Then\n\n$(1+pk)^{p}= 1+p(pk)+\\frac{p(p-1)}{2}(pk)^{2}+... \\equiv 1 \\bmod p^{2}$ [/hide]" } { "Tag": [ "topology" ], "Problem": "Def: A compact space X is maximal compact iff every strictly larger (finer) topology on X is noncompact.\r\n\r\nProblem:\r\n1) A compact space X is maximal compact iff every compact subset is closed.\r\n\r\nFor one direction of (1), I can argue as follows:\r\n\r\nProof: Suppose that every compact subset of X is closed with respect to $ \\tau_{1}$. Suppose on the contrary there exists a $ \\tau_{2}$ that is strictly finer than $ \\tau_{1}$ such that $ (X, \\tau_{2})$ is compact. Let C be a closed subset of $ (X, \\tau_{2})$. Then since every closed subset of a compact space is compact, C must be compact in $ (X, \\tau_{2})$. But then C is compact in $ (X, \\tau_{1})$ because $ \\tau_{1} \\subset \\tau_{2}$ and since every compact subset of $ ( X, \\tau_{1})$ is closed by the hypothesis, \r\nC is closed in $ (X, \\tau_{1})$, which means that $ \\tau_{2} \\subset \\tau_{1}$, a contradiction. Hence if every compact subset of X is closed then X is maximal compact. \r\n\r\nThe other direction is the one I would need help on.", "Solution_1": "For that other direction, use the contrapositive. Suppose $ K$ is a non-closed compact set. Define a new finer topology $ \\tau'$ on $ X$, generated by the open sets of the original topology $ \\tau$ and $ P=X\\setminus K$. Every open set in $ \\tau'$ is of the form $ (U\\cap P)\\cup V$ for some open sets $ U$ and $ V$ in $ \\tau$.\r\nNow, take an open (in $ \\tau'$) cover of $ X$ $ \\{(U_i\\cap P)\\cup V_i\\}$. Split it into $ \\{U_i\\cap P\\}\\cup \\{V_i\\}$; the subdivided open cover has a finite subcover if and only if the original did. Now, $ \\{V_i\\}$ is an open cover of $ K$, which has a finite subcover $ \\{G_j\\}$. Also, $ \\{U_i\\}\\cup \\{V_i\\}$ is an open cover of $ X$ in $ \\tau$, which has a finite subcover ${ \\{H_k\\}\\cup\\{G_l}$ (splitting among the two parts). The combination $ \\{H_k\\cap P\\}\\cup \\{G_j\\}\\cup \\{G_l\\}$ is a finite open cover of $ X$ with respect to $ \\tau'$, and is a subcover of our original (subdivided) cover.\r\nThis could be done for any open cover, so the finer topology $ \\tau'$ is compact. Hence, $ X$ is not maximal compact." } { "Tag": [ "\\/closed" ], "Problem": "I noticed the Resources page has lots of problems from previous years of math contests. I was just wondering, is there a page with answers to these problems?", "Solution_1": "I'm not sure what you're referring to. Our resources page from the left sidebar? Where are the problems?", "Solution_2": "You mean the Contests section, right? We are still working on those. We're more focused on gettting the problems up right now. There are still solutions linked to some of them. For example, http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2002 ...the S's on the right side of the page link to the solutions. The process of putting all of the problems and solutions into the database is a long and slow one. Bear with us :)", "Solution_3": "The resources pages will be upgraded at some point to provide easier means to input solutions. For now you can click on the problem's number to discuss it on the forum." } { "Tag": [ "geometry", "rectangle", "calculus", "calculus computations" ], "Problem": "Find the solutions $u(x,y)$ to\r\n\\[ u\\frac{\\partial^2 u}{\\partial x\\partial y} = \\frac{\\partial u}{\\partial x}\\frac{\\partial u}{\\partial y} \\]", "Solution_1": "Very nice indeed.\r\n\r\n[hide=\"Idea\"]\n\n$u\\frac{\\partial^2 u}{\\partial x\\partial y} = \\frac{\\partial u}{\\partial x}\\frac{\\partial u}{\\partial y}$\n$\\Rightarrow \\frac1u\\left(u\\frac{\\partial^2 u}{\\partial x\\partial y} - \\frac{\\partial u}{\\partial x}\\frac{\\partial u}{\\partial y}\\right)=0$\n$\\Rightarrow \\frac{\\partial }{\\partial x}\\left(\\frac1u\\frac{\\partial u}{\\partial y}\\right)=0$ or $\\frac{\\partial }{\\partial y}\\left(\\frac1u\\frac{\\partial u}{\\partial x}\\right)=0$\n\n[/hide]\r\n\r\nYou can complete from here!", "Solution_2": "Another idea, which turns this nonlinear equation into a truly simple linear PDE.\r\n\r\n[hide=\"hint\"]Let $u=e^v.$ (Deal with the possibility that $u$ changes signs later)[/hide]\n\n[hide=\"answer\"]$u(x,y)=f(x)g(y)$[/hide]", "Solution_3": "[quote=\"jmerry\"] answer $u(x,y)=f(x)g(y)$[/quote]\r\nThat is indeed true if $u$ doesnt vanish anywhere. But you may have terribly complicated solutions looking like countably many different islands of this kind supported by rectangles floating in the ocean where $u=0$..." } { "Tag": [], "Problem": "Who do you think will be really good next year in the NFL?", "Solution_1": "Carolina Panthers\r\n\r\nI hope patriots go 0-16.", "Solution_2": "philly!!! come back and repeat as NFC champs, but this time, win the Super Bowl :D", "Solution_3": "As long as Belicheck is still coach, New England Patriots are the team to beat.", "Solution_4": "this will not happen, BUT GO COWBOYS!!!!", "Solution_5": "[quote=\"GoBraves\"]I hope patriots go 0-16.[/quote]\nI hope the Patsies play the Steelers in New England for the AFC Championship, the Steelers are supposed to lose by 20, and the Patriots are up by 6 with 10 seconds to play, Brady fumbles it, and the Steelers run it back for a touchdown!(and win)\n\n[quote=\"plokoon51\"]philly!!! come back and repeat as NFC champs, but this time, win the Super Bowl [/quote]\r\nI live near Philly, so I want them to win the NFC, but not the Superbowl. That spot is reserved for the STEELERS :D :D :D :D :D :D :D :D :D :D", "Solution_6": "[quote=\"ln(dx/dy)\"][quote=\"GoBraves\"]I hope patriots go 0-16.[/quote]\nI hope the Patsies play the Steelers in New England for the AFC Championship, the Steelers are supposed to lose by 20, and the Patriots are up by 6 with 10 seconds to play, Brady fumbles it, and the Steelers run it back for a touchdown!(and win)[/quote]\r\n\r\nLol, very entertaining story. \r\n\r\nToo bad there's a such thing as a \"quarterback kneel\".", "Solution_7": "the raiders are gonna be killin it next year.... :lol: anyways.... now they got randy moss, and trade for micheal vick.... tom brady ain't nothin!\r\n\r\nby the way, hope the patriots suck next year!!! 0-16 :D :D :D", "Solution_8": "[quote=\"KrNboy10\"]the raiders are gonna be killin it next year.... :lol: anyways.... now they got randy moss, and trade for micheal vick.... tom brady ain't nothin!\n\nby the way, hope the patriots suck next year!!! 0-16 :D :D :D[/quote]\r\n\r\nLOL, RAIDERS??? They still don't have any defense at all!", "Solution_9": "mark my words, the $Minnesota Vikings - Randy Moss = Vikings^{2}$\r\n\r\nThis year $Vikings \\geq Any Past Superbowl Winning Team$", "Solution_10": "Amen to that!", "Solution_11": "NFL, is that sth about the SUPER BOWL?", "Solution_12": "[quote=\"shobber\"]NFL, is that sth about the SUPER BOWL?[/quote]\r\n\r\nwhat is 'sth'? I think that what has been posted pertains to next year: aka Year of the Dynasty.", "Solution_13": "Who do you think the 49ers will draft with their #1 pick? I like them but there is consensus on the best player this year.... A bunch of good players but no superstar....", "Solution_14": "[quote=\"Jhyon8149\"]Who do you think the 49ers will draft with their #1 pick? I like them but there is consensus on the best player this year.... A bunch of good players but no superstar....[/quote]\r\n\r\nthey (ESPN) say they'll go after a QB like Alex Smith of Utah or Aaron Rodgers of Cal. but i thought they still had Ken Dorsey, who should be a pretty able (future) QB...\r\n\r\nthen again, the 49ers aren't really my all-time favorite team in the world", "Solution_15": "Patriots! Yeah! They're gonna win again! 4 in 5 years!\r\n\r\n...erm, I could swear I remember someone saying something similar before :lol:", "Solution_16": "[b][size=200][color=green]Eagles!![/color][/size][/b]", "Solution_17": "[quote=\"ankur87\"][b][size=200][color=green]Eagles!![/color][/size][/b][/quote]\r\n\r\nankur87, i like your thinking :D", "Solution_18": "[color=red][size=200]BUCCANEERS!!!!!!!![/size][/color]\r\n\r\nthey happen to be the team that won the super bowl between all the patriot's wins. nobody seems to realize that. and they'll do it again this year.", "Solution_19": "Go Kansas City Royals!!!,\r\ner, Chiefs!! (whichever one plays football)", "Solution_20": "[quote=\"miraculouspostmaster\"]Go Kansas City Royals!!!,\ner, Chiefs!! (whichever one plays football)[/quote]\r\n\r\nLOL. i can tell you are a real sports fan.", "Solution_21": "[quote=\"ln(dx/dy)\"]Who do you think will be really good next year in the NFL?[/quote]\r\n\r\nAll Philadelphia Super Bowl would be amazing. If not, Steelers making to the Super Bowl, which is WAY overdue. \r\n\r\nI agree with GoBraves....0-16 for the Patriots would be pretty cool. :P", "Solution_22": "[color=green][size=200][u][b]EAGLES![/b][/u][/size][/color]", "Solution_23": "[quote=\"plokoon51\"][quote=\"ankur87\"][b][size=200][color=green]Eagles!![/color][/size][/b][/quote]\n\nankur87, i like your thinking :D[/quote]\r\n\r\nAs do I! :D", "Solution_24": "[quote=\"Dcadu9lvr\"][quote=\"plokoon51\"][quote=\"ankur87\"][b][size=200][color=green]Eagles!![/color][/size][/b][/quote]\n\nankur87, i like your thinking :D[/quote]\n\nAs do I! :D[/quote]\r\n\r\nAnd I admire your guys' ability to recognize good thinking.", "Solution_25": "[quote=\"ankur87\"][/quote][quote=\"Dcadu9lvr\"][quote=\"plokoon51\"][quote=\"ankur87\"][b][size=200][color=green]Eagles!![/color][/size][/b][/quote]\n\nankur87, i like your thinking :D[/quote]\n\nAs do I! :D[/quote][quote=\"ankur87\"]\n\nAnd I admire your guys' ability to recognize good thinking.[/quote]\r\n\r\nand i scorn ur inability to recognize good thinking....\r\n[b]COWBOYS!!![/b]", "Solution_26": "Up until the late 90s, your \"team\" was consisted mainly of jailbirds and problem-players. \r\n\r\n\r\nAnd Michael Irvin is a terrible TV personality :P", "Solution_27": "lol, who cares they are AMERICA's TEAM" } { "Tag": [ "quadratics", "algebra" ], "Problem": "\\[ \\frac{1}{{p \\plus{} q \\plus{} x}} \\equal{} \\frac{1}{p} \\plus{} \\frac{1}{q} \\plus{} \\frac{1}{x}\r\n\\]", "Solution_1": "hello, we want to solve the equation with respect to $ x$. The range of definition is $ p\\neq0,q\\neq0,x\\neq0$ and $ p\\plus{}q\\neq\\minus{}x$. Further we have\r\n$ \\frac{1}{p\\plus{}q\\plus{}x}\\equal{}\\frac{x(p\\plus{}q)\\plus{}pq}{pqx} \\Leftrightarrow pqx\\equal{}(p\\plus{}q\\plus{}x)(pq\\plus{}x(p\\plus{}q))$. Simplifying this we get the equation\r\n$ 0\\equal{}(x^2\\plus{}x(p\\plus{}q)\\plus{}pq)(p\\plus{}q)$. If $ p\\plus{}q\\equal{}0$ then solves $ x \\in \\mathbb{R}, x\\neq0$ the equation. Assuming $ p\\plus{}q\\neq0$ and solving the quadratic equation we get $ x_1\\equal{}\\minus{}q$ or $ x_2\\equal{}\\minus{}p$.\r\nSonnhard." } { "Tag": [ "trigonometry" ], "Problem": "Given a triangle $ABC$ let \\begin{eqnarray*} x &=& \\tan\\left(\\dfrac{B-C}{2}\\right) \\tan \\left(\\dfrac{A}{2}\\right) \\\\ y &=& \\tan\\left(\\dfrac{C-A}{2}\\right) \\tan \\left(\\dfrac{B}{2}\\right) \\\\ z &=& \\tan\\left(\\dfrac{A-B}{2}\\right) \\tan \\left(\\dfrac{C}{2}\\right). \\end{eqnarray*} Prove that $x+ y + z + xyz = 0$.", "Solution_1": "We now by trigonometry \\[ \\displaystyle tan(p)tan(q)=\\dfrac{cos{p-q}-cos(p+q)}{cos(p-q)+cos(p+q)} \\] So $x=\\dfrac{cos(B)-cos(C)}{cos(B)+cos(C)}$ and so on. If we replace these values in the equality it is equivalent to$\\Sigma{(cos(B)-cos(C))(cos(A)+cos(B))(cos(A)+cos(C))}=0$. The sum of these three is:\\[ \\Sigma{cos^2(A)cos(B)-\\Sigma{cos(A)cos^2(B)}=0 }\\]. The first side is \\[ (cos(A)-cos(B))(cos(B)-cos(C))(cos(C)-cos(A)) \\]. I don't know if I am wrong or your problem is not correct.", "Solution_2": "[b]Lemma[/b]: $\\tan{\\frac{(C-A)}{2}}\\tan{\\frac{(B-C)}{2}}\\tan{\\frac{(A-B)}{2}}=\\tan{\\frac{(C-A)}{2}}+\\tan{\\frac{(B-C)}{2}}+\\tan{\\frac{(A-B)}{2}}$.\r\n[u]Proof[/u]: The statement of the lemma is equivalent to $\\tan{\\frac{(C-A)}{2}}=-\\frac{\\tan{\\frac{(B-C)}{2}}+\\tan{\\frac{(A-B)}{2}}}{1-\\tan{\\frac{(B-C)}{2}}\\tan{\\frac{(A-B)}{2}}}=-\\tan{\\frac{(A-C)}{2}}$. But this equality is obvious since $\\tan{x}=-\\tan{-x}$.\r\n\r\n[color=red]Proof[/color]: Put the given expressions of $x,y,z$ into the identity we need to prove to obtain $\\tan{\\frac{B-C}{2}}\\tan{\\frac{A}{2}}+\\tan{\\frac{C-A}{2}}\\tan{\\frac{B}{2}}+\\tan{\\frac{A-B}{2}}\\tan{\\frac{C}{2}}+\\tan{\\frac{A}{2}}\\tan{\\frac{B}{2}}\\tan{\\frac{C}{2}}\\tan{\\frac{(C-A)}{2}}\\tan{\\frac{(B-C)}{2}}\\tan{\\frac{(A-B)}{2}}=0$ \r\n\r\nUsing the statement of the lemma this identity is equivalent to \\[ \\tan{\\frac{B-C}{2}}\\tan{\\frac{A}{2}}+\\tan{\\frac{C-A}{2}}\\tan{\\frac{B}{2}}+\\tan{\\frac{A-B}{2}}\\tan{\\frac{C}{2}}+\\tan{\\frac{A}{2}}\\tan{\\frac{B}{2}}\\tan{\\frac{C}{2}}(\\tan{\\frac{(C-A)}{2}}+\\tan{\\frac{(B-C)}{2}}+\\tan{\\frac{(A-B)}{2}})=0 \\]\r\n$\\Longleftrightarrow\\tan{\\frac{B-C}{2}}\\tan{\\frac{A}{2}}(1+\\tan{\\frac{B}{2}}\\tan{\\frac{C}{2}})+\\tan{\\frac{C-A}{2}}\\tan{\\frac{B}{2}}(1+\\tan{\\frac{A}{2}}\\tan{\\frac{C}{2}})+\\tan{\\frac{A-B}{2}}\\tan{\\frac{C}{2}}(1+\\tan{\\frac{A}{2}}\\tan{\\frac{B}{2}})=0$\r\n$\\Longleftrightarrow\\sum_{cyc}\\tan{\\frac{B-C}{2}}\\tan{\\frac{A}{2}}(1+\\frac{\\cos{\\frac{B-C}{2}}}{\\cos{\\frac{B}{2}}\\cos{\\frac{C}{2}}})=0$\r\n$\\Longleftrightarrow\\sum_{cyc}\\sin{\\frac{B-C}{2}}\\sin{\\frac{A}{2}}=0$\r\n$\\Longleftrightarrow\\sum_{cyc}\\frac{1}{2}(\\cos{\\frac{B-C-A}{2}}-\\cos{\\frac{A+B-C}{2}})=0$\r\nBut the last identity is trivial since $\\cos{-x}=\\cos{x}$.", "Solution_3": "it's well known that in a triangle (a+b)/(a-b)=tan(A/2+B/2) / tan(A/2-B/2)\r\n\r\nsince tan (A/2)= tan(90-B/2-C/2)=ctg(B/2+C/2)=1/tan(B/2+C/2)\r\n\r\nthen we have those three eqautions\r\n\r\nx=(b-c)/(b+c)\r\ny=(c-a)/(c+a)\r\nz=(a-b)/(a+b)\r\n\r\nwe can put now \r\ne=a+b\r\nf= b+c\r\ng=c+a\r\n\r\nand a-b= g-f, b-c= e-g and c-a= f-e\r\n\r\nand then the eqution is pretty easy to prove..", "Solution_4": "[hide=\"Here is a similar proof with the nice Felixx-s proof.\"][u]Theorem of the tangent[/u] ([b]Mollweide[/b]) $x\\equiv \\tan \\frac{B-C}{2}\\cdot \\tan \\frac A2=\\frac{b-c}{b+c}$ a.s.o. $\\Longrightarrow$\n\n$\\{ \\begin{array}{cc} 0\\cdot a+(x-1)\\cdot b+(x+1)\\cdot c=0\\\\ (y+1)\\cdot a+0\\cdot b+(y-1)\\cdot c=0\\\\ (z-1)\\cdot a+(z+1)\\cdot b+0\\cdot c=0\\end{array}\\Longrightarrow | \\begin{array}{ccc}0 & x-1 & x+1\\\\ y+1 & 0 & y-1\\\\ z-1 & z+1 & 0\\ \\end{array}|=0\\Longrightarrow$\n\n$(1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)\\Longrightarrow x+y+z+xyz=0\\ .$[/hide]", "Solution_5": "If we notice carefully , we dont even need the assumption that $A,B,C$ is angles in a triangle . :) The result remains true for any real $A,B,C$ ;)", "Solution_6": "Simply use Napier's Analogy, which gives x = (b-c)/(b+c), etc. So, x + y + z = -(a-b)(b-c )(c-a )/[(a+b)(b+c)(c+a)] = -xyz. Hence proved. \n\n", "Solution_7": "A similar question from KVS-JMO 2001 paper:\n\nSimplify:\n(a-b )/(a+b) + (b-c )/(b+c) + (c-a )/(c+a) + [(a-b )(b-c )(c-a )]/[(a+b)(b+c)(c+a)]", "Solution_8": "Given,\n$x=tan(B-C/2)tan(A/2)=tan(B-C/2)tan(90^{\\circ{}}-(B+C/2))=tan(B-C/2)cot(B+C/2)=tan(B-C/2)/tan(B+C/2)=\\frac{sin(B-C/2)cos(B+C/2)}{sin(B+C/2)cos(B-C/2)}=2\\frac{2sin(B-C/2)cos(B+C/2)}{2sin(B+C/2)cos(B-C/2)}=\\frac{sinB-sinC}{sinB+sinC}$\nSimilarly, \n$y=tan(C-A/2)tan(B/2)=tan(C-A/2)/tan(C+A/2)=\\frac{sinC-sinA}{sinC+sinA}$\n$z=tan(A-B/2)tan(C/2)=tan(A-B/2)/tan(A+B/2)=\\frac{sinA-sinB}{sinA+sinB}$\nSo,\n$\\frac{sinB-sinC}{sinB+sinC}+\\frac{sinC-sinA}{sinC+sinA}+\\frac{sinA-sinB}{sinA+sinB}\n+\\frac{sinB-sinC}{sinB+sinC}\\frac{sinC-sinA}{sinC+sinA}\\frac{sinA-sinB}{sinA+sinB}$\n$((sinB-sinC)(sinA+sinB)(sinC+sinA) + (sinB+sinC)(sinA-sinB)(sinC+sinA)+(sinB+sinC)(sinA+sinB)(sinC-sinA)+(sinB-sinC)(sinA-sinB)(sinC-sinA))/((sinB+sinC)(sinA+sinB)(sinC+sinA))=0$\n$x+y+z+xyz=0$\n", "Solution_9": "[hide=a problem of worth!] so we have $x=\\tan{(\\frac{B-C}{2})}\\cdot \\tan{\\frac{A}{2}}$\\\\\n$A=90^{\\circ}-(\\frac{B+C}{2})$\\\\\n$x=\\frac{\\tan\\left(\\dfrac{B-C}{2}\\right)}{\\tan\\left(\\dfrac{B+C}{2}\\right)}$ $\\implies x=\\frac{\\sin\\left(\\dfrac{B-C}{2}\\right)\\cdot \\cos\\left(\\dfrac{B+C}{2}\\right)}{\\sin\\left(\\dfrac{B+C}{2}\\right)\\cdot \\cos\\left(\\dfrac{B-C}{2}\\right)} $\\\\\nwe get $x=\\frac{\\sin{B}-\\sin{C}}{\\sin{B}+\\sin{C}}\\implies \\frac{1-x}{1+x}=\\frac{\\sin{C}}{\\sin{B}}$\\\\\nsimilarly compute $\\frac{1-y}{1+y}$ and $\\frac{1-z}{1+z}$ to get \\\\\n$\\frac{1-x}{1+x}\\cdot \\frac{1-y}{1+y}\\cdot \\frac{1-z}{1+z}=1\\implies (1-x)(1-y)(1-z)=(1+x)(1+y)(1+z)$\\\\\nconsider a cubic equation in $\"t\"$ with roots $x,y,z$ s.t.\\\\\n$P(t)=t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz$\\\\\nwe have $P(1)=-P(-1) \\implies \\boxed{x+y+z+xyz=0} \\blacksquare$[/hide]" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Six circles are tangent to the seventh one, such that each of them is tangent to its two neighbours. Prove, that three lines joing opposite points of tangency are concurrent in a point.\r\n\r\nIt is also on [url=http://mathworld.wolfram.com/SevenCirclesTheorem.html]mathworld.[/url]", "Solution_1": "Dear Mathlinkers,\r\n a nice proof can be find on the website of Stanley Rabinowitz \r\nhttp://www.mathpropress.com/stan/\r\nSincerely\r\nJean-Louis" } { "Tag": [ "calculus", "integration", "superior algebra", "superior algebra unsolved" ], "Problem": "need to find orthogonal base of {1 ,sinx ,cosx} x (o or 2p) iner product = 0 integral 2p (f(x)g(x))dx [/img]", "Solution_1": "Exactly where do you get stuck?\r\n\r\nAlso, this belongs in the linear algebra forum.", "Solution_2": "i traid to find Eigenvector no success", "Solution_3": "Please be more specific. I'll give you one eigenvector which I found by inspection, $ (1,1,1)$." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "show An is only subgroup of index 2 of Sn.", "Solution_1": "It was posted before that $A_{n}$ is the only nontrivial normal subgroup if $n>4$ (can be seen by intersecting with the simple group $A_{n}$).\r\nBut every subgroup of index $2$ is normal -> done.\r\n\r\nAnother argument: the problem is equivalent to determining a surjective homomorphism onto $\\mathbb{Z}/(2)$. But that requires us to send at least one transposition to $1 \\mod 2$ (they generate $S_{n}$), thus they require us to send all transpositions there (conjugate them).\r\nThus the kernel will always be $A_{n}$.", "Solution_2": "will you please give details of An is only nontrivial normal subgroup of Sn where n>4", "Solution_3": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=116164 ." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "a,b,c are positive real numbers, such that you can't construct with them a triangle.\r\n$Prove$ $that$\r\n$\\sum {\\frac{a}{b} + \\sum {\\frac{b}{a} \\ge 7} }$.", "Solution_1": "[quote=\"vardanch\"]a,b,c are positive real numbers, such that you can't construct with them a triangle.\n$Prove$ $that$\n$\\sum {\\frac{a}{b} + \\sum {\\frac{b}{a} \\ge 7} }$.[/quote]\r\nLet $a\\geq b+c.$\r\nSince $\\frac{b}{c}+\\frac{c}{b}\\geq2$ remain to prove that\r\n$a\\left(\\frac{1}{b}+\\frac{1}{c}\\right)+\\frac{b+c}{a}\\geq5\\Leftrightarrow$\r\n$\\Leftrightarrow(b+c)a^2-5bca+(b+c)bc\\geq0.$\r\nLet $f(x)=(b+c)x^2-5bcx+(b+c)bc.$\r\nBut $\\frac{5bc}{2(b+c)}\\leq b+c$ and $f(b+c)\\geq0.$ \r\nHence, $f(a)\\geq f(b+c)\\geq0.$ :)", "Solution_2": "$PROBLEM$\r\n[quote=\"vardanch\"]a,b,c are positive real numbers, such that you can't construct with them a triangle.\n$Prove$ $that$\n$\\sum {\\frac{a}{b} + \\sum {\\frac{b}{a} \\ge 7} }$.[/quote]\r\n$SOLUTION$\r\n\r\nLet $a\\leq b ab, \\quad a \\ne b$", "Solution_1": "Axiomatic proof ???? :|", "Solution_2": "[quote=\"Ramanujan\"]Axiomatic proof ???? :|[/quote]\r\n\r\nHOW ??", "Solution_3": "$1)$ $\\boxed{a\\geq b \\Longrightarrow a+c \\geq b+c}$ is an axiom at an ordered field like $\\mathbb{R}$.\r\nNow since $a+c \\geq b+c$ adding $(-c)$ we take $a+c \\geq b+c \\Longrightarrow a+c +(-c) \\geq b+c +(-c) \\Longrightarrow a \\geq b$\r\n\r\n$2)$ $a \\geq b \\Longrightarrow a + (-b) \\geq b+ (-b) \\Longrightarrow a-b \\geq 0$\r\nNow another axiom say that $\\boxed{a \\geq 0 \\, \\wedge b \\geq 0 \\Longrightarrow a \\cdot b \\geq 0}$.\r\nWe have $a-b \\geq 0$ and $c \\geq 0$ so $c \\cdot (a-b) \\geq 0 \\Longrightarrow ca-cb\\geq 0 \\Longrightarrow ca-cb +cb \\geq cb \\Longrightarrow ca \\geq cb$.\r\n\r\n$3)$ For this i am not sure that it needs a proof. Maybe it comes from $x^2 \\geq 0 , \\, \\forall x \\in \\mathbb{R}$.\r\nHere is an idea: If $x \\geq 0$ then $x^2 \\geq 0$ using the second axiom. If $x \\leq 0$ then $- x \\geq 0$ and again $(- x)^2 \\geq 0$ or $x^2 \\geq 0$.\r\nNow take $x= \\frac{a-b}{2}$ and bring it to the form you want using the axioms . I think that is. :)\r\n\r\nI am not sure why $(- x)^2 = x^2$ . Anyone help ?? :D", "Solution_4": "Thank you for help. I think I found the 3rd one:\r\n\r\n$a \\ne b \\Longleftrightarrow a>b \\vee b>a$\r\n\r\n$i) a > b \\Longleftrightarrow a-b>0 \\Longleftrightarrow (a-b)^2>0\\Longleftrightarrow a^2 +b^2 -2ab>0\\Longleftrightarrow a^2+b^2+2ab>4ab \\Longleftrightarrow \\boxed{\\left (\\frac{a+b}{2}\\right)^2>ab}$\r\n\r\n$ii) b>a \\Longleftrightarrow b-a>0 \\Longleftrightarrow (b-a)^2>0\\Longleftrightarrow b^2+a^2-2ab>0 \\Longleftrightarrow b^2+a^2+2ab>4ab \\Longleftrightarrow \\boxed{\\left (\\frac{a+b}{2}\\right)^2>ab}$\r\n\r\n$\\therefore \\boxed{\\left (\\frac{a+b}{2}\\right)^2>ab, \\quad a\\ne b}$" } { "Tag": [ "MIT", "college", "Harvard", "Princeton", "Yale" ], "Problem": "I am applying to some colleges for Fall 2008 and wanted to know whether I must send 2 high school teacher recommendations apart from the counselor recommendation. Can't I send 1 high school recommendation and 1 from a university mathematics professor? I think that a weak recommendation from school would further weaken my already weak chances of getting in. Besides, I live in India and people in my high school know almost nothing about writing recommendations. I am applying only to need-blind colleges like Northwestern, Boston College, Cornell, Stanford. Could anyone guide me in this? I am posting this here because those colleges don't reply until weeks after I send them any e-mail.", "Solution_1": "I myself am using a recommendation from a teacher at a university and another from a retired teacher. It's my understanding that as long as the teachers taught classes that you took while in high school, this is valid.", "Solution_2": "The important thing is to find a teacher who knows you well enough to mention specific information about you and above to find a teacher who likes you and will give you a strong recommendation. If your most recent math teacher teaches at a college rather than at a high school, that should be fine.", "Solution_3": "Thanks for all the info. My low Gpa in 10th grade and lack of international contests like IMO make it necessary for me to have a strong recommendation, which is why I want to send one from him. The people at the high school I go to do not know me well enough to give a strong recommendation.", "Solution_4": "Are you applying to the institutions that are need-blind for international students? They are as follows:\r\n\r\n 1. MIT, Massachusetts Institute of Technology in Massachusetts\r\n 2. Harvard University in Massachusetts\r\n 3. Princeton University in New Jersey\r\n 4. Yale University in Connecticut\r\n 5. Williams College in Massachusetts\r\n 6. Middlebury College in Vermont", "Solution_5": "Actually, I am not an international student since I am a US citizen. However, I have been living outside USA for 16 years. Hence, all states will treat me as a domestic but out-state applicant. That is why I cannot apply to state colleges which might be easier to get into." } { "Tag": [ "limit", "LaTeX", "calculus", "real analysis" ], "Problem": "Can someone explain the concept of infinity?\r\n\r\nQUESTIONS:\r\n\r\n(1) What is infinity divided by infinity? Explain.\r\n\r\n(2) Why is any number over x = 0? \r\n\r\nSAMPLES: \r\n\r\n(1/x) = 0\r\n\r\n(2/x) = 0\r\n\r\n(3/x) = 0, etc.....", "Solution_1": "[quote=\"Interval\"]Can someone explain the concept of infinity?\n\nQUESTIONS:\n\n(1) What is infinity divided by infinity? Explain.\n\n(2) Why is any number over x = 0? \n\nSAMPLES: \n\n(1/x) = 0\n\n(2/x) = 0\n\n(3/x) = 0, etc.....[/quote]\r\nPlease don't start this debate again.", "Solution_2": "$\\frac{\\infty}{\\infty}$ is what is called an indeterminate form:\r\n\r\n [url]http://en.wikipedia.org/wiki/Indeterminate_form[/url].\r\n\r\nNow for your second question, think of the limit $\\lim_{x\\to\\infty}\\frac{a}{x},\\ a\\in\\mathbb{R}_{+}$.", "Solution_3": "You said:\r\n\r\ninfinity/infinity is called an indeterminate form.\r\n\r\nWhy is it INDETERMINATE?\r\n\r\nWhat about the other question?", "Solution_4": "Well one way to think about it is that you don't know the relative size of the infities. Consider $\\lim_{x\\to\\infty}\\frac{e^{x}}{x}$ this gives you $\\frac{\\infty}{\\infty}$, but the limit is $\\infty$ because the infinity in the numerator is greater. \r\n\r\nYou really have to analyze the situation from which the infities arose.", "Solution_5": "[quote=\"Interval\"]Can someone explain the concept of infinity?\n\nQUESTIONS:\n\n(1) What is infinity divided by infinity? Explain.\n\n(2) Why is any number over x = 0? \n\nSAMPLES: \n\n(1/x) = 0\n\n(2/x) = 0\n\n(3/x) = 0, etc.....[/quote]\r\n\r\nthere are different infinities. the lim x-> infinity of x/x is 1 while the lim x-> infinity of x^2/x is infinity.\r\n\r\nalso, nobody said that any number over x is 0 :maybe: :huh:", "Solution_6": "[quote=\"jli\"][quote=\"Interval\"]Can someone explain the concept of infinity?\n\nQUESTIONS:\n\n(1) What is infinity divided by infinity? Explain.\n\n(2) Why is any number over x = 0? \n\nSAMPLES: \n\n(1/x) = 0\n\n(2/x) = 0\n\n(3/x) = 0, etc.....[/quote]\n\nthere are different infinities. the lim x-> infinity of x/x is 1 while the lim x-> infinity of x^2/x is infinity.\n\nalso, nobody said that any number over x is 0 :maybe: :huh:[/quote]\r\n\r\nI think he meant for the x to represent infinity... in which case, Interval, use the latex code [code]\\infty[/code] to use the symbol for infinity.\r\n\r\nAnyways, imagine we have the fraction $\\frac1x$. Then, say that $x$ gets really big. Not just big; HUGE. That is to say, it approaches infinity. Then, think about what happens when the denominator gets large. The fraction gets smaller. One hundredth is smaller than one tenth; one trillionth is alot smaller than one hundredth. But no, infinity is alot bigger than a trillion. Thus, it will continue to shrink, and shrink, and shrink, until it becomes so infinitesimally small that it approaches zero. Even if we multiply it by some number, ANY constant number $k$, it will still shrink until it approaches zero. \r\n\r\nTranslated to math, this means that for $k \\in \\mathbb{R}$ we have that $\\lim_{x \\rightarrow \\infty^{+}}\\left( \\frac{k}x \\right) =0$.", "Solution_7": "[quote=\"Interval\"](1) What is infinity divided by infinity? [/quote]\n\nI think a better question to ask would be\n\n[quote=\"Interval\"](0) [b]What is infinity?[/b][/quote]\r\n\r\nGood question!\r\n\r\nFirst of all, it's very important to realize that [b]infinity is not a number in the normal sense[/b]. It's just a concept. Operations involving infinity are not necessarily well-defined. \r\n\r\nA rigorous treatment of the infinite begins in calculus (real analysis), where we consider [b]limits.[/b] Rather than naively \"plugging in infinity,\" we say that\r\n\r\n$\\lim_{x \\to \\infty}f(x) = L$\r\n\r\nIf $| f(x)-L |$ can be made arbitrarily small. This is, effectively, $f(\\infty)$, but keep in mind that [b]sometimes limits don't exist.[/b]", "Solution_8": "[quote=\"Interval\"]You said:\n\ninfinity/infinity is called an indeterminate form.\n\nWhy is it INDETERMINATE?\n\nWhat about the other question?[/quote]\r\n\r\nSo (roughly speaking) 0/0 might mean 1, or it might mean 0 (and in fact, by appropriate examples, it can be made to mean anything); so it is indeterminate.", "Solution_9": "[quote=\"moogra\"][quote=\"Interval\"]You said:\n\ninfinity/infinity is called an indeterminate form.\n\nWhy is it INDETERMINATE?\n\nWhat about the other question?[/quote]\n\nSo (roughly speaking) 0/0 might mean 1, or it might mean 0 (and in fact, by appropriate examples, it can be made to mean anything); so it is indeterminate.[/quote]\r\n\r\nYes; here's something of a demonstration.\r\n\r\nLet's say that $\\frac{0}{0}= a$. What does this mean? Well, by the definition of division it means \r\n\r\n$0 = 0 \\cdot a$\r\n\r\nBut this is true for [b]any finite a.[/b] Hence $\\frac{0}{0}$ is [b]not well-defined.[/b]" } { "Tag": [ "inequalities", "algorithm", "function", "floor function", "algebra", "polynomial" ], "Problem": "Does there exist any general algorithm with which one can find the number of solutions to an inequality of the form $ \\sum^{n}_{k\\equal{}1}c_{k}x_{k} \\leq c_{n\\plus{}1}$\r\nwhere all the $ c_{i}$ are given constants and all the $ x_{i}$ are positive integers?", "Solution_1": "Use generating functions:\r\n$ \\displaystyle{(1 \\plus{} a^{c_1} \\plus{} a^{c_1^2} \\plus{} \\cdots a^{c_1^{\\lfloor log_{c_1}{c_{n \\plus{} 1}} \\rfloor}})(1 \\plus{} a^{c_2} \\plus{} a^{c_2^2} \\plus{} \\cdots a^{c_2^{\\lfloor log_{c_2}{c_{n \\plus{} 1}} \\rfloor}})\\cdots(1 \\plus{} a^{c_n} \\plus{} a^{c_n^2} \\plus{} \\cdots a^{c_n^{\\lfloor log_{c_n}{c_{n \\plus{} 1}} \\rfloor}})}$\r\nThe answer will be the sum of the coefficients of $ a^{z} : 0 \\le z \\le c_{n\\plus{}1}$.\r\n\r\nSorry for butchering the latex. :oops:", "Solution_2": "Generalizing Pick's theorem to higher dimensions and then considering the right simplex might be the best way, though all the $ \\gcd$'s involved to compute boundary points will make things hairy. [url=http://mathworld.wolfram.com/EhrhartPolynomial.html]Ehrhart polynomials[/url] are this generalization; that link shows that even doing this for a 3-term sum on the left side is not pretty." } { "Tag": [ "MATHCOUNTS", "search", "AwesomeMath", "summer program", "FTW" ], "Problem": "The winners from this year's national Mathcounts competition were recognized at the White House on Tuesday. The top four kids and their coaches along with the 1st place team were given an hour and a half behind the scene tour of the White House. They got a guided tour by a secret service agent who told them all sorts of interesting stories about things in every room that they went into. They got to go into the library and the china room and the reception room that is used to welcome visiting heads of state. They also went into the green room, the blue room and the main ball room. \r\n\r\nAfter the tour they were taken to the West Wing entrance and were escorted into the Roosevelt room where they waited for the President. Several got to sit in the Presidential chair at the conference table. The President spent around 20-30 minutes with the entire group in the Oval Office where he told them about the history of the Presidential desk (Old Resolute) and how it was presented to the United States fby the Queen of England for their efforts in saving an English ship that had become trapped in the ice in the search for the Northwest Passage. \r\n\r\nAfterwards he presented every member of the group with either a stick pin for the ladies or a tie clip to the gentlemen. They all had the United States seal and the President's signature on them. They also received a silver round page marker that had the Presidential seal on it. They also had pictures taken by the White House photographer that will be mailed to them in the next few weeks. The entire group then had their pictures taken by the national press corp for the local papers. If you go to the White House web site their is a picture of the entire group with the President. \r\n\r\nIt was a really cool afternoon. :lol: :lol: :lol:", "Solution_1": "I looked at the website and the picture; why wasn't Bobby there? :(", "Solution_2": "Bobby was missing because he was representing the USA at the World Primary Math Contest in Hong Kong. He could not get back in time to join us at the White House. I was able to get a set of the presents for him from the President. \r\n\r\nThe world Primary Math Contest is an intermational math contest that usually has around 40 countries involved. The problems are a lot like Mathcounts problems but much harder. He was on the team last year and he had a perfect score and his team won the award for the best non-Asian team at the competition. I have not talked to him yet to see how he and the team did this year. I think he gets back today.", "Solution_3": "Wait, doesn't Kevin Tian also do that? Or did he just miss that contest and go to the White House? Or did Kevin not make the PMWC?", "Solution_4": "From what I understood, Kevin qualified to be on the team but declined for some reason. He is at Awesome Math camp right now.", "Solution_5": "[quote=\"JBoyd\"]From what I understood, Kevin qualified to be on the team but declined for some reason. He is at Awesome Math camp right now.[/quote]Oh, so I guess he went to AwesomeMath immediately after the White House ceremony?", "Solution_6": "Yes, all three of the other Texas kids went to camp right after the Presidential trip. Kevin did compete in the World Primary Math contest last year when they also won the award for the top non-Asian team in the competition.", "Solution_7": "Oh, I see. He's beast :D \r\n\r\nAnd why isn't Bobby going to AwesomeMath? :(", "Solution_8": "wow.... meeting the president. I guess that is another good reason for preparing for mathcounts. But then again, Bush will be out of office so we never know how Obama and Mccain think about math. Hopefully they love it as much as we do. :lol:", "Solution_9": "[quote=\"JBoyd\"]Yes, all three of the other Texas kids went to camp right after the Presidential trip. Kevin did compete in the World Primary Math contest last year when they also won the award for the top non-Asian team in the competition.[/quote]\r\n\r\nHahaha, they award awards for the top non-asian team? Sounds like a beastly competition.", "Solution_10": "And I supposed the teams from china, japan, indonesia, etc. pwn everybody?", "Solution_11": "Sure. The Asians probably do, so that's why there exists an award for the top non asian team", "Solution_12": "In Asia math is probably so much more publicized and accepted then in the US. It sourta sucks here...", "Solution_13": "[quote=\"RussianRocket\"]In Asia math is probably so much more publicized and accepted then in the US. It sourta sucks here...[/quote] [b] Asia ftw![/b]", "Solution_14": "[quote=\"RussianRocket\"]In Asia math is probably so much more publicized and accepted then in the US. It sourta sucks here...[/quote]\r\nYes I agree. The average fifth grader in China has the mathematical mind of an average 7TH GRADER IN THE USA!", "Solution_15": "And that's just the [b]average[/b] fifth grader...\r\n\r\nHeh, in my school kids think I'm a \"genius\" because I skipped two grades of math, but in China I'd just be average.", "Solution_16": "[quote=\"gauss1181\"][quote=\"RussianRocket\"]In Asia math is probably so much more publicized and accepted then in the US. It sourta sucks here...[/quote]\nYes I agree. The average fifth grader in China has the mathematical mind of an average 7TH GRADER IN THE USA![/quote]\r\n\r\nNah... more like the average high-schooler...", "Solution_17": "Only the best AoPS like Bobby Shen compare to the Chinease people.\r\n\r\nMaybe it is also so much more accepted in China because there are so many people and not that many job slots. \r\nI think if you are in the top 1% of smart people then you are accepted into college for free.", "Solution_18": "That means you got a scholarship if you are that smart...", "Solution_19": "[quote=\"isabella2296\"]And that's just the [b]average[/b] fifth grader...\n\nHeh, in my school kids think I'm a \"genius\" because I skipped two grades of math, but in China I'd just be average.[/quote]\r\n\r\ni skipped two grades too...and we wouldn't be average we'd be way below...\r\n\r\nin 1st grade they do long multiplication and division :(\r\n\r\nits scary", "Solution_20": "I did multiplication and long division in the summer before 1st grade....", "Solution_21": ":( i learned it in 1st grade, but they didnt teach it in school till late 3rd and 4th.", "Solution_22": "Yes, that's true. My brother taught me all that stuff.", "Solution_23": "i learned by myself with my brothers textbooks o_o\r\nthis is going off-topic.", "Solution_24": "They have textbooks on multiplication/long division?!?", "Solution_25": "Well in the 4th grade glencoe books...yes..theres stuff about it", "Solution_26": "4th grade? Woah...that's late...", "Solution_27": "multiplication in 3rd, division in early 4th", "Solution_28": "[quote=\"AIME15\"]multiplication in 3rd, division in early 4th[/quote]\r\n\r\nI think I learned that in first grade.\r\n\r\nAnyways, this is getting out-of-hand.\r\n\r\nI spam. A lot.", "Solution_29": "i did too. but in our school system thats when you learn it :P" } { "Tag": [ "geometry", "rectangle" ], "Problem": "The points of this 3-by-3 grid are equally spaced\nhorizontally and vertically. How many different\nsets of three points of this grid can be the three\nvertices of an isosceles triangle?\n\n[asy]size(50);\ndot((0,0));\ndot((10,0));\ndot((20,0));\ndot((0,10));\ndot((10,10));\ndot((20,10));\ndot((0,20));\ndot((10,20));\ndot((20,20));[/asy]", "Solution_1": "If we count the different possibilities, we find there are $ \\boxed{4}$.", "Solution_2": "Err... I count at least 24.\r\n\r\nIf you consider the four corners of a square, you can group any three to make a 45-45-90 isosceles triangle. In this way we may form 4 triangles for each small square, and 4 for the large square.\r\n\r\nBeyond that, you may select any two adjacent corners of the big square and connect them with the midpoint of the side opposite, and that accounts for 4 more triangles.\r\n\r\nThat's how I got 24, lemme know if I forgot any.\r\n\r\nEDIT: Argh, yes, there are a lot more! :P See Kiseki's note below.... you get 4 more by pairing any two corners with the center, 4 more by grouping any set of three of the edge midpoints, and 4 more by grouping two adjacent midpoints with the far corner.", "Solution_3": "Do the triangles need to be distinct or can they be congruent? That makes a difference.\r\n\r\nEDIT: Then I am incorrect - I believe 24 is the right answer.", "Solution_4": "they can be congruent\r\nall the question says is to choose three points for the triangle\r\nso as long as the triangle is not in the exact same place, it counts", "Solution_5": "ummm...the answer wuz 36...but i counted 24 too...\r\n\r\n1-1-sqrt2 * 16\r\n2-2-2sqrt2 * 4\r\n2-sqrt2-sqrt2 * 8\r\n2-sqrt5-sqrt5 * 4\r\nsqrt2-sqrt5-sqrt5 * 4\r\n\r\nHey, thats 36!", "Solution_6": "For small right triangles in a 1*1 square, there are 16\nFor right triangles in a 1*2 square, there are 16 (running count 32)\nFor right triangles in the large square there are 4 (running count 36)\nFor non-right triangles in a 1*2 rectangle, there are 8 (running count 44)\nfor non right triangles in the large square, there are 8 (running count 52)\n\nDid i miss any? I think there are 52 triangles, much more than 36.", "Solution_7": "I think right triangles in 1*2 square are not \"isoceles\". So the answer is really 36... thank you very much :)" } { "Tag": [ "algebra", "polynomial", "Vieta" ], "Problem": "Let $ x;y$ be positive integer numbers satisfying $ xy|x^2\\plus{}y^2\\plus{}2$\r\nCaculate $ A\\equal{}\\frac{x^2\\plus{}y^2\\plus{}2}{3}$", "Solution_1": "Given $ xy|x^2\\plus{}y^2\\plus{}2$, therefore $ \\frac{x^2\\plus{}y^2\\plus{}2}{xy}$ is and integer. Given that 3 can only be formed of positive integers by $ 3.1$. Without loss of generality, let $ x\\equal{}3,y\\equal{}1$\r\n\r\n$ A\\equal{}\\frac{x^2\\plus{}y^2\\plus{}2}{xy}$\r\n$ A\\equal{}\\frac{3^2\\plus{}1^2\\plus{}2}{3}$\r\n$ A\\equal{}4$", "Solution_2": "[quote=\"SimonM\"]Given $ xy|x^2 \\plus{} y^2 \\plus{} 2$, therefore $ \\frac {x^2 \\plus{} y^2 \\plus{} 2}{xy}$ is and integer. Given that 3 can only be formed of positive integers by $ 3.1$. Without loss of generality, let $ x \\equal{} 3,y \\equal{} 1$\n\n$ A \\equal{} \\frac {x^2 \\plus{} y^2 \\plus{} 2}{xy}$\n$ A \\equal{} \\frac {3^2 \\plus{} 1^2 \\plus{} 2}{3}$\n$ A \\equal{} 4$[/quote]\r\nNo one is telling you that xy=3, you have no guarantee that A is an integer.\r\n[hide=\"idea\"]Root flipping[/hide]", "Solution_3": "[quote=\"SimonM\"]Given $ xy|x^2 \\plus{} y^2 \\plus{} 2$, therefore $ \\frac {x^2 \\plus{} y^2 \\plus{} 2}{xy}$ is and integer. Given that 3 can only be formed of positive integers by $ 3.1$. Without loss of generality, let $ x \\equal{} 3,y \\equal{} 1$\n\n$ A \\equal{} \\frac {x^2 \\plus{} y^2 \\plus{} 2}{xy}$\n$ A \\equal{} \\frac {3^2 \\plus{} 1^2 \\plus{} 2}{3}$\n$ A \\equal{} 4$[/quote]\r\n\r\ngood solution.", "Solution_4": "Let $ \\frac{x^2\\plus{}y^2\\plus{}2}{xy}\\equal{}k$,hence \r\n$ x^2\\minus{}x(ky)\\plus{}y^2\\plus{}2\\equal{}0$ $ (1)$.\r\nLet us consider 2 cases:\r\n1 case:\r\n$ x\\equal{}y\\Rightarrow x^2|2$,thus $ x\\equal{}y\\equal{}1$. So $ A\\equal{}\\frac{4}{3}$.\r\n2 case:$ x>y>1$.\r\nLet $ x_0,y$,be a solution of the $ (1)$,such that $ x_0\\plus{}y$ is less than any other $ x\\plus{}y$,where $ (x,y)$ is a solution of $ (1)$ too.\r\nAlso we know that there exist one more solution of $ (1)$,we call it $ x_1$.\r\nThen from Vieta:\r\n$ x_0\\cdot x_1\\equal{}y^2\\plus{}2$.Since $ y^2\\plus{}2,x_0>0$,hence $ x_1>0$.And \r\n$ x_1\\equal{}\\frac{y^2\\plus{}2}{x_0}\\leq x_0$.Equality doesn't occurs,because $ x_0^2\\equal{}y^2\\plus{}2$,has no solutions in positive integers.\r\nHence we have a contradiction to the min. of $ x_0\\plus{}y$.Thus $ y\\equal{}1$.\r\n$ (1)\\equal{}x\\plus{}\\frac{3}{x}\\in\\mathbb{N}$.Hence $ x\\equal{}1$ or $ x\\equal{}3$.\r\nIf $ x\\equal{}1$ then as we've already known $ A\\equal{}\\frac{4}{3}$.\r\nIf $ x\\equal{}3$ then $ A\\equal{}4$.", "Solution_5": "thats the general idea of these problems...comment:\r\n\r\ni) there is not necessarily a unique pair x,y so x+y is min\r\n\r\nwhat if (4,3), (5,2) were solutions...", "Solution_6": "[quote=\"Altheman\"]thats the general idea of these problems...comment:\n\ni) there is not necessarily a unique pair x,y so x+y is min\n\nwhat if (4,3), (5,2) were solutions...[/quote]\r\nOK,i think we must suppose that $ x$ is a minimal solution to this equation,or not?" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Is it true that every distribution in the linear transformation sense ( http://en.wikipedia.org/wiki/Distribution_(mathematics) ) \r\nis also a distribution in the probabilistic sense, and, if so, do the notions of weak convergence match?", "Solution_1": "[quote=\"mli\"]Is it true that every distribution in the linear transformation sense ( http://en.wikipedia.org/wiki/Distribution_(mathematics) ) \nis also a distribution in the probabilistic sense, and, if so, do the notions of weak convergence match? [/quote] \r\n\r\nNo, these are two unrelated uses of the word \"distribution\".", "Solution_2": "A particular example on $ \\mathbb{R}:$\r\n\r\n$ \\delta',$ defined so that its action on a test function $ \\varphi$ produces the number $ \\minus{} \\varphi'(0),$ is a perfectly reasonable distribution in the Schwartz or generalized function sense. \r\n\r\nI don't see any way that $ \\delta'$ makes any probabilistic sense at all.\r\n\r\nI think Schwartz made an infelicitous choice of terminology when he chose the word \"distribution.\" We might be better off if \"generalized function\" or something like that had become the more widely accepted term. What is the usual term in Russian?\r\n\r\n[Spelling note: yes, there is a \"t\" in this Schwartz. I'm referring to L. Schwartz (20th century, French), not H.A. Schwarz (19th century, German), whose name you see in a number of other places.]", "Solution_3": "I'm sorry...I misphrased that question. \r\n\r\nIs it true that every distribution can be interpreted as a signed measure on Lebesgue-measurable sets?\r\n\r\nerr...never mind. The above is still a good example.", "Solution_4": "A signed measure is a linear functional acting on a space of continuous functions. A distribution is a linear functional acting on a space of $ C^{\\infty}$ functions. Hence you can always differentiate a distribution, whereas you can't count on being able to differentiate a signed measure.\r\n\r\nDistributions were built to be differentiated - they're associated with the study of differential equations." } { "Tag": [ "geometry", "3D geometry", "algebra unsolved", "algebra" ], "Problem": "a is positive integer. For all positive integer n the number 4(a^n+1) =b^3 for any positive integer b. Prove that a=1", "Solution_1": "Hint:\r\n$ (a\\plus{}1)(a^2\\plus{}1)(a^3\\plus{}1)$ would have to be a cube :wink:", "Solution_2": "ok, (a+1)(...)() is cube, why a=1?", "Solution_3": "If $ a>1$ then the above expression can be placed strictly between two consecutive cubes.", "Solution_4": "very good, my solution 2e^3=a^9+1=(a^3+1)(a^6-a^3+1)=2k^3*m, m=a^6-a^3+1. We have, that m=t^3, but (a^2-1)^31" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "The circles S and S' meet at A and B. A line through A meets S, S' again at C, D respectively. M is a point on CD. The line through M parallel to BC meets BD at K, and the line through M parallel to BD meets BC at N. The perpendicular to BC at N meets S at the point E on the opposite side of BC to A. The perpendicular to BD at K meets S' at F on the opposite side of BD to A. Show that \u2220EMF = 90o. ;)", "Solution_1": "See:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=22165 \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=22201 \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=17322 \r\nhttp://www.kalva.demon.co.uk/short/soln/sh02g8.html \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=15588 post #2 file IMO_2002_shortlist.pdf page 21 .\r\n\r\n Darij" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Prove or disprove:\r\nEvery [b]directed[/b] graph with no leaves[i](vertex of degree 1)[/i] has two vertices of equal degree.", "Solution_1": "This holds only for finite graphs.\r\nMoreover, I think you have to be more precise about the degrees in your statement. Are you talking two times about indegrees? Outdegrees? the sum of them?\r\n\r\nPierre.", "Solution_2": "I mean by the degree of a vertex that $ deg(v)\\equal{} indeg(v) \\minus{} outdeg(v)$\r\nand also for finite directed graphs.\r\nHow con I prove that?", "Solution_3": "[quote=\"fesghelbahal\"]I mean by the degree of a vertex that $ deg(v) \\equal{} indeg(v) \\minus{} outdeg(v)$\n[/quote]\r\n\r\nThen it is not true.\r\nLet $ A,B,D$ be the vertices with directed egdes $ A \\minus{} > B,A \\minus{} > D,B \\minus{} > D,$ :wink: \r\n\r\nPierre.", "Solution_4": "Oops, You are right. :oops:\r\nI forgot an assumption: suppose that the degree of each vertex is nonzero!", "Solution_5": "Still not true...\r\nConsider vertices $ A,B,C,D,E$ and oriented edges $ A \\to E,B \\to E,C \\to E,D \\to E,A \\to D,B \\to D,C \\to D,A \\to C$.\r\nThe respective degrees of $ A,B,C,D,E$ are $ \\minus{} 3, \\minus{} 2, \\minus{} 1,2,4$.\r\n\r\nPierre.", "Solution_6": "Pierre, \\to makes a much nicer-looking arrow in LaTeX.", "Solution_7": ":blush: \r\n\r\nPierre.", "Solution_8": "My fault.\r\nI think It's better to write the whole question here again:\r\n\r\nProve or disprove:\r\nEvery finite [b]directed[/b] graph with $ | \\delta | \\geq2$ has two vertices of equal degree.\r\nWhere $ deg(v)\\equal{}indeg(v)\\minus{}outdeg(v)$\r\n\r\n[hide]PS: the assumption $ | \\delta | \\geq2$ is equal to that the graph has no leaves and no vertex of zero degree![/hide]", "Solution_9": "Seems still untrue...\r\nConsider graph with vertices $ A_1,A_2,A_3,B_1,B_2,B_3$ and oriented edges $ B_j \\to A_i$ for $ i,j \\equal{} 1,2,3$, and $ A_{i \\plus{} 1} \\to A_i$ for $ i \\equal{} 1,2$, and $ B_3 \\to B_2$.\r\nThe respective degrees of $ A_1,A_2,A_3,B_1,B_2,B_3$ are $ 4,3,2, \\minus{} 3, \\minus{} 2, \\minus{} 4$. :wink: \r\n\r\nPierre." } { "Tag": [ "trigonometry" ], "Problem": "If \\[x \\equal{} \\frac { \\minus{} 1 \\plus{} i\\sqrt3}{2}\\qquad\\text{and}\\qquad y \\equal{} \\frac { \\minus{} 1 \\minus{} i\\sqrt3}{2},\\] where $ i^2 \\equal{} \\minus{} 1$, then which of the following is [i]not[/i] correct?\n\n$ \\textbf{(A)}\\ x^5 \\plus{} y^5 \\equal{} \\minus{} 1 \\qquad \\textbf{(B)}\\ x^7 \\plus{} y^7 \\equal{} \\minus{} 1 \\qquad \\textbf{(C)}\\ x^9 \\plus{} y^9 \\equal{} \\minus{} 1$\n$ \\textbf{(D)}\\ x^{11} \\plus{} y^{11} \\equal{} \\minus{} 1 \\qquad \\textbf{(E)}\\ x^{13} \\plus{} y^{13} \\equal{} \\minus{} 1$", "Solution_1": "We have $ x\\equal{}e^{i \\frac{2 \\pi}{3}}$, $ y\\equal{}e^{\\minus{}i\\frac{2 \\pi}{3}}$\r\n\r\nNow $ x^n\\plus{}y^n\\equal{}e^{i \\frac{2n \\pi}{3}}\\plus{}e^{\\minus{}i\\frac{2n \\pi}{3}}\\equal{}2 Re(e^{i \\frac{2n \\pi}{3}})\\equal{}2 cos (\\frac{2n \\pi}{3})\\equal{}1$\r\nSo $ cos (\\frac{2n \\pi}{3})\\equal{}\\frac{1}{2}$ , checking $ n(mod \\;\\; 3)$ we know $ n\\equal{}1, 2(mod \\;\\; 3)$ is the solution. So the answer is $ C$", "Solution_2": "Did you used Muavre's theorem for power of complex number ?\r\n I'd like to know more for this theorem", "Solution_3": "He used Euler's formula more than DeMoivre.\r\n\r\n$ e^{i\\theta}\\equal{}\\cos\\theta\\plus{}i\\sin\\theta$\r\n\r\nAnyway, DeMoivre's theorem is simple: $ (\\cos\\theta\\plus{}i\\sin\\theta)^n\\equal{}\\cos n\\theta\\plus{}i\\sin n\\theta$, which follows from Euler's formula.", "Solution_4": "So DeMoivre's theorem is just a case of Euler's theorem ?", "Solution_5": "[quote=\"stephencheng\"]We have $ x \\equal{} e^{i \\frac {2 \\pi}{3}}$, $ y \\equal{} e^{ \\minus{} i\\frac {2 \\pi}{3}}$\n\nNow $ x^n \\plus{} y^n \\equal{} e^{i \\frac {2n \\pi}{3}} \\plus{} e^{ \\minus{} i\\frac {2n \\pi}{3}} \\equal{} 2 Re(e^{i \\frac {2n \\pi}{3}}) \\equal{} 2 cos (\\frac {2n \\pi}{3}) \\equal{} 1$\nSo $ cos (\\frac {2n \\pi}{3}) \\equal{} \\frac {1}{2}$ , checking $ n(mod \\;\\; 3)$ we know $ n \\equal{} 1, 2(mod \\;\\; 3)$ is the solution. So the answer is $ C$[/quote]\r\nI think it should be $ x^n \\plus{} y^n \\equal{} 2\\cos(2n\\pi/3) \\equal{} \\minus{} 1$ or $ \\cos[2\\pi/3 \\plus{} 2(n \\minus{} 1)\\pi/3] \\equal{} \\minus{} 1/2\\implies n\\equiv \\pm 1\\mod 3$.\r\n\r\nEDIT: fixed the $ \\pm$ sign...", "Solution_6": "[quote=\"enndb0x\"]So DeMoivre's theorem is just a case of Euler's theorem ?[/quote]\r\n\r\nSort of, yes. Euler's formula was discovered after DeMoivre's theorem, so the earlier proof of DeMoivre's theorem involves binomial expansion or something... but now DeMoivre's theorem is in a sense generalized by Euler's formula" } { "Tag": [ "factorial", "number theory", "prime numbers", "number theory proposed" ], "Problem": "Hi everybody,\r\n\r\nI have found a way to write factorial (p-1)! where p is prime \r\n\r\n(p-1)! = (a(1)p-1)*(a(2)p-1)*(a(3)p-1)*....*(a(k)p-1)\r\n\r\na(i) is an integer >0 \r\n\r\nI want to build some table of the coefficients a(i) to understand how it works.\r\nIf we can predict the values of a(i) it will be very helpful.\r\nAny idea will be helpful.\r\n\r\nExample :\r\n\r\n12!= (1*13-1)*(2*13-1)*(5*13-1)*(6*13-1)*(25*13-1)\r\n\r\na(1)=1\r\na(2)=2\r\na(3)=5\r\na(4)=6\r\na(5)=25\r\n\r\nThank you for any comment", "Solution_1": "Have you ever looked at the proof of Wilson's theorem?", "Solution_2": "[quote=\"fedja\"]Have you ever looked at the proof of Wilson's theorem?[/quote]\r\n\r\nYes I did.", "Solution_3": "Here is an algo to find the a(i)'s\r\n\r\n1. For p create a list L={p-1,p-2,p-3,.....,p-(p-2),p-(p-1)}\r\n\r\nExample : p=23\r\n\r\nList L={22,21,20,19,.....,4,3,2,1}\r\n\r\n2. From left to the right and for each number of the list solve ((p-i)*x) mod p = p-1 \r\nx is the first number solving the equation\r\n\r\nExample : \r\n\r\n22x mod 23 = 22 x=1 then store a(1)=(22*1 + 1)/23=1 remove 22 from the list L\r\n21x mod 23 = 22 x=12 then store a(2)=(21*12 + 1)/23=11 remove 21 and [u]12[/u] from the list L (edited because of mistake)\r\n20x mod 23 = 22 x=8 then store a(3)=(20*12 + 1)/23=7 remove 20 and [u]8[/u] from the list L (edited because of mistake)\r\n19x mod 23 = 22 x=6 then store a(4)=(19*6 + 1)/23=5 remove 19 and 6 from the list L\r\n\r\nand so on until the last couple of the list L\r\n\r\n3. Sort the a(i) (ascending order) and print the new list of a(i) ordered with new indexes a(j)\r\n\r\nMy goal is not to compute (p-1)!\r\nMy goal is to see if there is some formula to write directly the a(i)'s \r\nSomething like a Pascal triangle.", "Solution_4": "I forgot something.\r\nIf someone can program it I will thank him a lot if he can give me the list of the a(i) for the prime numbers between 100 and 120.\r\n\r\nThank you for any help" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Let $ a,b,c,d\\in\\mathbb{N}$ such that $ \\frac{a}{c}\\equal{}\\frac{b}{d}\\equal{}\\frac{ab\\plus{}1}{cd\\plus{}1}$\r\nProve that $ a\\equal{}c$", "Solution_1": "We have:\r\n\\[ \\frac{ab}{cd}\\equal{}\\left(\\frac{ab\\plus{}1}{cd\\plus{}1}\\right)^2\\Leftrightarrow \\frac{(ab\\plus{}1)^2}{ab}\\equal{}\\frac{(cd\\plus{}1)^2}{cd}\\Leftrightarrow ab\\plus{}2\\plus{}\\frac{1}{ab}\\equal{}cd\\plus{}2\\plus{}\\frac{1}{cd}\\Leftrightarrow ab\\plus{}\\frac{1}{ab}\\equal{}cd\\plus{}\\frac{1}{cd}\\]\r\nThen either $ ab\\equal{}cd$ or $ ab\\equal{}\\frac{1}{cd}$. But $ ab$, $ cd$ - integer numbers, so in the second case $ ab\\equal{}cd\\equal{}1$, and we have $ ab\\equal{}cd$ anyway. Then \r\n\\[ \\frac{a}{c}\\equal{}\\frac{ab\\plus{}1}{cd\\plus{}1}\\equal{}1\\]", "Solution_2": "Let $ \\frac{x}{y}\\equal{}\\frac{a}{c}$ where $ (x,y)\\equal{}1$. Then $ x|a,x|b$ implies that $ ab\\plus{}1\\equiv1\\pmod{x}$. Since we must have $ x|ab\\plus{}1$ also, $ x\\equal{}1$. Similarly, we find $ y\\equal{}1$ and so $ a\\equal{}c$." } { "Tag": [ "algebra", "polynomial", "function", "quadratics", "complex numbers", "algebra proposed" ], "Problem": "Find all real variables $ a$, for which the sum of all real roots of the equation $ x^4\\minus{}5x\\plus{}a\\equal{}0$ is $ a$.", "Solution_1": "[hide]If there are no real roots (if this case is allowed), we must have an empty sum for $ a\\equal{}0$, but this has a real root of course. Similarly, if there are four real roots, the sum of the real roots is the sum of all roots, also 0. But the polynomial $ x^4\\minus{}5x$ only has two real roots.\n\nSo we reduce to the case of two real roots and a complex conjugate pair $ z,\\overline{z}$ such that $ z\\plus{}\\overline{z}\\equal{}\\minus{}a,z\\minus{}\\overline{z}\\neq 0$. Subtract the equations for $ z,\\overline{z}$ to get\n\\[ 0\\equal{}z^4\\minus{}\\overline{z}^4\\minus{}5(z\\minus{}\\overline{z})\\equal{}(z\\minus{}\\overline{z})((z\\plus{}\\overline{z})(z^2\\plus{}\\overline{z}^2)\\minus{}5)\\]\nThis gives $ a\\neq 0$, $ z^2\\plus{}\\overline{z}^2\\equal{}\\minus{}\\frac{5}{a}$, and $ |z|^2\\equal{}\\frac{1}{2}a^2\\minus{}\\frac{5}{2a}$.\n\nNow add the equations for $ z,\\overline{z}$ to get\n\\[ 0\\equal{}2a\\minus{}5(\\minus{}a)\\plus{}z^4\\plus{}\\overline{z}^4\\equal{}7a\\plus{}(z^2\\plus{}\\overline{z}^2)^2\\minus{}2|z|^4\\]\n\\[ \\equal{}7a\\plus{}\\frac{25}{a^2}\\minus{}\\frac{1}{2}\\left( a^2\\minus{}\\frac{5}{a}\\right)^2\\]\n\\[ \\Rightarrow 0\\equal{}14a^3\\plus{}50\\minus{}(a^3\\minus{}5)^2\\equal{}25\\plus{}4a^3\\minus{}a^6\\]\n\\[ \\Rightarrow a\\equal{}\\boxed{\\sqrt[3]{2\\pm \\sqrt{29}}}\\][/hide]", "Solution_2": "[quote=\"scorpius119\"]\n\\[ a \\equal{} \\boxed{\\sqrt [3]{2\\pm \\sqrt {29}}}\n\\]\n[/quote]\r\n\r\nThe sum of the roots is always $ > 0$. It is easy to see from the graphic of the function $ f(x) \\equal{} x^4 \\minus{} 5x$. So $ a \\equal{} \\sqrt [3]{2 \\plus{} \\sqrt {29}}$.", "Solution_3": "Yes, I should have checked $ |z|^2\\geq 0$ to get $ a^3\\geq 5$. At least for $ a\\equal{}\\sqrt[3]{5\\plus{}\\sqrt{29}}$, we can actually find $ z,\\overline{z}$ with sum $ \\minus{}\\sqrt[3]{5\\plus{}\\sqrt{29}}$ and product $ \\frac{\\sqrt{29}}{2\\sqrt[3]{5\\plus{}\\sqrt{29}}}$ because the discriminant of the resulting quadratic is negative." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "If a positive integer is even, divide it by two. It it is odd, subtract one from it. Repeat this until the number is $ 0$. Let $ s(n)$ be the number of subtractions when input number is $ n$. How to compute $ \\sum_{i \\equal{} 1}^ns(i)$ effectively, in the cases $ 1\\leq n\\leq 10^{20}$?", "Solution_1": "[hide=\"Hint\"]Use base $ 2$[/hide]\n\n[hide=\"Formula for s(n)\"]Consider $ n$ in base $ 2$ then $ s(n)$ is the number of ones on that base[/hide]\n\n[hide=\"Formula for the sum for special n\"]\nConsider $ n\\equal{}2^k$, consider all the $ i$ numbers on base $ 2$ with $ k\\plus{}1$ digits there are in total $ 2^k$ numbers and $ (k\\plus{}1)\\cdot 2^k$ digits in that base and half of them are $ 1$ then that sum $ \\equal{}(k\\plus{}1)\\cdot 2^{k\\minus{}1}$[/hide]\r\n\r\nIs it true?", "Solution_2": "This is what I found too. But if $ 2^k < n < 2^{k \\plus{} 1}$ for some $ k\\in\\mathbb{Z}^\\plus{}$, how to evaluate the sum $ \\sum_{i \\equal{} 2^k \\plus{} 1}^n s(i)$?", "Solution_3": "[quote=\"MeKnowsNothing\"]This is what I found too. But if $ 2^k < n < 2^{k \\plus{} 1}$ for some $ k\\in\\mathbb{Z}^ \\plus{}$, how to evaluate the sum $ \\sum_{i \\equal{} 2^k \\plus{} 1}^n s(i)$?[/quote]\r\n\r\n[hide=\"Hint\"]Find a recursive formula, notice that for all $ i$ they always have the same amount of digits in base $ 2$ and start with a $ 1$[/hide]\n[hide=\"Hint2\"]$ n \\equal{} 2^k \\plus{} m$ with $ 0 < m < 2^k$[/hide]\n[hide=\"Hint3\"]Apply several times that recursive formula as the amount of $ 1$ on $ n$'s base $ 2$ expresion[/hide]" } { "Tag": [ "trigonometry", "geometry", "ratio", "geometry unsolved" ], "Problem": "Can anybody draw picture to this problem\r\n\r\nLet a convex quadrilateral $ APQC$ have its sides $ AP$, $ PQ$ and $ QC$ tangent to a minor circular arc $ ABC$ at the points $ A$,$ B$ and $ C$, respectively.\r\nLet $ E$ be the projection of $ B$ onto $ AC$. Let semicircle with $ PQ$ as diameter cut $ AC$ at $ H$ and $ K$, with $ H$ between $ A$ and $ K$.\r\n\r\nNow problem comes:\r\nWithout using triginometry, prove that $ BE$ bisects $ \\angle PEQ$ and that $ PH$ bisects $ \\angle APB$.", "Solution_1": "[quote=\"canada\"]Can anybody draw picture to this problem\n\nLet a convex quadrilateral $ APQC$ have its sides $ AP$, $ PQ$ and $ QC$ tangent to a minor circular arc $ ABC$ at the points $ A$,$ B$ and $ C$, respectively.\nLet $ E$ be the projection of $ B$ onto $ AC$. Let semicircle with $ PQ$ as diameter cut $ AC$ at $ H$ and $ K$, with $ H$ between $ A$ and $ K$.\n\nNow problem comes:\nWithout using triginometry, prove that $ BE$ bisects $ \\angle PEQ$ and that $ PH$ bisects $ \\angle APB$.[/quote]", "Solution_2": "Thank you very much man\r\n\r\nMay you take gold this year at IMO!!! :first:", "Solution_3": "Denote: $ D=AP\\cap CQ,\\,J=AC\\cap PQ$\r\nWe have $ DB,\\,PC,\\,QA$ are concurrent, so $ (J,\\,B,\\,P,\\,Q)=-1\\implies H(J,\\,B,\\,P,\\,Q)=-1$\r\nOn the other hand, $ HB\\perp HJ,$ hence: $ \\angle PHB=\\angle QHB$", "Solution_4": "[quote=\"canada\"][color=darkred]Let a convex quadrilateral $ APQC$ have its sides $ AP$, $ PQ$ and $ QC$ tangent to a minor circular arc $ ABC$ at the points $ A$,$ B$ and $ C$, respectively.\nLet $ E$ be the projection of $ B$ onto $ AC$. Let semicircle with $ PQ$ as diameter cut $ AC$ at $ H$ and $ K$, with $ H$ between $ A$ and $ K$.\nProve that $ BE$ bisects $ \\angle PEQ$ and that $ PH$ bisects $ \\angle APB$.[/color][/quote]\n[color=darkblue][b]I don't like this bushy enunciation ![/b] I\"ll present an equivalent enunciation.[/color]\n[quote=\"Virgil Nicula\"][color=darkred]Let $ w=\\mathcal C (O)$ be the circumcircle of the triangle $ ABC$. Denote the intersections $ \\{\\begin{array}{c}E\\in AC\\ ,\\ BE\\perp AC\\\\\\ J\\in AC\\cap BB\\\\\\ P\\in AA\\cap BB\\ ,\\ Q\\in CC\\cap BB\\end{array}$.\nThe circle with the diameter $ [PQ]$ cuts the line $ AC$ in the points $ H$, $ K$, where $ H\\in (AK)$. Denote $ R\\in PK\\cap QH$.\nProve that the ray $ [EB$ is the bisector of the angle $ \\widehat{PEQ}$ and $ O\\in PH\\cap QK\\cap BR$.[/color][/quote]\n[color=darkblue] Upon a time I posted the following very usual lemmas :[/color]\n\n[color=darkred][b]Lemma I.[/b] Let $ ABCD$ be a convex quadrilateral for which $ \\{\\begin{array}{c}BA=BC\\\\\\ \\widehat{ADB}\\equiv\\widehat{CDB}\\end{array}$. Then $ DA=DC$ or the given quadrilateral is cyclically.\n\n[b]Lemma II.[/b] Let $ A,C,B,D$ be four collinear points in thie mentioned order and let $ P$ be a point which doesn't belong to the line $ AB$.\nProve that if two from the following sentencies are truly, then and another sentence is truly : $ \\{\\begin{array}{cc}1\\blacktriangleright & \\frac{CA}{CB}=\\frac{DA}{DB}\\\\\\ 2\\blacktriangleright & PA\\perp PB\\\\\\ 3\\blacktriangleright & \\widehat{CPB}\\equiv\\widehat{DPB}\\end{array}$.\n[b]Remark.[/b] The first sentence states that the division $ (A,B,C,D)$ is harmonically, i.e. the points $ C$, $ D$ are conjugate w.r.t. the points $ A$, $ B$.[/color]\n\n[color=darkblue][b]Proof.[/b] Denote the intersection $ D\\in AA\\cap CC$. Apply the [b]Menelaus' theorem[/b] to the transversal $ \\overline{ACJ}$ and the triangle $ PDQ$ :\n\n$ \\frac{JP}{JQ}\\cdot\\frac{CQ}{CD}\\cdot\\frac{AD}{AP}=1$ $ \\implies\\over{(CD=AD)}$ $ \\frac{JP}{JQ}=\\frac{AP}{CQ}$ $ \\implies\\over{\\|\\begin{array}{c}PA=PB\\\\\\ QB=QC\\end{array}\\|}$ $ \\frac{JP}{JQ}=\\frac{BP}{BQ}$, i.e. the point $ B$ is the conjugate of the point $ J$ w.r.t. the points $ P$, $ Q$.\nApply the [b]lemma II[/b] : $ \\{\\begin{array}{ccc}HP\\perp HQ & \\Longrightarrow & \\widehat{AHP}\\equiv\\widehat{BHP}\\\\\\\\ EB\\perp EJ & \\Longrightarrow & \\boxed{\\widehat{PEB}\\equiv\\widehat{QEB}}\\\\\\\\ KP\\perp KQ & \\Longrightarrow & \\widehat{BKQ}\\equiv\\widehat{JKQ}\\end{array}$. If denote the point $ R\\in PK\\cap QH$, then $ BR\\perp PQ$ because $ \\{\\begin{array}{c}HQ\\perp HP\\\\\\ KP\\perp KQ\\end{array}$\nand the division $ (P,Q,B,J)$ is harmonically (the point $ R$ is the orthocenter of the triangle $ PSQ$, where $ S\\in PH\\cap QK$). Thus, $ \\boxed{O\\in BR}$.\nApply the [b]lemma I[/b] to the convex quadrilaterals $ APBH$ and $ CQBK$ :\n$ 1\\blacktriangleright \\{\\begin{array}{c}PA=PB\\\\\\ \\widehat{AHP}\\equiv\\widehat{BHP}\\end{array}\\|$ $ \\implies$ $ HA=HP$ because contrary, the quadrilateral $ APBH$ is inscribed in the circle with the diameter $ [PR]$, what is falsely.\n\n$ 2\\blacktriangleright \\{\\begin{array}{c}QC=QB\\\\\\ \\widehat{BKQ}\\equiv \\widehat{CKQ}\\end{array}\\|$ $ \\implies$ $ KC=KB$ because contrary, the quadrilateral $ CQBK$ is inscribed in the circle with the diameter $ [BR]$, what is falsely.\nIn conclusion, $ \\{\\begin{array}{c}HA=HB\\\\\\ KC=KB\\end{array}\\|$ $ \\implies$ $ \\boxed{O\\in PH\\cap QK}$. Accurately, $ O\\in PH\\cap QK\\cap BR$.\n\n[b]Remark.[/b] Here is a nice related problem about which I think that it is the [b]\"spring\"[/b] of the proposed problem.[/color]\n\n[quote=\"Virgil Nicula\"][color=darkred]\n\n$ \\mathrm{IN}\\blacktriangleright$ The incircle $ w=\\mathcal C(I)$ of the triangle $ ABC$ touches it in the points $ \\{\\begin{array}{c}M\\in BC\\\\\\ N\\in CA\\\\\\ P\\in AB\\end{array}$.\nDenote $ \\{\\begin{array}{c}U\\in IB\\cap NP\\\\\\ V\\in IC\\cap NP\\\\\\ R\\in BV\\cap CU\\end{array}$. Prove that the points $ U,V$ belong to the circle with the diameter $ [BC]$ and $ R\\in IM$.\n$ \\mathrm{EX}\\blacktriangleright$ The $ A$- exincircle $ w=\\mathcal C(I_{a})$ of the triangle $ ABC$ touches it in the points $ \\{\\begin{array}{c}M\\in BC\\\\\\ N\\in CA\\\\\\ P\\in AB\\end{array}$.\nDenote $ \\{\\begin{array}{c}U\\in I_{a}B\\cap NP\\\\\\ V\\in I_{a}C\\cap NP\\\\\\ R\\in BV\\cap CU\\end{array}$. Prove that the points $ U,V$ belong to the circle with the diameter $ [BC]$ and $ R\\in I_{a}M$.[/color][/quote]", "Solution_5": "I have solved first part of problem with similarity and some angle chasing \r\nI think that it is a nice and simple solution and I put it as attachment.\r\n\r\nDoes anybody have solution to second part of problem with using only similarity, Menelaus ( I mean without this ratios, isogonal conjugate points etc. becous it is a little bit difficult to understand it)", "Solution_6": "[quote=\"April\"]Denote: $ D=AP\\cap CQ,\\,J=AC\\cap PQ$\nWe have $ DB,\\,PC,\\,QA$ are concurrent, so $ (J,\\,B,\\,P,\\,Q)=-1\\implies H(J,\\,B,\\,P,\\,Q)=-1$\nOn the other hand, $ HB\\perp HJ,$ hence: $ \\angle PHB=\\angle QHB$[/quote]\r\n\r\n[b] Figure[/b]", "Solution_7": "[b]Stergiu[/b] what did you want to say with above post." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "$\\forall n\\in \\mathbb{N}$ Define $x_{n+1}=x_{n}(1-x_{n})$ which $0 0$ and $a+b+c=1$. Prove that\r\n\r\n\\[\\sqrt{\\frac{ab}{c}+1}+\\sqrt{\\frac{bc}{a}+1}+\\sqrt{\\frac{ca}{b}+1} \\geq 2(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}).\\]", "Solution_1": "I don't think it's true: consider $a=2,b=3,c=6$\r\n\r\n$(2)(3)+(3)(6)+(6)(2) = (2)(3)(6)$\r\n\r\n$\\sqrt{2}+\\sqrt{10}+\\sqrt{5} \\not \\ge 2(\\sqrt{2}+\\sqrt{3}+\\sqrt{6})$.", "Solution_2": "Haha yeah, I wrote the wrong condition--it's edited now. :)", "Solution_3": "If I made no mistake, it's minkowski + cauchy.", "Solution_4": "Can you elaborate? I'm not seeing it.", "Solution_5": "I don't think it needs the equals sign. AM-GM seems to give the result... (two variable)", "Solution_6": "[quote=\"blahblahblah\"]Can you elaborate? I'm not seeing it.[/quote]\r\nI forgot my solution too... anyway, i find this simpler\r\n\\[\r\n\\sqrt {\\frac{{ab}}\r\n{c} + 1} = \\sqrt {\\frac{{(c + a)(c + b)}}\r\n{c}} \\geqslant \\sqrt {\\frac{{c\\left( {\\sqrt a + \\sqrt b } \\right)^2 }}\r\n{c}} = \\sqrt a + \\sqrt b \r\n\\]\r\nby cauchy.\r\nThe result follows.", "Solution_7": "[quote=\"siuhochung\"][quote=\"blahblahblah\"]Can you elaborate? I'm not seeing it.[/quote]\nI forgot my solution too... anyway, i find this simpler\n\\[\n\\sqrt {\\frac{{ab}}\n{c} + 1} = \\sqrt {\\frac{{(c + a)(c + b)}}\n{c}} \\geqslant \\sqrt {\\frac{{c\\left( {\\sqrt a + \\sqrt b } \\right)^2 }}\n{c}} = \\sqrt a + \\sqrt b \n\\]\nby cauchy.\nThe result follows.[/quote]\r\n\r\nsame result here...", "Solution_8": "And I find this even easier:\r\n\r\n$\\frac {ab}{c}+a+b+c\\geq a+b+2\\sqrt{ab}=(\\sqrt{a}+\\sqrt{b})^2$\r\n\r\nI was just asking because I couldn't see the Minkowski (but then again, I never see when Minkowski can be applied, unless it just slaps me in the face)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "How do you show that the average number of representations of a natural number as a sum of two squares is $ \\pi$?", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=150965 ." } { "Tag": [ "LaTeX", "Ring Theory", "superior algebra", "superior algebra solved" ], "Problem": "Let P be a Prime Ideal of the commutative ring R. Prove that if I and J are Ideals of R and I $intersection$ J $subset$ P, then either I $subset$ P or J $subset$ P\r\n\r\nP.S. I tried unsing Latex but I guess it didn't work out. Sorry :blush:", "Solution_1": "hint: try showing that $IJ\\subset I\\cap J$ . once u've showed this, use the definition of a prime ideal (namely that if P is a prime ideal and whenever $ab\\in P$ then one of a or b is in P) to get what you wanted. i hope this is correct :| \r\n\r\ngreets", "Solution_2": "Assume $I \\cap J \\subseteq P$, but neither $I \\subseteq P$ nor $J \\subseteq P$. Then there exist $x \\in I \\backslash P,~y \\in J \\backslash P$. But $xy \\in I \\cap J \\subseteq P$ implies $x \\in P$ or $y \\in P$, a contradiction.", "Solution_3": "[quote=\"mikejones\"]\nP.S. I tried using Latex but I guess it didn't work out. Sorry :blush:[/quote]\r\n\r\nUsually latex just comes down to putting between dollar signs.\r\nBut for symbols like intersection you have put a \\ before those words(intersection doesn't exist, cap does however) :\r\n\r\nLet P be a Prime Ideal of the commutative ring R. Prove that if I and J are Ideals of R and I $\\cap$ J $\\subset$ P, then either I $\\subset$ P or J $\\subset$ P\r\n\r\nYou can 'steal' :wink: other people's code by clicking the quote button on their messages\r\n\r\n\r\nMaybe this list can help you too :\r\n\r\n[url]http://www.mathlinks.ro/LaTeX/AoPS_L_GuideSym.php[/url]" } { "Tag": [ "induction" ], "Problem": "Let $a$ and $b$ be positive integers greater than $2$. Prove that there exists a positive integer $k$ and a finite sequence $n_1$, $\\cdots$, $n_k$ of positive integers such that $n_1 =a$, $n_k =b$, and $n_i n_{i+1}$ is divisible by $n_{i}+n_{i+1}$ for each $i$ $(1 \\le i \\le k)$.", "Solution_1": "[b]Definition[/b] We will denote by $ a\\rightarrow b$ the fact that there exists a positive integer $ k$ and a finite sequence $ n_{1},\\cdots, n_{k}$ of positive integers such that $ n_{1}\\equal{} a$, $ n_{k}\\equal{} b$, and $ n_{i}n_{i\\plus{}1}$ is divisible by $ n_{i}\\plus{}n_{i\\plus{}1}$ for each $ i$ ($ 1\\le i\\le k$).\r\n[b]Observation 1[/b] If we reverse the sequence of $ k$ positive integers such that $ n_{1}\\equal{} a$, $ n_{k}\\equal{} b$, and $ n_{i}n_{i\\plus{}1}$ is divisible by $ n_{i}\\plus{}n_{i\\plus{}1}$ for each $ i$ ($ 1\\le i\\le k$), we get a sequence of $ k$ positive integers such that $ n_{1}\\equal{} b$, $ n_{k}\\equal{} a$, and $ n_{i}n_{i\\plus{}1}$ is divisible by $ n_{i}\\plus{}n_{i\\plus{}1}$ for each $ i$ ($ 1\\le i\\le k$). Therefore, $ a\\rightarrow b\\Leftrightarrow b\\rightarrow a$.\r\nAlso, note that if $ (x\\plus{}y)|xy$ then $ x\\rightarrow y$ and $ y\\rightarrow x$.\r\n[b]Lemma 1[/b] $ a\\rightarrow a(a\\minus{}1)$.\r\n[b]Proof[/b] $ a\\plus{}a(a\\minus{}1) \\equal{} a^{2}|a^{2}(a\\minus{}1) \\equal{} (a)(a(a\\minus{}1))\\Rightarrow (a\\plus{}a(a\\minus{}1))|(a)(a(a\\minus{}1))$ so $ a\\rightarrow a(a\\minus{}1)$ and the lemma is proved.\r\n[b]Lemma 2[/b] If $ a$ is odd, then $ a(a\\minus{}1)\\rightarrow a(a\\plus{}1)$.\r\n[b]Proof[/b] $ a(a\\minus{}1)\\plus{}a(a\\plus{}1) \\equal{} 2a^{2}|a^{2}(a\\minus{}1)(a\\plus{}1) \\equal{} (a(a\\minus{}1))(a(a\\plus{}1))$ because $ a\\minus{}1$ is even as $ a$ is odd. So, $ (a(a\\minus{}1)\\plus{}a(a\\plus{}1))|(a(a\\minus{}1))(a(a\\plus{}1))$ and the result of lemma 2 follows.\r\n[b]Lemma 3[/b] $ a\\rightarrow b, b\\rightarrow c\\Rightarrow a\\rightarrow c$.\r\n[b]Proof[/b] We can just take the finite sequence $ n_{1},\\cdots, n_{k}$ of positive integers such that $ n_{1}\\equal{} a$, $ n_{k}\\equal{} b$, and $ n_{i}n_{i\\plus{}1}$ is divisible by $ n_{i}\\plus{}n_{i\\plus{}1}$ for each $ i$ ($ 1\\le i\\le k$) and the a finite sequence $ m_{1},\\cdots, m_{l}$ of positive integers such that $ m_{1}\\equal{} b$, $ m_{l}\\equal{} c$, and $ m_{i}m_{i\\plus{}1}$ is divisible by $ m_{i}\\plus{}m_{i\\plus{}1}$ for each $ i$ ($ 1\\le i\\le l$), and the sequence $ a \\equal{} n_{1}, n_{2},...,n_{k}\\equal{} b \\equal{} m_{1}, m_{2},...,m_{l}\\equal{} c$ will satisfy the fact that the sum of very two consecutive terms in the sequence will divide their product, so $ a\\rightarrow c$ and lemma 3is proved.\r\n[b]Lemma 4[/b] For any positive integer $ a\\geq 3$, $ 3\\rightarrow a$.\r\n[b]Proof[/b] We will prove the result using mathematical induction.\r\n[b]Base Step[/b] If $ a \\equal{} 3$ then $ a\\rightarrow 3\\Rightarrow 3\\rightarrow 3$ which is true.\r\n[b]Induction step[/b] Assume, for $ i \\equal{} 3,4,...,k$ we have $ 3\\rightarrow i$. Let us prove $ 3\\rightarrow k\\plus{}1$.\r\nIf $ k\\plus{}1$ is even, $ k$ is odd, so by lemmas 1 and 2, have $ k\\plus{}1\\rightarrow k(k\\plus{}1)\\rightarrow k(k\\minus{}1)\\rightarrow k$ ($ k(k\\minus{}1)\\rightarrow k$ because of observation 1 and $ k\\rightarrow k(k\\minus{}1)$ by lemma 1). Hence, by lemma 3 $ k\\plus{}1\\rightarrow k$, so $ k\\rightarrow k\\plus{}1$. But $ 3\\rightarrow k, k\\rightarrow k\\plus{}1$ so by lemma 3 $ 3\\rightarrow k\\plus{}1$ and the result is true for $ a \\equal{} k\\plus{}1$.\r\nIf $ k\\plus{}1$ is odd, by lemmas 1 and 2 have $ k\\plus{}1\\rightarrow k(k\\plus{}1)\\rightarrow (k\\plus{}2)(k\\plus{}1)\\rightarrow k\\plus{}2$ so by lemma 3, $ k\\plus{}1\\rightarrow k\\plus{}2$ so $ k\\plus{}2\\rightarrow k\\plus{}1$. Because $ k\\plus{}1$ is odd, $ k\\plus{}2$ is even, let $ k\\plus{}2 \\equal{} 2l$, then $ k \\equal{} 2l\\minus{}2$. Now, $ (2l\\plus{}2l(l\\minus{}1))|(2l*2l(l\\minus{}1))\\Rightarrow 2l(l\\minus{}1)\\rightarrow 2l \\equal{} k\\plus{}2$. Also, $ (2l(l\\minus{}1)\\plus{}2(l\\minus{}1)(l\\minus{}2)) \\equal{} 4(l\\minus{}1)^{2}|(2l(l\\minus{}1))(2(l\\minus{}1)(l\\minus{}2))\\Rightarrow 2(l\\minus{}1)(l\\minus{}2)\\rightarrow 2l(l\\minus{}1)$. Finally, $ (2(l\\minus{}1)\\plus{}2(l\\minus{}2)(l\\minus{}1)) \\equal{} 2(l\\minus{}1)^{2}|(2(l\\minus{}1)*2(l\\minus{}2)(l\\minus{}1))\\Rightarrow 2(l\\minus{}1)\\rightarrow 2(l\\minus{}2)(l\\minus{}1)$.\r\nBecause $ k\\plus{}2 \\equal{} 2l$, then $ k \\equal{} 2l\\minus{}2 \\equal{} 2(l\\minus{}1)$.\r\nSo, $ k \\equal{} 2l(l\\minus{}1)\\rightarrow 2(l\\minus{}2)(l\\minus{}1)\\rightarrow 2l(l\\minus{}1)\\rightarrow 2l \\equal{} k\\plus{}2\\rightarrow k\\plus{}1\\Rightarrow k\\rightarrow k\\plus{}1$, but because $ 3\\rightarrow k$ by lemma 3 it follows that $ 3\\rightarrow k\\plus{}1$ and the result is true for $ a \\equal{} k\\plus{}1$.\r\nHence, in both cases we have proved the result for $ a \\equal{} k\\plus{}1$ and by mathematical induction it follows that $ 3\\rightarrow a$ for any positive integer $ a\\geq 3$ and lemma 4 is proved.\r\n\r\nNow, for any two positive integers $ a,b$ greater than $ 2$, by lemma 4, $ 3\\rightarrow a\\Leftrightarrow a\\rightarrow 3$ and $ 3\\rightarrow b$. So, $ a\\rightarrow 3$ and $ 3\\rightarrow b$ so by lemma 3, $ a\\rightarrow b$ and the result of the problem follows.", "Solution_2": "First, we can see that if there is a finite sequence satisfying the problem requirements such that $n_{1} = a$ and $n_{k} = b$, then we can make a sequence that satisfies the problem requirements such that $m_{1} = b$ and $m_{k} = a$ with the simple mapping of $m_{j} = n_{k-j+1}$. Also, note that if we have a term X in our sequence and D is a divisor of X then (D-1)X may be the next term in the sequence because $\\frac{(D-1)X^2}{DX}$ is an integer. Thus if X is a term in our sequence, then we can make X! a term in our sequence as well. If we want to construct a sequence that takes a to b, we can assume that a is less than b. Now, we can construct a sequence that takes a to a! and we now want to construct a sequence that takes a! to b. Let us consider the case in which b is odd. We can see that b+1 is now even and thus divisible by 2. Also, let us note that we can take a! to (2a)! by simply taking a! to 2*a! and noting that this is divisible by 2a which lets us take it to (2a)!. If $a < \\frac{b+1}{2} \\rightarrow 2a < b+1$. We can now take a! to $(2^k*a)!$ such that $\\frac{b+1}{2} \\le 2^k*a < b+1$. We can now note that 2 and $\\frac{b+1}{2}$ both divide $(2^k*a)!$ thus we can take $(2^k*a)!$ to $b*(2^k*a)!$ and take this to b!. However, we can take b! to b because we can take b to b!. Now let us consider the case in which b is even. We can take b to b(b-1) to (b-2)(b-1) to b-1 which is odd and thus we can construct a path from a to b-1 to b. Thus we can always construct a path from a to b for any pair of integers (a, b)." } { "Tag": [ "function", "geometry", "analytic geometry", "algebra", "domain", "functional equation", "algebra proposed" ], "Problem": "Find all continuous functions $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ satisfying\r\n$ f(f(x)y \\minus{} xf(y)) \\equal{} yf(f(x)) \\minus{} f(x)f(y)$.\r\n\r\n\r\nI don't know whether the condition continuous is necessary.\r\nMy interest is not finding the solution but how such functions satisfy this equation.. After you solve this, I think you would feel same as me...", "Solution_1": "Take $ x \\equal{} y \\equal{} 0$ then $ f(0)^2 \\plus{} f(0) \\equal{} 0$ \r\nTherefore $ f(0) \\equal{} 0$ or $ f(0) \\equal{} \\minus{} 1$\r\nCase 1 : $ f(0) \\equal{} 0$\r\nTake $ x \\equal{} y$ then $ f(x)^2 \\equal{} xff((x))$\r\nWe will prove that if exist a real number $ x_0$ different from 0 and $ f(x_0) \\equal{} 0$ then $ f(x) \\equal{} \\equiv 0$\r\nTake $ y \\equal{} x_0$ then $ f(x_0f(y)) \\equal{} x_0f(f(y))$ (1)\r\nTake $ x \\equal{} x_0$ then $ f( \\minus{} x_0f(y)) \\equal{} 0$\r\nTake $ x \\equal{} \\minus{} x_0f(y)$ then $ f(x_0f(y)^2) \\equal{} 0$\r\nTake $ y \\equal{} 1$ then we have $ f(x_0f(f(1)) \\equal{} 0$\r\nTake $ x \\equal{} f(1)$ in (1) then $ f_3(1) \\equal{} 0$\r\nEasy to check that $ f_3(1) \\equal{} \\frac {f(1)^3}{1^2}$ so $ f(1) \\equal{} 0$\r\nTake $ x \\equal{} 1$ then $ f( \\minus{} f(x)) \\equal{} 0$ therefore $ f(f(x)f(y)) \\equal{} 0$\r\nIf exist a real a such that $ f(f(a))$ is different from 0 then we have : \r\n$ f(f(a)y \\minus{} af(y)) \\plus{} f(a)f(y) \\equal{} xf(f(a))$\r\nSo there exist at least of $ f(f(a)y \\minus{} af(y)),f(y) > 1$ or exist a real r such that $ f(r) > 1$\r\nConsider $ g(x) \\equal{} f(x) \\minus{} 1$\r\n$ G(0) \\equal{} \\minus{} 1 ,f(r) > 0$ and because g continuous so exist $ r'$ such that $ f(r') \\equal{} 1$\r\nTake $ x \\equal{} r'$ then $ f(f(x)) \\equal{} 0,\\forall x$\r\nBut from $ xf(f(x)) \\equal{} f(x)^2$ so $ f(x)\\equiv 0 ,\\forall x$ ,it gives contradiction . \r\nNow consider problem when has no real different from 0 such that $ f(x_0) \\equal{} 0$\r\nFrom condition $ xf(f(x)) \\equal{} f^2(x)$ it shows that f is an inject function . \r\n$ xf(f(x)y \\minus{} yf(x)) \\equal{} f(x)(yf(x) \\minus{} xf(y))$\r\nTake $ y\\to x,x\\to y$ then \r\n$ xf(yf(x) \\minus{} xf(y)) \\plus{} yf(xf(y) \\minus{} yf(x)) \\equal{} 0$\r\nTake $ y\\to \\minus{} x$ and from f is inject then \r\n$ f(x) \\plus{} f( \\minus{} x) \\equal{} 0$ so f is an odd function . \r\nTherefore $ (x \\minus{} y)(f(f(x)y \\minus{} xf(y)) \\equal{} 0$ \r\nIt gives that for all $ (x,y)$ then $ f(x)y \\minus{} yf(x) \\equal{} 0$ or $ f(x)\\equiv ax$\r\nCase 2 :$ f(0) \\equal{} \\minus{} 1$\r\nTake $ y \\equal{} 1,x \\equal{} 0$ then $ f(0)f(1) \\equal{} 0$ and because $ f(0) \\equal{} \\minus{} 1$ so $ f(1) \\equal{} 0$\r\nTake $ x\\to y$ then $ xf(f(x)) \\equal{} f(x)^2 \\minus{} 1$\r\nTake $ x \\to 1 then$ then $ f( \\minus{} f(y)) \\equal{} \\minus{} y$\r\nTherefore take $ x\\to \\minus{} f(x)$ then \r\n$ f(x)f( \\minus{} x) \\equal{} 1 \\minus{} x^2$\r\nMore than take $ x\\to 0$ then $ f( \\minus{} y) \\equal{} yf( \\minus{} 1) \\plus{} f(y)$\r\nThen we have recursion $ f(x)f( \\minus{} x) \\equal{} 1 \\minus{} x^2,f(x) \\minus{} f( \\minus{} x) \\equal{} xf( \\minus{} 1)$\r\nThen easy to check that $ f(x) \\equal{} x \\minus{} 1$ is solution \r\n1.$ f(x)\\equiv 0$\r\n2.$ f(x)\\equiv ax$\r\n3. $ f(x)\\equiv x \\minus{} 1$\r\nI think we don't need condition ''continuous '' ,but now i can not find out a complete solution :)", "Solution_2": "[quote=\"TTsphn\"]Take $ x \\equal{} y$ then $ f(x)^{2} \\equal{} xf(x)$[/quote]\r\nWhy?? We just get $ f(0) \\equal{} xf(f(x)) \\minus{} f(x)^{2}$ so $ f(x)^{2} \\equal{} xf(f(x))$...\r\nBecause $ f(x) \\equal{} ax$ is one of its roots, your equation can't be hold...Try more.", "Solution_3": "I have proved that $ f(0)\\equal{}0$ . Take $ x\\equal{}y$ then $ xf(f(x))\\equal{}f(x)^2$", "Solution_4": "What do you mean? You wrote \"$ f(x)^{2}\\equal{}xf(x)$\".\r\nThis problem is not that easy I think.....", "Solution_5": "I'm sorry ,it is only a typo . I just have repair its .", "Solution_6": "[quote=\"TTsphn\"]\nTake $ y \\equal{} x_0$ then $ f(x_0f(y)) \\equal{} x_0f(f(x))$ (1)[/quote]\r\nIt should be $ f(x_{0}f(x)) \\equal{} x_{0}f(f(x))$ and you can't do anything more.\r\nabout Case 2 : $ x \\minus{} 1$ is solution, too. So your argument is false.....\r\nPlease check your whole solution and edit it...", "Solution_7": "[quote=\"Little Gauss\"][quote=\"TTsphn\"]\nTake $ y \\equal{} x_0$ then $ f(x_0f(y)) \\equal{} x_0f(f(x))$ (1)[/quote]\nIt should be $ f(x_{0}f(x)) \\equal{} x_{0}f(f(x))$ and you can't do anything more.\nabout Case 2 : $ x \\minus{} 1$ is solution, too. So your argument is false.....\nPlease check your whole solution and edit it...[/quote]\r\nI will complete solution ,it is case 2 of my problem . \r\n$ f(1)\\equal{}0,f(0)\\equal{}\\minus{}1$", "Solution_8": "You still have a problem in Case1.\r\nFor the final step of Case 2, you can't say like that. try $ f(x)\\equal{}2x\\minus{}\\sqrt{3x^{2}\\plus{}1}$.", "Solution_9": "[hide][quote=\"TTsphn\"]Take $ x \\equal{} y \\equal{} 0$ then $ f(0)^2 \\plus{} f(0) \\equal{} 0$ \nTherefore $ f(0) \\equal{} 0$ or $ f(0) \\equal{} \\minus{} 1$\nCase 1 : $ f(0) \\equal{} 0$\nTake $ x \\equal{} y$ then $ f(x)^2 \\equal{} xff((x))$\nWe will prove that if exist a real number $ x_0$ different from 0 and $ f(x_0) \\equal{} 0$ then $ f(x) \\equal{} \\equiv 0$\nTake $ y \\equal{} x_0$ then $ f(x_0f(y)) \\equal{} x_0f(f(y))$ (1)\nTake $ x \\equal{} x_0$ then $ f( \\minus{} x_0f(y)) \\equal{} 0$\nTake $ x \\equal{} \\minus{} x_0f(y)$ then $ f(x_0f(y)^2) \\equal{} 0$\nTake $ y \\equal{} 1$ then we have $ f(x_0f(f(1)) \\equal{} 0$\nTake $ x \\equal{} f(1)$ in (1) then $ f_3(1) \\equal{} 0$\nEasy to check that $ f_3(1) \\equal{} \\frac {f(1)^3}{1^2}$ so $ f(1) \\equal{} 0$\nTake $ x \\equal{} 1$ then $ f( \\minus{} f(x)) \\equal{} 0$ therefore $ f(f(x)f(y)) \\equal{} 0$\nIf exist a real a such that $ f(f(a))$ is different from 0 then we have : \n$ f(f(a)y \\minus{} af(y)) \\plus{} f(a)f(y) \\equal{} xf(f(a))$\nSo there exist at least of $ f(f(a)y \\minus{} af(y)),f(y) > 1$ or exist a real r such that $ f(r) > 1$\nConsider $ g(x) \\equal{} f(x) \\minus{} 1$\n$ G(0) \\equal{} \\minus{} 1 ,f(r) > 0$ and because g continuous so exist $ r'$ such that $ f(r') \\equal{} 1$\nTake $ x \\equal{} r'$ then $ f(f(x)) \\equal{} 0,\\forall x$\nBut from $ xf(f(x)) \\equal{} f(x)^2$ so $ f(x)\\equiv 0 ,\\forall x$ ,it gives contradiction . \nNow consider problem when has no real different from 0 such that $ f(x_0) \\equal{} 0$\nFrom condition $ xf(f(x)) \\equal{} f^2(x)$ it shows that f is an inject function . \n$ xf(f(x)y \\minus{} yf(x)) \\equal{} f(x)(yf(x) \\minus{} xf(y))$\nTake $ y\\to x,x\\to y$ then \n$ xf(yf(x) \\minus{} xf(y)) \\plus{} yf(xf(y) \\minus{} yf(x)) \\equal{} 0$\nTake $ y\\to \\minus{} x$ and from f is inject then \n$ f(x) \\plus{} f( \\minus{} x) \\equal{} 0$ so f is an odd function . \nTherefore $ (x \\minus{} y)(f(f(x)y \\minus{} xf(y)) \\equal{} 0$ \nIt gives that for all $ (x,y)$ then $ f(x)y \\minus{} yf(x) \\equal{} 0$ or $ f(x)\\equiv ax$\nCase 2 :$ f(0) \\equal{} \\minus{} 1$\nTake $ y \\equal{} 1,x \\equal{} 0$ then $ f(0)f(1) \\equal{} 0$ and because $ f(0) \\equal{} \\minus{} 1$ so $ f(1) \\equal{} 0$\nTake $ x\\to y$ then $ xf(f(x)) \\equal{} f(x)^2 \\minus{} 1$\nTake $ x \\to 1 then$ then $ f( \\minus{} f(y)) \\equal{} \\minus{} y$\nTherefore take $ x\\to \\minus{} f(x)$ then \n$ f(x)f( \\minus{} x) \\equal{} 1 \\minus{} x^2$\nMore than take $ x\\to 0$ then $ f( \\minus{} y) \\equal{} yf( \\minus{} 1) \\plus{} f(y)$\nThen we have recursion $ f(x)f( \\minus{} x) \\equal{} 1 \\minus{} x^2,f(x) \\minus{} f( \\minus{} x) \\equal{} xf( \\minus{} 1)$\nThen easy to check that $ f(x) \\equal{} x \\minus{} 1$ is solution \n1.$ f(x)\\equiv 0$\n2.$ f(x)\\equiv ax$\n3. $ f(x)\\equiv x \\minus{} 1$\n\nI think we don't need condition ''continuous '' ,but now i can not find out a complete solution :)[/quote][/hide]I don't see how \"Take $ y \\equal{} x_0$ then $ f(x_0f(y)) \\equal{} x_0f(f(y))$ (1).\" Perhaps I'm just stupid.", "Solution_10": "It is true ,because $ f(x_0)\\equal{}0$ \r\nTake $ y\\to x_0$ then $ f(x_0f(x))\\equal{}x_0f(f(x))$\r\nIt is also true for any real y .", "Solution_11": "Oh yeah, I forgot you stated that $ f(x_0) \\equal{} 0$. I always make this kind of stupid mistakes of forgetting the previously stated conditions.", "Solution_12": "[quote=\"TTsphn\"]\n$ xf(f(x)y \\minus{} yf(x)) \\equal{} f(x)(yf(x) \\minus{} xf(y))$\nTake $ y\\to x,x\\to y$ then \n$ xf(yf(x) \\minus{} xf(y)) \\plus{} yf(xf(y) \\minus{} yf(x)) \\equal{} 0$[/quote]\r\nI can't understand this part... It can be deduced for only case $ f(x)\\equal{}f(y)$... isn't it?\r\n\r\nActually you used the fact \"f is continuous\"!", "Solution_13": "Case 1: $ f(0) \\equal{} \\minus{}1$.\r\n\r\nThen, we have $ f(x) \\equal{} kx\\minus{}\\sqrt{(k^2\\minus{}1)x^2\\plus{}1}$ for some $ k \\geq 1$.\r\n\r\nCase 2: $ f(0) \\equal{} 0$.\r\n\r\n2.1 $ f(t) \\equal{} 0$ for some $ t \\neq 0$.\r\n Then, for some $ k \\geq 0$, $ f(x) \\equal{} 0$ if $ x \\geq 0$ and $ f(x) \\equal{} kx$ if $ x<0$.\r\n\r\n2.2 $ f(t) \\equal{} 0$ if and only if $ t\\equal{}0$.\r\n I didn't complete this case, but I believe the solutions are in the form $ f(x) \\equal{} kx$ for some $ k \\neq 0$.\r\n\r\nI did finish Cases 1 and 2.1, but my solution is lengthy (and brutally messy). Do you have any short solution for this problem?", "Solution_14": "My solution is also long, messy. It seems to be impossible to show that above root of your Case 1 is really root without substituting and calculating both side.\r\nI have solution for $ f(0) \\equal{} 0$ using area of triangle in coordinate plane.. how did you do 2.1?\r\n\r\nAnyway, It is very curious. How such irregular function satisfy this tidy(?) equation?\r\nI think there are deeper reasons of it.", "Solution_15": "Case 2.1: $ f(0) \\equal{} 0$ and $ f(t) \\equal{} 0$ for some $ t \\neq 0$\r\n\r\nI borrowed a lot of ideas from TTsphn.\r\n\r\nI will use his results that\r\n(1) $ f(1) \\equal{} 0$,\r\n(2) $ xf\\left(f(x)\\right) \\equal{} \\left(f(x)\\right)^2$ for all $ x$,\r\n(3) $ f\\left( \\minus{} f(y)\\right) \\equal{} 0$ for all $ y$, and\r\n(4) $ f\\left(f(x)f(y)\\right) \\equal{} 0$ for all $ x$ and $ y$.\r\n\r\nLet $ S: \\equal{} f^{ \\minus{} 1}\\left(\\mathbb{R} \\minus{} \\{0\\}\\right)$. If $ S \\equal{} \\emptyset$, then $ f$ is the zero function. Suppose $ S$ is nonempty and let $ F$ be the restriction of $ f$ on $ S$. With slight effort, we get $ F$ is injective.\r\n\r\nSuppose that there exists $ a \\neq 0$ such that $ f(a) > 0$. Thus, $ f\\left(f(a)\\right) \\equal{} \\frac {\\left(f(a)\\right)^2}{a} \\neq 0$. TTsphn suggests that for all $ y$,\r\n\\[ yf\\left(f(a)\\right) \\equal{} f\\left(f(a)y \\minus{} af(y)\\right) \\plus{} f(a)f(y)\\,.\r\n\\]\r\nHence, we can choose $ y$ positively large or negatively large enough that $ yf\\left(f(a)\\right) \\gg \\max\\left\\{f(a),1\\right\\}$. Thus, the above equation implies that either\r\n\\[ f\\left(f(a)y \\minus{} af(y)\\right) \\gg 1\r\n\\]\r\nor\r\n\\[ f(y) \\gg 1\\,.\r\n\\]\r\nIn any case, using the continuity of $ f$, there must be a number $ u \\neq 0$ such that $ f(u) \\equal{} 1$. Consequently,\r\n\\[ 0 \\equal{} uf(1) \\equal{} uf\\left(f(u)\\right) \\equal{} \\left(f(u)\\right)^2 \\equal{} 1\r\n\\]\r\nis a blatant contradiction. Therefore, $ f(x) \\leq 0$ for all $ x$.\r\n\r\nNow, if $ f(p) < 0$ and $ f(q) < 0$ for some $ p > 0$ and $ q < 0$, then the injectivity of $ F$ is violated (argued by continuity of $ f$). Thus, either \r\n(i) $ f(x) \\equal{} 0$ for all $ x \\geq 0$ and $ f(x) \\leq 0$ for $ x < 0$, or\r\n(ii) $ f(x) \\equal{} 0$ for all $ x \\leq 0$ and $ f(x) \\leq 0$ for $ x > 0$.\r\nThe latter case cannot happen; otherwise, there would exist $ s > 0$ such that $ f(s) < 0$ and $ f\\left(f(s)\\right) \\equal{} \\frac {\\left(f(s)\\right)^2}{s}$ would be positive.\r\n\r\nHence, $ f(x) \\equal{} 0$ if $ x \\geq 0$ and $ f(x) \\leq 0$ for $ x < 0$. Now, assume that $ f(r) < 0$ for some $ r < 0$. Thus, for all $ y \\geq 0$,\r\n\\[ f\\left(f(r)y\\right) \\equal{} f\\left(f(r)y \\minus{} rf(y)\\right) \\equal{} yf\\left(f(r)\\right) \\minus{} f(r)f(y) \\equal{} y\\frac {\\left(f(r)\\right)^2}{r} \\,,\r\n\\]\r\nor $ f\\left(f(r)y\\right) \\equal{} k\\left(yf(r)\\right)$, where $ k: \\equal{} \\frac {f(r)}{r} > 0$. Since $ y$ is arbitrary, we then prove that $ f(x) \\equal{} kx$ for all negative $ x$.\r\n\r\nWe can easily check that both the zero function and the functions in the form discussed above satisfy the condition. Therefore, for any $ k \\geq 0$, $ f(x) \\equal{} 0$ if $ x \\geq 0$ and $ f(x) \\equal{} kx$ for $ x < 0$ is a solution to the functional equation.\r\n\r\n\r\nPS:\r\nCan anyone show me how to deal with the case where $ f(t) \\equal{} 0$ if and only if $ t \\equal{} 0$, please?", "Solution_16": "Case 1: $ f(0) \\equal{} \\minus{} 1$\r\n\r\nLet $ k : \\equal{} \\minus{} \\frac {f( \\minus{} 1)}{2}$. We know that\r\n(1) $ xf\\left(f(x)\\right) \\equal{} \\left(f(x)\\right)^2 \\minus{} 1$ for all $ x$,\r\n(2) $ f( \\minus{} y) \\equal{} \\minus{} 2ky \\plus{} f(y)$ for all $ y$ (replacing $ x$ by $ 0$ in the original equation), \r\n(3) $ f(1) \\equal{} 0$ (replacing $ y$ by $ 1$ in (2)), and\r\n(4) $ f\\left( \\minus{} f(y)\\right) \\equal{} \\minus{} y$ (replacing $ x$ by $ 1$ in the original equation).\r\nReplace $ y$ by $ \\minus{} f(y)$ in (2) and use (4) to get\r\n\\[ f\\left(f(y)\\right) \\equal{} 2kf(y) \\minus{} y\\,.\r\n\\]\r\nFrom (1) and the above,\r\n\\[ \\left(f(x)\\right)^2 \\minus{} 1 \\equal{} xf\\left(f(x)\\right) \\equal{} x\\left(2kf(x) \\minus{} x\\right)\\,.\r\n\\]\r\nThis shows that $ f(x) \\equal{} kx \\pm \\sqrt {\\left(k^2 \\minus{} 1\\right)x^2 \\plus{} 1}$ for each $ x$. By continuity of $ f$ and the fact that $ f(0) \\equal{} \\minus{} 1$, we conclude that\r\n\\[ f(x) \\equal{} kx \\minus{} \\sqrt {\\left(k^2 \\minus{} 1\\right)x^2 \\plus{} 1}\\,.\r\n\\]\r\nNow, $ f(1) \\equal{} 0$ implies that $ k$ is nonnegative. Since the domain of $ f$ is $ \\mathbb{R}$, then $ k \\geq 1$.\r\n\r\nYou should check that any $ f$ in the above form satisfies the condition. (This is not an easy task, but I confirm it is true.) Hence, all functions $ f$ satisfying the condition with $ f(0) \\equal{} \\minus{} 1$ are\r\n\\[ f(x) \\equal{} kx \\minus{} \\sqrt {\\left(k^2 \\minus{} 1\\right)x^2 \\plus{} 1}\\,,\r\n\\]\r\nwhere $ k \\geq 1$. (For $ k \\equal{} 1$, we get $ f(x) \\equal{} x \\minus{} 1$, for example.)", "Solution_17": "Case 2.2: $ f(t) = 0$ iff $ t = 0$.\r\n\r\nBorrowed ideas from TTsphn are\r\n(1) $ f$ is injective and\r\n(2) $ xf(f(x)y - xf(y)) = f(x)(f(x)y - xf(y))$ for all $ x$ and $ y$.\r\n\r\nFor each $ x$, define $ S_x : = \\{f(x)y - xf(y): y \\in \\mathbb{R}\\}$ (of course, $ 0 \\in S_x$ for each $ x$). Moreover, if $ x \\neq 0$, $ m_x : = \\frac {f(x)}{x}$. Also, let $ T$ denote the set $ \\{x \\in \\mathbb{R}: S_x \\neq \\{0\\}\\}$. Thus, for all $ x \\in \\mathbb{R}$ and $ y \\in S_x$,\r\n\\[ f(y) = m_xy\\,.\r\n\\]\r\nIf $ T = \\emptyset$, then there exits $ k\\in\\mathbb{R}$ such that $ f(x) = kx$ for all $ x$.\r\n\r\nAssume that $ T \\neq \\emptyset$ (but keep in mind that $ 0 \\notin T$). Hence, whenever $ a$ and $ b$ are in $ T$, $ S_a$ and $ S_b$ contain a common nonzero element or else, one of them contains only nonnegative elements and the other contains only nonpositive elements. Thus,\r\n\\[ T = T^ + \\cup T^ - \\,,\r\n\\]\r\nwhere $ T^ + = \\{x \\in T: S_x \\subseteq [0, + \\infty)\\}$ and $ T^ - = \\{x \\in T: S_x \\subseteq ( - \\infty,0]\\}$. Hence, $ m_x$ are equal for all $ x \\in T^ +$; let say that they are all $ m_ +$ (if $ T^ + = \\emptyset$, let just say $ m_ + = 0$). The number $ m_{ - }$ is defined similarly.\r\n\r\nNow, if $ x \\in \\mathbb{R} \\setminus T$ and $ y \\in T$, we have $ S_x = \\emptyset$, or\r\n\\[ f(x)y - xf(y) = 0\\,.\r\n\\]\r\nSince $ f(y) = m_{\\pm}y$, we then have\r\n\\[ f(x) = m_{\\pm}x\r\n\\]\r\nas well. This shows that $ f(x) = m_{\\pm}x$ for all $ x \\in \\mathbb{R}$.\r\n\r\nBy continuity of $ f$, for all $ x \\geq 0$, $ f(x) = k_{ + }x$ for some $ k_ + \\in \\{m_{\\pm}\\}$. Similarly, for all $ x \\leq 0$, $ f(x) = k_{ - }x$ for some $ k_ - \\in \\{m_{\\pm}\\}$.\r\n\r\n(i) If $ k_ + > 0$ and $ k_ - > 0$, then for $ x > 0$ and $ y < 0$,\r\n\\[ f((k_ + - k_ - )xy) = f(f(x)y - f(x)y) = yf(f(x)) - f(x)f(y)\\,.\r\n\\]\r\nThis shows that $ f((k_ + - k_ - )xy) = k_ + ((k_ + - k_ - )xy)$ or $ 0 < k_ + < k_ -$.\r\n\r\n(ii) If $ k_ + > 0$ and $ k_ - < 0$, then for $ x > 0$ and $ y < 0$,\r\n\\[ f((k_ + - k_ - )xy) = f(f(x)y - f(x)y) = yf(f(x)) - f(x)f(y)\\,.\r\n\\]\r\nHence, $ f((k_ + - k_ - )xy) = k_ + ((k_ + - k_ - )xy)$. However, since $ (k_ + - k_ - )xy < 0$, $ f((k_ + - k_ - )xy) = k_ - ((k_ + - k_ - )xy)$. Therefore, $ k_ + = k_ -$, which is absurd.\r\n\r\n(iii) If $ k_ + < 0$, then for $ x > 0$ and $ y < 0$,\r\n\\[ f((k_ + - k_ - )xy) = f(f(x)y - f(x)y) = yf(f(x)) - f(x)f(y)\r\n\\]\r\nimplies $ f((k_ + - k_ - )xy) = 0$. Thus, $ k_ + = k_ -$. This contradicts the fact that $ T$ is nonempty.\r\n\r\nTherefore, the solutions are\r\n(1) $ f(x) = kx$ for all $ x$, and\r\n(2) $ f(x) = k_ + x$ if $ x\\geq0$ and $ f(x) = k_ - x$ if $ x\\leq0$, where $ 0 < k_ + < k_ -$.\r\nCombining all Cases I have writen, the solutions to this functional equations are\r\n(1) $ f(x) = kx$ for all $ x$, where $ k$ is an arbitrary constant,\r\n(2) $ f(x) = k_ + x$ if $ x\\geq0$ and $ f(x) = k_ - x$ if $ x\\leq0$, where $ 0 \\leq k_ + < k_ -$, and\r\n(3) $ f(x) = kx - \\sqrt {(k^2 - 1)x^2 + 1}$, where $ k \\geq 1$.\r\n\r\nPS: This is the weirdest (and toughest) functional equation I have ever seen. By the way, there is no need to deal with Case 2.1 at all. I think Case 2.2 can be made to cover 2.1." } { "Tag": [ "vector", "topology", "linear algebra", "linear algebra unsolved" ], "Problem": "Let X be a real vector space and let M:X->X be a map such that 2M( (x+y)/2 ) =M(x)+M(y) for any x,y in X. Deduce that M is linear.\r\n\r\n(May need to assume that M is continuous).", "Solution_1": "[quote=\"chungyc\"]\n(May need to assume that M is continuous).[/quote]\r\n\r\nWhat is X? A subset of $\\mathbb{C}$?", "Solution_2": "X is just a vector space over $\\mathbb{R}$. However, I suppose we need a topology on X to talk about continuity. Let's assume that there is a norm defined on X." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "If $ A\\subseteq\\mathbb{R}$, and $ g: \\mathbb{A}\\to\\mathbb{R}$ is a lower semi-continuous function, prove that for each $ f: \\mathbb{A}\\to\\mathbb{R}$ with $ f(x)\\geq g(x)$ for every x in A, we have $ \\liminf_{x\\rightarrow y}{f(x)}\\geq g(y)$ for every y in A.", "Solution_1": "Split it into two parts: $ \\liminf_{x\\rightarrow y}{g(x)}\\geq g(y)$ and $ \\liminf_{x\\rightarrow y}{f(x)} \\ge \\liminf_{x\\rightarrow y}{g(x)}$." } { "Tag": [ "geometry", "inradius" ], "Problem": "Prove that the inradius of a right angled triangle with integer sides is an integer.", "Solution_1": "We have a right triangle $\\Delta = s \\cdot r = a \\cdot b / 2$\r\n\r\n$a \\cdot b = (a+b+c)(a+b-c)$\r\n\r\nSo $r = (a+b-c)/2$\r\n\r\nSo we have to prove that $2 \\mid a+b-c$ but since $a^2 + b^2 = c^2$, $a+b$ has same parity as $c$.", "Solution_2": "Use \\cdot instead of \\times or \"X\" next time, ok? ;)", "Solution_3": "[quote=\"Valentin Vornicu\"]Use \\cdot instead of \\times or \"X\" next time, ok? ;)[/quote]\r\n\r\nOK ......... [hide]but why ?[/hide]", "Solution_4": "Because \\times is used either for vectorial product, or for sizes of matrices, arrays and tabulars. \\cdot (which means \"central dot\") is for multiplication." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "From a circular piece of paper with radius BC, Jeff removes a sector of the circimference. Using the large sector left, he joins edge BC to edge BA(no overlap) to form a cone of radius 12 centimeters and of volume 432$\\pi$ cubic centimeters. What is the number of degrees in the measure of angle ABC of the sector that is not used?\r\nEnjoy... :D", "Solution_1": "[hide]You find the height to be 9 if you use the volumn and the radius.\n You can then easily tell that the slant height or the radius of circle B is 15 using the Pythagorean Theorem.\nTherefore the circumference of circle B is 30$\\pi$\nYou find the circumference of the cone is 24$\\pi$\n\n$360(\\frac{30-24}{30})=\\boxed{72}$[/hide]", "Solution_2": "This question is actually number 29, by the way.", "Solution_3": "[hide]h=height of cone, r=radius of circle=slant height of cone, x=degree of arc\n\n$\\frac{12^2\\times h}{3}=432\\pi \\implies h=9$.\nSo $r=15$. Then $x=360-360\\times\\frac{2\\times 12\\pi}{2\\times 15\\pi}=72$.[/hide]" } { "Tag": [ "algorithm", "number theory", "relatively prime" ], "Problem": "prove that any two consecutive integers are relativley prime.", "Solution_1": "[hide]Let be the smaller of the two integers, and be the larger. Thus, to see if they are relatively prime, a fraction containing them must not reduce. This fraction is or its inverse. Expanding this fraction, it gives . Since x is an integer, is always not an integer, making the two numbers relatively prime.\n\n\n\nThere are 2 special cases to this:\n\n, in which case 1 and 2 are relatively prime\n\n or , so that x or x+1 is 0, therefore making relative primality trivial.[/hide]", "Solution_2": "mrblueskies wrote:prove that any two consecutive integers are relativley prime.\n\n[hide]we could use euclids algorithm which completely trivializes it...[/hide]", "Solution_3": "Oh really? I was thinking of somthing along the lines of towbomb. What is Euclid's algorithim and how do you use it?", "Solution_4": "Suppose the integer [i]n[/i] is divisible by any number of facors {a,b,c...}. \r\n\r\nn must be congruent to all of these integers mod 0.\r\n\r\nn + 1 must be congruent to all of these integers mod 1 . Consequently, n and n + 1 have no factors in common (except 1) and are therefore relatively prime. [/i]", "Solution_5": "(n+1)-n=1. QED." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "x_1, x_2,..., x_5 are pisitive reals such that 1/(1+x_1)+1/(1+x_2)+...+1/(1+x_5)=1.\r\nProve that 1/(4+x_1^2)+1/(4+x_2^2)+...+1/(4+x_5^2)<=1", "Solution_1": "Posted here:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3048[/url]\r\n\r\nAnd please choose better titles for your posts; the title \"Inequalities\" is too generic." } { "Tag": [ "topology", "function", "advanced fields", "advanced fields unsolved" ], "Problem": "$E$ the set of function from $[0;1]$ to $\\mathbb{R}$. Does there exist a topology on $E$ that generates the simple convergence.", "Solution_1": "It's the topological space $\\mathbb R^{[0,1]}$, i.e. the topological product of $c$ copies of $\\mathbb R$ ($c$ being the continuum cardinality).", "Solution_2": "in fact this question was a particular case of the following question. On a topological space $E$, if one define a kind of convergence, (i.e. if we know all convergent sequence) does there always exists a topology that generates this convergence.\r\n\r\nAt the beginning we wanted to define the set of closed sets of $E$ as follows:\r\nLet $C$ the set of all convergent sequences of $E$. To each $U \\in C$, let $L(U)$ the limit of $U$.\r\nOne says that $F$ is a closed subset if any sequence $U$ of element of $F$ that belongs to $C$ satisfies $L(U) \\in F$.\r\nBut it does not work because maybe there is more convergent sequence wiht this new topology. So, is the general statement true ? but maybe this is a very stupid question ..", "Solution_3": "dear Grobber, I think i've not perfectly understand your example. Could you precise it ? because it seems it doesn't work isn't it ?", "Solution_4": "The functions from $[0,1]$ to $\\mathbb R$ can be regarded simply as elements of $\\mathbb R^{[0,1]}$. A net (generalized sequence) of functions $(f_{\\alpha})_{\\alpha}$ from $[0,1]$ to $\\mathbb R$ converges pointwisely to a function $f: [0,1]\\to\\mathbb R$ iff the corresponding elements form a net converging to the element corresponding to $f$. There really isn't anything to prove. This is how you define the product topology.", "Solution_5": "I thought the product topology on $\\mathbb{R}^{[0;1]}$ was the topology generated by th sets of the form :\r\n$\\Pi_{x \\in F} w_x$ where $F$ is a finite set of $[0;1]$ and $w_x$ an open set of $\\mathbb{R}$ for any $x \\in F$. Is it ?", "Solution_6": "And why would you think that this is different from the topology of pointwise convergence? :)", "Solution_7": "because i was tired :D", "Solution_8": "consider the topology generated by the sets of the form :\r\n$\\Pi_{x \\in S} w_x$ where $S$ is an arbitrary set of $[0;1]$ and $w_x$ an open set of $\\mathbb{R}$ for any $x \\in S$. What is now the convergence ? it is much stronger than uniform convergence ..", "Solution_9": "There are several topologies you can put on an (infinite) cartesian product.\r\nwhat you need for simple convergence is the smallest topology for which the following functions are continuous:\r\n$ev_x: E\\to\\mathbb R$, $ev_x(f)=f(x)$, for $x\\in [0,1]$. \r\nIn other words a basis of open sets is \r\n$\\bigcap_{i=1}^nev_{x_i}^{-1}(G_i)$\r\nwhere $x_1,\\dots, x_n$ are finitely many points and $G_i$ are open in $\\mathbb R$.\r\nYour example is not of this kind.", "Solution_10": "yep, I got this (this is exactly the usual product topology). But I still wonder what kind of convergence we obtain with the new topology ?", "Solution_11": "$f_n\\to 0$ if for any positive function $\\phi: [0,1]\\to (0,\\infty)$, there exists $N(\\phi)$ such that\r\n$|f_n(x)|\\leq\\phi(x)$, $\\forall x\\in [0,1]$, $\\forall n\\geq N(\\phi)$. \r\n\r\nexample: if $f_n(x)=0$ for $x\\neq 1/2$ and $f_n(1/2)=1/n$, then $f_n\\to 0$." } { "Tag": [ "percent" ], "Problem": "What common fraction is equal to half of 80%?", "Solution_1": "[hide]\n80% = 8/10\n\n40% = 4/10\n4/10 = 2/5\n\n2/5 is the common fraction[/hide]", "Solution_2": "[hide]$\\frac{80}{100}\\cdot\\frac{1}{2}=\\frac{2}{5}$[/hide]", "Solution_3": "[hide][quote=\"anirudh\"]What common fraction is equal to half of 80%?[/quote]\n\n80%=4/5\nhalf of that is 2/5[/hide]", "Solution_4": "Wasn't this a state countdown problem from this year?", "Solution_5": "[hide]$\\frac{4}{10}=\\boxed{\\frac{2}{5}}$[/hide]", "Solution_6": "[hide]$80\\%=\\frac{80}{100}=\\frac{8}{10}=\\frac{4}{5}$\nHalf of this is $\\frac{2}{5}$[/hide]", "Solution_7": "A good way to do this problem, or any problem which has a % sign, is to think of % as a variable which stands for 1/100. Then just work with the resulting fraction.", "Solution_8": "[hide]80% is 4/5, half of which is 2/5[/hide]", "Solution_9": "Percent = percentage/100, therefore 80/100 * 1/2", "Solution_10": "[quote=\"bpms\"]Wasn't this a state countdown problem from this year?[/quote]\r\n\r\nYes it was (2006 State Countdown #12). If you want to see all of the problems, click [url=http://www.mathcounts.org/webarticles/articlefiles/510-06SCountdown.pdf]here[/url].", "Solution_11": "[hide]$\\frac{80\\%}{2}= 40\\%$\n$40\\% = \\frac{40}{100}$\n\n$\\frac{\\frac{40}{10}}{\\frac{100}{10}}= \\frac{4}{10}$\n\n$\\frac{\\frac{4}{2}}{\\frac{10}{2}}= \\boxed{\\frac{2}{5}}$[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $\\large a;b;c > 0$.\r\n$\\large \\sum \\sqrt{\\frac{a}{a^2+bc}} \\leq\\ \\sum \\frac{1}{\\sqrt{a+b}}$", "Solution_1": "Yes it is true. Put $x=\\frac1a$ etc., rename and see\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=67457\r\nIt is the $n=\\frac12$ case of (2). (posts #4 and #5 prove it for all positive $n$) :D", "Solution_2": "Thanks [b]spanferkel[/b]" } { "Tag": [ "AMC" ], "Problem": "Sooo, who got a perfect? There have got to be some on the forum.", "Solution_1": "Thazn or whatever his username is. :blush:", "Solution_2": "...woot :)", "Solution_3": "That 's so wonderful. Congratulations, ThAnz1!" } { "Tag": [ "calculus", "integration", "limit", "trigonometry", "algebra", "function", "domain" ], "Problem": "I'm trying to solve some problems from the Measure & Integral class and I'm having difficulties with the following problems:\r\n\r\n1. Calculate\r\n\r\n$ \\lim_{n\\to\\infty}\\sum_{m \\equal{} 0}^{\\infty}\\left(\\frac {1}{2}\\right)^{nm}\\cos{(nm)}$\r\n\r\nand justify your calculations.\r\n\r\n2. Let $ f_n(x) \\equal{} \\left(1 \\minus{} \\left(1 \\minus{} \\frac {1}{x^2}\\right)^2\\right)^n$ be a sequence of functions. Calculate\r\n\r\n$ \\lim_{n\\to\\infty}\\int_{2}^{ \\plus{} \\infty}f_n(x)dx$\r\n\r\nand justify your calculations.\r\n\r\nFor the first problem, I don't know how to approach it. I've solved some problems where an integral sign is placed instead of the sum by using the LDCT (Lebesgue's Dominant Convergence Teorem) but I don't know if something similar applies when there is a sum.\r\n\r\nFor the second problem, if I take the domain of $ f_n$ to be $ X \\equal{} \\left[1, \\plus{} \\infty\\right\\rangle$, than the limit $ \\lim_{n\\to\\infty}f_n(x)$ equals the indicator function $ \\textbf{1}_{\\{1\\}}$ so to apply the LDCT I need to find an integrable function $ g\\colon X \\to \\left[0, \\plus{} \\infty\\right]$ such that $ \\forall n\\in\\mathbb{N}\\quad |f_n(x)|\\leq g(x) \\quad (\\mu \\minus{} a.e.)$ on $ X$, but I'm having difficulties with finding $ g$.\r\n\r\nThanks for your help. :)", "Solution_1": "For the first one, I'm thinking Weierstrass M-test. Fix $ m$ so that we get $ \\frac {\\cos(mn)}{2^{mn}}\\leq\\frac {1}{2^{mn}} \\equal{} \\frac {1}{(2^{m})^{n}} \\equal{} M_{n}$. Then $ \\sum_{n \\equal{} 1}^{\\infty}M_{n} \\equal{} \\sum_{n \\equal{} 1}^{\\infty}(2^{ \\minus{} m})^{n} \\equal{} \\frac {1}{1 \\minus{} 2^{ \\minus{} m}} \\equal{} \\frac {1}{1 \\minus{} \\frac {1}{2^{m}}} \\equal{} \\frac {2^{m}}{2^{m} \\minus{} 1}$, which converges to 1. This means that $ f_{m}(n) \\equal{} \\frac {\\cos(mn)}{2^{mn}}$ converges uniformly. So you can swap limits and bring $ \\lim_{n\\rightarrow\\infty}$ inside the summation. That is, $ \\sum_{m \\equal{} 0}^{\\infty}\\lim_{n\\rightarrow\\infty}\\frac {\\cos(mn)}{2^{mn}} \\equal{}1\\plus{} \\sum_{m \\equal{} 1}^{\\infty}0 \\equal{}1\\plus{} 0\\equal{}1$. I'm not sure if this works.\r\n\r\nFor the second one, as far as I can tell, that function is monotone on your interval.", "Solution_2": "The DCT does apply to sums. A sum [i]is[/i] an integral over the integers (or some subset of the integers) with counting measure - that is, the measure of each integer point is 1. You could argue that the in the first problem, the terms tend pointwise to the the function $ (1,0,0,\\dots)$ - that is, the $ m\\equal{}0$ term tends to $ 1$ while every other term tends to $ 0.$ All terms are bounded above by $ 2^{\\minus{}m},$ which is summable (that is, integrable), so $ 2^{\\minus{}m}$ serves as the dominating function. So the limit is $ 1\\plus{}0\\plus{}0\\plus{}\\cdots \\equal{}1.$\r\n\r\nBut it's not a problem that showcases the need for the DCT. You can justify the interchange of sum and limit by more elementary means, as JRav showed. (Even if he did overlook the fact that the $ m\\equal{}0$ term is different.)\r\n\r\nThe second problem is also not a particularly convincing display of the need for the DCT or MCT. Note, by the way, that having a decreasing positive monotone sequence of functions doesn't guarantee anything - just consider $ \\int_0^{\\infty}\\frac{x}{n}\\,dx$ as $ n\\to\\infty.$ To get anything out of that, you have to have some function be integrable.\r\n\r\nHow I'd argue that: start with $ 0\\le1\\minus{}\\left(1\\minus{}\\frac1{x^2}\\right)^2\\le\\frac{2}{x^2}$ just by direct computation.\r\n\r\nThen $ \\int_2^{\\infty}f_n(x)\\,dx\\le\\int_2^{\\infty}\\frac{2^n}{x^{2n}}\\equal{}\\left.\\minus{}\\frac{2^n}{(2n\\minus{}1)x^{2n\\minus{}1}}\\right|_2^{\\infty}\\equal{}\\frac1{(2n\\minus{}1)2^{n\\minus{}1}}.$\r\n\r\nAnd that goes to zero. The argument is just calculus estimation and didn't need any big theorems. Yes, you could call it DCT using $ f_1$ as the dominating function - but that seems like overkill.\r\n\r\nOr is the in integral for the second problem actually from $ 1$ to $ \\infty$ rather than starting at $ 2?$ Because then we need a little more. If we are integrating on $ [1,\\infty),$ we see that $ f_n(x)\\to 0$ for $ x\\in (0,\\infty)$ but $ f_n(1)\\equal{}1.$ That's still going to zero almost everywhere. Then we can use that $ f_n(x)\\le f_1(x)\\le\\frac2{x^2}$ and that's an integrable function. So the limit is zero by the DCT.", "Solution_3": "Yes, I forgot to include that $ m\\equal{}0$ term in my sum :). I've edited it." } { "Tag": [ "inequalities", "trigonometry", "inequalities open" ], "Problem": "I am recalling trigonometric identities. (For inequalities)\r\n\r\nI know that if I have $ a\\plus{}b\\plus{}c\\equal{}abc$ then $ a\\equal{}\\tan{\\alpha},b\\equal{}\\tan{\\beta},c\\equal{}\\tan{\\gamma}$ with $ \\alpha\\plus{}\\beta\\plus{}\\gamma\\equal{}\\pi$ may work. (there are more, but my main question is:)\r\n\r\nIs there any similar trig substitution for $ ab\\plus{}bc\\plus{}ca\\equal{}abc$?", "Solution_1": "In such situation you can always change variables to $ x\\equal{}\\frac{1}{a}$, $ y\\equal{}\\frac{1}{b}$, $ z\\equal{}\\frac{1}{c}$ and then $ x\\plus{}y\\plus{}z\\equal{}1$.", "Solution_2": "You can try :\r\n\r\n$ a\\equal{} tan X tan Y$\r\n$ b\\equal{} tan Y tan Z$\r\n$ c\\equal{} tan Z tan X$\r\n\r\nFor example, prove that :\r\n\r\n$ \\frac{3}{2}\\sqrt{x\\plus{}y\\plus{}z\\minus{}1} \\ge \\sqrt{x\\minus{}1}\\plus{}\\sqrt{y\\minus{}1}\\plus{}\\sqrt{z\\minus{}1}$\r\n\r\nwhere $ xy\\plus{}yz\\plus{}zx\\equal{}xyz$ and $ x,y,z>1$", "Solution_3": "[quote=\"Nemion\"] Is there any similar trig substitution for $ ab \\plus{} bc \\plus{} ca \\equal{} abc$?[/quote]\r\n\r\nThis is not trigonometic, but it works in many inequalities. It's easy to check that for any $ x$, $ y$ and $ z$:\r\n\r\n\\[ {\\frac{3}{1\\plus{}x\\minus{}y} \\cdot \\frac{3}{1\\plus{}y\\minus{}z} \\plus{} \\frac{3}{1\\plus{}y\\minus{}z} \\cdot \\frac{3}{1\\plus{}z\\minus{}x} \\plus{} \\frac{3}{1\\plus{}z\\minus{}x} \\cdot \\frac{3}{1\\plus{}x\\minus{}y} \\equal{} \\frac{3}{1\\plus{}x\\minus{}y} \\cdot \\frac{3}{1\\plus{}y\\minus{}z} \\cdot \\frac{3}{1\\plus{}z\\minus{}x}}\\]\r\n\r\nSo we can substitute $ a\\equal{}\\frac{3}{1\\plus{}x\\minus{}y}$, $ b\\equal{}\\frac{3}{1\\plus{}y\\minus{}z}$ and $ c\\equal{}\\frac{3}{1\\plus{}z\\minus{}x}$.\r\n\r\nSergey Markelov" } { "Tag": [ "limit", "integration", "trigonometry" ], "Problem": "Using Riemann sums, find\r\n\r\n$\r\n\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\sin(k/n)\\cdot\\sin(k/n^2)\r\n$", "Solution_1": "[hide]$\\int_{0}^{1}x\\sin xdx$[/hide]", "Solution_2": "hmmm, if you are right please explain why..\r\n\r\n$\\lim \\sum k/n\\sin(k/n)=\\int_0^1 x\\sin x\\mathrm{d}x$\r\n\r\nif your result is true, then then we should prove\r\n\r\n$\\lim \\sum (k/n-\\sin(k/n^2)\\cdot\\sin(k/n)=0$\r\n\r\n\r\nwhy?", "Solution_3": "Well as k/n^2 goes to 0 we have sin(k/n^2) = k/n^2...", "Solution_4": "[quote=\"probability1.01\"]Well as k/n^2 goes to 0 we have sin(k/n^2) = k/n^2...[/quote]that is not a valid proof for it's summation being zero ;)", "Solution_5": "[quote=\"sebas\"]... please explain\n\n[/quote]\r\n\r\n\r\nUse\r\n\r\n$x-\\frac{x^3}{6}\\leq \\sin x \\leq x$ for $x\\geq 0$", "Solution_6": "this is what I tried\r\n\r\n$\r\n\\sin(k/n)\\cdot\\sin(k/n^2)\\leq k/n^2\\cdot\\sin(k/n)\r\n$\r\n\r\nthen\r\n\r\n$\r\nS_n=\\sum_{k=1}^n \\sin(k/n)\\cdot\\sin(k/n^2)\\leq\r\n\\sum_{k=1}^n k/n^2\\cdot\\sin(k/n)\r\n\\to\\int_0^1 t\\sin t\\mathrm{d}t\r\n$\r\n\r\nSo we get the bounds\r\n\r\n$\r\n0\\leq S_n\\leq \\int_0^1 t\\sin t\\mathrm{d}t\r\n$\r\n\r\nSimilarly\r\n\r\n$\r\n\\sin(k/n)\\cdot\\sin(k/n^2)\\geq \\bigl(\\sin(k/n^2)\\bigr)=\r\n\\frac{1-\\cos(2k/n^2)}{2}\r\n$\r\n\r\nthen\r\n\r\n$\r\nS_n\\geq \\sum_{k=1}^n \\frac{1-\\cos(2k/n^2)}{2}\r\n$\r\n\r\nI've found the sum of the right side, made $n\\to\\infty$ and the limit is different from $\\int_0^1 t\\sin t\\mathrm{d}t$", "Solution_7": "$\\sin(k/n)\\cdot\\sin(k/n^2)\\geq (k/n^2-k^3/(6n^6))\\sin(k/n)$\r\n\r\n$S_n\\geq 1/n\\sum_{k=1}^{n}k/n\\sin(k/n) - \\sum_{k=1}^{n}k^3/(6n^6)\\sin(k/n)$\r\n\r\nThe second sum in RHS tends to 0" } { "Tag": [ "geometry", "rectangle" ], "Problem": "Given a rectangle $ABCD$ with $AB > BC$, pick an arbitrary point $P$ in the plane. Prove that for any $P$, one of the two expressions:\r\n\\[m= | [APD]+[BPC] | , \\ \\ \\ n = |[APD]-[BPC] | \\]\r\nremains constant, and find that constant.\r\n\r\n$( [XYZ]$ denotes the area of triangle $XYZ$.$)$", "Solution_1": "[quote=\"cincodemayo5590\"]Given a rectangle $ABCD$ with $AB > BC$, pick an arbitrary point $P$ in the plane. Prove that for any $P$, one of the two expressions:\n\\[m= | [APD]+[BPC] | , \\ \\ \\ n = |[APD]-[BPC] | \\]\nremains constant, and find that constant.\n\n$( [XYZ]$ denotes the area of triangle $XYZ$.$)$[/quote]\r\n\r\nUm...\r\n[hide]\nIf we draw the altitudes from P to AD and BC...[APD]+[BPC] is $\\frac{BC}{2}(PE+PF)$ if E and F are the points of intersection of the altitude...but PE+PF is constant? and BC is constant as well...or is that not it... :maybe: [/hide]", "Solution_2": "[hide=\"hint\"] Consider the two cases: (1) $P$ is between lines $AD$ and $BC$, and (2) $P$ is not between them. Figure out which case corresponds to which expression being constant. [/hide]", "Solution_3": "[quote=\"paladin8\"][hide=\"hint\"] Consider the two cases: (1) $P$ is between lines $AD$ and $BC$, and (2) $P$ is not between them. Figure out which case corresponds to which expression being constant. [/hide][/quote]\n\nWhoops. \n[hide]\nSo then if P is between AD and BC then it's the addition one, and wouldn't it be basically the same if P isn't between AD and BC, except you subtract it instead...since if P were outside AD and P was closer to AD than BC you'd get PF-PE as a constant where F is the altitude to BC and E the altitude to AD...and same if P were closer to BC?[/hide]" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "What is the order of the group $ G$ with presentation\r\n$ \\left < a,b|a^{16} \\equal{} b^6 \\equal{} 1, bab^{ \\minus{} 1} \\equal{} a^3 \\right >$\r\n\r\nNo hint given :(\r\n\r\nAlso, can someone reccomend me a book that discusses free groups and finitely presented groups in a \"teaching\" way (i.e. not for researchers, but for students). I find the material we have is not enough.", "Solution_1": "$ a \\equal{} b^6 a b^{ \\minus{} 6} \\equal{} a^{3^6} \\equal{} a^9$, I think you can do the rest.\r\n\r\n[hide=\"solution\"]Thus it's $ \\langle a,b : a^8 \\equal{} b^6 \\equal{} 1 , b^{\\minus{}1} a b \\equal{} a^3 \\rangle \\cong \\mathbb{Z}/6 \\rtimes \\mathbb{Z}/8$ (remark $ gcd(3,8)\\equal{}1$ and $ 3^6 \\equiv 1$ mod $ 8$).[/hide]", "Solution_2": "Ok, thanks! Did you write something else at first? Because from $ a^8 \\equal{} 1$ I have also found the order to be $ 48$, but I thought you wrote $ Z_3 \\rtimes Z_8$ so I have been thinking about if perhaps I made a mistake in my reasoning.\r\n\r\nAnother question: You write $ \\langle a,b : a^8 \\equal{} b^6 \\equal{} 1 , b^{ \\minus{} 1} a b \\equal{} a^3 \\rangle \\cong \\mathbb{Z}/6 \\rtimes \\mathbb{Z}/8$. Can you really write isomorphism here? I thought the second expression doesn't uniquely determine a group unless both subgroups are normal, although it does give the order and many other properties. \r\n\r\nWhat about book advice?", "Solution_3": "yes I edited the post a couple of times. it's virtually impossible to determine the exact order if you don't give a more concrete representation of this group. so you should review your own proof. from the relations alone you just get an upper bound.\r\n\r\nin the context it is clear that the semidirect product is given by $ \\mathbb{Z}/6 \\to Aut(\\mathbb{Z}/8), 1 \\mapsto (z \\mapsto 3z)$.", "Solution_4": "I did give a concrete operation to the group by defining $ (a^xb^y)(a^ib^j)$ by looking at $ ba \\equal{} a^3b$, and then checking that the operation works ($ 3^6 \\equal{} 1$ mod 8 comes into play as you wrote). I'm not too sure about what the semi-direct product stuff means, but I'll try to read a little about it and think.", "Solution_5": "ok you probably worked out the semidirect product in this special case. you may have a look at http://en.wikipedia.org/wiki/Semidirect_product. but one important thing is missing there (maybe I'll add it): if $ \\alpha : G \\to Aut(N)$ is given, $ G \\rtimes_{\\alpha} N$ (in the article $ N \\rtimes_{\\alpha} G$) is the free group with generators of $ G$ and $ N$, relations of $ G$ and $ N$ as well as $ g^{ \\minus{} 1} n g \\equal{} \\alpha(g)(n)$ for all $ g \\in G, n \\in N$." } { "Tag": [ "summer program", "PROMYS", "Ross Mathematics Program", "number theory" ], "Problem": "Hallo! I am wavering between CTY Number Theory and Promys. Which one would I learn more at?\r\n\r\nThanks a muffinload,\r\n\r\n05.1 Logic\r\nDavid/Johnny", "Solution_1": "[quote=\"DavidL\"]Hallo! I am wavering between CTY Number Theory and Promys. Which one would I learn more at?\n\nThanks a muffinload,\n\n05.1 Logic\nDavid/Johnny[/quote]\r\n\r\nAs a PROMYS alumni, I would have to say that I'd be surprised if a high school student could learn that much NT in one summer anywhere else... I haven't heard much about CTY's NT course, but I'd bet PROMYS is much more rigorous and challenging, not to mention more selective (I think).", "Solution_2": "You would learn much much much much ... much more at PROMYS. CTY NT is pretty basic. ;)", "Solution_3": "Errr yeah CTY NT is pretty trivial. They're just like hmmm let's learn how to count I mean induct. This is true of CTY courses in general, I think. Of course, PROMYS may require a certain amount of mathematical experience (correct me if I'm wrong), but even so, you probably satisfy the requirements. It's always a good idea to do something challenging. You should ask beta, who went to PROMYS. OK, now I'm expecting that muffin: http://www.artofproblemsolving.com/Forum/weblog.php?w=46", "Solution_4": "This thread would certainly be on topic in, and might get more replies in, the [url=http://www.artofproblemsolving.com/Forum/index.php?f=136]Other U.S. Contests and Programs Forum[/url]. \r\n\r\nI have read in online discussion from one alumnus of Ross Program (similar to PROMYS) and the CTY Number Theory that two weeks of Ross >>>>> three weeks of CTY." } { "Tag": [ "number theory solved", "number theory" ], "Problem": "$p_{1}$,$p_{2}$,\u2026\u2026$p_{n}$ are primes,and $p_{i}<>p_{j},i<>j$\r\nprove :sqrt: $p_{1}$+ :sqrt:$ p_{2}$+\u2026\u2026+ :sqrt:$ p_{n}$can't be an integer", "Solution_1": "Have a look here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=rational&t=4496\r\n\r\nPierre." } { "Tag": [], "Problem": "How many equilateral triangles in the plane have two vertices in the set $ \\{(0, 0), (0, 1), (1, 0), (1, 1)\\}$?", "Solution_1": "The four points make up a square. There are 4 eq. triangle that share a side of the square and lies outside the square. Similarly, there are 4 eq. triangles that share a side and is inside the square. There are also 4 eq. triangles that use the diagonals as a side. Thus, there are a total of 12 different eq. triangles.", "Solution_2": "Choose 2 points from the given 4 in 4C2 ways = 6\nFrom 2 points, 2 different triangles are possible. Hence 12." } { "Tag": [ "inequalities", "function", "inequalities unsolved" ], "Problem": "Prove that\r\n\r\n\\[ (x + y + z)^4 \\ge 16(x^2 y^2 + y^2 z^2 + x^2 z^2 ) \\]", "Solution_1": "$(x+y+z)^2\\ge 3(xy+yz+zx)$\r\n$(x+y+z)^4\\ge 9(xy+yz+zx)^2$, but\r\n$\\frac{(xy+yz+zx)}{3}\\cdot\\frac{(xy+yz+zx)}{3}\\underbrace{\\ge}_{Chebishov} \\frac{x^2y^2+y^2z^2+z^2x^2}{3}$.\r\nHence $(xy+yz+zx)^2\\ge 3(x^2y^2+y^2z^2+z^2x^2)$\r\nSo $(x+y+z)^4\\ge 9(xy+yz+zx)^2\\ge 27(x^2y^2+y^2z^2+z^2x^2)$", "Solution_2": "[quote=\"vardanch\"]Prove that\n\n\\[ (x + y + z)^4 \\ge 27(x^2 y^2 + y^2 z^2 + x^2 z^2 ) \\][/quote]\r\n\r\nThis is wrong; take x = 2, y = 2, z = 1.\r\n\r\nBeat, you applied Chebyshev the wrong way round...\r\n\r\n Darij", "Solution_3": "It must be\r\n\\[ (x+y+z)^4\\geq16(x^2y^2+y^2z^2+z^2x^2). \\]\r\nThe proof involves AG, and derivative.", "Solution_4": "[quote=\"pvthuan\"]It must be\n\\[ (x+y+z)^4\\geq16(x^2y^2+y^2z^2+z^2x^2). \\]\nThe proof involves AG, and derivative.[/quote]\r\n\r\n I just curious to see the nice proof for this one. \r\nI have done it by substitute $y=x+a$, $z=x+a+b$ with $x,a,b \\geq 0$. After expanding, :blush: I get all of terms are positive. So this inequality is perfectly correct.", "Solution_5": "See Theorem 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=316023#p316023]http://www.mathlinks.ro/Forum/viewtopic.php?t=43904 post #13[/url] (if anyone can find the topic Harazi is speaking about, I would be grateful).\r\n\r\n Darij", "Solution_6": "Problem. Prove that if $a,b,c\\geq 0$, then $(a+b+c)^4\\geq16(a^2b^2+b^2c^2+c^2a^2)$. \r\n\r\nSolution. Without loss of generality, we can suppose that $a+b+c=1$, and $a\\geq b\\geq c$. It follows that $0\\leq c\\leq\\frac13$. Let $f=a^2b^2+b^2c^2+c^2a^2$, by AG inequaliy\r\n\\[ f\\leq \\frac{(a+b)^4}{4^2}+c^2(a+b)^2=\\frac{(1-c)^4}{16}+c^2(1-c)^2. \\]\r\nConsider function $f(c)=\\tfrac1{16}(1-c)^4+c^2(1-c)^2$ over $[0,\\tfrac13]$, differentiating $f(c)$ gives \r\n\\[ f'(c)=-\\frac14(1-c)(17c^2-10c+1). \\]\r\nWe have $f'(c)=0$ when $c=1$, or $c=\\tfrac{5-\\sqrt8}{17}$. Fuction $f(c)$ is decreasing on $[1,\\tfrac{5-\\sqrt8}{17}]$, increasing on $[\\tfrac{5-\\sqrt8}{17},\\tfrac13]$. It follows tha\r\n\\[ f(c)\\leq\\max\\{f(0),f(\\tfrac13)\\}=\\frac1{16}. \\]\r\nEquality holds when $(a,b,c)$ is some permutation of $(0,\\tfrac12,\\tfrac12)$.\r\n :lol:", "Solution_7": "@darij: All I could find is http://www.mathlinks.ro/Forum/viewtopic.php?t=2848 . Is this what you were looking for?\r\n\r\nBTW, is your avatar supposed to represent a version broken-line version of Pappus' Theorem?", "Solution_8": "I will prove the inequality by using Mixing Variables.\r\n\r\n[quote] Let $a,b,c$ be nonnegative reals. Prove that:\n\\[ (x+y+z)^4 \\ge 16(x^2y^2+y^2z^2+z^2x^2) \\] [/quote]\r\nLet $f(x,y,z)=(x+y+z)^4-16(x^2y^2+y^2z^2+z^2x^2)$.\r\nWlog, assume $0 \\le z \\le y \\le x$.\r\n\\begin{eqnarray*} f(x,y,z)-f\\left(x,\\frac{y+z}{2},\\frac{y+z}{2}\\right) &=& 16\\left[x^2\\left(\\frac{(y+z)^2}{2}-y^2-z^2\\right)+\\left(\\left(\\frac{y+z}{2}\\right)^2-yz\\right) \\left(\\left(\\frac{y+z}{2}\\right)^2+yz\\right) \\right] \\\\ &=& 4(y-z)^2\\left(\\left(\\frac{y+z}{2}\\right)^2+yz-2x^2\\right) \\\\ & \\ge & 0 \\\\ \\implies f(x,y,z) \\ge f\\left(x,\\frac{y+z}{2},\\frac{y+z}{2}\\right) \\end{eqnarray*}\r\nNormalize choice y letting $x+y+z=2$,that follows $x \\in \\left[0,\\frac{2}{3}\\right]$.\r\nNow:\r\n\\begin{eqnarray*} f(x,y,z) \\ge f\\left(x,\\frac{y+z}{2},\\frac{y+z}{2}\\right) \\\\ &=& 16-16\\left(2x^2\\left(\\frac{y+z}{2}\\right)^2+\\left(\\frac{y+z}{2}\\right)^4\\right) \\\\ &=& 16-16\\left(2x^2\\left(\\frac{2-x}{2}\\right)^2+\\left(\\frac{2-x}{2}\\right)^4 \\right) \\\\ &\\ge& 0 \\end{eqnarray*}\r\nThis is obviously true, because:\r\n\\[ 2x^2\\left(\\frac{2-x}{2}\\right)^2+\\left(\\frac{2-x}{2}\\right)^4 \\le 1 \\\\ \\iff x\\left(9x^3-40x^2+56x-32\\right) \\le 0\\ \\text{for all}\\ x\\in\\left[0,\\frac{2}{3}\\right] \\]\r\nDone.", "Solution_9": "Lovasz, it is nice too. Have you got this kind of inequality in four numbers?", "Solution_10": "Thanks, pvthuan.\r\nI think this inequality is true, it maybe trivial.\r\n\r\nLet $x,y,z,t \\ge 0$. Show that:\r\n\\[ (x+y+z+t)^4 \\ge 16\\left(x^2y^2+y^2z^2+z^2t^2+t^2x^2 \\right) \\]\r\n\r\nIt's easy. Right?\r\nI had another one but I'm not sure it's true.", "Solution_11": "Yes, it is easy. I solved it.", "Solution_12": "Can you post it?\r\nI wonder if our solution is same or different.", "Solution_13": "[quote=\"Lovasz\"]Can you post it?\nI wonder if our solution is same or different.[/quote]\r\nOk, let's wait me one day.", "Solution_14": "my solution:\r\nfor the orginal problem,it is equal to:$\\sum_{cyc}x^4+4\\sum_{sym}x^3y+6\\sum_{cyc}x^2y^2+12\\sum_{cyc}x^2yz \\ge 16\\sum_{cyc}x^2y^2$\r\njust use schur's:$\\sum_{cyc}x^4+\\sum_{cyc}x^2yz \\ge \\sum_{sym}x^3y$\r\nand $\\sum_{sym}x^3y \\ge 2\\sum_{cyc}x^2y^2$\r\n\r\n\r\nfor the n variables:assume x be the minimal one:\r\nwe can prove:$(z+t)^2x^2+y^2(z+t)^2 \\ge y^2z^2+z^2 t^2+t^2x^2$\r\nit reduce to $n-1$ variables..", "Solution_15": "[quote=\"zhaobin\"]my solution:\nfor the orginal problem,it is equal to:$\\sum_{cyc}x^4+4\\sum_{sym}x^3y+6\\sum_{cyc}x^2y^2+12\\sum_{cyc}x^2yz \\ge 16\\sum_{cyc}x^2y^2$\njust use schur's:$\\sum_{cyc}x^4+\\sum_{cyc}x^2yz \\ge \\sum_{sym}x^3y$\nand $\\sum_{sym}x^3y \\ge 2\\sum_{cyc}x^2y^2$\n\n\nfor the n variables:assume x be the minimal one:\nwe can prove:$(z+t)^2x^2+y^2(z+t)^2 \\ge y^2z^2+z^2 t^2+t^2x^2$\nit reduce to $n-1$ variables..[/quote]\r\n\r\nOh... master of expanding messy expression. I'm not brave enough like you...\r\nIndeed, I don't really like yours, zb.", "Solution_16": "[quote=\"Lovasz\"]I will prove the inequality by using Mixing Variables.\n\n[quote] Let $a,b,c$ be nonnegative reals. Prove that:\n\\[ (x+y+z)^4 \\ge 16(x^2y^2+y^2z^2+z^2x^2) \\] [/quote]\nLet $f(x,y,z)=(x+y+z)^4-16(x^2y^2+y^2z^2+z^2x^2)$.\nWlog, assume $0 \\le z \\le y \\le x$.\n\\begin{eqnarray*} f(x,y,z)-f\\left(x,\\frac{y+z}{2},\\frac{y+z}{2}\\right) &=& 16\\left[x^2\\left(\\frac{(y+z)^2}{2}-y^2-z^2\\right)+\\left(\\left(\\frac{y+z}{2}\\right)^2-yz\\right) \\left(\\left(\\frac{y+z}{2}\\right)^2+yz\\right) \\right] \\\\ &=& 4(y-z)^2\\left(\\left(\\frac{y+z}{2}\\right)^2+yz-2x^2\\right) \\\\ & \\ge & 0 \\\\ \\implies f(x,y,z) \\ge f\\left(x,\\frac{y+z}{2},\\frac{y+z}{2}\\right) \\end{eqnarray*}\nNormalize choice y letting $x+y+z=2$,that follows $x \\in \\left[0,\\frac{2}{3}\\right]$.\nNow:\n\\begin{eqnarray*} f(x,y,z) \\ge f\\left(x,\\frac{y+z}{2},\\frac{y+z}{2}\\right) \\\\ &=& 16-16\\left(2x^2\\left(\\frac{y+z}{2}\\right)^2+\\left(\\frac{y+z}{2}\\right)^4\\right) \\\\ &=& 16-16\\left(2x^2\\left(\\frac{2-x}{2}\\right)^2+\\left(\\frac{2-x}{2}\\right)^4 \\right) \\\\ &\\ge& 0 \\end{eqnarray*}\nThis is obviously true, because:\n\\[ 2x^2\\left(\\frac{2-x}{2}\\right)^2+\\left(\\frac{2-x}{2}\\right)^4 \\le 1 \\\\ \\iff x\\left(9x^3-40x^2+56x-32\\right) \\le 0\\ \\text{for all}\\ x\\in\\left[0,\\frac{2}{3}\\right] \\]\nDone.[/quote]\r\nsorry,I prefer zhaobin's solution to yours.", "Solution_17": "But I don't think you want to expand the 4-variable inequality?", "Solution_18": "[quote=\"Lovasz\"]But I don't think you want to expand the 4-variable inequality?[/quote]\r\nbut I wrote a method to reduce 4-varialbes to 3-varialbes :) \r\n\r\n\r\nThanks,Hawk Tiger", "Solution_19": "Problem: Let $a,b,c,d\\geq0$, prove that $(a+b+c+d)^4\\geq16(a^2b^2+b^2c^2+c^2d^2+d^2a^2)$.\r\n\r\nSolution: WLOG, we suppose that $0\\leq a\\leq b\\leq d\\leq c$ and $a+b+c+d=1$. The right hand reads\r\n\\[ f=(a^2+c^2)(b^2+d^2). \\]\r\nWe need to show that $f\\leq 1/16$. Note that\r\n\\begin{eqnarray*} f&=&\\{(a+c)^2-2ac\\}\\{(b+d)^2-2bd\\}\\\\ &\\leq& (a+c)^2\\{(b+d)^2-2bd\\}\\\\ &=&(1-s)^2(s^2-2p). \\end{eqnarray*}\r\nwhere $s=b+d$, $p=bd$. Consider $f(p)=-2p(s-1)^2+s^4-2s^3+s^2$ over $0\\leq p\\leq s^2/4$. We have $f(p)$ is decreasing over its interval of definition. So that $f(p)\\leq f(0)$. Moreover, $f(0)=s^2(1-s)^2$. \r\n\r\nBy AG inequality, we have\r\n\\[ s^2(1-s)^2\\leq\\Bigl(\\frac{s+1-s}2\\Bigl)^4=\\frac1{16}. \\]\r\n\r\n\r\npvthuan", "Solution_20": "Problem: Let $a,b,c,d \\ge 0$, prove that $(a+b+c+d)^4 \\ge 16 (a^2b^2+b^2c^2+c^2d^2+d^2a^2)$. \r\n\r\nThere is another straightforward proof: \r\n\r\n$(a+b+c+d)^4 \\ge 16(a+c)^2(b+d)^2 \\ge 16(a^2 b^2 + b^2c^2 + c^2d^2 + d^2a^2)$.\r\n\r\nThe first inequality is by AM-GM, the second one is obvious. :)", "Solution_21": "Yes, that is good solution.", "Solution_22": "[quote=\"darij grinberg\"]See Theorem 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=316023#p316023]http://www.mathlinks.ro/Forum/viewtopic.php?t=43904 post #13[/url] (if anyone can find the topic Harazi is speaking about, I would be grateful).\n\n Darij[/quote]\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=%28ab%29%5En&t=2934\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=%28ab%29%5En&t=2070\r\n\r\n\r\nSo, I think that the more general case also holds:\r\n\r\nIf $a,b,c\\geq 0$ and $n \\in \\mathbb N$ with $n\\geq 2$ then\r\n\r\n$(a+b+c)^{2n}\\geq 4^n(a^nb^n+b^nc^n+c^na^n)$\r\n\r\nStelios...", "Solution_23": "Even the following is true\r\n\\[ (a+b+c)^4\\geq16(a^2b^2+b^2c^2+c^2a^2)+11abc(a+b+c). \\]", "Solution_24": "What about this one?\r\n[hide=\"Click to reveal hidden content\"]\n[quote]\nLet $a,b,c,d$ be positive reals. Prove that:\n\\[ (a+b+c+d)^5 \\ge 256(a^2b^2c+b^2c^2d+c^2d^2a+d^2a^2b) \\]\n[/quote] [/hide]\r\nIt seems to be quite ugly.", "Solution_25": "[quote=\"pvthuan\"]Even the following is true\n\\[ (a+b+c)^4\\geq16(a^2b^2+b^2c^2+c^2a^2)+11abc(a+b+c). \\][/quote]\r\nYes, it is true for all non-negative real $a,$ $b$ and $c.$", "Solution_26": "[quote=\"Lovasz\"]What about this one?\nLet $a,b,c,d$ be positive reals. Prove that:\n\\[ (a+b+c+d)^5 \\ge 256(a^2b^2c+b^2c^2d+c^2d^2a+d^2a^2b) \\]\nIt seems to be quite ugly.[/quote]\r\nNo, it is very nice, but wrong. Tty $a=b=c=1$ and $d\\rightarrow0^{+}.$ ;)" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Didn't know where to post it, move it if needed.\r\n\r\nSo basicly you have to prove the equations [b]using only \"Algebra\" of sets, dont use the term x[/b], choose one side of the eauation and develop it to the other side while using known and proved sentences.\r\n\r\n[b]In the second equation, A' or B' means [u]Complement[/u], the wierd plus with the circle around him in means [u](A-B)U(B-A)[/u], the unit of the sets without there cut.[/b]\r\n\r\n\r\n[img]http://img80.imageshack.us/img80/7541/qaaub7.jpg[/img]\r\n\r\nTHANKS", "Solution_1": "PLEASE HELP", "Solution_2": "Let me try the second: $ A \\oplus B \\equal{} A' \\oplus B'$.\r\n\r\nThe operation $ A \\oplus B$ (which was also represented as $ A \\Delta B$) is (or was) called [i]symetric difference[/i], and equals $ A \\oplus B \\equal{} (A \\cup B) \\minus{} (A \\cap B)$. So we are to prove:\r\n\r\n$ (A \\cup B) \\minus{} (A \\cap B) \\equal{} (A' \\cup B') \\minus{} (A' \\cap B')$.\r\n\r\nBut we have:\r\n\r\n$ A' \\oplus B' \\equal{} (A' \\cup B') \\minus{} (A' \\cap B') \\equal{} (A' \\cup B') \\cap (A' \\cap B')'$\r\n\r\n$ \\equal{} (A' \\cup B') \\cap (A \\cup B)$\r\n\r\n$ \\equal{} (A \\cap B)' \\cap (A \\cup B)$\r\n\r\n$ \\equal{} (A \\cup B) \\cap (A \\cap B)'$\r\n\r\n$ \\equal{} (A \\cup B) \\minus{} (A \\cap B) \\equal{} A \\oplus B$\r\n\r\nand so our equality is proved.", "Solution_3": "Great solution, i managed to solve it in another way. \r\nfor each A,B, A-B= A$ \\cap$B' and then we get A+B= [(A') $ \\cap$ (B')'] $ \\cup$ [(A')' $ \\cap$ (B')]= A'+B'\r\nBut i can't solve the first, please help!", "Solution_4": "For the first equality, let's deal with the second member:\r\n\r\n$ (A_1 \\minus{} B_1) \\cup (A_1 \\minus{} B_2) \\equal{} (A_1 \\cap B_1') \\cup (A_1 \\cap B_2') \\equal{} A_1 \\cap (B_1' \\cup B_2') \\equal{} A_1 \\cap (B_1 \\cap B_2)'$\r\n\r\nwhere I have used the definition of difference, the distributive law, and Morgan's law. In a similar way:\r\n\r\n$ (A_2 \\minus{} B_1) \\cup (A_2 \\minus{} B_2) \\equal{} (A_2 \\cap B_1') \\cup (A_2 \\cap B_2') \\equal{} A_2 \\cap (B_1' \\cup B_2') \\equal{} A_2 \\cap (B_1 \\cap B_2)'$\r\n\r\nand so\r\n\r\n$ (A_1 \\minus{} B_1) \\cup (A_1 \\minus{} B_2) \\cup (A_2 \\minus{} B_1) \\cup (A_2 \\minus{} B_2) \\equal{} [A_1 \\cap (B_1 \\cap B_2)'] \\cup [A_2 \\cap (B_1 \\cap B_2)']$\r\n\r\n$ \\equal{} (A_1 \\cup A_2) \\cap (B_1 \\cap B_2)' \\equal{} (A_1 \\cup A_2) \\minus{} (B_1 \\cap B_2)$\r\n\r\nwhere I have used the distributive law and the definition of difference. So, the equality is proved." } { "Tag": [ "logarithms", "integration", "inequalities proposed", "inequalities" ], "Problem": "Let $H_n=\\sum\\limits_{k=1}^{n}\\frac{1}{k}$ and $A_n=H_n-\\frac{1}{2^n}\\sum\\limits_{k=1}^{n}{n \\choose k}H_k\\; .$ [color=green][b] Prove [/b][/color]\\[ \\begin{array}{|c|} \\hline \\\\ \\displaystyle \\frac{1}{(n+2)2^{n}} < \\displaystyle \\ln{2} -A_n < \\displaystyle\\frac{1}{(n+1)2^{n}}\\\\ \\\\ \\hline \\end{array} \\; , \\; (n\\in {\\mathbb N}^* ). \\]", "Solution_1": "[quote=\"flip2004\"]Let $H_n=\\sum\\limits_{k=1}^{n}\\frac{1}{k}$ and $A_n=H_n-\\frac{1}{2^n}\\sum\\limits_{k=1}^{n}{n \\choose k}H_k\\; .$ Prove that $\\frac{1}{(n+2)2^{n}} < \\ln{2} -A_n < \\frac{1}{(n+1)2^{n}}\\; \\; ,\\; \\; (n\\in {\\mathbb N}^* ).$[/quote]\r\nSome hints for possible complete solutions:\r\n[i]First solution:[/i] $\\ln{2}-A_n=\\frac{1}{2^n}\\sum\\limits_{k=0}^{\\infty} \\frac{(-1)^k}{(k+1){k+n+1\\choose n}}\\; .$ Further use properties of alternating series.\r\n[i]Second solution:[/i] If $S_n: =\\sum\\limits_{k=1}^n {n \\choose k}H_k ,$ then $S_n=2S_{n-1}+\\frac{1}{n}\\sum\\limits_{k=1}^n {n\\choose k} .$ Thus $S_n= 2^n\\left(H_n-\\sum\\limits_{k=1}^n \\frac{1}{k2^k}\\right)$ , $A_n=\\sum\\limits_{k=1}^{n} \\frac{1}{k2^k}$ and $\\ln{2}- A_n= \\frac{1}{2^n}\\int\\limits_{0}^{1}\\frac{x^n}{2-x}\\; dx .$" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "i dont understand proof of $Inn(G) \\unlhd Aut(G)$ where $Inn(G)$ interior automorphisms of G", "Solution_1": "Let $\\sigma \\in \\mbox{Aut} ( G )$. And $c_{x} \\in \\mbox{Inn} ( G )$ defined by $c_{x} ( a ) = xax^{- 1}$ for all $a \\in G$ and an $x \\in G$.\r\nThen for all $a \\in G$,\r\n\\begin{eqnarray*} \\left( \\sigma c_{x} \\sigma^{- 1} \\right) ( a ) & = & \\left. \\sigma \\left( c_{x} ( \\sigma^{- 1} ( a ) \\right) \\right)\\\\ & = & \\left. \\sigma ( x \\sigma^{- 1} ( a ) x^{- 1} \\right)\\\\ & = & \\sigma ( x ) \\sigma \\left( \\sigma^{- 1} ( a ) \\right) \\sigma ( x )^{- 1}\\\\ & = & \\sigma ( x ) a \\sigma ( x )^{- 1}\\\\ & = & c_{\\sigma ( x )} \\end{eqnarray*}\r\nSince $\\sigma$ is an automorphism, as $x$ goes through $G$, $\\sigma ( x )$ goes through $G$.\r\nHence $\\sigma \\mbox{Inn} ( G ) \\sigma^{- 1} = \\left\\{ \\sigma c_{x} \\sigma^{- 1} | x \\in G \\right\\} = \\left\\{ c_{\\sigma ( x )} | x \\in G \\right\\} = \\left\\{ c_{x} | x \\in G \\right\\} = \\mbox{Inn} ( G )$ for all $\\sigma \\in \\mbox{Aut}(G)$.\r\nIf anything is not clear, just tell me.", "Solution_2": "so how does the last thing\r\n$\\sigma \\mbox{Inn} ( G ) \\sigma^{- 1}=\\mbox{Inn} ( G )$ \r\nproove that innG is normal subgroup of G ?", "Solution_3": "Well isn't that the definition of normality ?\r\nLet $H$ be a subgroup of $G$.\r\n$H$ is normal in $G$ if and only for all $x \\in G$, we have $xHx^{-1}=H$.\r\nCorrect ?", "Solution_4": "[quote=\"spx2\"]so how does the last thing\n$\\sigma \\mbox{Inn} ( G ) \\sigma^{- 1}=\\mbox{Inn} ( G )$ \nproove that innG is normal subgroup of G ?[/quote]\r\nYou of course ment normal subgroup of $AutG$", "Solution_5": "ok...i understood.\r\ncould i get some bibliography that i can download from the internet ?\r\ni found something wrote by beachy http://www.math.niu.edu/~beachy/abstract_algebra/study_guide/contents.html\r\nbut i dont seem to have all proofs there.\r\ni need also a book with allot of exercices and examples", "Solution_6": "[url]http://www.jmilne.org/math/CourseNotes/math594g.html[/url]", "Solution_7": "amfulger thank you for the resource.\r\nnow comes a problem about that course of j.s. milne.\r\ni started reading it,i find that some of the theorems used are said\r\nto have been taught in a different course.\r\nwhere might that be?\r\nalso i understand that it is math594 - the name of that course\r\nwhat does this exactly mean ? that number 594 corresponds to some branch of\r\nmathematics and how advanced is the course.\r\ni think i read somewhere that courses (outside of romania ofcourse) have\r\nthese numbers attached to them..." } { "Tag": [], "Problem": "during a super collision ,as in the case of neutrons what is the external force that provides an increase in velocity after collision?", "Solution_1": "What? You mean like in fission? In fission the neutrons that hit the nuclei disturb the strong nuclear force holding the particles, which liberates the potential energy in the protons held together.", "Solution_2": "serialk11r what is the generic term of that force was my question .your answer was for a specific case" } { "Tag": [], "Problem": "(1) What is the best way to master, if you will, solving word problems?\r\n\r\n(Please, don't answer the obvious like PRACTICE, PRACTICE, PRACTICE, etc.)\r\n\r\n(2) Why do college textbooks call word problems APPLICATIONS?\r\n\r\n(3) Why do elementary school teachers call word problems MATH STORIES?\r\n\r\nThanks", "Solution_1": "[quote=\"Interval\"](1) What is the best way to master, if you will, solving word problems?\n\n(Please, don't answer the obvious like PRACTICE, PRACTICE, PRACTICE, etc.)\n\n(2) Why do college textbooks call word problems APPLICATIONS?\n\n(3) Why do elementary school teachers call word problems MATH STORIES?\n\nThanks[/quote]\r\n\r\n(1) PRACTICE, PRACTICE, PRACTICE, etc.\r\nSeriously, though, learn to look for the key information in a problem. Translate some of the relationships into mathematical expressions. Etc.\r\n\r\n(2) Because college students are mainly focusing on how math can be \"applied\" to whatever their major is. So the word problems are \"applications\" of the math concepts to the real world.\r\n\r\n(3) Because the teachers don't want to scare the little kids away from math by calling them \"applications\".", "Solution_2": "(1)\r\nLearn the lingo. You already know 'of' means '$*$' and stuff, but word problems are written in words, and you need to be able to translate any words you find into math. \r\n\r\n(2) \r\nmathnerd314 said it\r\n\r\n(3)\r\nbecause at the elementary school level they are stories. Bob the clown has 7 balloons.... that's a story!", "Solution_3": "For #3, the kids most likely don't know a lot of vocabulary but they probably know the words \"math\" and \"story\"." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "inequalities", "geometry", "function", "algebra" ], "Problem": "What general subjects do you hope each problem will cover?\r\n\r\nMy dream USAMO ( :P ):\r\n\r\n1. Inequality\r\n2. Geometry\r\n3. Number theory/divisibility\r\n\r\n4. Game theory\r\n5. Number theory/general\r\n6. Algebra/polynomial", "Solution_1": "1. Number theory\r\n2. Inequality or something bashable\r\n3. Don't care\r\n\r\n4. Geometry\r\n5. Functional equation\r\n6. Don't care", "Solution_2": "Last year's USAMO was as close to my dream as it'll get; I think the main reason I did well is that my solution to #4 was more rigorous than most. I seem to be better at those kinds of problems than at problems that require large theorems, for example.", "Solution_3": "1. Geometry\r\n2. Geometry\r\n3. Geometry\r\n\r\n4. Geometry\r\n5. Geometry\r\n6. Geometry\r\n\r\nNotice a trend here? They're the only things I can do xD", "Solution_4": "[quote=\"t0rajir0u\"]Last year's USAMO was as close to my dream as it'll get; I think the main reason I did well is that my solution to #4 was more rigorous than most. I seem to be better at those kinds of problems than at problems that require large theorems, for example.[/quote]\n\nLast year's day 1 was pretty sweet; I got all 3. Then day 2 pwn'd me; I only got #5 and the answer (un-rigorous proof) to #4.\n\nOf course, last year was the year I didn't qualify for the USAMO :(\n\n[quote=\"mathcrazed\"]1. Geometry\n2. Geometry\n3. Geometry\n\n4. Geometry\n5. Geometry\n6. Geometry\n\nNotice a trend here? They're the only things I can do xD[/quote]\r\n\r\nI'd say that's my best too, but I tried to be more realistic :D", "Solution_5": "Here's my hopes:\r\n\r\n1. Not Geometry\r\n2. Not Geometry\r\n3. Not Geometry\r\n\r\n4. Not Geometry\r\n5. Not Geometry\r\n6. Not Geometry\r\n\r\nSorry, mathcrazed <_<", "Solution_6": "Mine are \r\n\r\n1.Logic\r\n2. Algebra\r\n\r\n\r\n4. Number Theory\r\n5. Inequality\r\n\r\nDon't care about others...", "Solution_7": "1. combo\r\n2. geo\r\n3. ...\r\n\r\n4. alg/ineq\r\n5. geo\r\n6. ...\r\n\r\nI don't really plan on working on more than 2 problem per day.", "Solution_8": "NOT inequalities. I epic fail those.\r\n\r\n1. Geo\r\n2. Number Theory\r\n3. \r\n\r\n4. Combo\r\n5. Geo\r\n6. \r\n\r\nI have [i]never[/i] solved a 3/6, so I don't really care what goes there. I probably have a best shot if it's number theory or geo...\r\n\r\nand as I was writing this, I realized I really have no idea what my strengths/weaknesses are...\r\n\r\n[quote=\"nr1337\"]Last year's day 1 was pretty sweet; I got all 3. Then day 2 pwn'd me; I only got #5 and the answer (un-rigorous proof) to #4.\n\nOf course, last year was the year I didn't qualify for the USAMO :([/quote]\r\n\r\nWow, that really stinks", "Solution_9": "[quote=\"tjhance\"]NOT inequalities. I epic fail those.[/quote]\r\n\r\nMy theory is that if there's an inequality for #1, it will be a relatively straightforward application of AMGM/Cauchy/Jensen/etc. I'm not all that great with inequalities, but I'm sure I could do a #1.", "Solution_10": "[quote=\"nr1337\"][quote=\"tjhance\"]NOT inequalities. I epic fail those.[/quote]\n\nMy theory is that if there's an inequality for #1, it will be a relatively straightforward application of AMGM/Cauchy/Jensen/etc. I'm not all that great with inequalities, but I'm sure I could do a #1.[/quote]\r\n\r\nTrue, but if there is an inequality, it's probably either going to be geometric, or just one of those freaky weird ones that's asymmetric and involve weird functions, and not a #1.\r\n\r\nI hope I'm wrong.", "Solution_11": "[quote=\"Hamster1800\"]Here's my hopes:\n\n1. Not Geometry\n2. Not Geometry\n3. Not Geometry\n\n4. Not Geometry\n5. Not Geometry\n6. Not Geometry\n\nSorry, mathcrazed <_<[/quote]\r\nQFAAAAAAAAT\r\n\r\nAnyway, \r\n1. Combo\r\n2. Algebra\r\n3. Game Theory\r\n4. Combo\r\n5. Algebra\r\n6. Game Theory", "Solution_12": "I think all inequalities on the USAMO are meant to be applications of those basic 3 inequalities (plus rearrangement), just probably not always straight forward (though the last one we had was a #1, and it was really straight forward...)", "Solution_13": "I just hope that\r\n\r\nPlease let geo on the test is purely synthetic, and doesn't involved much length and/or angle measurement. I can't stand lengthy equations in geometry problems.\r\nNot more than one inequality. And a nice one, please.\r\nMore problems like #3 last year, pl0x. Gimme good combo.\r\nNT: Meh, as long as it's not a factorization like last year's #5.\r\nAlgebra: Please go easy on the polynomial stuff, it seems to suddenly be a weakness of mine\r\nYay game theory.\r\n\r\nIn general, nothing bash, because that's something I've avoided in problems ever since MOP, and I think it would be hard to go back to it.", "Solution_14": "Hmmm I'm not really sure...\r\n\r\nBest Performance:\r\n\r\n1. nothing weird\r\n2. geometry\r\n3. combinatorics that lends itself to analyzing small cases\r\n\r\n4. nothing weird\r\n5. combinatoricsy NT\r\n6. geometry\r\n\r\nMost Exciting:\r\n\r\n1. ...\r\n2. any combinatorics\r\n3. polynomials\r\n\r\n4. ...\r\n5. something unconventional\r\n6. combinatorial geometry\r\n\r\nMost Annoying:\r\n\r\n1. casework\r\n2. a \"does there exist\" problem where the answer is yes\r\n3. 3 variable inequality\r\n\r\n4. casework\r\n5. diophantine equation\r\n6. functional equation with lots of variables", "Solution_15": "1.number theory\r\n2.graph theory\r\n3.geometry(something like USAMO 2005 #3)\r\n\r\n4.inequality\r\n5.sequence\r\n6.geometry(something like USAMO 2006 #6)\r\n :D", "Solution_16": "1/4 anything besides geometry\r\n2/5 game theory, number theory, inequalities\r\n3/6 geometry. eww.\r\n\r\nAs you can see, I don't like geometry very much. :P", "Solution_17": "Hmm I know I can't get in on pure skill...\r\n\r\n1. Game theory\r\n2. random creativity problem\r\n3. a collection of 100 mathcounts state sprint problems\r\n\r\n4. Game theory\r\n5. random creativity problem\r\n6. a collection of 300 mathcounts national countdown problems\r\n\r\nok realistically.\r\n\r\n1. Geometry\r\n2. NT involving prime numbers\r\n3. Dont care won't get it\r\n\r\n4. Game theory / random creativity\r\n5. Algebra\r\n6. Dont care wont get it.", "Solution_18": "hopefully number 3/6 is a geo, cuz those are the only 3/6s i have any remote chance at getting\r\n\r\nif at least three of the remaining four problems are bashable, i'll be VERY happy (and i'll be seeing you all at MOP ;p )", "Solution_19": "[quote=\"K81o7\"][quote=\"cyberspace\"]Game Theory is pretty much all im good at and thats just because i get bored in 4.5 hrs and i dont have anything else to do except play the game :)[/quote]\n\nErgh...I lost...\n\nSure Anaxerzia...but you can't possibly suck as much as I do. I'm going to end up getting a negative zero.\n\n\nArgh...can't stand the tension waiting for the problems! No matter what problems come up, there's some luck involved...[/quote]\r\n\r\nYeah you suck so abysmally you shot be shot... -_-\r\n\r\nSeriously this self-depreciation helps us feel better about ourselves, and your just mocking it by being a USAMO winner an all...\r\n\r\n\r\nbah.", "Solution_20": "[quote=\"Anaxerzia\"][quote=\"K81o7\"][quote=\"cyberspace\"]Game Theory is pretty much all im good at and thats just because i get bored in 4.5 hrs and i dont have anything else to do except play the game :)[/quote]\n\nErgh...I lost...\n\nSure Anaxerzia...but you can't possibly suck as much as I do. I'm going to end up getting a negative zero.\n\n\nArgh...can't stand the tension waiting for the problems! No matter what problems come up, there's some luck involved...[/quote]\n\nYeah you suck so abysmally you shot be shot... -_-\n\nSeriously this self-depreciation helps us feel better about ourselves, and your just mocking it by being a USAMO winner an all...\n\n\nbah.[/quote]\r\n\r\nIn case anyone didn't quite catch it, the negative zero is a joke.", "Solution_21": "[quote=\"Hamster1800\"][/quote][quote=\"Anaxerzia\"][quote=\"K81o7\"][quote=\"cyberspace\"]Game Theory is pretty much all im good at and thats just because i get bored in 4.5 hrs and i dont have anything else to do except play the game :)[/quote]\n\nErgh...I lost...\n\nSure Anaxerzia...but you can't possibly suck as much as I do. I'm going to end up getting a negative zero.\n\n\nArgh...can't stand the tension waiting for the problems! No matter what problems come up, there's some luck involved...[/quote]\n\nYeah you suck so abysmally you shot be shot... -_-\n\nSeriously this self-depreciation helps us feel better about ourselves, and your just mocking it by being a USAMO winner an all...\n\n\nbah.[/quote][quote=\"Hamster1800\"]\n\nIn case anyone didn't quite catch it, the negative zero is a joke.[/quote]\r\n\r\nIn case anyone didn't quite catch it, mine was only 3/17ths serious :P", "Solution_22": "1. Functions\r\n2. Algebra/Sets\r\n3. Geometry (no inequalies/equalities in it)\r\n4. Number theory\r\n5. Game theory/logic\r\n6. Polynomials (dont really care since ill get a 0 on it)\r\n\r\nthe only 3/6 i can get are geometries and only with like the ones with angle chasing/concyclicity", "Solution_23": "[quote=\"Anaxerzia\"]\nYeah you suck so abysmally you shot be shot... -_-\n\nSeriously this self-depreciation helps us feel better about ourselves, and your just mocking it by being a USAMO winner an all...\n\nbah.[/quote]\r\n\r\nI'm trying to make a point here. Everyone says they will do so badly...\r\n\r\nI was taking the USAMO for the first time last year, and said the same thing. And look at what happened.", "Solution_24": "As someone who has experienced the [b]Korean Mathmatical Olympiads.....[/b]\r\n\r\nI daresay the USAMO is about the same as the Final Round of the KMO series...Hmm.\r\n\r\nAs both tests select the 12-member national team~\r\n\r\nMy ideal USAMO:\r\n\r\n1 - Something like Easy Algebra or Easy Combinatorics (But Alas, expecting an EASY on the USAMO is stupid :| )\r\n\r\n2 - A \"sensible\" Geometry/ Number Theory\r\n\r\n3 - I'd really like some Algebra here...something I can tackle :P \r\n\r\n4 - Number Theory/Geometry\r\n\r\n5 - Algebra\r\n\r\n6 - Don't care that much...I just wish I'd be able to UNDERSTAND the question and write a few lines... :P \r\n\r\n\"Hey, it's TODAY! Just a few more hours and we'll see for OURSELVES :lol: \"", "Solution_25": "[quote=\"Hamster1800\"]Here's my hopes:\n\n1. Not Geometry\n2. Not Geometry\n3. Not Geometry\n\n4. Not Geometry\n5. Not Geometry\n6. Not Geometry\n\nSorry, mathcrazed <_<[/quote]\r\nAgreed!\r\nExcept no inequalities either.", "Solution_26": "I got a 15... probably 7/1/0/7/0/0", "Solution_27": "[quote=\"DonkeyKingKong\"]I got a 15... probably 7/1/0/7/0/0[/quote]\r\n\r\nYou got a 3 on the AIME per your post http://www.artofproblemsolving.com/Forum/viewtopic.php?t=203500 :wink:", "Solution_28": "obviously donkeykingkong is full of crap :D", "Solution_29": "[quote=\"DonkeyKingKong\"]I got a 15... probably 7/1/0/7/0/0[/quote]\r\n\r\nThat's bullshit\r\nYou're bragging about your 3 on AIME and you're trying to skip to MIT" } { "Tag": [ "limit" ], "Problem": "Let $ S \\equal{}{99}!^{\\frac{1}{99}}$ and $ T \\equal{}{100!}^{\\frac{1}{100}}$\r\na) without using a calculator, which number is larger, $ S$ or $ T$?\r\nb) Which is larger: $ \\frac{S}{99}$ or $ \\frac{T}{100}$?\r\nc) Limit as $ n$ tends to infinity of $ \\frac{{n!}^{\\frac{1}{n}}}{n}$", "Solution_1": "Your question b) is exactly equivalent to a).\r\nLooking at c), I think that question b) should be:\r\nWhich is larger: $ \\frac{S}{99}$ or $ \\frac{T}{100}$?\r\nc) The limit is $ \\frac{1}{e}$, using Stirling's approximation:\r\n[url]http://en.wikipedia.org/wiki/Stirling's_approximation[/url]", "Solution_2": "[quote=\"lordWings\"]Your question b) is exactly equivalent to a).\nLooking at c), I think that question b) should be:\nWhich is larger: $ \\frac{S}{99}$ or $ \\frac{T}{100}$?\nc) The limit is $ \\frac{1}{e}$, using Stirling's approximation:\n[url]http://en.wikipedia.org/wiki/Stirling's_approximation[/url][/quote]\r\n\r\nAny takers for (a) or (b)?", "Solution_3": "I might be wrong, but can we say that $ \\frac{1}{e}\\approx\\frac{99!^{\\frac{1}{99}}}{99}$\r\nand therefore $ \\frac{99}{e}\\approx 99!^{\\frac{1}{99}}$?\r\n\r\nSame process for 100\r\n\r\nAnd we have that $ \\frac{100}{e}>\\frac{99}{e}$\r\n\r\nI'm not sure how to find the error term on that expression though.", "Solution_4": "a) is much simpler than that. [hide=\"All you have to say is...\"] $ S$ is the geometric mean of the set $ \\{ 1, 2, ... 99\\}$. The geometric mean will stay unchanged if we toss in the mean itself, which is clearly less than $ 99$. Hence tossing in the extra value $ 100$ increases the value of the new geometric mean $ T$. [/hide]", "Solution_5": "Or you can just clear the exponents and substitute $ n \\equal{} 99!$ to compare\r\n\r\n$ n^{100}\\quad\\rho\\quad (100n)^{99}$\r\n\r\nwhere $ \\rho$ is the desired relation. This simplifies to\r\n\r\n$ n\\quad\\rho\\quad 100^{99}$\r\n\r\nand $ \\rho$ is obviously $ <$, since all the factors on LHS are less than $ 100$.", "Solution_6": "great solution Farenhajt :)", "Solution_7": "$ T \\equal{} 100!^{\\frac{1}{100}}$\r\n\r\n$ T \\equal{} (100\\cdot 99!)^{\\frac{1}{99}\\cdot\\frac{99}{100}}$\r\n\r\n$ T \\equal{} 100^{\\frac{1}{100}}\\cdot (99!^{\\frac{1}{99}})^{\\frac{99}{100}}$\r\n\r\n$ T \\equal{} 100^{\\frac{1}{100}}\\cdot S^{\\frac{99}{100}}$\r\n\r\n$ T \\equal{}\\frac{100^{\\frac{1}{100}}}{S^{\\frac{1}{100}}}\\cdot S$\r\n\r\n$ T \\equal{}\\left(\\frac{100}{S}\\right)^{\\frac{1}{100}}\\cdot S$\r\n\r\nsince S is clearly less than 100, the left factor is greater than 1, so $ T > S$", "Solution_8": "Another solution for $ c)$\r\n\r\nLet $ a_{n}\\equal{}\\frac{n!}{n^{n}}$.\r\nWe have to compute $ \\lim_{n\\to\\infty}\\sqrt[n]{a_{n}}$\r\n\r\nFrom Cauchy-d'Alembert criteria we know that $ \\lim_{n\\to\\infty}\\sqrt[n]{a_{n}}\\equal{}\\lim_{n\\to\\infty}\\frac{a_{n\\plus{}1}}{a_{n}}\\equal{}\\lim_{n\\to\\infty}\\left(1\\minus{}\\frac{1}{n\\plus{}1}\\right)^{n}\\equal{}e^{\\lim\\minus{}\\frac{n}{n\\plus{}1}}\\equal{}e^{\\minus{}1}$" } { "Tag": [ "calculus" ], "Problem": "hi evryone i am new and i wich that i became more good in math", "Solution_1": "Can the moderator please add the subforum for the Linear Algebra, Complex Analysis, Calculus and Real Analysis", "Solution_2": "[quote=\"Qruni\"]Can the moderator please add the subforum for the Linear Algebra, Complex Analysis, Calculus and Real Analysis[/quote]\r\n\r\n...but they already exist, under college playground...", "Solution_3": "Can I know how to be a moderator here ??? :roll:" } { "Tag": [ "function", "inequalities", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Prove that there's no function $f: \\mathbb{R}^+\\rightarrow\\mathbb{R}^+$ such that $f(x)^2\\ge f(x+y)\\left(f(x)+y\\right)$ for all $x,y>0$.", "Solution_1": "Note that the inequality can be rewritten as $ f(x)(f(x) \\minus{} f(x\\plus{}y)) \\geq f(x\\plus{}y)f(y)$ so it is clear that $ f$ is strictly decreasing.\r\nThere are two possibilities: $ \\lim_{x\\to 0^{\\plus{}}} f(x) \\equal{} M$ where $ M$ is positive real number or $ \\lim_{x\\to 0^{\\plus{}}} f(x) \\equal{} \\plus{}\\infty$.\r\n\r\nSuppose that the first holds. Then consider the sequence $ (x_n)$ given by $ x_{n\\plus{}1}\\equal{}\\frac{x_n}{2}$ and $ x_0\\equal{}x$. Put $ x\\equal{}y\\equal{}x_{n\\plus{}1}$ in the original equation. We get\r\n$ (f(x_{n\\plus{}1}))^2 \\geq f(x_n)(f(x_{n\\plus{}1}) \\plus{} f(x_{n\\plus{}1}))$\r\nTaking the limit with $ n \\to \\infty$ we obtain that $ M^2 \\geq 2M^2$ which is a contradiction since $ M > 0$.\r\n\r\nSo let's assume that $ \\lim_{x\\to 0^{\\plus{}}} f(x) \\equal{} \\plus{}\\infty$. If $ y f(2x)f(x) \\plus{} f(2x)f(y)$, thus\r\n$ (f(x))^2 > f(2x)f(y)$\r\ntaking limit with $ y \\to 0^{\\plus{}}$ we obtain a contradiction.", "Solution_2": "TomciO, you solved $ f(x)^2\\ge f(x \\plus{} y)\\left(f(x) \\plus{} f(y)\\right)$ but the problem asked for $ f(x)^2\\ge f(x \\plus{} y)\\left(f(x) \\plus{} y\\right)$ :P .\r\n\r\nHere is a solution:\r\n\r\nSubstituting $ g(x) \\equal{} \\frac {f(x)}{x}$ we get: $ g(x)^2\\geq g(x \\plus{} y)\\left( 1 \\plus{} \\frac {y}{x}\\right) \\left( g(x) \\plus{} \\frac {y}{x} \\right) > g(x)g(x \\plus{} y)$ so $ g$ is strictly decreasing and $ l \\equal{} \\lim_{x\\to 0}g(x)$ exists.\r\nPutting $ x \\equal{} y$ we get $ g(x)^2\\geq 2g(2x)\\left( g(x) \\plus{} 1 \\right)$. For $ x\\to 0$ we get $ l^2\\geq 2l(l \\plus{} 1)$, a contradiction because $ l > 0$.", "Solution_3": "Why would g decreasing imply that the limit exists? 1/x?\r\n\r\nIn the case limit does not exist, we can follow the earlier suggestion. y > x means g(x) > g(y) so take limit as y tends to +0", "Solution_4": "Sorry, I rushed. I will try to fix it." } { "Tag": [ "limit", "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "If $f$ is $C^{2}$ on an interval prove that\r\n\r\n$\\lim_{h\\to 0}\\frac{f(x+h)-2f(x)+fx-h)}{h^{2}}=f''(x).$", "Solution_1": "By L'Hopital, we have\r\n$\\lim_{h\\to 0}\\frac{f(x+h)-2f(x)+f(x-h)}{h^{2}}$\r\n$= \\lim_{h\\to 0}\\frac{f'(x+h)-f'(x-h)}{2h}$\r\n$= \\lim_{h\\to 0}\\frac{f'(x+h)-f'(x)+f'(x)-f'(x-h)}{2h}$\r\n$= \\lim_{h\\to 0}\\frac{f'(x+h)+f'(x)}{2h}-\\lim_{h\\to 0}\\frac{f'(x-h)+f'(x)}{2h}$\r\n$= \\frac{1}{2}\\lim_{h\\to 0}\\frac{f'(x+h)+f'(x)}{h}+\\frac{1}{2}\\lim_{h\\to 0}\\frac{f'(x-h)+f'(x)}{-h}$\r\n$= \\frac{f''(x)+f''(x))}{2}= f''(x).$", "Solution_2": "yeah... why didn't i think of l'hopital... \r\n\r\nthanks for the help.", "Solution_3": "Alternatively, since $f\\in C^{2},$ we can say:\r\n\r\n$f(x+h)=f(x)+f'(x)h+\\frac12f''(x)h^{2}+o(h^{2})$\r\n\r\n$f(x-h)=f(x)-f'(x)h+\\frac12f''(x)h^{2}+o(h^{2})$\r\n\r\nAdd those together, subtract $2f(x),$ and divide by $h^{2}$ to get\r\n\r\n$\\frac{f(x+h)-2f(x)+f(x-h)}{h^{2}}=f''(x)+o(1).$\r\n\r\n[The next line is of interest to numerical analysts.]\r\n\r\nIf we assume more smoothness, say $f\\in C^{4},$ then we can include the $h^{3}$ and $h^{4}$ terms of the Taylor polynomials, the $h^{3}$ terms cancel, and the estimate becomes\r\n\r\n$\\frac{f(x+h)-2f(x)+f(x-h)}{h^{2}}=f''(x)+O(h^{2}).$\r\n\r\n----\r\n\r\nThe L'H\u00f4pital computation works, but for robustness, flexibility, and the possible devising of other similar formulas (hence my reference to numerical analysis), I would prefer to work with the Taylor polynomials, exactly as above. As an example: can you find $a,b,$ and $c$ such that\r\n\r\n$\\frac{af(x)+bf(x+h)+cf(x+2h)}{h}=f'(x)+o(h^{2})\\ ?$" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "let a1,a2,......,an n ordered positive integers \r\nprove that there exists i and j where 1<=i 0 and < 1000 that are divisible by 7: 1000 / 7 = 142 Remainder 6. Every 4 + 10i, i=0,1,2,3,... of those end in digit 8. 142 / 10 = 14 remainder 2. Perhaps that is what the previous reply said, more elegantly. I will try to learn proper notation for future replies." } { "Tag": [ "calculus", "algebra", "polynomial", "geometry solved", "geometry" ], "Problem": "Is there a (finite) formula for sin(x^2) and is it possible to construct the square of a given angle?", "Solution_1": "[quote=\"clinkmath\"]Is there a (finite) formula for sin(x^2)?[/quote]\nYes, indeed. For example: $\\sin(x^{2})=\\sqrt{1-cos(x^{2})^{2}}$ :D \n\n[quote=\"clinkmath\"]is it possible to construct the square of a given angle?[/quote]\r\n\r\nSquares often have perpendicular sides... :rotfl:", "Solution_2": "Disregarding the post above me :roll: \r\n\r\nNo, there isn't any nice synthetic geometrical method of finding squares of an angle as such, since there is no operation which multiplies angles (this would be absurd, also, due to units).\r\n\r\nOf course using calculus we can solve problems involving polynomials of angles although it is messy with Taylor polynomials.\r\n\r\nFurthermore, there might be interesting non standard geometries which involve angle multiplication (maybe in a distance formula or something)" } { "Tag": [ "AMC", "AIME", "geometry" ], "Problem": "So, any of you guys going to state ciphering at the Univ of AL at Tuscaloosa?", "Solution_1": "what is that and when", "Solution_2": "its the Alabama statewide ciphering event. Your team had to go to the alabama state math test, and the comprehensive team had to place top 16 in each division. (so if you went, you pretty much qualify for ciphering). Its this weekend on saturday at tuscaloosa (univ of al).", "Solution_3": "yay it does not coincide with my piano competition. Thus I may go. I hope the questions are aime level as usual (as in i always hope the questions are hard), but that is a ridiculous hope.", "Solution_4": "But it does coincide with State science olympiad... oh well I suck at math anyway", "Solution_5": "Psh, xinke, dont get your hopes up too high.... :wink: \r\n\r\nState ciphering never has been difficult in the past, at least since I went to it in 8th grade. The questions are hardly AMC 12a level, nevertheless AIME. its funny, 99% of the people who go to ciphering cant make higher than a 120 on the AMC 12. \r\n\r\nPlus, only 4 ppl get to cipher, so taking an entire team is useless. (but there is a team ciphering, so who knows :P )", "Solution_6": "u could go in 8th grade?!?!?", "Solution_7": "no, it starts in 9th grade, but our high school wanted to take me b/c i was also taking geometry courses in high school, while in middle school. And, i was on the math team and was pretty good at it at the time :P so i went\r\n :ninja:\r\n\r\n\r\nso, no YOU cant go in 8th grade, but if youre straight up baller like i am, then you can go :wink: (jk) :rotfl:" } { "Tag": [], "Problem": "consider the sequence $ S_1 $, $ S_2 $, ...$ S_r $, ... where each $ S_i $ is itself a sequence of the natural numbers 1, 2, 3, 4, ...\r\n\r\nlet the $ n $th sequence $ S_n $ be denoted by $ u_1 $, $ u_2 $, $ u_3 $, ... and $ S_{n+1} $ = $ u'_1 $, $ u'_2 $, $ u'_3 $, ...\r\n\r\nFor each $ n $, the terms of the sequence $ S_{n+1} $ are derived from those of the sequence $ S_n $ as follows:\r\n\r\n$ u'_i $ = $ u_i $ + 1 if $ n $ is a divisor of $ u_i $\r\n $ u_i $, if otherwise.\r\n\r\nDetermine those positive integers $ k $ for which the first $ k-1 $ terms of $ S_k $ are all equal to the number $ k $, but for which the $ k $th term is not.", "Solution_1": "Source: AMOC extension programme 4 Set 1 problem 6\r\n\r\ntest the first 10 cases (i know.. but it isn't too hard) and it should be really very apparent. \r\n\r\nThe proof itself was quite annoying... proving the k-1th term isn't too much effort.. trying to prove the rest.. well..." } { "Tag": [ "geometry", "3D geometry", "analytic geometry", "geometry unsolved" ], "Problem": "$ABCDA_{1}B_{1}C_{1}D_{1}$ is a cube with side length $4a$. Points $E$ and $F$ are taken on $(AA_{1})$ and $(BB_{1})$ such that $AE=B_{1}F=a$. $G$ and $H$ are midpoints of $(A_{1}B_{1})$ and $(C_{1}D_{1})$, respectively.\r\nFind the minimum value of the $CP+PQ$, where $P\\in[GH]$ and $Q\\in[EF]$.", "Solution_1": "I really hate when I see a post with you do that and that and rest\r\nis easy but in fact, the rest is what makes the problem hard. This\r\nis such post, but really, the rest of it is as not so important as\r\nthe first part. Maybe I had made a mistake in calculation somewhere,\r\nbut problem can be solved like this for sure.\r\n\r\nPut the origin of Cartesian system in $A$ such that\r\n$A(0,0,0),B(4,0,0),C(4,4,0)...$ (we put $a=1$). Then it's not hard\r\nto se that $P(2,y,4)$ and $Q(x,0,\\frac{x+2}2)$ where $x,y\\in [0,4]$.\r\nSo $CP+PQ=\\sqrt{20+(y-4)^{2}}+\\sqrt{(x-2)^{2}+(\\frac{x+2}2-4)^{2}+y^{2}}$. We want to calculate the minimum of this expression when\r\n$x,y\\in [0,4]$. The minimum of $(x-2)^{2}+(\\frac{x+2}2-4)^{2}$ can\r\neasily be calculated, and if I am not wrong it is $\\frac65$ and it\r\nis obtained for $x=\\frac{16}5\\in [0,4]$ (you see that only this part\r\ndepends on $x$, so we can calculate its minimum and use it in\r\nfurther solving). Now we should calculate minimum of $\\sqrt{(y-4)^{2}+20}+\\sqrt{y^{2}+\\frac{6}5}$. If you put $M(4,\\sqrt{20})$ and\r\n$N(0,\\sqrt{\\frac{6}{5}})$, then $\\sqrt{(y-4)^{2}+20}+\\sqrt{y^{2}+\\frac{6}5}$ is $YM+YN$ where $Y(y,0)$. This is well known problem,\r\nand solution is (again if I am not wrong, i really calculated all\r\nthis in maximum speed) is $y=\\frac{4\\sqrt{6}}{10+4\\sqrt{5}}$. This\r\nis all for $a=1$, but only multiply everything whit $a$, and that's\r\nit.", "Solution_2": "I didn't even think about solving this with coordinates! I guess you made some mistaken calculations... The answer is $\\sqrt{\\frac{276}{5}}a$. I took it \"easier\". I expressed $CP+PQ$ by the right triangles $CHP$ and $QGP$. Setting $HP=x$, we get $CP+PQ=\\sqrt{CH^{2}+x^{2}}+\\sqrt{QG^{2}+(4a-x)^{2}}$ and we use Minkovski. So, to attain the minimum, $QG$ must be firstly minimal, that is $Q$ is the projection of $G$ on $EF$.\r\nThen we apply Minkovski to \"get rid\" of $x$. So minimum is $\\sqrt{(CH+QG)^{2}+(4a)^{2}}$. Well, I don't make all the calculations. It is easy to get $CH$ from right triangle $CHP$, and $QG$ as being altitude in $EGF$.", "Solution_3": "Well, I like coordinate geometry, and this is the first time I use it on 3D that's why I was so exciting, and didn't look for other \"easier\" solutions :).\r\nI usually use it in 2D like in http://www.mathlinks.ro/Forum/viewtopic.php?t=136680v \r\nBye", "Solution_4": "[quote=freemind]I didn't even think about solving this with coordinates! I guess you made some mistaken calculations... The answer is $\\sqrt{\\frac{276}{5}}a$. I took it \"easier\". I expressed $CP+PQ$ by the right triangles $CHP$ and $QGP$. Setting $HP=x$, we get $CP+PQ=\\sqrt{CH^{2}+x^{2}}+\\sqrt{QG^{2}+(4a-x)^{2}}$ and we use Minkovski. So, to attain the minimum, $QG$ must be firstly minimal, that is $Q$ is the projection of $G$ on $EF$.\nThen we apply Minkovski to \"get rid\" of $x$. So minimum is $\\sqrt{(CH+QG)^{2}+(4a)^{2}}$. Well, I don't make all the calculations. It is easy to get $CH$ from right triangle $CHP$, and $QG$ as being altitude in $EGF$.[/quote]\n\nI think apply Minkowski is wrong because your equation never attain $\\sqrt{\\frac{276}{5}}a$. I think to solve your equation need derivative equalize to zero and $x$ will be nearly $2.38$ and answer is nearly $7.47938$ " } { "Tag": [ "calculus", "integration", "logarithms", "ratio", "AMC", "USA(J)MO", "USAMO" ], "Problem": "The increasing geometric sequence $x_{0},x_{1},x_{2},\\ldots$ consists entirely of integral powers of $3.$ Given that \\[\\sum_{n=0}^{7}\\log_{3}(x_{n}) = 308\\qquad\\text{and}\\qquad 56 \\leq \\log_{3}\\left ( \\sum_{n=0}^{7}x_{n}\\right ) \\leq 57,\\] find $\\log_{3}(x_{14}).$", "Solution_1": "i need help with this one too. this is what i had so far...\r\n\r\n[hide]\nif you express the sequence in the form $3^{a}, 3^{b}, 3^{c},\\ldots$, then you know a, b, c, ... are in an arithmetic sequence.\n\nthe first part basically tells you that the sum of the first eight powers ($a+b+c+...+h$) is 308, right?\n\nthe second part tells you that the product of those eight powers is between 56 and 57 inclusive i think. \n\nthat's where i got stuck. i don't see how that's possible.[/hide]", "Solution_2": "[hide=\"I can not believe I missed this one too\"]\nSince the series $x_{0},x_{1},\\ldots$ is geometric and consists of only powers of 3, let the series be $3^{a}, 3^{a+k},3^{a+2k},\\ldots$.\n\nThe first sum equation tells us that $a+(a+k)+...(a+7k)=8a+28k=308$ because $x_{7}$ is the [b]eighth[/b] term, not the seventh :wallbash: \n\nThe second sum tells us that these numbers have a sum betweemn $3^{56}$ and $3^{57}$. Let's say $a+7k<56$. if this is true, there is no way to possibly satisfy the second sum, and if $a+7k>57$, there is no way to satisfy the second sum. So $a+7k=56$.\n\nWe now have\n$8a+28k=308$ and $a+7k=56$. Multiplying the second equation by 8 and subtracting the first yields $28k=140\\rightarrow{k}=5$\n\nOur target is $a+14k=56+7(5)=\\boxed{091}$\n[/hide]\r\nBut leave it to me to miss the $x_{0}$ part and probably miss the floor because of it.", "Solution_3": "i liked this problem!", "Solution_4": "[quote=\"shake9991\"]i liked this problem![/quote]\r\n\r\nSame here :)", "Solution_5": "But it definetely was to easy for a number 12.", "Solution_6": "I would agree.... the first equation told you the sum of an arithmetic series and the second one told you one of the terms... makes it into a simple algebra problem", "Solution_7": "[quote=\"sapphyre571\"][hide=\"I can not believe I missed this one too\"]\nSince the series $x_{0},x_{1},\\ldots$ is geometric and consists of only powers of 3, let the series be $3^{a}, 3^{a+k},3^{a+2k},\\ldots$.\n\nThe first sum equation tells us that $a+(a+k)+...(a+7k)=8a+28k=308$ because $x_{7}$ is the [b]eighth[/b] term, not the seventh :wallbash: \n\nThe second sum tells us that these numbers have a sum betweemn $3^{56}$ and $3^{57}$. Let's say $a+7k<56$. if this is true, there is no way to possibly satisfy the second sum, and if $a+7k>57$, there is no way to satisfy the second sum. So $a+7k=56$.\n\nWe now have\n$8a+28k=308$ and $a+7k=56$. Multiplying the second equation by 8 and subtracting the first yields $28k=140\\rightarrow{k}=5$\n\nOur target is $a+14k=56+7(5)=\\boxed{091}$\n[/hide]\nBut leave it to me to miss the $x_{0}$ part and probably miss the floor because of it.[/quote]\r\n\r\nwow. I actually did miss the floor because of it.", "Solution_8": "[quote=\"sapphyre571\"]\nThe second sum tells us that these numbers have a sum betweemn $ 3^{56}$ and $ 3^{57}$. Let's say $ a \\plus{} 7k < 56$. if this is true, there is no way to possibly satisfy the second sum, and if $ a \\plus{} 7k > 57$, there is no way to satisfy the second sum. So $ a \\plus{} 7k \\equal{} 56$..[/quote]\r\n\r\nWhat is the \"second sum\" you are referring to? Can someone elaborate this part of the solution? It seems to be the key to solving this problem.", "Solution_9": "Tachyon:\r\n$ 56\\leq\\log_{3} (\\sum_{n=0}^{7}x_{n} )\\leq 57$\r\nExponentiating with a base of 3 yields:\r\n$ 3^{56}\\leq \\sum_{n=0}^{7}x_n \\leq 3^{57}$\r\nTry it now.", "Solution_10": "I thought of that specific part this way:\r\n[hide]\nAs in sapphyre's post, let $ x_0 \\equal{} 3^a$, and the common ratio be $ k$.\n$ \\log_{3}\\left (\\sum_{n \\equal{} 0}^{7}x_{n}\\right ) \\equal{} \\log_{3}(3^a \\plus{} 3^{a \\plus{} k} \\plus{} ... \\plus{} 3^{a \\plus{} 7k})$\n\n$ \\log_{3}(3^a \\plus{} 3^{a \\plus{} k} \\plus{} ... \\plus{} 3^{a \\plus{} 7k}) \\equal{} a \\plus{} 7k \\plus{} \\log_{3}(3^{ \\minus{} 7k} \\plus{} 3^{ \\minus{} 6k} \\plus{} ... \\plus{} 3^{ \\minus{} k} \\plus{} 1)$\n\nSince $ k$ has to be positive, $ 3^{ \\minus{} 7k} \\plus{} 3^{ \\minus{} 6k} \\plus{} ... \\plus{} 3^{ \\minus{} k} < \\frac {1}{2}$. Therefore,\n\n$ 0 < \\log_{3}(3^{ \\minus{} 7k} \\plus{} 3^{ \\minus{} 6k} \\plus{} ... \\plus{} 3^{ \\minus{} k} \\plus{} 1) < 1$,\nso the only integral value of $ a \\plus{} 7k$ that will place the sum between the right values is $ 56$. The solution follows from there.\n[/hide]\r\n\r\nI wish the average #12 was this easy. That would make USAMO placement a reasonable goal for me...", "Solution_11": "[quote=\"pkothari13\"]The increasing geometric sequence $ x_{0},x_{1},x_{2},\\ldots$ consists entirely of integral powers of $ 3.$ Given that\n\n$ \\sum_{n \\equal{} 0}^{7}\\log_{3}(x_{n}) \\equal{} 308$ and $ 56 \\leq \\log_{3}\\left ( \\sum_{n \\equal{} 0}^{7}x_{n}\\right ) \\leq 57,$\n\nfind $ \\log_{3}(x_{14}).$[/quote]\r\n\r\nDenote $ x_0, x_1, x_2 \\dots x^7$ as $ 3^p, 3^{p \\plus{} q} \\dots$\r\n\r\nThen, $ p \\plus{} (p \\plus{} q) \\plus{} \\dots \\plus{} (p \\plus{} 7q) \\equal{} 308 \\implies 8p \\plus{} 28q \\equal{} 308 \\text{ \\minus{} Equation I}$.\r\n\r\nAlso, $ 3^{56} \\leq \\sum_{n \\equal{} 0}^{7}x_{n} \\leq 3^{57}.$ (This is obtained from raising 3 to the power of both sides of the inequality.)\r\n\r\nThis means, $ 3^{56} \\leq 3^p(1 \\plus{} 3^q \\plus{} 3^{2q} \\plus{} \\dots \\plus{} 3^{7q}) \\leq 3^{57} \\text{ \\minus{} Equation II.}$\r\n\r\nThe problem tells us that the powers of 3 are integral. We will consider the last power, namely $ 3^{a \\plus{} 7k}.$ \r\n\r\nBy $ \\text{Equation II}$, we find that the last power cannot be 57 or greater nor less than 56. Therefore, the only integral power of $ p \\plus{} 7q$ that is remotely possible is $ 3^{p \\plus{} 7q} \\equal{} 3^{56}.$ \r\n\r\nTaking $ \\log_{3}$ of both sides, we find that $ p \\plus{} 7q \\equal{} 56.$\r\n\r\nNow we have:\r\n$ 8p \\plus{} 28q \\equal{} 308$ along with $ p \\plus{} 7q \\equal{} 56.$ Solving for $ q$ and $ p$, we get that $ p \\equal{} 21$ and that $ q \\equal{} 5.$\r\n\r\nTherefore, $ a_{14}$, namely $ p \\plus{} 14q \\equal{} 21 \\plus{} 14(5) \\equal{} \\boxed{091.}$", "Solution_12": "[quote=\"rpond\"]\n\n$ \\log_{3}(3^a \\plus{} 3^{a \\plus{} k} \\plus{} ... \\plus{} 3^{a \\plus{} 7k}) \\equal{} a \\plus{} 7k \\plus{} \\log_{3}(3^{ \\minus{} 7k} \\plus{} 3^{ \\minus{} 6k} \\plus{} ... \\plus{} 3^{ \\minus{} k} \\plus{} 1)$\n\n[/quote]\n\n\nNice solution, however I'm confused on how you got the above equation. Thanks :).", "Solution_13": "Just factor: \\[ 3^a + 3^{a+k} + \\cdots + 3^{a+7k} = 3^{a+7k}(3^{-7k} + 3^{-6k} + \\cdots + 1), \\] then take the base-3 logarithms of both sides.", "Solution_14": "@Maxima, I liked your solution but there\u2019s just one question I still have- why can\u2019t p+7q be less than or greater than 56? I get that it can\u2019t be equal to 57 because then that would not satisfy the RHS of the inequality, but why can\u2019t it be greater than or less than 56? If the last term had a power of 3 that was less than 56, there\u2019s still 6 more terms to go! So maybe, just maybe, their sum might still be bigger than or equal to 3^56? How do we know this won\u2019t happen? Also, I don\u2019t understand why p+7q can\u2019t be more than 56\u2026how can we be sure that there won\u2019t be any geometric sequence that are powers of 3 that satisfy all the given constraints but has a power of 3 that\u2019s greater than 56? ", "Solution_15": "The first equation gives us $x_0x_1 \\dots x_7 = 3^{308}$. Since all $x_0, x_1, \\dots$ are power of three we get $x_0 = 3^r$. So this means $3^{8r + 28s} = 3^{308}$ which means $2r + 7s = 77$. Since $r, s$ are integers we know that $2r = 7(11 - s)$ which implies $s$ is odd. The second equation gives $3^{56} \\le x_1 + x_2+ \\dots + x_7 \\le 3^{57}$. Replacing with $r ,s$ we know that $3^{56} \\le 3^{r} + 3^{r+s} + \\dots + 3^{r + 7s} \\le 3^{57}$. Using the geometric series formula we get $3^{56} \\le \\frac{3^r(3^{8s} - 1)}{3^s - 1 } \\le 3^{57}.$ Now we can just casework on $s$. \n\nCase 1: $s = 1$\nThis means $r = 35$. So we have $3^{21} \\le \\frac{3^8 - 1}{3-1} \\le 3^{22}$. Clearly this isn't true. \n\nCase 2: $s = 3$\nThis means $r = 28$. So we have $3^{28} \\le \\frac{3^{24} - 1}{3^3 -1} \\le 3^{29}$. Clearly this can't be true. \n\nCase 3: $s = 5$\nThis means $r = 21$. So we have $3^{35} \\le \\frac{3^{40} - 1}{3^5 - 1} \\le 3^{36}$. Clearly $\\frac{3^{40} - 1}{3^5 - 1} \\le 3^{36}$. Multiplying by $3^5 - 1$ we get $3^{40} - 3^{35} \\le 3^{40} - 1$ and this is clearly true. Thus our answer is $21 + 14(5) = 091$. " } { "Tag": [ "search", "combinatorics proposed", "combinatorics" ], "Problem": "Several knights participated in tournament .Knights fought with each other.We will call $ A$ a king,if for every knight $ B$, either $ A$ win $ B$,or there exist $ C$,such that $ A$ won $ C$ and $ C$ won $ B$.Prove that \r\n1.There exist a king \r\n2.If each knight has lost at least one battle,then there exist at least three kings.", "Solution_1": "read the hidden section in my post here:\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1779208502&t=159625[/url]\r\n\r\nfor the first part.\r\n\r\nand for the second part use the following lemma:\r\n\r\n[b]let $ T$ be a tournament having no vertex with indegree $ 0$.prove that if $ x$ is a king in $ T$,then $ T$ has another king in $ N^\\minus{} (x)$[/b]\r\n\r\n(here $ N^\\minus{} (x)$ means the in-neighborhood of vertex $ x$,i.e. the set $ \\{y\\in V(G): y\\to x\\}$)", "Solution_2": "[quote=\"BaBaK Ghalebi\"]read the hidden section in my post here:\n\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1779208502&t=159625[/url]\n\nfor the first part.\n\nand for the second part use the following lemma:\n\n[b]let $ T$ be a tournament having no vertex with indegree $ 0$.prove that if $ x$ is a king in $ T$,then $ T$ has another king in $ N^ \\minus{} (x)$[/b]\n\n(here $ N^ \\minus{} (x)$ means the in-neighborhood of vertex $ x$,i.e. the set $ \\{y\\in V(G): y\\to x\\}$)[/quote]\r\nOk,i agree that the first part is very easy,it is just enough to consider the vertex with maximum degree and the second part is very simple too...but please post a complete proof,for those who didn't solve this problem.", "Solution_3": "ok,the proof of the first part is given in the link above,for the second part,first of all we will prove the lemma I mentioned in my last post:\r\n\r\n[b]lemma.[/b]let $ T$ be a tournament having no vertex with indegree $ 0$.now let $ x$ be a king in $ T$,then $ T$ has another king in $ N^ \\minus{} (x)$.\r\n\r\nthe proof of this lemma can be done in a similar way as part 1.because note that the vertices of $ T$ can be partitioned into three subsets $ \\{x\\},N^ \\plus{} (x),N^ \\minus{} (x)$.\r\nnow note that the set $ N^ \\minus{} (x)$ is a tournament itself,so it has a king,let it be $ v$.\r\nnow we claim that $ v$ is also a king in $ T$.\r\nnote that $ v\\in N^ \\minus{} (x)$ hence there is an edge in $ T$ from $ v$ to $ x$.\r\nand also for every $ y\\in N^ \\plus{} (x)$ consider the path $ vxy$ which is a path with length $ 2$ from $ v$ to $ y$.so every vertex in $ N^ \\plus{} (x)$ can be reached from $ v$ by a path with length at most $ 2$,also $ v$ was a king in $ N^ \\minus{} (x)$,so every vertex in $ N^ \\minus{} (x)$ can be reached from $ v$ by a path with length at most 2.\r\nhence $ v$ is a king in $ T$,as wanted.\r\n\r\nnow use this lemma another time for $ v$.so there exists a vertex $ u$ in $ N^ \\minus{} (v)$ which is a king in $ T$.now note that we had $ v\\in N^ \\minus{} (x)$,hence $ u\\neq x$,thus there are at least three different kings (x,v,u) in $ T$.", "Solution_4": "All parts have appeared before and are linked to from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162646" } { "Tag": [ "logarithms", "inequalities", "inequalities unsolved" ], "Problem": "Show that for any positive reals $ x,y,z,t$ is true that : $ (x^xy^yz^zt^t)^4 \\ge (xyzt)^{(x\\plus{}y\\plus{}z\\plus{}t)}$.", "Solution_1": "[quote=\"Adriana N.\"]Show that for any positive reals $ x,y,z,t$ is true that : $ (x^xy^yz^zt^t)^4 \\ge (xyzt)^{(x \\plus{} y \\plus{} z \\plus{} t)}$.[/quote]\r\n$ \\Leftrightarrow x\\ln x\\plus{}y\\ln y \\plus{} z \\ln z \\plus{}t\\ln t \\ge \\frac{1}{4}(x\\plus{}y\\plus{}z\\plus{}t)(\\ln x \\plus{} \\ln y \\plus{} \\ln z \\plus{}\\ln t )$ - true by Chebyshev", "Solution_2": "nice...Why I dindn't think it?! :maybe: :maybe:" } { "Tag": [ "number theory", "least common multiple", "prime factorization" ], "Problem": "What is the least common multiple of the numbers 1332 and\n888?", "Solution_1": "Factoring the numbers, we get $ 6\\times222$ and $ 4\\times222$. Now we just need to find $ LCM(4,6)$ which is $ 12$. The answer is thus $ 12\\times222\\equal{}\\boxed{2664}$", "Solution_2": "ill answer this in a series of steps\n\n[hide=\"ANSWER\"]1.our best chance of success is to first is to factor the #'s 1332 and 888\n\n2.when factoring these #'s we get :$ 6\\times222 $ and $ 4\\times222 $\n\n3.now that we have simplified our question we can just now find the LCM (6,4) \n\n4.doing so, we get an answer of 12\n\n5.and back to the prime factorization we know the #'s 222 repeat themselves so we multiply 222*12=2664\n\n6.[hide]2664[/hide][/hide]\n\n[hide=\".\"][size=200][color=#BF0000]PLEASE RATE MY POST THANK YOU SO MUCH!![/color][/size][/hide]\n\n[size=50][b][i][u]please be fair when you're rating my post and please PM me if there is a better way to explain this thank you so much!![/u][/i][/b][/size]" } { "Tag": [ "geometry", "rectangle", "geometric transformation", "rotation", "function", "analytic geometry", "combinatorics proposed" ], "Problem": "Give a one-line argument showing that it is impossible to dissect a $1 \\times 2$ rectangle into finitely many smaller rectangles and reassemble the pieces (rotations are allowed) to form a $\\sqrt{2} \\times \\sqrt{2}$ rectangle. \r\n\r\n ;) \r\n\r\n--Vesselin", "Solution_1": "Let $f: \\mathbb R\\to\\mathbb R$ be an additive function which vanishes precisely on $\\mathbb Q$, and to each rectangle $\\mathcal R$ with axes parallel to the coordinate axes and sides $x,y$ assign $T(\\mathcal R)=f(x)f(y)$. Clearly, $T$ is additive, and it vanishes on a $1\\times 2$ rectangle, but not on a $\\sqrt 2\\times\\sqrt 2$ one.\r\n\r\nThese are $3$ lines, but now we're just splitting hairs :)." } { "Tag": [ "geometry", "function", "trigonometry", "inequalities", "circumcircle", "cyclic quadrilateral", "geometry unsolved" ], "Problem": "$A,B,C,D,E,F$ are on the circle $O$ counterclockwise. $BD$ is a diameter of O and it lies at right angles to $CF$. Also, $CF, BE, AD$ are concurrent. The $M$ is a foot of perpendicular of $B$ on $AC$, and $N$ is a foot of perpendicular of $D$ on $CE$. Prove that\r\n\\[S[MNC]\\leq\\frac{S[ACEF]}{2}\\]\r\n(Function $S$ is the area function.)", "Solution_1": "Quadrilateral ACEF is cyclic, hence $\\angle F = 180^\\circ-\\angle C,$ $\\sin F = \\sin C.$ The quadrilateral area is\r\n\r\n$S(ACEF) = \\frac{1}{2}\\ CA \\cdot CE\\ \\sin C+\\frac{1}{2}\\ FE \\cdot FA\\ \\sin F =$\r\n\r\n$= \\frac{1}{2}\\ (CA \\cdot CE+FE \\cdot FA) \\sin C$\r\n\r\nOn the other hand, area of the triangle $\\triangle MCN$ is\r\n\r\n$S(MCN) = \\frac{1}{2}\\ CM \\cdot CN\\ \\sin C$\r\n\r\nThe proposed inequality $S(MCN) < \\frac{1}{2}\\ S(ACEF)$ is actually sharp. Thus we have to show that\r\n\r\n$(?)\\ \\ \\ 2\\ CM \\cdot CN < CA \\cdot CE+FE \\cdot FA$\r\n\r\nSince $CF \\perp BD,$ B, D are midpoints of the circumcircle arcs CF containing A, E, respectively. By the problem conditions, C lies on the opposite side of the circumcircle diameter BD than A, E, F, hence $CA > CB = FB > FA$ and $CE > CD = FD > FE.$ Since in addition $BM \\perp CA,\\ DN \\perp CE,$ by Archimedes theorem (see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=115281]External bisector[/url]), M cuts in half the polygonal path CAF, while N cuts in half the polygonal path CEF. Thus $2\\ CM = CA+FA,$ $2\\ CN = CE+FE.$ Substituting this to the equation in question,\r\n\r\n$(?)\\ \\ \\ (CA+FA) \\cdot (CE+FE) < 2\\ (CA \\cdot CE+FE \\cdot FA)$\r\n\r\nAs a result, we no longer have to consider the triangle $\\triangle MCN.$ Multiplying the left side out and canceling equal terms on both sides, this becomes\r\n\r\n$(?)\\ \\ \\ CA \\cdot FE+CE \\cdot FA < CA \\cdot CE+FE \\cdot FA$\r\n\r\nRearranging and factoring,\r\n\r\n$(?)\\ \\ \\ (CA-FA)(CE-FE) > 0$\r\n\r\nSince $CA > FA,\\ CE > FE$ (see above), the last inequality trivially holds.\r\n\r\n\r\n[b]Comment:[/b] The concurrence of CF, BE, DA is not necessary and it was not used. Since B, D are midpoints of the circumcircle arcs CF containing A, E, respectively, AD bisects $\\angle A,$ while EB bisects $\\angle E$ of the cyclic quadrilateral ACEF. Assuming CF, BE, DA concur at P, we have $\\frac{CA}{FA}= \\frac{CP}{FP}= \\frac{CE}{FE}$ and it follows that $MN \\parallel AE.$ But again, it is not necessary and it was not used." } { "Tag": [ "vector", "geometry", "parallelogram", "analytic geometry", "graphing lines", "slope" ], "Problem": "This is problem 158 chapter 10 vol 2 aops. \r\nThree vertices of parallelogram PQRS are p(-3,-2), q(1,-5), r(9,1) with P and Q diagonally opposite. What is the S coordinate?\r\n\r\nFirst, P and Q being diagonally opposite, does that mean they are opposite angles of the parallelogram?\r\n\r\nIf my above question is yes, then i dont see how the book solution (i think it said (5,4), but i dont have the solution guide with me.) is wrong i think. \r\n\r\nThis is what i did:\r\nI set it up as two circles, one \r\n(x+3)^2 + (y+2)^2 = 100.\r\nThe other is (x-1)^2 + (y+5)^2 = 153. The first one is centered at P and the second at Q and this would determine the intersection line of the two.\r\n\r\nSolving, I get -4x + 3y = 20. Now I need to find another equation of a line to find where point S is. I know that from point Q, there is going to be a 1/4 slope till point S because the slope will be equal to the slope of R to P. \r\ny = x/4 + b. Plugging Q in, -5 = 1/4 + b, and b = -21/4.\r\nso y = x/4 -21/4, and -x + 4y = -21. \r\nSolving both equations found, i get the intersection at (-11,-8). First, I see no reason at all for this answer to be positive like the book says. And second, I see no reason why I would be wrong. \r\nHelp is appreciated, and I thank you in advance.", "Solution_1": "Gah - you caught us on a typo :oops: . Kind of surprised no one has mentioned before. Problem (and first line of solution) should have had P and R diagonally opposite.\r\n\r\n(Yet another reason we're glad we started this forum... Can clear up typos...)\r\n\r\nGo back through the problem (as stated in the book) and do it using the vector approach (as described in the solutions, but for the case of P and R being diagonally opposite). It's much faster. You'll know when you've done it right." } { "Tag": [ "calculus" ], "Problem": "Find the Difference Quotient of f, that is, find\r\nf(x + h) - f(x)/h, where h does not = 0, for each function. Be sure to simplify.\r\n\r\nNOTE: This is a precalculus question. Please, do not answer the question using calculus concepts.\r\n\r\n(1) f(x) = -3x + 1\r\n\r\n(2) f(x) = 1/(x + 3)", "Solution_1": "Post some progress or an attempt first, then we'll help you. We can help with understanding concepts related to homework, but we won't do it." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "trigonometry", "calculus", "geometry", "linear algebra" ], "Problem": "What type of calculator restrictions are there for the USAMO and the IMO??? I want to start looking more seriously into these tests and I want to know what instruments I can use.", "Solution_1": "wait, are you even allowed calculators on the USAMO?\r\ni thought you weren't...", "Solution_2": "that would be an answer :\\, i didn't state there was.. for all I know, you aren't allowed to use one, this is what I am trying to figure out.", "Solution_3": "No calculators.\r\n\r\n[quote]No notes, books, headphones, slide rules, mathematical tables, calculators or calculator watches are allowed during the exam. [/quote]\r\n\r\n(from http://www.unl.edu/amc/d-publication/d-1-pubarchive/2002-3pub/03usamo/html/04-03usamo-text.html )", "Solution_4": "You can not even use calculators on the AIME.(Although you probably already knew that)", "Solution_5": "yup, that's wat i thought...", "Solution_6": "No calculators are allowed for the IMO ... :D", "Solution_7": "I think the only famous competitions where calculators allowed are AMC and part of MC...", "Solution_8": "Can you use a calculator in the AMC ? Didn't know that :?", "Solution_9": "Yes you can. You can use any calculator permitted in SAT, so you could use TI-83 se, which is pretty useful when it comes to logs and trigs and such.\r\nI'm not sure about 89 though.", "Solution_10": "Do some of you know the TI 200 ?\r\nI won a TI 200 with our national olympiad and it's really an incredible thing :D", "Solution_11": "200!?!?!?Three d-d-digits :!: :?: \r\n\r\n... :)", "Solution_12": "Is it a Voyager 200?\r\n\r\nYou can use the 89 on the AMC", "Solution_13": "I hadn't heard of the TI-200 until you mentioned it, Arne. Would you mind telling us about it?", "Solution_14": "It's a \"voyager\" indeed :D There are lots of interesting programs on it such as mathcad, derive, cabri, geometer's sketchpad, ... It has a special calculus & linear algebra program. ... and lots of memory !", "Solution_15": "[color=cyan]So it's really a computer, not a calculator? QWERTY keyboard and such?[/color]", "Solution_16": "How much does it cost?", "Solution_17": "[url=http://education.ti.com/us/product/tech/89/features/features.html]Explanation of Voyage 200[/url]\r\n[url=http://epsstore.ti.com/webs/catlist.asp?deptid=609&catid=902]$200@TI site[/url]\r\n\r\n[url=http://www.artofproblemsolving.com/wforum/viewtopic.php?p=14418#14418]My attempt at making a post entirely linked :) (Even the emoticon's linked)[/url]", "Solution_18": "[quote=\"Tare\"]Yes you can. You can use any calculator permitted in SAT, so you could use TI-83 se, which is pretty useful when it comes to logs and trigs and such.\nI'm not sure about 89 though.[/quote]\r\n\r\nFor AMC,\r\n\r\n89 = yes\r\nVoyager/200 = no\r\n\r\nIf I remember correctly (yeah, I sound like a really old, old person...), the Voyager isn't allowed because it has a QWERTY keypad. This is how they distinguish between \"calculator\" and \"computer\" - that is, you can't use a laptop, either.\r\n\r\nBut the '89 is allowed. I believe the AMC allows the same calculators as the SAT in order to be consistent.", "Solution_19": "Voyage 200 does have a lot of memory for programs, but the bad thing is... It's probably not allowed in most competitions. They won't let you use anything with a qwerty keyboard. Voyage 200 has that. Most of the apps that runs on the voyage 200 will run on TI-89. So I suggest getting a TI-89. It's very helpful. I also heard some HP graphing calculators is even more powerful then a TI-89.\r\n\r\nReally? Calculators aren't allowed in IMO? I can see how calculators are problems. For example if they want you to calculate the digits in for example 299!; you can enter it on your TI-89 and get the exact digits.", "Solution_20": "Well IMO's are like proofs (and correct me if I'm wrong) and hard geometry problems so calculation might be not required at all/required very little.\r\nAnyway if a person is good enough to go to the IMO I'll bet that anyone of them could do long division :)", "Solution_21": "That isn't necessarily true. There are myriads of anecdotes about the greatest mathematicians in history who were unable to do basic arithmetic. If you're proving difficult theorems, you might not need to know basic arithmetic." } { "Tag": [ "modular arithmetic", "induction", "number theory" ], "Problem": "Okay I just learned this cool modular arithmetic trick from ernie on the post titled remainder, and I want to see if I'm doing this correctly. If you know how to solve the problem, then please tell me if I did it right. \r\n\r\nProblem; \r\nWhat is the remainder when [125^(625)]/6\r\n\r\n\r\nMy Solution; \r\n\r\n125^625 (mod 6) = [125^(104 + 1)] (mod 6) \r\n125^105 (mod 6) = [125^(17 + 3)] (mod 6) \r\n125^20 (mod 6) = [125^(3 + 2)](mod 6) \r\n125^5 (mod 6) = 5^15 (mod 6) = [5^(2 + 3)] (mod 6)\r\n5^5 (mod 6) = 3125 (mod 6) \r\n5 (mod 6)\r\n\r\nSo the remainder is 5.", "Solution_1": "That's correct.\r\n[hide=\"A faster way\"]$ 125^{625} \\pmod 6 \\equiv (\\minus{}1)^{625}\\pmod 6 \\equiv \\minus{}1\\pmod 6 \\equiv \\boxed{5}\\pmod 6$.[/hide]", "Solution_2": "Oh, that's a really cool solution.", "Solution_3": "[hide=\"This also works, I think?\"]\n\\[ 625\\equiv{1}\\pmod{6}\\\\\n125\\equiv{5}\\pmod{6}\\\\\n5^1\\equiv{\\fbox{5}}\\pmod{6}\\\\\n\\]\nBut I'm new to modular arithmetic so I'm not sure.\nAnyway, RisingFlame's solution is probably the easiest one.\n[/hide]", "Solution_4": "[quote=\"grn_trtle\"][hide=\"This also works, I think?\"]\n\\[ 625\\equiv{1}\\pmod{6} \\\\\n125\\equiv{5}\\pmod{6} \\\\\n5^1\\equiv{\\fbox{5}}\\pmod{6} \\\\\n\\]\nBut I'm new to modular arithmetic so I'm not sure.\nAnyway, RisingFlame's solution is probably the easiest one.\n[/hide][/quote]\r\nNo, that doesn't work. Sorry.\r\nBasically, you can't take the mod off of an exponent...\r\nThere are ways to do this, but it uses Fermat's little theorem.", "Solution_5": "[quote=\"grn_trtle\"][hide=\"This also works, I think?\"]\n\\[ 625\\equiv{1}\\pmod{6} \\\\\n125\\equiv{5}\\pmod{6} \\\\\n5^1\\equiv{\\fbox{5}}\\pmod{6} \\\\\n\\]\nBut I'm new to modular arithmetic so I'm not sure.\nAnyway, RisingFlame's solution is probably the easiest one.\n[/hide][/quote]\r\nit's not right.", "Solution_6": "[quote=\"Brut3Forc3\"][quote=\"grn_trtle\"][hide=\"This also works, I think?\"]\n\\[ 625\\equiv{1}\\pmod{6} \\\\\n125\\equiv{5}\\pmod{6} \\\\\n5^1\\equiv{\\fbox{5}}\\pmod{6} \\\\\n\\]\nBut I'm new to modular arithmetic so I'm not sure.\nAnyway, RisingFlame's solution is probably the easiest one.\n[/hide][/quote]\nNo, that doesn't work. Sorry.\nBasically, you can't take the mod off of an exponent...\nThere are ways to do this, but it uses Fermat's little theorem.[/quote]\r\n\r\n:lol: Didn't think so.\r\nWhat about this:\r\n\\[ 125\\equiv{5}\\pmod{15} \\\\\r\n5^{625}\\equiv{25^{312}*5}\\pmod{15} \\\\\r\n25\\equiv{10}\\pmod{15} \\\\\r\n10^{312}\\equiv{10}\\pmod{15} \\\\\r\n10*5\\equiv{50}\\equiv{\\fbox{5}}\\pmod{15}\r\n\\]\r\nWhich sort of uses my newly invented conjecture that $ 10^n\\equiv{10}\\pmod{15}, {n\\in{\\mathbb{Z}}}$ :o", "Solution_7": "[quote=\"grn_trtle\"]\nWhich sort of uses my newly invented conjecture that $ 10^n\\equiv{5}\\pmod{15}, {n\\in{\\mathbb{Z}}}$ :o[/quote]\r\n:) Did you check $ n \\equal{} 2$?", "Solution_8": "Sorry, sorry, typo.\r\nI meant $ 10^n\\equiv{10}\\pmod{15}$, not $ \\equiv{5}$", "Solution_9": "How can we prove that $ 10^n \\minus{}10 \\equiv 0 \\mod 15$?\r\n[hide=\"hint\"]Consider that it must be divisible by both 5 and 3.[/hide]", "Solution_10": "I'm going to try it without looking at the hint :)\r\nIt suffices to show that $ 15|(10^n - 10), \\forall{n}\\in\\mathbb{Z}^ +$\r\nLet $ P(n) = 10^n - 10$\r\n$ P(1) = 10 - 10 = 0$, and $ 15|P(1)$ because everything divides 0.\r\n$ P(2) = 100 - 10 = 90$, and $ 15*6 = 90$, so this is true.\r\nAnd, making the assumption that $ P(k)$ is true, we can show that $ P(k)$ implies $ P(k + 1)$\r\n\\begin{eqnarray*} P(k + 1) & = & 10^{k + 1} - 10 \\\\\r\n& = & 10(10^k - 1) \\\\\r\n& = & 10(P(k) + 9) \\\\\r\n& = & 10*P(k) + 90 \\end{eqnarray*}\r\nNow, because we assumed that $ 15|P(k)$, we know that $ 15|(10*P(k))$. We also know that $ 15|90$ by considering $ P(2)$. Adding a multiple of 15 to another multiple of 15 results in a multiple of 15.\r\n\r\nThus, by induction, $ 15|(10^n-10), \\forall{n}\\in{\\mathbb{Z}^ + }$, Q.E.D.\r\n\r\n(And yes, I'm sure there's a much more elegant way to do it than induction.)", "Solution_11": "It's correct, but you used $ P(n)$ both as a number and as a statement.\r\n(in the line \"And, making the assumption that $ P(n)$ is true...)", "Solution_12": "Thanks for pointing that out :). Taken care of.", "Solution_13": "i think this is simpler.\r\npowers of 10 give remaindre 10 u can observe.\r\nso, 10^n = 10(mod 15)\r\nimplies, 10^n-10=0(mod 15)\r\nhence, proved.", "Solution_14": "[quote=\"analytic\"]\npowers of 10 give remaindre 10 u can observe.[/quote]\r\n\r\nI think archimedes1 was trying to get me to prove that $ 10^n\\equiv{10}\\pmod{15}$.\r\nTo prove that $ 10^n\\minus{}10\\equiv{0}\\pmod{15}$ you must prove that $ 10^n\\equiv{10}\\pmod{15}$... you can't just \"observe\" or it wouldn't be a proof so much as a hypothesis.", "Solution_15": "remainder 10^1 =10(mod 15)\r\nremainder 10^2= 10*10=100=10(mod 15)\r\nthis chain will continue till infinity.just like u have non terminating recurring decimals.", "Solution_16": "Yes, the premise behind my proof by induction was to show that that chain continues infinitely.", "Solution_17": "Wickedestrjr will you teach me that method of Ernies? Now well be even because ive helped you.", "Solution_18": "What about this for a proof?\r\n\r\n[hide]We need to show that $ 10^n - 10 \\equiv 0 \\mod 15$\n\nSince $ 15 = 3 \\times 5$ , and 3 and 5 are both prime, it is sufficient to show that $ 10^n - 10 \\equiv 0 \\mod 3$ and $ 10^n - 10 \\equiv 0 \\mod 5$.\n\n$ 10^n - 10 \\equiv 1^n - 1 \\mod 3$\n$ 10^n - 10 \\equiv 0 \\mod 3$\n\n$ 10^n - 10 \\equiv 0^n - 0 \\mod 5$\n$ 10^n - 10 \\equiv 0 \\mod 5$\n\nTherefore, $ 10^n - 10 \\equiv 0 \\mod 15$ and $ 10^n \\equiv 10 \\mod 15$.\n\nBut about the $ {n\\in{\\mathbb{Z}}}$, I'm not too sure...[/hide]\r\n\r\nAnd Wickedestjr, can you post the link with the modular arithmetic trick?\r\nThanks", "Solution_19": "nice proof :)", "Solution_20": "Nice proof :lol: I would have done something more creative if I hadn't been so tired.\r\n$ n\\in\\mathbb{Z}$ simply means $ n$ in the set of all integers. (Basically, defines $ n$ as an integer).", "Solution_21": "N is the set of Natural numbers. \r\nZ is the set of Integers. \r\nQ is the set of Rational Numbers. \r\nR is the set of Real Numbers. \r\n\r\nFor those that don't know.", "Solution_22": "[quote=\"RisingFlame\"]That's correct.\n[hide=\"A faster way\"]$ 125^{625} \\pmod 6 \\equiv ( \\minus{} 1)^{625}\\pmod 6 \\equiv \\minus{} 1\\pmod 6 \\equiv \\boxed{5}\\pmod 6$.[/hide][/quote]\r\ni am not very good with modular arithmetic so, could you please explain your solution?", "Solution_23": "[quote=\"mathemonster\"]Wickedestrjr will you teach me that method of Ernies? Now well be even because ive helped you.[/quote]\r\n\r\nWhy learn it from Wickedestjr, when you could learn it from ernie himself!\r\n\r\nOk, sample problem.\r\n\r\nWhat is $ 6^{25432} \\text{ mod } 7$?\r\n\r\nFirst, you take $ 25432$, and divide it by $ 7$, to get $ 3633 \\frac {1}{7}$. Take $ 3633 \\plus{} 1$.\r\n\r\nWhat is $ 6^{3634} \\text{ mod } 7$?\r\n\r\nTake $ 3634$ and divide it by $ 7$, to get $ 519 \\frac {1}{7}$. Take $ 519 \\plus{} 1$.\r\n\r\nWhat is $ 6^{520} \\text{ mod } 6$?\r\n\r\nTake $ 520$ and divide it by $ 7$, to get $ 74 \\frac {2}{7}$. Take $ 74 \\plus{} 2$.\r\n\r\nWhat is $ 6^{74} \\text{ mod } 7$?\r\n\r\nTake $ 74$ and divide it by $ 7$, to get $ 10 \\frac {4}{7}$. Take $ 10 \\plus{} 4$.\r\n\r\nWhat is $ 6^{14} \\text{ mod } 7$?\r\n\r\nTake $ 14$ and divide it by $ 7$, to get $ 2 \\frac {0}{7}$. Take $ 2 \\plus{} 0$.\r\n\r\nWhat is $ 6^2 \\text{ mod } 7$?\r\n\r\nSince it is obviously cannot be simplified any more, take $ 36 \\text{ mod } 7$, to get the answer of $ \\boxed{1}$.\r\n\r\nThere are many easier ways. xD", "Solution_24": "[quote=\"Poincare\"][quote=\"RisingFlame\"]That's correct.\n[hide=\"A faster way\"]$ 125^{625} \\pmod 6 \\equiv ( \\minus{} 1)^{625}\\pmod 6 \\equiv \\minus{} 1\\pmod 6 \\equiv \\boxed{5}\\pmod 6$.[/hide][/quote]\ni am not very good with modular arithmetic so, could you please explain your solution?[/quote]\r\ndivide 125 by 6 , y'll get 5 or -1 in terms of 6. replace 125 by -1 and rest is simple.", "Solution_25": "[quote=\"ernie\"][quote=\"mathemonster\"]Wickedestrjr will you teach me that method of Ernies? Now well be even because ive helped you.[/quote]\n\nWhy learn it from Wickedestjr, when you could learn it from ernie himself!\n\nOk, sample problem.\n\nWhat is $ 6^{25432} \\text{ mod } 7$?\n\nFirst, you take $ 25432$, and divide it by $ 7$, to get $ 3633 \\frac {1}{7}$. Take $ 3633 \\plus{} 1$.\n\nWhat is $ 6^{3634} \\text{ mod } 7$?\n\nTake $ 3634$ and divide it by $ 7$, to get $ 519 \\frac {1}{7}$. Take $ 519 \\plus{} 1$.\n\nWhat is $ 6^{520} \\text{ mod } 6$?\n\nTake $ 520$ and divide it by $ 7$, to get $ 74 \\frac {2}{7}$. Take $ 74 \\plus{} 2$.\n\nWhat is $ 6^{74} \\text{ mod } 7$?\n\nTake $ 74$ and divide it by $ 7$, to get $ 10 \\frac {4}{7}$. Take $ 10 \\plus{} 4$.\n\nWhat is $ 6^{14} \\text{ mod } 7$?\n\nTake $ 14$ and divide it by $ 7$, to get $ 2 \\frac {0}{7}$. Take $ 2 \\plus{} 0$.\n\nWhat is $ 6^2 \\text{ mod } 7$?\n\nSince it is obviously cannot be simplified any more, take $ 36 \\text{ mod } 7$, to get the answer of $ \\boxed{1}$.\n\nThere are many easier ways. xD[/quote]\r\n\r\nWould if it is something like; \r\n\r\nx^y (mod z), when they are all very large numbers like 37, 81, and 93. \r\n\r\n\r\nSolve this; \r\n\r\n37^81 (mod 93) That procedure doesn't always work.", "Solution_26": "Why does the ernie method work?\r\nYes it can't be used on $ 37^{81}\\pmod{93}$\r\nhow would u solve this one?", "Solution_27": "This is exactly what I couldn't figure out in my other modular arithmetic post." } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "I know I may break some rules when I post this post,so if necessaryly medortor can delete it.\r\nIt is a solution to crux 2950,the problem appears in crux 2005.4 again,it is unsolved now.\r\nBut when I sent my solution to the editor,he said he can't understand what I wrote.I am so sorry,and may there is something unclear.I don't what can I do with it.So I post it here,I hope somebody can help me :)", "Solution_1": "I'm sorry but I don't understand even the first line of the problem: let $ABC$ be a triangle whose largest angle does not exceed , for , Can you please write more clearly what you are going to say.", "Solution_2": "Sorry,I forget \"$\\frac{2\\pi}3$ there,now I've edited my file.\r\nThank you", "Solution_3": "Hi,you have a nice solution. :) \r\nDid you think about generalizations? :P \r\nCompare $\\sum sin\\frac{A}{2}.sin\\frac{B}{2}$ and $x+y.\\sum cos\\frac{A}{2}.cos\\frac{B}{2}$" } { "Tag": [ "function", "calculus", "derivative", "limit", "absolute value", "real analysis", "real analysis unsolved" ], "Problem": "Let $f: R^2 \\to R$ be defined by:\r\n\r\n[img]http://img211.imageshack.us/img211/3874/image0102lr.gif[/img]\r\n\r\nShow that $f$ is not differentiable at $(0,0)$\r\n\r\nHow would you do this question? I am having trouble dealing with the $|y|$ and showing that the function is indeed not differentiable at $(0,0)$", "Solution_1": "The partial derivatives are zero, so if it were differentiable the derivative would be zero and $\\frac{f(x,y)}{\\|(x,y)\\|}=\\frac{x|y|}{x^2+y^2}$ would tend to zero at zero.\r\n\r\nThe absolute value is just there so that the directional derivatives exist.", "Solution_2": "I am quite confused about what you wrote, sorry. So the function IS differentiable at 0? Or were you just suggesting to use a proof by contradiction showing that $f$ is not differentiable at 0?", "Solution_3": "$\\frac{\\partial f}{\\partial y}$ and $\\frac{\\partial f}{\\partial x}$ exist and are zero at $x=0$. So if $f$ were differentiable, then we would have to have\r\n\r\n$\\lim_{\\|(x,y)\\|\\to 0} \\frac{f(x,y)-f(0,0)}{\\|(x,y)\\|}=0$\r\n\r\nBut this is not true (take $x=y$ and let $x,y\\to 0$)...", "Solution_4": "Oh, I get it now. Thanks a lot." } { "Tag": [ "arithmetic sequence", "geometric sequence", "geometric series", "arithmetic series" ], "Problem": "Would someone please refer some sites or ebooks where I can read information about sequences in general and about geometic and arithmetic progressions? My portuguese textbooks are too cold, if I make me understand...\r\n\r\nThank you... :lol:", "Solution_1": "Always start at Wikipedia or MathWorld. :D\r\n[url=http://en.wikipedia.org/wiki/Arithmetic_progression]Arithmetic Progression - Wikipedia[/url]\r\n[url=http://en.wikipedia.org/wiki/Geometric_progression]Geometric Progression - Wikipedia[/url]\r\n[url=http://mathworld.wolfram.com/GeometricSeries.html]Geometric Series - MathWorld[/url]\r\n[url=http://mathworld.wolfram.com/ArithmeticSeries.html]Arithmetic Series - MathWorld[/url]" } { "Tag": [ "MATHCOUNTS" ], "Problem": "[quote=\"aznness\"]Phelpedo, who's your brother?\n\nAlex is right, i did exaggerate a bit. but hey, we are still going to own you guys! And whats with Texas. Just because bush won the election doesnt mean texas owns the world. CAlifornia has the most electoral votes. EAT THAT!![/quote]\r\n\r\nPlease note that bush is not capitalized", "Solution_1": "Does 'bush' have to be capitalized?", "Solution_2": "[quote=\"NoSoupForYou\"]Does 'bush' have to be capitalized?[/quote]\r\n\r\nJoseph Chu was capitalized... while bush wasn't... does that tell you something?", "Solution_3": "not really", "Solution_4": "[quote=\"NoSoupForYou\"]not really[/quote]\r\n\r\nJoseph Chu>bush", "Solution_5": "$x>bush \\text{ where } [x|x\\text{ is any human, some animals, and most inanimate objects}]$", "Solution_6": "[quote=\"NoSoupForYou\"]$x>bush \\text{ where } [x|x\\text{ is any human, some animals, and most inanimate objects}]$[/quote]\r\nnice one :D \r\nwow a lot of people here dont' like bush, myself included.", "Solution_7": "agreed :)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $ x_n$ such that\r\n$ x_n\\equal{}\\sum\\limits_{k\\equal{}1}^{n} \\frac{1}{(k!)^2}$\r\nProve that a) this sequence has lim ( it is easy)\r\n b) This lim is not a rational number (it is hard)\r\nHelp help", "Solution_1": "a) note that the sequence is increasing and is bounded(for example by $ e$). this implies that ti has a limit.\r\nb) suppose it is equal to $ x\\equal{}\\frac{p}{q}$\r\n$ x(q!)^2\\equal{}n\\plus{}\\left(\\frac{1}{(q\\plus{}1)^2}\\plus{}\\frac{1}{(q\\plus{}1)^2(q\\plus{}2)^2}\\plus{}\\cdots\\right)$\r\nwhere $ n\\in \\mathbb{Z}$ \r\nbut $ 01 there are positive integers x and y such that x^2 - y^2 = m^3.\r\nb) Find all pairs of positive integers (x,y) such that x^6 = y^2 +127\r\n\r\n2002 5)\r\nEach of the first 13 letters of the alphabet is written on the back of a card and the 13 cards are placed in a row in the order\r\nA,B,C,D,E,F,G,H,I,J,K,L,M\r\nThe cards are then turned over so that the letters are face down. The cards are rearranged and again placed in a row, but of course they may be in a different order. They are rearranged and placed in a row a second time and both rearrangements were performed exactly the same way. When the cards are turned over the letters are in the order\r\nB,M,A,H,G,C,F,E,D,L,I,K,J\r\nWhat was the order of the letters after the cards were rearranged the first time?\r\n\r\n2000 5)\r\nThe Olive View Airline in the remote country of Kuklafrania has decided to use the following rule to establish its air routes: If A and B are two distinct cities, then there is to be an air route connecting A with B either if there is no city closer to A than B or if there is no city closer to B than A. No further routes will be permitted. Distances between Kuklafranian cities are never equal. Prove that no city will be connected by air routes to more than ve other cities.\r\n\r\n1998 1)\r\nAn organization decides to raise funds by holding a $60 a plate dinner. They get prices from two caterers. The first caterer charges $50 a plate. The second caterer charges according to the following schedule: $500 set-up fee plus $40 a plate for up to and including\r\n61 plates, and $2500 log (p/4) for p > 61 plates.\r\na) For what number of plates N does it become at least as cheap to use the second caterer as the first?\r\nb) Let N be the number you found in a). For what number of plates X is the second caterer's price exactly double the price for N plates?\r\nc) Let X be the number you found in b). When X people appear for the dinner, how much profit does the organization raise for itself by using the second caterer?", "Solution_1": "1. [hide]a) x=(m+m:^2:)/2, y=(m:^2:-m)/2. Note that those will indeed be integers as m and m:^2: have the same parity. b)(x:^3:-y)(x:^3:+y)=127 (prime), so x:^3:-y=1 and x:^3:+y=127=> x:^3:=64, y=63 => (x,y)=(4,63)[/hide]\n\n\n\n3. [hide]Assume that city A is connected to six cities, named B, C, D, E, F, and G. Connect A to these 6 cities. By pigeonhole, one of the angles between two cities and A (such as BAC, CAD, etc.) will be :le:60. Let the other two cities at the endpoints of this angle be B and C wlog. Then either ABC>60 or ABC<60. ABC cannot be equal to 60 b/c then the triangle would be equilateral, violating the assumption that all distances between cities are different. In the first case above, there should not be a road between A and C b/c AB500:le:10n=>n:ge:50, so the smallest integral n is 50.\n\nb)ok, we want the x for which g(x)=2*(2500)=5000. g(61)=2940, so as g(n) is an increasing function of n, we do have to worry about that freaky log part. okay, if y=x-61, g(x)=2940+2500 log (y/4). if this is to equal 5000, 2940+2500 log (y/4)=5000=> log (y/4)=.824 => y/4=10^(.824)=6.668 => y=27.672 okay so the smallest integral x is thus 89.\n\n\n\nc) we want 60*89-g(89)=5340-2940-2500log(7)=2400=$287.25[/hide]", "Solution_4": "2. [hide]1 a 3 b 6 c 7 d 5 e 8 f 4 || g 9 h 11 i 12 j 10 k 13 l 2 m. in the above string of letters and numbers, the numbers refer to the places of the letters in the stack of cards, while the letters are variables. for example, the first three characters represent that the card in the first position is mapped to something in the \"a\"th position, which will then be mapped to the third position. Notice that we have 26 characters up there, but we only have cycles of 13 characters. Note also that the double vertical lines separate the two sets of 13 characters. Thus, they must be the same cycle, i.e. 1=g, a=9, 3=h, etc. This makes the cycle as follows:\n\n\n\n1 9 3 11 6 12 7 10 5 13 8 2 4, meaning that ABCDEFGHIJKLM is mapped to DHIBJKLMAGCFE. You can indeed check that after another reiteration of this cycle, DHIBJKLMAGCFE is mapped to BMAHGCFEDLIKJ[/hide]", "Solution_5": "Mysticterminator gets all 4.", "Solution_6": "fun contest. when do we get a second round?", "Solution_7": "How did you think to do that for number 2? That is a very interesting method. Have you seen that used somewhere before?", "Solution_8": "no, it just seemed logical to do it that way. I am not entirely sure that it is rigorous, as I never explicitly stated (proved) that the cycles could not be of length <13, which would obviously detract from the method, but besides that little thing, it just seems logical to write out the sequence like that.", "Solution_9": "Second round will be up as soon as I get the problems together. I'm starting now." } { "Tag": [ "ratio", "Pythagorean Theorem", "geometry" ], "Problem": "In a tri. ABC, ang.B = 90 deg. ; ang.A = 30 and if the hypotenuse AC = 10 cm then find the other angles.", "Solution_1": "Are you serious?", "Solution_2": "im trying to prevent myself from thinking that this is a trick question.\r\n\r\n[hide]sum of angles in triangle = 180\n90 + 30 + x = 180\nx = 60[/hide]", "Solution_3": "60. This barely qualifies as a math problem, certainly not one for competitions.", "Solution_4": "i am sorry ......the question is find the remaining sides... :(", "Solution_5": "This is a famous \"30-60-90\" triangle, whose ratio of side lengths is $1: \\sqrt{3}: 2$. To prove this, consider an equilateral triangle, and cut it into two 30-60-90 triangles. (You need to use Pythagorean Theorem to finish the proof.)\r\n\r\nSo the remaining side lengths are $AB=5\\sqrt{3}$ and $BC=5$.\r\n\r\nI think this thread should be moved to middle school forum." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Assumue $x_1=\\frac{1}{3}$, $x_{n+1}=x_n(1-x_n)$ for $n>1$.Find $\\lim_{n\\to\\infty}{x_n}$. :)", "Solution_1": "$x_1=\\frac 13\\ ,\\ x_2=\\frac 12\\ ,\\ x_3=1\\ ,\\ x_4=\\ \\frac 10\\ ?$ You are cunning !\r\nIn my opinion this sequence hasn't sense. Can you think that $x_n=\\frac{1}{4-n},\\ \\left(\\forall\\right)\\ n\\in N^*,\\ n\\ne 4\\ !$ Maybe possible, because from a sequence we can eliminate a finite number of terms. In this case, $x_n\\nearrow 0\\ .$", "Solution_2": "What do you see, Virgil? I see a perfectly normal decreasing positive sequence $x_{n+1}=x_n\\cdot(1-x_n)$, which clearly converges to zero.", "Solution_3": "Just drawing lines $y=x$ and $y=x(1-x)$ and bouncing from one to another gives an answer in those cases :)" } { "Tag": [ "probability" ], "Problem": "Ben rolls four fair 20-sided dice, and each of the dice has faces numbered from 1 to 20. What is the probability that exactly two of the dice show an even number?", "Solution_1": "$ \\binom{4}{2} \\times \\frac{1}{16}\\equal{}\\boxed{\\frac{3}{8}}$.", "Solution_2": "Shouldn't it be $\\frac{4!}{2!2!} \\cdot \\frac{1}{16}$?", "Solution_3": "[quote=SlurpBurp]Shouldn't it be $\\frac{4!}{2!2!} \\cdot \\frac{1}{16}$?[/quote]\n\n$\\binom42=\\frac{4!}{2!(4-2)!}.$", "Solution_4": "[quote=SlurpBurp]Shouldn't it be $\\frac{4!}{2!2!} \\cdot \\frac{1}{16}$?[/quote]\n\nThat's the same thing as the post above you just in different notation.", "Solution_5": "[quote=kittenwarrior][quote=SlurpBurp]Shouldn't it be $\\frac{4!}{2!2!} \\cdot \\frac{1}{16}$?[/quote]\n\n$\\binom42=\\frac{4!}{2!(4-2)!}.$[/quote]\n\nFTFY", "Solution_6": "@above lol, thanks! I don't know why I even typed a $3\\dots $ :P", "Solution_7": "[hide=\"Sol\"]\nIt is $\\left(\\frac12\\right)^4=\\frac{1}{16}$. However, they can be rearranged in $\\frac{4!}{2!2!}=6$ ways. So it is $\\frac{6}{16}=\\boxed{\\frac38}$\n[/hide]", "Solution_8": "Wow we can literally just let them be fair coins because it only cares about parity.", "Solution_9": "[hide=pog // 896 posts]\nNote that each die has a probability of $\\dfrac{1}{2}$ of getting an even number or odd number, so we have $\\left(\\dfrac12\\right)^2 \\cdot \\left(\\dfrac12\\right)^2 = \\dfrac{1}{16}.$ Also, we must choose $2$ of the die to even, so we have $\\dfrac{4 \\cdot 3}{2}$ ways (choosing die $1$ and $2$ is the same as choosing die $2$ and $1.$) Thus we have $6 \\cdot \\dfrac{1}{16} = \\dfrac{6}{16} = \\boxed{\\dfrac{3}{8}}.$\n[/hide]", "Solution_10": "[hide=Solution]The number of ways to choose the dice that are even is $\\binom{4}{2} = 6$. The rest must be odd. \nHence, the probability that Ben rolls an even number is \n$6 \\cdot (\\frac{1}{2})^2 \\cdot (\\frac{1}{2})^2 = \\frac{6}{16} = \\boxed{\\frac{3}{8}}$.\u0010[/hide]" } { "Tag": [ "complex analysis" ], "Problem": "[img]http://i42.tinypic.com/2n3fy0.jpg[/img]\r\n\r\nIs there a mistake in the solution?\r\n\r\nThe line $ V_y\\equal{}U_x$. Should that not be $ V_y\\equal{}\\frac{\\minus{}1}{y}{U_x}$ and the proceed from there?\r\n\r\nthanks,\r\nbos1234", "Solution_1": "no, that's true, why should it be $ v_y \\equal{} \\frac { \\minus{} 1}{y}u_x$? I think you've confused with the polar version of Cauchy-Riemann equations." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all triples of prime numbers $ (p,q,r)$ such that $ p^q\\plus{}p^r$ is a perfect square.", "Solution_1": "$ q < r$ $ \\implies$ $ p^q \\plus{} p^r \\equal{} p^q.(1 \\plus{} p^{q \\minus{} r}) \\equal{} a^2$\r\n$ q > 2$ $ \\implies$ $ p| 1 \\plus{} p^{q \\minus{} r}$ contradiction!\r\n$ q \\equal{} 2$ $ \\implies$ $ p^2 \\plus{} p^r \\equal{} a^2$ $ \\implies$ $ (a \\minus{} p)(a \\plus{} p) \\equal{} p^r, a \\minus{} p \\equal{} p^x,a \\plus{} p \\equal{} p^y,2p \\equal{} p^y \\minus{} p^x$\r\n$ x > 1$ $ \\implies$ $ p|2$ $ \\implies$ $ p \\equal{} 2,2^y \\minus{} 2^x \\equal{} 4,y \\equal{} 3,x \\equal{} 2, r \\equal{} 5,q \\equal{} 2,p \\equal{} 2$\r\n$ x \\equal{} 1$ $ \\implies$ $ a \\equal{} 2p$ $ \\implies$ $ 3p \\equal{} p^y$ $ \\implies$ $ p \\equal{} 3. a \\equal{} 6, a^2 \\minus{} p^2 \\equal{} p^r$\r\n$ 36 \\minus{} 9 \\equal{} 3^3$ $ \\implies$ $ r \\equal{} 3,p \\equal{} 3,q \\equal{} 2$\r\n$ x \\equal{} 0$ $ \\implies$ $ a \\equal{} p \\plus{} 1,2p \\plus{} 1 \\equal{} p^y$ Contradiction!\r\n$ q \\equal{} r$ $ \\implies$ $ 2p^q \\equal{} a^2,p \\equal{} 2,q > 2 (2,q,q)$" } { "Tag": [ "Support" ], "Problem": "pick it\r\n\r\nIts the rules we'll play with for the MARCH tournament", "Solution_1": "our rules are like the same tho.", "Solution_2": "wats with 2 of the same names?\r\nR u all crazy?", "Solution_3": "personally, i really don't care (both seem fine to me)\r\n\r\nbut seeing as verizon's on my team, i'll support his rules\r\n\r\nGO BOYZ OF DARKNESS!!!!!", "Solution_4": "I voted verizonwires. Otherwise we will have no countdowns which means no fun. :(", "Solution_5": "Wait a second. The normal round was for the first round. I am thinking about a way to have a countdown for the second round and then a long normal game for the last round, which will be loads of fun!\r\nThats why I said in my Private Message that I sent to everyone, that those rules [b]only[/b] apply to the first round. It will get better and better each round making people want to win!", "Solution_6": "Uhh, wasn't that what I originally said on my very first post? I'm pretty sure there was a reason we decided not to do it that way :D .\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=187175", "Solution_7": "dunno cause when I read your rules, I was under the impression that all that would be done in EACH round. Isn't that why you only had two types of competition? Otherwise you would've had three. Either way, it was a bit unclear and verizonwires/minicon restated the rules so it was crystal. At least in a relative sense XD\r\n\r\nlet's just go with minicon's - further changes will only confuse everyone." } { "Tag": [], "Problem": "A swarm of evil bunnies that were being breeded by Drunken_Math contains 1000 evil bunnies. \r\n\r\nThe evil bunnies can do this: Every minute, each evil bunny can bite one person, and that person will instantly turn into an evil bunny, but that human who just turned into an evil bunny can't bite anyone until the next minute. For example, if one evil bunny bites someone at 12:38, the person who just turned into an evil bunny can't bite anyone until 12:39.\r\n\r\nAnother thing about the evil bunnies, after an evil bunny bites 100 people, it shrivels up and turns into a human. \r\n\r\nIf my swarm of 1000 bunnies attack the earth, in how long will the entire Earth be made completely into evil bunnies (round to the nearest day)? Assume the earth has 10 billion people.\r\n\r\nPS: I tried to do this problem and it got really messy!", "Solution_1": ":rotfl: Haha! This is a really well made problem. I'll post my solution if nobody else posts.", "Solution_2": ":o this problem is crazy! I could solve it if you gave me like 3 hours of adding. I bet you can make it algebra though where did you come up with this? Please show us how 4everwise :D It has a pattern 2,4,8,16,32,64,128,256,512,1024,2048,4096,8192 ok i know how to do it! Yay. 2^100 is more than a billion. It will take the bunnies 30 minutes to destroy the world, if you do not count the time it will take to travel all over the world. :D" } { "Tag": [ "Support", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "I don't have a very deep understanding of the concepts $ Ass$,$ Supp$,etc.. (yet). Can someone clear up this line.\r\n\r\nIf $ I$ is an ideal of $ A$ then $ Supp_A(A/I)$ is the set of prime ideals containing $ I$.", "Solution_1": "So we look for $ p \\in Spec(A)$ such that the localization of the $ A$-module $ A/I$ is non-zero\r\n\r\nso $ (A/I)_p$ as $ A_p$-module may not be zero\r\n\r\nequivalently\r\n\r\n$ (A_p/IA_p)$ as $ A_p$-module may not be zero\r\n\r\nthis is non-zero as long $ p \\supset I$\r\n\r\nelse if $ p \\subset I$, then $ IA_p \\equal{} A_p$ and $ A_p/IA_p \\equal{} 0$\r\n\r\n\r\nI've been learning these things way to fast. Apparently I now pay the price for this. I have a very chaotic understanding of these abstract concepts. \r\n\r\nDoes the above make any sense ?", "Solution_2": "the first part is ok. but in the second part you are very inaccurate and you assume that every two sets are comparable by inclusion ..\r\n \r\n$ I A_p \\equal{} A_p$ is equivalent to: there is $ i \\in I$ and $ s \\in A \\backslash p$ such that $ i/s \\equal{} 1$ in $ A_p$, i.e. there is $ t \\in A \\backslash p$ such that $ t(i \\minus{} s) \\equal{} 0$. it follows $ i \\minus{} s \\in p$ and then $ i \\notin p$. conversely, every $ i \\in I \\backslash p$ yields $ 1 \\equal{} i/i \\in I A_p$. thus: $ supp_A(A/I) \\equal{} \\{p \\in Spec(A) : A_p \\neq I A_p\\} \\equal{} V(I)$.", "Solution_3": "Is this why the name \"prime divisor\" is used? I never read any motivation actually\r\n\r\nSo $ P$ is a prime divisor of $ I$ if $ P \\in Ass(A/I)$.\r\n\r\nBecause $ Ass_A(A/I) \\subset Supp_A(A/I) \\equal{} V(I)$, the elements in $ Ass_A(A/I)$ are prime ideals containing $ I$ too.\r\n\r\nThen we can use the same terminology as in $ \\bf Z$, i.e. if $ (a) \\subset (b)$ one writes $ b \\mid a$ \r\n\r\nso $ P \\in Ass(A/I)$ means $ I \\subset P$ means \"$ P$ divides $ I$\" means \"$ P$ is a prime divisor of $ I$\"", "Solution_4": "An ideal $ J$ divides an ideal $ I$ means $ J\\supseteq I$. Unfortunately, \"prime divisor\" is not equivalent to prime + divisor. Any minimal prime over some ideal $ I$ (i.e., a minimal element w.r.t. inclusion of the set of primes containing $ I$) is a prime divisor, but there may or may not be more. I am not quite sure whether this answers your question, or what your question was in the first place. owk" } { "Tag": [ "geometry", "perimeter", "geometric transformation", "reflection", "area of a triangle", "Heron\\u0027s formula", "inequalities theorems" ], "Problem": "How to show that if $n\\geq 3$ is a given integer and $s>0$ is also given, then there is an $n$-gon with perimeter $s$ which has a maximal area? Furthermore, show that the maximal area is achieved if and only if the polygon is regular.", "Solution_1": "Okay, I have heard a sketch of the proof: First you have to prove that the polygon is convex. This is easy, since we can always find two points in a non-convex polygon where the line segment goes also outside the polygon. Then we just reflect the outer part to get a bigger polygon with the same perimeter. Then we prove that if the polygon has different length of sides, we can step by step change the polygon so that finally the polygon is regular. By Heron's formula one can show that the area won't get smaller. From the isoperimetric theorem, it follows that a polygon with fixed side lengths, the largest one is cyclic. Finally, by compactness argument one can show that the largest polygon exists. \r\n\r\nBut I think that my formulation is not correct. In my opinion, there are non-regular cyclic quadrilaterals with the same area and the same perimeter as a regular one.", "Solution_2": "No you are correct the maximal polygon is cycliq and has all equal sides so it will be regular :wink:", "Solution_3": "[quote=\"Albanian Eagle\"]No you are correct the maximal polygon is cycliq and has all equal sides so it will be regular :wink:[/quote]\r\nWhat part of my sketch is not true?", "Solution_4": "you don't have any mistake in your sketch,\r\nyou have actually shown that in the maximal polygon all sides are equal(by mixing the sides+Heron formula), also that it will be cyclic so it must be regular. Therefor your statement was correct since the beginning :)" } { "Tag": [ "vector", "geometry", "parallelogram" ], "Problem": "I need a solution using vectors to this problem, BAEDC is a pentagon, with vector BC equal and parallel to vector ED. X Y and Z are midpoints of respective vectors; BA, AE and DC respectively. Prove that XYZC is a parallelogram.", "Solution_1": "Why you don't want a solution of pure geometry? :mad: \r\n\r\nIn the triangle $ABE$, the segment $XY$ is joining the midpoints of $AB,AE$, so $XY \\parallel BE$ and $XY=\\frac{BE}{2}$\r\n\r\nBut $BCDE$ is a parallelogram, so $CD \\parallel = BE \\Rightarrow CZ \\parallel = XY\\Rightarrow CZYX$ parallelogram\r\n\r\n($\\parallel =$ means parallel and equal) \r\n\r\n\r\n\r\nOk here is a solution with vectors:\r\n\r\n$\\overline{BC}=\\overline{ED}\\Rightarrow BCDE$ parallelogram $\\Rightarrow \\overline{BE}=\\overline{CD}$\r\n\r\n$\\overline{BE}=\\overline{BA}+\\overline{AE}=2\\overline{XA}+2\\overline{AY} = 2(\\overline{XA}+\\overline{AY}) = 2\\overline{XY}$\r\n\r\n$\\overline{CD}=2\\overline{CZ}$\r\n\r\n$\\overline{BE}=\\overline{CD}\\Rightarrow \\overline{CZ}=\\overline{XY}\\Rightarrow XYZC$ parallelogram" } { "Tag": [ "geometric sequence", "number theory", "modular arithmetic" ], "Problem": "Hi, im not sure how to approach this problem. Any help would be appreciated, thanks.\r\n\r\nFor how many natural numbers n =<12 is 4^n + 5^n + 6^n + 7^n a multiple of 11", "Solution_1": "Other questions that have been bugging me: why is it that in most textbooks the sum of a geometric sequence is shown as a(1-r^n)/(1-r)? Since this is used for |r|>1, 1-r^n will be negative, as will 1-r. So why isnt -1 factored out so it is a(r^n-1)/(r-1)?\r\nAlso, if a question asks for the sum of the 1st n powers of 2, the answer is 2^(n+1)-1. This means that it goes from 2^0...2^n. Why does it start at 2^0 instead of 2^1? Thanks.", "Solution_2": "The formula for the summation of a geometric sequence can be written as a(1-r^n)/(1-r) or a(r^n-1)/(r-1); they're just different forms of the same formula.\r\n\r\nFor the powers of two, I assume that n starts from 0 since it is the smallest natural number.\r\n\r\n\"For how many natural numbers n =<12 is 4^n + 5^n + 6^n + 7^n a multiple of 11\"\r\n\r\nThis problem can probably be solved with modular arithmetic. Using the fact that a number is divisible by 11 iff the alternate sum of its digits is divisible by 11 (i.e. ex. 121 :equiv: 1-2+1 (mod 11) :equiv: 0 (mod 11).\r\n\r\n-interesting_move", "Solution_3": "I realize that they're different forms of the same equation, but what's the purpose of having the first form if you're going to get a negative numerator and denominator and you're going to have to cancel them? In most textbooks it's written the first way, but the 2nd way seems more practical to me.\r\nIf you count to 20 you start at 0?\r\nIn order to solve this problem by using the divisibility rules for 11, you have to actually sum the numbers, so it becomes tedious after n=2, unless if there is a pattern or something, but im not sure if we can find a pattern by calculating for n=1,2?", "Solution_4": "wow, how systematic hgmm76.\r\n\r\n-interesting_move", "Solution_5": "[quote=\"interesting_move\"]For the powers of two, I assume that n starts from 0 since it is the smallest natural number.[/quote]\r\n\r\nNo, 0 is not the smallest natural number. It isn't a natural number. \r\n\r\nI quote from The Big Book (AoPS #1, pg. 99): \"We start with the natural numbers: 1,2,3,...\"\r\n\r\nIt's easy to confuse natural numbers and whole numbers. One way to remember it is this: for whole numbers, think \"hole\". When you think \"hole\" you think of the number that looks like a hole - 0! So whole numbers include 0, as well as the positive integers. Whereas natural numbers don't.", "Solution_6": "[quote=\"interesting_move\"]\"For how many natural numbers n =<12 is 4^n + 5^n + 6^n + 7^n a multiple of 11\"\n\nThis problem can probably be solved with modular arithmetic. Using the fact that a number is divisible by 11 iff the alternate sum of its digits is divisible by 11 (i.e. ex. 121 :equiv: 1-2+1 (mod 11) :equiv: 0 (mod 11).[/quote]\r\n\r\nYes, you said you could use modular arithmetic. But you don't specify how - you just informed hgmm76 about the sum/difference of digits of multiples of 11. How is [i]saying[/i] that you could use modular arithmetic going to help him?", "Solution_7": "[color=cyan]Look at the cycles of things mod 11. For example:\n4^0 :equiv: 1 (11)\n4^1 :equiv: 4 (11)\n4^2 = 16 :equiv: 5 (11)\n4^3 = 4*4^2 :equiv: 4*5 = 20 :equiv: 9 (11)\n4^4 :equiv: 3\n4^5 :equiv: 1\nand then it has to repeat. You can do the same with the others. Also, you might note that\n4 :equiv: -7 (11) and 5 :equiv: -6 (11), so that [i]all[/i] odd powers will work, since the four and seven will cancel each otehr, as will the 5 and 6. The even powers, you will have to check by hand or some other clever method. Also, it has to be cyclic every tenth power, by Fermat's Little Theorem.\n\nAs for the other question, you start with 2^0 because then the formula looks nice; you use 1-r^n because the most interesting thing geometric sequences can do happens when |r| < 1, when this will be positive. I mean, those aren't real answers, but the real answer is \"by convention.\" Someone started doing it, and everyone else picked up on it. That's all.[/color]", "Solution_8": "thanks for the responses.", "Solution_9": "[quote=\"mathfanatic\"]\n\nNo, 0 is not the smallest natural number. It isn't a natural number. \n\n[/quote]\r\n\r\nThere is no convention on if to include 0 to the natural numbers. Sometimes one includes sometimes not. So to avoid ambiguousness I usually write Z+ to denote the positive integers and Z* the nonnegative integers.", "Solution_10": "Yep.\r\n\r\n[url]http://mathworld.wolfram.com/NaturalNumber.html[/url]", "Solution_11": "Here's something I just discovered that applies to this problem:\r\n\r\nLet x and y be positive integers such that x+y=11\r\n\r\nThen for all odd positive n, x^n + y^n mod 11 :equiv: 0\r\n\r\nThat is, x^n + y^n is a multiple of 11.\r\n\r\nI haven't been able to prove this, but it seems to work. Any ideas on a proof? It would probably involve modular arithmetic of some sort.\r\n\r\nIn this problem, we have 4^n + 5^n + 6^n + 7^n \r\nwhich can be written as (4^n + 7^n) + (5^n + 6^n).\r\n\r\nThese two pairs are examples of x^n + y^n such that x + y = 11 (4+7=5+6 = 11); hence the above statement applies and for odd n, 4^n + 5^n + 6^n + 7^n is a multiple of 11. This is for n = 1, 3, 5, 7, 9, 11, ...; that is, [b]6[/b] values for n :le: 12.", "Solution_12": "Sure. Rewrite y as 11-x, or -x (mod 11). Then x^n+(-x)^n=0 if n is odd because (-x)^n=(-1)^n*x^n=-x^n, so adding them gives x^n + - x^n = 0.", "Solution_13": "Here's some modular arithmetic that tells something about the previous statement that for x+y=11 and odd n, x^n + y^n mod 11 :equiv: 0\r\n\r\n(n=1)\r\n4 mod 11 = 4\r\n7 mod 11 = 7\r\n4+7 = 11\r\n\r\n(n=2) (Even power; shouldn't work)\r\n16 mod 11 = 5\r\n49 mod 11 = 5\r\n5 + 5 \u2260 11\r\n\r\n(n=3)\r\n64 mod 11 = 9\r\n343 mod 11 = 2\r\n9+2 = 11\r\n\r\n(n=4) (shouldn't work)\r\n256 mod 11 = 3\r\n2401 mod 11 = 3\r\n3 + 3 \u2260 11\r\n\r\n(n=5)\r\n1024 mod 11 = 1\r\n16807 mod 11 = 10\r\n1 + 10 = 11\r\n\r\n(n=6)\r\n4096 mod 11 = 4\r\n117649 mod 11 = 4\r\n4 + 4 \u2260 11\r\n\r\nAnother startling discovery (again, I can't prove it but it'll definitely have something to do with modular arithmetic) is:\r\n\r\nFor x+y=11 and even n, x^n mod 11 = y^n mod 11\r\n\r\nAny ideas? This confirms our suspicion that even powers of n won't ever work, because x^n mod 11 + y^n mod 11 will always produce an even mod 11 number (not 11)", "Solution_14": "Again, that's right. Let us write x=2k. Then x^n+(-x)^n=x^(2k)+(-x)^2k=(x^2)^k+(x^2)^k=2(x^2)^k, and since x^2 is not congruent to 0 (mod 11), 2(x^2)^k will be twice a number not congruent to 0 (mod 11), so it will not be divisible by 11. (Specifically, 11/2:equiv: 0 (mod 11), if you are used to seeing fractions with modulos.)", "Solution_15": "Oh! After reading ComplexZeta's proof of my first statement, I can prove the second one.\r\n\r\nAgain, by writing y as 11-x, we have:\r\ny^n mod 11 = (11-x)^n mod 11 = (-x)^n mod 11\r\nSince n is EVEN, (-x)^n = x^n\r\nSo y^n mod 11 = x^n mod 11.\r\n\r\nThis is why the mod 11 of x^n and y^n for even n is equal.\r\n\r\nHere's as summary of the weird 11-things:\r\nFor all positive integers such that x + y = 11,\r\nIf n is odd, then x^n + y^n mod 11 = 0\r\nIf n is even, then x^n mod 11 = y^n mod 11", "Solution_16": "Ah, we can generalize this statement for all numbers (since ComplexZeta's proof can be applied to other values other than 11; i.e. if x+y=6, then you could write y as 6-x, then apply the same logic for mod 6)\n\n\n\nSuppose x + y = Z\n\nThen for odd n, (x^n + y^n) mod Z :equiv: 0\n\nFor even n, x^n mod Z = y^n mod Z\n\n\n\nFor instance, if x + y = 21,\n\nThen if n is 5, x^5 + y^5 mod 21 :equiv: 0\n\nIf n is 6, x^5 mod 21 :equiv: y^5 mod 21\n\n\n\nA problem similar to hgmm76's original one is this:\n\nFor how many values n:le:100 is 2^n + 3^n + 6^n + 7^n divisible by 9?\n\n\n\n[hide]The answer is, of course, 50, since for all the odd ns, 2^n + 9^n and 3^n + 6^n will be divisble by 9. For all the even ns, 2^n mod 9 :equiv: 7^n mod 9, etcetera.[/hide]\n\n\n\nSo that's the easy way to do problems like this: by knowing this special formula or discovering it through experimentation.[/hide]", "Solution_17": "I don't quite see your logic for your problem, mathfanatic. Your answer is right though. But here's how I did it. We use the results from before to write for even n 2(2^n+3^n) (mod 9). 3^n will be divisible by 9, so we are left with 2*2^n (mod 9), to which there are clearly no zero solutions.", "Solution_18": "Here is the original problem:\r\n\r\n[quote]For how many natural numbers n =<12 is 4^n + 5^n + 6^n + 7^n a multiple of 11?[/quote]\n\nHere's how I solved it. \n\nI noticed that 4 + 7 = 5 + 6 = 11\nThen I noticed that 4^3 + 7^3 mod 11 :equiv: 5^3 + 6^3 mod 11:equiv: 0\nSimilarly, 4^5 + 7^5 mod 11 :equiv: 5^5 + 6^5 mod 11:equiv: 0\nAnd 4^7 + 7^7 mod 11 :equiv: 5^7 + 6^7 mod 11:equiv: 0\n\nSo I figured out that if x + y = 11 (just as 4&7, 5&6 do), then x^n + y^n mod 11 :equiv: 0 for odd n only.\n\nEventually, I expanded this to read:\n\n[quote]Suppose x + y = Z \nThen for odd n, (x^n + y^n) mod Z :equiv: 0 \nFor even n, x^n mod Z :equiv: y^n mod Z [/quote]\r\n\r\nYou can apply this rule to the new problem, \"For how many values n:le:100 is 2^n + 3^n + 6^n + 7^n divisible by 9?\"\r\n\r\nIn this case, 2+7 = 3+6 = 9. If we split the terms up into:\r\n2^n + 7^n\r\nand 3^n + 6^n\r\nWe find that each one works with the above rule. In the first pair, 2^n + 7^n mod 9 :equiv: 0 for all odd n. In the second pair, 3^n + 6^n mod 9 :equiv: 0 for all odd n. Because 2^n + 3^n + 6^n + 7^n is just the sum of two things that are each divisible by 9 when n is odd, the four-term sum is also divisible by 9 when n is odd. \r\n\r\nHence, there are 50 values for n that are less than 100 which satisfy the condition that 2^n + 3^n + 6^n + 7^n is divisible by 9.", "Solution_19": "thanks everyone, esp. mathfanatic." } { "Tag": [ "algebra", "polynomial", "LaTeX" ], "Problem": "Simply the following multiplication:\r\n$(x+z+y)^-1(x^-1+z^-1+y^-1)(xy+yz+xz)[(xy)^-1+(yz)^-1+(zx)^-1]$\r\n\r\n\r\nSimply the following:\r\n$(\\sqrt[6]{27}-\\sqrt{6\\frac{3}{4}})^2$\r\n\r\nWhen$x^5,x+\\frac{1}{x},$ $and$ $1+\\frac{2}{x}+\\frac{3}{x^2}$\r\nare multiplied, what is the degree of the resulting polynomial?", "Solution_1": "[hide=\"2\"]Square it to get: \n$\\sqrt[3]{27}-2(\\sqrt[6]{27})(\\frac{\\sqrt27}{\\sqrt4})+\\frac{27}{4}$ \n\nLook at the second term. \n\n$2(27^{\\frac{1}{6}})(\\frac{27^{\\frac{1}{2}}}{2})=2(\\frac{27^{\\frac{2}{3}}}{2})=2(\\frac{9}{2})=9$. \n\n$3-9+\\frac{27}{4}=\\frac{3}{4}$? [/hide]", "Solution_2": "1.[hide]\n$\\displaystyle\\frac{1}{x+y+z}(\\frac{1}{x}+\\frac1y+\\frac1z)(xy+yz+xz)(\\frac{1}{xy}+\\frac{1}{yz}+\\frac{1}{xz})$\n$\\displaystyle\\frac{1}{x+y+z}(\\frac{yz+xz+xy}{xyz})(xy+yz+xz)\\frac{xyz^2+x^2yz+y^2xz}{x^2y^2z^2}$\n$\\displaystyle\\frac{1}{x+y+z}(\\frac{yz+xz+xy}{xyz})(xy+yz+xz)\\frac{(xyz)(x+y+z)}{x^2y^2z^2}$\nNow we cancel:\n$\\displaystyle\\frac{(xy+yz+xz)^2}{x^2y^2z^2}\\Rightarrow (\\frac{xy+yz+xz}{xyz})^2$\n[/hide]\n\n2. [hide]\nWe'll start with simplification of the powers:\n$\\displaystyle(\\sqrt[6]{3^3}-\\sqrt{\\frac{3^3}{2^2}}\\Rightarrow (3^{\\frac12}-\\frac{3^{\\frac32}}{2})^2$\n\nLet y be $3^{\\frac12}$. Then we have the equation $\\displaystyle(\\frac{y-y^3}{2})^2\\Rightarrow (\\frac{y(y-1)(y+1)}{2})^2$. Now we expand: $\\displaystyle\\frac{y^2(y^2-1)^2}{4}=\\frac{3(3-1)^2}{4}=\\frac{3(4)}{4}=\\box{3}$\n[/hide]\r\n\r\nBilly", "Solution_3": "You guys made 2 a lot harder than it needed to be...\r\n$\\sqrt[6]{3^3}-\\sqrt{\\frac{3^3}{2^2}}= \\sqrt 3 - \\frac{3\\sqrt{3}}{2} = \\frac{-\\sqrt{3}}{2}$. Square to get $\\frac 3 4$", "Solution_4": "Anyone get 3?", "Solution_5": "Isa thea thirda ona 6?a", "Solution_6": "[quote=\"solafidefarms\"]\n2. [hide]\nWe'll start with simplification of the powers:\n$\\displaystyle(\\sqrt[6]{3^3}-\\sqrt{\\frac{3^3}{2^2}}\\Rightarrow (3^{\\frac12}-\\frac{3^{\\frac32}}{2})^2$\n\nLet y be $3^{\\frac12}$. Then we have the equation $\\displaystyle(\\frac{y-y^3}{2})^2\\Rightarrow (\\frac{y(y-1)(y+1)}{2})^2$. Now we expand: $\\displaystyle\\frac{y^2(y^2-1)^2}{4}=\\frac{3(3-1)^2}{4}=\\frac{3(4)}{4}=\\box{3}$\n[/hide]\n\nBilly[/quote]\r\n\r\nThere is a problem in the section that states $\\displaystyle(\\frac{y-y^3}{2})^2$; it is supposed to be $\\displaystyle(3^{\\frac12}-\\frac{3^{\\frac32}}{2})^2\\Rightarrow (y-\\frac{y^3}{2})^2=(\\frac{2y-y^3}{2})^2$...\r\n\r\nIn any case, this problem is easier than the amount of work you've put into it.\r\n\r\nP.S. I have currently typed my first Latex code by finding a pattern in other people's codes; hence, I may have some typos in there too.\r\n\r\nMasoud Zargar", "Solution_7": "[hide=\"3\"]We want the highest degree term. So $x^5\\cdot x\\cdot 1=x^6$, so the answer is $6$.[/hide]" } { "Tag": [], "Problem": "\u00a31000 is placed in a bank account at the start of each year, beginning in the year 2000.\r\nthe account pays 5% interest per year at the end of each year.\r\nhow much is the account after 15 years?", "Solution_1": "[hide]He'll have 2078.93 in the bank account after 15 years.\n\n1000 (1+5%)^15\n\nInterest= Principal x Rate.\n\nSo after 15 years, he'll have 2078. 93 (a profit of 1078.93!).[/hide]", "Solution_2": "Is it simple interest or compound interest?", "Solution_3": "Assuming compound interest, the way I read your problem is that we have 15 total payments (one at the start of each year), but we valuate one year after the last payment. Giving : [hide]$1000(1.05)^{15}+1000(1.05)^{14}+...+1000(1.05)^{1}=1000(1.05) \\frac{(1.05^{15}-1)}{(1.05-1)}=22657.49$[/hide]", "Solution_4": "what I would do is make a chart...\r\n\r\nif it takes too long..\r\n\r\nI'll choose the longer way! :D \r\n\r\nmake a graph! :P" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "The following two statements are easy to \"sense\" that they're true, but I found it hard to write a formal proof.\r\n\r\n1) Let $f: R^{+}\\rightarrow R^{+}$ be a strictly decreasing function. Show that there exists $x$ such that $f(x)=x$\r\n\r\n2) Let $f: R^{+}\\rightarrow R^{+}$ be a strictly increasing function. Show that $\\lim_{x\\rightarrow a^{-}}f(x)$ and $\\lim_{x\\rightarrow a^{+}}f(x)$ exist for all $a\\in R^{+}$", "Solution_1": "1) I'm assuming it's continuous (it's not implied by strictly decreasing, right?). Consider $g(x) = f(x)-x$. Since $\\lim_{x \\rightarrow-\\infty}g(x) = \\infty$ and $\\lim_{x \\rightarrow \\infty}g(x) =-\\infty$ so by the intermediate value theorem there exists an $x_{0}$ such that $g(x_{0}) = 0 \\Rightarrow f(x_{0}) = x_{0}$.", "Solution_2": "2) Let f be strictly increasing. Fix a in R+. Let xn be a increasing sequence converging to a-. Then f(xn) is an increasing sequence in R+ bounded above by f(a) thus it coverges in R+ to the limit. The limit exists since one sequence implies all sequences will converge, easy in reals.", "Solution_3": "Without the assumption of continuity, #1 is false. Example: $f(x)=\\begin{cases}1-x&x\\le0\\\\-1-x&x>0\\end{cases}$", "Solution_4": "[quote=\"paladin8\"] Since $\\lim_{x \\rightarrow-\\infty}g(x) = \\infty$ [/quote]\r\n\r\nBut I defined $f: R^{+}\\to R^{+}$...", "Solution_5": "[quote=\"Jumbler\"][quote=\"paladin8\"] Since $\\lim_{x \\rightarrow-\\infty}g(x) = \\infty$ [/quote]\n\nBut I defined $f: R^{+}\\to R^{+}$...[/quote]\r\n\r\nOops. Sorry. Then you just have $g(0) = f(0) > 0$ instead." } { "Tag": [], "Problem": "The president of the company has decided to use a new packaging scheme. she wants to have the same number of red and blue candies as are in the original package (10 red, 30 blue) but now she wants to make the same number of white candies 50% of the total number of candies in the package.\r\n\r\n-how many white candies would need to be added to make then 50% of the total mixture?\r\n-Explain\r\n-How many candies would NOW be put in each package?\r\n\r\nTHANK YOU :roll:", "Solution_1": "so right now there are 40 candles, and we want to add $x$ white so that white will be half\r\nfor white to be half, it has to = the rest.\r\nSo since there are 40 right now, we must add 40 white.\r\nSo 40 white candles is half of the package, which means 80 total.", "Solution_2": "[quote=\"helpmex55\"]The president of the company has decided to use a new packaging scheme. she wants to have the same number of red and blue candies as are in the original package (10 red, 30 blue) but now she wants to make the same number of white candies 50% of the total number of candies in the package.\n\n-how many white candies would need to be added to make then 50% of the total mixture?\n-Explain\n-How many candies would NOW be put in each package?\n\nTHANK YOU :roll:[/quote]\r\n\r\n[hide=\"te prueba es muy facil\"]I guessed and checked. Seemed faster that way...\n\nOriginal$=10 R, 30 B$\n\n$a)$ Add $40$ white candies. You get:\n $10 R+30 B+40 W=80 RBW$, where the white candy is 50% of total\n$b)$ It would be a total of 80 candies[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "How do you have braces facing up with text under it like this?\r\n\r\n111...111\r\nThen you have $\\}$ opening up\r\nand under it says 100 1s", "Solution_1": "Do you mean like this $\\underbrace{111...111}_{100}$ which is given by \\underbrace{111...111}_{100}?", "Solution_2": "[quote=\"stevem\"]Do you mean like this $\\underbrace{111...111}_{100}$ which is given by \\underbrace{111...111}_{100}?[/quote]\r\n\r\nYeah, thanks." } { "Tag": [ "geometric series" ], "Problem": "Prove that the sum of an infinite convergent geometrical series of rational numbers is rational.", "Solution_1": "[hide=\"Quick Attempt\"]\nThe sum of an infinite geometric series is $ \\frac {x}{1 \\minus{} r}$. Since the statement says the terms of the series are rational, that makes the first term rational. For the rest of the terms to be rational the rate, $ r$ , must be rational. This makes the [i]fraction[/i] a rational number.\n[/hide]", "Solution_2": "Nice job, though maybe you should explicitly mention closure on the operation of multiplication in the field of rationals. :)", "Solution_3": "Yeah, that would make it a bit more complete. :P" } { "Tag": [ "logarithms" ], "Problem": "solve for x:\r\n\\[{{3\\log_{2}x+8}\\over{2\\log_{2}x+2}}+{{5\\log_{2}x-4}\\over{\\log_{2}x}}= 6\\]", "Solution_1": "[quote=\"akimatsu\"]solve for x:\n\\[{{3\\log_{2}x+8}\\over{2\\log_{2}x+2}}+{{5\\log_{2}x-4}\\over{\\log_{2}x}}= 6 \\]\n[/quote]\r\n[hide]Let's simplify this by eliminating the fractions.\n$\\frac{3\\log_{2}x+8}{2\\log_{2}x+2}+5-4\\log_{x}2=6$\n$3\\log_{2}x+8+10\\log_{2}x+10-8-8\\log_{x}2=12\\log_{2}x+12$\nBy simplyfying and applying the substitution $\\log_{2}x=a$ We get $a-2-\\frac{8}{a}=0$\n$a^{2}-8a-2=0$\n$a=4+3\\sqrt{2}$ \nSo $x=2^{3\\sqrt{2}+4}$[/hide]", "Solution_2": "[quote=\"bpms\"]$a-2-\\frac{8}{a}=0\\implies a^{2}-8a-2=0$[/quote]\r\n\r\nDo recheck this step, please.", "Solution_3": "[quote=\"akimatsu\"]solve for x:\n\\[{{3\\log_{2}x+8}\\over{2\\log_{2}x+2}}+{{5\\log_{2}x-4}\\over{\\log_{2}x}}= 6 \\]\n[/quote]\r\n\r\n[hide]Let $a=\\log_{2}x$\n\n$\\frac{3a+8}{2a+2}+\\frac{5a-4}{a}=6$\n$(3a+8)(a)+(5a-4)(2a+2)=6(a)(2a+2)$\n$13a^{2}+10a-8=12a^{2}+12a$\n$a^{2}-2a-8=0\\implies a=4.-2$\n\nSo $x=16,\\frac14$[/hide]", "Solution_4": "That makes much more sense as teh answer..." } { "Tag": [], "Problem": "Let be $ a>0$ . Solve in the set of the real numbers this system :\r\n\r\n$ \\{\\begin{aligned}(x_1+a)(ax_1+1)&=(a+1)^2x_{2008}\\\\\r\n(x_2+a)(ax_2+1)&=(a+1)^2x_1\\\\\r\n(x_3+a)(ax_3+1)&=(a+1)^2x_2\\\\\r\n...............................................\\\\\r\n(x_{2008}+a)(ax_{2008}+1)=(a+1)^2x_{2007}\r\n\\end{aligned}$", "Solution_1": "[hide=\"Solution\"]\nFrom AM-GM,\n\n$ a(x_k^2 \\plus{} 1)\\ge2ax_k\\implies ax_k^2 \\plus{} a^2x_k \\plus{} x_k \\plus{} a\\ge a^2x_k \\plus{} 2ax_k \\plus{} x_k$\n\n$ \\implies(x_k \\plus{} a)(ax_k \\plus{} 1)\\ge(a \\plus{} 1)^2x_k\\implies\\prod_{k \\equal{} 1}^{2008}(x_k \\plus{} a)(ax_k \\plus{} 1)\\ge\\prod_{k \\equal{} 1}^{2008}(a \\plus{} 1)^2x_k$.\n\nIt is required by multiplying the given equations for\n\n$ \\prod_{k \\equal{} 1}^{2008}(x_k \\plus{} a)(ax_k \\plus{} 1) \\equal{} \\prod_{k \\equal{} 1}^{2008}(a \\plus{} 1)^2x_k$, so by the equality condition of AM-GM,\n\n$ x_k \\equal{} \\boxed1$.\n[/hide]", "Solution_2": "Equality occurs when $ x_k^2\\equal{}1$ so couldn't $ x_k$ be $ \\minus{}1$, too? But nice solution anyways...", "Solution_3": "If you plug in $ x_k\\equal{}\\minus{}1$, it doesn't work. But it's not a valid candidate in the first place because it's really AM-GM on $ x_k$ and $ 1$ ($ (x_k)^2\\plus{}1\\ge2(x_k)(1)$).", "Solution_4": "[quote=\"math154\"]If you plug in $ x_k \\equal{} \\minus{} 1$, it doesn't work. But it's not a valid candidate in the first place because it's really AM-GM on $ x_k$ and $ 1$ ($ (x_k)^2 \\plus{} 1\\ge2(x_k)(1)$).[/quote]\r\nShouldn't it be $ x_k^2 \\plus{} 1\\ge 2\\sqrt {x_k^2*1} \\equal{} 2|x_k|$?\r\nEDIT: hmm actually we require $ x_k\\equal{}|x_k|$ in order to factor that out...", "Solution_5": "Yeah, sorry; I only solved it for nonnegative $ x$. Now I'm not really sure how to solve it...", "Solution_6": "[hide=\"Straightforward\"]Expand the LHS's, sum up all the equations, and denote $ S_2: \\equal{} \\sum x_k^2, S_1: \\equal{} \\sum x_k$ to get\n\n$ aS_2 \\plus{} (a^2 \\plus{} 1)S_1 \\plus{} 2008a \\equal{} (a \\plus{} 1)^2S_1\\iff S_2 \\minus{} 2S_1 \\plus{} 2008 \\equal{} 0$\n\nHowever, $ S_2 \\minus{} 2S_1 \\plus{} 2008 \\equal{} \\sum(x_k \\minus{} 1)^2$, hence the only solution is $ x_1 \\equal{} x_2 \\equal{} \\dots \\equal{} x_{2008} \\equal{} 1$\n\nIt actually works for $ a<0$ too, hence $ a\\neq 0$ is sufficient.[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$x,y,z$ are positive prove that \r\n$x.(x-z)^2+y(y-z)^2\\ge (x-z)(y-z)(x+y-z)$", "Solution_1": ":rotfl:\r\n\r\n$x\\left(x-z\\right)^2+y\\left(y-z\\right)^2-\\left(x-z\\right)\\left(y-z\\right)\\left(x+y-z\\right)$\r\n$=x^3+y^3+z^3+3xyz-y^2z-z^2y-z^2x-x^2z-x^2y-y^2x\\geq 0$\r\n\r\nby the Schur inequality; thus,\r\n\r\n$x\\left(x-z\\right)^2+y\\left(y-z\\right)^2\\geq\\left(x-z\\right)\\left(y-z\\right)\\left(x+y-z\\right)$.\r\n\r\n Darij", "Solution_2": "i wrote that it is easy inequality :D ;)" } { "Tag": [], "Problem": "Charlene's calculator displays the digits 0, 1, 6, 8 and 9 so that, when the calculator is held upside-down, these digits appear to be 0, 1, 9, 8 and 6 respectively. How many three-digit numbers look the same upside-down and right-side-up when using these five possible digits? (A three-digit string that begins with either 0 or 00 is not considered a three-digit number.)", "Solution_1": "The numbers must be of the form $ \\overline{aba}$, where $ a,b\\neq6,9$ and $ a\\neq0$, or $ \\overline{6a9}$ or $ \\overline{9a6}$, where $ a\\neq6,9$, for a total of $ 2\\cdot3 + 2\\cdot3 = \\boxed{12}$.", "Solution_2": "Why not 6 or 9?", "Solution_3": "9 and 6 look different when upside-down.", "Solution_4": "$\\text{This was not a good problem.}$" } { "Tag": [ "rotation", "combinatorial geometry", "geometry solved", "geometry" ], "Problem": "In the plane, n circles of unit radius are drawn with different centers. Of course, overlapping circles partly cover each other's circumferences. A given circle could be so overlaid that any uncovered parts of its circumference are all quite small; that is, it might have no sizable uncovered arcs at all. However, this can't be true of every circle; prove that some circle must have a continuously uncovered arc which is at least 1/n th of its cirumference.", "Solution_1": "Clearly, we have to assume that the circles are pairwise distincts...\r\nSo let C be the convex hull of the set of the n circles. It is the union of segments and circular arcs from the circles. As we rotate to 360 when travelling along the boundary of C, it follows that the total length of the circular arcs used in C is 2 \\pi . Since there are n circles, there are no more than n circular arcs so that the average length of these circular arcs is at least 2 \\pi /n, and we are done.\r\n\r\nPierre." } { "Tag": [], "Problem": "If $ [a,b][c,d] \\equal{} [0,0]$, prove that $ ac\\minus{}bd \\equal{} 0$ and $ bc\\plus{}ad\\equal{}0$, and hence that $ (a^{2}\\plus{}b^{2})(c^{2}\\plus{}d^{2})\\equal{}0$. Deduce that either $ [a,b] \\equal{} [0,0]$ or $ [c,d]\\equal{}[0,0]$\r\n\r\nHow to do the second part and third part?", "Solution_1": "Hmm, do you mean $ [a,b] \\equal{} a\\plus{}bi$? [hide=\"Then\"] $ 0 \\equal{} [0,0] \\equal{} [a,b][c,d] \\equal{} (a\\plus{}bi)(c\\plus{}di) \\equal{} (ac\\minus{}bd)\\plus{}(ad\\plus{}bc)i$. So $ ac\\minus{}bd \\equal{}\\Re([a,b][c,d]) \\equal{} 0$ and $ ad\\plus{}bc \\equal{}\\Im([a,b][c,d]) \\equal{} 0$. Now one remarks that $ (a^{2}\\plus{}b^{2})(c^{2}\\plus{}d^{2}) \\equal{} (ac\\minus{}bd)^{2}\\plus{}(ad\\plus{}bc)^{2}\\equal{} 0$. And this means $ 0 \\equal{} |a\\plus{}bi|\\cdot|c\\plus{}di|$, so either $ |a\\plus{}bi| \\equal{} 0$, i.e. $ [a,b] \\equal{} [0,0]$ or $ |c\\plus{}di| \\equal{} 0$, i.e. $ [c,d] \\equal{} [0,0]$.[/hide]", "Solution_2": "ok thanks i understand now\r\n\r\nyes thats what i meant $ [a,b] \\equal{}\\Rightarrow a\\plus{}bi$" } { "Tag": [ "trigonometry", "inequalities", "LaTeX", "geometry", "geometric transformation", "inequalities proposed" ], "Problem": "Let ABC be a triangle .Prove that\r\n $\\frac{1]{cos\\frac{B-C}{2}cos\\frac{C-A}{2}cos\\frac{A-B}{2}} + 1 \\geq \\frac{3}{cosA+cosB+cosC}$\r\n I posted this problem a long time ago, but finally it wasn't completely proved. \r\nAt that time, my inequality was for acute triangle.tranminhhoang, who is my friend, did tell me this one was also right for any triangle. So can anyone help me?\r\nAny solution you can, even if very very \"buffalic\" :D", "Solution_1": "1/{cos(B-C)/2cos(C-A)/2cos(B-C)/2} +1 >= 3/{cosA+cosB+cosC}.\r\n I 'm afraid you can't see my latex. :?", "Solution_2": "Here is the LaTeX:\r\n\r\n[quote=\"treegoner\"]Let ABC be a triangle. Prove that\n$\\displaystyle \\frac{1}{\\cos\\frac{B-C}{2}\\cos\\frac{C-A}{2}\\cos\\frac{A-B}{2}} + 1 \\geq \\frac{3}{\\cos A+\\cos B+\\cos C}$[/quote]\n\nYou can always go back to edit your original post instead of double posting.\n\nOK, I did a translation, no idea whether it's helpful or not.\n\n\\[\n\\frac{(abc)^2}{(a+b)(b+c)(c+a)(s-a)(s-b)(s-c)} + 1 \\geq \\frac{3abc}{abc + 4(s-a)(s-b)(s-c)} \n\\]\n\nLet $x=s-a ,y =s-b, z=s-c$, then $a=y+z, b= z+x, c=x+y$, and it's translated into:\n[quote] If $x,y,z>0$, then\n\\[\n\\frac{[(x+y)(y+z)(z+x)]^2}{(x+y+2z)(x+2y+z)(2x+y+z)xyz} + 1 \\geq \\frac{3(x+y)(y+z)(z+x)}{(x+y)(y+z)(z+x) + 4xyz}.\n\\]\n[/quote]", "Solution_3": "[quote=\"treegoner\"]Let ABC be a triangle. Prove that\n$\\displaystyle \\frac{1}{\\cos\\frac{B-C}{2}\\cos\\frac{C-A}{2}\\cos\\frac{A-B}{2}} + 1 \\geq \\frac{3}{\\cos A+\\cos B+\\cos C}$[/quote]\r\nWe can easily prove these two equalities \r\n${\\\\cos\\frac{B-C}{2}\\cos\\frac{C-A}{2}\\cos\\frac{A-B}{2}} = \\frac{p^2 +2Rr + r^2}{8R^2} $\r\n$\\cos A+\\cos B+\\cos C = 1 + \\frac{r}{R} $\r\nSo treegoner 's inequality is equivalent to this one \r\n$\\displaystyle \\frac{8R^2(R+r)}{2R-r} -r^2-2Rr \\geq \\ p^2 $\r\n\r\nBecause treegoner did prove for acute triangles, now I prove for obstute triangles.\r\n[b]Lemma[/b] : for obstute triangles we have $\\ p \\leq\\ 2R+r $\r\n\r\nApply this lemma, all we have to do now is to prove the inequality\r\n$\\displaystyle \\frac{8R^2(R+r)}{2R-r} -r^2-2Rr \\geq \\ (2R+r)^2 $\r\n\r\nBut this one is equivalent to $\\ 4R^2 \\geq\\ 4R^2 - r^2 $ (obviously true)\r\nSo treegoner 's inequality is true for all triangles. \r\nAlso use this method we can prove for acute triangles but we must use another lemma which is rather complicated. Therefore, treegoner should post his solution for this case. \r\nAfter all, I hope my solution is not very buffalic :D", "Solution_4": "Hey, but I don't even know how to prove the inequality for acute triangle.\r\n Or I should try proving it one more time. Maybe I have luck this time. :) \r\n But can you send your solution in this topic, tranminhhoang ? :cool:", "Solution_5": "Hm treegoner, I really do not want to post such a complex solution. \r\n\r\nThe lemma which I talked about is that\r\n[b]Lemma[/b] For all triangles we have\r\n$ p^2 \\leq \\ 2R^2 + 10Rr - r^2 +2(R-2r)\\sqrt{R(R-2r)} $\r\n\r\nUse this lemma we have to prove\r\n$\\displaystyle \\frac{8R^2(R+r)}{2R-r} -r^2-2Rr \\geq \\ 2R^2 + 10Rr - r^2 +2(R-2r)\\sqrt{R(R-2r)} $\r\n\r\nDenote $ t= \\frac{R}{r} $ the inequality becomes\r\n$ (t-2)t(2t-3) \\geq \\ (t-2)(2t-1)\\sqrt{t(t-2)} $\r\n\r\nThis one is easy because $ t \\geq \\ 2 $\r\n\r\nThe lemma is very strong and the proof is too long for me to write. So... I must stop here :D", "Solution_6": "Don't worry, tranminhhoang. I know this lemma. Nice solution. :)" } { "Tag": [ "quadratics", "function", "floor function", "algebra", "algebra unsolved" ], "Problem": "Let $ f(x)\\equal{}a_1 x^2 \\plus{}b_1 x \\plus{} c_1$ $ g(x)\\equal{}a_2 x^2 \\plus{}b_2x \\plus{}c_2$ , $ a_1 \\neq 0$ ,$ a_2 \\neq 0$ quadratic functions such that $ [f(x)]\\equal{}[g(x)]$ ,$ x \\in R$.Prove that $ f(x)\\equal{}g(x)$ \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n___________________________\r\nAzerbaijan Land of the Fire :ninja: :roll:", "Solution_1": "do you mean $ \\lfloor f(x) \\rfloor \\equal{} \\lfloor g(x) \\rfloor$?", "Solution_2": "where $ x\\leq [x] < x\\plus{}1$ :ninja:", "Solution_3": "coefficients are integer,real?....", "Solution_4": "[quote=\"Pirkuliyev Rovsen\"]where $ x\\leq [x] < x \\plus{} 1$ :ninja:[/quote]\r\n\r\nthis isn't well-defined. I think what you mean to say is greatest integer function, i.e. floor function.", "Solution_5": "coefficients are real !!!", "Solution_6": "[quote=\"Pirkuliyev Rovsen\"]coefficients are real !!![/quote]\r\n\r\nyes, and?", "Solution_7": "We may assume $ a_1,a_2>0$\r\n\r\nThen $ f,g \\rightarrow \\infty$ as $ x\\rightarrow \\infty$.\r\n\r\nHence f(x) and g(x) take on infinitely many integer values i.e. f(x) = g(x) at infinitely many values of x\r\n\r\nSince $ f(x)\\minus{}g(x)$ is of degree utmost 2 we have $ f(x)\\minus{}g(x) \\equiv 0$.\r\n\r\nHence $ f(x) \\equal{} g(x)$", "Solution_8": "It is good :D :lol:" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that for all natural n, $\\frac{1}{2n}<$ {$n\\sqrt{7}$} $<1-\\frac{1}{6n}$", "Solution_1": "Is it hard? or just very easy.i cannot solve it :(", "Solution_2": "One part is solved here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=74032[/url]" } { "Tag": [], "Problem": "Prove that the equation\r\n\r\n$x^{2}+x+1=py$ has integer solutions $(x,y)$ for infinitely many primes $p$.\r\n\r\nHere's a hint\r\n\r\n[hide]Another interpretation of the problem is that inifnitely number of primes occur for among the factors for numbers in the form of $x^{2}+x+1$.[/hide]", "Solution_1": "[hide]\nSuppose there are only finite $p$ such that $p|(x^{2}+x+1)$ for some integer $x$. Let the set of all such $p$ be $S = \\{p_{1}, p_{2}, \\ldots, p_{k}\\}$. Then take\n\n$x = p_{1}p_{2}\\cdots p_{k}\\Rightarrow x^{2}+x+1 = p_{1}p_{2}\\cdots p_{k}(p_{1}p_{2}\\cdots p_{k}+1)+1$.\n\nThis, however, means that $x^{2}+x+1$ is not divisible by $p_{1}, p_{2}, \\ldots, p_{k}$ which implies that it is divisible by some other prime $p_{k+1}$, contradicting the fact that $S$ was the set of all primes dividing $x^{2}+x+1$ for any integer $x$. Hence $S$ is infinite. [/hide]" } { "Tag": [ "calculus", "integration", "inequalities", "symmetry" ], "Problem": "1)Find the number of $non \\ negative$ integral solution to:-\r\n\r\n$2x+y+z \\leq 19$\r\n\r\n2)Find the number of $non \\ negative$ integral solution to:-\r\n\r\n$x+y+z=18$,where $x0,x-1<0$, $2x+2=x+1$, $x=-1$\n\n$x+3>0,x-1>0$, $x=3$\n\n$x+3<0,x-1<0$, $x=-5$\n\n$x+3=0$ and $x-1=0$ don't work, $x=\\boxed{-1,3,-5}$[/hide]", "Solution_2": "This is trivial. Just consider 3 cases, and you to get 3 solutions.\r\n\r\nEDIT: The above is the correct solution." } { "Tag": [ "analytic geometry", "geometry", "geometric transformation", "reflection", "parallelogram", "geometry proposed" ], "Problem": "[b][size=100][color=DarkBlue]Let $ \\bigtriangleup A'BC,\\ \\bigtriangleup B'AC,\\ \\bigtriangleup C'AB$ be, three similar isosceles triangles with bases the side-segments $ BC,\\ AC,\\ AB$ respectively of a given triangle $ \\bigtriangleup ABC,$ erected outwardly to it. Prove that the line segments $ AA',\\ BB'\\ CC',$ where $ A',\\ B',\\ C'$ are the midpoints of the segments $ EF,\\ DF,\\ DE$ respectively, are concurrent at one point.[/color][/size][/b]\r\n\r\nKostas Vittas.", "Solution_1": "Let $ \\phi$ be the angle base of these isosceles triangles. Using Conway's formula, the barycentric coordinates of $D,E,F$ are:\n\n$ D \\equiv ( \\minus{} a^2: S_C \\plus{} S_{\\phi}: S_B \\plus{} S_{\\phi})$\n\n$ E \\equiv (S_C \\plus{} S_{\\phi}: \\minus{} b^2: S_A \\plus{} S_{\\phi})$\n\n$ F \\equiv (S_B \\plus{} S_{\\phi}: S_A \\plus{} S_{\\phi}: \\minus{} c^2)$\n\nSince the sum of the coordinates of $D,E,F$ equals $ 2S_{\\phi},$ we compute easily the coordinates of the midpoints of segments $DE,EF,FD$ as:\n\n$ A' \\equiv (a^2 \\plus{} 2S_{\\phi}: S_{\\phi} \\minus{} S_C: S_{\\phi} \\minus{} S_B)$\n\n$ B' \\equiv (S_{\\phi} \\minus{} S_C: 2S_{\\phi} \\plus{} b^2: S_{\\phi} \\minus{} S_A)$\n\n$ C' \\equiv (S_{\\phi} \\minus{} S_B: S_{\\phi} \\minus{} S_A: 2S_{\\phi} \\plus{} c^2)$ \n\nTherefore, $ AA',BB',CC'$ concur at the inner Kiepert perspector relative to $\\phi$\n\n$K \\equiv \\left (\\frac {1}{ S_A \\minus{} S_{\\phi}}: \\frac {1}{ S_B \\minus{} S_{\\phi}}: \\frac {1}{ S_C \\minus{} S_{\\phi}} \\right )$", "Solution_2": "Reflect $\\triangle BCD$ about sideline $ BC$ into $\\triangle BCD'.$ Notice that $ \\angle FBD' \\equal{} \\angle ABC,$ since $ \\angle FBD'$ is obtained by rotating $\\angle ABC$ about $ B$ with angle +$\\phi.$ Further, $ \\triangle BFD'$ and $ \\triangle BAC$ are similar with similarity coefficient $\\frac{_{ BF}}{^{BA}}.$ But $ \\frac {_{BF}}{^{BA}} \\equal{} \\frac {_{AE}}{^{AC}}$ $ \\Longrightarrow$ $ FD' \\equal{} AE$ and analogously $ ED' \\equal{} AF,$ which implies that $ AFD'E$ is a parallelogram $\\Longrightarrow$ Line $ AA'$ goes through $ D'.$ By similar reasoning, we conclude that $ AA',BB',CC'$ concur at the inner Kiepert perspector relative to $\\phi.$" } { "Tag": [ "geometry", "geometric transformation", "reflection", "analytic geometry", "perimeter", "perpendicular bisector", "puzzles" ], "Problem": "say you have Rx=0: (x,y)--> (-x,y) \r\nand Ry=-x: (x,y)--> (-y,-x) \r\n\r\nwhat is the composition of the above two reflections: \r\nRy=-x o Rx=0. \r\nWhat single transformation will accomplish the same thing as the composition of these two reflection?", "Solution_1": "Rx=0: (x,y)--> (-x,y) \r\nThis means that the reflection of the point (x,y) over line x=0 (i.e. y axis) is (-x,y). \r\n\r\nRy=-x: (x,y)--> (-y,-x) \r\nThis means that the reflection of the point (x,y) over line y=-x is (-y,-x). \r\n\r\nThe composition of the above two reflections: Ry=-x o Rx=0 is (x,y)--> (-y,x) \r\n\r\nThat's all I could do.", "Solution_2": "This isn't so bad if you think about it.\r\n\r\nLet $R_1: (x,y)\\longrightarrow(-x,y)$,\r\nand let $R_2: (x,y)\\longrightarrow(-x,-y)$\r\n\r\nThe composition $R_1+R_2$ Will map $(x,y)\\longrightarrow(-x,y)\\longrightarrow(x,-y)$\r\n\r\nRemember that if $P$ reflects across a line to $Q$, then the reflection line is the perpendicular bisector of $PQ$ (draw the diagram in your head to convince yourself). So we are looking for the perpendicular bisector of the line that goes through $(x,y)$ and $(x,-y)$.\r\n\r\nThe midpoint of this line is quite obviously $(x,0)$. It is also easy to notice that since they have the same $x$-coordinates, they two points are on the same vertical line, meaning that their perpendicular bisector must be horizontal. The horizontal line that goes through $(x,0)$ is $f(x)=0$.\r\n\r\nAlso note that the reflections can be done in either order. $R_1+R_2=R_2+R_1$.\r\n\r\n\r\nHere's an problem I saw recently regarding reflections:\r\nYou have a point $P$ on the line $f(x)=x$, and a point $Q$ on the line $f(x)=0$. You have another point $R$ inbetween the these two lines. $P$, $Q$, and $R$ are all in the first quadrant. Defining $R$ as $(a,b)$, find points $P$ and $Q$ in terms of $a$ and $b$ such that the perimeter of triangle $\\Delta PQR$ is a minimum.", "Solution_3": "[quote=\"gopherhole112\"]Also note that the reflections can be done in either order. $R_1+R_2=R_2+R_1$.[/quote]\r\n\r\nBe careful, though -- this isn't a general truth, but works in this case just because the lines of reflection are perpendicular to each other." } { "Tag": [ "geometry", "3D geometry", "pyramid", "Pythagorean Theorem" ], "Problem": "Two regular square pyramids have all edges 12cm in length. The pyramids have parallel bases and parallel edges, and each has a vertex at the center of the other pyramid's base. What is the total number of cubic centimeters in the volume of the solid of intersection of the two pyramids? Express your answer in simplest radical form.", "Solution_1": "If you \"chop\" this combined figure in half, then there will be a tiny pyraimd inside of a frustrum. The base of the tiny pyramid is 1/4 the area of one of the original pyramids, while the height is 1/2 the original pyramid. THerefore, 1/2 of the intersection thing is 1/8 of one pyramid. So the entire intersection is 2 * 1/8 = 1/4 of the original pyramid's volume.\r\n\r\nUsing Pythagorean Theorem and stuff, we get the height of one pyramid is $ 6\\sqrt{2}$. (I can't really show how, but it's not too hard) Therefore, the entire pyramid's volume is $ \\frac{(12)^2\\cdot 6\\sqrt{2}}{3}\\equal{}288\\sqrt{3}$. Thus, the volume of the intersection is $ 72\\sqrt{3}$", "Solution_2": "[quote=\"dragon96\"]If you \"chop\" this combined figure in half, then there will be a tiny pyraimd inside of a frustrum. The base of the tiny pyramid is 1/4 the area of one of the original pyramids, while the height is 1/2 the original pyramid. THerefore, 1/2 of the intersection thing is 1/8 of one pyramid. So the entire intersection is 2 * 1/8 = 1/4 of the original pyramid's volume.\n\nUsing Pythagorean Theorem and stuff, we get the height of one pyramid is $ 6\\sqrt{2}$. (I can't really show how, but it's not too hard) Therefore, the entire pyramid's volume is $ \\frac{(12)^2\\cdot 6\\sqrt{2}}{3}\\equal{}288\\sqrt{3}$. Thus, the volume of the intersection is $ 72\\sqrt{3}$[/quote]\n\nHow to find the intersection, any picture or so?\nI don't see why you have $\\sqrt{3}$ from the first thing", "Solution_3": "Try to picture two pyramids stacked on top of each other, just that one pyramid's vertex is at the other's center.", "Solution_4": "[quote=\"professordad\"]Try to picture two pyramids stacked on top of each other, just that one pyramid's vertex is at the other's center.[/quote]\n\nIs the correct answer $72\\sqrt{2}$ than?\n( I got that if I understand your explaining)", "Solution_5": "yes, the answers say that 72 \\sqrt 2 is the answer." } { "Tag": [ "LaTeX", "linear algebra", "matrix" ], "Problem": "I don't know how to write matrices in latex, sorry, but let \r\n\r\n\"B = Matrix([[1, 3, 4, 1, 3], [3, 9, 8, 6, 5], [5, 8, 7, 5, 2], [7, 7, 6, 3, 9], [9, 6, 5, 7, 7]])\" (Maple notation).\r\n\r\nGiven that 13413, 39865, 58752, 77639, 96577 are all divisible by 17, show that the determinant of B is divisble by 17. As a hint it is given that $13413 = 1\\cdot 10^4 + 3\\cdot10^3 + 4\\cdot10^2 + 1\\cdot10 + 3$.", "Solution_1": "Replace the fifth column $C_5$ with $10^4C_1+10^3C_2+10^2C_3+10C_4+C_5$. This doesn't change the determinant, and the new fifth column is a multiple of 17." } { "Tag": [ "geometry" ], "Problem": "I'm going to india in a few days and I want to buy a few books from there. Anywhere from algebra, geometry, etc. to analysis and linear algebra. Any recommendations? Thin, and light books are very welcome since I can convince my mom easier that they won't be a hassle on the plane :)", "Solution_1": "Well [b]Challenges and Thrills in Pre - College Mathematics[/b] is a very nice book. The Authors are B.J.Venkatachala, K.N.Ranganathan, C.R.Pranesachar.\r\nAlso [b]An Excursion In Mathematics[/b] is very good.\r\n[b]Problem Primer For the Olympiad[/b] is also a nice book." } { "Tag": [ "analytic geometry", "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $E$ be a vectorial space with dimension $n$ and $\\{v_1,\\ldots,v_n\\}$ its base, and let $a_1,\\ldots,a_n$ be real numbers, not all zero. Take \r\n\r\n$F=\\{x=x_1v_1+\\ldots+x_nv_n \\in E|a_1x_1+\\ldots+a_nx_n=0\\}$\r\n\r\nHow can I show that the dimension of $F$ is $n-1$?\r\n\r\nthanks", "Solution_1": "Intuitively, since the $a_{k}$ are not all $0$ and because of the definition of $F$, each element of $F$ has one of its coordinates entirely determined by its $n-1$ other coordinates.", "Solution_2": "I feel retarded i couldn't see that. I thought of it as a given vector $(a_1,\\ldots,a_n)$, and F as the subspace containing all perpendicular vectors to $(a_1,\\ldots,a_n)$. For exemple, in R^3, it should be a plane passing through the origin containing the vectors perpendicular to $(a_1,a_2,a_3)$, and I knew that it had to loose one dimension...too geometric, less algebric :P" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Given $ a, b, c > 0$ and $ n$ is an integer satisfy $ n \\ge 4$. Prove that:\r\n\\[ \\frac{ab^{n\\minus{}1}}{c^n}\\plus{}\\frac{bc^{n\\minus{}1}}{a^n}\\plus{}\\frac{ca^{n\\minus{}1}}{b^n} \\ge \\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{a}\\]\r\n :)", "Solution_1": "[quote=\"nguoivn\"]Given $ a, b, c > 0$ and $ n$ is an integer satisfy $ n \\ge 4$. Prove that:\n\\[ \\frac {ab^{n \\minus{} 1}}{c^n} \\plus{} \\frac {bc^{n \\minus{} 1}}{a^n} \\plus{} \\frac {ca^{n \\minus{} 1}}{b^n} \\ge \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a}\\]\n:)[/quote]\r\n\r\nwe have \r\n\r\n$ \\sum_{cyc}\\left(\\frac {ab^{n \\minus{} 1}}{c^n} \\plus{} \\frac {c}{a}\\right)\\ge 2\\sum_{cyc}\\left(\\frac {b^{\\frac {n \\minus{} 1}{2}}}{c^{\\frac {n \\minus{} 1}{2}}}\\right)\\ge \\frac {2}{3}(\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a})(\\sum_{cyc}\\left(\\frac {b^{\\frac {n \\minus{} 1}{2} \\minus{} 1}}{c^{\\frac {n \\minus{} 1}{2} \\minus{} 1}}\\right) \\ge 2 \\text{RHS}$ \r\n\r\nDone", "Solution_2": "or we can use\r\n\r\n$ \\left(\\frac {ab^{n \\minus{} 1}}{c^n} \\plus{}\\frac{c}{a} \\plus{}n \\minus{} 3 \\plus{} \\frac {b}{c}\\right)\\ge n.\\frac {b}{c}$" } { "Tag": [ "number theory", "relatively prime", "prime numbers" ], "Problem": "The largest prime is said to have been found. But how can it exist?\r\n\r\n\r\nLet P1, P2,.....,Pn be the set of all the primes.\r\n\r\nThen (P1.P2....Pn)+1 will not be divisible by any of the primes.\r\n\r\nThis means that this itself is prime.\r\n\r\nBut this is greater than Pn.\r\n\r\nThus, we reach a contradiction.", "Solution_1": "Where did you hear that? In all likelihood, whatever source you read meant the largest [i]known[/i] prime.", "Solution_2": "From Elementary Number Theory by William Stein\r\n[quote]A Mersenne prime is a prime of the form $ 2q \\minus{} 1$. The largest known prime as of July $ 2004$ is the Mersenne prime $ p \\equal{} 2^{24036583} \\minus{} 1$, which has $ 7235733$ decimal digits.[/quote]", "Solution_3": "If you want to know how big can be such a number ,you can see that in this link : [url]http://www.math.utah.edu/~pa/math/largeprime.html[/url]", "Solution_4": "Impressive :D \r\nIs there anyone can memorize it :P", "Solution_5": "Because there is four topics about the same thing \"JRav\" seems not to notice this topic so I would like to group everything in one place.\r\n[quote=\"JRav\"]Indeed, there are an infinite number of primes. What I'm going to guess you're referring to is that some group of researchers have found a prime larger than all others previously discovered. It's not the largest prime (there is none), just the largest found so far.[/quote]", "Solution_6": "There is no such thing as a largest prime. Your proof that there are infinitely many primes is completely rigorous, and rigorous proofs tend to hold true...", "Solution_7": "As many of the people here have replied, you probably misread whatever you read and it actually said \"largest known prime.\"", "Solution_8": "It might be best to note that in Euclid's proof of the infinitude of the primes, the number (p0*p1*...*pn-1*pn) + 1, while guaranteed to be relatively prime to all numbers p0, p1 ..., pn, may not itself be prime.\r\n\r\nAs another proof of the infinitude of the primes, Furstenberg's illustrates the many ways in which it can be shown that the primes have no limit. \r\n\r\nhttp://primes.utm.edu/notes/proofs/infinite/topproof.html\r\n\r\nHeuristics arguments also hold in showing the infinitude of the primes (as well as the number of primes less than n).", "Solution_9": "[quote=\"cstan\"]It might be best to note that in Euclid's proof of the infinitude of the primes, the number (p0*p1*...*pn-1*pn) + 1, while guaranteed to be relatively prime to all numbers p0, p1 ..., pn, may not itself be prime.[/quote]\r\nThat's completely a matter of how you phrase the contradiction. [i]In the proof[/i], the contradiction is that $ p_0 p_1 ... p_n \\plus{} 1$ is not divisible by any smaller prime (by assumption), which is the definition of primality. On the other hand, it also can't itself be prime (by assumption). [i]Outside the proof[/i], either of these conclusions may or may not hold.", "Solution_10": "Just continuing from the other posts above, you can alternatively prove that for any $n$, there necessarily exists a prime $p$ such that $n0?$ Does it converge uniformly on $ [0,b]?$\r\n\r\nThe second question has a cheap answer: since each $ I_n$ is a continuous function but the limit is not continuous at $ 0,$ the convergence cannot be uniform on any interval containing zero.\r\n\r\nalekk's suggestion would give us\r\n\r\n$ I_n(x)\\equal{}\\int_0^{nt}\\frac{\\sin u}{u}\\,du.$ Integrate this by parts to get\r\n\r\n$ I_n(x)\\equal{}\\frac{1\\minus{}\\cos nt}{nt}\\plus{}\\int_0^{nt}\\frac{1\\minus{}\\cos u}{u^2}\\,du.$\r\n\r\nExamine that for uniform convergence or the lack thereof." } { "Tag": [ "function", "logarithms", "inequalities proposed", "inequalities" ], "Problem": "[b]PRO-1[/b]: Assume $0b^ac^bd^ca^d$ \r\n[b]PRO-2[/b]: Assume $a>b>c>d>e$ , show that:$ a^{e^b}\\cdot b^{e^c}\\cdot c^{e^d}\\cdot d^{e^a}\\ge b^{e^a}\\cdot c^{e^b}\\cdot d^{e^c}\\cdot a^{e^d}$ \r\nNOTE: e such that: $\\ln{e}=1$ :D ", "Solution_1": "It's solved on M&Y 1/05 :)", "Solution_2": "[quote=\"nttu\"]It's solved on M&Y 1/05 :)[/quote]\r\n\r\nyes,but only PRO-1 :D , and PRO-2 is mine.okie?" } { "Tag": [ "trigonometry", "geometry", "circumcircle", "angle bisector", "geometry proposed" ], "Problem": "In a triangle $\\triangle ABC$ prove that the internal angle bisector of $A$ passes through the center of the 9-point circle if and only if\r\n\r\n\\[\r\n\\left( {2\\cos A - 1} \\right)\\left( {\\cos B - \\cos C} \\right) = 0\r\n\\]", "Solution_1": "Let us show that the internal angle bisector passes through the center of the nine point circle iff $AB=AC$ or $\\angle{A}=60^{\\circ}$. Clearly, it solves the problem.\r\n\r\nLet ABC be an arbitrary triangle, H be its orthocenter, O be the center of the circumcircle, N be the center of the nine point circle.\r\n\r\nIt's obvious that if $AB=AC$ then N lies at the angle bisector of $A$. Let's prove the converse.\r\n\r\nThere are two well-known facts.\r\n\r\n(1) The center of the nine point circle is a midpoint of the segment OH.\r\n\r\n(2) The points O and H are isogonal conjugate.\r\n\r\nSuppose that N lies at the internal angle bisector of A. From (2) we immediately obtain that AN is an angle bisector in the triangle OAH. And from (1) we obtain that AN is a median in the triangle OAH. And from the last two facts we have that the triangle OAH is isosceles (or the points O, H and A are collinear, but then $AB=AC$ and we are done). It means that $AO=AH$. But AO is a circumradius and AH is a circumdiameter of the triangle $H_BAH_C$ (where $H_B$ and $H_C$ are the feet of altitudes of the original triangle). But it is well-known that the triangles $BAC$ and $H_CAH_B$ are similar with coefficient $\\cos{\\alpha}$, where $\\alpha=\\angle{BAC}$. And hence $AH=2AO \\cos{\\alpha}$. Since $AO=AH$, we have $\\cos{\\alpha}=1/2$. It can be true only if $\\alpha=60^{\\circ}$. So, we are done." } { "Tag": [ "trigonometry", "search", "complex numbers" ], "Problem": "A friend of mine came up with this question. We solved it different ways and got different answers. Any help?\r\n$ \\displaystyle\\sum^\\infty_{n\\equal{}1}\\dfrac{\\cos{(n\\cdot x)}}{3^n}$\r\n(Where cos is in radians obviously)", "Solution_1": "Things very much like this have popped up on the forum on several occasions, but I'm not sure what a good search would be to try. The easiest solution method starts by noting that [hide]$ \\cos t \\equal{} \\textrm{Re}(e^{i t})$.[/hide]", "Solution_2": "[hide=\"Solution\"]\n$ \\sum_{n\\equal{}1}^{\\infty}\\frac{\\cos nx}{3^n}\\equal{}\\text{Re}\\sum_{n\\equal{}1}^{\\infty}\\left(\\frac{e^{ix}}{3}\\right)^n$\n\nThe sum is $ \\frac{e^{ix}{3}}{1\\minus{}\\frac{e^{ix}}{3}}\\equal{}\\frac{e^{ix}}{3\\minus{}e^{ix}}\\equal{}\\frac{\\cos x\\plus{}i\\sin x}{3\\minus{}\\cos x\\minus{}i\\sin x}$.\n\n$ \\frac{\\cos x\\plus{}i\\sin x}{3\\minus{}\\cos x\\minus{}i\\sin x}\\cdot\\frac{3\\minus{}\\cos x\\plus{}i\\sin x}{3\\minus{}\\cos x\\plus{}i\\sin x}\\equal{}\\frac{3\\cos x\\plus{}3i\\sin x\\minus{}1}{10\\minus{}6\\cos x}$\n\nThe real part of that is, of course, $ \\boxed{\\frac{3\\cos x\\minus{}1}{10\\minus{}6\\cos x}}$.\n[/hide]", "Solution_3": "Oops Im sorry, I didn't think to search before I posted.\r\nThanks for the solution!", "Solution_4": "[quote=\"grn_trtle\"]\n\nThe sum is $ \\frac {e^{ix}{3}}{1 \\minus{} \\frac {e^{ix}}{3}}$\n[/quote]\r\n\r\nThis should be $ \\frac {\\frac{e^{ix}}{3}}{1 \\minus{} \\frac {e^{ix}}{3}}$." } { "Tag": [ "function", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Hello!\r\n\r\n\r\nEdit:\r\n\r\nI'll repost the problem:\r\n\r\nIf $f(x,y)$ is a continuous function in $(x,y)$ and is defined on $M=C_{1}\\times C_{2}$, where $C_{1}$ and $C_{2}$ are compact sets in the euclidean spaces $E^{m}$ and $E^{n}$, then $\\Psi(x)$ is a continuous function in $x$, where $\\Psi (x): = \\mathop{\\max }_{y \\in C_{2}}f(x,y)$\r\n\r\n/Edit\r\n\r\n\r\nWe have $f(x,y)$, $f$ is affine\r\n\r\nFurthermore is f continuous on a compact (even convex) set $M$\r\n\r\n$M: = C_{1}\\times C_{2},x \\in C_{1},y \\in C_{2}$\r\n\r\nThe mapping $x \\mapsto f(x,y)$ is affine\r\n\r\nLet $\\Psi (x): = \\mathop{\\max }_{y}f(x,y)$\r\n\r\n (The maximum of $f(x,y)$ upon $y$)\r\n\r\nHow do I show that $\\Psi (x)$ is continuous on $M$?\r\n\r\n$\\Pi\\Delta\\Phi$", "Solution_1": "$C_{1}$ and $C_{2}$ are subsets of what? $\\mathbb{R}?$ $\\mathbb{R}^{m}$ and $\\mathbb{R}^{n}?$ Some Banach space?", "Solution_2": "I will disregard convexity and linearity, and only assume that $C_{1}$ and $C_{2}$ are compact metric spaces. We know that $f$ is uniformly continuous on $M$. For any $\\epsilon$ let $\\delta$ be as in the definition of uniform continuity. Then $|\\Psi(x)-\\Psi(x')|\\le \\max_{y}|f(x,y)-f(x',y)|\\le \\delta$ whenever $|x-x'|<\\epsilon$, as desired.", "Solution_3": "Hi!\r\n\r\n$C_{1}$ and $C_{2}$ are indeed compact metric spaces.\r\n\r\nI'll repost the problem:\r\n\r\nIf $f(x,y)$ is a continuous function in $(x,y)$ and is defined on $M=C_{1}\\times C_{2}$, where $C_{1}$ and $C_{2}$ are compact sets in the euclidean spaces $E^{m}$ and $E^{n}$, then $\\Psi(x)$ is a continuous function in $x$, where $\\Psi (x): = \\mathop{\\max }_{y \\in C_{2}}f(x,y)$\r\n\r\nMy idea is the following (it seems that linearity and convexity is not needed):\r\n\r\nFor every $\\epsilon>0$ and $x' \\in C_{1}$ we have to show that we can find a $\\delta>0$ so that\r\n\r\n$\\left|{\\Psi (x)-\\Psi (x')}\\right| < \\varepsilon ,\\forall x \\in C_{1}$ with $\\left|{x-x'}\\right| < \\delta$\r\n\r\nAs mlok said, from what is given we have that $f(x,y)$ is uniformly continuous, so that:\r\n\r\n$\\begin{gathered}\\forall y,y' \\in C_{2}: \\hfill \\\\ \\left|{f(x',y)-f(x,y)}\\right| < \\varepsilon \\hfill \\\\ \\end{gathered}$\r\n\r\nAND\r\n\r\n$\\left|{f(x',y')-f(x,y')}\\right| < \\varepsilon$\r\n\r\nif $\\left|{x-x'}\\right| < \\delta$ for $\\delta>0$\r\n\r\nChoose $y$ and $y'$ so that: $f(x',y') = \\Psi(x')$ and $f(x,y)=\\Psi(x)$ for some $x$ with $\\left|{x-x'}\\right| < \\delta$ for $\\delta>0$\r\n\r\nOkay, now comes the problem:\r\n\r\nIf I can show that: $f(x',y) \\leqslant f(x',y')$ and $f(x,y') \\leqslant f(x,y)$ I would be basically finished.\r\n\r\nHow do I show this, or, why is it true?\r\n\r\n$\\Pi\\Delta\\Phi$", "Solution_4": "Your approach is correct through \"Choose $y$ ...\". However, you don't need $y'$, because there is no way to tell if $|y-y'|$ is small. Instead, observe that $\\Psi(x)-\\Psi(x')=f(x,y)-\\Psi(x')\\le f(x,y)-f(x',y)<\\delta$. Since $x$ and $x'$ are interchangeable, you also have $\\Psi(x')-\\Psi(x)<\\delta$, hence $|\\Psi(x)-\\Psi(x')|<\\delta$.", "Solution_5": "Hi!\r\n\r\nNice proof\r\n\r\n[quote=\"mlok\"]Since $x$ and $x'$ are interchangeable[/quote]\n\nIs this always true?\n\n[quote=\"mlok\"]However, you don't need $y'$, because there is no way to tell if $|y-y'|$ is small[/quote]\r\n\r\nUmmm...\r\nCheck this out (the proof for lemma 1):\r\n\r\n[img]http://img521.imageshack.us/img521/6133/77295079sw9.jpg[/img]\r\n\r\n[img]http://img524.imageshack.us/img524/6457/67605287xt1.jpg[/img]\r\n\r\n[img]http://img161.imageshack.us/img161/6024/32387299nz1.jpg[/img]\r\n\r\n\r\nThis is from SIAM Review, Vol. 7, No. 2. (Apr., 1965), pp. 181-188\r\nYou see the step where he gets two inequalities (Step 13)? I don't understand how he gets this result. \r\nThanks for your posts.\r\n\r\n$\\Pi\\Delta\\Phi$", "Solution_6": "The inequalities (12) and (13) hold by the definition of $\\psi$. This is essentially the same proof that I gave -- it would be hard to come up with something original here. The only difference is that I did not bother to write the second part, where the roles of $x$ and $x'$ are exchanged. \r\n[quote][quote]Since x and x' are interchangeable[/quote]\nIs this always true?[/quote] No. This is only true in a particular context.\r\n[My standard reply to that question]" } { "Tag": [], "Problem": "Sarah's mom wants to bake 4 meatballs, but only three of them fit in her pan at a time. It takes 5 minutes to bake one side of a meatball, and 5 minutes to bake the other side. Without cutting the meatballs, can all 4 of them be baked in less than 20 minutes?", "Solution_1": "Solution: [hide] First 3 meatballs: 5 + 5 = 10, Second meatball: 5 + 5 = 10\n\nTherefore, she cannot bake all the meatballs in less than 20 minute, but in exactly 20 minutes[/hide]", "Solution_2": "That's what I thought, but I have a feeling she can...\r\nwhat is the longest period of time each meatball can stay if every meatball can have an equal time in the pan?", "Solution_3": "bake first 3 for 5 minutes (one side)\r\n\r\ntake out first, put in fourth. bake for five minutes\r\n\r\ntake out second and third. Put in first. Bake for 5 minutes. \r\n\r\nThis takes 15 minutes so yes.", "Solution_4": "Well must it be baked for five minutes continously, or cummulative?", "Solution_5": "cumulative, at least I hope", "Solution_6": "realistically, i dont think that cumulative would work, because it would cool of, and then at least some of the time would be spent reheating, so the meatball would be just barely undercooked...but maybe im wrong...", "Solution_7": "i know I know, but this is an imaginative math world, that's why problem solving is an art in some ways, not only do you have to think of ideas, but you also have to imagine the situation :lol:", "Solution_8": "i think you contradiced yourself...\r\n\r\nfirst you said its imaginitive, then u said that you have to imagine the situation...which is right?", "Solution_9": "well, you get what I mean.", "Solution_10": "no...\r\nwhich side r u on?\r\n\r\ncontiuously or cumulative?", "Solution_11": "each side has to be baked continuously for 5 minutes but you can cook the two sides separately", "Solution_12": "Then it's easy.\r\n\r\nLabel the bottom half of a meatball with a -1 suffix, and the top half with a -2 suffix.\r\n\r\nFirst five minutes:\r\n\r\n1-1\r\n2-1\r\n3-1\r\n\r\nNext five minutes\r\n\r\n2-2\r\n3-2\r\n4-2\r\n\r\n2 and 3 are done.\r\n\r\nLast five minutes\r\n\r\n1-4\r\n4-1", "Solution_13": "that's what I said in post 4......", "Solution_14": "yeah but it wasn't really that hard of a question anyway", "Solution_15": "Thank you everyone, that problem just asked if she could cook the meatballs in less than 20 minutes, problem solved :10: !" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Is it true that if $x,y,z$ are three positive reals such that $xyz=1$, then \r\n$\\frac{1}{2(x+y)+1}+\\frac{1}{2(y+z)+1}+\\frac{1}{2(z+x)+1}\\le \\frac{1}{2}$", "Solution_1": "$x=y=z=1$ :?:", "Solution_2": "Beat,did you mean ...$\\leq$1?", "Solution_3": "No, zhaobin is right :( \r\n\r\nI saw this inequality in my notebook, maybe there is a mistake and the sign is reversed. But anyway since i`m not sure just leave this topic.\r\nAppologize to every one who saw this\r\n\r\nBut if you can prove it with $\\le 1$, phamduyhiep, I ll be very happy to see the solution.", "Solution_4": "it is easy to prove:\r\n$\\frac{1}{2(x+y)+1}+\\frac{1}{2(y+z)+1}+\\frac{1}{2(z+x)+1}\\le 1$\r\nbecause:\r\n$\\frac{1}{2(x+y)+1}+\\frac{1}{2(y+z)+1}+\\frac{1}{2(z+x)+1}\\le \\frac{1}{x+y+1}+\\frac{1}{y+z+1}+\\frac{1}{z+x+1}$\r\nand let:$x=a^3,y=b^3,z=c^3$\r\nthen we have\r\n$\\frac{1}{x+y+1}+\\frac{1}{y+z+1}+\\frac{1}{z+x+1}= \\frac{abc}{a^3+b^3+abc}+\\frac{abc}{b^3+c^3+abc}+\\frac{abc}{c^3+a^3+abc}$\r\nuse $a^3+b^3 \\ge ab(a^2+b^2)$ and similar.\r\nthen we have:\r\n$\\frac{1}{2(x+y)+1}+\\frac{1}{2(y+z)+1}+\\frac{1}{2(z+x)+1}\\le \\frac{1}{x+y+1}+\\frac{1}{y+z+1}+\\frac{1}{z+x+1} \\le 1$\r\nin other hand:let $x=y \\to 0$,we will see that 1 is the best constant :)", "Solution_5": ":o \r\nThat was USAMO. I didn`t saw that..\r\nThanks anyway\r\nMust go to school. Goodbye for now :bye:", "Solution_6": "[quote=\"zhaobin\"]it is easy to prove:\n$\\frac{1}{2(x+y)+1}+\\frac{1}{2(y+z)+1}+\\frac{1}{2(z+x)+1}\\le 1$\nbecause:\n$\\frac{1}{2(x+y)+1}+\\frac{1}{2(y+z)+1}+\\frac{1}{2(z+x)+1}\\le \\frac{1}{x+y+1}+\\frac{1}{y+z+1}+\\frac{1}{z+x+1}$\nand let:$x=a^3,y=b^3,z=c^3$\nthen we have\n$\\frac{1}{x+y+1}+\\frac{1}{y+z+1}+\\frac{1}{z+x+1}= \\frac{abc}{a^3+b^3+abc}+\\frac{abc}{b^3+c^3+abc}+\\frac{abc}{c^3+a^3+abc}$\nuse $a^3+b^3 \\ge ab(a^2+b^2)$ and similar.\nthen we have:\n$\\frac{1}{2(x+y)+1}+\\frac{1}{2(y+z)+1}+\\frac{1}{2(z+x)+1}\\le \\frac{1}{x+y+1}+\\frac{1}{y+z+1}+\\frac{1}{z+x+1} \\le 1$\nin other hand:let $x=y \\to 0$,we will see that 1 is the best constant :)[/quote]\r\nYes!It's my solution.Thanks,zhaobin!" } { "Tag": [ "LaTeX" ], "Problem": "What do you guys think of the equation editor in word 2007?\r\n\r\nI mean sure its probably not as good as LaTex if you've been using it for years and remember all the complex codes and stuff. \r\n\r\nBut for someone like me who started typing these equations with mathtype it seems like an improvement.\r\n\r\nAnd the new equation editor has latex like recognition capabilities so you can type on the keyboard without stopping anyway. just need to remember a new set of strings (and i think some of them were taken right off latex anyway)\r\n\r\nany opinions? im considering whether or not i should get office 2007. ive only tried it.", "Solution_1": "Could you please post a screenshot of some of those equations? I haven't tried out Word 2007 yet. Somehow I'm not optimistic, though...", "Solution_2": "1) If you're going to learn the codes for EE, why not just learn the codes for LaTeX?\r\n2) It is not, in fact, very difficult to learn TeX. You just have to keep using it, and asking questions when you can't figure things out.\r\n3) EE makes image files. Documents become hugely massive and burdensome. This really sucks.\r\n4) Everyone can read .pdf-s (and could read .dvi-s or .ps-s with one or two additional pieces of free software). Not everyone can read Word documents. Also, you should find out if this new EE will be backwards-compatible, i.e. if anyone without Word 2007 will be able to read your equations.", "Solution_3": "$\\LaTeX$ still is better. I don't want to re-learn a lot of commands...\r\n\r\nAnd you can't use EE codes on this forum :D", "Solution_4": "but dun u think its easier to see what ur getting than a gazillion codes on one page??\r\n\r\nLike you have to intertwine text with latex code and that just makes it really confusing for me", "Solution_5": "After using $\\LaTeX$ for some time, you get used to the coding and can see the final result in your mind's eye. You always have the option to preview your posts, or compile your code.\r\nI find this to be a lot less cumbersome than using the Equation editor, which has me open a small window and switch between mouse and keyboard to input symbols, and then repeat for more equations later on.", "Solution_6": "I'm sorry I'm writing in this topic. I just didn't want to create another similar topic.\r\n\r\nNow the question :\r\n\r\nI have two old files on Word 2003 with MathType equations. Now, I want to convert the files in Word 2007 format and I prefer the mathematical equation to be in Cambria Math style (MS Equation Editor 3.0), not as a MathType equation. I've searched in the net, but there isn't useful information. So, does anybody know how to convert MathType symbols into Cambria style :| ?", "Solution_7": "I, for one, use Equation Editor all the time! It's just so simple, so I love it.", "Solution_8": "I use $ LaTex$ on the forum, and subscript on word 2007 when I am doing challenge set problems or homework. It is just annoying having codes and the words for my proofs mixed up on a page. \r\n@Karamata there is [b]never[/b] a good reason for reviving a 2 year old topic", "Solution_9": "Oh, isn't it better than starting a new topic? I think less topics is better. :blush: \r\n\r\nBy the way, I found an interesting program - GrindEq Math Utilities - which can resolve the problem with MathType equations. If somebody don't like MathType and prefer MS office 2007's equations, this will convert them in Cambria style. The problem is in that the program isn't free and a registration code is wanted after 10 conversions. Fortunately, 10 attempts were enough for me." } { "Tag": [ "function", "induction", "Functional Equations" ], "Problem": "Find all functions $f: \\mathbb{Q}\\to \\mathbb{Q}$ such that for all $x,y \\in \\mathbb{Q}$: \\[f(x+y)+f(x-y)=2(f(x)+f(y)).\\]", "Solution_1": "[b]1.[/b] If we plug in $ x = a$ and $ y = 0$ we get $ 2f(a) = 2(f(a)+f(0))$ therefore $ f(0) = 0$.\r\n[b]2.[/b] If we plug in $ x = 0$ and $ y = a$ we get $ f(a)+f(-a) = 2(f(a)+f(0))$ so $ f(a)+f(-a) = 2f(a)$ hence $ f(a) = f(-a)$, therefore $ f(x) = f(-x)$ for any $ x\\in\\mathbb{Q}$.\r\n[b]3.[/b] Take any $ t\\in\\mathbb{N}$. Then, plugging into the original equation $ x = y =\\frac{1}{t}$ we obtain $ f(\\frac{2}{t})+f(0) = 2(f(\\frac{1}{t})+f(\\frac{1}{t}))$ hence $ f(\\frac{2}{t}) = 4f(\\frac{1}{t})$.\r\nLet us now prove by mathematical induction that for any $ n\\in\\mathbb{N}$ have $ f(\\frac{n}{t}) = n^{2}f(\\frac{1}{t})$.\r\n[b]Base Step[/b]: For $ n = 1,2$ the result is true -- we have proved $ f(\\frac{2}{t}) = 4f(\\frac{1}{t})$ in the first paragraph of 3, and also $ f(\\frac{1}{t}) = 1^{2}f(\\frac{1}{t})$\r\n[b]Induction Step[/b]: Let us assume the result is true for $ n = k,k-1$. Then, let us prove the result for $ n = k+1$. If we plug into the original equation $ x =\\frac{n}{t}$ and $ y =\\frac{1}{t}$. Then we get:\r\n$ f(\\frac{k+1}{t})+f(\\frac{k-1}{t}) = 2(f(\\frac{k}{t})+f(\\frac{1}{t}))$\r\n$ \\Leftrightarrow f(\\frac{k+1}{t}) = 2f(\\frac{k}{t})+2f(\\frac{1}{t})-f(\\frac{k-1}{t})$\r\n$ \\Leftrightarrow f(\\frac{k+1}{t}) = 2k^{2}f(\\frac{1}{t})+2f(\\frac{1}{t})-(k-1)^{2}f(\\frac{1}{t})$\r\n$ \\Leftrightarrow f(\\frac{k+1}{t}) = 2k^{2}f(\\frac{1}{t})+2f(\\frac{1}{t})-(k-1)^{2}f(\\frac{1}{t})$\r\n$ \\Leftrightarrow f(\\frac{k+1}{t}) = (k+1)^{2}f(\\frac{1}{t})$\r\nso the result is true for $ n = k+1$ and by mathematical induction it follows that for any $ n,t\\in\\mathbb{N}$ have $ f(\\frac{n}{t}) = n^{2}f(\\frac{1}{t})$.\r\n[b]4.[/b] Now, take any $ t\\in\\mathbb{N}$. From 3, we get $ f(1) = f(\\frac{t}{t}) = t^{2}f(\\frac{1}{t})$ hence $ f(\\frac{1}{t}) =\\frac{1}{t^{2}}f(1)$.\r\nTherefore, combining this with 3 we get for any $ n,t\\in\\mathbb{N}$ ${ f(\\frac{n}{t}) = (\\frac{n}{t})^{2}}f(1)$, and therefore for any $ x\\in\\mathbb{Q}$ with $ x > 0$ we get $ f(x) = x^{2}f(1)$.\r\n\r\nFinally because from 2, $ f(x) = f(-x)$, $ f(0) = 0$ and for any $ x\\in\\mathbb{Q}$ with $ x > 0$ we get $ f(x) = x^{2}f(1)$, then for any $ x\\in\\mathbb{Q}$ have $ f(x) = x^{2}f(1)$.\r\n\r\n[b]Solution:[/b] $ f(x) = x^{2}*c$ where $ c$ is any rational number.", "Solution_2": "Plugging $x=y=0$ yields $f(0)=0.$ By induction $f(xy)=y^2f(x).$ Noting that $f$ is even we get $f(x)=cx^2,$ for some rational constant $c$, it clearly works." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "The numbers $ x,y,z>0$ are so that $ x 3^{x\\plus{}1}\\plus{}y3^{y\\plus{}1}\\plus{}z3^{z\\plus{}1}\\equal{}(xy\\plus{}yz\\plus{}zx)^2$.Prove that $ x\\equal{}y\\equal{}z.$", "Solution_1": "NICE ,SKYLOVER\r\nLet $ LHS\\equal{}f(x).RHS\\equal{}g(x)$\r\nthen $ f'(x)\\equal{}g'(x0$,simialerly,with $ f(y).g(y)$\r\nWe have $ T(a)\\equal{}a3^{a\\plus{}1}$ is a crease function with$ a>0$\r\nSo if $ x>y thus f'(x)>f'(y) thus g'(x)>g'(y) hence y>x$ (it is easy to find)\r\nSo $ x\\equal{}y$,\r\nThen $ x\\equal{}y\\equal{}z$\r\nWe have done :lol:", "Solution_2": "Here is a another solution:\r\n[hide]We have $ 3^x\\geq x^3$ and the others, so $ (xy\\plus{}yz\\plus{}zx)^2\\geq 3(x^4\\plus{}y^4\\plus{}z^4)\\geq (x^2\\plus{}y^2\\plus{}z^2)^2\\geq\\ (xy\\plus{}yz\\plus{}zx)^2$ .\nThus, we have the equality case and the conclusion follows.[/hide]", "Solution_3": "[quote=\"skylover\"]Here is a another solution:\n[hide]We have $ 3^x\\geq x^3$ and the others, so $ (xy \\plus{} yz \\plus{} zx)^2\\geq 3(x^4 \\plus{} y^4 \\plus{} z^4)\\geq (x^2 \\plus{} y^2 \\plus{} z^2)^2\\geq\\ (xy \\plus{} yz \\plus{} zx)^2$ .\nThus, we have the equality case and the conclusion follows.[/hide][/quote]\r\nIt is wrong,try $ x\\equal{}2,8$, :wink: :wink: :wink:", "Solution_4": "Yeah, that ineq holds only for $ x\\geq 3$..:D..", "Solution_5": "Oh no,your proof is still true for $ 00\nApplying AM-GM we have: 4x^2+1>=4x\n 4y^2+1>=4y\n 4z^2+1>=4z\nTherefore, the LHS=0). That's the only solution to this system of equation(Q.E.D)[/hide]", "Solution_3": "I think you forgot that $ x \\equal{} y \\equal{} z \\equal{} 0$ is also true for all three equations ;)", "Solution_4": "2 - it is similar to Bulgaria 1996 - IV Round problem\r\n1 - Canadian Math Olympiad 1996 problem", "Solution_5": "[quote=\">Aver<\"]I think you forgot that $ x \\equal{} y \\equal{} z \\equal{} 0$ is also true for all three equations ;)[/quote]\r\nOh yeah you're right :blush: . I left out that simple case.", "Solution_6": "Someone solves the #2 and #3 for me, please! :(", "Solution_7": "one solution for #2 is $ p\\equal{}q\\equal{}3$ (because $ 3|5^x \\minus{} 2^x$ for all positive integers $ x$)", "Solution_8": "you can see the solution of problem#2 here\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1791960291&t=50838[/url]" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "Let $ A \\equal{} \\begin{pmatrix} 7 \\minus{} \\lambda & \\minus{} 12 & 6 \\\\\r\n10 & \\minus{} 19 \\minus{} \\lambda & 10 \\\\\r\n12 & \\minus{} 24 & 13 \\minus{} \\lambda\\end{pmatrix}$. Find $ \\operatorname{rank} A$", "Solution_1": "Of course, the rank depends on $ \\lambda.$ For all but at most three values of $ \\lambda,$ the rank is 3. The only $ \\lambda$ for which the question is interesting are the roots of the characteristic polynomial, which is to say the eigenvalues. Do you know what those are?", "Solution_2": "[hide]All lines add up to 1, so 1 is an eigenvalue ($ \\lambda_3 \\equal{} 1$). The trace equals 1, so $ \\lambda_1 \\plus{} \\lambda_2 \\plus{} 1 \\equal{} 1 \\equal{} > \\lambda_2 \\equal{} \\minus{} \\lambda_1$. On the other hand,\n\n$ \\left | \\begin{array}{ccc} 7 & \\minus{} 12 & 6 \\\\\n10 & \\minus{} 19 & 10 \\\\\n12 & \\minus{} 24 & 13 \\end{array} \\right| \\equal{} \\left | \\begin{array}{ccc} 7 & 1 & 6 \\\\\n10 & 1 & 10 \\\\\n12 & 1 & 13 \\end{array} \\right| \\equal{} \\left | \\begin{array}{ccc} 1 & 1 & 6 \\\\\n0 & 1 & 10 \\\\\n\\minus{} 1 & 1 & 13 \\end{array} \\right| \\equal{} \\left | \\begin{array}{ccc} 1 & 1 & 6 \\\\\n0 & 1 & 10 \\\\\n0 & 2 & 19 \\end{array} \\right|$\n\n$ \\equal{} \\left | \\begin{array}{ccc} 1 & 1 & 6 \\\\\n0 & 1 & 10 \\\\\n0 & 0 & \\minus{} 1 \\end{array} \\right| \\equal{} \\minus{} 1$\n\nand so $ \\minus{} \\lambda_1^2 \\equal{} \\minus{} 1 \\equal{} > \\lambda_1 \\equal{} 1$ and $ \\lambda_2 \\equal{} \\minus{} 1$.[/hide]", "Solution_3": "So when $ \\lambda\\equal{}\\minus{}1,$ the rank of $ A\\minus{}\\lambda I$ is 2. The only remaining question is the rank of $ A\\minus{}\\lambda I$ for $ \\lambda \\equal{}1.$ Carcul showed that that eigenvalue has algebraic multiplicity $ 2.$ Its geometric multiplicity has yet to be determined.", "Solution_4": "We are to solve the system\r\n\r\n$ \\left [\\begin{array}{ccc} 6 & \\minus{}12 & 6 \\\\\r\n10 & \\minus{}20 & 10 \\\\\r\n12 & \\minus{}24 & 12 \\end{array} \\right] \\left [\\begin{array}{c} x \\\\ y \\\\ z \\end{array} \\right] \\equal{} \\left [\\begin{array}{c} 0 \\\\ 0 \\\\ 0 \\end{array} \\right] <\\equal{}> \\left [\\begin{array}{ccc} 1 & \\minus{}2 & 1 \\\\\r\n1 & \\minus{}2 & 1 \\\\\r\n1 & \\minus{}2 & 1 \\end{array} \\right] \\left [\\begin{array}{c} x \\\\ y \\\\ z \\end{array} \\right] \\equal{} \\left [\\begin{array}{c} 0 \\\\ 0 \\\\ 0 \\end{array} \\right]$\r\n\r\n$ <\\equal{}> z \\equal{} \\minus{}x \\plus{} 2y$. \r\n\r\nTherefore, all eigenvectors associated with $ \\lambda \\equal{} 1$ have the form $ (x,y,\\minus{}x\\plus{}2y)^T$, i e., they belong to $ Span \\{(1,0,\\minus{}1)^T,(0,1,2)^T \\}$.", "Solution_5": "... which is to say that the eigenvalue $ \\lambda\\equal{}1$ has geometric multiplicity $ 2,$ which is to say that $ \\operatorname{rank}(A\\minus{}I)\\equal{}1.$" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "[b]Let[/b] $ A_{i} \\in S_{n} \\mathbb{(R)}$ ($ A_{i}=A_{i}^{T}$ if $ A \\in S_{n}(R)$) \r\n$ \\forall i \\in$ {1,2,...,m} , m>2 [b]such that [/b]: $ A_{i}A_{j}=0_{n}$ if $ i \\neq j$\r\n[b]Prove that :[/b] \r\n $ rank(A_{1})+rank(A_{2})+rank(A_{3})+....+rank(A_{m}) \\leq n$\r\n :blush:", "Solution_1": "[hide=\"Hint\"]The matrices are simultaneously diagonalizable (why?), hence you can assume WLOG that all matrices are diagonal. Now is should be easy.[/hide]" } { "Tag": [ "percent" ], "Problem": "A merchant bought some goods at a discount of $ 20\\%$ of the list price. He wants to mark them at such a price that he can give a discount of $ 20 \\%$ of the marked price and still make a profit of $ 20\\%$ of the selling price. The percent of the list price at which he should mark them is:\r\n\r\n$ \\textbf{(A)}\\ 20 \\qquad\\textbf{(B)}\\ 100 \\qquad\\textbf{(C)}\\ 125 \\qquad\\textbf{(D)}\\ 80 \\qquad\\textbf{(E)}\\ 120$", "Solution_1": "[hide=\"Click for solution\"]\nLet $ C$ be the cost, $ L$ the list price, $ S$ the selling price and $ M$ the marked price. $ C\\equal{}\\frac{4}{5}L$, $ S\\equal{}C\\plus{}\\frac{1}{5}S$, $ S\\equal{}\\frac{4}{5}M$. Thus, $ \\frac{4}{5} \\times \\frac{4}{5}M\\equal{}\\frac{4}{5}L \\implies M\\equal{}\\frac{5}{4}L$, so $ \\boxed{\\textbf{(C)}}$ is correct.\n[/hide]" } { "Tag": [], "Problem": "The numbers p and q are primes and p^2 + 1\u22610(mod q) and q^2-1\u22610(mod p).\r\nProve that p+q+1 is a composite.\r\n\r\nAll i could show is that q is either 2 or of the form 4t+1. If q=2 it follows trivially that p+q+1 is a composite but what if q=4t+1 ???", "Solution_1": "$p|q^2-1\\Rightarrow p|p^2+2pq+q^2-1=(p+q+1)(p+q-1)$. Assume $p+q+1$ is a prime. In this case, $p|p+q-1\\Rightarrow p|q-1\\Rightarrow q=mp+1,\\ m\\ge 1$. $q$ also divides $p^2+1$, so $\\frac{p^2+1}q=np+1,\\ n\\ge 0$. If $n>0$, then it's clear that $(mp+1)(np+1)>p^2+1$, which is false, so $n=0\\Rightarrow m=p$, so $q=p^2+1$, which means that $p+q+1=p^2+p+2$, which is even and $>2$, and thus a composite." } { "Tag": [ "floor function", "inequalities", "number theory proposed", "number theory" ], "Problem": "These three are my own problems, and actually i only know the solution of 2 and 3 but not 1. :oops: but i think its a nice problem and you can solve it... :D \r\n\r\n\r\n\r\n\r\n$1$.Find all positive integers $n>1$ such that there exist $2n$ distinct positive integers $a_{1},a_{2},\\cdots,a_{n},b_{1},b_{2},\\cdots,b_{n}$ such that:\r\n$\\sum^{n}_{i=1}a_{i}= \\sum^{n}_{i=1}b_{i}$\r\n$\\prod^{n}_{i=1}a_{i}= \\prod^{n}_{i=1}b_{i}$\r\n\r\n\r\n$2$.$n$ is a positive integer. Let $T(n)$ be the number of all positive integers which do not have any $0$ digit and the sum of their digits (in base $10$) is $n$. Prove that $T(n) \\leq 2^{n-1}$\r\n\r\n\r\n$3$.$n$ is a positive integer and for all reals ${x}$, we denote $\\left\\{x\\right\\}$ as $x-\\left\\lfloor x\\right\\rfloor$ . Assume that $A=\\sum^{6n-1}_{m=1}( \\left\\{\\frac{m}{6n}\\right\\}+\\left\\{\\frac{m}{3n}\\right\\})$\r\nFind the smallest prime number $p$ such that $\\left\\lfloor A \\right\\rfloor \\equiv 1 ( mod \\: p)$", "Solution_1": "#2: We have that $T(n)$ is the number of compositions of $n$ into positive integers from 1 to 9. But the number of compositions into positive integers is $\\binom{n-1}{0}+\\binom{n-1}{1}+...+\\binom{n-1}{n-1}=2^{n-1}$, and we are counting a subset of those compositions. Then the desired inequality follows. Was this your solution?", "Solution_2": "exactly k81o7, that was my solution too :D , don't you have a solution for 1 ? :maybe:", "Solution_3": "well , here is the solution of $\\#3$ : $A = \\sum^{6n-1}_{m=1}(\\frac{m}{6n}+\\frac{m}{3n})-\\sum^{6n-1}_{m=1}\\left( \\left\\lfloor \\frac{m}{6n}\\right\\rfloor+\\left\\lfloor \\frac{m}{3n}\\right\\rfloor \\right)$ \r\nnotice that: if $1 \\leq m \\leq 6n-1$ then $\\left\\lfloor \\frac{m}{6n}\\right\\rfloor = 0$\r\nand: if $1 \\leq m \\leq 3n-1$ then $\\left\\lfloor \\frac{m}{3n}\\right\\rfloor = 0$\r\n if $3n \\leq m \\leq 6n-1$ then $\\left\\lfloor \\frac{m}{3n}\\right\\rfloor = 1$\r\n\r\nAs a consequence we have : $\\sum^{6n-1}_{m=1}\\left( \\left\\lfloor \\frac{m}{6n}\\right\\rfloor+\\left\\lfloor \\frac{m}{3n}\\right\\rfloor \\right) = 3n$\r\nAlso its easy to prove: $\\sum^{6n-1}_{m=1}( \\frac{m}{6n}+\\frac{m}{3n}) = \\frac{3}{2}\\times ( 6n-1 )$\r\nThus $A = 12n-\\frac{3}{2}\\Rightarrow \\left\\lfloor A \\right\\rfloor = 12n-2$ , and the smallest possible value of $p$ is $3$ . :lol:", "Solution_4": "i still cant find any solution for $ 1$, it seems easy but a problem...! :huh:", "Solution_5": "$ 1)$ it is imposible for $ n=2$\r\nAssume it is possible:\r\n$ b_{1}=a_{1}+k$ with $ k$ integer $ \\ne 0$ then $ b_{2}=a_{2}-k$ \r\n$ a_{1}a_{2}= b_{1}b_{2}$ implies that $ 0=k (a_{2}-(a_{1}+k) )$ contradiction", "Solution_6": "Solution for $ n=3$\r\n\r\n$ (2,12,7)$ and $ ( 3, 14, 4)$\r\nSums 21\r\nProduct 168\r\n\r\n[hide=\" How to build a n+3 solution\"]Notice that if we multiply by a k a solution we get another one \nthen multiply by big number (for example the maximum plus one) and append previous solution[/hide]\r\nI try to post a solution for $ n=4$ and $ n=5$ and then it is posible for $ n\\ge 3$\r\nEdited: Solution for $ n=4$\r\n$ (2,7,180,13)$ and $ (3,5,182,12)$\r\nSolution for $ n=5$\r\n$ (2, 7, 11, 975, 28)$ and $ (3, 5, 13, 980, 22)$" } { "Tag": [], "Problem": "wasn't sure where to post it, so I decided to post here.\r\nI didn't know about the source of the problems soon enough last year, so had some problems with misunderstandings.\r\nAnyone who sees these questions (counting similar format with only a few names changed) that are obviously seeking aid, which is not permitted, please report/lock them until due date. These came out a few days ago from my knowledge.\r\n\r\nhttp://www.nzamt.org.nz/sites/cms/index.php?option=com_content&task=view&id=68&Itemid=82", "Solution_1": "Can we answer them yet?", "Solution_2": "yep. Selection for this problem has finished :D\r\nI'll post my solutions soon" } { "Tag": [ "inequalities" ], "Problem": "Prove that \r\n\r\n$ x^{n} \\plus{} x^{n\\minus{}2} \\plus{} x^{n\\minus{}4} \\plus{} ... \\plus{} \\frac{1}{x^{n\\minus{}4}} \\plus{} \\frac {1}{x^{n\\minus{}2}} \\plus{} \\frac{1}{x^{n}} \\ge n\\plus{}1$,\r\n\r\nfor all positive numbers $ x$ and for all $ n \\in \\mathbb{N}$.", "Solution_1": "[hide]For even $ n$ \n\n$ x^n \\plus{} \\frac{1}{x^n} > 2$ by AMGM\n\nThere are $ \\frac{n}{2}$ such pairs plus $ 1$\n\nTherefore $ L.H.S. > n\\plus{}1$\n\nFor odd $ n$ \n\nThere are $ \\frac{n\\plus{}1}{2}$ such pairs, \n\nTherefore $ L.H.S. \\ge n\\plus{}1$[/hide]", "Solution_2": "[quote=\"chrischris\"]$ x^n \\plus{} \\frac {1}{x^n} > 2$ by AMGM[/quote] Not quite.", "Solution_3": "For $ n\\equal{}1$, we have $ x\\plus{}\\frac{1}{x} \\ge 2$, which is true because $ (x\\minus{}1)^{2} \\ge 0$.\r\n\r\nReplacing $ x$ by $ x^{2}$, we get \r\n\r\n$ x^{2} \\plus{} \\frac{1}{x^{2}} \\ge 2$.\r\n\r\nAdding $ 1$ to both sides, we have $ x^{2} \\plus{} 1 \\plus{} \\frac{1}{x^{2}} \\ge 3$.\r\n\r\nTherefore $ x^{2} \\plus{} 2 \\plus{} \\frac{1}{x^{2}} \\ge 4 > 3$.\r\n\r\nTherefore the inequality is true for $ n\\equal{}1$ (1) and for $ n\\equal{}2$. (2)\r\n\r\nFor the inductive step, we show that \r\n\r\nif it is true for $ n\\equal{}k$ then it is true for $ n\\equal{}k\\plus{}2$:\r\n\r\nIf $ x^{k} \\plus{} x^{k\\minus{}2} \\plus{} ... \\plus{}\\frac{1}{x^{k\\minus{}2}} \\plus{} \\frac{1}{x^{k}} \\ge k\\plus{}1$\r\n\r\nthen adding $ x^{k\\plus{}2} \\plus{} \\frac{1}{x^{k\\plus{}2}} \\ge 2$, \r\n\r\nwe get the required result which with (1) implies that the inequality is true for all odd natural numbers and with (2) implies that it is true for all even natural numbers.", "Solution_4": "[quote=\"JBL\"][quote=\"chrischris\"]$ x^n \\plus{} \\frac {1}{x^n} > 2$ by AMGM[/quote] Not quite.[/quote]\r\n\r\n :?: I don't understand why not...?", "Solution_5": "$ \\ge$", "Solution_6": "(and in particular equality is obtained when $ x \\equal{} 1$)" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Find all real polynomials $p(x)$ such that $p(x) = a$ has $\\deg{p}$ real roots, for any $a > 1995$. Multiple roots are counted according to their multiplicities.", "Solution_1": ":huh: $p(x)=a$ has at most $2$ real roots for $a$ large enough.", "Solution_2": "[quote=\"olorin\"]:huh: $p(x)=a$ has at most $2$ real roots for $a$ large enough.[/quote]\r\nWhy? :maybe:", "Solution_3": "If $\\deg p$ is even, then the line $f(x)=a$ would intersect $p(x)$ at exactly two places for $a$ sufficiently large that $f(x)$ clears the tallest \"bump\" (local maximum) in the graph of $p(x)$. If $\\deg p$ is odd, then we will similarly have only one intersection.", "Solution_4": "Yes.\r\n$p$ has only finitely many local extremes $(x_{1},p(x_{1})),\\ldots,(x_{r},p(x_{r}))$.\r\nFor $a>\\max\\{|p(x_{1})|,\\ldots,|p(x_{r})|\\}$, then $p(x)=a$ has at most $2$ solutions.\r\n\r\nNot sure what you mean [b]N.T.TUAN[/b]??", "Solution_5": "Ok, you are right! Thanks olorin and mathisfun1. I edited it\r\n\r\n[quote=\"N.T.TUAN\"]Find all real polynomials $p(x)$ such that $p(x) = a$ has $\\deg{p}$ real roots for any $a > 1995$. Multiple roots are counted according to their multiplicities.[/quote]", "Solution_6": "$\\deg p\\leq 2$, so \r\n$p(x)=mx+n$ for any $m,n\\in\\mathbb{R},m\\not =0$\r\nor $p(x)=kx^{2}+mx+n$ for any $k,m,n\\in\\mathbb{R},k>0$ with $n-{m^{2}\\over 4k}\\leq 1995$." } { "Tag": [ "ARML", "geometry", "AMC", "AIME", "AMC 10" ], "Problem": "Maryland is going to field its first home school ARML team this year. It draws from all over the state, but with the requirement that students must be fully homeschooled according to the laws of Maryland. The state doesn't allow home school students to participate in extracurricular activities with the public schools, so the ARML administration is allowing us to field our own team. If you are interested in trying out for the team, contact\r\n\r\nCatherine Asaro, Coach\r\nHoward Area Homeschoolers\r\nasaro@sff.net\r\n\r\nIf you have scores from the AMC-10, AMC-12, AIME, Sigma League, or Mandelbrot competitions, please have those ready.\r\n\r\nWe had originally planned to take one team and a few alternates. However, we've had more people than expected try out. We may possibly field two teams; that remains to be seen. Selection of the team(s) and alternates will be made the last week in March, so if you are interested please contact me ASAP.\r\n\r\nAnd my thanks to the Art of Problem Solving for making these Forums available. :-)\r\n\r\nBest regards\r\nCatherine Asaro", "Solution_1": "You might also consider posting this in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=240]Maryland forum[/url].", "Solution_2": "Thanks. I didn't realize Maryland had a forum. I will go post over there.\r\n\r\nBest,\r\nCatherine" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AMC 12" ], "Problem": "So i started doing contest math problems last September, right now i barely get a 3 or 4 on the practice AIME and since i'm a sophomore i will be taking AMC 12 next year which obviously kills my chance for USAMO. I just wished that i knew the competition couple years earlier, so that i can be like most of you to qualify USAMO.", "Solution_1": "you are NOT a loser, i was in the same case you are currently in last year\r\ndon't give up, just keep doing problems and keep a positive attitude :D", "Solution_2": "USAMO qualifiers make up 400 of approximately 11,000 AIME qualifiers every year. Even on AoPS, I seriously don't believe that USAMO qualifiers make up the majority of the posters in the high school forums. Contest mathematics shouldn't be about being the best; it should be about being exposed to interesting mathematical ideas you wouldn't see any of in a classroom setting. Relax and have fun with it :)", "Solution_3": "Don't have such a negative attitude! I realize this sounds clich\u00e9d. Taking the AMC 12 doesn't kill your chances for USAMO - you'll be surprised how much you improve just by doing lots of problems.", "Solution_4": "I agree with t0rajir0u.\r\nThis is my first time taking AMC12, and I have learnt a lot through studying AoPS Volume 1. Like t0rajir0u said, I think you shouldn't limit yourself to doing well on the contest. Just enjoy problem solving!", "Solution_5": "advice:\r\nbut AoPS V1 and V2. They really help!" } { "Tag": [ "trigonometry", "function", "inequalities", "geometry unsolved", "geometry" ], "Problem": "Let $a,b$ be the smaller sides of a right triangle. Let $c$ be the hypothenuse and $h$ be the altitude from the right angle. Fint the maximal value of $\\frac{c+h}{a+b}$.", "Solution_1": "Set $ABC$ be a right traingle such that $BC=a,\\ CA=b,\\ AB=c.$ and $\\angle{BCA=90^\\circ}.$ Let $\\angle {ABC} =\\theta ,$ we have $a=c\\cos \\theta ,\\ b=c\\sin \\theta , h=a\\sin \\theta =c\\sin \\theta \\cos \\theta .$ Thus $\\frac{c+h}{a+b}=\\frac{c(1+\\sin \\theta \\cos \\theta)}{c(\\sin \\theta +\\cos \\theta)}.$ Let $x=\\sin \\theta +\\cos \\theta,$ by $(\\sin \\theta +\\cos \\theta)^2=1+2\\sin \\theta \\cos \\theta ,$ we have $\\sin \\theta \\cos \\theta =\\frac{x^2-1}{2}.$ so we can rewrite this equation as $\\frac{c+h}{a+b}=\\frac{1}{2}\\left(x+\\frac{1}{x}\\right).$ Now the range of $x$ is $1< x\\leq \\sqrt{2},$\r\nbecause we can write $x=\\sqrt{2}\\sin \\left(\\theta +\\frac{\\pi}{4}\\right),$ from $0<\\theta <\\frac{\\pi}{2}\\Longrightarrow \\frac{\\pi}{4} <\\theta +\\frac{\\pi}{4}<\\frac{3}{4}\\pi ,$ yielding $1=1.and then we have infinitely many solutions for n=1,2,4\r\n(i guess).ihavent worked the inequality out.ne ideas\r\ncheers", "Solution_2": "I've a soln\r\nHint: Prove n<=4\r\nChoose (ao,bo,co,do ) satisfy equa such that (ao+bo+co+do) minimum\r\nWLOG, suppose ao>=bo>=co>=do\r\nQuadratic equation:\r\n(x+b+c+d)^2=n^2.x.bcd has another root a1:\r\na1+ao=n^2bo.co.do-2(bo+co+do)(1)\r\na1*ao=(bo+co+do)^2(2)\r\nSince a1>=ao, (1)=> a1>=1/2.n*n*bo.co.do-(bo+co+do)\r\n(2)=> a1==(bo+co+do)^2/ao <= 3(bo+co+do)\r\nthen we get n*n*bo.co.do <= 8(bo+co+do)\r\nthen n<=4\r\ngood???" } { "Tag": [ "group theory", "abstract algebra", "number theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose $ G$ is a group such that $ |G|$ is coprime to 2. Further suppose that $ |Z(G)|$ is not divisible by 3. Show that $ G$ has no order 3 normal subgroups.", "Solution_1": "supppse that $ G$ has a normal subgroup of order $ 3$. this is cyclic generated by an element $ x$. because $ |Z(G)|$ is not divisible by $ 3$, $ x$ does not lie in the center. so take $ y$ such that $ y^{ \\minus{} 1} x y \\neq x$. since $ \\langle x \\rangle$ is normal, we find $ 0 \\leq i \\leq 2$ such that $ y^{ \\minus{} 1} x y \\equal{} x^i$. clearly we must have $ i \\equal{} 2$. then $ y^{ \\minus{} 2} x y^2 \\equal{} x^{2^2} \\equal{} x$, and inductively $ y^{ \\minus{} i} x y^i \\equal{} x$ for all even $ i$. since $ |G|$ is coprime to $ 2$, $ y$ has odd order, so $ y^i \\equal{} y$ for some even $ i$. this yields the contradiction $ y^{ \\minus{} 1} x y \\equal{} x$.", "Solution_2": "Thanks for the reply. After reading this i tried to use the same approach for when $ |G|$ is coprime to 4 and 5 doesn't divide $ |Z(G)|$ but ran into problems.\r\nI got to that $ yxy^{\\minus{}1}\\equal{}x^i$ where $ 0\\le i\\le 4$ but an having trouble seeing which value $ i$ must be.", "Solution_3": "what's exactly the new exercise?", "Solution_4": "my issue was that in this case you could choose $ 0\\le i\\le 4$ but 0 and 1 are not possible so you are left with 2,3,4.\r\nhowever i think i have managed to get around this. \r\nwhat i found was that $ y^nxy^{ \\minus{} n} \\equal{} x^{r^n}$ so then by Fermat's little theorem for all multiples of 4 $ y^nxy^{ \\minus{} n} \\equal{} x$. And then from here I basically use the same ideas as in your previous post." } { "Tag": [ "geometry" ], "Problem": "An $ 8$-foot by $ 10$-foot floor is tiled with square tiles of size $ 1$ foot by $ 1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $ 1/2$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?\n[asy]unitsize(2cm);\ndefaultpen(linewidth(.8pt));\n\nfill(unitsquare,gray);\nfilldraw(Arc((0,0),.5,0,90)--(0,0)--cycle,white,black);\nfilldraw(Arc((1,0),.5,90,180)--(1,0)--cycle,white,black);\nfilldraw(Arc((1,1),.5,180,270)--(1,1)--cycle,white,black);\nfilldraw(Arc((0,1),.5,270,360)--(0,1)--cycle,white,black);[/asy]$ \\textbf{(A)}\\ 80\\minus{}20\\pi \\qquad\n\\textbf{(B)}\\ 60\\minus{}10\\pi \\qquad\n\\textbf{(C)}\\ 80\\minus{}10\\pi \\qquad\n\\textbf{(D)}\\ 60\\plus{}10\\pi \\qquad\n\\textbf{(E)}\\ 80\\plus{}10\\pi$", "Solution_1": "[hide=\"Click for solution\"]\nThe four quarter circles have the same area as a whole circle of radius $ \\frac{1}{2}$ or $ \\frac{\\pi}{4}$ square feet. Thus the area of the shaded portion of each tile is $ 1\\minus{}\\frac{\\pi}{4}$ square feet. There are $ 8 \\times 10\\equal{}80$ tiles in the entire floor, so the area of the total shaded region is $ 80 \\left (1\\minus{}\\frac{\\pi}{4} \\right )\\equal{}80\\minus{}20\\pi$, or $ \\boxed{\\textbf{(A)}}$.\n[/hide]" } { "Tag": [ "induction", "algebra", "polynomial" ], "Problem": "How can I prove that there is a possible 2n+1 multiplications (when the operations are performed as indicated) using mathematical induction?\r\n\r\n$\\displaystyle\\sum_{i=0}^n a\\sub_i x^i = a\\sub_0 +a\\sub_1 x+a\\sub_2 x^2 +...+a\\sub_n x^n$\r\n\r\nThank you.", "Solution_1": "I'm not sure what you mean. Do you mean the total number of multiplications needed to evaluate such a sum? Because the total number of multiplications would be $\\frac{n(n+1)}{2}$... so I'm not sure what you mean.\r\n\r\nOr do you mean the minimum number of multiplications needed? Because this number is $n$... yet again, I'm not sure what you mean.", "Solution_2": "Can you prove, using mathematical induction that the total number of multiplications is [n(n+1)]/2?", "Solution_3": "Actually, i see a way to have $2n-1$ multiplications (if $n>0$) without converting to synthetic-division-ish form:\r\n\r\n[hide]Base case: $n=1$: 1 multiplication here. That is $2(1)-1$.\nInductive step: The polynomial $\\sum_{k=0}^n a_kx^k$ takes $2n-1$ multiplications according to inductive hypothesis. During these multiplications, the value of $x^n$ is determined (as shown below for $n>1$: for $n=1$ it is given).\n\nThen add on $a_{n+1}x^{n+1}$. First multiply $x^n$ by $x$ (now the value of $x^{n+1}$ has been determined). Then multiply by $a_{n+1}$. Two multiplications were required here, so the total is $(2n-1)+2=2(n+1)-1$. Done.[/hide]\n\nunless, of course, you wanted the multiplications to be done in left-to-right order: then it would be $\\frac{n(n+1)}{2}$: this is similar to above, except\n\n[hide]$n+1$ multiplications are needed for $a_{n+1}x^{n+1}$.[/hide]", "Solution_4": "[quote=\"scorpius119\"]Actually, i see a way to have $2n-1$ multiplications (if $n>0$) without converting to synthetic-division-ish form:\n\n[hide]Base case: $n=1$: 1 multiplication here. That is $2(1)-1$.\nInductive step: The polynomial $\\sum_{k=0}^n a_kx^k$ takes $2n-1$ multiplications according to inductive hypothesis. During these multiplications, the value of $x^n$ is determined (as shown below for $n>1$: for $n=1$ it is given).\n\nThen add on $a_{n+1}x^{n+1}$. First multiply $x^n$ by $x$ (now the value of $x^{n+1}$ has been determined). Then multiply by $a_{n+1}$. Two multiplications were required here, so the total is $(2n-1)+2=2(n+1)-1$. Done.[/hide]\n\nunless, of course, you wanted the multiplications to be done in left-to-right order: then it would be $\\frac{n(n+1)}{2}$: this is similar to above, except\n\n[hide]$n+1$ multiplications are needed for $a_{n+1}x^{n+1}$.[/hide][/quote]\r\n\r\nAre you proving there are 2n+1 or 2n-1 multiplications.. your last step confused me. Can you please prove via induction how there are $\\frac{n(n+1)}{2}$ multiplications?", "Solution_5": "[quote=\"shyce\"][quote=\"scorpius119\"]Actually, i see a way to have $2n-1$ multiplications (if $n>0$) without converting to synthetic-division-ish form:\n\n[hide]Base case: $n=1$: 1 multiplication here. That is $2(1)-1$.\nInductive step: The polynomial $\\sum_{k=0}^n a_kx^k$ takes $2n-1$ multiplications according to inductive hypothesis. During these multiplications, the value of $x^n$ is determined (as shown below for $n>1$: for $n=1$ it is given).\n\nThen add on $a_{n+1}x^{n+1}$. First multiply $x^n$ by $x$ (now the value of $x^{n+1}$ has been determined). Then multiply by $a_{n+1}$. Two multiplications were required here, so the total is $(2n-1)+2=2(n+1)-1$. Done.[/hide]\n\nunless, of course, you wanted the multiplications to be done in left-to-right order: then it would be $\\frac{n(n+1)}{2}$: this is similar to above, except\n\n[hide]$n+1$ multiplications are needed for $a_{n+1}x^{n+1}$.[/hide][/quote]\n\nAre you proving there are 2n+1 or 2n-1 multiplications.. your last step confused me. Can you please prove via induction how there are $\\frac{n(n+1)}{2}$ multiplications?[/quote]\n\nI proved that there were $2n-1$ (the induction assumes true for $m$, needs to be proven for $m+1$: when $n=m+1$ then $2m+1=2n-1$).\n\n[hide=\"solution for n(n+1)/2\"]Base case ($n=0$): no multiplication needed. $0=\\frac{0(0+1)}{2}$.\n\nInductive step: (assuming true for $m$) The polynomial $\\sum_{k=0}^m a_kx^k$ will require $\\frac{m(m+1)}{2}$ multiplications.\n\nTo find the number of multiplications for $\\sum_{k=0}^{m+1} a_kx^k$, add on the number of multiplications from the $x^{m+1}$ term. This is $a_{m+1}*\\underbrace{x*x*x*\\cdots*x}_{m+1}$. There is exactly one $*$ before each $x$ so $m+1$ multiplications.\n\nThe total for $m+1$ is then $\\frac{m(m+1)}{2}+(m+1)=\\frac{m(m+1)+2(m+1)}{2}=\\frac{(m+1)(m+2)}{2}=\\frac{([m+1])([m+1]+1)}{2}$. Done.[/hide]", "Solution_6": "Thank you very much" } { "Tag": [ "Pascal\\u0027s Triangle" ], "Problem": "How many ways are there to spell out the way MTHOMBA going from top to bottom below?\r\n[code]\n M\nTT\nHHH\nOOOO\nMMM\nBB\nA[/code]\r\n\r\nYou can only move to adjacent letters directly below.", "Solution_1": "[quote=\"13375P34K43V312\"]How many ways are there to spell out the way MTHOMBA going from top to bottom below?\n[code]\n M\nTT\nHHH\nOOOO\nMMM\nBB\nA[/code]\n\nYou can only move to adjacent letters directly below.[/quote]\r\nYou stole my question :P\r\n[hide]16[/hide]", "Solution_2": "can you post a solution?", "Solution_3": "[quote=\"davidli\"]can you post a solution?[/quote]\r\nOk.\r\n[hide]Notice that from the M on top, you have two choices. The same is for the Ts, and the same is for the Os. That is 8. Now it gets tricky, but notice that there are 8 ways from the bottom a to any o as well, so our answer is 8*8=64[/hide]", "Solution_4": "No, that's incorrect.\r\n\r\nThere was this dude named Pascal long ago...", "Solution_5": "hm wut about him?", "Solution_6": "[quote=\"ProtestanT\"]hm wut about him?[/quote]\r\nHe liked triangles.", "Solution_7": "[hide]At each letter I wrote how many ways there were to get to that letter by adding the 1 or 2 above it. The answer is 20. You can also draw a Pascal's triangle over it and the number that coincides with the A is a 20.[/hide]", "Solution_8": "\"You can only move to adjacent letters directly below.\"\r\n\r\nThis makes no sense." } { "Tag": [ "Support" ], "Problem": "I have never done chemistry before, and I was wondering a few things about atoms. \r\n\r\n*What do atoms have to do with cells? Do the atoms make up the cells? \r\n*If quarks make up atoms, then is there anything more basic than quarks?", "Solution_1": "Quarks r the most basic products known till now......\r\n\r\ncells r made up of atoms.....quarks never exist independently they always combine to make one or the other atom or structure\r\n\r\nso if we see the complete hierachy of life,atoms will be found at bottom--below cellular level", "Solution_2": "Did scientists just stop at quarks, because they did not have good enough technology to see anything smaller?", "Solution_3": "As far as I'm aware even the existence of quarks is debatable. They started off as useful theoretical tools but they make such good predictions that people generally believe they exist. It's not so much that we cannot see them (after all there are a lot of things we can't see) but that we have no need to postulate something more elementary yet.", "Solution_4": "[quote=\"BanishedTraitor\"]As far as I'm aware even the existence of quarks is debatable. They started off as useful theoretical tools but they make such good predictions that people generally believe they exist. It's not so much that we cannot see them (after there are a lot of things we can't see) but that we have no need to postulate something more elementary yet.[/quote]\r\n\r\nOkay, I get it. Thanks you two for your help.", "Solution_5": "and well i think the LHC proposes to see quarks(bosons)\r\n\r\nmore specifically, i think the scientists who 1st thought abt quarks just gave the name quark --to the smallest possible particle---and in future wat we call today as quarks may not be called so", "Solution_6": "[quote=\"rituraj007\"]and well i think the LHC proposes to see quarks(bosons)\n\nmore specifically, i think the scientists who 1st thought abt quarks just gave the name quark --to the smallest possible particle---and in future wat we call today as quarks may not be called so[/quote]\r\n\r\nYeah, because we don't have the technology, so they just kind of guess that there is something that makes up the atoms. Right? \r\n\r\nWhy do you say (bosons)?", "Solution_7": "they r trying to use the tech to prove themselves correct... :P \r\n\r\nwell LHC proposes to see bosons which r a type of elementary particle :)", "Solution_8": "LHC proposes to support the existence of Higgs Bosons (not just any bosons because we know they exist e.g. proton, neutron etc.).", "Solution_9": "Okay, thanks for your help guys." } { "Tag": [ "Euler" ], "Problem": "[b][size=100][color=DarkBlue]\u0394\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ \\bigtriangleup ABC$ \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ A',\\ B',\\ C',$ \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b5\u03c0\u03b1\u03c6\u03ae\u03c2 \u03c4\u03bf\u03c5 \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 $ (I),$ \u03c3\u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 $ BC,\\ AC,\\ AB,$ \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03af\u03c7\u03c9\u03c2. \u0395\u03ac\u03bd $ I_{a},\\ I_{b},\\ I_{c},$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b1 \u03ba\u03ad\u03bd\u03c4\u03c1\u03b1 \u03c4\u03c9\u03bd \u03c0\u03b1\u03c1\u03b5\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03c9\u03bd \u03ba\u03cd\u03ba\u03bb\u03c9\u03bd \u03c0\u03bf\u03c5 \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03b9\u03c7\u03bf\u03cd\u03bd \u03c3\u03c4\u03b9\u03c2 \u03c9\u03c2 \u03ac\u03bd\u03c9 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2, \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 $ I_{a}A',\\ I_{b}B',\\ I_{c}C',$ \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03ad\u03c3\u03c4\u03c9 $ P,$ \u03ba\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b1\u03c5\u03c4\u03cc \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 $ OI,$ \u03cc\u03c0\u03bf\u03c5 $ O,$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c0\u03b5\u03c1\u03af\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 $ \\bigtriangleup ABC$[/color][/size][/b].\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_1": "\u03a0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 ... \u039c\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9 \u03c4\u03bf \u03c3\u03ba\u03b5\u03c0\u03c4\u03b9\u03ba\u03cc \u03c4\u03b7\u03c2 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03ae \u03c4\u03bf\u03c5 \u0395\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03bf\u03cd \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd.\r\n\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ A'C'//I_aI_c$ ,$ B'C'//I_bI_c$, $ A'B'//I_aI_b$ \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 \u03c4\u03b1 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03b1 $ A'B'C'$ \u03ba\u03b1\u03b9 $ I_aI_bI_c$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03b9\u03cc\u03b8\u03b5\u03c4\u03b1\r\n\u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 \u03bf\u03b9 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 $ I_aA',\\ I_bB',\\ I_cC'$ \u03c3\u03c5\u03bd\u03c4\u03c1\u03ad\u03c7\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf $ P$ \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf \u03bf\u03bc\u03bf\u03b9\u03bf\u03b8\u03b5\u03c3\u03af\u03b1\u03c2 \u03c4\u03c9\u03bd \u03b4\u03cd\u03bf \u03b1\u03c5\u03c4\u03ce\u03bd \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03c9\u03bd.\r\n\u0391\u03bd \u03c4\u03ce\u03c1\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03bf\u03bc\u03bf\u03b9\u03bf\u03b8\u03b5\u03c3\u03af\u03b1 \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf\u03c5 $ P$ \u03c0\u03bf\u03c5 \u03c3\u03c4\u03ad\u03bb\u03bd\u03b5\u03b9 \u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ A'B'C'$ \u03c3\u03c4\u03bf $ I_aI_bI_c$ \u03b1\u03c5\u03c4\u03ae \u03b7 \u03bf\u03bc\u03bf\u03b9\u03bf\u03b8\u03b5\u03c3\u03af\u03b1 \u03b8\u03b1 \u03c3\u03c4\u03b5\u03af\u03bb\u03b5\u03b9 \u03c4\u03bf \u03ad\u03b3\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf $ I$ (\u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03af\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 $ A'B'C'$ ) \u03c3\u03c4\u03bf $ I'$ \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c0\u03b5\u03c1\u03af\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 $ I_aI_bI_c$ \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 \r\n\u03c4\u03b1 $ I,P,I'$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac . \u038c\u03bc\u03c9\u03c2 \u03ba\u03b1\u03b8\u03ce\u03c2 \u03c4\u03bf $ I$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03c1\u03b8\u03cc\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 $ I_aI_bI_c$ \u03b7 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 \u03c0\u03bf\u03c5 \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b1 $ I,I',P$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 Euler tou $ I_aI_bI_c$ . \u038c\u03bc\u03c9\u03c2 \u03c4\u03bf $ O$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c4\u03bf\u03c5 Euler sto $ I_aI_bI_c$ \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 $ I,I',P,O$ \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac and we are done . :)" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "how to do the following integration:\r\n\r\nt is a constant,\r\nIntegral from 0 to t of (y^2)(e^[(t^2)/(y^2)]) dy ?????????", "Solution_1": "I'm fairly sure that requires complex analysis. I don't think that's going to have a nice solution over the reals.", "Solution_2": "ok...then I think I approach this problem wrongly...\r\n\r\nthe original question is:\r\n\r\nlim (t->0) of [ (1/t) A]\r\n\r\nwhere A= double integral of e^[(tx)/(y^2)] dxdy over the region [0,t] x [0,t]\r\n\r\n\r\nIs there a theorem or a trick that I missed????", "Solution_3": "That question [i]again[/i]? What's the source?", "Solution_4": "And of course, it simply diverges. The value is $ \\plus{}\\infty.$ Rather than linking to where I said that, how about you show some initiative and find my post.", "Solution_5": "I've actually done something to it in the past few minutes..\r\n\r\nwhat about applying green's theorem to the double integral??\r\nSo integral of Pdx+Qdy over the boundry (square in our case)= double integral of dQ/dx-dP/dy over [0,t]x[0,t]\r\nset P=0 and dQ/dx=e^[(tx)/(y^2)]\r\nso Q=(y^2)[e^[(tx)/(y^2)] t^(-1)\r\n\r\nnow by setting C1 and C3 be the two horizontal boundry (where dy=0), and C2 and C4 be the two vertical boundry (where y is ranging from 0 to t (when x=t) and t to 0(when x=0))\r\n\r\nso the integral over the boundry over C1 and C3 vanishes.\r\nand the whole thing becomes:\r\n\r\nlim, as t->0, [ (integral of Q dy from 0 to t)- (t^2)/3 ] / t\r\nby L-hopital's rule and FTC,\r\nlim, as t->0, te-(2/3)t = 0\r\n\r\n\r\nIs it wrong?? And if yes, where is my mistake?" } { "Tag": [], "Problem": "This may be one of the senseless posts ... but I just looked at your picture, Valentin, and wondered, if you make gel in your hair or not. It looks a little bit wet, but decent (very nice). I for myself always use a little bit gel to style my hair in a similar way.\r\n\r\nI thought Romania was a poor country, so no one should use gel there :D", "Solution_1": "[quote=\"Daniel Gutekunst\"]This may be one of the senseless posts ... but I just looked at your picture, Valentin, and wondered, if you make gel in your hair or not. It looks a little bit wet, but decent (very nice). I for myself always use a little bit gel to style my hair in a similar way.\n\nI thought Romania was a poor country, so no one should use gel there :D[/quote]\r\nabout the last phrase, Romania is far away from a poor country. Just that a few people are rich, but those who are, they are really rich. Many people are poor, but for example, on my street I can see 3 mercedez, 2 BMWs, and a lot of other very expensive cars. :D\r\n\r\nAnyway, usually most young people in romania (and I might say Japan, because I know :D) use a looot of gel. I personally never use gel, so that's just what my hair looks like :D.", "Solution_2": "btw,how old are u in this pic of urs,valentin? :D (in india,it's considered bad manners if u ask sum1 elder than u his/her(more importantly) the\r\nage)\r\nbut i'll take the liberty coz i'm asking ur age in the pic,not ur present age :D :D", "Solution_3": "this pic is like 8 months old. so i'm 19 and 10 months old. :)", "Solution_4": "The $4$ oldest posts are locked, so this is the oldest unlocked post!" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Let $f_0=f_1=1$ and $f_{i+2}=f_{i+1}+f_i$ for all $n\\ge 0$. Find all real solutions to the equation \n\\[x^{2010}=f_{2009}\\cdot x+f_{2008}\\]", "Solution_1": "[hide=\"reduction\"]\n$ x^{2010} \\minus{} f_{2009} \\cdot x \\minus{} f_{2008} \\equal{} (x^2 \\minus{} x \\minus{} 1)(\\sum_{i \\equal{} 0}^{2008} f_{2008 \\minus{} i} x^i )$ by $ ij$\nTwo roots are given by the solutions to $ x^2 \\minus{} x \\minus{} 1 \\equal{} 0$; they are $ \\frac {1 \\pm \\sqrt {5}}{2}$.\n[/hide]", "Solution_2": "by calculating $(x^{2010}-f_{2009}x-f_{2008})''=2010*2009x^{2008}$\nwe know that $x^{2010}-f_{2009}x-f_{2008}$ is a convex function\nhence it has no more than $2$ zeroes on $R$\nwe can confirm that $x=\\frac{1+\\sqrt5}{2}$ and $\\frac{1-\\sqrt5}{2}$ are solutions,so they are the only roots." } { "Tag": [], "Problem": "hey folks can you tell me the TOP 5 rock gutarists you would choose as the best?\r\n\r\nhere's my list....\r\n\r\n1 SLASH(gnr)\r\n\r\n2 Mark knophler(dire straits)\r\n\r\n3 Blackmoor(deep purple)\r\n\r\n4 Kirk(metallica)\r\n\r\n5 jimi hendrix", "Solution_1": "I'm going to move this thread to Fun and Games based on a previous statement by AoPS management of what the topic scope of the Music and Art Forum is.", "Solution_2": "um, i don't know many rock gutarists but i think jimmy hendrix was good.", "Solution_3": "Carlos Santana", "Solution_4": "i would also have to say santana. his music has a great blend of classic rock and latin. jimi hendrix was a great guitarist as well.", "Solution_5": "need to add tom morello (rage against the machine/audioslave) to the top 5 list.", "Solution_6": "Ooooh Santana!!!!!", "Solution_7": "any other guitarists worth mentioning?", "Solution_8": "My two favorites: Jimmy Page and Edward Van Halen.", "Solution_9": "Here's mine...\r\n\r\n1# Carlos Santana\r\n\r\n2# Matt B------ (forgot his name, it's weird) From Muse\r\n\r\n3# Jack White... (white stripes)\r\n\r\nBut see, those are only the ones I could name, I have heard better. I just don't know their names. I can't remember Jimmy Hendrix playing guitar even though I know he does, I just remember his voice... I'm more of a voice person when it comes to rock! Thus my liking for White Stripes and MUSE", "Solution_10": "I dont have favorites, but all of the above mentioned are good, though I think Slash is a little bit overrated. He's good... hes not legend-type stuff but everyone seems to think hes a guitar god...He is good though." } { "Tag": [ "calculus", "integration", "function", "trigonometry", "analytic geometry", "parameterization", "algebra" ], "Problem": "Let $ C$ be the circle with radius 1 centered on the origin. Fix the endpoint of the string with length $ 2\\pi$ on the point $ A(1,\\ 0)$ and put the other end point $ P$ on the point $ P_0(1,\\ 2\\pi)$. From this situation, when we twist the string around $ C$ by moving the point $ P$ in anti clockwise with the string streched tightly, find the length of the curve that the point $ P$ draws from sarting point $ P_0$ to reaching point $ A$.", "Solution_1": "I am getting an answer of $ 3{\\pi}^2$", "Solution_2": "Regrettably that is incorrect.", "Solution_3": "Parametrize the movement of P by $ P_t: \\equal{}(x(t),y(t))$ where $ t$ is the polar angle of the point of tangency of the sting to C (in Radians). Then the functions $ x\\equal{}x(t)$ and $ y\\equal{}y(t)$ satisfy the system of equations:\r\n\r\n\\[ \\begin{array}{rr} (x\\minus{}\\cos t)^2\\plus{}(y\\minus{}\\sin t)^2 \\equal{} (2\\pi \\minus{} t)^2 \\\\ \\frac{y\\minus{}\\sin t}{x\\minus{}\\cos t}\\equal{}\\minus{}\\cot t \\end{array}\\] \r\n\r\nwhere the first equation was deduced from the length of the string and the second equation requires the sting be tangent to C when it \"leaves it\".\r\n\r\nThe required value is then\r\n\r\n\\[ S\\equal{}\\int_{0}^{2\\pi}\\sqrt{\\left(\\frac{dx}{dt}\\right) ^2\\plus{}\\left(\\frac{dy}{dt}\\right) ^2}\\, dt\\]\r\n\r\nSolving the system I get\r\n\r\n\\[ \\begin{array}{ll} x(t)\\equal{}\\cos t \\plus{} (2\\pi \\minus{}t) \\, |\\sin t| \\\\ y(t) \\equal{} \\minus{}\\cos t \\left( \\mbox{sgn} (\\sin t) (2\\pi \\minus{}t) \\plus{} \\cot t\\right) \\end{array}\\]\r\n\r\nAnd I'm going to smoke while mull over the nagging feeling that the is a clever way to formulate that integrand without solving the system explicitly...", "Solution_4": "[quote=\"kunny\"]Regrettably that is incorrect.[/quote]\r\nyeah sorry made a blunder.. will fix it and post asap!", "Solution_5": "Can anyone solve the problem?", "Solution_6": "We have to evaluate $ \\int_{0}^{2\\pi\\minus{}\\arctan{2\\pi}}\\sqrt{\\frac{(1\\plus{}\\varphi^2)^2\\plus{}\\varphi^2}{1\\plus{}\\varphi^2}}d\\varphi$.", "Solution_7": "Could you explain it more in detail?", "Solution_8": "Is the answer $ 2 \\pi ^2$ by any chance?", "Solution_9": "Yes.", "Solution_10": "I used the plane polar coordinate system. I let the origin of the polar coordinate system be the center of the circle $ C$, and the half-line $ \\theta\\equal{}0$ be the $ x$ axis. Then, a point in the curve can be denoted by $ \\bold{\\hat{r}}\\plus{}(2\\pi\\minus{}\\theta)\\bold{\\hat{\\theta}}$. Applying the formula for calculating length of a curve which is $ \\int_{\\theta_i}^{\\theta_f}\\sqrt{r^2\\plus{}(dr/d\\theta)^2}d\\theta$, we get the integral($ r\\equal{}\\sqrt{1^2\\plus{}(2\\pi\\minus{}\\theta)^2}$, $ \\theta_i\\equal{}\\arctan{2\\pi}$, $ \\theta_f\\equal{}2\\pi$).\r\n\r\n Something is wrong in my approach, if the answer is surely $ 2\\pi^2$. Mathematica says the value of my integral is $ 13.883$ approaximately.", "Solution_11": "The error was the meaning of $ \\theta$. $ \\theta$ in $ \\bold{\\hat{r}}\\plus{}(2\\pi\\minus{}\\theta)\\bold{\\hat{\\theta}}$ and in $ \\int_{\\theta_i}^{\\theta_f}\\sqrt{r^2\\plus{}(dr/d\\theta)^2}d\\theta$ is different. Let the latter $ \\theta$ be $ \\Theta$ and the former be $ \\theta$, the relation $ \\theta\\equal{}\\Theta\\minus{}\\arctan (2\\pi\\minus{}\\theta)$ holds because $ \\theta$ is the angle of the tangent point, not a point in the curve. \r\n\r\n Using polar coordinate system is useless. We can simply find an expression of $ P(x,y)$ in Cartesian coordinate system enough, using parameter $ \\theta$ which is the angle of the tangent point. $ x\\equal{}\\cos\\theta\\minus{}(2\\pi\\minus{}\\theta)\\sin\\theta$ and $ y\\equal{}\\sin\\theta\\plus{}(2\\pi\\minus{}\\theta)\\cos\\theta$ holds. Then, we can apply the formula $ L\\equal{}\\int_{\\theta_i}^{\\theta_f}\\sqrt{(dx/d\\theta)^2\\plus{}(dy/d\\theta)^2}d\\theta$($ L$ is, of course, the length of the curve between $ \\theta\\equal{}\\theta_i$ to $ \\theta_f$) based on the Phytagorian theorem. \r\n\r\nIt gives $ \\int_{0}^{2\\pi}\\sqrt{(\\minus{}\\sin\\theta\\minus{}2\\pi\\cos\\theta\\plus{}\\sin\\theta\\plus{}\\theta\\cos\\theta)^2\\plus{}(\\cos\\theta\\minus{}2\\pi\\sin\\theta\\minus{}\\cos\\theta\\plus{}\\theta\\sin\\theta)^2}d\\theta$\r\n $ \\equal{}\\int_{0}^{2\\pi}\\sqrt{4\\pi^2\\minus{}4\\pi\\theta\\plus{}\\theta^2}d\\theta\\equal{}\\int_{0}^{2\\pi}(2\\pi\\minus{}\\theta) d\\theta\\equal{}\\boxed{2\\pi^2}$.", "Solution_12": "[quote=\"kunny\"]Yes.[/quote]\r\nSadly I was rotating it in clock wise direction :(", "Solution_13": "[quote=\"J.Y.Choi\"]\n Using polar coordinate system is useless. We can simply find an expression of $ P(x,y)$ in Cartesian coordinate system enough, using parameter $ \\theta$ which is the angle of the tangent point. $ x \\equal{} \\cos\\theta \\minus{} (2\\pi \\minus{} \\theta)\\sin\\theta$ and $ y \\equal{} \\sin\\theta \\plus{} (2\\pi \\minus{} \\theta)\\cos\\theta$ holds. Then, we can apply the formula $ L \\equal{} \\int_{\\theta_i}^{\\theta_f}\\sqrt {(dx/d\\theta)^2 \\plus{} (dy/d\\theta)^2}d\\theta$($ L$ is, of course, the length of the curve between $ \\theta \\equal{} \\theta_i$ to $ \\theta_f$) based on the Phytagorian theorem. \n\nIt gives $ \\int_{0}^{2\\pi}\\sqrt {( \\minus{} \\sin\\theta \\minus{} 2\\pi\\cos\\theta \\plus{} \\sin\\theta \\plus{} \\theta\\cos\\theta)^2 \\plus{} (\\cos\\theta \\minus{} 2\\pi\\sin\\theta \\minus{} \\cos\\theta \\plus{} \\theta\\sin\\theta)^2}d\\theta$\n $ \\equal{} \\int_{0}^{2\\pi}\\sqrt {4\\pi^2 \\minus{} 4\\pi\\theta \\plus{} \\theta^2}d\\theta \\equal{} \\int_{0}^{2\\pi}(2\\pi \\minus{} \\theta) d\\theta \\equal{} \\boxed{2\\pi^2}$.[/quote]\r\n\r\nYes, so did I. :lol:" } { "Tag": [], "Problem": "http://en.wikipedia.org/wiki/Rebus\r\n\r\nRebus marathon? Solve the previous rebus and post a new one. Or, just post a new one, and start a queue of unsolved rebuses, but you don't get any points for doing this.\r\n\r\nThere are two classes of rebuses: Those that are tricky and those that are just unbearably weird.\r\n\r\nFor example.\r\n[img]http://www.webmikey.com/images/wallpaper/starfire.jpg[/img] + [img]http://www.x-mencharacters.com/images/Scott.gif[/img] + [img]http://media.gamersroundtable.com/images/naruto/naruto.gif[/img] + [img]http://www.mad-cow.org/00/turtle.gif[/img] = ??\r\n\r\n\r\nOr\r\n\r\n[code]\n\n\nCCCSAILINGCCCC\n\n[/code] = ?", "Solution_1": "I solved the second one:\r\nsailing in the seas, right?", "Solution_2": "[quote=\"Koda\"]I solved the second one:\nsailing in the seas, right?[/quote]\r\n\r\nNope, sailing in the seven seas :D", "Solution_3": "try this one\r\n\r\n[b]sta4nce[/b]", "Solution_4": "The first one is:\r\n\r\nTeenage Mutant Ninja Turtle", "Solution_5": "[quote=\"jhredsox\"]try this one\n\n[b]sta4nce[/b][/quote]\r\nThat is easy, [b][u]for instance [/u][/b]I can think of a harder one! \r\n\r\n\r\n[u]ONHOLEE[/u]", "Solution_6": "[quote=\"jackdillon\"][quote=\"jhredsox\"]try this one\n\n[b]sta4nce[/b][/quote]\nThat is easy, [b][u]for instance [/u][/b]I can think of a harder one! \n\n\n[u]ONHOLEE[/u][/quote]\r\nhole in one", "Solution_7": "YOUJUSTME\r\n\r\n[code]\n\n1D 5U\n2R 6L\n3A 7A\n4C\n\n\n[/code]\n\n15S68A20F9E41T28Y975\n\n\n[code] D R\nS I\n K N\nHOUSE[/code]", "Solution_8": "just between you and me.\r\n\r\nthe others i can't get the exact things:\r\ndracula's letters?\r\nsafety among the numbers? (as in crowd)\r\n\r\nno idea about the last one - how are drinks adn house related?", "Solution_9": "Just between you and me\r\n\r\n\r\n\r\nSafety in numbers\r\n\r\nDrinks all around on the house", "Solution_10": "Count Dracula.\r\n\r\n\r\nHERE'S A REALLY GOOD ONE\r\n\r\nGEGS", "Solution_11": "scrambled eggs :D", "Solution_12": "HIJKLMNO\r\n\r\n(message too small :roll: )", "Solution_13": "$\\text{H}_{2}\\text{O}$\r\n\r\nCount Dracula was great!" } { "Tag": [], "Problem": "Source: ACoPS\r\nIf $x^2+y^2+z^2=49$ and $x+y+z=x^3+y^3+z^3=7$, find $xyz$.", "Solution_1": "[hide]\n$x^2 + y^2 + z^2 = (x+y+z)(x+y+z)$\nExpanding right hand side, then simplifying, then multiplying by (x+y+z)\n$(x+y+z)(xy + yz + xz) = 0$\n$3xyz + x^2(y+z) + y^2(x+z) + z^2(y+x) = 0$\nx+y+z = 7, so we sub to simplify..\n$3xyz + x^2(7 - x) + y^2(7 - y) + z^2(7 - z) = 0$\nsimplifying, then subbing back in\n$3xyz + 7*49 - 7 = 0$\ntherefore, xyz = -112\n[/hide]", "Solution_2": "[hide=\"short and sweet\"]$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$\n\n$\\implies xy+yz+zx=0$\n\n$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$\n\nSubstitute values and get the answer[/hide]", "Solution_3": "[quote=\"shadysaysurspammed\"][hide=\"short and sweet\"]$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$\n\n$\\implies xy+yz+zx=0$\n\n$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$\n\nSubstitute values and get the answer[/hide][/quote]\r\n\r\nJajaja, short and sweet....... nice one!" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Given a,b,c>0\r\nprove", "Solution_1": "See here : http://www.mathlinks.ro/viewtopic.php?t=56071", "Solution_2": "[quote=\"duc 95\"]Given a,b,c>0\nprove[/quote]\r\n\r\n$ \\frac {a}{\\sqrt {a \\plus{} b}} \\plus{} \\frac {b}{\\sqrt {b \\plus{} c}} \\plus{} \\frac {c}{\\sqrt {c \\plus{} a}}\\geq \\frac {\\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c}}{\\sqrt{2}}$" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "[size=134]Find a splitting field $ K \\subseteq \\mathbb{C}$ for $ x^5\\minus{}1 \\in \\mathbb{Q}[x],$ and determine $ [K: Q]$[/size]", "Solution_1": "$ x^n \\minus{} 1 \\equal{} \\prod_{k\\equal{}0}^{n} (x \\minus{} e^{ \\frac{2 \\pi i k}{n} } )$ over $ \\mathbb{C}$, so the splitting field of $ x^n \\minus{} 1$ over the rationals is the [url=http://mathworld.wolfram.com/CyclotomicField.html]cyclotomic field[/url] $ \\mathbb{Q}[ e^{\\frac{2 \\pi i}{n} } ]$." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Give example of $f \\in L^2(\\mathbb{R}) - L^1(\\mathbb{R})$ such that $\\widehat{f} \\in L^1(\\mathbb{R})$. Under what circumstances can this happen ?", "Solution_1": "A function that has very fast decay but is not very smooth will have a Fourier transform that is quite smooth but very slow decay. \r\n\r\nSuppose $\\widehat{f} = \\chi_{[-1, 1]}$. Then $f$ is $\\frac{\\sin(2\\pi \\theta)}{\\pi \\theta}$, I believe. Does this work?" } { "Tag": [ "projective geometry", "geometry proposed", "geometry" ], "Problem": "Triangle $ABC$ is acute-angled.$M$ is a midpoint of bisector $AA_{1}$.$P$ and $Q$ are points on $MB$ and $MC$ respectively.$\\measuredangle{APC}=\\measuredangle{AQB}=\\frac{\\pi}{2}$.Prove that $A_{1}PMQ$ is cyclic.", "Solution_1": "[hide] This problem is hard. But if you make inversion in point A it will become an easy exersise for using symmetry. :wink: \nThere is other two provings.They didn't use inversion.\nOne is hard and long.\nSecond is long but not hard.Consider the line $l \\parallel AM, B \\in l$\nConsider the altitude from A to l and circle AQB.We get some eqeations between angles.And then a lot of calcules... \n[/hide]", "Solution_2": "let $l$ be a line through $B$ and parallel with $AM$. $CM$ meets $l$ at $D$\r\nWe have $MC/CD=MA_{1}/BD=AM/BD=EM/ED$ so $(C,E,M,D)=-1$\r\nBut $AM$ is the bisector of $\\angle EAC$ so $DA\\bot AM$\r\nOtherwise $BD//AM$ so $BD\\bot AD$\r\nThis means $A,D,B,Q$ are cyclic\r\nSo $\\angle MCA+\\angle MAC=\\angle A_{1}MC=\\angle BDC=\\angle BAQ=\\angle BAM+\\angle MAQ$\r\nso $\\angle MCA=\\angle MAQ$\r\nSo triangle $MAQ$ and $MCA$ are similar\r\nthus $MA_{1}^{2}=MA^{2}=MQ*MC$\r\nso triangle $MQA_{1}$ and $MA_{1}C$ are similar\r\nSo $\\angle MQA_{1}=\\angle MA_{1}C$\r\nSimilarly $\\angle MPA_{1}=\\angle MA_{1}B$ . But $\\angle MA_{1}B+\\angle MA_{1}C=180$\r\nso $\\angle MQA_{1}+\\angle MPA_{1}=180$\r\nThis means $M,Q,A_{1},P$ are cyclic $q.e.d$", "Solution_3": "[quote=\"Huy\u1ec1n V\u0169\"]let $l$ be a line through $B$ and parallel with $AM$. $CM$ meets $l$ at $D$\nWe have $MC/CD=MA_{1}/BD=AM/BD=EM/ED$ so $(C,E,M,D)=-1$\nBut $AM$ is the bisector of $\\angle EAC$ so $DA\\bot AM$\nOtherwise $BD//AM$ so $BD\\bot AD$\nThis means $A,D,B,Q$ are cyclic\nSo $\\angle MCA+\\angle MAC=\\angle A_{1}MC=\\angle BDC=\\angle BAQ=\\angle BAM+\\angle MAQ$\nso $\\angle MCA=\\angle MAQ$\nSo triangle $MAQ$ and $MCA$ are similar\nthus $MA_{1}^{2}=MA^{2}=MQ*MC$\nso triangle $MQA_{1}$ and $MA_{1}C$ are similar\nSo $\\angle MQA_{1}=\\angle MA_{1}C$\nSimilarly $\\angle MPA_{1}=\\angle MA_{1}B$ . But $\\angle MA_{1}B+\\angle MA_{1}C=180$\nso $\\angle MQA_{1}+\\angle MPA_{1}=180$\nThis means $M,Q,A_{1},P$ are cyclic $q.e.d$[/quote]\r\n\r\n[b] NOTE[/b]\r\n Please ,let me ''improve'' (make easier?) this [color=red]nice[/color] solution in the first step , avoinding the harmonic division:\r\n If line $BD$ meets line $CA$ at $Z$ , then since $BZ // AA_{1}$ we take that triangle $ABZ$ is isosceles and since $M$ is the mid point of $AA_{1}$, point $D$ will be the mid point of $BE$.Thus $AD \\perp BZ$.\r\n\r\n[u] Babis[/u]", "Solution_4": "OH! Your note is very nice stergiu. Thanks", "Solution_5": "[Hide=solutions]My solutions \n[url]https://artofproblemsolving.com/community/c6h1832889p12277985[url]\n\n$(MA_{1})^2=MA^2=MP.MB$ $\\rightarrow$ $\\angle{MA_{1}P}=\\angle{MBC}$\n$\\angle{APM}=\\angle{AQM}$$\\rightarrow$ $BPQC$ is concyclic $\\angle{MBC}=\\angle{PQM}$\n$\\angle{MA_{1}P}=\\angle{PQM}$ $\\rightarrow$ $MPA_{1}Q$ is cyclic" } { "Tag": [ "inequalities", "function", "inequalities open" ], "Problem": "Let $ a,b,c,d$ non-negative numbers such that $ a^{6}\\plus{}b^{6}\\plus{}c^{6}\\plus{}d^{6}\\equal{}1$, then\r\n\\[ \\sqrt{a^{4}\\plus{}b^{4}\\plus{}c^{4}\\plus{}d^{4}}\\geq\\sqrt[3]{a^{3}\\plus{}b^{3}\\plus{}c^{3}\\plus{}d^{3}}\\]", "Solution_1": "[quote=\"Sunjee\"]Let $ a,b,c,d$ non-negative numbers such that $ a^{6}+b^{6}+c^{6}+d^{6}= 1$, then\n\n$ \\sqrt{a^{4}+b^{4}+c^{4}+d^{4}}\\geq\\sqrt [3]{a^{3}+b^{3}+c^{3}+d^{3}}$\n\n[/quote]\r\n\r\nI have solved a same problem with $ 3$-variable.\r\n\r\n\r\n$ \\sqrt{a^{4}+b^{4}+c^{4}}\\leq\\sqrt [3]{a^{3}+b^{3}+c^{3}}$(*)\r\n\r\nWe will prove (*) inequality .\r\n\r\n(*)$ \\Leftrightarrow\\sum3a^{8}.b^{4}+6a^{4}b^{4}c^{4}-\\sum2a^{6}b^{6}-\\sum2a^{9}b^{3}-\\sum2c^{6}a^{3}b^{3})\\le 0$\r\n\r\nBy Muirhead 'inequality , $ \\sum a^{8}b^{4}+b^{8}a^{4}\\le\\sum a^{9}b^{3}$\r\n\r\n$ 6a^{4}b^{4}c^{4}\\le2\\sum a^{6}b^{3}c^{3}$.\r\n\r\nAnd $ \\sum a^{9}b^{3}-2\\sum a^{8}b^{4}+\\sum a^{6}b^{6}\\ge 0\\Leftrightarrow (a-b)^{2}(a^{4}-a^{2}b^{2}+b^{4})\\ge 0$\r\n\r\nWe're done.\r\nEqualiry occurs onliy if $ a = b = c$", "Solution_2": "[quote=\"Sunjee\"]Let $ a,b,c,d$ non-negative numbers such that $ a^{6}\\plus{}b^{6}\\plus{}c^{6}\\plus{}d^{6}\\equal{} 1$, then\n\\[ \\sqrt{a^{4}\\plus{}b^{4}\\plus{}c^{4}\\plus{}d^{4}}\\geq\\sqrt[3]{a^{3}\\plus{}b^{3}\\plus{}c^{3}\\plus{}d^{3}}\\]\n[/quote]\r\nI don't think it is true", "Solution_3": "[quote=\"Harry Potter\"][quote=\"Sunjee\"]Let $ a,b,c,d$ non-negative numbers such that $ a^{6}\\plus{}b^{6}\\plus{}c^{6}\\plus{}d^{6}\\equal{} 1$, then\n\\[ \\sqrt{a^{4}\\plus{}b^{4}\\plus{}c^{4}\\plus{}d^{4}}\\geq\\sqrt[3]{a^{3}\\plus{}b^{3}\\plus{}c^{3}\\plus{}d^{3}}\\]\n[/quote]\nI don't think it is true[/quote]\r\n\r\nuhm, for $ 3$-variable, Sunjee 'inequality is true ( follow my proof ) . \r\n\r\nAnd for $ 4$ - variable , i think, the inequality must be :\r\n\r\n$ \\sqrt{a^{4}\\plus{}b^{4}\\plus{}c^{4}\\plus{}d^{4}}\\le\\sqrt[3]{a^{3}\\plus{}b^{3}\\plus{}c^{3}\\plus{}d^{3}}$\r\n\r\nI am true ?? :blush:", "Solution_4": "The inequality inversely holds!\r\n\r\nsee:http://www.irgoc.org/bbs/dispbbs.asp?boardid=12&id=2696&star=2#8228", "Solution_5": "[quote=\"fjwxcsl\"]The inequality inversely holds!\n\nsee:http://www.irgoc.org/bbs/dispbbs.asp?boardid=12&id=2696&star=2#8228[/quote]\r\nSorry I dont understand this language...If its inversely holds,show me please example.", "Solution_6": "[quote=\"Sunjee\"][quote=\"fjwxcsl\"]The inequality inversely holds!\n\nsee:http://www.irgoc.org/bbs/dispbbs.asp?boardid=12&id=2696&star=2#8228[/quote]\nSorry I dont understand this language...If its inversely holds,show me please example.[/quote]\r\n\r\nI have root this solution from fjwxcsl'link . \r\n\r\n\r\n$ (x^{3}\\plus{}y^{3}\\plus{}z^{3})^{2}(x^{6}\\plus{}y^{6}\\plus{}z^{6})\\minus{}(x^{4}\\plus{}y^{4}\\plus{}z^{4})^{3}\\equal{}\\sum f(x,y,z)(x\\minus{}y)(y\\minus{}z)$\r\n\r\nwith $ f$ is a 3-variable function so that if $ x\\ge y\\ge z$ then: $ \\sqrt{f(x,y,z)}\\minus{}\\sqrt{f(z,x,y)}\\plus{}\\sqrt{f(y,x,z)}\\ge 0$\r\n\r\nthen after some argument, we have $ \\sum f(x,y,z)(x\\minus{}y)(y\\minus{}z)\\ge 0$.\r\n\r\nI have seen a same argument :\r\n\r\nLet $ f:\\mathbb{R^{\\plus{}}}\\to\\mathbb{R^{\\plus{}}}$ so that :\r\n\r\n$ \\sqrt{f(x)}\\plus{}\\sqrt{f(y)}\\ge\\sqrt{f(z)}$ with$ x\\ge y\\ge z > 0$ .\r\nThen :\r\n$ \\sum_{\\text{a,b,c}}f(a)(a\\minus{}b)(b\\minus{}c)\\ge 0$\r\n\r\nBut it is easier than own inequality . :blush:", "Solution_7": "hi Sunjee $ a\\equal{}b\\equal{}0$ $ c\\equal{}\\frac{1}{2}$ $ d\\equal{}\\frac{\\sqrt[6]{63}}{2}$ not true", "Solution_8": "Who can you solve the inequality by the way from the link :\r\n\r\nhttp://www.irgoc.org/bbs/dispbbs.asp?boardid=12&id=2696&star=2#8228\r\n\r\n :)", "Solution_9": "[quote=\"Sunjee\"]Let $ a,b,c,d$ non-negative numbers such that $ a^{6}\\plus{}b^{6}\\plus{}c^{6}\\plus{}d^{6}\\equal{} 1$, then\n\\[ \\sqrt{a^{4}\\plus{}b^{4}\\plus{}c^{4}\\plus{}d^{4}}\\geq\\sqrt[3]{a^{3}\\plus{}b^{3}\\plus{}c^{3}\\plus{}d^{3}}\\]\n[/quote] Sorry everybody,in my proof i mistyped,so \r\n\\[ \\sqrt{a^{4}\\plus{}b^{4}\\plus{}c^{4}\\plus{}d^{4}}\\leq\\sqrt[3]{a^{3}\\plus{}b^{3}\\plus{}c^{3}\\plus{}d^{3}}\\]\r\nin general case: if $ \\sum x^{6}_{i}\\equal{}1$ then\r\n\\[ \\sqrt{\\sum x^{4}_{i}}\\leq\\sqrt[3]{\\sum x^{3}_{i}}\\]\r\nthank you.", "Solution_10": "The inequality is equivalent to :\r\n\r\n$ (x^{4}\\plus{}y^{4}\\plus{}z^{4})^{3}\\le (x^{3}\\plus{}y^{3}\\plus{}z^{3})^{2}(x^{6}\\plus{}y^{6}\\plus{}z^{6})$.\r\n\r\nBy Holder , \r\n\r\n$ (x^{3}\\plus{}y^{3}\\plus{}z^{3})(x^{3}\\plus{}y^{3}\\plus{}z^{3})(x^{6}\\plus{}y^{6}\\plus{}z^{6})\\ge (\\sum\\sqrt [3]{x^{3}x^{3}x^{6}})^{3}\\equal{} (x^{4}\\plus{}y^{4}\\plus{}z^{4})^{3}$. :wink:", "Solution_11": "[quote=\"quangpbc\"]The inequality is equivalent to :\n\n$ (x^{4}\\plus{}y^{4}\\plus{}z^{4})^{3}\\le (x^{3}\\plus{}y^{3}\\plus{}z^{3})^{2}(x^{6}\\plus{}y^{6}\\plus{}z^{6})$.\n\nBy Holder , \n\n$ (x^{3}\\plus{}y^{3}\\plus{}z^{3})(x^{3}\\plus{}y^{3}\\plus{}z^{3})(x^{6}\\plus{}y^{6}\\plus{}z^{6})\\ge (\\sum\\sqrt [3]{x^{3}x^{3}x^{6}})^{3}\\equal{} (x^{4}\\plus{}y^{4}\\plus{}z^{4})^{3}$. :wink:[/quote]\r\nOk.If you used HOlder's inequality,equality occurs when $ (a,b,c)\\equal{}(t,t,t)$ here $ 3t^{6}\\equal{}1$...\r\nIn my inequality equality occurs if $ n\\equal{}3$ then $ (a,b,c)\\equal{}(t,t,t)$ and $ (a,b,c)\\equal{}(m,m,0)$ here $ 2m^{6}\\equal{}1$ and $ (a,b,c)\\equal{}(1,0,0)$\r\nso i think holder not work here. I want general solution...", "Solution_12": "[quote=\"Sunjee\"][quote=\"quangpbc\"]The inequality is equivalent to :\n\n$ (x^{4}\\plus{}y^{4}\\plus{}z^{4})^{3}\\le (x^{3}\\plus{}y^{3}\\plus{}z^{3})^{2}(x^{6}\\plus{}y^{6}\\plus{}z^{6})$.\n\nBy Holder , \n\n$ (x^{3}\\plus{}y^{3}\\plus{}z^{3})(x^{3}\\plus{}y^{3}\\plus{}z^{3})(x^{6}\\plus{}y^{6}\\plus{}z^{6})\\ge (\\sum\\sqrt [3]{x^{3}x^{3}x^{6}})^{3}\\equal{} (x^{4}\\plus{}y^{4}\\plus{}z^{4})^{3}$. :wink:[/quote]\nOk.If you used HOlder's inequality,equality occurs when $ (a,b,c) \\equal{} (t,t,t)$ here $ 3t^{6}\\equal{} 1$...\nIn my inequality equality occurs if $ n \\equal{} 3$ then $ (a,b,c) \\equal{} (t,t,t)$ and $ (a,b,c) \\equal{} (m,m,0)$ here $ 2m^{6}\\equal{} 1$ and $ (a,b,c) \\equal{} (1,0,0)$\nso i think holder not work here. I want general solution...[/quote]\r\n\r\n :D For $ n\\equal{}3$ i have posted a solution above by Muirhead , but the inequality occurs if $ a\\equal{}b\\equal{}c$ or or $ a\\equal{}b,c\\equal{}0$or $ b\\equal{}c\\equal{}0$ and cyclic . ( i have typed only $ a\\equal{}b\\equal{}c$. Sorry :blush: .Do you have a solution for $ n\\equal{}4$. I think it can be solve by Muirhead, ( simirlary for $ n\\equal{}3$ )", "Solution_13": "Holder works for this one, as well as the $ n$-variable case\r\n\\[ \\left(\\sum_{k\\equal{}1}^{n}a_{k}^{6}\\right)\\left(\\sum_{k\\equal{}1}^{n}a_{k}^{3}\\right)^{2}\\geq\\left(\\sum_{k\\equal{}1}^{n}a_{k}^{4}\\right)^{3}\\]\r\nas 6+2*3=3*4.\r\n\r\nThere's nothing wrong with the equality case, either. There will be equality as long as there exist nonzero constants $ x,y,z$ such that\r\n\\[ xa_{k}^{6}\\equal{}ya_{k}^{3}\\equal{}za_{k}^{4}\\]\r\nfor all $ k$, which not only occurs when all $ a_{k}$ are equal, but when all [i]nonzero[/i] $ a_{k}$ are equal. (For a simpler example, Cauchy, a special case of Holder, proves $ (a^{3}\\plus{}b^{3})(a\\plus{}b)\\geq (a^{2}\\plus{}b^{2})^{2}$ for nonnegative $ a,b$ even though there's equality when $ b \\equal{} 0$.)" } { "Tag": [ "inequalities" ], "Problem": "Let's $a,$ $b$ and $c$ are non-negative numbers and $f$ is increasing positive function.\r\nProve that $\\frac{ab+ac-2bc}{f(b)+f(c)}+\\frac{ab+bc-2ac}{f(a)+f(c)}+\\frac{ac+bc-2ab}{f(a)+f(b)}\\geq0.$", "Solution_1": "[hide]\nsince f is positive and increasing, let $f(a) \\ge f(b) \\ge f(c)$\nso \n$f(a)+f(b) \\ge f(b)+f(c)$\n$f(a)+f(b) \\ge f(a)+f(c)$\n\nand $\\frac{ab+ac-2bc}{f(b)+f(c)}+\\frac{ab+bc-2ac}{f(a)+f(c)}+\\frac{ac+bc-2ab}{f(a)+f(b)}\\ge$$\\frac{ab+ac-2bc}{f(a)+f(b)}+\\frac{ab+bc-2ac}{f(a)+f(b)}+\\frac{ac+bc-2ab}{f(a)+f(b)}$\nand $\\frac{1}{f(a)+f(b)}(2ab+2ac+2bc-2ab-2ac-2bc) \\ge 0 \\Longleftrightarrow 0 \\ge 0$[/hide]", "Solution_2": "[quote=\"hydro\"][hide]\nsince f is positive and increasing, let $f(a) \\ge f(b) \\ge f(c)$\nso \n$f(a)+f(b) \\ge f(b)+f(c)$\n$f(a)+f(b) \\ge f(a)+f(c)$\n\nand $\\frac{ab+ac-2bc}{f(b)+f(c)}+\\frac{ab+bc-2ac}{f(a)+f(c)}+\\frac{ac+bc-2ab}{f(a)+f(b)}\\ge$$\\frac{ab+ac-2bc}{f(a)+f(b)}+\\frac{ab+bc-2ac}{f(a)+f(b)}+\\frac{ac+bc-2ab}{f(a)+f(b)}$\nand $\\frac{1}{f(a)+f(b)}(2ab+2ac+2bc-2ab-2ac-2bc) \\ge 0 \\Longleftrightarrow 0 \\ge 0$[/hide][/quote]\r\nWhy $\\frac{ab+ac-2bc}{f(b)+f(c)}+\\frac{ab+bc-2ac}{f(a)+f(c)}\\ge\\frac{ab+ac-2bc}{f(a)+f(b)}+\\frac{ab+bc-2ac}{f(a)+f(b)}$ $?$ :wink:\r\nThank you." } { "Tag": [], "Problem": "Four tangent circles are centered on the x-axis. Circle O is centered at the origin. The radius of Circle A is twice the radius of Circle O. The radius of Circle B is three times the radius of Circle O. The radius of Circle C is four times the radius of Circle O. All circles have integer radii and the point (63,16) is on Circle C. What is the equation of Circle A?", "Solution_1": "Was not this answered in Basic?", "Solution_2": "nope. i was told to move it to intermediate.", "Solution_3": "[quote=\"7h3.D3m0n.117\"]\nThis should go into Intermediate :?:[/quote]\r\n\r\nOk, sorry.", "Solution_4": "Im just going to assume that they are in the order on the x axis, O, A, B ,C.\r\n\r\nlet the radius for O be $r$, this implies that the radius for A is $2r$, B is $3r$ and C is $4r$.\r\n\r\nThis means that the center of C is $r+2*2r+2*3r+4r = 15r$ from the origin\r\n\r\nSo we can set up an equation for circle C\r\n$(x-15r)^{2}+y^{2}= (4r)^{2}= 16r^{2}$\r\n\r\nSince C is on the point (63,16), we can plug in these x and y values into our equation and solve for r.\r\n\r\n$(63-15r)^{2}+16^{2}= 16r^{2}$\r\n$\\Rightarrow 63^{2}-2(63)(15r)+225r^{2}+256 = 16r^{2}$\r\n$\\Rightarrow 3969+256-1890r+225r^{2}= 16r^{2}$\r\n$\\Rightarrow 209r^{2}-1890r+4225 = 0$\r\n$\\Rightarrow r_{1,2}= \\frac{1890 \\pm 200}{418}$\r\n$r_{1}= 5$, $r_{2}= 4.04306....$\r\n$r_{2}$ is inadmissable since the radii are of integer lengths.\r\n\r\nSince we now know the radius, we can begin to construct an equation for A.\r\n\r\nCircle O is centered on the Origin with radius 5, thus its max point on x is 5. Circle A follows, but its radius is two times that of circle O, it follows that the center of circle A is on $x = 15$\r\nSo our equation for circle A is:\r\n$\\boxed{(x-15)^{2}+y^{2}= 100}$", "Solution_5": "[quote=\"sharkboy\"]nope. i was told to move it to intermediate.[/quote]\r\n\r\npsh by \"move\" i was talking to the mods. i didn't mean for u to make a new topic." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all $ f : \\mathbb R \\rightarrow \\mathbb R$ functions such that for all reals $ x$ and $ y$ we have:\r\n\r\n$ f(x^{3})\\minus{}f(y^{3})\\equal{}(x^{2}\\plus{}xy\\plus{}y^{2})(f(x)\\minus{}f(y))$", "Solution_1": "[color=green]This problem is easy. And the same problem is the problem below\n\nFind all functions $ f : \\mathbb R^ \\plus{} \\longrightarrow \\mathbb R^ \\plus{}$ such that for all positive reals $ x$ and $ y$ we have:\n\n$ f(x^{3}) \\minus{} f(y^{3}) \\equal{} (x^{2} \\plus{} xy \\plus{} y^{2})(f(x) \\minus{} f(y))$.[/color]", "Solution_2": "[quote=\"delegat\"]Find all $ f : \\mathbb R \\rightarrow \\mathbb R$ functions such that for all reals $ x$ and $ y$ we have:\n\n$ f(x^{3}) \\minus{} f(y^{3}) \\equal{} (x^{2} \\plus{} xy \\plus{} y^{2})(f(x) \\minus{} f(y))$[/quote]\r\nHere is my solution \r\n$ g(x) \\equal{} f(x) \\minus{} f(0)$ then \r\n$ g(x^3) \\minus{} g(y^3) \\equal{} (x^2 \\plus{} xy \\plus{} y^2)(g(x) \\minus{} g(y))$\r\nTake $ y \\equal{} 0$ then \r\n$ g(x^3) \\equal{} x^2g(x)$\r\nSo $ x^2g(x) \\minus{} y^2g(y) \\equal{} (x^2 \\plus{} xy \\plus{} y^2)(g(x) \\minus{} g(y))$\r\n$ \\Leftrightarrow (x \\plus{} y)(xg(y) \\minus{} yg(x)) \\equal{} 0$\r\nTake $ y \\equal{} 1$ then \r\n$ g(x) \\equal{} g(1)x$ for all x is different from $ \\minus{} 1$ (1)\r\nReplace to the equation we have \r\n$ g(1)x^3 \\minus{} g( \\minus{} 1) \\equal{} (x^2 \\minus{} x \\plus{} 1)(g(1)x \\minus{} g( \\minus{} 1))$\r\nSo $ g( \\minus{} 1) \\plus{} g(1) \\equal{} 0$ (2)\r\nFrom (1)+(2) then \r\n$ g(x) \\equal{} ax$\r\nIt follow that $ f(x) \\equal{} ax \\plus{} b$", "Solution_3": "[color=green]For the second problem in this topic, who can solve it? :) [/color]", "Solution_4": "Set $ g: \\mathbf{R}^ \\plus{} \\to \\mathbf{R}$, $ g(x) \\equal{} f(x) \\minus{} f(1)$. Hence $ g(1) \\equal{} 0$ and $ g$ satisfy same equation. Setting $ y \\equal{} 1$ gives $ g(x^3) \\equal{} g(x)(x^2 \\plus{} x \\plus{} 1)$. This implies $ g(x)(x \\plus{} 1 \\minus{} xy \\minus{} y^2) \\equal{} g(y)(y \\plus{} 1 \\minus{} xy \\minus{} x^2)$, which gives $ g(x)(1 \\minus{} y) \\equal{} g(y)(1 \\minus{} x)$ . Therefore $ g$ is linear function and so it is $ f$." } { "Tag": [ "algebra", "polynomial", "modular arithmetic", "quadratics", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ p$ be a primer number. Determine the number of points in the quadric of $ \\mathbb{Z}_{p}^{2}$ given by: $ x^{2}\\plus{}y^{2} \\equal{} a$, $ a\\in \\mathbb{Z}_{p}$.", "Solution_1": "any solution?", "Solution_2": "Assume $ p > 2, a \\neq 0$. Then the associated projective variety is isomorphic to $ \\mathbf{P}^1_{\\mathbf{F}_p}$ as it has a rational point (counting argument). Now count the points at infinity.", "Solution_3": "something more elementary? this problem is desinged to be in a undergraduate contest", "Solution_4": "Compute $ \\sum_{x\\plus{}y\\equal{}a}\\left(1\\plus{}\\left(\\frac{x}{p}\\right)\\right)\\left(1\\plus{}\\left(\\frac{y}{p}\\right)\\right)$", "Solution_5": "why is this useful? can you write down a complete proof pleasE?", "Solution_6": "$ \\ldots \\equal{} p \\plus{} \\sum_{x\\equal{}0}^{p\\minus{}1}\\left(\\left(\\frac{x}{p}\\right) \\plus{} \\left(\\frac{a\\minus{}x}{p} \\plus{} \\left(\\frac{x(a\\minus{}x)}{p}\\right)\\right)\\right) \\equal{} \\sum_{x\\equal{}1}^{p\\minus{}1}\\left(\\frac{x^{\\minus{}1}}{p}\\right)^2\\left(\\frac{x(a\\minus{}x)}{p}\\right) \\equal{} \\ldots$", "Solution_7": "let's work on that computation", "Solution_8": "Note that the number of solutions in $ \\mathbf{F}_p$ to $ z^2 \\equal{} x$ is equal to $ 1\\plus{}\\left(\\frac{x}{p}\\right)$.", "Solution_9": "I don't know what is [b]$ F_{p}$[/b] please, can you write down all details? it would be great ... :lol:", "Solution_10": "[quote=\"Lousin Garckz\"]I don't know what is [b]$ F_{p}$[/b] please, can you write down all details? it would be great ... :lol:[/quote]\r\n\r\n$ F_p$ is the field on $ p$ elements. Its another way to write $ \\mathbb{Z}_p$ if you don't want to confuse it with the $ p$-adics and if you want to emphasize that it's a field.\r\n\r\nThere's not many details to write down there: he just wrote that there are two solutions if $ x$ is a square mod $ p$, and zero otherwise.\r\n\r\nEDIT: Or 1, if the RHS is zero.", "Solution_11": "i can't get it :(", "Solution_12": "If $ F$ is a field of characteristic different from 2, so that $ 2a \\equal{} 0 \\implies a \\equal{} 0$, then $ z^2 \\equal{} x$ has either 0 or 2 roots -- 1 iff $ x \\equal{} 0$. It can't have more than two roots, since a polynomial of degree $ n$ has as most $ n$ roots over a field, and if $ r^2 \\equal{} x$ then $ ( \\minus{} r)^2 \\equal{} x$, so a root $ r$ implies the existence of a second roots $ \\minus{} r$, provided $ r$ is nonzero.\r\n\r\nDoes that make sense?", "Solution_13": "yes, I understand it, but I don't understand how is related it with the topic, I mean, how can I use it to find the number of pairs $ (x,y)$ such that $ x^{2}\\plus{}y^{2} \\equal{} a$.", "Solution_14": "[quote=\"myan\"]Compute $ \\sum_{x \\plus{} y \\equal{} a}\\left(1 \\plus{} \\left(\\frac {x}{p}\\right)\\right)\\left(1 \\plus{} \\left(\\frac {y}{p}\\right)\\right)$[/quote]\r\n\r\nThis computes the number of ways to write $ a \\equal{} x \\plus{} y$, and to furthermore find $ z$ and $ w$ such that $ z^2 \\equal{} x$ and $ w^2 \\equal{} y$. So, it counts all solutions to $ z^2 \\plus{} w^2 \\equal{} a$.", "Solution_15": "I got that part, but, actually what is $ (\\dfrac{x}{p})$ is this the Jacobi symbol? can you compute the sum please?", "Solution_16": "It's the Legendre symbol (since $ p$ is prime). myan started expanding the sum. You should give it a try first :).", "Solution_17": "ok I'll try it, i'm really sorry about my poor knowledge in number theory and been so insistent, I will try!..", "Solution_18": "Nukular, i can't get the result :( .... help :blush:", "Solution_19": "This is just an outline... if you need more, I can flesh out the details.\r\n\r\nYou should be able to reduce it to a calculation of $ \\sum_{x \\equal{} 1}^{p \\minus{} 1} \\left( \\frac {x(a \\minus{} x)}{p} \\right)$. Then you can replace that with $ \\sum_{x\\equal{}1}^{p\\minus{}1} \\left( \\frac{ax^{\\minus{}1} \\minus{} 1}{p} \\right)$, factoring out the $ x^2$. \r\n\r\nYou get two cases: either $ a \\equiv 0 \\pmod{p}$ or not. The first case should be relatively easy. I'll assume we're not there for now.\r\n\r\nThen $ ax^{\\minus{}1}$ runs over all nonzero residues mod $ p$; therefore $ ax^{\\minus{}1} \\minus{} 1$ runs from $ 0,1,\\dots,p\\minus{}2$. So then you compute $ \\sum_{y \\equal{} 0}^{p\\minus{}2} \\left( \\frac{y}{p} \\right) \\equal{} \\minus{}\\left(\\frac{\\minus{}1}{p}\\right) \\equal{} \\minus{}(\\minus{}1)^{(p\\minus{}1)/2}$.\r\n\r\nI think the answer then ends up being something like $ p \\minus{} (\\minus{}1)^{(p\\minus{}1)/2}$, but I might be off by one or something.\r\n\r\n(Sorry, written in haste and organized for crap. If you need more help I'll clean it up a bit and take care of the constants that fall out.)", "Solution_20": "nop, i have problem manipulating the Legendre symbol, i don't know why i can't do nothing :(", "Solution_21": "Re-check my previous post, I edited it while you were posting to have some more helpful details. \r\n\r\nUseful facts:\r\n\r\n$ \\left(\\frac{x}{p}\\right)\\left(\\frac{y}{p}\\right) \\equal{} \\left(\\frac{xy}{p}\\right)$\r\n$ \\sum_{i\\equal{}0}^p \\left(\\frac{i}{p}\\right) \\equal{} 0.$\r\n$ \\left(\\frac{\\minus{}1}{p}\\right) \\equal{} (\\minus{}1)^{(p\\minus{}1)/2}.$\r\n\r\nAnd the ever-useful (though not useful here) Quadratic Reciprocity Law, which you should learn at some point: for odd primes $ p$ and $ q$:\r\n$ \\left(\\frac{p}{q}\\right)\\left(\\frac{q}{p}\\right) \\equal{} (\\minus{}1)^{(p\\minus{}1)(q\\minus{}1)/4}.$\r\n\r\nThose should get you started.", "Solution_22": "OK, let's see: Using Nukular arguments and recopilating to write more clear:\r\n\r\n$ \\sum_{x\\plus{}y\\equal{}a}(1\\plus{}(\\dfrac{x}{p}))(1\\plus{}(\\dfrac{y}{p})) \\equal{} \\sum_{x\\plus{}y\\equal{}a}1\\plus{}(\\dfrac{y}{p})\\plus{}(\\dfrac{x}{p})\\plus{}(\\dfrac{xy}{p}) \\equal{} \\sum_{x\\equal{}0}^{p\\minus{}1}1\\plus{}(\\dfrac{a\\minus{}x}{p})\\plus{}(\\dfrac{x}{p})\\plus{}(\\dfrac{x(a\\minus{}x)}{p}) \\equal{} \\sum_{x\\equal{}0}^{p\\minus{}1}1\\plus{}(\\dfrac{x(a\\minus{}x)}{p}) \\equal{} p \\plus{} \\sum_{x\\equal{}0}^{p\\minus{}1}(\\dfrac{x(a\\minus{}x)}{p}) \\equal{} 1 \\plus{} p \\plus{}\\sum_{x\\equal{}1}^{p\\minus{}1}(\\dfrac{x(a\\minus{}x)}{p}) \\equal{} 1 \\plus{} p \\plus{} \\sum_{x\\equal{}1}^{p\\minus{}1}(\\dfrac{xa^{\\minus{}1}\\minus{}1}{p})$. As nukular said, let's think about this by part. If $ a \\equiv 0 (p) \\Rightarrow ax^{\\minus{}1}\\minus{}1 \\equiv \\minus{}1 (p) \\Rightarrow (\\dfrac{ax^{\\minus{}1}\\minus{}1}{p}) \\equal{} (\\minus{}1)^{\\dfrac{(p\\minus{}1)}{2}}$ and the final result is $ 1 \\plus{} p \\plus{} p (\\minus{}1)^{\\dfrac{(p\\minus{}1)}{2}}$. By the other hand, if $ a \\neq 0 (p)$ then $ ax^{\\minus{}1}$ runs over all non-zero residues mod $ p$ then $ \\sum_{x\\equal{}1}^{p\\minus{}1} (\\dfrac{ax^{\\minus{}1}\\minus{}1}{p}) \\equal{} \\sum_{y\\equal{}0}^{p\\minus{}2} (\\dfrac{y}{p}) \\equal{} \\minus{}(\\dfrac{\\minus{}1}{p}) \\equal{} \\minus{}(\\minus{}1)^{\\dfrac{(p\\minus{}1)}{2}}$ and the final result is $ p\\plus{}1\\minus{}(\\minus{}1)^{\\dfrac{(p\\minus{}1)}{2}}$. Is all this right?\r\n\r\nP/D: Nukular, about what you wrote in other post of mine, believe me that I'm working so hard on all that problems, but I have pretty much work to do and is for tomorrow, and i Can't, i mean, I know that this is not of your interest but at least i want to give an explanation.", "Solution_23": "It looks fine, except that $ \\left(\\frac{0}{p}\\right) \\equal{} 0$, not $ 1$.", "Solution_24": "[quote=\"Lousin Garckz\"]\nP/D: Nukular, about what you wrote in other post of mine, believe me that I'm working so hard on all that problems, but I have pretty much work to do and is for tomorrow, and i Can't, i mean, I know that this is not of your interest but at least i want to give an explanation.[/quote]\r\n\r\nOh, if it's for homework, then, good luck with it! I know the stress of a problem set due the next day firsthand :).", "Solution_25": "all right, and thank you very very much Nukular!!!! you are great!" } { "Tag": [ "analytic geometry", "function", "algebra solved", "algebra" ], "Problem": "We have a curve in space (you may consider it to be a polygonal line). Prove that there are 4 coplanar on this curve which divide it in 4 parts of equal lengths.", "Solution_1": "First of all, I only solved the problem for a CLOSED CURVE of FINITE LENGTH. If it wasn't closed then the four points would divide the curve into five parts. If the curve didn't have a finite length, then there is no point in defining 'equal lengths' for the pieces. Assume that the curve has length L>0. We fix a point O on the curve and we say that a point X has coordinate 0<=u<=L and we write X(u) if the length of the arc of the curve from O to X when going in the clockwise direction is u.\r\n\r\nAlso by convention O(0) and the point with coordinate u+L has coordinate u, for all real u. \r\n\r\nNow, take an x>0 , 0<=x<=L and consider the points A(x) , B(x+L/4) , C(x+2L/4) , D(x+3L/4). If they lie on the same plane, we are done. Let us define the function f(x) as follows : if A(xa,ya,za) , B(xb,yb,zb) , C(xc,yc,zc) , D(xd,yd,zd), then \r\n\r\n|xa ya za 1|\r\n|xb yb zb 1| = f(x)\r\n|xc yc zc 1|\r\n|xd yd zd 1|\r\n\r\nIt's easy to see that f(x) = 0 <=> A,B,C,D lie on the same plane. Now we have :\r\n\r\n|xd yd zd 1|\r\n|xa ya za 1| = f(x+3L/4) = - f(x)\r\n|xb yb zb 1|\r\n|xc yc zc 1|\r\n\r\nf is obviously continuous and f(x) + f(x+3L/4) = 0 and f(x) is not 0 => f(x)f(x+3L/4) < 0 => there is an x0 in (x,x+3L/4) such that f(x0)=0 , QED.", "Solution_2": "Of course! :D Sorry about the incomplete text: the curve must indeed be closed and it must have finite length." } { "Tag": [ "conics", "parabola" ], "Problem": "Hi,\r\n\r\nI am kind of confused about the derivations of the equations: $ y\\equal{}a(x\\minus{}h)^2\\plus{}k; x\\equal{}a(y\\minus{}k)^2\\plus{}h.$ I am finding myself basically memorizing these 'formulas' and I can't really understand them. Any explanation would be highly appreciated. \r\n\r\nI am also confused as to why $ y\\equal{}ax^2\\plus{}bx\\plus{}c, x\\equal{}ay^2\\plus{}by\\plus{}c$ are useful and why they work. Maybe it is just that I am misunderstanding some basic concept. \r\n\r\nThank you very much for the help :)", "Solution_1": "Not sure if this is gonna help but:\r\nA most primitive parabolic equation is: $ y\\equal{}x^2$\r\n\r\nNow, if I alter the equation by adding any number (constant) to the end, to form $ y\\equal{}x^2 \\plus{} k$, all the values for [i]y[/i] in this equation will be [i]k[/i] more than the values in the first equation for the respective [i]x[/i] value. Basically a shift upward of [i]k[/i].\r\nExample: $ y\\equal{}x^2$ vs $ y\\equal{}x^2\\plus{}2$\r\nIf we plug in $ x\\equal{}1$ then $ y\\equal{}1$ for the first, but $ y\\equal{}1\\plus{}2\\equal{}3$ in the second.\r\n\r\nNow, about the [i]h[/i]. Let's look back at the example of $ y\\equal{}x^2$. Using an example, say $ y\\equal{}(x\\plus{}2)^2$.\r\nIf plugging in values, notice that for the second equation, if $ x\\equal{}1$, then the result is $ y\\equal{}(1\\plus{}2)^2$ which is $ y\\equal{}3^2$. (Basically the same result as $ y\\equal{}x^2$ if we had plugged in \"3\" for x).\r\nThus, the [i]y[/i] value we obtain from the second equation by plugging in [i]x[/i] is actually the same as the y value from the first equation if we plug in $ x\\plus{}2$.\r\nThen entire graph of $ y\\equal{}x^2$ will be reproduced two units to the left because each y value of $ y\\equal{}(x\\plus{}2)^2$ is the one from $ y\\equal{}x^2$ that is 2 to the right (or, I mean, +2 units on the x-axis). So the result is a shift of the graph to the left. Which is why the form is $ y\\equal{}(x\\minus{}k)^2\\plus{}h$ instead of $ y\\equal{}(x\\plus{}k)^2\\plus{}h$. We want positive [i]h[/i] to correspond with a shift of the graph to the right, for some reason.\r\n\r\nI hope it was at least a bit helpful.", "Solution_2": "Thank you for the reply, dChampionLance4. That basically cleared up my conception about the $ y\\equal{}(x\\minus{}k)^{2}\\plus{}h$ forms. That was really helpful. That was my main point of confusion. I kept confusing $ ax^2\\plus{}bx\\plus{}c \\equal{} 0$ with something like $ y\\equal{}ax^{2}\\plus{}bx\\plus{}c, x\\equal{}ay^{2}\\plus{}by\\plus{}c ,$ and that was after all a pretty big misconception. Your example cleared that up for me :) (I probably wasn't paying that much attention when the book was explaining that gulping down a pound and a half of Nachoz ;))\r\n\r\nThanks again,\r\nlimac" } { "Tag": [], "Problem": "hey guys i need a little help\r\n\r\n1. what is the molarity of a solution prepared by dissolving 1.25 mol of AgNO3 in enough water to make 625ml of solution. Please show details\r\n\r\n2. How many ml of water should be added to 50.0ml of a 15.0 M H2SO4 soultion to give a final concentration of 0.300 M\r\n\r\n3. How Many ML of 0.150 M NaOH are needed to neutralize 50.00 ml of a 0.120 M solution of H2SO4\r\n\r\nPls show explanation trying to prepare for a test next week and i seem not to know how to work these type of questions\r\n\r\nThanks in advance", "Solution_1": "Molarity is concentration in mol/L. So:\r\n\r\n1) $[AgNO_{3}] = \\frac{1.25mol}{0.625L}= 2.0M$\r\n\r\n2) The number of moles of \"H2SO4\" thal will be present in the final 0.3 M solution is the same that exists in the initial solution (50 mL of 15.0 M). As \r\n\r\nnumber of moles = concentration$\\times$volume,\r\n\r\nthen,\r\n\r\n$15.0M\\times50mL = 0.30M \\times V_{final}\\Rightarrow V_{final}= 2500mL$\r\n\r\nand so we must add 2450 mL of water.\r\n\r\n3) To answer this problem correctly, we need to calculate the $[H_{3}O^{+}]$ in the 0.120 M solution of sulphuric acid. Now, we must pay attention because of a special property of this acid: its the only (common) polyprotic acid for which the first ionization is complete [i]and[/i] the conjugate base is a moderatly strong acid, whose acid-base equilibrium cannot be ignored. So we have\r\n\r\n$H_{2}SO_{4}(aq)+H_{2}O(l) \\longrightarrow HSO_{4}^{-}(aq)+H_{3}O^{+}(aq)$\r\n$HSO_{4}^{-}(aq)+H_{2}O(l) \\rightleftharpoons SO_{4}^{2-}(aq)+H_{3}O^{+}(aq), K$,\r\n\r\nwhere K is the acidity constant of hydrogensulphate ion. Now, using this information it can be proved that for a solution of sulphuric acid of concentration C, the concentration of H+ is given by\r\n\r\n$[H_{3}O^{+}] = \\frac{\\sqrt{C^{2}+6KC+K^{2}}+C-K}{2}$\r\n\r\n(I will leave the proof as an exercise).\r\n\r\nNow, knowing that K = 0.012, and in our case C = 0.120 M, we have by substitution is the above expression\r\n\r\n$[H_{3}O^{+}] = 0.13 M$.\r\n\r\nThe number of moles of H+ that exists in the 50.0 mL of H2SO4 solution is then\r\n\r\n$n(H_{3}O^{+}) = 0.13M\\times0.050L = 6.5\\times10^{-3}mol$.\r\n\r\nTo neutralize them we will need the same number of moles of hydroxide ions, and so:\r\n\r\n$0.150M\\times V_{NaHO}= 6.5\\times10^{-3}mol \\Rightarrow V_{NaHO}= 43.3mL$.\r\n\r\nThe volume of sodium hydroxide needed is thus 43.3 mL." } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "if $\\;x_{n}\\;=\\;\\cos\\;\\frac{\\pi}{2^{n}}\\;+\\;\\imath\\;\\sin\\;\\frac{\\pi}{2^{n}}\\;$,\r\n\r\nfind an expression for\r\n\r\n$\\;\\prod_{k=1}^{\\infty}\\;x_{k}\\;$\r\n\r\n :D", "Solution_1": "We have $x_{1}.x_{2}...x_{n}=\\cos (\\pi (\\frac{1}{2}+\\frac{1}{2^{2}}+...+\\frac{1}{2^{n}}))+i\\sin (\\pi (\\frac{1}{2}+\\frac{1}{2^{2}}+...+\\frac{1}{2^{n}}))$. Now $\\sum_{i=1}^{\\infty}\\frac{1}{2^{i}}=1$.\r\n\r\nFinal, that product is equal to $-1$ and Thanks :D", "Solution_2": "$\\prod_{k=1}^{\\infty}x_{k}= \\prod_{n=1}^{\\infty}\\left(\\cos(\\frac{\\pi}{2^{k}})+i\\sen(\\frac{\\pi}{2^{k}})\\right) = \\prod_{k=1}^{\\infty}e^{i\\frac{\\pi}{2^{k}}}= e^{i\\pi\\sum_{k=1}^{\\infty}\\frac{1}{2^{k}}}= e^{i\\pi}$\r\n$= \\cos(\\pi)+i\\sin(\\pi) =-1$." } { "Tag": [ "inequalities", "number theory proposed", "number theory" ], "Problem": "Does there exist a natural number $z$ which can be represented in the form $z=x!+y!,$ where $x$ and $y$ are natural numbers satisfying the inequality $x\\leq y,$ in (at least) two different ways ?", "Solution_1": "small values (below 3) obviously don't give solutions\r\n\r\nn! + (n-1)! is strictly smaller than (n+1)! so in the place where the lowest y is, we have that the other part isn't able of getting the same level (always above) :)" } { "Tag": [ "induction", "inequalities", "algebra proposed", "algebra" ], "Problem": "Show that there exists a subset $A$ of the set $\\left\\{ 1,2,\\ldots,2^{1996}-1\\right\\} $ having the following properties: \r\na) $1$ and $2^{1996}-1$ belong to $A;$\r\nb) every element of $A$ except $1$ is the sum of two (not necessarily distinct) elements of $A;$\r\nc) the number of elements of $A$ does not exceed $2012.$\r\n\r\nI have a solution which shows that 2010 is enough. I wonder if it is the best estimate.", "Solution_1": "You can show us your solution... ;)", "Solution_2": "JURY SOLUTION\r\nLet $f\\left( n\\right) $ be the least number of elements of a set $A\\subset\\left\\{ 1,2,\\ldots,2^{n}-1\\right\\} $ satisfying conditions a) and b). For instance, $f\\left( 3\\right) \\leq5,$ since the set\r\n\\[\r\nA=\\left\\{ 1,2,3,6,7\\right\\} \\subset\\left\\{ 1,2,3,4,5,6,7\\right\\}\r\n\\]\r\nsatisfies the required conditions. Thus, we have to prove that $f\\left(\r\n1996\\right) \\leq2012.$\r\n\r\nFirst, observe that\r\n\\[\r\nf\\left( n+1\\right) \\leq f\\left( n\\right) +2(*)%\r\n\\]\r\n\r\nIndeed, if $A\\subset\\left\\{ 1,2,\\ldots,2^{n}-1\\right\\} $ satisfies a) and\r\nb), so does\r\n\\[\r\nB=A\\cup\\left\\{ 2^{n+1}-2,2^{n+1}-1\\right\\} ,\r\n\\]\r\nsince\r\n\\[\r\n2^{n+1}-2=2\\left( 2^{n}-1\\right)\r\n\\]\r\nand\r\n\\[\r\n2^{n+1}-1=1+(2^{n+1}-2).\r\n\\]\r\nAlso, notice that\r\n\\[\r\nf\\left( 2n\\right) \\leq f\\left( n\\right) +n+1.(**)%\r\n\\]\r\nThis is because\r\n\\[\r\nB=A\\cup\\left\\{ 2\\left( 2^{n}-1\\right) ,2^{2}\\left( 2^{n}-1\\right)\r\n,\\ldots,2^{n}\\left( 2^{n}-1\\right) ,2^{2n}-1\\right\\}\r\n\\]\r\nsatisfies a) and b):\r\n\\[\r\n2^{k+1}\\left( 2^{n}-1\\right) =2\\cdot2^{k}\\left( 2^{n}-1\\right) ,\r\n\\]\r\nfor $k=0,1,\\ldots,n-1$ and\r\n\\[\r\n2^{2n}-1=2^{n}\\left( 2^{n}-1\\right) +\\left( 2^{n}-1\\right) .\r\n\\]\r\nNow, we can obtain an upper estimate of $f\\left( m\\right) $ using\r\nsuccessively either (*) or (**), according to the parity of the argument of\r\n$f.$ To shorten the computation, observe that (1) and (2) imply\r\n\\[\r\nf\\left( 2n+1\\right) \\leq f\\left( n\\right) +n+3,\r\n\\]\r\nand a simple induction shows that\r\n\\[\r\nf\\left( 2^{k}\\left( 2n+1\\right) \\right) \\leq f\\left( n\\right)\r\n+k+m+3+\\left( 2n+1\\right) \\left( 2^{k}-1\\right) .\r\n\\]\r\nSince $1996=2^{2}\\cdot499=2^{2}\\left( 2\\cdot249+1\\right) ,$ we have\r\n\\[\r\nf\\left( 1996\\right) \\leq f\\left( 249\\right) +1751.\r\n\\]\r\nIn a similar way, we obtain\r\n\\[\r\nf\\left( 249\\right) \\leq f\\left( 124\\right) +127,\\\\\r\n\\]\r\n\\[\r\nf\\left( 124\\right) \\leq f\\left( 15\\right) +113,\\\\\r\n\\]\r\n\\[\r\nf\\left( 15\\right) \\leq f\\left( 7\\right) +10,\\\\\r\n\\]\r\n\\[f\\left( 7\\right) \\leq f\\left( 3\\right) +6,\r\n\\]\r\nand, as we saw,\r\n\\[\r\nf\\left( 3\\right) \\leq5.\r\n\\]\r\nAdding up yields\r\n\\[\r\nf\\left( 1996\\right) \\leq2012,\r\n\\]\r\nas needed.\r\n\r\n\r\nTHE RESULT CAN BE IMPROVED\r\nIndeed, I claim that if we impose to the set $A$ the additional condition $7\\in A,$ then the following inequality holds:\r\n\\[\r\nf\\left( n+3\\right) \\leq f\\left( n\\right) +4.(***)%\r\n\\]\r\nTo prove this, observe that if $A\\subset\\left\\{ 1,2,\\ldots,2^{n}-1\\right\\} $\r\nsatisfies a) and b) and $7\\in A$, so does\r\n\\[\r\nB=A\\cup\\left\\{ 2^{n+1}-2,2^{n+2}-4,2^{n+3}-8,2^{n+3}-1\\right\\} .\r\n\\]\r\nWe now have (using only (**) and (***))\r\n\\[\r\nf\\left( 1996\\right) \\leq f\\left( 499\\right) +1499\\\\\r\nf\\left( 499\\right) \\leq f\\left( 496\\right) +4\\\\\r\nf\\left( 496\\right) \\leq f\\left( 31\\right) +469\\\\\r\nf\\left( 31\\right) \\leq f\\left( 28\\right) +4\\\\\r\nf\\left( 28\\right) \\leq f\\left( 7\\right) +23,\r\n\\]\r\nand, finally,\r\n\\[\r\nf\\left( 7\\right) \\leq11\r\n\\]\r\nas before. Adding up, we obtain $f\\left( 1996\\right) \\leq2010.$\r\n(It is easy to see that $7$ belongs to $A$).\r\n\r\nFURTHER IMPROVEMENTS?", "Solution_3": "Thank you, enescu!\r\nI am interested in lower bound." } { "Tag": [ "geometry", "trigonometry", "linear algebra", "matrix" ], "Problem": "If [b]O[/b] be the origin and if [b]P1(x1,y1) [/b]and [b]P2(x2,y2)[/b] be two points, then [b]OP1.OP2.cos(angle [/b][b]P1OP2)[/b] is equal to \r\n\r\n[b](a) x1y2+x2y1\n(b)(x1^2+y1^2)(x2^2+y2^2)\n(c)(x1-x2)^2 + (y1-y2)^2\n(d) x1x2+y1y2[/b]\r\n\r\nIts answer is [b](d)[/b].Please give its solution as more explanation as you can.Thanks a lot for your help.", "Solution_1": "Do you know inner product?", "Solution_2": "hello, after the formula\r\n$ \\vec{a}\\circ\\vec{b}\\equal{}|\\vec{a}\\parallel{}\\vec{b}|\\cos(\\angle(\\vec{a},\\vec{b}))$ we get\r\n$ \\cos(\\angle(\\vec{OP_1},\\vec{OP_2}))\\equal{}\\frac{x_1x_2\\plus{}y_1y_2}{\\sqrt{x_1^2\\plus{}y_1^2}\\sqrt{x_2^2\\plus{}y_2^2}}$\r\nSonnhard.", "Solution_3": "Let $ \\mathbf{a}\\equal{}\\begin{pmatrix}a_{1}\\\\ a_{2}\\\\ \\vdots\\\\ a_{n}\\end{pmatrix}, \\mathbf{b}\\equal{}\\begin{pmatrix}b_{1}\\\\ b_{2}\\\\ \\vdots\\\\ b_{n}\\end{pmatrix} \\in \\mathbb{R}^n$,\r\n\r\n$ (\\mathbf{a}, \\mathbf{b}) \\equal{} \\|\\mathbf{a}\\| \\ \\|\\mathbf{b}\\| \\cos \\theta$,\r\n$ (\\mathbf{a}, \\mathbf{b}) \\equal{} \\sum_{i\\equal{}1}^n a_ib_i$" } { "Tag": [ "geometry", "geometric transformation", "function", "trigonometry" ], "Problem": "bonjour, :D \r\nJ'ai un puzzle pour toi \r\nte donner quatre z\u00e9ros: 0,0,0,0\r\npouvez vous les transformer en 24 ?\r\n\r\nExemple : 1,2,3,4\r\n\r\n1*2*3*4=24 ,\r\n\r\nnow 0,0,0,0 , comment \u00e0 :?: \r\nwelcome,", "Solution_1": "Unless I'm missing something (which is quite possible given my poor french), I think the answer is no, since\r\n$0+0=0-0=0*0=0$, $\\frac{0}{0}$ and $0^{0}$ are undefined, though possibly we can use $0!=1$?", "Solution_2": "[hide]$(0!+(0!+(0!+(0!))))!$[/hide]", "Solution_3": "[quote=\"jjsneaker\"]bonjour, :D \nJ'ai un puzzle pour toi \nte donner quatre z\u00e9ros: 0,0,0,0\npouvez vous les transformer en 24 ?\n\nExemple : 1,2,3,4\n\n1*2*3*4=24 ,\n\nnow 0,0,0,0 , comment \u00e0 :?: \nwelcome,[/quote]\r\nTranslation: (I hope this is correct :roll: )\r\nHello. :D \r\nI have a puzzle for you.\r\nYou are given four zeros (0,0,0,0).\r\nCan you transform them into 24?\r\n\r\nExample: (1,2,3,4)\r\n$1 \\cdot 2 \\cdot 3 \\cdot 4 = 24$\r\n\r\nNow, how would you transform (0,0,0,0)?\r\n\r\n[hide]I think the only way you could do this is to use $0!$, since every other \"function\" of 0 just produces 0 again. I'm guessing that it's \\[(0!+0!+0!+0!)!\\] [/hide]", "Solution_4": "[hide=\"facile\"]$(0!+0!+0!+0!)!=(1+1+1+1)!=4!=24$ :?:\nEste-ce que vous avez un autre solution? [/hide]\r\n\r\nTHis should be moved thought :wink:", "Solution_5": "[hide=\"You could also have\"]\n$(\\cos 0+\\cos 0+\\cos 0+\\cos 0)!$\nif trigonometry was allowed.[/hide]\r\n\r\n\r\nBut yes, it should probably be moved to puzzles and brainteasers.", "Solution_6": "[quote=\"jjsneaker\"]bonjour, :D \nJ'ai un puzzle pour toi \nte donner quatre z\u00e9ros: 0,0,0,0\npouvez vous les transformer en 24 ?\n\nExemple : 1,2,3,4\n\n1*2*3*4=24 ,\n\nnow 0,0,0,0 , comment \u00e0 :?: \nwelcome,[/quote]\r\n\r\n[hide]\nJ'ai pens\u00e9 de ceci, et la seule r\u00e9ponse que je pourrais penser \u00e0 est\n$(0!+0!+0!+0!)!=4!=24$.\n\nLe seul moyen de transformer c'est en changant le z\u00e9ro \u00e0 z\u00e9ro factorielle[/hide]", "Solution_7": "[quote=\"anirudh\"][quote=\"jjsneaker\"]bonjour, :D \nJ'ai un puzzle pour toi \nte donner quatre z\u00e9ros: 0,0,0,0\npouvez vous les transformer en 24 ?\n\nExemple : 1,2,3,4\n\n1*2*3*4=24 ,\n\nnow 0,0,0,0 , comment \u00e0 :?: \nwelcome,[/quote]\n\n[hide]\nJ'ai pens\u00e9 de ceci, et la seule r\u00e9ponse que je pourrais penser \u00e0 est\n$(0!+0!+0!+0!)=4!=24$.\n\nLe seul moyen de transformer c'est en changant le z\u00e9ro \u00e0 z\u00e9ro factorielle[/hide][/quote]\r\nYou forgot a factorial.", "Solution_8": ":) :) \r\nohhhh,so clever,\r\nGood idea, I see the difference between us ,I am Chinese,\r\nI must learn more !!!\r\nThanks everyone.\r\nI am happy \r\noh, sorry for my poor english,I can't show my feeling,", "Solution_9": "[quote=\"i_like_pie\"][quote=\"anirudh\"][quote=\"jjsneaker\"]bonjour, :D \nJ'ai un puzzle pour toi \nte donner quatre z\u00e9ros: 0,0,0,0\npouvez vous les transformer en 24 ?\n\nExemple : 1,2,3,4\n\n1*2*3*4=24 ,\n\nnow 0,0,0,0 , comment \u00e0 :?: \nwelcome,[/quote]\n\n[hide]\nJ'ai pens\u00e9 de ceci, et la seule r\u00e9ponse que je pourrais penser \u00e0 est\n$(0!+0!+0!+0!)=4!=24$.\n\nLe seul moyen de transformer c'est en changant le z\u00e9ro \u00e0 z\u00e9ro factorielle[/hide][/quote]\nYou forgot a factorial.[/quote]\r\n\r\nI fixed it.\r\nJe l'ai r\u00e9par\u00e9" } { "Tag": [ "geometry", "trapezoid", "circumcircle", "radical axis", "geometry unsolved" ], "Problem": "Trapezoid $ ABCD$, with $ AB\\parallel{}CD$, is inscribed in circle $ \\omega$ and point $ G$ lies inside triangle $ BCD$. Rays $ AG$ and $ BG$ meet $ \\omega$ again at points $ P$ and $ Q$, respectively. Let the line through $ G$ parallel to $ AB$ intersects $ BD$ and $ BC$ at points $ R$ and $ S$, respectively. \r\nProve that quadrilateral $ PQRS$ is cyclic if and only if $ BG$ bisects $ \\angle{CBD}$.", "Solution_1": "$ BB, QQ$ are tangents of the circle $ \\omega \\equiv \\odot(ABCD)$ at $ B, Q.$ By Pascal theorem for $ ABBPQQ,$ the intersections $ G \\equiv AP \\cap BQ, H \\equiv AB \\cap QQ, K \\equiv BB \\cap PQ$ are collinear.\r\n\r\nAssume $ BGQ$ bisects $ \\angle CBD.$ Then $ QQ \\parallel DC \\parallel AB$ $ \\Longrightarrow$ $ H$ is at infinity, $ GK \\parallel AB$, $ R, S \\in GK.$ $ \\triangle BSR \\sim \\triangle BCD$ are centrally similar with center $ B,$ $ KB$ is the single common tangent of their circumcircles, $ \\overline{KS} \\cdot \\overline{KR} \\equal{} \\overline{KB}^2 \\equal{} \\overline{KP} \\cdot \\overline{KQ}$ $ \\Longrightarrow$ $ PQRS$ is cyclic.\r\nAssume $ PQRS$ is cyclic, $ K \\equiv BB \\cap PQ \\cap RS$ is radical center of the circles $ \\odot(PQRS), \\odot(ABCD), \\odot (BSR)$ $ \\Longrightarrow$ $ QQ \\parallel GK \\parallel AB \\parallel DC$ $ \\Longrightarrow$ $ BGQ$ bisects $ \\angle CBD.$", "Solution_2": "For me, this is easier than P1.\r\n :o" } { "Tag": [ "search" ], "Problem": "Hey my name's Ian, I'm 15 yr old high school freshman, and I just found this forum, so I thought I'd join it! I won't be talking about high school math or contests though, I like working with infinity, number patterns, making up my own useful equations, and stuff like that. I could post stuff like that here right?", "Solution_1": "Welcome to AoPS!\r\n\r\nIf by \"here\" you mean the Round Table forum, I would suggest instead that you try the Other Problem Solving Topics forum. The Round Table is really for non-mathematical discussion and OPST is a good catch-all forum if what you're posting about doesn't fit neatly into one of the other forums. That said, I suggest you poke around the other parts of the high school section just to see if anything catches your eye. (If you have particular topics in mind, you might also try a search of this site to see if other people have discussed them before.)" } { "Tag": [ "geometry", "perimeter" ], "Problem": "The medians to the legs of a right triangle are $ 12$m and $ 16$m in length. Find the perimeter of the triangle.", "Solution_1": "Let the legs be $ a$ and $ b$. Then we have $ a^2 \\plus{} \\frac {b^2}4 \\equal{} 144$ and $ \\frac {a^2}4 \\plus{} b^2 \\equal{} 256$, so\r\n\r\n$ \\frac54a^2 \\plus{} \\frac54b^2 \\equal{} 400\\implies a^2 \\plus{} b^2 \\equal{} c^2 \\equal{} 320$. This also gives\r\n\r\n$ (a^2 \\plus{} b^2) \\minus{} \\left(a^2 \\plus{} \\frac {b^2}2\\right) \\equal{} \\frac34b^2 \\equal{} 176\\implies b^2 \\equal{} \\frac {704}3 \\equal{} \\frac {2112}9$ and\r\n\r\n$ (a^2 \\plus{} b^2) \\minus{} \\left(b^2 \\plus{} \\frac {a^2}2\\right) \\equal{} \\frac34a^2 \\equal{} 64\\implies a^2 \\equal{} \\frac {256}3 \\equal{} \\frac {768}9$.\r\n\r\nFinally, $ a \\plus{} b \\plus{} c \\equal{} \\boxed{\\frac {16\\sqrt3 \\plus{} 24\\sqrt5 \\plus{} 8\\sqrt {33}}3}$." } { "Tag": [], "Problem": "1 1/2 + 3 1/6 +5 1/12 + 7 1/20 + 9 1/30 + 11 1/42 = ?", "Solution_1": "[quote=\"jsc\"]1 1/2 + 3 1/6 +5 1/12 + 7 1/20 + 9 1/30 + 11 1/42 = ?[/quote]\r\n[hide]$1\\frac{1}{2}+3\\frac{1}{6}+5\\frac{1}{12}+7\\frac{1}{20}+9\\frac{1}{30}+11\\frac{1}{42}=$\n$36+\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\frac{1}{3\\times 4}+\\frac{1}{4\\times 5}+\\frac{1}{5\\times 6}+\\frac{1}{6\\times 7}=36+\\displaystyle\\left(1-\\frac{1}{7}\\right)=\\boxed{36\\frac{6}{7}}$[/hide]" } { "Tag": [ "linear algebra", "matrix", "vector", "inequalities" ], "Problem": "$A=((i+j))_{1\\leq i,j \\leq n}$ \r\n\r\n1/ Find $rank(A)$ ? IMC 2005\r\n\r\n-----------------------------------------------------\r\nContinuation\r\n\r\n2/ Find the eigenvalues of $A$ ?", "Solution_1": "Assume $n\\ge2$ for the sake of nontriviality.\r\n\r\nThe rank is 2. The column space is spanned by $\\begin{bmatrix}1\\\\2\\\\ \\vdots\\\\n\\end{bmatrix}$ and $\\begin{bmatrix}1\\\\1\\\\ \\vdots\\\\1\\end{bmatrix}.$\r\n\r\nI'm not done with the eigenvalues, but I know this much: there are $n-2$ zero eigenvalues (that much just from the rank), one positive, and one negative eigenvalue. The nonzero eigenvalues are of the form\r\n\r\n$\\frac{n(n+1)}2\\pm\\sqrt{m}$ where $m>\\left(\\frac{n(n+1)}2\\right)^2.$", "Solution_2": "Let $u$ and $v$ be two linearly independent column vectors. The marix $A=uv^T+vu^T$ is similar to $\\begin{bmatrix}v^Tu&v^Tv&0&\\cdots&0\\\\u^Tu&u^Tv&0&\\cdots&0\\\\0&0&0& \\cdots&0\\\\\\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\0&0&0&\\cdots&0\\end{bmatrix}$, using a basis of $u,v$, and other vectors orthogonal to both. The eigenvalues are clearly $\\langle u,v\\rangle\\pm\\sqrt{\\langle u,u\\rangle\\cdot\\langle v,v\\rangle}$ and many copies of 0.\r\n\r\nThe relation to the original problem should be obvious.\r\nKent's inequality is simply Cauchy-Schwarz (and the expression has a typo)." } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Prove that the sum of a square number's digits is never equal to 5\r\n$x^2=\\overline{a_na_{n-1}a_{n-2}....a_2a_1}$\r\nProve that :\r\n$a_n + a_{n-1} + .... + a_2 + a_1 \\neq 5$", "Solution_1": "$x^2$ mod $9$ $\\in \\{0,1,4,7\\}$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Suppose $ k$ is some integer. Show that for any integer $ m$, if $ k | m$, then $ k | m_r$, where $ m_r$ is $ m$ with its digits reversed. For example, $ 123_r \\equal{} 321$. \r\n\r\nEDIT: Oh now I realize another reason why this problem is unclear. I OMITTED A HALF OF THE PROBLEM STATEMENT. \r\n\r\nShow that if $ k$ satisfies this, then $ k$ must be a divisor of 99. \r\n\r\nSource: Problem-Solving Strategies.", "Solution_1": "[quote=\"Zhero\"]Suppose $ k$ is some integer. Show that for any integer $ m$, if $ k | m$, then $ k | m_r$, where $ m_r$ is $ m$ with its digits reversed. For example, $ 123_r \\equal{} 321$. \n\nSource: Problem-Solving Strategies.[/quote]\r\n\r\nI dont understand this problem.\r\n\r\nDo you mean \"show that $ \\forall k\\in\\mathbb N$, $ \\forall m\\in\\mathbb N$ : $ k | m$ $ \\implies$ $ k | m_r$\"?. If so, it's wrong : choose $ k \\equal{} 2$ and $ m \\equal{} 12$, for example.\r\n\r\nDo you mean \"show that $ \\exists k\\in\\mathbb N$ such that $ \\forall m\\in\\mathbb N$ : $ k | m$ $ \\implies$ $ k | m_r$\"?. If so, it's trivial : choose $ k \\equal{} 1$ or $ k\\equal{}3$, or $ k\\equal{}9$, or $ k\\equal{}11$...\r\n\r\nDo you mean \"Find all $ k\\in\\mathbb N$ such that $ \\forall m\\in\\mathbb N$ : $ k | m$ $ \\implies$ $ k | m_r$\"?. \r\n\r\nDo you mean something else ?", "Solution_2": "Yes, I mean \"Find all $ k \\in \\mathbb{N}$ such that $ \\forall m \\in \\mathbb{N} : k | m \\implies k | m_r$. \r\n\r\nI was going at \"Show that for any integer $ k$, $ k | m \\implies k| m_r$ implies that $ k | 99$\", though on hindsight even that is not so clear. \r\n\r\nI apologize for the lack of clarity." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "geometry unsolved" ], "Problem": "Let be given a tetrahedron $ ABCD$ such that triangle $ BCD$ equilateral and $ AB \\equal{} AC \\equal{} AD$. The height is $ h$ and the angle between two planes $ ABC$ and $ BCD$ is $ \\alpha$. The point $ X$ is taken on $ AB$ such that the plane $ XCD$ is perpendicular to $ AB$. Find the volume of the tetrahedron $ XBCD$.", "Solution_1": "Solution posted at \n\nhttp://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.VMO1962Problem4\n\nby Vo Duc Dien" } { "Tag": [ "geometry", "similar triangles" ], "Problem": "\u0393\u03b9\u03b1 \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd \u03bc\u03b5\u03c3\u03b1 \u03c3\u03c4\u03b9\u03c2 \u03b5\u03be\u03b5\u03c4\u03b1\u03c3\u03b5\u03b9\u03c2 \u03b8\u03b5\u03bb\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03b4\u03b9\u03b1\u03bb\u03b5\u03b9\u03bc\u03bc\u03b1 \u03c3\u03c4\u03bf \u03b4\u03b9\u03b1\u03b2\u03b1\u03c3\u03bc\u03b1, \u03b2\u03b1\u03b6\u03c9 \u03c4\u03b7\u03bd \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7:\r\n\r\n\u0395\u03c3\u03c4\u03c9 $ ABC$ \u03bf\u03c1\u03b8\u03bf\u03b3\u03c9\u03bd\u03b9\u03bf \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 $ C\\equal{}90$. \u0395\u03c3\u03c4\u03c9 $ CK$ \u03c5\u03c8\u03bf\u03c2 \u03ba\u03b1\u03b9 $ CE$ \u03b4\u03b9\u03c7\u03bf\u03c4\u03bf\u03bc\u03bf\u03c2 \u03c4\u03b7\u03c2 $ ACK$ \u03c0\u03bf\u03c5 \u03bf\u03c1\u03b9\u03b6\u03b5\u03b9 \u03c3\u03b7\u03bc\u03b5\u03b9\u03bf $ E$ \u03c3\u03c4\u03b7\u03bd $ AB$. \u03a6\u03b5\u03c1\u03bd\u03c9 \u03b1\u03c0\u03bf \u03c4\u03bf $ B$ \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03b7 \r\n\u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7\u03bd $ CE$ \u03b7 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03c4\u03b5\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd $ CK$ \u03c3\u03c4\u03bf $ F$. \u0394\u03b5\u03b9\u03be\u03c4\u03b5 \u03bf\u03c4\u03b9 \u03b7 $ FE$ \u03c3\u03c5\u03bd\u03b1\u03bd\u03c4\u03b1 \u03c4\u03b7\u03bd $ CA$ \u03c3\u03c4\u03bf \u03bc\u03b5\u03c3\u03bf \u03c4\u03b7\u03c2.", "Solution_1": "\u03a3\u03c5\u03bd\u03bf\u03c0\u03c4\u03b9\u03ba\u03cc\u03c4\u03b1\u03c4\u03b1:\r\n\r\nUsing Menelaus theorem on $ AKC$ and line $ MF$ where $ M$ is the intersection of $ AC$ and $ EF$ it is sufficient to prove that:\r\n\r\n$ AE/EK\\equal{}FC/FK$\r\n\r\nUsing the parallels it is sufficient to prove that:\r\n\r\n$ AE/EK\\equal{}BE/BK$\r\n\r\nUsing the bisector theorem it is sufficient to prove that:\r\n\r\n$ AC/CK\\equal{}BE/BK$\r\n\r\nUsing similar triangles $ ACK$ and $ CBK$ it is sufficient to prove that:\r\n\r\n$ CB/BK\\equal{}BE/BK$\r\n\r\nHence we only need to prove that $ BCE$ is isosceles which is trivial with some angles-chasing.", "Solution_2": "\u0391\u03c2 \u03b4\u03ce\u03c3\u03c9 \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1, \u03c7\u03c9\u03c1\u03af\u03c2 \u03c7\u03c1\u03ae\u03c3\u03b7 \u03bf\u03bc\u03bf\u03b9\u03bf\u03c4\u03ae\u03c4\u03c9\u03bd \u03ae \u03ac\u03bb\u03bb\u03c9\u03bd \u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ce\u03bd \u03c3\u03c7\u03ad\u03c3\u03b5\u03c9\u03bd!\r\n\r\n\u03a6\u03ad\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0' \u03c4\u03bf $ C$ \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03c3\u03c4\u03b7 $ CE$ \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd $ AB$ \u03c3\u03c4\u03bf $ L$.\r\n\u0395\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b7 $ BC\\perp FL$ \u03b1\u03c6\u03bf\u03cd $ B$ \u03c4\u03bf \u03bf\u03c1\u03b8\u03cc\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 $ CFL$ ($ LB\\perp FC$, $ CL\\perp CE//FB$).\r\n \u0391\u03c1\u03b1 $ FL//AC$\r\n\u03a4\u03ce\u03c1\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 A,E,K,L \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ac \u03c3\u03c5\u03b6\u03b7\u03b3\u03ae \u03b1\u03c6\u03bf\u03cd \u03c3\u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ AKC$ \u03bf\u03b9 CE, CL \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ae/ \u03b5\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ae \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf\u03b9 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1!\r\n\r\nE\u03c4\u03c3\u03b9 \u03c4\u03ce\u03c1\u03b1 $ F(A, E,K,L)$ \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ae \u03b4\u03ad\u03c3\u03bc\u03b7! \u03ba\u03b9 \u03b1\u03c6\u03bf\u03cd $ FL//AC$ \u03b7 $ FE$ \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd $ AC$ \u03c3\u03c4\u03bf \u03bc\u03ad\u03c3\u03bf \u03c4\u03b7\u03c2!!\r\n\r\nQED :)", "Solution_3": "\u039c\u03b9\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b5 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ad\u03c2 \u03b4\u03ad\u03c3\u03bc\u03b5\u03c2 \u03ad\u03c8\u03b1\u03c7\u03bd\u03b1 \u03ba\u03b9 \u03b5\u03b3\u03ce \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03ad\u03c6\u03b5\u03c1\u03b1 \u03c4\u03b7 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b7 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7... \u0388\u03c6\u03b5\u03c1\u03b1 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03b1\u03c0\u03cc \u03c4\u03bf C \u03c3\u03c4\u03b7\u03bd AC \u03ba\u03b1\u03b9 \u03c0\u03ae\u03b3\u03b1 \u03ad\u03c4\u03c3\u03b9 \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5 \u03b2\u03b3\u03ae\u03ba\u03b5 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03bb\u03ad\u03c9 \u03bd\u03c4\u03ac\u03be\u03b5\u03b9 \u03c0\u03ac\u03bc\u03b5 \u03c0\u03af\u03c3\u03c9 \u03c3\u03c4\u03bf \u039c\u03b5\u03bd\u03ad\u03bb\u03b1\u03bf... :) \u03a9\u03c1\u03b1\u03af\u03bf\u03c2 \u03bc\u03b9\u03bc.", "Solution_4": "[quote=\"Protonios\"]\u039c\u03b9\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b5 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ad\u03c2 \u03b4\u03ad\u03c3\u03bc\u03b5\u03c2 \u03ad\u03c8\u03b1\u03c7\u03bd\u03b1 \u03ba\u03b9 \u03b5\u03b3\u03ce \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03ad\u03c6\u03b5\u03c1\u03b1 \u03c4\u03b7 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b7 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7... \u0388\u03c6\u03b5\u03c1\u03b1 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03b1\u03c0\u03cc \u03c4\u03bf C \u03c3\u03c4\u03b7\u03bd AC \u03ba\u03b1\u03b9 \u03c0\u03ae\u03b3\u03b1 \u03ad\u03c4\u03c3\u03b9 \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5 \u03b2\u03b3\u03ae\u03ba\u03b5 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03bb\u03ad\u03c9 \u03bd\u03c4\u03ac\u03be\u03b5\u03b9 \u03c0\u03ac\u03bc\u03b5 \u03c0\u03af\u03c3\u03c9 \u03c3\u03c4\u03bf \u039c\u03b5\u03bd\u03ad\u03bb\u03b1\u03bf... :) \u03a9\u03c1\u03b1\u03af\u03bf\u03c2 \u03bc\u03b9\u03bc.[/quote]\r\n\r\n\u039a\u03b1\u03c4' \u03b1\u03c1\u03c7\u03ac\u03c2 \u03b7 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03b1\u03c0\u03cc \u03c4\u03bf C \u03c3\u03c4\u03b7\u03bd AC \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 AB! :P \u03bc\u03ae\u03c0\u03c9\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2 \u03ba\u03ac\u03c4\u03b9 \u03ac\u03bb\u03bb\u03bf?\r\n\r\n\u039a\u03b1\u03c4\u03ac \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf\u03bd \u03bd\u03b1 \u03b5\u03be\u03b7\u03b3\u03ae\u03c3\u03c9 \u03c4\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03b5\u03af \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7:\r\n\u039f\u03c4\u03b1\u03bd \u03b2\u03bb\u03ad\u03c0\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ad\u03bd\u03b1 \u03c4\u03bc\u03ae\u03bc\u03b1 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03ac\u03bb\u03bb\u03bf \u03c3\u03c4\u03b7 \u03bc\u03ad\u03c3\u03b7 \u03bc\u03af\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03bc\u03b5\u03b8\u03cc\u03b4\u03bf\u03c5\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03bb\u03ae\u03bc\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03bb\u03ad\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03bc\u03b9\u03b1 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ae \u03b4\u03ad\u03c3\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03c6\u03ad\u03c1\u03b5\u03b9\u03c2 \u03c0\u03b1\u03c1\u03ac\u03bb\u03bb\u03b7\u03bb\u03b7 \u03c0\u03c1\u03c2 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03c4\u03b7\u03c2 \u03b4\u03ad\u03c3\u03bc\u03b7\u03c2, \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c5\u03c4\u03ae \u03b8\u03b1 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b9\u03c2 \u03ac\u03bb\u03bb\u03b5\u03c2 3 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03c3\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 \u03c4\u03bf \u03ad\u03bd\u03b1 \u03bd\u03b1 '\u03bd\u03b1\u03b9 \u03bc\u03ad\u03c3\u03bf \u03c4\u03c9\u03bd \u03ac\u03bb\u03bb\u03c9\u03bd.\r\n\r\n\u03a4\u03ce\u03c1\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b4\u03ad\u03c3\u03bc\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c6\u03b1\u03bd\u03b5\u03c1\u03cc \u03cc\u03c4\u03b9 \u03b8\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf $ F$ \u03c9\u03c2 \u03ba\u03bf\u03c1\u03c5\u03c6\u03ae \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c2 $ FA,FE, FC$ \u03c9\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03c4\u03b7\u03c2 \u03b4\u03ad\u03c3\u03bc\u03b7\u03c2! \u0397 4\u03b7 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03ac\u03bb\u03bb\u03b7\u03bb\u03b7 \u03b1\u03c0' \u03c4\u03bf $ F$ \u03c3\u03c4\u03b7\u03bd $ AC$. E\u03c4\u03c3\u03b9 \u03c6\u03ad\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03ac\u03bb\u03bb\u03b7\u03bb\u03b7 \u03ba\u03b1\u03b9 \u03b2\u03bb\u03ad\u03c0\u03bf\u03c5\u03bc\u03b5 \u03c0\u03bf\u03c5 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd AB! \u0395\u03c0\u03b5\u03b9\u03c4\u03b1 \u03b8\u03ad\u03bb\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03c4\u03b9 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2... \u03cc\u03c4\u03b9 $ A,E,K,L$ \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ac! \u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03cc\u03c4\u03b9 $ CE\\perp CL$ (\u03b5\u03b4\u03ce \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03b6\u03c9 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bb\u03af\u03b3\u03bf \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03b1 \u03b1\u03c0\u03bf \u03cc\u03c4\u03b9 \u03c3\u03c4\u03bf \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03c0\u03bf\u03c3\u03c4). \u039f\u03bc\u03c9\u03c2 \u03b1\u03c5\u03c4\u03cc \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03ac\u03bc\u03b5\u03c3\u03b1 \u03b1\u03c6\u03bf\u03cd \u03c0\u03bb\u03ad\u03bf\u03bd [i]\u03bf\u03b4\u03b7\u03b3\u03b5\u03af\u03c3\u03b1\u03b9 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ac[/i] \u03c3\u03c4\u03bf \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 B \u03bf\u03c1\u03b8\u03cc\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf!", "Solution_5": "\u0398\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bb\u03af\u03b3\u03bf \u03c0\u03b9\u03bf basic. \u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03ad\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 \u039c \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03bc\u03ae\u03c2 AC \u03ba\u03b1\u03b9 FE. E\u03c0\u03b5\u03b9\u03b4\u03ae BF//CE \u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 BFC=ECK=ACE. \u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03b3\u03c9\u03bd\u03af\u03b1 \u039aCB=CAB. \u0386\u03c1\u03b1 \u03c4\u03b1 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03b1 \u0391CE \u03ba\u03b1\u03b9 CFB \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bc\u03bf\u03b9\u03b1. \u03a6\u03ad\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf L \u03c4\u03b7\u03c2 CF. \u0391\u03bd \u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 LBF \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd MEC \u03c4\u03b5\u03bb\u03b5\u03b9\u03ce\u03c3\u03b1\u03bc\u03b5. \u03a0\u03c1\u03bf\u03b5\u03ba\u03c4\u03b5\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd BL \u03ba\u03b1\u03c4\u03ac \u03c4\u03bc\u03ae\u03bc\u03b1 LB' \u03af\u03c3\u03bf \u03bc\u03b5 \u03c4\u03b7\u03bd \u0392L. \u0388\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03bf\u03c4\u03b9 \u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 MEC \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 B'EF \u03ac\u03c1\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 B'EBF \u03b5\u03b3\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03bf \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1\u03c4\u03b9 \u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 B'FB \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd ECB(\u03b1\u03c0\u03bf \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03cc\u03b3\u03c1\u03b1\u03bc\u03bc\u03bf) \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c3\u03b5\u03b9\u03c1\u03ac \u03c4\u03b7\u03c2 \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 CEB(EAC+ACE=KCB+ECK=ECB) DONE :)", "Solution_6": "\u039c\u03b9\u03bc \u03b5\u03bd\u03bd\u03bf\u03ce \u03b1\u03c0\u03cc \u03c4\u03bf B \u03c3\u03c4\u03b7 BC. \u039d\u03b1\u03b9 \u03b1\u03c2 \u03c0\u03bf\u03cd\u03bc\u03b5 \u03b5\u03b3\u03ce \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03b4\u03ad\u03c3\u03bc\u03b7 BC, BM, BA, \u03ba\u03b1\u03b9 \u03b7 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03c6\u03b5\u03c1\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ae \u03ba\u03b1\u03b9 \u03c0\u03ae\u03c1\u03b1 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03cc\u03c4\u03b9 \u03bc\u03b9\u03b1 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b7 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b4\u03b1 \u03c4\u03bf\u03bc\u03ce\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ae \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03cc \u03bb\u03bf\u03b3\u03b9\u03ba\u03ac \u03b4\u03b5\u03bd \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03ac\u03bc\u03b5\u03c3\u03b1 \u03cc\u03c0\u03c9\u03c2 \u03c4\u03bf \u03b4\u03b9\u03ba\u03cc \u03c3\u03bf\u03c5... :)", "Solution_7": "\u0397 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b1\u03c5\u03c4\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03ba\u03c1\u03b9\u03b2\u03c9\u03c2 \u03b9\u03b4\u03b9\u03b1 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03b7\u03bd \u03c4\u03bf\u03c5 \u039a\u03c9\u03c3\u03c4\u03b1.\r\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03c9 \u03bf\u03bc\u03c9\u03c2 \u03c4\u03bf\u03bd \u0394\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b5\u03be\u03b1\u03b9\u03c1\u03b5\u03c4\u03b9\u03ba\u03b7 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03c5\u03c8\u03b7\u03bb\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03b5\u03b4\u03bf\u03c5 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b1 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03b1\u03c2 \u03c0\u03bf\u03c5 \u03bc\u03b1\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c6\u03b5\u03c1\u03b5\u03b9. :)", "Solution_8": "[quote=\"Sakis\"]\u0397 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b1\u03c5\u03c4\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03ba\u03c1\u03b9\u03b2\u03c9\u03c2 \u03b9\u03b4\u03b9\u03b1 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03b7\u03bd \u03c4\u03bf\u03c5 \u039a\u03c9\u03c3\u03c4\u03b1.\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03c9 \u03bf\u03bc\u03c9\u03c2 \u03c4\u03bf\u03bd \u0394\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b5\u03be\u03b1\u03b9\u03c1\u03b5\u03c4\u03b9\u03ba\u03b7 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03c5\u03c8\u03b7\u03bb\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03b5\u03b4\u03bf\u03c5 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b1 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03b1\u03c2 \u03c0\u03bf\u03c5 \u03bc\u03b1\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c6\u03b5\u03c1\u03b5\u03b9. :)[/quote]\r\n\r\n\u03a3' \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03ba\u03b1\u03bb\u03ac \u03c3\u03bf\u03c5 \u03bb\u03cc\u03b3\u03b9\u03b1 \u03a3\u03ac\u03ba\u03b7... \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bc\u03c9\u03c2 \u03b4\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03c7\u03b1\u03c1\u03ac \u03bd\u03b1 \u03b2\u03bf\u03b7\u03b8\u03ce \u03cc\u03c3\u03bf \u03bc\u03c0\u03bf\u03c1\u03ce \u03c3\u03b5 \u03b1\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7\u03bd \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae \u03ba\u03bf\u03b9\u03bd\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03c4\u03cc\u03c3\u03b1 \u03bc\u03bf\u03c5 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03ac\u03b8\u03b5\u03b9 :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Hi all,this is my 1st post :lol: \r\n\r\n$a,b,c$ are postive reals,Prove that \r\n$\\frac{1}{a(2b+1)} + \\frac{1}{b(2c+1)} + \\frac{1}{c(2a+1)} \\geq \\frac{6}{1+8abc}$", "Solution_1": "Brute force \r\n\r\nit is equipvelent to show that\r\n\r\nsigma cyclic(1+2a)ac(4bc-1)^2 >=0 (equaloty holds iff a=b=c=0.5)", "Solution_2": "[quote=\"IneqLover\"]$ a,b,c$ are postive reals,Prove that \n$ \\frac{1}{a(2b+1)}+\\frac{1}{b(2c+1)}+\\frac{1}{c(2a+1)}\\geq \\frac{6}{1+8abc}$[/quote]\r\n\r\nThis is actually the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 , applied to the numbers 2a, 2b, 2c.\r\n\r\n darij" } { "Tag": [ "calculus", "integration", "real analysis", "function", "vector", "limit", "inequalities" ], "Problem": "These are two definitions of Reimann integration I've come acrooss and I'm unable to prove them equivalent.\r\n(note: $ S(f,\\dot{P})$,$ L(f,P)$ and $ U(f,P)$ are Reimann, Lower Reimann and Upper Reimann sums repectively on an interval$ [a,b]$ and tagged partition $ \\dot{P}$)\r\n\r\n(1) (Bartle & Sherbert) $ f$ is Reimann Integrable on $ [a,b]$ iff $ \\exists$ $ L\\in R$ such that $ \\forall \\epsilon >0$ $ \\exists \\delta >0$ with $ |S(f,\\dot{P})\\minus{}L| < \\epsilon$ $ \\forall$ Tagged Partitions with $ \\|\\dot{P}\\| < \\delta$\r\n\r\n(2) (Baby Rudin) $ f$ is Reimann Integrable iff $ \\inf U(f,P) \\equal{} \\sup L(f,P)$, where the $ \\inf$ and the $ \\sup$ are taken over all $ P$.\r\n\r\n(1) $ \\implies$ (2) is pretty easy, using refined partitions and the fact that $ L$ and $ U$ are special cases of $ S$ but I am stuck with the converse because (1) is basically distinguishing between different partitions whereas (2) does not mention the nature of the Partitions at all. I've been able to show that $ f$ is integrable according to (1) implies $ |S(f,\\dot{P}_1)\\minus{}L|<|S(f,\\dot{P}_2)\\minus{}L|$ if $ \\| P _1\\| < \\| P_2\\|$ ... (3) \r\nIs it possible to arrive at this step using (2), or is it possible to do the converse without this step?", "Solution_1": "1 implies 2 is actually pretty hard; my book said it was \"obvious\" and left it at that. 2 implies 1 is a lot easier; you can use the squeeze theorem knowing that the Riemann sums are sandwiched between the Darboux sums and that you want to take the limit as the mesh of the partitions goes to 0.", "Solution_2": "Proof:\r\n($ \\implies$) Let $ \\epsilon >0$ be given. (1)$ \\implies$ $ \\exists$ partition $ \\dot{P}$ s.t. $ |S(f,\\dot{P})\\minus{}L|<\\epsilon$\r\n\r\nLet $ P^*$ be a refinement of $ P$. Consider partition $ P^*$ a) with tags $ u_i| f(u_i)\\equal{}\\sup f(x), x\\in [a_i,a_{i\\plus{}1}]$ and b) with tags $ l_i| f(l_i)\\equal{}\\inf f(x), x\\in [a_i,a_{i\\plus{}1}]$\r\n\r\nfrom (1), The Reimann Sums for both are within $ \\epsilon$ of $ L$. But we can see that a and b are $ U$ and $ L$ for the partition $ P^*$ respectively. Hence, we have that $ \\forall$ $ \\epsilon >0$ $ \\exists$ Partitions $ P^*$ such that $ U(f,P^*)\\minus{}L(f,P^*) < \\epsilon$ which implies (2).\r\n\r\nCould you elaborate on your method? The \"limit\" of the Reimann sum is not the limit we usually define (it is over the norm of the partition) and I don't know if the usual results like Squeeze Theorem apply.", "Solution_3": "Well I don't know if you know what a filter base is (you can google it). It's a general method used to classify limits. Examples of bases that you are familiar with are the base $ n \\to \\infty$ for sequences and the bases $ x \\to a$, $ x \\to 0 \\plus{} a$, and $ x \\to \\minus{} \\infty$ for functions on R. These are not 'just symbols'; they are all instances of the same concept of a filter base. There is indeed a base that goes by the name 'as the mesh/norm of the partition goes to 0' and it has its own symbol too.\r\n\r\nIn this case the Riemann sum is our function and it is defined on a certain base over the set of all partitions of [a,b]\r\nThe utility of defining the integral in terms of bases is that you can easily define the integral of a complex valued function, or a vector valued function. In fact, this definition in terms of bases is set up deliberately so you can analogously define the integral of a function with values in some complete normed space.\r\n\r\nOf course, the definition is equivalent to the one in terms of upper and lower Darboux sums but there's something nice about having everything you know about limits (Cauchy sequences, limits of functions, integrals, etc) tied into one arching concept. :)\r\n\r\nHere is a [url=http://books.google.com/books?id=qA5FTMT7HE4C&pg=PA131&lpg=PA131&dq=vladimir+zorich+base+limit&source=web&ots=to5sUhZnq6&sig=dI1P6XtL3ZTmvEa-Yjdvarx3PIk#PPA128,M1]link[/url]to google books if you're interested; it's the portion of the book I'm using that introduces bases.", "Solution_4": "Thanks for the help. The Filter Base sounds like a powerful notion. Is this how we go about it?\r\n\r\nWe consider set $ X$ to be the set of all partitions on $ [a,b]$ and Base $ \\textsl{\\textbf{B}}$ the set of all $ U_{\\delta}\\subset X$ where $ U_{\\delta}\\equal{} \\{P: \\|P\\| < \\delta\\}$\r\nand require that $ \\lim_{\\textsl{\\textbf{B}}} S(f,P)\\equal{} L$\r\n\r\nNow we first want to prove the squeeze theorem;\r\nIf $ L(f,P)\\leq S(f,P)\\leq U(f,P)$ $ \\forall P$ ...(1) \r\nand $ \\lim_{\\textsl{\\textbf{B}}} L(f,P)\\equal{}L\\equal{}\\lim_{\\textsl{\\textbf{B}}} U(f,P)$ ...(2) \r\nthen $ \\lim_{\\textsl{\\textbf{B}}} S(f,P)\\equal{} L$\r\n\r\n(2)$ \\implies$ $ \\forall \\epsilon>0$, $ \\exists$ $ U_{\\delta}$ and $ U'_{\\delta}$ such that both $ L(f,P)$ and $ U(f,P)$ $ \\in$ $ V_{\\epsilon}(L)$ $ \\forall$ $ P \\in U_{\\delta}$ and $ P \\in U'_{\\delta}$ respectively.\r\nAs $ \\textsl{\\textbf{B}}$ is a base, $ \\exists$ $ V\\in \\textsl{\\textbf{B}}$ with $ V$ $ \\subset$ $ U_{\\delta}\\cap U'_{\\delta}$\r\n$ V$ satisfies the required condition.\r\n\r\nSo this approach is basically introducing ordering of the partitions", "Solution_5": "That's about right. :)\r\n\r\nThe base is such a set of partitions, except you forgot to mention that each partition in any $ U_\\delta$ has intermediate points associated with it. I think you also meant if P is in $ U_\\delta$ and P' is in $ U_{\\delta '}$. But once we have found V, how do we know that every partition with mesh less than that of V satisfies the same property (remember, that is what the limit over our base requires)?\r\n\r\nIf the condition in 2 holds then by the squeeze theorem the condition in 1 holds. Of course, it's not obvious that the limit of lower sums over the base is the same as sup (all P) L(f,P) so that's something you have to verify as an intermediate step to. As for the other direction I've already mentioned what I think you omitted. The way I proved that was through the triangle inequality.", "Solution_6": "I should've put that more clearly.\r\nLet $ \\delta_1$ and $ \\delta_2$ be respective upper norms for $ U_{\\delta}$ and $ U'_{\\delta}$, and $ \\delta\\equal{}\\inf\\{ \\delta_1, \\delta_2\\}$\r\n$ V\\subset U_{\\delta}\\cap U'_{\\delta}$ and $ V\\in \\textsl{\\textbf{B}}$ $ \\implies$ $ \\forall P$ $ \\in$ $ V$, $ \\| P\\|<\\delta$\r\n\r\nAs for the intermediate step, is it not true that $ \\forall P\\in X$, $ P\\in U_{b\\minus{}a}$, and hence, $ \\sup_{\\textsl{\\textbf{B}}}L(f,P)$ $ \\equal{}$ $ \\sup_{X}L(f,P)$?\r\n\r\nI did not get the problem in the other direction, could you repeat it?", "Solution_7": "$ \\forall \\varepsilon > 0\\exists \\delta > 0(\\parallel{}P\\parallel{} < \\delta \\Rightarrow |I \\minus{} S(f,P,\\alpha )| < {\\textstyle{\\varepsilon \\over 6}})$\r\n$ U(f,P) \\equal{} \\sup _\\beta S(f,P,\\beta ) \\Rightarrow \\exists \\alpha '(0 \\le U(f,P) \\minus{} S(f,P,\\alpha ') < {\\textstyle{\\varepsilon \\over 3}})$\r\n\r\nThe triangle inequality and the first statement show that $ |S(f,P,\\alpha ) \\minus{} S(f,P,\\alpha ')| < {\\textstyle{\\varepsilon \\over 3}}$\r\n\r\nCombining these inequalities and using the triangle inequality gives $ |I \\minus{} U(f,P)| < \\varepsilon$\r\n\r\n(The alpha's and the betas and the alpha prime's refer to the distinguished points of the partition P)\r\n\r\nThe book I'm using left this as an \"it's obvious from the definition\"..", "Solution_8": "Yes, but if you use Refined partitions, $ U$ and $ L$ can be seen as special cases of $ S$ and as integrability implies $ |S(f,\\dot{P})\\minus{}L|<\\epsilon$ for every refinement of the original partition, we get that $ U(f,P)\\minus{}L(f,P)<\\epsilon$ which can be easily shown to be equivalent to the Darboux Integral.", "Solution_9": "I'm not sure what you mean by U and L are special cases of S. A function may not attain its maximum on an interval.", "Solution_10": "Yes, you are right. I was subconsciously assuming that $ f$ is continous.\r\n\r\nHowever, I have a more serious problem. \r\nHow do we know that if $ f$ is Darboux Integrable ((2)) then $ \\lim_{\\mathcal{B}}U(f,P)\\equal{}\\lim_{\\mathcal{B}}L(f,P)$ considering $ \\mathcal{B}$ as I have defined it?\r\nSeems like begging the question.", "Solution_11": "I think this should do it.\r\n\r\nLet $ f$ be Darboux integrable on $ [a,b]$\r\nDefine function $ \\varepsilon (\\delta)\\equal{}\\sup\\{ |L\\minus{}S(f,\\dot{P})|: \\|\\dot{P}\\|\\equal{}\\delta\\}$\r\n\r\nWe have to show that, \r\n$ \\varepsilon$ is a non-decreasing function, i.e., if $ \\delta_1 <\\delta_2$ then $ \\varepsilon (\\delta_1)<\\varepsilon (\\delta_2)$...(1)\r\n\r\nLet $ T$ be set of tagged partitions $ \\dot{P}$ such that $ |L\\minus{}S(f,\\dot{P})|<\\varepsilon(\\delta_1)$. Suppose, $ \\dot{P'}\\subset\\dot{P}\\in T$ are such that $ \\|P'\\|\\equal{}\\delta_2 >\\delta_1$ \r\nWe see that \\sup$ |L\\minus{}S(f,\\dot{P'})|\\geq\\varepsilon(\\delta_1)$ and hence, $ \\varepsilon (\\delta_1)<\\varepsilon (\\delta_2)$ with $ \\delta_1 <\\delta_2$\r\nAs $ \\dot{P}$ is arbitrary, we have (1).\r\n\r\n$ f$ is Darboux Integrable implies $ \\exists$ partition $ P: L(f,P) 0$, there exists a $ \\delta > 0$ such that $ U(f; \\mathcal{P}) \\minus{} L(f; \\mathcal{P}) < \\epsilon$, for all partitions $ \\mathcal{P}$ of mesh less than $ \\delta$.\r\n\r\n[b]Proof:[/b]\r\n\r\n($ \\Leftarrow$) Trivial.\r\n\r\n($ \\Rightarrow$) We may suppose that $ \\vert f \\vert \\leq M$, for some $ M > 0$.\r\n\r\nLet $ \\mathcal{P}$ be any partition of $ [a,b]$. Consider the effect of dropping a new point $ p$ in $ \\mathcal{P} \\equal{} \\{a \\equal{} x_0 < x_1 < \\ldots < x_n \\equal{} b\\}$. If the new point $ p$ equals $ x_i$ for some $ i$, then the upper sum does not change. If not, then $ p \\in (x_{i \\minus{} 1}, x_i)$ for some $ i$. Label the new refined partition $ \\mathcal{P}'$. Then $ 0 \\leq U(f; \\mathcal{P}) \\minus{} U(f; \\mathcal{P}')$ = $ \\sup_{[x_{i \\minus{} 1}, x_i]} f \\cdot (x_i \\minus{} x_{i \\minus{} 1}) \\minus{} \\sup_{[x_{i \\minus{} 1},p]} f \\cdot (p \\minus{} x_{i \\minus{} 1}) \\minus{} \\sup_{[p,x_i]} f \\cdot (x_i \\minus{} p)$ = $ (\\sup_{[x_{i \\minus{} 1}, x_i]}f \\minus{} \\sup_{[x_{i \\minus{} 1},p]}f)(p \\minus{} x_{i \\minus{} 1}) \\plus{} (\\sup_{[x_{i \\minus{} 1}, x_i]}f \\minus{} \\sup_{[p,x_i]}f)(x_i \\minus{} p)$ $ \\leq 2M(p \\minus{} x_{i \\minus{} 1}) \\plus{} 2M(x_i \\minus{} p) \\leq 2M \\cdot$ mesh $ \\mathcal{P}$.\r\n\r\nThus, if we drop $ k$ points in $ \\mathcal{P}$, the upper sum does not decrease by more than $ 2kM \\cdot$ mesh $ \\mathcal{P}$. (apply the above $ k$ times). A similar result holds for lower sums.\r\n\r\nLet $ \\epsilon > 0$ be given. Now, since $ f$ is given to be Darboux integrable on $ [a,b]$, there exists a partition $ \\mathcal{Q}$ such that $ U(f; \\mathcal{Q}) \\minus{} L(f; \\mathcal{Q}) < \\frac {\\epsilon}{2}$. Suppose that $ \\mathcal{Q}$ has $ k$ inner partitioning points.\r\n\r\nSet $ \\delta \\equal{} \\frac {\\epsilon}{8kM}$, and let $ \\mathcal{P}$ be a partition with mesh $ \\mathcal{P} < \\delta$. Let $ \\mathcal{R}$ be a common refinement of $ \\mathcal{P}, \\mathcal{Q}$. Observe that $ \\mathcal{R}$ is obtained by adding $ \\mathcal{Q}$'s $ k$ points in $ \\mathcal{P}$. Therefore, $ 0 \\leq U(f; \\mathcal{P}) \\minus{} U(f; \\mathcal{R}) \\leq 2kM \\cdot$ mesh $ \\mathcal{P} < 2kM \\delta \\Rightarrow U(f; \\mathcal{P}) < U(f; \\mathcal{R}) \\plus{} 2kM \\delta$. Similarly, $ L(f; \\mathcal{P}) > L(f; \\mathcal{R}) \\minus{} 2kM \\delta$. Hence, $ U(f; \\mathcal{P}) \\minus{} L(f; \\mathcal{P}) < U(f; \\mathcal{R}) \\minus{} L(f; \\mathcal{R}) \\plus{} 4kM\\delta$ $ \\leq U(f; \\mathcal{Q}) \\minus{} L(f; \\mathcal{Q}) \\plus{} 4kM\\delta$ $ < \\frac {\\epsilon}{2} \\plus{} 4kM \\cdot \\frac {\\epsilon}{8kM} \\equal{} \\epsilon$, and the claim follows.\r\n\r\n[b]Main Claim[/b]\r\nA bounded function is Riemann integrable if and only if it is Darboux integrable.\r\n\r\n[b]Proof:[/b]\r\n\r\n($ \\Rightarrow$) $ f$ is Riemann-integrable. Let $ \\epsilon > 0$ be given. Since $ f$ is Riemann-integrable, there exists a $ \\delta > 0$ such that $ \\left\\vert S(f; \\mathcal{P}) \\minus{} \\int_{[a,b]} f \\right\\vert < \\frac {\\epsilon}{4}$ for every partition with mesh $ \\mathcal{P} < \\delta$, and any corresponding set of sampling points. Fix such a partition $ \\mathcal{P} \\equal{} \\{a \\equal{} x_0 < x_1 < \\ldots < x_n \\equal{} b\\}$. Observe that there exist $ t_i \\in [x_{i \\minus{} 1},x_i]$ (for every $ i$) such that $ f(t_i) > \\sup_{[x_{i \\minus{} 1},x_i]} f \\minus{} \\frac {\\epsilon}{4(b \\minus{} a)}$. Thus $ U(f; \\mathcal{P}) \\equal{} \\sum_{i \\equal{} 1}^n \\sup_{[x_{i \\minus{} 1},x_i]} f \\cdot (x_i \\minus{} x_{i \\minus{} 1}) < \\sum_{i \\equal{} 1}^n (f(t_i) \\plus{} \\frac {\\epsilon}{4(b \\minus{} a)})(x_i \\minus{} x_{i \\minus{} 1})$ = $ S(f; \\mathcal{P}) \\plus{} \\frac {\\epsilon}{4} < \\int_{[a,b]} f \\plus{} \\frac {\\epsilon}{2}$. Similarly, $ L(f; \\mathcal{P}) > \\int_{[a,b]} f \\minus{} \\frac {\\epsilon}{2}$. Thus, $ U(f; \\mathcal{P}) \\minus{} L(f; \\mathcal{P}) < \\epsilon$, so $ f$ is Darboux-integrable.\r\n\r\n($ \\Leftarrow$) $ f$ is Darboux-integrable. Let $ \\epsilon > 0$ be given. Choose $ \\delta$ as in the sideclaim above, and let $ \\mathcal{P}$ be any partition with mesh $ \\mathcal{P} < \\delta$. Then $ L(f; \\mathcal{P}) \\leq \\int_{[a,b]} f \\leq U(f; \\mathcal{P}) < L(f; \\mathcal{P}) \\plus{} \\epsilon$. Also, trivially, $ L(f; \\mathcal{P}) \\leq S(f; \\mathcal{P}) \\leq U(f; \\mathcal{P})$, for any set of sampling points for the riemann sum $ S$. Hence $ \\minus{} \\epsilon < L(f; \\mathcal{P}) \\minus{} \\int_{[a,b]} f \\leq S(f; \\mathcal{P}) \\minus{} \\int_{[a,b]} f \\leq U(f; \\mathcal{P}) \\minus{} \\int_{[a,b]} f < \\epsilon$, so $ \\left\\vert S(f; \\mathcal{P}) \\minus{} \\int_{[a,b]} f \\right\\vert < \\epsilon$, so $ f$ is Riemann-integrable.", "Solution_13": "I think cmad gave the standard argument. It's not all that easy to see how to generalize it to multiple integrals though. At least I didn't see it at first. :blush:", "Solution_14": "[quote=\"\u00a7outh\u00a7tar\"]I think cmad gave the standard argument. It's not all that easy to see how to generalize it to multiple integrals though. At least I didn't see it at first. :blush:[/quote]\r\n\r\nThe main claim works almost as it is for multiple integrals, just rename $ [a,b] \\equal{} R$ (some compact rectangle) and pick $ t_I \\in I$, instead, for all subrectangles $ I \\in \\mathcal{P}$.\r\n\r\nThe only real difference lies in the sideclaim. Observe that inserting a new partition point affects the difference of the upper sums (resp. lower sums) in a compact rectangle of dimensions at most $ W \\times \\cdots \\times W \\times$ mesh $ \\mathcal{P}$, where $ W$ is the max-width of $ R$. Adding a factor $ W^{n\\minus{}1}$ in the denominator of our choice of $ \\delta$ takes care of that.", "Solution_15": "What year are you cmad?\r\n\r\n(I meant to say in the earlier post that I didn't see the proof you now give immediately although I did later on)" } { "Tag": [ "LaTeX" ], "Problem": "How did the administrators of this site get this TeX functionality into this website? I wouldn't mind being able to make TeX gif's for myself every now and again.\r\n\r\nThanks,\r\nJames", "Solution_1": "Do you have the ability to install LaTeX and ImageMagick on your server?\r\n\r\nIf so, [url=http://www.mayer.dial.pipex.com/tex.htm#latexrender]try this[/url]. If not, there are instructions there for how to do it without LaTeX/ImageMagick, but I haven't tried them." } { "Tag": [ "geometry", "trapezoid", "parallelogram", "factorial", "calculus", "similar triangles", "power of a point" ], "Problem": "two segments $AB,CD$ intersects each other at $O$,with the property of;$\\frac{OB}{OD}=\\frac{OC}{OA}$\r\n\r\nwhich is always true,about quadrilateral $ACBD$?\r\n\r\n$1)$cyclic\r\n\r\n$2)$trapezoid\r\n\r\n$3)$parallelogram\r\n\r\n$4)$square\r\n\r\n$5)$none of the above", "Solution_1": "[hide][size=200]2[/size][/hide]", "Solution_2": ":mad: \r\n\r\nu are wrong...\r\n\r\n[hide]$3$[/hide]", "Solution_3": "[quote=\"Ashegh\"]:mad: \n\nu are wrong...\n\n[hide]$3$[/hide][/quote]\r\n\r\nwe can get from this that:\r\n\r\n1.Your geometry is weak \r\n\r\nor \r\n\r\n2.you are not good at English.", "Solution_4": "[hide=\"to end \"][hide=\"this controversy\"][hide=\"i say\"][size=150]5!![/size][/hide][/hide][/hide]\r\nPeople, let's be nice now. :P", "Solution_5": "[hide]\nNot sure if converse power of a point is true, but I'd guess [size=200]1[/size] :D\n[/hide]", "Solution_6": "[quote=\"ccy\"][hide]\nNot sure if converse power of a point is true, but I'd guess [size=200]1[/size] :D\n[/hide][/quote]\r\n\r\nofcourse it is ;)", "Solution_7": "[quote=\"easyas3.14159...\"][hide=\"to end \"][hide=\"this controversy\"][hide=\"i say\"][size=150]5!![/size][/hide][/hide][/hide]\nPeople, let's be nice now. :P[/quote]\r\n\r\nhow could u find that the answer is $5$??????", "Solution_8": "isnt the ansswer[hide]$4$[/hide]?", "Solution_9": "[quote=\"britney spears\"]isnt the ansswer[hide]$4$[/hide]?[/quote]\r\n\r\nno it isnt... :rotfl:", "Solution_10": "how would you do this problem? i am very confused.", "Solution_11": "I think it is [hide=\"...\"]2\nthere are two similar triangles\nalternate interior angles theorem states that the segments are parallel\ntherefore, trapeoid[/hide]", "Solution_12": "I find two pairs of alternate interior angles, stating that two pairs of lines are parallel, so it's a parallelogram?\r\n\r\nI'm not sure about the trapezoid...", "Solution_13": "[quote=\"Ashegh\"]two segments $AB,CD$ intersects each other at $O$,with the property of;$\\frac{OB}{OD}=\\frac{OC}{OA}$\n\nwhich is always true,about quadrilateral $ACBD$?\n\n$1)$cyclic\n\n$2)$trapezoid\n\n$3)$parallelogram\n\n$4)$square\n\n$5)$none of the above[/quote]\r\n\r\n[hide]The answer is 2. It is only sometimes a parallelogram because it never said that AO=BO. You just know that their proportions to the other sides are same. If it isn't a || gram, it can't be a square :) I don't know what a cyclic is so I will not guess that. \n\nBut still I'm not sure[/hide]", "Solution_14": "HAHAHA\r\nall the answers have been guessed\r\nbut i choose[hide=\"...\"]4[/hide]", "Solution_15": "WHOOPS! i mean [hide=\"...\"]3[/hide]", "Solution_16": "[quote=\"polarpan523\"]WHOOPS! i mean [hide=\"...\"]3[/hide][/quote]\r\n\r\n\u00bf?", "Solution_17": "[quote=\"236factorial\"]I find two pairs of alternate interior angles, stating that two pairs of lines are parallel, so it's a parallelogram?\n\nI'm not sure about the trapezoid...[/quote]\r\n\r\nWho says there are two pairs of parallel lines?", "Solution_18": "Well, here's my perspective shown in the picture below. Maybe I'm assuming something incorrectly. If angle ODA = angle OCB, wouldn't AC be parallel to BD?", "Solution_19": "dear friends : i dont know why do we have so many different answers here,for this problem....\r\n\r\ni think $ACBD$ is always cyclic,because:\r\n\r\ntriangles $AOC$ and $ODB$are simmilaar to each other;and:\r\n\r\n$\\angle ACO=\\angle OBD$ and $\\angle CAO=\\angle ODB$\r\n\r\nthen the quadrilateral is cyclic...\r\n\r\nis it wrong????...\r\n\r\ni will wait for ure ideas... ;)", "Solution_20": "why isnt the answer trapezoid?\r\n\r\ni really want to know", "Solution_21": "because the answer is $1$.\r\n\r\ndidnt u see my answer?", "Solution_22": "[quote=\"Ashegh\"]because the answer is $1$.\n\ndidnt u see my answer?[/quote]\r\n\r\nPlease, try to be more polite.", "Solution_23": "I think that the figure is cyclic. Apparently, I assumed the corresponding angles incorrectly. \r\n\r\nFor further reading about cyclic quadrilaterals, visit [url=http://mathworld.wolfram.com/CyclicQuadrilateral.html]this[/url]\r\n\r\nFor now, I'll leave some of the spam (because I'm so nice :P :roll: ), but be aware of this... don't pick 5 answer choices at once... :D", "Solution_24": "[quote=\"Jos\u00e9\"][quote=\"Ashegh\"]because the answer is $1$.\n\ndidnt u see my answer?[/quote]\n\nPlease, try to be more polite.[/quote]\r\n\r\nits none of u bissines... :mad: \r\n\r\nAshegh helped me alot.\r\n\r\nu cant talk with him ,this way.is it clear?", "Solution_25": "to jose:\r\n\r\n :mad: which part of my sentenses was impolitely? \r\n\r\nto 236 factorial:\r\n\r\ndear 236 factorial, i dont alow jose, writting these rubish sentenses. \r\n\r\nand we should talk here only about math and math...\r\n\r\nif he again does it, i will be angry...\r\n\r\n\r\nu know,i dont have time to answer stupid boys... ;)", "Solution_26": "[quote=\"Ashegh\"]to jose:\n\n :mad: which part of my sentenses was impolitely? \n\nto 236 factorial:\n\ndear 236 factorial, i dont alow jose, writting these rubish sentenses. \n\nand we should talk here only about math and math...\n\nif he again does it, i will be angry...\n\n\nu know,i dont have time to answer stupid boys... ;)[/quote]\r\n\r\nHmph, how rude and degenerate of you. Your impoliteness stems from the fact that you seem to hold a sense of self-superiority and therefore do not accept criticism or disagreement.", "Solution_27": "Can I change my answer?I think it's \r\n[hide]a trapezoid[/hide] since triangle OAD is similar to triangle OCB", "Solution_28": "Im pretty sure that its a trapezoid, but is it an isosceles trapazoid? Otherwise it wouldnt be cyclic...", "Solution_29": "This topic is really going out of hand. I'm not a moderator here, so I have no control over this. \r\n\r\nPlease stop using chat language and please stick to correct spelling and grammar (at least try to). That way, the sentence will be less \"impolite\" (because there wasn't anything too impolite about what Ashegh said; it was just presented poorly). \r\n\r\nBack to the topic... \r\n\r\nI had a diagram in one of my posts on the first page. It's labeled incorrectly, right?", "Solution_30": "[quote=\"ccy\"]Im pretty sure that its a trapezoid, but is it an isosceles trapazoid? Otherwise it wouldnt be cyclic...[/quote]\r\n\r\ndear $ccy$,could u plz say,why the answer is trapezoid.???\r\n\r\nor u can say, why is my answer wrong...", "Solution_31": "Perhaps I can clearly explain an answer\r\n[hide]\nI believe that the only true statement is 1.\nIf we cross multiply our two fractions we find that $OA*OB=OC*OD$. If the quadrilateral is inscibed in a circle (cyclic) then regardless of the point we choose we can see this equality holds by power of a point. It is certainly possible for the quadrilateral to be a parallelogram, a trapezoid, or a square but this is not what the question is asking. What we want to know is what MUST be true about the quadrilateral. Certainly one could choose 4 random points on a circle that do not form any of these three shapes and the equation would still hold true.\n[/hide]", "Solution_32": "Let's clear this up. \r\n\r\nWhy it is cyclic\r\n[hide]\nThink about chords in a circle.\nsee http://mathworld.wolfram.com/Chord.html\nBy power of a point, we know that because the segments divide each other in this certain way, they are chords or a circle and cyclic.\n[/hide]\n\nWhy the answer can't be a trapezoid\n[hide]\nWe know our figure is cyclic by the reasoning above. It's not too hard to select 4 points on a circle such that there are no paralell sides. If there are no paralell sides, it is not a trapezoid.\n[/hide]\r\n\r\nSimilar reasoning finds flaws in the other choices as well.\r\nTherefore, the answer is 1, Cyclic.\r\n\r\nEdit: AHHH! Sniped! <_<;;", "Solution_33": "Why are you guys wasting your time. Using power of a point, you know that $OA\\cdot OB=OC\\cdot OD\\implies \\frac{OB}{OD}=\\frac{OC}{OA}$. So the figure is cyclic for sure.\r\n\r\nMasoud Zargar", "Solution_34": "Really children, you've blown this topic out of proportion. If you are not willing to consider the ideas and views expressed by other members of the Art of Problem Solving community, frankly, you should not be posting. It is completely obvious, then, that mockery of another user's post or answer will in no way be tolerated. If you think so and so is a witless fool, do not share it with us, a simple \"I disagree\" is sufficient. Personal attacks and pokes at ones mathematical abilities are completely inappropriate, these are manner's one is taught in kindergarten. I will keep the topic open for the sole sake of mathematical discussion, it appears the answer to this problem has received much controversy. If the tone of the topic persists, however, one of the moderators will be forced to lock it. A thanks to the courteous members who have attempted to sort out the matter.", "Solution_35": "I's not sure what the trouble is, but I'm pretty sure, the person who said the answer was $5$ is correct. If $OB=OD$ and $O$ is not the midpoint of the segments, $AO$ and $CO$ are also equal, but with a different measure, and it is an isosceles trapezoid. If $OB$ and $OD$ are not equal, then $OC=OD$ and $OA=OB$ and it is a parallelogram. If they are equal and $O$ is the midpoint, then obviously, it is a square, and in the case of either the trapezoid or the square, it is cyclic, but not for the parallelograms that aren't squares. Therefore, any of the four answers are sometimes true, but none of them always are, and the answer is $5$", "Solution_36": "[quote=\"DanK\"]If $OB$ and $OD$ are not equal, then $OC=OD$ and $OA=OB$ [/quote]\r\n\r\n?\r\n\r\nWe were given: $\\frac{OB}{OD}=\\frac{OC}{OA}$ so making your substitutions, $\\frac{OA}{OD}=\\frac{OD}{OA} =\\implies OA=OD=OB=OC$ but you just said that $OB\\ne OD$...contradiction\r\n\r\nThe correct answer is that this is a cyclic figure. The people that are getting trapezoid, be very careful about how you are labeling your congruent angles, the pair of angles that are congruent are not alternate interior angles, but they are $\\angle CAB=\\angle CDB$.", "Solution_37": "[quote=\"G-UNIT\"]Really children, you've blown this topic out of proportion. If you are not willing to consider the ideas and views expressed by other members of the Art of Problem Solving community, frankly, you should not be posting. It is completely obvious, then, that mockery of another user's post or answer will in no way be tolerated. If you think so and so is a witless fool, do not share it with us, a simple \"I disagree\" is sufficient. Personal attacks and pokes at ones mathematical abilities are completely inappropriate, these are manner's one is taught in kindergarten. I will keep the topic open for the sole sake of mathematical discussion, it appears the answer to this problem has received much controversy. If the tone of the topic persists, however, one of the moderators will be forced to lock it. A thanks to the courteous members who have attempted to sort out the matter.[/quote]\r\n\r\nUncle G-Unit speaks right.Reading through this topic was nonsense,so keep your ego to yourselves whoever this implies to.\r\nP.S-No offense G-Unit,I was praising you :)", "Solution_38": "[quote=\"boxedexe\"]Why are you guys wasting your time. Using power of a point, you know that $OA\\cdot OB=OC\\cdot OD\\implies \\frac{OB}{OD}=\\frac{OC}{OA}$. So the figure is cyclic for sure.\n\nMasoud Zargar[/quote]\r\n\r\nQuite nice, but keep in mind that everything on the first and half the stuff on the second page were from middle school students, and middle schoolers are not expected to know neither power of a point nor cyclic quadrilaterals. \r\n\r\nAnd why does that necessarily imply that the quadrilateral is cyclic?", "Solution_39": "[quote=\"236factorial\"]\n\nQuite nice, but keep in mind that everything on the first and half the stuff on the second page were from middle school students, and middle schoolers are not expected to know neither power of a point nor cyclic quadrilaterals. \n\nAnd why does that necessarily imply that the quadrilateral is cyclic?[/quote]\r\n\r\nHmmm...I think if u read some of the above posts 236!,u'll understand otherwise read about this theorem once,it is pretty clear that it is a cyclic quad [b]bcos once you prove the two traingles similar(any 2),you will get two angles correspondingly equal which are angles in the same segment for a cyclic quad[/b].Hope this helps....\r\n\r\nI'm really bad at explaining.", "Solution_40": "Oops that was the stupidest question in the world... I even said so earlier :blush:", "Solution_41": "Ok, I see what I did wrong, especially now that I finally learned what power of a point actually is last night...\r\n\r\nHow come I didn't learn that theorem until now?", "Solution_42": "[quote=\"DanK\"]Ok, I see what I did wrong, especially now that I finally learned what power of a point actually is last night...\n\nHow come I didn't learn that theorem until now?[/quote]\r\n\r\ndank , after u get more power full in power of points...\r\n\r\ngo and learn pole and polar...\r\n\r\nit is really enjoyable... ;)", "Solution_43": "Actually, we're doing polar stuff in my pre-calculus class right now..." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Supposed $ A$ is a complex matrices so that $ A^2\\equal{}A^T$, compute all possible eigenvalues of $ A$. Thank you for the solution before.", "Solution_1": "$ (A^T)^2 \\equal{} A \\implies A^4 \\equal{} A$, so $ A$ can have eigenvalues $ 0, 1, \\omega, \\omega^2$.", "Solution_2": "And all four are possible, even in the same matrix.\r\n\\[ \\begin{pmatrix}0&1&0&0\\\\0&0&1&0\\\\1&0&0&0\\\\0&0&0&0\\end{pmatrix}\\]", "Solution_3": "We can say quite a bit more: for one, $ A$ must be diagonalizable, since $ A^4\\equal{}A$." } { "Tag": [ "function", "floor function" ], "Problem": "How would one go about solving for this:\r\n\r\nFind the smallest $n$, a positive integer, that satisfies this equation:\r\n\r\n$\\lfloor\\frac{252n}{1000}\\rfloor - \\lfloor\\frac{251n}{1000}\\rfloor =1$", "Solution_1": "[hide]Notice that 252 is close to 250 and 1000/250 is 4.\nLet's try multiples of 4. Hopefully, we can find a point where 252x/1000 has a bigger unit digit and at multiples of 4, we have the greatest chance since there, we must change the unit digit.\nAt 4, we see that 1.008 has the same unit digit as 1.004. At 8, it's 2.016 and 2.008. The one for 252n/1000 grows by 1.008 for every multiple of 4 and for 251n/1000, the growth is 1.004 for every multiple of 4. We want to find when .008 equals 1. This is when x=125. However, remember that we add 1.008 at multiples of 4 so 125*4=500 is the smallest.[/hide]", "Solution_2": "Well, that IS a possibility for $n$, but it is not the smallest. \r\n\r\nThe answer is actually [hide]127[/hide]\r\n\r\nAnyone have any ideas?", "Solution_3": "It is the smallest value such as $1000*n/251 - 1000*n/252 > 2$ and indeed this is $127$", "Solution_4": ":blush: Would you mind explaining that a little more?", "Solution_5": "I see what I did wrong.\r\n127 is correct because I forgot to count the possitiblity that 4k+3 would cause a carryover effect.\r\n252*3/1000=.756. We want to find an whole number x that when multiplied with .008 (the growth rate of .252 for every 4 numbers since the one's digit is meaningless.) will equal or be very close to one. .756+.008x=1 so .244=.008x and the closest x is 30 which gives .240. But .240+.756=.996 so we need 31. But that's 31*4 or 124 and adding .756 is 3 more than that or 127." } { "Tag": [ "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $ M$ be a bounded set in $ C[a,b]$. Prove that: the set $ A$ of all functions is defined below is a compact set:\r\n$ y(t)\\equal{}\\int_{0}^{t}x(s)ds$, $ x\\in M$", "Solution_1": "I believe you have to use the Ascoli- Arzela theorem\r\n\r\nfirstly prove that A is equicontinuous . This is easy since $ M$ is bounded, then prove the image of each sequence $ $ in $ A$ is compact in $ \\mathbb{R}$ with the remark that in $ \\mathbb{R}$ compact = bounded + closed", "Solution_2": "What's the good definition of $ y$ because the function $ x$ don't define out of segment $ [a,b]$?", "Solution_3": "The function space is $ C[0,1]$, and $ t\\in[0,1]$. Sory for wrong typing", "Solution_4": "[quote=\"1234567a\"]I believe you have to use the Ascoli- Arzela theorem [/quote]\r\n\r\nBut the Ascoli- Arzela theorem is used to prove that $ A$ is relatively compact(ie $ \\overline{A}$ is compact in $ C[a,b]$) but not of $ A$" } { "Tag": [], "Problem": "The average January temperature in Charleston is $ 15^\\circ$C, and the average January temperature is $ \\minus{}8^\\circ$C in Salt Lake City. What is the positive difference of the average temperatures of these two cities?", "Solution_1": "15-(-8) = 15+8=23", "Solution_2": "how about a colorful solution:\n\n$ {\\colorbox{red}{\\colorbox{yellow}{\\colorbox{green}{\\colorbox{blue}{\\colorbox{red}{\\colorbox{yellow}{\\colorbox{green}{\\colorbox{blue}{\\colorbox{red}{\\colorbox{yellow}{\\colorbox{yellow}{\\text{23}}}}}}}}}}}} }$" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Two series\r\n\r\n$ \\sum_{n\\equal{}0}^{\\infty}\\frac{1}{(2n\\plus{}1)!!}$\r\n\r\nand\r\n\r\n$ \\sum_{n\\equal{}0}^{\\infty}\\frac{n!}{(2n\\plus{}1)!!}$\r\n\r\nwhere $ n!!\\equal{}n\\cdot(n\\minus{}2)\\cdot(n\\minus{}4)\\cdots$", "Solution_1": "For the first one, please refer to this [url]http://www.mathlinks.ro/viewtopic.php?p=1269448#1269448[/url].\r\n\r\nThe second one, \r\n\r\n$ \\sum_{n\\equal{}0}^{\\infty}\\frac{n!}{(2n\\plus{}1)!!} \\equal{} \\sum_{n\\equal{}0}^{\\infty}\\frac{1}{2^n}\\prod_{j\\equal{}1}^{n} \\left( 1 \\minus{} \\frac{1}{2j\\plus{}1} \\right) \\equal{} 2$" } { "Tag": [], "Problem": "According to the special theory of relativity, if we want to work with mass and time for something moving at a significant speed, we need to use the Lorent transformation $ t \\equal{} \\frac{t_0}{\\sqrt{1\\minus{}v^2/c^2}}$. Thus \"time moves more slowly\" for this object than on the earth. But the whole point of relativity is that the laws of physics are unchanged no matter what (inertial) reference frame we use. So lets say we take the object in question as the \"origin\" - then the calculations are exactly the same, and \"time moves more slowly\" on the earth than for this object. How does this work out?", "Solution_1": "It works out well. If an object and the earth are moving relative to each other, each one thinks the other has a slower clock. At first, this creates a lot of apparent paradoxes. However, you also have to take into account the fact that leading clocks lag and the relativity of simultaneity.\r\n\r\nWhen viewed as a whole, special relativity works incredibly well (though it feels at times it shouldn't) and is surprisingly uncomplicated." } { "Tag": [], "Problem": "What's your grade and how long time you spend to learn in your school in one day. :D \r\n\r\nRemember that learn by yourself is not learn in school! :)\r\n\r\nI'm 10th grade and learn 7-8 hrs a day.", "Solution_1": "none(I'm in 7th, homeschool and am practicing higher topics than the basic logs.)", "Solution_2": "[quote=\"nutz_for2.718281828\"]none(I'm in 7th, homeschool and am practicing higher topics than the basic logs.)[/quote]\r\nStrange for me. :D", "Solution_3": "I'm in college first year and am in college from 10 to 5 generally, with a recess from 1 to 2:15. Twice a week we start at 11, and thrice a week end at 4. Now calculate. :P", "Solution_4": "In GA, we are in school from 8:30 till 3:30. That's 7 hours.", "Solution_5": "[quote=\"Xantos C. Guin\"]In GA, we are in school from 8:30 till 3:30. That's 7 hours.[/quote]\r\n\r\n7 hours in Antarctica is hard work!", "Solution_6": "7 hours daily ... too much for me. :P", "Solution_7": "I don't learn anything at school :o everything's so easy(especially math)\r\nMaybe highschool will be different..", "Solution_8": "7 to 3.\r\n\r\n7 hours if you dont count breaks etc.\r\n8 if you do." } { "Tag": [ "probability", "Harvard", "college", "articles" ], "Problem": "It seems as though mathematicians have appeared in quite a few movies and TV shows recently. There's [i]NUMB3RS[/i], the TV show about a math genius who helps police solve crimes. [i]A Beautiful Mind[/i] portrayed the life of John Nash, the Nobel prizewinning mathematician who suffered from schizophrenia. Soon the movie [i]Proof[/i] will be released, starring Anthony Hopkins as a brilliant but mad mathematician, and Gwyneth Paltrow as his mathematician daughter. Even in [i]Fever Pitch[/i], the movie about a passionate Red Sox fan, the stars have math-related careers.\r\n\r\nDo these portrayals of mathematicians help or hurt? Are inaccurate portrayals better than no portrayals? Do many people associate genius with madness?", "Solution_1": "I am not sure, some do some don't probably. I think most people think math is too difficult for them to grasp, and only mad genius people can do it. LOL", "Solution_2": "[quote=\"wild octagon\"] [i]A Beautiful Mind[/i] portrayed the life of John Nash, the Nobel prizewinning mathematician who suffered from schizophrenia. [/quote]\r\n\r\nThe original book is great. The movie is somewhat twisted. The story about John Nash is real and probably pretty accurate. The sad part is that Nash was very immature, selfish (without realizing his selfish), and worst of all, his tendency to put other ppl down when he was young. Nash is at the extreme end amoung mathematicians (but not limit to mathematician group). I highly recommend the book.", "Solution_3": "[quote=\"wild octagon\"] There's [i]NUMB3RS[/i], the TV show about a math genius who helps police solve crimes.[/quote]\r\n\r\nI just heard about this recently. I'm looking forward to watching it on Friday :lol: What do people think of it who have seen it?", "Solution_4": "Somehow I think that some of those still present a somehow bad idea of the mathematician as a person. [b]A Beautiful Mind[/b] present a genius (with actually very nice papers published) but the main part is just showing that he was mentally ill and he could realize it and overcome such an illness. [b]Good Will Hunting[/b] is also a good movie, but any way it present an ill boy that is a genius in mathematics, but has a hard time overcoming his traumas from childhood. [b]Proof[/b], I hope the movie will be better than the book, because the book, has a few very good moments, but overall has a pretty weak argument, and again, present an ill mathematician who is dying, and his also somehow mentally ill dauther who also have good skills as a mathematician, even given her lack of formal instruction on the area.\r\n\r\nIf you are a mathematician, that means that you are different from the average person, but that do not make you a sick unstable person by default. I have met some mathematicians who have very turbulent lifes and who are somehow sick, but those are by far a minority, I think that most of the mathematicians live regular lifes, fighting with the dayly problems as everyone else, and overcoming most of them. \r\n\r\nAny way, it is just an opinion.", "Solution_5": "As I it at this point the media presents mathematicians mostly as ill, or mad minds (even if they are geniuses). This gives the general person a bad idea about mathematicians in general :(", "Solution_6": "Yeah, that's the way it is in the U.S. People have such a terrible atitude about math. Eventually, we won't be able to import so many mathematicians here because of our great universites, and that will suck for us. This country needs and education makeover, especially a math education makeover...(and maybe education reform should be done by an educated person...see quote at bottom of the page). I bet the attitude is much better in Romania, right Mr. Vornicu?", "Solution_7": "[i]NUMB3RS[/i], interestingly enough, portrays the mathematician in residence, Charlie, as... a really cool guy. I have been pleasantly surprised. It's basically a CSI+math show, but still interesting. Charlie is a grad student and his older brother is a federal officer. True, there is the typical \"girlfriendless geek\" vibe going on with Charlie but (as this is American television) there is a really pretty girl that he's been tutoring for a while that's into him. Basically, in every single show, he saves the day using some sort of mathematical probability formula. In a couple of them he's solving some sort of problem or something, but most of them use probability. Either way, it's still pretty nifty. Good light Friday night entertainment.", "Solution_8": "I just watched the last couple of episodes of NUMB3RS. It's pretty funny what TV script writers think mathematicians are like. Fortunately the main character, Charlie, is likable and not too nerdlike, though somewhat divorced from reality as he tries to understand the real world by searching for the right set of equations. Overall the shows are fun to watch if you don't take them too seriously as the plots are totally implausible. I think Charlie will help raise the status of young mathematicians if he doesn't get cancelled.", "Solution_9": "Two comments:\r\n1) Just a quick mention of a professor at SUNY Buffalo who is doing research at Harvard but who also runs a consulting company to help movie and TV writers portray math and science in their shows. (Spiderman 2 could have used his services ... electron volts, my ear.) Anyhow, an article from the Harvard Crimson on Tuesday about him can be found [url=http://www.thecrimson.com/article.aspx?ref=507510]here.[/url]\r\n\r\n2) Pendulum, seeing your signature, I note that you must have read Hamlet at some point. Have you read [u]Rosencrantz & Guildenstern Are Dead[/u], by Tom Stoppard? The first scene is absolutely fabulous, and it's worth purchasing the play just for that. (The rest of it isn't bad, either.) Also, apparently there's a movie version of it released 1990, about which I know nothing.", "Solution_10": "[quote=\"wild octagon\"]It seems as though mathematicians have appeared in quite a few movies and TV shows recently. There's [i]NUMB3RS[/i], the TV show about a math genius who helps police solve crimes. [i]A Beautiful Mind[/i] portrayed the life of John Nash, the Nobel prizewinning mathematician who suffered from schizophrenia. Soon the movie [i]Proof[/i] will be released, starring Anthony Hopkins as a brilliant but mad mathematician, and Gwyneth Paltrow as his mathematician daughter. Even in [i]Fever Pitch[/i], the movie about a passionate Red Sox fan, the stars have math-related careers.\n\nDo these portrayals of mathematicians help or hurt? Are inaccurate portrayals better than no portrayals? Do many people associate genius with madness?[/quote]\r\n\r\nI don't know that they necessarily hurt anything, though I have found [i]NUMB3RS[/i] extremely banal. David Krumholtz's Pinnochio incessantly being reminded of the rudiments of the history of math by Peter McNichol's Jimmy Cricket got stale by the end of the pilot episode. Furthermore, I am not entertained by endless re-solution of an inverse problem by a character that was intentionally written to be a rip-off of John Nash's real life. What are we being told? Quants have to be disturbed to be interesting?", "Solution_11": "[quote=\"paladin8\"][quote=\"wild octagon\"] There's [i]NUMB3RS[/i], the TV show about a math genius who helps police solve crimes.[/quote]\n\nI just heard about this recently. I'm looking forward to watching it on Friday :lol: What do people think of it who have seen it?[/quote]\r\n\r\nI have watched [i]NUMB3RS[/i] numerous times, from the premiere show. It is interesting, and like somebody have said before, it involves math and crime investigation. One thing I don't like is that the program just shows that he solves problem, but do not tell the watchers HOW he solves them.... Then of course, it will make it boring, so I dunno. But a great show.", "Solution_12": "Yup. I've watched the last couple of episodes. Very entertaining, but not enough math :D Perhaps that's asking too much from TV producers, though.", "Solution_13": "[quote=\"JBL\"]2) Pendulum, seeing your signature, I note that you must have read Hamlet at some point. [/quote]\r\nI chose the Hamlet quote because I liked the idea of adding a bit of Shakespeare to a math site. Hamlet's such an un-mathy kind of guy too. More like an English major or art major. No, I haven't read [i]Rosencrantz[/i] but I've heard it's very funny. I watched Ethan Hawke's [i]Hamlet[/i] recently. The modern setting was an interesting twist but the American actors -- except Liev Schreiber -- had trouble capturing the rhythm of Shakespearean speech.", "Solution_14": "[quote=\"bubala\"]Yeah, that's the way it is in the U.S. People have such a terrible atitude about math. Eventually, we won't be able to import so many mathematicians here because of our great universites, and that will suck for us. This country needs and education makeover, especially a math education makeover...(and maybe education reform should be done by an educated person...see quote at bottom of the page). I bet the attitude is much better in Romania, right Mr. Vornicu?[/quote]Well the general atitude about Math olympiad in Romania is OK as long as you're a medalist. After that, there's not too much respect for the work, and it certainly doesn't pay off like Computer Science, of Finance. So all people interested in math really leave Romania, so there's not much talking about the Mathematicians in the media (again outside of the IMO team when we return with medals :D)." } { "Tag": [ "probability", "trigonometry", "geometry", "MATHCOUNTS", "calculus", "articles" ], "Problem": "Hi, I'm an eight grader and I'm taking accelerated 1 (high school course) and i have additionally taken harder online courses for college algebra.\r\nProblem is, every time I go to a competition, I've solved harder problems before, but I can never get a perfect score etc...\r\nIS it because I don't focus on math competition problems?\r\nAnd how would you recommend preparing for a competition besides taking practice tests.\r\n\r\nFor example, I know things like exponential and logarithmic functions/conic sections/matrices/a lot of rules(Cramer/Gauss-Jordan), but I always mess up on simpler concepts like probability or word problems..", "Solution_1": "Because you have been focusing on upper level high school courses, and even some college algebra, you know many tricks that can make competitions easier. Some simple trig can never hurt on geometry problems, especially some of the trickier Mathcounts questions.\r\n\r\nI don't know how much free time you have or what activities you do, but I always find it helpful to discuss contest problems with a math coach or teacher. Even if you practice tests and competitions frequently, and even though you know all the principles of the problems, everyone makes the occasional error. Your mistakes may be very close to the actual answer, such as making simple addition errors or rounding to the wrong decimal place.\r\n\r\nThat being said, you always should read questions very carefully. Watch for clues about form of answer, e.g. \"to the nearest integer\" implies a non-terminating decimal. You may have done all the right work, but mistakenly answering 29.73 instead of 29.7, or $ 3\\pi$ instead of 9.4, can kill you.\r\n\r\nMy last bit of advice is to discuss the problem-solving process with peers. See how they solve problems, and even look at their notes/work. Discussing your mistakes with others will help you to understand your own thought process, as well as ways you can correct simple concepts and arithmetic errors.", "Solution_2": "The reason is that math taught in school does not prepare you for things like math competitions, it only prepares you for a test in that class.\r\n\r\n\r\nThere are two main differences between school math and competition math: material and method. \r\n\r\nA typical school curriculum goes through Pre-Algebra, Algebra 1, Geometry, Algebra 2, Trig/Precalc, Calculus. Since Calculus is at the end of the path, people think of it differently as they should. For example, see this article called the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php][b]Calculus Trap[/b][/url]. This leads people to think that this is all there is to math. Math competitions turn it the other way around and include [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_DiscreteMath.php][b]Discrete Math[/b][/url], such as probability, counting, and number theory. This kind of math requires more thought and creativeness than memorization.\r\n\r\nThe kind of math schools teach is based on memorization and formulas, while competition and leisure math is based on [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php][b]problem solving[/b][/url]. And there is a huge difference between the two.\r\n\r\n\r\nTo prepare and learn for math competitions, the best way is to do guided problems. What I mean by this is to do problems which lead you to understand and discover new things, rather than someone just telling you it. How you want to do that is up to you, but AoPS offers [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]books[/url] and [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]classes[/url] that do this.\r\n\r\nOne last thing: [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]Not all math competitions are good[/url].", "Solution_3": "Thanks for your replies.\r\nYes, I've already purchased a book from AoPS (The volume 1 of problem solving)\r\nI haven't even started it since I was too busy focused on online courses :( '\r\nI'll take your tips and start studying more on \"problem solving methods\".\r\n\r\n@Power of Pi\r\nI don't really solve problems in the math competition based on my knowledge from school math but rather other courses I've taken before. Also, I have a shipload of math stuff I have to look at since my dad bought them at Korea...The math there is more challenging than here though in my opinion :(" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that: \\[a^{3}b^{3}c^{3}\\left(a^{15}+b^{15}+c^{15}\\right)+4a^{6}b^{6}c^{6}\\left(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}\\right)+2a^{2}b^{2}c^{2}\\sum_{sym}a^{15}b^{3}\\geq \\\\ \\geq 9a^{8}b^{8}c^{8}+2\\left(a^{12}b^{12}+b^{12}c^{12}+c^{12}a^{12}\\right)+4a^{4}b^{4}c^{4}\\left(a^{12}+b^{12}+c^{12}\\right).\\] where $a,b,c>0$.\r\nArqady, what do you think?", "Solution_1": "[quote=\"Lovasz\"]Prove that: \\[a^{3}b^{3}c^{3}\\left(a^{15}+b^{15}+c^{15}\\right)+4a^{6}b^{6}c^{6}\\left(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}\\right)+2a^{2}b^{2}c^{2}\\sum_{sym}a^{15}b^{3}\\geq \\\\ \\geq 9a^{8}b^{8}c^{8}+2\\left(a^{12}b^{12}+b^{12}c^{12}+c^{12}a^{12}\\right)+4a^{4}b^{4}c^{4}\\left(a^{12}+b^{12}+c^{12}\\right).\\] where $a,b,c>0$.\nArqady, what do you think?[/quote]\r\nThis is wrong.\r\n$a\\rightarrow 0, b=c=1$", "Solution_2": "Sorry, it was not ht e inequality I wnated to say.\r\nPlease solve the following, \\[a^{15}b^{15}+b^{15}c^{15}+c^{15}a^{15}+4a^{8}b^{8}c^{8}\\left(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}\\right)+2a^{4}b^{4}c^{4}\\sum_{sym}a^{15}b^{3}\\\\ \\ge 9(abc)^{10}+2a^{2}b^{2}c^{2}\\left(a^{12}b^{12}+b^{12}c^{12}+c^{12}a^{12}\\right)+4(abc)^{6}\\left(a^{12}+b^{12}+c^{12}\\right).\\]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove that:$\\frac{1+a^{2}}{1+b+c^{2}}+\\frac{1+b^{2}}{1+c+a^{2}}+\\frac{1+c^{2}}{1+a+b^{2}}\\geq 2$\r\nwhere $a;b;c$ are real numbers satisfying $a;b;c \\geq-1$", "Solution_1": "see http://www.mathlinks.ro/Forum/viewtopic.php?p=20974#p20974" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": ":( :( :(", "Solution_1": "Through the point P draw a line L3 such that forms an isosceles triangle with the lines L1 and L2 .The base of this triangle is the loci. demostration is very elementary", "Solution_2": "don't understand :(" } { "Tag": [ "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "show :D\r\n\r\n$ \\int_{0}^{\\infty}\\;\\;\\frac{\\textbf dx}{\\left(\\;x\\plus{}\\sqrt{\\;1\\plus{}x^{2}\\;}\\;\\right)^{n}}\\;\\;\\equal{}\\;\\;\\boxed{\\frac{n}{n^{2}\\minus{}1}}$", "Solution_1": "Make the substitution $ u \\equal{} x\\plus{}\\sqrt{1\\plus{}x^{2}}.$ Solving for $ x,$ we get $ x \\equal{}\\frac{u^{2}\\minus{}1}{2u}\\equal{}\\frac{1}{2}\\left(u\\minus{}\\frac{1}{u}\\right).$\r\n\r\nThat gives $ dx \\equal{}\\frac{1}{2}\\left(1\\plus{}\\frac{1}{u^{2}}\\right)\\,du.$ The integral becomes\r\n\r\n$ \\frac{1}{2}\\int_{1}^{\\infty}u^{\\minus{}n}\\plus{}u^{\\minus{}n\\minus{}2}\\,du \\equal{}\\frac{1}{2}\\left(\\frac{1}{n\\minus{}1}\\plus{}\\frac{1}{n\\plus{}1}\\right)$\r\n\r\nwhich gives the desired result.\r\n\r\n[Required nothing beyond the first portion of a second-semester calculus course. But I'm not putting it on the exam I'm giving tomorrow.]", "Solution_2": "Another method:\r\n\r\nUse the substitution $ x\\equal{}\\sinh t$ and we get that\r\n\r\n$ I\\equal{}\\int_{0}^{\\infty}\\frac{\\cosh t}{e^{nt}}dt\\equal{}\\int_{0}^{\\infty}\\frac{e^{\\minus{}(n\\minus{}1)t}\\plus{}e^{(n\\plus{}1)t}}{2}dt$\r\n\r\n$ \\equal{}\\frac{1}{2}\\left(\\frac{1}{n\\minus{}1}\\plus{}\\frac{1}{n\\plus{}1}\\right)\\equal{}\\frac{n}{n^{2}\\minus{}1}$.\r\n\r\nOf course $ n\\neq 1$, otherwise the integral diverges." } { "Tag": [ "algebra open", "algebra" ], "Problem": "Two problems on sequences...\r\n1) We have the sequence {a_n} such that:\r\n[tex]a_0=a_1=1[/tex]\r\n[tex]a_{n+2}=a_{n+1}^2+a_n[/tex]\r\nShow that exist an m in N such that a_m ends with 2005 zeros i.e. 10^2005|a_m.\r\nCan we find such definited m if the sequence have [tex]a_{n+2}=a_{n+1}+a_n^2[/tex]\r\n\r\n2) Find a closed form for this sequence:\r\n[tex]x_0=x_1=1[/tex]\r\n[tex]x_{n+2}=(n+1)(x_{n+1}+x_n)[/tex]\r\nDoes exist a closed form if x_0=0 and x_1=1 ?", "Solution_1": "1) We can define recursively $(a_n)$ for negative numbers by setting $a_{n-2}=a_n-a_{n-1}^2$. Now let us examine $(a_n)$ modulo $10^{2005}$. Since $(a_n)$ is completely determined by a pairs of consecutive elements the sequence is periodic and we have $a_k \\equiv a_0 = 1 - 1 = 0\\ mod\\ 10^{2005}$.", "Solution_2": "$x_{n+2}=(n+1)(x_{n+1}+x_n)$\r\n$x_{n+2} - (n+2)x_{n+1} = - (x_{n+1} - (n+1)x_n)$\r\n$y_n = x_{n+1} - (n+1)x_n$\r\n$y_n = -y_{n+1}$\r\n\r\nSo y oscillates.\r\n\r\n$y_0 = x_1 - 2x_0 = 1 - 2 = -1$\r\n\r\nSo $y_{2n} = -1, y_{2n+1} = 1$\r\nThus, $x_{n+1} - (n+1)x_n = (-1)^{n+1}$\r\n\r\n...", "Solution_3": "[quote=\"kueh\"]$x_{n+2}=(n+1)(x_{n+1}+x_n)$\n$x_{n+2} - (n+2)x_{n+1} = - (x_{n+1} - (n+1)x_n)$\n$y_n = x_{n+1} - (n+1)x_n$\n$y_n = -y_{n+1}$\n\nSo y oscillates.\n\n$y_0 = x_1 - 2x_0 = 1 - 2 = -1$\n\nSo $y_{2n} = -1, y_{2n+1} = 1$\nThus, $x_{n+1} - (n+1)x_n = (-1)^{n+1}$[/quote]\r\n\r\nDividing by $(n+1)!$,.....", "Solution_4": "Ahem... [tex]y_0=x_1-(1+0)x_0=0[/tex]....", "Solution_5": "oops! yes! so $x_n = n!$... and the second part yields $x_{n+1} - (n+1)x_n = (-1)^{n}$..." } { "Tag": [ "\\/closed" ], "Problem": "How can I write hidden solutions on forum? :?:", "Solution_1": "The best way to find out how to do something on a forum is sometimes by quoting a post that did this ;)\r\n\r\nI think you want something like:\r\n[hide=\"My hidden spam\"]\nSPAM!\nSPAM!\nSPAM!\n[/hide]\r\n\r\n[code][hide=My hidden spam]\nSPAM!\nSPAM!\nSPAM!\n[/hide][/code]", "Solution_2": "Thanks! :)", "Solution_3": "[quote=\"j555\"]Thanks! :)[/quote]\nI still don't get it!", "Solution_4": "[code][hide=My hidden spam]\nSPAM!\nSPAM!\nSPAM!\n[/hide][/code]\n\noutputs\n\n[hide=\"My hidden spam\"]\nSPAM!\nSPAM!\nSPAM!\n[/hide]\n\n(click on it to see what happens)", "Solution_5": "[quote=\"AkshajK\"][code][hide=My hidden spam]\nSPAM!\nSPAM!\nSPAM!\n[/hide][/code]\n\noutputs\n\n[hide=\"My hidden spam\"]\nSPAM!\nSPAM!\nSPAM!\n[/hide]\n\n(click on it to see what happens)[/quote]\n$[hide=\"solution\"]$1000$[/hide]$", "Solution_6": "See those green boxes with stuff inside them over your \"post reply\" box?\nclick on the one that says \"hide\". \nthen, type whatever you want to hide inside it.", "Solution_7": "How did you get the rate thingy?", "Solution_8": "[quote=\"roderickhuang\"]How did you get the rate thingy?[/quote]\n\nCould you be more clear? Are you talking about the squares on top of each post?", "Solution_9": "Also, if you want to test, there's a Test Forum. No need to test in a thread here.", "Solution_10": "I think he's wondering how my post rating shows. \n\nSorry, you have to figure out for yourself. I promised not to tell, even though I figured it out for myself too.", "Solution_11": "I don't think it is particularly difficult to see that it's listed as part of your location. (so much for \"not telling\")" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a,b,c,d>0$,and $ abcd\\equal{}1$.Prove that:\r\n$ 3(a^4\\plus{}b^4\\plus{}c^4\\plus{}d^4)\\plus{}24 \\ge (ab\\plus{}bc\\plus{}cd\\plus{}da\\plus{}ac\\plus{}bd)^2$\r\nI hope you will like it,my brothers :) :lol:", "Solution_1": "[quote=\"quykhtn-qa1\"]Given $ a,b,c,d > 0$,and $ abcd \\equal{} 1$.Prove that:\n$ 3(a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4) \\plus{} 24 \\ge (ab \\plus{} bc \\plus{} cd \\plus{} da \\plus{} ac \\plus{} bd)^2$\n[/quote]\r\nAfter using $ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4 \\plus{} 2abcd\\geq a^2b^2 \\plus{} a^2c^2 \\plus{} a^2d^2 \\plus{} b^2c^2 \\plus{} b^2d^2 \\plus{} c^2d^2$ we need to prove that\r\n$ d^2\\sum_{cyc}(a^2 \\minus{} ab) \\minus{} d\\sum_{cyc}(a^2b \\plus{} a^2c \\minus{} 2abc) \\plus{} \\sum_{cyc}(a^2b^2 \\minus{} a^2bc)\\geq0,$ for which we need to prove that\r\n$ \\left(\\sum_{cyc}(a^2b \\plus{} a^2c \\minus{} 2abc)\\right)^2 \\minus{} 4\\sum_{cyc}(a^2 \\minus{} ab)\\sum_{cyc}(a^2b^2 \\minus{} a^2bc)\\leq0,$\r\nwhich is equivalent to $ (a \\minus{} b)^2(a \\minus{} c)^2(b \\minus{} c)^2\\geq0.$", "Solution_2": "[quote=\"arqady\"][quote=\"quykhtn-qa1\"]Given $ a,b,c,d > 0$,and $ abcd \\equal{} 1$.Prove that:\n$ 3(a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4) \\plus{} 24 \\ge (ab \\plus{} bc \\plus{} cd \\plus{} da \\plus{} ac \\plus{} bd)^2$\n[/quote]\nAfter using $ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4 \\plus{} 2abcd\\geq a^2b^2 \\plus{} a^2c^2 \\plus{} a^2d^2 \\plus{} b^2c^2 \\plus{} b^2d^2 \\plus{} c^2d^2$ we need to prove that\n$ d^2\\sum_{cyc}(a^2 \\minus{} ab) \\minus{} d\\sum_{cyc}(a^2b \\plus{} a^2c \\minus{} 2abc) \\plus{} \\sum_{cyc}(a^2b^2 \\minus{} a^2bc)\\geq0,$ for which we need to prove that\n$ \\left(\\sum_{cyc}(a^2b \\plus{} a^2c \\minus{} 2abc)\\right)^2 \\minus{} 4\\sum_{cyc}(a^2 \\minus{} ab)\\sum_{cyc}(a^2b^2 \\minus{} a^2bc)\\leq0,$\nwhich is equivalent to $ (a \\minus{} b)^2(a \\minus{} c)^2(b \\minus{} c)^2\\geq0.$[/quote]\r\nNice proof,thanks arqady :lol:" } { "Tag": [], "Problem": "| | |x| + 1 | + 2 | = 10\r\n\r\n SOLUTION????? :blush:", "Solution_1": "[quote=\"Stitch\"]| | |x| + 1 | + 2 | = 10\n\n SOLUTION????? :blush:[/quote]\r\nWe have $ |x| \\plus{} 1 > 0$ so $ \\parallel{}|x| \\plus{} 1| \\plus{} 2| \\equal{} \\parallel{}x| \\plus{} 3|$\r\nNote that $ |x| \\plus{} 3 > 0$ so $ \\parallel{}x| \\plus{} 3| \\equal{} |x| \\plus{} 3$\r\nthen we have $ |x| \\plus{} 3 \\equal{} 10$<=>$ |x| \\equal{} 7$\r\nSo $ x \\equal{} \\pm 7$" } { "Tag": [ "number theory", "prime numbers" ], "Problem": "Emirps are defined as prime numbers, when reversed, they are still prime. What are the two largest emirps less than 1000? Solutions please.", "Solution_1": "[hide]Well, the largest prime less than 100 is 97, and 97 reversed is 79, so 97 works. The next largest is 89, but reversing 89 gets 98. In fact, all primes in the 80's will be composite when reversed (because they'll end in 8). The next largest number is 79. We already know 79 works, so we have our 2 numbers, 97 and 79.[/hide]", "Solution_2": "[hide]You can use trial and error to get the answer to be 79 and 97.[/hide]", "Solution_3": "[hide]Find large primes and test if they are 'emirp' s. The largest prime less than 100,[b]97[/b], is an 'emirp';the next largest emirp is not 87 nor 83, but [b]79[/b].[/hide]", "Solution_4": "I believe the question was \"the two largest emirps less than [b]1000[/b]\" ;)", "Solution_5": "[quote=\"mathnerd314\"]I believe the question was \"the two largest emirps less than [b]1000[/b]\" ;)[/quote]\r\nThe question was editted after everyone answered. :lol:", "Solution_6": ":roll:\r\n\r\nAnswer to the edited :mad: problem:\r\n\r\n[hide]999 is divisible by 3.\n997 is prime, but 799=47*17.\n993 is divisible by 3.\n991 and 199 are both primes, so 991 is one emirp.\n989 is 43*23 and composite.\n987 is 47*7*3 and composite.\n983 and 389 are both primes, so 983 is the other one.\n\n991 and 983[/hide]\r\n\r\nHey, it may have been cheap that I used my calculator's PFACTOR program, but...\r\n\r\nAnd mgao, your signature says $x^{2}-7+10=0$. It should be $x^{2}-7x+10=0$ for $x=2,5$. As it stands, $x=\\pm i\\sqrt{3}$.", "Solution_7": "more or less guess and check\r\n\r\n[hide]u get all the primes from 950-999. start with the top(999). and try to find emirps. u finally find that $\\boxed{991}$ $\\text{and}$ $\\boxed{983}$\n[/hide]" } { "Tag": [], "Problem": ":blush: Maybe this is funy query but my braing has just stopped. \r\nI need to find out: expl. How many sick days hours goes to 1 worker in month?; if, there are 3143 workers, works 8hours/day, 22 day/month. 548 of them were on sick days (absent); sick day is count also as 8 hours.\r\nThe answer shoult be something like: # hours goes to 1 worker", "Solution_1": "I'm afraid your problem confuses me. Are you trying to find out how many sick day hours go to one worker in a month? And what are you mening with the second part with the 548 absent?", "Solution_2": "This problem is a bit confusing, but here's what I interpreted of it:\r\n[hide]The number of hours worked in a month is 3143*8*22=553168. If 548 of them were absent for 8 hours each, then subtracting 548*8 from 553168 gives us 548784. Dividing this by the number of workes gets approximately 175 hours. This is where the problem is: each worker a month only works 176 hrs. (It would be nice if they got that much time off, though). :) Did I do something wrong in the problem?[/hide]\r\n\r\nBy the way, to both of you, :welcomeani: :welcome: to AoPS!!", "Solution_3": "[quote=\"easyas3.14159...\"]This problem is a bit confusing, but here's what I interpreted of it:\n[hide]The number of hours worked in a month is 3143*8*22=553168. If 548 of them were absent for 8 hours each, then subtracting 548*8 from 553168 gives us 548784. Dividing this by the number of workes gets approximately 175 hours. This is where the problem is: each worker a month only works 176 hrs. (It would be nice if they got that much time off, though). :) Did I do something wrong in the problem?[/quote]\n\n[hide]umm i THINK that since you subtracted 548*8(sick hours) from 553168(total work hours), then 548784 should be the total work time WITHOUT the sick hours. Then you divided by 3143, which gives you about 175. So I think that they WORK about 175 hours per month, not have 175 SICK hours. But then again, they only have 1 sick hour per month?? :huh: That is too unrealistic. So I think I just made the problem worse... :? [/hide][/hide]" } { "Tag": [ "AoPSwiki", "function", "Support", "parameterization" ], "Problem": "http://www.artofproblemsolving.com/Wiki/index.php/Template:Incomplete\r\n\r\nThe #ifexpr shows \"These\" regardless of whether variable {{{1}}} is 1 or not, as shown by http://www.artofproblemsolving.com/Wiki/index.php/User:Temperal/sandbox .\r\n\r\nIs it because != is not an operator on MediaWiki?", "Solution_1": "Have no idea. The main core, including the part you refer to, is the same with the specifications of the mediawiki software (http://www.mediawiki.org). If that doesn't work for some reason on the AoPSWiki, it's most likely it doesn't work on the Wikipedia as well.", "Solution_2": "[url=http://meta.wikimedia.org/wiki/ParserFunctions]Parser functions[/url] can only be used on MediaWiki version > 1.6.8. I remember seeing somewhere before that the AoPSWiki only supports MediaWiki 1.6, so none of the parser templates work on this wiki. The only special thing you can do with parameters is {{{param|}}}, which only shows it if the parameter exists. (so we can't evaluate expressions, use if statements, etc.)\r\n\r\nI think parser functions would save a lot of time and make templates more convenient.", "Solution_3": "Mmm, but didn't AoPS upgrade their version of MediaWiki recently?", "Solution_4": "[quote=\"azjps\"]I remember seeing somewhere before that the AoPSWiki only supports MediaWiki 1.6, so none of the parser templates work on this wiki.[/quote]That is wrong, AoPSWiki is currently based on the latest version of MediaWiki (1.11.0).", "Solution_5": "Yes, I thought so.\r\n\r\nAnyways, this is not working on Wikipedia either. http://en.wikipedia.org/wiki/User:Nousernamesleft/sandbox\r\n\r\nI think it's just a syntax error on my part, with nothing to do with support or no support. I'll try to get it to work; if I can't then I'll get rid of it.\r\n\r\nEDIT: It doesn't matter what I do on any wiki, it won't work. I'll revert.", "Solution_6": "[quote=\"Valentin Vornicu\"][quote=\"azjps\"]I remember seeing somewhere before that the AoPSWiki only supports MediaWiki 1.6, so none of the parser templates work on this wiki.[/quote]That is wrong, AoPSWiki is currently based on the latest version of MediaWiki (1.11.0).[/quote]\r\n\r\nI'm slightly behind times, thanks for correcting me.\r\n\r\nBut that's weird, none of the [url=http://www.artofproblemsolving.com/Wiki/index.php/User:Azjps/sandbox]parser function[/url] seem to work on the AoPSWiki ... :( \r\n\r\n@Temperal: The reason why your sandbox on Wikipedia wasn't working was because you had 3 \"{\"s instead of 2 [you fixed it on AoPS but not 'pedia] so it evaluated weirdly, it works now.\r\n\r\n[b]Edit[/b]: OH... [url=http://meta.wikimedia.org/wiki/ParserFunctions#Installation]it has to be installed[/url]. That would probably explain it!" } { "Tag": [ "AMC", "AIME", "search", "AMC 8", "AMC 10" ], "Problem": "I've just recently learned about the competition called AMC (Australian Mathematical Competition, not American Mathematical Competition). It seems to be a counterpart of AHSME in Asia. I know a little bit about the test; the format is similar to ASHME's. However, is the contest's difficulty comparable to AMC here? Can someone provide me more information about the contest? Just out of curiosity..................", "Solution_1": "Did you try to google it?", "Solution_2": "What type of info do you want?\r\nTheres 30 multichoice (5 options each) questions.. The scoring used to be 3, 4, and 5 marks respectively for a correct answer in the first, second, and third lot of 10 questions, 0 if you leave it blank, and you'd lose a quarter of the possible marks if you got it wrong, add all that with a starting score of 30 marks. But now its a bit different, I can't quite remember how much the first 20 are worth, but you lose no marks for wrong answers there, with the last ten being 8 for correct, 3 for blank and 0 for wrong if I remember correctly. As for difficulty, well, its hard to say in comparison with your tests, I haven't tried them properly, but the last 10 questions get pretty tricky. Only one person out of all of australia and new zealand (and a few people in other countries) got 30/30 last year. I was lucky enough to get 28 I think it was, and get flown to aussie to get a medal..", "Solution_3": "Stephen once posted a problem from that contest here, and in my opinion, it was more difficult than anything on the American MC or AHSME, but I'm not sure if that was an exceptionally difficult question from the Australian MC. (It was the one with the number of ones in 1+11+111+...+(2002 1s).)", "Solution_4": "Yes, that was a very nice question by the way.\r\n\r\nI suspect from that question, and from the rules for the scoring, that the level of difficulty probably ranges from AMC-12 difficulty for the first few, to AIME level for the last few questions. Sort of like the Canadian Open Mathematics Challenge or the BAMM Test of Ingenuity - where the questions start out like AMC-12 questions but have a much larger range of difficulty than the AMC-12.\r\n\r\nI think this is generally true of most other countries that have national exams; because most don't have an AIME-type exam between the general and the National Olympiad exams.\r\n\r\nTry a search though. I'm sure you can probably find some information or sample questions from the test.\r\n\r\nEdit: I did a search and it turns out they have a lot of separate tests, sort of like our AMC-8, AMC-10, and AMC-12, but ranging all the way from Elementary School to High School - with the toughest exam being called the Senior exam. http://www.amt.canberra.edu.au/amcfact.html\r\n\r\nI suspect most of us are referring to the Senior exam in terms of difficulty. TripleM probably knows more about this than any of the rest us though, being that he's from that region.", "Solution_5": "By the way, thats 1s problem was the final question in the exam. So yeah it was harder than the others :P I may have some old papers around here somewhere..", "Solution_6": "thanks for the help. I'm just curious about the exam and the objective views about it." } { "Tag": [ "inequalities" ], "Problem": "For all real $a,b,c$ prove that\r\n\r\n$a^2+2b^2+3c^2+4\\geq 2ab+2ac+4c$\r\n\r\n :)", "Solution_1": "[hide=\"Solution\"]\nWe have $(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc\\ge{0}$ so $a^2+b^2+c^2\\ge 2ab+2ac-2bc$.\nNow $a^2+2b^2+3c^2+4-2ab-2ac-4c$\n$\\ge b^2-2bc+2c^2-4c+4$\n$=(b-c)^2+(c-2)^2$\n$\\geq 0$, as desired :D [/hide]", "Solution_2": "[quote=\"Melissa\"]\nWe have $(a+b+c)^2=a^2+b^2+c^2-2(ab+bc+ac)\\ge{0}$ :( [/quote]", "Solution_3": "Oh...right :blush: Sorry, it's edited now.", "Solution_4": "[quote=\"Melissa\"][hide=\"Solution\"]\nWe have $(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc\\ge{0}$ so $a^2+b^2+c^2\\ge 2ab+2ac-2bc$.\nNow $a^2+2b^2+3c^2+4-2ab-2ac-4c$\n$\\ge b^2-2bc+2c^2-4c+4$\n$=(b-c)^2+(c-2)^2$\n$\\geq 0$, as desired :D [/hide][/quote]\r\n\r\n :D :D \r\n\r\nI also did almost the same . But im wondering whether theres other way to do it or not ? Maybe AM-GM or whatever :P", "Solution_5": "AM-GM only works with positive reals :P...", "Solution_6": "[hide]$a^2+2b^2+3c^2+4-2ab-2ac-4c=(a-b-c)^2+(b-c)^2+(c-2)^2\\geq 0$[/hide]", "Solution_7": "[quote=\"kunny\"][hide]$a^2+2b^2+3c^2+4-2ab-2ac-4c=(a-b-c)^2+(b-c)^2+(c-2)^2\\geq 0$[/hide][/quote]\r\n\r\n :D :D you got it kunny !", "Solution_8": "[quote=\"Melissa\"]AM-GM only works with positive reals :P...[/quote]\r\nBut AM-GM for 2 variables is true for all real numbers.\r\n$a^2+2b^2+3c^2+4=(\\frac12a^2+2b^2)+(\\frac12a^2+2c^2)+(c^2+4)\\geq 2ab+2ac+4c$", "Solution_9": "[quote=\"leepakhin\"][quote=\"Melissa\"]AM-GM only works with positive reals :P...[/quote]\nBut AM-GM for 2 variables is true for all real numbers.\n$a^2+2b^2+3c^2+4=(\\frac12a^2+2b^2)+(\\frac12a^2+2c^2)+(c^2+4)\\geq 2ab+2ac+4c$[/quote]\r\n\r\nHow do you prove $a^2+b^2\\geq 2ab$ in using AM-GM? :roll:", "Solution_10": "[quote=\"kunny\"]How do you prove $a^2+b^2\\geq 2ab$ in using AM-GM?[/quote]\r\n\r\n$\\sqrt{\\frac{a^2+b^2}{2}}\\geq \\sqrt{ab} \\Rightarrow a^2+b^2\\geq 2ab$ :D", "Solution_11": "maybe something like $|a|^2+|b|^2\\geq 2|a||b|\\geq 2ab$ :P", "Solution_12": "[quote=\"shyong\"]maybe something like $|a|^2+|b|^2\\geq 2|a||b|\\geq 2ab$ :P[/quote]\r\n\r\nThat's right! :lol:" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Prove that\r\n\r\n$ \\int_{0}^{1} y^{y} dy \\equal{} \\iint_{P} (xy)^{xy} dxdy$\r\n\r\nwhere P is a square, x=0, x=1, y=0, y=1", "Solution_1": "What is quadrate? Why do you have differential form dp?", "Solution_2": "its ok now" } { "Tag": [ "integration", "trigonometry", "vector", "geometry", "3D geometry", "sphere", "ratio" ], "Problem": "i have no clue how to do this. I know it has to do with kinetic and potential energy but the book is horendous, and i can't find anything helpful online\r\n\r\nAn electron is projected with an initial speed of 4.2 x 10^5 m/s directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to 3 times the initial value?\r\n\r\nfixed ^5\r\nand as a note the proton is fixed in place", "Solution_1": "[color=blue]I can't find any other solution to this problem without using kenertic and potential energy.\nAnd can you post the anser of this problem?[/color]", "Solution_2": "well i didn't say u can't use potential and kinetic energy. and i do not know the answer as it is a homework problem, i have some sort of idea, but not completely sure. Thx", "Solution_3": "$4.2 X 10^5$ or 4.2 X 105 ??\r\n\r\nAnd well, could you give an idea of what you intend to do using kinetic and potential energy ? In both cases, the energy of the electron, being either $mgh$ or $\\frac{1}{2}mv^2$ will be negligible since both mass and gravity are negligible quantities for an electron.", "Solution_4": "[color=blue]Using the law of conservation of energy we have\n$\\large \\frac{1}{2}m_ev_0^2=-\\frac{ke^2}{R^2}+\\frac{1}{2}m_e(3v_0)^2$\n\nwhere $v_0$ is the initial velocity of the electron\n\n$\\Longrightarrow R=\\frac{e}{2v_0}\\sqrt{\\frac{k}{m_e}}$\n\nI wonder if the solution is wrong[/color]", "Solution_5": "It looks OK to me, as long as the proton doesn't move.", "Solution_6": "Uhm... are you sure that expression for potential energy is right?", "Solution_7": "It should be $k\\frac{e^2}{R}$, right? ;)", "Solution_8": "Right! what an unpleasant mistake....", "Solution_9": "guys i also have another problem:\r\n\r\nSphere 1 with radius R1 has positive charge q. Sphere 2 with radius 3.0R1 is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to retain only negligible charge, what fraction of q ends up on (a) sphere 1 and (b) sphere 2? (c) What is the ratio of the final surface charge density of sphere 1 to that of sphere 2? (Answer with 2 significant figures.)", "Solution_10": "ok, ill attempt this even though i dont know electricity that well...\r\n\r\nthe current in the wire will cease when both spheres have the same potential. Taking the reference point to be infinity, we set up the equation\r\n\r\n$\\frac{kQ_1}{R_1}$ $=$ $\\frac{kQ_2}{3.0R_1}$\r\n\r\nthis simplifies to $Q_2 = 3.0Q_1$, which means that $\\frac{q}{4}$ ended up on the first sphere and $\\frac{3q}{4}$ ended up on the second sphere.\r\n\r\nNow we can calculate density by placing this amount over the surface area. If the charge density of Sphere $1$ is \r\n\r\n$\\frac{q}{4\\pi R^2}$\r\n\r\nthen the charge density of sphere $2$ is \r\n\r\n$\\frac{3q}{4\\pi(3R)^2}$\r\n\r\nand the final ratio of charge densities is $\\frac{1}{3}$", "Solution_11": "nice, i'm just not really getting this idea from bringing it in fron infinity", "Solution_12": "i tried calculating E at the center where with ds=1.5/2 but i am not sure. can someone check this prob for me.\r\n\r\nTwo large parallel metal plates are 1.5 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +6.1 V, what is the electric field in the region between the plates?", "Solution_13": "[color=blue]The electric field between the plates is uniform \n${\\large U=E\\frac{d}{2}\\Longrightarrow E=U\\frac{d}2}$[/color]", "Solution_14": "i also have another question on a problem requiring integration. How do i solve this:\r\n\r\nIn Fig. 24-38, a plastic rod having a uniformly distributed charge Q = -22.4 pC has been bent into a circular arc of radius 4.04 cm and central angle 120\u00b0. With V = 0 at infinity, what is the electric potential in volts at P, the center of curvature of the rod?\r\n\r\nI have tried using ds length of the segment as r*d(theta) but it doesn't seem to work. I would greatly appreciate some guidance here. Also does anyone know any good physics resources online?\r\n\r\nThanks again", "Solution_15": "[color=blue]I have tried with this problem and my answer is $\\large V=\\frac{kQ}{R}$\nIf it's the correct answer, I'll post my solution later [/color]", "Solution_16": "but does that take into account the angle? I mean i'm not exactly sure of all of the properties of potential difference. I would assume it doesn't matter but i would just like to get this confirmed.", "Solution_17": "Hm...electric field is $\\frac{kq}{r^2}$, right? Yeah...so we want to integrate the y-component of this as the x-component cancels out ... so we have something. Yeah. It's like $\\int_{-\\frac{\\pi}{3}}^{\\frac{\\pi}{3}} \\frac{k \\lambda \\cos \\theta}{r^2} (r d \\theta)$, huh. Well what does that come out to be. I think that is something like $\\frac{k \\lambda \\sqrt{3}}{r} = \\frac{kq 3\\sqrt{3}}{2 \\pi r}$ that's pretty crazy huh", "Solution_18": "darn it you wanted potential huh well that's pretty trivial then as potential's not even a vector sheesh", "Solution_19": "$\\large\\lambda=\\frac{3Q}{2\\pi R}$\r\n\r\n$\\large dQ=\\lambda Rd\\theta=\\frac{3Q}{2\\pi}d\\theta$\r\n\r\n$\\large dV=\\frac{kdQ}{R}=\\frac{3kQ}{2\\pi R}d\\theta$\r\n\r\n$\\large V=\\int_{-\\frac{\\pi}{3}}^{\\frac{\\pi}{3}}\\frac{3kQ}{2\\pi R}d\\theta=\\frac{kQ}{R}$", "Solution_20": "I have a couple more problems which i don't know how to do.\r\n\r\n1.\r\nA charged isolated metal sphere of diameter 14 cm has a potential of 11000 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.\r\n\r\n2.\r\nIn Fig. 25-30, the battery has a potential difference of V = 10.0 V and the five capacitors each have a capacitance of 9.90 \u00b5F. What is the charge in microcoulombs on (a) capacitor 1 and (b) capacitor 2?\r\n(see attached picture)\r\nI just need to figure out (b)\r\n\r\n3.\r\nA certain parallel-plate capacitor is filled with a dielectric for which k= 5.00. The area of each plate is 0.0678 m2, and the plates are separated by 2.32 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 206 kN/C. What is the maximum energy that can be stored in the capacitor?\r\n\r\nI would greatly appreciate suggestions on how to solve these problems.", "Solution_21": "3) The maximal $E$ is given, therefore the maximal voltage is $U = Ed$ where $d$ is the distance between the plates. The capacity is $C=k\\varepsilon_0\\frac{S}{d}$ and the energy it can store -- at the maximal voltage -- is $W=\\frac{CU^2}{2}=\\frac{1}{2}k\\varepsilon_0dSE^2$.\r\n\r\n2) The equivalent capacity of the capacitors to the left of the EMF is $\\frac{3}{5}C$ and the charge on them is $q=\\frac{3}{5}CV$. The charges on the single capacitor and the three capacitors (with $C_2$) are equal to $q$. The voltage on the three capacitors is $V_1=V-q/C=\\frac{2}{5}V$. This is the voltage on the two capacitors to the right; the voltage on $C_2$ is then $V_2 = V_1/2$. Thus, the charge on $C_2$ is $q_2=\\frac{CV}{5}$. Am I right?", "Solution_22": "what is the $S$ in (3)", "Solution_23": "for (2) how did u get $\\frac{3}{5}C$ ? I can only see $\\frac{2}{3}C$ for the path where $C_2$ is located.", "Solution_24": "Sorry I didn't make it clear...\r\n\r\n$S$ is the area of the plates.\r\n\r\nYou know that if two capacitors are serially connected the equivalent capacity is $\\frac{1}{C}=\\frac{1}{C_1}+\\frac{1}{C_2}$; for a parallel connection you just add the capacities. Now, we have $C_2$ serially connected to another capacitor, which gives us $C/2$; then this $C/2$ has a parallel connection to another capacitor of $C$ capacity and their equivalent is $3C/2$. Next, these are serially connected to yet another $C$ which gives $3C/5$. Check if you don't believe me :) BTW, I think that you got your $2C/3$ when you tried to connect the capacitors as you'd connect resistors or solenoids...", "Solution_25": "o ok i got it, stupid mistake on my part. Thanks a lot, makes sense now.", "Solution_26": "[quote=\"demji\"]1.\nA charged isolated metal sphere of diameter 14 cm has a potential of 11000 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.[/quote]\r\n[color=blue]We know that for a charge isolated sphere, the electric field and the potential outer the sphere is\n\n$\\large E=\\frac{kQ^2}{d}$\n\n$\\large V=\\frac{kQ}{d}$\n\n$\\large\\Longrightarrow E=QV$\n\nThe energy density of the electric field near the the surface of the sphere is ($d=R$)\n\n$\\large \\omega_E=\\frac{\\epsilon_0E^2}{2}=\\frac{\\epsilon_0Q^2V^2}{2}$\n\nwhere $\\epsilon_0=8.85\\times 10^{-12}C^2/Nm^2$.[/color]", "Solution_27": "but the problem doesn't give the value for $Q$ so some other method needs to be used. I just can't see that method.", "Solution_28": "Notice the line $V = \\frac{kQ}{R}$. You can calculate $Q$ from it since you know the potential $V$.", "Solution_29": "o yeah right, missed that. Thanks", "Solution_30": "since the book gives an answer of $.11 \\frac{J}{m^3}$ working backwards gives the formula:\r\n\r\n$u=\\frac{\\epsilon_0(\\frac{2V}{d})^2}{2}$\r\n\r\nBut shouldn't it just be $V$ and not $2V$?\r\n\r\nSomething screwy is going on here.", "Solution_31": "How ridiculous will I get? [b]phucnv87[/b]'s solution contains a tiny mistake :) and we've calculated everything according to his solution! Here's the real thing:\r\n\r\nThe electric field on the surface of a charged sphere is $E=\\frac{kQ}{R^2}$ (note the difference) and the potential is $V=\\frac{kQ}{R}$. Thus, $E = V/R=2V/d$!!! Now we use the fact that $w = \\frac{\\varepsilon_0E^2}{2}$ and case solved...", "Solution_32": "omg, i am so stupid, assumed the d was radius... oh well. Guess it pays to go through each variable. Thanks Djole" } { "Tag": [ "algebra", "function", "domain", "geometry", "3D geometry", "topology" ], "Problem": "Ok this is a question from munkres that should be obvious but I am not seeing the solution. I'm studying for an exam tomorrow so if anyone could explain the solution to me it would be greatly appreciated.\r\n\r\nIf n>1, show that every continuous map from S^n to S^1 is nulhomotopic.\r\n\r\nThanks for the help.", "Solution_1": "The universal cover of $ S^1$ is $ \\mathbb{R}$, so let $ q: \\mathbb{R}\\to S^1$ be the covering map. Let $ f: S^n\\to S^1$ be your continuous map.\r\n\r\nNow by the lifting criterion, this map lifts and hence factors through $ \\mathbb{R}$ if and only if the induced map $ f_*$ is trivial. But this is the case since your domain is just $ \\pi_1(S^n) \\equal{} 1$. Thus the lift exists, so since $ \\mathbb{R}$ is simply connected $ \\bar{f}$ ~ $ \\bar{c}$ where the bar indicates the lifted maps. So $ f$ is homotopic to a constant function by the Monodromy Theorem.\r\n\r\nHmm...On second thought, I'm not sure it has to be as complicated as factoring through the universal cover, since we already have $ \\pi_1(S^n) \\equal{} 1$...Any other suggestions?", "Solution_2": "i think the shortest (and more general) way is in fact proving the lemma that says that if $ p: Y\\to X$ is a covering, then $ \\pi_n(Y) \\equal{} \\pi_n(X)$.\r\nof course, with this in mind, the exercise gets kind of trivial :)\r\n\r\nthe proof of the lemma should go as follow: consider a map $ S^n\\to X$ as a map from $ I^n$ (the unit cube) such that the image of $ \\partial I^n$ is a fixed point. if you choose a basepoint in $ Y$, you can lift this map, and this lifting is still a map $ I^n\\to Y$ that factors through $ S^n\\to Y$, because $ \\partial I^n$ is connected.\r\nof course homotopies \"down\" lift up, and you have the induced map $ p_*$ that reverses your lifting, so you have an isomorphism.\r\nactually, we've proved something stronger, that $ p_*$ is an isomorphism $ \\pi_n(Y) \\to \\pi_n(X)$.." } { "Tag": [ "floor function" ], "Problem": "Compute $ [x]\\plus{}[1\\minus{}x], x\\in\\mathbb{R},$ where $ [x]$ is the floor function.", "Solution_1": "Is this $ 1$ for $ x \\in \\mathbb{Z}$ and $ 0$ otherwise?", "Solution_2": "hello, we get $ 1+\\lfloor[-x]+\\lfloor[x]$\r\nSonnhard.", "Solution_3": "Let $ x\\equal{}n\\plus{}\\alpha$ with $ 0\\le \\alpha<1$ and $ n\\in \\mathbb{Z}$\r\n\r\nIf $ \\alpha\\equal{}0$,\r\n\r\n$ \\lfloor x\\rfloor \\plus{} \\lfloor 1\\minus{}x\\rfloor\\equal{}n\\plus{}(1\\minus{}n)\\equal{}\\boxed{1}$\r\n\r\nIf $ \\alpha\\not\\equal{}0$ and $ n>0$,\r\n\r\n$ \\lfloor x\\rfloor \\plus{} \\lfloor 1\\minus{}x\\rfloor\\equal{}n\\plus{}(\\minus{}n)\\equal{}\\boxed{0}$\r\n\r\nIf $ \\alpha\\not\\equal{}0$ and $ n<0$,\r\n\r\n$ \\lfloor x\\rfloor \\plus{} \\lfloor 1\\minus{}x\\rfloor\\equal{}(n\\minus{}1)\\plus{}(1\\minus{}n)\\equal{}\\boxed{0}$\r\n\r\nIf $ \\alpha\\not\\equal{}0$ and $ n\\equal{}0$,\r\n\r\n$ \\lfloor x\\rfloor \\plus{} \\lfloor 1\\minus{}x\\rfloor\\equal{}\\boxed{0}$\r\n\r\nTherefore, as AndrewTom said, the answer is $ 1$ if $ n$ is an integer, or $ 0$ if $ n$ is not an integer." } { "Tag": [], "Problem": "Prove that $ \\forall a,b,c\\in\\mathbb{N}$,there exist $ x,y\\in\\mathbb{Z}$,$ x,y\\neq 0$,such that\r\n$ ax \\plus{} by$ is divisible by $ c$,and $ |x|,|y|\\leq[\\sqrt {c}]$.", "Solution_1": "See numbers $ ax\\plus{}by$ , here $ x,y\\in \\{\\frac{\\minus{}n}{2},...,\\frac{n}{2}\\}(n\\equal{}[\\sqrt{c}])$", "Solution_2": "[quote=\"N.T.TUAN\"]See numbers $ ax \\plus{} by$ , here $ x,y\\in \\{\\frac { \\minus{} n}{2},...,\\frac {n}{2}\\}(n \\equal{} [\\sqrt {c}])$[/quote]\r\nI know the proof,i think that you should explain your solution to the others :) \r\nP.S:\r\nTo post it here was not a good idea,could someone please move it to the pre-olympiad topic,thanks" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "hi everyone,\r\n\r\nhere is a simple pb:\r\nlet p>3 a prime number . Prove that there exists (x,y,k) so that 1<=2k<=p \r\nand kp+3=x^2+y^2 ..\r\n\r\nbye.", "Solution_1": "The sets $\\{x^2|x\\in \\mathbb Z_p\\}$ and $\\{-y^2+3|x\\in \\mathbb Z_p\\}$ each has $\\frac {p+1}2$ elements, so their intersection can't be void, so there exist $x,y\\in \\mathbb Z_p$ s.t. $x^2+y^2-3=0$. We can, of course, choose $x,y\\in [0,\\frac {p-1}2]$, and this gives $kp=x^2+y^2-3<\\frac {p^2}4+\\frac {p^2}4-3<\\frac {p^2}2\\Rightarrow k<\\frac p2$.\r\n\r\nThere might be some slight mistakes, but this is the main idea." } { "Tag": [], "Problem": "f(x)ra noghteie doreii migooiand agar:\r\nf(f(f(...(f(x)))...)=x\r\nsabet konid:\r\nf[m](x)=f[k](x) agar va faghat agar x ozve Q bashad.\r\n-----f[m](x)iani f(f(f(...(f(x))))...) m daf e.iani 'm' bar fof(x).----", "Solution_1": ":?: :?: :?: :?:" } { "Tag": [ "geometry", "geometric transformation", "reflection", "trapezoid", "perpendicular bisector", "geometry solved" ], "Problem": "The distinct points $X_1, X_2, X_3, X_4, X_5, X_6$ all lie on the same side of the line $AB$. The six triangles $ABX_i$ are all similar. Show that the $X_i$ lie on a circle.", "Solution_1": "do u know the source of this problem?", "Solution_2": "I am changing the notations (as usual ;) ):\r\n\r\n[color=blue][b]Problem.[/b] Let $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$, B, C be eight points in the plane such that the three triangles $A_1BC$, $CA_2B$ and $BCA_5$ are directly similar to each other, the three triangles $A_6CB$, $BA_3C$ and $CBA_4$ are directly similar to each other, and the former three triangles $A_1BC$, $CA_2B$ and $BCA_5$ are inversely similar to the latter three triangles $A_6CB$, $BA_3C$ and $CBA_4$. Then, prove that $A_1A_6\\parallel A_2A_3\\parallel A_4A_5\\parallel BC$, and that the points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ and $A_6$ lie on one circle.[/color]\r\n\r\n[i]Solution.[/i] (For an illustration, see http://www.pandd.demon.nl/lemoine/neuberg.htm#2 Stelling 3; note that the picture has a typo: A should be $A_6$.)\r\n\r\nWe will use directed angles modulo 180\u00b0.\r\n\r\nThe reflection with respect to the perpendicular bisector of the segment BC maps the point B to the point C and the point C to the point B. Let it map the point $A_1$ to a point $A_1^{\\prime}$. Then, since reflection in a line changes the sign of an angle, $\\measuredangle A_1^{\\prime}CB=-\\measuredangle A_1BC$. On the other hand, since the triangles $A_1BC$ and $A_6CB$ are inversely similar, we have $\\measuredangle A_6CB=-\\measuredangle A_1BC$. Thus, $\\measuredangle A_1^{\\prime}CB=\\measuredangle A_6CB$, so that the point $A_1^{\\prime}$ lies on the line $CA_6$. Similarly, the point $A_1^{\\prime}$ lies on the line $BA_6$. But the lines $CA_6$ and $BA_6$ have only one common point, namely the point $A_6$; thus, the point $A_1^{\\prime}$ coincides with the point $A_6$. Since the point $A_1^{\\prime}$ was defined as the image of the point $A_1$ under the reflection with respect to the perpendicular bisector of the segment BC, it follows that the point $A_6$ is the image of the point $A_1$ under the reflection with respect to the perpendicular bisector of the segment BC. Hence, the line $A_1A_6$ is perpendicular to the perpendicular bisector of the segment BC; thus, $A_1A_6\\parallel BC$. Similarly, $A_2A_3\\parallel BC$ and $A_4A_5\\parallel BC$. Hence, $A_1A_6\\parallel A_2A_3\\parallel A_4A_5\\parallel BC$ is proven. Remains to show that the points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ and $A_6$ lie on one circle.\r\n\r\nThis is pretty easy now: The point $A_6$ is the image of the point $A_1$ under the reflection with respect to the perpendicular bisector of the segment BC; similarly, the point $A_5$ is the image of the point $A_4$ under this reflection, and thus the quadrilateral $A_1A_6A_4A_5$ is an isosceles trapezoid, so that it has a circumcircle. In other words, the points $A_1$, $A_6$, $A_4$, $A_5$ lie on one circle k.\r\n\r\nSince the triangles $A_6CB$ and $BCA_5$ are inversely similar, we have $\\measuredangle A_6CB=-\\measuredangle BCA_5$. In other words, $\\measuredangle A_6CB=\\measuredangle A_5CB$. Hence, the points C, $A_5$, $A_6$ are collinear. Similarly, the points C, $A_2$, $A_4$ are collinear; so are the points C, $A_1$, $A_3$, the points B, $A_1$, $A_4$, the points B, $A_2$, $A_6$, and the points B, $A_3$, $A_5$.\r\n\r\nNow, since $A_4A_5\\parallel BC$, we have $\\measuredangle\\left(A_4A_5;\\;CA_6\\right)=\\measuredangle\\left(BC;\\;CA_6\\right)$; in other words, $\\measuredangle A_4A_5A_6=\\measuredangle BCA_6$. But on the other hand, since the triangles $A_6CB$ and $CA_2B$ are inversely similar, we have $\\measuredangle BCA_6=-\\measuredangle BA_2C$, so that $\\measuredangle A_4A_5A_6=-\\measuredangle BA_2C=\\measuredangle CA_2B=\\measuredangle A_4A_2A_6$. Hence, the points $A_4$, $A_6$, $A_5$, $A_2$ lie on one circle; thus, the point $A_2$ lies on the circle k which passes through the points $A_4$, $A_6$, $A_5$. Similarly, the point $A_3$ also lies on this circle k. Hence, all six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ and $A_6$ lie on one circle, namely on the circle k, and the problem is solved.\r\n\r\nThe author of the problem is Joseph J. B. Neuberg, who discovered it probably at the end of the 19th century.\r\n\r\n Darij", "Solution_3": "I remember this problem was probably at the Austria-Polish math competition few years ago." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "The cities in a country are connected with paths. It is known that two cities are connected with no more than one path(s) and each city is connected with not less than three paths. A traveller left some city must pass through at least six other cities (it is not allow passing a path more than once) before he goes back to its starting position. Prove that the country have at least $ 24$ cities.", "Solution_1": "[quote=\"Moonmathpi496\"]The cities in a country are connected with paths. It is known that two cities are connected with no more than one path(s) and each city is connected with not less than three paths. A traveller left some city must pass through at least six other cities (it is not allow passing a path more than once) before he goes back to its starting position. Prove that the country have at least $ 24$ cities.[/quote]\r\n\r\nI have result more than 24 :maybe:", "Solution_2": "[quote=\"SaYaT\"]I have result more than 24 :maybe:[/quote]\r\nThen post the solution, the problem asks to prove that [b]at least 24[/b] cities are there.", "Solution_3": "[quote=\"Moonmathpi496\"]The cities in a country are connected with paths. It is known that two cities are connected with no more than one path(s) and each city is connected with not less than three paths. A traveller left some city must pass through at least six other cities (it is not allow passing a path more than once) before he goes back to its starting position. Prove that the country have at least $ 24$ cities.[/quote]\r\n\r\nLet's consider $ m$ - the shortest route of traveller to return to his starting position (we searching $ m$ among all possible routes and starting position of traveller)(where $ m$ is the number of cities that he had passed to return including the city where he had started).\r\n\r\nIf $ m\\equal{}7$ then let's $ B_1B_2B_3B_4B_5B_6B_7$ is one of such possible routes(where $ B_i$ is one of the city of this country)\r\n Then we have that among this $ 7$ cities there are no other direct connection but $ B_iB_{i\\plus{}1}$(for $ 1 \\le i \\le 7$ and $ B_8\\equal{}B_1$),becouse otherwise $ m<7$ which make contradiction.\r\nEvery city have direct conection at least with $ 3$ other cities, so let's every $ B_i$ conected with $ C_i$. Now among this defined 14 cities every $ C_i$ is connected only with $ B_i$, becouse otherwise again we will have that $ m<7$ which is contradiction.\r\nSo every $ C_i$ connected with $ D_i$ and $ E_i$(becouse every city have at least direct connection with $ 3$ different cities). So we have that there are at least $ 28$ different cities in this country.\r\n\r\nIf $ m>7$ then in the similiar way we will find out that there are more than $ 28$ cities. \r\n\r\n[b]Remark:[/b] when i said [b] every $ C_i$ connected with $ D_i$ and $ E_i$[/b] i just defined cities $ D_i$ and $ E_i$ that connected with $ C_i$." } { "Tag": [ "algebra", "polynomial", "integration", "calculus", "induction", "calculus computations" ], "Problem": "The Hermite polynomials $H_{n}(x)$ can be defined by the Rodrigues formula where ($Df=f'$)\r\n\\[H_{n}(x) = (-1)^{n}e^{x^{2}}D^{n}(e^{-x^{2}}) \\]\r\na) show $H_{n}$, defined as above, is a polynomial of degree $n$. What is its leading coefficient?\r\n\r\nb) Show that the polynomials $H_{n}$ are orthogonal for the inner product\r\n\\[(f,g) = \\int_{\\mathbb{R}}f(x)g(x)e^{-x^{2}}dx \\]\r\nand compute $\\parallel H_{n}\\parallel^{2}=(H_{n},H_{n})$ Hint; if $m \\leq m$, use integration by parts n times.", "Solution_1": "a)\r\n\r\nInduction on $n$ for the first : \r\n\r\nAssume thet it's true for $n = k$ .\r\n\r\nThen $H_{k}(x) = p(x)$ a polynomial of degree $k$ .\r\n\r\nThen \r\n\r\n$\\\\ H_{k+1}(x) = (-1)^{k+1}e^{x^{2}}D^{k+1}(e^{x^{2}}) \\Rightarrow \\\\ \\\\ H_{k+1}(x) =-e^{x^{2}}D[e^{-x^{2}}(-1)^{k}e^{x^{2}}D^{k}(e^{x^{2}})] \\Rightarrow \\\\ \\\\ H_{k+1}(x) =-e^{x^{2}}D(e^{-x^{2}}p(x)) \\Rightarrow \\\\ \\\\ H_{k+1}(x) =-e^{x^{2}}(p'(x)e^{-x^{2}}-2xe^{-x^{2}}p(x)) \\Rightarrow \\\\ \\\\ H_{k+1}(x) = 2xp(x)-p'(x)$\r\n\r\npolynomial of degree $k+1$ .\r\n\r\n[b]Edit:[/b] I didnt noticed the question : What is its leading coefficient?\r\n\r\nBy induction and by using $H_{k+1}(x) = 2xH_{k}(x)-H'_{k}(x)$ \r\n\r\nand $H_{1}=2x$ it turns to be $2^{n}$ .", "Solution_2": "[quote=\"Ramanujan\"]a)\n\nInduction on $n$ for the first : \n\n...[/quote]\r\n\r\nThank you for your reply." } { "Tag": [ "calculus", "integration", "complex analysis", "trigonometry", "complex numbers", "complex analysis theorems" ], "Problem": "Let $f(z)=z+\\sum^{\\infty}_{n=2}a_nz^n$ a serie with the convergence radius $R\\geq 1$ and $a_n\\in R$.\r\nShow that if $f$ is injective on the disk of complex numbers with modulus < 1 then $|a_n|\\leq n$.", "Solution_1": "Maybe you could give the first question of the problem, Killer, since like this the problem is a little bit too difficult. It is a theorem of Dieudonne, if I remember correctly and even if the proof is elementary, it is very difficult (IMHO) to find it. By the way, the conclusion remains true if we consider just complex numbers, but the proof is unbelievable difficult.", "Solution_2": "Do we have to use Cauchy's integral formula?", "Solution_3": "The generalized case($a_i \\in \\mathbb{C}$) has been solved only recently (around 1985) told me a friend.", "Solution_4": "This is a hard problem. It is actually the real version of the Bieberbach conjecture. The full version was only proved in recent memory as alekk mentioned. A proof for the real version goes something like this:\r\n\r\nLet $f=u+iv$ and $00$ when $\\Im z>0$ and $v(z)<0$ when $\\Im z<0$).\r\n\r\nNow just let $r\\to 1^{-}$ and the result follows.\r\n\r\nFinally, this proof is not original to me. \r\n\r\nOral exam, eh? How did that go anyway? :rotfl:" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "USAMTS", "\\/closed" ], "Problem": "I thnk that we should hold an official AoPS Contest, aiming at High school students, not yet at the level of taking the MathLinks contest. I tried to take MathLinks test, but they are WAYYY too hard.", "Solution_1": "[quote=\"xxreddevilzxx\"]I thnk that we should hold an official AoPS Contest, aiming at High school students, not yet at the level of taking the MathLinks contest. I tried to take MathLinks test, but they are WAYYY too hard.[/quote]\r\nI think it will be very difficult to start a new nation-wide contest among high school students. Besides, you already have USAMO, AIME, and other contests...", "Solution_2": "first, of all, I think reddevilz wants to make a contest for AoPS members. The questions would be posted online and done in the honor system. I think it's a great idea--for me (at least for now) the MathLinks contest is waaaaay beyond me. \r\n\r\nsecond of all, I'd note that while difficult, it's not impossible to start a nationwide contest. Mr. Rusczyk and his old pals Sandor Lehoczky and Sam Vandervelde started the Mandelbrot only a little more than a decade back--I think. I could be wrong.", "Solution_3": "[quote=\"bubala\"]first, of all, I think reddevilz wants to make a contest for AoPS members. The questions would be posted online and done in the honor system. I think it's a great idea--for me (at least for now) the MathLinks contest is waaaaay beyond me. \n\nsecond of all, I'd note that while difficult, it's not impossible to start a nationwide contest. Mr. Rusczyk and his old pals Sandor Lehoczky and Sam Vandervelde started the Mandelbrot only a little more than a decade back--I think. I could be wrong.[/quote]\r\n\r\nThere's a ton of things that have to be done to make a contest. Mostly comimg up with problems, but also orginazation, planning it, administering it, a ton of other stuff. You might not mean it, but it sort of sounds like you think starting a nation-wide contest isn't too hard.\r\n\r\nIt might be cool, but it would probably depend a lot on the level of interest in the contest.", "Solution_4": "no no no, I never said it wouldn't be hard. It would be a gigantic undertaking. What I meant was it's not impossible. \r\n\r\nAt the beginning, I was saying that starting an AoPS contest wouldn't be terribly hard. Then, I went on to say that although extremely difficult, starting a nation-wide contest is not impossible. \r\n\r\nSOrry if that's unclear.", "Solution_5": "It's impossible at this point to have a contest like your suggesting. Mainly because we already are dealing with a few online contests (USAMTS, MathLinks Contest) and because of other projects that are underway.", "Solution_6": "oh well. I'll just keep improving my math skills and one day I'll be good enough for the MathLinks one.", "Solution_7": "[quote=\"bubala\"]no no no, I never said it wouldn't be hard. It would be a gigantic undertaking. What I meant was it's not impossible. \n\nAt the beginning, I was saying that starting an AoPS contest wouldn't be terribly hard. Then, I went on to say that although extremely difficult, starting a nation-wide contest is not impossible. \n\nSOrry if that's unclear.[/quote]\r\n\r\nMmm, gotcha. :) \r\n\r\nAlso, since you're a middle schooler, you haven't been exposed to a lot of math contests. I think you'll find that there are many more contests available once you're in high school, as long as you're part of your school's math team. Although this may vary from school to school.", "Solution_8": "There are already [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]many, many contests[/url] out there." } { "Tag": [ "integration", "algebra", "function", "domain", "advanced fields", "advanced fields unsolved" ], "Problem": "I'm trying to solve this ODE problem:\r\n\r\nlet $ x^{(1)} \\equal{} \\left( \\begin{array}{ccc}1\\\\1\\\\1\\end{array} \\right)$, $ x^{(2)} \\equal{} \\left( \\begin{array}{ccc}1\\\\t\\\\t^2\\end{array} \\right)$, $ x^{(3)} \\equal{} \\left( \\begin{array}{ccc}1\\\\t\\\\t^3\\end{array} \\right)$ \r\n\r\na) Find the Wronskian $ W(x^{(1)}, x^{(2)}, x^{(3)})$\r\n\r\nb) You are told that $ x^{(1)}, x^{(2)}, x^{(3)}$ are solutions of a \r\n\r\nlinear homogeneous system $ x' \\equal{} P(t)x$\r\n\r\ni) what can you say about the discontinuities of the coefficient $ p_{i,j}(t)$?\r\n\r\nii) Find $ p_{1,1}(t) \\plus{} p_{2,2}(t) \\plus{} p_{3,3}(t)$. (Hint: use your solution to part a).\r\n\r\n\r\n==============================================\r\n\r\nFor part a), the Wronskian is W = determinant of $ (x^{(1)}, x^{(2)}, x^{(3)})$, which I found to be $ t^4 \\minus{}2t^3 \\plus{} t^2$. I'm not sure how to do part b). What discontinuities is the question refering to, and how do I find them? How do I use my answer to part a) to solve b)ii)? Any help is appreciated!", "Solution_1": "1) $ W(t)=[x_1(t),x_2(t),x_3(t)]$; $ det(W(t))=det(W(t_0))e^{\\int_{t_0}^{t}trace(P(t))dt}$. Thus if $ det(W(t_0))\\not=0$ then $ \\forall{t}\\in{D},det(W(t))=(t-1)^2t^2\\not=0$ whith $ D$ = the domain where $ P$ is continuous.\r\nConclusion: $ P$ can't be continuous in $ t=1$ or $ t=0$.\r\n2) ${ \\int_{t_0}^{t}trace(P(t))dt}=ln(\\dfrac{t^4-2t^3+t^2}{t_0^4-2t_0^3+t_0^2})$ and $ trace(P(t))=\\dfrac{4t^3-6t^2+2t}{t^4-2t^3+t^2}$." } { "Tag": [ "geometry", "limit" ], "Problem": "A pirate ship spots, 10 nautical miles to the east, an oblivious caravel sailing 60 degrees south of west at a\r\nsteady 12 nm/hour. What is the minimum speed that the pirate ship must maintain at to be able\r\nto catch the caravel?\r\n\r\nNote: I think the answer is 12, but could someone confirm this?", "Solution_1": "[hide=\"solution\"]\nThe fastest rout between any two points is a straight line. So construct a triangle with one side length $ 10$ and the other length $ 12t$ and an angle between them of $ 120^o$\nThe distance that the pirate ship travels, $ c$ is $ c^2 \\equal{} 10^2 \\plus{} (12.t)^2 \\minus{} 120.t.cos(120)$\nSo the speed it is traveling is $ \\dfrac{c}{t}$\n$ \\displaystyle\\lim_{t\\to\\infty}\\dfrac{\\sqrt {144t^2 \\plus{} 60t \\plus{} 100}}{t} \\equal{} 12$\n\nSo $ 12$ m/h is correct[/hide]" } { "Tag": [ "calculus" ], "Problem": "m$(d^2 x$/$dt)$ - b$( dx/dt)$ = -a ( a and b and m are constants) \r\n\r\nI got the CF as x=A+Be^ (-bt/m)\r\nBut i do not know how to get the PI", "Solution_1": "Should this be in advanced physic ?I think it is under calculus .Anyway \r\n\r\nthe auxiliary equation is $mp^2-bp=0$ $\\implies p=0,b/m$ So the CF is $x=A+Be^{bt/m}$\r\n\r\nTo choose PI , we notice that it is $-a$ which is a constant . Hence we should also pick a constant for PI. But notice that in CF, we already have a constant , hence we have to multiply it with $t$ , so our PI is $x=ct$ for constant $c$ . Plugging back into the original equation and solve it to get $c=\\frac{b}a$ .\r\n\r\nSo our equation should looks like\r\n\r\n$x=A+Be^{bt/m}+\\frac{b}at$ where $A,B$ are arbitrary constant ." } { "Tag": [ "function", "algebra", "domain", "Functional Analysis", "calculus", "calculus computations" ], "Problem": "I am not really sure if this question belongs here. Here it goes:\r\n\r\nI was wondering what's the difference between the following two \"mapping\" symbols:\r\n$f: A \\to B$ vs $f: A \\mapsto B$\r\n\r\nAlso, in how many and in what ways can you read the following notation:\r\n$f \\circ g$", "Solution_1": "$\\to$ is used when the thing on the left is the domain and the thing on the right is (a set containing) the range. Thus, we say $f: \\bf R \\to C$ is a function from the reals to the complexes.\r\n\r\n$\\mapsto$ is used for particular values of a function. For instance, the function $f(x) = x^2 + i$ can be written $f: x \\mapsto x^2 + i$. If $x$ is only allowed to take real values, this function is a function $f: \\bf R \\to C$, as well.", "Solution_2": "So $\\mapsto$ is used to specify the rule itself whereas $\\to$ is used to specify the domain and co-domain?", "Solution_3": "Yeah, that's basically what he said. One thing though is that I don't think it should be called the co-domain as you just mentioned in your question. If I remember the terms correctly, the co-domain of $f: A\\rightarrow B$ is $\\textrm{Im}\\left(f\\right)$ and $\\textrm{Im}\\left(f\\right)$ is only equal to $B$ if $f$ is onto.", "Solution_4": "The notational distinction JBL is making is news to me. Many people use $f: A\\mapsto B$ just to show the domain and codomain.\r\n\r\nAs for swapnillium's second question, I would read $f\\circ g$ only as the composition of functions - that is, $f\\circ g(x)=f(g(x)).$\r\n\r\nThat's for writing functions on the left. If you for some reason write functions on the right, then you had better explain your notation at the beginning of what you're writing.", "Solution_5": "[quote=\"Kent Merryfield\"]\nAs for swapnillium's second question, I would read $f\\circ g$ only as the composition of functions - that is, $f\\circ g(x)=f(g(x)).$\n[/quote]\r\nI know that $f\\circ g$ means $f(g(x))$ but how do you read it? I mean, when I first saw it, I started calling it \"f dot g,\" but is this the way most people read this notation?", "Solution_6": "[quote=\"swapnillium\"][quote=\"Kent Merryfield\"]\nAs for swapnillium's second question, I would read $f\\circ g$ only as the composition of functions - that is, $f\\circ g(x)=f(g(x)).$\n[/quote]\nI know that $f\\circ g$ means $f(g(x))$ but how do you read it? I mean, when I first saw it, I started calling it \"f dot g,\" but is this the way most people read this notation?[/quote]\r\n\r\nI say \"f of g of x\" as if it was just written as $f(g(x))$.", "Solution_7": "[quote=\"paladin8\"]I say \"f of g of x\" as if it was just written as $f(g(x)).$[/quote]\r\nAbsolutely. Whatever you call it, you can't call it \"dot\" because it's not a dot. It's a circle (which is vaguely like an \"o\", which is the first letter of \"of\".\r\n\r\nIf you're feeling slightly more formal, say \"f composed with g at x\" or \"f compose g/\"", "Solution_8": "[quote=\"Kent Merryfield\"]The notational distinction JBL is making is news to me. Many people use $f: A\\mapsto B$ just to show the domain and codomain.[/quote]\r\n\r\nHmm, okay. I feel like that's the way all of my professors have used the two notations, but I suppose that could be not true or just a coincidence.\r\n\r\n\r\nFor the second question, \"f composed with g\" or \"f of g\" is how I would say that out loud. (I don't say \"at x\" unless there actually is an x there.)", "Solution_9": "[quote=\"amcavoy\"]Yeah, that's basically what he said. One thing though is that I don't think it should be called the co-domain as you just mentioned in your question. If I remember the terms correctly, the co-domain of $f: A\\rightarrow B$ is $\\textrm{Im}\\left(f\\right)$ and $\\textrm{Im}\\left(f\\right)$ is only equal to $B$ if $f$ is onto.[/quote]\r\n\r\nYou have the meaning of co-domain backwards. In $f: A \\rightarrow B$, $A$ is the [i]domain[/i] of $f$, $B$ is the [i]co-domain[/i] of $f$, and $\\textrm{Im}\\left(f\\right)$ is the [i]range[/i] of $f$. The co-domain is a [i]superset[/i] of the range.\r\n\r\nIf $\\textrm{Im}\\left(f\\right) = B$, then $f$ is said to be [i]onto[/i] (or surjective)", "Solution_10": "[quote=\"amcavoy\"]Yeah, that's basically what he said. One thing though is that I don't think it should be called the co-domain as you just mentioned in your question. If I remember the terms correctly, the co-domain of $f: A\\rightarrow B$ is $\\textrm{Im}\\left(f\\right)$ and $\\textrm{Im}\\left(f\\right)$ is only equal to $B$ if $f$ is onto.[/quote]\r\n\r\nWait, are you sure? I mean, for example, let $f(x)=x^2$ and we can describe the input and output of this function as $f: \\mathbb{R} \\to \\mathbb{R}$, right? Then, don't we then mean that $f: domain \\to codomain$ since $\\mathbb{R}$ is not really the range but the co-domain of $f$?", "Solution_11": "anyone else have anything to say in this matter?", "Solution_12": "[quote]Kent Merryfield wrote: \n[quote]The notational distinction JBL is making is news to me. Many people use just to show the domain and codomain.[/quote] \n\n\n\nHmm, okay. I feel like that's the way all of my professors have used the two notations, but I suppose that could be not true or just a coincidence. [/quote] \r\n\r\nI've seen this notational distinction explicitly stated in a book before. So it's a real thing, but maybe it hasn't become standard yet.", "Solution_13": "I think Stein uses them as JBL said, but that's the only text where I remember any kind of distinction. \r\n\r\nIt doesn't really matter so long as the author makes clear the meaning of his notation when he first uses it (and if it isn't too bizarre).", "Solution_14": "Well, that makes me feel better :)\r\n\r\n[quote=\"swapnillium\"]anyone else have anything to say in this matter?[/quote]\r\nRead the post above yours.", "Solution_15": "I am sorry for the confusion. Here's my main question:\r\nIn general when we write $f: domain \\to ?$, do we actually mean $f: domain \\to codomain$ or $f: domain \\to range$?", "Solution_16": "The co-domain is the set [i]into[/i] which a function maps its domain.\r\n\r\nThe range is the set [i]onto[/i] which a function maps its domain.", "Solution_17": "gauss202, you are not answering his question. If you see this $f: A\\to B$, would you interpret it as\r\n\r\nA) $f: \\mathrm{domain}\\to\\mathrm{codomain}$;\r\n\r\nB) $f: \\mathrm{domain}\\to\\mathrm{range}$;\r\n\r\nC) Both A and B\r\n\r\nD) None of the above.\r\n\r\nNo explanation, just [size=150][b][u]CHOOSE[/u][/b][/size] one please. Thank you.", "Solution_18": "swapnillium: you can delete that kind of post with the 'x' box in the corner. \r\n\r\n10000th user: I choose A. Often you want to discuss some kind of general function, such as an integrable function. Then you'll say, \"Suppose $f: R \\rightarrow R$ is integrable.\" I don't want to imply that the range of $f$ is all of $R$ -- I want $f$ to be as general as possible. Wouldn't it be a bit unwieldy if I always had to write \"$f: R \\rightarrow A$ with $A \\subset R$\" all the time?", "Solution_19": "[quote=\"Xevarion\"]swapnillium: you can delete that kind of post with the 'x' box in the corner. [/quote]\r\nI don't see an 'x.' Where exactly is it?\r\n\r\n\r\nBTW, I also agree with A). I mean, that's what I see my professors doing all the time. :lol:", "Solution_20": "[quote=\"10000th User\"]If you see this $f: A\\to B$, would you interpret it as\n\nA) $f: \\mathrm{domain}\\to\\mathrm{codomain}$;\n\nB) $f: \\mathrm{domain}\\to\\mathrm{range}$;\n\nC) Both A and B\n\nD) None of the above.\n[/quote]\r\n\r\n[b]A[/b]", "Solution_21": "[quote=\"swapnillium\"]I don't see an 'x.' Where exactly is it?[/quote]\r\n\r\nAt the bottom right of anyone else's post, you should see \"quote\" and nothing else. At the bottom right of your own posts, you should see \"quote\" \"edit\" \"[x]\" -- that third box is the one I'm talking about. \r\n\r\nSee MareleG's recent post in Unsolved and Proposed problems for an example of this notation in action (using interpretation A, as seems to be popular).", "Solution_22": "[quote=\"Xevarion\"]\nAt the bottom right of anyone else's post, you should see \"quote\" and nothing else. At the bottom right of your own posts, you should see \"quote\" \"edit\" \"[x]\" -- that third box is the one I'm talking about. [/quote]\r\nCall me delusional, but I just see two icons (\"quote\" and \"edit\") on the bottom right corner of my post. However, when I move my mouse over the \"edit\" icon, it does display \"Edit/Delete this post,\" but when I do click on the \"edit\" icon it just gives me the same old editing menu. :huh:", "Solution_23": "Oh, I get it! You can't freely delete it anymore if someone has replied, I think. But if you click edit, there's still nowhere to delete it? Let me post this and then check on an older message of mine.\r\n\r\nEdit: Sorry, you're right. I can't figure out how to delete the post if someone has already replied. I guess you're just supposed to replace the useless post with something witty. ;)", "Solution_24": "But moderators can delete any post, whether replied to or not. Since you're all fussing about this, I just did that to swapnillium's post.\r\n\r\nFairly often, when I notice a post with no content, I'll delete it. But I don't get to everything.", "Solution_25": "[quote=\"Kent Merryfield\"]The notational distinction JBL is making is news to me. Many people use $f: A\\mapsto B$ just to show the domain and codomain.[/quote]\r\nSurprising, it always seemed a common notation to me ! At last in Serge Lang's \"Algebra\" and \"Real and functional analysis\" the distinction (that of JBL) is clearly mentioned. Also French people standardly use this convention (For example, a french math teacher told me I've made a mistake when I wrote $f : A \\mapsto B$). This distinction is introduced almost as early as functions are in the French Educational system (functions are introduced in [i]troisieme[/i] and this notation comes a year later in [i]seconde[/i])." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "number theory theorems", "number theory" ], "Problem": "I want to know which theorems and fomulas are useful for math olympiad.Is there any body who can introduce me an electric book or pdf file or something like this to find them and their solutions.\r\nThank You.", "Solution_1": "Well, I'd be glad to see a recomendation as well.", "Solution_2": "Engles: \"Problem solving stratagies\" is really good, it'll show you all the tricks and has a mindboggaling amount of problems.\r\n\r\nZetiz: \"The Art and Craft of problem solving\" is very good aswell, it shows you whatg mindframe to be in and how to start looking at problems in the right light.\r\n\r\nThere the only 2 books i have, so i cant say much more!", "Solution_3": "I woudl say that mathematical olympiad chellenges is also an amazing book. It has theorems and many problems from USAMO, IMO, Journals, etc." } { "Tag": [ "linear algebra", "matrix", "function", "linear algebra unsolved" ], "Problem": "Let $ X$ be any set, and let $ f$ be any function from $ X$ into $ \\mathbb{C}$. Let $ x_1,x_2,\\cdots,x_n$ be any points of $ X$. Prove that the matrix $ A$ defined by $ A_{ij}\\equal{}f(x_i)\\overline{f(x_j)}$ is nonnegative definite.", "Solution_1": "[hide=\"hint\"]For $ \\xi \\in \\mathbb{C}^n$, $ \\xi^*A\\xi \\equal{} \\sum_{i \\equal{} 1}^n\\sum_{j \\equal{} 1}^nf(x_i)\\overline{f(x_j)}\\xi _i \\overline{\\xi _j} \\equal{} \\eta^*J\\eta$ where $ \\eta_i \\equal{} \\xi _i f(x_i)$ and $ J$ is the matrix of all ones. Hence to show $ A$ is positive semi-definite it suffices to show that $ J$ is, which is easy enough (show that $ J$ has $ n$ as an eigenvalue, and $ 0$ as an eigenvalue with multiplicity $ n \\minus{} 1$).[/hide]", "Solution_2": "thanks a lot.", "Solution_3": "It is a special kind of Gram matrix. Thus... :)", "Solution_4": "JoeBlow: Your $ J$ is [i]not[/i] the all-ones matrix. It is simpler than that ;)\r\n\r\n darij" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "( a+b+c)2 (a+b)2 (b+c)2 (c+a)2\r\n-------------- +3 > -------- + --------- + ---------\r\na2+b2+c2 ~ a2+b2 b2+c2 c2+a2\r\n\r\n2 means squared.\r\nProve it for a,b,c real", "Solution_1": "[quote=\"andi\"]( a+b+c)2 (a+b)2 (b+c)2 (c+a)2\n-------------- +3 > -------- + --------- + ---------\na2+b2+c2 ~ a2+b2 b2+c2 c2+a2\n\n2 means squared.\nProve it for a,b,c real[/quote]\r\nIf are you mean we need to prove that\r\n $ \\frac {(a \\plus{} b \\plus{} c)^2}{a^2 \\plus{} b^2 \\plus{} c^2} \\plus{} 3\\geq\\frac {(a \\plus{} b)^2}{a^2 \\plus{} b^2} \\plus{} \\frac {(b \\plus{} c)^2}{b^2 \\plus{} c^2} \\plus{} \\frac {(c \\plus{} a)^2}{c^2 \\plus{} a^2}$ then, it's wrong.\r\nTry $ a \\equal{} b \\equal{} c \\equal{} 0.$ :wink:\r\nFor $ a,$ $ b$ and $ c$ such that $ |ab|\\plus{}|ac|\\plus{}|bc|\\neq0$ we obtain:\r\n$ \\frac {(a \\plus{} b \\plus{} c)^2}{a^2 \\plus{} b^2 \\plus{} c^2} \\plus{} 3\\geq\\frac {(a \\plus{} b)^2}{a^2 \\plus{} b^2} \\plus{} \\frac {(b \\plus{} c)^2}{b^2 \\plus{} c^2} \\plus{} \\frac {(c \\plus{} a)^2}{c^2 \\plus{} a^2}\\Leftrightarrow$\r\n$ \\Leftrightarrow1\\plus{}\\sum_{cyc}\\frac{2ab}{a^2\\plus{}b^2\\plus{}c^2}\\geq\\sum_{cyc}\\frac{2ab}{a^2\\plus{}b^2}\\Leftrightarrow1\\geq\\sum_{cyc}\\frac{2abc^2}{(a^2\\plus{}b^2)(a^2\\plus{}b^2\\plus{}c^2)}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\left(c^2\\minus{}\\frac{2abc^2}{a^2\\plus{}b^2}\\right)\\geq0\\Leftrightarrow\\sum_{cyc}\\frac{c^2(a\\minus{}b)^2}{a^2\\plus{}b^2}\\geq0.$" } { "Tag": [ "analytic geometry", "geometry proposed", "geometry" ], "Problem": "Given to points $ A,B$, show that all points $ C$ with the ability $ 2AC \\equal{} BC$ lies on a circle.\r\n\r\nGeneralisation: Given to points $ A,B$, show that all points $ C$ with the ability $ \\frac {AC}{BC} \\equal{} k$ for a given $ k \\in \\mathbb{R}_ \\plus{}$, lies on a circle.", "Solution_1": "[hide=\"Easy, here is a hint\"]\nPut coordinate system\n[hide=\"Hint 2\"]\n such that A=(0,0) B=(0,b) C=(x,y)\n[hide=\"Final\"]\nNow use the condition to get the equation of the circle and that's it\n[/hide]\n[/hide]\n[/hide]", "Solution_2": "[quote=\"Bugi\"][hide=\"Easy, here is a hint\"]\nPut coordinate system\n[hide=\"Hint 2\"]\n such that A=(0,0) B=(0,b) C=(x,y)\n[hide=\"Final\"]\nNow use the condition to get the equation of the circle and that's it\n[/hide]\n[/hide]\n[/hide][/quote]\r\nOk :) I did it with complex numbers. Then $ 2|a\\minus{}z| \\equal{} |b\\minus{}z|$ simplifies to $ \\left | z \\minus{} \\frac{4a\\minus{}b}{3} \\right | \\equal{} \\frac{2}{3}|a\\minus{}b|$ :)\r\n\r\nAnd likewise $ |a\\minus{}z| \\equal{} k|b\\minus{}z| \\iff \\left | z \\minus{} \\frac{k^2b\\minus{}a}{k^2\\minus{}1} \\right | \\equal{} \\frac{k}{k^2\\minus{}1} |a\\minus{}b|$ .\r\n\r\nHow can this be proved synthetic?", "Solution_3": "[quote]Given to points $ A,B$, show that all points $ C$ with the ability $ 2AC \\equal{} BC$ lies on a circle.\n\nGeneralisation: Given to points $ A,B$, show that all points $ C$ with the ability $ \\frac {AC}{BC} \\equal{} k$ for a given $ k \\in \\mathbb{R}_ \\plus{}$, lies on a circle.[/quote]\r\n[b]Maybe this prove is synethic enough[/b], dear [b]Mathias_DK[/b]\r\nThe internal bisector of $ \\angle ACB$ wrt $ \\triangle ABC$ intersects $ BA$ at $ D$, and the outner bisector of $ \\angle ACB$ wrt $ \\triangle ABC$ intersects $ BA$ at $ E$. It is obvious that $ \\frac {CA}{CB} \\equal{} \\frac {DA}{DB} \\equal{} \\frac {EA}{EB} \\equal{} k \\equal{} const$, but $ E,D\\in AB$ wich is fixed $ \\Longrightarrow E,D$ are fixed. But $ CE\\bot CD$ at $ C$\r\nTherefore $ C\\in$ the circle with diameter $ ED$- wich is the fixed circle.\r\nThe result is lead as follow.", "Solution_4": "This is the famous Apollonius Circle" } { "Tag": [ "email", "LaTeX" ], "Problem": "Can someone pm or email me pre-1999 chapter tests? I would reaaaaaaly appreciate it. Pre-1996 state tests and answer keys to any nationals from 1991 and on would also be very appreciated.", "Solution_1": "Look at my last post here:[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78829]www.artofproblemsolving.com/Forum/viewtopic.php?t=78829[/url]I need to figure out how to work the tex center or get a scanner or something.", "Solution_2": "PM WHATEVER YOU HAVE AND I WILL MAKE IT INTO PDF LATEX OR WHATEVER :wow: :wow: YOU ARE MY HERO AND SAVIOR", "Solution_3": "I'll just type it up on that pdf program ccy gave. I'll post it in the public site thread.", "Solution_4": "youre gonna type up that much!?\r\nwoah. If you need any help let me know", "Solution_5": "[quote=\"biffanddoc\"]youre gonna type up that much!?\nwoah. If you need any help let me know[/quote]\r\nnah i'll prob end up getting a scanner. I'm a slow typer.", "Solution_6": "ok.\r\nare the tests online anywhere?", "Solution_7": "Can you PM anything before 1991 that you have. PLEASE!!!", "Solution_8": "[quote=\"xxm0rg07hxx\"]Can you PM anything before 1991 that you have. PLEASE!!![/quote]\r\nI infer from this that you have stuff from after 1991. Can you email/pm me stuff you have from 1991-1996 in states? thanks", "Solution_9": "I dont think he has a scanner." } { "Tag": [ "algebra", "polynomial", "function", "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ f$ be an irreducible polynomial over field $ K$ of degree p, p - prime. Prove that if the Galois group of this polynomial is solvable (i.e. Galois group of the extension of K to the splitting field of $ f$), then for every extension $ L$ of $ K$ $ f$ has exactly 0, 1 or p roots in $ L$.", "Solution_1": "Use this result due to Galois himself: \r\nIn order that an irreducible equation of prime degree be solvable by radicals, it is necessary and sufficient that all its roots be rational functions of two roots." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a;b;c$ be positive real numbers satisfying $ ab\\plus{}bc\\plus{}ca\\equal{}7$\r\nFind min of $ T\\equal{}\\frac{8a^4\\plus{}1}{a^2} \\plus{}\\frac{ 108b^5\\plus{}1}{b^2} \\plus{} \\frac {16c^6\\plus{}1}{c^2}$", "Solution_1": "[quote=\"chien than\"]Let $ a;b;c$ be positive real numbers satisfying $ ab \\plus{} bc \\plus{} ca \\equal{} 7$\nFind min of $ T \\equal{} \\frac {8a^4 \\plus{} 1}{a^2} \\plus{} \\frac { 108b^5 \\plus{} 1}{b^2} \\plus{} \\frac {16c^6 \\plus{} 1}{c^2}$[/quote]\r\nWhy title of the topic is ab+bc+ca=7abc?", "Solution_2": "[quote=\"N.T.TUAN\"][quote=\"chien than\"]Let $ a;b;c$ be positive real numbers satisfying $ ab \\plus{} bc \\plus{} ca \\equal{} 7$\nFind min of $ T \\equal{} \\frac {8a^4 \\plus{} 1}{a^2} \\plus{} \\frac { 108b^5 \\plus{} 1}{b^2} \\plus{} \\frac {16c^6 \\plus{} 1}{c^2}$[/quote]\nWhy title of the topic is ab+bc+ca=7abc?[/quote]\r\nif $ 7abc \\equal{} ab \\plus{} bc \\plus{} ca$ => $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\equal{} 7$\r\n$ T \\equal{} (8a^2 \\plus{} \\frac {1}{2a^2}) \\plus{} (54b^3 \\plus{} 54b^3 \\plus{} \\frac {2}{9b^2} \\plus{} \\frac {2}{9b^2} \\plus{} \\frac {2}{9b^2}) \\plus{} (16c^4 \\plus{} \\frac {1}{4c^2} \\plus{} \\frac {1}{4c^2}) \\plus{} (\\frac {1}{2a^2} \\plus{} \\frac {1}{3b^2} \\plus{} \\frac {1}{2c^2})\\ge 4 \\plus{} 10 \\plus{} 3 \\plus{} \\frac {1}{7}.(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})^2 \\equal{} 24$\r\n :D" } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "[b]1[/b] $\\mathop {\\lim }\\limits_{x \\to 1} \\frac{{x^{\\frac{p}{q}} - 1}}{{x^{\\frac{r}{s}} - 1}}$\r\n\r\n[b]2[/b] $\\mathop {\\lim }\\limits_{x \\to 0} \\frac{{1 - \\cos x \\cdot \\cos 2x \\cdot \\cos 3x}}{{x^2 }}$\r\n\r\n[b]3[/b] $\\mathop {\\lim }\\limits_{x \\to 0} \\frac{{\\sqrt[m]{{\\cos \\alpha x}} - \\sqrt[n]{{\\cos \\beta x}}}}{{x^2 }}$ $m,n \\in N$", "Solution_1": "1. $\\mathop {\\lim }\\limits_{x \\to 1} \\frac{{x^{\\frac{p}{q}} - 1}}{{x^{\\frac{r}{s}} - 1}}= \\frac{ps}{qr}$\r\n\r\n2. $\\mathop {\\lim }\\limits_{x \\to 0} \\frac{{1 - \\cos x \\cdot \\cos 2x \\cdot \\cos 3x}}{{x^2 }} = 7$\r\n\r\n3. $\\mathop {\\lim }\\limits_{x \\to 0} \\frac{{\\sqrt[m]{{\\cos \\alpha x}} - \\sqrt[n]{{\\cos \\beta x}}}}{{x^2 }} = \\frac{b^2m-a^2n}{2mn}$", "Solution_2": "For positive integers $n,$ we can generalize \r\n\\[ \\lim_{x\\rightarrow 0} \\frac{1-\\cos x\\cdot \\cos 2x\\cdots \\cdots\\cos nx}{x^2}=\\frac{1}{12}n(n+1)(2n+1). \\]", "Solution_3": "[quote=\"kunny\"]For positive integers $n,$ we can generalize \n\\[ \\lim_{x\\rightarrow 0} \\frac{1-\\cos x\\cdot \\cos 2x\\cdots \\cdots\\cos nx}{x^2}=\\frac{1}{12}n(n+1)(2n+1). \\][/quote]\r\nThis can be proved by induction.", "Solution_4": "[quote=\"kunny\"]For positive integers $n,$ we can generalize \n\\[ \\lim_{x\\rightarrow 0} \\frac{1-\\cos x\\cdot \\cos 2x\\cdots \\cdots\\cos nx}{x^2}=\\frac{1}{12}n(n+1)(2n+1). \\][/quote]\r\nThis can be proved by induction.", "Solution_5": "We can prove this directly. :D" } { "Tag": [], "Problem": "I have 6 friends and during a certain vacation, i met them during several dinners. I found that I dined with all the 6 on 1 day, with every 5 of them on 2 days, with every 4 of them on 3 days, with every 3 of them on 4 days, and with every 2 of them on 5 days. Further every friend was absent at 7 dinners, and every friend was present at 7 dinners. How many dinners did I have alone?\r\n\r\n :blush:", "Solution_1": "Let the friends be called $ A_i$, for $ i\\equal{}1, \\, 2,\\,\\ldots ,\\,6$.\r\nI had $ 1$ dinner with all of them $ A_1A_2A_3A_4A_5A_6$. \r\nWith every five of them, I should have exactly two dinners. All possible five of them are:\r\n$ A_1A_2A_3A_4A_5$, \r\n$ A_1A_2A_3A_4A_6$, \r\n$ A_1A_2A_3A_5A_6$, \r\n$ A_1A_2A_4A_5A_6$, \r\n$ A_1A_3A_4A_5A_6$, and \r\n$ A_2A_3A_4A_5A_6$\r\nWe see that each of the six groups of five friends were present when I had dinner with all six. So I ought to have one more dinner with each of the above groups to make the number of dinners to be two with every five of my friends. That makes the total number of dinner so far to be 7. \r\nFurther, we see that, by now, I already have three dinners with every four of my friends, four dinners with every three of them, and five dinners with every two of them. So the conditions given in the problem are all fulfilled, except the last condition which requires each friend to be present in exactly seven dinners and each of them to be absent in exactly seven of them.\r\nIf we again look at the above groupings, we see that $ A_1$ is so far present in exactly six of the dinners. In order for $ A_1$ to be present in exactly seven dinners, I must have one more dinner with him. In a similar way, I must have one dinner each with only $ A_2$, only $ A_3$, only $ A_4$, only $ A_5$, and only $ A_6$. So by now, I have a total of 13 dinners.\r\nFinally, by now we see that $ A_1$ is not preent in exactly six dinners. The same goes true for each friend so far. Thus, to make the number of absents for each friend to be exactly seven, [b]I must have one dinner alone.[/b]" } { "Tag": [ "email" ], "Problem": "the semifinalist list got posted today, if anyone here entered.\r\nhttp://www.siemens-foundation.org/", "Solution_1": "Congratulations to your son.", "Solution_2": "I entered, and I am a semifinalists also. I pray that I make it to Regional Finalist, though I was very surprised to make it to semifinalist in the first place.", "Solution_3": "Out of curiosity, how on earth are regional finalists supposed to prepare for the poster session (I am referring to those who have to do it on Nov 4), considering that regional finalists aren't even announced until Nov 1.", "Solution_4": "good question, i'd like to know this as well.\r\n\r\nby the way, anyone know many many semi's make it to regionals in each of the 6 regions???\r\n\r\nEDIT: to answer that first question, I'd guess S-W contacts the regional finalists BEFORE the nov. 1 notification (by phone, email? i dunno). that way, they have some advance notice. This is just a guess though, but it seems logical.", "Solution_5": "I'm sure that there are many people here on AoPS who have qualified for regional finalist or beyond. Does anyone know the answer? (are regional finalists notified before Nov 1?)", "Solution_6": "Out of curiousity, how do people get their research published? Do they just randomly find a scientific/mathematical journal and ask that their work be published? I mean, can high schoolers really do research that matches the work of people who spent the past thirty years of their life in some field? I would be rather depressed if I had a Phd in some field, and a high schooler like thirty years younger than me and without so much as a high school diploma published his/her research in the same journal that I was publishing mine.", "Solution_7": "[quote=\"**********\"]\nby the way, anyone know many many semi's make it to regionals in each of the 6 regions???\n[/quote]\r\n\r\nthere is a maximum of 5 individuals and 5 teams for each region. If there are 2 -3 people per team, that means that a maximum of 15-20 people advance in each region." } { "Tag": [ "geometry", "3D geometry", "probability" ], "Problem": "Express probability in a common fraction please!\r\n\r\n1. If a white 8x8x8 cube has 2 sides (non-adjacent) painted black and you cut the cube into 1x1x1 cubes and placed them into a bag, what is the probability that you will pick out one cube with one black face on it?\r\n\r\n2. If 2 [i]adjacent[/i] sides are painted black (on the same cube), what is the probability that you will pick out a cube with [i]exactly[/i] one face painted black?\r\n\r\n3. If [i]3[/i] sides ([i]all touching the same corner[/i]) are painted black (on the same cube), what is the probability that you will pick out a cube with [i]exactly[/i] one face painted black?", "Solution_1": "[hide=\"The Answers :rotfl: \"][hide=\"1)\"]1/4[/hide][hide=\"2)\"]7/32[/hide][hide=\"3)\"]147/512[/hide][/hide]", "Solution_2": "ugh can you please explain? I looked at number one and didn't feel like trying to even attempt it.", "Solution_3": "[hide=\"1\"]There are 128 painted faces and 512 cubes so the answer is 1/4 when you divide[/hide]", "Solution_4": "Wouldn't #3 be\r\n[hide]\n37/128\n\nBecause there are 170 cubes painted black if I'm not mistaken and 22 of those have 2 or 3 sides painted. 170 - 22 = 148 cubes painted 1 side\n\n148 / 512 = 37 / 128[/hide]", "Solution_5": "If you think about it...\r\n\r\n[hide]one face of an 8x8x8 cubes has 64 squares. For #3, you subtract two of the edges to get 7x7 squares = 49 squares then, you multiply that by 3 to get 147/512, which is in lowest terms, not 148/512 [/hide] :P", "Solution_6": "[quote=\"Quevvy\"]Express probability in a common fraction please!\n\n1. If a white 8x8x8 cube has 2 sides (non-adjacent) painted black and you cut the cube into 1x1x1 cubes and placed them into a bag, what is the probability that you will pick out one cube with one black face on it?\n\n2. If 2 [i]adjacent[/i] sides are painted black (on the same cube), what is the probability that you will pick out a cube with [i]exactly[/i] one face painted black?\n\n3. If [i]3[/i] sides ([i]all touching the same corner[/i]) are painted black (on the same cube), what is the probability that you will pick out a cube with [i]exactly[/i] one face painted black?[/quote]\r\n[hide=\"1\"]$p=\\frac{8\\times8\\times2}{8\\times8\\times8}=\\boxed{\\frac14}$[/hide]\n[hide=\"2\"]$p=\\frac{8\\times7\\times2}{8\\times8\\times8}=\\boxed{\\frac7{32}}$[/hide]\n[hide=\"3\"]$p=\\frac{7\\times7\\times3}{8\\times8\\times8}=\\boxed{\\frac{147}{512}}$[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ n \\geq 3$ be an odd integer. We denote by $ [\\minus{}n,n]$ the set of all integers greater or equal than $ \\minus{}n$ and less or equal than $ n$.\r\nPlayer $ A$ chooses an arbitrary positive integer $ k$, then player $ B$ picks a subset of $ k$ (distinct) elements from $ [\\minus{}n,n]$. Let this subset be $ S$.\r\nIf all numbers in $ [\\minus{}n,n]$ can be written as the sum of exactly $ n$ distinct elements of $ S$, then player $ A$ wins the game. If not, $ B$ wins.\r\nFind the least value of $ k$ such that player $ A$ can always win the game.", "Solution_1": "k>n+1, since if k\\widehat{MDE}\\equal{}\\widehat{FEC}$\r\nthat: $ \\widehat{FDM}\\equal{}\\widehat{MDE}$\r\n$ \\equal{}>\\widehat{FDM}\\equal{}\\widehat{FEC}$" } { "Tag": [ "geometry", "geometry solved" ], "Problem": "The convex quadrilateral ABCD is inscribed in the circle $S_1$. Let $O$ be the intersection of $\\overline{AC}$ and $\\overline{BD}$. Circle $S_2$ passes through $D$ and $O$, intersecting $\\overline{AD}$ and $\\overline{CD}$ at $M$ and $N$, respectively. $\\overline{OM}$ and $\\overline{AB}$ intersect at $R$, lines $ON$ and $BC$ intersect at $T$, and $R$ and $T$ lie on the same side of line $BD$ as $A$. Prove that $O, R, T,$ and $B$ are concyclic", "Solution_1": "That's soooo banal...\r\n\r\nUsing directed angles modulo 180, we have < OMD = < OND since the points O, M, N and D lie on one circle (namely the circle $S_2$), and we have < DAB = < DCB since the points A, B, C and D lie on one circle (namely the circle $S_1$). Hence,\r\n\r\n < ORB = < (OM; AB) = < (OM; AD) + < (AD; AB) = < OMD + < DAB\r\n = < OND + < DCB = < (ON; CD) + < (CD; BC) = < (ON; BC)\r\n = < OTB,\r\n\r\nwhat immediately yields that the points O, R, T and B are concyclic. The condition that the points R and T lie on the same side of line BD as the point A is not necessary, of course.\r\n\r\n Darij" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Show that the sequence $(a_{n})_{n\\geq 1}$ does not contain a square of a integer:\r\n$a_{n}=[n+\\sqrt{n}+\\frac{1}{2}]$", "Solution_1": "if $n$ is a square then $a_{n}=x(x+1)$ where $x^{2}=n$ so if n is a perfect sq. we are done\r\nlet n not be a perfect square then let $x^{2}< n < (x+1)^{2}$\r\n[b]case 1: [/b] $(\\sqrt{n})=\\sqrt{n}-[\\sqrt{n}]<\\frac{1}{2}$\r\nthen $a_{n}=n+x$\r\nthus $x^{2}< x^{2}+x x^{2}+x$\r\nso we are done", "Solution_2": "Let $f(x)=x+\\sqrt x+\\frac{1}{2}$.\r\nf(x) increase and $f(k^{2}-k)k^{2}+1$ it give contradition with $k^{2}\\le f(n) 0$ for all $ i$. Also, of course, the top expression isn't defined when $ x \\equal{} 0$.", "Solution_2": "We can use L'H\u00f4pital's rule here.\r\n\r\n$ \\ln M(x,\\mathbf{a}) \\equal{} \\frac {\\ln\\left(\\frac1n\\sum_{j \\equal{} 1}^n a_j^x\\right)}{x}$\r\n\r\nBy L'H\u00f4pital,\r\n\r\n$ \\lim_{x\\to 0}\\ln M(x,\\mathbf{a}) \\equal{} \\lim_{x\\to0}\\frac {\\frac1n\\sum_{j \\equal{} 1}^na_j^x\\ln a_j}{\\frac1n\\sum_{j \\equal{} 1}^na_j^x}$\r\n\r\nSince $ \\lim_{x\\to0}\\frac1n\\sum_{j \\equal{} 1}^na_j^x \\equal{} 1,$ this becomes\r\n\r\n$ \\lim_{x\\to 0}\\ln M(x,\\mathbf{a}) \\equal{} \\frac1n\\sum_{j \\equal{} 1}^na_j^0\\ln a_j \\equal{} \\frac1n\\sum_{j \\equal{} 1}^n\\ln a_j.$\r\n\r\nExponentiating this gives the desired result.\r\n\r\nOther things to prove include:\r\n\r\n$ M(x,\\mathbf{a})$ is an increasing function of $ x,$ and $ M(x,\\mathbf{a}) < GM(\\mathbf{a})$ for $ x < 0$ and $ M(x,\\mathbf{a}) > GM(\\mathbf{a})$ for $ x > 0.$ (This is assuming that $ a_j\\ne a_k$ for some $ j,k.$ Else replace these inequalities with equalities.)\r\n\r\n$ \\lim_{x\\to\\infty}M(x,\\mathbf{a}) \\equal{} \\max_{1\\le j\\le n}a_j.$\r\n\r\n$ \\lim_{x\\to \\minus{} \\infty}M(x,\\mathbf{a}) \\equal{} \\min_{1\\le j\\le n}a_j.$", "Solution_3": "Thanks Professor Merryfield!\r\n\r\nMissed the \"+\" :oops: , thanks JBL.", "Solution_4": "at first use asymptotical expression of $ a_{j}^{x}$ when $ x \\minus{} > 0$\r\nit's $ a_{j}^{x}$~$ 1 \\plus{} x\\ln a_{j}$\r\nand in seccond use the next fact :\r\n$ (1 \\plus{} \\frac {x \\ln \\prod_{1\\leq j \\leq n } a_{j}}{n})^{1/x} \\equal{} \\exp (\\frac {\\ln \\prod_{1\\leq j \\leq n } a_{j}}{n})$ when $ x \\minus{} > 0$\r\nor $ (1 \\plus{} xA)^{1/x} \\equal{} \\exp ( A)$ when $ x \\minus{} > 0$" } { "Tag": [ "linear algebra", "matrix", "calculus", "function", "number theory", "greatest common divisor", "least common multiple" ], "Problem": "Let $X_1,X_2,\\ldots,X_m$ a numbering of the $m=2^n-1$ non-empty subsets of the set $\\{1,2,\\ldots,n\\}$, $n\\geq 2$. We consider the matrix $(a_{ij})_{1\\leq i,j\\leq m}$, where $a_{ij}=0$, if $X_i \\cap X_j = \\emptyset$, and $a_{ij}=1$ otherwise. Prove that the determinant $d$ of this matrix does not depend on the way the numbering was done and compute $d$.", "Solution_1": "ok, let's try...\r\nfirst of all, by the (axiomatic) definition of determinant of a matrix, we have that if $d'$ is the determinant of a matrix with two exchanged rows or columns, with respect to a matrix with determinant $d$, we have $d' = -d$. (*)\r\nnow, instead of showing that any for any permutations of the $m$ sets $d$ remains constant, we can just show that for a transposition (i mean a permutation $\\sigma$ such that $\\sigma(X_i) = X_j$, $\\sigma(X_j) = X_i$, $\\sigma(X_k) = X_k \\forall k \\neq i,j$), we have $d_\\sigma = d$, where $d_\\sigma$ is the determinant of the matrix associated to the set after permutation $\\sigma$.\r\nbut this easily follows from the (*) above.\r\nnow, we're trying to show that $d_n = -1 \\forall n > 1$, and $d_1 = 1$.\r\nfirst of all, cases $n=1,2$ are trivial, and determinants are $d_1 = 1$, $d_2 = -1$ so the given formula seems to be working. ;)\r\nlet's call $M_{n-1}$ the matrix associated to a fixed numbering of the $2^{n-1} - 1$ subsets of $\\{1,\\cdots n-1\\}$.\r\nwe can get $M_n$ by adding $X_{2^{n-1}} = {n}$, and $X_{2^{n-1}+i} = {n} \\cup X_i$.\r\nnow, the matrix associated to this permutation gets obviously:\r\n\\[ \\left( \\begin{tabular}{ccc|c|ccc} & & & 0 & & & \\\\ & M_{n-1} & & \\vdots & & M_{n-1} & \\\\& & & 0 & & & \\\\ \\hline 0 & \\cdots & 0 & 1 & 1 & \\cdots & 1 \\\\ \\hline & & & 1 & 1 & \\cdots & 1 \\\\ & M_{n-1} & & \\vdots & \\vdots & \\ddots & \\vdots \\\\ & & & 1 & 1& \\cdots & 1 \\end{tabular}\\right) \\].\r\nby subtracting the $2^{n-1}$-th row, and applying Laplace's rule for the calculus of $d_n$ , we get that $d_n = (-1)^{2^{n-1}-1}(d_{n-1})^2$, because we can get to a matrix divided into three square boxes, the bottom-left of which is $0$. but, for inductive hypothesis, $ d_{n-1} = -1$ so we have our nice result.", "Solution_2": "It occurs to me to ask what happens if you replace the lattice of subsets of $[n]$ with a more general lattice with a bottom element $\\mathbf 0$? The definition of $a_{ij}$ would then become $a_{ij} = 0$ if $X_i \\wedge X_k = {\\mathbf 0}$, otherwise 1. (I don't have an answer BTW.)", "Solution_3": "I don't want to think about the determinant of an infinite matrix, so let's restrict to finite lattices.\r\n\r\nIf there is some pair $x,y$ such that for every $z$, either both or neither of $x\\wedge z$ and $y\\wedge z$ are $0$, then the determinant is zero. In the nontrivial case, each element is completely determined by the set of elements it has nonzero intersection with. We can extend this another step, by considering the set of minimal nonzero elements; $x$ and $y$ have nonzero intersection if and only if there is some minimal element $z$ with $z\\le x$ and $z\\le y$.\r\n\r\nThe nontrivial case then reduces to considering subsets of the power set of the minimal nonzero elements.\r\n\r\nThis still leaves room for variation; the lattice $0,a,b,c,abc$ gives a determinant of $-2$.", "Solution_4": "In Lovasz's book [i]Combinatorial Problems and Exercises[/i], one of the problems expresses the determinant of the lattice matrix in terms of the Mobius function of the lattice.", "Solution_5": "I believe this is called Smith's determinant, after the case(s) where $a_{ij}$ is the GCD or LCM of $i$ and $j$ (which Smith computed in the 1800's)." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that \r\n$ \\frac{a}{(b\\plus{}1)(c\\plus{}1)}\\plus{}\\frac{b}{(c\\plus{}1)(a\\plus{}1)}\\plus{}\\frac{c}{(a\\plus{}1)(b\\plus{}1)} \\ge \\frac{3}{4}$ \r\nwhere $ a,b$ and $ c$ are positive real numbers satisfying $ abc\\equal{}1$", "Solution_1": "[quote=\"Litlle 1000t\"]Prove that \n$ \\frac {a}{(b \\plus{} 1)(c \\plus{} 1)} \\plus{} \\frac {b}{(c \\plus{} 1)(a \\plus{} 1)} \\plus{} \\frac {c}{(a \\plus{} 1)(b \\plus{} 1)} \\ge \\frac {3}{4}$ \nwhere $ a,b$ and $ c$ are positive real numbers satisfying $ abc \\equal{} 1$[/quote]\r\n\r\nBy Cauchy Schwarz ineuqlaity we have \r\n\r\n\\[ \\left( {\\sum {\\frac{a}{{\\left( {b \\plus{} 1} \\right)\\left( {c \\plus{} 1} \\right)}}} } \\right)\\left( {\\sum {a\\left( {b \\plus{} 1} \\right)\\left( {c \\plus{} 1} \\right)} } \\right) \\ge \\left( {\\sum a } \\right)^2 \r\n\\]\r\n\r\nWe will prove \r\n\r\n\\[ 4\\left( {\\sum a } \\right)^2 \\ge 3\\left( {\\sum {a\\left( {b \\plus{} 1} \\right)\\left( {c \\plus{} 1} \\right)} } \\right)\r\n\\]\r\n\r\n\\[ \\Leftrightarrow 4\\left( {\\sum a } \\right)^2 \\ge 3\\left( {3abc \\plus{} 2\\sum {ab} \\plus{} \\sum a } \\right)\r\n\\]\r\n\r\ntrue since \r\n\r\n\\[ 3\\sum {a^2 } \\ge 3\\sum a \r\n\\]\r\n\r\n\\[ 2\\sum {ab} \\plus{} \\sum {a^2 } \\ge 9\r\n\\]", "Solution_2": "[quote=\"Litlle 1000t\"]Prove that \n$ \\frac {a}{(b \\plus{} 1)(c \\plus{} 1)} \\plus{} \\frac {b}{(c \\plus{} 1)(a \\plus{} 1)} \\plus{} \\frac {c}{(a \\plus{} 1)(b \\plus{} 1)} \\ge \\frac {3}{4}$ \nwhere $ a,b$ and $ c$ are positive real numbers satisfying $ abc \\equal{} 1$[/quote]\r\n\r\nThe above inequality equivalents to:\r\n$ 4\\sum a(a\\plus{}1) \\ge 3(a\\plus{}1)(b\\plus{}1)(c\\plus{}1)$\r\n\r\n$ \\Leftrightarrow 4(a^2\\plus{}b^2\\plus{}c^2) \\plus{}4(a\\plus{}b\\plus{}c) \\ge 3abc\\plus{}3(ab\\plus{}bc\\plus{}ca)\\plus{}3(a\\plus{}b\\plus{}c)\\plus{}3$\r\n\r\n$ \\Leftrightarrow 4(a^2\\plus{}b^2\\plus{}c^2) \\plus{}(a\\plus{}b\\plus{}c) \\ge 6\\plus{}3(ab\\plus{}bc\\plus{}ca)$\r\n\r\nWhich is clearly true by AM-GM inequality:\r\n$ a^2\\plus{}b^2\\plus{}c^2\\plus{}a\\plus{}b\\plus{}c \\ge 6$ and $ a^2\\plus{}b^2\\plus{}c^2 \\ge ab\\plus{}bc\\plus{}ca$", "Solution_3": "I think your inequality should be :\r\n\r\n$ \\sum \\frac{a}{(a\\plus{}1)(b\\plus{}1)}$ ??\r\n\r\nBecause $ abc\\equal{}1$\r\n\r\nWe may set \r\n$ a\\equal{}\\frac{x}{y}$\r\n$ b\\equal{}\\frac{y}{z}$\r\n$ c\\equal{}\\frac{z}{x}$\r\n\r\nSo we need to prove :\r\n\r\n$ \\sum \\frac{yz}{(y\\plus{}z)(z\\plus{}x)} \\ge \\frac{3}{4}$\r\n$ \\sum \\frac{yz(y\\plus{}z)}{(y\\plus{}z)(z\\plus{}x)(x\\plus{}y)} \\ge \\frac{3}{4}$\r\n\r\n$ 4 \\sum_{sym} x^2y \\ge 3 (\\sum_{sym} x^2y \\plus{}2xyz)$\r\n$ \\sum_{sym} x^2y \\ge \\sum_{sym} xyz$\r\n\r\nWhich is obvious.." } { "Tag": [], "Problem": "With this I start a series of motto's. They are my own creation and I post them as discution topics, essay topics...\r\n\r\nSo here it comes:\r\n\"Surprises only come to those not expecting them\"", "Solution_1": "that's absolutely true..\r\nif you expect it, it would become a suprise.... isn't it?", "Solution_2": "A surprise also comes when an expected surprise does not come! :)", "Solution_3": "you do not expect that what you expected won't come...\r\nso when it really does not come, it is a surprise to you.", "Solution_4": "The mottos are a very good idea [img]http://www.ectona.org/Mysmilies/Smilies/eclipsee_steering.gif[/img]", "Solution_5": "As liyi said, they wouldn't be considered surprises otherwise, right? This similar in concept with one of Murphy's laws: \"The place where yoyu find something is the last one you're searching.\" (or something like that :D). That's actually not surprising at all, since you obviously stop looking after you've found what you were looking for. \r\n\r\nI think these are truisms: of course it's like that, we don't really need to make mottos out of them... :D:D", "Solution_6": "In this motto I see a warning: If you don't want an unpleasant surprise spoil your fun, then you should better be cautious. \r\nThat's if you don't have a blind trust in life and think you will only be pleasantly surprised. Or if you are pessimist and think you can only get nice surprises if you think for the worst possible future.\r\n\r\nI am realistic and cautious.", "Solution_7": "16 year bump" } { "Tag": [], "Problem": "What is the value of \r\n(n) + (n) + ... + (n)\r\n(1) (2) (n)\r\n\r\nPlease give me an easy way to[b]solve[/b] the problem?\r\nI dont just need an answer.", "Solution_1": "i dont understand this question", "Solution_2": "It's #11 on the Mock Sprint. Don't worry, we helped him over in the Mock Results thread.", "Solution_3": "Two words.\r\nPascal's Triangle.", "Solution_4": "I don't understand....\r\nO\r\nwow...\r\nI was wondering what the parentheses were about.\r\nJust use binomial theorem." } { "Tag": [ "inequalities", "induction", "inequalities unsolved" ], "Problem": "$ a_{1},a_{2},....,a_{2009} > 0$ for $ 2\\leq i \\leq2009$ $ \\implies$ $ a_{i}\\geq a_{1} \\plus{} ... \\plus{} a_{i \\minus{} 1}$ $ \\implies$ find maximum value of that\r\n\r\n$ \\frac {a_{1}}{a_{2}} \\plus{} \\frac {a_{2}}{a_{3}} \\plus{} ... \\plus{} \\frac {a_{2008}}{a_{2009}}$", "Solution_1": "Problem is very easy to find the [b]minimum[/b] ,I can post my solution if you want.\r\n\r\nAnd for the maximum ,I'm not sure if it is bounded from above.\r\n\r\nExperimenting with numbers $ a_1 ,a_2,a_3,a_4$ ,we have \r\n\r\n$ a_1 \\leq a_2 \\implies \\frac {a_1}{a_2}\\leq 1$ \r\n\r\n$ a_1 \\plus{} a_2 \\leq a_3 \\implies \\frac {a_1}{a_3} \\plus{} \\frac {a_2}{a_3} \\leq 1$\r\n\r\n$ a_1 \\plus{} a_2 \\plus{} a_3 \\leq a_4 \\implies \\frac {a_1}{a_4} \\plus{} \\frac {a_2}{a_4} \\plus{} \\frac {a_3}{a_4} \\leq 1$\r\n\r\n$ \\left(\\frac {a_1}{a_3} \\plus{} \\frac {a_1}{a_4} \\plus{} \\frac {a_2}{a_4} \\right ) \\plus{} \\left(\\frac {a_1}{a_2} \\plus{} \\frac {a_2}{a_3} \\plus{} \\frac {a_3}{a_4}\\right) \\leq 4$\r\n\r\nTo maximise $ \\left(\\frac {a_1}{a_2} \\plus{} \\frac {a_2}{a_3} \\plus{} \\frac {a_3}{a_4}\\right)$ we need to minimise $ \\left(\\frac {a_1}{a_3} \\plus{} \\frac {a_1}{a_4} \\plus{} \\frac {a_2}{a_4} \\right )$ \r\n\r\nBy Am-Gm \r\n\r\n$ \\left(\\frac {a_1}{a_3} \\plus{} \\frac {a_1}{a_4} \\plus{} \\frac {a_2}{a_4} \\right ) \\geq 3 \\sqrt [3]{\\frac {a_1}{a_3} \\cdot \\frac {a_1}{a_4} \\cdot \\frac {a_2}{a_4}}$ \r\n\r\nand the equality occur if and only if $ \\frac {a_1}{a_3} \\equal{} \\frac {a_1}{a_4} \\equal{} \\frac {a_2}{a_4} \\iff a_1 \\equal{} a_2$ and $ a_3 \\equal{} a_4$ ,contradiction .\r\n\r\nSo I think we can not minimise $ \\left(\\frac {a_1}{a_3} \\plus{} \\frac {a_1}{a_4} \\plus{} \\frac {a_2}{a_4} \\right )$ \r\n\r\ntherefore we can not maximise $ \\left(\\frac {a_1}{a_2} \\plus{} \\frac {a_2}{a_3} \\plus{} \\frac {a_3}{a_4}\\right)$\r\n\r\n----------------------------------------------------\r\n\r\nAnother thing that makes me thinking that it is not bounded from above is this : \r\n\r\nBy induction we can easily show that \r\n\r\n$ a_k \\geq 2^{k\\minus{}2}a_1$\r\n\r\nLet $ S\\equal{}\\frac{a_1}{a_2} \\plus{}\\frac{a_2}{a_3} \\plus{}...\\plus{}\\frac{a_{2008}}{a_{2009}}$\r\n\r\nThen \r\n\r\n$ S\\leq\\frac{a_1}{a_1} \\plus{}\\frac{a_2}{2a_1} \\plus{}\\frac{a_3}{4a_1} \\plus{} ...\\plus{}\\frac{a_{2009}}{2^{2007}a_1} \\leq \\frac{a_{2009}}{a_1} \\left(1\\plus{}\\frac{1}{2} \\plus{}\\frac{1}{4} \\plus{} ...\\plus{}\\frac{1}{2^{2007}}\\right)$\r\n\r\nTo minimise We need to have $ \\frac{a_{2009}}{a_1} \\leq x$ ,not $ \\frac{a_{2009}}{a_1} \\geq 2^{2007}$", "Solution_2": "[quote=\"cnyd\"]$ a_{1},a_{2},....,a_{2009} > 0$ for $ 2\\leq i \\leq2009$ $ \\implies$ $ a_{i}\\geq a_{1} \\plus{} ... \\plus{} a_{i \\minus{} 1}$ $ \\implies$ find maximum value of that\n\n$ \\frac {a_{1}}{a_{2}} \\plus{} \\frac {a_{2}}{a_{3}} \\plus{} ... \\plus{} \\frac {a_{2008}}{a_{2009}}$[/quote]\r\nThe max is $ \\frac {2009}{2}$.\r\n$ \\sum_{i \\equal{} 1}^{2008} \\left ( \\frac {a_i}{a_{i \\plus{} 1}} \\right ) \\minus{} \\frac {n}{2} \\equal{}$\r\n$ \\frac {a_1}{a_2} \\minus{} 1 \\plus{} \\sum_{i \\equal{} 2}^{2008} \\left ( \\frac {a_i}{a_{i \\plus{} 1}} \\minus{} \\frac {1}{2} \\right ) \\equal{}$ $ \\frac {a_1 \\minus{} a_2}{a_2} \\plus{} \\sum_{i \\equal{} 2}^{2008} \\left ( \\frac {(a_1 \\plus{} a_2 \\plus{} .. \\plus{} a_i) \\minus{} a_{i \\plus{} 1}}{2a_{i \\plus{} 1}} \\minus{} \\frac {(a_1 \\plus{} a_2 \\plus{} .. \\plus{} a_{i \\minus{} 1}) \\minus{} a_i}{2a_{i \\plus{} 1}} \\right ) \\equal{}$\r\n$ \\frac {a_1 \\minus{} a_2}{2a_2} \\plus{} \\sum_{i \\equal{} 2}^{2008} (a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_{i \\minus{} 1} \\minus{} a_i) \\left ( \\frac {1}{2a_i} \\minus{} \\frac {1}{2a_{i \\plus{} 1}} \\right ) \\le 0$.\r\nEquality obtained when $ a_1\\equal{}1, a_k \\equal{} 2^{k\\minus{}2}, k \\ge 2$.", "Solution_3": "good solution ! :lol:", "Solution_4": "[quote=\"Mathias_DK\"]\n\n $ \\frac {a_1 \\minus{} a_2}{a_2} \\plus{} \\sum_{i \\equal{} 2}^{2008} \\left ( \\frac {(a_1 \\plus{} a_2 \\plus{} .. \\plus{} a_i) \\minus{} a_{i \\plus{} 1}}{2a_{i \\plus{} 1}} \\minus{} \\frac {(a_1 \\plus{} a_2 \\plus{} .. \\plus{} a_{i \\minus{} 1}) \\minus{} a_i}{2a_{i \\plus{} 1}} \\right ) \\equal{}$\n$ \\frac {a_1 \\minus{} a_2}{2a_2} \\plus{} \\sum_{i \\equal{} 2}^{2008} (a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_{i \\minus{} 1} \\minus{} a_i) \\left ( \\frac {1}{2a_i} \\minus{} \\frac {1}{2a_{i \\plus{} 1}} \\right ) \\leq0$.\n[/quote]\nI think, it should be\n$ \\frac {a_1 \\minus{} a_2}{a_2} \\plus{} \\sum_{i \\equal{} 2}^{2008} \\left ( \\frac {(a_1 \\plus{} a_2 \\plus{} .. \\plus{} a_i) \\minus{} a_{i \\plus{} 1}}{2a_{i \\plus{} 1}} \\minus{} \\frac {(a_1 \\plus{} a_2 \\plus{} .. \\plus{} a_{i \\minus{} 1}) \\minus{} a_i}{2a_{i \\plus{} 1}} \\right ) \\equal{}$\n$ \\frac {a_1 \\minus{} a_2}{2a_2} \\plus{} \\sum_{i \\equal{} 2}^{2008} (a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_{i \\minus{} 1} \\minus{} a_i) \\left ( \\frac {1}{2a_i} \\minus{} \\frac {1}{2a_{i \\plus{} 1}} \\right )+\\frac{\\sum\\limits_{i=1}^{2008}a_i-a_{2009}}{2a_{2009}}\\leq0$." } { "Tag": [ "geometry", "3D geometry", "sphere", "inequalities", "circumcircle", "analytic geometry", "graphing lines" ], "Problem": "can anyone post the answer for #22 and #25?", "Solution_1": "Tomorrow would probably be better to post answers, I got everything but #23,24 & 25, though I found 24 later. This should tie me in my school from who I've talked to, although it's not as good as I would have liked.", "Solution_2": "25 is D and 22 is C.\r\n\r\nFor 25, consider the aerial view of the whole structure and find the distance between the centres of the cone and the sphere. Then consider a side 2D view of one half-cone (right triangle) and the sphere (circle). Use similar triangles to knock off the problem.\r\n\r\nFor 22, it's just some sort of a greedy distribution into: 6x_1 + 5x_2 and 2y_1 + 3y_2. I got 9:43, I'm not sure if there's an earlier time though. It seems right ;) .", "Solution_3": "22. Quincy and Celine have to move 16 small boxes and 10 large boxes. The chart indicates the time that each person takes to move each type of box. They start moving the boxes at 9:00 a.m. The earliest time at which they can be finished moving all of the boxes is\r\n(A) 9:4a a.m. (B) 9:42 a.m. (C) 9:43 a.m. (D) 9:44 a.m. (E) 9:45 a.m.\r\n\r\n(A chart is given with the following information:\r\n- Celine moves a small box in 2 minutes and a large box in 6 minutes\r\n- Quincy moves a small box in 3 minutes and a large box in 5 minutes)\r\n\r\n[hide=\"Comments\"]I have a problem with the wording of the question. It leaves questions like \"Can more than one person move a box at the same time?\" and \"If only one person can move a box at a time, and Quincy finishes moving all of his boxes, and Celine has been moving a large box for three minutes, can Quincy take over to save 30 seconds?\" I assumed that each box had to be moved completely by one person, and the question should have contained this clause.[/hide]\n\n[hide=\"Solution\"]C is the answer. We will consider pairs (C,Q), where C is the time Celine takes, Q is the time Quincy takes. We are trying to minimize max(C,Q). We have the inequality max(C,Q) >= avg(C,Q). C+Q is minimized when Celine moves all the samll boxes, and Quincy moves all the large boxes; avg(32,50) = 41. Now, starting from this situation, consider \"switches\" of boxes, a switch being Quincy moving an additional small box, or Celine moving an additional large box. Each switch increases the average by 0.5. We can make 3 switches (1 small, 2 large), to get (42,43), with max=43. Can we do better? Not with more than 2 switches, since then avg(C,Q) >= 42.5, so max(C,Q) >= 43 (since the max is an integer). Thus we are limited to 1 switch or 2 switches. There are only 5 possibilities. Computing them we get (38,45) and (30,52) for 1 switch; (44,40), (36,48), and (28,56) for 2 swtiches. These yield max's 45, 52, 44, 48, 56. Thus 43 is the best we can do.[/hide]\n\n(The attachment is part of my solution, not part of the question.)\n\n25. Three identical cones each have a radius of 50 and a height of 120. The cones are placed so that their circular bases are touching each other. A sphere is placed so that it rests in the space created by the three cones. If the top of the sphere is level with the tops of the cones, then the radius of the sphere is closest to\n(A) 38.9 (B) 38.7 (C) 38.1 (D) 38.5 (E) 38.3\n\n[hide=\"Solution\"]Consider the three circular bases. The centres of the three circles form an equilateral with side length $100$. Let the circumcircle of this triangle have centre D. Its radius is $\\frac{100}{\\sqrt{3}}$.\n\nNow consider the attached diagram, where O is the centre of the sphere radius $r$, and Q is the point of contact of the sphere with one of the cones. D is the point of contact of the perpendicular from O to the plane containing the bases of the cones. BD = $\\frac{100}{\\sqrt{3}}$.\nAC has slope $-\\frac{12}{5}$, so QO has slope $\\frac{5}{12} \\Rightarrow OT = \\frac{5r}{13}, \\ QT = \\frac{12r}{13}.$\n\n$AP = SO + OT = \\frac{18r}{13}.$ By similar triangles, $\\frac{PQ}{AP} = \\frac{50}{120} \\Rightarrow PQ = \\frac{15r}{26}.$\n\n$PT = PQ + PT = \\frac{15r}{26} + \\frac{12r}{13} = \\frac{3r}{2}.$ But $PT = BD$, so\n\n$\\frac{3r}{2} = \\frac{100}{\\sqrt{3}} \\Rightarrow r = \\frac{200}{3\\sqrt{3}} \\approx 38.4900 \\Rightarrow \\boxed{D} \\ .$[/hide]", "Solution_4": "almost everyone in my class (even my teacher) got # 10 wrong...", "Solution_5": "Steven Karp is my hero.\r\nI left the last one blank .. :(. Pretty sure I got everything else though.\r\n\r\nYeah, #10 was a trickster, like 20, and 22.", "Solution_6": "What was the answer for #10? I put 8 intersections but then realized after it is probably 4. Similarly, my teacher thought the same thing.", "Solution_7": "why would it be 4?? Almost everyone at my school put 8, unless they're all wrong", "Solution_8": "[quote=\"kool_dudy\"]What was the answer for #10? I put 8 intersections but then realized after it is probably 4. Similarly, my teacher thought the same thing.[/quote]\r\n\r\nIt's 8. See AIME 1999 #4. I got pwned on this question. The funny thing is, I was working on this question the night before on the particular AIME. But nah...my brain decided to imagine that the answer was 4. :(\r\n\r\nEDIT: Pwnage repealed. Thank goodness for stupidity. :rotfl:", "Solution_9": "[quote=\"Effervescence\"]why would it be 4?? Almost everyone at my school put 8, unless they're all wrong[/quote]\r\nThat's exactly what I thought too! But, I think the answer is 4. Since it says \"[b]any two rectangles[/b]\" in the question, you have to take the minimum of all the possibilities. -> No matter what the sizes of the two rectangles are, you can always have four points of intersection at corner of the bigger rectangle (you place the smaller rectangle at of other corners of the big rectangle so that they have a common vertex and two sides touching each other [see attachment], and then you turn the small rectangle until you get fours points of intersection), and that's the maximum. You can have 8 points under ideal conditions, but it doesn't apply to [i]any two rectangles[/i].\r\n\r\nI am not sure if that's the right answer, so correct me if I am wrong. :P", "Solution_10": "Well it asks for the maximum possible points of intersection, and I think the \"any\" just means that the rectangles are not restricted to any conditions, meaning like, you take any rectangles of your choice and try to find a max number of points of intersection (I don't know if that makes sense, I'm explaining really poorly). It shouldn't make you think this much if it's only part A :huh:\r\n\r\nIf you argue it your way, wouldn't it be correct to say that the minimum number of intersections is 0, if they're not touching?, or 2, like they showed in the example.", "Solution_11": "[quote]If you argue it your way, wouldn't it be correct to say that the minimum number of intersections is 0, if they're not touching?, or 2, like they showed in the example.[/quote]\njust added a little bit more to my reply I poseted.. you can have a look again..\n(for the diagram on the contest, you can still make it to have 4 points of intersection...)\n\n[quote]It shouldn't make you think this much if it's only part A :huh:[/quote]\r\nI said that to myself too... :D", "Solution_12": "Now that I think about it, chilly_snow seems to have put up a valid argument. Even if we have say...a microscropic rectangle vs. a rectangle the size of Colorado, the smaller rectangle can wedge out 4 intersections from the corner of the larger one. Drawing a picture confirms this. However, 8 intersections can only occur if the rectangles are closer in size to each other...in ideal conditions as chill_snow puts it. Kinda hard to get 8 intersections with given a really tiny and a really huge rectangle. \r\n\r\n...Does this mean I got #10 right? Haha...man! :roll:", "Solution_13": "thanks a lot guys...it seems like i'm doing worse every year...for this year, i already messed up amc 12..and now fermat is just adding to the list...sigh..", "Solution_14": "can anyone post how to find # 23? the tom and jerry one?\r\ni totally guessed on it :?", "Solution_15": "[quote=\"tzhu\"]I just checked the CEMC website and the solutions to this year's Fermat is out. Question 10 turns out to be D, so it wasn't a trick question after all. :lol:[/quote]\r\nyaya... I just checked it.. the answer is 8..!!!\r\nthat means I did it right~!\r\nbut I still think the answer should be 4...\r\no well...", "Solution_16": "I owned(150). :D \r\nThe thing was such a joke: I had over 25 min for question 25. :rotfl: \r\nI think they are making contests easier and easier by the year.", "Solution_17": "[quote=\"rem\"]I think they are making contests easier and easier by the year.[/quote]This is definitely true - I have heard it directly from someone who formulates questions and decides which should be included on the contests.", "Solution_18": "[quote]I owned(150). \nThe thing was such a joke: I had over 25 min for question 25. \nI think they are making contests easier and easier by the year.[/quote]\r\n\r\nNot everyone is superman like Alex though. (This is Tai by the way :P ). Thinking back I probably should've been able to do 25. Had a brain fart with about 30 minutes left on 22-23, then in a flash Burns was like, \"5 minutes guys\".\r\nI think people are getting smarter too, or they need to make contest easier so more people can take it.", "Solution_19": "I can't believe thay still haven't posted the results. It shouldn't take too long to mark everything and to do the statistics. :mad:", "Solution_20": "[quote=\"Effervescence\"][quote]I owned(150). \nThe thing was such a joke: I had over 25 min for question 25. \nI think they are making contests easier and easier by the year.[/quote]\n\nNot everyone is superman like Alex though. (This is Tai by the way :P ). Thinking back I probably should've been able to do 25. Had a brain fart with about 30 minutes left on 22-23, then in a flash Burns was like, \"5 minutes guys\".\nI think people are getting smarter too, or they need to make contest easier so more people can take it.[/quote]\r\nit's just because human r getting smarter...Darwin was right hehe", "Solution_21": "may i ask you when the results will be posted??", "Solution_22": "The results should have been posted about a week ago. I have no idea why it is taking them so long. maybe it is because many more people wrote the contest compared to last year.", "Solution_23": "So finally the results book is out. http://cemc.uwaterloo.ca/english/contests/past_result/2006/pcf_results.pdf\r\n\r\nHaha my school came top 5 in Fermat. svg tech. My teacher will be happy.", "Solution_24": "[quote=\"carpo\"]So finally the results book is out. http://cemc.uwaterloo.ca/english/contests/past_result/2005/pcf_results.pdf\n\nHaha my school came top 5 in Fermat. svg tech. My teacher will be happy.[/quote]\r\nits the last year's result lol", "Solution_25": "[quote=\"nirvanatear\"][quote=\"carpo\"]So finally the results book is out. http://cemc.uwaterloo.ca/english/contests/past_result/2005/pcf_results.pdf\n\nHaha my school came top 5 in Fermat. svg tech. My teacher will be happy.[/quote]\nits the last year's result lol[/quote]They looked the same :? \r\n\r\nI fixed the link.", "Solution_26": "[quote=\"carpo\"]So finally the results book is out. http://cemc.uwaterloo.ca/english/contests/past_result/2006/pcf_results.pdf\n\nHaha my school came top 5 in Fermat. svg tech. My teacher will be happy.[/quote]\r\nHAHA my school got 1st in both Fermat and Pascal :)", "Solution_27": "In group 4 fermat, students names are written first name first last name last, while in the other groups it's the other way. :rotfl: :rotfl:", "Solution_28": "[quote=\"rem\"]In group 4 fermat, students names are written first name first last name last, while in the other groups it's the other way. :rotfl: :rotfl:[/quote]\r\n\r\n\"first name first last name last\"\r\n\r\nDo you actually expect a human to be able to parse that? It took me quite a few seconds before I actually realized what you were trying to say.", "Solution_29": "haha funny" } { "Tag": [], "Problem": "I don't understand Modular Arithmetic. Can someone explain it to me? And possibly give an example of how it can be used?\r\n\r\nThanks :lol:", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=modular+arithmetic&t=13194" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "let [tex]f: R^n \\rightarrow R^n [/tex] be differentiable, with [tex]\\sum_{i=1}^n (-1)^i \\frac{ \\partial{f_i}}{\\partial{x_i}} = 0 [/tex]. show that [tex]\\omega = \\sum_{i=1}^n f_i dx_1 \\Lambda ... \\Lambda \\hat{dx_i}\\Lambda ... \\Lamda dx_n [/tex]\r\nis exact.\r\n\r\n-------------------------------------\r\nhere's what i got so far:\r\n\r\n[tex]\\omega[/tex] is a n-1 form, since the \"^\" on the [tex]dx_i[/tex] indicates that this term is omitted. f is defined on [tex]R^n[/tex] which is an open, star-shaped set. because of this, we can use poincare's lemma, which states that on an open starshaped set, every closed form is exact. therefore we only need to show that [tex]\\omega [/tex] is closed. that is, dw = 0.\r\n\r\nwhen i write out dw, i get\r\n\r\n[tex]dw = \\sum_{i=1}^n \\sum_{\\alpha = 1}^n \\frac{ \\partial{f_i}}{\\partial{x_\\alpha}} dx_\\alpha \\Lambda dx_1 \\Lambda... \\Lambda \\hat{dx_i}\\Lambda ... \\Lamda dx_n [/tex]\r\n\r\nhow do i simplify this? how do i make this equal zero? if [tex]\\sum_{i=1}^n (-1)^i \\frac{ \\partial{f_i}}{\\partial{x_i}} = 0 [/tex], does this mean that the expression i wrote for dw equals zero? i don't really see the connection here, since [tex]\\sum_{i=1}^n (-1)^i \\frac{ \\partial{f_i}}{\\partial{x_i}} [/tex] has an extra factor of [tex](-1)^i[/tex]...i'm not sure about the subscripts either. i hope someone can help me with this.", "Solution_1": "Hi, man,\r\n\r\nI don't know how to write formulas, so sorry if you can't understans smth. \r\n\r\nLook at your formula for the dw. Do you understand that all if alpha is equal to any index except the index of the omitted member, the item is zero? And if it's equal to i than it will require exactly i transpositions to put x_alpha to the place of the omitted member, that's how you'll get (-1)^i. The rest is obvious. \r\n\r\nP.S. It would have been easier to understand if you started with 1-form on R^2 and have written it all without double sum formulas but as is.", "Solution_2": "here's what i got so far:\r\n\r\n[quote]when i write out dw, i get\n\n[tex]dw = \\sum_{i=1}^n \\sum_{\\alpha = 1}^n \\frac{ \\partial{f_i}}{\\partial{x_\\alpha}} dx_\\alpha \\Lambda dx_1 \\Lambda... \\Lambda \\hat{dx_i}\\Lambda ... \\Lamda dx_n [/tex][/quote]\r\n\r\nSince $dx_i\\wedge dx_i=0$, then\r\n\\begin{eqnarray*}\r\ndw &=& \\sum_{i=1}^n \\frac{\\partial{f_i}}{\\partial{x_i}}dx_i\\wedge dx_1\\wedge\\cdots \\hat{dx_i}\\wedge\\cdots\\wedge dx_n\\\\\r\n& = & \\left(\\sum (-1)^i \\frac{\\partial{f_i}}{\\partial{x_i}}\\right)dx_1\\wedge\\cdots \\wedge dx_i\\cdots\\wedge dx_n=0\r\n\\end{eqnarray*}" } { "Tag": [ "search", "geometry unsolved", "geometry" ], "Problem": "Let triangle ABC with each angle less than 90 degree. let p be the FERMAT point of triangle ABC .\r\nTHEN we will get three smaller triangles. prove that the RULER LINE of these three triangles meet within one point.\r\nAnd when the triangle has one angle larger than 90 degree .Does the problem be all right.", "Solution_1": "This problem has been post before\r\nsee http://www.mathlinks.ro/viewtopic.php?search_id=916513781&t=6188" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry", "geometric transformation", "reflection" ], "Problem": "Tiankai Liu\r\nMatt Ince\r\nAlison Miller\r\nAaron Pixton\r\nTony Zhang\r\nOleg Golberg\r\n\r\nAlternate: Anders Kaseorg\r\n2nd Alternate: Brian Lawrence (he actually started out in blue zuming picked him as alternate because he totally pwned everyone on the TST)", "Solution_1": "Congratulations to Alison, who has been working toward being on the team for YEARS. (Her mom is the very first person who ever told me about competition culture, back when my oldest son was first-grade age.) Matt Ince is a homeschooler too, so that's a pretty good representation of homeschoolers on this year's team, what with Anders Kaseorg being an alternate.", "Solution_2": "Congratulations to everyone on the team. And I'm looking forward to meeting you guys in Greece.", "Solution_3": "Interesting how half of the U.S. team is from the same school (Exeter) and only one contestant is from a public high school (Aaron Pixton). Congrats and good luck at IMO.", "Solution_4": "Oleg Golberg does not attend Exeter.", "Solution_5": "[quote=\"MysticTerminator\"]Oleg Goldberg does not attend Exeter.[/quote]\r\n\r\nIt says so on the AMC website.", "Solution_6": "Is one of those people you MysticTerminator? Are any of those people from AoPS?", "Solution_7": "[quote=\"JS1527\"]Is one of those people you MysticTerminator?[/quote]\n\nNo.\n\n[quote] Are any of those people from AoPS?[/quote]\r\n\r\nOccasionally Alison and Aaron.", "Solution_8": "At least four of the people on the team have accounts on AoPS.", "Solution_9": "Congratulations Alison!! Have a great time in Greece! We will all be cheering for you!\r\n\r\nGood luck to you also Bill!! We will be cheering for you too.\r\n\r\nOne good reason there are so many Exeter students on the Team this year is that Zuming Feng, the US Team Coach, is a teacher there. He has attracted a lot of very good students to Exeter recently; and I'm sure they get GREAT practice for Olympiads throughout the year.\r\n\r\nOleg Goldberg, who is a student there now, is a recent immigrant from Russia. He was on the Russian IMO Teams in 2002 and 2003, and received Gold Medals (with nearly perfect scores) in both years.\r\n\r\nNotice the team this year is also all seniors. So 6 completely new students will be competing next year.", "Solution_10": "didn't exeter also have like 3 people on last year's team?", "Solution_11": "[quote=\"yif man12\"]didn't exeter also have like 3 people on last year's team?[/quote]\r\n\r\nActually, I think there were none last year, heh. Although if I remember correctly, Tony was an alternate.", "Solution_12": "No. They didn't have any on last year's team. Tiankai decided to go to RSI instead.", "Solution_13": "oops, must've been something else i saw then. i don't follow this much and i just saw something about the IMO or something on a magazine some time ago.", "Solution_14": "[quote=\"JS1527\"]Is one of those people you MysticTerminator? Are any of those people from AoPS?[/quote]\r\n\r\nhehe not this year, but I know at least 2 of the people are active on AoPS.", "Solution_15": "[quote=\"Fierytycoon\"][quote=\"yif man12\"]didn't exeter also have like 3 people on last year's team?[/quote]\n\nActually, I think there were none last year, heh. Although if I remember correctly, Tony was an alternate.[/quote]\r\n\r\nActually, Alison was the alternate last year. Congrats to her and all the others who made the team this year.", "Solution_16": "I believe Alison wasn't a USAMO winner last year.", "Solution_17": "[quote=\"314159265358979323846\"]I believe Alison wasn't a USAMO winner last year.[/quote]\r\n\r\nI wouldn't bet my life on it, but I'm pretty still pretty sure she was the alternate.\r\n\r\nBeing an alternate, or even team member, does not require being a USAMO winner. USAMO winners = top 12 USAMO scores, including Canadians. But the candidates for the IMO team = top 12 people eligible to be on the US team only (citizens or green card holders). So however many Canadians there are, there will be that many extra US people eligible for the US IMO team. Anders wasn't a winner this year either, but he's the alternate.", "Solution_18": "Last year Canadians were ineligible, though Tiankai Liu (and possibly others) declined the invitation to MOsP, so at least one USAMO Honorable mention took the selection exams last year.", "Solution_19": "Again, I am not sure, but I believe that there were in fact two alternate last year because of a scoring tie. I'm pretty sure Tony was one of them, but Alison may have been the other.", "Solution_20": "I thought Boris Alexeev was an alternate last year.", "Solution_21": "USAMO perfect score but did not attend MOP (opted for RSI):\r\nTiankai Liu [Exeter]\r\n\r\nUS IMO team 2003:\r\nDani Kane\r\nAnders Kaseorg\r\nMark Lipson\r\nPo-Ru Loh [also perfect USAMO]\r\nAaron Pixton\r\nYan Zhang\r\n\r\nAlternates (tied by USAMO + TST, tie broken by MOP score):\r\nTony Zhang (1st alternate) [Exeter]\r\nBoris Alexeev (2nd alternate)\r\n\r\nRounding out the top 12:\r\nJae Bae\r\nPo-Ling Loh\r\nKwok Fung Tang [Exeter]\r\n\r\n\"Blue\" group at MOP (27 on the USAMO -- top 12 cutoff was 28):\r\nJongmin Baek\r\nRicky Biggs\r\nMatt Ince\r\nNate Ince\r\nAlison Miller\r\n\r\nAll of this should be correct, except at worst one mistake in \"Blue\". Note that indeed nobody from Exeter was on the team last year.", "Solution_22": "First of all, for those casually browsing this discussion and seeing that the team in my prior comment resembles last year's team ... that's because it is. The first comment, to the best of my knowledge, actually reflects this year's team.\r\n\r\n[quote=\"MCrawford\"]Last year Canadians were ineligible, though Tiankai Liu (and possibly others) declined the invitation to MOsP, so at least one USAMO Honorable mention took the selection exams last year.[/quote]\r\n\r\nLast year only the USAMO Top 12 were eligible for the US IMO team. Unfortunately, you can't extrapolate an exceptional situation from this year onto previous years (or even later years, although the AMC might now be compelled to use the rule from this year in the future).\r\n\r\nFor example, I'd be interested to learn whether any honorable mentions were eligible for the team the year Gabriel Carroll opted for RSI.", "Solution_23": "well, I'm not totally hallucinating. Alison was an alternate. But it was 2002, not 2003.", "Solution_24": "The second alternate this year is Brian Lawrence, a freshman who got a 13 on the USAMO but got the highest score (31?) on the TST.\r\n\r\nCrazy...\r\n\r\n-Oleg R", "Solution_25": "[quote=\"Celeborn\"]The second alternate this year is Brian Lawrence, a freshman who got a 13 on the USAMO but got the highest score (31?) on the TST.\n\nCrazy...\n\n-Oleg R[/quote]\r\n\r\nNo one scored higher than a 31 on the TST? I would have thought that there would have been higher scores, since students score even higher on the IMO.", "Solution_26": "How can one be an alternate without being a USAMO winner?", "Solution_27": "[quote=\"ComplexZeta\"]How can one be an alternate without being a USAMO winner?[/quote]\r\n\r\n(After edit:) Because the Canadians who were USAMO winners can't be on the USAMO team, they went deeper into the depth chart after USAMO, I think, to establish eligibility for the IMO team. [Checking this year's [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamowinners.html]USAMO winner's page on the Web[/url]:] Yes, even Anders wasn't a winner this year, although he was an honorable mention. But the name mentioned above, Brian Lawrence, as the second alternate doesn't seem to appear on the USAMO winner's page on the Web, although it does appear on the [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamoqualstate.html]USAMO qualifier's page[/url]. Maybe the team selection test is counting for more than it used to these days.", "Solution_28": "[quote=\"tokenadult\"]\nBecause the Canadians who were USAMO winners can't be on the USAMO team, they went deeper into the depth chart after USAMO, I think, to establish eligibility for USAMO. [/quote]\r\n\r\nNo, Brian's score on TST was exceptional, so Zuming made him alternate anyway; it had nothing to do with the Canadians.\r\n\r\nHowever, it is not going to matter (unless two people on the team get murderplied or something).", "Solution_29": "Yes, I was the alternate two years ago and did not place last year.\r\n\r\nBrian Lawrence's score was a 33. This year's TST was weird in that it was not really harder than the IMO, but one of the problems was computationally intensive, and a number of people misread another.", "Solution_30": "[quote=\"tokenadult\"]Congratulations to Alison, who has been working toward being on the team for YEARS. (Her mom is the very first person who ever told me about competition culture, back when my oldest son was first-grade age.) Matt Ince is a homeschooler too, so that's a pretty good representation of homeschoolers on this year's team, what with Anders Kaseorg being an alternate.[/quote]\r\n\r\nPerhaps the team will include 3 homeschoolers this year. Two team members are having difficulty obtaining visas to go to Greece at the moment and Anders is next in line.", "Solution_31": "btw, at the awards, aaron pixton got 2nd for black mop tests scoring like 20 pts higher than everyone and oleg golberg just pwned everyone (something like 106 points on 8 combined MOP tests) awesomeness. On a lighter note, I was \"most likely to be arrested\" and runner-up for something else I cannae remember. Dunbar was voted most late to class :) (after having been about an hour late to one and missing another entirely)", "Solution_32": "Are all these awards listed somewhere?", "Solution_33": "Probably not on the internet. The graders probably still have the list though.", "Solution_34": "moppers.kaseorg.com\r\n\r\ncheck there after a bit" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "The incircle $ (I)$ of triangle $ ABC$ touches side $ BC,CA,AB$ at $ A_1,B_1,C_1$ respectively.The circumcircle of $ ABC$ meets the lines $ AI,BI,CI$ at $ A_2,B_2,C_2$ respectively. Denote $ S,S_1,S_2$ be the areas of triangles $ ABC,A_1B_1C_1,A_2B_2C_2$ respectively. Prove that $ S^2\\equal{}4S_1S_2$", "Solution_1": "$ (O, R)$ is the circumcircle, $ (I, r)$ is the incircle. $ \\triangle A_1B_1C_1$ is pedal triangle of $ I$ WRT the $ \\triangle ABC,$\r\n\r\n$ S_1 \\equal{} \\frac {p(I, (O))}{4R^2}\\ S \\equal{} \\frac {R^2 \\minus{} IO^2}{4R^2}\\ S \\equal{} \\frac {R^2 \\minus{} (R^2 \\minus{} 2rR)}{4R^2}\\ S \\equal{} \\frac {r}{2R}\\ S.$\r\n\r\n(see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=38164[/url]). $ \\triangle A_1B_1C_1 \\sim \\triangle A_2B_2C_2$ with circumcircles $ (I), (O)$ are (centrally) similar, having parallel their vertex circumradii,\r\n\r\n$ \\frac {S_2}{S_1} \\equal{} \\frac {R^2}{r^2}\\ \\Longrightarrow\\ 4S_1S_2 \\equal{} \\frac {4R^2}{r^2}\\ S_1^2 \\equal{} S^2.$" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Hi Guys \r\nLet f be strictly increasing on the subset S of R\r\nIF f(S) is open then [b]prove f is continuous on S[/b]\r\nThank u in advance", "Solution_1": "Decompose $f(S)$ into a bunch of disconnected open intervals. Now consider $x\\in\\text{bdd}(S)$. For small enough $\\epsilon$, $f(x+\\epsilon)\\not\\in f(S)$. Then if $f(x)\\in f(S)$, $f(x)\\in\\text{bdd}(f(S))$. But this is absurd, because $f(S)$ is open and hence does not contain its boundary. So $S$ does not contain any of its boundary. In particular, that means that it is open." } { "Tag": [ "function", "algebra", "domain" ], "Problem": "1 Suppose that g(x)=x^2 +4x and that f(z) =z square root . Let h(x)=f(g(x)) Calculate h(2). Use two decimals, for example 6.21\r\n\r\n2 Which of the following functions will have an inverse?\r\nV\u00e4lj minst ett svar\r\n\t a. f(x) = x^2 with domain R\t\r\n\t b. f(x) = x^2 with domain x > 0\t\r\n\t c. f(x) = x^2 with domain x > 0 AND x = 0\t\r\n\t d. f(x) = ln(x) (x > 0)\t\r\n\t e. for domain R\t\r\n\t f. f(x) = x - ln(x) for x > 0\r\n\r\n3 Suppose that f(1) = 1, f(-1) = 2 and f(3) = 2. Let g(x) = (f(x) + f(-x))*f(3x). Calculate g(1).", "Solution_1": "What are you having trouble with specifically?", "Solution_2": "Need help to find solutions to all of them please", "Solution_3": "1.\r\n$ g(2)\\equal{}12$, I think that $ f(x)\\equal{}\\sqrt{x}$ right if so, $ \\sqrt{12}\\equal{}\\boxed{3.46}$" } { "Tag": [ "integration", "calculus", "Euler", "function", "limit", "calculus computations" ], "Problem": "$\\int_{a}^{b}\\frac{xdx}{\\sqrt{(x-a)(b-x)}}$ \r\n caculus intergal", "Solution_1": "Do the Euler substitution $\\sqrt{(x-a)(b-x)}= t(x-a)$, i.e. $t = \\sqrt{\\frac{b-x}{x-a}}, \\, x = \\frac{b+a t^{2}}{1+t^{2}}, \\, dx = \\frac{2(a-b)t}{\\left( t^{2}+1 \\right)^{2}}$!\r\n\r\nThus,\r\n\\begin{eqnarray*}I &=& \\int_{a}^{b}\\frac{x}{\\sqrt{(x-a)(b-x)}}\\, dx \\\\ \\ &=& \\int_\\infty^{0}\\frac{\\frac{b+a t^{2}}{1+t^{2}}}{\\frac{t(b-a)}{1+t^{2}}}\\cdot \\frac{2(a-b)t}{\\left( t^{2}+1 \\right)^{2}}\\, dt \\\\ \\ &=& 2 \\int_{0}^\\infty \\frac{b+a t^{2}}{\\left( t^{2}+1 \\right)^{2}}\\, dt \\\\ \\ &=& \\int_{-\\infty}^\\infty \\frac{b+a t^{2}}{\\left( t^{2}+1 \\right)^{2}}\\, dt .\\end{eqnarray*}\r\n\r\nConsider the function $f(z) = \\frac{b+a z^{2}}{\\left( z^{2}+1 \\right)^{2}}$. We integrate this from $-R$ to $R$ on the real line and then over a semicircle of radius $R$, centered at the origin, traced counter-clockwise. Call this contour $\\gamma(R)$.\r\n\r\nThe first part clearly goes to $\\int_{-\\infty}^\\infty \\frac{b+a t^{2}}{\\left( t^{2}+1 \\right)^{2}}\\, dt$ (which is finite) as $R \\to \\infty$. The second part can be written as $\\int_{0}^\\pi \\frac{b+a R^{2}\\exp \\left( 2 i \\theta \\right)}{\\left( 1+R^{2}\\exp \\left( 2 i \\theta \\right) \\right)^{2}}\\cdot i R \\exp \\left( i \\theta \\right) \\, d \\theta$. It should be pretty clear that this goes to $0$ as $R \\to \\infty$.\r\n\r\nHence, $I = \\lim_{R \\to \\infty}\\int_{\\gamma(R)}f(z) \\, dz$.\r\n\r\nWe can compute the last integral via the residue theorem. The only pole inside $\\gamma(R)$ is $i$. The residue there is\r\n\\begin{eqnarray*}\\textrm{Res}(f,i) &=& \\frac1{1!}\\cdot \\lim_{z \\to i}\\frac{ d \\left( \\frac{b+a z^{2}}{(z+i)^{2}}\\right)}{dz}\\\\ \\ &=& \\lim_{z \\to i}\\frac{2az(z+i)-2 \\left( b+a z^{2}\\right)}{(z+i)^{3}}\\\\ \\ &=& \\frac{a+b}{4i}.\\end{eqnarray*}\r\nHence, $\\int_{\\gamma(R)}f(z) \\, dz = 2 \\pi i \\cdot \\textrm{Res}(f,i) = \\frac{\\pi \\left( a+b \\right)}2$.\r\nTherefore, $I = \\frac{\\pi \\left( a+b \\right)}2$.\r\n\r\nPS: There are many other ways to compute this... and I think I chose the most tedious one.", "Solution_2": "i think the following may be an easier way\r\n\r\nput $x=acos^{2}t+bsin^{2}t$ with appropriate limits. sorry i dont have more time to work out the full solution", "Solution_3": "The easiest (???) solution is, I think, $y = x-\\frac{a+b}2$:\r\n\\[I = \\int_{\\frac{a-b}2}^{\\frac{b-a}2}\\frac{y+\\frac{a+b}2}{\\sqrt{\\left( \\frac{b-a}2 \\right)^{2}-y^{2}}}\\, dy = \\left. \\frac{a+b}2 \\cdot \\arcsin \\left( \\frac{2y}{b-a}\\right) \\right|_\\frac{a-b}2^\\frac{b-a}2 = \\frac{\\pi \\left( a+b \\right)}2 . \\]\r\n(The other part of the integral is $0$, since we're dealing with an odd function and the improper integral obtained is convergent.)" } { "Tag": [ "probability", "geometry", "ratio" ], "Problem": "A man arrives at an airport from 11:00 to 12:00, waits for his wife to pick him up for 20 minutes, and goes home if she doesn't. His wife arrives at 11:00 to 11:40, waits for him for 10 minutes, and drives home if he is not there after 10 minutes. What is the probability that the wife picks him up?", "Solution_1": "[hide]\n$\\frac{7\\sqrt{10}-1}{48}$\n[/hide]", "Solution_2": "let the times that they arrive be x, and y for the man and woman respectively\r\ny-x <20 or x-y<10, then draw a box on a cooridinate grid that has 0 to 60 for x, and 0-40 for y and find the ratio of the area that is shaded to the ratio of the whole box\r\nas this for being geometry, not really, it is pretty easy to find areas of these triangles, it is more algebra than anything, this is a lot like linear programming" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "In how many ways can we select 'r' objects from 'n' distinct objects placed in a circle such that no two of these selected objects are placed together.", "Solution_1": "cut the \"necklace\" of objects somewhere, say between consecutive objects $ A$ and $ B$, and make a line out of them: $ A, ..., B$. how many ways are there to choose $ r$ objects from this line so that no two are next to each other? well, $ \\binom{n \\minus{} r \\plus{} 1}{r}$. but now back to the necklace: it may be that we chose both $ A$ and $ B$ from the line, which is not allowed for the necklace. so the question is, how many ways are there to pick $ r$ objects from the line, two of which are $ A$ and $ B$, so that no two are next to each other? well, $ \\binom{n \\minus{} 4 \\minus{} (r \\minus{} 2 \\minus{} 1)}{r \\minus{} 2} \\equal{} \\binom{n \\minus{} r \\minus{} 1}{r \\minus{} 2}$. (take $ A$ and $ B$ and their neighbors out of the picture (4 objects in total), then use the same formula as above, with $ n$ replaced by $ n \\minus{} 4$ and $ r$ replaced by $ r \\minus{} 2$. so the answer is $ \\binom{n \\minus{} r \\plus{} 1}{r} \\minus{} \\binom{n \\minus{} r \\minus{} 1}{r \\minus{} 2}$.", "Solution_2": "Arithmetypo: $ \\minus{} 4 \\minus{} ( \\minus{} 2 \\minus{} 1) \\equal{} \\minus{} 1$ ;). \r\n\r\nAlternatively, choose the first element in $ n$ ways; then we must choose $ r \\minus{} 1$ elements from a line of $ n \\minus{} 3$. We can do this in $ \\binom{n \\minus{} 3 \\minus{} (r \\minus{} 1) \\plus{} 1}{r \\minus{} 1} \\equal{} \\binom{n \\minus{} r \\minus{} 1}{r \\minus{} 1}$ ways. However, this counts every choice $ r$ times, so we have in total $ \\frac {n}{r} \\binom{n \\minus{} r \\minus{} 1}{r \\minus{} 1}$ choices.", "Solution_3": "[quote=\"JBL\"]Arithmetypo: $ \\minus{} 4 \\minus{} ( \\minus{} 2 \\minus{} 1) \\equal{} \\minus{} 1$ [/quote]\r\n\r\nnow that's a word worthy of any linguist's lexicon ;)" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "For positive integers $ n,k$, define $ F(n,k)\\equal{}\\displaystyle\\sum_{r\\equal{}1}^{n}r^{2k\\minus{}1}$. Prove that: $ F(n,1)$ divides $ F(n,k)$.", "Solution_1": "[quote=\"moldovan\"]For positive integers $ n,k$, define $ F(n,k) \\equal{} \\displaystyle\\sum_{r \\equal{} 1}^{n}r^{2k \\minus{} 1}$. Prove that: $ F(n,1)$ divides $ F(n,k)$.[/quote]\r\n\r\n$ F(n,1)\\equal{}\\frac{n(n\\plus{}1)}{2}$\r\n\r\n[b]Case 1: [/b] $ n$ odd, then\r\n\r\n$ F(n,k) \\equiv [1^{2k\\minus{}1}\\plus{}(n\\minus{}1)^{2k\\minus{}1}]\\plus{}[2^{2k\\minus{}1}\\plus{}(n\\minus{}2)^{2k\\minus{}1}]\\plus{}...$\r\n$ \\plus{}[(\\frac{(n\\minus{}1)}{2})^{2k\\minus{}1}\\plus{}(\\frac{(n\\plus{}1)}{2})^{2k\\minus{}1}] \\equiv 0 \\pmod{n}$\r\n\r\n$ F(n,k) \\equiv [1^{2k\\minus{}1}\\plus{}n^{2k\\minus{}1}]\\plus{}[2^{2k\\minus{}1}\\plus{}(n\\minus{}1)^{2k\\minus{}1}]\\plus{}...$\r\n$ \\plus{}[(\\frac{n\\minus{}1}{2})^{2k\\minus{}1}\\plus{}(\\frac{n\\plus{}3}{2})^{2k\\minus{}1}]\\plus{}(\\frac{n\\plus{}1}{2})^2 \\equiv 0 \\pmod{n\\plus{}1}$\r\n\r\n$ \\Rightarrow F(n,1)|F(n,k)$\r\n\r\n[b]Case 2:[/b] $ n$ even, then\r\n\r\n$ F(n,k) \\equiv [1^{2k\\minus{}1}\\plus{}(n\\minus{}1)^{2k\\minus{}1}]\\plus{}[2^{2k\\minus{}1}\\plus{}(n\\minus{}2)^{2k\\minus{}1}]\\plus{}...$\r\n$ \\plus{}[(\\frac{n\\minus{}2}{2})^{2k\\minus{}1}\\plus{}(\\frac{n\\plus{}2}{2})^{2k\\minus{}1}]\\plus{}(\\frac{n}{2})^2 \\equiv 0 \\pmod{\\frac{n}{2}}$\r\n\r\n$ F(n,k) \\equiv [1^{2k\\minus{}1}\\plus{}n^{2k\\minus{}1}]\\plus{}[2^{2k\\minus{}1}\\plus{}(n\\minus{}1)^{2k\\minus{}1}]\\plus{}...$\r\n$ \\plus{}[(\\frac{n}{2})^{2k\\minus{}1}\\plus{}(\\frac{n\\plus{}2}{2})^{2k\\minus{}1}] \\equiv 0 \\pmod{\\frac{n\\plus{}1}{2}}$\r\n\r\n$ \\Rightarrow F(n,1)|F(n,k)$, done." } { "Tag": [ "NIMO", "AMC", "AIME", "MATHCOUNTS" ], "Problem": "This is pretty much where you make up your own awesome math competition, and then you make a cool acronym and stuff!", "Solution_1": "NIMO, all the way :D", "Solution_2": ":rotfl: , looks like this thread is failing... I would call mine: the Mathematics Elements Listing Examination(MELE) :D !\r\n\r\nIt would be like a qualifying exam, sorta like AIME!", "Solution_3": "The Critical Objects of Mathematical Information Contest (COMIC)", "Solution_4": "I would have to go with NIMO, y'know?", "Solution_5": "The International Math Competition for Elementary Schoolers, or IMCES for short. It would be sort of like Mathcounts, except I'd have 5 different rounds(and for once an elementary schooler would get a chance to participate in a real competition) The \"Speed\" round is where the competitors try to answer 250 word problems ranging from \"3rd grade ISTEP+ level\" to hard MC states\" level in 40 minutes(That way there shouldn't be so many ties). Then a \"Skills\" round where 20 problems are handed out in sets of fours with 4 minutes per set of problems(calculators allowed). Following that would be a \"Stretch\" round where the third through fifth graders group in teams of four and answer 20 pretty difficult questions in 25 minutes. Then the 50 highest scorers would participate in a \"Semi-Final\" round where all the students are(while on a stage), asked a problem and given 45 seconds to answer. Each student must write out his/her answer and if they get 2 questions wrong, they're out. Finally, the \"Oral Round\" is a bracket style tournament for the top 16 students very similar to an MC countdown round. And the best part is, the competition would take place in Athens, Greece!", "Solution_6": "Hm...tough, but I'd have to say \"NIMO\".", "Solution_7": "I can't think of anything, so I'll say NIMO. (pronounced NYE-moh)", "Solution_8": "All of you who said NIMO run NIMO. It is an actual competition that is run by you guys... Not bad... I'm thinking of doing something similar :wink:", "Solution_9": "Oh, come on, you didn't even say what NIMO meant (the National Internet Mathematical Olympiad).\r\n\r\nNo wonder this thread is failing." } { "Tag": [ "geometry" ], "Problem": "The sides of unit square $ ABCD$ have trisection points $ X$, $ Y$, $ Z$ and $ W$, as shown. If $ AX: XB \\equal{} BY: YC \\equal{} CZ: ZD \\equal{} DW: WA \\equal{} 2: 1$, what is the area of the shaded region? Express your answer as a common fraction.\n\n[asy]fill((.9,.3)--(2.7,.9)--(2.1,2.7)--(.3,2.1)--cycle,gray(.7));\ndraw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1));\ndraw((0,0)--(3,1),linewidth(1));\ndraw((3,0)--(2,3),linewidth(1));\ndraw((3,3)--(0,2),linewidth(1));\ndraw((0,3)--(1,0),linewidth(1));\nlabel(\"A\",(0,3),N);\nlabel(\"X\",(2,3),N);\nlabel(\"B\",(3,3),N);\n\nlabel(\"W\",(0,2),W);\nlabel(\"Y\",(3,1),E);\nlabel(\"D\",(0,0),W);\nlabel(\"Z\",(1,0),S);\nlabel(\"C\",(3,0),E);[/asy]", "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=252209]Click because I'm lazy[/url] :P", "Solution_2": "[quote=\"Kouichi Nakagawa\"][asy]fill((1,1)--(3,1)--(3,3)--(1,3)--cycle,grey);\n\nfor (int i = 0; i <= 4; ++i)\ndraw((i,0)--(i,4)^^(0,i)--(4,i));\n\ndraw((0,1)--(3,0)--(4,3)--(1,4)--cycle,linewidth(2));\n\ndraw((1,4)--(1,2/3),linewidth(1));\ndraw((0,1)--(3+1/3,1),linewidth(1));\ndraw((3,0)--(3,3+1/3),linewidth(1));\ndraw((4,3)--(2/3,3),linewidth(1));\n\nlabel(\"$A$\",(1,4), N);\nlabel(\"$B$\",(4,3), E);\nlabel(\"$C$\",(3,0), S);\nlabel(\"$D$\",(0,1), W);\nlabel(\"$X$\",(3, 3+1/3), NE);\nlabel(\"$Y$\",(3+1/3, 1), SE);\nlabel(\"$Z$\",(1, 2/3), SW);\nlabel(\"$W$\",(2/3, 3), NW);[/asy]\n\nThus, $ \\frac {2^2}{(\\sqrt {1 \\plus{} 3^2})^2} \\equal{} \\frac {4}{10} \\equal{} \\boxed{\\frac {2}{5}}$.[/quote]\r\n\r\nI put this here just because i think it's a beautiful solution, and so people don't have to go searching the other discussion for the solution" } { "Tag": [ "trigonometry", "AMC" ], "Problem": "[color=cyan]If 1 + sin(2x) = 2cos(2x), compute all possible values of tan(x)[/color]", "Solution_1": "would u use the LAw of tangents for this?,\r\n\r\nor sum of sine-cosine laws?", "Solution_2": "Transform this into sin (90-2x)/(1+cos (90-2x)) = 1/2, hence tan (45-x) = 1/sqrt(2). From this it is very, very easy to determine tan x, but I won't, because it's too hard!" } { "Tag": [ "pigeonhole principle", "combinatorics proposed", "combinatorics" ], "Problem": "In a school there are $ 10$ rooms. Each student from a room knows exactly one student from each one of the other $ 9$ rooms. Prove that the rooms have the same number of students (we suppose that if $ A$ knows $ B$ then $ B$ knows $ A$).", "Solution_1": "[quote=\"felixmann\"]In a school there are $ 10$ rooms. Each student from a room knows exactly one student from each one of the other $ 9$ rooms. Prove that the rooms have the same number of students (we suppose that if $ A$ knows $ B$ then $ B$ knows $ A$).[/quote]\r\n\r\n[hide]Suppose there are two rooms $ M$ and $ N$ such that $ M$ has more students than $ N$. Each student in $ M$ has exactly one friend in $ N$. But since $ \\# (M) > \\#(N)$, then by pigeonhole, there is at least two students in $ M$ who have the same friend in $ N$. However, this cannot be, because $ N$ would have more than one friend. Hence, all rooms must have an equal number of students.[/hide]" } { "Tag": [], "Problem": "Since I recently found out what the China Syndrome is, I asked a bunch of people if they knew what it was. No one did. So, I decided to move the poll here for fun. So, do you know what the China syndrome is?", "Solution_1": "Well, [url=http://en.wikipedia.org/wiki/China_Syndrome]now I do.[/url]", "Solution_2": "[quote=\"t0rajir0u\"]Well, [url=http://en.wikipedia.org/wiki/China_Syndrome]now I do.[/url][/quote]\r\n\r\nLol if you put \"Don't Google it\" people will Google.", "Solution_3": "I immediately thought of the China effect. Aww... well.", "Solution_4": "I thought of lead poisoning or something :wink: Sorry to you people in China.", "Solution_5": "Believe it or not, I immediately thought of chow mein ( :D ). This is because I got confused with something I read somewhere about some kind of Chinese food effect thingy, which is about how Chinese food is not a good idea for dieting.", "Solution_6": "Well yes, if I put \"don't Google it,\" people will probably Google it, but who cares. This is just for fun.", "Solution_7": "It was on the news, \"The China Syndrome\" about how they got into the car building business?", "Solution_8": "nope :huh:" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Can you draw the picture about the function : f-0(x)=e^x", "Solution_1": "I'm sorry . But I don't know how to input the expressions into the messeage ,could you tell me??", "Solution_2": "Put your expression between two $\\$$", "Solution_3": "$f-0(x)=e^x$ is this whatb u want ?", "Solution_4": "It's $f-0(x)=e^x$", "Solution_5": "Oh,I'm sorry .this is the first time that I write in the forum.Thanks for your help! $f_0(x)=e^x$", "Solution_6": "very good." } { "Tag": [ "geometry", "pigeonhole principle", "induction", "symmetry", "inequalities", "number theory" ], "Problem": "For the first time since I can remember, there is no discussion of National Camp on this forum. Is anyone here coming this year? July 6-12 at John Abbott College in Montreal - yay! :D", "Solution_1": "what (olympiad?) topics are we going to learn there?", "Solution_2": "We have two sessions on Number Theory (mods and diophantine equations), two on Geometry (circle geometry and triangle geometry), one on proofs and problem solving techniques (Pigeonhole, induction, wolog, etc), a session on Parity and Symmetry, a session on Combinatorics, and possibly a session on either Inequalities or something else. We focus on developing good-quality solution-writing, and help transition into being able to solve (more) olympiad-level problems. We also have \"fun talks\" about cool things in math that don't necessarily relate to competitions (past fun talk topics have included Ramsey Theory, Surreal Numbers, Cryptography, etc). And the whole experience is as much about meeting other math students across Canada as it is about learning math.\r\n\r\nIf you are Heinrich from Windsor (which I assume you are, because I don't know too many other Heinrichs), you should be able to ask numerous National Camp alums about what it's like. And while you're at it, send my regards to Mr. White!", "Solution_3": "haha...thanks for the info Taotao :D", "Solution_4": "ahhhh I want to go again. Pretty awesome camp there=)", "Solution_5": "I agree Jonathan ... very good camp", "Solution_6": "=)\r\n\r\nSafe to say it was a fun, successful year!!! Let's see how many national campers we bring to AoPS this year eh?", "Solution_7": "[quote=\"Sunny\"]Let's see how many national campers we bring to AoPS this year eh?[/quote]\r\nWell, you did convince me.....\r\nTook a look on here for the first time today\r\nIt's a forum.... but dedicated for nerdy, mathy people.... *gasp*\r\nThis is almost as exciting as that 755-page IMO book we pirated off the internet XD", "Solution_8": "[quote=\"richrichardw\"]\nIt's a forum.... but dedicated for nerdy, mathy people.... *gasp*\nThis is almost as exciting as that 755-page IMO book we pirated off the internet XD[/quote]\r\n\r\nHahaha, Hi Richie! You mean, the book you pirated when you were supposed to be sleeping? =)", "Solution_9": "everyone pirates that book lol", "Solution_10": "Huh? What book is this?", "Solution_11": "Probably the IMO Compendium from 1959 to 2004... correct me if I'm wrong. :D", "Solution_12": ":D It was a really fun camp!!:) \r\n\r\nI just want to remind everyone that there's a facebook group for the national camp, too! \r\n\r\nCome on up and join!!!! \r\n\r\nWe need more people :wink:" } { "Tag": [ "function", "inequalities", "inequalities proposed" ], "Problem": "Let $ x_i > 0$ be positive reals $ (i \\equal{} 1,2,\\cdots n)$ such that $ \\sum_{i \\equal{} 1}^{n} a_i \\equal{} n$. Find $ P_{min}$ where:\r\n$ P \\equal{} \\frac {x_1^k}{x_n \\plus{} x_1} \\plus{} \\frac {x_2^k}{x_1 \\plus{} x_2} \\plus{} \\cdots \\plus{} \\frac {x_n^k}{x_1 \\plus{} x_{n \\minus{} 1}}$.\r\n :blush: \r\nHope you will enjoy it! :)\r\nHappy Bengali New year to all mathlinkers!(It might be a bit late) :lol: \r\n[hide=\"P/s\"] I have to be offline for 7 days or so from today as my internet session is to expire today midnight[/hide]", "Solution_1": "what about $ k$ is it a natural number?", "Solution_2": "[quote=\"peine\"]what about $ k$ is it a natural number?[/quote]\r\nYeah, forgot to add: $ k\\in\\mathbb{N}$. :oops:", "Solution_3": "[quote=\"Potla\"]Let $ x_i > 0$ be positive reals $ (i \\equal{} 1,2,\\cdots n)$ such that $ \\sum_{i \\equal{} 1}^{n} a_i \\equal{} n$. Find $ P_{min}$ where:\n$ P \\equal{} \\frac {x_1^k}{x_n \\plus{} x_1} \\plus{} \\frac {x_2^k}{x_1 \\plus{} x_2} \\plus{} \\cdots \\plus{} \\frac {x_n^k}{x_1 \\plus{} x_{n \\minus{} 1}}$.\n :blush: \nHope you will enjoy it! :)\nHappy Bengali New year to all mathlinkers!(It might be a bit late) :lol: \n[hide=\"P/s\"] I have to be offline for 7 days or so from today as my internet session is to expire today midnight[/hide][/quote]\r\nI think $ \\sum_{i\\equal{}1}^{n}x_i\\equal{}n$.If so,use AM-GM or Holder :wink: :)", "Solution_4": "$ P \\equal{} \\frac {x_1^k}{x_n \\plus{} x_1} \\plus{} \\frac {x_2^k}{x_1 \\plus{} x_2} \\plus{} \\cdots \\plus{} \\frac {x_n^k}{x_1 \\plus{} x_{n \\minus{} 1}}$\r\n\r\nMaybe, we could use the convex function $ f(x)\\equal{}\\frac{1}{x}$ and Jensens ineq. \r\n\r\n$ P\\geq (\\sum {{x}_{i}}^{k})(\\frac{1}{\\frac{\\sum 2{x}_{i}}{n}})\\equal{}\\frac{ (\\sum {{x}_{i}}^{k})}{2}$\r\n \r\nIf k=1 this results in the minimum n/2. Assume that k>1 Now, using jensen inequality on the function $ f(x)\\equal{}{x}^{k}$\r\n\r\n $ \\frac{ (\\sum {{x}_{i}}^{k})}{2}\\geq \\frac{n}{2}$\r\n\r\n Which gives the minimum if all the variables are equal.", "Solution_5": "[quote=\"fredrik1232009\"]$ P \\equal{} \\frac {x_1^k}{x_n \\plus{} x_1} \\plus{} \\frac {x_2^k}{x_1 \\plus{} x_2} \\plus{} \\cdots \\plus{} \\frac {x_n^k}{x_1 \\plus{} x_{n \\minus{} 1}}$\n\nMaybe, we could use the convex function $ f(x) \\equal{} \\frac {1}{x}$ and Jensens ineq. \n\n$ P\\geq (\\sum {{x}_{i}}^{k})(\\frac {1}{\\frac {\\sum 2{x}_{i}}{n}}) \\equal{} \\frac { (\\sum {{x}_{i}}^{k})}{2}$\n \nIf k=1 this results in the minimum n/2. Assume that k>1 Now, using jensen inequality on the function $ f(x) \\equal{} {x}^{k}$\n\n $ \\frac { (\\sum {{x}_{i}}^{k})}{2}\\geq \\frac {n}{2}$\n\n Which gives the minimum if all the variables are equal.[/quote]\r\nThanks for your nice solution, but I think my solution is easier than this..... :oops: \r\nI do hope that Kunny will give us a solution....... :blush:", "Solution_6": "I think that the right enunciation of your problem is the following:\r\n\r\nLet $ x_i > 0$ be positive real numbers $ (i \\equal{} 1,2,...,n),$ where $ n\\in N, n\\geq 1$, such that $ \\sum_{i \\equal{} 1}^{n}x_{i} \\equal{} n$. \r\nFor $ k > 0$, find $ P_{min}$, where:\r\n$ P \\equal{} \\frac {x_{1}^{k}}{x_{n} \\plus{} x_{1}} \\plus{} \\frac {x_{2}^{k}}{x_{1} \\plus{} x_{2}} \\plus{} \\cdots \\plus{} \\frac {x_{n}^{k}}{x_{n \\minus{} 1} \\plus{} x_{n}}.$\r\n\r\n[u]My solution[/u]:\r\n\r\n$ P \\equal{} \\frac {x_{1}^{k}}{x_{n} \\plus{} x_{1}} \\plus{} \\frac {x_{2}^{k}}{x_{1} \\plus{} x_{2}} \\plus{} \\cdots \\plus{} \\frac {x_{n}^{k}}{x_{n \\minus{} 1} \\plus{} x_{n}}\\geq$\r\n$ \\geq \\frac {(x_1 \\plus{} x_2 \\plus{} \\cdots \\plus{} x_n)^k}{n^{k \\minus{} 2}\\cdot\\left[(x_n \\plus{} x_1) \\plus{} (x_1 \\plus{} x_2) \\plus{} \\cdots \\plus{} (x_{n \\minus{} 1} \\plus{} x_n)\\right]} \\equal{} \\frac {n^k}{n^{k \\minus{} 2}\\cdot2\\cdot n} \\equal{} \\frac {n}{2}$\r\nThus: $ P_{min} \\equal{} \\frac {n}{2}$\r\n\r\nI'm sure you know that:\r\n\r\nIf: $ a_i\\geq0,b_i > 0,i \\equal{} 1,2,...,n,$ and $ n\\in N,n\\geq1$, then, for $ k > 0,$ we have:\r\n$ \\frac {{a_1}^k}{b_1} \\plus{} \\frac {{a_2}^k}{b_2} \\plus{} \\cdots \\plus{} \\frac {{a_n}^k}{b_n}\\geq\\frac {(a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n)^k}{n^{k \\minus{} 2}\\cdot(b_1 \\plus{} b_2 \\plus{} \\cdots \\plus{} b_n)}$", "Solution_7": "[quote=\"marin.bancos\"]\nI'm sure you know that:\n\nIf: $ a_i\\geq0,b_i > 0,i \\equal{} 1,2,...,n,$ and $ n\\in N,n\\geq1$, then, for $ k > 0,$ we have:\n$ \\frac {{a_1}^k}{b_1} \\plus{} \\frac {{a_2}^k}{b_2} \\plus{} \\cdots \\plus{} \\frac {{a_n}^k}{b_n}\\geq\\frac {(a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n)^k}{n^{k \\minus{} 2}\\cdot(b_1 \\plus{} b_2 \\plus{} \\cdots \\plus{} b_n)}$[/quote]\r\nYes :rotfl: that is H\u00f6lder:\r\n\r\n$ \\left(\\frac {{a_1}^k}{b_1} \\plus{} \\frac {{a_2}^k}{b_2} \\plus{} \\cdots \\plus{} \\frac {{a_n}^k}{b_n}\\right)(b_1 \\plus{} b_2 \\plus{} \\cdots \\plus{} b_n)(1\\plus{}\\cdots\\plus{}1)^{n\\minus{}2}$ $ \\ge {(a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n)^k}$", "Solution_8": "You're right, [b]spanferkel[/b]!" } { "Tag": [ "calculus", "integration", "inequalities", "function", "logarithms", "calculus computations" ], "Problem": "From my Math Club:\r\n\r\nf(x) is a continuous positive function on [0..1] Prove that\r\n$\\int_{0}^{1}f(x) dx \\geq e^{\\int_{0}^{1}ln(f(x)) dx}$\r\nThanks", "Solution_1": "[quote=\"Gibbenergy\"]From my Math Club:\n\nf(x) is a continuous positive function on [0..1] Prove that\n$\\int_{0}^{1}f(x) dx \\geq e^{\\int_{0}^{1}ln(f(x)) dx}$\nThanks[/quote]\r\n\r\nI think it follows from $\\ln{x}$ being concave and the continuous version of Jensen's Inequality (I'm not sure of the details of this, though).", "Solution_2": "[quote=\"Gibbenergy\"]From my Math Club:\n\nf(x) is a continuous positive function on [0..1] Prove that\n$\\int_{0}^{1}f(x) dx \\geq e^{\\int_{0}^{1}ln(f(x)) dx}$\nThanks[/quote]\r\n\r\nBy Jensen inequality\r\n$\\frac{1}{n}{\\sum_{k=1}^{n}ln(f(\\frac{i}{n})) \\leq ln[ \\frac{1}{n}{\\sum_{k=1}^{n}f(\\frac{i}{n})]}}$\r\n\r\nLeft side approaches to:\r\n$\\int_{0}^{1}lnf(x) dx$\r\nand right side approaches to:\r\n$ln \\int_{0}^{1}f(x) dx$", "Solution_3": "Here's one way to phrase Jensen's inequality: Suppose that $\\varphi$ is a convex function on a certain interval, and suppose that $X$ is a random variable taking values in that interval. Suppose also that all of the expectations in this theorem are absolutely convergent. Then \r\n\r\n$E(\\varphi(X))\\ge\\varphi(E(X)).$\r\n\r\n(For a concave function, reverse the order of the inequality.)\r\n\r\nI like this form, as it tends to unify the various other ways of writing Jensen. The proof is essentially the same as for all of the others, just a little more streamlined. Here's the proof:\r\n\r\nLet $x_{0}=E(X).$ There is a \"supporting line,\" a function $L(x)=ax+b$ such that $L(x)\\le\\varphi(x)$ and $L(x_{0})=\\varphi(x_{0}).$\r\n\r\nThen $E(\\varphi(X))\\ge E(L(X))=L(E(X))=L(x_{0})=\\varphi(x_{0}) =\\varphi(E(X)).$\r\n\r\nThe key step, and the place where we need that this is a random variable (that is, that we are integrating over a set of total mass 1) is that $E(L(X))=L(E(X).$ In more detail, $E(L(X))=E(aX+b)=aE(X)+E(b)=aE(X)+b=L(E(X)).$\r\n\r\nFor the case of this problem, note that if $X$ is the random variable that is uniformly distributed on $[0,1]$ and $Y=f(X),$ then we are asked to show that $E(Y)\\ge e^{E(\\ln(Y)),}$ which is the same as $\\ln(E(Y))\\ge E(\\ln(Y))$ and is true by Jensen (with the sign reversed because $\\ln$ is concave.)" } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "Let x1,x2,...,xn in [0,1] and x1 + x2 + ... + xn = 1\r\n Find the maximum of : f(x1,x2,...,xn) = x1 2 + x2 2 + ... + xn 2 - x1 4 - x2 4 - ... - xn 4", "Solution_1": "If i hadn't made any mistakes, sum (x_i)^3-sum (x_i)^4=sum (1<=i $ ( \\frac{(2k\\minus{}1)!}{p} ) \\equal{} 1$ and $ (2k\\minus{}1)! \\equiv \\minus{}1 (\\mod p)$ <=> $ ( \\frac{(2k\\minus{}1)!}{p} ) \\equal{} \\minus{}1$\r\n\r\n(ii) $ (\\frac{(2k\\minus{}1)!}{p}) \\equal{} (\\minus{}1)^a$, where a is the number of quadratic nonresidues of p; (this is trivial)\r\n\r\n(iii) Observe that $ (\\frac{\\minus{}1}{p}) \\equal{} \\minus{}1$, and therefore $ (\\frac{\\minus{}k}{p}) \\equal{} \\minus{}(\\frac{k}{p})$\r\n\r\nThis implies that a number i=1,2,..,2k-1 is a quadratic nonresidue iff the number p-i is a quadratic residue.\r\n\r\nThis yields a=m, and this finishes the proof.\r\n(The number of quadratic nonresidues between 1 and $ \\frac{p}{2}$ is equal to the number of quadratic residues between $ \\frac{p}{2}$ and p.)" } { "Tag": [ "trigonometry" ], "Problem": "Find the exact value of $ \\csc{20^{\\circ}}\\minus{}\\cot{40^{\\circ}}$ without a calculator", "Solution_1": "[quote=\"tinytim\"]Find the exact value of $ \\csc{20^{\\circ}} \\minus{} \\cot{40^{\\circ}}$ without a calculator[/quote]\r\n\r\nI don't understand: $ \\boxed{\\csc{20^{\\circ}}}$", "Solution_2": "$ \\csc{20^{\\circ}} \\equal{} \\frac {1}{\\sin{20^{\\circ}}}$", "Solution_3": "csc=reciprocal of sin\r\n\r\nsec= reciprocal of cos\r\n\r\ncot=reciprocal of tan\r\n\r\nHint for my problem:\r\n\r\n$ \\csc{20^{\\circ}}\\minus{}\\cot{40^{\\circ}}\\equal{}\\frac{1}{\\sin{20^{\\circ}}}\\minus{}\\frac{\\cos{40^{\\circ}}}{\\sin{40^{\\circ}}}$", "Solution_4": "fine, ignore my problem.... :mad:", "Solution_5": "$ \\frac{1}{\\sin{20^{\\circ}}}\\minus{}\\frac{\\cos{40^{\\circ}}}{\\sin{40^{\\circ}}}$\r\n\r\n$ \\equal{}\\frac{2\\cos20^0\\minus{}\\cos40^0}{\\sin{40^{\\circ}}}$\r\n\r\n$ \\equal{}\\frac{(\\cos20^0\\minus{}\\cos40^0)\\plus{}\\cos20^0}{\\sin{40^{\\circ}}}$\r\n\r\n$ \\equal{}\\frac{\\minus{}2\\sin30^0\\sin(\\minus{}10^0)\\plus{}\\cos20^0}{\\sin{40^{\\circ}}}$\r\n\r\n$ \\equal{}\\frac{\\sin10^0\\plus{}\\cos20^0}{\\sin{40^{\\circ}}}$\r\n\r\n$ \\equal{}\\frac{\\cos80^0\\plus{}\\cos20^0}{\\sin{40^{\\circ}}}$\r\n\r\n$ \\equal{}\\frac{2\\cos50^0.\\cos30^0}{\\sin{40^{\\circ}}}$\r\n\r\n$ \\equal{}\\frac{2\\sin40^0.\\cos30^0}{\\sin{40^{\\circ}}}$\r\n\r\n$ \\equal{}2\\cos30^0 \\equal{} \\sqrt{3}$" } { "Tag": [ "calculus", "induction", "algebra unsolved", "algebra" ], "Problem": "Let $ (U_n)$ be a sequence given recursively as follows:\r\n\r\n$ \\left \\{\\begin{array}{l}U_0 \\equal{} 2 \\ ; \\ U_1 \\equal{} 10 \\\\\r\nU_{n \\plus{} 2} \\equal{} 8U_{n \\plus{} 1} \\minus{} 2U_n \\end{array} \\right.$\r\n\r\nDetermine the maximum possible value of n such that $ \\ U_n \\leq 2009^2$", "Solution_1": "[quote=\"mcdat\"]Let $ (U_n)$ be a sequence given recursively as follows:\n\n$ \\left \\{\\begin{array}{l}U_0 \\equal{} 2 \\ ; \\ U_1 \\equal{} 10 \\\\\nU_{n \\plus{} 2} \\equal{} 8U_{n \\plus{} 1} \\minus{} 2U_n \\end{array} \\right.$\n\nDetermine the maximum possible value of n such that $ \\ U_n \\leq 2009^2$[/quote]\r\n\r\nAccording to me, the simpliest methedod is rude calculus (though not very clever)\r\n\r\nIt's immediate to show with induction that $ u_n$ is increasing\r\nThen :\r\n$ u_2\\equal{}76< 4036081\\equal{}2009^2$\r\n$ u_3\\equal{}588< 4036081\\equal{}2009^2$\r\n$ u_4\\equal{}4552< 4036081\\equal{}2009^2$\r\n$ u_5\\equal{}35240< 4036081\\equal{}2009^2$\r\n$ u_6\\equal{}272816< 4036081\\equal{}2009^2$\r\n$ u_7\\equal{}2112048< 4036081\\equal{}2009^2$\r\n$ u_8\\equal{}16350752> 4036081\\equal{}2009^2$\r\n\r\nSo the required value is $ n\\equal{}7$", "Solution_2": "[quote=\"pco\"][quote=\"mcdat\"]Let $ (U_n)$ be a sequence given recursively as follows:\n\n$ \\left \\{\\begin{array}{l}U_0 \\equal{} 2 \\ ; \\ U_1 \\equal{} 10 \\\\\nU_{n \\plus{} 2} \\equal{} 8U_{n \\plus{} 1} \\minus{} 2U_n \\end{array} \\right.$\n\nDetermine the maximum possible value of n such that $ \\ U_n \\leq 2009^2$[/quote]\n\nAccording to me, the simpliest methedod is rude calculus (though not very clever)\n\nIt's immediate to show with induction that $ u_n$ is increasing\nThen :\n$ u_2 \\equal{} 76 < 4036081 \\equal{} 2009^2$\n$ u_3 \\equal{} 588 < 4036081 \\equal{} 2009^2$\n$ u_4 \\equal{} 4552 < 4036081 \\equal{} 2009^2$\n$ u_5 \\equal{} 35240 < 4036081 \\equal{} 2009^2$\n$ u_6 \\equal{} 272816 < 4036081 \\equal{} 2009^2$\n$ u_7 \\equal{} 2112048 < 4036081 \\equal{} 2009^2$\n$ u_8 \\equal{} 16350752 > 4036081 \\equal{} 2009^2$\n\nSo the required value is $ n \\equal{} 7$[/quote]\r\n\r\nThank :) :)" } { "Tag": [ "trigonometry", "function", "geometry", "trig identities", "Law of Cosines", "area of a triangle", "Heron\\u0027s formula" ], "Problem": "a triangle with sides of a,b,c in that order is a=75\r\n\r\nb=53\r\nc=47\r\n\r\n\r\nangle y is between the 75 and the 53\r\n while angle x is between 53 and 47\r\n\r\nShow your work and the formula u used\r\n\r\nfind y and x", "Solution_1": "What are you supposed to do? :huh:", "Solution_2": "[quote=\"CheeseIsGood\"]What are you supposed to do? :huh:[/quote]\r\n\r\nI think you just find the missing angles, since they are deemed variables.\r\n\r\nLaw of Cosines...?", "Solution_3": "i need to find the angles.....", "Solution_4": "[hide]The Law of Cosines is sqrt(a^2 + b^2 - 2abcosc), where a and b are the sides adjacent to an angle c in a triangle. It finds the side opposite from the angle. \nUsing the law, you can get this equation:\n75 = sqrt(47^2 + 53^2 - 2(47)(53)cosx)\nBy squaring, subtracting, and dividing, you can isolate it to cosine of X, which you can undo by using the cosecant function or cos^-1. \nUse the same method for Y. \nHope it works....[/hide]", "Solution_5": "you get rated..... im so stupid.....i didnt thought about that\r\n\r\nbut tell me the answer for x and y so i can check", "Solution_6": "can someone tell me te answeR?", "Solution_7": "Why can't you calculate it yourself?", "Solution_8": "Alternate way of getting the answer which may or may not be quicker...\r\n\r\nUse Heron's Formula to find the area, then use the formula $ \\frac{ab\\sin \\theta}{2}$, and equate it to what you got for Heron's and simply solve for $ \\theta$", "Solution_9": "[quote=\"ZzZzZzZzZzZz\"]Why can't you calculate it yourself?[/quote]\r\n\r\ni said for checking....i got the answer already :football:", "Solution_10": "Hmm. Your new name shall be Sharkman.\r\n\r\n[/spam]\r\n\r\nAoPS is not a place where you check your work. Really." } { "Tag": [], "Problem": "$ a, b, c$ real positive numbers, $ abc \\equal{} 1$. Prove that $ \\frac {a \\plus{} b}{2(a^7 \\plus{} b^7 \\plus{} c)}$ +$ \\frac {b \\plus{} c}{2(b^7 \\plus{} c^7 \\plus{} a)}$+ $ \\frac {c \\plus{} a}{2(c^7 \\plus{} a^7 \\plus{} b)}$<=1", "Solution_1": "[hide=\"An idea\"]\n\nCauchy-Schwarz: $ (a^7+b^7+c)(\\frac{1}{a^3}+\\frac{1}{b^3}+c^3) \\ge (a^2+b^2+c^2)^2$ so $ \\frac{1}{a^7+b^7+c} \\le \\frac{b^3c^3+a^3c^3+c^3}{(a^2+b^2+c^2)^2}$. Then, the problem would reduce to $ \\sum_{cyc} c^3(a+b)(a^3+b^3+1) \\le 2(a^2+b^2+c^2)^2$.[/hide]", "Solution_2": "$ a^7 \\plus{} b^7 \\ge a^2 b^2 (a^3 \\plus{} b^3 )$\r\nSo, $ \\sum\\limits_{cyc}^{a,b,c} {\\frac{{a \\plus{} b}}{{2(a^7 \\plus{} b^7 \\plus{} c)}}} \\le \\sum\\limits_{cyc}^{a,b,c} {\\frac{{a \\plus{} b}}{{2a^2 b^2 (a^3 \\plus{} b^3 \\plus{} c^3 )}}}$\r\n$ \\sum\\limits_{cyc}^{a,b,c} {\\frac{{a \\plus{} b}}{{2a^2 b^2 (a^3 \\plus{} b^3 \\plus{} c^3 )}}} \\equal{} \\frac{{(a \\plus{} b)c^2 \\plus{} (a \\plus{} c)b^2 \\plus{} (b \\plus{} c)a^2 }}{{2(a^3 \\plus{} b^3 \\plus{} c^3 )}}\\le 1$ :gathering:" } { "Tag": [ "function", "limit" ], "Problem": "Find the function f = f (x), defined and continuous on R+ = {x | 0 \u2264 x < \u221e}, that satisfies f (x+1) = f (x)+x on R+ and f (1) = 0.", "Solution_1": "From the equation $ f(x + 1) = f(x) + x$, we have that\r\n\\begin{align*}f(x) & = f(x - 1) + x - 1 \\\\\r\nf(x - 1) & = f(x - 2) + x - 2 \\\\\r\nf(x - 2) & = f(x - 3) + x - 3 \\\\\r\n\\dots \\\\\r\nf(2) & = f(1) + 1 = 1\\end{align*}\r\nSo we have\r\n\\begin{align*}f(x) & = f(x - 1) + x - 1 \\\\\r\n& = f(x - 2) + 2x - 3 \\\\\r\n& = f(x - 3) + 3x - 6 \\\\\r\n& = f(x - 4) + 4x - 10 \\\\\r\n\\dots \\\\\r\n& = f(x - (x - 1)) + (x - 1)x - \\frac {x(x - 1)}2 \\\\\r\n& = f(1) + \\frac {x(x - 1)}2 \\\\\r\n& = \\frac {x(x - 1)}2 \\end{align*}\r\nWe can check this:\r\n\\begin{align*}f(x + 1) & = \\frac {x(x - 1)}2 + x \\\\\r\n& = \\frac {x(x - 1) + 2x}2 \\\\\r\n& = \\frac {x(x + 1)}2 \\\\\r\n& = \\frac {((x + 1) - 1)(x + 1)}2, \\end{align*}\r\nAs desired.\r\n\r\n[color=darkred]Edit: Hmm, I seem to have only solved this for $ \\mathbb Z^ +$... oh well.[/color]", "Solution_2": "I think you can define an arbitrary continuous function $ g : [0,1] \\rightarrow [0,\\infty)$ with $ g(0) \\equal{} g(1) \\equal{} 0$, and extend it to an $ f$ that works. It's easy to show $ f$ is continuous inductively with the definition $ \\lim_{\\epsilon \\rightarrow 0} f(x\\plus{}\\epsilon) \\minus{} f(x) \\equal{} 0$.\r\n\r\nAlternately, let $ h(x) \\equal{} f(x) \\minus{} x(x\\minus{}1)/2$, and the given equation reduces to $ h$ having period $ 1$ (and some range oddities from $ f \\geq 0$), with no other information." } { "Tag": [ "trigonometry", "geometry", "geometric transformation", "reflection", "circumcircle", "power of a point", "radical axis" ], "Problem": "In a square $ABCD$, let $P$ and $Q$ be points on the sides $BC$ and $CD$ respectively, different from its endpoints, such that $BP=CQ$. Consider points $X$ and $Y$ such that $X\\neq Y$, in the segments $AP$ and $AQ$ respectively. Show that, for every $X$ and $Y$ chosen, there exists a triangle whose sides have lengths $BX$, $XY$ and $DY$.", "Solution_1": "Since both AP, AQ are different from the bisector AC of the right angle $\\angle DAB$, we have\r\n\r\n$BX < DX \\le DY + XY,\\ \\ DY < BY \\le BX + XY$\r\n\r\nIt remains to show that XY < BX + DY. Denote a = AB = BC = CD = DA the square side, 0 < p = BP = CQ < a, and $\\beta = \\angle PAB < 45^\\circ,\\ \\delta = \\angle QAD < 45^\\circ$.\r\n\r\n$\\tan \\beta = \\frac{BP}{AB} = \\frac p a,\\ \\ \\tan \\delta = \\frac{DQ}{DA} = \\frac{a - p}{a}$\r\n\r\n$\\tan(\\beta + \\delta) = \\frac{\\tan \\beta + \\tan \\delta }{1 - \\tan \\beta \\tan \\delta} = \\frac{\\frac p a + \\frac{a - p}{a}}{1 - \\frac{p(a - p)}{a^2}} = \\frac{1}{1 - \\frac{p(a - p)}{a^2}} > 1$\r\n\r\nConsequently, $\\beta + \\delta > 45^\\circ,\\ 2\\beta + 2\\delta > 90^\\circ.$ There are other ways to show this, for example, see the solution of problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=81900]Square[/url], but it is not worth the effort. Let B', D' be reflections of the square vertices B, D in the lines AP, AQ, respectively. The quadrilateral BD'B'D is cyclic with the circumcenter A, because the perpendicular bisectors AP, AQ, AC of BB', DD', BD meet at the vertex A. Since $\\angle BAB' = 2 \\beta,\\ \\angle DAD' = 2\\delta,$ we have $\\angle BAB' + \\angle DAD' > 90^\\circ$ and consequently, the points B, D', B', D follow on the minor arc BD of the circumcircle (A) in this order. As a result, any circle (X) centered on the segment (AP) and with the radius BX = B'X contains the point D' in its interior and any circle (Y) centered on the segment (AQ) and with radius DY = DY' contains the point B' in its interior. Thus any 2 such circles (X), (Y) intersect, which means that\r\n\r\n$XY < B'X + D'Y = BX + DY$", "Solution_2": "Let $ \\omega_1$ and $ \\omega_2$ be the circumferences centered at $ X,Y$ with radii $ XB$ and $ YD,$ respectively. It suffices to prove that $ \\omega_1$ and $ \\omega_2$ are secant. Let $ M \\equiv BQ \\cap AP$ and $ N \\equiv DP \\cap AQ.$ From $ \\triangle ABP \\cong \\triangle BCQ$ and $ \\triangle ADQ \\cong \\triangle DCP,$ it follows that $ AM \\perp BQ$ and $ AN \\perp DP$ $ \\Longrightarrow$ $ T \\equiv BQ \\cap DP$ is the orthocenter of $ \\triangle APQ.$ \n\n$ TP^2 \\minus{} TQ^2 \\equal{} AP^2 \\minus{} AQ^2 \\equal{} BP^2 \\minus{} DQ^2 \\Longrightarrow TX^2 \\minus{} XB^2 \\equal{} TY^2 \\minus{} YD^2$ \n\n$ \\Longrightarrow p(T,\\omega_1) \\equal{} p(T,\\omega_2),$\n\nwhich means that radical axis $ \\zeta$ of $ \\omega_1$ and $ \\omega_2$ passes through $ T.$ Since $ BQ > BC \\equal{} BA,$ then we have $ \\angle BAQ > \\angle AQB$ $ \\Longrightarrow$ $ \\angle DTQ > \\angle TDQ$ $ \\Longrightarrow$ $ QD > QT,$ $ YD > YT.$ Hence, $ T$ is always inside $ \\omega_2$ $\\Longrightarrow$ $ \\zeta$ cuts $ \\omega_2,$ say at $ U,V$ $ \\Longrightarrow$ $ \\omega_1$ and $ \\omega_2$ meet at $ U,V$ and proof is completed.", "Solution_3": "Here I have a super beautiful solution by the argentinian Carlos di Fiore. Enjoy:\n\nLet $ PB=QC=a $ and $ BA=CP=b $ (obviously $ BA=AD=a+b $).\nNow consider in $ \\mathbb{R}^3 $ the points $ A'=(a+b, 0, 0) $, $ P'=(0, a, 0) $ and $ Q'=(0, 0, b) $ (and $ O=(0,0,0) $).\nSince $ \\angle PBA = 90^\\circ = \\angle P'OA' $, $ BA=a+b=OA' $ and $ PB=a=P'O $ we have that $ \\triangle PBA \\equiv \\triangle P'OA' $. In the same way $ \\triangle QDA \\equiv \\triangle Q'OA' $ and $ \\triangle PCQ \\equiv Q'OP' $. From that we have $ PA=P'A' $, $ QA=Q'A' $ and $ PQ=P'Q' $ so $ \\triangle AQP \\equiv \\triangle A'Q'P' $.\nNow we consider $ X' $ and $ Y' $ in $ A'P' $ and $ A'Q' $ respectively such that $ AX=A'X' $ and $ AY=A'Y' $. Obviously $ \\triangle AXB \\equiv \\triangle A'X'O' $, $ \\triangle AYD \\equiv \\triangle A'Y'O $ and $ \\triangle AXY \\equiv \\triangle A'X'Y' $ so $ BX=OX' $, $ DY=OY' $ and $ XY=X'Y' $. But it's obvious that $ \\triangle X'OY' $ is a triangle, and has sides $ BX $, $ DY $ and $ XY $, hence the solution is complete $ \\Box $." } { "Tag": [ "blogs" ], "Problem": "[b][size=150][color=red]PA[/color][color=white]RA[/color][color=blue]GUAY[/size][/b][/color]\r\n\r\n[b][i][color=red]Tit[/color][color=white]ula[/color][color=blue]res[/color][/i][/b]\r\n1. Iv\u00e1n Hazevich\r\n2. Jes\u00fas Torres\r\n3. Rodrigo Benitez\r\n4. Mauricio Maluff\r\n\r\n[b][i][color=red]Sup[/color][color=white]len[/color][color=blue]tes[/color][/i][/b]\r\n1. Osmar Qui\u00f1\u00f3nez\r\n2. Nair Aguilera", "Solution_1": "[b][size=150][color=red]PE[/color][color=white]RU[/color][color=red] 2007[/color][/size][/b]\r\n\r\n[b]Titulares:[/b]\r\n\r\n- Angles Larico, Tom\u00e1s Miguel\r\n- Cuenca Lucero, Cesar Augusto\r\n- V\u00e9lez Lee, Luis Enrique\r\n- Rey Rodriguez, Diego Eduardo\r\n\r\n[b]Suplentes:[/b]\r\n\r\n(Primer suplente) Guerra Rios, Percy Augusto\r\n(Segundo suplente) Leiva Torres, Juan Luis\r\n\r\nSaludos,", "Solution_2": "[color=blue]Me extra\u00f1a que no se presenten los seleccionados de los dem\u00e1s pa\u00edses\u2026[/color]", "Solution_3": "[color=blue]Quiz\u00e1 por vez \u00fanica, una olimpiada del Cono Sur tiene sitio Web propio (blog, en este caso). La direcci\u00f3n es [url]http://www.conosur2007.org/[/url].\n\nLastimosamente, no aparecen ni los resultados ni los enunciados. Si alguien tiene alguno de ellos\u2026 ya sabe qu\u00e9 hacer (preferir\u00eda recibir los enunciados).\n\nSaludos a todos.[/color]", "Solution_4": "Los seleccionados chilenos son:\r\n\r\n * Jaime Soza Parra \r\n * Francisco Orlando Garrido Valenzuela \r\n * Sebastian Zu\u00f1iga Alterman \r\n * Cristobal Carvajal Hausdorf \r\n\r\nSuerte a ellos.\r\nDato obtenido de http://www.fmat.cl", "Solution_5": "[b][size=150][color=red]PE[/color][color=white]RU[/color][color=red] 2007[/color][/size][/b]\r\n\r\n[b]Titulares:[/b]\r\n\r\n- Angles Larico, Tom\u00e1s Miguel -> ORO 51 PUNTOS.\r\n- Cuenca Lucero, Cesar Augusto -> ORO 53 PUNTOS.\r\n- V\u00e9lez Lee, Luis Enrique -> ORO 50 PUNTOS.\r\n- Rey Rodriguez, Diego Eduardo -> ORO 50 PUNTOS.\r\n\r\nSaludos.", "Solution_6": "Ya pueden ver en mi blog\r\n[url]http://selectivos-peru.blogspot.com[/url]\r\nlas dos pruebas tomadas, y pr\u00f3ximamente con soluciones :D \r\n\r\n$Tipe$", "Solution_7": "se encuentran en algun lado los resultados? \r\nsupe que la delegacion Chilena ha salido quinta, pero no veo los resultados en ninguna parte.", "Solution_8": "[quote=\"tipe\"]Ya pueden ver en mi blog\n[url]http://selectivos-peru.blogspot.com[/url]\n[/quote]\r\n\r\nAh\u00ed tambi\u00e9n puse los resultados por pa\u00edses.\r\n\r\n$Tipe$", "Solution_9": "[quote=\"tipe\"][quote=\"tipe\"]Ya pueden ver en mi blog\n[url]http://selectivos-peru.blogspot.com[/url]\n[/quote]\n\nAh\u00ed tambi\u00e9n puse los resultados por pa\u00edses.\n\n$Tipe$[/quote]\r\n\r\nS\u00ed, gracias, de ah\u00ed obtuve la informaci\u00f3n, soloq ue me gustaria saber a nivel individual.\r\n\r\nSaludos" } { "Tag": [ "function" ], "Problem": "Give an explicit bijection between $ N$, the set of natural numbers, and $ N^{2}$.", "Solution_1": "does the word explicit have any significance?", "Solution_2": "I guess it means give your function explicitly in mathematical notation.", "Solution_3": "It means that, given any integer $ n$, I should be able to compute from your answer the pair $ (p, q)$ that it corresponds to, and vice-versa.\r\n\r\n\r\nHere's an example of a proof that those two are of the same cardinality without an explicit bijection:\r\n\r\nGiven an integer $ n \\in \\bf N$, assign it the ordered pair $ (n, 0)$. This is an injection from $ \\bf N$ to $ {\\bf N} \\times {\\bf N}$.\r\nGiven an ordered pair $ (p, q)$, assign it the integer $ 2^p 3^q$. This is an injection from $ {\\bf N} \\times {\\bf N}$ to $ \\bf N$.\r\nWhenever we have an injection from $ A$ to $ B$ and an injection from $ B$ to $ A$, there exists a bijection between $ A$ and $ B$. (I can't remember the name of this theorem.) Thus $ \\bf N$ and $ {\\bf N} \\times {\\bf N}$ are in bijection, as desired.", "Solution_4": "I'm actually having a hard time coming up with a non-constructive proof. Does anyone have a proof that doesn't actually hand you a bijection?", "Solution_5": "Heh, I was adding one in while you posted. What's the name of that theorem?", "Solution_6": "Cantor-Bernstein/Schroder Bernstein", "Solution_7": "But the only proof I know of CBS is constructive - given two injections, it explicitly constructs the corresponding bijection. Is the standard proof non-constructive?", "Solution_8": "Well, okay, I would say that invoking CBS gives a non-constructive proof but that any such proof can be \"constructivised\" by replacing CBS with the proof of CBS. This is obviously just a matter of personal taste, though. Does the proof of CBS actually work \"in practice,\" i.e., can you write down the bijection that results if you start with the two injections that I listed?" } { "Tag": [ "algebra", "polynomial", "induction", "complex numbers", "algebra solved" ], "Problem": "Let $\\, P(z) \\,$ be a polynomial with complex coefficients which is of degree $\\, 1992 \\,$ and has distinct zeros. Prove that there exist complex numbers $\\, a_1, a_2, \\ldots, a_{1992} \\,$ such that $\\, P(z) \\,$ divides the polynomial \\[ \\left( \\cdots \\left( (z-a_1)^2 - a_2 \\right)^2 \\cdots - a_{1991} \\right)^2 - a_{1992}. \\]", "Solution_1": "I was asked for a solution of this, so let me post one.\r\n\r\nSince the zeros $z_{1}$, $z_{2}$, ..., $z_{1992}$ of the polynomial P(z) are distinct, the polynomial P(z) divides the polynomial $Q\\left(z\\right) = \\left( ... \\left( \\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}...-a_{1991}\\right)^{2}-a_{1992}$ if and only if $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;...;\\;1992\\right\\}$. So it is enough to show that for any 1992 complex numbers $z_{1}$, $z_{2}$, ..., $z_{1992}$, there exist 1992 complex numbers $a_{1}$, $a_{2}$, ..., $a_{1992}$, such that the polynomial $Q\\left(z\\right) = \\left( ... \\left( \\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}...-a_{1991}\\right)^{2}-a_{1992}$ satisfies $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;...;\\;1992\\right\\}$. This can be generalized:\r\n\r\n[color=blue][b]Theorem.[/b] Let n be a positive integer, and $z_{1}$, $z_{2}$, ..., $z_{n}$ be n complex numbers. Then, there exist n complex numbers $a_{1}$, $a_{2}$, ..., $a_{n}$ such that the polynomial $Q\\left(z\\right) = \\left( ... \\left( \\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}...-a_{n-1}\\right)^{2}-a_{n}$ satisfies $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;...;\\;n\\right\\}$.[/color]\r\n\r\n[i]Proof.[/i] We use induction over n.\r\n\r\nFor n = 1, the theorem is trivial, since $Q\\left(z\\right)=z-a_{1}$, so we can take $a_{1}=z_{1}$, and then the polynomial $Q\\left(z\\right)=z-z_{1}$ clearly satisfies $Q\\left(z_{1}\\right)=0$.\r\n\r\nThe interesting part is the induction step: Let k be a positive integer. Assume that for n = k, the theorem is proven, i. e. for any k complex numbers $z_{1}$, $z_{2}$, ..., $z_{k}$, there exist k complex numbers $a_{1}$, $a_{2}$, ..., $a_{k}$ such that the polynomial $Q\\left(z\\right)=\\left(...\\left( \\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}...-a_{k-1}\\right)^{2}-a_{k}$ satisfies $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;...;\\;k\\right\\}$. We have to prove the theorem for n = k + 1, so we have to prove that for any k + 1 complex numbers $z_{1}$, $z_{2}$, ..., $z_{k+1}$, there exist k + 1 complex numbers $b_{1}$, $b_{2}$, ..., $b_{k+1}$ such that the polynomial $R\\left(z\\right)=\\left(...\\left( \\left(z-b_{1}\\right)^{2}-b_{2}\\right)^{2}...-b_{k}\\right)^{2}-b_{k+1}$ satisfies $R\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;...;\\;k+1\\right\\}$.\r\n\r\nIn fact, after our assumption, there exist k complex numbers $a_{1}$, $a_{2}$, ..., $a_{k}$ such that the polynomial $Q\\left(z\\right)=\\left(...\\left( \\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}...-a_{k-1}\\right)^{2}-a_{k}$ satisfies $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;...;\\;k\\right\\}$. We want to construct our polynomial R from this polynomial Q.\r\n\r\nIn fact, let $b_{i}=a_{i}$ for every $i\\leq k-1$, then let $b_{k}=a_{k}+\\frac12Q\\left(z_{k+1}\\right)$ and $b_{k+1}=\\frac14\\left(Q\\left(z_{k+1}\\right)\\right)^{2}$. Then,\r\n\r\n$R\\left(z\\right)=\\left(...\\left( \\left(\\left(z-b_{1}\\right)^{2}-b_{2}\\right)^{2}...-b_{k-1}\\right)^{2}-b_{k}\\right)^{2}-b_{k+1}$\r\n$=\\left(...\\left( \\left(\\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}...-a_{k-1}\\right)^{2}-\\left(a_{k}+\\frac12Q\\left(z_{k+1}\\right)\\right) \\right)^{2}-\\frac14\\left(Q\\left(z_{k+1}\\right)\\right)^{2}$\r\n$=\\left(\\left(...\\left( \\left(\\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}...-a_{k-1}\\right)^{2}-a_{k}\\right)-\\frac12Q\\left(z_{k+1}\\right)\\right)^{2}-\\frac14\\left(Q\\left(z_{k+1}\\right)\\right)^{2}$\r\n$=\\left(Q\\left(z\\right)-\\frac12Q\\left(z_{k+1}\\right)\\right)^{2}-\\frac14\\left(Q\\left(z_{k+1}\\right)\\right)^{2}=Q\\left(z\\right)\\left(Q\\left(z\\right)-Q\\left(z_{k+1}\\right)\\right)$.\r\n\r\nHence, for $i\\in\\left\\{1;\\;2;\\;...;\\;k\\right\\}$, we have $R\\left(z_{i}\\right)=Q\\left(z_{i}\\right)\\left(Q\\left(z_{i}\\right)-Q\\left(z_{k+1}\\right)\\right)=0$ because $Q\\left(z_{i}\\right)=0$, and $R\\left(z_{k+1}\\right)=Q\\left(z_{k+1}\\right)\\left(Q\\left(z_{k+1}\\right)-Q\\left(z_{k+1}\\right)\\right)=0$. Thus, $R\\left(z_{i}\\right)=0$ holds for every $i\\in\\left\\{1;\\;2;\\;...;\\;k+1\\right\\}$, and this proves the theorem for n = k + 1. Hence, the induction step is complete, and the Theorem is proven.\r\n\r\n Darij", "Solution_2": "Here's a different approach:\r\n\r\nLemma: Consider the equation $ p((f\\minus{}a)^{2}) \\equal{} 0$, where $ p$ is a polynomial, $ a$ is a constant, and $ f$ is a variable. Let's say we can choose the polynomial $ p$ to have any set of $ k\\minus{}1$ distinct roots for an integer $ k\\ge 2$ and that we can choose $ a$ to be any complex number. Then for any set of $ k$ distinct complex numbers $ f_{1}, ..., f_{k}$, we can choose a value of $ a$ and a set of $ k\\minus{}1$ distinct roots for $ p$ such that $ p((f\\minus{}a)^{2}) \\equal{} 0$, when viewed as a polynomial in $ f$, has $ f_{1},...,f_{k}$ as roots.\r\n\r\nChoose $ a \\equal{}\\frac{f_{1}\\plus{}f_{k}}{2}$ and let $ p$ have roots $ (f_{1}\\minus{}a)^{2}, ... , (f_{k\\minus{}1}\\minus{}a)^{2}$. Then $ p((f\\minus{}a)^{2})$ is clearly $ 0$ for $ f \\equal{} f_{1},...,f_{k\\minus{}1}$. For $ f_{k}$, we have $ p((f_{k}\\minus{}a)^{2}) \\equal{} p((\\minus{}(f_{1}\\minus{}a))^{2}) \\equal{} p((f_{1}\\minus{}a)^{2}) \\equal{} 0$, so $ f$ has all of the possible values $ f_{1},...,f_{k}$.\r\n[hide=\"Remark\"]\nGeometric intuition is key here. The idea is that if we can choose any $ k\\minus{}1$ possible values for $ (f\\minus{}a)^{2}$, that's the same as choosing $ 2k\\minus{}2$ complex numbers for $ (f\\minus{}a)$ that come in pairs such that the midpoint of each pair is the origin. Based on this, we can get any set of $ k\\minus{}1$ possible values of $ f$ by choosing each of the $ k\\minus{}1$ pairs to have a desired root. This, however, completely determines the other number in each pair. We can then choose $ a$ to move the midpoint of each pair from $ 0$ to $ a$. If we choose $ a$ to be at the midpoint of two of the desired possible values of $ f$, one of the $ k\\minus{}1$ pairs can produce two possible values of $ f$, leaving the other $ k\\minus{}2$ pairs to produce the other desired $ k\\minus{}2$ values for $ f$.\n[/hide]\r\n\r\n\r\nWe now use induction to show that $ (...((z\\minus{}a_{1})^{2}\\minus{}a_{2})^{2}...\\minus{}a_{n\\minus{}1})^{2}\\minus{}a_{n}\\equal{} 0$ can have any given $ n$ distinct roots for $ z$ if we appropriately choose $ a_{1},...,a_{n}$.\r\n\r\nFor $ n \\equal{} 1$ we choose $ a_{1}$ to be the one distinct root since our equation is $ z\\minus{}a_{1}\\equal{} 0$.\r\n\r\nThen let's assume it's true for $ n\\minus{}1$. Consider $ (...((z\\minus{}a_{1})^{2}\\minus{}a_{2})^{2}...\\minus{}a_{n\\minus{}1})^{2}\\minus{}a_{n}\\equal{} 0$. By the induction hypothesis we can get any $ n\\minus{}1$ distinct possible values for $ (z\\minus{}a_{1})^{2}$ with an appropriate choice of $ a_{2},...,a_{n}$. By the lemma, we can thus choose $ a_{1}$ so that we have the desired $ n$ possible values of $ z$, and the induction is complete.\r\n\r\nIf we replace $ n$ with $ 1992$, we see that we can achieve any desired set of $ 1992$ roots in $ z$, and if we choose those to be the roots of $ P(z)$, it follows that $ P(z)$ must divide this polynomial.\r\n\r\nQED", "Solution_3": "What was Darij's motivation behind letting:\n$b_k=a_k+1/2Q(a_k+1)$", "Solution_4": "Let $z_1, z_2, \\cdots$ be the roots of $P$. We prove the case for general $n$ via induction. The base case is trivial; take $a_1 = z_1$.\n\nNow suppose that there exist constants $b_1, b_2, \\cdots, b_n$ such that setting $a_i = b_i$ for each $i$ creates a valid polynomial. Then, this means that$$(\\dotsb ((z_i - b_1)^2 - b_2)^2 \\dots - b_{n-1})^2 = k$$for some constant $k$ for each $1 \\leq i \\leq n$. Furthermore, let the value of this expression for $z_{n+1}$ be $r$. Then, we need to pick $b_n, b_{n+1}$ such that$$(k-b_n)^2 = (r - b_n)^2 = b_{n+1}.$$As the roots of the quadratic$$(x-a)^2 = b$$can be any pair of complex numbers by choosing suitable $a$ and $b$, such $b_n$ and $b_{n+1}$ clearly exist, as required." } { "Tag": [ "function", "trigonometry", "geometry", "geometric transformation", "reflection", "analytic geometry", "graphing lines" ], "Problem": "Hello everyone,\r\n\r\nI can express $ \\tan (90 \\minus{} x)$ as $ \\frac {1}{\\tan x}$ using algebra, but how would I figure out $ \\tan (90 \\minus{} x)$ out using the unit circle and comparing the two points to the tangent line?\r\n\r\nThank you. My diagram is below (Grey lines mark the x and y axis):\r\n\r\n---\r\n\r\n[img]http://img178.imageshack.us/img178/3271/tan90xfy2.jpg[/img]", "Solution_1": "The line with angle $ 90^{\\circ} \\minus{} x$ is the reflection of the line with angle $ x$ across $ y \\equal{} x$ (the line with angle $ 45^{\\circ}$). Another way to phrase this reflection is the transformation $ (x, y) \\mapsto (y, x)$. What are the slopes of these lines?", "Solution_2": "Alternatively, you can use the properties of triangles. You know that the two right triangles (formed by connecting each of the points on your diagram to the origin and dropping a perpendicular to the x-axis) are congruent because they have the same hypotenuse (the radius of the circle) and angles (they each have angles of measure $ 90$, $ x$, and $ 90 \\minus{} x$). From there, you can compute the tangent of each in terms of length and height.", "Solution_3": "Thank you very much for your replies. I understand the concept now." } { "Tag": [ "LaTeX", "inequalities solved", "inequalities" ], "Problem": "If $x_{k}, y_{k} \\in R, k=1,n$ then we have:\r\n\r\n$4( \\sum_{k=1}^{n} x_{k}^2)( \\sum_{k=1}^{n} y_{k}^2) + 4 ( \\sum_{k=1}^{n} x_{k}y_{k})( \\sum_{k=1}^{n} x_{k}) + 4n( \\sum_{k=1}^{n} x_{k}^2) \\\\ > 4 ( \\sum_{k=1}^{n} x_{k}y_{k})^{2} +( \\sum_{k=1}^{n} x_{k})^{2}+4( \\sum_{k=1}^{n} x_{k}^2)( \\sum_{k=1}^{n} y_{k}) $\r\n\r\ncheers! :D :D", "Solution_1": "there was a mistake here and I have fixed it, try this ineq now! :D \r\n\r\ncheers! :D :D :cool:", "Solution_2": "just a small latex tip: you need not use $ x_k * y_k $ now, just use $x_ky_k$ or $x_k \\cdot y_k$ the central dot's command is \\cdot." } { "Tag": [ "pigeonhole principle" ], "Problem": "Suppose a musical group has 11 weeks to prepare for opening night, and they intend to have at least one rehearsal each day. However, they decide not to schedule more than 12 rehearsals in any 7-day period, to keep from getting burned out. Prove that there exists a sequence of successive days during which the band has exactly 21 rehearsals.", "Solution_1": "[quote=\"mdk\"]Suppose a musical group has 11 weeks to prepare for opening night, and they intend to have at least one rehearsal each day. However, they decide not to schedule more than 12 rehearsals in any 7-day period, to keep from getting burned out. Prove that there exists a sequence of successive days during which the band has exactly 21 rehearsals.[/quote]\r\n\r\n[hide=\"Solution\"]\nLet $ a_{i}$ mean the number of total rehersals upto the $ i$-th day.\n\nConsider the numbers,\n$ a_{1},a_{2},...,a_{77},a_{1}+21,a_{2}+21,...,a_{77}+21$ ... $ (*)$\n\nNow the most number of rehershals possible is $ 11\\cdot 12 = 132$.\nSo, the numbers $ a_{i}$ and $ a_{i}+21$ are among $ \\{1,2,...,143\\}$.\n\nBy pigeonhole (*) has at least two elements being equal. Now it cant be $ a_{i}=a_{j}$ for $ i\\not = j$ because they place [b]at least one[/b] rehershal each day. And likewise it cant be $ a_{i}+21 = a_{j}+21$ for $ i\\not =j$. So it must mean that $ a_{i}= a_{j}+21$. And so $ a_{i}-a_{j}= 21$. Which means from $ j+1,..,i$ the band played exactly $ 21$ games.\n[/hide]" } { "Tag": [], "Problem": "Un n\u00famero $p$ se dice perfecto si la suma de sus divisores, exceptuando al propio $p$, da como resultado $p$. Sea $f$ una funci\u00f3n tal que:\r\n\r\n$f(n) = 0$ si $n$ es perfecto\r\n$f(n) = 0$ si la cifra de las unidades de $n$ es $4$\r\n$f(a.b) = f(a) + f(b)$\r\n\r\nCalcular $f(1988)$\r\n\r\nLo he resuelto pero no me sirve el primer dato, lo cual me hace pensar q me he equivocado :blush:", "Solution_1": "$f: \\mathbb{N} \\to \\mathbb{N}$ ? :)", "Solution_2": "[quote=\"mathmanman\"]$f: \\mathbb{N} \\to \\mathbb{N}$ ? :)[/quote]\r\n\r\nsi", "Solution_3": "Ok, que piensas de eso? :\r\n$f(1988)+f(8) = f(15904) \\Longrightarrow f(1988)=0$\r\n\r\n\r\nO, por ejemplo : \r\n$f(2006)+f(4) = f(8024) \\Longrightarrow f(2006)=0$\r\n\r\n:)\r\n\r\n\r\nPienso que podriamos modificar la tercera condicion :\r\n$f(a)+f(b)=f(a \\cdot b)$. $\\forall a$, podemos hallar $b$ tal que el cifro de las unidades de $a \\cdot b$ sea 4. Entonces $f(a)=0\\ \\ \\forall a$.", "Solution_4": "correccion:\r\n[quote=\"mathmanman\"]\n... podemos hallar $b$ tal que el cifro de las unidades de $a \\cdot b$ sea 4... [/quote]\n[quote=\"yo\"]\n... podemos hallar $b$ tal que [b]la cifra[/b] de las unidades de $a \\cdot b$ sea 4... [/quote]\r\n----------\r\nrespecto a la idea.........\r\nesto ya esta dada en la condicion...\r\n\r\nCarlos Bravo ;)\r\nLima - PERU", "Solution_5": "Ok, gracias [b]carlosbr[/b] por la correccion! :D \r\n\r\n\r\nIt's cool that you correct me so I can improve my poor spanish. ;)", "Solution_6": "Entonces mi respuesta estaba bien.:lol:", "Solution_7": "S\u00ed, el resultado es por supuesto 0, pero es un poco raro que se haga f\u00e1cilmente el problema sin hacer referencia a n\u00fameros perfectos.", "Solution_8": "[quote=\"mathmanman\"]Ok, que piensas de eso? :\n$ f(1988) \\plus{} f(8) \\equal{} f(15904) \\Longrightarrow f(1988) \\equal{} 0$\n\n\nO, por ejemplo : \n$ f(2006) \\plus{} f(4) \\equal{} f(8024) \\Longrightarrow f(2006) \\equal{} 0$\n\n:)\n\n\nPienso que podriamos modificar la tercera condicion :\n$ f(a) \\plus{} f(b) \\equal{} f(a \\cdot b)$. $ \\forall a$, podemos hallar $ b$ tal que el cifro de las unidades de $ a \\cdot b$ sea 4. Entonces $ f(a) \\equal{} 0\\ \\ \\forall a$.[/quote]\r\nhay un peque\u00f1o error... toca calcular f(8) para poder hallar f(1988) de esa forma.\r\nSin embargo esto es muy facil:\r\nf(8)=f(4)+f(2)\r\n4 termina en 4, 2 es perfecto, entonces f(8)=0, entonces f(1988)=0\r\nah\u00ed se usa la primera condici\u00f3n :)\r\n\r\nEn cuanto a el problema con la condici\u00f3n modificada: \r\nf(k)=f(k)+f(2)=f(2k)\r\nf(2k)=f(2k)+f(2)=f(4k)\r\n.\r\n.\r\n.\r\n$ f(k) \\equal{} f(2^n k)$\r\neventualmente, para algun n, $ f(2^n k)$ va a terminar en 4, a menos que 5 divida a n; esto es f\u00e1cil de comprobar por tanteo.", "Solution_9": "$ 2$ es perfecto?\r\nPero $ \\sigma(2)\\equal{}1\\plus{}2\\equal{}3$.", "Solution_10": ":fool:\r\n\r\nf(1988)+f(28)=f(1988\u00b728) \r\n1988\u00b728 termina en 4 y 28 es perfecto, entonces f(1988)=0 :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ a,b,c$ are positive numbers, then\r\n\r\n$ \\frac a{\\sqrt{2a+b}}+\\frac b{\\sqrt{2b+c}}+\\frac c{\\sqrt{2c+a}}\\le \\sqrt{a+b+c}\\le \\frac a{\\sqrt{a+2b}}+\\frac b{\\sqrt{b+2c}}+\\frac c{\\sqrt{c+2a}}$.", "Solution_1": "Dear VASC! The right part is an exercises of secrets in Inequality. We use Holder to prove it. The left part follows Cauchy and the famous one\r\n\\[ \\sum \\frac{a}{2a+b}\\le 1. \\]", "Solution_2": "[quote=\"hungkhtn\"]The right part is an exercises of secrets in Inequality. We use Holder to prove it. The left part follows by Cauchy and the famous one\n\\[ \\sum \\frac{a}{2a+b}\\le 1. \\]\n[/quote]\r\nIndeed, these are the solutions. The left inequality can be generalized as follows:\r\n\r\nIf $ a_{1},a_{2},...,a_{n}$ are positive numbers, then\r\n\r\n\\[ \\frac{a_{1}}{\\sqrt{(n-1)a_{1}+a_{2}}}+\\frac{a_{2}}{\\sqrt{(n-1)a_{2}+a_{3}}}+\\frac{a_{n}}{\\sqrt{(n-1)a_{n}+a_{1}}}\\le \\sqrt{a_{1}+a_{2}+...+a_{n}}. \\]", "Solution_3": "[quote=\"Vasc\"]If $ a,b,c$ are positive numbers, then\n\n$ \\frac a{\\sqrt{2a+b}}+\\frac b{\\sqrt{2b+c}}+\\frac c{\\sqrt{2c+a}}\\le \\sqrt{a+b+c}$.[/quote]\r\nThis one for $ a,b,c>0$ and $ k\\ge 2$ holds too:\r\n\\[ \\frac a{\\sqrt{ka+b}}+\\frac b{\\sqrt{kb+c}}+\\frac c{\\sqrt{kc+a}}\\le \\sqrt{\\frac{3(a+b+c)}{k+1}}\\]" } { "Tag": [ "function", "logarithms", "calculus", "derivative", "inequalities", "calculus computations" ], "Problem": "Is (1+x/n)^n an increasing function of n for x real and n>0. If not, what if we restrict n>1.", "Solution_1": "Take logarithm and differentiate it wrt n. The derivative is $ \\log(1\\plus{}x/n)\\minus{}\\frac{x/n}{1\\plus{}x/n}$. So we need the inequality $ \\log(1\\plus{}t)\\ge \\frac{t}{1\\plus{}t}$, $ t\\ge 0$. This runs in the opposite direction to the familiar $ \\log(1\\plus{}t)\\le t$, but it's easy to turn a logarithm around: $ \\log(1\\plus{}t)\\equal{}\\minus{}\\log \\frac{1}{1\\plus{}t}\\equal{}\\minus{}\\log\\left(1\\minus{}\\frac{t}{1\\plus{}t}\\right)\\ge \\frac{t}{1\\plus{}t}$." } { "Tag": [], "Problem": "Solve: $(x^{3}+1)+(x^{2}+1)+3x\\sqrt{x+1}>0$", "Solution_1": "Condition: $x\\ge-1$\r\nPut $t=x\\sqrt{x+1}$ then the inequation becomes:\r\n$t^{2}+3t+2>0\\Leftrightarrow t<-2$ or $t>-1$\r\nFrom that we can easy to continue with variable $x$" } { "Tag": [ "geometry", "perimeter", "ratio", "LaTeX" ], "Problem": "[b]1.[/b] An equilateral regions of side length $ 2cm$ is cut from a corner of an equilateral triangular region with side length $ 6cm$. What is the number of centimeters in the perimeter of the resulting solid? \r\n\r\n\r\n[b]2.[/b] In the figure below, the smaller circle has the radius of two feet and the larger circle has the radius of four feet. What is the total area of the four shaded regions? Express your answer in terms of[i] pi[/i]. \r\n\r\n[b]3. [/b]Joe purchases a mirror that fits exactly in the frame. The outer perimeter of the frame measures $ 60cm$ by $ 80cm$. The width of the frame measures 10cm. What is the area of the mirror?\r\n\r\n[b]4.[/b] In square $ ABCD$, point $ M$ is the midpoint of the side $ AB$ and point $ N$ is the midpoint of side $ BC$. What is the ratio of the area of the triangle $ AMN$ to the are of the square $ ABCD$? Express your answer as a common fraction.", "Solution_1": "[hide=\"Question 1\"]The resulting polygon is a regular hexagon with sides two, because each side is subtracted 4. Perimeter is, thus, [b]12[/b].[/hide]\n\nWhere is the picture for 2?\n\n[hide=\"Question 3\"]If the frame is 10 cm thick, then 20 is subtracted from the length and height which would be 40cmx60cm mirror. The are of the mirror is [b]2400cm^2[/b].[/hide]\n\n[hide=\"Question 4\"]I do not how to explain this, and I wish I have LaTeX, but AMN/ABCD = [b]1/8[/b]. I kind of need a diagram to explain it, so sorry.[/hide]", "Solution_2": "[quote=\"garfielddisco123\"][hide=\"Question 1\"]The resulting polygon is a regular hexagon with sides two, because each side is subtracted 4. Perimeter is, thus, [b]12[/b].[/hide]\n\nWhere is the picture for 2?\n\n[hide=\"Question 3\"]If the frame is 10 cm thick, then 20 is subtracted from the length and height which would be 40cmx60cm mirror. The are of the mirror is [b]2400cm^2[/b].[/hide]\n\n[hide=\"Question 4\"]I do not how to explain this, and I wish I have LaTeX, but AMN/ABCD = [b]1/8[/b]. I kind of need a diagram to explain it, so sorry.[/hide][/quote]\r\n\r\nOh, my bad. Here, omit #3, will you? \r\n\r\nSorry! :blush:", "Solution_3": "Well I can't take it out because I can't edit my post anymore so... yeah." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "[b]Suppose that $n$ is a given integer ( $n\\ge 2$ ). When the number $a^{n}+b^{n}$ may be written as a product of two distinct integers $\\ge 2$ ?[/b]\r\n [color=green][b]It's possible to find all such pairs $(a,b), a\\ne b ,$ with $a^{n}+b^{n}=cd$ where $c,d$ , $c\\ne d$ , are in $\\{2,3,...\\}$ ?[/b][/color]\r\nFor instance, in case $n=2$ ; \\[\\begin{array}{lcl}2^{2}+5^{2}&=& 29 \\\\ 3^{2}+5^{2}&=& 2\\times 17 \\\\ 4^{2}+5^{2}&=& 41\\\\ \\vdots &\\vdots & \\vdots \\\\ \\end{array}\\]", "Solution_1": "If n had odd divisor d (n=md) then $a^{n}+b^{n}=(a^{m}+b^{m})(a^{m(d-1)}-a^{m(d-2)}b^{m}+...+b^{(d-1)m}$.", "Solution_2": "If $n$ has an odd prime divisor we can use $a^{qm}+b^{qm}=(a^{m}+b^{m})(a^{(m-1)q}-a^{(m-2)q}b+-...+b^{(m-1)q})$ for odd $q$.\r\n\r\nFor the case of $n=2^{k}$, we can reduce it to $k=1$.\r\nIn this case, exactly the primes $\\equiv 1 \\mod 4$ are possible to be written as sum of two squares, so this are all cases.", "Solution_3": "If $n=2^{k}, k\\ge 2$ $a^{n}+b^{n}=(a,b)^{n}D,D=a_{1}^{n}+b_{1}^{n},a_{1}=a/(a,b),b_{1}=b/(a,b)$. If p is odd prime divisors D, then $p=1(mod \\ 2^{k+2}).$" } { "Tag": [], "Problem": "Each successive term of the sequence 1, 3, 7, 15, 31, ... is obtained\nby adding one to twice the previous term. The 10th term is 1023.\nWhat is the value of the 9th term?", "Solution_1": "If the $ \\ 10th$ term is $ \\ 1023$, then the $ \\ 9th$ term must be $ \\ 1023 \\minus{} 1 \\equal{} 2x$\r\n$ \\ x \\equal{} \\boxed{511}$", "Solution_2": "Note that the $ n^\\text{th}$ term is $ 2^n\\minus{}1$, so the $ 9^\\text{th}$ term is $ 2^9\\minus{}1\\equal{}\\boxed{511}$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let $ a$ , $ b$ , $ c$ , $ d$ $ \\in$ $ {\\mathbb R}^{\\plus{}}$ , prove that :\r\n\r\n$ (a\\plus{}b\\plus{}c\\plus{}d)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\plus{}\\frac{1}{d})$ $ \\geq$ $ \\frac{a\\plus{}b}{c\\plus{}d}\\plus{}\\frac{b\\plus{}c}{d\\plus{}a}\\plus{}\\frac{c\\plus{}d}{a\\plus{}b}\\plus{}\\frac{d\\plus{}a}{b\\plus{}c}$", "Solution_1": "hi , if we expand the first member i think it becomes easy to see :\r\nFirst member = 4 + a/b + a/c + a/d + b/a + b/c + b/d + c/a + c/b + c/d + d/a + d/b + d/c\r\nbut for any x,y,z and t > 0 :\r\nx/z + y/t >= x+y / z+t (difference positive)\r\nhence :\r\na/c + b/d >= a+b / c+d\r\nb/a + c/d >= b+c / a+d\r\nc/a + d/b >= c+d / a+b\r\nand\r\nd/c + a/b >= d+a / b+c\r\nthe result follows if no mistake.", "Solution_2": "thank's , but i think that you prof is wrong", "Solution_3": "[quote=\"y-a-s-s-i-n-e\"]thank's , but i think that you prof is wrong[/quote]\r\ni dont think so yassin!; why you think it wrong?", "Solution_4": "When are LHS and RHS equal?\r\nI don't know.", "Solution_5": "This is a trivial and not a good inequality.\r\nYou have in $ RHS$ $ 16$ terms of degree $ 0$ and in $ LHS$ only $ 4$ terms of degree $ 0$.As Huynsoo Park saw, we have no equality case.If we equal terms on both sides, the ineq. should be:\r\n$ (a\\plus{}b\\plus{}c\\plus{}d)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\plus{}\\frac{1}{d})\\geq4(\\frac{a\\plus{}b}{c\\plus{}d}\\plus{}\\frac{b\\plus{}c}{d\\plus{}a}\\plus{}\\frac{c\\plus{}d}{a\\plus{}b}\\plus{}\\frac{d\\plus{}a}{b\\plus{}c}).$\r\nBut this is not true.Try $ a\\equal{}b\\equal{}c\\equal{}1, d\\equal{}5$.\r\nSo,y-a-s-s-i-n-e, sorry,but you have to improve your inequality. :(", "Solution_6": "why it isn't true with $ a \\equal{} b \\equal{} c \\equal{} 1$ and $ d \\equal{} 5$ you find after the calculation $ 19.2 \\geq 6.666666667$\r\n\r\ntry to start from $ \\frac {a \\plus{} b}{c \\plus{} d} \\plus{} \\frac {b \\plus{} c}{d \\plus{} a} \\plus{} \\frac {c \\plus{} d}{a \\plus{} b} \\plus{} \\frac {d \\plus{} a}{b \\plus{} c}$\r\n\r\nhint\r\nprove that \r\n$ \\frac {a \\plus{} b \\plus{} c \\plus{} d}{a} \\geq \\frac {a \\plus{} b}{c \\plus{} d}$", "Solution_7": "[quote=\"y-a-s-s-i-n-e\"]\nhint\nprove that \n$ \\frac {a \\plus{} b \\plus{} c \\plus{} d}{a} \\geq \\frac {a \\plus{} b}{c \\plus{} d}$[/quote]\r\nI think, your hint is confusing. \r\nTry $ c \\equal{} d\\rightarrow0^ \\plus{} .$ :wink:\r\nBy the way, the kprepaf's proof is true and nice.", "Solution_8": "i didn't understand any thing in kperpaf's proof can you explaine please ?\r\n\r\nis my inequality true ?", "Solution_9": "[quote=\"y-a-s-s-i-n-e\"]i didn't understand any thing in kperpaf's proof can you explaine please ?\n\nis my inequality true ?[/quote]\r\nYour inequality is true and nice.\r\nMaybe kprepaf meant the following reasoning: \r\n$ \\frac{x}{z} \\plus{}\\frac{ y}{t}\\geq\\frac{ x\\plus{}y}{ z\\plus{}t}$ is true for positive $ x,$ $ y,$ $ z$ and $ t$ since,\r\n$ \\frac{x}{z} \\plus{}\\frac{ y}{t}\\equal{}\\frac{x^2}{xz}\\plus{}\\frac{y^2}{yt}\\geq\\frac{(x\\plus{}y)^2}{xz\\plus{}yt}\\geq\\frac{ x\\plus{}y}{ z\\plus{}t}.$ :wink:", "Solution_10": "thank's arqady \r\n sorry but if we assume we have : \r\n\r\n$ \\frac{2a}{c}\\plus{}\\frac{2c}{a}\\plus{}\\frac{2b}{d}\\plus{}\\frac{2d}{b}$ $ \\geq$ $ \\frac {a \\plus{} b}{c \\plus{} d} \\plus{} \\frac {b \\plus{} c}{d \\plus{} a} \\plus{} \\frac {c \\plus{} d}{a \\plus{} b} \\plus{} \\frac {d \\plus{} a}{b \\plus{} c}$\r\n\r\nplease explaine :?:", "Solution_11": "[quote=\"y-a-s-s-i-n-e\"]thank's arqady \n sorry but if we assume we have : \n\n$ \\frac {2a}{c} \\plus{} \\frac {2c}{a} \\plus{} \\frac {2b}{d} \\plus{} \\frac {2d}{b}$ $ \\geq$ $ \\frac {a \\plus{} b}{c \\plus{} d} \\plus{} \\frac {b \\plus{} c}{d \\plus{} a} \\plus{} \\frac {c \\plus{} d}{a \\plus{} b} \\plus{} \\frac {d \\plus{} a}{b \\plus{} c}$\n\nplease explaine :?:[/quote]\r\nBut kprepaf explained it already.", "Solution_12": "yes \r\n\r\ni did a little mistake when i assume \r\n\r\nsorry kprepaf :ush: thank's a lot you proof is very nice :D" } { "Tag": [ "probability" ], "Problem": "A bag have 9 chips, 3 tan, 2 pink and 4 violet, if drawn randomly with out replacement, what is the probability in such way that 3 tan chips are consecutively, 2 pink consecutively and 4 violet consecutively, but not necessary in the tan-pink-violet order?", "Solution_1": "Find the normal probability without replacement then multiply by 3! for the 6 different order of the colors.\r\n\r\n3!(3/9)(2/8)(1/7)(2/6)(1/5)(4/4)(3/3)(2/2)(1/1)\r\n\r\nYou can work it out to get the final answer.", "Solution_2": "thank you!", "Solution_3": "Why some time ask 8 red marble, 5 blue marble, the probability of getting 3 red marble should like this:\r\n\r\n[(8*7*6)/3!]/[13*12*11/3!)\r\n\r\nCan someone tell me why this needs to divide 3!, but the problem do not need to divide byanything?\r\n\r\nThank you!", "Solution_4": "[quote=\"Big M\"]Why some time ask 8 red marble, 5 blue marble, the probability of getting 3 red marble should like this:\n\n[(8*7*6)/3!]/[13*12*11/3!)\n\nCan someone tell me why this needs to divide 3!, but the problem do not need to divide byanything?\n\nThank you![/quote]\r\n\r\nActually, you are multiplying by $3!$:\r\n\r\n\\[\\frac{\\frac{8*7*6}{3!}}{\\frac{13*12*11}{3!}}\\]\r\n\\[=\\frac{8*7*6}{3!}\\cdot\\frac{3!}{13*12*11}\\]", "Solution_5": "[hide]I got\n\n\\[\\frac{(3!)(2!)(4!)}{9!}*3!=\\[\\frac{1}{210}\\][/hide]" } { "Tag": [], "Problem": "I also find this a little difficult\r\n* means exponent\r\n-2x*3 + 3x*2 + 5x - 6 > 70 ;)", "Solution_1": "this means exponent ^ \r\nfor example 2^3\r\nis $2^3$", "Solution_2": "Factor the expression and those are your answers", "Solution_3": "$-2x^3 + 3x^2 +5x - 6 > 70$" } { "Tag": [ "Putnam", "calculus", "derivative", "conics", "hyperbola", "inequalities", "college contests" ], "Problem": "Find the minimum calue of \r\n$(u-v)^{2}+\\left(\\sqrt{2-v^{2}}-\\frac{9}{v}\\right)^{2}$\r\n\r\nfor $00$\r\n\r\nwell obviously taking the derivative isnt going to help too much. :P \r\nso i am think to try and relate it to points on the plane but that isnt going to well.\r\nso far i have that $016->37->58->89->145->42->20->4." } { "Tag": [ "Pythagorean Theorem", "geometry" ], "Problem": "A pilot flew from the equator always on a bearing North East. How far had he traveled when he reached the north pole? Take the radius of the earth to 4000 miles", "Solution_1": "[hide]\nnortheast = north + east.\nnorth=4000 mi.\ntherefore east =4000 mi.\ntherefore traveled = 4000 \\sqrt{2} mi.\n\n(right?)\n[/hide]", "Solution_2": "radius = 4000 miles\r\ncircumference = 8000 pi miles\r\n\r\nIf the plot flew directly North, then he would have to travel 1/4 of the circumference, or 2000 pi miles, to reach the pole.\r\n\r\nFlying North - East means for every mile he flies north, he also flies exactly one mile east. Note that flying east does not get him any closer to the pole. So in the end he still has to fly 2000 pi miles to the North, in addition to 2000 pi miles to the East.\r\n\r\nUse Pythagorean theorem to find the total distance:\r\nsqrt((2000 pi miles to North)^2 + (2000 pi miles to East)^2) = 2000 sqrt(2) pi miles" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "This is my thought, but have no idea whether it is true :(\r\nConsider any real sequence that is convergent, is any rearrangement of this sequence convergent too? \r\n\r\nI think it is correct and it has the same limit as the original sequence. I mean I see that no matter how we arrange some infinite numbers of the rearrangement will lie around the limit. But I think it is a delicate topic. Some body might create a monster sequence that does not agree to me \r\n\r\nAny ideas appreciated,\r\nThanks :lol:", "Solution_1": "It is true, just apply the definitions (although you did not provide your rearrangement definition), e.g., let $ x$ be the limit of the sequence $ (x_n)_{n\\equal{}1,2,\\ldots}$ and $ (y_n)_{n\\equal{}1,2,\\ldots}$ be a [i]rearrangement[/i], then we have the following:\r\n\r\n$ \\forall \\epsilon >0: \\exists N_{\\epsilon}$ such that $ \\forall n>N_{\\epsilon}: |x\\minus{}x_n|<\\epsilon$ holds. Now for a fixed arbitrary $ \\epsilon >0$ determine $ y_{k_1},\\ldots,y_{k_{N_{\\epsilon}}}$ in the rearrangement, which correspond to $ x_1,\\ldots,x_{N_{\\epsilon}}$. Let $ M_{\\epsilon}\\equal{}\\max\\{k_1,\\ldots,k_{N_{\\epsilon}}\\}$. Then we have $ \\forall n>M_{\\epsilon}: |x\\minus{}y_n|<\\epsilon$ and since $ \\epsilon$ was arbitrary $ y_n\\rightarrow x$", "Solution_2": "I had the same idea, but i thought I might be missing something very minute in the argument. Since in series, a conditionally convergent series behaves strangely, I thought a sequence might suffer correspondingly. But my argument showed otherwise, so wanted to clarify :P \r\nThanks :lol:" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "if a+b=1 ,a,b >0\r\nprove\r\n( a+(1\\a ) )^2 + ( b+(1\\b) )^2 >= 25\\2", "Solution_1": "Just note that from the ineq. $2(x^{2}+y^{2})\\geq (x+y)^{2}$=> \\[2((a+\\frac{1}{a})^{2}+(b+\\frac{1}{b})^{2})\\geq ((a+b)+(\\frac{1}{a}+\\frac{1}{b}))^{2}\\geq (a+b+\\frac{4}{a+b})^{2}=25\\]" } { "Tag": [ "AMC" ], "Problem": "We all know our instructors are never wrong, so these might as well be the official answers from U of N in Lincoln.\n\n\n\nThis is for those who missed the Math Jam \n\n[hide](I'm not too great at taking notes so you might want to check the Math Jam transcripts anyways.)[/hide]\n\n1. 592\n\n2. 441\n\n3. 384\n\n4. 927\n\n5. 766\n\n6. 408\n\n7. 293\n\n8. 054\n\n9. 973\n\n10. 913\n\n11. 625\n\n12. 134\n\n13. 484\n\n14. 108\n\n15. 593", "Solution_1": "isn't R on the AMC Advisory Panel?? Doesn't he have hte answers????", "Solution_2": "I dropped off all test-related duties when I started this site, so I don't have access to the tests anymore until they're administered." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "let a_n be a sequence of real number and we define the sequence u_n by:\r\nu_(n+1)=u_n+sqrt(u_n^2 + a_n)\r\nProve that the sequence u_n converges iff the series sum(a_n,n) converges.", "Solution_1": "Hmmm... Am I missing something? What if $a_n=0$ and $u_0=1$?. $u_n$ clearly doesn't converge... Or does it? Another thing that should be cleared up: is it $u_{n^2}$ or $(u_n)^2$?" } { "Tag": [ "function", "linear algebra", "matrix", "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Firstly, it's not a problem, just some dicussions. \r\nReally this is very known but I' fond at its ideals. All we know that an ODE like as: $\\frac{dy}{dt}=A(t)y$, $y(0)=y_{0}$ where $A$ is function on an interval $I$ with $n\\times n$ matrix-values. If we could find a bound $M$ of $A$, namely $\\sup|A(t)|<+\\infty$ (sup takes on $t\\in I$) then the local solution can be extended to all $I$. Now let me assume that $\\sup|A(t)|=+\\infty$,\r\na) Is there such $A$ so that all local solution can be extend to all $I$?\r\nb) Is there such $A$ so that all local solution can not be extend to all $I$?", "Solution_1": "It could go either way. At least for scalars, there's really very little difference between this and the equation $y'=A(t)$. If the integral of $A$ converges, solutions can be extended. If it doesn't, solutions can't be extended.[/i]", "Solution_2": "All right, here's a problem:\r\n\r\nConstruct an explicit matrix-valued function $A(t)$ on an interval $I$ such that\r\n(1) The (improper) integral $\\int_{I}A(t)\\,dt$ converges.\r\n(2) Some solutions to the differential equation $y'(t)=A(t)y(t)$ on subintervals of $I$ cannot be extended to all of $I$.\r\n\r\n[hide=\"Hints\"]If $\\int_{I}\\|A(t)\\|\\,dt$ converges, this is impossible. We need conditional convergence to make this work.[/hide]" } { "Tag": [], "Problem": "$ ABCD$ is convex quadrilateral with $ [AB]$ is not parallel to $ [CD]$.\r\n$ E$ and $ F$ be the midpoints of $ [AD]$ and $ [BC]$ respectively. $ |CD|\\equal{}12$ , $ |AB|\\equal{}22$ and $ |EF|\\equal{}x$.\r\nWhat are the possible integer values of $ x$ ?", "Solution_1": "We can use coordinates.\r\nWe have A(a,b) B(c,d) C(12,0) D(0,0)\r\nLet $ x \\equal{} d \\minus{} b$, $ y \\equal{} c \\minus{} a$\r\nFrom the given, we have\r\n$ x^2 \\plus{} y^2 \\equal{} 484$\r\nWe have to find the possible values of the square root of\r\n$ \\frac {x^2 \\plus{} (y \\plus{} 24)^2}{4} \\equal{} 265 \\plus{} 12y$\r\nBut y can range from 0 to 22, so the possibilities are 265-520, the integer square roots ranging from 17-22." } { "Tag": [ "topology", "function", "vector", "advanced fields", "advanced fields unsolved" ], "Problem": "What are the extreme points of the unit ball of $ C(X)$, where $ X$ is a compact Hausdorff space ?\r\nWhat about $ l^p$.", "Solution_1": "The extreme points are the functions such that the $ |f(x)|\\equal{}1$ (identically). In the real (connected) case, this is not very interesting. In the complex case however, it turns out that these functions are dense in the unit ball. \r\n\r\nFor $ l^p$, the extremal boundary of the unit ball is the topological boundary: every unit vector is an extreme point.", "Solution_2": "[quote=\"phils\"]these functions are dense in the unit ball.[/quote]\nDo you mean: the convex combinations of these functions are dense?\n\n[quote=\"phils\"]For $ l^p$, the extremal boundary of the unit ball is the topological boundary: every unit vector is an extreme point.[/quote]\r\nUnless $ p\\equal{}1$ or $ p\\equal{}\\infty$.", "Solution_3": "Thanks for the corrections, mlok. I'm embarrassed by my semi-literate guesses - it's been a while since I've last seen that stuff." } { "Tag": [], "Problem": "Samantha deposited $ \\$100$ in a bank account that has earned $ 5\\%$ interest during this year. Audrey deposited $ \\$50$ more than Samantha deposited but earned only $ 4\\%$ interest during this year. Now, Audrey has how much more money than Samantha?", "Solution_1": "Samantha gets 5 dollars of interest, while Audrey gets 6 dollars of interest. So Audrey has 51 more dollars than Samantha." } { "Tag": [ "inequalities", "geometry unsolved", "geometry" ], "Problem": "Let $ABC$ be a triangle with $BC>CA>AB$ and let $G$ be the centroid of the triangle. Prove that \\[ \\angle GCA+\\angle GBC<\\angle BAC<\\angle GAC+\\angle GBA . \\]\r\n[i]Dinu Serbanescu[/i]", "Solution_1": "[quote=\"indybar\"]Let $ABC$ be a triangle with $BC>CA>AB$ and let $G$ be the centroid of the triangle. Prove that\n\\[\\angle GCA+\\angle GBC<\\angle BAC<\\angle GAC+\\angle GBA . \\]\n[i]Dinu Serbanescu[/i][/quote]\r\n\r\nit is false?? :?:" } { "Tag": [ "inequalities", "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $H(x)$ be a differentiable function on $R$ , $H(x_0) = 0$ and\r\n\r\n $H'(x) > 2cxH(x)$, $c >0$ for all $x > x_0$.\r\n\r\nProve that $H(y)$ is not equal to 0 for all $y>x_0$", "Solution_1": "Multiply by $e^{-cx^2}$ of both sides." } { "Tag": [ "linear algebra", "matrix", "algebra", "binomial theorem", "superior algebra", "superior algebra unsolved" ], "Problem": "$ \\mathcal R$ is a commutative ring with $ 1$ in which $ 2$ has inverse. Prove that for each $ n\\geq 2$, and $ A\\in\\mathcal M_n(\\mathcal R)$, there are 3 matrices $ A_1,A_2,A_3$ such that\r\n\\[ A \\equal{} A_1^2 \\plus{} A_2^2 \\plus{} A_3^2\r\n\\]", "Solution_1": "I have the solution (found by my father, actually), but right now I can only provide a \r\n\r\n[hide=\"rough sketch.\"]\nFirst, for the $ n\\equal{}2$ case, use\n\n\\[ \\begin{pmatrix} 1\\plus{}a & x \\\\ 0 & 1\\minus{}a \\end{pmatrix}^2 \\plus{} \n \\begin{pmatrix} 1\\minus{}b & 0 \\\\ y & 1\\plus{}b \\end{pmatrix}^2 \\plus{} \n \\begin{pmatrix} 0 & \\minus{}(1\\minus{}a)^2\\minus{}(1\\minus{}b)^2 \\\\ 1 & 0 \\end{pmatrix}^2 \\equal{} \n \\begin{pmatrix} 4a & 2x \\\\ 2y & 4b \\end{pmatrix}\n\\]\n\nand observe that since 2 is invertible, the result can be made anything we want.\n\nFor $ n \\geq 3$, first consider matrices with 0 on the main diagonal, 1's below the diagonal, and variables above the diagonal. Square, and prove that you can make the main diagonal of the result anything you want (disregarding the results outside of the main diagonal).\n\nNext, consider an upper triangular matrix with $ 1/2$ on the main diagonal, 0 below the diagonal, and variables $ x_{ij}$ above the diagonal. Square, and you'll notice that each entry in the result looks like $ x_{ij} \\plus{}$ stuff involving variables on lower diagonals. Conclude that the square of such a matrix can be fully controlled above the main diagonal, while the main diagonal is of course forced to be $ 1/4$ and anything below it remains 0.\n\nNow do the same for lower triangular. \n\nThe last step is easy: start with any matrix, use the first step to transform the main diagonal into all $ 1/2$, and then use the second step to find suitable upper and lower triangular matrices that can fix the result off the main diagonal.\n[/hide]", "Solution_2": "Wow, that's an impressive idea, though I don't understand what you are doing for $ n \\equal{} 3$.\r\nHere is a complete proof based on your idea.\r\n\r\nFirst I'll need the two formulas\r\n$ \\begin{pmatrix}{x \\minus{} y \\plus{} 1\\over 2} & 2y \\minus{} {(x \\minus{} y \\minus{} 1)^2\\over 2} \\\\\r\n{1\\over2} & {x \\minus{} y \\minus{} 1\\over 2}\\end{pmatrix}^2 \\equal{} \\begin{pmatrix}x & * \\\\\r\n* & y\\end{pmatrix}$\r\n$ \\begin{pmatrix}0 & {x \\plus{} y \\minus{} z\\over 2} & {x \\minus{} y \\plus{} z\\over 2} \\\\\r\n1 & 0 & { \\minus{} x \\plus{} y \\plus{} z\\over 2} \\\\\r\n1 & 1 & 0\\end{pmatrix}^2 \\equal{} \\begin{pmatrix}x & * & * \\\\\r\n* & y & * \\\\\r\n* & * & z\\end{pmatrix}$\r\nfor any $ x,y,z\\in\\mathcal{R}$.\r\n\r\nThen take any $ n\\times n$-matrix $ A$ over $ \\mathcal{R}$ with $ n\\ge 2$. \r\nUsing the above formulas we can find some $ n\\times n$-matrix $ A_1$, s.t. $ A_1^2$ and $ A \\minus{} 2\\cdot I_n$ have the same diagonal,\r\nby putting some $ 2\\times 2$ and $ 3\\times 3$-matrices as blocks on the diagonal of $ A_1$ and otherwise $ 0$.\r\n\r\nThis gives $ A \\minus{} 2\\cdot I_n \\minus{} A_1^2 \\equal{} N_1 \\plus{} N_2$ with some nilpotent triangular matrices $ N_1,N_2$ (upper and lower),\r\nwhich leaves to show, that $ I_n \\plus{} N$ is a square for any nilpotent $ n\\times n$-matrix $ N$ over $ \\mathcal{R}$.\r\n\r\nBut using [url=http://en.wikipedia.org/wiki/Binomial_theorem]Newton's generalized binomial theorem[/url] we get\r\n$ M^2 \\equal{} I_n \\plus{} N$ with $ M: \\equal{} \\sum_{k\\ge 0}\\binom{1/2}{k}N^k$, where $ N^k \\equal{} 0_n$ for large $ k$\r\nand the binomial coefficient $ \\binom{1/2}{k}: \\equal{} {{1\\over 2}({1\\over 2} \\minus{} 1)\\cdots ({1\\over 2} \\minus{} k \\plus{} 1)\\over k!}$ is well-defined in $ \\mathcal{R}$ for all $ k\\ge 0$\r\nusing $ \\binom{1/2}{0} \\equal{} 1,\\binom{1/2}{1} \\equal{} {1\\over 2}$ and the recursive formula $ \\binom{1/2}{k} \\equal{} \\minus{} {1\\over 2}\\sum_{0 < l < k}\\binom{1/2}{l}\\binom{1/2}{k \\minus{} l}$ for all $ k\\ge 2$,\r\nwhich comes from comparing coefficients of $ 1 \\plus{} X \\equal{} (\\sum_{k\\ge 0}\\binom{1/2}{k}X^k)^2$ in $ \\mathbb{Z} [ [X] ]$.\r\n\r\nPS this shows, that $ 4^kA$ is a sum of three squares over any ring with $ 1$ (not just commutative) and $ k\\in\\mathbb{N}$ large enough.\r\nPPS explicitly we get $ 4^{4n \\minus{} 5}A$ a sum of three squares over any ring with $ 1$ and any $ n\\times n$-matrix $ A$ with $ n\\ge 2$.", "Solution_3": "[quote=\"olorin\"]$ 1 \\plus{} X \\equal{} (\\sum_{k\\ge 0}\\binom{1/2}{k}X^k)^2$ in $ \\mathbb{Z} [ [X] ]$[/quote]\r\nSorry, I meant in $ \\mathbb{Q} [ [X] ]$ the ring of [url=http://en.wikipedia.org/wiki/Power_series_ring#The_ring_of_formal_power_series]formal power series[/url] over $ \\mathbb{Q}$." } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "function", "calculus computations" ], "Problem": "Evaluate $ \\int_{\\frac{\\pi}{2}}^{\\pi} x^{\\sin x}(1\\plus{}x\\cos x\\cdot\\ln x \\plus{}\\sin x)\\ dx$.", "Solution_1": "The first guess worked:\r\n$ (x^{\\sin x\\plus{}1})'\\equal{}(e^{(\\sin x\\plus{}1)\\ln x)})'\\equal{}((\\sin x\\plus{}1)\\ln x)' e^{(\\sin x\\plus{}1)\\ln x}\\equal{} (\\cos x \\ln x \\plus{} \\frac{\\sin x \\plus{}1}{x}) x^{\\sin x\\plus{}1}\\equal{}x^{\\sin x}(1\\plus{}x\\cos x\\cdot\\ln x \\plus{}\\sin x)$\r\nHence, the integral equals to $ \\pi ^{\\sin \\pi \\plus{}1} \\minus{} (\\frac{\\pi}{2})^{\\sin\\frac{\\pi}{2}\\plus{}1}\\equal{}\\frac{4\\pi \\minus{}\\pi ^{2}}{4}$", "Solution_2": "That's right! :)", "Solution_3": "[hide]$ I \\equal{} \\int_{\\pi/2}^{\\pi}x^{\\sin x}(1 \\plus{} x \\cos x \\ln x \\plus{} \\sin x)\\,dx \\equal{} \\int_{\\pi/2}^{\\pi}\\left(x^{\\sin x \\plus{} 1}\\right)'\\,dx \\equal{} \\left[x^{\\sin x \\plus{} 1}\\right]_{\\pi/2}^{\\pi}$\n\n$ \\equal{} \\pi \\minus{} \\left(\\frac{\\pi}{2}\\right)^2$.[/hide]", "Solution_4": "Carcul, your solution is same as bambaman's. :wink:", "Solution_5": "[quote=\"bambaman\"]The first guess worked:\n$ (x^{\\sin x \\plus{} 1})'$[/quote]\r\nWhat made you think about it? Just a trial and error way of resolution or there were some \"clues\" in the integral? Your solution starts with a function that seems completely pointless in order to calculate the integral and then gets a very simple way to solve it.\r\n\r\nGoodbye\r\nGM" } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "For any straight line, let $m$ and $b$ represent its slope and y-intercept, respectively. Consider all lines having the property that $2m+b = 2000$. These lines all have the specific point $(x_1,y_1)$ in common. Find this common point $(x_1,y_1)$.\r\n\r\nSolve this problem without use of calculator.", "Solution_1": "[hide]\nIsolating, we get:\n\n$m=\\frac{2000-b}{2}$. In solpe intercept point:\n\n$y=\\frac{2000-b}{2} x+b$. The only value (x,y) which is constant regaudrless of b is $(2,2000)$[/hide]", "Solution_2": "I hid your answer.\r\n\r\nGuys, please hide your answer. ;)" } { "Tag": [ "calculus", "integration", "function", "algebra", "domain", "LaTeX", "complex analysis" ], "Problem": "Hi all,\r\n\r\nThis is a basic theoretical question I am not quite sure about.\r\n\r\nSuppose you have an analytic function F defined on some domain contained in the open upper half plane. You then decide to define say F* on the \"conjugate\" domain D* in the lower half plane, as F*(x) = conjugate(F(conjugate(x))\r\n\r\nNow I want to integrate F* around some oriented curved contained in D*, can I say that the integral is equal to the conjugate of the integral of F over the reversely-oriented conjugate curve? This seems obvious... yet I am a bit concerned regarding the umm symantics of proving this because when I want to make the substitution u = conjugate(z), but then I am concerned that the dz becomes d(conjugate(u)), which I don't know how to deal with, and I want to integrate wrt to u not its conjugate...\r\n\r\nAnyways hope that makes sense, sorry I don't know latex to make that clearer, but if you can drop some knowledge regarding this I would greatly appreciate it, I am sure this is pretty basic, just not sure how to get around the issue of integrating wrt. the conjugate. Thanks.", "Solution_1": "Conjugation is not a conformal map, so $ f(z) \\equal{} \\bar{z}$ is not complex-differentiable." } { "Tag": [], "Problem": "Suppose that $a+b+c=0$. Prove that:\r\n\\[ \\frac{a^3+b^3+c^3}{3}\\cdot\\frac{a^4+b^4+c^4}{2}=\\frac{a^7+b^7+c^7}{7}=\\frac{a^2+b^2+c^2}{2}\\cdot\\frac{a^5+b^5+c^5}{5} \\]", "Solution_1": "My advice: Brute-force it. But first, check your posting. Should it be (a^4 + b^4 + c^4) / 4, NOT divided by 2? I'm just observing the pattern.", "Solution_2": "Teki-Teki: No, $\\frac{a^4+b^4+c^4}{2}$ is right. Try $(1,2,-3)$.", "Solution_3": "[quote=\"Teki-Teki\"]My advice: Brute-force it. But first, check your posting. Should it be (a^4 + b^4 + c^4) / 4, NOT divided by 2? I'm just observing the pattern.[/quote]\r\n\r\nDid you try to brute-force it? :-)", "Solution_4": "This question is not very difficult to deal with by brute force.\r\nIf you don't like brute force, here is an alternative.\r\n\r\nDenote $A = ab + bc + ca, B = abc$. \r\nThen $a, b, c$ are the roots of $x^3+Ax-B =0$.\r\n\r\nDefine $S_n=a^n+b^n+c^n$ for $n = 0, 1, 2, 3,......$\r\nThen $S_0=3$, $S_1=0$,$S_2=(a+b+c)^2-2(ab+bc+ca)=-2A$.\r\n\r\nNote that $S_{n+3}=-AS_{n+1}+BS_n$ for $n = 0, 1, 2, .....$\r\n$S_3=-AS_1+BS_0=3B$,\r\n$S_4=-AS_2+BS_1=2A^2$,\r\n$S_5=-AS_3+BS_2=-5AB$, and\r\n$S_7=-AS_5+BS_4=7A^2B$.\r\n\r\nDirect verification shows that $\\frac{S_3} 3 \\cdot \\frac{S_4} 2 = \\frac{S_7} 7 = \\frac{S_2} 2 \\cdot \\frac{S_5} 5$, which is equivalent to what we need to show." } { "Tag": [ "algebra", "system of equations" ], "Problem": "Does anyone know this?\r\n\r\nSolve using elimination method x + 2y = -17, and 2x + 1y = -16", "Solution_1": "$x+2y=-17$\r\n$2x+y=-16$\r\n\r\nthe first equation can be written as $2x+4y=-34$\r\nsubtracting the second equation from this gives\r\n$3y=-18$, so $y=-6$. we can plug this into any of the equations to find $x=-5$.", "Solution_2": "$x+2y =-17$\r\n$2x+1y =-16$ \r\n\r\nEliminate either y or x from either equation. Let's choses to eliminate $x$ from the second equation. This can be done by adding to the second equation -2 time th e first equation to get a new second equation. \r\n\r\n $2x+1y =-16$\r\n(+) $-2x-4y = 34$\r\n----------------------------\r\n $-3y = 18$\r\n $y =-6$ \r\n\r\nTry finding $x$ now =P", "Solution_3": "Daermon,\r\n\r\nThanks for your help :)", "Solution_4": "[quote=\"Imperial Effect\"]\n $-4y = 18$\n[/quote]\r\ncheck your subtraction again :wink:\r\nalso, $-2\\cdot4\\ne18$", "Solution_5": "Daermon,\r\n\r\nI just pm'd you? Is that alright?", "Solution_6": "a general method when given a system of equations \r\n$a_{1}x+b_{1}y=w\\\\ a_{2}x+b_{2}y=z$ is to multiply the first equation by $a_{2}$ and the second by $-a_{2}$, then adding the equations together. The coefficient of $x$ in the new equation will be $a_{1}a_{2}-a_{1}a_{2}=0$, so we will only have one variable to deal with: $y$ :D", "Solution_7": "Another (relatively) painless method with linear systems is [url=http://en.wikipedia.org/wiki/Cramer%27s_rule]Cramer's Rule[/url], which involves determinants of matrices (which aren't that hard to do by hard). This method is also a bit easier to do than the substitution method for a larger system of equations.", "Solution_8": "[quote=\"jenny221\"]Does anyone know this?\n\nSolve using elimination method x + 2y = -17, and 2x + 1y = -16[/quote]\r\n[hide]\nelimination is getting rid of a variable. An easy variable to get rid of is x (though y is just as easy).\n2x+4y=-34\n2x+y=-16\nsubtract them\n3y=-18\ny=-6\nx=-5[/hide]", "Solution_9": "[hide=\"formula\"]\n\nwe have\n\n$ax+by=c$\n$dx+ey=f$\n\n$aex+bey=ce$\n$dbx+eby=fb$\n____________\n$(ae-db)x=(ce-fb)$\n\n$x=\\frac{ce-fb}{ae-db}$\n\n$ax+by=c$\n$dx+ey=f$\n\n$adx+bdy=cd$\n$adx+aey=af$\n_____________\n$(bd-ae)y=(cd-af)$\n\n\n$y=\\frac{cd-af}{bd-ae}$\n\n$(x,y)=\\left(\\frac{ce-fb}{ae-db},\\frac{cd-af}{bd-ae}\\right)$[/hide]\r\n\r\nI recommend to NOT memorize the formula, but to see why it works.", "Solution_10": "I need help with the following problem.\r\n\r\n2/3x-3/5y=-17/15\r\n8/5x-7/6y=-3/10\r\n\r\nPlease solve by elimination and give the steps. Thanks![/list]", "Solution_11": "[quote=\"Trisha\"]I need help with the following problem.\n\n2/3x-3/5y=-17/15\n8/5x-7/6y=-3/10\n\nPlease solve by elimination and give the steps. Thanks![/list][/quote]\r\nPlease do not revive a 2 month old topic." } { "Tag": [ "inequalities", "algebra unsolved", "algebra" ], "Problem": "Give a,b,c are real numbers.\r\nProve that : a(a+b)^3+b(b+c)^3+c(c+a)^3 >= 0", "Solution_1": "This problem was created by vasile Cartoaje long time ago and it eas already posted on this forum.", "Solution_2": "I want to know how is solution??????", "Solution_3": "Move it from Algebra to Inequalities. Maybe some one will give us a proof :)", "Solution_4": "A stronger inequality was posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=113218[/url]." } { "Tag": [ "integration", "logarithms", "calculus", "calculus computations" ], "Problem": "Help me to check the solution, thanks.\r\n\r\n$ \\text{* }I\\equal{}\\int\\limits_{1}^{e}{\\frac{\\log \\left( 1\\plus{}\\log x \\right).dx}{x}}$\r\n \r\n$ \\text{Since for }x\\in \\left[ 1,e \\right]\\text{, we have }0\\le \\log x\\le 1\\Rightarrow$\r\n \r\n$ I\\equal{}\\int\\limits_{1}^{e}{\\frac{dx}{x}.\\sum\\limits_{n\\equal{}1}^{\\plus{}\\infty }{\\frac{{{\\left( \\minus{}1 \\right)}^{n\\plus{}1}}.{{\\left( \\log x \\right)}^{n}}}{n}}}\\equal{}\\sum\\limits_{n\\equal{}1}^{\\infty }{\\frac{{{\\left( \\minus{}1 \\right)}^{n\\plus{}1}}}{n}}\\int\\limits_{1}^{e}{{{\\left( \\log x \\right)}^{n\\minus{}1}}.d\\left( \\log x \\right)}\\text{, }$\r\n \r\n$ \\int\\limits_{1}^{e}{{{\\left( \\log x \\right)}^{n\\minus{}1}}.d\\left( \\log x \\right)}\\equal{}\\left[ \\frac{{{\\left( \\log x \\right)}^{n}}}{n} \\right]_{1}^{e}\\equal{}\\frac{1}{n}\\Rightarrow I\\equal{}\\sum\\limits_{n\\equal{}1}^{\\infty }{\\frac{{{\\left( \\minus{}1 \\right)}^{n\\plus{}1}}}{{{n}^{2}}}}\\equal{}\\frac{1}{2}\\zeta \\left( 2 \\right)\\equal{}\\frac{{{\\pi }^{2}}}{12}$\r\nand\r\n \r\n$ \\text{Put }y\\equal{}1\\plus{}\\log x\\Rightarrow dy\\equal{}\\frac{dx}{x}\\Rightarrow I\\equal{}\\int\\limits_{1}^{2}{\\log y.dy}\\equal{}\\left[ y\\left( \\log y\\minus{}1 \\right) \\right]_{1}^{2}$\r\n \r\n$ \\equal{}2.\\left( \\log 2\\minus{}1 \\right)\\plus{}1\\equal{}2.\\log 2\\minus{}1$", "Solution_1": "what a pity!\r\nThe first answer is wrong,and the second is right.\r\nyou wrong at second \"=\"\r\nit shoud be (ln x)^n,not (ln x)^(n-1)." } { "Tag": [ "probability", "number theory", "prime numbers" ], "Problem": "An random integer is choosen from the first 500 natural numbers. What is the probability that the integer picked is such that there is a 1/4 chance of picking any one of its positive, proper divisors?", "Solution_1": "Is the proper divisor including or not including 1?\r\nAlso, I can't exactly understand what you're asking about 1/4 of the proper divisors.\r\nDoes this mean that we're looking for numbers that have 4 proper divisors?", "Solution_2": "[quote=\"pianoforte\"]An random integer is choosen from the first 500 natural numbers. What is the probability that the integer picked is such that there is a 1/4 chance of picking any one of its positive, proper divisors?[/quote]\r\n\r\nHm. Interesting problem. :P \r\n[hide]I think the answer is $\\frac{179}{500}$. I found that only numbers that can be expressed in the form $x\\cdot y$ where $x$ and $y$ are prime numbers have a $1/4$ chance of picking any one of its positive, proper divisors. If $x=2$, there are 53 primes that can replace y and still be under 500. I did that until 500/x1} (-2^{k_i}+2^{k_i+l_i+1} ) . \\]" } { "Tag": [], "Problem": "if a and b are two positive integers,\r\nWhat is the smallest possible value of a+b given that $6a^3 = 5b^5$ ?\r\nCorny Title, I know", "Solution_1": "Quote:If a and b are two positive integers,\nWhat is the smallest possible value of a+b given that ?[hide]\n\nLet a = 2p*3q*5r, and b = 2s*3t*5u.\n\n\n\nThen we have: (2)(3)(2p*3q*5r)3 = 5(2s*3t*5u)5\n\n\n\nAnd: 23p+1*33q+1*53r = 25s*35t*55u+1\n\n\n\nHence: 3p + 1 = 5s, 3q + 1 = 5t, 3r = 5u+1\n\n\n\nFor the smallest values: p = 3, s = 2, q = 3, t = 2, r = 2, u = 1\n\n\n\nTherefore: a = 23*33*52 = 5400, b = 22*32*51 = 180 . . . and a + b = 5580[/hide]" } { "Tag": [ "national olympiad" ], "Problem": "These are the questions of Iranian pre-TST first round\r\nQuestion 1- http://www.mathlinks.ro/Forum/viewtopic.php?p=185850#p185850\r\nQuestion 2- http://www.mathlinks.ro/Forum/viewtopic.php?t=30002\r\nQuestion 3- http://www.mathlinks.ro/Forum/viewtopic.php?p=185853#185853\r\nAnd the PDF version of questions:", "Solution_1": "Thank you - can you tell us when the next rounds will take place? :D And how many rounds are there ?", "Solution_2": "Well there have been 3 exams week ago, 2 weeks ago, 3 weeks ago. I was in Tehran for the IMO Preparation program. Now because of Iraninan new year that is 4 days later we have vacation. I'm home in Isfahan and I have time for posting the questions. I'll post the second pre-TST soon", "Solution_3": "Then post the link here, please!", "Solution_4": "Which link ?", "Solution_5": "For the second pre-TST.", "Solution_6": "Questions of second Round:\r\n1- http://www.mathlinks.ro/Forum/viewtopic.php?p=187969#187969\r\n2- http://www.mathlinks.ro/Forum/viewtopic.php?p=187974#187974\r\n3- http://www.mathlinks.ro/Forum/viewtopic.php?p=187975#187975\r\nAnd PDF version:", "Solution_7": "Omid Hatami: When you will post the third test? :)", "Solution_8": "Well for some reasons I couldn't send the problems of 3rd exam, But these are the problems of 4th exam that are from Silk Road Math Olympiad 2005\r\nProblem 1- http://www.mathlinks.ro/Forum/viewtopic.php?p=205676#205676\r\nProblem 2- http://www.mathlinks.ro/Forum/viewtopic.php?p=205678#205678\r\nProblem 3- http://www.mathlinks.ro/Forum/viewtopic.php?p=205679#205679\r\nProblem 4- http://www.mathlinks.ro/Forum/viewtopic.php?t=33037" } { "Tag": [], "Problem": "I propose a contest where each person makes a picture. The winner will be the one who has the most creative drawing. Although it is not a very exciting prize the winner will be awarded the first consideration for any other contest I run. Now here are some restrictions:\r\n\r\nThere are up to 31 players(I will join in too).\r\nThere must be at least 8 players \r\nIf any new player wants to join the game he/she will post on the forum and be considered.\r\nAt the beginnig of the game 5 judges will be elected to work out the winner(after the game) .\r\nThe picture can be made however you like.\r\nIt is not permitted to copy someone else. If this happens then the person who posted the copy will be thrown out.\r\nthere will be three rounds.\r\n\r\nLet the tournament begin!!!\r\n :starwars:", "Solution_1": "I will join\r\nI drew my avatar." } { "Tag": [ "MATHCOUNTS" ], "Problem": "1. What is the average of all [i]4[/i] digit numbers with distinct digits fomred by choosing out of the digits 1,3,4,6.\r\n\r\n2. What is the average of all [i]3[/i] digit numbers with distinct digits fomred by choosing out of the digits 1,3,4,6.\r\n\r\n3. What is the average of all 3 digit [b]even[/b] numbers with distinct digits formed by choosing out of the digits 1,3,4,6.", "Solution_1": "i dont know of any easy way to do these type of problems so i would love to learn one cause i see them all of the time on mathcounts and such.\r\n\r\n#1\r\n[hide]Ok so there are 24 numbers, and every number is in every position 6 times. so we have \n\n 1(6) 1(6) 1(6) 1(6)\n 3(6) 3(6) 3(6) 3(6) \n 4(6) 4(6) 4(6) 4(6)\n +6(6) 6(6) 6(6) 6(6)\n____________________\n 9 3 3 2 4 \n\nWhen this is divided by 24 to get the averege we have $ \\boxed{3888.5}$[/hide]", "Solution_2": "#2. \r\nSimilarly, we have the average is 388,5.\r\n#3.\r\nthey are 134, 164, 314, 364, 614, 634, 136, 146, 316, 346, 416, 436.\r\nHence, the average is 335." } { "Tag": [ "symmetry", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "One day a hunter lost in the forest.He doesn't know where the road is,but he knows that the road is somewhere on a distance equal to $ 1$,from the place where he initially stands.Find the trajectory in which he must go to find the road,so that the length of this trajectory would be minimized.Also find the length of this trajectory.", "Solution_1": "making a lot of assumptions here.. most importantly\r\n\r\n1) said hunter cant \"see\" the road from any distance except when he stands on it?\r\n\r\nthen any path he takes out to the circle of radius 1 will be of length 1. then he would walk around in either direction the circumfrence of 2 pi. (at most)\r\n\r\nthe symmetry of the unknowns seems to make this the only choice right?", "Solution_2": "Do we know anything about the road? For example, if the road is known to be perfectly straight, then a path of length $ 2 \\plus{} \\sqrt {2} \\plus{} \\pi < 1 \\plus{} 2\\pi$ is possible (maybe even shorter). If at least the road is not 'too wavy' in some sense (e.g. the curve is smooth and its curvature is sufficiently small), it may be possible to get a bound below $ 1 \\plus{} 2\\pi$. But if we know nothing about the road then it's effectively like searching for a point, so kenn4000's observation would seem to be correct.", "Solution_3": "[quote=\"kenn4000\"]making a lot of assumptions here.. most importantly\n\n1) said hunter cant \"see\" the road from any distance except when he stands on it?\n\nthen any path he takes out to the circle of radius 1 will be of length 1. then he would walk around in either direction the circumfrence of 2 pi. (at most)\n\nthe symmetry of the unknowns seems to make this the only choice right?[/quote]\r\nNope,there exist a shorter way.\r\nAnswering to JoeBlow's question : Yes,we know that the road is a straight line.\r\nIf he could see the road from some distance then I would definitely mention it in the statement,so we may assume that our hunter is blind. :)\r\nWhat do you mean by \"symmetry of the unknowns\"?\r\nIt seems to me that $ \\sqrt {2} \\plus{} 2 \\plus{} \\pi$ is correct answer,but I am not sure,maybe you would be so kind to post the image or anything else,for us to check it out.(at least a description)", "Solution_4": "Red is the set of points where the road may be, green is the hunter's path.\r\n[asy]draw(Circle((0,0),10),red+linewidth(1));\ndraw(Arc((0,0),10,0,180),green+linewidth(1));\ndraw((-10,0)--(-10,-10),green+linewidth(1));\ndraw((10,0)--(10,-10),green+linewidth(1));\ndraw((0,0)--(-10,-10),green+linewidth(1));\nlabel(\"$\\sqrt{2}$\",(-5,-5),SE);\nlabel(\"$1$\",(-10,-5),W);\nlabel(\"$1$\",(10,-5),E);\nlabel(\"$\\pi$\",(0,10),N);[/asy]\r\nI feel fairly confident this is the shortest path, but not sure how one might go about proving that.", "Solution_5": "Your example is almost correct,but not completely...", "Solution_6": "You're right, looks like I messed up the calculus. $ \\sec \\theta \\plus{} \\tan \\theta \\plus{} \\frac {\\pi}{2} \\minus{} \\theta$ has a minimum at $ \\frac {\\pi}{6}$, not at $ \\frac {\\pi}{4}$. Here's the picture\r\n[asy]draw(Circle((0,0),10),red+linewidth(1));\n draw(Arc((0,0),10,0,210),green+linewidth(1));\n draw((-8.66025,-5)--(-5.7735,-10),green+linewidth(1));\n draw((10,0)--(10,-10),green+linewidth(1));\n draw((0,0)--(-5.7735,-10),green+linewidth(1));\n draw((0,0)--(-8.66025,-5),dashed);\n draw((-5.7735,-10)--(10,-10),dashed);\n draw((0,0)--(0,-10),dashed);\n label(\"$\\frac{2}{\\sqrt{3}}$\",(-2.88675,-5),SE);\n label(\"$\\frac{1}{\\sqrt{3}}$\",(-7.21688,-7.5),SW);\n label(\"$1$\",(10,-5),E);\n label(\"$\\frac{7\\pi}{6}$\",(0,10),N);\n label(\"$\\frac{\\pi}{6}$\",(-2,-2),SW);[/asy]\r\nSo the length of the path is $ \\sqrt {3} \\plus{} 1 \\plus{} \\frac {7\\pi}{6}$. Any other way to shorten it?", "Solution_7": "Now it seems to be correct.\r\nI have no idea how it can be proved,but I'm pretty sure that this is the only right example.\r\nI doubt if there exist any easy solution,I think its proof kinda messy and uses a lot of calculus." } { "Tag": [ "function", "combinatorics unsolved", "combinatorics" ], "Problem": "A class of 25 students and 2 teachers collect money for charity. Each of the teachers donates 0, 3 or 6 coin, and each student 0 or 1. Let $a_{n}$ be the number of ways in which n coin can be collected\r\n\r\na) Find generating function $G_{\\bar{a}}(x)$ for the sequence $\\bar{a}$.\r\n\r\nb)Determine $a_{20}$.", "Solution_1": "About the part b) it could be 25,584,855.\r\n\r\nThe teachers can collect 0, 3 or 6 coins, so there are 9 different ways:\r\n\r\n0, 0\r\n0, 3\r\n0, 6\r\n3, 0\r\n3, 3\r\n3, 6\r\n6, 0\r\n6, 3\r\n6, 6\r\n\r\nAnd the students have to collect the other coins:\r\n\r\n20 coins\r\n17\r\n14\r\n17\r\n14\r\n11\r\n14\r\n11\r\n8\r\n\r\nSo the different ways to collect 20 coins are\r\n$(25,20)+2*(25,17)+3*(25,14)+2*(25,11)+(25,8)$" } { "Tag": [ "probability", "analytic geometry", "graphing lines", "slope" ], "Problem": "As the diagram shown , the gird-table is formed by 25 small squares.\r\n\r\nWhen we get any $ 2$ points from the grids beside point $ A$ , what is the \r\n\r\nprobabilty of forming a straight line linked with $ A$?\r\n\r\n :D", "Solution_1": "Just a junior problem-solver. I used organized brute force (though you were probably looking for something more creative) [hide]sum of all possible cases where three points are collinear/total no. of possible choices=8/119, or maybe I screwed up, I am so tired![/hide] If anyone finds a way other than organized brute force and casework, please tell me.", "Solution_2": "[hide=\"Solution\"]On the first diagonal, $ \\binom{5}{2}$\n\nOn the other, $ \\binom{4}{2}$.\n\nOn the two vertical lines, $ \\binom{5}{2}$\n\nSumming, $ 36$ ways.\n\nAnswer: $ \\frac{36}{\\binom{24}{2}}\\equal{}\\boxed{\\frac{3}{23}}$[/hide]\r\n\r\nI can't think of a better way to do it than those cases.", "Solution_3": "[quote=\"grn_trtle\"][hide=\"Solution\"]On the first diagonal, $ \\binom{5}{2}$\n\nOn the other, $ \\binom{4}{2}$.\n\nOn the two vertical lines, $ \\binom{5}{2}$\n\nSumming, $ 36$ ways.\n\nAnswer: $ \\frac {36}{\\binom{24}{2}} \\equal{} \\boxed{\\frac {3}{23}}$[/hide]\n\nI can't think of a better way to do it than those cases.[/quote]\r\n\r\nthere are 4 more lines, forming the slope of 2,-2,1/2,-1/2 around point A\r\n:)\r\nand the total # of ways of picking 2 points should be 35*34/2! i think", "Solution_4": "Wait, the method you guys are using doesn't take into account the [u]order[/u] of the different ways through which two points may be collinear with $ A$. For example, if $ X$ and $ Y$ are collinear to A, there are actually $ 2$ ways to pick points $ X$ and $ Y$ (if order counts), not 1 way as $ 2!/2!0!\\equal{}1$ shows. Or perhaps I'm wrong and order isn't taken into account at all...", "Solution_5": "It doesn't matter so long as you use the same method when counting the number of \"good\" ways and the number of total ways. (Switching between the two methods is the same as multiplying by $ \\frac{2!}{2!}$.)", "Solution_6": "I don't see the other 4 lines... I looked for other lines, but...\r\nA line with slope of 2 can't fit three collinear points in that square, can it?", "Solution_7": ":huh: Let B be the point two up and one right from A and let C be the point two down and one left from A. Then CAB is a straight line.\r\n\r\nWe're not choosing squares, we're choosing vertices of squares." } { "Tag": [], "Problem": "Given an ordered sequence of N elements, what is the minimum number of random relations of the form \"A is before B\" that will completely fix the order of the sequence, no matter what that order is?\r\n\r\nThe relations are required to be distinct, i.e. no two relations can deal with the same two elements of the sequence.\r\n\r\nI don't have a solution either, so I'm opening it to general discussion.", "Solution_1": "[hide=\"I think this works\"]\nIn order to completely fix the order of the set, I believe the minimum number would have to concern every element's relation to two other elements unlessthe element lies at the beginning or end, in which case only one relation is noted, so, if given the correct relations, the order of the set can be defined by $n-1$ relations. However, the problem says \"random relations\" so it may mean that for any two randomly selected elements, the relation is given. I believe the maximum to define the order is $n$ choose $2$, but the problem is somewhat unclear to me...[/hide]", "Solution_2": "If we know the relation of every number to each other except the biggest two: that gives us $1$ less than the maximum and you still can't determine the order, so you need all of them to fix it.", "Solution_3": "Actually, you don't.\r\n\r\nSay the order is A, B, C, D, E,...\r\n\r\nThen the subset \"A is before B,\" \"B is before C,\" \"C is before D,\" etc. of the set of all (n choose 2) possible relations is obviously sufficient to determine the order. There are only n - 1 relations in this set, and the rest are all unnecessary. \r\n\r\nThe problem is, of course, that if you're randomly given n - 1 different relations, they're likely not to be exactly those, as sapphyre pointed out...", "Solution_4": "By the way, here's a clearer way of putting the problem:\r\n\r\nGiven an ordered sequence with N elements, whose order you know nothing about. How many different questions of the form \"Does element A come before element B?\" would you need to ask before you're [i]guaranteed[/i] to know the order (no matter what that order is)?", "Solution_5": "[quote=\"Laplace's_Demon\"]Actually, you don't.\n\nSay the order is A, B, C, D, E,...\n\nThen the subset \"A is before B,\" \"B is before C,\" \"C is before D,\" etc. of the set of all (n choose 2) possible relations is obviously sufficient to determine the order. There are only n - 1 relations in this set, and the rest are all unnecessary. \n\nThe problem is, of course, that if you're randomly given n - 1 different relations, they're likely not to be exactly those, as sapphyre pointed out...[/quote]\r\n\r\nI thought the problem was talking about the worst-case scenario (i.e. the maximum number of of relations such that the order is not fixed). So you need all of them to guarantee you can order the sequence.", "Solution_6": "Sorry if my problem statement remains somewhat unclear. \r\n\r\nIt asks for the [i]minimum[/i] number of relations (given at random) that you would need before you could determine the order of any N-element sequence with certainty. So the \"worst-case scenario\" is where you have the largest possible set of [i]logically independent[/i] relations that still doesn't fix the order. If you're given (say) \"A is before B\" and \"B is before C,\" then \"A is before C\" would [i]not[/i] be in this set because that can be inferred from the given relations.\r\n\r\nAs you see, for an ordered sequence A, B, C, D,... with N elements, you would have to be given at least N-1 relations (these being specifically \"A is before B,\" \"B is before C,\" etc.) before you can determine the order. But we assume you're given these relations randomly, so the \"worst-case\" set of relations--the one we're interested in--would have quite a bit more than N-1 relations. But it is not the set of all (N choose 2) relations, because those include precisely the N-1 we considered earlier that fix the order of the sequence on their own. \r\n\r\nTherefore the worst-case set of relations, which \r\n\r\n1) are logically independent of each other (none of them can be inferred from the others),\r\n2) do not fix the order of the sequence, and\r\n3) form the largest set that satisfies (1) and (2),\r\n\r\nhas strictly between N - 1 and (N choose 2) elements. If we add any relation to this set that cannot be inferred from those already in the set, then the order of the sequence must be fixed. The number of relations in this last set, in terms of N, is the answer to the question." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Consider two distinct positive integers $a$ and $b$ having integer arithmetic, geometric and harmonic means. Find the minimal value of $|a-b|$. \r\n\r\n[i]Mircea Fianu[/i]", "Solution_1": "Clearly $a$ and $b$ must have the same parity so let $a = k+m$ and $b = k - m$ for integers $k > m > 0$. We have $|a-b| = 2m$. Note that $ab = k^2 - m^2 = n^2$ is a perfect square. Thus, $(n,m,k)$ is a Pythagorean triple.\r\n\r\nNow, $\\left( \\frac{1}{2a} + \\frac{1}{2b} \\right)^{-1} = \\left( \\frac{k}{k^2-m^2} \\right)^{-1} = \\frac{k^2-m^2}{k}$. Thus, $k | m^2$ and $k | n^2$. Starting with a primitive Pythagorean triple $(r,s,t)$ we want $t | r^2$ and $t | s^2$ so we must scale everything by $t$ since $(r,s,t) = 1$. From this idea, we use $(20, 15, 25) = (n,m,k)$.\r\n\r\nFinally, $a = 40, b = 10$ and $|a-b| = 30$. We can check $\\frac{a+b}{2} = 25$, $\\sqrt{ab} = 20$, and $\\frac{2ab}{a+b} = 16$!", "Solution_2": "Edit: ThAzN1, your solution agrees with mine but you seem to have a slightly wrong formula for harmonic mean. :)\r\n\r\n\r\n\r\n$A = \\frac{a+b}{2} \\in \\mathbb{N} \\implies a \\equiv b \\bmod 2$\r\n\r\n$G = \\sqrt{ab} \\in \\mathbb{N} \\implies ab = G^2$. Suppose WLOG that $a > b$. Then we can write $a = kG, b = \\frac{G}{k}$ for some integer $k | G$, and hence $a = bk^2$. Furthermore, $A = \\frac{b(k^2+1)}{2}$ and $G = bk$. \r\n\r\n$H = \\frac{2ab}{a+b} \\in \\mathbb{N} \\implies Hb(k^2 + 1) = 2b^2 k^2 \\implies H = \\frac{2bk^2}{k^2 + 1}$. Since $k^2, k^2 + 1$ share no common divisors, it follows that $(k^2 + 1) | 2b$. \r\n\r\nWe know that $k \\neq 1$. Now, substitute $c = k^2 - 1$. We are trying to minimize the product $|bc| = |a-b|$, and we can do this by rewriting in terms of $b$ and $c$:\r\n\r\n$A = \\frac{b(c+2)}{2}$\r\n$G = b \\sqrt{c+1}$\r\n$H = \\frac{2b(c+1)}{c+2}$\r\n\r\nWe divide into two cases.\r\n\r\nCase 1: $c$ is even. Write $c = 2d$. We have the condition $2d+2 | 2b \\implies b = e(d+1)$. Now, $c = 3, 8, 15, 24, ...$, so $d = 4, 12, ...$ will guarantee that $G \\in \\mathbb{N}$. Because $c+2 = 2(d+1)$ is even, we have guaranteed that $A \\in \\mathbb{N}$, and because $c+2 | 2b$ we have guaranteed that $H \\in \\mathbb{N}$. The minimum value here is therefore attained when $b = e(d+1) = 1(4+1) = 5$, and $bc = \\boxed{40}$.\r\n\r\nCase 2: $c$ is odd. We require that $c+2 | 2b$ - hence, $b = f(c+2)$. In order that $A \\in \\mathbb{N}$ we require $f = 2g$, so $b = 2g(c+2)$ guarantees $A, G, H \\in \\mathbb{N}$. The minimum value here is therefore attained when $b = 2g(c+2) = 2(3+2) = 10$ and $bc = \\boxed{30}$.", "Solution_3": "OH oops yeah I'll edit it.. good thing that factor of 2 doesn't change anything :)", "Solution_4": "A dumb solution: Let $ab=k^2$ and $a+b=2\\ell$. Then $(a-b)^2 = (a+b)^2 - 4ab=4 \\left( \\ell^2 - k^2 \\right)$, so we have to minimize $\\left| \\ell^2 - k^2 \\right|$. Now, the dirty work: we check some cases, which will lead to the correct answer, I think.", "Solution_5": "[quote=\"t0rajir0u\"]$G = \\sqrt{ab} \\in \\mathbb{N} \\implies ab = G^2$. Suppose WLOG that $a > b$. Then we can write $a = kG, b = \\frac{G}{k}$ for some integer $k | G$,[/quote]\n\nAre you sure ? Counter example : $225 = 9 \\times 25$. But, $\\frac{15}{9}$ is not integer.", "Solution_6": "Now, that this thread has been revived ... true! a rare occasion t0rajir0u has erred. Notice also that post #5 leads verbatim to the solution of post #2; the best shortcut, in my view. From $ab=k^2$ and $a+b=2\\ell$, we need minimize $|a-b|$, or equivalently, minimize $(a-b)^2 = (a+b)^2 - 4ab = 4(\\ell^2 - k^2)$. But $a,b$ are the roots of $\\lambda^2 - 2\\ell\\lambda + k^2$, so $a,b = \\ell \\pm \\sqrt{\\ell^2 - k^2}$, whence we need $\\ell^2 = k^2 + m^2$, a Pythagorean triple. This yields all instances where $\\frac {a+b} {2}$ and $\\sqrt{ab}$ are integer, with the minimum for $|a-b|$ equal to $6$, given for $5^2 = 4^2 + 3^2$, when $\\{a,b\\} = \\{2,8\\}$. The [b]third condition [/b]on $\\frac {2ab} {a+b} = \\frac {k^2} {\\ell}$ to be integer now forces $\\ell \\mid k^2$ and $\\ell \\mid m^2$, thus the minimum for $|a-b|$ equal to $30$, given for $(5\\cdot 5)^2 = (5\\cdot 4)^2 + (5\\cdot 3)^2$, when $\\{a,b\\} = \\{10,40\\}$.", "Solution_7": "Set $a=da_1$ and $b=db_1$ with $(a_1,b_1)=1$. Since $\\sqrt{ab}\\in\\mathbb{N}$, we obtain $a_1,b_1$ are perfect squares, set $a_1=u^2$ and $b_1=v^2$. With this, $(a+b)/2\\in\\mathbb{N}$ implies $2\\mid d(u^2+v^2)$. Furthermore, the harmonic mean of $a$ and $b$ arranges to\n\\[\n\\frac{2}{\\frac{1}{du^2}+\\frac{1}{dv^2}} = \\frac{2du^2v^2}{u^2+v^2}.\n\\]\nAs $(u,v)=1$, we have $(u^2v^2,u^2+v^2)=1$, yielding $u^2+v^2\\mid 2d$. Now if $u^2+v^2$ is odd, then $d=2s(u^2+v^2)$ for some arbitrary $s\\in\\mathbb{N}$, yielding $|a-b| = 2s|u^4-v^4|$, whose smallest value (under $u,v$ being odd) is clearly obtained for $\\{u,v\\}=\\{1,2\\}$ and $s=1$, that is $d=10$ and $(a,b)=(10,40)$. The second case, namely $u^2+v^2$ being even, is handled analogously; hence the smallest value is $|a-b|=|10-40|=30$. " } { "Tag": [ "function", "LaTeX", "search" ], "Problem": "I posted this once but forgot and can't find the old topic.\r\n\r\nHow do I do floor and ceiling function in $ \\text{\\LaTeX}$?", "Solution_1": "The topic is right [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=278238405&t=298226]here[/url]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "If $a_{1},a_{2}, \\ldots ,a_{n}>0$ then:\r\n\r\n$ \\frac{ \\sqrt[k]{a_{1}a_{2} \\ldots a_{k}} + \\ldots + \\sqrt[k]{a_{n-k+1} \\ldots a_{n}}}{ \\binom{n}{k}} \\leq \\sqrt[k]{ \\frac{ a_{1}a_{2} \\ldots a_{k} + \\ldots + a_{n-k+1} \\ldots a_{n}}{ \\binom{n}{k}}}$.\r\n\r\ncheers! :D :D", "Solution_1": "In TeX writeup:\r\n\r\n[tex]\\large \\frac{\\sqrt[k]{a_{1}a_{2}...a_{k}}+\\sqrt[k]{a_{n-k+1}a_{n-k+2}...a_{n}}}{\\binom{n}{k}}\\leq \\sqrt[k]{\\frac{a_{1}a_{2}...a_{k}+a_{n-k+1}a_{n-k+2}...a_{n}}{\\binom{n}{k}}}[/tex].\r\n\r\nIn order to have the inequality valid you need the extra condition 0 < k < n.\r\n\r\nThen, the inequality is trivial: Call $\\sqrt[k]{a_{1}a_{2}...a_{k}}=a$ and $\\sqrt[k]{a_{n-k+1}a_{n-k+2}...a_{n}}=b$; then we have to prove\r\n\r\n[tex]\\large \\frac{a+b}{\\binom{n}{k}}\\leq \\sqrt[k]{\\frac{a^{k}+b^{k}}{\\binom{n}{k}}}[/tex].\r\n\r\nBut since 0 < k < n, we have $\\binom{n}{k}\\leq 2$, so that\r\n\r\n[tex]\\large \\frac{2}{\\binom{n}{k}}\\leq 1[/tex]\r\n\r\nand\r\n\r\n[tex]\\large \\frac{2}{\\binom{n}{k}}\\leq \\sqrt[k]{\\frac{2}{\\binom{n}{k}}}[/tex],\r\n\r\nand finally\r\n\r\n[tex]\\large \\frac{a+b}{\\binom{n}{k}}=\\frac{a+b}{2}\\cdot \\frac{2}{\\binom{n}{k}}\\leq \\frac{a+b}{2}\\cdot \\sqrt[k]{\\frac{2}{\\binom{n}{k}}}\\leq \\sqrt[k]{\\frac{a^{k}+b^{k}}{2}}\\cdot \\sqrt[k]{\\frac{2}{\\binom{n}{k}}}=\\sqrt[k]{\\frac{a^{k}+b^{k}}{\\binom{n}{k}}}[/tex],\r\n\r\nwhere we have used the Power Mean inequality. Proof complete.\r\n\r\nBut I guess you meant a different inequality:\r\n\r\n[tex]\\large \\frac{\\sqrt[k]{a_{1}a_{2}...a_{k}}+...+\\sqrt[k]{a_{n-k+1}a_{n-k+2}...a_{n}}}{\\binom{n}{k}}\\leq \\sqrt[k]{\\frac{a_{1}a_{2}...a_{k}+...+a_{n-k+1}a_{n-k+2}...a_{n}}{\\binom{n}{k}}}[/tex]\r\n\r\n(summation over all possible products of k of the n numbers). This one is, however, even more trivial using power means :-)\r\n\r\n Darij" } { "Tag": [ "MATHCOUNTS" ], "Problem": "This is another one of those problems that is easy, but I forgot how to do it...\r\n\r\nWhen Kelly is not on the moving sidewalk, she can walk the length of the sidewalk in 3 minutes. If she stands on the sidewalk as it moves, she can travel the length in 2 minutes. If Kelly walks on the sidewalk as it moves, how many minutes will it take her to travel the same distance? Assume she always walks at the same speed.\r\n\r\n :oops:", "Solution_1": "Kelly's rate $ \\equal{} \\ r_1$ and sidewalk rate $ \\equal{} \\ r_2$\r\nDistance of sidewalk $ \\equal{} \\ d$. Recall $ dr \\equal{} t$ where $ d$ is distance, $ r$ is rate, and $ t$ is time.\r\nYou're given $ 3 r_1 \\equal{} d$ and $ 2r_2 \\equal{} d$. You want to find some number $ t$ that makes $ t(r_1 \\plus{} r_2) \\equal{} d$ true.\r\n\r\nMultiply the first equation by $ 2$ and the second by $ 3$ to get: $ 6r_1 \\equal{} 2d$ and $ 6r_2 \\equal{} 3d$ and then add them to get \r\n$ 6r_1 \\plus{} 6r_2 \\equal{} 5d \\Rightarrow 6(r_1 \\plus{} r_2) \\equal{} 5d \\Rightarrow \\frac {6}{5}(r_1 \\plus{} r_2) \\equal{} d \\Rightarrow \\boxed{1 \\ minute \\ 12 \\ seconds}$", "Solution_2": "Thanks... But I solved a bit differently...\r\n\r\n :thumbup: :blush: :play_ball:", "Solution_3": "this problem is from mathcounts, and should not be posted in the high school basics forum. in the future, post mathcounts problems in the [b]mathcounts forum[/b]." } { "Tag": [], "Problem": "Derek just went out and bought a 4 GB (gigabyte) iPod Nano for 250 dollars while Eustace bought a 512 MB (half of a gigabyte) iPod Shuffle for a certain amount of money. Given that Eustace's 512 MB iPod Shuffle cost 2.24 times as much per gigabyte than Derek's iPod Nano, how much did Eustace's iPod Shuffle cost? \r\n\r\n[hide=\"Answers\"] [b]70[/b] dollars. By the way, these prices are the exact same current retail prices, so if you've been to your local electronics store and seen the \"Retail Price: 69 dollars\" sign on a 512 GB iPod shuffle, this problem should be pretty easy :p .\n[/hide]\r\n\r\nOn a side note, Iron Fist has a 4 GB iPod Mini. It's just in the middle - too small to have a lot of memory and too big to fit in an Orbit gum pack. >_>", "Solution_1": "[hide]$70$[/hide]\r\nIs there anyone else here besides me who is music deprived?\r\nI have no MP3 Player and crappy internet and only like 3 albums on an itunes that freezes up my computer when played too long.", "Solution_2": "I have nothing :P", "Solution_3": "[hide]$1/8*250=125/4*2.24=125*14/25=70$[/hide]", "Solution_4": "I have a nano. But i just got it and there isn't much on it. I'm bringing it to nats anyway.", "Solution_5": "[quote=\"ehehheehee\"]I have nothing :P[/quote]\r\n\r\nMe too.\r\nThen again, I have an alarmclock/radio.", "Solution_6": "[hide][i]4000MB : $250 512MB : 1/2 * 2.24 *$250/4\n=[b]70[/b][/i] :lol:[/hide]", "Solution_7": "I have TWO ipods.", "Solution_8": "[quote=\"mbabbitt2003\"]I have TWO ipods.[/quote]\r\n\r\nThat's great, if you give me half of an iPod we'll both have the same amount.", "Solution_9": "If YOU give me a half of an ipod, I will have 5 times as many as you.", "Solution_10": "Haha.\r\n\r\nIf mbabbit gives treething 1/2 of a jPod, they will have the same amount.\r\nIf treething gives mbabbit 1/2 of a jPod, mbabbit will have 5 times as many jPods as treething.\r\nHow many jPods does...[hide]Klebian[/hide] have?", "Solution_11": "[hide]Klebian[/hide] has 0 [b]j[/b]pods.", "Solution_12": "woah, you're good.\r\nand yes, i did jpod on purpose, in case you were \"subtly\" implying something.", "Solution_13": "No, I was just pointing out the j to people who didn't notice.", "Solution_14": "um i got $1400.00 anyone have any ideas as to how that happened because im severly confused \r\n :huh:", "Solution_15": "[quote=\"MonkeyBoy\"]um i got $\\$1400.00$ anyone have any ideas as to how that happened because im severly confused \n :huh:[/quote]\r\nFirst, I thought you were talking about real life, as in \"I just got a whole bunch of money and I don't know how I got it.\" :rotfl: And sorry, I don't know how you got 1400. It should be $\\frac{.5}{4}*2.24*250=\\boxed{70}$." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "[b]Let[/b] $ A,B,C \\in S_{n}(R)$ [b]such that [/b]: $ AB\\equal{}BA,BC\\equal{}CB,CA\\equal{}AC$\r\n[b]Slove the system of matrix equations :[/b]\r\n $ \\begin{cases} e^{A}\\minus{}e^{A\\minus{}B}\\equal{}B \\\\ e^{B}\\minus{}e^{B\\minus{}C}\\equal{}C \\\\ e^{C}\\minus{}e^{C\\minus{}A}\\equal{}A \\end{cases}$\r\n[b]Here[/b] $ S_{n}(R)$ [b]are set real symmetric nxn matrices .[/b]\r\n------------------------------------------------\r\nDVG : OOO :wink:", "Solution_1": "One more time: Note that the matrices involved are simultaneously diagonalizable. \r\n\r\nps: Is that homework of a linear algebra class? I must admit that it smells like ... :P" } { "Tag": [], "Problem": "If $p$ and $q$ are the roots of the equation $(x-a)(x-b)-2x+1=0$, then what are the roots of the equation $(x-p)(x-q)+2x-1=0$?\r\n\r\n(a)$a$, $b$\r\n(b)$-a$, $-b$\r\n(c)$\\displaystyle\\frac{1}{a}$, $\\frac{1}{b}$\r\n(d)$-\\frac{1}{a}$, $-\\frac{1}{b}$\r\n(e) None of the above", "Solution_1": "$p+q=b+a+2$\r\n\r\nand for the second equation, the sum of the roots is $q+p-2=b+a$, so the roots are $a,b$", "Solution_2": "That seems reasonable. Could you please explain to me how you got $p+q=b+a+2$ though?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a, b, c$ be positive real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove o disprove that\r\n\r\n$ \\sum\\frac{4ab \\plus{} 3}{c \\plus{} 1}\\le 8$ .\r\n\r\n__________\r\n :roll:", "Solution_1": "[quote=\"Ligouras\"]Let $ a, b, c$ be positive real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove o disprove that\n\n$ \\sum\\frac {4ab \\plus{} 3}{c \\plus{} 1}\\le 8$ .\n\n__________\n :roll:[/quote]\r\n\r\nMy solution. :)", "Solution_2": "ineq $ \\Leftrightarrow\\sum(4ab\\plus{}3)(b\\plus{}1)(a\\plus{}1)\\leq 8\\prod(a\\plus{}1)\\Leftrightarrow 4\\sum a^2b^2\\plus{}3\\sum ab\\leq 20abc\\plus{}1\\Leftrightarrow 4q^2\\plus{}3q\\leq 28r\\plus{}1$\r\nCase $ 0\\leq q\\leq\\frac{1}{4}\\Rightarrow LHS\\leq 1;RHS\\geq 1$\r\nCase $ \\frac{1}{4}\\leq q\\leq\\frac{1}{3}:$\r\n$ 4q^2\\plus{}3q\\minus{}28r\\minus{}1\\leq 4q^2\\minus{}\\frac{85}{9}q\\plus{}\\frac{19}{9}\\equal{}(q\\minus{}\\frac{1}{4})(4q\\minus{}\\frac{76}{9})\\leq 0$\r\nWe are done :)", "Solution_3": "After expand, it's becomes:\r\n$ 4\\sum\\ a^2b^2 \\plus{} 4\\sum\\ ab(a\\plus{}b) \\leq\\ (a\\plus{}b\\plus{}c)^3\\plus{}4\\sum\\ ab\\plus{}8abc$\r\nObviously because:\r\n$ (a\\plus{}b\\plus{}c)^3\\plus{}8abc \\geq\\ 4\\sum\\ ab(a\\plus{}b)$ (Schur)\r\nAnd $ 4 \\sum\\ a^2b^2 \\leq\\ \\sum\\ ab(a\\plus{}b)^2 \\leq\\ \\sum\\ ab(a\\plus{}b\\plus{}c)^2 \\equal{} ab\\plus{}bc\\plus{}ca$\r\nWe have done. :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that in any triangle $ABC$ the following inequality holds:\r\n\r\n $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\leq\\frac{3}{2}\\cdot\\frac{s}{r(4R+r)}$ \r\n\r\n standard notations.", "Solution_1": "We have : $ ab+ab+bc = s^2+4Rr+r^2 ; abc =4sRr $ \r\n Change the problem into the expression of $ s,R,r $ and use the famous ineq : $ s^2 \\geq 16Rr -5r^2 $ ;)" } { "Tag": [ "inequalities" ], "Problem": "\u0393\u03b5\u03b9\u03ac \u03c3\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03bc\u03ad\u03bd\u03b1. \u03a3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03ae\u03c1\u03b9\u03b1 \u03c3\u03b5 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c0\u03bf\u03c5 \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03cd\u03bd\u03c4\u03b1\u03b9 \u03c0\u03b1\u03b8\u03b9\u03b1\u03c3\u03bc\u03ad\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac, \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03c0\u03b9\u03b8\u03b1\u03bd\u03cc\u03c4\u03b1\u03c4\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03b9 \u03b1\u03c0\u03cc \u03b5\u03c3\u03ac\u03c2 \u03c3\u03c4\u03bf \u03bc\u03ad\u03bb\u03bb\u03bf\u03bd \u03b8\u03b1 \u03b5\u03af\u03c3\u03c4\u03b5 \u03b5\u03c1\u03b5\u03c5\u03bd\u03b7\u03c4\u03ad\u03c2 \u03c0\u03c1\u03ce\u03c4\u03b7\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae\u03c2. \u0398\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce \u03cc\u03c0\u03bf\u03c4\u03b5 \u03b2\u03c1\u03af\u03c3\u03ba\u03c9 \u03c4\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf (\u03b5\u03bb\u03ad\u03c9 \u03c6\u03c1\u03bf\u03bd\u03c4\u03b9\u03c3\u03c4\u03b7\u03c1\u03af\u03c9\u03bd) \u03bd\u03b1 \u03c0\u03bf\u03c3\u03c4\u03ac\u03c1\u03c9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2. \u039e\u03b5\u03ba\u03b9\u03bd\u03ce \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b7\u03bd \u03c3\u03bd\u03bf\u03bc\u03c0\u03ac\u03c1\u03bf\u03c5\u03bd \u03bf\u03b9 \u03ad\u03bc\u03c0\u03b5\u03b9\u03c1\u03bf\u03b9 \u03bb\u03cd\u03c4\u03b5\u03c2 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03ae\u03c4\u03c9\u03bd \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03b3\u03bd\u03c9\u03c3\u03c4\u03bf\u03cd \u03bb\u03ae\u03bc\u03bc\u03b1\u03c4\u03bf\u03c2.\r\n\r\n\u0391\u03bd a, b, c \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03ba\u03b1\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 abc=1 \u03c4\u03cc\u03c4\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9:\r\n\r\n $ \\frac {1}{a^{6}} \\plus{} \\frac {1}{b^{6}} \\plus{} \\frac {1}{c^{6}} \\geq a^{3} \\plus{} b^{3} \\plus{} c^{3}$\r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03b1\u03bd a, b, c \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af \u03c4\u03cc\u03c4\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03b5\u03af\u03c4\u03b5 \u03c4\u03b7\u03bd \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03b7 \u03c4\u03b9\u03bc\u03ae \u03c4\u03bf\u03c5 \u03c0\u03c1\u03ce\u03c4\u03bf\u03c5 \u03bc\u03ad\u03bb\u03bf\u03c5\u03c2.", "Solution_1": "[quote=\"KostasGT\"]\u0393\u03b5\u03b9\u03ac \u03c3\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03bc\u03ad\u03bd\u03b1. \u03a3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03ae\u03c1\u03b9\u03b1 \u03c3\u03b5 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c0\u03bf\u03c5 \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03cd\u03bd\u03c4\u03b1\u03b9 \u03c0\u03b1\u03b8\u03b9\u03b1\u03c3\u03bc\u03ad\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac, \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03c0\u03b9\u03b8\u03b1\u03bd\u03cc\u03c4\u03b1\u03c4\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03b9 \u03b1\u03c0\u03cc \u03b5\u03c3\u03ac\u03c2 \u03c3\u03c4\u03bf \u03bc\u03ad\u03bb\u03bb\u03bf\u03bd \u03b8\u03b1 \u03b5\u03af\u03c3\u03c4\u03b5 \u03b5\u03c1\u03b5\u03c5\u03bd\u03b7\u03c4\u03ad\u03c2 \u03c0\u03c1\u03ce\u03c4\u03b7\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae\u03c2. \u0398\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce \u03cc\u03c0\u03bf\u03c4\u03b5 \u03b2\u03c1\u03af\u03c3\u03ba\u03c9 \u03c4\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf (\u03b5\u03bb\u03ad\u03c9 \u03c6\u03c1\u03bf\u03bd\u03c4\u03b9\u03c3\u03c4\u03b7\u03c1\u03af\u03c9\u03bd) \u03bd\u03b1 \u03c0\u03bf\u03c3\u03c4\u03ac\u03c1\u03c9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2. \u039e\u03b5\u03ba\u03b9\u03bd\u03ce \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b7\u03bd \u03c3\u03bd\u03bf\u03bc\u03c0\u03ac\u03c1\u03bf\u03c5\u03bd \u03bf\u03b9 \u03ad\u03bc\u03c0\u03b5\u03b9\u03c1\u03bf\u03b9 \u03bb\u03cd\u03c4\u03b5\u03c2 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03ae\u03c4\u03c9\u03bd \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03b3\u03bd\u03c9\u03c3\u03c4\u03bf\u03cd \u03bb\u03ae\u03bc\u03bc\u03b1\u03c4\u03bf\u03c2.\n\n\u0391\u03bd a, b, c \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03ba\u03b1\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 abc=1 \u03c4\u03cc\u03c4\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9:\n\n $ \\frac {1}{a^{6}} \\plus{} \\frac {1}{b^{6}} \\plus{} \\frac {1}{c^{6}} \\geq a^{3} \\plus{} b^{3} \\plus{} c^{3}$\n\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03b1\u03bd a, b, c \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af \u03c4\u03cc\u03c4\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03b5\u03af\u03c4\u03b5 \u03c4\u03b7\u03bd \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03b7 \u03c4\u03b9\u03bc\u03ae \u03c4\u03bf\u03c5 \u03c0\u03c1\u03ce\u03c4\u03bf\u03c5 \u03bc\u03ad\u03bb\u03bf\u03c5\u03c2.[/quote]\r\n\r\nKanoume omonuma ta klasmata kai pollaplasiazoume me 2 kai pairnoume oti $ \\sum_{sym}a^{6}b^{6}\\geq \\sum_{sym}a^{3}$. Pollaplasizaoume kathe oro tou deuterou melous me $ a^{3}b^{3}c^{3}$ kai paratiroume oti isxuei apo Muirhead me triades tis $ (6,6,0)\\succ (6,3,3)$", "Solution_2": "you killed it! ego tha ekana to eksis $ LHS\\equal{}\\sum{\\frac{b^3c^3}{a^3}}$ kai meta $ {\\frac{b^3c^3}{a^3}}\\plus{}{\\frac{a^3c^3}{b^3}}\\geq 2c^3$ prostheoume kai telos. to extremum pou zitas einai apli efarmogi am-gm.", "Solution_3": "\u039c\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7: \u0391\u03c0\u03cc \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7, $ \\frac{1}{a^6} \\plus{} \\frac{1}{b^6} \\plus{} \\frac{1}{c^6} \\geqslant \\frac{1}{a^3b^3} \\plus{} \\frac{1}{b^3c^3} \\plus{} \\frac{1}{c^3a^3} \\equal{} c^3 \\plus{} a^3 \\plus{} b^3$", "Solution_4": "[quote=\"Nick Rapanos\"]you killed it! ego tha ekana to eksis $ LHS \\equal{} \\sum{\\frac {b^3c^3}{a^3}}$ kai meta $ {\\frac {b^3c^3}{a^3}} \\plus{} {\\frac {a^3c^3}{b^3}}\\geq 2c^3$ prostheoume kai telos. to extremum pou zitas einai apli efarmogi am-gm.[/quote]\r\n\r\ntelika i lusi einai sosti i oxi? mepiase anxos tora :P", "Solution_5": "\u03a0\u03bf\u03bb\u03b5\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2, \u03bf\u03c1\u03b9\u03c3\u03c4\u03b5 \u03b1\u03bb\u03bb\u03b7 \u03bc\u03b9\u03b1:\r\n\r\n$ LHS \\equal{} \\sum{a^6b^6} \\equal{} \\sum{(b^3a^3)^2} \\geq \\sum{a^3b^3b^3c^3} \\equal{} a^3b^3c^3\\sum{a^3} \\equal{} \\sum{a^3} \\equal{} RHS$", "Solution_6": "[quote=\"\u0393\u03b9\u03ce\u03c1\u03b3\u03bf\u03c2\"][quote=\"Nick Rapanos\"]you killed it! ego tha ekana to eksis $ LHS \\equal{} \\sum{\\frac {b^3c^3}{a^3}}$ kai meta $ {\\frac {b^3c^3}{a^3}} \\plus{} {\\frac {a^3c^3}{b^3}}\\geq 2c^3$ prostheoume kai telos. to extremum pou zitas einai apli efarmogi am-gm.[/quote]\n\ntelika i lusi einai sosti i oxi? mepiase anxos tora :P[/quote]\r\n\r\nxaxax i didnt go through the calculations but i trust you! :P", "Solution_7": "\u0398\u03b1\u03c5\u03bc\u03ac\u03c3\u03b9\u03b5\u03c2 \u03c3\u03ba\u03ad\u03c8\u03b5\u03b9\u03c2 \u03cc\u03bb\u03b5\u03c2. \u0397 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ac\u03c3\u03c4\u03b7\u03ba\u03b5 \u03bc\u03b5 \u03c4\u03bf \u03c3\u03ba\u03b5\u03c0\u03c4\u03b9\u03ba\u03cc \u03c4\u03bf\u03c5 NickNafplio, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b2\u03b1\u03c3\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf:\r\n\r\n$ a^{2} \\plus{} b^{2} \\plus{} c^{2} \\geq ab \\plus{} bc \\plus{} ca$\r\n\r\n\u039a\u03b1\u03bb\u03cc \u03b2\u03c1\u03ac\u03b4\u03c5 \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2." } { "Tag": [ "function", "trigonometry", "logarithms", "calculus", "integration", "algebra", "floor function" ], "Problem": "Suppose that $ b, X $ are positive integers $(b>2)$. Assuming that $b, X, j , (j \\in \\{0,1,...,b-1\\}), $ are given, [b] how many times the digit $ j $ appears the representation of $ X $ in base $b$ ?[/b] Let $ N_j=N_j(X; b) $ be this number, e.g. $ N_0(20052006; 10)=4\\; . $ Using only floor function (integer part), it's possible to find a closed form for $ N_j $ (perhaps as a sum of a convergent series) ?\r\nLet $[ \\cdot ] $ denote the integer part function and define $ G_b:{\\mathbb R}\\to {\\mathbb R} $ by $ \\displaystyle G_b(t)= \\left\\{\\begin{array}{lcl} \\displaystyle \\frac{1}{b^2}\\left( \\frac{\\sin{\\pi t} }{\\sin{\\frac{\\pi t}{b}}}\r\n\\right)^2 & , & t\\ne b\\cdot p\\\\ 1 &, & t=b\\cdot p \\end{array}\\right.\\; ,\\; p\\in {\\mathbb Z} \\; .$\r\nTry to prove/disprove that \r\n\\[\r\nN_j(X;b)=\\sum\\limits_{k=0}^{\\left[\\log_b{X}\\right]}G_b\r\n\\left(\\left[\\frac{X}{b^k}-j\\right]\\right)\\; \\; \\; \\; \\; \\;\r\n\\mbox{(according to Robert Israel).}\r\n\\]", "Solution_1": "[quote=\"flip2004\"]Suppose that $b, X$ are positive integers $(b>2)$. Assuming that $b, X, j , (j \\in \\{0,1,...,b-1\\}),$ are given, [b] how many times the digit $j$ appears the representation of $X$ in base $b$ ?[/b] Let $N_j=N_j(X; b)$ be this number, e.g. $N_0(20052006; 10)=4\\; .$ Using only floor function (integer part), it's possible to find a closed form for $N_j$ (perhaps as a sum of a convergent series) ?\nLet $[ \\cdot ]$ denote the integer part function and define $G_b: {\\mathbb R}\\to {\\mathbb R}$ by $\\displaystyle G_b(t)= \\left\\{\\begin{array}{lcl} \\displaystyle \\frac{1}{b^2}\\left( \\frac{\\sin{\\pi t} }{\\sin{\\frac{\\pi t}{b}}} \\right)^2 & , & t\\ne b\\cdot p\\\\ 1 &, & t=b\\cdot p \\end{array}\\right.\\; ,\\; p\\in {\\mathbb Z} \\; .$\nTry to prove/disprove that \n\\[ N_j(X;b)=\\sum\\limits_{k=0}^{\\left[\\log_b{X}\\right]}G_b \\left(\\left[\\frac{X}{b^k}-j\\right]\\right)\\; \\; \\; \\; \\; \\; \\mbox{(according to Robert Israel).} \\][/quote]\r\nI (re-)discovered that \r\n\\[ \\begin{array}{|c|} \\hline \\\\ \\displaystyle N_j(X,b)=\\displaystyle \\sum\\limits_{k=1}^{\\infty}\\left( \\left[\\frac{X}{b^k} -\\frac{j}{b}\\right] - \\left[\\frac{X}{b^k} - \\frac{j+1}{b} \\right] \\right)\\\\ \\\\ \\hline \\end{array}\\; , \\]\r\nwhere $[\\; \\cdot\\; ] \\;$ denotes integral part.\r\n[b]Reference:[/b]\r\nGazeta Matematica (Seria B), anul XCIV (1989),nr.1,p.35,[i] Problem C :572[/i].,proposed by Alex. L ." } { "Tag": [ "function", "limit", "calculus", "calculus computations" ], "Problem": "Suppose $ f(x) \\equal{} \\begin{cases} 0 \\ \\ \\ \\text{if} \\ x \\notin \\mathbb{Q} \\\\\r\n1/q \\ \\ \\ \\text{if} \\ x \\in \\mathbb{Q} \\ \\text{and} \\ x \\equal{} p/q \\ \\text{in lowest terms} \\end{cases}$\r\n\r\n1. Fix $ t \\in \\mathbb{R}$ and $ n \\in \\mathbb{N}$. Show that $ f$ maps finitely many elements of $ (t \\minus{} 1/2, t \\plus{} 1/2)$ to $ 1/n$.\r\n\r\nSo the length of the interval is $ 1$. I think its $ n \\minus{} 1$. But that is intuition. How do you make it rigorous?\r\n\r\n\r\n2. Prove that $ f$ is continuous at every irrational number. So $ \\forall \\epsilon > 0$, $ \\exists \\delta > 0$ such that $ |x \\minus{} a| < \\delta \\implies |f(x)| < \\epsilon$. Do we want elements that dont map to $ 1/n$?", "Solution_1": "[quote=\"lordmahone123\"]I think it's $ n\\minus{}1$.[/quote]It's not. It's [i]at most[/i] $ n\\minus{}1$. How many fractions with denominator $ \\frac1n$ are there in that range? Some of them won't be in lowest terms; why isn't that a problem?\r\n\r\n2: The essential rule for finding $ \\delta$ here is to look for the closest bad point $ y$ at which $ |f(y)|\\ge\\epsilon$, and choose $ \\delta\\equal{}|y\\minus{}x|$. How can you show that there is a closest bad point?", "Solution_2": "[quote=\"jmerry\"][quote=\"lordmahone123\"]I think it's $ n \\minus{} 1$.[/quote]It's not. It's [i]at most[/i] $ n \\minus{} 1$. How many fractions with denominator $ \\frac1n$ are there in that range? Some of them won't be in lowest terms; why isn't that a problem?\n\n2: The essential rule for finding $ \\delta$ here is to look for the closest bad point $ y$ at which $ |f(y)|\\ge\\epsilon$, and choose $ \\delta \\equal{} |y \\minus{} x|$. How can you show that there is a closest bad point?[/quote]\r\n\r\n\r\nBecause the function is defined so that $ x$ is in lowest terms. You can show there is a closest bad point by using the Well Ordering Property? But dont you want to pick $ \\delta$ to be the minimum of 2 values?\r\n\r\nSo you end up with $ 1\\plus{}1\\plus{}2\\plus{}3\\plus{}5\\plus{}7\\plus{} \\ldots$.", "Solution_3": "That doesn't give me any confidence that you understood what I said.\r\n\r\nAlso, it's not about primes.", "Solution_4": "2)\r\nLet x be irrational.\r\nDefine g(n) as follows. Consider the fractions in lowest terms with denominator $ \\leq$ n. They are bounded away from $ x$ for each n. Let g(n) be the smallest distance between one of those fractions and $ x$. Then, $ \\lim_{n\\to\\infty}g(n) \\equal{} 0$, but $ g(n)\\neq0$ for all $ n$, and $ g$ is decreasing. So if we want to find some $ \\delta$ such that for all rationals $ q$ in $ (x \\minus{} \\delta,x \\plus{} \\delta)$, $ f(q) < \\epsilon$. So take $ n$ to be the least positive integer such that $ g(n) < \\epsilon$, then $ \\delta \\equal{} 1/n$. Since all fractions within $ \\epsilon$ of x have denominators $ \\geq n$ by choice of n, this works I think.", "Solution_5": "[quote=\"jmerry\"][quote=\"lordmahone123\"]I think it's $ n \\minus{} 1$.[/quote]It's not. It's [i]at most[/i] $ n \\minus{} 1$. How many fractions with denominator $ \\frac1n$ are there in that range? Some of them won't be in lowest terms; why isn't that a problem?\n\n2: The essential rule for finding $ \\delta$ here is to look for the closest bad point $ y$ at which $ |f(y)|\\ge\\epsilon$, and choose $ \\delta \\equal{} |y \\minus{} x|$. How can you show that there is a closest bad point?[/quote]\r\n\r\nBasicaly we are trying to show that $ f$ maps only finitely many rational numbers in lowest terms in this interval to $ 1/n$. Let $ I \\equal{} (t \\minus{} 1/2, t \\plus{} 1/2)$. Let $ x \\equal{} p/n \\in I$ in lowest terms. Then $ p$ is coprime to $ n$. Can we somehow use this to show the maximum number is $ n \\minus{} 1$? E.g. $ \\varphi(n) \\equal{} n\\minus{}1$?", "Solution_6": "Don't worry about that stuff. How many fractions $ \\frac{k}{n}$, not necessarily in lowest terms, are in that interval?", "Solution_7": "[quote=\"jmerry\"]Don't worry about that stuff. How many fractions $ \\frac {k}{n}$, not necessarily in lowest terms, are in that interval?[/quote]\r\n\r\n$ n \\minus{} 1$ by inspection. I let $ t \\equal{} 0$ and $ n \\equal{} 3$. Thus number in lowest terms is $ \\leq n\\minus{}1$.", "Solution_8": "You don't get to choose $ t$ and $ n$. This should apply for all choices. Also, your count is wrong; $ \\frac03$ is a fraction with denominator 3.", "Solution_9": "You fix $ t$ and $ n$. So the length of the interval really only matters. So number of fractions is $ n$. Thus number of fractions in lowest terms is $ \\leq n$.\r\n\r\n$ p/n < t\\plus{}1/2$\r\n\r\nSo $ p < n(t\\plus{}1/2)$." } { "Tag": [ "calculus", "vector", "search", "geometry", "linear algebra", "graph theory", "real analysis" ], "Problem": "This semester I have found a new love for pure mathematics. Having gone to study actuarial science in university on a hunch rather than having any previous math interest, I currently lack much of the skills that the pure math majors currently have. I am switching from the honors math courses (linear algebra 1 and calculus 2) into advanced linear algebra 2 and advanced calculus 3.\r\n\r\nI am asking the people here that have done the advanced stream what I can do to ease the transition? I have a 4 month work term now and have a lot of free time studying.\r\n\r\nHere is my current book list to catch up:\r\n\r\nRudin's Introduction to Mathematical Analysis (Not sure if I should start with this for analysis, perhaps spivak's calculus?)\r\nHow to Solve It and How to Prove it (Going through these now, very good for proofs/problem solving)\r\n\r\nI still need a good linear algebra book.", "Solution_1": "I'd say starting with Spivak is a good idea. I've also heard Wade's \"introduction to analysis\" covers the material in a way very similar to 147/148 (although it would probably be a bit dry compared to Spivak). As for linear algebra, the standard 146/245 textbook is Linear Algebra by (oh god... the author escapes me right now but I'll find it in the morning). I'd say that as far as the transition goes, 247 will be a much bigger jump than 245 because I found 146 to be a bit closer to 136 (although everything was catered more to abstract vector spaces and not just a select few finite dimensional spaces) whereas 147/148 differed significantly in content on some topics when compared to 137/138.", "Solution_2": "To follow up, the book is Linear Algebra by Friedberg, Insel and Spence.", "Solution_3": "Thank you very much, your help is much appreciated. To my luck, there is an international version.", "Solution_4": "Very cool, I did the same thing. \r\nI went from regular lin alg 1 and calc 2 to adv lin alg 2 and calc 3...\r\nIf you can get a hold of notes and assignments from a previous term, that would help you a lot. \r\nWhen I did calc 3 there were course notes, not a textbook, so you can try and see if Graphics is still selling the coursenotes from the Winter term.", "Solution_5": "So, the last post was a while ago, but I was wondering what you guys think of U Waterloo, what you like, what you don't. Do yuou get lots of time with math profs if you need it?\r\n\r\nI'm seriously looking at U Waterloo for BMath next year (started the Ontario Application thing today) and then doing the teaching option... I've talked to a bunhc of people, but no math majors... care to shed some light?", "Solution_6": "I would suggest you nose around the UW website and see if you can find faculty members that can answer your questions about Teaching option. As for UW, the most valuable experience would be the COOP program, although I am not sure how that plays with students with teaching options. And yes, you can walk into any math prof's office and they are usually very nice and would be more than happy to chat with you (although I've occasionally heard stories of hard nosed profs)", "Solution_7": "If anyone is in the double degree program (BMath/BBA) at WLU and UW, how hard is it? Is the Co-op good or a hindrance, and is the business part boring? Also, what kinds of things would you learn in the business courses?", "Solution_8": "Good questions. I can't answer anything about the Teaching Option or the Double Degree option, but I can talk about UWaterloo in general. I'd be happy to answer anything I can :) \r\n\r\nI love UW. I find the advanced math courses interesting, in general the math professors are really really good. They are friendly, not intimidating, and always willing to help. I'm not sure what you guys would like to know on the side...\r\n\r\nI'm in the co-op program here at Waterloo, I'm actually starting my 4th of 6 work terms in January. I think it's a terrific program. I must admit we're not fond of Jobmine (the job search system students use), and interviews can come at some stressful times. If you're in high demand too, it's hard to juggle all your interviews and class times. Co-op is great because you get more than a year's worth of real world job experience before you graduate, which makes for a very attractive resume when you do enter the real world. Not only that, but with 6 possible co-op terms, it allows you the flexibility to try different areas and see what tickles your fancy. More often than not you get introduced into fields of work that you never thought you'd be interested in. It's also relatively common that the place where you do your last co-op term will extend an offer for permanent placement upon graduation. If you're looking at a Business degree, I highly recommend co-op, it will let you develop real business skills early on. There are so many companies that higher Waterloo co-ops, some would argue Google, Amazon, and Microsoft being among the more attractive companies. \r\n\r\nAnother thing I love about UW is the people. There are tons of clubs and events where you can meet tons of great people, always willing to help out students or just relax between assignments. For instance, if you're in math, the PMAMC&OC (Pure math, applied math, combinatorics and optimization club) is one club full of math students who like to hang out, or do math, or take a break and play frisbee. I hear rumors that some other university math departments can get pretty competitive, and it's not really the case in Waterloo. \r\n\r\nDownsides....first of all, the town of Waterloo itself doesn't have much to do. It's kind of a bubble in fact. There's movie theaters and bars and stuff, but if you're with a group of friends on a friday night, sometimes it just seems like there's nothing to do, Waterloo is pretty small. You get used to it after a while though. And I usually console myself with trips into Toronto (particularly if I want a good mall, being a girl who likes to shop, heh). \r\n\r\nSecond downside is sometimes you can get stuck with a crappy co-op job. In particular this seems to happen to first years, where you don't do too much and seem like a glorified secretary. Don't worry, it does get better as long as you don't get bad evaluations. \r\n\r\nAll in all, university is what you make of it. Make sure you take everything into consideration though...even the town it's in. Though it might not seem like it, it can make a difference. =) \r\nI hope that helped...if anyone wants anything more specific, ask away, or message me!", "Solution_9": "Less than a month till my coop term is over and back to school!\r\n\r\nHas anyone taken CO 350, linear optimization and Math 249 Adv Combinatorics?\r\n\r\nHow are they as courses, in terms of interestingness and difficulty?\r\n\r\nAlso, how are the upper year pure math courses in terms of difficulty? How many is it recommended to take per term? 4?", "Solution_10": "As far as I know CO355 is the preferred course in terms of intensity, and for the pure math courses there are two distinct levels of intensity as well.", "Solution_11": "@samspotting\r\n\r\nWell now it is the new term. Are you in Math 249? If so I can't add detail, but if not I can tell you how it has been so far (I am in it this winter term :) )", "Solution_12": "Yes, I am having an absolute blast. Honours math was ok... but I hated how the profs didnt give a rats ass about the proofs and a lot of the time it became a repeating computational exercise.\r\n\r\nAdvanced courses so far are manageable. I am finding advanced lin alg is easier than advanced calc, but that might change. I have mixed feelings about advanced combinatorics, I don't like how half the stuff we are doing there's no explanation other than it just works.", "Solution_13": "Yes that, and I am finding it rather difficult =) Although it is enlightening.", "Solution_14": "Ok things have changed lol\r\n\r\nAdvanced calc is easy, advanced lin alg is really hard :( almost failed the midterm.\r\n\r\nThe advanced combinatorics midterm was really hard as well.", "Solution_15": "Ehh how are you finding this graph theory stuff? I prefer the enumeration ^_^ It was easier.\r\nAnd yea it was difficult. This is my first advanced class and it is definitely a bump up :(\r\nMarcoux has a reputation for getting very hard. Just pray you don't have him for Real Analysis :P (PMath 351). Lawrence teaches it this summer, might be a good time to take it instead.", "Solution_16": "graph theory is ok, what i found really really helped was that I did a book on proofs that focused on relations and set theory before the course, so my mastery of proofs involving relations and sets is pretty good.\r\n\r\nThe midterm was brutal though, I didnt get the last question and the conway's checkers game (grr 2/3 theorem). Gona have to pick it up in the final.\r\n\r\nHis third assignment was ridiculous, had so much long computation omg. Im not made of time lol. Spent 3 hours correcting myself on the first question.", "Solution_17": "I love reading your conversations about classes - i'm really not sure what to expect next year in terms of academic stuff. You hear all about rez and food and life, but as for the actual courses I don't know what to expect. \r\n\r\nI'm registered right now for Honors mathematics co-op and then the teaching option. But I'm not sure about the honors part. I'm taking (and doing well in) AP Calculus right now, but my math 12 background was a little sketchy, so Euclid was definitely a challenge. \r\n- Someone mentioned that they were in the honors program - what is the difference between that and regular non-honors math? Who would you suggest the honors courses to? \r\n\r\nThanks,", "Solution_18": "Honours math is the normal program at Waterloo. Everyone starts in the honours math program, and the people who are unable to maintain a sufficient average are forced to drop into the 3-year general program. Also, I think you are no longer eligible for co-op once you are in the general program, and can't take many of the courses offered by the faculty of math.", "Solution_19": "Don't worry about it, I came to waterloo having not taken math for a year (finished the math stream in gr 11) and failed the math assesment test with something like 3/15. \r\n\r\nI got excellent standing in both my first and second semester (>80%). Honours math is easy.", "Solution_20": "Ok, that makes a lot of sense. and now I feel a whol lot better about the whole math thing. Thanks a lot:)", "Solution_21": "Are you going to be in Waterloo this fall, andrea.lee?", "Solution_22": "Yes, I have a conditional acceptance based on my mark in Calculus (i'm finished now), so UW is my plan for September. I'm really excited - I can't wait. Are you there/planning on going?", "Solution_23": "I'm going to Waterloo Honours Math Co-op in September too!!! :lol:", "Solution_24": "Ahh actually I will be third year come this fall. I am also an Orientation Leader so maybe we'll see each other around, both of you ;)" } { "Tag": [ "\\/closed" ], "Problem": "When I logged in today, I noticed that the way the forums were display was not as it has been. For example, I have the High School forums showing usually but, today it was missing. Also the Contests & Programs section has new sections that are uncollapsable. Though, after I went to my preferences and reapplied them, it is back to normal except for those sections under Contests & Programs.\r\nSo, it is not really a problem, just thought I say something. :)", "Solution_1": "We had a small problem when uploading some updated files. Problems should be fixed now though.", "Solution_2": "Edit: if you have any problems please erase your cookies (or go to the Preferences page, hide forums and click submit).", "Solution_3": "Problem is fixed." } { "Tag": [ "limit", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "I can not understand the line denoted by red line .\r\n can you help me ? \r\nthanks", "Solution_1": "I think like that .\r\nBecause $ f(x_n)\\geq x_n$, then $ \\lim_{n\\to\\infty} f(x_n) \\geq x_0$.\r\nBecause $ f$ is increasing, $ f(x_n)\\leq f(x_0)0$. Since $ x_0\\plus{}\\epsilon\\not\\in A$, $ f(x_0\\plus{}\\epsilon)0$ $ \\exists N$ such that $ \\forall n>N$\r\nwe have $ |f(\\minus{}x)\\minus{}L|<\\epsilon$", "Solution_1": "[quote=\"azatkgz\"]Is just $ \\lim_{x\\rightarrow \\minus{} \\infty}f(x) \\equal{} \\lim_{x\\rightarrow \\minus{} \\infty}f( \\minus{} x)$[/quote]\n\nIncorrect!\n\n[quote=\"azatkgz\"]Given a function $ f: R\\rightarrow R$ and a number L,write down a definition of the statement\n\n$ \\lim_{x\\rightarrow \\minus{} \\infty}f(x) \\equal{} L$[/quote]\r\n\r\nThe definition is as follows: for $ \\forall \\epsilon > 0$ $ \\exists N>0$ such that $ \\forall x <\\minus{} N$ we have $ |f(x) \\minus{} L| < \\epsilon$" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Show that if $ G$ is a group and $ H$ is a subgroup of $ G$ with index $ 2$, then $ H$ is a normal subgroup of $ G$. \r\n\r\nI just need a hint, can anyone help?", "Solution_1": "If $ g \\notin H$ then $ gH \\equal{} H'$ where $ H' \\equal{} G\\minus{}H$\r\n\r\nSimilarly $ Hg \\equal{} H'$\r\n\r\nHence gH = Hg making H a normal subgroup" } { "Tag": [ "inequalities", "triangle inequality" ], "Problem": "A quadrilater has side lengths 3,3,x, and y. We choose x randomly, and then we choose y (x and y are integers). Let N be the number of possible y after x is chosen. What is the maximum of N.", "Solution_1": "[hide=\"Solution\"]$4^{9}+9^{4}= 2^{18}+3^{8}$, which from memorization of powers of two and three is \n$262144+6561 = 268705$\nWhich is divisible by 5:\n$5*53741$\n\nAfter some ugly prime checking, I see that 61 divides it.\nExpression $= 5 \\cdot 61 \\cdot 881$, making 881 the largest prime divisor.\n[/hide]\r\nHow can one do this nicely?", "Solution_2": "Uhoh. I posted a reply to the problem what is the largest prime divisor of $9^{4}+4^{9}$, not this one.\r\nSomething odd is going on.", "Solution_3": "I thought the solution looked \"ackward\" for this problem :wink: .", "Solution_4": "[hide=\"hint\"]\n$6+x>y$\n$6+y>x$[/hide]", "Solution_5": "[quote=\"DiscreetFourierTransform\"]$4^{9}+9^{4}= 2^{18}+3^{8}$, which from memorization of powers of two and three is \n$262144+6561 = 268705$\nWhich is divisible by 5:\n$5*53741$\n\nAfter some ugly prime checking, I see that 61 divides it.\nExpression $= 5 \\cdot 61 \\cdot 881$, making 881 the largest prime divisor.\n\nHow can one do this nicely?[/quote]\r\n\r\nYea for that 4^9+9^4 thing..its just 2^18+3^8, which is just (2^9)^2+(3^4)^2, which is a^2+b^2, which is (a+b)^2-2ab, so its (2^9+3^4)-2*2^9*3^4, which is (2^9+3^4)-2^10*3^4, which is a^2-b^2. So we have (2^9+3^4)-(2^5*3^2)^2, which is (2^9+3^4+2^5*3^2)(2^9+3^4-2^5*3^2), 2^9+3^4+2^5*3^2=881. Now that is the largest possible value that can work, but still check if it works, apparently it does so 881.\r\n\r\nFor the quadrilateral, it requires some triangle inequality. But just using diophantient thing, you can get 11 solutions for y after x is chosen (x-5 to x+5)", "Solution_6": "[quote=\"DiscreetFourierTransform\"]Uhoh. I posted a reply to the problem what is the largest prime divisor of $9^{4}+4^{9}$, not this one.\nSomething odd is going on.[/quote]\r\nDiscreetFourierTransform said that the solution was not for this thread.", "Solution_7": "[quote=\"AstroPhys\"][quote=\"DiscreetFourierTransform\"]Uhoh. I posted a reply to the problem what is the largest prime divisor of $9^{4}+4^{9}$, not this one.\nSomething odd is going on.[/quote]\nDiscreetFourierTransform said that the solution was not for this thread.[/quote]\r\n\r\nhe wanted a better way to do it.", "Solution_8": "[hide=\"Solution\"]Per the triangle inequality theorem extended to quadrilaterals, $6+x>y$ and $6+y>x$.\nSo we have\n$6+y>x$ which can also be written $y>x-6$, combined with $x+6>y$\nand therefore $x-6 2. \r\n\r\nApplying the same idea to the lower limit does not work. We need something stronger. We may continue (*) inductively to obtain f(2n+1) = f(2n) + f(2n - 1) + ... + f(3) + f(2) + f(1) + 1. (**) We now use the following lemma: \r\n\r\n f(1) + f(2) + ... + f(2r) \u2265 2r f(r) \r\n\r\nWe group the terms on the lhs into pairs and claim that f(1) + f(2r) \u2265 f(2) + f(2r-1) \u2265 f(3) + f(2r-2) \u2265 ... \u2265 f(r) + f(r+1). If k is even, then f(k) = f(k+1) and f(2r-k) = f(2r+1-k), so f(k) + f(2r+1-k) = f(k+1) + f(2r-k). If k is odd, then f(k+1) = f(k) + f((k+1)/2) and f(2r+1-k) = f(2r-k) + f((2r-k+1)/2), but f is monotone so f((k+1)/2) \u2264 f((2r+1-k)/2) and hence f(k) + f(2r+1-k) \u2265 f(k+1) + f(2r-k), as required. \r\n\r\nApplying the lemma to (**) gives f(2n+1) > 2n+1f(2n-1). This is sufficient to prove the lower limit by induction. It is true for n = 1. Suppose it is true for n. Then f(2n+1) > 2n+12(n-1)2/4 = 2(n2-2n+1+4n+4)/4 > 2(n+1)2/4, so it is true for n+1.", "Solution_2": "The link in chien than's answer no longer works. Also, his proof can be a bit hard to read because copying exponents went wrong apparently. Here is my version.\n\nWe will first look at the function $f(m)$, where $m$ can be any non-negative integer and is not required to be a power of two itself. Define $f_k(m)$ as the number of ways of representing $m$ as a sum of powers of two where we use $1$ in the sum exactly $k$ times. With the definition of $f_k(m)$, we clearly get $f(m) = \\displaystyle \\sum_{k=0}^m f_k(m)$. We also have $f_k(m) = f_0(m-k)$ and, if $m$ is odd, $f_0(m) = 0$, while if $m$ is even, $f_0(m) = f\\left(\\frac{m}{2}\\right)$. The latter equality can be seen by noticing that one can put the expressions for $m$ that do not contain a $1$ in a one-to-one correspondence with all expressions for $\\frac{m}{2}$, by simply dividing every term by two. Applying these properties one can obtain the following way to recursively calculate $f(m)$:\n\n\\begin{align}\nf(m) &= \\sum_{k=0}^m f_k(m) \\nonumber \\\\\n&= \\sum_{k=0}^m f_0(m-k) \\nonumber \\\\\n&= \\sum_{k=0}^m f_0(k) \\nonumber \\\\\n&= \\sum_{k=0}^{\\left \\lfloor \\frac{m}{2} \\right \\rfloor} f_0(2k) \\nonumber \\\\\n&= \\sum_{k=0}^{\\left \\lfloor \\frac{m}{2} \\right \\rfloor} f(k)\n\\end{align}\n\nWe will now start our proof of the upper bound. With equality (1) it is straightforward to check $f(8) = f(0) + f(1) + f(2) + f(3) + f(4) = 1 + 1 + 2 + 2 + 4 = 10$, so that the inequality $f(2^n) < 2^{n^2/2}$ is seen to hold for $n = 3$. Let us therefore assume $n \\ge 4$ and by induction we may assume $f(2^{n-1}) < 2^{(n-1)^2/2}$. Furthermore note that $f$ is a monotonically increasing function.\n\n\\begin{align*}\nf(2^n) &= \\sum_{k=0}^{2^{n-1}} f(k) \\\\\n&= 1 + 1 + \\sum_{k=2}^{2^{n-1}} f(k) \\\\\n&< f(2^{n-1}) + \\sum_{k=2}^{2^{n-1}} f(2^{n-1}) \\\\\n&= 2^{n-1} f(2^{n-1}) \\\\\n&< 2^{n-1} 2^{(n-1)^2/2} \\\\\n&= 2^{n^2/2 - n + 1/2 + n - 1} \\\\\n&= 2^{n^2/2 - 1/2} \\\\\n&< 2^{n^2/2}\n\\end{align*}\n\nFor the lower bound, we first need to show a little lemma.\n\n[b]Lemma.[/b]\nFor all $n \\ge 2$ and all $i$ with $1 \\le i \\le 2^{n-2}$ we have the following inequality:\n\n$$\nf(i) + f(2^{n-1} - i + 1) \\ge 2f(2^{n-2})\n$$\n\n[i]Proof.[/i] We will actually prove $f(i) + f(2^{n-1} - i + 1) \\ge f(i + 1) + f(2^{n-1} - i)$, which implies $f(i) + f(2^{n-1} - i + 1) \\ge f(2^{n-2}) + f(2^{n-2} + 1) = 2f(2^{n-2})$ by iteration. To show $f(i) + f(2^{n-1} - i + 1) \\ge f(i + 1) + f(2^{n-1} - i)$, note that when $i$ is even we have $f(i) = f(i+1)$ and $f(2^{n-1} - i + 1) = f(2^{n-1} - i)$, so that we actually have equality. When $i$ is odd, we have $f(i+1) - f(i) = f\\left(\\frac{i+1}{2}\\right)$ and $f(2^{n-1} - i + 1) - f(2^{n-1} - i) = f\\left(\\frac{2^{n-1} - i + 1}{2}\\right)$, both by equation (1), so that:\n\n\\begin{align*}\nf(i) + f(2^{n-1} - i + 1) &= f(i+1) - f\\left(\\frac{i+1}{2}\\right) + f(2^{n-1} - i) + f\\left(\\frac{2^{n-1} - i + 1}{2}\\right) \\\\\n&= f(i+1) + f(2^{n-1} - i) + \\left(f\\left(\\frac{2^{n-1} - i + 1}{2}\\right) - f\\left(\\frac{i+1}{2}\\right)\\right) \\\\\n&\\ge f(i+1) + f(2^{n-1} - i)\n\\end{align*}\n\nWhere the final inequality follows from that fact that $f$ is monotonically increasing and $\\frac{2^{n-1} - i + 1}{2} \\ge \\frac{i+1}{2}$ since $i \\le 2^{n-2}$. This finishes the proof of the lemma.\n\nNow we can finally prove the lower bound. Since $f(2) = 2$, the inequality $2^{n^2/4} < f(2^n)$ is seen to hold for $n = 1$, so assume $n \\ge 2$. By applying equation (1) and the above lemma we get:\n\n\\begin{align*}\nf(2^n) &= \\sum_{k=0}^{2^{n-1}} f(k) \\\\\n&> \\sum_{i=1}^{2^{n-2}} \\left(f(i) + f(2^{n-1} - i + 1) \\right) \\\\\n&\\ge 2^{n-1} f(2^{n-2}) \\\\\n&> 2^{n-1} 2^{(n-2)^2/4} \\\\\n&= 2^{n^2/4 - n + 1 + n - 1} \\\\\n&= 2^{n^2/4}\n\\end{align*}" } { "Tag": [ "search", "Diophantine equation", "number theory proposed", "number theory" ], "Problem": "Let $ m$ and $ n$ natural numbers such that:\r\n$ \\frac{m\\plus{}1}{n}\\plus{}\\frac{n\\plus{}1}{m}\\equal{}k$, when k is a natural number too(not constant).\r\nProve that there are infinitely pairs $ (m,n)$.", "Solution_1": "[hide=\"Hint\"]Use Pell equation,we prove there are infinitely pairs $ (m,n)$ such that\n$ \\frac {m\\plus{}1}{n}\\plus{}\\frac{n\\plus{}1}{m}\\equal{}3$.It is quite obivious\n[/hide]", "Solution_2": "http://www.mathlinks.ro/viewtopic.php?search_id=772572583&t=210528" } { "Tag": [ "group theory", "abstract algebra", "number theory", "relatively prime", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello,\r\n\r\nLet G be a finite group and H a normal subgroup of G such that card(H) = n, and n is relatively prime to card(G).\r\nShow that H is the only subgroup of G with cardinal n.\r\n\r\nI have a solution using Sylow subgroups but I believe there is something more elementary. Any suggestion ?", "Solution_1": "I assume you meant that $n$ is coprime to $[G:H]$, right?\r\n\r\nLat $K\\le G$ be another subgroup of $G$ of order $n$. Then $H$ is a normal subgroup of $KH$, and $KH/H$ is a subgroup of $G/H$, meaning that $[KH:H]$ divides $[G:H]$. However, $KH/H$ is isomorphic to $K/(K\\cap H)$, meaning that, in particular, they have the same order, which must then divide $|K|=n$. We have shown that the order of $KH/H$ divides both $n$ and $[G:H]$, so it must be $1$, or, in other words, $H=KH\\Rightarrow K\\le H\\Rightarrow K=H$.", "Solution_2": "Very nice Grobber ! \r\n\r\nI had as a proof: every p-Sylow of H is a p-Sylow of G ( since gcd(n,[G:H])=1 ) and reciprocally every p-Sylow of G (where p divides n) must be conjugated with a p-Sylow of H. But then it must be contained in H (since H is normal). That means every p-Sylow of H' (H' is any group of G of cardinal n) must be contained in H. In particular if n = p_1^k_1* ... *p_n^k_n, then each p_i^k_i divides $H \\cap H'$ ... which I am not sure it's correct, but it is definitely not elementary.\r\n\r\nbye", "Solution_3": "Hi,\r\n\r\nA more direct proof:\r\nLet F be the morphism from G to G/H and let K be a subgroup of G with |K|=|H|.\r\nThen F(K) is a subgroup of G/H and |F(K)| is a divisor of |K| and of |G/H|. Thus |F(K)|=1 and K=H.\r\n\r\nFriendly,\r\nGeorges", "Solution_4": "Hello Georgres,\r\n\r\nThat's a very short proof indeed; glad to see you eventually decided to have a look at the forum and become a member !!!\r\n\r\n@+\r\n\r\nJulien" } { "Tag": [ "probability", "calculus", "symmetry", "algebra unsolved", "algebra" ], "Problem": "You have two jars, 50 red marbles, 50 blue marbles. You need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) You can arrange the marbles however you like, but each marble must be in a jar.\r\n\r\nIt is some what intuitive that if we put just one red marble in one of the jars and remaining 49 red and 50 blue marbles in the other jar, we get the maximum probability \\[ \\frac{1}{2}*(\\frac{1}{1} \\plus{} \\frac{49}{99})\\]. Can some one prove this claim or give a proof of optimal case.", "Solution_1": "Did you try to use calculus? Set $ f(k,l)$ to be the probability you want to maximize in terms of \r\n$ k$, the number of red marbles in the first jar, and $ l$ the number of blue marbles in the first jar. \r\nYou can do it with $ n$ instead of 50 marbles, and assume by symmetry that $ k \\le n/2$. \r\nIt shouldn't be too difficult to maximize $ f$ for fixed $ k$ with respect to $ l$, and afterwards \r\nwith respect to $ k$ with optimal $ l_k$. Maybe there is some ugly casework that has to be done, \r\nbut in principle it should work fine.", "Solution_2": "In one of the jars, the probability of pick a red marble will be smaller than $ 1/2$, since the number of red marbles is equals to the number of blue marbles (both probabilities will be $ 1/2$ if the number of blue marbles is equals to the number of red marbles in each of the jars, but this case gives probability $ 1/2$, which is smaller than our conjectured answer). Being smaller than $ 1/2$, the maximum possible value of this probability is $ 49/99$. The other probability clearly has maximum $ 1$." } { "Tag": [ "inequalities", "trigonometry", "geometry proposed", "geometry" ], "Problem": "[color=purple] Let triangle $ ABC$ in plane and M is an arbitrary point.Denote $ r_{a},r_{b},r_{c}$ are three exradii. Prove that :\n $ \\frac{MA}{r_{a}}\\plus{}\\frac{MB}{r_{b}}\\plus{}\\frac{MC}{r_{c}}\\ge 2$\n Easy or not ? :D :) [/color]", "Solution_1": "[hide=\"maybe a start\"] $ \\frac{MA}{\\tan \\frac{A}{2}}+\\frac{MB}{\\tan \\frac{B}{2}}+\\frac{MC}{\\tan \\frac{C}{2}}\\geq 2p$ $ \\implies$ $ \\frac{MA\\tan \\frac{B}{2}\\tan \\frac{C}{2}+MB\\tan\\frac{A}{2}\\tan\\frac{C}{2}+MC\\tan\\frac{A}{2}\\tan\\frac{B}{2}}{\\tan\\frac{B}{2}\\tan\\frac{A}{2}\\tan\\frac{C}{2}}\\geq 2p$ and it is know that\n\n $ \\|\\begin{array}{cccc}\\tan\\frac{B}{2}\\tan\\frac{A}{2}\\tan\\frac{C}{2}=\\frac{r}{p}\\\\\n\\\\\n\\tan \\frac{B}{2}\\tan \\frac{C}{2}+\\tan\\frac{A}{2}\\tan\\frac{C}{2}+\\tan\\frac{A}{2}\\tan\\frac{B}{2}=1\\end{array}$[/hide]" } { "Tag": [ "trigonometry", "geometry" ], "Problem": "How can I prove that $x=30$ \"exactly\" is the solution of the following equation?\r\n\r\n$\\frac{\\sin(54-x)}{sinx}={\\frac{\\sin54}{sin84}}$\r\n\r\nAll I've been able to find is $x=arctan{\\frac{\\sin84sin54}{sin54+sin84cos54}}{\\simeq}30$\r\n\r\nThanks and regards.", "Solution_1": "You want to prove that $\\sin24 \\sin 84=\\sin 30 \\sin 54$ and this is equivalent to $2 \\sin24 \\sin 84=\\sin 54$ or $\\cos 60-\\cos 108= \\sin54$. \r\nNow $\\frac{1}{2}= 1-2 \\sin^{2}{54}+\\sin54$.\r\nThe equation $2x^{2}-x-\\frac{1}{2}= 0$ has for positive solution $x=\\frac{\\sqrt5+1}{2}$ and this is exactly $\\sin54=\\sin \\frac{3\\pi}{10}$.\r\nI tried to solve the original equation but failed. Maybe you should post it this way ( I mean Solve )\r\nAlso such type of equations appear when trying to solve a geometry problem with sine theoreme. If this is the case post the geometry problem.", "Solution_2": "I believe you should go with delta's method.\r\nI got the relation,\r\ncot x - cot 54 = cosec 84\r\n\r\ndidn't follow up, see if it gets you anywhere.", "Solution_3": "It is $\\cos 60^\\circ-\\cos108^\\circ=\\sqrt{3}\\sin54^\\circ\\implies \\frac{1}{2}+\\cos 72^\\circ = \\sqrt{3}\\sin54^\\circ$\r\n\r\nNote that: $54 = 36+18, 72 = 2\\cdot36$ and that $\\cos 18^\\circ = \\frac{\\sqrt{10+2\\sqrt{5}}}{4}, \\sin 18^\\circ=\\frac{\\sqrt{5}-1}{4}$\r\n\r\nThe equation could be written like that:\r\n\r\n$\\frac{1}{2}+\\cos (2\\cdot36^\\circ)=\\sqrt{3}\\sin (36^\\circ+18^\\circ)\\implies\\ldots$\r\n\r\nAll the trigonometric numbers are known, so it's a matter of calculations now..", "Solution_4": "Thanks for the answers and please excuse me for my late reply. Happy new year." } { "Tag": [ "induction", "real analysis", "function" ], "Problem": "Solve the recurence \\[a_{n+1}= \\sum_{k = 0}^{n}{\\binom{n}{k}}{a_{k}}a_{n-k}\\] depending on $a_{0}$", "Solution_1": "1) This has nothing to do with complex analysis.\r\n2) This is not, in fact, very hard: calculating the first 4 terms makes the general form clear, and induction polishes it off.", "Solution_2": "JBL I undersand that for you is easy ,but for me ,how I said is hard ,so would someone be so kindly to show me how can it be solved.Thanks", "Solution_3": "[quote=\"silviu\"]would someone be so kindly to show me how can it be solved.[/quote]\n\n[quote=\"JBL\"]calculating the first 4 terms makes the general form clear, and induction polishes it off.[/quote]\r\n\r\n\r\nIf a moderator could move this to a more appropriate forum (Pre-Olympiad? Intermediate?), that would be nice.", "Solution_4": "I might as well add:\r\n\r\nThere are other ways to look at this. In particular, the expression on the right is the binomial convolution of $\\{a_{i}\\}$ with itself, which suggests that we might define the exponential generating function\r\n\r\n$A(x) = \\sum_{n = 0}^\\infty \\frac{a_{n}x^{n}}{n!}$,\r\n\r\nand then the given relation implies $A'(x) = A(x)^{2}$, which leads to $A(x) = \\frac{c}{1-cx}$ and thus to the result." } { "Tag": [ "geometric transformation", "geometry", "homothety", "cyclic quadrilateral", "geometry proposed" ], "Problem": "The circles $W_1, W_2, W_3$ in the plane are pairwise externally tangent to each other. Let $P_1$ be the point of tangency between circles $W_1$ and $W_3$, and let $P_2$ be the point of tangency between circles $W_2$ and $W_3$. $A$ and $B$, both different from $P_1$ and $P_2$, are points on $W_3$ such that $AB$ is a diameter of $W_3$. Line $AP_1$ intersects $W_1$ again at $X$, line $BP_2$ intersects $W_2$ again at $Y$, and lines $AP_2$ and $BP_1$ intersect at $Z$. Prove that $X, Y$, and $Z$ are collinear.", "Solution_1": "Let $ B',A'$ denote the 2nd intersections of $ BP_1$ and $ AP_2$ with $ \\omega_1$ and $ \\omega_2.$ Since $ P_1$ and $ P_2$ are insimilicenters of $ \\omega_1 \\sim \\omega_3$ and $ \\omega_2 \\sim \\omega_3,$ it follows that $ AB \\parallel XB' \\parallel YA'.$ If $ a,b$ denote the tangents of $ \\omega_3$ at $A,B$ and $a',b'$ the tangents of $\\omega_2,\\omega_1$ at $ A',B',$ we have $ a \\parallel a' , b \\parallel b'.$ Since $ a \\parallel b,$ then $ a' \\parallel b'$ $\\Longrightarrow$ $ b'$ is image of $ a'$ through the negative homothety that takes $ \\omega_1$ into $\\omega_2$ $ \\Longrightarrow$ $ A',B',P_3$ are collinear. $ B'X,A'Y$ are diameters of $ \\omega_1$ and $ \\omega_2$ $\\Longrightarrow$ $X,Y$ lie on perpendicular to $ A'B'$ through $ P_3.$ From cyclic quadrilateral $ ABP_2P_1,$ we get\n\n$ \\angle ABP_1 \\equal{} \\angle BB'X \\equal{} \\angle AP_2P_1 \\ , \\ \\angle BAP_2 \\equal{} \\angle AA'Y \\equal{} \\angle BP_1P_2$\n\nOn the other hand, from the cyclic quadrilaterals $ P_1P_3XB',$ $ P_2YP_3A'$ we get $ \\angle P_1P_3Y \\equal{} \\angle BB'X,$ $\\angle P_2P_3Y \\equal{} \\angle AA'Y $ $ \\Longrightarrow$ $ \\angle P_2P_3P_1 \\equal{} \\angle AP_2P_1 \\plus{} \\angle BP_1P_2$ $\\Longrightarrow$ $ZP_2P_3P_1$ is cyclic. Since $ \\angle P_2P_3Y \\equal{} \\angle ZP_1P_2,$ then $ YX$ is identical with the diagonal $ ZP_3$ of the quadrilateral $ ZP_2P_3P_1,$ i.e. $ X,Y,Z$ are collinear." } { "Tag": [ "ratio", "quadratics", "algebra", "quadratic formula" ], "Problem": "1)Solve for x: 12abx2 \u2014 (9a2 \u2014 8b2) x - 6ab = 0\r\n2)\r\n$ {\\rm{If the ratio of the roots of equation, }}lx^2 {\\rm{ }} \\plus{} {\\rm{ }}mx{\\rm{ }} \\plus{} {\\rm{ }}m{\\rm{ }} \\equal{} {\\rm{ }}0{\\rm{, is }}p: q{\\rm{ prove that: }}\\sqrt {\\frac{p}{q}} \\plus{} \\sqrt {\\frac{q}{p}} \\plus{} \\sqrt {\\frac{m}{l}} \\equal{} 0$\r\n3)Q.4 Using the quadratic formula, solve the equation:\r\n a2b2x2 \u2013 (4b4 \u2013 3a4)x \u2013 12a2b2 = 0\r\n4)\r\n\\[ \\left( {x \\minus{} \\frac{1}{x}} \\right)^2 \\plus{} 8\\left( {x \\plus{} \\frac{1}{x}} \\right) \\minus{} 29 \\equal{} 0\r\n\\]", "Solution_1": "[quote=\"sanbittu\"]1)Solve for x: 12abx^2 \u2014 (9a^2 \u2014 8b^2) x - 6ab = 0\n2)\n$ {\\rm{If the ratio of the roots of equation, }}lx^2 {\\rm{ }} \\plus{} {\\rm{ }}mx{\\rm{ }} \\plus{} {\\rm{ }}m{\\rm{ }} \\equal{} {\\rm{ }}0{\\rm{, is }}p: q{\\rm{ prove that: }}\\sqrt {\\frac {p}{q}} \\plus{} \\sqrt {\\frac {q}{p}} \\plus{} \\sqrt {\\frac {m}{l}} \\equal{} 0$\n3)Q.4 Using the quadratic formula, solve the equation:\n a^2b^2x^2 \u2013 (4b^4 \u2013 3a^4)x \u2013 12a^2b^2 = 0\n4)\n\\[ \\left( {x \\minus{} \\frac {1}{x}} \\right)^2 \\plus{} 8\\left( {x \\plus{} \\frac {1}{x}} \\right) \\minus{} 29 \\equal{} 0\n\\]\n[/quote]\r\n1.$ \\Delta_x \\equal{} (9a^2 \\minus{} 8b^2)^2 \\plus{} 4.12.6ab \\equal{} (9a^2 \\plus{} 8b^2)^2$\r\nIt follows that:\r\n$ x \\equal{} \\frac {9a^2 \\minus{} 8b^2 \\plus{} 9a^2 \\plus{} 8b^2}{24ab} \\equal{} \\frac {3a}{4b}$\r\nor $ x \\equal{} \\frac {9a^2 \\minus{} 8b^2 \\minus{} (9a^2 \\plus{} 8b^2)}{24ab} \\equal{} \\frac { \\minus{} 2b}{3a}$\r\n3.Similarly, we have: $ x \\equal{} \\frac {4b^2}{a^2}$ or $ x \\equal{} \\frac { \\minus{} 3a^2}{b^2}$", "Solution_2": "4)\r\n$ \\left({x\\minus{}\\frac{1}{x}}\\right)^2\\plus{}8\\left({x\\plus{}\\frac{1}{x}}\\right)\\minus{}29\\equal{}0 \\Rightarrow x^2\\minus{}2\\plus{}\\frac{1}{x^2}\\plus{}8x\\plus{}\\frac{8}{x}\\minus{}29\\equal{}0$\r\nMultily by $ x^2$\r\n\r\n$ x^4\\plus{}8x^3\\minus{}31x^2\\plus{}8x\\plus{}1\\equal{}0$\r\n$ (x^2\\minus{}3x\\plus{}1)(x^2\\plus{}11x\\plus{}1)\\equal{}0$\r\n$ x^2\\minus{}3x\\plus{}1\\equal{}0 \\Rightarrow \\boxed {x\\equal{}\\frac{3\\pm \\sqrt{5}}{2}}$\r\n$ x^2\\plus{}11x\\plus{}1\\equal{}0 \\Rightarrow \\boxed {x\\equal{}\\frac{\\minus{}11\\pm 3\\sqrt{13}}{2}}$" } { "Tag": [], "Problem": "For what integers $ n\\ge 3$ is it possible to accommodate, in some order, the numbers $ 1,2,\\cdots, n$ in a circular form such that every number divides the sum of the next two numbers, in a clockwise direction?", "Solution_1": "[hide]$ n\\equal{}3$ obviously works for any permutation we choose. We now show no other $ n$ can work.\n\nLet us draw out sections of the circular form linearly, with the understanding that the rightward direction corresponds to a clockwise direction on the circle, and that E represents an even number, while O represents an odd number.\n\nSuppose there were two even numbers next to each other: EE_\nThen the next number (clockwise) could not be odd (EEO), for we can not have an even number divide the sum of an even number and an odd number. Thus, the next number must be even. (EEE) Similarly, the following number must also be even, and so on (EEEE...), giving a clear contradiction.\n\nNow consider an even number. It must be surrounded by odds on both sides: OEO\nThe arrangment EOE is impossible, since the sum of the two rightmost numbers is odd, and an even number cannot divide it. Thus, we cannot have E on the left or the right of OEO; that is, it must be arranged OOEOO, so the two number preceeding and succeding an even number must be odd.\n\nFor any even number, consider how many odd numbers are between that number and the next even number (clockwise). Our analysis shows that this number is at least 2. Thus, there are at least twice as many odds as there are evens. This is impossible for any $ n$ except 3, so we are done.[/hide]", "Solution_2": "[size=100]See that we can't have two evens together because in that case they all will have to be even, so if we have an even it w'd have to be [color=#FF0000]odd[/color]-[color=#0000FF]even[/color]-[color=#FF0000]odd[/color] and from this we can deduce the parity of the next and the previous, something like this:\n[color=#FF0000]odd[/color]-[color=#FF0000]odd[/color]-[color=#0000FF]even[/color]-[color=#FF0000]odd[/color]-[color=#FF0000]odd[/color] from this we conclude that for every even we will have at least $2$ odds, if $n=2d$ there are $d$ odds and $d$ evens, so $n$ can't be even, if $n=2d+1$ there are $d$ evens and $d+1$ odds, from this last conclutions we have $d+1 \\ge 2d$ thus $1 \\ge d$ so $d=1$ and $n=3$.[/size]" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Is there $f$ defined on $[0,1]$ with real numbers such that the preimage of any compact interval is never measurable?", "Solution_1": "Here's a possible construction:\r\n\r\nLet $H$ be a Hamel basis of $\\mathbb R$ over $\\mathbb Q$, and fix some $h_{0}\\in H$. Let $V$ be $\\mathbb Q$-subspace of $\\mathbb R$ with basis $H-\\{h_{0}\\}$, and let $W$ be its complement in $\\mathbb R$ with basis $h_{0}$. Define $f$ to be the projection on $V$ along $W$ (in other words, if $x=\\sum_{h\\in H}x_{h}h,\\ x_{h}\\in\\mathbb Q$, then $f(x)=\\sum_{h\\in H-\\{h_{0}\\}}x_{h}h$).\r\n\r\nIf $I$ is a bounded interval, then the countable union $\\bigcup_{p,q\\in\\mathbb Q}(ph_{0}+qf^{-1}(I))$ is $\\mathbb R$, so $f^{-1}(I)$ can't have measure zero $(*)$. On the other hand, let's look at the set $f^{-1}(I)+f^{-1}(I)$. Its image through $f$ is contained in $I+I$ and hence is bounded, but it's easy to see that the image through $f$ of an interval is dense in $\\mathbb R$. This means that $f^{-1}(I)+f^{-1}(I)$ cannot contain an interval, so it's not measurable with positive measure. Together with $(*)$, this proves that $f$ does indeed have the required property." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $\\epsilon$ be in $(0,1)$ and $n$ in $N^{*}$. We consider $n$ postive ral numbers $x_{1}, \\ldots ,x_{n}$ such that:\r\n\r\n$\\forall (i,j) \\in{1,\\ldots ,n}^{2}, x_{i}x_{j}\\leq \\epsilon^{|i-j|}.$\r\n\r\nShow that:\r\n\r\n$\\sum_{i=1}^{n}{x_{i}}\\leq\\frac{1}{1-\\sqrt{\\epsilon}}$", "Solution_1": "Any Idea? :maybe:", "Solution_2": "Hint:\r\nFor every integer $k$,there is at most $k$ numbers of $x_{1},x_{2},\\cdots,x_{n}$ $\\ge$ $(\\sqrt{\\epsilon})^{k}$ :D", "Solution_3": "Thanks. I got it.", "Solution_4": "thxthxthxthx" } { "Tag": [ "modular arithmetic", "number theory" ], "Problem": "These are just cool problems we came across in the Introductory Number Theory final challenge for 2006:\r\n\r\n1) Prove that with fourteen integers, $a_{1}, a_{2}, a_{3}, ..., a_{14}$, that $a_{1}^{4}+a_{2}^{4}+a_{3}^{4}+...+a_{14}^{4}$ cannot ever equal 9999.\r\n\r\n2) Prove that with integers, a, b, and c, that $a^{2}+b^{2}+c^{2}$ cannot equal $a^{2}c^{2}$.", "Solution_1": "[hide=\"1\"]\n$a_{1}^{4}+\\cdots+a_{14}^{4}=9999$\n$a_{1}^{4}+\\cdots+a_{14}^{4}=15\\mod16$\nbut all fourths are $=0,1\\mod16$ so contradiction[/hide]\n[hide=\"2, actually it can be equal, but they would all be 0.\"]\nWLOG they are all positive\n$a^{2}+b^{2}+c^{2}=a^{2}c^{2}\\mod4$\n$x^{2}$ is 1 mod 4 when x is odd, and 0 mod 4 when even.\nif all are odd, $1+1+1=1$, contradiction. so one is even\nWLOG let a=even.\n$0+b^{2}+c^{2}=0\\mod4$ so all must be even\nso substitute, $a=2a_{1},b=2b_{1},c=2c_{1}$\nso $4a_{1}^{2}+4b_{1}^{2}+4c_{1}^{2}=16a^{2}_{1}c^{2}+1$\n$a_{1}^{2}+b^{2}_{1}+c^{2}_{1}=4a^{2}_{1}c^{2}_{1}$\n$a_{1}^{2}+b^{2}_{1}+c^{2}_{1}=0\\mod4$\nso again they are all even.\nthis repeats on and on, and the $a_{i}$s get smaller and smaller infinitly, so they can't be integers. so contradiction, and $(a,b,c)=(0,0,0)$ is onkly solution\n[/hide]", "Solution_2": "[quote=\"junggi\"] this repeats on and on, and the $a_{i}$s get smaller and smaller infinitly, so they can't be integers[/quote]\r\n\r\nI'm pretty sure this situation is called \"Infinite Descent\". (right?) :)", "Solution_3": "Yep.\r\n\r\n[hide=\"Alternate solution\"] $(a^{2}-1)(c^{2}-1) = b^{2}+1$\n\nIf $b$ is even, then $a, c$ are even so of $a-1, a+1, c-1, c+1$ two must be $\\equiv 3 \\bmod 4$, but such numbers cannot divide $b^{2}+1$, contradiction.\n\nIf $b$ is odd, then one of $a, c$ is odd so the LHS is $\\equiv 0 \\bmod 4$ but the RHS is $\\equiv 1 \\bmod 4$, contradiction. [/hide]", "Solution_4": "[quote=\"junggi\"][hide=\"1\"]\n$a_{1}^{4}+\\cdots+a_{14}^{4}=9999$\n$a_{1}^{4}+\\cdots+a_{14}^{4}=15\\mod16$\nbut all fourths are $=0,1\\mod16$ so contradiction[/hide]\n[hide=\"2, actually it can be equal, but they would all be 0.\"]\nWLOG they are all positive\n$a^{2}+b^{2}+c^{2}=a^{2}c^{2}\\mod4$\n$x^{2}$ is 1 mod 4 when x is odd, and 0 mod 4 when even.\nif all are odd, $1+1+1=1$, contradiction. so one is even\nWLOG let a=even.\n$0+b^{2}+c^{2}=0\\mod4$ so all must be even\nso substitute, $a=2a_{1},b=2b_{1},c=2c_{1}$\nso $4a_{1}^{2}+4b_{1}^{2}+4c_{1}^{2}=16a^{2}_{1}c^{2}+1$\n$a_{1}^{2}+b^{2}_{1}+c^{2}_{1}=4a^{2}_{1}c^{2}_{1}$\n$a_{1}^{2}+b^{2}_{1}+c^{2}_{1}=0\\mod4$\nso again they are all even.\nthis repeats on and on, and the $a_{i}$s get smaller and smaller infinitly, so they can't be integers. so contradiction, and $(a,b,c)=(0,0,0)$ is onkly solution\n[/hide][/quote]\r\nfor Q1, how do you get all fourths are $=0,1\\mod16$ so contradiction? (without checking all the possibilites?)", "Solution_5": "[quote=\"srulikbd\"]without checking all the possibilites?[/quote]\r\nIt isn't hard. Notice that fourth powers of even numbers are divisible by $16$ and for any $x$ we have $x^{4}\\equiv (16-x)^{4}\\pmod{16}$, so we only have to check $1,3,5,7$ and it's easy.", "Solution_6": "[quote=\"srulikbd\"][quote=\"junggi\"][hide=\"1\"]\n$a_{1}^{4}+\\cdots+a_{14}^{4}=9999$\n$a_{1}^{4}+\\cdots+a_{14}^{4}=15\\mod16$\nbut all fourths are $=0,1\\mod16$ so contradiction[/hide]\n[hide=\"2, actually it can be equal, but they would all be 0.\"]\nWLOG they are all positive\n$a^{2}+b^{2}+c^{2}=a^{2}c^{2}\\mod4$\n$x^{2}$ is 1 mod 4 when x is odd, and 0 mod 4 when even.\nif all are odd, $1+1+1=1$, contradiction. so one is even\nWLOG let a=even.\n$0+b^{2}+c^{2}=0\\mod4$ so all must be even\nso substitute, $a=2a_{1},b=2b_{1},c=2c_{1}$\nso $4a_{1}^{2}+4b_{1}^{2}+4c_{1}^{2}=16a^{2}_{1}c^{2}+1$\n$a_{1}^{2}+b^{2}_{1}+c^{2}_{1}=4a^{2}_{1}c^{2}_{1}$\n$a_{1}^{2}+b^{2}_{1}+c^{2}_{1}=0\\mod4$\nso again they are all even.\nthis repeats on and on, and the $a_{i}$s get smaller and smaller infinitly, so they can't be integers. so contradiction, and $(a,b,c)=(0,0,0)$ is onkly solution\n[/hide][/quote]\nfor Q1, how do you get all fourths are $=0,1\\mod16$ so contradiction? (without checking all the possibilites?)[/quote]\r\n\r\nOr if you know the fermat-Eular theorem...", "Solution_7": "[quote=\"seamusoboyle\"]Or if you know the fermat-Eular theorem...[/quote]\r\nDoesn't suffice.\r\nBut it's always good to know the structure of prime residue classes, not depending on such weak theorems.", "Solution_8": "Well if you assume its obvious for even numbers...", "Solution_9": ":?:", "Solution_10": "Sorry, my bad." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "[color=red][b] True or false ?[/b][/color]\r\n\r\n If function $ f$ is continuous on $ [a,b]$ with $ f(a) \\equal{} f(b) \\equal{} 0$ and between any two roots of $ f$ , there is at least one more root of $ f$, then $ f$ is constant on the interval $ [a,b]$.\r\n \r\n[i]Please give an answer as much more simple as possible(for school use, if possible).[/i] \r\nThanks in advance - Babis", "Solution_1": "[quote=\"stergiu\"][color=red][b] True or false ?[/b][/color]\n\n If function $ f$ is continuous on $ [a,b]$ with $ f(a) \\equal{} f(b) \\equal{} 0$ and between any two roots of $ f$ , there is at least one more root of $ f$, then $ f$ is constant on the interval $ [a,b]$.\n \n[i]Please give an answer as much more simple as possible(for school use, if possible).[/i] \nThanks in advance - Babis[/quote]\r\nI will give it a shot :)\r\nAssume there exists $ c,d \\in [a;b]$ such that $ f(x) \\neq 0$ when $ x \\in (c;d)$. But then $ f(c) \\equal{} f(d) \\equal{} 0$, and there is exists $ r \\in (c;d)$ such that $ f(r) \\equal{} 0$. So there is no open interval $ (c;d)$ such that $ f(x) \\neq 0, \\forall x \\in (c;d)$. And so $ f(x) \\equal{} 0, \\forall x \\in [a;b]$ because of the contiunity. QED", "Solution_2": "Dear Mathias, thank you for trying the problem. But I'm afraid that your solution needs more details. I think that we must begin with the asumption : Let$ m$ with $ f(m)$ different from $ 0$ and to find an absurd.But this is not done in the above solution\r\n Babis", "Solution_3": "Assume that there exists a real number $ x_0$ such that: $ f(x)\\neq 0$ (Assume that f(x)>0)\r\n\r\nBecause of the continuity of f(x) there exists an interval $ [c;d]$ such that $ f(x) > 0\\forall x\\in [c;d]$ (1)\r\n\r\nDefine $ A \\equal{} \\{ x\\in [a;b]|f(x) \\equal{} 0 ; x < c\\}$ ; $ B \\equal{} \\{ x\\in [a;b]|f(x) \\equal{} 0 ; x > d\\}$\r\n\r\nSo there exists $ \\sup A \\equal{} c_1$(2) and $ \\inf B \\equal{} d_1$ (3): $ f(c_1) \\equal{} f(d_1) \\equal{} 0$. By the hypothesis, there exists $ m\\in (c_1;d_1)$ such that: f(m)=0\r\n\r\nm cannot be in ][c;d] because (1) and it cannot be in [c1;c) and (d;d1] because (2) and (3) -> contraction", "Solution_4": "[quote=\"stergiu\"]Dear Mathias, thank you for trying the problem. But I'm afraid that your solution needs more details. I think that we must begin with the asumption : Let$ m$ with $ f(m)$ different from $ 0$ and to find an absurd.But this is not done in the above solution\n Babis[/quote]\r\nDear Babis,\r\nI will try to clarify :)\r\nAssume there exists $ r \\in [a;b]$ such that $ f(r) \\neq 0$. Then there would exist a interval $ (c;d)$ (with $ c,d \\in [a;b]$) such that $ f(x) \\neq 0 \\forall x \\in (c;d)$ (Because of continuity) But we have proven that no such interval exists! Contradiction. So our assumption was wrong and there exist no $ r$ such that $ f(r) \\neq 0$ when $ r \\in [a;b]$. (Or $ f(x) \\equal{} 0 \\forall x \\in [a;b]$)", "Solution_5": "[quote=\"marshell\"]Assume that there exists a real number $ x_0$ such that: $ f(x)\\neq 0$ (Assume that f(x)>0)\n\nBecause of the continuity of f(x) there exists an interval $ [c;d]$ such that $ f(x) > 0\\forall x\\in [c;d]$ (1)\n\nDefine $ A \\equal{} \\{ x\\in [a;b]|f(x) \\equal{} 0 ; x < c\\}$ ; $ B \\equal{} \\{ x\\in [a;b]|f(x) \\equal{} 0 ; x > d\\}$\n\nSo there exists $ \\sup A \\equal{} c_1$(2) and $ \\inf B \\equal{} d_1$ (3): $ f(c_1) \\equal{} f(d_1) \\equal{} 0$. By the hypothesis, there exists $ m\\in (c_1;d_1)$ such that: f(m)=0\n\nm cannot be in ][c;d] because (1) and it cannot be in [c1;c) and (d;d1] because (2) and (3) -> contraction[/quote]\r\n\r\n Thank you :) ! \r\nThis was also my first approach , but I had - and have still - this question.: Why must we have , for example , \r\n\r\n$ f(c_1) \\equal{} 0$ ? Has $ \\sup A$ the same property with $ x's$ in $ A$ ? But I'm thinking - by looking just now at your \r\n\r\nsolution carefully -that we have two possibilities : If $ A$ is finite or if $ c_1 \\in A$, then it is clear that $ f(c_1) \\equal{} 0$. If $ A$ is infinite and does not contain $ c_1$, then \r\n\r\n$ c_1$must be an acumulation point of $ A$.But then we can find a sequence $ a_n$ in $ A$ with $ \\lim {a_n} \\equal{} c_1$.Since \r\n\r\n$ f$ is continuous, we must have that $ \\lim f({a_n}) \\equal{} f(c_1)$ But\r\n\r\n $ f(a_n) \\equal{} 0$ , so $ f(c_1) \\equal{} 0$.\r\n\r\n Well , if the above steps are correct, then I think that the solution is complete .But is really $ c_1$ a point of acumulation of $ A$ ?My intuition says YES !\r\n \r\n Babis", "Solution_6": "The set of zeroes of a continuous function is closed. ($ f^{\\minus{}1}$ takes closed sets to closed sets, and $ \\{0\\}$ is closed.) Thus its complement is open and so is a union of open intervals or empty. Suppose not empty. Then choose one of these intervals, say $ (m, n)$ for some $ m$ and $ n$. But then $ m, n$ are zeroes of $ f$ with no zero between them, a contradiction, so the complement of the zeroes must actually be empty and $ f$ is the zero function." } { "Tag": [ "function", "limit", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "If $f: (a,b)\\longrightarrow \\mathbb R$ is a function s.t. the limit $\\lim_{x\\nearrow x_{0}}\\frac{f(x)-f(x_{0})}{x-x_{0}}$ exists and is finite $\\forall x_{0}\\in (a,b)$ and the function $g: (a,b)\\longrightarrow \\mathbb R$, $g(y)=\\lim_{x\\nearrow y}\\frac{f(x)-f(y)}{x-y}$ is increasing, prove that $f$ is convex.", "Solution_1": "No answer? :( Is it too easy? Or the problem is incorrectly stated?", "Solution_2": "A counterexample: $f(x)=x^{2}$ for $x\\le 0$, $f(x)=x^{2}+1$ for $x>0$. \r\n\r\nLet's assume in addition that $f$ is continuous. If $f$ is not convex, then there exist numbers $a0$, it follows that $\\tilde f$ attains its maximum at some point $d\\in(a,b)$. Clearly $\\tilde g(d)\\ge 0$, which implies $\\tilde g(x)\\ge 0$ for all $x\\ge d$. This will be hard to reconcile with the inequality $\\tilde f(b)<\\tilde f(d)$." } { "Tag": [ "MATHCOUNTS" ], "Problem": "I'm training for MathCounts, and I kinda want to come in the top 12 in Nationals. So, I'll be posting like crazy here. I'll definitely still spoiler post, obviously, but you'll just see pages saying \"Last post by: Treething\".\r\n\r\nI'm just saying that I'm not trying to spam to increase my post count.\r\n\r\n\r\nOther middle-schoolers are encouraged to stop me from dominating *cough* be active here too :P .", "Solution_1": "[quote=\"Treething\"]Other middle-schoolers are encouraged to stop me from dominating *cough* be active here too :P .[/quote]I also encourage this :)", "Solution_2": "[quote=\"Treething\"]I'm training for MathCounts, and I kinda want to come in the top 12 in Nationals. So, I'll be posting like crazy here. [/quote]\r\n\r\nAre you sure your way \"posting like crazy\" will work (make you in the top12)? Could you please PM me after May this year if it works. I will do the same thing (posting like crazy) next year. :D", "Solution_3": "it may work\r\ntake myself as example, I do want to reply every problem I see, but 80% of them are too difficult to me, so I don't have the Ability to post like crazy, since he has that ability, he must be good at math\r\nif he wins some awards, even interviewed by some media, he should promote our AoPS then.", "Solution_4": "[quote=\"paganinio\"]it may work\ntake myself as example, I do want to reply every problem I see, but 80% of them are too difficult to me, so I don't have the Ability to post like crazy, since he has that ability, he must be good at math\nif he wins some awards, even interviewed by some media, he should promote our AoPS then.[/quote]\r\n\r\nLol :P .\r\n\r\nGeneral Pinochet comes to mind, but that's a little too severe ;P .", "Solution_5": "[quote=\"cj\"][quote=\"Treething\"]I'm training for MathCounts, and I kinda want to come in the top 12 in Nationals. So, I'll be posting like crazy here. [/quote]\n\nAre you sure your way \"posting like crazy\" will work (make you in the top12)? Could you please PM me after May this year if it works. I will do the same thing (posting like crazy) next year. :D[/quote]\r\n\r\n\r\nAnu reminded me about this post.\r\n\r\n\r\nIt did.", "Solution_6": "Okay.\r\n\r\nI will also be posting like crazy, except not in this forum.\r\n\r\nBut yes, good luck Treething.", "Solution_7": "[quote=\"Treething\"][quote=\"cj\"][quote=\"Treething\"]I'm training for MathCounts, and I kinda want to come in the top 12 in Nationals. So, I'll be posting like crazy here. [/quote]\n\nAre you sure your way \"posting like crazy\" will work (make you in the top12)? Could you please PM me after May this year if it works. I will do the same thing (posting like crazy) next year. :D[/quote]\n\n\nAnu reminded me about this post.\n\n\nIt did.[/quote]\r\n\r\nHe said \"this year\" it is already \"next year\"", "Solution_8": "Good luck making the top twelve at nats :roll:", "Solution_9": "Speaking of New Jersey, anyone from New Jersey have a chance?", "Solution_10": "[quote=\"mathgeniuse^ln(x)\"]Speaking of New Jersey, anyone from New Jersey have a chance?[/quote]\r\nWho was speaking of new jersey?", "Solution_11": "NO one was, but treething is from New Jersey.", "Solution_12": "Er, this is slightly old. Locked." } { "Tag": [], "Problem": "How many different ways can 36 identical chairs be placed in rows \nif all rows have the same number of chairs, each chair is in exactly\none row, and no row has more than 20 chairs or less than 3 chairs?", "Solution_1": "This problem wants the number of factors of 36, such that the factor is greater than 3 and less than 20. Counting you get 4 ways", "Solution_2": "HUh, you forgot to count 3, so the answer is 5.", "Solution_3": "[hide]I get $\\boxed{6}$: $3, 4, 6, 9, 12,$ and $18.$[/hide]", "Solution_4": "@AIME15- I think that you forgot 6" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve that equation of system:\r\n\r\n$ \\left\\{\\begin{array}{l} x \\plus{} \\frac {{2xy}}{{\\sqrt [3]{{x^2 \\minus{} 2x \\plus{} 9}}}} \\equal{} x^2 \\plus{} y \\\\\r\ny \\plus{} \\frac {{2xy}}{{\\sqrt [3]{{y^2 \\minus{} 2y \\plus{} 9}}}} \\equal{} y^2 \\plus{} x \\\\\r\n\\end{array} \\right.$", "Solution_1": "[quote=\"L_Euler\"]Solve that equation of system:\n\n$ \\left\\{\\begin{array}{l} x \\plus{} \\frac {{2xy}}{{\\sqrt [3]{{x^2 \\minus{} 2x \\plus{} 9}}}} \\equal{} x^2 \\plus{} y \\\\\ny \\plus{} \\frac {{2xy}}{{\\sqrt [3]{{y^2 \\minus{} 2y \\plus{} 9}}}} \\equal{} y^2 \\plus{} x \\\\\n\\end{array} \\right.$[/quote]\r\nCan you prove that $ x \\equal{} y$!! :wink:", "Solution_2": "[quote=\"L_Euler\"]\nCan you prove that $ x \\equal{} y \\equal{} z$!! :wink:[/quote]\r\n\r\nSeems difficult since there is no $ z$... :wink: \r\n\r\nPierre.", "Solution_3": "[quote=\"pbornsztein\"][quote=\"L_Euler\"]\nCan you prove that $ x \\equal{} y \\equal{} z$!! :wink:[/quote]\n\nSeems difficult since there is no $ z$... :wink: \n\nPierre.[/quote]\r\nOh, I'm sorry, there isn't $ z$! :wink:", "Solution_4": "[quote=\"L_Euler\"]Solve that equation of system:\n\n$ \\left\\{\\begin{array}{l} x \\plus{} \\frac {{2xy}}{{\\sqrt [3]{{x^2 \\minus{} 2x \\plus{} 9}}}} \\equal{} x^2 \\plus{} y \\\\\ny \\plus{} \\frac {{2xy}}{{\\sqrt [3]{{y^2 \\minus{} 2y \\plus{} 9}}}} \\equal{} y^2 \\plus{} x \\\\\n\\end{array} \\right.$[/quote]\r\n\r\n\r\n$ (\\plus{})\\minus{}\\minus{}\\minus{}\\minus{}>$ $ x\\plus{}y\\plus{}\\frac{2xy}{\\sqrt[3]{x^{2}\\minus{}2x\\plus{}9}}\\plus{}\\frac{2xy}{\\sqrt[3]{y^{2}\\minus{}2y\\plus{}9}}\\equal{}x^{2}\\plus{}y^{2}\\plus{}x\\plus{}y$---->\r\n$ 2xy(\\frac{1}{\\sqrt[3]{(x\\minus{}1)^{2}\\plus{}8}}\\plus{}\\frac{1}{\\sqrt[3]{(y\\minus{}1)^{2}\\plus{}8}})\\equal{}x^{2}\\plus{}y^{2}$\r\n\r\n$ \\frac{1}{\\sqrt[3]{(x\\minus{}1)^{2}\\plus{}8}}\\plus{}\\frac{1}{\\sqrt[3]{(y\\minus{}1)^{2}\\plus{}8}}\\geq 0$ and $ x^{2}\\plus{}y^{2}\\geq 0$----> $ 2xy\\geq 0$-----> \r\n\r\n$ x\\not \\equal{}y\\not \\equal{} 0$\r\n\r\n$ \\frac{1}{\\sqrt[3]{(x\\minus{}1)^{2}\\plus{}8}}\\plus{}\\frac{1}{\\sqrt[3]{(y\\minus{}1)^{2}\\plus{}8}}\\equal{}\\frac{x^{2}\\plus{}y^{2}}{2xy}$\r\n\r\n\r\n$ \\frac{1}{\\sqrt[3]{(x\\minus{}1)^{2}\\plus{}8}}\\plus{}\\frac{1}{\\sqrt[3]{(y\\minus{}1)^{2}\\plus{}8}}\\leq\\frac{1}{\\sqrt[3]{8}}\\plus{}\\frac{1}{\\sqrt[3]{8}}\\equal{}1\\leq\\frac{x^{2}\\plus{}y^{2}}{2xy}$------> $ x\\equal{}y\\equal{}1$ and $ x\\equal{}y\\equal{}0$", "Solution_5": "I think we should find $ x\\equal{}0$ or $ y\\equal{}0$ the first, and we are the root $ (0;0)$. \r\n\r\nThank for solving!! ^^" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find $ x,y\\in\\mathbb{R}$ such that\r\n$ x|x|\\plus{}y|y|\\equal{}1$ and $ [[x]]\\plus{}[[y]]\\equal{}1$", "Solution_1": "What did you meant by the double square brackets $ [ [ \\cdot ] ]$ ?", "Solution_2": "[quote=\"pqrs\"]Find $ x,y\\in\\mathbb{R}$ such that\n$ x|x| \\plus{} y|y| \\equal{} 1$ and $ [[x]] \\plus{} [[y]] \\equal{} 1$[/quote]\r\n\r\ni believe the[[.]] were for misleading only :wink: :wink: :lol: but the problem can be easily solved by plotting graphs.", "Solution_3": "$ [[x]]$ is a integer such that $ x\\ge[[x]]>x\\minus{}1$", "Solution_4": "I use [] for this.... $ x\\ge[x]>x\\minus{}1$, $ [x]$ is integer of cource", "Solution_5": "$ x\\ge y$\r\n$ 1)$\r\nIf $ y\\ge 0$ that $ 2>x\\ge 1$ and $ 1>y\\ge 0$($ [x]\\plus{}[y]\\equal{}1$)\r\nAnd\r\n$ x|x|\\plus{}y|y|\\equal{}x^2\\plus{}y^2\\equal{}1$, but $ x\\ge 1$. We have $ x\\equal{}1$ and $ y\\equal{}0$\r\n$ 2)$\r\n$ y<0$\r\n$ [y]\\equal{}\\minus{}k (k>0)$\r\n$ [x]\\equal{}k\\plus{}1$\r\n{y}$ \\equal{}a$($ 1>a\\ge 0$)\r\n{x}$ \\equal{}b$($ 1>b\\ge 0$)\r\n$ (\\minus{}k\\plus{}a)(k\\minus{}a)\\plus{}(k\\plus{}1\\plus{}b)^2\\equal{}1$, $ 1\\plus{}b\\equal{}s$, $ 2>s\\ge 1$\r\n$ \\minus{}k^2\\plus{}2ka\\minus{}a^2\\plus{}k^2\\plus{}2ks\\plus{}s^2\\equal{}1$\r\n$ 2(a\\plus{}s)k\\plus{}s^2\\minus{}a^2\\equal{}1$\r\n$ (a\\plus{}s)(2k\\plus{}s\\minus{}a)\\equal{}1$\r\n$ a\\plus{}s\\ge 1$\r\n$ 2k\\plus{}s\\minus{}a<\\equal{}1$\r\n$ 1\\ge 2k\\plus{}s\\minus{}a\\ge 2k$\r\nBut it's not true, because $ k$ is integer and $ k\\ge 1$\r\nWe have one solution $ (x,y)\\equal{}(1,0)$, $ x\\ge y$" } { "Tag": [ "blogs", "\\/closed" ], "Problem": "also, you used to be able to get to somebody's blog by viewing their profile, but now you can't. will this be changed?", "Solution_1": "[quote=\"perfectnumber628\"]also, you used to be able to get to somebody's blog by viewing their profile, but now you can't. will this be changed?[/quote]You can see the blog when you are viewing the topic, I forgot about the profile. I will fix it.", "Solution_2": "Speaking of the profile, could you add the online/offline status to a person's profile?", "Solution_3": "yes that would be a good idea. Also I noticed the online status thing does not always work properly, ie it says a user is online but he/she is actually not online...", "Solution_4": "[quote=\"rem\"]Also I noticed the online status thing does not always work properly, ie it says a user is online but he/she is actually not online...[/quote]\r\n\r\nI think that's because it's based on the last 5 minutes thing.", "Solution_5": "or the user is hidden", "Solution_6": "Added the new info to the profile of the users :)" } { "Tag": [ "geometry", "search", "trigonometry" ], "Problem": "I came up with this when doing another problem. It's almost definitely not new, but I couldn't find it with the search function.\r\n\r\nProve that $ [ABC]\\equal{}\\dfrac{(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}{r}$.", "Solution_1": "[hide=\"solution\"]It is well known that $ [ABC] \\equal{} \\sqrt{s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}$ (1) and that $ [ABC]\\equal{}rs$ (2). So square (1) and divide by (2) and the desired result is obtained.[/hide]", "Solution_2": "hmmmm I didn't realize it would be that trivial...my solution was to note that $ \\dfrac{s\\minus{}a}{r}\\equal{}\\tan{\\dfrac{A}{2}}$ (drop perpendiculars from I to the three sides, and draw AI,BI,CI) then to use the fact that $ \\displaystyle\\sum\\tan{\\dfrac{A}{2}}\\tan{\\dfrac{B}{2}}\\equal{}1$ (proven easily by just substituting C=180-A-B). This gives $ r^{2}s\\equal{}(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)$, so $ r[ABC]\\equal{}(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)$, etc.\r\n\r\nI guess this leads to a nicer proof of the aformentioned identity, which was the problem I mentioned in the original post." } { "Tag": [ "function", "number theory", "prime numbers", "algebra proposed", "algebra" ], "Problem": "A function $ f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfies:\r\n$ (a)$ $ f(ab)\\equal{}f(a)f(b)$ whenever $ a$ and $ b$ are coprime;\r\n$ (b)$ $ f(p\\plus{}q)\\equal{}f(p)\\plus{}f(q)$ for all prime numbers $ p$ and $ q$.\r\nProve that $ f(2)\\equal{}2,f(3)\\equal{}3$ and $ f(1999)\\equal{}1999.$", "Solution_1": "[quote=\"moldovan\"]A function $ f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfies:\n$ (a)$ $ f(ab) \\equal{} f(a)f(b)$ whenever $ a$ and $ b$ are coprime;\n$ (b)$ $ f(p \\plus{} q) \\equal{} f(p) \\plus{} f(q)$ for all prime numbers $ p$ and $ q$.\nProve that $ f(2) \\equal{} 2,f(3) \\equal{} 3$ and $ f(1999) \\equal{} 1999.$[/quote]\r\n\r\nLet $ p$ an odd prime : $ 2,p$ coprime $ \\implies$ $ f(2p)\\equal{}f(2)f(p)$ but $ p$ prime $ \\implies$ $ f(p\\plus{}p)\\equal{}f(p)\\plus{}f(p)$ and so $ f(2)\\equal{}2$\r\n\r\n$ f(2\\plus{}2)\\equal{}2f(2)$ and so $ f(4)\\equal{}4$\r\n$ f(5)\\equal{}f(3\\plus{}2)\\equal{}f(3)\\plus{}f(2)\\equal{}f(3)\\plus{}2$\r\n$ f(7)\\equal{}f(5\\plus{}2)\\equal{}f(5)\\plus{}f(2)\\equal{}f(3)\\plus{}4$\r\n\r\n$ f(12)\\equal{}f(3)f(4)\\equal{}4f(3)$\r\n$ f(12)\\equal{}f(5\\plus{}7)\\equal{}f(5)\\plus{}f(7)\\equal{}2f(3)\\plus{}6$\r\n\r\nSo $ 4f(3)\\equal{}2f(3)\\plus{}6$ and $ f(3)\\equal{}3$ and so $ f(5)\\equal{}5$ and $ f(7)\\equal{}7$\r\n\r\n$ 3$ and $ 5$ coprime $ \\implies$ $ f(15)\\equal{}f(3\\times 5)\\equal{}f(3)f(5)\\equal{}15$\r\n$ 13$ and $ 2$ primes $ \\implies$ $ f(15)\\equal{}f(13\\plus{}2)\\equal{}f(13)\\plus{}f(2)$ $ \\implies$ $ f(13)\\equal{}13$\r\n$ 11$ and $ 2$ primes $ \\implies$ $ f(13)\\equal{}f(11\\plus{}2)\\equal{}f(11)\\plus{}f(2)$ $ \\implies$ $ f(11)\\equal{}11$\r\n$ 11$ and $ 3$ coprime $ \\implies$ $ f(33)\\equal{}f(3\\times 11)\\equal{}f(3)f(11)\\equal{}33$\r\n$ 31$ and $ 2$ primes $ \\implies$ $ f(33)\\equal{}f(31\\plus{}2)\\equal{}f(31)\\plus{}f(2)$ $ \\implies$ $ f(31)\\equal{}31$\r\n$ 29$ and $ 2$ primes $ \\implies$ $ f(31)\\equal{}f(29\\plus{}2)\\equal{}f(29)\\plus{}f(2)$ $ \\implies$ $ f(29)\\equal{}29$\r\n$ 13$ and $ 2$ coprime $ \\implies$ $ f(26)\\equal{}f(2\\times 13)\\equal{}f(2)f(13)\\equal{}26$\r\n$ 23$ and $ 3$ primes $ \\implies$ $ f(26)\\equal{}f(23\\plus{}3)\\equal{}f(23)\\plus{}f(3)\\equal{}f(23)\\plus{}3$ $ \\implies$ $ f(23)\\equal{}23$\r\n$ 23$ and $ 29$ coprime $ \\implies$ $ f(667)\\equal{}f(23\\times 29)\\equal{}f(23)f(29)\\equal{}667$\r\n$ 667$ and $ 3$ coprime $ \\implies$ $ f(2001)\\equal{}f(3\\times 667)\\equal{}f(3)f(667)\\equal{}2001$\r\n$ 1999$ and $ 2$ primes $ \\implies$ $ f(2001)\\equal{}f(1999\\plus{}2)\\equal{}f(1999)\\plus{}2$ $ \\implies$ $ f(1999)\\equal{}1999$" } { "Tag": [ "Harvard", "college", "AMC", "AIME", "AMC 8", "AMC 10" ], "Problem": "Hello, I was referred to this site by someone in my school, who said he visited here to sharpen and improve his problem solving skills. I didn't really care for it at the time, but now that school is about to start, I'm beginning to take things more seriously.\r\n\r\nMy parents usually push me to enter math contests and the like, but I haven't really had the confidence to enter them. I've tried the AMC 10, but I didn't do so well (I guessed on too many problems, which is a bad habit I have). On the last SAT practice test I took (8th grade), I scored a 580 in math. I'm hoping to fix all this by working much harder on my math skills.\r\n\r\nI also am starting to feel that I may be beginning to struggle in math; most people in my class have also felt this way, and they all say its due to our merciless teacher. Whether this is the truth or not, I don't enjoy the feeling of struggling, so I've taken it to myself to improve, hopefully with the help of this site and books.\r\n\r\nI am not very far ahead in the math classes at my school; I'm ahead only by one year. By my school district I'm considered gifted and above average, but I have a feeling that my district seems to be pretty lenient on those sort of titles. I'm going into 10th grade at my high school, which, at my school, means Algebra 2. My parents don't have the money to spend on a summer courses, but this site, as well as the Art of Problem Solving books, sound like a perfect alternative. As another alternative, I'm beginning to look into online courses.\r\n\r\nI've looked at a few of the problems in this forum, and I've had difficulty solving a few of them, but I think it is because I haven't learned some of the necessary material. That will probably change as my high school education progresses.\r\n\r\nThe point to all this is that hopefully I will be able to enter math competitions and perform well in them, as well as do better in class, and maybe even on the SAT (College entrance exams). Do you think I'll be able achieve all this through visiting this site often for problems and tips, as well as reading through the Art of Problem Solving?\r\n\r\nThanks", "Solution_1": "[quote=\"Startup\"]Do you think I'll be able achieve all this through visiting this site often for problems and tips, as well as reading through the Art of Problem Solving?[/quote]\r\n\r\nNo doubt about it. :) I haven't ever seen anything like AoPS before... It truly is amazing! BTW, the classes here are excellent, too!", "Solution_2": "THis is definitely a great place to be. There are a lot of things happening on this site. By the sounds of it you are perfect for the getting started forum. I also suggest that you participate in Math Chat's. Recently I have hosted them. Usually they are on Friday's. You need Aol INstant Messenger. I have a thread on this in this forum... you can find more info there.", "Solution_3": "[quote=\"Startup\"]Hello, I was referred to this site by someone in my school, who said he visited here to sharpen and improve his problem solving skills.[/quote]\n\nWelcome aboard. This site is made for people like you and your classmate. \n\n[quote=\"Startup\"]I've tried the AMC 10, but I didn't do so well (I guessed on too many problems, which is a bad habit I have).[/quote]\n\nA key \"secret\" for scoring better on the AMC 10 is to do exactly and ONLY the questions you are quite sure about. The penalty for guessing on the AMC 10 is severe, as is also true for the AMC 12. My son has told the story here on this site about following advice from JBL, another forum participant who is now a math major at Harvard, to do enough problems to score decently and then leave all other problems blank. My son did the first eighteen problems, all correctly, on the AMC 10 last year, and completely disregarded the last seven problems. His score (18*6 + 7*2.5=125.5) was enough to qualify him for the AIME and make him the state winner in Minnesota on the AMC 10B. It's important not to guess on the AMC 10 or AMC 12. \n\n[quote=\"Startup\"]I also am starting to feel that I may be beginning to struggle in math . . . . I don't enjoy the feeling of struggling, so I've taken it to myself to improve, hopefully with the help of this site and books.[/quote]\n\nIf you're willing to take the advice of a forty-six-year-old fuddy-duddy, I can advise that that feeling of struggling can be your friend, if you take it as a clue to not give up. Often people reach learning plateaus just before they break through to a higher level of understanding. I can't comment at all on how good your school math program is (and I homeschool my own children because I hardly trust any school program), but your plan now to add other sources of information to your set of learning resources is a VERY good idea. You may continue to struggle a while, but if you persist, then I think you will break through and reach a new, higher level of understanding, and enjoy the concepts that now cause your struggling. I learned this from my wife, a pianist and piano teacher, who often appeared to hit brick walls years ago while practicing piano but who persisted and now plays MUCH better than she did when I first met her. Don't give up. \n\n[quote=\"Startup\"]By my school district I'm considered gifted and above average, but I have a feeling that my district seems to be pretty lenient on those sort of titles. I'm going into 10th grade at my high school, which, at my school, means Algebra 2.[/quote]\n\nI was considered \"gifted\" at a school in suburban Minneapolis that also offered algebra 2 as a tenth-grade class for its most advanced students. Later I met kids in Taiwan, and I mean AVERAGE kids in Taiwan, who learned that math at younger ages. In the United States, algebra 2 counts as a somewhat advanced math class for a tenth grader, but certainly not internationally. \n\n[quote=\"Startup\"]My parents don't have the money to spend on a summer courses, but this site, as well as the Art of Problem Solving books, sound like a perfect alternative. As another alternative, I'm beginning to look into online courses.[/quote]\n\nWe hammered our family budget this year by sending our son to a [url=http://www.jhu.edu/gifted/summer/catalogs/osmath.html]CTY summer program[/url], but no pain, no gain. My son also got a full scholarship to another summer program, but even the minimal out-of-pocket expenses we had from that program hurt our budget. If your family has a typical middle-class income, as mine does, you should qualify for SUBSTANTIAL financial aid from the [url=http://epgy.stanford.edu/courses/math/]EPGY distance learning programs[/url], which I highly recommend. \n\n[quote=\"Startup\"]I've looked at a few of the problems in this forum, and I've had difficulty solving a few of them, but I think it is because I haven't learned some of the necessary material. That will probably change as my high school education progresses.[/quote]\n\nYes, and we are also encouraging everyone to post more EASY problems on all the forums here, as there are many beginners (of various ages) who need more easy problems for practice. \n\n[quote=\"Startup\"]The point to all this is that hopefully I will be able to enter math competitions and perform well in them, as well as do better in class, and maybe even on the SAT (College entrance exams). Do you think I'll be able achieve all this through visiting this site often for problems and tips, as well as reading through the Art of Problem Solving?[/quote]\r\n\r\nUsing this site regularly, and reading the varied advice it has on study tips and taking tests, will surely be helpful. The Art of Problem Solving books are very good resources for a second, deeper look at most secondary school mathematics topics. \r\n\r\nEnjoy the site. Good luck in your studies.", "Solution_4": "I'm actually in the same boat as you. I've never taken math seriously until now and am also going into the 10th grade. I hope to aim for the AIME as well.\r\n\r\nI wish you the best of luck!\r\n\r\nJust note that 15 years isn't too late; the mathematician Lagrange only took math seriously at 15 and I know one physicist who took it seriously only at age 20.", "Solution_5": "[quote=\"countersixte\"][quote=\"Startup\"]Do you think I'll be able achieve all this through visiting this site often for problems and tips, as well as reading through the Art of Problem Solving?[/quote]\n\nNo doubt about it. :) I haven't ever seen anything like AoPS before... It truly is amazing! BTW, the classes here are excellent, too![/quote]\r\n\r\nAbout several months ago, I did see something like AoPS, Mathlinks! Perhaps you could call the site AoPS-Mathlinks? :)", "Solution_6": "Hi!\r\n[quote=\"Startup\"]My parents usually push me to enter math contests and the like,[/quote]\r\nMy parents encouraged me to learn math ever since I first started school, and I think it really helped to get me where I am now. I scored well on the AMC8 and got invited to Mathpath. I think math contests and stuff is a good way to practice for future contests. Even if you don't do well you'll do better next time. Good Luck with math!! :)" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Congratulations to all those who competed in the 2009 State Competition.\r\n\r\nThe WV State Winners / 2009 Team Members were:\r\n\r\nBenjamin Taylor - 7 Ayscue Taylor Home School\r\nJennifer Yu - 8 Suncrest Middle School\r\nShiv Sunil - 8 Suncrest Middle School\r\nMichelle Ma - 8 Suncrest Middle School\r\n\r\nCoach: Lois Swineford Suncrest Middle School\r\n\r\nGood Luck in Florida!!!", "Solution_1": "Do you know the scores? Cause my teacher ain't telling me.", "Solution_2": "[quote=\"O M G\"]Do you know the scores? Cause my teacher ain't telling me.[/quote]\r\n\r\nI do not know the scores, I got the names off of the MATHCOUNTS site. I am not of MATHCOUNTS level anymore, but I was a previous competitor for our state. Actually, I'm the only competitor from the last few years that didn't come from Morgantown.\r\n\r\nI don't know who you could contact about the scores, sorry...", "Solution_3": "Got any advice for me next year for mathcounts? Got 7th....", "Solution_4": "I'm guessing you are from Suncrest, so ask Ms. Swineford for all the help you can get and maybe even get some books from this site.\r\n\r\nIf you aren't from Suncrest, then ask your teacher for all the State or Nationals rounds he/she has.\r\n\r\nReading this, you can see how much it pays to go to Suncrest...lol\r\n\r\nI rly didint \"study\" for the comp, idk how i made nats...gl tho next yr", "Solution_5": "ok i'll try that", "Solution_6": "So... you saw my post of some of the scores right?" } { "Tag": [ "inequalities" ], "Problem": "Find all natural numbers $n$ such that the equation \r\n\r\n$a_{n+1}x^2 - 2x\\sqrt{a^2_{1} + a^2_{2} + ... + a^2_{n+1}} + a_{1} + a_{2} + ... + a_{n} = 0$\r\n\r\nhas real solutions for all real numbers $a_{1}, a_{2}, ..., a_{n+1}$.", "Solution_1": "There is no real solution if $4(a_1^2+a_2^2+...+a_{n+1}^2)-4(a_{n+1})(a_1+a_2+...+a_n)<0$\r\nThis can be expressed as $a_1^2-a_{n+1}a_1+a_2^2-a_{n+1}a_2+...+a_n^2-a_{n+1}a_n+a_{n+1}^2<0$\r\nEach $a_i^2-a_{n+1}a_i$ is lowest when $a_i=\\frac{a_{n+1}}{2}$. \r\nThen $a_i^2-a_{n+1}a_i$ comes out to be $-\\frac{a_{n+1}^2}{4}$. Now it's clear that n=1, 2, 3, 4." } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "prove that sqrt((a1-b1)^2+..+(an-bn)^2)+sqrt((b1-c1)^2+..+(bn-cn)^2))>=sqrt((c1-a1)^2+..+(cn-an)^2)) holds for all an,bn,cn reals n>=2\r\n\r\n[i]***[/i]", "Solution_1": "[b]suggestion[/b]\r\nuse ineq Minkowski\r\n( \\sum (a_k-b_k) 2 ) 0.5 +( \\sum (b_k-c_k)) 0.5 \\geq ( \\sum (c_k-a_k)) 0.5", "Solution_2": "In fact that is the solution:\r\n\r\napply Minkowsky to the cuples: (a1-b1,a2-b2,..,an-bn) and (b1-c1,..,bn-cn)\r\n\r\ncheers! :D" } { "Tag": [], "Problem": "Prove why does the equations n^2+n+41 and n^2-n+41?\r\nnot always give you a prime number??please show the steps", "Solution_1": "[hide=\"hint\"]substitute 41.[/hide]", "Solution_2": "This is actually true of any integer-valued polynomial. The proof is only slightly harder.", "Solution_3": "[quote=\"pythag011\"][hide=\"hint\"]substitute 41.[/hide][/quote]\r\n\r\nThis is the whole answer. A counterexample is enough in questions like this.", "Solution_4": "Please don't double post http://www.mathlinks.ro/viewtopic.php?p=1261945" } { "Tag": [ "quadratics", "LaTeX", "algebra" ], "Problem": "If you want to find [tex]\\sqrt{a + b\\sqrt{c}}[/tex], for any real values of a, b and c (including negative for c), does the single method I read before work all the time?\r\n\r\nIt goes like\r\n\r\n[tex]\r\n\\sqrt{a + b \\sqrt{c})} = (r+s)\r\n\r\n(a + b \\sqrt{c}) = (r+s) = r + 2rs + s\r\n\r\nso a = r + s\r\n\r\nbc = rs\r\n[/tex]\r\n\r\nAnd solve with r and s roots of the quadratic equation\r\nx + 2a + bc = 0 ?\r\n\r\nI think... does this work for every real value of a, b and c?\r\nAnd if I have this wrong, please tell me.. its new to me and I only remembered half the steps so I guestimated the rest. :lol:", "Solution_1": "eum could you please translate this in english? :?\r\n\r\nwhat is ? and what is ?? :?\r\n\r\nJust displaying a square here...", "Solution_2": "I think this is a great problem in itself - trying to work out what it all means, so here goes:\r\nClue 1: ? and only ever occur together so could stand for one symbol\r\nClue 2: \"including negative for c\" is relevant\r\nClue 3: ?(a + b?c) = (r+s) and then writes (a + b?c) = (r+s) \r\nSo I reckon ? is just :sqrt: which means that the question is:\r\n[quote]If you want to find [tex]\\sqrt{a + b\\sqrt{c}}[/tex], for any real values of a, b and c (including negative for c), does the single method I read before work all the time?[/quote]\r\nNow it's up to mikewd to say if I am right or wrong :)\r\n\r\nPS Next problem: work out how ? came to be used", "Solution_3": "[quote=\"stevem\"]PS Next problem: work out how ? came to be used[/quote]\r\n\r\nDifferent character system/encoding (unicode vs. ANSI, maybe)?", "Solution_4": "therefor we got latex: $\\sqrt{perfect\\ square}=integer$", "Solution_5": "Oops! You are correct - it's a square root sign. Copied and pasted from microsoft word. It worked on another forum. Sorry for the confusion, I'll edit it.\r\n\r\nedit (after correcting): latex is pretty handy actually!", "Solution_6": "[quote=\"mikewd\"]$\\sqrt{a + b \\sqrt{c})} = (r+s)$\nso $a = r + s$ \nand $bc = rs$[/quote]\r\n\r\nno clue where you get that non-sense: $a=1, b=4, c=4, r=3/2, s=3/2$\r\n\r\nthe condition $\\sqrt{1 + 4 \\sqrt{4})} = ( \\frac32 + \\frac32 )$ is fulfilled\r\n\r\nbut $1 = \\frac{18}4$ and $64 = \\frac{81}{16}$ are luckily false :)\r\n\r\nIt would be true though IF $a,b,c,r,s$ are integers and $c$ is not a perfect square...", "Solution_7": "I think wanting to solve [tex]\\sqrt{a+b\\sqrt{c}} = r + s\\sqrt{c}[/tex] That makes a bit more sense, and also gives [tex]a = r^2 + cs^2, b = 2rs[/tex], which looks like what mikewd was meaning.", "Solution_8": "Still, you're in trouble when c is a perfect square. And still, they need to be integers." } { "Tag": [], "Problem": "I don't understand the first part of their solution.\r\nthe first pic is the solution so be careful!!", "Solution_1": "They just applied the conjugation in $\\mathbb{Z}\\left[\\sqrt{2}\\right]$ to both sides." } { "Tag": [], "Problem": "Who made it on AoPS?\r\n\r\nI just got my letter today. Woot! See you guys next school year!", "Solution_1": "i wish i could go there next year as a freshman...too bad i like like 2.5 hrs away :( \r\n\r\nBut oh well, if i went there, i wouldnt be considered \"smart\" since everyone's smart at TJ...but here, there arnt a lot of prob-solving smart people... :roll:" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME" ], "Problem": "I just thought I'd get it started early this year, and go ahead and ask if they've graded the tests yet.....\r\nNot actually.\r\nBut- is there any way to get scores on individual problems when they do release the scores? As far as I can tell, there isn't. Is there a philosophical reason for this? Does the CAMC just not want people second guessing them or posting scanned solutions and saying \"I only got a 5 I can't believe it I should've made MOP\"? Or is it something else?\r\nThanks", "Solution_1": "I don't believe so. However, it's generally pretty easy to tell what you got on a problem, since you'll almost always get a 0, 1, 6, or 7 on any given problem.", "Solution_2": "Hamster is going to get 7 on all problems and be the 2007 IMO perfect scorer!", "Solution_3": "[quote=\"mrosett\"]I just thought I'd get it started early this year, and go ahead and ask if they've graded the tests yet.....\nNot actually.[/quote]\r\n\r\nhaha nice\r\n\r\nalso delong wtf?", "Solution_4": "Dang, I got a 3 last year...and i only answered one correctly another one had a completely incorrect answer...which i dont think they gave credit for or do they give credit for incorrect/incomplete solutions.", "Solution_5": "[quote=\"MysticTerminator\"][quote=\"mrosett\"]I just thought I'd get it started early this year, and go ahead and ask if they've graded the tests yet.....\nNot actually.[/quote]\n\nhaha nice\n\nalso delong wtf?[/quote]\r\n\r\nHe made this prediction at mathcamp...apparently I'm also supposed to prove the Riemann Hypothesis.", "Solution_6": "In regard to your second question, which was kind of mixed up with the first, I quote the 2007 USAMO Teacher's Manual:\r\n\r\nUSAMO Winners and all those invited to the MOSP will be notified by telephone shortly after the [b]USAMO grading on May 4,5, & 6, 2007[/b]. Information regarding the MOSP will be sent to those selected beginning on [b]May 8th[/b]. A copy of the USAMO with solutions and results [including the score of your student(s)] will be mailed to you on [b]Wednesday, May 10rd[/b]. Also included will be a USAMO Certificate of Participation to be presented to your student(s). Students and teachers should not inquire about results before [b]Friday, May 11th[/b].\r\n\r\nI like the \"Wednesday, May 10rd\" part. Just try saying that. It's quite amusing. I also like \"the MOSP\". Somehow \"MOSP\" never seems to need an article.", "Solution_7": "Isn't \"May 10rd\" a Thursday anyway? unless they are trying to trick us.", "Solution_8": "lol ya it is. I wonder where they think Thursday goes. Maybe we'll get warped into the 7th dimension and it will change all of our days around. Then by Friday it will be back to normal. (wait then I won't be able to fly to nats!)darn.", "Solution_9": "wow - that is kinda dumb. i missed something else which was obvious:\r\n\r\nWednesday, May 10rd ... Friday May 11th\r\n\r\nprobably it is meant to be thursday may 10th, but they changed it from previous years and somebody at CAMC didn't check their grammar - that's why we're here though.\r\n\r\nhey mr. dunbar (or whoever reads this first and has access to the AMCDirector account) care to comment?\r\n\r\nEDIT: Also to AMCDirector - will this forum be closed for the USAMO? I think I saw this asked somewhere else (maybe it was for AIME) but I assume it will...", "Solution_10": "[quote=\"Teki-Teki\"]Also to AMCDirector - will this forum be closed for the USAMO? I think I saw this asked somewhere else (maybe it was for AIME) but I assume it will...[/quote][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=818495#818495[/url]", "Solution_11": "Luckily, since the USAMO is administered at the same time the forum doesn't [i]have [/i]to be closed until like 12:25PM EST tomorrow.\r\n\r\n\r\n\r\nOlympiad scoring arithmetic is so awesome\r\n\r\n----\r\n\r\nA small, non-trivial correct step in a problem: 1 point\r\nTwo small non-trivial correct steps in a problem: 1 point\r\n\r\n1 + 1 = 1\r\n\r\n----\r\n\r\nA correct solution: 7\r\nA nearly correct solution with something small missing (like a trivial equality case): 6\r\nHaving only the said trivial equality case: 0\r\n\r\n7 - 0 = 6\r\n\r\n----" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "For a,b,c positive reals\r\n\r\n\r\n$\\frac {a^5+b^5+c^5}{5}+a^2bc^2\\geq ab^3c$", "Solution_1": "Assume that $ac=1$, then\r\n\\[\r\na^5+b^5+c^5+5a^2bc^2-5 ab^3c \\\\\r\n= (a^5+c^5) + b^5 -5b^3 +5 b \\\\\r\n \\geq b^5 - 5b^3 + 5 b + 2 \\\\\r\n = (b+2)(b^2-b-1)^2 \\\\\r\n \\geq 0 \\\\\r\n\\]" } { "Tag": [ "number theory open", "number theory" ], "Problem": "Find the smallest integer n with following property: do not exist prime with n digits.", "Solution_1": "No such $ n$ exists by [url=http://en.wikipedia.org/wiki/Bertrand%27s_postulate]Bertrand's postulate[/url]." } { "Tag": [ "AMC", "AIME" ], "Problem": "If A is the sequence a1, a2, a3, ... , define \u0394A to be the sequence a2 - a1, a3 - a2, a4 - a3, ... . If \u0394(\u0394A) has all terms 1 and a19 = a92 = 0, find a1.", "Solution_1": "Well, I just found this site a week or so ago, and after getting a good amount of AIME practice due to your posting problems, fiery, I guess I will post one of my solutions. (which is probably wrong anyway)\n\n\n\n[hide]\n\nFirst, notice that the difference of the sequence :tri: A is 1, so we can rewrite\n\n :tri: A_1=d\n\n :tri: A_2=d+1 ...etc\n\nnow since :tri: A_n is the difference between the terms A_n and A_n+1, we can write the sequence:\n\n\n\nA_1=A_1\n\nA_2=A_1+d\n\nA_3=A_1+2*d+1\n\nA_4=A_1+3*d+2+1\n\n\n\nA_n=A_1+(n-1)*d+(n-2)(n-1)/2\n\n\n\nSetting this value equal for A_19 and A_92, D was equal to -54, and then putting this back into our equation for A_n for A_19, and then solving for A_1\n\n\n\nA_19 = 0 = A_1+18*-54+(18)(17)/2\n\nA_1 = 819\n\n\n\nI can sort of see some ways of getting A_1 without solving for d (or :tri: A_1), but this is the way in which i can \"see\" the solution best, where I am not doing any manipulations that I don't fully understand.\n\n[/hide]\n\n\n\nTo everyone, thanks for posting all the practice AIME problems (and hints+solutions) in the last few days, they have been very helpful, and I will most definetly be back after the AIME tomorrow. GL", "Solution_2": "Welcome to the forum GeoMar. Good job, your solution is correct. If you hang around here a lot, you oughta post a bit more. Its fun, and I've learned so much stuff on here just by being around smart people." } { "Tag": [ "logarithms", "search", "inequalities", "analytic geometry" ], "Problem": "1.Solve the following inequality. \\[2\\log_{a}(x-4)>\\log_{a}(3x-8)\\ (a>0,\\ a\\neq 1).\\] 2. Explalin the meanimg of the real roots of the following equation. \\[x^{2}=2^{x}.\\] 3.Explain the meaning of the real roots of the following equation. \\[(x-1)^{3}+1-a=0.\\]", "Solution_1": "[hide=\"1\"]\n$log((x-4)^{2})>log(3x-8)$\n$x^{2}-8x+16>3x-8$\n$x^{2}-11x+24>0$\n$(x-8)(x-3)>0$\nThink about the graph\n\nAnswer: x<3 or x>8\n\n[/hide]\nThanks, SplashD. The roots are 2, 4, and a negative irrational.\n[/hide]\n[hide=\"3\"]\n$(x-1)^{3}= a-1$\n$x-1 = (a-1)^\\frac{1}{3}$\n$x = 1+(a-1)^\\frac{1}{3}$\nAgain, don't know the meaning.\n[/hide]", "Solution_2": "For number 2, x can also equal to 4 as well as a negative irrational that I don't know the exact value of", "Solution_3": "[quote=\"vvr1590\"]\n$log((x-4)^{2})>log(3x-8)$\n[/quote]\n\nwhat if $\\log a<0$?\n\n[quote=\"kunny\"]...\n3.Explain the meaning of the real roots of the following equation.\n[/quote]\r\n\r\nI don't understand what you want when you say: \"meaning.\"", "Solution_4": "[quote=\"kunny\"]2. Explalin the meanimg of the real roots of the following equation. \\[x^{2}=2^{x}.\\] [/quote]\r\nWe can see the obvious integer solutions, but I'm guessing the next best way to find the values of $x$ which satisfy this would be to set $f\\left(x\\right)=x^{2}-2^{x}$ and use Newton's Method to approximate the roots.", "Solution_5": "[quote=\"Altheman\"][quote=\"vvr1590\"]\n$log((x-4)^{2})>log(3x-8)$\n[/quote]\n\nwhat if $\\log a<0$?\n\n[quote=\"kunny\"]...\n3.Explain the meaning of the real roots of the following equation.\n[/quote]\n\nI don't understand what you want when you say: \"meaning.\"[/quote]\r\n\r\nSorry for my poor english, I ask you that what is the solution of equation on graphically.\r\n\r\nkunny", "Solution_6": "[quote=\"vvr1590\"][hide=\"1\"]\n$log((x-4)^{2})>log(3x-8)$\n$x^{2}-8x+16>3x-8$\n$x^{2}-11x+24>0$\n$(x-8)(x-3)>0$\nThink about the graph\n\nAnswer: x<3 or x>8\n\n[/hide][/quote]\n\n[hide] $x>4$ [/hide]", "Solution_7": "Number 2 was discussed many times before, please search for it (I've searched for it, but without result, maybe the search tool is not working well)", "Solution_8": "#1\r\n\r\n[hide]\n\nRemember that you cannot take the log of a negative number, so x-4>0 and 3x-8>0, the intersection of these two inequalities being x>4. \n\nFor a>1:\nx>8\n\nFor 0 c,\\ b \\plus{} c > a,\\ c \\plus{} a > b$.\r\n\r\n(1) Prove that $ a > 0,\\ b > 0,\\ c > 0$.\r\n\r\n(2) Prove that $ |b \\minus{} c| < a,\\ |c \\minus{} a| < b,\\ |a \\minus{} b| < c$.\r\n\r\n(3) Prove that $ 1\\leq \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} < 2.$\r\n\r\n\r\n$ \\boxed{2}$ Let $ S$ the area of the right triangle with the perimeter 1 and the hypotenuse $ t$ in length.\r\n\r\n(1) Let $ a,\\ b$ be the other two side lengths. Find the values of $ a \\plus{} b,\\ ab$ in terms of $ t$.\r\n\r\n(2) Let $ t$ vary to move, find the range of $ S$ for which $ S$ can be valued.", "Solution_1": "For #1 part 1:\r\nSuppose only c is negative. Then we have c>a-b and c>b-a, but one of a-b,b-a is positive (or 0), contradiction.\r\nSuppose b and c are negative. Then b+c>a, contradiction.\r\nSuppose all three are negative. Adding all three inequalities, we get a+b+c>0, contradiction.\r\n\r\nFor part 2, assume $ a \\ge b \\ge c$, then it is obvious, as the three are equivalent to the original three.\r\n\r\nFor part 3, it is equivalent to\r\n$ 0 \\le (a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2 \\le a^2 \\plus{} b^2 \\plus{} c^2$\r\nThe LHS is obvious. But for the RHS, we have -ca\\plus{}c\\implies 2b>0\\implies b>0$. Similarly for $ a>0, c>0$\n\n(2) From the first inequality, $ a>c\\minus{}b$, and from the third, $ a>b\\minus{}c$. Therefore $ a>|b\\minus{}c|$. Similarly for the remaining two.\n\n(3) Use $ (a\\minus{}b)^2\\plus{}(b\\minus{}c)^2\\plus{}(c\\minus{}a)^2\\geqslant 0$, expand and rearrange to get $ a^2\\plus{}b^2\\plus{}c^2\\geqslant ab\\plus{}bc\\plus{}ca$. Then use the result (2) to write $ (a\\minus{}b)^2\\plus{}(b\\minus{}c)^2\\plus{}(c\\minus{}a)^2t$, also\r\n$ \\frac{a^2\\plus{}b^2}{2} \\ge \\frac{(a\\plus{}b)^2}{4} \\Leftrightarrow \\sqrt{2}t \\ge a\\plus{}b$\r\n$ \\Leftrightarrow (\\sqrt{2}\\plus{}1)t \\ge 1$\r\n\r\nSo $ \\frac{1}{2}>t \\ge \\frac{1}{\\sqrt{2}\\plus{}1}\\equal{}\\sqrt{2}\\minus{}1$ is the range of $ t$\r\n\r\n$ \\frac{1}{2}\\minus{}(\\sqrt{2}\\minus{}1)\\equal{}\\frac{3}{2}\\minus{}\\sqrt{2} \\ge ab >0$\r\n$ \\Leftrightarrow \\frac{3\\minus{}2\\sqrt{2}}{4} \\ge S >0$ is the range of $ S$", "Solution_4": "Yeh, the problem has been solved. :lol:", "Solution_5": "[color=blue]1st place: Mnjm :winner_first: :first:\n2nd place: Iijm :winner_second: \n3rd place: Ski :winner_third: [/color]\r\n\r\n\r\n\r\n$ \\displaystyle \\displaystyle \\displaystyle \\begin{tabular}{|c|c|c|c|c|c\\parallel{}c|}\\hline Name & Problem 1 & Problem 2 & Total Score \\\\\r\n\\hline Mnjm & 50 & 33 & 83 \\\\\r\n\\hline Iijm & 30 & 46 & 76 \\\\\r\n\\hline Ski & 50 & 25 & 75 \\\\\r\n\\hline Average & 20.8 & 18.8 & 39.6 \\\\\r\n\\hline\\end{tabular}$\r\n\r\nEntry 55", "Solution_6": "sorry, i will not show the details because i have to be fast.\r\n$ 1$:\r\n1:\r\nsuppose it is not true and try arrive to an absurd.or use these: $ a\\plus{}b>{c}$and $ b\\plus{}c>{a}$.we\u00b4ll have that $ 2b\\plus{}c>c$,so $ b>0$.\r\n\r\n2:i think this is literally direct from the question :) \r\n\r\n3:$ ab\\plus{}bc\\plus{}ca0$\r\n\r\nThis is true since we can assume $ a,b,c$ are the sides of a triangle.. :)" } { "Tag": [], "Problem": "Suppose that $ n$ is a positive integer. Prove that there is a positive integer k for which $ (\\sqrt{n}-1)^{n}=\\sqrt{k}-\\sqrt{k-1}$", "Solution_1": "I will disprove the problem statement.\r\n\r\nLets assume that the problem statement is true:\r\n\r\nLet $ n = 1$\r\n\r\n$ \\Rightarrow (\\sqrt{n}-1)^{n}= (\\sqrt{1}-1)^{1}= 0$\r\n\r\n$ \\Rightarrow \\sqrt{k}-\\sqrt{k-1}= 0$\r\n\r\n$ \\Rightarrow \\sqrt{k}= \\sqrt{k-1}$\r\n\r\n$ \\Rightarrow k = k-1$\r\n\r\n$ \\Rightarrow 0 =-1$\r\n\r\nThis is a contradiction therefore the problem statement is false.", "Solution_2": "i think the problem is true for $ n>1$ though. but i don't have a solution yet.", "Solution_3": "Really? How so? Am I not understanding the problem?\r\n\r\nNote that the LHS side increases without bound and the RHS starts at 1 and converges to zero...\r\n\r\nIn fact, let $ n = m^{2}, \\; m \\in \\mathbb{Z}$. Then, the LHS is equal to $ (m-1)^{m^{2}}$, an integer.\r\n\r\n$ \\sqrt{k}-\\sqrt{k-1}$ is an integer iff. $ k=1$, in which case it equals $ 1$." } { "Tag": [ "induction" ], "Problem": "Hey all,\r\n\r\nI'm new to the board and have a quick question.\r\n\r\nWhere would I post anything related to Mathematical Reasoning, Induction, and Recursion?\r\n\r\nThank you,\r\nShamrock Hoax", "Solution_1": ":welcome:\r\n\r\nDon't quote me on this, but I think probably High School Basics or Intermediate Topics would be your best bet.", "Solution_2": "Hello and welcome :wink: to AOPS/MathLinks.\r\n\r\nIf you would like to see an overview of all the subforums here, click:\r\n[url=http://www.mathlinks.ro/Forum/index.php]this link[/url]", "Solution_3": "Hello! I didn't want to make myself a new topic so are you okay iff i post this:\r\n[url]http://www.claymath.org/millennium/[/url]\r\nare there anyother things you can be on this forum?\r\nthis is sort of another newbie thing but i am not a newbie so i hope you are okay with me posting it....i second the, \"Don't quote me on this, but I think probably High School Basics or Intermediate Topics would be your best bet.\" but cross out the don't quote me part..." } { "Tag": [ "geometry", "perimeter", "function", "inequalities", "number theory proposed", "number theory" ], "Problem": "1) For any natural number n, (n \\geq 3), let f(n) denote the number of congruent integer sided triangles with perimeter n\r\n (eg f(3)=1, f(4)=0,f(7)=2 ) .Show that\r\n a) f(1999) > f(1996)\r\n b) f(2000) = f(1997)\r\n \r\n2) Show that there do not exists integers m and n such that \r\n m/n + n+1/m =4", "Solution_1": "For (2):\r\n\r\nWe have m/n+n+1/m=4 iff m^2/mn+mn^2/mn+n/mn=4 iff m^2+mn^2+n=4mn iff m(m+n^2)=n(4m-1), thus m|n because (m,4m-1)=1. We have n=km, so m(m+k^2*m^2)=km(4m-1) iff m(1+mk^2)=k(4m-1), so m|k because (m,4m-1)=1, so k=am, so m(1+a^2*m^3)=am(4m-1) iff 1+a^2*m^3=a(4m-1), so a is 1, so 1+m^3=4m-1 iff 4m-m^3=2, so m|2, so m can only be 1, -1, 2, or -2. By checking all of these we see that there are no solutions.\r\n\r\nMaybe there's something wrong here, because it seems too easy :)", "Solution_2": "For 1) :\r\nMore generally, we prove that f(n+3) = f(n) if n is odd, and f(n+3) > f(n) if n is even (and the triangles are supposed to be nondegenerated).\r\n\r\nLet E(n) = {(a,b,c) / a,b,c integers, a \\leq b \\leq c, a+b > c}.\r\nThen f(n) = |E|.\r\n\r\nLet F the function defined on E(n) by F(a,b,c) = (a+1,b+1,c+1).\r\nClearly, F(a,b,c) belongs to E(n+3) and F is injective.\r\nThus equality or inequality between f(n) and f(n+3) will follows from that F is surjective or not.\r\n\r\n- If n = 4q :\r\nThen n+3 = 4q+3.\r\nIt's easy to see that (x,y,z) = (q+1, q+1, 2q+1) is in E(n+3). But (x-1,y-1,z-1) = (q,q,2q) is not in E(n). Thus, (x,y,z) is not in F(E(n)), which proves that F is not surjective.\r\n\r\n- If n = 4q+2 :\r\nThen n+3 = 4q+5, and we get the same conclusion as above with (x,y,z) = (q+1, q+2, 2q+2).\r\n\r\nThus f(n+3) > f(n) if n is even.\r\n\r\n- if n is odd :\r\nLet (x,y,z) in E(n+3). Let (a,b,c) = (x-1,y-1,z-1).\r\nClearly, a,b,c are integers such that a \\leq b \\leq c and a+b+c = n.\r\nSince x+y > z and x,y,z are integers, we have x+y \\geq z+1.\r\nIf x+y = z + 1 then n+3 = x+y+z = 2z+1, which is impossible since n+3 is even. Then x+y > z+1, which leads to a+b > c.\r\nIt remains to prove that a \\geq 1, that is x \\geq 2. Now suppose that x = 1. Then, from the triangular inequality, we have y > z - 1, that is y \\geq z, since we are dealing with integers. Since y \\leq z, we then have y = z. Thus n+3 = x + y + z = 1 + 2z, and we get the same contradiction as above. Thus x \\geq 2, and (a,b,c) is in E(n), and F(a,b,c) = (x,y,z), which proves that F is surjective. Thus F is bijective from E(n) onto E(n+3) and f(n) = f(n+3).\r\n\r\nPierre." } { "Tag": [ "MATHCOUNTS", "calculus", "integration" ], "Problem": "How many negative bases exists in which all positive integral numbers can be expressed? (Negative bases work the same way as positive bases. For example, 562 in base -7 is 5*(-7):^2:+6*(-7)^1+2*(-7)^0=5*(49)+6*-7+2*1=205)\r\n\r\nEasier Question: How many negative bases exists in which all integral numbers can be expressed?", "Solution_1": "Negative bases don't work very well in general because the representations for numbers aren't unique.", "Solution_2": "Yup, but that shouldn't be a problem...or will it?", "Solution_3": "Well, it can be. If I tell you to convert 43 to base -10, do I want you to write 163 or -57? It's not clear at all.", "Solution_4": "While it may be true (and I'm not quite sure since I think -57 in base -10 is just 57 in base 10), it shouldn't be a problem in this problem.", "Solution_5": "No. -57 in base -10 is -(5*-10+7)=-(-43)=43.", "Solution_6": "Oh, I thought it was (-5)7, not -(57), but oh well. Again, since the problem only ask if it [b]can [/b]be expressed, this shouldn't be a problem." } { "Tag": [ "floor function", "function" ], "Problem": "A set $M$ of positive integers is called [i]connected[/i] if for any element $x\\in M$ at least one of the numbers $x-1,x+1$ is in $M$. Let $U_n$ be the number of the connected subsets of $\\{1,2,\\ldots,n\\}$. \r\n\r\na) Compute $U_7$;\r\n\r\nb) Find the smallest number $n$ such that $U_n \\geq 2006$.", "Solution_1": "[hide]In other words, no number can be \"isolated\" (like in {2,3,5,8,9}, 9 is isolated.)\nSo U_n must consist of at least 2 elements.\n\nFor U7 there can be 2,3,4,5,6 or 7 elements.\n\n2 - each one must be a pair, there are 6 sets.\n3 - no isolation-- there must be a triplet, there are 5 sets.\n4 - we can either have a string of 4 (4 sets), or two pairs (10 sets) = 14 sets.\n5 - we can either have a 3-string and 2-string (6 sets), or a 5-string (3 sets) = 9 sets.\n6 - if 2 or 6 were missing we would have isolations, this leaves us with 5 sets.\n7 - 1 set.\n\n6+5+14+9+5+1 = [b]40 sets. [/b]\n\nDon't know #2 though :blush: [/hide]", "Solution_2": "[hide]Actually the answer is 37: there are only 6 connected subsets of ${1, \\dots, 7}$ that consist of 2 separate pairs, but you forgot to count the null set, which is connected.\n\nTo give this post some semblance of constructivity, I'll remark that the smallest $n$ in the second part is 15; if I'm right in my computation, $U_{14} = 1897$ but $U_{15} = 3329$.[/hide]", "Solution_3": "perhaps create a recursion for $U_n$", "Solution_4": "It doesn't right from my point of view.\r\nWhen you disscus about the numbers of 4-element sets, the two situations have something in common :!: \r\n\r\nAnd I think the correct recursion should be Un=Un-1 +Un-2 +2\r\n\r\nI got the answer U7=32, and 16 is the smallest number satisfies the request.", "Solution_5": "No.Here's an indication.Let's consider M={a,b,c,d} whith a=2006, which gives n>=66.The answer is n=66.", "Solution_6": "Valentin, is the question perhaps misstated? The previous poster solves a question you didn't ask, but one which is suggested by the title of the topic, namely where $U_n$ is the number of connected [i]4 element[/i] subsets of $\\{1, 2, \\ldots, n\\}$.\r\n\r\n(So far, by the way, Saul is the only person to give a correct calculation of $U_7$ as it is defined in the question.)", "Solution_7": "I don't think the Saul's solution is correct.How is Valentin? :?", "Solution_8": "Saul is definitely correct, given the current statement of the question.\r\n\r\n\r\nIn particular, the formula for the thing that the question currently asks is given by:\r\n\r\n$1+\\sum_{n=0}^m\\sum_{i=0}^{\\lfloor \\frac n2\\rfloor} {n-i-1 \\choose i-1}\\cdot{m-n+1 \\choose i}$.\r\n\r\n\r\nThe process to get to this is to define a function $P_i(n)$, the number of ordered partitions of $n$ into $i$ parts, each at least size 2, which works out to be $n - i - 1 \\choose i$, and the function $Q_i(k)$, the number of ordered partitions of $k$ into $i$ non-empty parts, which is $k+i-1 \\choose i-1$. The result comes from counting the subsets based on the shape and arrangement of their connected components.", "Solution_9": "I don't think so!I was at this olympiad and I took 7p for 7p(maximum) for this problem.Let's ask Valentin...", "Solution_10": "Read the question that Valentin has written here -- it is not the same question that you have solved in this thread!", "Solution_11": "Yo say about the problem, no?I've solved directly b)-find the smallest n such that U(n)>=2006 and this is n=66.", "Solution_12": "[quote=\"Altheman\"]perhaps create a recursion for $U_n$[/quote]\r\n\r\n[i]Is there anyone working along this line of attack?[/i]\r\n\r\nLet M be any connected subset of {1,2,3....,n}\r\n\r\nCase 1 $n \\notin M$\r\nThen M is a connected subset of {1,2,3,...,n-1}\r\nThere are $U_{n-1}$ cases.\r\n\r\nCase 2 $n \\in M$\r\nThen it is necessary that $n-1 \\in M$ also, for otherwise the number n will be isolated.\r\nBasically, I think we need to consider a subset of the form $M=N \\cup \\{n-1,n\\}$\r\nNow, either $N=P \\cup \\{n-2 \\}$ with P being a connected subset of {1,2,...,n-4}, if N is [b]disconnected[/b], or else $N$ is any [b]connected[/b] subset of {1,2,...,n-2}.\r\nHence, there are $U_{n-4}+U_{n-2}$ ways of forming $M$ in this case.\r\n\r\nCombine the results, $U_n =U_{n-1}+U_{n-2}+U_{n-4}$ for n >4.\r\n\r\nLet us work out some cases for n:\r\n$U_1=1$,\r\n$U_2=2$,\r\n$U_3=4$,\r\n$U_4=7$,\r\n$U_5=U_4+U_3+U_1=7+4+1=12$,\r\n$U_6=U_5+U_4+U_2=12+7+2=21$, and\r\n$U_7=U_6+U_5+U_3=21+12+4=37$ which is the same as Saul's result.\r\n\r\nIf this recurrsion is correct, then we may solve the problem in an easier way. :idea:\r\n\r\nWith this recurrsion, I found that $U_{14}=1896$ and $U_{15}=3329$.\r\nTherefore, the answer of this problem should be $n = 15$.\r\n\r\nIs there anyone who knows the answer for checking?", "Solution_13": "@stancioiu sorin:\r\nDid you re-read the question as written? Here is the definition of $U_n$ in the first post in this thread:\r\n[quote=\"Valentin Vornicu\"] Let $U_n$ be the number of the connected subsets of $\\{1,2,\\ldots,n\\}$. [/quote]\n\nYou are solving a question which uses the following definition:\n[quote=\"Not Valentin Vornicu\"]Let $U_n$ be the number of [i]4-element[/i] connected subsets of $\\{1,2,\\ldots,n\\}$. [/quote]\r\n\r\nDo you understand?\r\n\r\n\r\n@Luimichael:\r\nLooks good to me! Except that you wrote down 1896 when you should have written 1897. These are also the values Saul gave and which my, much uglier, solution gives.", "Solution_14": "JBL,\r\nThank you for your confirmation. \r\nThe value 1896 is a typing mistake. hehehe :lol:", "Solution_15": "[quote=\"JBL\"]@stancioiu sorin:\nDid you re-read the question as written? Here is the definition of $U_n$ in the first post in this thread:\n[quote=\"Valentin Vornicu\"] Let $U_n$ be the number of the connected subsets of $\\{1,2,\\ldots,n\\}$. [/quote]\n\nYou are solving a question which uses the following definition:\n[quote=\"Not Valentin Vornicu\"]Let $U_n$ be the number of [i]4-element[/i] connected subsets of $\\{1,2,\\ldots,n\\}$. [/quote]\n\nDo you understand?\n\n\n@Luimichael:\nLooks good to me! Except that you wrote down 1896 when you should have written 1897. These are also the values Saul gave and which my, much uglier, solution gives.[/quote]\r\n Yes, JBL, I've understood the mistake.But Vlaentin has changed the problem.When I solved this problem at the District Olympiad, M had this forme: M={a,b,c,d}-four integer numbers.That is[b] sure![/b]But, how I said, V.V has changed the problem;or he didn't read it correctly." } { "Tag": [ "Euler", "IMO", "IMO 2009" ], "Problem": "a^n+b^n=c^n isnt true for all even integers,say n=2p,then (a^p,b^p,c^p)is a pythagorean triple which are same exponent of different numbers which is impossible,can you prove this for odd numbers,then it will be proved totally", "Solution_1": "[quote=\"masum billal\"]a^n+b^n=c^n isnt true for all even integers,say n=2p,then (a^p,b^p,c^p)is a pythagorean triple which are same exponent of different numbers which is impossible,can you prove this for odd numbers,then it will be proved totally[/quote]\r\n\r\nWhat about n = 2, im sure $ 3^2 \\plus{} 4^2 \\equal{} 5^2$.\r\n\r\nAnd i think that if it really was that simple to prove, then it would have been solved by Euler in the 18th century ;)", "Solution_2": "1) This isn't the forum for this\r\n\r\n2) If the proof was that easy, Andrew Wiles wouldn't write hundreds of pages\r\n\r\n3) There were many guesses and tries to prove it\r\n\r\n4) kurt.math, I think you know that the LFT is for $ n>2$ :)", "Solution_3": "[quote=\"Bugi\"]kurt.math, I think you know that the LFT is for $ n > 2$ :)[/quote]\r\n\r\nI know :D lol" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Find all monic polynomials P(x) and Q(x) with integer coefficients such that Q(0)=0 and P(Q(x))=(x - 1)(x - 2)(x - 3) ... (x - 15).", "Solution_1": "Hey! This is still unsolved. what's the matter?\r\n Ashwath", "Solution_2": "degP+degQ odd then\r\n1, degP odd, degQ even. therefore the range of PQ is not R, the range of (x-1)..(x-15) is R\r\n2. similarly.\r\n=> there dont exist P and Q", "Solution_3": "Isn't $Q(X)=X$ and $P(x)=(x - 1)(x - 2)(x - 3) ... (x - 15)$ a solution nmtruong1986 ?\r\n\r\n\r\nI guess this is the only one ;)", "Solution_4": "My proposal (a bit complicated) :\r\n\r\nObviously $d \\textdegree Q \\geq 1$. \r\nLet's suppose $d \\textdegree Q > 1$ and then $d \\textdegree P \\leq 13$. \r\n\r\nAll values of the $Q(k)$, $k=1..15$ can't be different otherwise $P$ would have $> 13$ roots which is impossible ($P \\neq 0$).\r\n\r\nWe must have $Q(k_1) = Q(k_2)$ and $Q(k_3) = Q(k_4)$ or $Q(k_1) = Q(k_2)= Q(k_3)$, $k_i$ being in $1..15$ and different.\r\n\r\nLet's look at first case, wlog, $k_1< k_2 < k_3 < k_4$ :\r\n\r\n$Q(X) = A + (X-k_1)(X-k_2)R_1(X)$ \r\n$Q(X) = B + (X-k_3)(X-k_4)R_2(X)$ \r\n\r\nFirst note that $A$ and $B$ are not null otherwise $P(Q(X))$ would not be divisible by $(X-k_1)(X-k_2)$ or\r\n$(X-k_3)(X-k_3)$ \r\n\r\n$Q(X) = 0 \\Rightarrow k_1*k_2 | A$\r\n$Q(X) = 0 \\Rightarrow k_3*k_4 | B$ \r\nSince $1 \\leq A,B \\leq 15$ ($A$ is a root of $(x - 1)(x - 2)(x - 3) ... (x - 15)$) the only possibility is $k_1=1$, $k_2$ = 2, $k_3=3$, $k_4=4$ and $A=2$, $B=12$\r\n\r\n$Q(X) = 2 + (X-1)(X-2)R_1(X)$ \r\n$Q(X) = 12 + (X-3)(X-4)R_2(X)$ \r\n\r\nBut then $P$ is then given only even numbers, which prevent odd roots to be reachable.", "Solution_5": "Nice solution tptp\r\n Ashwath!" } { "Tag": [ "vector", "function", "linear algebra", "linear algebra unsolved" ], "Problem": "Let V be the subspace of the vector space of all real-valued functions that is spanned by the set S={(cos(x)^2), (sin(x)^2), cos(2x)}. Show that V is isomorphic to R^2.\r\n\r\nI have to prove this. This is my logic... since cos(2x) is a linear combination of (cos(x)^2) and (sin(x)^2), then I can form a new set S'={(cos(x)^2), (sin(x)^2)} which also spans the vector space of all real-valued functions. Obvisously, S' is linearly independent. (I used the Wronskian to show this.) Ok, so then S' forms a basis for the vector space of all real-valued functions. \r\n\r\nNow, since V is a subspace of the vector space of all real-valued functions, then S' also forms a basis for V. Now, the dim of V is 2 since the basis S' has 2 vectors. \r\n\r\nObviously dim(R^2) is 2. So, there is a theorem that two vector spaces are isomorphic to each other if and only if they have the same dimension. So, since dim(V)=dim(R^2)=2, then V is isomorphic to R^2.\r\n\r\nIs my logic and thus proof correct? I would appreciate it if someone could verify this. Thanks!", "Solution_1": "I deem it correct yet do not understand what you mean by \"S' forms a basis for the vector space of all real-valued functions. \". You mean by this V?" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ n \\geq 2$ be an integer. Find the largest real number $ \\lambda$ such that the inequality \\[ a^2_n \\geq \\lambda \\sum^{n\\minus{}1}_{i\\equal{}1} a_i \\plus{} 2 \\cdot a_n.\\] holds for any positive integers $ a_1, a_2, \\ldots a_n$ satisfying $ a_1 < a_2 < \\ldots < a_n.$", "Solution_1": "I will show that $MAX(\\lambda)=\\frac{2n-4}{n-1}$\n[hide=\"Proof\"]\nNow $1\\leq a_1a_n-n\\geq 0$ so $(a_n-n+2)(a_n-n)\\geq 0$\nSo $a_n^2-2a_n\\geq (n-2)(2a_n-n)$\nSo $a_n^2\\geq \\frac{2n-4}{n-1}\\sum^{n-1}_{i=1}a_i +2a_n$\n\nSo $MAX(\\lambda)\\geq\\frac{2n-4}{n-1}$\n\nConsider $a_m=m$ where $1\\leq m\\leq n$\nThen $n(n-2)=n^2-2n\\geq \\lambda \\cdot \\frac{n(n-1)}{2}$\nThen $\\lambda \\leq\\frac{2n-4}{n-1}$\n\nSo $MAX(\\lambda)\\leq\\frac{2n-4}{n-1}$\n\nHence $\\frac{2n-4}{n-1}\\leq MAX(\\lambda)\\leq\\frac{2n-4}{n-1}$\nHence $MAX(\\lambda)=\\frac{2n-4}{n-1}$\n[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that for every positive $ n$, the set of integers {$ 1, 2, 3, ... , 2n\\minus{}1, 2n$} can be partitioned into $ n$ pairs:\r\n\r\n{$ a_1, b_1$}, {$ a_2, b_2$}, ... , {$ a_n, b_n$}\r\n\r\nso that $ a_i \\plus{} b_i$ is prime for every $ i$.", "Solution_1": "Please wait for a while :\r\nhttp://mathcentral.uregina.ca/mp/current/\r\n\r\nPierre." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a, b, c \\geq\\ 0$ satisfy $ a + b + c = 5$. Prove that: ${ 10 + ab^2 + bc^2 + ca^2 \\geq\\ \\frac {7}{8}(a^2b + b^2c + c^2a)}$\r\nWhen does equality occurs ? \r\nWe can prove it easily by Am-Gm :)", "Solution_1": "Case 1: $ a \\geq c \\geq b$ then $ LHS\\minus{}RHS\\equal{}80\\plus{}7(a\\minus{}b)(b\\minus{}c)(c\\minus{}a)\\plus{}ab^2\\plus{}bc^2\\plus{}ca^2 \\geq 0$ Obvious\r\nCase 2: $ a \\geq b \\geq c$ then \r\n$ LHS\\minus{}RHS \\geq 80 \\minus{}7ab(a\\minus{}b)\\plus{}ab^2 \\equal{}80\\minus{}\\frac{ 5a.20b(7a\\minus{}8b)}{100} \\geq 80 \\minus{}\\frac{(12a\\plus{}12b)^3}{27.100} \\geq 0$.\r\nEquality occurs if and only if: $ a\\equal{}4,b\\equal{}1,c\\equal{}0$", "Solution_2": "[quote=\"Hong Quy\"]Case 1: $ a \\geq c \\geq b$ then $ LHS \\minus{} RHS \\equal{} 80 \\plus{} 7(a \\minus{} b)(b \\minus{} c)(c \\minus{} a) \\plus{} ab^2 \\plus{} bc^2 \\plus{} ca^2 \\geq 0$ Obvious\nCase 2: $ a \\geq b \\geq c$ then \n$ LHS \\minus{} RHS \\geq 80 \\minus{} 7ab(a \\minus{} b) \\plus{} ab^2 \\equal{} 80 \\minus{} \\frac { 5a.20b(7a \\minus{} 8b)}{100} \\geq 80 \\minus{} \\frac {(12a \\plus{} 12b)^3}{27.100} \\geq 0$.\nEquality occurs if and only if: $ a \\equal{} 4,b \\equal{} 1,c \\equal{} 0$[/quote]\r\nVery nice,Nam.Here is my proof:\r\nI prove that\r\n\\[ f(a,b,c)\\equal{} 80 \\plus{} 7(a \\minus{} b)(b \\minus{} c)(c \\minus{} a) \\plus{} ab^2 \\plus{} bc^2 \\plus{} ca^2 \\geq 0\r\n\\]\r\nNow,Assume $ c\\equal{}min(a,b,c)$;$ t$ is a nonnegative number such that $ t\\le c$\r\n.It is obivious to see \r\n\\[ f(a,b,c)\\ge f(a\\minus{}t;b\\minus{}t;c\\minus{}t)\r\n\\]\r\nHence we only need to prove that the inequality when $ c\\equal{}0$\r\nNow is the time for AMGM inequality,which is very easy as nguoivn said :lol:" } { "Tag": [ "algorithm", "geometry" ], "Problem": "I would like to know the answer to the question below.\r\n\r\nWhat is the best way to help students who post questions here? \r\n\r\nI notice that some of you answer my questions but do not provide an easy explanation or detail for your answers.\r\n\r\nWhy?\r\n\r\nTake this sample:\r\n\r\nx + 4 = 20\r\n\r\nMost of you say: x = 16 and NOTHING MORE. \r\n\r\nVery few of you take time to make sure that students fully grasp what is going on.\r\n\r\nWhat makes a good math tutor or teacher?\r\n\r\nAlso, what about [color=red]HINTS? [/color]\r\n\r\nSome of the hints are worst than the language used in my textbook.\r\n\r\nIsn't the goal to make things clear for people who are obviously lost in mathematics.\r\n\r\nI expect many replies here.\r\n\r\nSome of you complain that students that do not show an effort when posting questions. What most of you math tutors do not seem to understand is the fact that there are questions that are simply too complex to even begin an approach. \r\n\r\nI cannot speak for other students. In MY case, the questions posted here are questions that I personally tried answering for over 30 minutes. There is a small quiz at the end of the study book that I own with 25 questions. \r\n\r\nI simply post questions after getting most of the quiz questions right. The questions here are the ones I considered hard and thus, should be approached by people who are gifted in math or better prepared for upper level high school mathematics.\r\n\r\nWhat is your perspective? \r\n\r\nThanks!\r\ninterval", "Solution_1": "Some people come to this site for problems that they can't solve or for practice for a test or just for a challange! But I think that people should show all their steps. If someone needs help with a problem, they should post some things that they tried. One problem I posted got to $x^{3}-6x-9=0$ and I didn't know how to factor it or solve it and then the next step after that was $x=3$ (the only real number that solves it) and then I had to say, how did you get that? and even then, I didn't know how to solve it...", "Solution_2": "EDIT: what if the problems require multiple steps? or are questions that the posters haven't learned yet?", "Solution_3": "[b]we are not math tutors[/b]...where did you get that from??? we are just kids who are interested in math\r\n\r\ni am against your posting of problems...these types of problems don't belong on the forum because they follow a very simple pattern, there is no thinking involved, there is no unique twist to the problems, if you look at the amc problems, they are not just the run of the mill problems that you are posting, \r\n\r\nif you spend 30 mins on some of these problems, you are not approaching your study correctly, part of mathematics is a way of thinking, you have to learn to find patterns and algorithms to solve the problems, you learn these processes from an example in a book, then you use the practice tests to test if you know the process or not,", "Solution_4": "Yeah-why should we waste time posting trivial steps for trivial problems?", "Solution_5": "I'm a bit skeptical of some of the things you're saying. First, you refer to \"tutors.\" The Art of Problem Solving is not a tutoring site, it is a site for discussion of interesting problems (that have to do with areas of problem solving, not textbook problems). Yes, if you spend a lot of time on a problem and want a hint, posting a problem is the thing to do. But you ask for hints, many people post hints very often. Also, you always complain about problems that do not have a sufficient explanation. However, the solutions written are very well-explained, you don't seem to be trying hard enough to understand them. Yes, if they are using theorems or concepts that you don't know much about, it would be the right thing to do to ask about them. But many of the solutions that are posted to your problems use very elementary methods that are easy to understand. They are explained fine. Yes, it is annoying to just see an answer to a problem you don't know how to approach, but almost always when you request a better explained solution, the solution is self-explanatory. I've been through this experience before, mind you, when I first read AoPS Volume 2. I thought the solutions to their examples were way too vague, and I never picked them up at first. But I realized what I was doing wrong, I was just trying to read. They leave \"in-between,\" more trivial steps out, because the reader should fill them in themselves. This is a good tactic, I believe Zeitz calls it \"active reading.\"\r\n\r\nAs a side note, I have noticed a pattern in your posts. After people post solutions that you like for a problem, you almost always make a post like \"I like the replies,\" \"Thanks for the tips,\" \"Very well explained,\" or \"Nice work.\" As I have notified you before via pm, this is generally spam. If you have a compliment for someone, by all means give it, but not in a thread, but via pm.\r\n\r\n(edit) man, Altheman beat me with what I wanted to say.", "Solution_6": "[quote=\"Altheman\"][b]we are not math tutors[/b]...where did you get that from??? we are just kids who are interested in math\n[/quote]\r\nyes, this isn't for tutoring people-if i wanted to tutor i'd go over to the library and charge people lots of money per hour to be tutored \r\n\r\nthe lowest level here is middle school math and..at least from where i am...middle schoolers would understand a x-4=16 or whatever problem\r\nperhaps not quadratics(though most do!) but we shouldn't have to cover each trivial step if we post a solution(sorry if others feel differently)\r\nread every step and see how it works...research easy things yourself\r\n\r\n\r\npersonally, most teachers(and tutors) can easily, very easily be replaced now\r\nwhat makes a good teacher, tutor, etc. is that they help but don't give you the answer-you have to discover some things for your self", "Solution_7": "I also strongly suggest that you move your posts to the Classroom Math forum, many people are getting annoyed with your posts clogging the forums when they want to solve challenging problems for fun.", "Solution_8": "These questions are ridiculously easy, even for Classroom Math.\r\n\r\nNo offense.\r\n\r\nAnd yeah, if you think this is a tutoring, you are sadly mistaken.\r\n\r\nForums are meant to talk about things, not necessarily tutors.", "Solution_9": "Yes, I think that is something that you need to realize.\r\nWe are NOT OBLIGATED to answer your questions. We come here for fun because we enjoy solving problems that are challenging. I think most of your questions could be solved by consulting an algebra textbook, no offense meant.\r\nMost of us don't show work because most of the problems you post are considered trivial by most of AoPS.", "Solution_10": "Take it easy, guys.\r\n\r\nInterval, if you are looking for a math tutoring site, here are a couple that I found:\r\n\r\nhttp://mathforum.org/dr.math/\r\nhttp://www.fliegler.com/mathman.htm\r\n\r\nI hope I have been of assistance.", "Solution_11": "[quote=\"mathnerd314\"]Take it easy, guys.\n\nInterval, if you are looking for a math tutoring site, here are a couple that I found:\n\nhttp://mathforum.org/dr.math/\nhttp://www.fliegler.com/mathman.htm\n\nI hope I have been of assistance.[/quote]\r\nI think most people are tired of him flooding the forum even though several people have said that he should mostly post in the classroom math forum. \r\n\r\nThis could discourage a new member who is interested in solving challenging and unique contest problems, but then sees these posts and thinks AoPS is a website for school math problems.", "Solution_12": "[quote=\"bpms\"][quote=\"mathnerd314\"]Take it easy, guys.\n\nInterval, if you are looking for a math tutoring site, here are a couple that I found:\n\nhttp://mathforum.org/dr.math/\nhttp://www.fliegler.com/mathman.htm\n\nI hope I have been of assistance.[/quote]\nI think most people are tired of him flooding the forum even though several people have said that he should mostly post in the classroom math forum. \n\nThis could discourage a new member who is interested in solving challenging and unique contest problems, but then sees these posts and thinks AoPS is a website for school math problems.[/quote]\r\n\r\n\r\nhe should either:\r\n\r\n\r\npost in Classroom Math \r\n\r\n\r\nor: don't post in this HSB\r\n\r\n\r\nit's getting spammy sort of full of easy problems" } { "Tag": [ "FTW" ], "Problem": "One-half of the people in a room have two brown eyes, one-fourth\nhave two blue eyes and one-sixth have two green eyes.\nThe other 9 people in the room have two hazel eyes. How\nmany people are in the room?", "Solution_1": "$ \\frac{6x}{12}$ = brown\r\n$ \\frac{3x}{12}$ = blue\r\n$ \\frac{2x}{12}$ = green\r\n\r\ntherefore $ \\frac{x}{12}$ = hazel = 9\r\n\r\nx=9*12=108\r\n\r\nanswer : 108", "Solution_2": "LOL i like how they specify they have $ 2$ hazel eyes. What if someone had one of each???", "Solution_3": "[b]Another way to solve this problem:[/b]\r\n\r\n$ \\frac{1}{2} \\plus{} \\frac{1}{4} \\plus{} \\frac{1}{6} \\equal{} \\frac{11}{12}$\r\n\r\n\r\nThe rest of the people have hazel eyes, and there's $ 9$ of them, so that means that $ 9$ is $ \\frac{1}{12}$ of the total number of people. So we divide:\r\n\r\n$ 9 \\div \\frac{1}{12} \\equal{} 9 \\times 12 \\equal{} 108$\r\n\r\nThere are $ \\boxed {108 \\text{ people}}$", "Solution_4": "A 3rd way of solving the problem. Some people could have more than 2 eyes, and as such, since 9 have 2 hazel eyes and we know about at least half of all people, then $ 2\\times{9} = 18$ for a minimum of 18, therefore:\r\n\r\n$ 18\\LEQ{\\text{number of people in the room}}\\leq{108}$ As it wants a set number of people, we average these two and find the average number of people in the room is $ \\frac {18 + 108}{2} = 63$ Unfortunately, we are overlooking a few problems, such as \"What defines people?\" and \"What constitutes the difference between brown and hazel?\" Also, FTW does not accept 63 as the answer, so I must assume that this problem is glitched with a \"corrcect\" answer of $ 108$", "Solution_5": "[quote=\"mewto55555\"]A 3rd way of solving the problem. Some people could have more than 2 eyes, and as such, since 9 have 2 hazel eyes and we know about at least half of all people, then $ 2\\times{9} \\equal{} 18$ for a minimum of 18, therefore:\n\n$ \\geq{18}\\text{number of people in the room}\\leq{108}$ As it wants a set number of people, we average these two and find the average number of people in the room is $ \\frac {18 \\plus{} 108}{2} \\equal{} 63$ Unfortunately, we are overlooking a few problems, such as \"What defines people?\" and \"What constitutes the difference between brown and hazel?\" Also, FTW does not accept 63 as the answer, so I must assume that this problem is glitched with a \"corrcect\" answer of $ 108$[/quote]\r\n\r\n :rotfl: :rotfl: :rotfl:" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x,y,z \\in \\mathbb{R}$ such that $ x \\plus{} y \\plus{} z \\equal{} 0$. Knowing there exists positive reals $ a,b,c \\in (0;1)$ such that $ a \\plus{} b \\plus{} c \\equal{} 1.5$ and $ ax \\plus{} by \\plus{} cz \\equal{} 1.5$ then prove that:\r\n\\[ p^x \\plus{} p^y \\plus{} p^z \\ge \\frac {3}{2}(p \\plus{} p^{ \\minus{} 1}), \\quad \\forall p \\in \\mathbb{R}_ \\plus{}\r\n\\]", "Solution_1": "If it's too easy try the generalized version:\r\nLet $ x_1,...,x_n \\in \\mathbb{R}$ such that $ x_1 \\plus{} ... \\plus{} x_n \\equal{} 0$. Knowing there exists positive reals $ a_1,a_2,...,a_n \\in (0;1)$ such that $ a_1 \\plus{}... \\plus{} a_n \\equal{} \\frac{n}{2}$ and $ a_1x_1 \\plus{} ... \\plus{} a_nx_n \\equal{} \\frac{n}{2}$ then prove that:\r\n\\[ p^{x_1} \\plus{} ... \\plus{} p^{x_n} \\ge \\frac {n}{2}(p \\plus{} p^{\\minus{}1}), \\quad \\forall p \\in \\mathbb{R}_ \\plus{}\r\n\\]", "Solution_2": "[hide=\"Strong hint\"]Use weighted AM-GM!![/hint][/hide]", "Solution_3": "Since no one could solve it, I'd better provide a solution :)\r\n\r\nFirst by weighted AM-GM:\r\n$ \\frac{a_1p^{x_1} \\plus{} ... \\plus{} a_np^{x_n}}{a_1 \\plus{} ... \\plus{} a_n} \\ge \\sqrt[a_1 \\plus{} ... \\plus{} a_n]{p^{a_1x_1 \\plus{} ... \\plus{} a_nx_n}} \\equal{} p$\r\nBut also:\r\n$ \\frac{(1\\minus{}a_1)p^{x_1} \\plus{} ... \\plus{} (1\\minus{}a_n)p^{x_n}}{n\\minus{}a_1 \\plus{} ... \\plus{} a_n} \\ge \\sqrt[n\\minus{}a_1 \\plus{} ... \\plus{} a_n]{p^{(1\\minus{}a_1)x_1 \\plus{} ... \\plus{} (1\\minus{}a_n)x_n}} \\equal{} p^{\\minus{}1}$\r\n\r\nAddition and multiplying by $ \\frac{n}{2}$ gives the desired. QED" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be nonnegative real numbers such that $ a^2\\plus{}b^2\\plus{}c^2\\plus{}abc\\equal{}4$, then\r\n\\[ a\\plus{}b\\plus{}c \\ge abc\\plus{}2\\]\r\n:)", "Solution_1": "Let me think! :lol: \r\nOk! This is the answer for your problem:\r\nBy squaring both sides, we get\r\n$ \\left(a\\plus{}b\\plus{}c\\right) \\geq \\left(abc\\plus{}2\\right)$\r\n$ \\leftrightarrow a^2\\plus{}b^2\\plus{}c^2 \\plus{} 2(ab\\plus{}bc\\plus{}ca)\\geq (abc)^2\\plus{}4abc\\plus{}4$\r\n$ \\leftrightarrow 2(ab\\plus{}bc\\plus{}ca) \\geq (abc)^2\\plus{}4abc$\r\nUsing Cauchy- Schwartz, we get\r\n$ 2(ab\\plus{}bc\\plus{}ca) \\geq 6\\sqrt[3]{a^2b^2c^2}$\r\nFrom the condition given, we easily deduct the following inequality\r\n$ abc \\leq 1$\r\nThat is what we need!" } { "Tag": [ "geometry", "parallelogram", "angle bisector", "geometry unsolved" ], "Problem": "I am a little stumped on this problem. any help is appreciated.\r\n\r\nTwo circles X and Y meet at point P. \r\nLet tangent line to X at P...meet Y at point B.\r\nLet tangent line to Y at P...meet X at point A.\r\nLet circle Z pass through P,B,A. \r\nLet tangent line to Z at P...meet X at C and Y at D.\r\n\r\nProve that PC=PD.\r\n(hint: draw lines joining p and centers of three circles. look for a parallelogram).", "Solution_1": "It's rather simple. Denote O,O',E be the centers of circles X,Y,Z. \r\n$ OE\\perp PA\\Rightarrow OE\\parallel{}O'P$, $ O'E\\perp PB\\Rightarrow O'E\\parallel{}OP$, thus OPO'E is a parallelogram.\r\nDraw the diameters PM,PN of (O),(O') respectively. Then E is the midpoint of MN.\r\n$ MC\\perp CD, ND\\perp CD, EP\\perp CD, EM\\equal{}EN$, hence PC=PD.\r\nP.s. Your diagram maybe rather bad :| .", "Solution_2": "\u25b3AO1P\u223d\u25b3PO2B => \u25b3PO1O3\u223d\u25b3APB\r\n\u2220O3PO2+\u2220O2PD=90=\u2220PBD+\u2220O2PD => \u2220PBD=\u2220O3PO2=\u2220ABP\r\nAnalogously, \u2220PAC=\u2220O1PO3=\u2220PAB\r\nUsing sin law \r\nPC=sin\u2220PAB*AP/sin\u2220ACP\r\nPD=sin\u2220PBA*BP/sin\u2220BDP\r\n=> PC=PD", "Solution_3": "Using the above notations ( mr.dan's): after having seen that POEO\u2019 was a parallelogram, hence $ \\triangle POE \\sim \\triangle APB$, we shall get $ \\triangle ACP\\sim \\triangle PDB\\sim \\triangle APB$, that is, $ \\angle ACP \\equal{} \\angle BDP$ ( 1 ), $ \\angle CAP \\equal{} \\angle BAP$ ( 2 ) and $ \\angle ABP \\equal{} \\angle DBP$ ( 3 ).\r\nIf {F} = AC\u2229BD, from (1) we get $ \\triangle CDF$ isosceles, while from (2) and (3) P is the F excircle of $ \\triangle ABF$, i.e. PF is the angle bisector of $ \\angle CFD$, hence the result.\r\n\r\nBest regards,\r\nsunken rock" } { "Tag": [ "rotation", "geometry", "geometric transformation", "function", "modular arithmetic", "combinatorics proposed", "combinatorics" ], "Problem": "A circular disk is partitioned into $ 2n$ equal sectors by $ n$ straight lines through its center. Then, these $ 2n$ sectors are colored in such a way that exactly $ n$ of the sectors are colored in blue, and the other $ n$ sectors are colored in red. We number the red sectors with numbers from $ 1$ to $ n$ in counter-clockwise direction (starting at some of these red sectors), and then we number the blue sectors with numbers from $ 1$ to $ n$ in clockwise direction (starting at some of these blue sectors).\r\n\r\nProve that one can find a half-disk which contains sectors numbered with all the numbers from $ 1$ to $ n$ (in some order). (In other words, prove that one can find $ n$ consecutive sectors which are numbered by all numbers $ 1$, $ 2$, ..., $ n$ in some order.)\r\n\r\n[hide=\"Problem 8 from CWMO 2007\"]$ n$ white and $ n$ black balls are placed at random on the circumference of a circle.Starting from a certain white ball,number all white balls in a clockwise direction by $ 1,2,\\dots,n$. Likewise number all black balls by $ 1,2,\\dots,n$ in anti-clockwise direction starting from a certain black ball.Prove that there exists a chain of $ n$ balls whose collection of numbering forms the set $ \\{1,2,3\\dots,n\\}$.[/hide]", "Solution_1": "Here's an idea that might (I'm not sure) work:\r\n\r\nWe regard the sectors as $2n$ colored and numbered points. The problem asks us to show there is a line which contains points numbered $1,2,..,n$ on each of its sides.\r\n\r\nWe draw $n$ segments, each having $2$ points numbered the same as its extremities. It's obvious that no $2$ such segments intersect. I think it's easy to derive from here that there must be a line cutting all $n$ segments, and that solves the problem.\r\n\r\nThink it's Ok?", "Solution_2": "I do not see why two segments won't intersect :( \r\n\r\nPierre.", "Solution_3": "You are right: it is possible for them to intersect :). As I said in that message, it just seemed Ok. Sorry.", "Solution_4": "Now that you broughtit up, does anyone have a really nice and short solution?\r\n\r\nI think I managed to solve it, but I doubt ever having the patience to write it here, since it's a bit hard to explain (even though it's not long). I have to define all sorts of orientations and stuff :(. A diagram would help a lot, but I don't know what to use to draw one which is fairly accurate.", "Solution_5": "I have one i found when the tst was pusblished but i think its the same as you, because it is imposible to write, lot of considerations and things, i will try to do it later", "Solution_6": "True source of this problem:\r\n\r\n20th Tournament of Towns, Spring 1999, Juniors, hard variant, problem 4.\r\n\r\nAuthor of the problem: V. Proizvolov.\r\n\r\nSee here:\r\n\r\nIn Russian: http://www.turgor.ru/20/turnir20.php .\r\n\r\nIn English: http://servus.matematik.su.se/matcir/turgor/tg20vo.htm .", "Solution_7": "Is the official solution not so ugly to be given?\r\n\r\nPierre.", "Solution_8": "Ok Pierre, I will post the official solution soon.", "Solution_9": "I don't know the official solution, but here my solution is.\r\n\r\nLet us consider two auxiliary disks consisting of n sectors, one over another. One for blue sectors (we will call it \"blue disk\") and another for red sectors (we will call it \"red disk\").\r\n\r\nConsider some half-disk on the original disk. There are some blue sectors in it and some red sectors. Consider corresponding sequences of sectors on the blue and red disks. If we rotate the half-disk there are 4 cases of behaviour these sequences:\r\n\r\n1. Red sequence rotates clockwise by one sector.\r\n2. Blue sequence rotates counter-clockwise by one sector.\r\n3. One sector from the end of red sequence moves to the head of blue sequence.\r\n4. One sector from the end of blue sequence moves to the head of red sequence.\r\n\r\nObserve that in all cases the distance between the heads of sequnces always decreases by 1 or allways increases by 1. (Distance is a number of sectors between these heads.) WLOG we may assume that in decreases (otherwise we may consider the distance between the ends of the sequences). Distance is a number from 0 to n-1. Therefore after at most n rotations we will have that such a distance will be 0. It is easy to see that this case corresponds the case that the half-disk contains all numbers 1, ..., n.", "Solution_10": "It looks pretty much like what I did :). I guess it wasn't as nasty as I thought :?.", "Solution_11": "Ah... :? I didn't understand what you meant with the blue/red circles and what is a head and the distance, and how it uses the values... :( \r\nHope Valentin's proposal would be clearer... :blush: \r\n\r\nPierre.", "Solution_12": "Hi!!\r\nMaybe the easiest consideration (also the official one :D) is to consider the 2 numbers that are equal, of different colors, and closest to each other among all other such pairs. Some checking will help you see that somewhere there lies the sequence we are looking for! :)", "Solution_13": "Ok, I think I got it. :)\r\nThanks Dust.\r\n\r\nPierre.", "Solution_14": "Here is my version: \r\nIf all the blue sectors lie in a demi-disk, then all the integers are also in it. \r\nSuppose that there is no demi-disk with only one color, and suppose that all the integers of the blue sectors are placed. The position of the integers on red sectors is only determined by the position of 1. So, there are n ways to place the integers on red sectors in the counter clockwise direction. \r\nNow, I define a bijection between the n diameters, and the n possible positions for the 1 on red sector. \r\nA diameter separates the disk in 2 demi-disks, and there is only one position for the red 1 so that all the integers from 1 to n are on these 2 demi-disks (here we use the hypothesis that there is no demi-disk with only one color). It is quite easy to see that this function is an injection (this is again due to the fact that there is no unicolor demi-disk). As the two sets have the same number of elements, the function is also a surjection, which means that for every position of 1, there is a diameter which splits the disk in two demi-disks which contain all the numbers from 1 to n.", "Solution_15": "[quote] \nThere is a circle, which is divided into 2n sectors. n of them are black and n are white. Now you have to number the white ones consecutively clockwise and the black ones counterclockwise. \n \nYou have to proof that there is always a way to divide the circle into to halves so you have the numbers from 1 to n in each of them, no matter in which way the black and white sectors are arranged and where you start to number them.[/quote]\r\n\r\nDoes this work? \r\n \r\nFirst of all suppose that there is no line that divides the circle in a black and a white half; in that case that line would verify the assertion. Then order the sectors clockwise and call f(i) the number of sector i, where 0<=i<=2n-1 and \"i\" is taken modulo 2n and f(i) modulo n. Define also g(i) belonging to {0,1} as the colour of the corresponding sector, where g(i)=0 means \"white\" and g(i)=1 means \"black\". Define the k-th \"half of the circle\" ( k modulo 2n) as the one made up by the n sectors k+1, k+2,... k+n. Consider the sequence f(k+1), f(k+2).. f(k+n); it can be partitioned in 2 nonempty (by the initial assumption) subsequences, the one made up by the f(i) for which g(i)=0, and the one made up by the f(i) for which g(i)=1. The first subsequence is an increasing sequence of consecutive residues modulo n ( in the sense a(i+1)=a(i)+1 (mod n) ) which has as its initial and bottom element f(j) for the \"leftmost\" (in the sense \"conterclockwise-most\") j in the k-th half of the circle such that g(j)=0. We will call that element W(k). The second subsequence, on the contrary, is a decreasing sequence of consecutive residues modulo n which has as its initial term, which we will call B(k), f(j) for the \"leftmost j\" in the k-th half of the circle such that g(j)=1. \r\nThe fact that the k-th half contains all the n numbers is then equivalent to B(k)=W(k)-1 or D(k)=W(k)-B(k)=1 (mod n).When we pass from k-th half to the k+1-th one. There are 2 possibilities: for the leftmost j in the k-th half g(j)=0 and therefore W(k+1)=W(k)+1 and B(k+1)=B(k), or g(j)=1 and therefore W(k+1)=W(k) and B(k+1)=B(k)-1. In both cases D(k+1)=D(k)+1 (all mod n). Now it becomes obvious thar proceeding clockwise we will reach a k such that D(k)=1, and we are done. \r\n \r\n \r\nSorry for any mistakes.", "Solution_16": "Just an addendum: it's problem 1 of 2. round of Bundeswettbewerb Mathematik 2006, too.", "Solution_17": "I hope that this works.\n\n[b]Solution to CWMO Problem:[/b] Number the balls in clockwise order on the circle as $\\omega_1, \\omega_2, \\dots, \\omega_{2n}$ where $\\omega_1$ is chosen arbitrarily. All indices of balls will be taken in $\\pmod{2n}$. Let $a_{i}$ denote the number on the white ball either equal to $\\omega_i$ or closest to $\\omega_i$ in a clockwise direction around the circle. Let $b_{i}$ denote the number on the black ball either equal to $\\omega_i$ or closest to $\\omega_i$ in a clockwise direction around the circle. Let $c_i = a_i - b_i$. Now, considering the cases corresponding to the colours of consecutive balls, it follows that $c_{i+1} \\equiv c_{i} + 1 \\pmod{n}$. This implies that varying the ball $\\omega_i$ causes $c_i$ to assume all values in $\\pmod{n}$. Hence there exists a ball $\\omega_{j}$ such that $c_{j} \\equiv 1 \\pmod{n}$ which implies that $a_j = b_j +1=m$ for some $m$. Now consider the balls $\\omega_j, \\omega_{j+1}, \\dots, \\omega_{j+n-1}$. If this set of balls contains $k$ white balls, then these have numbers $m, m+1, \\dots, m+k-1 \\pmod{n}$ and the blacks balls have numbers $m-1, m-2, \\dots, m+k-n \\pmod{n}$. Since this forms a complete residue set in $\\pmod{n}$, the numbers on the balls $\\omega_j, \\omega_{j+1}, \\dots, \\omega_{j+n-1}$ are $\\{ 1, 2, \\dots, n \\}$ in some permutation.", "Solution_18": "Label the sectors $A_0, A_2, \\dots A_{2n-1}$ in circular order. Take the integer $i$ such that the two sectors with number $i$ are closest together over all choices of $i.$ WLOG, the two sectors in question are $A_0$ and $A_k$ where $k \\le n.$ Sectors $A_1, A_2, \\dots A_{k-1}$ must be all the same color. Suppose WLOG sectors $A_0$ through $A_{k-1}$ are one color and $A_k$ is another, then choosing $A_1, A_2, \\dots A_{n}$ works." } { "Tag": [], "Problem": "How many integers from 1 to 2003 inclusive have an odd number of distinct positive divisors?", "Solution_1": "Only perfect squares have odd number of divisors.", "Solution_2": "That's a good hint Myth, thanks. I also think that the trick in the question is the word 'distinct'.", "Solution_3": "[quote]Only perfect squares have odd number of divisors.[/quote]\r\n\r\nIs it possible to prove it?\r\n\r\nI just want to know how you know it. :)", "Solution_4": "[quote=\"Silverfalcon\"][quote]Only perfect squares have odd number of divisors.[/quote]\nIs it possible to prove it?[/quote]\nNo, it is impossible. It is an open question, and if you prove it you will win $10^6$ dollars :D\n\n[quote]I just want to know how you know it.[/quote]\r\nI think it is not my business to solve problems from Getting Started Section ;)", "Solution_5": "[quote=\"Myth\"]\nNo, it is impossible. It is an open question, and if you prove it you will win $10^6$ dollars :D\n[/quote]\r\n\r\nMath humor is funny.", "Solution_6": "Just wondering: why is it impossible to prove?\r\nWouldn't it be enough to say that if there are an odd number of divisors, then there exists a factor $x$ such that $x^2$ is equal to the number, making it a perfect square? If there are an even number of divisors, then each factor $a$ corresponds to another factor $b$ such that $a$ and $b$ are distinct and $ab=c$. Or is that not a very rigorous proof? This problem seems quite interesting.", "Solution_7": "Well, strictly speaking, every natural number can be written in the form:\r\n\r\n$\\displaystyle\\prod_{p_i\\in\\mathbb{P}} {p_i}^{r_i}$, where the $p_i$'s are prime.\r\n\r\nNow, the number of divisors of such a number is given by:\r\n\r\n$\\displaystyle\\prod (r_i+1)$\r\n\r\nA number is a perfect square iff all of the $r_i$'s are even. If one of the $r_i$'s is odd, then $\\displaystyle\\prod (r_i+1)$ will contain a factor of two and hence be even. Hence a number contains an odd number of divisors iff it is a perfect square.", "Solution_8": "[quote=\"nolachrymose\"]Just wondering: why is it impossible to prove?[/quote]\r\nIt was a joke!", "Solution_9": "[hide]answer is 44. [/hide]", "Solution_10": "[quote=\"blahblahblah\"]Well, strictly speaking, every natural number can be written in the form:\n\n$\\displaystyle\\prod_{p_i\\in\\mathbb{P}} {p_i}^{r_i}$, where the $p_i$'s are prime.\n\nNow, the number of divisors of such a number is given by:\n\n$\\displaystyle\\prod (r_i+1)$\n\nA number is a perfect square iff all of the $r_i$'s are even. If one of the $r_i$'s is odd, then $\\displaystyle\\prod (r_i+1)$ will contain a factor of two and hence be even. Hence a number contains an odd number of divisors iff it is a perfect square.[/quote]\r\n\r\nWHAZZZAH???? please explain in further detail!! im not THAT smart \"1+1=3 yepthatzme\" get the picture?? lol just kiddin!", "Solution_11": "done how do you knwo the answer is 44 did you use a math program or some sort or equation?", "Solution_12": "Well, we know that only perfect squares have an odd number of divisors. Therefore, all he had to do was find the largest perfect square less than or equal to 2003. Since 45^2=2025, he knows the largest square is 44^2, so the answer is 44.", "Solution_13": "exactly that.", "Solution_14": "i guess i learned something new. Thanks for that idea", "Solution_15": "[quote=\"1+1=3 yepthatzme\"][quote=\"blahblahblah\"]Well, strictly speaking, every natural number can be written in the form:\n\n$\\displaystyle\\prod_{p_i\\in\\mathbb{P}} {p_i}^{r_i}$, where the $p_i$'s are prime.\n\nNow, the number of divisors of such a number is given by:\n\n$\\displaystyle\\prod (r_i+1)$\n\nA number is a perfect square iff all of the $r_i$'s are even. If one of the $r_i$'s is odd, then $\\displaystyle\\prod (r_i+1)$ will contain a factor of two and hence be even. Hence a number contains an odd number of divisors iff it is a perfect square.[/quote]\n\nWHAZZZAH???? please explain in further detail!! im not THAT smart \"1+1=3 yepthatzme\" get the picture?? lol just kiddin![/quote]\r\n\r\nI ll try to explain those in layman's terms.\r\n\r\nFirst $\\prod$ means the product of a sequence. It is used just like $\\sum$ but you multiply everything. For example $\\prod_1^4{x}=1\\cdot 2\\cdot 3\\cdot 4=24.$\r\n\r\n$\\prod p_i^{r_i}$ where the $p_i$ are primes.\r\n\r\nThis means that every number can be prime factorized. It can also be written as $p_1^{r_1}\\cdot p_2^{r_2}\\cdots p_n^{r_n}.$\r\n\r\n\r\nNext the number of divisors of that number is given by the formula:\r\n$\\prod (r_i+1)$.\r\n\r\nSo if we wanted the number of divisors of $150$ we say that $150=2\\cdot3\\cdot5^2.$ So the number of divisors is $(1+1)(1+1)(2+1)=12$.\r\n\r\n(You can use a counting argument to prove this...try it)\r\n\r\n\r\nNext, he basically says that a number is a perfect square if and only if all the exponents when the number is factorized are even.\r\n\r\nHope this helps!", "Solution_16": "[quote=\"Done\"]That's a good hint Myth, thanks. I also think that the trick in the question is the word 'distinct'.[/quote]I am just curious why do you think that the trick in the question is the word 'distinct'?\r\nIt seems to me that the trick was knowing that the only numbers with an odd number of divisors are perfect squares (we are assuming here that we are talking about positive divisors), so the word distinct is superfluos in the statement.", "Solution_17": "Take for example number 4.\r\n\r\n4 = 2 * 2.\r\n\r\nWe say that number 4 has a total of 4 factors: 1, 4, and two prime factors. Or, it has 3 factors: 1, 4, and a single distinct prime factor.\r\n\r\nSo the number of factors differs, depending whether we are talking about total factors or total distinct factors.\r\n\r\nSame applies for, say, 25.\r\n\r\nThis is why the question specifically includes 'distinct'.", "Solution_18": "[quote=\"Done\"]Take for example number 4.\n\n4 = 2 * 2.\n\nWe say that number 4 has a total of 4 factors: 1, 4, and two prime factors. Or, it has 3 factors: 1, 4, and a single distinct prime factor.\n\nSo the number of factors differs, depending whether we are talking about total factors or total distinct factors.\n\nSame applies for, say, 25.\n\nThis is why the question specifically includes 'distinct'.[/quote]You are wrong. First of all if we are talking about decompositions into prime factors the correct form is $ 4 =2^2$, or $36= 2^2\\cdot 3^2$, and so on. There is only one prime factor in the decomposition of 4, namely 2. There are 2 prime factors in the decomposition of 36, namely 2 and 3. And so on. \r\n\r\nFurther more, I hope that you weren't thinking that when we say the sum of divisors of 4 we mean $1+2+2+4 = 9$ (notice that I haven't used the distinct attribute), because it's wrong. In fact saying distinct divisors is superfluous. Just if you are not convinced yet, think about what sense does the \"non-distinct\" divisors expression have? ;)", "Solution_19": "I wonder how we should approach this question then,\r\n\r\nFor any positive integer n, n>1, the \"length\" of n is the number of its positive primes whose product is n. \r\n\r\nWhat is the greatest possible length of a positive integer less than 1000?\r\n\r\na) 10\r\nb) 9\r\nc) 8\r\nd) 7\r\ne) 6\r\n\r\nnote that I haven't used the word 'distinct' here.", "Solution_20": "[quote=\"Done\"]I wonder how we should approach this question then,\n\nFor any positive integer n, n>1, the \"length\" of n is the number of its positive primes whose product is n. \n[/quote]The definition of the \"length\" is ambiguous, due to a gramatical flaw. Please either correct it or rephrase the question :)" } { "Tag": [ "inequalities", "inequalities proposed", "highschoolmath" ], "Problem": "Let $ a, b, c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that: $ \\frac {1 \\plus{} a}{1 \\minus{} a} \\plus{} \\frac {1 \\plus{} b}{1 \\minus{} b} \\plus{} \\frac {1 \\plus{} c}{1 \\minus{} c} \\leq 2(\\frac {b}{a} \\plus{} \\frac {c}{b} \\plus{} \\frac {a}{c})$", "Solution_1": "CBS = Cauchy-Bunyakovski-Schwarz! ^^", "Solution_2": "$ S\\equal{}\\frac {1 \\plus{} a}{1 \\minus{} a} \\plus{} \\frac {1 \\plus{} b}{1 \\minus{} b} \\plus{} \\frac {1 \\plus{} c}{1 \\minus{} c}\\equal{}\\frac {2a\\plus{}b\\plus{}c}{b\\plus{}c} \\plus{} \\frac {2b\\plus{}a\\plus{}c}{a\\plus{}c} \\plus{} \\frac {2c\\plus{}a\\plus{}b}{a\\plus{}b}\\equal{}3\\plus{}2\\sum_{cyc}\\frac{a}{b\\plus{}c}$\r\nwe have $ (a\\plus{}b)^2\\geq4ab$\r\n$ \\leftrightarrow \\frac{2}{a\\plus{}b}\\leq\\frac{1}{2}\\left(\\frac{1}{a}\\plus{}\\frac{1}{b}\\right)$\r\n$ \\leftrightarrow \\frac{2c}{a\\plus{}b}\\leq\\frac{1}{2}\\left(\\frac{c}{a}\\plus{}\\frac{c}{b}\\right)$\r\ntherefore:\r\n$ 2\\sum_{cyc}\\frac{a}{b\\plus{}c}\\leq\\left(\\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{a} \\right)$\r\nthen: $ S\\leq 2\\left(\\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{a} \\right)$", "Solution_3": "Thank you but i don't understand your solution\r\n[quote=\"peine\"]$ S \\equal{} \\frac {1 \\plus{} a}{1 \\minus{} a} \\plus{} \\frac {1 \\plus{} b}{1 \\minus{} b} \\plus{} \\frac {1 \\plus{} c}{1 \\minus{} c} \\equal{} \\frac {2a \\plus{} b \\plus{} c}{b \\plus{} c} \\plus{} \\frac {2b \\plus{} a \\plus{} c}{a \\plus{} c} \\plus{} \\frac {2c \\plus{} a \\plus{} b}{a \\plus{} b} \\equal{} 3 \\plus{} 2\\sum_{cyc}\\frac {a}{b \\plus{} c}$\nwe have $ (a \\plus{} b)^2\\geq4ab$\n$ \\leftrightarrow \\frac {2}{a \\plus{} b}\\leq\\frac {1}{2}\\left(\\frac {1}{a} \\plus{} \\frac {1}{b}\\right)$\n$ \\leftrightarrow \\frac {2c}{a \\plus{} b}\\leq\\frac {1}{2}\\left(\\frac {c}{a} \\plus{} \\frac {c}{b}\\right)$\ntherefore:\n[color=red](Here!!!!!!!!!)[/color] [color=red]$ 2\\sum_{cyc}\\frac {a}{b \\plus{} c}\\leq\\left(\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\right)$\nthen: $ S\\leq 2\\left(\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\right)$[/color][/quote]\r\nCan you show me how you do it?\r\n\r\nSorry for my bad English!! ^^\r\nBest Regarts!", "Solution_4": "Let $ a, b, c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that: $ \\frac {1 \\plus{} a}{1 \\minus{} a} \\plus{} \\frac {1 \\plus{} b}{1 \\minus{} b} \\plus{} \\frac {1 \\plus{} c}{1 \\minus{} c} \\leq 2(\\frac {b}{a} \\plus{} \\frac {c}{b} \\plus{} \\frac {a}{c})$\r\n\r\nI think apply Bunyacovski inequality is the best way to solute this inequality!\r\n\r\n$ RHS \\equal{} 3\\plus{}2\\sum_{a,b,c}\\frac{a}{1\\minus{}a} \\leq 2(\\frac{b}{a}\\plus{}\\frac{c}{b}\\plus{}\\frac{a}{c})\r\n\\Leftrightarrow 2\\sum_{a,b,c}a(\\frac{1}{c}\\minus{}\\frac{1}{1\\minus{}a}) \\geq 3\r\n\\Leftrightarrow \\sum_{a,b,c}a\\frac{1\\minus{}a\\minus{}c}{c(1\\minus{}a)} \\geq \\frac{3}{2}\r\n\\Leftrightarrow \\sum_{a,b,c}\\frac{ab}{c(b\\plus{}c)} \\geq \\frac{3}{2}$\r\nApply Bunyacovski inequality we have: \r\n$ (\\sqrt{\\frac{ab}{c}}\\plus{}\\sqrt{\\frac{bc}{a}}\\plus{}\\sqrt{\\frac{ca}{b}})^2 \\equal{}\r\n(\\sqrt{\\frac{ab}{c(b\\plus{}c)}(b\\plus{}c)}\\plus{}\\sqrt{\\frac{bc}{a(c\\plus{}a)}(c\\plus{}a)}\\plus{}\\sqrt{\\frac{ca}{b(a\\plus{}b)}(a\\plus{}b)})^2 \\leq\r\n2.\\sum_{a,b,c}\\frac{ab}{c(b\\plus{}c)}$\r\nBecause $ (a\\plus{}b\\plus{}c)^2 \\geq 3(ab\\plus{}bc\\plus{}ca)$\r\n$ \\Rightarrow (\\sqrt{\\frac{ab}{c}}\\plus{}\\sqrt{\\frac{bc}{a}}\\plus{}\\sqrt{\\frac{ca}{b}})^2 \\geq 3(a\\plus{}b\\plus{}c) \\equal{} 3$\r\nCompleted!\r\n^^", "Solution_5": "I have two solution apply CBS for this unit but i don't know a different solution! Can you show me! Thank you! \r\nLet $ a, b, c>0$. Prove that: $ \\frac{a}{bc(c\\plus{}a)}\\plus{}\\frac{b}{ca(a\\plus{}b)}\\plus{}\\frac{c}{ab(b\\plus{}c)} \\geq \\frac{27}{2(a\\plus{}b\\plus{}c)^2}$", "Solution_6": "I think this is an JBMO Problem, isn't it ?\r\nMy solution just use AM-GM", "Solution_7": "[quote=\"Ronald Widjojo\"]I think this is an JBMO Problem, isn't it ?\nMy solution just use AM-GM[/quote]\r\n^^ No! It's from MO Romanian 2004. Can you show your solution?", "Solution_8": "my solution\r\nlet $ abc=1$\r\nwe must prove $ \\frac{a}{bc(c+a)}+\\frac{b}{ca(a+b)}+\\frac{c}{ab(b+c)}>=\\frac{27}{2(a+b+c)^2<->\\frac{a^2}{(c+a)}+\\frac{b^2}{(a+b)}+\\frac{c^2}{(b+c)}>=\\frac{27}{2(a+b+c)^2$\r\nCBS $ \\frac{a^2}{(c+a)}+\\frac{b^2}{(a+b)}+\\frac{c^2}{(b+c)}>=\\frac{a+b+c}{2}>=\\frac{27}{(a+b+c)^2}$", "Solution_9": "[quote=\"D.I.Culianop\"]I have two solution apply CBS for this unit but i don't know a different solution! Can you show me! Thank you! \nLet $ a, b, c > 0$. Prove that: $ \\frac {a}{bc(c \\plus{} a)} \\plus{} \\frac {b}{ca(a \\plus{} b)} \\plus{} \\frac {c}{ab(b \\plus{} c)} \\geq \\frac {27}{2(a \\plus{} b \\plus{} c)^2}$[/quote]\r\nthis is my solution \r\nwe have with AM-GM:\r\n$ S\\equal{}\\frac {a}{bc(c \\plus{} a)} \\plus{} \\frac {b}{ca(a \\plus{} b)} \\plus{} \\frac {c}{ab(b \\plus{} c)} \\geq 3\\sqrt[3]{\\frac{1}{abc(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}}$\r\n$ \\leftrightarrow S\\geq3\\sqrt[3]{\\frac{1}{\\left(\\frac{a\\plus{}b\\plus{}c}{3}\\right)^3.\\left(\\frac{2a\\plus{}2b\\plus{}2c}{3}\\right)^3)}}$\r\nthen: $ S\\geq\\frac{27}{2(a\\plus{}b\\plus{}c)^2}$", "Solution_10": "[quote=\"gagaga\"]my solution\nlet $ abc \\equal{} 1$\n\nwe must prove \n\n$ \\frac{a}{bc(c \\plus{} a)} \\plus{} \\frac{b}{ca(a \\plus{} b)} \\plus{} \\frac{c}{ab(b \\plus{} c)} \\geq \\frac{27}{2(a \\plus{} b \\plus{} c)^2}$\n\n\n$ \\Leftrightarrow$ $ \\frac{a^2}{c \\plus{} a} \\plus{}$ $ \\frac{b^2}{a \\plus{} b} \\plus{}$ $ \\frac{c^2}{b \\plus{} c}$ $ \\geq$ $ \\frac{27}{2(a \\plus{} b \\plus{} c)^2}$\n\n\n CBS $ \\frac{a^2}{(c \\plus{} a)} \\plus{} \\frac{b^2}{(a \\plus{} b)} \\plus{} \\frac{c^2}{(b \\plus{} c)} \\geq \\frac{a \\plus{} b \\plus{} c}{2} \\geq \\frac{27}{(a \\plus{} b \\plus{} c)^2}$[/quote]\r\n^^ Thank you gagaga. I think we don't need let $ abc \\equal{} 1$. You use Schwarz first and than we can use Cauchy to sovle. It's a beautiful solution!", "Solution_11": "[quote=\"peine\"][quote=\"D.I.Culianop\"]I have two solution apply CBS for this unit but i don't know a different solution! Can you show me! Thank you! \nLet $ a, b, c > 0$. Prove that: $ \\frac {a}{bc(c \\plus{} a)} \\plus{} \\frac {b}{ca(a \\plus{} b)} \\plus{} \\frac {c}{ab(b \\plus{} c)} \\geq \\frac {27}{2(a \\plus{} b \\plus{} c)^2}$[/quote]\nthis is my solution \nwe have with AM-GM:\n$ S \\equal{} \\frac {a}{bc(c \\plus{} a)} \\plus{} \\frac {b}{ca(a \\plus{} b)} \\plus{} \\frac {c}{ab(b \\plus{} c)} \\geq 3\\sqrt [3]{\\frac {1}{abc(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}}$\n$ \\leftrightarrow S\\geq3\\sqrt [3]{\\frac {1}{\\left(\\frac {a \\plus{} b \\plus{} c}{3}\\right)^3.\\left(\\frac {2a \\plus{} 2b \\plus{} 2c}{3}\\right)^3)}}$\nthen: $ S\\geq\\frac {27}{2(a \\plus{} b \\plus{} c)^2}$[/quote]\r\n\r\nThank you very much, peine! ^^ :oops: You're intelitence! I think your solution with AG is very nice. This is my solution:\r\n\r\n$ (\\sqrt {\\frac {a}{bc}} \\plus{} \\sqrt {\\frac {b}{ca}} \\plus{} \\sqrt {\\frac {c}{ab}})^2 \\equal{} (\\sqrt {\\frac {a}{bc(c \\plus{} a)}(c \\plus{} a)} \\plus{} \\sqrt {\\frac {b}{ca(a \\plus{} b)}(a \\plus{} b)} \\plus{} \\sqrt {\\frac {c}{ab(b \\plus{} c)}(b \\plus{} c)})^2 \\leq 2(a \\plus{} b \\plus{} c)(\\sum_{a,b,c}\\frac {a}{bc(c \\plus{} a)}) \\equal{} 2(a \\plus{} b \\plus{} c)M$\r\nBecause $ (a \\plus{} b \\plus{} c)^2 \\geq 3(ab \\plus{} bc \\plus{} ca)$\r\nWe have: $ (\\sqrt {\\frac {a}{bc}} \\plus{} \\sqrt {\\frac {b}{ca}} \\plus{} \\sqrt {\\frac {c}{ab}})^2 \\geq 3(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})$\r\n$ \\rightarrow M \\geq \\frac {3(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})}{2(a \\plus{} b \\plus{} c)} \\geq \\frac {27}{2(a \\plus{} b \\plus{} c)^2}$\r\n$ \\leftrightarrow \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\geq \\frac {9}{a \\plus{} b \\plus{} c}$ (true by Schwarz) :D", "Solution_12": "^^ Let $ x, y, z > 0$ and $ \\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=4$. Prove that: \r\n$ \\frac{1}{2x+y+z} + \\frac{1}{x+2y+z} + \\frac{1}{x+y+2z} \\leq 1$", "Solution_13": "[quote=\"D.I.Culianop\"]^^ Let $ x, y, z > 0$ and $ \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\equal{} 4$. Prove that: \n$ \\frac {1}{2x \\plus{} y \\plus{} z} \\plus{} \\frac {1}{x \\plus{} 2y \\plus{} z} \\plus{} \\frac {1}{x \\plus{} y \\plus{} 2z} \\leq 1$[/quote]\r\nFrom CS; \\[ LHS \\le \\sum{\\frac{1}{4}(\\frac{1}{x\\plus{}y}\\plus{}\\frac{1}{y\\plus{}z})} \\equal{} \\frac{1}{2}(\\sum{\\frac{1}{x\\plus{}y}}) \\leq \\frac{1}{2}\\sum\\frac{1}{4}(\\frac{1}{x}\\plus{}\\frac{1}{y}) \\equal{} 1\\]", "Solution_14": "[quote=\"apollo\"][quote=\"D.I.Culianop\"]^^ Let $ x, y, z > 0$ and $ \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\equal{} 4$. Prove that: \n$ \\frac {1}{2x \\plus{} y \\plus{} z} \\plus{} \\frac {1}{x \\plus{} 2y \\plus{} z} \\plus{} \\frac {1}{x \\plus{} y \\plus{} 2z} \\leq 1$[/quote]\nFrom CS;\n\\[ LHS \\le \\sum{\\frac {1}{4}(\\frac {1}{x \\plus{} y} \\plus{} \\frac {1}{y \\plus{} z})} \\equal{} \\frac {1}{2}(\\sum{\\frac {1}{x \\plus{} y}}) \\leq \\frac {1}{2}\\sum\\frac {1}{4}(\\frac {1}{x} \\plus{} \\frac {1}{y}) \\equal{} 1\n\\]\n[/quote]\r\nYour solution apply Schwarz inequality ! ^^ Perfect! Thank you!", "Solution_15": "Back to the original problem, the following inequalities hold for postive reals $ a,\\ b,\\ c$.\r\n\\[ \\frac {b \\plus{} c}{a \\plus{} b} \\plus{} \\frac {c \\plus{} a}{b \\plus{} c} \\plus{} \\frac {a \\plus{} b}{c \\plus{} a}\\leq \\frac {b}{a} \\plus{} \\frac {c}{b} \\plus{} \\frac {a}{c}\r\n\\]\r\n\r\n\\[ \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b}\\leq \\frac {b}{a} \\plus{} \\frac {c}{b} \\plus{} \\frac {a}{c}\r\n\\]" } { "Tag": [ "calculus", "integration", "topology", "invariant", "analytic geometry" ], "Problem": "problem in Algebraic Topology\r\nHey guys, i am studying cohomology by hatcher's. Could anyone provide me some ideas on these problems? Thank you all!\r\n\r\nLet f : S2n-1 -> Sn denote a continuous map. Let Xf = D2n union f Sn be the space obtained by attaching a 2n-dim cell to Sn using the map f.\r\ni). Calculate the integral homology of Xf .\r\nii). Find an n and an f such that H*(Xf ;Z=2) has a non-trivial product. Justify it.", "Solution_1": "I'm going to assume for now that $ n > 1$.\r\n\r\ni) You have a single 0-cell, a single n-cell and a single (2n)-cell. As $ n > 1$ you can't have any boundary homomorphisms, and so you get homology $ \\mathbb{Z}$ in dimensions 0,n,2n; no other homology exists.\r\nii) Since the homology is torsion-free you get the same groups appearing as cohomology; $ \\mathbb{Z}$ in 0,n,2n; nothing else. If $ x \\in H^n(Xf)$ is the nontrivial generator, you may get a cup product $ x \\smile x \\in H^{2n}(Xf);$ it will be a multiple $ h$ of the generator in that dimension - $ h$ is called the Hopf invariant of $ f$. \r\n\r\nA simple example: let $ Xf \\equal{} \\mathbb{C}P^2$, which is formed with attaching map $ f \\equal{} \\eta$ (the Hopf map -- same Hopf, no surprise.) Then since $ H^*(\\mathbb{C}P^2) \\equal{} \\mathbb{Z}[x]/(x^3)$ we get that nontrivial product $ x^2$ in dimension four, and the map has hopf invariant one. The multiplication will be nontrivial no matter what coefficient group you are using, as squaring in this cases gives an isomorphism $ H^2(\\mathbb{C}P^2) \\to H^4(\\mathbb{C}P^2)$.\r\n\r\nTaking care of the case $ n \\equal{} 1$ shouldn't be too hard, since then the map is $ f : S^1 \\to S^1$, which is entirely classified up to homotopy by degree, and the homotopy type of $ Xf$ only on the homotopy class of $ f$.", "Solution_2": "Thank you very much! hopf map is so hard. Do you mean that CP2=D6 Uf S3 with hopf f? Do you have an easy example? Thanks anyway.", "Solution_3": "No, I mean $ \\mathbb{C}P^2 \\equal{} S^2 \\cup_\\eta D^4;$ the space has cells in even dimensions, and the hopf map is $ \\eta : S^3 \\to S^2$. It's very easy to write down, since $ S^2 \\equal{} \\mathbb{C}P^1$ canonically, and $ S^3 \\subset \\mathbb{R}^4 \\equal{} \\mathbb{C}^2$. The hopf map is given by $ \\eta(z,w) \\equal{} [z,w] \\in \\mathbb{C}P^1$.", "Solution_4": "sorry, i have no idea about this operation", "Solution_5": "S^3={(X_1,X_2,X_3,X_4):X_1^2+X_2^2+X_3^2+X_4^2=1}, \t\r\n\r\nS^2={(x_1,x_2,x_3):x_1^2+x_2^2+x_3^2=1}. \t\r\n\r\nx_1\t=\t2(X_1X_2+X_3X_4)\t\r\n(3)\r\nx_2\t=\t2(X_1X_4-X_2X_3)\t\r\n(4)\r\nx_3\t=\t(X_1^2+X_3^2)-(X_2^2+X_4^2).\r\n\r\nyes?", "Solution_6": "Oh sorry $ [x,y]$ is just the homogeneous coordinate in $ \\mathbb{C}P^1$. The space $ \\mathbb{C}P^1$ consists of all pairs $ [a,b]$ where $ a,b \\ne 0$, under the equivalence relation $ [a,b] \\sim [c,d]$ whenever $ ad \\minus{} bc \\equal{} 0$. You can identify such a point $ [a,b]$ with $ b/a \\in \\mathbb{C} \\cup \\{\\infty\\}$ (in fact $ \\mathbb{C}P^1 \\equal{} S^2$ is the Riemann sphere.)" } { "Tag": [], "Problem": "", "Solution_1": "" } { "Tag": [], "Problem": "which expression is the same as (6+4) X 8?\r\na) 8 X 6 + 8 X 4 \r\nb) 8 X 6 + 4 \r\nc) 8 X 4 + 6 \r\nd) 8 X 4 X 8 X 6", "Solution_1": "This is known as the distributive property which says:\r\n\r\n$ a(b \\plus{} c) \\equal{} ab \\plus{} ac$\r\n\r\nI think you can figure out the rest, and once again wrong forum.", "Solution_2": "I really never went to school and now I'm trying hard to help my son \r\nwe moved recently to the US \r\n\r\ncould you also tell me the right forum I was supposed to post to \r\n\r\nhave a great evening \r\nEmanuele", "Solution_3": "Classroom Math, it is the first forum on the forums page, its in the middle school section." } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Find all the ploynominals $p(x)$ satisfying :\r\n$(\\forall x \\in \\mathbb R) : (x-8)p(2x)=8(x-1)p(x)$", "Solution_1": "Well, puting $x=1$ we obtain $p(2)=0$, now put $x=2$, and we have $p(4)=0$. It is obvious that $p(8)=0$. Hence we can write $p$ as $p(x)=(x-2)(x-4)(x-8)q(x)$ for some polinomial $q$. Replacing with this in the given identity yields $8(x-1)(x-2)(x-4)(x-8)q(2x) = 8(x-1)(x-2)(x-4)(x-8)q(x)$ therefor for all $x \\notin {1,2,4,8}$ we have that $q(2x)=q(x)$, and if $\\deg{ q} >0$ then for some $k$ we will have $|q(2x)|>|q(x)|$ $ \\forall x>k$. So $q$ is constant, and it is easy to see that the polinomial $p(x)=(x-2)(x-4)(x-8)k$ works for any $k \\in \\mathbb R$" } { "Tag": [ "geometry", "factorial", "number theory", "prime factorization" ], "Problem": "Do you know any useful formulas for MOEMS that you want to share?\r\n\r\nHere's one: To find the number of positive factors of a number, take the prime factorization of a number, add 1 to each of the exponents of each prime, and multiply these new exponents together. For example, take 150. The prime factorization is $2^13^15^2$. Add 1 to each exponent to get $2^23^25^3$. Multiply all the exponents together and get 2*2*3=12 factors of 150.", "Solution_1": "Also, make sure you know how to manipulate and work with area and perimeter.", "Solution_2": "Also, to determine the number of zeroes at the end of the factorial of a number (let's say N) take N and divide it by powers of 5 starting from 1 and take the integer part and finally summing it all up.", "Solution_3": ":huh: looks complicated. I'd better give an example. let's say 200!\r\n\r\n200/5=40, 200/25=8, 200/125=1.6\r\n\r\nSumming up all the integer parts: 40+80+1=49 zeroes at the end.", "Solution_4": "$1+2+3+4+5+.... +n=\\frac{n(n+1)}{2}$. \r\n\r\n$2+4+6+8+... +2n=n(n+1)$\r\n\r\n$1+3+5+7+...+(2n-1)=n^2$. \r\n\r\nI hope I didn't screw up....", "Solution_5": "236factorial's reminded me of:\r\n\r\n$1^2+2^2+3^2+...+n^2=\\frac{n(n+1)(2n+1)}{6}$ \r\n\r\n$1^3+2^3+3^3+...+n^3=(\\frac{n(n+1)}{2})^2$", "Solution_6": "$1^3+2^3+...+n^3=(1+2+...+n)^2$\r\n$1+2^1+...+2^{m-1}=2^m-1$", "Solution_7": "just to let you guys know, this topic is really helping me so keep posting formulas!!!!! :lol: :lol:", "Solution_8": "[quote=\"math92\"]just to let you guys know, this topic is really helping me so keep posting formulas!!!!! :lol: :lol:[/quote]\r\nLearning the technique/method is much better than memorizing formulas. Maybe some come in handy for timed competitions but you should spend time learning the technique rather than using the time to learn the formulas. Sorry to break the fun out :( Maybe this didn't help, I am not familiar with MOEMS...\r\n[Edit]: Ack, I posted in Middle School...", "Solution_9": "These formulas are given to you with a warning from me. Be careful on how you use them, as it is the application of these formulas which is essential. Often times, you will have to manipulate one of these formulas to solve a problem, and without understanding where a formula is from, it is very difficult to think in that sense. So, take heed, and happy problem solving." } { "Tag": [ "geometry", "rectangle" ], "Problem": "A semicircle with center O has a radius of 9cm. What is the number of centimeters in the length of $ \\overline{RQ}$, a diagonal of the rectangle shown?\n[asy]pair O,R,Q;\nO = (0,0); R = (-3,0); Q = (0,4);\ndraw((0,5)--Q--R--Q+R--Q--O); draw((-5,0)--(5,0));\ndraw(arc(O,5,0,180));\nlabel(\"O\",O,S); label(\"R\",R,S); label(\"Q\",Q,E);[/asy]", "Solution_1": "Well, since the figure is a rectangle, the diagonals are equal. One can notice that if the other diagonal is drawn, it's a radius. So the answer is $ \\boxed{9}$." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Let be given seven circles: $(O), (O_{1}),(O_{2}),(O_{3}),(O_{4}),(O_{5}),(O_{6})$ \r\nConsider that: \r\n*$(O_{1}),(O_{2}),(O_{3}),(O_{4}),(O_{5}),(O_{6})$ are inside $(O)$\r\n*$(O_{1}),(O_{2}),(O_{3}),(O_{4}),(O_{5}),(O_{6})$ tangent internally $(O)$ at $A_{1},A_{2},A_{3},A_{4},A_{5},A_{6}$ respectively\r\n*$(O_{1})$ tangents externally $(O_{2})$,$(O_{2})$ tangents externally $(O_{3})$,,$(O_{3})$ tangents externally $(O_{4})$, $(O_{4})$ tangents externally $(O_{5})$,$(O_{5})$ tangents externally $(O_{6})$,$(O_{6})$ tangents externally $(O_{1})$ \r\nProve that :$A_{1}A_{4},A_{2}A_{5},A_{3}A_{6}$ are concurrent", "Solution_1": "Discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=5190 , also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=60519 .\r\n\r\n darij", "Solution_2": "ha ha ha ha...\r\n\r\nagain inversion shows its power...\r\n\r\nthe problem will be solved very nice by inversion... ;)" } { "Tag": [ "rotation", "ratio", "calculus", "integration" ], "Problem": "Hi, have a fixed pulley and a string in tension placed over it in contact with the pulley over an angle A. Coefficient of friction between the pulley and string is u. The pulley is then rotated and slips with respect to the string, show that the ration of the tensions in the two parts of the string is e^(uA). Assuming theres an integral to find but where do you get it from?", "Solution_1": "I don't think this is the correct forum for that question.", "Solution_2": "What forum should I move this to? Is it physics maybe?", "Solution_3": "Yea physics I think.", "Solution_4": "perhaps you should try \r\nhttp://www.physicsforums.com. I am quite hopeful that it would be a helpful place, I know that personally. :)", "Solution_5": "Thanks, sorry wasn't sure. I suppose was off a physics training sheet.", "Solution_6": "Now that this is in the correct part of the forums can anyone be of assistance?", "Solution_7": "I know this sounds dumb but what does uA stand for :?:", "Solution_8": "[quote=\"Inspired By Nature\"]I know this sounds dumb but what does uA stand for :?:[/quote]\r\n\r\n$u$ is coefficient of friction and $A$ is the angle" } { "Tag": [ "videos", "geometry", "rotation", "geometric transformation", "induction", "puzzles" ], "Problem": "Some medic I use damned God as sad Ogden made suicide memos.\r\n\r\n This sentence highlights a mthematical principle. What is it?", "Solution_1": "palindrome! That's a good one... never seen it.", "Solution_2": "There are only so many mathematical principles that can be represented by a bunch of strange words... ;) Nice one, anyway.", "Solution_3": "its really good ,thanks for it,I am going to give it right away for the bulletein board \r\nOfcource I could not crack it", "Solution_4": "no sir, away a papaya war is on.\r\nlol, that's longest palidrome i know.", "Solution_5": "A man a plan a canal panama", "Solution_6": "Look at this palindrome. It's so long! :o \r\n\r\n[b]Star? Not I! Movie \u2013 it too has a star in or a cameo who wore mask \u2013 cast are livewires. \n\nSoda-pop straws are sold, as part-encased a hot tin, I saw it in mad dog I met. Is dog rosy? Tie-dye booths in rocks. \n\nAll ewes lessen ill. I see sheep in Syria? He, not I, deep in Syria, has done. No one radio drew old one. \n\nMany moths \u2013 I fondle his; no lemons are sold. Loot delis, yob, moths in a deli bundle his tin. Pins to net a ball I won \u2013 pins burst input. I loot to get a looter a spot paler. Arm a damsel \u2013 doom a dam. Not a base camera was in a frost, first on knees on top spot. Now a camera was a widened dam. \n\nAsk: Cold, do we dye? No, hot \u2013 push tap, set on to hosepipe. Nuts in a pod liven. \n\nA chasm regrets a motto of a fine veto of wars. Too bad \u2013 I all won. A sadist sent cadets \u2013 a war reign a hero derides. A bad loser, a seer, tossed a cradle \u2013 he begat to cosset \u2013 a minaret for Carole, Beryl, Nora. We\u2019re not as poor to self. \n\nI risk cold as main is tidal. As not one to delay burden, I don\u2019t set it on \u201chot\u201d. A foot made free pie race losses runnier. As draw won pull, eye won nose. Vile hero saw order it was in \u2013 even a moron saw it \u2013 no, witnessed it: Llama drops \u2013 ark riots. Evil P.M. in a sorer opus enacts all laws but worst arose. Grab a nosey llama \u2013 nil lesser good, same nicer omen. \n\nIn pins? No, it is open. If a top spins, dip in soot. \n\nMadam, as I desire, dictates: Pull aside, damsels, I set a rag not for a state bastion. A test I won e.g. a contest I won. \n\nKidnap, in part, an idle hero. Megastars, red, rosy, tied no tie. Blast! A hero! We do risk a yeti\u2019s opposition! \n\n \n\n\nHe too has a wee bagel still up to here held. \n\nDemigods pack no mask, cap nor a bonnet, for at last a case is open \u2013 I left a tip \u2013 it wets. A dog wets too. Radios to help pay my tip, pull a tip. \n\nAle, zoo beer, frets yon animal. Can it? New sex arose but, we sots, not to panic \u2013 it\u2019s ale \u2013 did I barrel? Did I lose diadem, rare carrot in a jar of mine? Droop as tops sag \u2013 unseen knots. \n\nA cat ate straw as buck risk cud; evil foe, nil a red nag ate? Bah! Plan it \u2013 silage. Model foot in arboreta. \n\nI, dark Satanist, set fire \u2013 voodoo \u2013 to slat. I design a metal as parrot, I deem it now. One vast sum is no ten in set \u2013 amen! Indeed, nine drag a yam, nine drag a tie. Dame nabs flower; can we help man? Woman is worse nob. \n\nMud level rose, so refill a rut. A nag of iron I made to trot I defied \u2013 I risk leg and its ulnae. Can a pen I felt to bid dollar or recite open a crate, open a cradle, his garret? \n\nSample hot Edam in a pan. I\u2019m a rotten digger \u2013 often garden I plan, I agreed; All agreed? Aye, bore ensign; I\u2019d a veto \u2013 I did lose us site. Wool to hem us? No, cotton. Site pen in acacias or petals a last angel bee frets in. \n\nI met a gorilla (simian); a mate got top snug Noel fire-lit role. Manet, Pagnol, both girdle his reed bogs. \n\nFlan I reviled, a vet nods to order it, Bob, and assign it. Totem users go help mates pull as eye meets eye. Son \u2013 mine \u2013 pots a free pie, yes? No. Left a tip? Order a dish to get. A ring is worn \u2013 it is gold. Log no Latin in a monsignor, wet or wise. Many a menu to note carrot. \n\nCat in a boot loots; As I live, do not tell! A bare pussy, as flat on fire, I know loots guns, fires a baton, nets a hero my ale drop made too lax. \n\nIf it is \n\n \n\nto rain, a man is a sign; I wore macs, no melons rot. I use moths if rats relive, sir, or retire. \n\nVendor pays: I admire vendee, his pots net roe. Nine dames order an opal fan; I\u2019ll ask cold log fire vendor to log igloo frost. Under Flat Six exist no devils. \n\nMarxist nods to Lenin. To Lenin I say: \u201cMama is a deb, besides a bad dosser.\u201d \n\nGen it up to get \u201cova\u201d for \u201cegg\u201d. I recall a tarot code: yell at a dessert side-dish sale. Yes/nos a task cartel put correlate: E.S.P. rocks a man. I am a man, am no cad, I\u2019m aware where it\u2019s at! \n\nFire! Its an ogre-god to help, man, as I go. Do not swap; draw, pull a troll! \n\nIt\u2019s not a cat I milk \u2013 calf, for a fee, sews a button - knit or tie damsel over us. Mined gold lode I fill until red nudes I met in a moor-top bar can. I sit, I fill a diary \u2013 trap nine men in ten-part net \u2013 oh, sir, I ask, cod nose? No, damp eel. \n\nSo, to get a name! I say, Al! I am Al! Last, I felt, to breed, deer begat. \n\nTo can I tie tissue \u2013 damp \u2013 or deliver Omani artist \u2013 a man of Islam. \n\nIn a den mad dogs lived on minis a signor who lived afore targets in at. As eremites pull, I, we, surf, fantasise, mend a bad eye. No hero met satyr; Tony, as I stressed, won\u2019t, so cosset satyr. \n\nA vet on isles made us sign it, a name. Foe man one sub. \n\nAside no dell I fret a wallaby; metal ferrets yodel, like so. On a wall I ate rye. Bored? No, was I rapt! One more calf? O.K., calf, one more, bossy! No! Lock cabin, rob yam, sip martini. Megastar was in a risk. \n\nCat? No, I\u2019m a dog; I\u2019m a sad loyal pet. A design I wore \u2013 kilts (a clan); if net drawn, I put it up. Royal spots snag \u2013 royal prevents rift. \n\nComposer, good diet, are both super, God \u2013 label it a love of art, lustre. Video bored, no wise tale e.g. a mini tale \u2013 no sagas seen. Knack: cede no foes a canal. \n\nPay \u2013 as I sign I lie; clear sin it is; e.g. \u201cAmadeus\u201d sign I \u2013 lira for ecu, decimal \u2013 sin as liar. \n\nTrad artistes pull a \n\n \n\ndoom, a drawer won\u2019t. \n\nIs it sold loot? No, I suffered loss. A man is god; Amen! I came nice Tahiti (sic). \n\nIt\u2019s ale for a ban if for a fast \u2013 is role to help mash turnip? Use zoo? No - grasp order \u2013 use no zoos. Warts on time did sag. \n\nNo grade \u201cX\u201d \u201cA\u201d Level? Oh, \u201cA\u201d! I\u2019d a \u201cB\u201d or a \u201cC\u201d. So \u2013 pot? No, we lop. Date? Take no date! Bah! Play L.P. \n\nMiss (a lass, all right?) flew to space in NASA era. Rose no (zero) cadets ate raw. As a wise tart I fined rags red Lenin, we help pay bet \u2013 a risk \u2013 cash to Brian. I put a clam in a pool \u2013 a pool wets. \n\nMahdi puts a stop to harem \u2013 miss it in one vote, lost in one, veto of none. Post-op, no tonsil; I ate; no tastier, eh? We sleep at noon time so I dare not at one; no time stops as I time tides. A bed: under it, roll; in a mania, panic! \n\nIn a pond I did as Eros as Lee felt tenrec. \u201cInk\u201d \u2013 list it under \u201cI\u201d. Termites put pen in a way. Democrats wonder, I too. To slay moths a dog did. \n\nI saw elf; elf, far now, is a devilish taboo, rag-naked. I hid a bootleg disc. I, saboteur, toss it in. Oops! No legs! Laminated, a cask, conker in it, negates all if it is simple. \n\nHot pages are in a mag, nor will I peer, familiar tat, so lewd, native rot. Toner, ewe wore no trace; vagabond ewes do. Oh, Ada! Have pity! A pitiable eel \u2013 \u201cOh wet am I!\u201d - to save, note: bite gill as I do. \n\nCall a matador minor, eh? As I live, don\u2019t! Is torero no rigid animal debaser if tipsy? Ale drew esteem in a matador. A bolero, monks I rate play or go dig rocks; a can I step on. \n\nGo! Gas \u2013 it evades a bedsit \u2013 set a roost on fire. Boss sent a faded eclair to green imp or dog, I\u2019d don a belt to boot it; if Ada hid a boot, panic. \n\nI mock comic in a mask, comedian is a wit if for eventide. Vole no emu loved is not a ferret, so pet or witness a weasel if not. I hired less, am not so bossy, as yet amateur. \n\nTo stir evil, Edna can impugn a hotel: bad loos, hot on Elba: I may melt. Tart solicits it rawer, gets it rare. Push crate open; I ram buses, use no trams. \n\n \n\n\nDid I say, not to idiot nor a bare ferret, to trap rat, strap loops rat? Stewpot was on. Hot? I was red! Lessen it! Fine man on pot? No, pen inside by a bad law. So I made rips \u2013 nine delays. \n\nSome Roman items in a.m. ordered \u201cIs room for a ban?\u201d \u201cIt is,\u201d I voted: I sat pews in aisle. Beryl, no tiro to my burden, made off for a contest, I won kiss. I may raid fine dales. I raid lochs if I to help am. \n\nForecast for Clare v. Essex: If no rain, a man is ref. Fusspots net foxes. \n\nSenor is a gnome, latinos\u2019 bad eyesore. Help misses run to border, Casanova, now, or drab hotel. \n\nMa has a heron; I sleep, pet\u2019s on nose, sir! Rev. I rag loved art live \u2013 fine poser. Ultra-plan: I feign, I lie: cedar to disperse \u2013 last one? No, last six. Enamel bonnet for a dark car to toss a snail at. In it all, Eve lost; Seth\u2019s a hero slain on a trap \u2013 Rise, Sir Ogre Tamer. \n\nUpon Siamese box I draw design. I, knight able to help, missed an alp seen in Tangier of fine metal pots. Tin I mined rages \u2013 order nine, melt ten. Tone radios; tones are not to concur. Ten-tone radar I bomb \u2013 best fire-lit so hostel side meets eerie mini red domicile. A gulf to get is not a rare tale; no time to nod. \n\nRow on, evil yobs, tug, pull. If dogs drowse, fill a rut. An era\u2019s drawers draw. Put in mid-field in a band I dig a tub deep. Staff on a remit did refill a minaret. \n\nSam\u2019s a name held in a flat, or, sir, bedsit. I wonder, is it illicit ore? No ties? A bit under? Retarded? Is \u2018owt amiss? I\u2019m on pot; not so Cecil, a posh guy a hero met. A red date was not to last so Cecil sat. \n\nTip? An iota to pay, a dot; sad, I drop item. I\u2019d ask, call, Odin, a Norseman\u2019s god: \u201cPay payee we owe radio dosh o.n.o.\u201d I to me? No, I to media. \n\nPeril in golf \u2013 is ball a \u201cfore\u201d? K.O.! \n\nVexed I am re my raw desires. Alto has eye on nose but tone-muser pianist is level-eyed. I lost a tie. Blast! In uni no grades are musts. Avast! Never port! Sea may be rut. \n\nPart on rose? - It\u2019s a petal. Define metal: \n\n \n\nTin is \u2026 (I gulp!) can! \n\nI am a fine posse man, I pull a ton. Ron, a man I put on, I made suffer of evil emu\u2019s sadism. Leo\u2019s never a baron - a bad loss but evil \u2013 topple him, Leo\u2019s lad. Assign a pen, can I? A pal is note decoding. \n\nIs damp mule tail-less? No, ill; I breed for its tone. Radio speed, to grower, grew. Open a lot? No, stamp it; if for a free peso \u2013 not ecu -deign it. Times ago stone rates, e.g. at Scilly, display a wont. \n\nNo wish to get a design I, Sir Des, I\u2019ve let? No bus sees Xmas fir. O.K. \u2013 cab \u2013 tart it up; tie lots \u2013 diamond, log or tinsel; first end errata edit. So \u201cle vin (A.C.)\u201d, Martini, Pils lager, one tonic. \n\nI pegged a ball up to here when I got a top star role, Beryl. Gun is too big \u2013 won\u2019t I menace? Yes? No? \n\nIll? A cold? Abet icecap\u2019s nip. U.S.A. meets E.E.C. inside tacit sale \u2013 see! Beg a cotton tie, ma! No trial, so dodo traps exist. Arabs under-admire card label good hood stole. \n\nIn rage erupted Etna. Will a rotunda, bare villa, to tyro. Lack car? Non-U! Get a mini! My, my, Ella, more drums per gong; get a frog \u2013 nil less. Rod, never ever sneer. Got to? \n\nI disperse last pair of devils (ah!) here today or else order cash to breed emus. Said I: \u201cAre both superlative?\u201d C.I.D. assign it lemon peel still. I wore halo of one bottle from a ref (football) \u2013 a tip; so hit last ego slap a mate got. \n\nLate p.m. I saw gnu here (non-a.m.) or an idea got a dog to nod \u2013 I made felt to boot. \n\nFill in a lad? Nay, not all, Edna \u2013 lash to buoy. Did you biff one Venus? Not I! \u201cBroth, girl!\u201d ladies ordered \u2013 \u201cNo, with gin!\u201d \u2013 a fine plate, maybe suet; no carton I made rots in it. \n\nMed: a hill, Etna, clears in it. Ali, Emir, to slap in/slam in. All in all I made bad losers sign it \u2013 alibi. Set a lap for a level bat. \n\nA bed, sir, eh? To put cat now? Drat! Such an idyll of a dog\u2019s lair! That`s it, open it \u2013 a cage! Big nit sent rat! Some day (A.D.) send ewe. No, draw a pot now, do! Of wary rat in a six ton tub. \n\n \n\nEdna, ask satyr: \u201cTel. a.m.?\u201d No, tel. p.m.; Israeli tuner is damp. Use item: \u201cAnna Regina\u201d. No! Dye main room (\u201csalle\u201d) red! \n\nNice caps for a sea cadet in U.S.A. \u2013 Now I, space cadet, am it, sea vessel rep. Pin it on Maria, help Maria fondle her fine hotpot. No! Meet; set up to net, avoid a lesion. Set acid arena: Bruno one, Reg nil. Like it to sign in? Even I am nine-toed! I vote votes. \n\nOh, can a nose-rut annoy? No, best is Dorset. I know, as liar, to snoop, malign. \u201cI\u2019ll order it to get a bedroom door,\u201d began a miser I fed. \n\nAm I to peer, fan? Is a door by metal? Ere sun-up, drowse, nod, lose magnet. Food? Buns? I\u2019ll ask. Corn? I\u2019ll ask. Corn \u2013 I snack. Cats snack (cold rat). Sum for a bag: nil. First, is remit \u201ctraps in net\u201d? Yes, on a par. Coots yell over a dam I made. Bared nudist went a foot, I made roots. I tip a canon: \u201cRow, sir, at same tide; man one: row tug.\u201d \n\nSewer of denim axes a wide tail \u2013 a terror recipe to hero made manic. I, to resign? I ? Never! \n\n\u201cOFT I FELT ITS SENSUOUSNESS\u201d \u2013 title fit for evening is erotic; I named a more hot epic \u2013 error retaliated \u2013 I was examined for ewe\u2019s gut, wore no named item. \n\nA star is worn on a cap, it is too red. Am I too fat? Newts I\u2019d under a bed. Am I mad? Are volleys too crap? A nosey tennis part-timer sits rifling a bar of mustard. \n\nLock cans, stack cans in rocks, all in rocks, all I snub. Do often games, old ones, word-pun use; relate, my brood, as in a free pot I made fires, I manage brood. Moor debate got tired rolling, I lampoon, so trail saw on kites. \n\nRod sits, ebony on nature, so Nana chose to veto video. Ten in main evening is O.T.T. i.e. killing; Ere noon, urban eradicates noise, lad, I ovate not. Put esteem on top (to hen, if reheld). \n\nNo fair ample hair \u2013 am not I nipper-less? Eva estimated ace caps I won as united. A Caesar of space, Cinderella\u2019s moor, Niamey Don (a Niger-an name), ties up mad sire, nut! I, Lear, simpleton male, try tasks \u201cA\u201d and \u201cE\u201d \n\n \n\nbut not \u201cXI\u201d. Sanitary raw food won top award one Wednesday \u2013 a demo. \n\nStart nesting, I beg a cat. I? Nepotist? Ah, trials, God! A folly, Dinah, custard won\u2019t act up; other is debatable. Velar: of palate; sibilating is \u201cs\u201d. \n\nResold: a bed, a mill, an ill animal \u2013 snip, also trim. Eilat in Israel can tell I had \u2018em. Tin I stored (am I not raconteuse?) by a metal pen. If a night, I wondered, rose, I\u2019d all right orbit on sun, even off. \n\nI buoy, did you? Both Sal and Ella, Tony and Alan (\u201cIll if too bottle-fed, am I?\u201d) do not. God! A toga! Ed in a Roman one, rehung! Was I, M.P. et al., to get a map? Also get salt? I, hospital lab to offer, am, or felt to be, no fool \u2013 a hero. \n\nWill it sleep? No, melting is sad ice. Vital re-push to be raid, I assume. Deer, both sacred roes, Leroy (a doter, eh?) has lived for. I, apt sales rep\u2019s idiot to greens, revere vendors selling or fat egg-nog reps. \n\nMurder O\u2019Malley, my mini mate \u2013 gun on rack. Calory total: liver, a bad nut or all I wanted (\u201cet puree garnie\u201d): lots. \u201cDo, oh do, ogle bald racer,\u201d I\u2019m dared \u2013 N.U.S. bar at six. \n\nEsparto, dodo\u2019s lair to name it, not to cage bees, elasticated, is nice. Esteem, as up in space, cite bad local lions, eye can emit now. G.I. boots in ugly rebel or rat\u2019s potato gin (eh?) were hot. Pull a bad egg \u2013 epic, I note, no regal slip in it. Ram can \u2026 (I\u2019ve lost idea!) \n\nTarred nets, rifles, nitro, gold \u2013 no maid stole it. Put it, rat, back or if Sam (\u201cX\u201d) sees sub on televised rising, I sedate Goths. I won\u2019t \u2013 no way. \n\nAlps, idyllic stage set, are not so gas-emitting, I educe. To nose, peer, far off, I tip mats onto lane. Power grew or got deep so I dare not stir. Of deer, billions sell. I ate lump \u2013 mad sign, I do cede \u2013 tonsil a pain, acne pang is sad also. Elm I help pot, live \u2013 tub\u2019s sold; a ban or a bar, even so, elms, I\u2019d assume, live for. Effused am I not, up in a manor, not all up in a mess. \n\nOpen if a main A.C. plug is in it. \n\n \n\nLate men I fed late \u2013 pasties or not. \u201cRapture\u201d by a maestro prevents a vast sum erased. \n\nArgon in units, albeit at solid eye level, sits in a \u2026 (I presume not) \u2026 tube, son. No eyes: a hot laser \u2013 is Ed wary? \n\nMermaid, ex- evoker of all A.B.s, I flog. Nil I repaid. Emotion! Emotion, oh so do I dare, woe! \n\nWee yap-yap dog\u2019s name\u2019s Ron. An idol lacks a dime tip, or did, as today a potato in a pitta slice costs a lot \u2013 tons. A wet adder ate more hay. Ugh! So, pal, ice cost on top? No, miss, I\u2019m a two-sided rat, erred nut, I base it on erotic ill; It is I, red now; it is debris, rot. \n\nAlf, an idle he-man as \u201cmaster animal lifer\u201d did time, ran off at speed, but a G.I. did nab an idle if dim nit. Upwards rewards are natural life\u2019s words, God. Fill up guts, boy, live now or do not emit one later. A rat on site got flu. \n\nGaelic, I\u2019m odd Erin, I\u2019m Eire, esteemed islet. So hostile rifts ebb. Mob, I.R.A., dare not net R.U.C. \u2013 no cotton. Erase not, so I dare not nettle men in red rose garden \u2013 I\u2019m in it. \n\nStop late men if foreign at nine. Esplanades, simple hotel, bath, gin \u2013 king is Edward IX; obese; Ma is no pure mater. Go! Rise, sir; part anon. \n\nI also rehash tests \u2013 \u2018O\u2019 Level Latin, Italian. S.A.S., so, to track radar. Often nobleman exists alone \u2013 not sales reps \u2013 I do. Trade ceiling, i.e. final part, lures open if evil trade. \n\nVolga River rises on no steppe. Elsinore has a hamlet \u2013 Oh, Bard, row on Avon! \n\nA sacred robot nurses simple hero\u2019s eye; dabs on it a lemon. Gas, iron, Essex often stops, suffers in a mania. Ron fixes several crofts, acer of maple. Hot, I fish; cold, I arise laden; if diary amiss, I know it set no car off. Foe-damned ruby motor, it only rebels. \n\nIan I swept aside to visit, in a bar of moorside red, Romanis met in a more mossy ale den. Inspired am I, Oswald. A bay bed is nine p on top. No name, niftiness- elder saw it. Oh no! Saw top wet star\u2019s pool \u2013 part star, part otter. Refer a baron to idiot, Tony, as I did. \n\n \n\nSmart ones use submarine. \n\nPoet, arch-super-artiste, grew artistic. I lost rattle; my amiable, not oh so old, able to hang up, mina, can deliver it, so true. \u201cTa, matey!\u201d \u2013 says so Boston (Mass.) elder I hit. \n\nOn file S.A.E. was sent \u2013 I wrote poster re fat on side, volume one \u2013 loved it, never off it, I was in. Aide mocks a manic; I mock comic, I nap: too bad I had a fit, I too. Bottle ban odd, I go drop mine, ergo trial ceded a fatness, sober if not so, or a test is debased. \n\nA vet is agog \u2013 no pet\u2019s in a cask \u2013 corgi dog, royal pet, a risk no more. \n\nLob a rod at a man I meet. Sewer delays pit fires \u2013 a bedlam in a dig \u2013 iron ore rots it. No devil is a hero \u2013 Nimrod. \n\nAt a mall a cod is all I get. I bet on Eva, so Tim ate whole eel bait, I pay tip, Eva had a hood sewed. No B.A. gave car to Nero, we were not to rev it and we lost a trail; I\u2019m a free pill, I wrong a man. I erase gap; to help miss it, I fill a set. A gent in ire knocks a cadet. \n\nAnimals\u2019 gel on spoon \u2013 it is so true to basics \u2013 I\u2019d gel; too bad I hide kangaroo baths \u2013 I lived as I won raffle, flew as I did go, dash, to my, also too tired now, star comedy: A wan, inept, upset I\u2019m retired, nut; its ilk, nicer. Nettle feels a sore; sad, I did no panic in a pain, am an ill or tired, nude, based item; it is a spot. \n\nSemitone, not a tone, radios emit; no, on tape; elsewhere it\u2019s a tone. \n\nTail is not on; pots open on foot, even on it, so let oven (on, it is) simmer \u2013 a hotpot\u2019s a stupid ham stew. \n\nLoop a loop, animal \u2013 cat up in air. \n\nBoth sacks I rate by apple hewn in elder\u2019s garden if it rates, I was aware \u2013 tasted a core. \n\nZones or areas, Annie, cap, so twelfth girl, lass, alas, simply (alpha beta) done, Kate. Tadpole won top Oscar, Obadiah, \u201cO\u201d Level axed. \n\nArgon gas did emit no straw, so ozone sure drops argon, oozes up in Ruth\u2019s ample hotel or sits afar off in a bar \u2013 of elastic, is it? \n\nI hate cinema; cinema dogs in a mass. Older effusion to old \u2013 lost, is it now? Reward: a mood. \n\n \n\nAll upsets it. \n\nRadar trails an Islamic educer of a riling issue, damages it in Israel. Ceiling is, I say, a plan, a case of one deck. Can knees sag as one Latin image elates, I wonder? \n\nOboe diverts ultra foe, volatile bald ogre \u2013 push to berate; I\u2019d do, ogre. So, p.m., Oct. first, never play organ\u2019s stops \u2013 lay or put it up in ward ten. \n\nFinal cast like rowing \u2013 I sedate play, old as am I, God! Am I! On tacks I ran; I saw rats. A Gemini tramp is May born. \n\nI back colony\u2019s sober omen of lack of lace. Rome, not Paris, a wonder. \n\nObey retail law \u2013 a noose killed oyster. Reflate my ball, a water-filled one. Disabuse no name of emanating issue. \n\nDamsels, I note, vary tastes so cost now desserts. I say no! Try taste more honeyed. A bad nemesis at naff ruse will upset. I, mere Satanist, e.g. rater of a devil \u2013 (Oh wrong is a sin!) \u2013 I\u2019m no devil\u2019s god, damned. \n\nAnimals, if on a mat, sit. Rain, a more vile drop, made us site it in a cottage. Breed deer \u2013 bottle fits a llama. \n\nI lay, as I emanate, go to sleep, mad ones on docks \u2013 air is hot. Entrap, net, nine men in party raid - all if it is in a crab-pot room, an itemised, under-lit, nullified old log den \u2013 I\u2019m sure voles made it rot in knot. \n\nTubas we see far off lack limit. A cat on still or tall upward paws to no dog is an ample hot-dog, ergo nastier if tastier, eh? We, raw amid a conman, a mama in a mask, corpse et al., err. \n\nOctuple tracks at a son\u2019s eyelash side distressed a tall eye doctor, a tall ace, rigger of a vote: got put in egress; odd, abased, is ebbed, as I am, Amy, asinine lot! Nine lots! Don\u2019t six rams live? Don\u2019t six exist? \n\nAlfred, nuts or fool gigolo, trod never if gold locks all in a flap on a red rose; made nine or ten stops. \n\nI heed never, I\u2019m Daisy, a prod never, I terrorise viler starfish. To me suitors, no lemons, came rowing. Is a sin a mania? Rot! \n\n \n\nSit! I fix a looted amp or delay more, hasten not. A baser if snug stool, wonkier, if not - Alf says - super, a ballet to no devil, is a stool too. Ban it, actor, race to no tune. \n\nMay names I wrote wrong (Is no man in it, a long old log?) sit in row, sign irate Goths; I dare drop it. At felon\u2019s eye I peer, fast open \u2013 I\u2019m nosey, esteem eyes. All upset, ample hogs resume totting. Is sad nabob tired? Roots don\u2019t evade liver in Alf\u2019s gob. \n\nDeers I held right; oblong, apt enamel or tile rifle on gun spot to get a man \u2013 aim is all. I rogate, minister. Feeble gnats, alas late, prosaic, a canine pet is not to consume hot. \n\nLoo, wet, issues old idiot; evading, I sneer, obey a deer, gall a deer, gain alpine dragnet for egg I\u2019d net to ram in a pan I made to help master. Rags I held, arcane poet, arcane poetic error, all odd; I bottle fine panacean lust. I\u2019d nag elks I ride if editor toted a minor. I fog a natural life. \n\nRoses, or level dumb ones \u2013 rows in a mown, ample, hewn acre. Wolfsbane made it a garden in May, a garden indeed. \n\nNine mates, nine tons I must save now on time \u2013 editor raps a late man. G.I.s edit also, too. Do over if tests in a task radiate. Rob ran; I, too, fled. \n\n\u201cOmega\u201d \u2013 list in alphabet. \n\nA gander, a line of live ducks, irk cubs. A wart, set at a cast on knee, snug as spots. \n\nA poor denim for a janitor, racer, armed aide, solid idler \u2013 rabid; I\u2019d elastic in a pot, tons to sew. \n\nTubes or axes went in a clam, in an oyster. Free booze \u2013 lap it all up. Pity, my apple hot, so I\u2019d a root stew. God, a stew! Tip it at feline! Posies, a cat\u2019s altar often, no baron packs. A monk caps dog \u2013 I meddle here \u2013 hot? Pull its leg! A bee was a hoot, eh? \n\n \n\nNo, it is opposite. Yaks I rode wore hats, albeit on deity\u2019s orders. Rats age more held in a trap, nip and I know it \u2013 set no cage now. \n\nIt\u2019s eta; no, it\u2019s a beta \u2013 Tsar of Tonga rates isles. Mad Ed is all upset at cider, is Ed? Is a madam too? Snip? I\u2019d snip, spot a fine position, snip nine more cinemas. \n\nDo ogres sell in a mall? Yes, on a barge so rats row tubs. \n\nWall last canes up or Eros, an imp, lives to irk, rasp or dam all tides sent. I won\u2019t \u2013 I was no Roman \u2013 even I saw tired row \u2013 a sore. He lives on. \u201cNo!\u201d we yell. \n\nUp, now! Wards are in nurses\u2019 sole care. I, peer, fed, am too fat? Oh, not I, test no dined ruby ale; dote not on salad it\u2019s in \u2013 I am sad. \n\nLocks I rifle so troops atone re war. Only rebel or a crofter animates so cottage beheld arcades, so trees are sold, abased. I redo, rehang, I err \u2013 a wasted act; nests I\u2019d \u2013 as an owl \u2013 laid. A boot\u2019s raw foot, even if a foot to master, germs (ah!) can evil do. \n\nPan is tune-pipe \u2013 so hot notes, paths up to honeydew. \n\nOdd locks, a maddened (I was aware) macaw on top, spot no seen knots, rifts or fan, I saw. Are maces a baton, madam? Oodles, madam? Rare laptops are too late \u2013 got too lit up. \n\nNits rub \u2013 snip now, I\u2019ll abate, not snip, nits I held. \n\nNubile Danish tomboys I led to old loser as no melons I held; no fish to my name. Nod lower, do I dare? No, one nods a hairy snipe. (Edit: one hairy snipe, eh?) See silliness, else we\u2019ll ask cornish to obey deity\u2019s or god\u2019s item. I, God, damn it! I was in it! To Hades, acne trap, sad loser! As warts pop, a dosser I \u2013 we \u2013 vile rat, sack! Same row, oh woe! Macaroni, rats, as a hoot, tie. I vomit on rats. [/b]", "Solution_7": "Wow thats amazing!!! :bruce: :bruce: :bruce: :bruce:", "Solution_8": "Does anyone want to check whether this is indeed a palindrome? :D", "Solution_9": "this, indeed, is a palindrome, if we dont count all the quotations, exclamation/question marks, periods, etc.", "Solution_10": "OMG :o :o :o :o :rotfl:", "Solution_11": "wow :10:", "Solution_12": ":wacko: :wacko: :wacko: :wacko: :wacko: \r\n\r\nwhere did you get that one? or did you make it up? which i highly doubt (no offense).", "Solution_13": "He got it from: \"Source: http://www.palindromelist.com/longest.htm\", as he stated above. And if you really think about, this is just numerous palindromes piled together, they really have no flow, or mutual topic....", "Solution_14": "[quote=\"gr8sk8r\"]:wacko: :wacko: :wacko: :wacko: :wacko: \n\nwhere did you get that one? or did you make it up? which i highly doubt (no offense).[/quote]\r\nNone taken. The person who made this must been really bored. :rotfl:", "Solution_15": "For $x^3-23x+2=0$\r\n\r\n$x^3+2=23x$\r\n\r\nNow do you see it?", "Solution_16": "[quote=\"mathnerd314\"]And, of course, the old joke that fear of palindromes is\n\n[size=200]AIBOHPHOBIA[/size][/quote]\r\n\r\nnow I know why it's called that :D", "Solution_17": "[b]A man, a plan, a canal, Panama.[/b]", "Solution_18": "How about birthdays? 09/11/90 woohoo!", "Solution_19": "kayak, radar\r\nTen animals I slam in a net :!:", "Solution_20": "All forgot about the original question, or what? \r\nI feel its The Principle of Mathematical Induction??\r\nSanjith", "Solution_21": "When x = 9...\r\n\r\n$2x+x^2 = 18 + 81$", "Solution_22": "A palindrome in [url=http://users.utu.fi/mitaja/pitkpali.txt]Finnish[/url]. ;)", "Solution_23": "[quote=\"Sathej\"]All forgot about the original question, or what? \nI feel its The Principle of Mathematical Induction??\nSanjith[/quote]\r\n\r\nWhat? The answer to the original question was that it's a palindrome.", "Solution_24": "Does the finnish palindrome actually make sense? If it does, that is so cool! :)", "Solution_25": "Do you people know about sentences like this: \"In this sentence, there are three letters \"i\", three letters \"n\"\", etc., except the fact that the sentence describes the number of letters for each letter in the alphabet!\r\nI know some in Russian. they are really cool.\r\nIt's actually very hard to come up with those.", "Solution_26": "The palindrome has no proper meaning. For example, here is a part of the palindrome and its translation. It's little hard to translate it to English since the palindrome is not good Finnish. Eino and Riku are Finnish male names as well as Veli-Matti. And Vellu is a nickname of Veli-Matti. \r\n\r\nKaskelotti: ovi eh\u00e4tti nen\u00e4\u00e4n. Ei, Eino! Poni,\r\nponi vei Vellun. Is\u00e4s s\u00e4tti k\u00e4si\u00e4, Riku mielisteli. -Enon akku katosi,\r\nmuksusi sanoi.\r\n\r\nA sperm whale: the door had time to go to nose. Eino, no! A pony, pony took Vellu. Your father was scloding hands, Riku flattered. -Uncle's battery disappeared said your kid.", "Solution_27": "111111111 * 111111111= 12345678987654321= 111111111 * 111111111\r\n :D", "Solution_28": "When I was younger my teacher had a book called\r\n\"Go hang a salami! I'm a lasagna hog!\"\r\nwhich is a book of illustrated palindromes (which is always funny, as they aren't meant to be comprehended)", "Solution_29": "[quote=\"me@home\"]When I was younger my teacher had a book called\n\"Go hang a salami! I'm a lasagna hog!\"\nwhich is a book of illustrated palindromes (which is always funny, as they aren't meant to be comprehended)[/quote]\r\n\r\nI have that book!! :D I loved the illustrations..." } { "Tag": [ "search", "combinatorics proposed", "combinatorics" ], "Problem": "Noah has 8 species of animals to fit into 4 cages of the ark. He plans to put species in each cage. It turns out that, for each species, there are at most 3 other species with which it cannot share the accomodation. Prove that there is a way to assign the animals to their cages so that each species shares with compatible species.", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=907693445&t=212214\r\n\r\nI searched for \"noah cage\" to find that thread. The problem was also apparently reused at the 2005 Pan African Olympiad." } { "Tag": [ "LaTeX" ], "Problem": "Is it possible to have a comment environment in $ \\text{\\LaTeX}$? Such as the code tags are on this forum.\r\n\r\nFor example instead of \r\n\r\n[code]% bla bla\n% asd asd\n% tt tt\n[/code]\n\none would use \n\n[code]\\begin{comment}\nbla bla\nasd asd\ntt tt\n\\end{comment}\n[/code]", "Solution_1": "You can have [i]exactly[/i]\r\n[code]\\begin{comment} \nbla bla \nasd asd \ntt tt \n\\end{comment}[/code] All you need is to add the comment package with \\usepackage{comment} in the preamble.", "Solution_2": "Awesome! Thanks for the quick reply Steve." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c>0$ that satisfy $abc=1$. Prove that:\r\n\r\n\\[ 2\\cdot \\frac{(a+b+c)^2}{\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}\\geq \\sqrt{a}+\\sqrt{b}+\\sqrt{c}+3 \\]", "Solution_1": "[quote=\"Ezbakhe Yassin\"]Let $a,b,c>0$ that satisfy $abc=1$. Prove that:\n\n\\[ 2\\cdot \\frac{(a+b+c)^2}{\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}\\geq \\sqrt{a}+\\sqrt{b}+\\sqrt{c}+3 \\][/quote]\r\n\r\nHi Yassin!\r\n\r\nCauchy's ineq yields $3(a+b+c)\\geq (\\sqrt{a}+\\sqrt{b}+\\sqrt{c})^2$\r\n\r\nso $\\text{LHS}\\geq 2 \\frac{(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})^3}{9}$\r\n\r\nCalling $\\sqrt{a}+\\sqrt{b}+\\sqrt{c}=A$ we just need to prove that \r\n\r\n$2A^3\\geq 9A+27$ which is equivalent to $(A-3)(2A^2+6A+9)\\geq 0$\r\n\r\nobviously true due to AM-GM: $\\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\geq 3\\sqrt[6]{abc}=3$\r\n\r\nHope I'm right and unterstandable :) :D" } { "Tag": [ "probability", "function", "topology", "real analysis", "real analysis theorems" ], "Problem": "a) Since there are infinite many reals and rationals, is it true that there are \"more\" reals than rationals? and how do you prove that?\r\n\r\nb) From Archimedean property of real - for every real number x, there exists a natural number n, where n>x.\r\nI know both reals and natural numbers are NOT bounded above, but am I correct to interpret the above as this - [i]no matter how large a real number is selected, I can always find a larger natural number, hence the largest natural number is bigger than the largest real number (I know both do not exist) [/i]?\r\n\r\nThanks for the help.", "Solution_1": "\"More\" in what sense?\r\n\r\nThere are three senses for what you might mean by \"more\".\r\n\r\n1. The crudest version of the word \"more\" is given by cardinality. Two sets have the same cardinality if and only if they they can be put into a one-to-one correspondence with each other. By this criterion, by cardinality, there are exactly as many even integers are there are integers. In fact, there are exactly as many rational numbers as there are integers. In fact, there are exactly as many algebraic numbers as there are integers. (Examples of algebraic numbers include $ \\sqrt{2},\\sqrt[5]{8\\plus{}\\sqrt[3]{17}},$ the real root of $ x^7\\plus{}x^3\\plus{}6\\equal{}0,$ and so on.) All of these sets are countably infinite - all that means is that they can be put into one-to-one correspondence with $ \\mathbb{N},$ the set of natural numbers.\r\n\r\nOn the other hand, the real numbers are uncountable - which means they cannot be put into a one-to-one correspondence with $ \\mathbb{N}.$ And that means that the irrational numbers are also uncountable - that there as many of them as the real numbers. This cardinality - the cardinality of the reals, the cardinality of the continuum - is strictly greater than the countable cardinality. So there are, by cardinality, more irrationals than rationals.\r\n\r\nFor a proof of the uncountability of $ \\mathbb{R},$ look up \"Cantor's diagonal argument\" or something like that. For a theorem that guarantees that you can actually compare cardinalities, see the \"Cantor-Schroeder-Bernstein Theorem\" (or any two of those names).\r\n\r\n2. Or you can compare sets by measure. Probability is a specific version of measure. Choose a real number in $ [0,1]$ randomly according to the uniform distribution on that interval. The probability that the number you choose is rational is zero. The probability that the number you choose is irrational is one. By that criterion, there are more irrationals than rationals.\r\n\r\n3. Baire category. The rationals can be written as a countable union of nowhere dense sets. The real numbers, being a complete metric space, cannot be written as a countable union of nowhere dense sets, meaning that the irrationals cannot be written as a countable union of nowhere dense sets. So by Baire category, there are more irrationals than rationals.\r\n\r\nAs for what you said in part (b): there is no such thing as the largest real number. Or the largest natural numbers. Given any real, I can find a larger integer. But given any integer I can find a larger real. Any set which is not bounded above has no largest element. But even some sets that are bounded above, such as the open interval $ (0,1),$ also do not have a largest element.", "Solution_2": "As always, I appreciate your detail response, Prof. Merryfield.\r\n\r\na) For \"more\", I am referring to Cardinality. Your explanation is clear, I will have to do some reading on countability.\r\n\r\nb) On my other question, I knew something was wrong, I am aware that they are not bounded above and there is no largest element but I couldn't escape my fallacy from misinterpreting the \"n>x\" in Archimedean property. Furthermore, what you said \r\n[quote]... But given any integer I can find a larger real...[/quote]\r\nimplies that given any n, there exists a real number x, where x>n. How can both \"n>x\" and \"x>n\" be true? or perhaps, it is more like \"n>x\" and \"y>n\" where x,y are real numbers. I am confused. \r\n\r\nSorry for another push for detail :)", "Solution_3": "They are indeed not the same instances of those variable names. We reuse names all the time.\r\n\r\nIt looks like you're not really getting the quantifiers; in \"for any $ n$, there exists $ x$ such that...\", that $ x$ doesn't exist independently of $ n$, and actually choosing one would be a function $ x(n)$.", "Solution_4": "Yes, I see my problem now, hopefully it sticks, thanks for pointing that out :)" } { "Tag": [ "search", "\\/closed" ], "Problem": "what do bots do exactly? i've looked at--for example--MSNBot's profile and they never posted a thing!! is there a point to them?", "Solution_1": "Well, I know GameBot posts people's questions when you click \"Ask on Forum\". As for the other bots, no clue.", "Solution_2": "Please use the following links:\r\nhttp://www.mathlinks.ro/viewtopic.php?search_id=1205226179&t=99267\r\nhttp://en.wikipedia.org/wiki/Internet_bot", "Solution_3": "thanks guyz" } { "Tag": [ "inequalities", "algebra", "polynomial", "algebra unsolved" ], "Problem": "schow that if the polynomiale $ax^{3}+bx^{2}+cx+d$ have three positif roots then the condition $bc\\leq3ad$ is nessec\u00e9re.", "Solution_1": "$a\\not =0$, as $bx^{2}+cx+d$ has no three roots.\r\n$x_{1},x_{2},x_{3}>0$ distinct roots of $ax^{3}+bx^{2}+cx+d$, then\r\nThen [url=http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas]Vi\u00e8te's formulas[/url] gives\r\n$x_{1}+x_{2}+x_{3}=-{b\\over a}$,\r\n$x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}={c\\over a}$ and\r\n$x_{1}x_{2}x_{3}=-{d\\over a}$.\r\nThe [url=http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means]AM-GM inequality[/url] then gives\r\n$-{b\\over 3a}={x_{1}+x_{2}+x_{3}\\over 3}>\\sqrt[3]{x_{1}x_{2}x_{3}}=\\sqrt[3]{-{d\\over a}}$ and\r\n${c\\over 3a}={x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}\\over 3}>\\sqrt[3]{x_{1}x_{2}x_{1}x_{3}x_{2}x_{3}}=\\sqrt[3]{(-{d\\over a}})^{2}$,\r\nand multiplying those inequalities gives\r\n$-{bc\\over 9a^{2}}>-{d\\over a}$.\r\nMultiplying with $-9a^{2}<0$ gives $bc<9ad$.\r\nNow $ad=-x_{1}x_{2}x_{3}a^{2}<0$, so $9ad<3ad$ and the result follows.", "Solution_2": "really good \r\nthinks olorin :thumbup:", "Solution_3": "Now you mentioned roots of cubics here a very simple but quite useful result.\r\n\r\nExcluding the cubic $f(x)=x^{3}$.\r\n\r\nLet $y_{1}, y_{2}$ be the $y-value$ of the two stationary points of a cubic polynomial\r\n\r\n1. If $y_{1}y_{2}< 0$ then the cubic has $3$ distinct real roots\r\n\r\n2. If $y_{1}y_{2}= 0$ then the cubic has $2$ real roots where one is a root of multiplicity 2 (i.e a double root)\r\n\r\n3. If $y_{1}y_{2}> 0$ then the cubic has $1$ real root\r\n\r\nThese can be proved graphically." } { "Tag": [], "Problem": "prove that there is not n that:\r\n[b]7l2^n+1[/b]", "Solution_1": "there are three expressions for each positive integer like $n$:\r\n$n = 3k \\rightarrow 2^{3k}+1 \\equiv 8^{k}+1 \\equiv 2 \\;( mod\\;7 )$\r\n$n = 3k+1 \\rightarrow 2^{3k+1}+1 \\equiv 8^{k}\\times 2+1 \\equiv 3 \\;( mod\\;7 )$\r\n$n = 3k+2 \\rightarrow 2^{3k+2}+1 \\equiv 8^{k}\\times 4+1 \\equiv 5 \\;( mod\\;7 )$\r\nAs a consequence there's no positive integer $n$ that $7 | 2^{n}+1$ .", "Solution_2": "Essentially the same thing, but $2^{n}\\equiv 1, 2, 4 \\mod{7}$, and an $n$ which satisfies the condition required would require $2^{n}\\equiv 6\\mod{7}$. Where is this from? I don't think this would be an olympiad problem." } { "Tag": [], "Problem": "Hey i found this problem and i was trying to brute force it but it wouldn't come out so yea please help\r\n\r\nFind all real triples (a,b,c) of real numbers such that x is a real number that satisfies\r\n\r\n$ \\sqrt {2x^2 + ax + b} > x-c $\r\n\r\n$ x\\le 0$ and $x> 1$\r\n\r\nyea i got it down to something but there were so many combinations so im not sure how to get this", "Solution_1": "I think you switched around the greater than and less than signs.", "Solution_2": "yea probably otherwise i dun see how u could get a finite amount of (a,b,c)\r\noh well i found it on this kid's site\r\nhttp://www.xanga.com/osdes\r\nso maybe he wrote it wrong", "Solution_3": "But its a 2002 MSOP homework problem :? \r\n\r\nIf and only if x<=0 and x>1\r\nedit: oops, yeah If and only if x<=0 or x>1\r\n\r\nAnd I have a question. I know it means x can't be between 0 and 1, but does it also mean that x [i]must[/i] be true for [i]all[/i] values {x: x<=0 :cup: x>0}", "Solution_4": "[quote=\"djshowdown2\"]But its a 2002 MSOP homework problem :? \n\nIf and only if x<=0 and x>1[/quote]\r\n\r\nMaybe you mean 'or' instead of 'and' :)" } { "Tag": [ "inequalities" ], "Problem": "Solve this absolute value inequality: \r\n\r\n$|x+3|<|x-8|$.", "Solution_1": "[hide]\nby a simple graphic interpretation, $x<\\frac{5}{2}$, finding this value from the equation $-x+8=x+3$\n[/hide]", "Solution_2": "Or if this was harder, i.e it had $2x$ or something in it, just square both sides to make them positive, and solve normally.\r\n$x<\\frac{5}{2}$", "Solution_3": "$|x+3|<|x-8|\\Longleftrightarrow |x+3|^2<|x-8|^2$, yielding $x<\\frac{55}{22}.$ The answer is $x<\\frac{5}{2}.$", "Solution_4": "[quote=\"kunny\"]yielding $x<\\frac{55}{22}.$ The answer is $x<\\frac{5}{2}.$[/quote]\r\n\r\nHow did you make that transition?", "Solution_5": "You can reduce by 11.", "Solution_6": "[quote=\"kunny\"]You can reduce by 11.[/quote]\r\n\r\nOh :rotfl: that was stupid of me. \r\n\r\n\r\nOn another thought, can you use the theorem \r\n\r\n$|x| \\frac{1}{\\tau}$.", "Solution_2": "Sorry - should be $\\tau = \\cos^{-1}{\\left(1-\\epsilon\\right)}$ (won't let me edit)", "Solution_3": "sorry but i am unable to follow.", "Solution_4": "Teki-teki said, roughtly this:\r\nPlot the cos(x) function.\r\nNow plot the line y=1-e.\r\nThe two plots intersect at infinitely many points.\r\nThe distance between two points consecutive of intersection is fixed (the points should be such that for every point x between them cos(x)-(1-e)>0). And your question follows immediately form the irrationality of pi (the period between the intervals)", "Solution_5": "so basically you are using the fact that between any two numbers $a,b\\in\\mathbb{R}$ we know $\\exists{c}\\in\\mathbb{Q}$ such that $a \\equal{} 0$pou profanws einai to $ M \\equal{} 0$.\r\nP.S. opou dio payles ennow to apolito. :oops:", "Solution_2": "[quote=\"Athinaios\"]$ \\left|\\left|\\sum_{cyc}ab(a^{2} - b^{2})\\right|\\leq \\left|\\sum_{cyc}\\frac {\\left(a^{2} - b^{2}\\right)\\left(a^{2} + b^{2}\\right)}{2\\right|} = \\left|\\sum_{cyc}\\frac {a^{4} - b^{4}}{2}\\right| = 0$ \u03b1\u03c1\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03bf \u03b5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf $ M$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ M(a^{2} + b^{2} + c^{2})^{2}\\geq 0\\Longrightarrow M = 0$ [/quote]\r\n\r\n\u0395\u03ac\u03bd \u03b8\u03ad\u03c3\u03b5\u03b9\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03cc\u03c0\u03bf\u03c5 $ M = 0$ \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ \\left| ab(a^{2} - b^{2}) + bc(b^{2} - c^{2}) + ca(c^{2} - a^{2})\\right|\\leq 0$ \u03cc\u03c0\u03bf\u03c5 \u03b5\u03b4\u03ce \u03c4\u03ce\u03c1\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b7 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9 \u03b7 \u03b1\u03bd\u03af\u03c3\u03c9\u03c3\u03b7. :!:\r\n\r\n[hide]\u0397 \u03c3\u03c9\u03c3\u03c4\u03ae \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 $ M = \\frac {9}{32}\\sqrt {2}$[/hide]", "Solution_3": "[quote=\"\u0393\u03b9\u03ce\u03c1\u03b3\u03bf\u03c2\"]\u0395\u03c0\u03b5\u03b9\u03b4\u03ae \u03b4\u03b5 \u03b2\u03bb\u03ad\u03c0\u03c9 \u03bd\u03b1 \u03bc\u03c0\u03b1\u03af\u03bd\u03bf\u03c5\u03bd related \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03b1\u03c0\u03cc I.M.Os \u03c3\u03ba\u03ad\u03c6\u03c4\u03b7\u03ba\u03b1 \u03bd\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03b5\u03b3\u03ce \u03ad\u03bd\u03b1...\n\n\u039a\u03b1\u03b8\u03bf\u03c1\u03af\u03c3\u03c4\u03b5 \u03c4\u03bf\u03bd \u03b5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc $ M$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03bd \u03ce\u03c3\u03c4\u03b5 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 $ \\left| ab(a^{2} \\minus{} b^{2}) \\plus{} bc(b^{2} \\minus{} c^{2}) \\plus{} ca(c^{2} \\plus{} a^{2})\\right|\\leq M(a^{2} \\plus{} b^{2} \\plus{} c^{2})^{2}$ \u03bd\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03cc\u03bb\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 $ a,b,c$.[/quote]\r\n\r\n\u0395\u03c7\u03b5\u03b9\u03c2 \u03b5\u03bd\u03b1 \u03bc\u03b9\u03ba\u03c1\u03bf typo... \u03c4\u03bf \u03c3\u03c9\u03c3\u03c4\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9\r\n\r\n$ \\left| ab(a^{2} \\minus{} b^{2}) \\plus{} bc(b^{2} \\minus{} c^{2}) \\plus{} ca(c^{2} \\minus{} a^{2})\\right|\\leq M(a^{2} \\plus{} b^{2} \\plus{} c^{2})^{2}$", "Solution_4": "\u03a4\u03b9 \u03ad\u03b3\u03b9\u03bd\u03b5 \u03c1\u03b5 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac? \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1\u03c2? :o", "Solution_5": "\u0393\u03b9\u03ce\u03c1\u03b3\u03bf \u03b4\u03b5\u03bd \u03b1\u03c3\u03c7\u03bf\u03bb\u03ae\u03b8\u03b7\u03ba\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03c7\u03b5 \u03c3\u03c5\u03b6\u03b7\u03c4\u03b7\u03b8\u03b5\u03af \u03ba\u03b1\u03b9 \u03c0\u03b1\u03bb\u03b1\u03b9\u03cc\u03c4\u03b5\u03c1\u03b1 \u03c3\u03c4\u03bf \u03b1\u03b3\u03b3\u03bb\u03b9\u03ba\u03cc Forum (\u03c3\u03c4\u03bf section \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2) \u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u0395\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03cc. \u03a3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ad\u03bc\u03c0\u03c9 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03c3\u03c4\u03b9\u03c2 \u03b5\u03be\u03ae\u03c2 \u03c3\u03b5\u03bb\u03af\u03b4\u03b5\u03c2:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=571945#571945\r\n\r\n\u03ba\u03b1\u03b8\u03ce\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd http://www.mathlinks.ro/Forum/viewtopic.php?t=101316 \u03cc\u03c0\u03bf\u03c5 \u03bf \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03cc\u03c2 \u03b5\u03af\u03c7\u03b5 \u03b2\u03ac\u03bb\u03b5\u03b9 \u03bc\u03af\u03b1 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03bb\u03cd\u03c3\u03b7.\r\n\r\n\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "For $-1\\leq x\\leq 1$ and $n\\in\\mathbb N$ define $T_{n}(x)=\\frac{1}{2^{n}}[(x+\\sqrt{1-x^{2}})^{n}+(x-\\sqrt{1-x^{2}})^{n}]$.\r\na)Prove that $T_{n}$ is a monic polynomial of degree $n$ in $x$ and that the maximum value of $|T_{n}(x)|$ is $\\frac{1}{2^{n-1}}$.\r\nb)Suppose that $p(x)=x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}\\in\\mathbb{R}[x]$ is a monic polynomial of degree $n$ such that $p(x)>-\\frac{1}{2^{n-1}}$ forall $x$, $-1\\leq x\\leq 1$. Prove that there exists $x_{0}$, $-1\\leq x_{0}\\leq 1$ such that $p(x_{0})\\geq\\frac{1}{2^{n-1}}$.", "Solution_1": "If $T_{n}(x)=\\frac{1}{2^{n}}[(x+\\sqrt{1-x^{2}})^{n}+(x-\\sqrt{1-x^{2}})^{n}]$, then\r\n\\[a_{n}=\\frac{1}{2^{n-1}}\\sum_{k=0}^{[n/2]}C_{n}^{2k}(-1)^{k}=\\frac{2^{n/2}}{2^{n-1}}cos\\frac{\\pi n}{4}\\]\r\n.\r\nI Think $T_{n}(x)=\\frac{1}{2^{n}}[(x+\\sqrt{x^{2}-1})^{n}+(x-\\sqrt{x^{2}-1})^{n}]$. In these case\r\n\\[a_{n}=\\frac{1}{2^{n-1}}\\sum_{k=0}^{[n/2]}C_{n}^{2k}=1. \\]", "Solution_2": "But then $x^{2}-1\\leq 0$ with $-1\\leq x\\leq 1$.", "Solution_3": "Let $-1\\le x\\le 1$, then exist y, suth that $x=cos y, \\ \\sqrt{1-x^{2}}=siny$.\r\nIt give $T_{n}(x)=\\frac{1}{2^{n}}[(cosy+isiny)^{n}+(cosy-isiny )^{n}]=\\frac{cos(ny)}{2^{n-1}}.$ Therefore $T_{n}(x)=\\frac{1}{2^{n-1}}cos(n arccos x).$" } { "Tag": [ "geometry", "3D geometry", "algebra", "factorization", "difference of cubes", "special factorizations", "number theory unsolved" ], "Problem": "Find all pair integer numbers $ (x,y)$ satisfy equation: $ x^3\\minus{}y^3\\equal{}2xy\\plus{}8$", "Solution_1": "Multiply through by 27 and use $ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{}3abc \\equal{} (a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2 \\minus{}ab\\minus{}bc\\minus{}ca)$", "Solution_2": "[quote=\"KevinB\"]Multiply through by 27 and use $ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca)$[/quote]\r\n\r\nI don't suppose the result from multiplying by 27 would yield the form $ a^3\\plus{}b^3\\plus{}c^3\\minus{}3abc$...", "Solution_3": "$ x^3 \\minus{} y^3 \\equal{} 2xy \\plus{} 8 \\implies x^3 \\minus{} y^3 \\minus{} 1 \\minus{} 2xy \\equal{} 7 \\implies (3x)^3 \\plus{} ( \\minus{} 3y)^3 \\plus{} ( \\minus{} 3)^3 \\minus{} 2(3x)( \\minus{} 3y)( \\minus{} 3) \\equal{} 7\\cdot3^3$", "Solution_4": "[quote=\"KevinB\"]$ x^3 \\minus{} y^3 \\equal{} 2xy \\plus{} 8 \\implies x^3 \\minus{} y^3 \\minus{} 1 \\minus{} 2xy \\equal{} 7 \\implies (3x)^3 \\plus{} ( \\minus{} 3y)^3 \\plus{} ( \\minus{} 3)^3 \\minus{} 2(3x)( \\minus{} 3y)( \\minus{} 3) \\equal{} 7\\cdot3^3$[/quote] Sorry ignore this, its wrong.", "Solution_5": "Seems like you can limit the range of values for x,y by analyzing the difference of cubes on the left with the expression on the right.\r\n\r\nI.e. Consider the case when the cubes on the left are consecutive, x=y+1\r\n(y+1)^3-y^3 = 2y(y+1)+8\r\n\r\nThis has roots around y=-3.1926 and y=2.1926 which suggests for y>2.192 and y<-3.1926 the left side is greater than the right.\r\n\r\nWith that knowledge, you should only have to test y=2,1,0,-1,-2,-3 and evaluate in the original equation if x can be integer.\r\n\r\nThis occurs for \r\ny=0, x=2\r\ny=-2, x=0" } { "Tag": [ "geometry", "circumcircle", "incenter" ], "Problem": "I am finding many probems that are using some geometric terminology I do not yet know, especially related to triangles. So I'm deciding to start a thread asking about the definition of these terms, such as: orthocenter, circumcenter, centroid, incenter,... there might be others I don't know.", "Solution_1": "Here are some of the common ones:\r\n\r\n$\\text{centroid}$ - $\\text{where the three medians of a }\\triangle\\text{ meet}$\r\n$\\text{circumcenter}$ - $\\text{ where the three }\\perp\\text{ bisectors of a }\\triangle\\text{ meet}$\r\n$\\text{incenter}$ - $\\text{ where the three }\\angle\\text{ bisectors of a }\\triangle\\text{ meet}$\r\n$\\text{orthocenter}$ - $\\text{ where the three altitudes of a }\\triangle\\text{ meet}$", "Solution_2": "Note that the orthocenter and the circumcenter may occur outside of the triangle.", "Solution_3": "Also note that the centroid is sometimes referred to the 'center of mass', I believe. \r\n\r\nThe incenter is also the center of the inscribed circle; similarly, the circumcenter is the center of the circumscribed circle.", "Solution_4": "Does the perpendicular referred to in the definition of the circumcenter have to come from the opposite vertex.", "Solution_5": "[quote=\"mna851\"]Does the perpendicular referred to in the definition of the circumcenter have to come from the opposite vertex.[/quote]\r\nNo." } { "Tag": [ "probability", "function", "probability and stats" ], "Problem": "Suppose someone put popcorn in a microwave.\r\nThe minimum time to a popcorn explode is $ a$ and the maximum time is $ b$. This can be found empirically.\r\n\r\nI wanted to build a density of probability function for this model. After $ t$ times what is the probability of a popcorn explode?\r\n\r\nThere is $ n$ ways to make this function. I thought of one that I think is simple.\r\n\r\nYou put the popcorn in the microwave and record the sound. :)\r\n\r\nBut the next step; how can I make this function knowing only the Sound?\r\n\r\nI wanted to build a simple program to compute this function empirically. So this program computes the maximums of intensity of the sound and return the time of this peak.\r\n\r\nAnd in the final I have the time of each explosion (to make it by hand would be problematical and imprecise). This is simple.\r\n\r\nKnowing this final result, we can say that this is a function. At each time it return 1 for explosion and 0 for non-explosion.\r\n\r\nBut the Time is continuous but the number of explosions is discrete. This is my problem.\r\n\r\nIf someone has another method to find this density of probability, I would be happy to learn.", "Solution_1": "you should google \"maximum likelihood\". This is a usual way to deal with such kind of problems, once you know a little more what kind of density you are looking for.", "Solution_2": "Er... I saind I wanted to \"convert\" the sound in probability and asked a method to do this.\r\n\r\nAnd I asked how to make a simple program that do that.\r\n\r\nI had a idea of how to do this, but is fairly complicated.\r\n\r\nOn internet I doubt that there is something like what I wanted.\r\n\r\nIf I had the DATA I wouldn't be asking here." } { "Tag": [ "trigonometry" ], "Problem": "[b]If $ A,B,C$ are the angles of any triangle, prove that \n$ 3(\\sin^2A\\plus{}\\sin^2B\\plus{}\\sin^2C)\\minus{}2(\\cos^3A\\plus{}\\cos^3B\\plus{}\\cos^3C)\\leq 6$[/b]", "Solution_1": "can anybody provide a hint or start?", "Solution_2": "^The LHS will be maximized when $ A\\equal{}B\\equal{}C$?" } { "Tag": [ "calculus", "integration", "function", "trigonometry", "calculus computations" ], "Problem": "$ g(x)$ is a differentiable function for $ 0\\leq x\\leq \\pi$ and $ g'(x)$ is a continuous function for $ 0\\leq x\\leq \\pi$.\r\nLet $ f(x) \\equal{} g(x)\\sin x.$ Find $ g(x)$ such that $ \\int_0^{\\pi} \\{f(x)\\}^2dx \\equal{} \\int_0^{\\pi}\\{f'(x)\\}^2dx$.", "Solution_1": "First:\r\n$ \\int_{0}^{\\pi}{\\{f(x)\\}^2 dx}\\equal{}\\int_{0}^{\\pi}{g^2 (x) \\sin^2(x) dx}\\equal{}\\frac{\\int_{0}^{\\pi}{\\sin^2 (x)dx}}{\\pi \\minus{} 0}\\int_{0}^{\\pi}{g^2 (x) dx}\\equal{}$\r\n$ \\equal{}\\frac{1}{2}\\int_{0}^{\\pi}{g^2 (x) dx}$\r\nSecond:\r\n$ \\int_{0}^{\\pi}{\\{f'(x)\\}^2 dx}\\equal{}\\int_{0}^{\\pi}{\\left(g'(x)^2 \\sin^2 (x) \\plus{} g(x)g'(x)\\sin(2x) \\plus{} g(x)^2 \\cos^2 (x)\\right) dx}\\equal{}$\r\n$ \\equal{}\\frac{1}{2}\\int_{0}^{\\pi}{g'(x)^2 dx}\\plus{}0\\plus{}\\frac{1}{2}\\int_{0}^{\\pi}{g^2 (x) dx}$\r\nSo, we will have:\r\n$ \\frac{1}{2}\\int_{0}^{\\pi}{g'(x)^2 dx}\\equal{}0$\r\nThis means that: $ g(x)\\equal{}c$, where $ c$ is a constant.", "Solution_2": "Your answer is correct, Yuriy Solovyov. :)" } { "Tag": [], "Problem": "Well, as the thread's title says, I'm planning to make the sixth installation of The Mole in AoPS!\r\nAlthough my first attempt failed, I think that this time I'll make it better. I would like to rather have a co/backup/replacement mod, given that I'll be insanely busy due to university... offers to the charge can be PM'd to me.\r\nAbout the setup... I've already made about what, 20% of the setup. So, if I'm lucky (and university isn't so much work, and I get a good backup mod), the plot will be complete before end of January, signups would start at early or mid-february, and the game would be underway around March.\r\nSo, comments come here, and co/backup mod offers go to my PM box. \r\n\r\n[size=75]If any mod thinks this is spam, they can lock the thread if wished... this is just for information, and as I thought this would rather belong to FF, I posted here.[/size]", "Solution_1": "Judging by my game, the mole has become relatively unpopular now.", "Solution_2": "Well, maybe given that the inmediately previous game imploded due to me losing internet access, that wasn't much popular.\r\nAnd, if I'm not wrong, the Mole 5 exploded due to [i]you[/i] being inactive :wink:\r\nAnd, after all, I posted this in order to get a co-mod (in case I don't get computer acces on time, or in case the university makes my brain explode due to excessive homework >.<)", "Solution_3": "I'll play, but co-modding is out of my time... :\\", "Solution_4": "I'll play, with a possibility of co-modding.", "Solution_5": "I'll play, but no co-modding for me...", "Solution_6": "I'd like in.\r\nNo co-modding for me.\r\nI'm lazy at the moment.", "Solution_7": "I want to play.\r\n\r\nNo co-modding for me.", "Solution_8": "is pacman still checking this?", "Solution_9": "[quote=\"pacman2812\"][b] I would like to rather have a co/backup/replacement mod, given that I'll be insanely busy due to university... offers to the charge can be PM'd to me.\n (and university isn't so much work,if I get a good backup mod), the plot will be complete before end of January, signups would start at early or mid-february, and the game would be underway around March.\nSo, comments come here, and co/backup mod offers go to my PM box. [/b][/quote]", "Solution_10": "ah, right.\r\nI should read more often.", "Solution_11": "I'll play, but can't mod due to creating a game that's also The Mole plus a bunch of other elements :O", "Solution_12": "I'll play, but please let it have an end :)", "Solution_13": "[quote=\"pacman2812\"]Well, maybe given that the inmediately previous game imploded due to me losing internet access, that wasn't much popular.\nAnd, if I'm not wrong, the Mole 5 exploded due to [i]you[/i] being inactive :wink:\nAnd, after all, I posted this in order to get a co-mod (in case I don't get computer acces on time, or in case the university makes my brain explode due to excessive homework >.<)[/quote]\r\n\r\nME being inactive? I tried to restart it, prodded everyone at least 6 times, and waited for something good to happen. Which it didn't.\r\n\r\nMaybe I can co-mod, I don't know yet.", "Solution_14": "well, it seems that I'll be OK with access from February-...\r\nThe problem is, the plot I have is HUGE, and the game would end around October or November :what?: :wacko:", "Solution_15": "whoa.", "Solution_16": "can I join? and how do you play?", "Solution_17": "@musicnmath One way to find the rules would be to wait till the game starts, another is to study a previous game 2-3 game pages back.", "Solution_18": "Plot status: 55% complete.\r\nEstimated start for signups: February 1.\r\n\r\n\r\nNote: the signups are made by applications, via PM. At the date above, I'll create a thread where all the regulations are exposed, then you can start signing up.\r\nAnyone interested should look at the prior The Mole games (preferred games 1-3, as 4-5 were abandoned)", "Solution_19": "I shall play.\r\n\r\nMwahahaha.", "Solution_20": "Finally, a Mole game run by someone who will be active and knows how to make real games without throwing in random dumb twists b/c they got tired of modding!", "Solution_21": "Now how far are you into the storyline?\r\n\r\n2 weeks ago=55%\r\n\r\njust wondering.....", "Solution_22": "Around 90%. Only one task (the final task), details on other tasks, some quizzes and some flavor to go.\r\nWhenever I get a computer (currently from internet cafe), I'll start the applications.\r\nI'm still accepting offers for co-mod." } { "Tag": [ "number theory", "greatest common divisor" ], "Problem": "For distinct natural numbers $a,b$, prove that $gcd(a,b) \\leq \\frac{a+b}{3}$", "Solution_1": "Ooh...I'm not good with these kind of problems...can someone please help me with these.... :blush:", "Solution_2": "We know that if $a>b$ then the gcd. is not more than $\\frac{a}{2}$", "Solution_3": "I sort of have an idea, but I'm not sure if it's rigorous enough. It follows what you said. I'm sorry if it's too wordy or if it's wrong. I'm not really good at proofs\r\n\r\n[hide]\nI thought that since the numbers are distinct, we can assume without loss of generality that a>b. The maximum the gcd could possibly be would then be b. But any common denominator would be a factor of both numbers, so if b is a factor of a then so is a/b. Since factors must be integers, the smallest a/b can be is 2 (1 wouldn't work because then a=b), which would give us the maximum value of b as a/2. So gcd(a,b) has to be less than or equal to a/2, or (a+b)/3.\n[/hide]\r\n\r\nEwww.... I definitely need latex.", "Solution_4": "hopefully i got it...\r\n[hide]we can let $g=\\gcd(a,b)$ and so we can write $a=m*g$ and $b=n*g$.\nim going to pick $a3$ since the terms tend to $\\infty$. The only question that remains is whether the series converges or diverges at each point $x$ with $|x|=3$.\r\n\r\nIf $x$ is a real variable, the only points to check are $x=3$ and $x=-3$. When $x=3$, the series becomes $\\sum_{n=1}^\\infty \\frac1{3n}$, which is a multiple of the harmonic series and divergent. When $x=-3$, the series becomes $\\sum_{n=1}^\\infty \\frac{(-1)^{n-1}}{3n}$ which is a conditionally convergent alternating series.\r\n\r\nIf $x$ is a complex variable, we need more complicated methods to deal with the rest of the circle. The series does converge for all $x$ with $|x|=3$ except $3$ itself, and we can calculate the exact values.", "Solution_4": "ok\u00e9 I see but for calculating the convergentia area I use always x=1 is this ok\u00e9? why can I don't use for example 2 or 4 ect...?\r\n\r\n[quote](Use \\infty for that upper limit) \n[/quote]\r\n\r\nWhat is the different?\r\n\r\nGreets thanks a lot.", "Solution_5": "the difference is that $\\propto$ is a completely different sign than $\\infty$, it's like using $\\lhd$ for $<$ because it looks like it...\r\n\r\nBut I don't understand your question on $|x|=1$... on what mathematical grounds would you [i]choose[/i] the convergence radius? :? Usually it has to follow from the problem, and you don't have the freedom to 'take' it $|x|=1$ or something.", "Solution_6": "[quote]But I don't understand your question on ... on what mathematical grounds would you choose the convergence radius? Usually it has to follow from the problem, and you don't have the freedom to 'take' it or something.\n[/quote]\r\n\r\nI know but the question is just calculate the convergence radius I don't know who to do that if I use x=1 then I find the rigth answer but I don't know why I need to choice x=1?\r\n\r\nGreets.", "Solution_7": "I don't understand what you mean by \"if I use x=1 then I find the rigth answer\"? :?", "Solution_8": "first of all I need to find the convergence radius but I don't know who so I go look into my solutions in the back of my book and then I see that I do the next calculation that I get 3 as the answer.\r\n\r\nSo $\\sum^{%Error. \"intify\" is a bad command.\n}_{n=1}\\frac{x^{n-1}}{n3^n}$\r\n\r\nthen I put 1 in my exprsession so $\\sum^{%Error. \"intify\" is a bad command.\n}_{n=1}\\frac{1^{n-1}}{ n3^n}$ and if n become infinite then we get ooh sorry I can't get the right value\r\n\r\nBut can someone tell me who I need to do that? Greets.", "Solution_9": "In fact, I still can't follow you... what's the problem with jmerry's solution?", "Solution_10": "[quote]In fact, I still can't follow you... what's the problem with jmerry's solution?\n[/quote]\r\n\r\nNothing but I can not find the convergention radius on a correct way that's the problem.", "Solution_11": "Let $S_n(p)=\\sum_{k=1}^n \\frac{1}{k\\cdot p^k}\\ (p>2),$ we have $\\lim_{n\\to\\infty} S_n(p)=\\ln \\frac{p}{p-1}.$\r\n\r\nHere is my proof, but this may be a famous.\r\n\r\nFor $x\\neq 1,$ let $\\ f_n(x)=\\sum_{k=1}^n x^{k-1},$ we have $\\frac{1}{1-x}-f_n(x)=\\frac{x^n}{1-x}.$ Thus for $0\\leq x\\leq \\frac{1}{p}\\ (p>2),$ we have $1\\leq \\frac{1}{1-x}\\leq \\frac{p}{p-1},$ yielding\r\n$0\\leq \\frac{x^n}{1-x}\\leq \\frac{p}{p-1}x^n.$ Integrate with respect to $x$ for the interval $0\\leq x\\leq \\frac{1}{p},$ we have $0<\\int_0^{\\frac{1}{p}} \\left\\{\\frac{1}{1-x}-f_n(x)\\right\\}dx<\\frac{p}{p-1}\\int_0^{\\frac{1}{p}} x^ndx.$\r\n\r\n$\\therefore 0<\\ln \\frac{p}{p-1}-\\sum_{k=1}^n \\frac{1}{k\\cdot p^k}<\\frac{p}{p-1}\\cdot \\frac{1}{(n+1)p^{n+1}}.$\r\n\r\nLet $n$ approach infinity, $\\lim_{n\\to\\infty} \\left(\\ln \\frac{p}{p-1}-\\sum_{k=1}^n \\frac{1}{k\\cdot p^k}\\right)=0.$\r\n\r\nTherefore $\\sum_{n=1}^{\\infty} \\frac{1}{n\\cdot p^n}=\\ln \\frac{p}{p-1}.$", "Solution_12": "briljant proof but can I stil find the convergention radius?", "Solution_13": "[quote=\"Bert\"]Hello,\n\nGiven is $\\sum^{\\propto}_{n=1}\\frac{x^{n-1}}{n3^n}$ I can find that the convergention region is 3 but what can I say about the points 3 and -3 ?\n\nGreets.[/quote]\r\n\r\n You can try ratio test and find that the radius of convergence is 3. About the end point, jmeery has answered you already.\r\n Ratio test stated that : If $lim_{n \\to \\infty } abs ({\\frac{a_{n+1}}{a_n})=L <1}$ then the series $\\sum_{n=1}^{\\infty } a_n$ is absolutely converge and therefore it is converge.\r\n\r\n BTW, Kunny sols for $x=1$ is very nice.", "Solution_14": "In fact, it can be done much easier than Kunny's proof...\r\n\r\nTermswise differentiation gives $\\sum -\\frac{1}{p^{k+1}} = -\\frac{1}{p(p-1)}$, integrating again gives $\\ln(p) - \\ln(p-1)$, which is the same. ;)\r\n\r\nJust as trivial we have that the original sum is $x^{n-1} \\ln\\left(\\frac{p}{p-1}\\right)$.", "Solution_15": "About finding that radius of convergence:\r\n\r\nThe ratio test and/or the root test compare the series to a geometric series. In practice, they are used to find what it would take to balance out everything that grows or shrinks geometrically fast or faster, while rendering invisible everything that grows or shrinks slower than geometrically fast.\r\n\r\nThe powers of the variable in a power series are always geometric, which is why the root and/or ratio tests are always in order. On the other hand, once you're used to this, there are many cases (including this one) in which you can determine the radius of convergence \"by inspection.\"\r\n\r\nLook at what you have: you have $x^{n-1},$ which is geometric, but whose exact behavior depends on $x.$ You have $\\frac1{3^n}$ which is geometric - and you can see that setting $|x|=3$ would exactly balance it out. And you have $\\frac1n$ which shrinks and a slower-than-geometric race, and hence has no influence on the radius of convergence. If you go ahead and use the root or ratio test, then the $\\frac1n$ will contribute a factor of $1$ to the limiting root or limiting ratio - that's what I meant above by being \"rendered invisible.\" So, for the radius of convergence, once we see that $|x|=3$ will balance the geometric things we have and there's nothing else that matters, then the radius of convergence has to be $3.$\r\n\r\nDetermining the radius of convergence is quite different than determining the behavior at the boundary points. If $|x|=3$ then all the geomtric things balance out perfectly (at least in size) and all that's left is that \"slower\" object, the $\\frac1n,$ along with whatever $\\pm$ signs we have. Follow up on that thinking and you have jmerry's first post in this thread.\r\n\r\n[Note: the limsup version of the root test is theoretically superior to any version of the ratio test - you will see it used to give a formula for the radius of convergence. That doesn't matter in this problem.]", "Solution_16": "thank you for you explanation." } { "Tag": [ "vector", "advanced fields", "advanced fields unsolved" ], "Problem": "Hello folks! Here another one that I could not handle!\r\nThankl for any help!\r\n\r\nLet $E$ and $F$ normed vector spaces. Suppose that $T : E \\to F$ is a closed linear mapping :\r\n\r\na) Let $K$ be a compact subset of $E$. SHow that $T(K)$ is a closed subset of $F$.\r\n\r\nb) If $K$ is compact subset of $F$, show that $T^-{1}(K)$ (the inverse image) is a closed subset of $E$", "Solution_1": "a) Assume $(T(x_{n}))$ is a sequence of $T(K)$ converging to a point $e \\in F$. By compacity of $K$, there is a subsequence $(x'_{n})$ of $(x_{n})$ which converges to a point $x' \\in K$. But then $(x'_{n},T(x'_{n}))$ is converging to $(x',e)$, and since the graph of $T$ is closed, $e=T(x') \\in T(K)$. This proves that $T(K)$ is closed in $F$.\r\n\r\nb) Assume $(x_{n})$ is a sequence of $T^{-1}(K)$ converging to $x \\in E$. Then $(T(x_{n}))$ is a sequence of $K$ from which you can extract a subsequence $(T(x'_{n}))$ converging to $y \\in K$. Now since $T$ is closed, $(x'_{n},T(x'_{n}))$ converges to $(x,y)$ and $y=T(x)$, meaning $x \\in T^{-1}(K)$. Whence $T^{-1}(K)$ is closed.", "Solution_2": "Thank you Julien!" } { "Tag": [ "quadratics", "LaTeX", "algebra", "quadratic formula", "calculus", "calculus computations" ], "Problem": "I have problems finding the limits of this fraction:\r\n\r\n$f(x)=\\frac{x^2+3x}{x+4}$\r\n\r\n$x=\\frac{-b+/-\\sqrt{b^2-4ac}}{2a}$\r\n\r\n$x=0;-3$\r\n\r\n$f(x)=\\frac{(x+0)(x+3)}{x+4}$\r\n\r\n$f(x)=\\frac{x(x+3)}{x+4}$\r\n\r\nBut my calculator doesn't agree. What am i doing wrong?\r\n\r\n[u]EDIT:[/u] My calculator says there's one more around -3,9. How do i find that one?\r\n\r\n(For my own usage: 322,1)", "Solution_1": "what do you mean with\r\n\r\nEDIT: My calculator says there's one more around -3,9. How do i find that one? \r\n\r\nAnyhow \r\n \r\nif x->-4(-) [respectively x->-4(+)] then f->-inf [respectively +inf]\r\n\r\nif x -> +inf [-inf] then f ->+ inf [-inf]", "Solution_2": "Well, on my calculator it looked like it was $-3,9$, but i figured it was exactly $-4$.\r\n\r\nOk, so i have to take each factor i get:\r\n\r\n$(x+0);(x+3);(x+4)$\r\n\r\nand from there i can find the limits..\r\n\r\nOf course, i knew that. How am i stupid :lol: \r\n\r\nWelcome to the forum BTW", "Solution_3": "And here are two other ones. I only need help to find the full fraction from which i can do the factorizing.\r\n\r\n1) $\\frac{1}{x}+\\frac{5}{x-2}=0$\r\n\r\n2) $\\frac{x+6}{x+1}=2$\r\n\r\nI've read this earlier:\r\n\r\n$f(x)=x+\\frac{x-4}{x+2}$\r\n$f(x)=\\frac{x(x+2)}{x+2}+\\frac{x-4}{x+2}$\r\n$f(x)=\\frac{x^2+3x-4}{x+2}$\r\n\r\nI can follow the direction of the book, but now that we have a fraction + another fraction i get confused.\r\n\r\nThx in advance. BTW, thx to you, Pic, with the first problem.", "Solution_4": "[quote=\"Sporally\"]I have problems finding the limits of this fraction:\n\n$f(x)=\\frac{x^2+3x}{x+4}$\n\n$x=\\frac{-b+/-\\sqrt{b^2-4ac}}{2a}$\n\n$x=0;-3$[/quote]\r\n\r\nHehe, you don't need the quadratic formula there :) Just take out a common factor of $x$ from $x^2+3x$, and you get $(x)(x+3)$, which is the same thing you had, but a thousand times easier. As my math teacher tells me almost daily, \"Common factoring is the #1 skill in math.\"", "Solution_5": "Ah, of course. I have done that several times now, i don't know why i didn't do it this time though. Can you help me with my number 1 and 2 in the last post?", "Solution_6": "[quote=\"Sporally\"]And here are two other ones. I only need help to find the full fraction from which i can do the factorizing.\n\n1) $\\frac{1}{x}+\\frac{5}{x-2}=0$\n\n2) $\\frac{x+6}{x+1}=2$[/quote]\r\nBasically, you need to take the common denominator of the two fractions, just like when you calculate $\\frac{1}{2}+\\frac{1}{3}$.\r\n\r\n[hide=\"Number 1\"]\n$\\frac{1}{x}+\\frac{5}{x-2}=0$\n\n$\\frac{x-2}{x(x-2)}+\\frac{5x}{x(x-2)}=0$\n\n$\\frac{x-2+5x}{x(x-2)}=0$\n\n$\\frac{6x-2}{x(x-2)}=0$[/hide]\r\n\r\nFor #2, move 2 (on the right side) to the left side of the equation and then take the common denominator.", "Solution_7": "I have a suggestion: Every time you don't know how to do a problem, try manipulating the equation a little and see what you can come up with. Just because it's not exactly like what you've seen before doesn't mean you can't figure it out.", "Solution_8": "I figured i had to find a common denominator, it was more like i didn't know how to get it like that. But it seems like i can do it now, thx to you. Regarding nr. 2 i knew that i could move 2 over to the other side as -2, but didn't know what to do from there.\r\n\r\nI usally try and manipulating with the fractions, but just couldn't find a way that my calculator would show the graph the same way everytime.", "Solution_9": "BTW: Regarind LaTex, how do i make the +/- symbol so that i can write\r\n\r\n$x=\\frac{-b+/-\\sqrt{b^2-4ac}}{2a}$\r\n\r\nproperly?", "Solution_10": "Use \\pm ;)", "Solution_11": "Yeah, got the same info from the LaTex-subforum. Found it just a little after i've posted the last post in this thread. But thx for the info anyway.\r\n\r\n\r\n\r\nAnd just as you think you understand it all in this chapter you encounter another problem.\r\n\r\nOk, i have this fraction which i first of rewrite, from which i must find the numbers that can not be used (can't remember the english word for it, i hope you get it anyway once you see the fraction):\r\n\r\n$\\frac{1}{x}+\\frac{5}{x-2}=0$\r\n\r\n$\\frac{6x-2}{x(x-2}=0$\r\n\r\nIf i want to find the numbers that i can't used for this fraction i have to find the numbers for x that makes the denominator equal 0 as you can't divide with 0. But on the other hand you used the denominator when you want to find the limits of the fraction, where you will see what numbers makes either the nominator and the denominator equal 0.\r\n\r\n$x=0$\r\n\r\nand\r\n\r\n$x-2=0$\r\n\r\n$x=2$\r\n\r\nSo these numbers are both used as the limits for the denominator and the numbers which can't be used due to the law that you can't divide by 0.\r\n\r\nI believe i have misunderstood something about the law of dividing with 0, so can anyone guide me, plz?" } { "Tag": [], "Problem": "From my homework, I have no idea:\r\n\r\nBoron occurs naturally as 10B and 11B. Which isotope is most abundant? (Hint: Refer to the Periodic Table.) \r\nA) boron-5 \tB) boron-6 \tC) boron-10 \tD) boron-11 \tE) none of the above", "Solution_1": "[quote=\"Gibbenergy\"](Hint: Refer to the Periodic Table.) [/quote]\r\n\r\nMight not be a bad idea.\r\n\r\nLook up the atomic weight of Boron. Is it closer to $10$ or $11$?" } { "Tag": [ "function", "calculus", "logarithms", "derivative" ], "Problem": "I'm not sure if this problem is supposed to be stated in this subforum, but here it goes:\r\n\r\nAs you probably all know, exponential functions of the form $ a^{x}$ (whereby $ a$ is a constant), all have inverse functions of the form $ log_{a}(x)$. When graphed, it becomes evident that some of these functions and their inverses (those with large values of $ a$) don't intersect with one another, while others do so twice (with smaller $ a$ values);\r\n\r\nTry getting a graphics calculator and graphing $ e^{x}$ and $ ln(x)$. It's evident that these don't intersect, and if you graph something like $ 1.1^{x}$ and $ log_{1.1}(x)$, you can see that they do twice.\r\n\r\nTherefore, there must be a value for $ a$ in $ a^{x}$ which only intersects once with its inverse, $ log_{a}(x)$ . The problem is, what is it?", "Solution_1": "[hide=\"a little calculus\"]\nAt any place where the graphs of $ a^{x}$ and $ \\log_{a}x$ intersect, $ x$ does too. So we want the value of $ a$ such that $ a^{x}=x$ has a unique solution.\nNotice that if this is the case, then at the solution $ x$ to this equation, the derivative wrt $ x$ of $ a^{x}$ must be $ 1$. The derivative wrt $ x$ of this function is $ a^{x}\\ln{a}$. So we have\n\n\\[ a^{x}=x\\]\n\\[ a^{x}\\ln{a}=1\\]\n\nOkay, so clearly $ x=\\frac{1}{\\ln{a}}$. So then\n\n$ a^{\\frac{1}{\\ln{a}}}=\\frac{1}{\\ln{a}}$.\nBut $ a^{\\frac{1}{\\ln{a}}}=e^{\\frac{\\ln{a}}{\\ln{a}}}=e$\n$ \\implies e=\\frac{1}{\\ln{a}}\\implies a=\\boxed{e^{\\frac{1}{e}}}$[/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find every pair of positive real numbes $ (\\alpha,\\beta)$ such that for all sequence $ (u_n)$ which satisfies $ \\exists K: |u_n|\\leg K,\\forall n$ and $ \\alpha u_{n + 1} + \\beta u_n \\geq u_{n - 1}$ then $ (u_n)$ converges.", "Solution_1": "[quote]Find every pair of positive real numbes such that for all sequence which satisfies $ \\exists K: |u_n|\\leg K,\\forall n$ and $ \\alpha u_{n + 1} + \\beta u_n \\geq u_{n - 1}$ then converges.\n[/quote]\r\nDo you mean $ K > 0$ and $ |u_n|\\le K$?", "Solution_2": "If $ \\alpha +\\beta \\ge 1$ exist sequence ${ u_{3n}=1,u_{3n+1}=a,u_{3n+2}=b, \\{a,b\\}\\not=\\{1,1}$, tytrefore nonconverged, satisfyed $ \\alpha b +\\beta a\\ge 1,\\alpha a+\\beta \\ge b, \\alpha +\\beta b\\ge a$.\r\nIf $ \\alpha +\\beta <1$, then $ \\alpha (u_{n+1}-u_n)+(\\alpha+\\beta )(u_n-u_{n-1})\\ge (1-\\alpha +\\beta )u_{n-1}$ give, that $ u_n$ converged." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ \\bigstar$ [b]Problem:[/b]\r\nLet $ a,b,c,d$ be positive real numbers. Prove that:\r\n\\[ \\frac {a^3}{b^3 \\plus{} a^2b} \\plus{} \\frac {b^3}{c^3 \\plus{} b^2c} \\plus{} \\frac {c^3}{d^3 \\plus{} c^2d} \\plus{} \\frac {d^3}{a^3 \\plus{} d^2a}\\ge 2\r\n\\]", "Solution_1": "[quote=\"zaizai-hoang\"]$ \\bigstar$ [b]Problem:[/b]\nLet $ a,b,c,d$ be positive real numbers. Prove that:\n\\[ \\frac {a^3}{b^3 \\plus{} a^2b} \\plus{} \\frac {b^3}{c^3 \\plus{} b^2c} \\plus{} \\frac {c^3}{d^3 \\plus{} c^2d} \\plus{} \\frac {d^3}{a^3 \\plus{} d^2a}\\ge 2\n\\]\n[/quote]\r\n\r\nPut $ x\\equal{}\\frac ba$ etc. We have to prove $ \\sum_{cyc}\\frac1{x^3\\plus{}x}\\ge2$ for $ xyzt\\equal{}1$.\r\n\r\nNow, $ \\frac1{u^6\\plus{}u^2}\\plus{}\\frac1{v^6\\plus{}v^2}\\ge\\frac2{u^3v^3\\plus{}uv}$ (easy by AM-GM), so we can prove it by mixing variables. :lol:", "Solution_2": "Can you explain clearly ? :) The result of n-variables is also correct :D", "Solution_3": "What is not clear? My proof applies to $ n$ variables, too. :)", "Solution_4": "$ LHS\\equal{}\\sum\\frac{\\frac{a}{b}(a^2b\\plus{}b^3)\\minus{}ab^2}{b^3\\plus{}a^2b}\\equal{}\\sum\\frac{a}{b}\\minus{}\\sum\\frac{ab^2}{b^3\\plus{}a^2b}\\geq 2$\r\nI think this proof can solve n-variables :)" } { "Tag": [ "geometry", "inequalities unsolved", "inequalities" ], "Problem": ":oops: a,b,c,A,B,C are nonnegative numbers satisfying a+A=b+B=c+C=k\r\nSo prove that aB+bC+cA$ \\le k^2$", "Solution_1": "brian_gold, why do you need to post the same problem again?\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=219077[/url]", "Solution_2": "great,kunny!but use geometry is much nicer and I think have a ineq tronger than bri_an gold 's problem!\r\nI'm trying it! :blush:" } { "Tag": [], "Problem": "I'm a college freshman right now, trying to figure out what to do this summer. I've identified some REU's (NSF summer research programs) that I plan to apply for, but the problem is that many of them are geared towards more advanced students (sophomores or juniors), so I definitely need a backup plan. I am several years advanced in math courses, so it's not prereqs holding me back so much as \r\n\r\nLong term, I'm basically interested in pure mathematics, most likely as a research mathematician. Barring that, I basically want to be thinking and solving problems. So short term I'm looking for something that will both allow me to solve problems, keep me from doing nothing all summer, and maybe give me some useful experience and money on the side.\r\n\r\nWhere would be a good place to look for interesting internships or other opportunities? I imagine I will need to start applying very soon.", "Solution_1": "I'm in a similar situation and would appreciate any advice. Like eugevir, the internships I've seen are geared towards older students (sophomores, juniors), but I would like to get some experience this year.", "Solution_2": "REUs (NSF or otherwise) generally don't mind if you're a frosh as long as you've had the prereqs. Don't let that hold you back---apply to at least 6 - 8 programs, though, to increase your chances of getting in. (I'm saying that because my students have recently been getting into one or two of the 6 - 8 they apply for... they're often alternates at a couple other REUs.)\r\n\r\nI worked for a professor the summer after my first year of college, doing very low-level research. (I looked for papers in the library and ran data analyses against some of their theories.) You might ask around to see if anyone at your institution has that kind of funding." } { "Tag": [], "Problem": "What number should be added to both the numerator and the denominator of $ \\frac{1}{5}$ to get a fraction equivalent to $ \\frac{4}{5}$?", "Solution_1": "$ \\frac{4}{5}\\equal{}\\frac{8}{10},\\frac{12}{15},\\frac{16}{20}$\r\n\r\n$ \\frac{16\\minus{}1}{20\\minus{}5}\\equal{}\\frac{15}{15}$\r\n\r\nIf you add 15 to the numerator and denominator of $ \\frac{1}{5},$ you get $ \\frac{16}{20}\\equal{}\\frac{4}{5}$", "Solution_2": "instead of guessing and checking, which may take a while if the numbers were larger, we use an equation. Let $ x\\equal{}\\text{the number added}.$ We have \\[ \\frac{1\\plus{}x}{5\\plus{}x}\\equal{}\\frac{4}{5}\\implies 5x\\plus{}5\\equal{}4x\\plus{}20\\implies x\\equal{}\\boxed{15}\\]" } { "Tag": [], "Problem": "how does one thermodynamically define freezing point and in particu;ar why is there 'depression of freezing point' in a 'impure' solution", "Solution_1": "In the freezing point of a pure solvent, the rate at which the solvent molecules stick together to form the solid is equal to the rate at which solvent molecules come out of the solid to return to the liquid phase. However, when a solute is present (volatile or not) its molecules \"get in the way\" of the solvent molecules and so there are fewer solvent molecules in contact with the solid solvent. However, the rate at which solvent molecules leave the solid to go into the solution is unaffected, and so there is a net flow of solvent molecules from the solid to the liquid, and solidification in therefore not possible. Only if the temperature is lowered, in order for the solid -> liquid flow of solvent molecules to be slowed, will there be a net flow of molecules from the liquid to the solid and so solidification can now occur.", "Solution_2": "thanks :lol:" } { "Tag": [], "Problem": "Triangle $ ABC$ is isosceles with angle $ A$ congruent to angle $ B$.\nThe measure of angle $ C$ is 30 degrees more than the measure of\nangle $ A$. What is the number of degrees in the measure of angle $ C$?", "Solution_1": "call angle A x\r\n\r\nthen angle B is also x\r\n\r\nand angle C is x+30\r\n\r\nsince they must be equal to 180;\r\n\r\nx+x+x+30=180\r\n\r\n3x+30=180\r\n\r\n3x=150\r\n\r\nx=50; angle C=50+30=80" } { "Tag": [ "geometry", "circumcircle", "power of a point", "radical axis", "geometry proposed" ], "Problem": "11.4 Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$. \r\n([i]L. Emel'yanov[/i])", "Solution_1": "Denote $H'$ intersection of $CH$ and $A'B'$, $O'$ intersection of $CO$ and $A_1B_1$\r\n\r\nIt is well known that $ CH'\\perp A'B'$ and $CO'\\perp A_1B_1$. So $CC'O'H'$ is cyclic.\r\n\r\nConsider inversion with th pole $C$ and power $k = CA'*CA_1=CB'*CB_1$.\r\n\r\nLine $A'H'B'$ transforms in $(A_1HB_1C)$\r\nLine $A_1O'B_1$ transforms in $(A'OB'C)$\r\nIt can easily to see $I(O')= O, I(H')= H$\r\n\r\nSo circle $(CC'O'H')$ transforms in line $OH$ which is perpendicular to diameter $(CC'O'H')$ - $CC'$. \r\n\r\nSo $OH\\perp CC'$.", "Solution_2": "Here's another solution:\r\n\r\nSince $A_1A'B'B_1$ is cyclic, $C'A'\\cdot C'B'=C'A_1\\cdot C'B_1$, so $C'$ lies on the radical axis of the circles $(CA_1B_1),(CA'B')$. This means that $CC'$ is perpendicular to the line of centers of these two circles (call this $(*)$). However, the center of $(CA_1B_1)$ is the midpoint of $CH$, while the center of $(CA'B')$ is the midpoint of $OH$, so $OH$ is parallel to the line of centers, and thus perpendicular to $CC'$, according to $(*)$.", "Solution_3": "The official solution is the same as grobber's one.\r\nIt is quiate unusual, since they needed to present a notion of radical axe which is out of school education.", "Solution_4": "[quote=\"Myth\"]The official solution is the same as grobber's one.\nIt is quiate unusual, since they needed to present a notion of radical axe which is out of school education.[/quote]\r\nMaybe L. Emel'yanov was expecting the contestants to deduce the radical axe idea by themselves :D", "Solution_5": "Very funny!", "Solution_6": "[quote=\"Myth\"]11.4 Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$.[/quote]\r\n\r\nMore general (but equally trivial):\r\n\r\nLet P and Q be two isogonal conjugate points with respect to triangle ABC. Let X and Y be the orthogonal projections of the point P on the sides BC and CA of triangle ABC, and let X' and Y' be the orthogonal projections of the point Q on these sides. The lines XY and X'Y' intersect each other at the point C'. Prove that $CC^{\\prime}\\perp PQ$.\r\n\r\nThe above problem 11.4 is a particular case of this if P = H and Q = O.\r\n\r\nBoth Prowler's and Grobber's solutions of problem 11.4 work similarly for the general version (one just has to apply the well-known result that the points X, Y, X', Y' are concyclic).\r\n\r\n darij", "Solution_7": "[quote=\"darij grinberg\"][quote=\"Myth\"]11.4 Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$.[/quote]\n\nMore general (but equally trivial)...[/quote]\r\nSay it only for yourself. If this problem appeared as 11.4 it means that the jury had some reasons for that.", "Solution_8": "[quote=\"grobber\"]The center of $(CA'B')$ is the midpoint of $OH$.[/quote]\r\n\r\nSmall typo here: it should be the midpoint of $OC$. A very nice solution, grobber!" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Let P(x) be a polynomial with real coefficients. Prove that there exists a nonzero polynomial Q(x) with real coefficients such that P(x)Q(x) has terms which are all of a degree divisible by 10^9.", "Solution_1": "Try $Q(x)=P(z_1x)P(z_2x)\\dots P(z_{n-1}x)$, where $z_k$ are the roots of $z^m=1$ different from $1$. Here $m=10^9$." } { "Tag": [ "inequalities", "geometry solved", "geometry" ], "Problem": "P is in Triangle ABC and makes =(PA+PB+PC)^2", "Solution_1": "Nice problem!\r\n\r\n[color=blue][b]Problem.[/b] Let P be a point inside a triangle ABC such that < PBC = < PCA = < PAB. Prove that $a^2+b^2+c^2\\geq\\left(PA+PB+PC\\right)^2$.[/color]\r\n\r\n[i]Solution.[/i] Let the lines AP, BP, CP meet the sides BC, CA, AB of triangle ABC at the points X, Y, Z. Then, by the Gergonne theorem (see http://cut-the-knot.com/triangle/Gergonne.shtml or http://www.mathlinks.ro/Forum/viewtopic.php?t=31501 property ii) ), we have $\\frac{PX}{AX}+\\frac{PY}{BY}+\\frac{PZ}{CZ}=1$.\r\n\r\nNow, remember that < PBC = < PCA; this can be rewritten as < YBC = < YCP. Also, trivially, < BYC = < CYP. Thus, the triangles YBC and YCP are similar; hence, $\\frac{CY}{BY}=\\frac{PC}{CB}$ and $\\frac{PY}{CY}=\\frac{PC}{CB}$. Hence, $\\frac{PY}{BY}=\\frac{PY}{CY}\\cdot\\frac{CY}{BY}=\\frac{PC}{CB}\\cdot\\frac{PC}{CB}=\\left(\\frac{PC}{CB}\\right)^2=\\left(\\frac{PC}{a}\\right)^2$. Similarly, $\\frac{PZ}{CZ}=\\left(\\frac{PA}{b}\\right)^2$ and $\\frac{PX}{AX}=\\left(\\frac{PB}{c}\\right)^2$. Hence, the equation $\\frac{PX}{AX}+\\frac{PY}{BY}+\\frac{PZ}{CZ}=1$ becomes $\\left(\\frac{PB}{c}\\right)^2+\\left(\\frac{PC}{a}\\right)^2+\\left(\\frac{PA}{b}\\right)^2=1$. Now, by the Cauchy-Schwarz inequality,\r\n\r\n$\\left(a^2+b^2+c^2\\right)\\left(\\underbrace{\\left(\\frac{PC}{a}\\right)^2+\\left(\\frac{PA}{b}\\right)^2+\\left(\\frac{PB}{c}\\right)^2}_{=\\left(\\frac{PB}{c}\\right)^2+\\left(\\frac{PC}{a}\\right)^2+\\left(\\frac{PA}{b}\\right)^2=1}\\right)\\geq\\left(\\underbrace{a\\cdot\\frac{PC}{a}+b\\cdot\\frac{PA}{b}+c\\cdot\\frac{PB}{c}}_{=PA+PB+PC}\\right)^2$;\r\n\r\nin other words,\r\n\r\n$a^2+b^2+c^2\\geq\\left(PA+PB+PC\\right)^2$,\r\n\r\nand the problem is solved.\r\n\r\n[i]Note.[/i] The point P satisfying < PBC = < PCA = < PAB is called the [i]first Brocard point[/i] of triangle ABC, and the angle < PBC = < PCA = < PAB is referred to as the [i]Brocard angle[/i] of triangle ABC.\r\n\r\n darij" } { "Tag": [ "number theory", "relatively prime" ], "Problem": "For how many integers $ n$ is $ \\frac{n}{20\\minus{}n}$ the square of an integer?\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)}\\ 10$", "Solution_1": "[quote=\"worthawholebean\"]For how many integers $ n$ is $ \\frac {n}{20 \\minus{} n}$ the square of an integer?\n\n$ \\textbf{(A)}\\ 1 \\qquad \\textbf{(B)}\\ 2 \\qquad \\textbf{(C)}\\ 3 \\qquad \\textbf{(D)}\\ 4 \\qquad \\textbf{(E)}\\ 10$[/quote]\r\n\r\nThrough brute force, one can easily tell it is 3.\r\n$ N>10$\r\nN works when $ N\\equal{}10, 16, 18$\r\nC", "Solution_2": "Doesn't n=0 also work? Because it doesn't say it has to be positive.\r\nSo the answer is D.", "Solution_3": "[hide]Let the square be $ k^2$. Then, $ \\frac{n}{20\\minus{}n}\\equal{}k^2$, and solving this for $ n$, we have\n\\[ n\\equal{}\\frac{20k^2}{k^2\\plus{}1}\\]Because $ n$ is an integer, the fraction must be reducible, and since $ k^2$ and $ k^2\\plus{}1$ are relatively prime, $ k^2\\plus{}1$ must divide $ 20$, and that is true only when $ k\\equal{}0, 1, 2, 3$. The corresponding values of $ n$ are $ 0, 10, 16, 18$, and the answer is $ 4 \\Rightarrow \\boxed{D}$.[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "solve equation :$x^2+y^2+1=xyz$ where $x,y,z \\in Z$", "Solution_1": "I didn't you can post it here...in wrong part..." } { "Tag": [ "real analysis", "function", "integration", "real analysis unsolved" ], "Problem": "Prove or disprove:\r\n\r\nLet $ f: \\mathbb{R}\\to\\mathbb{R}$ be L-measurable.\r\n\r\n$ \\int f\\,dx \\equal{}\\sup\\{\\int s \\,dx |0\\le s\\le f\\}$\r\n\r\nwhere s is a step function", "Solution_1": "False.\r\n\r\nOn $ [0,1]$ let $ f(x)\\equal{}\\begin{cases}1,&x\\in\\mathbb{R}\\setminus \\mathbb{Q}\\\\ 0,&x\\in\\mathbb{Q}\\end{cases}.$\r\n\r\nThen $ \\sup\\int_{[0,1]}s\\equal{}0$ where the supremum is taken over step functions $ s\\le f,$ while $ \\int_{[0,1]}f\\equal{}1.$" } { "Tag": [ "AMC 12 A", "AMC 12 B" ], "Problem": "My school is offering the AMC 12 A, but I want to take both tests... I was wondering if there is anywhere in Maryland where I can take the AMC 12 B test this year. Thanks in advance!", "Solution_1": "Yes, dunno how close it is to you.\r\n\r\nhttp://www.unl.edu/amc/b-registration/b1-archive/2009-2010/2010-HS-CU-list.shtml", "Solution_2": "thank you for the link... i will check it out!" } { "Tag": [ "geometry", "3D geometry", "sphere", "power of a point", "radical axis", "geometry solved" ], "Problem": "Point is a circle or not?", "Solution_1": "\"A circle is the set of points in a plane that are equidistant from a given point $O$\"\r\n\r\nFor any two points $A,B$, their distance is $\\geq 0$\r\n(it is $0$ iff $A \\equiv B$)\r\n\r\n\r\nSo, a zero-distance is allowed. \r\n\r\nSo, we have \"The set of points in the plane that have distance $0$ from the point $O$ \" is the point $O$ itself\r\n\r\n\r\nFurther more, $O$ is a sphere with a zero-radius, or even an n-dimensional sphere", "Solution_2": "[quote=\"jensen\"]Point is a circle or not?[/quote]\r\n\r\nWell, this depends on how you interpret the word \"circle\". The same holds for your question whether a line is a circle, which you have posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=65374 . Mathematicians have noted that under some circumstances, points and lines behave similarly to circles, so they have found it useful to consider points and lines as \"degenerate\" cases of circles. The exact definitions are the following:\r\n\r\n- A point P can be identified with a \"degenerate\" circle with center P and radius 0; this circle is called the [b]point-circle[/b] P. This point-circle passes through one single point, namely P.\r\n- A line g can also be identified with a \"degenerate\" circle, a so called [b]line-circle[/b]. The center of this line-circle is the infinite point where all lines perpendicular to g intersect. The radius of this circle is undefined (you could say it is $\\infty$, but this is misleading, since usually, a circle is uniquely determined by its center and its radius, but in this case this would not be true, as there would be lots of circles all centered at one infinite point and having the same radius $\\infty$).\r\n\r\nActually, both notions of point-circles and line-circles are suggested by limiting observations: Imagine a fixed line t, a fixed point P on the line t, and a variable point M on the line t. Consider the circle with center M and radius MP. When the point M comes closer and closer to P, the radius MP tends to zero, and the circle resembles more and more to the single point P (point-circle). When the point M moves farther and farther from P, the radius MP tends to infinity, and the circle resembles more and more to a line, namely the perpendicular to the line t at P (line-circle).\r\n\r\nOf course, in order to ensure that these \"definitions\" are more than empty words, we have to extend some possible properties of circles to these \"degenerate\" circles. For instance, when are two degenerate circles, or a non-degenerate and a degenerate circle orthogonal? Well, the usual definition of orthogonality of two circles is: Two circles $k_1$ and $k_2$ are called orthogonal if $O_1O_2\\ ^2=r_1^2+r_2^2$, where $O_1$ and $O_2$ are the centers and $r_1$ and $r_2$ are the radii of the circles $k_1$ and $k_2$. This definition works for non-degenerate circles, and it also works for point-circles. However, since a point-circle has radius 0, the equation $O_1O_2\\ ^2=r_1^2+r_2^2$ simplifies to $O_1O_2\\ ^2=r_1^2$ when the circle $k_2$ is a point-circle, and this obviously means that the point $O_2$ lies on the circle $k_1$. Hence, we see that a point-circle P is orthogonal to a circle k if and only if the point P lies on the circle k. As a consequence of this, two point-circles are orthogonal if and only if they coincide. Well, so far we have defined orthogonality for point-circles. Defining orthogonality for line-circles is a bit harder, since they have no radii. But actually, line-circles are lines; so it is suggesting to say that two line-circles are orthogonal when they are perpendicular lines. Furthermore, a line-circle is said to be orthogonal to a point-circle when it passes through the point, and a line-circle is said to be orthogonal to a non-degenerate circle if the center of the non-degenerate circle lies on the line. (Both of these postulations are not arbitrary, but suggested by a limiting argument.)\r\n\r\nNow we can note that lots of theorems about circles continue to hold if some of the circles \"degenerate\" to point-circles or line-circles. Particularly in inversive geometry, you will find lines and circles having more or less equal rights. You can also define the radical axis of a circle and a point-circle, for instance,[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=249398#p249398]and use it in some problems[/url] (however, as usual, degenerate cases should be treated with care - e. g., you cannot define radical axes for line-circles). This is the main reason why the notions of point-circles and line-circles are used.\r\n\r\n Darij", "Solution_3": "point is a circle with a very very small radius. ;)", "Solution_4": "Darij grinberg, thanks a lot for your answer, I enjoyed reading it! :)\r\nThat was really interesting." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "The sequence $(a_n)$ is given by $a_1=x\\in R$ and $3a_{n+1}=a_n+1$ for $n\\geq 1$. Set \r\n$A=\\sum_{n=1}^\\infty [a_n-\\frac 16]$ and $B=\\sum_{n=1}^\\infty [a_n+\\frac 16]$. (where $[x]$ denote the greatest integer less that or equal to $x$).Compute the sum $A+B$ in terms of $x$. :lol:", "Solution_1": "$a_{n+1}=\\frac{1}{3}a_n+\\frac{1}{3}\\Longleftrightarrow a_{n+1}-\\frac{1}{2}=\\frac{1}{3}\\left(a_n-\\frac{1}{2}\\right)$, thus $a_n=\\left(\\frac{1}{3}\\right)^{n-1}\\left(x-\\frac{1}{2}\\right)+\\frac{1}{2}$.\r\n\r\nFrom $a_n-\\frac{1}{6}=\\left(\\frac{1}{3}\\right)^{n-1}\\left(x-\\frac{1}{2}\\right)+\\frac{1}{3},\\ a_n+\\frac{1}{6}=\\left(\\frac{1}{3}\\right)^{n-1}\\left(x-\\frac{1}{2}\\right)+\\frac{2}{3}$. If $[x]=m$,then $x-1<[x]\\leq x$,so \r\n\r\nsince $\\left(a_n-\\frac{1}{6}\\right)-1<\\left[a_n-\\frac{1}{6}\\right]\\leq a_n-\\frac{1}{6}$ and $\\left|\\frac{1}{3}\\right|<1$, we have $\\sum_{n=1}^{\\infty}\\left( a_n-\\frac{1}{6}\\right)=\\frac{x-\\frac{1}{2}}{1-\\frac{1}{3}}-\\infty=-\\infty$, similarly \r\n\r\n$ \\sum_{n=1}^{\\infty}\\left( a_n+\\frac{1}{6}\\right)=\\frac{x-\\frac{1}{2}}{1-\\frac{1}{3}}-\\infty=-\\infty$,... :?" } { "Tag": [ "number theory", "relatively prime", "Divisibility Theory" ], "Problem": "Prove that among any ten consecutive positive integers at least one is relatively prime to the product of the others.", "Solution_1": "Clearly the only common prime factors amongst 10 consecutive positive integers will be 2,3,5,7.\r\n\r\n5 of them will be divisible by 2, at least one of which must also be divisible by 3 and at least one of which must also be divisible by 5, and, if two of the numbers are divisible by 7, one of them will be even.\r\n\r\nThis leaves two multiples of 3, one multiple of 5, and one multiple of 7 unaccounted for, enough factors to dish out to 4 more of our integers.\r\n\r\nBut this still leaves one integer not divisible by 2,3,5,7, and therefore co-prime with all other integers of the set, and so coprime with their product.", "Solution_2": "Can we replace $ 10$ by $ n$?\r\n\r\nI have found that this fails for $ 25$, although I am not sure of the other values. Can anyone give anything?", "Solution_3": "Can you post your example for 25? Maybe that gives us a clue.", "Solution_4": "I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.", "Solution_5": "[quote=\"peter\"]Can you post your example for 25? Maybe that gives us a clue.[/quote]\n\nWell my example is the natural numbers from $ 1$ to $ 25$.\n\n[quote=\"MayankM\"]\t\nI have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.\n[/quote]\r\n\r\nThe source of the problem said [b]IHH pp.211[/b] and so I took that out and I found what [b]MayankM[/b] had siad to be true.", "Solution_6": "[quote=\"manjil\"][quote=\"peter\"]Can you post your example for 25? Maybe that gives us a clue.[/quote]\n\nWell my example is the natural numbers from $ 1$ to $ 25$.[/quote]Really? I think $ 23$ is relatively prime to the product of the others though... :wink:", "Solution_7": "[quote=\"MayankM\"]I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.[/quote]\r\nit's true. 16 can be done and 17 can't. the counterexample for $ n\\equal{}17$ are consecutive numbers of which the first equals $ 27830$.\r\n(in fact the first can be $ 27830\\plus{}30030k$ for a natural $ k$, mybe there exists a smaller example that i missed but the point is that there are infinitely many)\r\nu can proove it for $ n\\equal{}16$ basicaly same as ilthigore has done for $ n\\equal{}10$ above only a bit more complicated.\r\n\r\n$ n\\equal{}17$\r\nlet $ a_1,...a_{17}$ be these consecutive numbers. if we set $ a_1$ to be devisable by $ 2,5,11$, $ a_2$ devisable by $ 3$, $ a_3$ by $ 7$ and $ a_4$ by $ 13$. it's easy to check thet none of them is than relatively prime to all others. we can calculate $ 27830$ with china reminder theorem now.\r\n\r\nwhat is left to do is to show that if there is a counter wxample for $ n$ than we can also find a counter example for $ n\\plus{}1$. im working on this now.", "Solution_8": "[hide=drawn-out solution]Let any set of $10$ numbers be defined as $S=\\{n,n+1,n+2,n+3, \\dots ,n+9\\}$. \n\n[b]Lemma $1$[/b]: The only prime factors that the numbers in $S$ can have in common are $2,3,5$ and $7$. \nWe can take the set $S$ modulo $2,3,5,$ and $7$, and since there must contain at least two $0$'s in $S$ (by spacing of the modulo), there are two numbers with the modulo as their factors. However, if we take the modulo of any prime greater than $7$, there is at most one $0$ contained in the set. (Also, due to spacing) \nAlso, note that this argument makes it so that the only divisors that the elements in $S$ can have in common are the ones described above, due to euclidian algorithm (For example, $\\gcd(n,n+5)=\\gcd(n,5)$) $\\square$\n\nThere can be a maximum of $5$ factors of $2$ contained in $S$, $4$ factors of $3$, $2$ factors of $5$, and $2$ factors of $7$. Using a spacing argument, there must exist $2$ factors of $2$ and $3$; $1$ factor of $2$ and $5$; and $1$ factor of $2$ and $7$.Thus, to account for the overcounting, we have to subtract $4$, and we can now simply sum the numbers and subtract our overcounting factor to get $5+4+2+2-4=9$, and so there must exist some number that is relatively prime to the rest of the numbers.[/hide]", "Solution_9": "As mentioned above, the statement is true when 'ten' is replaced by any positive integer $m \\le 16$. It is false, however, if one replaces 'ten' by any integer $m > 16$. References for this fact:\n\nIn a paper that you can find here: [url]https://www.ias.ac.in/article/fulltext/seca/011/01/0006-0012[/url], Pillai proved that it is true for $m \\le 16$ and it is false for all $m$ with $17 \\le m \\le 430$.\n\nIn a paper that you can find here: [url]https://www.ams.org/journals/bull/1941-47-04/S0002-9904-1941-07455-0/S0002-9904-1941-07455-0.pdf[/url], Brauer proves that it is false for all $m \\ge 300$, thereby finishing the problem." } { "Tag": [ "geometry", "3D geometry", "number theory proposed", "number theory" ], "Problem": "These questions are almost classical rigth now but...\r\n\r\nWhich is the smallest positive integer that could be written as the sum of two (positive) squares in two different ways?\r\nWhich is the smallest positive integer that could be written as the sum of two (positive) squares in three different ways (three distinct pairs - any two elements are not equal)?\r\nWhich is the smallest positive integer that could be written as the sum of two (positive) cubes in two different ways?\r\nThere is an integer such that is the smallest positive integer that could be written as the sum of two (positive) cubes in three different ways? \r\nWhich is the smallest positive integer that could be written as the sum of two (positive) fourth powers in two different ways?", "Solution_1": "1. From $(a^2 + b^2)(c^2 + d^2) = (ac \\pm bd)^2 + (ad \\mp bc)^2$, we want the smallest number expressible as a product of two distinct sums of squares, which would be $(1^2 + 2^2)(1^2 + 3^2) = \\boxed{50} = 5^2 + 5^2 = 1^2 + 7^2$.\r\n\r\n2. Similar logic gives $(1^2 + 2^2)(1^2 + 3^2)(2^2 + 3^2) = \\boxed{650} = 5^2 + 25^2 = 11^2 + 23^2 = 17^2 + 19^2$. I think.\r\n\r\n3. Ramanujan's number! $\\boxed{1729} = 1^3 + 12^3 = 10^3 + 9^3$ :D", "Solution_2": "[quote=\"t0rajir0u\"]3. Ramanujan's number![/quote]\r\n\r\nAlso called taxicab number ;)", "Solution_3": "fOR THE LAST also Ramanajuan reply to this at 5 minutes after the question without using a computer and being on hospital that the smallest is\r\n\r\n$635318657=133^4+134^4=158^4+59^4$ :)", "Solution_4": "[quote=\"silouan\"]without using a computer[/quote]\r\nThere was no computers at that time ;)", "Solution_5": "Maybe in India was be :D", "Solution_6": "Sorry for spamming but Ramanujan was a computer himself :D \r\ni think noone has doubt for this." } { "Tag": [ "function", "integration", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "show :D \r\n\r\n$ \\int_{0}^{\\frac{\\pi}{2}}\\;\\;\\cos^{m}x\\;\\cdot\\;\\sin^{n}x\\;\\;\\textbf dx\\;\\;\\equal{}\\;\\;\\boxed{\\frac{(m\\plus{}n\\plus{}2)}{(m\\plus{}1)(n\\plus{}1)}\\;\\;\\cdot\\;\\;\\prod_{k\\equal{}1}^{\\infty}\\;\\;\\left[\\;\\frac{\\left(1\\plus{}\\frac{m\\plus{}n\\plus{}2}{2k}\\right)}{\\left(1\\plus{}\\frac{m\\plus{}1}{2k}\\right)\\cdot\\left(1\\plus{}\\frac{n\\plus{}1}{2k}\\right)}\\;\\right]}$", "Solution_1": "Let $ m\\equal{}2a\\minus{}1$ and $ n\\equal{}2b\\minus{}1$ and so,\r\n$ \\int_{0}^{\\pi/2}\\sin^{2a\\minus{}1}x\\cos^{2b\\minus{}1}x dx \\equal{}\\int_{0}^{\\pi/2}\\sin^{2a\\minus{}1}x\\cos^{2b\\minus{}2}x\\cos xdx$\r\nLet $ y\\equal{}\\sin x$ to get,\r\n$ \\int_{0}^{1}y^{2a\\minus{}1}(1\\minus{}y^{2})^{b\\minus{}1}dy$\r\nLet $ z\\equal{}y^{2}$ to get,\r\n$ \\frac{1}{2}\\int_{0}^{1}z^{a\\minus{}1}(1\\minus{}z)^{b\\minus{}1}dz \\equal{}\\frac{1}{2}\\mbox{B}(a,b) \\equal{}\\frac{1}{2}\\cdot\\frac{\\Gamma (a)\\Gamma(b)}{\\Gamma(a\\plus{}b)}$\r\nSo, subsitute back,\r\n$ \\frac{1}{2}\\frac{\\Gamma\\left(\\frac{n\\plus{}1}{2}\\right)\\cdot\\Gamma\\left(\\frac{m\\plus{}1}{2}\\right)}{\\Gamma\\left(\\frac{n\\plus{}m\\plus{}2}{2}\\right)}$\r\nApply the infinite product formulas for Gamma functions ... and it should work out. :maybe:" } { "Tag": [ "modular arithmetic" ], "Problem": "Do all squares differ by odd integers (e.g. 4, 9, 16, 25, etc.)?\r\n\r\nAre all squares perfect squares?", "Solution_1": "No, because $n^2\\equiv 0,1\\pmod 4$, so that two non-adjacent ones can be $1-1=0\\pmod 4$, and therefore not an odd integer.", "Solution_2": "[quote=\"erik-the-red\"]Are all squares perfect squares?[/quote]\r\nUhh, what do you mean?", "Solution_3": "well let x be an integer, then x+1 is the next integer...so the difference is\r\n$(x+1)^{2} - x^{2}$\r\n$2x + 1$\r\nso yes ...since $2x +1$ for any integer x is odd", "Solution_4": "[quote=\"Elemennop\"]No, because $n^2\\equiv 0,1\\pmod 4$, so that two non-adjacent ones can be $1-1=0\\pmod 4$, and therefore not an odd integer.[/quote]But that's assuming that there are two adjacent squares that are both $\\equiv 1\\pmod 4$...it turnes out they alternate between $0$ and $1$. If the square root is even then the square is $\\equiv 0\\pmod 4$, if the square root is odd then the square is $\\equiv 1\\pmod 4$.", "Solution_5": "He never said anywhere they had to be adjacent, plus I specified specifically that they have an even difference only if they are not adjacent.", "Solution_6": "[quote=\"Elemennop\"]He never said anywhere they had to be adjacent, plus I specified specifically that they have an even difference only if they are not adjacent.[/quote]Oh good point, I probably just assumed he meant consecutive squares.", "Solution_7": "No, not all squares are perfect squares. A perfect square is a positive integer which is the square of some other positive integer. A square can be one of many things. It can be any number. Or, it could be a regular quadrilateral.", "Solution_8": "Thanks for clarifying that to me.", "Solution_9": "Perfect squares are the square of an integer.\r\n\r\nOnly squares of integers differing by an odd number will differ by an odd number. For example: $n^2$ and $(n-3)^2=n^2-6n+9$ differ by an odd number. However, squares of integers differing by an even number will differ by an even number." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "We make n\u00b2 equal squares inside of a square and we write on each square a real number.\r\n\r\n$ \\forall 1 \\le i \\le n$ we denote $ r_i$ the sum of the numbers on the [u]row[/u] i and $ c_i$ the sum of the numbers on the [u]column[/u] i \r\n(Thank you JBL for correcting my english)\r\n\r\nprove that there exists i and j such that:\r\n$ (r_i \\minus{} c_i)(r_j \\minus{} c_j) \\le 0$\r\n\r\nWhere was this problem proposed please?\r\n\r\nThanks !", "Solution_1": "[quote=\"FOURRIER\"]... [u]line[/u] ... [u]array[/u] ...[/quote]\r\n\r\n\"Row\" and \"column,\" respectively. (Easy to remember in this instance, because of the $ r$ and the $ c$ ;).)", "Solution_2": "Indeed very easy:\r\nIf $ (r_i \\minus{} c_i)(r_j \\minus{} c_j) > 0$ for all $ i$ and $ j$ then\r\n$ r_i > c_i\\ \\ \\forall i$ (or $ (r_i < c_i\\ \\ \\forall i$ wich gives the same)\r\nthen $ r_1 \\plus{} r_2 \\plus{} ... \\plus{} r_n > c_1 \\plus{} c_2 \\plus{} ... \\plus{} c_n$ which contradicts $ r_1 \\plus{} r_2 \\plus{} ... \\plus{} r_n \\equal{} c_1 \\plus{} c_2 \\plus{} ... \\plus{} c_n$\r\n\r\nbtw no idea where it was proposed..." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c>0$ such that $ abc\\equal{}1$. Find the minimum value of the following inequality:\r\n\\[ \\frac{1}{{\\sqrt {1 \\plus{} a^2 } }} \\plus{} \\frac{1}{{\\sqrt {1 \\plus{} b^2 } }} \\plus{} \\frac{1}{{\\sqrt {1 \\plus{} c^2 } }}\r\n\\]", "Solution_1": "1 by using mixing variables?", "Solution_2": "I think it is correct :lol: \r\nInequality holds for a->0 and b,c--->infinity.\r\nI think for these kind of problem(only one inflection point in f(e^x)) it is easy because\r\nyou can put a=b=t, c=1/t^2 and differentiate.", "Solution_3": "How to solve it? This is the most important things in this topic :) Can anyone post solution?", "Solution_4": "[quote=\"ywl\"]\nyou can put a=b=t, c=1/t^2[/quote]\r\n\r\nI think you can't\r\n\r\nonly if you have shown : $ f(a,b,c)\\geq f(\\sqrt{ab},\\sqrt{ab},c)$\r\n :wink:", "Solution_5": "[quote=\"FOURRIER\"][quote=\"ywl\"]\nyou can put a=b=t, c=1/t^2[/quote]\n\nI think you can't\n\nonly if you have shown : $ f(a,b,c)\\geq f(\\sqrt {ab},\\sqrt {ab},c)$\n :wink:[/quote]\r\n\r\n\r\nI think I have shown. Have you tried the convexity of f(e^x) with f(x)=1/(root(1+x^2)? It has one inflection point \r\nso that you can put a=b=t and c=1/t^2 This follows from applications of Karamata and Jensen \r\n(See post in Theorems and Formulas sections by Zhaobin regarding the \"Half convex, half concave theorem) :)", "Solution_6": "$ S(a,b,c)\\equal{}\\frac{1}{\\sqrt{1\\plus{}a^{2}}}\\plus{}\\frac{1}{\\sqrt{1\\plus{}b^{2}}}\\plus{}\\frac{1}{\\sqrt{1\\plus{}c^{2}}}$. Let's prove that $ \\inf{S(a,b,c)}\\equal{}1$ (It hasn't got a minimum value) with $ abc\\equal{}1$. It's clear that if $ a>\\equal{}1,b<\\equal{}1,c<\\equal{}1$ , we have $ S(a,b,c)>\\frac{1}{\\sqrt{1\\plus{}b^{2}}}\\plus{}\\frac{1}{\\sqrt{1\\plus{}c^{2}}}>\\equal{}2\\cdot\\frac{1}{2}\\equal{}1$ , so $ a>\\equal{}1, b>\\equal{}1, c<\\equal{}1$. Then $ b<\\equal{}\\frac{1}{c}$. So we have $ S(a,b,c)>\\frac{1}{\\sqrt{1\\plus{}b^{2}}}\\plus{}\\frac{1}{\\sqrt{1\\plus{}c^{2}}}>\\equal{}\\frac{c\\plus{}1}{\\sqrt{1\\plus{}c^{2}}}>1$\r\nSo $ S(a,b,c)>1$. But choosing $ a\\equal{}b\\equal{}x$, $ c\\equal{}\\frac{1}{x^{2}}$ we have $ S(a,b,c)\\equal{}\\frac{x\\plus{}2}{\\sqrt{1\\plus{}x^{2}}}\\rightarrow 1$ with $ x\\rightarrow \\infty$. So $ \\inf{S(a,b,c)}\\equal{}1$", "Solution_7": "[i]I am a member of LeHongPhong highschool so i can say that the problem is \" find maximum of the expression\"[/i]\r\n[i]very beautiful problem\n\n But it's origin is very , very similiar to erveryone :)\n [/i] Denote $ x = 1 , y = \\frac {1}{a^2} , z = \\frac {1}{(ab)^2}$\r\n => $ a^2 = \\frac {x}{y} , b^2 = \\frac {y}{z}, c^2 = \\frac {z}{x}$\r\n [i]We have to prove with x,y,z be positive real numbers we have[/i] B=$ \\sum\\limits_{cyc} \\sqrt {\\frac {2x}{x + y}}$ $ \\leq$ 3\r\n Deduce B= $ \\sum\\limits_{cyc}\\sqrt {\\frac {2x(y + z)}{(x + y)(y + z)}}$\r\n \r\n => $ \\ B^{2}$ $ \\leq$(BCS) A= 4( $ xy + yz + zx$).($ \\frac {1}{(x + y)(y + z)} + \\frac {1}{(y + z)(z + x)} + \\frac {1}{(z + x)(x + y)}$)\r\n = 8$ \\frac {(xy + yz + zx)(x + y + z)}{(x + y)(y + z)(z + x)}$\r\n \r\n[i] From a known ineq[/i]\r\n $ (x + y)(y + z)(z + x)$$ \\geq$ $ \\frac {8}{9}$ $ (x + y + z)(xy + yz + zx)$ :lol: \r\n[i] ( Cauchy with 6 positive real numbers)[/i] \r\n [i] We have[/i] [size=150]A[/size] $ \\leq$ 9\r\n => LHS $ \\leq$ 3 ( QED)\r\n [i] Hero TV\u01a0 Y\u00eau An m\u00e3i m\u00e3i[/i]", "Solution_8": "[quote=\"zaizai-hoang\"]Let $ a,b,c > 0$ such that $ abc \\equal{} 1$. Find the minimum value of the following inequality:\n\\[ \\frac {1}{{\\sqrt {1 \\plus{} a^2 } }} \\plus{} \\frac {1}{{\\sqrt {1 \\plus{} b^2 } }} \\plus{} \\frac {1}{{\\sqrt {1 \\plus{} c^2 } }}\n\\]\n[/quote]\r\nAssume $ ab \\le 1$, then\r\n$ \\frac {1}{{\\sqrt {1 \\plus{} a^2 } }} \\plus{} \\frac {1}{{\\sqrt {1 \\plus{} b^2 } }} \\plus{} \\frac {1}{{\\sqrt {1 \\plus{} c^2 } }} \\\\ \\ge \\frac {1}{{\\sqrt {1 \\plus{} a^2 } }} \\plus{} \\frac {1}{{\\sqrt {1 \\plus{} b^2 } }} \\\\ \\ge \\sqrt{\\frac{1}{1\\plus{}a^2}\\plus{}\\frac{1}{1\\plus{}b^2}} \\\\\r\n\\ge 1$.\r\n\r\nWe know the general inequality\r\n\\[ \\min \\left(1;\\frac{n}{2^k} \\right) \\le \\frac{1}{(1\\plus{}a_1)^k}\\plus{}\\frac{1}{(1\\plus{}a_1)^k}\\plus{}...\\plus{}\\frac{1}{(1\\plus{}a_1)^k} \\le \\max \\left(n\\minus{}1;\\frac{n}{2^k} \\right)\r\n\\]\r\nwhere $ a_1,a_2,...,a_n > 0$ such that $ a_1a_2...a_n \\equal{}1$ and $ k$ is a positive real number." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c\\ge 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$,prove that\\[ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\ge \\frac {1}{4}\\]", "Solution_1": "[quote=\"skywalkerJ.L.\"]Let $ a,b,c\\ge 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$,prove that\n\\[ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\ge \\frac {1}{4}\\]\n[/quote]\r\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\ge \\frac {1}{4}\\Leftrightarrow4(a^3\\plus{}b^3\\plus{}c^3)\\plus{}24abc\\geq(a\\plus{}b\\plus{}c)^3,$ which is obviously true by Schur.", "Solution_2": "[quote=\"skywalkerJ.L.\"]Let $ a,b,c\\ge 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$,prove that\n\\[ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\ge \\frac {1}{4}\\]\n[/quote]\r\n\r\n\r\n$ \\Longleftrightarrow a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6abc\\ge \\frac {1}{4}(a\\plus{}b\\plus{}c)^3$\r\n\r\n\r\n$ \\Longleftrightarrow \\frac{3}{4}\\sum{a(a\\minus{}b)(a\\minus{}c)}\\plus{}\\frac{9}{4}abc\\geq 0$" } { "Tag": [ "absolute value", "number theory unsolved", "number theory" ], "Problem": "Prove that the number $[(5+\\sqrt{35})^{2n-1}] $ is divisible by $ 10^n$ for each $n\\in N$\r\n(where $ [x]$ denote the greatest integer less than or equal to $x$). :huh: :)", "Solution_1": "Define $x_0:=2 ; x_1:=10 ; x_{n+2} := 10( x_{n+1} + x_{n} )$, then every odd term of the sequence will be identically what your term gives (simply solve the recursion, you get some term $5-\\sqrt{35}$ that has absolute value $<1$ and so can be neglected when taking the integer part) and conclusion (even more) follows." } { "Tag": [], "Problem": "$ f\\left( x \\right) \\equal{} \\frac {x^5}{5x^4 \\minus{} 10x^3 \\plus{} 10x^2 \\minus{} 5x \\plus{} 1}$. \n$ \\sum_{i \\equal{} 1}^{2009} f\\left( \\frac {i}{2009} \\right) \\equal{} ?$\n\n$\\textbf{(A)}\\ 1000 \\qquad\\textbf{(B)}\\ 1005 \\qquad\\textbf{(C)}\\ 1010 \\qquad\\textbf{(D)}\\ 2009 \\qquad\\textbf{(E)}\\ 2010$", "Solution_1": "I guess the first term in the denominator should be $ 5x^4$?", "Solution_2": "absolutely right. edited.", "Solution_3": "$ f\\left( x \\right) \\equal{} \\frac {x^5}{5x^4 \\minus{} 10x^3 \\plus{} 10x^2 \\minus{} 5x \\plus{} 1} \\equal{} \\frac {x^5}{5(x^3 \\minus{} x^2 \\plus{} x)(x \\minus{} 1) \\plus{} 1}$. \r\n\r\nAnd it's 1005, but I don't know how to get there from this...", "Solution_4": "$ f(x) \\equal{} \\frac {x^5}{x^5 \\plus{} (1 \\minus{} x)^5}$\r\n$ f(x) \\plus{} f(1 \\minus{} x) \\equal{} \\frac {x^5}{x^5 \\plus{} (1 \\minus{} x)^5} \\plus{} \\frac {(1 \\minus{} x)^5}{x^5 \\plus{} (1 \\minus{} x)^5} \\equal{} 1$\r\n\r\nHence, all terms pair up except $ f(\\frac {2009}{2009}) \\equal{} f(1) \\equal{} \\frac {1}{1 \\plus{} 0} \\equal{} 1$, giving the sum to have value $ 1004 \\plus{} 1 \\equal{} 1005$\r\n\r\nSo B is the correct answer.\r\n\r\nThe trick was seeing that you could write the denominator like that.", "Solution_5": "[quote=\"rofler\"]The trick was seeing that you could write the denominator like that.[/quote]\r\n\r\nAh... I factored it wrong.\r\n\r\nBleh >_<. Mind's not working. Thank you!" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a,b,c,d are postive real numbers whose sum is 4, prove\r\n$ \\prod_{cyc}{(a^{3}+1)}\\geq 16$", "Solution_1": "[quote=\"hanschinaman\"]a,b,c,d are postive real numbers whose sum is 4, prove\n$ \\prod_{cyc}{(a^{3}+1)}\\geq 16$[/quote]\nThe following inequality is also true.\nLet $a$, $b$, $c$, $d$, $e$, $f$ and $g$ be positive numbers such that $a+b+c+d+e+f+g=7$. Prove that:\n\\[(a^3+1)(b^3+1)(c^3+1)(d^3+1)(e^3+1)(f^3+1)(g^3+1)\\geq128\\]" } { "Tag": [ "inequalities", "calculus", "inequalities proposed" ], "Problem": "SORRY - WRONG ONE", "Solution_1": "And,what is the right one? :D", "Solution_2": ":D i realized my mistake in calculus, so i erased it." } { "Tag": [ "probability", "geometry", "3D geometry", "calculus", "integration", "quadratics" ], "Problem": "I have n sticks, of length 1, 2, 3, ..., n.\r\nI choose three sticks at random. Whats the probability that they can form a triangle?\r\n\r\nI have an answer, hopefully it agrees with your ones.", "Solution_1": "Haven't tried it yet, but a possibly interesting variation: Given three real numbers between 0 and 1, with all triples equally likely to be chosen, what is the probability that a triangle can be formed with sides of those lengths?", "Solution_2": "Well, Joel's variant can be solved by considering a 1*1*1 cube and finding the volume that makes it work. Stephen's one seems like a very difficult problem.", "Solution_3": "Hmm, hadn't thought about that. Makes sense. It should also be the limiting case of the solution to Steven's problem, which I now attempt to give.\n\n\n\nHmm, I realize that I did a different problem than the one you gave -- you question didn't allow repeats, while my solution allowed for them. I'll try and do it your way later, but here's what I did first:\n\n[hide]\n\nOkay, so, denominator first: C(n, 3) ways to pick all different, C(n, 1) ways to pick all the same, 2C(n, 2) ways to pick two the same and one different, so (C(n, 3) + 2C(n, 2) + C(n, 1)) = (n3 + 3n2 + 2n)/6 total picks.\n\n\n\n\n\nNow, for the numerator. Let M(i) be the number of triangles with integral sides between 1 and i with largest side i. Then the numerator should be :Sigma:M(i) from i = 1 to n.\n\n\n\nI wish to generate M(i) recursively. I claim M(i) = M(i - 1) + [i/2], where [x] is the greatest integer less than or equal to x.\n\nProof:\n\nConsider any triple (a, b, i) that makes a triangle, with b > a. Then (a, b - 1, i - 1) is a triple that makes a triangle, with b - 1 :ge: a. That gives us M(i - 1) such triples. The only ones we've missed are those in which a = b, of which we have [i/2] which make triangles. Thus, we have the relation I claimed. \n\n\n\nSo, if we just look at what that gives us, we have M(1) = 1, M(2) = 2, M(3) = 3, M(4) = 5, M(5) = 7, M(6) = 10, M(7) = 13, and so on. This is roughly quadratic -- for odd i, it gives us (i2 + 3)/4, and for even i it gives us (i2/4 + 1), which can be encapsulated nicely in the formula ([i2/4] + 1).\n\nThen we have to sum that up from i = 1 to n.\n\n[/hide]\n\nI have to go, so I can't finish this now -- I will try to finish my way and adjust for the way I was supposed to do it later. Anyone, feel free to jump off of this.", "Solution_4": "My solution to Joel's problem:\n\n[hide]\n\nWell, I couldn't figure out how to get integral signs, so S_a^b means the integral from a to b:\n\n\n\nS_0^1 S_0^1 (min(1,x+y)-|x-y|) dy dx =\n\nS_0^1 S_0^1 min(1,x+y) - S_0^1 S_0^1 |x-y| dy dx =\n\nS_0^1 S_0^(1-x) x+y dy dx + S_0^1 S_0^(1-x) 1 dy dx - S_0^1 S_0^x (x-y) dy dx + S_0^1 S_x^1 (y-x) dy dx =\n\nS_0^1 (-1/2 x^2 + 1/2) dx + S_0^1 x dx - S_0^1 1/2 x^2 dx - S_0^1 (1/2x^2-x+1/2) dx = \n\nS_0^1 (-3/2 x^2 + 2x) dx =\n\n1/2\n\n[/hide]\n\n\n\nThe fact that the answer has such a simple form makes me think there must be a more elegant solution. Any ideas for say [hide]a bijection between triples that form a triangle and triples that don't[/hide]?\n\n\n\n--Dan", "Solution_5": "Well heres an outline of how I did it anyway. \n\n[hide]\n\nWe count the possible number of groups of 3 that work. First we consider the ones with largest stick of the 3 being the one length n:\n\nNow if we take n-1, we can choose one of (2,3,...,n-2) ie n-3 ways.\n\nIf we take n-2, we choose one of (3,4,..,n-3) ie n-5 ways.\n\netc etc\n\n\n\nSo the number of ways in this case is 0+2+4+..+(n-3) if n is odd, 1+3+5+..+(n-3) if n is even.\n\n\n\nWork out those sums, (the formulas end up differing by 1/4). So to find the total number of ways, you sum the (higher) formula from 3 to n, subtract off [(n-1)/2] quarters, and get an answer (after dividing by nC3).\n\n[/hide]", "Solution_6": "Okay, yeah, that was exactly what I was trying to do, except that I was working on it when I wasn't at the computer so I got that one part (about repeatability) wrong." } { "Tag": [ "inequalities", "calculus", "calculus computations" ], "Problem": "Does any one have good problems involving greatest integer and fractional part ?", "Solution_1": "Any problems please", "Solution_2": "do it by yourself.\r\n\r\ngo and use google, c'mon, you can't be that lazy.\r\n\r\nyou can even look at the forum.", "Solution_3": "Hi. A book which has problems on this is \"Equations and Inequalities\", pages 251 to 257. See here: http://www.muni.cz/research/publications/313831/" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "S is the complexe of the numbers in form p+a^p which a is a fix integer and p is a prime number and suppose that k is also an integer.is it true that we can find two member from the complexe like x and y so that (x,y)>k\r\nthank you very much.", "Solution_1": "Take some prime $q>k$ not dividing $a$. Then lets set $s \\equiv 1 \\mod (q-1)$ and $\\equiv -a \\mod q$ since this gives $s+a^s \\equiv s+a \\equiv 0 \\mod q$.\r\nBy Dirichlets theorem on primes in arithmetic progressions, there are two different primes $x,y \\equiv s \\mod q\\cdot(q-1)$, fulfilling $\\gcd(x+a^x , y+a^y) =q \\cdot \\mathrm{SPAM} \\geq q >k$." } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "just checking if i did this right\r\n$\\phi:H\\rightarrow H$, where $H$ is any group, and $\\phi(x)=x^{-1}$. Show that $\\phi$ is a homomorphism.\r\n\r\n\r\n$\\phi(x+y)=(x+y)^{-1}$, $\\phi(x) + \\phi(y)=x^{-1} + y^{-1}$ , $\\phi(x+y) \\neq \\phi(x) + \\phi(y)$ , so $\\phi$ wouldnt be a homomorphism", "Solution_1": "Are you sure that's + and not $\\times$ ?", "Solution_2": "If we set * the \"law\" (is it the good word in english ?) on H, we get :\r\n$\\phi(x) * \\phi(y)=x^{-1} * y^{-1}$, $\\phi(x*y)=(xy)^{-1}=y^{-1} * x^{-1}$\r\n$\\phi$ is an homomorphism if H is an abelian group", "Solution_3": "[quote=\"Rodolphe2005\"]If we set * the \"law\" (is it the good word in english ?) on H, we get :\n[/quote]\r\noperation.", "Solution_4": "would kern $\\phi = \\left\\{e\\right\\}$ ?", "Solution_5": "No, for example take $( \\mathbb{Z}_2 ; + )$", "Solution_6": "$\\phi:H\\rightarrow H$, where $H$ is any group, and $\\phi(x)=x^{-1}$. Show that $\\phi$ is a homomorphism.\r\n\r\n\r\nThis holds if and only if $H$ is abelian? Thus statement is not true :?", "Solution_7": "[quote=\"chenwb\"]$\\phi:H\\rightarrow H$, where $H$ is any group, and $\\phi(x)=x^{-1}$. Show that $\\phi$ is a homomorphism.\n\n\nThis holds if and only if $H$ is abelian? Thus statement is not true :?[/quote]\r\n\r\nTotally right, because by hypothesis, for every $x,y\\in H$ you'll have $\\phi(xy)=\\phi(x)\\phi(y)$, but therefore $(xy)^{-1}=y^{-1}x^{-1}=x^{-1}y^{-1}$, hence, $H$ is abelian!" } { "Tag": [ "geometry", "angle bisector", "Pythagorean Theorem" ], "Problem": "Right triangle ABC is divided, as shown, such that angle BAC is bisected by AD. What is the length of AD? Express your answer as a decimal to the nearest hundredth.\r\n\r\n--> Click on PDF attachment for diagram.\r\n\r\n[hide=\"ANSWER\"]10.06[/hide]", "Solution_1": "[hide]Recognizing Pythagorean Triplets tells us that BC=12.\nLet BD=x so CD=12-x.\nThe Angle Bisector Theorem tells us that $ \\frac{9}{x}=\\frac{15}{12-x}$ or\n$ 108-9x=15x$\n$ x=\\frac{9}{2}$ Applying the Pythagorean Theorem to triangle ABD gives that AD is approximately 10.06[/hide]" } { "Tag": [ "category theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $R$ a ring, $I$ ideal of $R$, $S$ a multiplicatively closed subset of $R$. Prove that $S^{-1}I=S^{-1}R$ iff $I\\cap S\\neq\\phi$.", "Solution_1": "Is $R$ commutative and what does the notation $S^{-1}$ stand for ?", "Solution_2": "$S^{-1}R$ is as always the localization of $R$ at $S$.", "Solution_3": "Here is my proof:\r\n\r\n$M \\mapsto S^{-1}M$ is an exact functor because $S^{-1}M = S^{-1}R \\otimes_{R}M$ and $S^{-1}R$ is a flat R-module.\r\n\r\nWe have the short exact sequence $0 \\to I \\to R \\to R/I \\to 0$, and therefore the short exact sequence $0 \\to S^{-1}I \\to S^{-1}R \\to S^{-1}(R/I) \\to 0$.\r\n\r\nHere $S^{-1}I \\to S^{-1}R$ is surjective if and only if the kernel of $S^{-1}R \\to S^{-1}(R/I)$ is all of $S^{-1}R$. But this map is also surjective, so this happens if and only if $S^{-1}(R/I) = 0$. \r\n\r\nNow $S^{-1}(R/I) \\simeq \\bar{S}^{-1}(R/I)$ where $\\bar{S}$ is the image of S in R/I. And this is 0 if and only if $0 \\in \\bar{S}$ which happens when and only when $S \\cap I \\neq \\emptyset$.", "Solution_4": "[quote=\"henrik\"]$M \\mapsto S^{-1}M$ is an exact functor because $S^{-1}M = S^{-1}R \\otimes_{R}M$ and $S^{-1}R$ is a flat R-module[/quote]\r\n\r\nUsually, this implication is vice-versa ;)", "Solution_5": "I'm not quite sure about that! May I see a proof? :maybe:", "Solution_6": "you've already got a proof (although, as always, you did not show any cooperation, also in the case of these trivial exercises ...)! what's unclear about it?" } { "Tag": [ "probability" ], "Problem": "A woman wants to open a lock. She has 10 keys. She tries them one by one, setting them aside if the lock does not open. Find the chance\r\nthat the seventh key opens the lock.", "Solution_1": "For the seventh key to open the lock, she must have picked six wrong keys first.\r\n\r\nI think that there is a probability of $ \\frac{9}{10}$ of picking an incorrect key the first time, then $ \\frac{8}{9}$ and so on, giving \r\n\r\n$ \\frac{9}{10} \\cdot \\frac{8}{9} \\cdot \\frac{7}{8} \\cdot \\frac{6}{7} \\cdot \\frac{5}{6} \\cdot \\frac{4}{5} \\cdot \\frac{1}{4} \\equal{} \\frac{1}{10}$.", "Solution_2": "Or you could think of it as \"Of the letters A B C D E F G H I J, what is the probability that A is 7th?\" In this case, A is the key that unlocks the lock and B C D E..... are all the other keys. A has an equal probability of being in any of the 10 spots: aka 1/10 probability of being first, 1/10 probability of being 2nd, 1/10th probability of being 3rd and so on. So A has a 1/10 probability of being 7th. \r\n\r\nYet another way to do this is find the total number of ways to arrange the keys (10!) and then the number of ways to arrange the keys with A being 7th (9!) and divide the two to get 1/10" } { "Tag": [ "articles", "MIT", "college", "Princeton", "Harvard", "Yale", "Stanford" ], "Problem": "I've read some articles saying that the gap between the quality of education at schools ranked say, #1-20 is very minute. Meaning, top schools (MIT, Princeton, Harvard, etc.) compared with good, but not elite schools (Michigan, Minnesota, Rice, Texas, etc.).\r\n\r\nWhat do you guys think about this? How big really is the gap between these schools? And if this gap isn't that big, do you think it's really worth it to travel long distances and pay big bucks for schools like Harvard? A lot of people have told me its much more important to go to a good school for a Graduate/Professional degree than for undergraduate.\r\n\r\nAs someone living in Austin, I will most likely go to Texas (UT Austin), or possibly Rice/A&M, which are among those second-tier schools I mentioned.", "Solution_1": "The general definition of \"second tier\" is very different from yours. Rice and Michigan are in no way second tier schools. What's your definition of first and second tier?", "Solution_2": "Sorry if my meaning is unclear. I mean to compare the very elite, ivy league schools, (ie. Harvard) with good but not elite schools (as I said before, schools like Michigan and Rice).\r\n\r\nJust wondering, what is the general definition of first and second tier?", "Solution_3": "'Lots' of people consider Harvard, Princeton, Yale, MIT, Stanford, and Caltech to be the definitive first tier schools (inasmuch as undergrad is concerned). I personally concur with this, although I'm sure others do not." } { "Tag": [ "group theory", "abstract algebra", "modular arithmetic", "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "This is a very beautiful problem: \r\n a) Let $K$ be a finite field and let $G$ be the set of linear transformations of the form $f(x,y)=(ax+by, y)$ with $a,b\\in K$, $b\\ne 0$. Prove that $G$ has a subgroup of order $n$ if and only if $n$ can be written $NM$ with $N\\parallel K|$, $M\\parallel K|-1$ and $M=1\\pmod N$. \r\n b) If $n$ has the property that any group of order divisible by $n$ has a subgroup of order $n$, then $n$ is a power of a prime. Is the converse true?", "Solution_1": "There some things I don't quite get:\r\n\r\nFirst of all, I think $a$ should be non-zero, not $b$, since $G$ is just the group of matrices of the form $\\begin{pmatrix}a&b\\\\0&1\\end{pmatrix}$. \r\n\r\nThat's is not so important though. What I'm having trouble with is this: the order of $G$ is $|K|\\cdot(|K|-1)$. If we take $n=|K|\\cdot(|K|-1)$ then of course, $G$ has subgroups of order $n$. The only candidates for $N,M$, however, are $N=|K|$ and $M=|K|-1$, and clearly, $M\\ne 1\\pmod N$..:? Should it be the other way around (that is, $N=1\\pmod M$)?\r\n\r\nSorry if I'm missing something obvious.", "Solution_2": "I'm sorry, you are right for the two remarks.", "Solution_3": "Let $K$ be a field with $q=p^{k}$ elements for some prime $p$ and some positive integer $k$. \r\n\r\n\r\n(a)\r\n\r\nLet $H\\le G$ be a subgroup of $G$. Since $G$ has $q(q-1)$ elements, the order $n$ of $H$ is indeed of the form $n=uv,\\ u|q,\\ v|q-1$. We want to prove that $v|u-1$. Notice that the subgroup $P$ of $H$ consisting of matrices of the form $\\begin{pmatrix}1&*\\\\0&1\\end{pmatrix}$ is the unique Sylow $p$-subgroup of $H$ (and so it has cardinality $u$). The action of $\\begin{pmatrix}a&*\\\\0&1\\end{pmatrix}\\ (\\#)$ by conjugation on $P$ depends only on $a$, as does the equivalence class of $(\\#)$ modulo $P$. Moreover, conjugation by $(\\#)$ sends $\\begin{pmatrix}1&b\\\\0&1\\end{pmatrix}$ to $\\begin{pmatrix}1&ab\\\\0&1\\end{pmatrix}$. This tells us what the orbits of this action look like: the orbit of the identity is, of course, trivial, and all the other orbits have precisely $\\frac{|H|}{|P|}=v$ elements. This shows that $P\\setminus\\{1\\}$ is partitioned into $v$-element subsets, so $v|u-1$, as desired.\r\n\r\nConversely, let $n=uv,\\ u|q,\\ v|q-1$ and $v|u-1$. Suppose $u=p^{l},\\ l\\le k$. Since $v$ divides both $p^{k}-1$ and $p^{l}-1$, it divides $p^{d}-1$, where $d=(k,l)$. The roots of the polynomial $x^{p^{d}}-x$ form a subfield $K'$ of $K$ with $p^{d}$ elements. Let $V$ be a vector subspace of $K$ over $K'$ of dimension $\\frac{l}{d}$ (and hence of cardinality $u=p^{l}$), and let $a$ be a primitive $v$\u2019th root of $1$ contained in $K'$. The subgroup of $\\mbox{Gl}_{2}(K)$ generated by the groups $\\left\\{\\begin{pmatrix}a^{n}&0\\\\0&1\\end{pmatrix}\\ |\\ n\\in\\mathbb Z\\right\\}$ and $\\left\\{\\begin{pmatrix}1&b\\\\0&1\\end{pmatrix}\\ |\\ b\\in V\\right\\}$ has the properties we\u2019re looking for.\r\n\r\n\r\n(b)\r\n\r\nThis is a direct consequence of (a). Let $n=p_{1}^{a_{1}}\\cdot\\ldots\\cdot p_{t}^{a_{t}},\\ t\\ge 2$, and assume that $p_{1}^{a_{1}}$ is the smallest among all $p_{i}^{a_{i}},\\ i=\\overline{1,t}$. Then we can find a field $K$ with $p^{k}=p_{1}^{k}$ elements for some large $k$ such that $n$ divides the order of the group $G$ defined as above, but $n$ does not satisfy the condition proved necessary for $G$ to have a subgroup of order $n$ (this condition is $\\frac n{p_{1}^{a_{1}}}\\ |\\ p_{1}^{a_{1}}-1$, which is clearly impossible because of the minimality of $p_{1}^{a_{1}}$ among $p_{i}^{a_{i}}$).\r\n\r\n\r\nP.S.\r\n\r\nThe converse of (b) is true and well-known :)." } { "Tag": [ "Support", "abstract algebra", "calculus", "integration" ], "Problem": "Me and my mate have been trying to solve this question for 2 weeks now and we haven't made any progress. We were given it for a challenge by our maths teacher and he knows we won't be able to solve it, he just likes toying with us, but i want to rpove him wrong...with your help :P\r\n\r\nThe question is as follows:\r\n\r\nThe pilot of a small plane is attempting to land on a small airstrip, and the wheels hit the runway only 1000 metres short of a large group of trees at the end of the runway. The plane\u2019s speed on the tarmac is modelled by the equation v = A \u2013 10^kt, where v the plane\u2019s velocity in m/s, t is the time taken in seconds to come to a complete halt, and A and k are constants.\r\n\r\n\r\nb) Investigate at least 3 different values of k between (but NOT including) 0.01 and 0.2 and for each value determine whether the plane comes to a complete stop before the end of the runway. Use any graphs and/or diagrams, any numerical techniques and any technology to support your arguments, justify your procedures. What physical conditions could make k vary? List any assumptions you have made. \r\n\r\nWould there be any values of k (not just the values between 0.01 and 0.2) for which this equation could not be used to model the velocity of a plane landing on a tarmac? \r\n\r\nMy friend does the hardest maths module at my high-school and yet still cannot get this question out, neither can i. I know it is a lost of physics as well but i think there may be some integration needed to solve. Any help would be appreciated.\r\n\r\nCheers", "Solution_1": "Hi, welcome to the forum.\r\nI've got two questions for you.\r\n1. Do you know the value of $A$? I don't think you can solve the problem without it.\r\n2. Can you integrate simple expressions like $10^{kt}$? Is it at your level?", "Solution_2": "\"t is the time taken in seconds to come to a complete halt\"\r\nso, is t a variable? I'll assume it is.\r\n\r\na) find the t value when v is zero, that would be the time it takes for a complete stop. integrate v from 0 to this value, that is the stopping distance (in terms of k). Then see when the distance is less than 1000m\r\n\r\nb) k cannot be negative or zero, since the plane is supposed to slow down, not speed up. If you want to be more precise, the model isn't a very good model at all, since either case, (k>0 or not), the limit of v as t goes to infinity is non-zero, unless A is zero and k is negative." } { "Tag": [ "function", "induction", "algebra proposed", "algebra" ], "Problem": "Find all $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ which satisfies\r\n$ f(xy \\plus{} f(x \\plus{} y)) \\equal{} xy \\plus{} f(x) \\plus{} f(y)$\r\nfor all $ x,y \\in \\mathbb{R}$.", "Solution_1": "Here's my progress (Still not coming up with the solution. :( )\r\n\r\nSubstitute $ y \\leftarrow 0$, $ f(f(x)) \\equal{} f(x) \\plus{} f(0)$. Denote by $ a$ the term $ f(0)$. This means $ f(y) \\equal{} y \\plus{} a \\forall y \\in Im(f)$.\r\n\r\nConsider any $ y \\in Im(f)$, substituting $ x \\rightarrow a$ yields $ f(ay \\plus{} f(a \\plus{} y)) \\equal{} ay \\plus{} f(a) \\plus{} f(y)$ (1).\r\n\r\nNote that $ a \\plus{} y \\equal{} f(y) \\in Im(f)$ and $ f(a) \\equal{} 2a$, $ (1) \\Rightarrow f(ay \\plus{} y \\plus{} 2a) \\equal{} ay \\plus{} y \\plus{} 2a$\r\n$ \\Rightarrow ay \\plus{} y \\plus{} 2a \\in Im(f) \\Rightarrow f(ay \\plus{} y \\plus{} 2a) \\equal{} ay \\plus{} y \\plus{} 3a$\r\nSo, $ ay \\plus{} y \\plus{} 2a \\equal{} ay \\plus{} y \\plus{} 3a$ or $ a \\equal{} 0$.\r\n\r\nFor now, we have: $ f(0) \\equal{} 0$ and $ f(x) \\equal{} x \\forall x \\in Im(f)$.\r\n\r\nHow do we proceed ? Could we prove that $ f$ is surjective ?", "Solution_2": "[quote=\"mathangel\"]\nNote that $ a \\plus{} y \\equal{} f(y) \\in Im(f)$ and $ f(a) \\equal{} 2a$, $ (1) \\Rightarrow f(ay \\plus{} y \\plus{} 2a) \\equal{} ay \\plus{} y \\plus{} 2a$[/quote]\r\n\r\nI think there is a $ a$ missing in the final RHS, since $ f(a)\\plus{}f(y) \\equal{} 2a\\plus{}y\\plus{}a$.\r\nIt follows that we don't have $ a\\equal{}0$, which is confirmed by the fact that any function of the form $ f(x)\\equal{}x\\plus{}c$, where $ c$ is a constant, is a solution of the problem.\r\n\r\nPierre.", "Solution_3": "Anyone with a complete solution?", "Solution_4": "[b]Step 1.[/b] If $y=0$ in the original equation, then\n\\[f(f(x))=f(x)+f(0),\\]\nso if $x$ is in the image of $f$, then $f(x)=x+f(0)$.\n\n[b]Step 2.[/b] If $x=1+f(x)$, $y=-1$ in the original equation, then\n\\[f(-1-f(x)+f(f(x)))=-1-f(x)+f(1+f(x))+f(-1),\\]\nso by step 1, \n\\[f(-1+f(0))=-1-f(x)+f(1+f(x))+f(-1).\\]\nIf $x=1$, $y=-1$ in the original equation, then\n\\[f(-1+f(0))=-1+f(1)+f(-1).\\]\nBy comparing these two equations, we get\n\\[f(1+f(x))=f(1)+f(x).\\]\n\n[b]Step 3.[/b] If $x=y=1$ in the original equation, then\n\\[f(1+f(2))=1+2f(1),\\]\nso by step 2,\n\\[f(1)+f(2)=1+2f(1),\\]\nso $f(2)=1+f(1)$. Then by step 1,2,\n\\[f(0)+f(2)=f(f(2))=f(1+f(1))=f(1)+f(1),\\]\nso $f(1)=f(0)+1$.\n\n[b]Step 4.[/b] Assume $f(x)=x+f(0)$. If $x=x+1$, $y=-1$ in the original equation, then by the proof of step 2,\n\\[-1+f(1)+f(-1)=f(-1+f(0))=f(-x-1+f(x))=-x-1+f(x+1)+f(-1),\\]\nso $f(x+1)=x+f(1)=x+1+f(0)$ by step 3. Also, $f(0)=0+f(0)$, so by induction, we get $f(n)=n+f(0)$ for all [i]nonnegative integers[/i] $n$, and $f(f(x))=f(x)+f(0)$, so by induction, we get $f(f(x)+n)=f(x)+n+f(0)$ for all [i]nonnegative integers[/i] $n$.\n\n[b]Step 5.[/b] Let $m$ be an integer greater than $2$. If $x=m$, $y=-m$ in the original equation, then\n\\[f(-m^2+f(0))=-m^2+f(m)+f(-m)=-m^2+m+f(0)+f(-m)\\]\nby step 4. If $x=m^2-2$, $y=-2$ in the original equation, then\n\\[f(-m^2+f(0))=f(-2m^2+4+f(m^2-4))=-2m^2+4+f(m^2-2)+f(-2)=-m^2+2+f(-2)+f(0)\\]\nby step 4. By comparing these two equations, we get\n\\[f(-m)=f(-2)-m+2.\\]\nTherefore, $f(-3)=f(-2)-1$. If $x=-2$, $y=-1$ in the original equation, then\n\\[f(0)+f(-2)+1=f(1)+f(-2)=f(1+f(-2))=f(2+f(-3))=2+f(-2)+f(-1),\\]\nby step 2,3, so $f(-1)=f(0)-1$. If $x=-1$, $y=-1$ in the original equation, then\n\\[f(1)+f(-2)=f(1+f(-2))=1+f(-1)+f(-1)\\]\nby step 2, so $f(-2)=1+2f(-1)-f(1)=f(0)-2$ by step 3. Then\n\\[f(-m)=f(-2)-m+2=f(0)-m.\\]\nTherefore, by step 4, $f(n)=n+f(0)$ for all [i]integers[/i] $n$.\n\n[b]Step 6.[/b] If $x=f(x)+2$, $y=-2$ in the original equation, then\n\\[f(-2f(x)-4+f(f(x)))=-2f(x)-4+f(f(x)+2)+f(-2).\\]\nBy step 4,5, we get\n\\[f(-f(x)-4+f(0))=-2f(x)-4+f(x)+2+f(0)-2+f(0)=-f(x)-4+2f(0).\\]\nIf $x=f(x)-4+f(0)$ in this equation, then\n\\[f(-(-f(x)-4+2f(0))-4+f(0))=-(-f(x)-4+2f(0))-4+2f(0),\\]\ni.e., \n\\[f(f(x)-f(0))=f(x).\\]\nTherefore, if $x$ is in the image of $f$, then $f(x-f(0))=x$.\n\n[b]Step 7.[/b] If $y=-1$ in the original equation, then\n\\[f(-x+f(x-1))=-x+f(x)+f(-1)=-x+f(x)-1+f(0)\\]\nby step 5, so $-x+f(x)-1+f(0)$ is in the image of $f$. By step 6,\n\\[f(-x-1+f(x))=-x+f(x)-1+f(0).\\]\nIf $x=x+1$, $y=-1$ in the original equation, then\n\\[f(-x-1+f(x))=-x-1+f(x+1)+f(-1)=-x-1+f(x+1)-1+f(0)\\]\nby step 5. By comparing these two equations, we get\n\\[f(x+1)=f(x)+1.\\]\n\n[b]Step 8.[/b] If $y=1$ in the original equation, then\n\\[1+f(x+f(x))=f(x+f(x+1))=x+f(x)+f(1)=x+f(x)+1+f(0)\\]\nby step 7, so $f(x+f(x))=x+f(x)+f(0)$.\n\n[b]Step 9.[/b] If $x=x+1$, $y=f(x)-1$ in the original equation, then\n\\[f((x+1)(f(x)-1))+f(x+f(x)))=(x+1)(f(x)-1)+f(x+1)+f(f(x)-1)),\\]\nso by step 8, we get\n\\[f(xf(x)+2f(x)+f(0)-1)=xf(x)+f(x)-x-1+f(x+1)+f(f(x)-1),\\]\nand then by step 1,7, we get\n\\[f(xf(x)+2f(x)+f(0))=xf(x)+3f(x)-x+f(0).\\]\nIf $y=f(x)$ in the original equation, then\n\\[f(xf(x)+x+f(x)+f(0))=f(xf(x)+f(x+f(x)))=xf(x)+f(x)+f(f(x))=xf(x)+2f(x)+f(0)\\]\nby step 1,8, so $xf(x)+2f(x)+f(0)$ is in the image of $f$. Then by step 1, we get\n\\[f(xf(x)+2f(x)+f(0))=xf(x)+2f(x)+2f(0).\\]\nTherefore,\n\\[xf(x)+3f(x)-x+f(0)=xf(x)+2f(x)+2f(0),\\]\nso $f(x)=x+f(0)$.\n\nIt is easy to verify that this satisfies the original equation. Q.E.D.", "Solution_5": "What is $Im(f)$?", "Solution_6": "[quote=Ryoma.Echizen]What is $Im(f)$?[/quote]\nimage/range of $f$, $f(\\mathbb{R})$" } { "Tag": [ "AMC", "AIME", "geometry", "email", "AIME II" ], "Problem": "hi guys\r\n\r\nIm in the toronto/scarborough area, my school just told me that there have been a little problem in ordering AIME II --they were asked to send in money before AIME but our school doesnt allow that, somehow. so the math department emailed AIME but no reply yet.\r\n\r\nSo i'm wondering if any of your schools, if, in that area, offers AIME to students who took AMC in their own school?\r\n\r\n\r\nThanks!", "Solution_1": "[quote=\"sansan\"]they were asked to send in money before AIME but our school doesnt allow that, somehow. [/quote]\r\n\r\nDoesn't allow?\r\nWhat a [b][color=red]-censored-[/color][/b] school policy\r\n\r\n\r\n(censored on my own)", "Solution_2": "LOl I don't know\r\ni think because the money has to go through our school board, that's why\r\nthe school board has some policies regarding prepayments...they somehow want to receive the exams before they pay.....can i somehow write it though?", "Solution_3": "First of all, you should ask someone from your school to [b]call[/b] the AMC to figure out what's going on (the number is on the inside cover of the AIME Teachers' Manual).\r\n\r\nThe fee is only 25 dollars US. (Maybe more for Canada.) It's absurd to me that a school won't fork over such a small amount of money (for a school) without having to go through layers of bureaucracy. The AMC does accept credit cards so maybe your parents can just pay the fee directly (if they are so inclined).\r\n\r\nThe problem with taking it at a different school is that for the AIME II, each school generally only has exactly enough copies of the contest for their own students, and no extras. You might be able to arrange to take it at a different school if their AIME exams haven't been shipped yet.\r\n\r\nBut I would start by having your teacher call the AMC and seeing what your options are. Good luck!", "Solution_4": "HI\r\n\r\nMy teacher has emailed the coordinator but I asked this morning, he said there's still no reply yet.\r\n\r\ndoes anyone know if the email they provide on the manual is active? people do use that right?\r\n\r\nthanks!", "Solution_5": "The most obvious solution is to ask your parents, or some other sympathic adult, to write the check. It isnt that much money.", "Solution_6": "well thanks everyone\r\n\r\nI've tried to ask my teacher about it, but he said he would take care of it, so i don't want to sound/act as if I can't trust him.\r\n\r\neven though i really want to pay the 25 dollars on my own, i dont know how to ask him...\r\n\r\nshould i keep waiting or what?\r\n\r\nthanks!", "Solution_7": "I would just recommend that you talk to whoever runs the competitions and pay.\r\n\r\nLast year, I had to pay for the exam, as my school would never imagine paying for anything like that. (In fact, we have to pay 3 dollars each for the AIME.) Their philosophy is that if they didn't force you to do the competition, they won't pay for it. Anyway, all I'm saying is that it's certainly possible for you to pay, as I did last year.", "Solution_8": "hi calc\r\n\r\nhow did you pay last year?\r\ndid you write a checque and mail it to AMC organizers?\r\nwhen did you pay, did you pay before your AIME? do you still know the deadline?\r\n\r\n\r\nthanks!", "Solution_9": "[quote=\"sansan\"]hi calc\n\nhow did you pay last year?\ndid you write a checque and mail it to AMC organizers?\nwhen did you pay, did you pay before your AIME? do you still know the deadline?\n\n\nthanks![/quote]\r\n\r\nI think all I did was give 25 dollars to the math department head at my school, though I don't exactly remember.", "Solution_10": "25 dollars in checque right?\r\n\r\nalso, did you pay before AIME ?\r\n\r\ndo you think i still have enough time...? i should pay it tomorrow then!", "Solution_11": "Yes, the email is active. It is checked and replied to daily, provided the mail was sent to the address on the contest booklets. (We have other addresses used for other purposes, and those may not be answered daily as workflow increases or decreases.)\r\n\r\nThis is a simple problem that can be easily resolved between the school and the AMC office. We have business procedures that allow a teacher to order with out prepayment in cash. The best procedure is for the teacher to call the AMC office and make arrangements.\r\n\r\n It does not make sense of a number of non-involved students to attempt to resolve the issue here without the teacher and without the AMC office. Please allow teachers and the AMC office to conduct regular business with our regular procedures. Some trust in your teachers would also be well-placed. \r\n\r\nSteve Dunbar\r\nDirector, American Mathematics Competitions", "Solution_12": "ok thanks!" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "MATHCOUNTS", "search" ], "Problem": "i wasnt sure where to put this post, sorry if its in the wrong place. i was wondering for all you amc, aime, & usamo people out there, did you attend any kind of math program or participate in some workshop, or read some book to improve your scores? thanks in advance :lol:", "Solution_1": "You've come to the right place :D\r\n\r\nMath Programs: Tons of them out there, you can find info about them in the \"Other US Contests & Programs\" forum.\r\nClasses: AoPS offers classes ranging from MathCounts to Olympiad level. Go to the home page and click on the \"Classes\" link.\r\nBooks: AoPS v1 and v2 are very useful for AMC/AIME and perhaps even USAMO. There's a lot of talk about other books randomly in the forum, so you can search them up." } { "Tag": [], "Problem": "Yann and Camille go to a restaurant. If there are 10 items on the menu, and each orders one dish, how many different combinations of meals can Yann and Camille order if they refuse to order the same dish?", "Solution_1": "Yann can order 1 of 10 dishes and Camille can order 1 of 9 dishes (since they don't order the same one)\r\n\r\ntherefore, there are 90 possibilities", "Solution_2": "Note that vallon got this by doing $ 10 \\times 9 \\equal{} 90$.", "Solution_3": "Since I hace 10 things and the friends has 9 the total is just 90. (remember you don't eivide by two since we have distinct people." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Hey I have a question that I can't seem to figure out. I have to find the limit as n goes to infinity of:\r\n\r\nlim (X^n/(1+X^n)) \r\n\r\nAnd this should equal 0, but I keep getting 1. Any help would be greatly appreaciated!\r\n\r\nThanks", "Solution_1": "Set $f_n(x)=\\frac{x^n}{1+x^n}$. Then \\[ \\lim_{n\\rightarrow\\infty}f_n(x)=\\begin{cases}1,& x>1\\\\ \\frac{1}{2},& x=1\\\\ 0,&1\\ge x\\ge 0\\end{cases} \\] I think the limit does not exist for $x<0$." } { "Tag": [ "trigonometry", "calculus", "conics", "parabola", "analytic geometry", "graphing lines", "slope" ], "Problem": "A certain football quarterback can throw a football a maximum range of 80 meters on level ground. What is the highest point reached by the football if thrown this maximum range? Ignore air friction. \r\n\r\na) 10m\r\nb) 20m\r\nc) 30m\r\nd) 40m\r\ne) 50m\r\n\r\nHere's my long-winded [hide=\"solution\"] \n\nLet $ v_{o}$ be the initial speed, $ t$ the total time in flight and $ h$ the maximum height.\n\nWe have 3 equations:\n\n1) Height and time: $ h \\equal{} v_{o}sin\\theta \\frac{t}{2} \\minus{} \\frac{1}{2}a(\\frac{t}{2})^2$\n\n2) Range and time: $ R \\equal{} v_{o}cos\\theta t$ or $ t \\equal{} \\frac{R}{v_{o}cos\\theta}$\n\n3) Conservation of energy: $ \\frac{1}{2}mv_{o}^2 \\equal{} \\frac{1}{2}m(v_{o}cos\\theta)^2 \\plus{} mgh$ or equivalently $ v_{o}^2 \\equal{} \\frac{gh}{\\frac{1}{2} \\minus{} \\frac{1}{2}cos^2\\theta}$\n\nNow to plug E2 into E1:\n\n$ h \\equal{} v_{o}sin\\theta \\frac{R}{2v_{o}cos\\theta} \\minus{} \\frac{1}{2}a(\\frac{R}{2v_{o}cos\\theta})^2 \\equal{} \\frac{rtan\\theta}{2} \\minus{} \\frac{aR^2}{8v_{o}^2cos^2\\theta}$\n\nPlugging E3 into the above equation:\n\n$ h \\equal{} \\frac{rtan\\theta}{2} \\minus{} \\frac{aR^2(\\frac{1}{2} \\minus{} \\frac{1}{2}cos^2\\theta)}{8ghcos^2\\theta}$\n\nSince the maximum range is achieved only when $ \\theta \\equal{} \\frac{\\pi}{4}$:\n\n$ h \\equal{} \\frac{r}{2} \\minus{} \\frac{aR^2(\\frac{1}{2} \\minus{} \\frac{1}{2}(\\frac {\\sqrt{2}}{2})^2)}{8gh(\\frac {\\sqrt{2}}{2})^2} \\equal{} \\frac{R}{2} \\minus{} \\frac{aR^2}{16gh}$\n\nThus, $ h \\equal{} 20$ [/hide]\r\n\r\nWith that ugly mess written, can someone suggest an easier method?", "Solution_1": "This one can be done without using energy.\r\n\r\n[hide=\"solution\"]$ v_0 \\sin(\\theta) t \\equal{} 80$ [i][b](1)[/b][/i]\n$ v \\equal{} v_0 \\plus{} at \\implies 2v_0 \\cos\\theta \\equal{} gt \\implies t \\equal{} \\frac {2v_0\\cos\\theta}{g}$ [i][b](2)[/b][/i]\n\nPlug [i][b](2)[/b][/i] back in to [i][b](1)[/b][/i] to get $ \\frac {2v_0^2\\sin\\theta\\cos\\theta}{g}$. And since $ \\sin\\theta \\equal{} \\cos\\theta$ when $ \\theta \\equal{} \\frac {pi}{4}$, this simplifies to $ \\frac {2v_0^2\\cos^2\\theta}{g}$[i][b] (3)[/b][/i]\n\nSo, for the vertical motion, $ v \\equal{} v_0^2 \\plus{} 2at$ and plugging in values will get $ 0 \\equal{} (v_0\\cos\\theta)^2 \\minus{} 2gx \\implies x \\equal{} \\frac {(v_0\\cos\\theta)^2}{2g}$. [i][b](4)[/b][/i]\n\nPlug the above equation [i][b](3) [/b][/i]into [i][b](4)[/b][/i] to get $ x \\equal{} \\frac {40g}{2g} \\equal{} \\boxed{20m}$.[/hide]", "Solution_2": "[quote=\"pascal12\"]This one can be done without using energy.\n\n[hide=\"solution\"]$ v_0 \\sin(\\theta) t \\equal{} 80$\n$ v \\equal{} v_0 \\plus{} at \\implies 2v_0 \\cos\\theta \\equal{} gt \\implies t \\equal{} \\frac {2v_0\\cos\\theta}{g}$\n\nPlug it back in to get $ \\frac {2v_0^2\\sin\\theta\\cos\\theta}{g}$. And since $ \\sin\\theta \\equal{} \\cos\\theta$ when $ \\theta \\equal{} \\frac {pi}{4}$, this simplifies to $ \\frac {2v_0^2\\cos^2\\theta}{g}$\n\nSo, for the vertical motion, $ v \\equal{} v_0^2 \\plus{} 2at$ and plugging in values will get $ 0 \\equal{} (v_0\\cos\\theta)^2 \\minus{} 2gx \\implies x \\equal{} \\frac {(v_0\\cos\\theta)^2}{2g}$. \n\nPlug the above equation to get $ x \\equal{} \\frac {40g}{2g} \\equal{} \\boxed{20m}$.[/hide][/quote]\r\n\r\nThanks. Your \"it\"s and \"plug the equation\" without references to the target were kinda confusing, but I understand that the three main equations that you used are:\r\n\r\n1) $ v_{o}sin\\theta t \\equal{} 80$\r\n\r\n2) $ v \\equal{} v_{o} \\plus{} at$\r\n\r\n3) $ 0 \\equal{} (v_{o}sin\\theta)^2 \\minus{} 2gx$\r\n\r\nI guess your method is shorter and involves less algebra because you used different equations from what I used.", "Solution_3": "Oh yeah, sorry, I should have made that clearer. Ok, I edited it and hopefully it should make more sense.\r\n\r\nAlso, I generally try to avoid the equation $ x \\equal{} vt \\plus{} \\frac {1}{2}at^2$ because unless $ v \\equal{} 0$, it usually leads to something really messy.", "Solution_4": "Have you taken calculus? If so, you will know some facts (unrelated to physics) that will make this problem doable with almost no calculation.\r\n\r\n[hide=\"Comments\"][b]Facts:[/b]\n- The ball follows a parabolic path\n- All parabolas are similar\n- The distance is maximized at an angle of 45 degrees, which corresponds to a slope of magnitude 1.\n\nI just want the ratio of the height to the distance for this throw. I can use any parabola I want, since they are all similar! I'll use $ y \\equal{} x^2$. The slope has magnitude one at $ x \\equal{} \\pm 1/2$. The height of the portion of the parabola within these bounds is $ \\frac{1}{4}$, and he width is just the difference in $ x$ values, or $ 1$, so the height is 1/4 of the width.[/hide]\r\n\r\nWhile I certainly am not encouraging anyone to use methods like that to solve problems, it can be helpful when there is little time left, or for checking answers in a different way.", "Solution_5": "That's a really ingenious way to solve the problem. I'll definitely look out for these mathematical methods from now on. Thanks!", "Solution_6": "It's useful to know the range and height formulas for projectile motion:\r\n\r\n$ R \\equal{} \\frac{v_0^2 \\sin(2\\theta)}{g}$\r\n\r\n$ H \\equal{} \\frac{v_0^2 \\sin^2(\\theta)}{2g}$", "Solution_7": "let: y=-(x/a)^2+(x/b)\r\n\r\nfrom dy/dx = 1 at x=0 we get b = 1\r\n\r\nhence range is a (the other solution -(x/a)^2+(x) = 0)\r\n\r\nheight is y at a/2." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Mykolka the numismatist possesses $241$ coins, each worth an integer number of turgiks. The total value of the coins is $360$ turgiks. Is is necessarily true that the coins can be divided into three groups of equal total value ? :maybe:", "Solution_1": "[quote=\"RDeepMath91\"]Mykolka the numismatist possesses $241$ coins, each worth an integer number of turgiks. The total value of the coins is $360$ turgiks. Is is necessarily true that the coins can be divided into three groups of equal total value ? :maybe:[/quote]\r\n\r\nLet $a_{1}$ the number of coins whose value is exactly $1$ turgik. Considering that all the other coins have a value of at least $2$, we have $a_{1}+2(241-a_{1})\\leq 360$, and so $a_{1}\\geq 122$\r\nSince we have $241$ coins whose total value is $360$, the biggest coin value possible is $120$\r\n\r\nNow split all the $241-a_{1}$ coins whose values are greater than $1$ turgik in two groups A and B whose difference of values is minimum, say $B=A+\\Delta$. Take then one coin in B (any coin). Let $m$ its value. Obviously, moving this coin from group B to group A may not reduce the difference between A and B (since $\\Delta$ is supposed the minimum one) and so $m\\geq \\Delta$.\r\nSince $A+\\Delta+A+a_{1}=360$ and since $a_{1}\\geq 122$, we have $A=180-\\frac{a_{1}}{2}-\\frac{\\Delta}{2}< 120$\r\n\r\nConsider now the 3 groups :\r\n$F$ is the group of $a_{1}$ $1$-turgik coins plus the selected $m$-turgiks coin : Total is $a_{1}+m$\r\n$G$ is the group $A$ : Total is $A=180-\\frac{a_{1}}{2}-\\frac{\\Delta}{2}<120$\r\n$H$ is the group $B$ minus the selected $m$-turgiks coin. Total is $A+\\Delta-m=180-\\frac{a_{1}}{2}+\\frac{\\Delta}{2}-m\\leq A<120$\r\n\r\nIn order to complete group $G$ up to $120$, we need $C_{G}=120-A=\\frac{a_{1}}{2}+\\frac{\\Delta}{2}-60$\r\nIn order to complete group $H$ up to $120$, we need $C_{H}=\\frac{a_{1}}{2}-\\frac{\\Delta}{2}+m-60$\r\n\r\nAnd $C_{G}+C_{H}=a_{1}+m-120$ and, since no coin may have a value $>120$, $C_{G}+C_{H}\\leq a_{1}$\r\n\r\nHence, we have enough $1$-turgik coins to move from $F$ to groups $G$ and $H$ to value them at $120$ (and group $F$ will then also have value of $120$).\r\n\r\nAnd it is true that the coins can be divided into three groups of equal total value.\r\n\r\n-- \r\nPatrick", "Solution_2": "That's a nice solution.. Thx, pco :)" } { "Tag": [ "inequalities", "trigonometry", "geometry", "3D geometry" ], "Problem": "SHow that \r\n\r\n\\[ 6n+5 < 3(\\sqrt[3]{n^2(n+2)} + \\sqrt[3]{n(n+2)^2}) < 6n+6 \\]", "Solution_1": "\\[ 3 \\sqrt[3]{n\\cdot n \\cdot (n+2)} \\leq 3n+2 \\]\r\n\\[ 3 \\sqrt[3]{n (n+2)(n+2)} \\leq 3n+4 \\].\r\nCombine it (equality can't occur it $n \\in N^*$) we get:\r\n\\[ 3(\\sqrt[3]{n^2(n+2)} + \\sqrt[3]{n(n+2)^2}) < 6n+6 \\]", "Solution_2": "I'm afraid the left side is only true for $n > \\frac{25}{12}$...\r\nTake for example $n=1$ or $n=0$, and the left side doesn't hold.\r\n\r\nNow if we assume that $n > \\frac{25}{12}$, it is easy to prove:\r\n\r\nBy AM-GM:\r\n\r\n$3(\\sqrt[3]{n^2(n+2)} + \\sqrt[3]{n(n+2)^2}) \\geq 6 \\sqrt{n(n+2)}$\r\n\r\nNow, because $n > \\frac{25}{12}$, it's true that \\[ \\frac{n}{3} > \\frac{25}{36} \\]\r\nSo \\[ n^2+2n > n^2 + \\frac{5n}{3} + \\frac{25}{36} \\]\r\nAnd taking the square root of both sides yields \\[ \\sqrt{n(n+2)} > n + \\frac{5}{6} \\]\r\n\r\nPlugging this in the original inequality gives us \\[ 3(\\sqrt[3]{n^2(n+2)} + \\sqrt[3]{n(n+2)^2}) \\geq 6 \\sqrt{n(n+2)} > 6n + 5 \\]\r\n\r\nSo it works in case $n > \\frac{25}{12}$", "Solution_3": "Notice that between the inequality sines, we have something in the form $3a^2b + 3ab^2$, though cube roots are involved. Does that bring any methods ideas to mind?" } { "Tag": [], "Problem": "If a curve with a radius of 88m is perfectly banked for a car traveling 75 km/hr, what must be the coefficient of static friction for a car not to skid when traveling at 95 km/hr.", "Solution_1": "I got [hide=\"ans\"]0.304.[/hide]\r\n\r\n\r\nclarification: when the road is \"perfectly banked,\" does that mean no static friction is needed to keep the car on the road?", "Solution_2": "yes, and your answer is wrong : (\r\n\r\nI'll post a solution tomorrow if you need it.", "Solution_3": "whoops, forgot to base my $ \\mu_s$ on the normal force.\r\n\r\n[hide]\n.352?\n[/hide]", "Solution_4": "noope =[\r\n\r\nhmm i don't feel like posting the solution\r\nmaybe this will get you to the answer\r\nhttp://en.wikipedia.org/wiki/Banked_curve#Banked_turns_with_friction", "Solution_5": "actually, that basically gives the answer to the solution... :|\r\n\r\nbleh, now I'm too lazy.\r\n\r\n\r\nhad my centripetal force going the wrong direction.\r\nit's been too long...", "Solution_6": "Yeah, the most difficult part (for me at least) was realizing that the frictional force does not go towards the center of the circle but rather vertically below that center and down the incline.", "Solution_7": "Friction force is parallel to contact surfaces." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "How many permutations $ (w_1, \\cdots, w_n)$ of $ (1,...,n)$ such that $ w_i \\not\\equal{} i$ or $ i\\minus{}1$?", "Solution_1": "The sequence is [url=http://www.research.att.com/~njas/sequences/A000271]A000271[/url] in the OEIS, which has a simple recursion that one could probably prove similarly to the proof for derangements. Similar to [url=http://www.research.att.com/~njas/sequences/A000179]A00179[/url]." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let a, b, c and\r\n[img]http://www.mathlinks.ro/Forum/latexrender/pictures/edf22308435cfdd8054e75beb6ea7eeb.gif[/img]\r\nProve that\r\n[img]http://www.mathlinks.ro/Forum/latexrender/pictures/45564c1892677ae2016aba9cdd9a50ff.gif[/img]\r\n :(", "Solution_1": "Please do not post the same topic in different forums." } { "Tag": [ "geometry", "circumcircle", "similar triangles", "geometry proposed" ], "Problem": "Let a triangle $ ABC$ inscribed in the circle $ (O)$. The bisectors of angles $ ABC$, $ ACB$ meet $ (O)$ at again $ M$, $ N$ respectively. Let $ D$ on the opposite ray of the ray $ BC$ and $ E$ on the opposite ray of the ray $ CB$ such that $ BD \\equal{} BA$, $ CE \\equal{} CA$. Denote $ O_1,O_2$ be the circumcenter of triangles $ NBD$, $ MCE$ respectively. Prove that $ \\angle O_1AB \\equal{} \\angle O_2AC$.\r\n\r\nP.s. Keepsake for 2 years on MathLinks. :lol: \r\n26th January, 2010. :)", "Solution_1": "My solution", "Solution_2": "We will prove the proposed problem, in its more general formulation a follows :\r\n\r\n[b][size=100][color=DarkBlue] A triangle $ \\bigtriangleup ABC$ is given and let $ \\bigtriangleup ABD,\\ \\bigtriangleup AMC$ be, two similar triangles with $ \\angle BAD \\equal{} \\angle MAC$ and $ \\angle ADB \\equal{} \\angle ACM,$ erected on its side-segments $ AB,\\ AC$ respectively, outwardly to it. Also, let $ \\bigtriangleup ABN,\\ \\bigtriangleup AEC$ be, two other similar triangles with $ \\angle BAN \\equal{} \\angle EAC$ and $ \\angle ABN \\equal{} \\angle AEC,$ erected by the same way. Prove that $ \\angle O_{1}AB \\equal{} \\angle O_{2}AC,$ where $ O_{1},\\ O_{2}$ are the circumcenters of the triangles $ \\bigtriangleup BDN,\\ \\bigtriangleup CEM$ respectively.[/color][/size][/b]\r\n\r\n( In the schema of the solution by [b]livetolove212[/b], we remind that $ \\angle BAD \\equal{} \\angle ADB \\equal{} \\frac {\\angle B}{2} \\equal{} \\angle CAM \\equal{} \\angle ACM$ and $ \\angle ABN \\equal{} \\angle BAN \\equal{} \\frac {\\angle C}{2} \\equal{} \\angle CAE \\equal{} \\angle AEC$ ).\r\n\r\n[b]PROOF OF THE GENERALIZED PROBLEM.[/b] - From the similar triangles $ \\bigtriangleup ABD,\\ \\bigtriangleup AMC,$ we have that $ \\frac {AD}{AC} \\equal{} \\frac {AB}{AM}$ $ ,(1)$\r\n\r\nFrom the similar triangles $ \\bigtriangleup ABN,\\ \\bigtriangleup AEC,$ $ \\Longrightarrow$ $ \\frac {AN}{AC} \\equal{} \\frac {AB}{AE}$ $ ,(2)$\r\n\r\nFrom $ (1),\\ (2)$ $ \\Longrightarrow$ $ \\frac {AD}{AE} \\equal{} \\frac {AN}{AM}$ $ ,(3)$\r\n\r\nFrom $ (3)$ and because of $ \\angle DAN \\equal{} \\angle EAM,$ we conclude that the triangles $ \\bigtriangleup AND,\\ \\bigtriangleup AME$ are similar and so, we have that $ \\angle ADN \\equal{} \\angle AEM$ $ ,(4)$\r\n\r\n$ \\bullet$ We denote the points $ G\\equiv (O_{1})\\cap AD,\\ D'\\equiv (O_{1})\\cap AN$ and then, we have also the similarity of the triangles $ \\bigtriangleup AGD',\\ \\bigtriangleup AME$. \r\n\r\nWe denote the point $ H\\equiv (O_{2})\\cap AE$ and it is easy to show that $ \\angle AGN \\equal{} \\angle DBN \\equal{} \\angle ABD \\minus{} \\angle ABN \\equal{} \\angle AMC \\minus{} \\angle AEC$ \r\n\r\n$ \\Longrightarrow$ $ \\angle AGN \\equal{} \\angle AMC \\minus{} \\angle HMC \\equal{} \\angle AMH$ $ ,(5)$\r\n\r\nFrom $ (5),$ in the configuration of the similar triangles $ \\bigtriangleup AGD',\\ \\bigtriangleup AME,$ we have the circles $ (O_{1}),\\ (O_{2}),$ as the circumcircles of the triangles $ \\bigtriangleup GD'N,\\ \\bigtriangleup MEH$ respectively, the vertices of which are homologous points.\r\n\r\n$ \\bullet$ So, we conclude that $ \\angle O_{1}AG \\equal{} \\angle O_{2}AM$ $ ,(6)$ because these angles are formed by the lines connecting homologous points ( = are formed by homologous lines ), of the similar triangles $ \\bigtriangleup AGD',\\ \\bigtriangleup AME.$\r\n\r\nHence, from $ (6)$ and because of $ \\angle BAD \\equal{} \\angle CAM,$ we conclude that $ \\angle O_{1}AB \\equal{} \\angle O_{2}AC$ and the proof is completed.\r\n\r\nKostas Vittas.", "Solution_3": "Two interesting solutions, livetolove212 and Kostas Vittas. :wink: \r\nCongrats! :lol:" } { "Tag": [], "Problem": "There are three piles of matches on the table: one with $ 100$ matches, one with $ 200$, and one with $ 300$. Two players play the following game. They play alternatively, and a player on turn takes one of the piles and divides each of the remaining piles into two nonempty piles. The player who cannot make a legal move loses. Who has a winning strategy?", "Solution_1": "May not belong in Pre-Olympiad.\r\n\r\n[hide=\"solution\"]Note that each game will last the same number of moves, as each move increases the number of piles by one, and we need $ 100\\plus{}200\\plus{}300\\equal{}600$ piles. The game will end after $ 600\\minus{}3 \\equal{} 597$ moves. Thus the first player to move wins.[/hide]", "Solution_2": "[quote=\"leoxnlin\"] as each move increases the number of piles by one, [/quote]\r\n\r\nNot true - read the problem carefully :wink:", "Solution_3": "the second player has a winning strategy.\r\nThe player who start is $ \\alpha$ the other is $ \\beta$.\r\nThe strategy for $ \\beta$ is to move always in the same pile of $ \\alpha$." } { "Tag": [ "geometry", "rectangle", "geometry proposed" ], "Problem": "[img]http://img204.imageshack.us/img204/1460/janyooj3.png[/img]\r\n\r\nSquares $ABDE$ and $BCFG$ are constructed on the sides $\\overline{AB}$ and $\\overline{BC}$ of a $\\triangle ABC$, $O$ is the intersection of $\\overline{AF}$ and $\\overline{CE}$. Prove that $\\overline{BO}\\perp \\overline{AC}$.", "Solution_1": "Also true, for similar rectangles instead of squares. The ''isogonic'' theorem is coming back... \r\n\r\nKostas Vittas." } { "Tag": [ "LaTeX", "induction", "inequalities", "inequalities theorems" ], "Problem": "For any positive reals $x_1,x_2,..,x_n$ we have\r\n\r\n$(n-1)(x_1^n+\\cdots+x_n^n)+nx_1\\cdots x_n \\geq (x_1+\\cdots+x_n)(x_1^{n-1}+\\cdots+x_n^{n-1}) $\r\n\r\nFor $n=3$ it is Shur.\r\n\r\nTo define beautiful this one is not so much !!", "Solution_1": "I guess I know a little better LaTeX than you. :) Use \\ldots or \\cdots instead of \"...\" I learned a lot of LaTeX tricks by simply looking at the quotes of other people's posting (I am still learning :) ). \r\n\r\nFor any positive reals $x_1,x_2,\\ldots,x_n$ we have \r\n$(n-1)(x_1^n+\\cdots+x_n^n)+nx_1\\cdots x_n \\geq (x_1+\\cdots+x_n)(x_1^{n-1}+\\cdots+x_n^{n-1}). $\r\n\r\nCheers! :D :D", "Solution_2": "Thank you very much Fuzzylogic for your advice.", "Solution_3": "Well, the only solution I know for this one is using induction, but the induction step is highly nontrivial. I'm in Athens now and I hope I will post the solution next week, when I'm home.", "Solution_4": "Have you come back home,Harazi? :)", "Solution_5": "Oh, yes, I apologise, Im home for a lot of time, but I simply forgot about the proof of this inequality. I will put a Word file with the solution, since its too long to write here.", "Solution_6": "[quote=\"harazi\"]Oh, yes, I apologise, I\ufffdm home for a lot of time, but I simply forgot about the proof of this inequality. I will put a Word file with the solution, since it\ufffds too long to write here.[/quote]\r\nThe proof is false, because we must prove that\r\n\\[(\\sum x_{i})(\\sum x_{i}^{n-1})-%Error. \"sumx\" is a bad command.\n_{i}^{n}+n(n-1)(n-2)\\prod x_{i}\\geq \\prod x_{i}(n+(n-2)(\\sum x_{i})(\\sum x_{i}^{n-1})) \\]\r\nSuranyi's inequality not works here... :mad: Please check it.", "Solution_7": "Sorry I have mistake... :blush:", "Solution_8": "Hey guys, could anyone explain me this? I dont understand this step :blush: \n\n[hide][img]http://i.gyazo.com/e868d4360b36b66e3f4eada661fef77e.png[/img][/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all integer solutions $(x^{2}+y^{2})^{2}=(x+y)^{3}$", "Solution_1": "Let $D: (x^{2}+y^{2})^{2}=(x+y)^{3}$\r\n\r\nD is $D \\subset (-1, 3)\\times (-1, 3)$\r\n(Proof is very easy)\r\n\r\nIf $x=0$ then $(0^{2}+y^{2})^{2}=(0+y)^{3}\\iff y^{3}(y-1) = 0 \\iff y=0, 1$\r\nIf $x=1$ then $(1^{2}+y^{2})^{2}=(1+y)^{3}\\iff y(y^{3}-y^{2}-y-3) = 0 \\Rightarrow y=0$\r\nIf $x=2$ then $(2^{2}+y^{2})^{2}=(2+y)^{3}\\iff (y-2)(y^{3}+y^{2}+4y-4) = 0 \\Rightarrow y=2$\r\n\r\nThus, $(x, y)=(0, 0), (0, 1), (1, 0), (2, 2)$", "Solution_2": "2nd solution :) \r\nCleraly $x+y\\geq 0$ and we can write that $x+y=a^{2},x^{2}+y^{2}=a^{3}(a\\in Z^{+}\\cup \\{ 0\\})$.So, $y=a^{2}-x$.\r\n\r\n$x^{2}+(a^{2}-x)^{2}=a^{3}\\Longrightarrow 2x^{2}-2a^{2}x+a^{4}-a^{3}=0$ Let us examine the discriminant:\r\n$\\bigtriangleup =4a^{4}-8(a^{4}-a^{3})=4a^{3}(2-a)\\geq 0$.Hence, $a$ may equal to $0,1,2$.\r\nIf $a=0\\Longrightarrow \\left\\{\\begin{array}{cc}& x^{2}+y^{2}=0^{3}\\\\ & x+y=0^{2}\\end{array}\\right.\\Longrightarrow (x,y)=(0,0)$\r\nIf $a=1\\Longrightarrow \\left\\{\\begin{array}{cc}& x^{2}+y^{2}=1^{3}\\\\ & x+y=1^{2}\\end{array}\\right.\\Longrightarrow (x,y)=(1,0),(0,1)$\r\nIf $a=2\\Longrightarrow \\left\\{\\begin{array}{cc}& x^{2}+y^{2}=2^{3}\\\\ & x+y=2^{2}\\end{array}\\right.\\Longrightarrow (x,y)=(2,2)$\r\n\r\nLokman G\u00d6K\u00c7E" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Given three external circles $ C_a,C_b,C_c$ with three external common tangents of them $ d_a,d_b,d_c$,respectively, prove that exist four circles tangent to $ d_a,d_b,d_c$ and show the contruction of them!", "Solution_1": "Sorry but how is this an extension of incircle/excircle ? \r\n\r\nAren't we just constructing circles tangent to 3 given lines, which is just the incircle and the 3 excircles of the formed triangle?", "Solution_2": "ohh yes of course, there are some my ideas in my thingking, only but you are right, it is only a circle tagent to three lines, but my idea, when three circle have the radius zero then it is the incircle or excircles!" } { "Tag": [ "function", "integration", "algebra", "domain", "search", "real analysis", "real analysis unsolved" ], "Problem": "We know :\r\n\r\n$ \\left(L^1[0,1],\\|\\cdot\\|_1\\right)$ is a normed space , where :\r\n\r\n$ L^1 \\cong$ Collection of equivalence classes according to the relation R , we say $ f \\ R \\ g$ iff $ f = g , a.e$ .\r\n\r\nand $ \\parallel{} f\\parallel{}_1 = \\int\\limits_{0}^1 |f(t)| dt$ . (Leb Int )\r\n\r\nAlso , $ \\left(L^\\infty [0,1],\\|\\cdot\\|_\\infty\\right)$ is a normed space , where :\r\n\r\n$ \\parallel{}f\\parallel{}_\\infty = \\mathop {essup }\\limits_{t\\in [0,1]} \\ |f(t)|$ , (Essential Supremum ) \r\n\r\nNote that : $ f = [f]$ , the class of functions which they equal a.e ( for both previous spaces .)\r\n\r\nThe Super question is :\r\n\r\nfor any measurable function $ g$ on $ [0,1]$ , then $ g\\in L^\\infty[0,1] \\Longleftrightarrow g.L^{1}[0,1] \\subseteq L^{1}[0,1]$\r\n\r\nIN FACT I NEED THE DIRECTION :\"$ \\Longleftarrow$\" \r\n\r\n :)", "Solution_1": "Consider the contrapositive: If $ g\\not\\in L^{\\infty},$ then there exists $ f\\in L^1$ such that $ fg\\not\\in L^1.$\r\n\r\nLet's look at a plan of attack. The first thing we can do is to assume WLOG that $ g\\ge 0$ and that the $ f$ we will pick is also nonnegative. We can do that by choosing the size of $ f$ and then multiplying that by $ \\operatorname{sgn}(g).$ (Or $ \\overline{\\operatorname{sgn}(g)}$ if we're using complex-valued functions.)\r\n\r\nDivide the domain into disjoint sets on which $ k\\le g<2^{k\\plus{}1}.$ Or maybe $ 2^k\\le g<2^{k\\plus{}1}$ - maybe that would be easier to manipulate. Choose a (constant) value for $ f$ on each of these sets. Can we arrange it so that $ \\int f<\\infty$ but $ \\int fg\\equal{}\\infty?$ Fiddle around with it until you can make it work.", "Solution_2": "Related to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1662488565&t=177843]this.[/url]", "Solution_3": "what do you mean by the size of $ f$? :| \r\n\r\nI couldn't do it yet :huh: \r\n\r\nAny help will be appreciated ." } { "Tag": [], "Problem": "Would someone mind giving me a list of every1 who made countdown and also possibly their AOPS account?", "Solution_1": "I've posted the first. The latter is up to those participants; revealing it without their consent is against privacy rules." } { "Tag": [ "trigonometry", "parameterization", "function", "Pythagorean Theorem", "geometry" ], "Problem": "Which of the following rectangular equations is equivalent to the parametric equations $ x\\equal{}sintcost$, $ y\\equal{}cos^{2}(2t)$?\r\nA) $ y\\equal{}\\sqrt{1\\minus{}x^{2}}$\r\nB) $ y\\equal{}1\\minus{}4x^{2}$\r\nC) $ y^{2}\\equal{}4\\minus{}4x^{2}$\r\nD) $ y\\equal{}2x^{2}$\r\nE) NOTA=None of the Above\r\n\r\nNote: I have absolutely no idea what parametric equations are so can someone clarify what this means?", "Solution_1": "$ x\\equal{}\\frac{1}{2}\\sin{2t}$ and $ y\\equal{}\\cos^{2}{2t}$. Then $ (2x)^{2}\\plus{}y^{2}\\equal{}1\\equal{}4x^{2}\\plus{}y^{2}$", "Solution_2": "[quote=\"JRav\"]$ x \\equal{} \\frac {1}{2}\\sin{2t}$ and $ y \\equal{} \\cos^{2}{2t}$. Then $ (2x)^{2} \\plus{} y^{2} \\equal{} 1 \\equal{} 4x^{2} \\plus{} y^{2}$[/quote]I would think then that\r\n$ 2x \\equal{} \\sin{2t}$\r\n$ 4x^2 \\equal{} \\sin^{2}{2t}$\r\n$ 4x^2 \\plus{} y \\equal{} \\sin^{2}{2t} \\plus{} \\cos^{2}{2t}$\r\n$ 4x^2 \\plus{} y \\equal{} 1$\r\n$ y \\equal{} 1 \\minus{} 4x^2$\r\nbut with so many $ 2$'s dancing in front of our eyes, we may see more or less of them.\r\n\r\nMy idea of parametric equations is that they are equations that give us the variables we care about (x, y, ..) in terms of one variable that we call a parameter (t) because we are not particularly interested in it at the time. For example, we may want to know an equation describing the trajectory of a projectile shot at a certain angle. We go through some parametric equations to get to the answer we want. We can somewhat easily determine the (parametric) functions relating the height (y) and the horizontal distance covered (x) to the time since launching (t). From those parametric equations, it takes some algebra to get y as a function of x to determine the shape of the trajectory. Occasionally, we may not be able to get a reasonable y=f(x) function and we may have to keep the parametric equations.", "Solution_3": "Basically you define x and y in terms of some other variable t.", "Solution_4": "[quote=\"JRav\"]$ x \\equal{} \\frac {1}{2}\\sin{2t}$ and $ y \\equal{} \\cos^{2}{2t}$. Then $ (2x)^{2} \\plus{} y^{2} \\equal{} 1 \\equal{} 4x^{2} \\plus{} y^{2}$[/quote]\r\n\r\nAlso, if the above confused you, it was a mistake. Since $ y$ was defined as $ \\cos^2 2t$, we should have $ (2x)^2 \\plus{} y \\equal{} 1 \\implies y \\equal{} 1 \\minus{} 4x^2$.", "Solution_5": "hello, we have $ x \\equal{} 1/2\\cdot \\sin(2t)$ and from here $ 4x^2 \\equal{} \\sin^2(2t)$ and by the Pythagorean theorem we get $ 4x^2 \\plus{} y\\equal{} 1$.\r\nSonnhard.", "Solution_6": "[quote=\"Dr Sonnhard Graubner\"]hello, we have $ x \\equal{} 1/2\\cdot \\sin(2t)$ and from here $ 4x^2 \\equal{} \\sin^2(2t)$ and by the Pythagorean theorem we get $ 4x^2 \\plus{} y^2 \\equal{} 1$.\nSonnhard.[/quote]\r\n\r\nYou also made the same mistake as JRav. $ y$ is defined as $ \\cos^2 2t$, so it should be $ 4x^2 \\plus{} y \\equal{} 1$.", "Solution_7": "hello, yes, you have right, i have corrected it.\r\nThank you very much,Sonnhard.", "Solution_8": "[quote=\"Yongyi\"]Also, if the above confused you, it was a mistake.[/quote]\r\nOops. I squared the $ y$-term by accident." } { "Tag": [ "inequalities", "geometry", "circumcircle", "trigonometry", "geometry proposed" ], "Problem": "Given triangle $ ABC$ and $ O$ is circumcenter $ P$ is an arbitrary point in plane prove that\r\n\\[ \\frac{\\sin A}{PA}+\\frac{\\sin B}{PB}+\\frac{\\sin C}{PA}\\ge 4\\sqrt{\\frac{OP}{PAPBPC}}\\sin A\\sin B\\sin C \\]\r\nEquality iff $ ABC$ is equilateral and $ P\\to\\infty$.\r\nGook luck all mathlinkers in IMO.", "Solution_1": "Just a question regarding this inequality. Is it correct the way it is written? \r\n\r\nIs it $ \\frac{\\sin C}{PA}$ or $ \\frac{\\sin C}{PC}$?", "Solution_2": "Sorry you are right it must be\r\nGiven triangle $ ABC$ and $ O$ is circumcenter $ P$ is an arbitrary point in plane prove that\r\n\\[ \\frac{\\sin A}{PA}+\\frac{\\sin B}{PB}+\\frac{\\sin C}{PC}\\ge 4\\sqrt{\\frac{OP}{PAPBPC}}\\sin A\\sin B\\sin C \\]\r\nEquality iff $ ABC$ is equilateral and $ P\\to\\infty$." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "is there a coloring of $\\mathbb{R}^2$ with countable color for the angel $\\theta$ such that there are not any $A,B,C$ st $\\widehat{ABC}=\\theta $", "Solution_1": "what is your idea guys?", "Solution_2": ":( no idea?" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Prove that for every polynomial $P(x)$ with real coefficients there exist a positive integer $m$ and polynomials $P_{1}(x),\\ldots , P_{m}(x)$ with real coefficients such that \n\\[P(x) = (P_{1}(x))^{3}+\\ldots +(P_{m}(x))^{3}\\]", "Solution_1": "$x=(x+\\frac 1{\\sqrt 6})^{3}+(x-\\frac 1{\\sqrt 6})^{3}-2x^{3}$ .. and so on :cool:", "Solution_2": "Wow! That's cool :)" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let \r\n \r\n M = X/(Y + aZ) + Y/(Z+aX) + Z/(X+aY),\r\n\r\nFind the minimun value of M. \r\n If X,Y,Z are positive real numbers and \"a\" a real number.", "Solution_1": "I have found a partial solution for this one:\r\nfirst we write:[tex] \r\nM= \\frac{x^2}{xy+azx} + \\frac{y^2}{yz+axy} + \\frac{z^2}{zx+ayz}[/tex]\r\nand by cauchy's inequality:\r\n[tex] (xy+yz+zx)(a+1)M \\ge (x+y+z)^2 \\ge 3(xy+yz+zx)[/tex]\r\nthen [tex] M \\ge \\frac{3}{1+a} [/tex] \r\nif [tex] a \\succ -1[/tex]" } { "Tag": [ "geometry", "3D geometry", "superior algebra", "superior algebra unsolved" ], "Problem": "Let E1/F and E2/F be Galois extensions. If E1 and E2 are finite, I know how to show that the composite field E1E2 is finite and Galois. What about when E1/F and E2/F are infinite? \r\n\r\nAlso, a classmate conjectured that if E1/F and E2/F are infinite and Galois, then E1E2 = E1 or E2. Is this true?", "Solution_1": "The composition field is clearly seperable (it's generated by such elements) and normal, the latter since being true by the following definition of normal (equivalent to the others): $E|F$ is normal iff every homomorphism to an algebraic closure is already an automorphism/a homomorphism into $E$ itself.\r\nAnd it's also infinite since a infinite subfield is contained in it\r\n\r\nNo, the conjecture is clearly false, e.g. take those extensions of $\\mathbb{Q}$ by addjoining square roots (giving $E_{1}$) and adjoining cube roots (also the complex ones; giving $E_{2}$).", "Solution_2": "Thanks. I couldn't get the normal part." } { "Tag": [ "induction", "algebra", "polynomial", "algebra proposed" ], "Problem": "here is my solution:\r\nlet $x_0=1,x_(n+1)=x_n^2+1$, then by induction we obtain $p(x_n)=x_n$,by polynomial identity theorem, we obtain $p(x)=x$.", "Solution_1": "Your solution to what? :?: ;) \r\n\r\nPierre.", "Solution_2": "[quote=\"pbornsztein\"]Your solution to what? :?: ;) \n\nPierre.[/quote]\r\nHe must have clicked \"New topics\" instead of \"Post reply\" when he was typing his solution... :D" } { "Tag": [ "calculus", "integration", "parameterization", "function", "vector", "absolute value", "calculus computations" ], "Problem": "Can arc length of a curve be negative?", "Solution_1": "I suppose it could be if you use a reversed parametrization.", "Solution_2": "If by \"arc length\" you mean \"the length of an arc\" then the answer is obviously \"no,\" as lengths are nonnegative. If by \"arc length\" you mean \"value of an integral meant to give arc length with respect to some parametrization,\" then the answer is \"yes, if you have the bounds of integration in an unusual order.\"", "Solution_3": "There are two attitudes about integrals here.\r\n\r\nOne is the attitude, strongest in first-semester calculus, that emphasizes the algebra of antiderivatives, and has - and needs - the theorem $ \\int_a^bf(x)\\,dx=-\\int_b^af(x),\\,dx.$\r\n\r\nAn example of this thinking: do the integral $ \\int_0^2x\\sqrt{4-x^2}\\,dx$ by making the substitution $ u=4-x^2.$ That gives us $ du=-2x\\,dx$ and leaves us with\r\n\r\n$ -\\frac12\\int_4^0\\sqrt{u}\\,du$ which we may then turn around to $ \\frac12\\int_0^4\\sqrt{u}\\,du=\\frac83.$\r\n\r\nJRav's post is rooted in this thinking.\r\n\r\nThere's a second attitude, which starts to really take root around the first multivariable calculus course. De-emphasize the algebraic and antiderivative aspects, and emphasize that we integrate functions over [i]sets.[/i] The notion of a multivariable change of variables, with its attendant Jacobian, forces us to think that way. With this attitude, return to the same problem.\r\n\r\n$ \\int_0^2x\\sqrt{4-x^2}\\,dx$ is the integral of a positive function, hence positive. (Well, the function is positive except for two points, but two points don't count.)\r\n\r\nMake the substitution $ u=4-x^2.$ We note that the mapping $ x\\mapsto 4-x^2$ sends the set $ [0,2]$ to the set $ [0,4].$ Sure, it flips the set around (because that's a decreasing function), but we don't care. As for the Jacobian: we always take the absolute value of that anyway. Hence $ du=2x\\,dx.$ The integral become\r\n\r\n$ \\frac12\\int_0^4\\sqrt{u}\\,du=\\frac83.$\r\n\r\nNow, I'm going to take that second attitude - the attitude that integrals are over sets, the vector calculus-analysis attitude - and I'll answer ThetaPi's question. \r\n\r\nNo, an arclength can't be negative. It's given by the integral over a set (and it's not an oriented integral) of a nonnegative function, and the result will be nonnegative. (In fact, positive except for trivial \"curves\" that consist of a single point.)" } { "Tag": [ "geometry" ], "Problem": "When each side of a square is decreased by $25\\%$, the area of the square decreases by 7 square meters. What is the number of square meters in the area of the original square?", "Solution_1": "[hide]$\\frac{7}{2}= 3.5$\n\n$\\frac{3.5}{4}= 0.875$\n\n$3.5+0.875 = 4.375$\n\n$4.375 * 4.375 = \\boxed{19.140625m^{2}}$[/hide]", "Solution_2": "[hide]\nOriginal side length was $x$.\n\nWe have \\[\\frac{3x}{4}^{2}+7 = x^{2}\\]\\[\\frac{9x^{2}}{16}+7 = x^{2}\\]\\[7 = \\frac{7x^{2}}{16}\\]\\[x^{2}= 16\\]\\[x = 4\\]\n\nSo the original area was $\\boxed{16}$.[/hide]", "Solution_3": "Yes this is a different solution\r\n\r\n[hide]chances are they give you integers. Doesn't hurt to try the two that differ by 7, 3&4. Almost instant answer of $4^{2}= 16$[/hide]", "Solution_4": "my anwer was 16\r\n\r\nI built an equation\r\n\r\nx^2-7=(.75x)^2\r\nx^2-9x^2/16=7\r\n7x^2/16=7\r\nx^2=16\r\nx=4", "Solution_5": "I know that ur new, but please hide solutions. This is done by surrounding the solution w/ hide tags.", "Solution_6": "[hide]\n$x^{2}-(\\frac{3}{4}x)^{2}=7$\n\ndon't know how to solve that, but since it's obvious that 8 doesn't work, 4 does (maybe 6, but chances are that 4 works)\n\n16-9=7\n\nanswer is 16[/hide]\r\n\r\njorian" } { "Tag": [ "number theory", "number theory proposed" ], "Problem": "We define a primitive solution to $ x^2 \\plus{} y^2 \\equal{} z^2$ such that $ (x,y,z) \\equal{}1$. Further more two solutions is essentialy alike if we can get from one to the other by changing signs of $ x,y,z,$ or by changing $ x$ and $ y$.\r\n\r\nGiven $ z \\in \\mathbb{Z}$ Prove that there are either $ 0$ or $ 2^n$ essentialy different solutions for some $ n \\in \\mathbb{N}_0$ to the equation $ x^2 \\plus{} y^2 \\equal{} z^2$ where $ x,y \\in \\mathbb{Z}$", "Solution_1": "Generally, this is not true. The smallest counterexample is $ z\\equal{}25$ for which there are 3 essentially different solutions:\r\n$ 0^2\\plus{}25^2\\equal{}25^2$\r\n$ 7^2\\plus{}24^2\\equal{}25^2$\r\n$ 15^2\\plus{}20^2\\equal{}25^2$\r\n\r\nIn general, the number of solutions is given by the [url=http://mathworld.wolfram.com/SumofSquaresFunction.html]formula (17)[/url].", "Solution_2": "The number of solutions with coprime $ x,y,z$ is a power of $ 2$ (or $ 0$). Seeing this is rather easy using arithmetic of $ \\mathbb Z[i]$. In fact, one already gets the some result for the solutions of $ x^2 \\plus{} y^2 \\equal{} n$ and the idea is that the number of ways to write a number as product of coprime numbers is a power of $ 2$.\r\nFormula 17 is in my oppinion not the nicest way to state the result ;)\r\n(and well, due to excessive errors, often very stupid ones, I've some bad attitude towards MathWorld in general; that formula is correct, but I wouldn't trust them on anything I don't have a second source or proof for)", "Solution_3": "[quote=\"maxal\"]Generally, this is not true. The smallest counterexample is $ z \\equal{} 25$ for which there are 3 essentially different solutions:\n$ 0^2 \\plus{} 25^2 \\equal{} 25^2$\n$ 7^2 \\plus{} 24^2 \\equal{} 25^2$\n$ 15^2 \\plus{} 20^2 \\equal{} 25^2$\n\nIn general, the number of solutions is given by the [url=http://mathworld.wolfram.com/SumofSquaresFunction.html]formula (17)[/url].[/quote]\r\nRemember that $ (x,y,z) \\equal{} 1$. Indeed $ (0,25,25) \\equal{} 25 \\neq 1$ and $ (15,20,25) \\equal{} 5 \\neq 1$.\r\n\r\nI have too proved it in $ \\mathbb{Z}[i]$ so I was just looking for a proof using classic number theory :)" } { "Tag": [], "Problem": "Four positive integers have a product of 504 and a sum of 20. What is their arithmetic mean?", "Solution_1": "Since the integers have a sum of 20, and there are 4 integers, the arithmetic mean is $ 20 \\div 4 \\equal{} 5$." } { "Tag": [ "function", "inequalities", "algebra unsolved", "algebra" ], "Problem": "determine all functions f defined from the set of positive real numbers to itself such that\r\n$ f(g(x))\\equal{} \\frac{x}{xf(x)\\minus{}2}$ and $ g(f(x))\\equal{} \\frac{x}{xg(x)\\minus{}2}$ for all positive real numbers x :roll:", "Solution_1": "Let's denote : $ u(x)\\equal{}xf(x) ,\\ v(x)\\equal{}xg(x)$ . Manifestly $ u(x)>2 ,\\ v(x)>2$ and $ u(g(x))\\equal{}\\frac{v(x)}{u(x)\\minus{}2}>2,\\ v(f(x))\\equal{}\\frac{u(x)}{v(x)\\minus{}2}>2$ therefore \r\n$ v(x)>2u(x)\\minus{}4,\\ u(x)>2v(x)\\minus{}4$, so $ u(x)\\plus{}v(x) < 8$ and it's easy to see that $ u(x),v(x)$ are bounded.\r\nLet's denote : $ U_i \\equal{} \\inf (u(x)),U_s\\equal{}\\sup (u(x)) ,\\ V_i \\equal{} \\inf (v(x)),V_s\\equal{}\\sup (v(x))$ the inferior,superior limit bounds of the functions.\r\n$ u(g(x))\\equal{}\\frac{v(x)}{u(x)\\minus{}2}\\ \\Longrightarrow U_i\\leq \\frac{v(x)}{u(x)\\minus{}2}\\leq U_s$ $ \\ \\Longrightarrow \\frac {V_s}{U_i}\\geq u(x)\\minus{}2 \\geq \\frac {V_i}{U_s} \\ \\Longrightarrow \\frac {V_s}{U_i}\\plus{}2 \\geq U_s\\geq U_i\\geq \\frac {V_i}{U_s}\\plus{}2$ $ \\Longrightarrow V_s\\plus{}2U_i\\geq U_iU_s\\geq V_i\\plus{}2U_s$\r\nIn the same way, from $ v(g(x))\\equal{}\\frac{u(x)}{v(x)\\minus{}2}$ we obtain $ U_s\\plus{}2V_i\\geq U_i\\plus{}2V_s$\r\nNow , by adding the two inequalities, we obtain : $ V_s\\plus{}2U_i\\plus{} U_s\\plus{}2V_i\\geq V_i\\plus{}2U_s \\plus{} U_i\\plus{}2V_s\\ \\Longrightarrow\\ 0\\geq (U_s\\minus{}U_i)\\plus{}(V_s\\minus{}V_i)$\r\nBecause of $ U_s\\geq U_i,\\ V_s\\geq V_i$ we can write $ U_s\\equal{}U_i\\equal{}U ,\\ V_s\\equal{}V_i\\equal{}V$ and $ U\\equal{}\\frac VU\\plus{}2,\\ V\\equal{}\\frac UV\\plus{}2$\r\nFinally we obtain $ U\\equal{}V\\equal{}3$ so $ f(x)\\equal{}g(x)\\equal{}\\frac 3x$\r\n :cool:" } { "Tag": [], "Problem": "solve the diophantine equation\r\n$(x-y-z)(x-y+z)(x+y-z)=8xyz$\r\nwhere x,y,z are relatively prime", "Solution_1": "${(x-y-z) (x-y+z) (x+y-z)=8 x y z}$\r\n${x=s-a}$\r\n${y=s-b}$\r\n${z=s-c}$\r\n${s=a+b+c}$\r\n\r\n${-a b c=(a+b) (a+c) (b+c)}$\r\n${(a+b+c) (a b+c b+a c)=0}$\r\n\r\n${a=-(b+c)}$\r\n\r\n${-a=\\frac{1}{c}+\\frac{1}{b}}$\r\n\r\nThe last equation is integer only if ${b^{2}=c^{2}}$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ be positive real numbers.Prove that the following inequality occurs: \\[\\sum\\frac{b+c}{a}\\geq4\\sum_{cyc}\\frac{a}{a+b}\\]", "Solution_1": "$4\\sum_{cyc}\\frac{a}{a+b}=2\\sum_{cyc}\\frac{2}{\\frac{b}{a}+1}\\le 2\\sum_{cyc}\\frac{\\frac{a}{b}+1}{2}=3+\\sum_{cyc}\\frac{a}{b}\\le \\sum_{cyc}\\frac{b}{a}+\\sum_{cyc}\\frac{a}{b}=\\sum_{cyc}\\frac{b+c}{a}$ :)", "Solution_2": "Wow :) \r\nCool solution :) \r\nI knew it was kind of weak,because in my solution I used a very weak Am-Gm.\r\nI have a soln with Cauchy for this ineq..", "Solution_3": "it is very similar with Levinson's inequality :)", "Solution_4": "Also ways by rearrangement:\r\n$\\sum_{cyc}\\frac{b+c}{a}\\geq4\\sum_{cyc}\\frac{a}{b+c}\\geq4\\sum_{cyc}\\frac{a}{a+b}$ \r\n $\\sum_{cyc}\\frac{b+c}{a}\\geq\\sum_{cyc}\\frac{a+b}{b}\\geq4\\sum_{cyc}\\frac{a}{a+b}$", "Solution_5": "[quote=\"andyciup\"]Let $a,b,c$ be positive real numbers.Prove that the following inequality occurs: \\[\\sum\\frac{b+c}{a}\\geq4\\sum_{cyc}\\frac{a}{a+b}\\] [/quote]\r\n\r\napply:\r\n\r\n$\\frac{1}{a}+\\frac{1}{b}\\geq \\frac{4}{a+b}$ (cauchy)\r\n\r\nso we have: \r\n\r\n$\\sum{\\frac{c}{a}+\\frac{c}{b}}\\geq \\sum{\\frac{4c}{a+b}}$\r\n\r\n$\\Rightarrow 2\\sum{\\frac{c+b}{a}}\\geq 2\\sum{\\frac{4c}{a+b}}$, and now rearrangement.", "Solution_6": "You might want to look up one post :)", "Solution_7": "The square root inequality is familiar with somebody. But it is still new with some others. So I post it (just for fun). \\[\\sqrt{b+c \\over a}+\\sqrt{c+a \\over b}+\\sqrt{a+b \\over c}\\ge 2\\left(\\sqrt{a \\over a+b}+\\sqrt{b \\over b+c}+\\sqrt{c \\over c+a}\\right).\\]", "Solution_8": "\\[\\sum\\sqrt{\\frac{b+c}{a}}\\geq 2\\sum\\sqrt{\\frac{a}{b+c}}\\geq2\\sum\\sqrt{\\frac{a}{a+b}}\\]", "Solution_9": "Here is my solution for the initial problem:\r\nBy AM-GM we have $a+b\\geq2\\sqrt{ab}$ and $c^{2}+ab\\geq2c\\sqrt{ab}$,which means that \\[(a+b)(c^{2}+ab)\\geq4abc\\] ,which is equivalent to \\[\\frac{b}{a+b}\\leq\\frac{c^{2}+ab}{4ac}\\] ,which means that \\[\\frac{c}{a}+\\frac{b}{c}\\geq4\\frac{b}{a+b}\\] Summing up the other two cyclic ones gives us the requested inequality.\r\n\r\nAlso,here is a proof with cauchy,but it`s not a big deal:\r\nBy cauchy we have $(a+b)(ab+c^{2})\\geq b(a+c)^{2}$,which means that \\[\\frac{b}{a+b}\\leq\\frac{c^{2}+ab}{(a+c)^{2}}\\] But $(a+c)^{2}\\geq4ac$,which means that $\\frac{b}{a+b}\\leq\\frac{c^{2}+ab}{4ac}$,which is what we were looking for..", "Solution_10": "http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=511112&p=3272156#p3272156\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=429236&p=3556281#p3556281]Middle European Mathematical Olympiad 2011[/url]" } { "Tag": [], "Problem": "and the penguin comes back :D:D:D with enhanced features this time :D", "Solution_1": "i got exactly 750.... :D\r\nand the best record is 1156.x after 1 min....\r\nnow the lastest record is 1215.8?", "Solution_2": "I'm not so sure about enhanced features...it's still \"click, click again when you want to hit the penguin\", right?\r\n\r\nI think the distances were scaled so 320 in the previous version should be around 1200 or so. :) [to reorientate ourselves to the record to beat] :D", "Solution_3": "previous record: 321.7\r\n Now: 1203.3. I aggree with Valiowk.", "Solution_4": "yeah but I love to hit it to go further than 320 :) \r\nmore of the penguin coming soon :D", "Solution_5": "first try 928.5\r\nbest record:1224.1", "Solution_6": "take a look at these :)", "Solution_7": "I love those pics :D", "Solution_8": "This is more fun :D .Best rec:1215 after 2 mins", "Solution_9": "Which is the direct link to the swf?\r\n\r\nEdit:Nevermind.It is:\r\n[url]http://www.mathlinks.ro/files/pingu3.swf[/url]", "Solution_10": "A friend played it a few hours and got 4 times 1224.1 (I think thats the max possible achievable distance)", "Solution_11": "wt's the shortest distance you can get?\r\nI get -56.3 :D , it's -56.3", "Solution_12": "Yup PDP, 1224.1 on several occasions - Not a tenth higher. Why have you done this to us Valentin?!?!?!?!?!?!", "Solution_13": "[quote=\"tripleJ\"]Why have you done this to us Valentin?!?!?!?!?!?![/quote]\r\nwhat, what, what have I done? :)", "Solution_14": "[quote]what, what, what have I done?[/quote]\r\nThis is what god said when he created Adam [img]http://www.smiliedb.de/s/sdb49878.gif[/img]", "Solution_15": "Here's my record after a few minutes of playing:)\r\n\r\nIndeed, what have you done to us ? :D", "Solution_16": "Maybe it's time to go back to good ol' penguin v. 1 (and try cracking Valentin's 327.1, because I don't think any of us can do the 383...) :D", "Solution_17": "the penguin got annoyed :D", "Solution_18": "hehe hehehe, my hand hurts now :D", "Solution_19": "Ahahahaha!I was thinking,what the heck,the poor pinguin gets kicked with the baseball bat,then poked...but NO,the pinguing strikes back.GO GO Pinguin! [img]http://community.the-underdogs.org/smiley/happy/thumb.gif[/img]", "Solution_20": "[quote=\"PiDeltaPhi\"]Ahahahaha!I was thinking,what the heck,the poor pinguin gets kicked with the baseball bat,then poked...but NO,the pinguing strikes back.GO GO Pinguin! [img]http://community.the-underdogs.org/smiley/happy/thumb.gif[/img][/quote]\r\nspeaking of strike-ing have fun with the new version of pingu. \r\n\r\nit's [url=http://www.mathlinks.ro/fun/pingslap.swf]HERE!!!![/url] :D", "Solution_21": "damn. I've played to much Pingu' and my comps' starting getting weird errors :?. The last one I've got looks like this:", "Solution_22": "[url]http://www.mathlinks.ro/fun/pingslap.swf[/url]\r\n\r\nThe link doesnt work,sorry", "Solution_23": "[quote=\"Valentin Vornicu\"]damn. I've played to much Pingu' and my comps' starting getting weird errors :?. The last one I've got looks like this:[/quote]\r\nsome bugs in your picture.\r\nthe fonts in the buttons are different...\r\nheh, you need to use \"MS Sans Serif\" for the two rightmost buttons.", "Solution_24": "the link is fixed. liyi there is no problem with the fonts, it was just a joke :D.", "Solution_25": "1197.5 :D", "Solution_26": "1199 is my new score.This is really nonsense..." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Given $ a_i\\ge0$ and $ \\sum_{i \\equal{} 1}^n a_i \\equal{} 1$.Find the maximum value of\r\n\r\n$ P \\equal{} \\sum_{i \\equal{} 1}^n \\frac1{\\sqrt {1 \\plus{} a_i^3}}$", "Solution_1": "[quote=\"dduclam\"]Given $ a_i\\ge0$ and $ \\sum_{i \\equal{} 1}^n \\equal{} 1$.Find the maximum value of\n\n$ P \\equal{} \\sum_{i \\equal{} 1}^n \\frac1{\\sqrt {1 \\plus{} a^3}}$[/quote]\r\n\r\n\r\nHi. [b]dduclam[/b] $ \\sum_{i \\equal{} 1}^n \\equal{} 1$ ? $ \\sum_{i \\equal{} 1}^n{a_{i}} \\equal{} 1$ or ...", "Solution_2": "[quote=\"zaya_yc\"][quote=\"dduclam\"]Given $ a_i\\ge0$ and $ \\sum_{i \\equal{} 1}^n \\equal{} 1$.Find the maximum value of\n\n$ P \\equal{} \\sum_{i \\equal{} 1}^n \\frac1{\\sqrt {1 \\plus{} a_i^3}}$[/quote]\n\n\nHi. [b]dduclam[/b] $ \\sum_{i \\equal{} 1}^n \\equal{} 1$ ? $ \\sum_{i \\equal{} 1}^n{a_{i}} \\equal{} 1$ or ...[/quote]\r\n\r\nYes, $ \\sum_{i \\equal{} 1}^n{a_{i}} \\equal{} 1$, i edited my post,sorry :(" } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Disculpen pero alguien podria postear toda la prueba de la OMR 2006, \u00f3 pasarme un link para ver la prueba. por favor.\r\nchau", "Solution_1": "tengo los dos primeros problemas de nivel 2 \r\nel primero esta en el topic de geo : \r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=123246\r\n\r\n\r\nel segundo problema es : \r\ncarlitos escribio todos los subconjuntos de ${1,2,...,2006}$ en los que la diferencia entre la cantidad de numeros pares y de numeros impares es multiplo de 3 \r\n\u00bfcuantos subconjuntos escribio carlitos?", "Solution_2": "que paso son las soluciones :blush:", "Solution_3": "P_3) se han elegido varios segmentos en una recta (con posibles superposiciones) si a cada segmento se le colorea de azul sus dos terceras partes de la izquierda, el conjunto de los puntos azules es un segmento de longitud 31. si a cada segmento se le colorea de rojo sus dos terceras partes de la derecha, el cunjunto de los puntos rojos es un segmento de longitud 23. sean $M$ y $m$ la maxima y minima longitud de un segmento de la coleccion, respectivamente.\r\n\u00bfcual es el valor minimo que puede tener $M-m$ ?", "Solution_4": "Ya que soy muy malo contando tengo una solucion horrible como esta.\r\n\r\nPrimero notemos que si un subconjunto tiene $a$ elementos pares y $b$ elementos impares, es necesario y suficiente para que Carlitos lo escriba que:\r\n$a \\equiv b (mod 3)$.\r\nEntonces si Carlitos escribio $n$ subconjuntos tenemos que:\r\n$\\\\ n= \\big[ \\binom{1003}{0}+\\binom{1003}{3}+...+\\binom{1003}{1002}\\big]^{2}+\\big[ \\binom{1003}{1}+\\binom{1003}{4}+...+\\binom{1003}{1003}\\big]^{2}+\\big[\\binom{1003}{2}+\\binom{1003}{5}+...+\\binom{1003}{1001}\\big]^{2}= a^{2}+b^{2}+c^{2}$\r\n\r\nPorque primero escogemos la cantidad de pares $a$ y luego la de impares $b$ que sea congruente con $a$ modulo 3.\r\n\r\nConsideremos a $\\omega \\not= 1$ tal que $\\omega^{3}-1=0$, tenemos que:\r\n$\\\\ \\omega^{2}+\\omega+1=0 \\implies \\\\ \\omega^{2i}+\\omega^{i}+1=0 \\forall i: 3\\not|i$ (*)\r\nAdem\u00e1s $(1+\\omega)(1+\\omega^{2})=1$ y $\\omega^{a}= \\omega^{b}\\iff a\\equiv b (3)$\r\n\r\nAhora consideremos $f(x)=(1+x)^{1003}$ expandimos y nos queda:\r\n$f(x)= \\sum_{i=0}^{1003}\\binom{1003}{i}x^{i}$\r\nAhora usemos las propiedades de $\\omega$ para hallar que:\r\n$f(1)+f(\\omega)+f(\\omega^{2}) = \\sum_{i=0}^{1003}\\binom{1003}{i}(1+\\omega^{i}+\\omega^{2i}) = 3a$ pues se cancelan todos los terminos que no tengan a $i$ mulitplo de 3.\r\nAdem\u00e1s:\r\n$f(1)+\\omega f(\\omega)+\\omega^{2}f(\\omega^{2}) = \\sum_{i=0}^{1003}\\binom{1003}{i}(1+\\omega^{i+1}+\\omega^{2(i+1)})= 3b$\r\n\r\n$\\\\ f(1)+\\omega^{2}f(\\omega)+\\omega f(\\omega^{2})= \\\\ f(1)+\\omega^{2}f(\\omega)+\\omega^{4}f(\\omega^{2})= \\sum_{i=0}^{1003}\\binom{1003}{i}(1+\\omega^{i+2}+\\omega^{2(i+2)})= 3c$\r\n\r\nEsto quiere decir que:\r\n$\\\\ n= a^{2}+b^{2}+c^{2}= \\\\ \\frac{[f(1)+f(\\omega)+f(\\omega^{2})]^{2}+[f(1)+\\omega f(\\omega)+\\omega^{2}f(\\omega^{2})]^{2}+[f(1)+\\omega^{2}f(\\omega)+\\omega f(\\omega^{2})]^{2}}{9}$\r\nExpandimos la expresion anterior y obtenemos despues de agrupar convenientemente algunos terminos:\r\n\r\n$n= \\frac{3f^{2}(1)+(f^{2}(\\omega)+f^{2}(\\omega^{2})+2f(1)f(\\omega)+2f(1)f(\\omega^{2})) (1+\\omega+\\omega^{2})+3\\omega^{3}f(\\omega)f(\\omega^{2})}{9}=$\r\n$\\frac{3f^{2}(1)+3f(\\omega)f(\\omega^{2})}{9}=$\r\n$3\\frac{[(1+1)^{1003}]^{2}+2(1+\\omega)^{1003}(1+\\omega)^{1003}}{9}=$\r\n$\\frac{2^{2006}+2}{3}$\r\n\r\nCon lo que queda terminado el problema.\r\n\r\nComo veran es una solucion muy tecnica, pero el truco de $\\omega$ es muy bueno." } { "Tag": [ "number theory", "least common multiple" ], "Problem": "Simplify and express your answer as a common fraction:\n$ \\frac{2\\cdot(3)^{\\minus{}2}\\plus{}3\\cdot(2)^{\\minus{}3}}{2\\cdot(3)^{\\minus{}2}\\minus{}3\\cdot(2)^{\\minus{}3}}$", "Solution_1": "First we simplify the top:\r\n\r\n$ 2 \\cdot (3)^{\\minus{}2} \\plus{} 3 \\cdot (2)^{\\minus{}3}$\r\n\r\n$ 2 \\cdot \\frac{1}{9} \\plus{} 3 \\cdot \\frac{1}{8}$\r\n\r\n$ \\frac{2}{9} \\plus{} \\frac{3}{8}$\r\n\r\n$ \\frac{16}{27}$\r\n\r\nAnd then the bottom:\r\n\r\n$ 2 \\cdot (3)^{\\minus{}2} \\minus{} 3 \\cdot (2)^{\\minus{}3}$\r\n\r\n$ 2 \\cdot \\frac{1}{9} \\minus{} 3 \\cdot \\frac{1}{8}$\r\n\r\n$ \\frac{2}{9} \\minus{} \\frac{3}{8}$\r\n\r\n$ \\frac{\\minus{}11}{72}$\r\n\r\nNow we divide:\r\n\r\n$ \\frac{16}{27} \\div \\frac{\\minus{}11}{72}$\r\n\r\n$ \\boxed{\\minus{}\\frac{43}{11}}$", "Solution_2": "Check out this alternate method of getting rid of \"fractions within factions\":\r\n\r\n$ \\frac{2\\cdot(3)^{\\minus{}2}\\plus{}3\\cdot(2)^{\\minus{}3}}{2\\cdot(3)^{\\minus{}2}\\minus{}3\\cdot(2)^{\\minus{}3}}$\r\n\r\nMultiply the top and bottom by $ 3^2$ and $ 2^3$, distribute and cancel:\r\n\r\n$ \\frac{3^2\\cdot 2^3\\cdot(2\\cdot(3)^{\\minus{}2}\\plus{}3\\cdot(2)^{\\minus{}3})}{3^2\\cdot 2^3\\cdot (2\\cdot(3) ^{\\minus{}2}\\minus{}3\\cdot(2)^{\\minus{}3})}\\equal{}\\frac{2^4 \\plus{}3^3}{2^4\\minus{}3^3}\\equal{}\\frac{16\\plus{}27}{16\\minus{}27}\\equal{}\\frac{43}{\\minus{}11}\\equal{}\\minus{}\\frac{43}{11}$", "Solution_3": "Would that be finding the LCM of the numerator and denominator?\r\n\r\nNice trick, btw. :wink:", "Solution_4": "Yeah ernie, that's right... the LCM of the denominators you want to wipe out.\r\n\r\nSo in \r\n$ \\frac{\\frac{1}{3}\\plus{}\\frac{5}{6}}{\\frac{3}{4}\\minus{}\\frac{2}{3}}$\r\n\r\nyou could multiply top and bottom by 12, reducing the problem to\r\n\r\n$ \\frac{4\\plus{}10}{9\\minus{}8}\\equal{}14$\r\n\r\nStandard algebra trick :)", "Solution_5": "Yeah.\r\n\r\n[quote=\"RightAnnihilator\"]\"fractions within factions\"[/quote]\r\n\r\nComplex fractions. :wink:" } { "Tag": [ "linear algebra" ], "Problem": "if A and B are two nxn matrices such that $ AB \\equal{} I_{n}$ then prove that $ BA \\equal{} I_{n}$ and prove that $ B \\equal{} A^{ \\minus{} 1}$", "Solution_1": "We have that $ BA\\equal{}(BA)(BB^{\\minus{}1})\\equal{}B(AB)B^{\\minus{}1}\\equal{}BI_{n}B^{\\minus{}1}\\equal{}I_{n}$.", "Solution_2": "So... how do you know that $ B$ has an inverse? You can't prove this by just manipulating operations, since it's not true in a general ring.\r\n\r\nSee also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=277173]this thread[/url].", "Solution_3": "He said to prove that $ B\\equal{}A^{\\minus{}1}$ so I assumed the existence of inverses. But if this isn't a division ring, then the statement shouldn't be true in general.", "Solution_4": "Surprise: the statement becomes truer if you actually read it.\r\n\r\n darij" } { "Tag": [ "LaTeX" ], "Problem": "Hi, when I run LaTeX on an emacs text file, I get this error: \r\n\r\nblazes > latex chap3.tex\r\nThis is pdfeTeX, Version 3.141592-1.21a-2.2 (Web2C 7.5.4)\r\nentering extended mode\r\n(./chap3.tex\r\nLaTeX2e <2003/12/01>\r\nBabel and hyphenation patterns for american, french, german, ngerman, b\r\nahasa, basque, bulgarian, catalan, croatian, czech, danish, dutch, esperanto, e\r\nstonian, finnish, greek, icelandic, irish, italian, latin, magyar, norsk, polis\r\nh, portuges, romanian, russian, serbian, slovak, slovene, spanish, swedish, tur\r\nkish, ukrainian, nohyphenation, loaded.\r\n! Undefined control sequence.\r\nl.1 \\section\r\n {Introduction}\r\n? \r\n\r\n\r\nI'm new to LaTeX. I have checked the syntax, which seems to be correct, and as far as I know the correct packages have already been installed for me, so what could be causing this error? Any advice appreciated, Thanks.", "Solution_1": "If \\section is indeed line 1 of your document (as your error message implies), then that's your problem. The first line should be the \\documentclass line, and before you can start doing stuff like \\section you have to have \\begin{document}", "Solution_2": "chap3.tex is part of a number of chapters in which are all put into one document, main.tex, when I run latex main.tex.\r\n\r\nIn main.tex I have \\begin{document} and \\end{document}. This should not be required again in the individual chapters, as I used the same format last year and I didn't have this problem. (My supervisor gave me the code) So do you think this problem might be with the package? Thank you.\r\n\r\nSee below:\r\n\r\n\r\n\\begin{document}\r\n\r\n\r\n\\title{} \r\n\r\n\\author{}\r\n\\date{September 2004}\r\n\\maketitle\r\n\r\n\\chapter*{Declaration}\r\n\r\nI hereby declare that this thesis is entirely my own work and that it\r\nhas not been submitted as an exercise for a degree at any other\r\nuniversity.\r\n\r\n\\begin{center}\r\n\\vspace*{2in}\r\n\r\n\\underline{\\hspace*{3in}} \\today\r\n\r\nMyriam\r\n\r\n\r\n\\end{center}\r\n\r\n\r\n\\chapter*{Acknowledgements}\r\n\\input{ack}\r\n\r\n\r\n\\pagestyle{headings}\r\n\\tableofcontents\r\n\\listoftables\r\n\r\n\r\n\\begin{abstract}\r\n\\noindent\r\n\r\n\\end{abstract}\r\n\r\n\\chapter{Introduction}\r\n\\include{chap1}\r\n\r\n\\chapter{}\r\n\\include{chap2}\r\n\r\n\\chapter{Programs Utilised for Empirical Research}\r\n\\include{chap4}\r\n\r\n\\chapter{}\r\n\\include{chap5}\r\n\r\n\\chapter{}\r\n\\include{chap6}\r\n\r\n\\chapter{}\r\n\\include{chap7}\r\n\r\n\\chapter{Conclusion}\r\n\\include{chap8}\r\n\r\n\r\n\r\n\\bibliography{bibliography}\r\n\\bibliographystyle{egapa}\r\n\r\n\\appendix\r\n\r\n\\chapter{}\r\n\\include{}\r\n\r\n\\chapter{}\r\n\\include{}\r\n\r\n\\chapter{}\r\n\\include{}\r\n\r\n\\chapter{}\r\n\\include{}\r\n\r\n\\chapter{}\r\n\\include{}\r\n\r\n\\chapter{}\r\n\\include{} \r\n\r\n\r\n\\end{document}", "Solution_3": "But the output you posted above (in your first message) was for the command \"latex chap3.tex\", not \"latex main.tex\"...maybe that's your problem?\r\n\r\nAnd \\documentclass is still required somewhere, before \\begin{document}.", "Solution_4": "[65]\r\nUnderfull \\hbox (badness 10000) in paragraph at lines 221--222\r\n\r\n[66]) [67] [68]\r\nChapter 7.\r\n[69] (./chap8.tex [70]) [71] (./main.bbl [72]) [73]\r\nAppendix A.\r\n[74]latex: Not writing to .aux (openout_any = p).\r\n\r\n! I can't write on file `.aux'.\r\n\\@include ...\\immediate \\openout \\@partaux #1.aux \r\n \\immediate \\write \\@partau...\r\nl.155 \\include{}\r\n \r\nPlease type another output file name:" } { "Tag": [], "Problem": "Korisni linkovi:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=70684[/url]\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=27757[/url]\r\n\r\nNapomena moderatora: Dio diskusije je preseljen ovdje sa teme Off-off-Broadway, radi lakseg snalazenja.\r\n\r\nP. S. Otvaranje besmislenih tema se ne tolerise, timmysa.", "Solution_1": "Da bi ova tema dobila neki smisao, evo ne\u0161to \u0161to me zanima (to sam se danas na tulumu zapitao): ima li $ \\frac {a^p \\minus{} 1}{a \\minus{} 1} \\equal{} p^k$ rje\u0161enja gdje su $ a, p, k$ prirodni brojevi, $ p \\ge 3$ prost broj i $ a \\equal{} 1$ $ (mod$ $ p)$.", "Solution_2": "[url]http://www.math.olympiaadid.ut.ee/eng/archive/prob0607.pdf[/url]\r\n\r\nEto vidi tu na stranici 12 imas taj 3 zadatak koji mi se donekle cini slicnim ovom tvom pa ti to mozda da neku ideju.", "Solution_3": "Jedino mislim da se tu dokazuje obrnuti smjer:\r\n\r\nAko je izraz potencija prostog broja ---> p je prost.", "Solution_4": "Ma to je trivijalno ako znas sledecu teoremu:\r\n\r\nAko je $ p>2$ prost tada $ p^k\\parallel a\\minus{}1$ ako i samo ako $ p^{k\\plus{}1}\\parallel a^p\\minus{}1$ ($ k\\geq 1$). ($ p^k\\parallel x$ ako i samo ako $ p^k\\mid x$ i $ p^{k\\plus{}1}\\nmid x$)", "Solution_5": "To zna\u010di da jednad\u017eba nema rje\u0161enja, dakle zadatak s tuluma je rije\u0161en!\r\n\r\nGobline hvala ti. Je li te\u0161ko taj teorem dokazati?", "Solution_6": "Nije tesko... Ali to i nije teorema bas , to je vise lema ...", "Solution_7": "U vezi teoreme.. Recim da sam je ja koristila na izbornom bez dokaza i da mi nisu oduzeli poene, ali su nam rekli da bi to trebalo dokazati na medjunarodnom taknicenju, jer ocigledno nije bas toliko poznato, a i nema ime ...", "Solution_8": "[quote=\"hana\"]U vezi teoreme.. Recim da sam je ja koristila na izbornom bez dokaza i da mi nisu oduzeli poene, ali su nam rekli da bi to trebalo dokazati na medjunarodnom taknicenju, jer ocigledno nije bas toliko poznato, a i nema ime ...[/quote]\r\n\r\nMa, ta teorema ili lema, kako god, je toliko poznata (svaka malo ozbiljnija knjiga iz teorije brojeva je mora imati, jer se preko nje dokazuje par mnogo bitnih teorema), da bih bio veoma iznenadjen ako bi vam na bilo kom ozbiljnijem takmicenju oduzeli poene. Kako nema ime mozda je treba navesti, tacnije napisati nesto u stilu koristicemo sledecu teoremu, i tada skoro sigurno nece biti nikakvih problema.", "Solution_9": "Sto se naziva tice ovde imate nesto http://www.mathlinks.ro/viewtopic.php?t=70684 .", "Solution_10": "Ovo je znano kao Hensels lemma dokazuje se indukcijom, a ba\u0161 ne znam ili ovo lahko na natjecanju samo napi\u0161e\u0161 da je teorem.", "Solution_11": "Otkrio sam i ja ovaj podforum :D\nAli kad sam znao da mornik i hsiljak moraju biti ovde, kao i hana kad se cima oko nekih gluposti :P\n\nTakodje korisna lema:\n\n-postoji prost broj q>p, gde je p takodje prosto, tako da q deli a^p-1, a ne deli a-1." } { "Tag": [ "geometry", "trigonometry" ], "Problem": "\u4eca\u5929\u6211\u7684\u6570\u5b66?\u5e08\u95ee\u6211\u4eec:\"?\u77e5?\u4e3a\u4ec0\u4e48?\"\u8bf1\u5bfc\u516c?\"?\"\u6211\u4eec\u6447\u5934,\u7b49\u5f85\u7b54\u6848,\u4e8e\u662f,?\u5e08?\u7d27?\u6162\u5730\u8bf4:\"\u6211\u4e5f?\u77e5?......\"\r\n\r\n\u54ea?\u80fd\u5e2e\u5e2e\u5fd9,\u544a\u8bc9\u6211\u5230\u5e95\u4e3a\u4ec0\u4e48?\u8bf1\u5bfc\u516c?? :?", "Solution_1": "maybe because you are adding the period, so its adding or inducing.", "Solution_2": "in what context? what area of math are you doing in class?", "Solution_3": "I've only heard \"\u8bf1\u5bfc\u516c?\" in trignometry...\r\nwhich is something like $\\sin(x+n\\frac{\\pi}{2})=\\pm\\sin x(\\cos x)$...", "Solution_4": "[quote=\"liyi\"]I've only heard \"\u8bf1\u5bfc\u516c?\" in trignometry...\nwhich is something like $\\sin(x+n\\frac{\\pi}{2})=\\pm\\sin x(\\cos x)$...[/quote]\r\n\r\n\u5c31\u662f\u4e09\u89d2\u91cc\u7684\u90a3\"\u4e00\u5927\u7823\"\u516c?\u5440......" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Hello,\r\n\r\nWhat is your favorite pet? Mine is either dogs or cats, but I voted for dogs. I tried to think in all the options I could, but it seems like it only accept up to 10 options.\r\n\r\nBest,", "Solution_1": "Little kids [img]http://forums.finalhit.org/html/emoticons/whistling.gif[/img] .", "Solution_2": "hmmm...between a cat and a dog, i'd have to go with...a dog. I never really had a pet before but my neighbors were pet crazy. They probably had every kind of pet listed in the survey. The cats were real nice and they can think for themselves. Dogs will follow you around EVERYWHERE. You can play catch with them. We taught his dog how to play soccer, but then it popped my soccer ball :(", "Solution_3": "[quote=\"Treething\"]Little kids [img]http://forums.finalhit.org/html/emoticons/whistling.gif[/img] .[/quote]\r\n\r\nLittle kids can be pretty nice. However, one time, i was taking care of my neighbors and their son (who's real little) started \"cutting\" me up with his toy lightsaber. Thats freaky!!! :P", "Solution_4": "[quote=\"Treething\"]Little kids [img]http://forums.finalhit.org/html/emoticons/whistling.gif[/img] .[/quote]\r\n\r\ngosh darnit...u beat me to it. i was going to say my little brother *evil grin*", "Solution_5": "Pet rocks, anyone?", "Solution_6": "[quote=\"miraculouspostmaster\"]Pet rocks, anyone?[/quote]\r\n\r\nRight here!", "Solution_7": "[quote=\"yif man12\"][quote=\"miraculouspostmaster\"]Pet rocks, anyone?[/quote]\n\nRight here![/quote]\r\n\r\nLol... Rocks. I chose a dog because my friend has a dog that plays with us all the time.", "Solution_8": "Clearly dogs are the best pets ever :P ... I am amazed that Gyan hasn't yet replied in this topic :)", "Solution_9": "[quote=\"Valentin Vornicu\"]Clearly dogs are the best pets ever :P ... I am amazed that Gyan hasn't yet replied in this topic :)[/quote]\r\n\r\nI do think that dogs are great pets; very loyal, usually they like to play. But also they require a lot of time, at least if you really want to have a happy dog. On the other hand, if you don't have too much time, usually a cat is great. They don't need that much attention, are very independent, but they are less prone to play with you.\r\n\r\nBest,", "Solution_10": "[quote=\"DarkKnight\"][quote=\"yif man12\"][quote=\"miraculouspostmaster\"]Pet rocks, anyone?[/quote]\n\nRight here![/quote]\n\nLol... Rocks. I chose a dog because my friend has a dog that plays with us all the time.[/quote]\r\n\r\nCool! My friend has a pet rock and he plays with us all the time too! :D ;)", "Solution_11": "I have a pet snake, houdina, and she is great. She is a corn snake that is somewhat olive colored. She is pretty big, about 1.5 meters, and as the name suggests, she is an escape artist. :lol: \r\n\r\nI used to have another snake, snowpowder, but she contracted cancer and it spread. A very, sad and painful death :( \r\nbut u have to move on", "Solution_12": "pet rocks are cool.... i know someone that got bit by one once", "Solution_13": "What's a pet rock anyway? (I only know that rock is a kind of stone (or vice-versa) or a musical genre - and an adverbe meaning something in the lines of \"cool\" - e.g. \"they rock\") :? :?", "Solution_14": "cats are better than dogs!!!!!!!! but... my parents hate the \"mess\" pets make so i don't have one.\r\n\r\nelvenchamp777", "Solution_15": "[quote=\"Valentin Vornicu\"]What's a pet rock anyway? (I only know that rock is a kind of stone (or vice-versa) or a musical genre - and an adverbe meaning something in the lines of \"cool\" - e.g. \"they rock\") :? :?[/quote]\r\n\r\nthere really is nothing special about a pet rock. it's basically a stone that sometimes gets features glued or drawn onto it and usually has a name.", "Solution_16": "ok so it is a [b]stone[/b] in the end :) ... I'm not sure, but last time I checked, pets ment [b]live[/b] animals, right? :? :)", "Solution_17": "pet rocks were a huge fad in the seventies. it is what it says it is- a pet ROCK. that is, a rock that someone has painted and is now a pet. you dont have to feed them, walk them, or clean up after them.\r\n\r\nbut anywho. you need to add an \"all of the above\" option to the poll.", "Solution_18": "Turtles are coo./ :)", "Solution_19": "gosh, school was great when we had class pets. \"sigh\", now all the science classes has fake human skeletons. Turtles are supercool!!!", "Solution_20": "[quote=\"Valentin Vornicu\"]ok so it is a [b]stone[/b] in the end :) ... I'm not sure, but last time I checked, pets ment [b]live[/b] animals, right? :? :)[/quote]\r\n\r\nOh, you should meet my pet rock. :D", "Solution_21": "does it (for lack of better word) have a name?", "Solution_22": "Chickens :D . Not only are they fun to watch run around their little coop, but they lay eggs which you can sell for big *inster your currency symbol here*. Though when it comes to a house pet, nothing beats a good dog.", "Solution_23": "My math book...", "Solution_24": "dogs are the best", "Solution_25": "[quote=\"filletwho\"]My math book...[/quote]\r\n\r\ndid you name it? :D", "Solution_26": "\"Mathybooky\"", "Solution_27": "[quote=\"mcalderbank\"]does it (for lack of better word) have a name?[/quote]\r\n\r\nHis name is Bill.", "Solution_28": "[quote=\"filletwho\"]My math book...[/quote]\r\n\r\nI beg your pardon, but I think it is worst than having a Pet Rock. C'mon, I love math, but everybody needs a life too.\r\n\r\nBest,", "Solution_29": "Cats are awesome! \r\n\r\nI have 3 cats, one of them is a ghost cat, hehehe. Its white with blue eyes that glow red. She actually doesn't show up in any pictures we took of her, only as an extremely faint outline if you look for it. Its really kinda scary. \r\n\r\nEver heard the quote by Churchill? Not about pets, but close enough. \"I like Pigs. Dogs look up to us. Cats look down on us. Pigs treat us as equals.\"", "Solution_30": "I voted for dogs. :) . The one pictured on the left is actually a graduate of the Oregon campus of the famous .[url=http://www.guidedogs.com/]Guide Dogs.[/url] and some of you may know him from [url=http://www.thefacebook.com/]\" The Face Book \"[/url] as a graduate student in MIT. (That explains why he spends most of his spare time socialising with his friends online) Besides, he is my friend and a family member so he is more than a pet.:) (His eating habits does inspire mathcounts problems: eg: [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25885]this ..:)[/url]", "Solution_31": "I have a pet rock named Mike the Mineral\r\n\r\nalso, i feed my pet cat-flavored dog food", "Solution_32": "You totally forgot about stuff like [hide]crawfish[/hide]", "Solution_33": "It seems like no body likes birds. Nobody have voted for parrots or singing birds :(", "Solution_34": "[quote=\"djimenez\"]It seems like no body likes birds. Nobody have voted for parrots or singing birds :([/quote]\r\n\r\nBirds are great! I used to own one and it's actually a very good pet. The only annoying part is shutting it up when you're trying to go to sleep, but it didn't disturb my sleeping after about a week. \r\n\r\nOn those lines, I think passenger pigeons would be very cool. But the coolest thing to do with a bird is to teach it to poop on other people. That would be pretty funny, but a little mean too.", "Solution_35": "I luv little bunnies. Dwarf rabbits are cool. Mine will respond to name and seems to like drinking tomato juice. I dunno why", "Solution_36": "mine may be a horse or an eagle, i love both!", "Solution_37": "My math book can show you tricks that a dog would never master...", "Solution_38": "[quote=\"filletwho\"]My math book can show you tricks that a dog would never master...[/quote]\r\n\r\nmy pet rock (mike the mineral) can eat your magical math book.", "Solution_39": "I voted for dogs but I really like all of them.\r\nAfter all I do have:\r\n4 dogs\r\n4 cats\r\n2 hedgehogs :D \r\n1 cockateel\r\nand soon to be...\r\n2 horses\r\n3 more dogs\r\n2 more cats\r\nand possibly a snake.\r\nFor a grand total of 19 pets. :D :D", "Solution_40": "[quote=\"jackroland\"]I voted for dogs but I really like all of them.\nAfter all I do have:\n4 dogs\n4 cats\n2 hedgehogs :D \n1 cockateel\nand soon to be...\n2 horses\n3 more dogs\n2 more cats\nand possibly a snake.\nFor a grand total of 19 pets. :D :D[/quote]\r\n\r\nthats a lot...my neighbors have...\r\n\r\n1 dog\r\n3-6 puppies\r\n3 cats\r\n5 fish\r\n4 ducks\r\n4 chickens\r\nmaybe more...", "Solution_41": "[quote=\"plokoon51\"]\n3-6 puppies\n[/quote]\r\n\"3-6\"? does that mean it changes?\r\nWe used to have chickens but they were all eaten by various other animals.\r\n(and I'm sure not complaining! ;) )", "Solution_42": "[quote=\"jackroland\"][quote=\"plokoon51\"]\n3-6 puppies\n[/quote]\n\"3-6\"? does that mean it changes?\nWe used to have chickens but they were all eaten by various other animals.\n(and I'm sure not complaining! ;) )[/quote]\r\n\r\nwell, i don't know how many exactly. they (neighbors) might've given some away, or kept them, i'm not quite sure.\r\n\r\non the other hand, the puppies like to deceive people by changing their numbers :D" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "I've seen these type of problems and I'm posting in Pre-Olympiad forum because I want to know the general way to solve this. (I consider Intermediate to find the answer rather than regular way)\r\n\r\nLike how woud you estimate the value of $\\sqrt[3] 7$? How about $\\sqrt[3] 4$? Or $\\sqrt[8] 91$? Also, how would you find the ones like this $\\sqrt[3] 2$ and $\\sqrt[3] 3$ where they must be very close.. \r\n\r\nHow would you do those estimations?\r\n\r\nI mean without using calculator.\r\n\r\nThank you.", "Solution_1": "How close do you want your estimation to be? Anyways the best way is usually just to multiply the number by 1000 or something and guess and check.", "Solution_2": "There is a way to do it...but it will be somewhat hard with cube roots\r\nWrite out the number.\r\nPlace lines every three places (for instance, place one between the hundreds and thousands, the hundred thousands and millions, etc. You may want to write some zeros to the right of the decimal point and place lines every three for those as well.)\r\nNow draw a radical sign over the number. Look at the group of digits on the very left (could be 1, 2 or 3 digits.). Write down the largest number whose cube is less that the cube of that number. Subtract the cube of the number from the number formed by the group. Now bring down the next three digits. If the first digit you wrote on top was n, then you must now find the greatest x such that\r\n300xn^2+30nx^2+x^3\r\nis less than this number. Subtract. Repeat. This basically uses the principle that (a+b)^3=a^3+3ba^2+3ab^2+b^3. As you get larger numbers, the process will become tedious, but you can get an exact number to as many decimal places as you wish." } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Let the Taylor series $ e^{e^{x}}\\equal{}\\sum_{n \\equal{} 1}^{\\infty}a_{n}x^{n}$\r\nProve:$ a_{n}\\ge e (\\gamma\\ln n)^{\\minus{}n}$\r\nwher $ \\gamma$ is a constant number greater than $ e$.\r\nI find $ a_{n}\\equal{}\\frac{1}{n!}\\sum_{k\\equal{}0}^{\\infty}\\frac{k^{n}}{k!}$\r\nwhich is related the sum of the Stirling Number of second kind.\r\nBut I don't know how to do next.", "Solution_1": "No one?It is from an entering eximine.", "Solution_2": "No one take care of this problem? :(", "Solution_3": "Hi \r\nyour question is equivalent to the next question :\r\nwe must limit $ \\sum\\frac{k^{n}}{k!}$ or find asymptotical expression for this.\r\nI am not sure but I think that $ a_{n}> a_{n\\minus{}1}$ if it is true then your problem solved. \r\nin this forum was posted some particular case of this sum \r\n$ \\sum\\frac{k^{n}}{k!}$\r\nto be exact this sum $ \\sum\\frac{k^{4}}{k!}$\r\nThe idea was in next :\r\n$ k^{n}\\equal{} Ak(k\\minus{}1)..1\\plus{}B(k\\minus{}1)(k\\minus{}2)..\\plus{}..\\plus{}Z$\r\nwe find this coeficient and after can calculate this sum\r\nor \r\nyour sum is equal to \r\n$ x(...(xe^{x}\\minus{}x)'...)'$ in point zero.\r\nI think somewhere must be calculated this sum \r\n$ \\sum\\frac{k^{n}}{k!}$ or asymptotical expression", "Solution_4": "The coefficients are (up to a multiple of $ e$) the Bell numbers. The [url=http://www.research.att.com/~njas/sequences/A000110]OEIS[/url] has a lot of information on them, as does [url=http://mathworld.wolfram.com/BellNumber.html]mathworld[/url]. I don't know if any of it is useful for solving your question, however.", "Solution_5": "Thank you,two.But I don't think it always be true that $ a_{n}>a_{n\\minus{}1}$\r\nIn fact you can try some small number you will find it is wrong.\r\nYes,it is really the Bell numbers,but it appeared in an eximine,so I think there may exist an acute solution.", "Solution_6": "Yes.\r\nI find some information about Bell number and I also calculate that when $ n\\to+oo$ then\r\n$ a_{n}$~${ Cn!e^{1+n(\\frac{1}{\\ln n}-\\ln\\ln n)}}$\r\nwhere $ C$ some constant" } { "Tag": [], "Problem": "Patrick drove to work on Monday at an average speed of 40 mph and arrived one minute late. He left at the same time on Tuesday, drove an average speed of 45 mph, and arrived one minute early. How many miles does he drive to work each day?", "Solution_1": "\\[ 40t\\equal{}45(t\\minus{}2)\\] \\[ d\\equal{}40t\\] \\[ d\\equal{}\\boxed{8}\\]", "Solution_2": "Watch the units :wink: \r\n\r\n\\begin{align*}\r\n40\\left(t+\\frac{1}{60}\\right)=45\\left(t-\\frac{1}{60}\\right) &\\Rightarrow 40t+\\frac{40}{60}=45t-\\frac{45}{60} \\\\\r\n&\\Rightarrow 5t=\\frac{85}{60} \\\\\r\n&\\Rightarrow t=\\frac{17}{60} \\\\\r\n&\\Rightarrow d=40\\left(\\frac{17}{60}+\\frac{1}{60}\\right)=\\boxed{12} \r\n\\end{align*}" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Show that 2 <= (1 - x2)2 + (1 - y2)2 + (1 - z2)2 <= (1 + x)(1 + y)(1 + z) \r\nfor non-negative reals x, y, z with sum 1.", "Solution_1": "For the problem with x+y+z=1 etc.:\r\n\r\nFor the first part:\r\n(1-x^2)^2+(1-y^2)^2+(1-z^2)^2>=2 iff x^4+y^4+z^4+1 >= 2(x^2+y^2+z^2) iff x^4+y^4+z^4+(x+y+z)^4 >= 2(x+y+z)^2*(x^2+y^2+z^2) iff x^4+y^4+z^4+2(xy+yz+zx)(x+y+z)^2 >= (x+y+z)^2*(x^2+y^2+z^2) iff x^4+y^4+z^4+4(xy+yz+zx)^2 >= (x^2+y^2+z^2)^2 iff 2(xy+yz+zx)^2 >= x^2*y^2+y^2*z^2+z^2*x^2, which is obviously true, with equality iff x=y=0 (or y=z=0 or z=x=0).\r\n\r\nThe second part:\r\n(1-x^2)^2+(1-y^2)^2+(1-z^2)^2 <= (x+1)(y+1)(z+1) iff x^4+y^4+z^4+1 <= 2(x^2+y^2+z^2)+xy+yz+ zx+xyz iff x^4+y^4+z^4+(x+y+z)^4 <= 2(x+y+z)^2*(x^2+y^2+z^2)+(xy+yz+zx)(x+y+z)^2+ xyz(x+y+z). We keep simplifying this (by subtracting common terms from both sides until we are left with 2xyz(x+y+z) <= x^3*(y+z)+y^3*(x+z)+z^3*(x+y). For proving this we just add the inequalities of the type x^3*y+y^3*x >= 2x^2*y^2 and then we use the fact that x^2*y^2+y^2*z^2+z^2*x^2 >= xyz(x+y+z)." } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Problem 2:\r\nLet $ n$ be a natural number such that $ n \\equal{} a^{2}\\plus{}b^{2}\\plus{}c^{2}$, for some natural numbers $ a, b, c$. Prove that\r\n$ 9n \\equal{} (p_{1}a \\plus{} q_{1}b \\plus{} r_{1}c)^{2} \\plus{} (p_{2}a \\plus{} q_{2}b \\plus{} r_{2}c)^{2} \\plus{} (p_{3}a \\plus{} q_{3}b \\plus{} r_{3}c)^{2}$,\r\nwhere $ p_{j} 's, q_{j} 's, r_{j} 's$ are all nonzero integers. Further, of $ 3$ does not divide at least one of $ a, b, c$, prove that $ 9n$ can be expressed in the form $ x^{2} \\plus{}y^{2} \\plus{}z^{2}$, where $ x, y, z$ are natural numbers none of which is divisible by $ 3$.", "Solution_1": "my solution isn't complete but...here is what i did:\r\npart 1-the answer is $ (p_1,p_2,p_3)\\equal{}(2,2,\\minus{}1),(q_1,q_2,q_3)\\equal{}(\\minus{}1,2,2),(r_1,r_2,r_3)\\equal{}(2,\\minus{}1,2)$\r\npart 2-if two of them(i mean $ a,b,c$) are divisable by $ 3$ then the answer will be the same, \r\nit remains two cases:\r\nnone of them is divisable by $ 3$\r\none of them is divisable by $ 3$\r\nif none of them is divisable by three and we want to present \r\n$ 9n \\equal{} (p_1a\\plus{}q_1b\\plus{}r_1c)^2\\plus{}(p_2a\\plus{}q_2b\\plus{}r_2c)^2\\plus{}(p_3a\\plus{}q_3b\\plus{}r_3c)^2$ but this time we r allowed to choose the coefficients between integers, then u can prove that none of them is $ 0$ because if it happens then one of the coefficients must be $ 3$ and...(the rest is trivial).", "Solution_2": "http://www.mathlinks.ro/viewtopic.php?search_id=551070467&t=131843" } { "Tag": [], "Problem": "What is the value of $ 99^3 \\plus{} 3(99^2) \\plus{} 3(99) \\plus{} 1$?", "Solution_1": "$ a^3\\plus{}3a^2b\\plus{}3ab^2\\plus{}b^3\\equal{}(a\\plus{}b)^3$.\r\nThus, $ 99^3 \\plus{} 3(99^2) \\plus{} 3(99) \\plus{} 1 \\equal{} (99\\plus{}1)^3\\equal{}100^3\\equal{}\\boxed{1000000}$." } { "Tag": [], "Problem": "Does it cost AoPS money to create new communities? If it does, I see why there wouldn't be a community for every state. However, if it does not, what are the disadvantages of having 50 state communities, besides a cluttered interface?", "Solution_1": "It costs them server space (I doubt it's substantial, though)... not sure who'd they pay the money too.... PhpBB is a open-source, free, forum package. (Run-on sentences for the win!)", "Solution_2": "Clutter, and time. Also, we'd have to police forums that don't have moderators, which we don't want to do. So, we only create forums when there's someone who wants to take care of them." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Assume $x,y>0, t\\in (0,1)$. Prove that\r\n$a) t/x+(1-t)/y \\geq 1/(x^{t}+y^{1-t})$ and\r\n$b) 1/(tx+(1-t)y) \\leq 1/(x^{t}\\cdot y^{1-t}) \\leq t/x+(1-t)/y$", "Solution_1": "[quote=\"puuhikki\"]Assume $x,y>0, t\\in (0,1)$. Prove that\n$a) t/x+(1-t)/y \\geq 1/(x^{t}+y^{1-t})$ and\n$b) 1/(tx+(1-t)y) \\leq 1/(x^{t}\\cdot y^{1-t}) \\leq t/x+(1-t)/y$[/quote]\r\n\r\na) isn't true, put $x=y=100$ and $t=1/2$.\r\n\r\nb) Jensen, of course." } { "Tag": [ "number theory", "Diophantine equation" ], "Problem": "For each positive integer $ {m}>{1}$, let P(m) denote the greatest prime factor of m. For how many positive integers n is it true that both P(n)=$ \\sqrt{n}$ and P(n+48)=$ \\sqrt{n\\plus{}48}$?\r\n\r\nThanks for the help!", "Solution_1": "[hide=\"Solution\"]$ P(n) \\equal{} \\sqrt {n}$ implies $ n$ is a perfect square. So, $ n$ and $ n \\plus{} 48$ are perfect squares. We want to solve the diophantine equation $ x^2 \\minus{} y^2 \\equal{} 48$, namely $ (x \\plus{} y)(x \\minus{} y) \\equal{} 48$. There are several cases.\n\nCase 1: $ x \\plus{} y \\equal{} 48$ and $ x \\minus{} y \\equal{} 1$. No solution.\n\nCase 2: $ x \\plus{} y \\equal{} 24$ and $ x \\minus{} y \\equal{} 2$. $ x \\equal{} 13$ and $ y \\equal{} 11$.\n\nCase 3: $ x \\plus{} y \\equal{} 16$ and $ x \\minus{} y \\equal{} 3$. No solution.\n\nCase 4: $ x \\plus{} y \\equal{} 12$ and $ x \\minus{} y \\equal{} 4$. $ x \\equal{} 8$ and $ y \\equal{} 4$.\n\nCase 5: $ x \\plus{} y \\equal{} 8$ and $ x \\minus{} y \\equal{} 6$. $ x \\equal{} 7$ and $ y \\equal{} 1$. However, $ y \\neq 1$, so this doesn't count.\n\nTherefore, there are $ \\boxed{2}$ such integers.[/hide]", "Solution_2": "[hide=\"Small change to above\"]\n$ P(n) \\equal{} \\sqrt{n}$ does imply that $ n$ is a perfect square, but more: that $ n$ is the square of a prime. (Note: it says the greatest prime factor.) So unfortunately Case 4 above doesn't work: $ P(16) \\neq \\sqrt{16}$, which means only one solution left.\n[/hide]" } { "Tag": [ "search" ], "Problem": "Just to let everyone know (I hope this hasn't already been posted), but the relly awards (where we find out if kevin chen won) will be on regis and kelly this friday", "Solution_1": "Klebian, I don't think this topic should be posted. There already has been many other topics about this Kevin Chen thing.\r\n\r\nMods please lock this.", "Solution_2": "He's just telling us when it will be. There's nothing wrong with that.", "Solution_3": "It is useful to know; however, there is already a topic about this still on the front page. Please post there :) .\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=166113\r\n\r\nBtw, if you don't know if a topic is already posted, searching the forum with the website's search is very easy." } { "Tag": [ "analytic geometry", "vector", "Euler", "geometry", "trigonometry", "trig identities", "Law of Cosines" ], "Problem": "I know that all AMC questions can be solved without a calculator, but there are just some that either you have to (1) think of an insanely simple and intelligent way or (2) brute force it taking like 20 minutes on one problem or (3) know how to add, divide, square root everything perfectly in a very short amount of time.\r\n\r\nfor example this problem: http://www.mathlinks.ro/viewtopic.php?p=383159#383159\r\nI can easily do it a simple way--assigning coordinates to the triangle and then using Hebron's formula/law/whatever by finding the length of each side and then calculating sqrt.S(S-A)(S-B)(S-C). but that's pretty much impossible in the short, short amount of time that you get, plus a little carelss mistake will kill the answer. now on my calculator i got it in like 10 seconds! so what now, is the new AMC going to have problems with messy work like this ? (yes, i know that there's probably some easy peasy way to do it, so...is the reason why we don't have calculators so we all have to find the intelligent way? :( )\r\n\r\nand yes, i'm probably stupid, but how was i supposed to know, in one of the problems, that 29 was a factor of 667 without a calculator? =P", "Solution_1": "[quote=\"Narcissa\"]is the reason why we don't have calculators so we all have to find the intelligent way? :( [/quote]\n\nYes. Think of this as incentive to improve! I'm sure you feel better when you find a quick way to solve a problem than when you brute-force it for 20 minutes. (Coordinate bashing is [i]never[/i] a good idea, and personally I don't find it any fun at all.)\n\n[quote=\"Narcissa\"]and yes, i'm probably stupid, but how was i supposed to know, in one of the problems, that 29 was a factor of 667 without a calculator? =P[/quote]\r\n\r\nI don't usually give this advice, but memorizing the prime factorizations of the year numbers is a good idea for competitions.", "Solution_2": "[quote]I don't usually give this advice, but memorizing the prime factorizations of the year numbers is a good idea for competitions.\n[/quote]\r\n\r\nthat will probably be a very smart thing to do =O", "Solution_3": "[quote=\"t0rajir0u\"]I don't usually give this advice, but memorizing the prime factorizations of the year numbers is a good idea for competitions.[/quote]\r\nHeh, I do this too.", "Solution_4": "[quote=\"t0rajir0u\"](Coordinate bashing is [i]never[/i] a good idea, and personally I don't find it any fun at all.)[/quote]\r\nSays you :P", "Solution_5": "I distinguish \"coordinate bashing\" from \"using vectors,\" if that's what you mean. :?:", "Solution_6": "[quote=\"t0rajir0u\"]I distinguish \"coordinate bashing\" from \"using vectors,\" if that's what you mean. :?:[/quote]\r\nNo, that isn't. I mean giving every point a coordinate, then using the distance formula/shoelace excessively :D", "Solution_7": "So will the problems be slightly different when compared to previous years (more clever solutions; not as much calculation), or will they write the problems to be just as they were before?", "Solution_8": "[quote=\"mwpl11\"]So will the problems be slightly different when compared to previous years (more clever solutions; not as much calculation), or will they write the problems to be just as they were before?[/quote]\r\n\r\n\r\nIf its just as before then it shouldnt be any difference because the best solutions as the years before required no calculator (very limited calculation). Only those \"out of the box\" solutions required a calculator.", "Solution_9": "[quote=\"firecricket91\"]Only those \"out of the box\" solutions required a calculator.[/quote]\r\n\r\nI guess you're right. :blush:\r\n\r\nBut by doing that, they're basically restricting people to solving the problems in a very limited number of ways (\"the best solutions\"). That's no fun.", "Solution_10": "If you like computational methods of solving problems, try the USACO or [url=http://projecteuler.net/]Project Euler[/url]. But the AMC is about mathematics - [i]good mathematics[/i] - and the best scores should go to the best solutions.", "Solution_11": "I think you can expect to see some problems in a slightly different style now, problems that weren't used before because calculators rendered them trivial. Take a look at pre-1994 (or is it 1995?) AHSME's to see what I mean. You'll see problems on there that are definitely interesting if you don't have calculators allowed, but trivial with them. As I understand it, including problems like that was one of the big motivations for removing calculators in the first place.", "Solution_12": "[quote=\"t0rajir0u\"]If you like computational methods of solving problems, try the USACO or [url=http://projecteuler.net/]Project Euler[/url]. But the AMC is about mathematics - [i]good mathematics[/i] - and the best scores should go to the best solutions.[/quote]\r\n\r\nI just have to say, USACO usually requires more thought than \"co-ordinate bashing\" on this problem.", "Solution_13": "Well, yes; I just meant that the USACO has a different focus compared to the (competitions leading up to) USAMO.", "Solution_14": "[quote=\"Sly Si\"]I think you can expect to see some problems in a slightly different style now, problems that weren't used before because calculators rendered them trivial. Take a look at pre-1994 (or is it 1995?) AHSME's to see what I mean. You'll see problems on there that are definitely interesting if you don't have calculators allowed, but trivial with them. As I understand it, including problems like that was one of the big motivations for removing calculators in the first place.[/quote]\r\n\r\nI agree. That's probably going to be the only major change; a question that occurred on any previous AMC could still occur now even without calculators (except, obviously, they'd be different questions -- you know what I mean). The question # might be slightly different, but that would be a minor change that no one would notice anyways.", "Solution_15": "[quote=\"t0rajir0u\"](Coordinate bashing is [i]never[/i] a good idea, and personally I don't find it any fun at all.)[/quote]\r\n\r\nSo, is there always a better solution than coordinate geometry, or can that sometimes be the most elegant method?", "Solution_16": "Like I said, I distinguish coordinate bashing from the use of vectors. While the use of vectors can often be quite illuminating (because we set certain quantities as vectors without computing them), the use of coordinate bashing in particular requires calculations that are never (in my experience) the most elegant way of doing a problem. At least try Law-of-Cosines bashing first, since that [i]can[/i] occasionally be a decent way of doing a problem.\r\n\r\nEdit: I'm talking about geometry problems that aren't stated in terms of coordinates. Even when problems are stated in terms of coordinates, coordinate-bashing (except as far as Pick's Theorem or the Shoelace Theorem are concerned) is rarely the best method.", "Solution_17": "someone suggested memorizing \"prime factors of the year numbers\" -- can someone tell me what is meant by that? By Prime factors of the year? Thanks", "Solution_18": "Memorizing the prime factorization of the year - for example, in year 2002, competitors may have memorized $ 2002 \\equal{} 2\\cdot7\\cdot11\\cdot13$, so that if the problem \"find the sum of the factors of $ 2002$\" came up at a competition, they could find the answer more quickly.", "Solution_19": "I don't know if someone already said this, but I like to split up things like $ 796^2$ into 2 binomials if you will.\r\n$ 796^2\\equal{}(800\\minus{}4)(800\\minus{}4)\\equal{}640000\\minus{}6400\\plus{}16\\equal{}633616$\r\n\r\nI actually did that one in my head today. I suggest not using a calculator for a bit, and that should help you adjust to manual arithmetic.", "Solution_20": "My math teacher says that because of the no calculator rule, there will be more clever-based questions that can be solved very quickly by paper, but require some knowledge of, say, correct formulas. But hey don't worry: if this new rule is gonna hurt you, then it's gonna hurt us all! *gulp*", "Solution_21": "[quote]but require some knowledge of, say, correct formulas.[/quote]GULP! don't know any formulas =O", "Solution_22": "[quote=\"AoPS book\"]\nIt is not important to remember formulas, but how they are derived. If you happen to forget that formula you can prove it.\n[/quote]", "Solution_23": "[quote=\"hunter34\"][quote=\"AoPS book\"]\nIt is not important to remember formulas, but how they are derived. If you happen to forget that formula you can prove it.\n[/quote][/quote]\r\nOr you could take the other route and just not forget formulas ;)", "Solution_24": "Yeah that works pretty well for me.", "Solution_25": "[quote=\"zephyredx\"]but require some knowledge of, say, correct formulas.[/quote]\r\n[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]This site's webmaster disagrees.[/url]\r\n\r\nBetter to say is \"but require some mathematical ingenuity.\" Granted, formulas do work on AMCs, but that approach won't get you much farther.", "Solution_26": "[quote=\"MellowMelon\"][quote=\"zephyredx\"]but require some knowledge of, say, correct formulas.[/quote]\n[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]This site's webmaster disagrees.[/url]\n\nBetter to say is \"but require some mathematical ingenuity.\" Granted, formulas do work on AMCs, but that approach won't get you much farther.[/quote]\r\n\r\nI didn't mean \"memorize formulas right-on\" i meant \"know them well\", which would imply knowing to derive them, and also know what kind of problem they appear in. I actually don't trust many of the formulas I can't derive unless it's an absolute emergency.", "Solution_27": "^whoa, yeah i've seen that before but not used to it. i guess i'll have to do more problems that way!! thanks for sharing" } { "Tag": [ "articles", "number theory", "complex analysis" ], "Problem": "So, if anyone has heard of those famous open problems (Millennium Problems) that includes: Riemann Hypothesis; P vs NP; Yang-Mills Theory etc.\r\n\r\nI have a query pertaining to these problems, and it is: Why are these problems so difficult to solve, secondly what does one require to solve these problems, is it advanced knowledge in their respective fields?", "Solution_1": "Try solving one of them and you'll see. You're not solving them without some really deep knowledge of the fields and very deep insight.", "Solution_2": "What does insight mean with reference to problem-solving? Is it intuition or something else? :maybe:", "Solution_3": ":o !!!! Insight is the most basic concept of problem solving!!!! It is what separates the US (and probably most other places) public school mathematics education from mathematics competitions.\r\n\r\nInsight is that feeling when you (often suddenly) understand or see through something, when you suddenly can view a problem or something else in a new way.\r\n\r\nThus to solve these problems, one not only needs knowledge in the advanced, technical fields in which they are posed (the tools to solve the problem), but also experience and insight (how to use those tools). Both of these are very difficult, since the subject matter is often quite abstract, but the insight is much more valuable and desired. Any old computer can be programmed the tools of mathematics, but what they lack is insight.\r\n\r\nAs for the relation with intuition, when solving a difficult problem, one first develops an intuition, which may be vague, and then insight to verify and sharpen the idea proposed by the intuition.", "Solution_4": "I have no such resolutions of solving any of those open problems, but one can get tempted by the reward one may get for solving any one of those open problems, but I sometimes feel it is not worth the effort; On other times, I feel maybe one should pursue those problems (any one of them) in earnest :D ,but even when someone pursues them in earnest the problems are not solved (consider the instance of 'The Riemann Hypothesis' which has remained an \"open problem\" for the last 150 years.\r\n\r\n\r\nI don't know -----what these problems demand from a human being?", "Solution_5": "They demand insight.\r\n\r\nUsually good insight comes with experience. As one becomes familiar with a subject one can more easily get insight. Thus the problem demands experience (i.e. years of work) from a human being.\r\n\r\nOn a side note: sometimes it is said that a problem is not likely to be solved by an already well-known, experienced mathematician, because those ideas ingrained in his/her head have already been tested, and this mathematician may not have the fresh, new mind and imagination to come up with the novel ideas that may be needed to solve a problem. In that case, talent is demanded. (The talent/work debate is one that often seems to come up.)", "Solution_6": "But, I feel the person who is going to solve the Riemann Hypothesis , that is , provided it is solvable ( mostly, it is going to be proved, rather than being disproved) will cover himself with glory, I wonder who that person might be. \r\n\r\nIf the biography \"A Beautiful Mind\" is taken into consideration, it is stated in it that John Nash has indeed tried his hand at conquering the 'Riemann Hypothesis' but due to some reason, he couldn't make further progress.\r\n\r\nHmm, coming to imagination, how much does role does this play in solving these problems ( or solving problems for that matter). :lol:", "Solution_7": "in general, it's kind of hard to explain why a problem is hard without already having some reasonable knowledge of the field in question, so I'm probably not going to be able to do that. Terry Tao can, though! Here's a link to his thoughts on why the Navier-Stokes problem is hard -- see what you can get out of it: http://terrytao.wordpress.com/2007/03/18/why-global-regularity-for-navier-stokes-is-hard/\r\n\r\nArguments I've seen for why P vs NP is hard (as presented in most introductory computational complexity books) usually cite that diagonalization arguments cannot possibly work by talking about oracles and whatnot, so this here is a very specific tactic that can be shown to fail, but I'm sure there's a much firmer understanding of the problem that I can't provide. \r\n\r\nI don't really know much about any of the other millennium problems (besides poincare), but Tao's article might give you some ideas.\r\n\r\nIt's a good question, certainly -- understanding why a problem is hard and what makes it hard is always a good thing to be thinking about at any level, both in terms of possible tactics to approaching the problem if you end up working on it and to make sure your eventual proof bypasses all the initial difficulties instead of having a hole in it somewhere.", "Solution_8": "The one I'm particularly interested in (and who isn't?) is Riemann. I would love to understand why this problem is so difficult but my knowledge of number theory and complex analysis is rather limited.", "Solution_9": "[quote=\"JRav\"]The one I'm particularly interested in (and who isn't?) is Riemann. I would love to understand why this problem is so difficult but my knowledge of number theory and complex analysis is rather limited.[/quote]\r\n\r\neh why what is so exciting about RH", "Solution_10": "I agree. The Hodge Conjecture is certainly the most interesting. :P", "Solution_11": "A few days ago there was published a proof of Goldbach Conjecture. It's being tested now. What do you think aobut it?\r\n\r\nhttp://arxiv.org/pdf/0812.0930", "Solution_12": "I don't understand anything in that paper but it would be cool if it turned out to be correct.", "Solution_13": "Is it humanely possible to solve not one , but two of these problems? ( A two-million dollar prize in the offing. :lol: ) \r\n\r\nWhat if Dr Perelman manages to solve the Navier-Stokes problem? It would be great, ain't it. \r\n\r\nI think they demand knowledge of vast amount of literature across different fields of Mathematics in order to solve two of these Millennium problems.\r\n\r\nAnd by the way, It's almost 150 years since the 'Riemann Hypothesis' was proposed by Riemann himself. \r\n\r\nI also personally feel P!=NP ( '!=' denotes not equal to), but I might be mistaken.\r\n\r\nI think if one could come up with an instance that shows-- A problem that is easy to check (NP) is also easy to solve (P), and Eureka!, P=NP problem is all but solved, but I don't think it is possible to come up with such an instance.", "Solution_14": "Is it possible? Yes. Is it likely? Not even remotely.", "Solution_15": "Firstly, I would like to apologize to JBL for reviving this thread after such a long time (I just completely forgot about this thread, so I couldn't ask this particular question at that time)\r\n\r\nI mean-- Is anyone on AoPS trying to solve any of those millennium problems? (I think one of them is resolved by Dr Perelman)", "Solution_16": "Yes, Grisha Perelman already solved Poincare, but by his own admission, he's completely withdrawn himself from mathematics. I have no idea if he has really stopped math altogether, but that's what he's claimed.\r\n\r\nIf someone on this site is working on one of them, they haven't done a lot of advertising." } { "Tag": [ "Support", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $ A, B\\in M_n(\\emph{C})$ be positive definite matrices, and $ p\\ge1$. Prove or disprove that \\[ Tr[(A\\plus{}B^{\\frac{1}{2}}AB^{\\frac{1}{2}})^p]\\le Tr[(A\\plus{}A^{\\frac{1}{2}}BA^{\\frac{1}{2}})^p].\\]", "Solution_1": "Can you do p=1?", "Solution_2": "[quote=\"transseries\"]Can you do p=1?[/quote]\r\nWhen $ p\\equal{}1$, the equality holds. It is not difficult to see the case when $ p\\equal{}2$.", "Solution_3": "I think the equality always holds. Since $ \\text{Tr}(AB)\\equal{}\\text{Tr}(BA)$, we expand $ (A\\plus{}A^{1/2}BA^{1/2})^p$, group the terms that contains $ i$ numbers of $ A$ and $ p\\minus{}i$ numbers of $ A^{1/2}BA^{1/2}$, take trace and change the order of multiplication, so to obtain $ \\text{Tr}[(A\\plus{}A^{1/2}BA^{1/2})^p]\\equal{}\\sum_i\\binom{p}{i}\\text{Tr}[A^i (A^{1/2}BA^{1/2})^{p\\minus{}i}]\\equal{}\\sum_i\\binom{p}{i}\\text{Tr}(A^p B^{p\\minus{}i})$. Similarly the right-hand side has the same expression.", "Solution_4": "[quote=\"sofeshue\"]I think the equality always holds. Since $ \\text{Tr}(AB) \\equal{} \\text{Tr}(BA)$, we expand $ (A \\plus{} A^{1/2}BA^{1/2})^p$, group the terms that contains $ i$ numbers of $ A$ and $ p \\minus{} i$ numbers of $ A^{1/2}BA^{1/2}$, take trace and change the order of multiplication, so to obtain $ \\text{Tr}[(A \\plus{} A^{1/2}BA^{1/2})^p] \\equal{} \\sum_i\\binom{p}{i}\\text{Tr}[A^i (A^{1/2}BA^{1/2})^{p \\minus{} i}] \\equal{} \\sum_i\\binom{p}{i}\\text{Tr}(A^p B^{p \\minus{} i})$. Similarly the right-hand side has the same expression.[/quote]\r\n\r\n$ \\text{Tr}[(A \\plus{} A^{1/2}BA^{1/2})^p] \\equal{} \\sum_i\\binom{p}{i}\\text{Tr}[A^i (A^{1/2}BA^{1/2})^{p \\minus{} i}]$ is not correct.", "Solution_5": "Numerical experiment supports this inequality. It is really difficult to give a proof." } { "Tag": [ "MATHCOUNTS", "FTW" ], "Problem": "Are there any events associated with pi day?\r\n\r\nDoes anyone actually \"celebrate\" it?", "Solution_1": "can u gimme any link where i can get information about thhis pi day. i am hearing about it for the very first time. :!:", "Solution_2": "Today, 3/14, is Pi Day because of the date. $ \\pi \\approx 3.14$. Of course, it won't really be Pi Day until 3/14/15 at 9:26 AM and 54 seconds. :P", "Solution_3": "hello, here is a link for you\r\nhttp://www.exploratorium.edu/pi/\r\nSonnhard.", "Solution_4": "[quote=\"isabella2296\"]Today, 3/14, is Pi Day because of the date. $ \\pi \\approx 3.14$. Of course, it won't really be Pi Day until 3/14/15 at 9:26 AM and 54 seconds. :P[/quote]\r\n\r\ncouldn't it also be at 9:26 PM? :)\r\naw dang\r\nit's only 7:27 *sits and waits*", "Solution_5": "9:26PM=21:26 in military time\r\n\r\n\r\nwe had pie today", "Solution_6": "I had MATHCOUNTS on Pi day, and all the coaches wore t-shirts that said stuff like \"Happy Pi Day, 3/14/09\" on them.", "Solution_7": "[quote=\"zamboni34\"]I had MATHCOUNTS on Pi day, and all the coaches wore t-shirts that said stuff like \"Happy Pi Day, 3/14/09\" on them.[/quote]\r\n\r\nHello zamboni, and welcome to AoPS. I see you are new. \r\n\r\nIn the future, I'd recommend trying to refrain from \"reviving\" old topics. It's commonly regarded on AoPS as \"spam\".", "Solution_8": "well, you just do alot of math, go on FTW, and eat pie!!! :lol: :icecream: \r\n (sorry it's icecream couldn't find pi)", "Solution_9": "Well, IN State Mathcounts was hosted on Pi Day(didn't give me any extra luck for my 6th grade year at doing well. I got 63rd. :() But according to my sister, at our local high school, some people give teachers pies. That's about it though." } { "Tag": [ "geometry", "rectangle", "ratio", "invariant", "geometry solved" ], "Problem": "We have a rectangle ABCD, and M and N are the midpoints of AD and BC (respectively!!!). Extend CD through D to any old point P. If Q is the intersection of PM and AC, prove that NM bisects angle QNP.", "Solution_1": "[b]EDIT:[/b] I have moved your two other problems into separate topics:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=141467\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=141464\r\n\r\nNice problem...\r\n\r\n[i]Solution:[/i]\r\n\r\nLet the lines NQ and AB meet at T. By Menelaos' theorem, applied to triangle ACD with the collinear points Q, M and P, we have\r\n\r\n$\\frac{AQ}{QC}\\cdot \\frac{CP}{PD}\\cdot \\frac{DM}{MA}=-1$,\r\n\r\nor simply\r\n\r\n$\\frac{AQ}{QC}\\cdot \\frac{CP}{PD}=-1$,\r\n\r\nsince DM / MA = 1. On the other hand, by Menelaos' theorem, applied to triangle ABC with the collinear points Q, N and T, we have\r\n\r\n$\\frac{AQ}{QC}\\cdot \\frac{CN}{NB}\\cdot \\frac{BT}{TA}=-1$,\r\n\r\nso that\r\n\r\n$\\frac{AQ}{QC}\\cdot \\frac{BT}{TA}=-1$,\r\n\r\nbecause CN / NB = 1. Comparing this with\r\n\r\n$\\frac{AQ}{QC}\\cdot \\frac{CP}{PD}=-1$,\r\n\r\nwe imply that\r\n\r\n$\\frac{BT}{TA}=\\frac{CP}{PD}$.\r\n\r\nIn other words, the points T and P divide the segments BA and CD in the same ratio. Now, since these two segments BA and CD are equal in length, we can conclude BT = CP. Together with BN = CN and < TBN = < PCN = 90\ufffd, this leads to the conclusion that the triangles TBN and PCN are congruent, so that < BNT = < CNP. Now, the line NM is perpendicular to BC (remember that ABCD is a rectangle!); therefore, < TNM = 90\ufffd - < BNT and < PNM = 90\ufffd - < CNP, so that from < BNT = < CNP we obtain < TNM = < PNM, and the line NM bisects the angle QNP.\r\n\r\n Darij", "Solution_2": "Here's a cute approach to the problem:\r\n\r\nThe lines $CD,CM,CA,CB$ form a harmonic cross-ratio, so the same can be said about the lines $NP,NM,NQ,NB$. This means that the map $NP\\leftrightarrow NQ$ is an involution on the pencil of lines through $N$. Such an involution is defined by two pairs of corresponding lines, so take $P\\to \\infty,P=D$, and the property we want to prove holds in these two cases, so it always holds.", "Solution_3": "[quote=\"grobber\"]Here's a cute approach to the problem:\n\nThe lines $CD,CM,CA,CB$ form a harmonic cross-ratio, so the same can be said about the lines $NP,NM,NQ,NB$. This means that the map $NP\\leftrightarrow NQ$ is an involution on the pencil of lines through $N$. Such an involution is defined by two pairs of corresponding lines, so take $P\\to \\infty,P=D$, and the property we want to prove holds in these two cases, so it always holds.[/quote]\r\n\r\nhmm. I am very much afraid that I only understood \"Here's a cute approach to the problem:\".\r\n\r\n- what is a harmonic cross ratio? Kedlaya's packet says that for four concurrent lines, the cross ratio can be defined as the cross ratio for the four points another line intersects the four concurrent lines at, and it is harmonic if it has ratio -1, but I don't see how you find that the ratio is -1...we don't know where line PMQ intersects CB\r\n\r\n- I think I understand how you deduce the second part of the first sentence, i.e. that NP, NM, NQ, NB are harmonic cross ratio, because we can draw the same line PMQ and use the definition of cross ratio on that (it being the same as the cross ratio of the previous four lines CD, CM, CA and CB)\r\n\r\n- I don't understand the rest very much. Mathworld defines line involution as \"pairs of points on a line, the product of whose distances from a fixed point is a given constant\", which I think is the definition you are using, but I don't quite understand how that follows from NP, NM, NQ, NB being a harmonic cross-ratio (because that only relates the distances PQ, PB, etc., not the distances involving N ?). \r\n\r\nCould you explain your solution or recommend me to a resource that explains all this please ?", "Solution_4": "If you cut four lines with another line, then the cross-ratio of the four points on that line will be invariant, so you can define the cross-ratio of the four lines. The lines $CD,CM,CA,CB$ have a harmonic cross-ratio because if you cut these four lines with $AD$, you get $A,D,M$ and the point at infinity, and these four form a harmonic cross-ratio (this really is well-known). Now cut the four lines with the line $PQ$. The points of intersection are $P,M,Q$ and $CB\\wedge PQ\\in NB$, so the lines $NP,NM,NQ,NB$ also form a harmonic cross-ratio because they pass through the four points.\r\n\r\nAn involution is, more generally, a transformation between points on the same line (or, as in this example, between lines through the same point), which conserves the cross-ratio and which is, of course, an involution :) ($f(f(d))=d$). The map I defined is an involution. Try to figure out why an involution is uniquely determined by two pairs of lines which correspond to each other." } { "Tag": [ "geometry", "incenter", "circumcircle", "geometric transformation", "reflection", "geometry solved" ], "Problem": "give $ABC$triangle. and $AA'$, $BB'$ , $CC'$ intersectin point $o$.prove:\r\n$A''$, $B''$ and $C''$ intersect in some common point .", "Solution_1": "Do you mean: $AA''$, $BB''$ and $CC''$ intersect in some common point :?:", "Solution_2": "[quote=\"M4RI0\"]Do you mean: $AA''$, $BB''$ and $CC''$ intersect in some common point :?:[/quote]\r\n\r\nProbably. But I think that this concurrency is true, only when the segment lines $A'A'',$ $B'B'',$ $C'C'',$ are concurrent, where $B'',$ $C'',$ are the orthogonal projections of $O,$ on $A'C',$ $A'B',$ respectively $($ opposite than in [b][size=109]Extremal's[/size][/b] figure $).$\r\n\r\nThe segment lines $A'A'',$ $B'B'',$ $C'C''$ now, are concurrent only in some special cases of $O,$ with respect to the triangle $\\bigtriangleup A'B'C'.$\r\n\r\nAs I know this concurrency is true, when the point $O,$ is coincided with the orthocenter $H',$, or the incenter $I'$, or the circumcenter $O'$ of $\\bigtriangleup A'B'C'$ and also with the points $H'',$ $I'',$ as the isotomic conjugates of $H',$ $I'$ respectively, with respect to the triangle $\\bigtriangleup A'B'C'$ $($ easy to prove $).$\r\n\r\nI am interesting if someone knows other points in the plane of a reference triangle, whose their pedal triangle is also a cevian triangle and what is the locus of such that points.\r\n\r\nKostas Vittas.", "Solution_3": "[quote=\"M4RI0\"]Do you mean: $AA''$, $BB''$ and $CC''$ intersect in some common point :?:[/quote]\r\nYes, $AA''$, $BB''$ and $CC''$ intersect in some common point.\r\nto vittasko\r\nTrust me it's true for all point's in traiangle. I have solved it, My sulution is very big. therefore maybe someone can solve it more best way than my.\r\nWhen $o=H$ this is particular case of my fact :wink:", "Solution_4": "[quote=\"Extremal\"][quote=\"M4RI0\"]Do you mean: $AA''$, $BB''$ and $CC''$ intersect in some common point :?:[/quote]\nYes, $AA''$, $BB''$ and $CC''$ intersect in some common point.\nto vittasko\nTrust me it's true for all point's in traiangle. I have solved it, My sulution is very big. therefore maybe someone can solve it more best way than my.\nWhen $o=H$ this is particular case of my fact :wink:[/quote]\r\n\r\nSorry, but I doupt that this result is true generally. Please see the figure t=144160(a).\r\n\r\nHowever in the figure t=144160(b), we have the configuration of the particular case when the concurrency point of $AA',$ $BB',$ $CC',$ is the incenter $I'$ of the triangle $\\bigtriangleup A'B'C'$ and, as well known, the segment lines $A'A'',$ $B'B'',$ $C'C'',$ are concurrent at the [b][size=109]Gergone's[/size][/b] point of $\\bigtriangleup A'B'C'.$\r\n\r\nHence, based on the Lemma I mentioned at http://www.mathlinks.ro/Forum/viewtopic.php?t=131476 , we conclude that the segment lines $AA'',$ $BB'',$ $CC'',$ are concurrent at one point, so be it $P.$\r\n\r\nThe proof of the above Lemma I have in mind, is not so difficult and as I have been promised, I will post it here soon.\r\n\r\nKostas Vittas.", "Solution_5": "I agree with vittasko; in general, the three lines are not concurrent.\r\n\r\n[quote=\"Extremal\"][quote=\"M4RI0\"]Do you mean: $AA''$, $BB''$ and $CC''$ intersect in some common point :?:[/quote]\nYes, $AA''$, $BB''$ and $CC''$ intersect in some common point.\nto vittasko\nTrust me it's true for all point's in traiangle. I have solved it, My sulution is very big. [/quote]", "Solution_6": "I'll try post my sulution very soon :wink:", "Solution_7": "[quote=\"Extremal\"]I'll try post my sulution very soon :wink:[/quote]\r\n\r\nOK. I am looking forward to your approach.\r\n\r\nKostas vittas.", "Solution_8": "There is no post after 1 year :) \r\nI think the first problem is not true, but if we denote $ A''$ be the reflection of $ O$ through $ B'C'$ ( instead of projection, as initial problem), the new problem is true, and I think it's nice :)", "Solution_9": "Indeed, PDatK40SP's version is true. This is called the Begonia theorem, and it is due to Floor van Lamoen. \r\nFor a proof using inversion, see Darij Grinberg, Begonia points and coaxal circles, available at [url]http://de.geocities.com/darij_grinberg/[/url]." } { "Tag": [ "Alcumus", "probability", "Support" ], "Problem": "im planning on not buying the intro to counting and probability book and just doing alcumus to learn it. is this a good idea? i have volume 1, so i have some background on counting and probability, so, yea...", "Solution_1": "It really depends on how you learn. I think Alcumus does cover lots of the Intro to C&P book, but the book has a lot more details.", "Solution_2": "I recommend you buy the book, because it contains lots of details and strategies, and then focus on the problems in Alcumus. Alcumus has over 1000 challenging problems, and it is a very good practice.", "Solution_3": "Don't really on Alcumus to learn, but it is good practice. You should buy the book though.", "Solution_4": "I agree with remy1140 and AwesomeToad. The book actually [i]teaches[/i] you the topics. First get the book, then do Alcumus for extended practice (1000+ [more] problems). And whenever you can't do a problem, review the topic covered and look at the solution, re-solving it in great detail.", "Solution_5": "Personally, i have to admit, i am not bragging. I am a lot smarter than i was before i did alcumus. I also have the book. They are both excellent. Combined, i could be be the smartest peson in the world. :rotfl:", "Solution_6": "Lol, but remember that there's intermediate series and [i]beyond[/i] (such as olympiad). But I agree that AoPS has the best resources for math competitions.", "Solution_7": "Probability used to be my worst subject of math. I did Alcumus. Now it's my best. \r\n\r\nNo, I don't mean to sound like an advertisement. I'm not kidding.", "Solution_8": "Actually, me too. Once I took a counting/probability quiz in school and I got about 10/30 correctly, and my teacher was pretty frustrated. Over the summer I worked on Intro to CP and I learned things I would never have learned anywhere else. Combinatorics is so fun now (and that quiz I failed seems too easy now that I am almost finishing Intermediate CP.)", "Solution_9": "There's also a lot of stuff you can find in Alcumus you never learn in school. Personally, I have found Geometric Probability most fascinating and helpful, while at the same time elegant.", "Solution_10": "I use alcumus to help me with my advanced 9th grade math! It rocks. I really like how it tracks your progress and the program won't give you a problem out of your range.", "Solution_11": "Many of us are probably attempting to unearth the secrets in the special recipe for Alcumus score, maybe? I am (failing at it).", "Solution_12": "The Alcumus score is a number calculated by looking at relative performance of everyone, isn't it?", "Solution_13": "[quote=\"AdithyaGanesh\"]The Alcumus score is a number calculated by looking at relative performance of everyone, isn't it?[/quote]\r\n\r\nYes. Passing and mastery is not relative to others, but your overall Alcumus score is. So, you can boost your Alcumus score by getting members of your state legislature to use it!", "Solution_14": "Alcumus is a great learning tool, but you should at least try the book. Maybe a friend of yours has it?", "Solution_15": "Alcumus was intended as a supplement to the book, and I believe it makes a very good supplement. After you get the book and you do some Alcumus, it's really only a matter of practice left until mastery of probability.", "Solution_16": "I think Alcumus is very helpful, but only as a supplement for the book. It covers other problems and acts like a review test for those who have finished the Counting and Probability book.", "Solution_17": "It isn't really a test as in a sort of assessment thing, in my opinion; the main focus is for additional practice, extending ideas, with just the \"progress report\" as an additional bonus." } { "Tag": [ "search", "number theory", "relatively prime", "number theory proposed" ], "Problem": "Find all positive integers $ n$ such that $ \\phi(n)\\plus{}\\delta(n)\\equal{}2n$ , where $ \\delta(n)$ denotes sum of positive divisors of $ n$ and $ \\phi(n)$ number of positive integers less then $ n$ and relatively prime to $ n$.", "Solution_1": "This problem is already posted [hide=\"here\"][url]http://www.mathlinks.ro/viewtopic.php?search_id=14841832&t=276352&start=40[/url][/hide]" } { "Tag": [], "Problem": "The number $916238457$ is an example of a nine-digit number which conatins each of the digits $1$ to $9$ exactly once. It also has the property that the digits $1$ to $5$ occur in their natural order, while the digits $1$ to $6$ do not. How many such numbers are there?", "Solution_1": "[hide=\"duno\"]nvm, i'll ponder this [/hide]", "Solution_2": "No, because you haven't taken into account that the digits 1-5 must be in the right order (or that 1-6 aren't)", "Solution_3": "Well, we know that there are 9! total possibilities.\r\nThe six simply must go before the five for the number to be valid.\r\nOf course, the number is clearly less than 9!, but I don't know by how much.", "Solution_4": "[hide=\"a lot less than 9!\"]\nWe begin with 9 ways to place the 9, 8 ways to place the 8, and 7 ways to place the 7. Then there are only 5 ways to place the 6 (anywhere except the rightmost remaining spot) and the order of 1-5 is fixed. So the answer is just $9\\cdot 8\\cdot 7\\cdot 5=\\boxed{2520}$.\n[/hide]", "Solution_5": "[hide]There are ${9\\choose 6}$ ways to choose where the digits 1-6 go. From these, given the placement of 6, the placements of 1-5 are determined. 6 can go in any of the six places, except the last, so $\\times 5$. There are $3!$ ways to place the remaining 7-9.\n\n${9\\choose 6}\\times 5\\times 3!=2520$.[/hide]", "Solution_6": "[hide]First place the 9,8,7, no restrictions on that, so $9\\cdot8\\cdot7=504$. Then, we have 6 spots left. There are 5 places to put the 6, since we can't put it in the last spot, since it would imply that 1-6 are in order. Thus, there are 5 more places, then the 1, 2, 3, 4, 5 are forced, so the final answer is $504\\cdot5=2520$. [/hide]" } { "Tag": [ "geometry", "perimeter", "ratio", "Pythagorean Theorem" ], "Problem": "Each right triangle has a hypotenus 7 cm and the shortest side 3 cm. Determine the perimeter of the figure.\r\n\r\n[hide=\"view image\"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=545[/img][/hide]\r\n\r\na) (28 + 2\u221a10 - 3) cm\r\nb) (28 + 8\u221a10 - 3) cm\r\nc) (28 + 2\u221a10) cm\r\nd) (28 + 8\u221a10) cm\r\ne) (16 + 8\u221a10) cm", "Solution_1": "[quote=\"ch*yeuk\"]Each right triangle has a hypotenus 7 cm and the shortest side 3 cm. Determine the perimeter of the figure.\n\n[hide=\"view image\"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=545[/img][/hide]\n\na) (28 + 2\u221a10 - 3) cm\nb) (28 + 8\u221a10 - 3) cm\nc) (28 + 2\u221a10) cm\nd) (28 + 8\u221a10) cm\ne) (16 + 8\u221a10) cm[/quote]\n[hide]\nE. there are four 7s so it's 28. then there is the small portions of $\\sqrt{49-9}=\\sqrt{40}$ and the portion of $\\sqrt{40}$ is 3 so the length that is in the perimeter is $\\sqrt{40}-3$, and that is there four times so the perimeter of the whoel thing is $28+4(\\sqrt{40}-3)$ or $16+4\\sqrt{40}=\\boxed{16+8\\sqrt{10}}$[/hide]", "Solution_2": "[quote=\"ch1n353ch3s54a1l\"][quote=\"ch*yeuk\"]Each right triangle has a hypotenus 7 cm and the shortest side 3 cm. Determine the perimeter of the figure.\n\n[hide=\"view image\"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=545[/img][/hide]\n\na) (28 + 2\u221a10 - 3) cm\nb) (28 + 8\u221a10 - 3) cm\nc) (28 + 2\u221a10) cm\nd) (28 + 8\u221a10) cm\ne) (16 + 8\u221a10) cm[/quote]\n[hide]\nE. there are four 7s so it's 28. then there is the small portions of $\\sqrt{49-9}=\\sqrt{40}$ and the portion of $\\sqrt{40}$ is 3 so the length that is in the perimeter is $\\sqrt{40}-3$, and that is there four times so the perimeter of the whoel thing is $28+4(\\sqrt{40}-3)$ or $16+4\\sqrt{40}=\\boxed{16+8\\sqrt{10}}$[/hide][/quote]\r\n\r\nAhhh so you have to use Pythagorean Theorem first! I tried the 30 60 90 theory first.. ahh. Thanks!", "Solution_3": "the 30 60 90 ratios of a triangle only works if the other angles in that triangle are 30 and 60, but they are not. so ya, you gotta use pythagorean theorem......", "Solution_4": "[quote=\"The Master\"]the 30 60 90 ratios of a triangle only works if the other angles in that triangle are 30 and 60, but they are not. so ya, you gotta use pythagorean theorem......[/quote]\r\n\r\nYeah I noticed after trying to figure it out from that theory.. ^_^;;", "Solution_5": "Option E is the correct one." } { "Tag": [ "trigonometry" ], "Problem": "If $ \\theta$ is a constant such that $ 0 < \\theta < \\pi$ and $ x \\plus{} \\frac{1}{x} \\equal{} 2\\cos{\\theta}$. then for each positive integer $ n$, $ x^n \\plus{} \\frac{1}{x^n}$ equals\r\n\r\n$ \\textbf{(A)}\\ 2\\cos{\\theta}\\qquad \r\n\\textbf{(B)}\\ 2^n\\cos{\\theta}\\qquad \r\n\\textbf{(C)}\\ 2\\cos^n{\\theta}\\qquad \r\n\\textbf{(D)}\\ 2\\cos{n\\theta}\\qquad \r\n\\textbf{(E)}\\ 2^n\\cos^n{\\theta}$", "Solution_1": "[hide=\"Fairly Large Hint\"]Note that a solution is $ x \\equal{} \\text{cis } \\theta$.[/hide]", "Solution_2": "[hide]$ \\implies x^2\\minus{}2x\\cos(\\theta)\\plus{}1\\equal{}0 \\implies x \\equal{} \\cos\\theta\\plus{}i\\sin\\theta \\equal{} e^{i\\theta}$, or $ x^n \\equal{} e^{ni\\theta}$\n\nIf $ f(e^{i\\theta}) \\equal{} 2\\cos\\theta$, then $ f(e^{ni\\theta}) \\equal{} 2\\cos(n\\theta)$, D.[/hide]" } { "Tag": [ "geometry", "LaTeX" ], "Problem": "Farz koni $D$ noghtei dar mosalase haddatozavieye $ABC$ bashad agar \r\n$DA.DB.AB+DB.DC.BC+DC.DA.CA=AB.BC.CA$ basahd mahalle noghteye $D$ ra biabid.\r\n[color=red][Moderator edit : LaTeX Formula edited][/color]", "Solution_1": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?&t=36551]Inja[/url] ra bebinid", "Solution_2": "merc, kheili mamnoon :D" } { "Tag": [ "function", "calculus", "inequalities proposed", "inequalities" ], "Problem": "With $ a,b,c > 0$. Prove that \r\n 1. $ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{c \\plus{} a} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\sqrt {\\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}} \\geq 5/2$\r\n 2. $ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{c \\plus{} a} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\sqrt {\\frac {2(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2}} \\geq 3$", "Solution_1": "(1) is obviously trues by CS and Am-Gm\r\n(2) is only a case in my old problem: (see it here) :) \r\nhttp://www.mathlinks.ro/viewtopic.php?t=279762\r\nWill your way work for this general problem, 2424? :)", "Solution_2": "in (1) the best constant that can replace the exponent of $ \\frac {ab \\plus{} ac \\plus{} bc}{a^2 \\plus{} b^2 \\plus{} c^2}$ is $ k_0 \\equal{} 0.9568...$ :wink: , we can obtain it by noting that the expression in function of $ abc$ is increasing , with the same way we can obtain the best value in (2) :)", "Solution_3": "[quote=\"anas\"]in (1) the best constant that can replace the exponent of $ \\frac {ab \\plus{} ac \\plus{} bc}{a^2 \\plus{} b^2 \\plus{} c^2}$ is $ k_0 \\equal{} 0.9568...$ :wink: , we can obtain it by noting that the expression in function of $ abc$ is increasing , with the same way we can obtain the best value in (2) :)[/quote]\r\nOf course, if use uvw of pqr or ABC... we only need to check these ineqs with $ a \\equal{} b$ and $ c \\equal{} 0$.\r\nBut I think the case $ a \\equal{} b$ is not nice because we'll have to use calculus.\r\nExample, with the general problem:\r\nGiven $ a, b, c > 0, k \\ge 1, n > 0$. Prove that:\r\n$ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{c \\plus{} a} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} 2\\sqrt [n]{2}k.\\sqrt [n]{\\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}} \\ge 2(k \\plus{} 1)$\r\nWe also can solve it easily by Am-Gm and not to use any calculus :)\r\n@To2424: The trues problem is: \r\n$ \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b}\\plus{}2\\sqrt{\\frac{2(ab\\plus{}bc\\plus{}ca)}{a^{2}\\plus{}b^{2}\\plus{}c^{2}}}\\geq 4$\r\nAnd $ \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b}\\plus{}\\sqrt{\\frac{2(ab\\plus{}bc\\plus{}ca)}{a^{2}\\plus{}b^{2}\\plus{}c^{2}}}\\geq \\frac{3}{2}\\plus{}\\sqrt{2}$ \r\nIt's also my mistake when posted the general problem. I edited it :oops:" } { "Tag": [], "Problem": "In a paralelipiped filled with water up to $H=250cm$ we make two holes $50cm$ apart (they are one under the other). Compute the distance between the surface of the water and hole $1$ s.t. the water reaches the ground in the same place.", "Solution_1": "[quote=\"perfect_radio\"]In a paralelipiped filled with water up to $H=250cm$ we make two holes $50cm$ apart (they are one under the other). Compute the distance between the surface of the water and hole $1$ s.t. the water reaches the ground in the same place.[/quote]\r\n\r\nCan you explain the problem?", "Solution_2": "you have something which holds water ( paralelipiped form, or whatever). On the side we make two holes through which water comes out. (they are on the same line - $\\perp$ to the earth) We must find the placement of these holes in terms of height such that the water which comes out of them reaches the ground in the same place (not necesarilly in the same time)\r\n\r\nsorry if you don't understand. i'm not good at explaining things :(", "Solution_3": "Enjoy.\r\n\r\n[url=http://img421.imageshack.us/my.php?image=pradio9bd.jpg][img]http://img421.imageshack.us/img421/2677/pradio9bd.th.jpg[/img][/url]", "Solution_4": "Let us consider a vessel filled with a liquid to a height $H$ and a hole in it a a depth $h$ from the liquid surface. \r\nThen, by Bernoulli's Principle the speed of the liquid coming out from the orifice $v = \\sqrt{2gh}$. let the horizontal range of the liquid coming out from the orifice be $d$. Let the time taken by the liquid to reach the bottom be $t$.\r\nThe $\\frac 1{2}gt^2 = H-h$\r\n$i.e, t = \\sqrt{\\frac{2(H-h)}{g}}$\r\nSo, $d = v.t$\r\n$i.e, d = 2\\sqrt{h(H-h)}$\r\nFor d to be maximum or minimum \r\n$\\frac{d [h(H-h)]}{dh} = 0$ \r\n$i.e, h = \\frac H{2}$\r\nalso $\\frac{d^2 [h(H-h)]}{dh^2} = -2$\r\nSo, $d$ is maximum at $h = \\frac H{2}$\r\nLet us now find the value of $d$ for an orifice located a distance $k$ to the top and bottom of the centre of the portion filled with liquid i.e, $h = \\frac H{2}$.\r\n$d_{upper} = 2\\sqrt{[(\\frac H{2} - k)(H - (\\frac H{2} - k))]}$\r\n$i.e, d_{upper} = 2\\sqrt{[(\\frac H{2} - k)(\\frac H{2} + k)]}$\r\nand, \r\n$d_{lower} = 2\\sqrt{[(\\frac H{2} + k)(H - (\\frac H{2} + k))]}$\r\n$i.e, d_{lower} = 2\\sqrt{[(\\frac H{2} + k)(\\frac H{2} - k)]}$\r\nSo, $d_{upper} = d_{lower}$\r\n\r\nSo, you can remember it as a result.\r\ni.e, the horizontal range first increases to become maximum at$h = \\frac H{2}$ and then decreases. Also, the points having the same horizontal range are symmetrically located about $h = \\frac H{2}$.\r\n\r\nAccording to given problem the holes are separated by a distance of $50 cm$. Maximum horizontal range of the liquid is at a distance $\\frac{250}{2} cm = 125 cm$ from top as well as bottom. So, the position of the holes is $(1.25 - 0.25)m$ and $(1.25 + 0.25)m$ from top as well as bottom the liquid surface.\r\nThus, the first hole should should be at a distance of $1m$ from top of the liquid surface.", "Solution_5": "That's a very very useful result!!!\r\n\r\nHowever, in a more classic way:\r\n\r\n$v_1=\\sqrt{2gh} ; v_2=\\sqrt{2g(h+l)} ; t_1=\\sqrt{\\frac{2(H-h)}{g}} ; t_2=\\sqrt{\\frac{2(H-h-l)}{g}}$\r\n\r\nSolving the equation $v_1 t_1 = v_2 t_2$:\r\n\r\n$h=\\frac{H - l}{2}$ therefore $h = 1 m$." } { "Tag": [ "inequalities" ], "Problem": "Let $a,b,c$ be three positive reals such that $abc=2$.\r\nProve that \\[a^{3}+b^{3}+c^{3}\\geq a \\sqrt{b+c}+b \\sqrt{c+a}+c \\sqrt{a+b}.\\]", "Solution_1": "By Cauchy Chebychev and again AM-GM we have \r\n$a\\sqrt{b+c}+b\\sqrt{c+a}+c\\sqrt{a+b}\\leq\\sqrt{2(a^{2}+b^{2}+c^{2})(a+b+c)}\\leq\\sqrt{6(a^{3}+b^{3}+c^{3})}\\leq a^{3}+b^{3}+c^{3}$\r\n\r\n :wink:", "Solution_2": "nice solution , silouan !\r\n\r\ni have an other one , but almost based on the same ideas:\r\n\r\n\r\nWe have:\r\n\r\n$3(a^{2}+b^{2}+c^{2}) \\geq (a+b+c)^{2}$\r\n\r\nand \r\n\r\n$(a+b+c)(a^{3}+b^{3}+c^{3}) \\geq (a^{2}+b^{2}+c^{2})^{2}$\r\n\r\n\r\nCombining these two , we have :\r\n\r\n$a^{3}+b^{3}+c^{3}\\geq \\frac{(a^{2}+b^{2}+c^{2})(a+b+c)}{3}$\r\n\r\n\r\n<=> $a^{3}+b^{3}+c^{3}\\geq \\frac{(a^{2}+b^{2}+c^{2})(b+c+a+c+a+b)}{6}$\r\n\r\nbut from Cauchy:\r\n\r\n$(a^{2}+b^{2}+c^{2})((b+c)+(a+c)+(a+b)) \\geq (a \\sqrt{b+c}+b \\sqrt{c+a}+c \\sqrt{a+b})^{2}$\r\n\r\nSo:\r\n\r\n$a^{3}+b^{3}+c^{3}\\geq \\frac{(a \\sqrt{b+c}+b \\sqrt{c+a}+c \\sqrt{a+b})^{2}}{6}$(*)\r\n\r\nbut from AM-GM:\r\n\r\n$a \\sqrt{b+c}+b \\sqrt{c+a}+c \\sqrt{a+b}\\geq 6$ (we replace this ineq in (*)\r\n\r\nSo: $a^{3}+b^{3}+c^{3}\\geq a \\sqrt{b+c}+b \\sqrt{c+a}+c \\sqrt{a+b}$\r\n\r\n\r\nThe end ;)", "Solution_3": "This problem is from JBMO Shortlist 2002 and problem 9 in \"Old and new inequalities\"", "Solution_4": "[quote=\"pohoatza\"]\nbut from AM-GM:\n$a \\sqrt{b+c}+b \\sqrt{c+a}+c \\sqrt{a+b}\\geq 6$\n[/quote]\r\n\r\nIs there a shorter way to arrive at this than:\r\n\r\n$abc\\sqrt{b+c}\\sqrt{c+a}\\sqrt{a+b}= \\sqrt{a^{2}b^{2}c^{2}(b+c)(c+a)(a+b)}\\geq \\sqrt{8a^{3}b^{3}c^{3}}=8$. I feel like I must be missing something more obvious." } { "Tag": [], "Problem": "What day of the year will it be 1500 days from now, as of Sept.7 2005?", "Solution_1": "[hide=\"click text\"]October 16 2009?[/hide][/hide]", "Solution_2": "[quote=\"pianoforte\"]What day of the year will it be 1500 days from now, as of Sept.7 2005?[/quote]\r\n\r\n[hide]$\\frac{1500}{365}=4$ remainder $40$ But in those $4$ years, there will be $1$ leap year, $2008$.\n\nSo we subtract $1$ from the remainder, making it $39$\n\nSo it would be $39$ days after September $7$ in the year $2009$.\n\n39 days after it would be October $16$.\n\nSo it would be October $16$, $2009$[/hide]", "Solution_3": "Yep. That's what I got.", "Solution_4": "[hide]\n$\\frac{1500}{365}=4 r 40$\nThere will be one leap year also, 2008.\nSo, you have 39 days after September 7th:\nOctober 16th\n(2009)[/hide]", "Solution_5": "[hide]you take 1500 and divide it by 365. You find that its 4r40. You know that every 4 years theres a leap year and there has to be 1 in there so you subtract a day. So find whats 39 days after september 7th, and you get October 16th, then add in the 4 yeras you got to get October 16th, 2009.[/hide]", "Solution_6": "[hide]10/16/2009[/hide]", "Solution_7": "1500 days from now...hmm...let me check the calendar...\r\n\r\n[hide]Just kidding. What day of the week was September 7? It was Wednesday, okay.\n\n[hide]We divide 1500 days by 7 to find out how many weeks (and thus how many Wednesdays) passed in that 1500 days. Since 1500 doesn't divide 7 equally, we have a remainder. The remainder is the number of days we must count forward to get our answer.\n\n1500/7= 214 r2\n\nSo there are 215 Wednesdays in the 1500 days. We then count forward to[/hide] [hide]Friday.[/hide][/hide]" } { "Tag": [ "calculus", "integration", "function", "real analysis", "probability", "special factorizations", "real analysis unsolved" ], "Problem": "Hello,\r\n\r\nI'm trying to solve this integral:\r\n\r\n$ \\int^{\\infty}_{ \\minus{} \\infty} e^{ \\minus{} ax^2} e^{\\left( \\minus{} \\frac {(x \\minus{} \\mu)^2}{2\\sigma^2}\\right)}dx$\r\n\r\nWould it be fine to integrate by parts, with $ u(x) \\equal{} e^{ \\minus{} ax^2}$, $ du \\equal{} \\minus{} 2ax e^{ \\minus{} ax^2}dx$, \r\n$ dv \\equal{} e^{\\left( \\minus{} \\frac {(x \\minus{} \\mu)^2}{2\\sigma^2}\\right)}dx$, and $ v \\equal{} \\sqrt {2\\pi\\sigma^2}$?\r\n\r\nI tried to do that and I got \r\n\r\n$ \\int^{\\infty}_{ \\minus{} \\infty} e^{ \\minus{} ax^2} e^{\\left( \\minus{} \\frac {(x \\minus{} \\mu)^2}{2\\sigma^2}\\right)}dx \\equal{} 2a^{ \\minus{} 1}\\sqrt {2\\pi\\sigma^2}$\r\n\r\nWhich appears to be incorrect.", "Solution_1": "$ \\int_{-\\infty}^\\infty \\frac{1}{\\sqrt{2\\pi \\sigma^2}}e^{-ax^2}e^{\\frac{-(x-\\mu)^2}{2\\sigma^2}dx}$\r\n\r\n$ =\\int_{-\\infty}^\\infty \\frac{1}{\\sqrt{2\\pi \\sigma^2}}e^{-\\frac{3a\\sigma^2+1}{2\\sigma^2}x^2+\\frac{2\\mu}{2\\sigma^2} x-\\frac{\\mu^2}{2\\sigma^2}}dx$\r\n\r\n$ y=\\sqrt{\\frac{3a\\sigma^2+1}{2\\sigma^2}}x, dy=\\sqrt{\\frac{3a\\sigma^2+1}{2\\sigma^2}}dx$\r\n\r\n$ -y^2+\\frac{2\\mu}{\\sqrt{2\\sigma^2}\\sqrt{3a\\sigma^2+1}}y-\\frac{\\mu^2}{2\\sigma^2}$\r\n\r\n$ =-y^2+\\frac{2\\mu}{\\sqrt{2\\sigma^2}\\sqrt{3a\\sigma^2+1}}y-\\frac{\\mu^2}{2\\sigma^2(3a\\sigma^2+1)}+\\frac{\\mu^2}{2\\sigma^2}(\\frac{1}{3a\\sigma^2+1}-1)$\r\n\r\n$ =-(y-\\frac{2\\mu}{\\sqrt{2\\sigma^2}\\sqrt{3a\\sigma^2+1}})^2 - \\frac{3a\\mu^2}{2(3a\\sigma^2+1)}$\r\n\r\nso we get $ \\int_{-\\infty}^\\infty \\frac{1}{\\sqrt{2\\pi \\sigma^2}}e^{-(y-\\frac{2\\mu}{\\sqrt{2\\sigma^2}\\sqrt{3a\\sigma^2+1}})^2 - \\frac{3a\\mu^2}{2(3a\\sigma^2+1)}}\\frac{dy}{\\sqrt{\\frac{3a\\sigma^2+1}{2\\sigma^2}}}$\r\n\r\n$ =e^{- \\frac{3a\\mu^2}{2(3a\\sigma^2+1)}}\\frac{2\\sigma^2}{3a\\sigma^2+1}\\frac{1}{\\sqrt{2\\pi \\sigma^2}} \\int_{-\\infty}^\\infty e^{-(y-\\mu')^2}dy=\\frac{2\\sigma}{3a\\sigma^2+1}e^{- \\frac{3a\\mu^2}{2(3a\\sigma^2+1)}}$\r\n\r\nthat seems a bit ugly", "Solution_2": "[quote=\"kenn4000\"]$ \\int_{ - \\infty}^\\infty \\frac {1}{\\sqrt {2\\pi \\sigma^2}}e^{ - ax^2}e^{\\frac { - (x - \\mu)^2}{2\\sigma^2}dx}$\n\n$ = \\int_{ - \\infty}^\\infty \\frac {1}{\\sqrt {2\\pi \\sigma^2}}e^{ - \\frac {3a\\sigma^2 + 1}{2\\sigma^2}x^2 + \\frac {2\\mu}{2\\sigma^2} x - \\frac {\\mu^2}{2\\sigma^2}}dx$\n\n$ y = \\sqrt {\\frac {3a\\sigma^2 + 1}{2\\sigma^2}}x, dy = \\sqrt {\\frac {3a\\sigma^2 + 1}{2\\sigma^2}}dx$\n\n$ - y^2 + \\frac {2\\mu}{\\sqrt {2\\sigma^2}\\sqrt {3a\\sigma^2 + 1}}y - \\frac {\\mu^2}{2\\sigma^2}$\n\n$ = - y^2 + \\frac {2\\mu}{\\sqrt {2\\sigma^2}\\sqrt {3a\\sigma^2 + 1}}y - \\frac {\\mu^2}{2\\sigma^2(3a\\sigma^2 + 1)} + \\frac {\\mu^2}{2\\sigma^2}(\\frac {1}{3a\\sigma^2 + 1} - 1)$\n\n$ = - (y - \\frac {2\\mu}{\\sqrt {2\\sigma^2}\\sqrt {3a\\sigma^2 + 1}})^2 - \\frac {3a\\mu^2}{2(3a\\sigma^2 + 1)}$\n\nso we get $ \\int_{ - \\infty}^\\infty \\frac {1}{\\sqrt {2\\pi \\sigma^2}}e^{ - (y - \\frac {2\\mu}{\\sqrt {2\\sigma^2}\\sqrt {3a\\sigma^2 + 1}})^2 - \\frac {3a\\mu^2}{2(3a\\sigma^2 + 1)}}\\frac {dy}{\\sqrt {\\frac {3a\\sigma^2 + 1}{2\\sigma^2}}}$\n\n$ = e^{ - \\frac {3a\\mu^2}{2(3a\\sigma^2 + 1)}}\\frac {2\\sigma^2}{3a\\sigma^2 + 1}\\frac {1}{\\sqrt {2\\pi \\sigma^2}} \\int_{ - \\infty}^\\infty e^{ - (y - \\mu')^2}dy = \\frac {2\\sigma}{3a\\sigma^2 + 1}e^{ - \\frac {3a\\mu^2}{2(3a\\sigma^2 + 1)}}$\n\nthat seems a bit ugly[/quote]\r\n\r\nThank you for your help.\r\n\r\nHowever, are you sure this is correct? I ran a simulation script in R to test your equation and the results do not appear to match:\r\n\r\n[code]\nsigma = 1 \nmu = 0\na = 1\n\nX = rnorm(1000000, mean=mu, sd=sigma)\nY = exp(-a * X^2)\n\nmean(Y) #Returns 0.5771073\n(2*sigma /(3 * a * sigma^2 + 1)) * exp((-3 * a * mu^2)/(6 * a * sigma^2 + 2)) #Returns .5\n[/code]\n\nEDIT: According to [url]http://www.mombu.com/science/mathematics/t-expectation-value-of-exponential-function-3094939.html[/url], completion of the square can be used to solve the integral.\n\nEDIT2: I think I got it! By completing the square, I got \n$ E[e^{ - a X^2}] = \\frac {e^{\\frac { - \\mu^2}{2\\sigma^2}}e^{\\frac {\\mu^2}{(2a\\sigma^2 + 1) (2\\sigma^2)}}}{\\sqrt {2a\\sigma^2 + 1}}$\n\nAnd according to simulation script, this is correct,\n\n[code]\nsigma = 3\nmu = -2\na = 1\n\nX = rnorm(1000000, mean=mu, sd=sigma) \nY = exp(-a * X^2) \n\nmean(Y) \nexp(-mu^2 / (2 * sigma^2)) * exp(mu^2 / ((2*a*sigma^2 + 1)*(2 * sigma^2))) / sqrt(2 * a * sigma^2 + 1)\n[/code]", "Solution_3": "yeah i dont know why i had a three there, its obviously a two from the first time it appears. just sloppy.", "Solution_4": "That's a convolution. You can make the fact that it's a convolution clearer by writing it as \r\n\r\n$ \\int_{-\\infty}^{\\infty}e^{-ax^2}e^{-\\frac{(\\mu-x)^2}{2\\sigma^2}}\\,dx$\r\n\r\nAnd with that being a convolution (we'll get a function of $ \\mu$), there's a very strong incentive to use Fourier transform method.\r\n\r\nGiven the obvious probabilistic context, I'll call use the so-called \"characteristic function\" from probability theory: $ \\chi(t)=E(e^{itX}).$\r\n\r\nMultiply by the appropriate constants to make the problem\r\n\r\n$ \\frac1{\\sqrt{2\\pi}\\sigma}\\cdot\\frac{\\sqrt{a}}{\\sqrt{\\pi}}\\int_{-\\infty}^{\\infty}e^{-ax^2}e^{-\\frac{(\\mu-x)^2}{2\\sigma^2}}\\,dx$\r\n\r\nLet $ X$ be the random variable whose density is $ \\frac{\\sqrt{a}}{\\sqrt{\\pi}}e^{-ax^2}$ and let $ Y$ be the random variable whose density is $ \\frac1{\\sqrt{2\\pi}\\sigma}e^{-\\frac{y^2}{2\\sigma^2}}.$\r\n\r\n$ Y$ is a normal random variable with mean zero and standard deviation $ \\sigma.$ It's fairly well known that its characteristic function is $ \\chi(t)=e^{-\\frac{\\sigma^2t^2}2.}$ \r\n\r\nThe fundamental fact there is that the Fourier transform of a Gaussian is a Guassian - we just have to work out the constant.\r\n\r\nThen $ X$ is a normal random variable with mean zero and standard deviation $ \\frac1{\\sqrt{2a}}.$ That means that its characteristic function is $ \\chi(t)=e^{-\\frac{t^2}{4a}}.$\r\n\r\nNow, suppose $ X$ and $ Y$ are independent. What is the distribution of $ X+Y?$ One way to answer that is to find the convolution of the densities of $ X$ and of $ Y$ - and that's precisely the integral we've been asked to find. The other think that you can say is that the characteristic function of $ X+Y$ is the pointwise product of the characteristic functions of $ X$ and $ Y.$ (Again, that's fundamental Fourier analysis: convolution on one side corresponds to pointwise multiplication on the other side.)\r\n\r\nThat gives the characteristic function of $ X$ and $ Y$ as $ \\exp(-\\frac{\\sigma^2}2+\\frac1{4a})t^2).$\r\n\r\nThat's the characteristic function of another normal distribution. In other words, the sum of independent normal distributions is a normal distribution. The variance of the sum is the sum of the variances; in this case $ \\sigma^2+\\frac1{2a}.$\r\n\r\nFrom that I get the following:\r\n\r\n$ \\frac1{\\sqrt{2\\pi}\\sigma}\\cdot\\frac{\\sqrt{a}}{\\sqrt{\\pi}}\\int_{-\\infty}^{\\infty}e^{-ax^2}e^{-\\frac{(\\mu-x)^2}{2\\sigma^2}}\\,dx=\\frac1{\\sqrt{2\\pi}\\sqrt{\\sigma^2+\\frac1{2a}}}\\exp(-\\frac{\\mu^2}{2(\\sigma^2+\\frac1{2a})}).$\r\n\r\nUndoing the multiplication by constant, I have that\r\n\r\n$ \\int_{-\\infty}^{\\infty}e^{-ax^2}e^{-\\frac{(\\mu-x)^2}{2\\sigma^2}}\\,dx=\\frac{\\sqrt{2\\pi}\\sigma}{\\sqrt{2a\\sigma^2+1}}\\exp(-\\frac{\\mu^2}{2(\\sigma^2+\\frac1{2a})}).$", "Solution_5": "[quote=\"Kent Merryfield\"]That's a convolution. You can make the fact that it's a convolution clearer by writing it as \n\n$ \\int_{ - \\infty}^{\\infty}e^{ - ax^2}e^{ - \\frac {(\\mu - x)^2}{2\\sigma^2}}\\,dx$\n\nAnd with that being a convolution (we'll get a function of $ \\mu$), there's a very strong incentive to use Fourier transform method.\n\nGiven the obvious probabilistic context, I'll call use the so-called \"characteristic function\" from probability theory: $ \\chi(t) = E(e^{itX}).$\n\nMultiply by the appropriate constants to make the problem\n\n$ \\frac1{\\sqrt {2\\pi}\\sigma}\\cdot\\frac {\\sqrt {a}}{\\sqrt {\\pi}}\\int_{ - \\infty}^{\\infty}e^{ - ax^2}e^{ - \\frac {(\\mu - x)^2}{2\\sigma^2}}\\,dx$\n\nLet $ X$ be the random variable whose density is $ \\frac {\\sqrt {a}}{\\sqrt {\\pi}}e^{ - ax^2}$ and let $ Y$ be the random variable whose density is $ \\frac1{\\sqrt {2\\pi}\\sigma}e^{ - \\frac {y^2}{2\\sigma^2}}.$\n\n$ Y$ is a normal random variable with mean zero and standard deviation $ \\sigma.$ It's fairly well known that its characteristic function is $ \\chi(t) = e^{ - \\frac {\\sigma^2t^2}2.}$ \n\nThe fundamental fact there is that the Fourier transform of a Gaussian is a Guassian - we just have to work out the constant.\n\nThen $ X$ is a normal random variable with mean zero and standard deviation $ \\frac1{\\sqrt {2a}}.$ That means that its characteristic function is $ \\chi(t) = e^{ - \\frac {t^2}{4a}}.$\n\nNow, suppose $ X$ and $ Y$ are independent. What is the distribution of $ X + Y?$ One way to answer that is to find the convolution of the densities of $ X$ and of $ Y$ - and that's precisely the integral we've been asked to find. The other think that you can say is that the characteristic function of $ X + Y$ is the pointwise product of the characteristic functions of $ X$ and $ Y.$ (Again, that's fundamental Fourier analysis: convolution on one side corresponds to pointwise multiplication on the other side.)\n\nThat gives the characteristic function of $ X$ and $ Y$ as $ \\exp( - \\frac {\\sigma^2}2 + \\frac1{4a})t^2).$\n\nThat's the characteristic function of another normal distribution. In other words, the sum of independent normal distributions is a normal distribution. The variance of the sum is the sum of the variances; in this case $ \\sigma^2 + \\frac1{2a}.$\n\nFrom that I get the following:\n\n$ \\frac1{\\sqrt {2\\pi}\\sigma}\\cdot\\frac {\\sqrt {a}}{\\sqrt {\\pi}}\\int_{ - \\infty}^{\\infty}e^{ - ax^2}e^{ - \\frac {(\\mu - x)^2}{2\\sigma^2}}\\,dx = \\frac1{\\sqrt {2\\pi}\\sqrt {\\sigma^2 + \\frac1{2a}}}\\exp( - \\frac {\\mu^2}{2(\\sigma^2 + \\frac1{2a})}).$\n\nUndoing the multiplication by constant, I have that\n\n$ \\int_{ - \\infty}^{\\infty}e^{ - ax^2}e^{ - \\frac {(\\mu - x)^2}{2\\sigma^2}}\\,dx = \\frac {\\sqrt {2\\pi}\\sigma}{\\sqrt {2a\\sigma^2 + 1}}\\exp( - \\frac {\\mu^2}{2(\\sigma^2 + \\frac1{2a})}).$[/quote]\r\n\r\nGreat way to solve it using Fourrier transform! Thank you!\r\n\r\nP.S. the answer should be $ \\frac {1}{\\sqrt {2a\\sigma^2 + 1}}\\exp( - \\frac {\\mu^2}{2(\\sigma^2 + \\frac1{2a})}).$" } { "Tag": [ "vector", "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $T$ be an endomorfisma of vector space $V$, prove that:\r\n$T$ is self adjoint iff all of the eigenvalues of $T$ is positive.", "Solution_1": "I thinks its false..\r\nunless we add condition of $T$ must be positive definite..", "Solution_2": "Assume that $T$ is positive, since $T$ is self-adjoint the eigenvalues are all real numbers, and for any vectorspace $v_{i}$ we have $\\langle{T(v_{i}),v_{i}}\\rangle >0$ and so $\\langle{\\lambda_{i}v_{i},v_{i}}\\rangle >0 \\Longrightarrow \\lambda_{i}\\langle{v_{i},v_{i}}\\rangle >0$ this conclude $\\lambda_{i}> 0$.\r\n\r\nFor the converse, i thinks we can use spectral theorem, but maybe there are another elementary proof.", "Solution_3": "The matrix $\\begin{pmatrix}-1 &0 \\\\ 0 &-1\\end{pmatrix}$ is self-adjoint and has negative eigenvalues.", "Solution_4": "I thinks he forgot the condition that $T$ must be positive (see my first post in this thread)", "Solution_5": "The matrix $\\begin{bmatrix}1&3\\\\0&2\\end{bmatrix}$ has positive eigenvalues and is most assuredly not self-adjoint.\r\n\r\nThe problem as stated is very strange - so false as to be irrelevant. Being self-adjoint and having positive eigenvalues have very little relationship to each other. \r\n\r\n\"Positive eigenvalues\" can be defined with no mention of either a basis or an inner product; to define \"self-adjoint\" requires either an inner product or an agreed-upon standard basis.\r\n\r\nMy suspicion is that bolpoin has misquoted his source in some way.", "Solution_6": "The problem say:\r\n\r\nLet $T$ be a self adjoint endomorfisma, prove that:\r\n\r\n$T$ is positive iff all of eigenvalues of $T$ is positive.\r\n\r\njust like Beginner said.", "Solution_7": "OK. That's straightforward in finite dimensional spaces - just use the orthogonal (or unitary) diagonalization result. In an infinite-dimensional space, it's a little trickier, and may require us to talk about the spectrum rather than the eigenvalues.", "Solution_8": "So we really need spectral theorem here.\r\n\r\nLet $y \\in V$ then by spectral theorem since T is self- adjoint \r\n$V$ is the direct sum of eigenspace of $T$, say $K(\\lambda_{i})$, for different eigenbalues $\\lambda_{i}$.\r\n\r\nthen $y=\\sum v_{i}$ with $v_{i}\\in K(\\lambda_{i})$ so\r\n\\[\\langle{ T(y),y\\rangle}= \\sum_{i}\\sum_{j}\\lambda_{i}\\langle{ v_{i}, v_{j}\\rangle}\\]\r\nsince $\\langle{v_{i},v_{j}\\rangle}=0$ for $i \\not= j$, and $\\lambda_{i}\\langle{v_{i},v_{i}\\rangle}> 0$ the conclusion follow." } { "Tag": [ "floor function", "inequalities" ], "Problem": "Let $ S$ be a set of $ n$ elements. Prove that if $ n$ is even, the only clutter of size $ \\binom{n}{\\lfloor\\frac{n}{2}\\rfloor}$ is the clutter of all $ \\frac{n}{2}$-combinations.", "Solution_1": "I don't know if everyone here knows what a \"clutter\" is - the more standard terminology as far as I know is \"antichain\", specifically an antichain of subsets of $ S$: this is a family of subsets none of which contains another. \r\n[hide=\"Hint\"]\nThis follows directly from one of the standard proofs of [url=http://en.wikipedia.org/wiki/Sperner%27s_theorem]Sperner's Theorem[/url], namely the one proving the stronger [url=http://en.wikipedia.org/wiki/Lubell-Yamamoto-Meshalkin_inequality]LYM inequality[/url]. \n[/hide]\r\nIt is also true that when $ n\\equal{}2m\\plus{}1$ is odd, the largest clutters are the one formed by all the $ m$-sets and the one formed by all the $ m\\plus{}1$-sets. This requires a slightly more detailed analysis." } { "Tag": [ "ratio", "AMC" ], "Problem": "In the figure, triangle ABC has angle A=45 and angle B=30. A line DE, with D on AB and angle ADE=60, divides triangle ABC into two pieces of equal area. (Note: the figure may not be accurate; perhaps E is on CB instead of AC.) The ratio AD/AB is:\r\n\r\n(A) $ \\frac {1}{\\sqrt{2}}$\r\n(B) $ \\frac {2}{2\\plus{}\\sqrt{2}}$\r\n(C) $ \\frac {1}{\\sqrt{3}}$\r\n(D) 1/cube root of 6\r\n(E) 1/ fourth root of 12\r\n\r\n[url]http://www.math.ksu.edu/main/events/hscomp/worksheets/amc12/worksheet11/Worksheet11.pdf[/url]\r\n\r\nhere is url, go to page 5 for diagram\r\n\r\ncan someone pm me on how to do a diagram WITHOUT geogebra on here?\r\nI am having major problems with it.\r\n\r\n\r\nFor this problem, if someone can give some hints, that would be great.\r\n\r\nThanks in advance!", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225961", "Solution_2": "oh, i usually dont check that forum, thanks for the link", "Solution_3": "can people post some hints?\r\ni really dont want to look at the solutions on that link, but im at my wit's end", "Solution_4": "drop altitudes", "Solution_5": "are there any other ways to do this?" } { "Tag": [ "geometry" ], "Problem": "Equilateral triangles are constructed on each side of a right triangle. The areas of the triangles constructed on the legs of the right triangle are 36 $ \\text{cm}^{2}$ and 64 $ \\text{cm}^{2}$. What is the number of square centimeters in the area of the triangle constructed on the hypotenuse?", "Solution_1": "[quote=\"GameBot\"]Equilateral triangles are constructed on each side of a right triangle. The areas of the triangles constructed on the legs of the right triangle are 36 $ \\text{cm}^{2}$ and 64 $ \\text{cm}^{2}$. What is the number of square centimeters in the area of the triangle constructed on the hypotenuse?[/quote]\r\n\r\n$ 36\\plus{}64\\equal{}100$.", "Solution_2": "[geogebra]9e88dcb65ce53a4df07faf38134828383ed34e98[/geogebra]", "Solution_3": "Its equilateral triangles, not squares." } { "Tag": [], "Problem": "I was wondering, does anyone know if the criteria for being selected for the US physics Team varies depending on grade. I know for MOP, this is true since Red and Blue MOP are much easier to get into than Black MOP. Judging by the distribution of the past participants of te US Physics Team, I'm inclined to say that the team is chosen solely based on your score on the semifinal round, but I just wondered if anyone knew for sure.", "Solution_1": "[quote=\"psychicalchemist\"]I was wondering, does anyone know if the criteria for being selected for the US physics Team varies depending on grade. I know for MOP, this is true since Red and Blue MOP are much easier to get into than Black MOP. Judging by the distribution of the past participants of te US Physics Team, I'm inclined to say that the team is chosen solely based on your score on the semifinal round, but I just wondered if anyone knew for sure.[/quote]\r\nI have been wondering about the same thing. Where did you find the statistics/distribution of the past participants? Unlike AMC, there are hardly any info about the participants of the physics teams and training camp, although there are some info on the traveling team. Does anyone know how many students take part in the first round of the test each year on average? In other words, how many students are the 200 semifinalists selected from?", "Solution_2": "yeah, that was my question too!\r\n\r\nalso one more thing, what is the average score needed for being a semi-finalist?" } { "Tag": [ "function", "limit", "inequalities", "logarithms", "integration", "real analysis", "real analysis unsolved" ], "Problem": "\u0001\u0084\u009c\u0082Find all continuous functions $ f: \\mathbb{R}\\to [1,\\infty)$ for wich there exists $ a\\in\\mathbb{R}$ and a positive integer $ k$ such that \\[ f(x)f(2x)\\cdot...\\cdot f(nx)\\leq an^k\\] \r\nfor all real $ x$ and all positive integers $ n$.\r\n\r\n[i]author :Radu Gologan[/i]", "Solution_1": "First of all easy to check that $ \\lim_{n\\to\\infty }f(x)=1$\r\nSuppose exist $ a$ such that $ f(a)>1$\r\nThen exist an interval $ [a,b)$ such that $ f(x)>1,\\forall x\\in [a,b]$\r\n$ M=inf f(x),x\\in [a,b]$\r\nFor each $ x_0$ there are at least for every $ k\\in [\\frac{a}{x+0}]-1,\\frac{b}{x_0}]$ then $ kx_0\\in [a,b]$\r\nSo $ f(x)....f(nx)\\geq M^{n-k}$\r\nBut $ \\lim_{n\\to\\+\\infty}\\frac{a^n}{x^m}=+\\infty$ so the inequalites doesn't hold for all n , \r\nIt mean that $ f(x)\\equiv 1$", "Solution_2": "A slightly different take:\r\n\r\nTake logarithms: $ \\sum_{j \\equal{} 1}^n\\ln (f(jx))\\le \\ln a \\plus{} k\\ln n$. Dividing by $ n$, $ \\frac1n\\sum_{j \\equal{} 1}^n\\ln (f(jx))\\le\\frac {\\ln a}{n} \\plus{} k\\frac {\\ln n}{n}$. Set $ x \\equal{} \\frac {B}{n}$, and let $ n\\to\\infty$; we get $ \\int_0^B \\ln(f(t))\\,dt\\le \\lim_{n\\to\\infty}\\frac {\\ln a}{n} \\plus{} k\\frac {\\ln n}{n} \\equal{} 0$.\r\nSince $ \\ln f(t)$ is nonnegative and continuous, it must be identically zero on $ [0,B]$. $ B$ was arbitrary, and $ f$ is identically $ 1$." } { "Tag": [ "geometry", "inradius", "circumcircle", "inequalities proposed", "inequalities" ], "Problem": "Prove that\r\n\r\n$ \\frac{l_ah_a\\minus{}1}{l_a^2}\\plus{}\\frac{l_bh_b\\minus{}1}{l_b^2}\\plus{}\\frac{l_ch_c\\minus{}1}{l_c^2}\\geq\\frac{12r^3\\plus{}3r^2R\\minus{}R}{3Rr^2}$", "Solution_1": "[quote=\"Ligouras\"]Prove that\n\n$ \\frac {l_ah_a \\minus{} 1}{l_a^2} \\plus{} \\frac {l_bh_b \\minus{} 1}{l_b^2} \\plus{} \\frac {l_ch_c \\minus{} 1}{l_c^2}\\geq\\frac {12r^3 \\plus{} 3r^2R \\minus{} R}{3Rr^2}$[/quote]\r\n\r\nthat's not a geometric inequality...\r\n\r\nand, what are the $ l_a,l_b,l_c$ mean?", "Solution_2": "[quote=\"kuing\"][quote=\"Ligouras\"]Prove that\n\n$ \\frac {l_ah_a \\minus{} 1}{l_a^2} \\plus{} \\frac {l_bh_b \\minus{} 1}{l_b^2} \\plus{} \\frac {l_ch_c \\minus{} 1}{l_c^2}\\geq\\frac {12r^3 \\plus{} 3r^2R \\minus{} R}{3Rr^2}$[/quote]\n\nthat's not a geometric inequality...\n\nand, what are the $ l_a,l_b,l_c$ mean?[/quote]\r\n\r\nMy Friend\r\n\r\n: Let $ h_a, h_b, h_c$ be the heights, $ l_a, l_b, l_c$ the bisectors, $ r$ the inradius, and $ R$ the circumradius of a triangle $ ABC$." } { "Tag": [ "geometry", "geometric transformation", "rotation", "analytic geometry", "linear algebra", "matrix", "trigonometry" ], "Problem": "I'm not always familiar with rotation of both points and line.\r\n\r\nLet's say we have a point $ (1,2)$ in the line $ y \\equal{} x\\plus{}1$. Here are some things I want to do:\r\n\r\n1) Rotate point $ (1,2)$ by 60 degree\r\n2) Rotate point $ (i,2)$ using complex set? If so, how would this work for rotating by 60 degrees? $ n$ degrees?\r\n3) Rotate line $ y \\equal{} x\\plus{}1$ by origin.\r\n4) Rotate line $ y \\equal{} x\\plus{}1$ by x-axis/y-axis.\r\n5) Rotate line $ y \\equal{} x\\plus{}1$ by $ n$ degrees.\r\n\r\nThanks a lot! I could never really get the concept of rotation (like how to do it..).", "Solution_1": "Two common ways are to use [url=http://mathworld.wolfram.com/RotationMatrix.html]rotation matrices[/url] or [url=http://mathworld.wolfram.com/PolarCoordinates.html]polar coordinates[/url].", "Solution_2": "[b]The Rotation Matrix:[/b] The rotation of a point $ \\binom{x}{y}$ through an angle of $ \\theta$ about the origin in $ \\mathbb{R}^2$ is given by:\r\n\r\n$ \\left( \\begin{array}{cc} \\cos \\theta & - \\sin \\theta & \\sin \\theta & \\cos \\theta \\end{array} \\right) \\binom{x}{y}$\r\n\r\nwhere the result is computed through standard matrix multiplication and the point $ (x,y)$ and it's image are expressed as column vectors.\r\n\r\n\r\n[b]Basic Usage:[/b] Suppose we want to rotate $ (1,2)$ by 60 degrees. Just plug in:\r\n$ \\left( \\begin{array}{cc} 1/2 & - \\sqrt {3}/2 & \\sqrt {3}/2 & 1/2 \\end{array} \\right) \\binom{1}{2} = \\binom{\\frac {1}{2} - \\sqrt {3}}{1 + \\frac {\\sqrt {3}}{2}}$\r\n\r\nSo, the resulting point is $ \\left(\\frac {1}{2} - \\sqrt {3} , \\; 1 + \\frac {\\sqrt {3}}{2}\\right)$\r\n\r\n\r\n[b]Rotating a Parametrizable Curve:[/b] Suppose we want to rotate the whole line $ y = x + 1$ by some angle -- let's do 30 degrees. The first step is to parametrize this in the minimum possible number of variables. For example, we can parametrize this as $ (x,y) = (t,t+1)$\r\n\r\nNext, we apply the rotation matrix to the parametrized form: \r\n$ \\left( \\begin{array}{cc} \\sqrt {3}/2 & - 1/2 & 1/2 & \\sqrt {3}/2 \\end{array} \\right) \\binom{t}{t + 1} = \\left( \\begin{array}{c} \\left( \\frac {\\sqrt {3} - 1}{2}\\right) t - \\frac {1}{2} & \\left( \\frac {\\sqrt {3} + 1}{2}\\right) t + \\frac {\\sqrt {3}}{2} \\end{array} \\right)$\r\n\r\nA parametrization of the points on the rotated line is thus:\r\n$ x = \\left( \\frac {\\sqrt {3} - 1}{2}\\right) t - \\frac {1}{2}$\r\n$ y = \\left( \\frac {\\sqrt {3} + 1}{2}\\right) t + \\frac {\\sqrt {3}}{2}$\r\n\r\nFinally, if you want an explicit solution (as in this case), eliminate the parameter from the system through algebraic manipulation. In this case, we get $ y = (2 + \\sqrt {3} ) x + (1 + \\sqrt {3})$\r\n\r\n\r\n[b]Alternate Solution:[/b] While the above is more representative of how to rotate a function in general, it is not the ideal method for rotating a line. We don't even need a rotation matrix.\r\n\r\n[b]Background/Preparation:[/b] Recall that a non-vertical line in $ \\mathbb{R}^2$ is uniquely defined by two parameters, which I will define in the most convenient way:\r\n1. Slope: The appropriately-signed tangent of the angle the line makes with the positive x-axis\r\n2. y-intercept: The unique value of $ r$ such that the polar point $ (r , 90^{\\circ})$ is on the line.\r\n\r\n[b]Procedure:[/b] The slope of $ y = x + 1$ is $ 1$, thus the tangent of the angle it makes with the x-axis is $ 1$. The slope of the new line will be the tangent of the angle 30 degrees more than the original angle it makes with the x-axis. Thus, by the tangent-addition formula, the slope of the new line is:\r\n\r\n$ \\frac {1 + \\tan 30^{\\circ}}{1 - 1 \\cdot \\tan 30^{\\circ}} = \\frac {1 + \\frac {1}{\\sqrt {3}}}{1 - \\frac {1}{\\sqrt {3}}} = 2 + \\sqrt {3}$\r\n\r\nRecall the definition of the y-intercept. The y-intercept of the new point takes the form $ (r, 90^{\\circ})$, so its pre-image takes the form $ (r, 60^{\\circ})$, where $ r$ is the same before and after the rotation. Since the pre-image has $ \\theta = 60^{\\circ}$, $ y = \\sqrt {3} \\cdot x$.\r\n\r\nWe solve $ \\sqrt {3} x = x + 1$ to find the pre-image of the y-intercept point. We get $ x = \\frac {1 + \\sqrt {3}}{2} \\implies y = \\frac {3 + \\sqrt {3}}{2}$. By the Pythagorean theorem, the $ r$ value of this point is $ 1 + \\sqrt {3}$.\r\n\r\nThus, the new line is $ y = (2 + \\sqrt {3} ) x + (1 + \\sqrt {3})$", "Solution_3": "Thanks, TZF! That is exactly what I needed... It's been really long time since I saw matrice so thanks for refreshing it! :) :)" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "$f(x_{1}+x_{2})=f(x_{1})+f(x_{2})$ for evry real $x_{1}$ and $x_{2}$\r\nf:R->R\r\nf(x) is continuous. Prove that f(x)=ax.\r\nPlease excuse my bad english, and try to explain me this problem. Thank you..", "Solution_1": "From the initial equation derive\r\n\r\n$f(\\underbrace{x+x+\\dots+x}_{m})=\\underbrace{f(x)+f(x)+\\dots+f(x)}_{m}\\iff f(mx)=mf(x)\\qquad(*)$\r\n\r\nThen putting $x=1$ we get $f(m)=mf(1)=am$ for integer $m$, where $a=f(1)$\r\n\r\nNow put $x={k\\over m}, k,m\\in\\mathbb{Z}$ into $(*)$ to obtain\r\n\r\n$f(k)=mf({k\\over m})\\iff f({k\\over m})={f(k)\\over m}=a{k\\over m}$\r\n\r\nsince for integer $k$ we have $f(k)=ak$.\r\n\r\nTherefore we've proven $f(x)=ax$ for all rational $x$.\r\n\r\nNow use the continuity of the function and Dedekind-cuts representation of irrationals to expand that result to any real $x$, and the proof is complete." } { "Tag": [], "Problem": "[url]http://www.msnbc.msn.com/id/26884523?GT1=43001[/url]\r\n\r\nPlease discuss the house voting no on the largest government intervention since the depression.\r\n\r\nPersonally, I thought the package would easily pass...but I think part of the reason it didn't was the idea that people in congress were afraid of having something like this on their record, in case the plan fails tremendously. Again, I really think this is incredible - this was not supposed to happen at all.", "Solution_1": "I think this is another example of the struggles between Republicans and Democrats. Republicans even said that it was Pelosi's speech about bipartisan work that \"killed\" the bill.\r\nWhat is Congress thinking? In this economy in the brink of another Depression, they refuse to pass this bill?\r\nI think this is utter madness and that Congress is forcing the American economy to \"commit suicide.\"\r\nLast time I checked, the Dow was down 600 points.", "Solution_2": "[quote]Despite pressure from supporters, not enough members were willing to take the political risk just five weeks before an election.[/quote]\r\nIt's sad that the representatives care more about their own re-election than the American economy.", "Solution_3": "The vote here was less partisan than many; there were a lot of Democrats voting against it, and a lot of Republicans voting for it. The strongest indicator for voting against it was having a competitive reelection race, with nearly all such representatives voting no.\r\n\r\nThe public outrage is very real; the people aren't convinced that the plan is actually good for the economy. The stock and bond markets are not \"the economy\".\r\n\r\nWhat went wrong? This plan started with an outrageously bad proposal from the Bush administration, which was then modified by the congressional Democrats. It would probably be a better idea to throw that out and start from scratch with a proposal from independent economists. That would include such radical ideas as reinstating the regulations that were dropped ten years ago, and allowed the mess.", "Solution_4": "Also, noting that fact that the House has a Democrat majority, one realizes that if the economy collapses, it would be blamed on Bush, severely hampering McCain.", "Solution_5": "that's not a very useful note. compare the percentage of democratic representatives who voted yes to that of the republicans. analyzing this on a strictly partisan level doesn't seem likely to help.", "Solution_6": "[quote=\"mewto55555\"]Also, noting that fact that the House has a Democrat majority, one realizes that if the economy collapses, it would be blamed on Bush, severely hampering McCain.[/quote]\r\n\r\nin addition to the obvious point that the bill was supported by the majority of democrats, allowing the economy to collapse to win an ground in an election is analogous to cutting off your nose to spite your face or however that stupid saying goes.", "Solution_7": "We still have to remember that the democrats didn't vote in 2:1 favor of it either...\r\n[code]\n Yes\t No\nDemocrats\t 140\t 95\nRepublicans\t 65\t 133\nTotal\t 205\t 228[/code]\r\n\r\nSo what do you think congress ought to do now - start from scratch, or revise this bill?", "Solution_8": "Just let it fail. Then start from scratch.\r\n\r\n(This can be interpreted as refering to either the economy or the bill...I don't know which one I'm refering to more...I think I'm refering to both.)", "Solution_9": "It seems the senate is trying to force the House to approve it, what do you guys think about this? They just approved it 75-24 I believe, and sweetened the deal, hoping to attract those republican votes they didn't get last time", "Solution_10": "[quote=\"ProtestanT\"]Just let it fail. Then start from scratch.\n\n(This can be interpreted as refering to either the economy or the bill...I don't know which one I'm refering to more...I think I'm refering to both.)[/quote]\r\n\r\nSo you wouldn't mind living your entire life in an economic depression, not know whether the world would ever recover? Notice how, in the words of Steven Colbert, we [b]are[/b] world leaders. Look at how everyone else's markets are following ours!!! And I'm sure those who [b]DIDN'T[/b] overextend themselves and [b]SAVED[/b] money would be glad that their lifetime of work amounted to nothing.\r\n\r\nThe only thing that saved us from the First Great Depression of the '30s (as it will probably be known as soon) was WWII. FDR did tons to help people, yet still, he didn't remove the national debt. \r\n\r\nWhat do you suppose people do in the transition state between the crash and rebuilding of the economy? A barter system? \r\n\r\nWhat about retired people? \"Oh, sorry sir, but you'll have to starve or beg from your family. You don't have any family left? I'm sorry sir, there's no welfare anymore, you'll just have to grow your own food. What? You're in a wheelchair? You'll have to beg.\"" } { "Tag": [], "Problem": "At $ 2: 15$ o'clock, the hour and minute hands of a clock form an angle of:\r\n\r\n$ \\textbf{(A)}\\ 30^{\\circ} \\qquad\\textbf{(B)}\\ 5^{\\circ} \\qquad\\textbf{(C)}\\ 22\\frac {1}{2}^{\\circ} \\qquad\\textbf{(D)}\\ 7\\frac {1}{2} ^{\\circ} \\qquad\\textbf{(E)}\\ 28^{\\circ}$", "Solution_1": "[hide=\"Click for solution\"]\nIn $ 15$ minutes the hour hand moves through an angle of $ \\frac{1}{4} \\cdot 30^\\circ\\equal{}\\frac{15}{2}^\\circ$. Therefore, the angle between the hour hand and the minute hand is $ 22 \\frac{1}{2}$, or $ \\boxed{\\textbf{(C)}}$.\n[/hide]" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all $ f: Z\\minus{}>Z$ such that:\r\n$ f(a^3\\plus{}b^3\\plus{}c^3)\\equal{}f(a)^3\\plus{}f(b)^3\\plus{}f(c)^3$", "Solution_1": "no one like solve it? :maybe:", "Solution_2": "[quote=\"Math pro\"]Find all $ f: Z \\minus{} > Z$ such that:\n$ f(a^3 \\plus{} b^3 \\plus{} c^3) \\equal{} f(a)^3 \\plus{} f(b)^3 \\plus{} f(c)^3$[/quote]\r\nHere is a problem of Vietnam TST 2005." } { "Tag": [ "group theory", "abstract algebra", "function", "Euler", "number theory", "totient function", "superior algebra" ], "Problem": "Suppose we have a bijection between two finite groups $ f: G \\rightarrow H$ such that $ \\forall g \\in G \\: |f(g)|\\equal{}|g|$ is $ f$ an isomorphism?\r\n\r\n($ \\, |g|$ is the order of the subgroup which is generated by $ g$)", "Solution_1": "No. Consider a cyclic group of order five; any of the 24 permutations of the non-identity elements satisfies your condition, but only four of them are group homomorphisms.\r\n\r\nAlso, it's possible to construct non-isomorphic groups with the same numbers of elements of each order; I believe the smallest examples are 16-element groups.", "Solution_2": "oh ok, pretty obvious actually.\r\n\r\nThinking about partitions of groups, is the equivalence relation $ a\\sim b$ iff $ \\equal{}$ useful for much? I remember that it provides an identity with the eular-totient function $ n\\equal{} \\sum_{d|n} \\varphi(d)$ where $ \\varphi(d)$ is the Euler totient function. What else is this useful for? if anything.." } { "Tag": [ "factorial", "floor function", "number theory", "prime factorization" ], "Problem": "For how many positive integers $ k$ does the ordinary decimal representation of the integer $ k!$ end in exactly $ 99$ zeros?", "Solution_1": "[hide]If it is possible for $ n!$ to end in 99 trailing zeros, then there are exactly 5 such numbers $ n$, which take the form $ 5k, \\; 5k\\plus{}1, \\; 5k\\plus{}2, \\; 5k\\plus{}3, \\; 5k\\plus{}4$ for some $ k$.\n\n[hide=\"Justification\"]The number of trailing zeros is the number of factors of $ 10$, or the lower of the number of factors of $ 2$ and the number of factors of $ 5$. The latter is necessarily lower, so we just want to count the number of factors of $ 5$.\n\n$ (5k\\minus{}1)!$ clearly has fewer factors of $ 5$ than $ (5k)!$, since $ 5k$ is divisible by $ 5$. Similarly, $ 5k\\plus{}1,2,3,4$ all have no factors of $ 5$, so multiplying them in to $ (5k)!$ to create their respective factorials will not add any factors of $ 5$. However, $ (5k\\plus{}5)$ is again divisible by $ 5$, therefore $ (5k\\plus{}5)!$ has at least one more factor of $ 5$ than $ (5n)!$.[/hide]\n\nWe must now determine if there is a number $ n$ such that $ n!$ ends in 99 trailing zeros.\n\nAs detailed in the justification, the number of trailing zeros equals the exponent on $ 5$ in the prime factorization of $ n!$. Therefore, the number of trailing zeros is $ z(n)\\equal{}\\sum_{j\\equal{}1}^{\\infty} \\left \\lfloor \\frac{n}{5^j} \\right \\rfloor$.\n\nObserve that $ z(n) \\leq \\frac{n}{5^j} \\equal{} \\frac{n}{4}$, so a good place to start is $ n\\equal{} 4 \\times 99 \\equal{} 396$. However, this has the same number of trailing zeros as $ 395!$, which is less than $ n/4$, so this must have fewer than $ 99$ trailing zeros (it actually has 97). So, we try the next multiple of $ 5$, which is $ 400$.\n\nIndeed, $ 400!$ ends in $ 99$ zeros, hence there are $ \\boxed{5}$ such numbers $ n$, and they are $ \\boxed{400,401,402,403,404}$.[/hide]", "Solution_2": "$ \\boxed{for\\ a \\ prime\\ p: \\ v_p(n) \\equal{} a\\iff p^{a}\\parallel{}n\\iff p^{a}|n\\ and\\ p^{a \\plus{} 1}\\not|n}$\r\n\r\nwe have $ v_2(k!) < v_5(k!)$ so $ k!$ end with exactly $ v_5(k!)$ z\u00e9ro in dicimal representation,\r\nhere we have $ v_5(k!) \\equal{} \\sum_{i \\equal{} 1}^{ \\plus{} \\infty}[k/5^i] \\equal{} 99$\r\n$ k < 5^{4}$ becsause $ v_5(5^{4}) > 99$ so we can write $ k \\equal{} a \\plus{} 5b \\plus{} 25c \\plus{} 125d$ where $ a,b,c,d\\in\\{0,1,2,3,4\\}$\r\nand we have $ v_5(k!) \\equal{} [k/5] \\plus{} [k/25] \\plus{} [k/125] \\equal{} b \\plus{} 6c \\plus{} 31d \\equal{} 99$\r\nand it's easy to find that the only solution is $ (b,c,d) \\equal{} (0,1,3)$\r\nthen $ k \\equal{} a \\plus{} 25 \\plus{} 125\\times 3$ where $ a\\in\\{0,1,2,3,4\\}$\r\nthere are $ 5$ solutions" } { "Tag": [], "Problem": "Let A={1,2,3,...,30}.Find the number of subset of A that has 3-element such that sum of it's elements can be divisible with 5.", "Solution_1": "I suggest first taking $(mod5)$ of all of the integers in your set, and then considering which combinations of 3 can sum to 0(mod5):\r\n\r\n0,0,0\r\n0,1,4\r\n0,2,3\r\netc.\r\n\r\nThen use combinatorics to choose from the equivalence classes!", "Solution_2": "there are not so many combinations: \r\n$(0,0,0),(0,1,4),(0,2,3),(1,2,2),(1,3,1)$\r\nSo then i find\r\n$6\\cdot 5 \\cdot 4+6\\cdot 6\\cdot 6 + 6\\cdot 6\\cdot 6 +6\\cdot 6\\cdot 5 +6\\cdot 6\\cdot 5=912$ sets. \r\n\r\nCould that be correct?", "Solution_3": "Is it true?912 or 812?" } { "Tag": [ "geometry", "3D geometry", "sphere", "combinatorial geometry", "geometry proposed" ], "Problem": "Finitely many points are given on the surface of a sphere, such that every four of them lie on the surface of open hemisphere. Prove that all points lie on the surface of an open hemisphere.", "Solution_1": "Let $S$ be the finite set of these points, situated on the surface of a sphere of centre $O$. Consider their convex hull $\\operatorname{conv}(S)$. It is easy to see that $O\\not \\in \\operatorname{conv}(S)$ if and only if $S$ is contained on the surface of an open hemisphere (by well-known results on separability of a point from a convex set).\n\nAssume $O\\in \\operatorname{conv}(S)$. By the Caratheodory theorem, there exists a subset $T$ of $4$ points in $S$, such that $O\\in \\operatorname{conv}(T)$. By the given condition we however have that $T$ is contained on the surface of an open hemisphere, thus $O\\not \\in \\operatorname{conv}(T)$, contradiction. Therefore $O\\not \\in \\operatorname{conv}(S)$, and so by the previous remark $S$ is contained on the surface of an open hemisphere.\n\nNotice the result stands no more for any infinite set of points, even a countable one. Say one point in $S$ lies on the equator, and each other point of the equator does not lie in $S$, but is an accumulation point for $S$. Then $S$ is not contained on the surface of an open hemisphere, but any $4$ points of it are." } { "Tag": [ "function", "calculus", "integration", "calculus computations" ], "Problem": "is the greatest integer function non differentiable at integral points", "Solution_1": "Yes the gretest integer function is non differentiable at integral points as they are not continous at those points . For an integer $ N$ $ lim_{x \\to 0^\\minus{}} {[x]} \\equal{} N\\minus{}1$ while $ lim_{x \\to 0^\\plus{} } { [x] } \\equal{} N$ while $ f(N) \\equal{} N$ , hence we see gthat it is not continous at integral points and hence not differentiable ." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that in any triangle $ABC$ the following inquality holds:\r\n\r\n \\[ ab+bc+ca\\geq 4S\\sqrt{3} +\\frac{1}{2}((a-b)^{2}+(b-c)^{2}+(c-a)^{2}) \\]", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=31598 .\r\n\r\n Darij" } { "Tag": [ "geometry", "inequalities", "triangle inequality" ], "Problem": "Heres a problem I cam across in one of my old geometry texts.\r\n\r\n Let $X$ and $Y$ be any two points contained in the disc of radius $r$ around $P$. Using the traingle inequality, prove that\r\n\r\n$d(X, Y) \\le 2r$\r\n\r\n Two notes.\r\n\r\n- $d(X, Y)$ means the distance from point $X$ to point $Y$.\r\n\r\n-\"disk\" is defined as follows: We define disk of center $P$ and radius $r$ to be the set of all points whose distance from $P$ is $\\le r$.", "Solution_1": "[hide=\"solution\"]From the triangle inequality (a property of all metric spaces) we have $d(X,Y)\\le{d}(X,P)+d(P,Y).$ But from the definition of the disc, the distances from $P$ to $X$ and $Y$ are $\\le{r}$. So $d(X,P)+d(P,Y)\\le2r$.[/hide]", "Solution_2": "[hide=\"No triangle inequality...\"]\nThe greatest possible distance between two points in a circle is the diameter, or $2r$. Therefore, the distance between any two points $X$ and $Y$ must be less than or equal to $2r$.\n[/hide]\r\n\r\nI don't think that's rigorous, is it? :(" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "The numbers from 1 to 9 are written in an arbitrary order. A move is to reverse the order of any block of consecutive increasing or decreasing numbers. For example, a move changes 916532748 to 913562748. Show that at most 12 moves are needed to arrange the numbers in increasing order.", "Solution_1": "any solution??????? :!:", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?p=124454", "Solution_3": "Is it 12 needed for some permutation? or can always be done with 11 or even lower?" } { "Tag": [ "geometry", "circumcircle", "trigonometry", "combinatorics unsolved", "combinatorics" ], "Problem": "a) Suppose that three congruent disks with radius $R$ have their centers in the vertices of a given triangle which they cover. Find the smallest value of $R$ for a given triangle.\r\n\r\nb) (this problem just asks itself to generalize :P) Suppose that $n$ congruent disks with radius $R$ have their centers in the vertices of a given convex $n$-gon which they cover. Find the smallest value of $R$ for a given convex $n$-gon.", "Solution_1": "for a), isn't the answer just the circumradius of the triangle itself? (at least for an acute triangle) because if R were to be smaller than the circumradius, then the circumcentre will never be covered by the disks....and i think for an obtuse triangle, min(R) is 1/2 of max{a,b,c}. :)", "Solution_2": "Your answer is correct for acute triangle but wrong for obtuse where it is $\\frac{a}{2 \\cos {\\beta}}$ where $a$ is the second longest side and $\\beta$ is the smallest angle in the given triangle. :)", "Solution_3": "ahh i see, it's because if BC is the largest side, have to make sure that R extended from A covers the midpoint of BC. okay, thanx! :D" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "How many positive integers less than 500 can be written as the sum of two positive perfect cubes?", "Solution_1": "It would be a great help if you knew the Ramanujan-Hardy number, $ 1729$ which is the first natural number that can be expressed as the sum of two distinct cubes in two different ways. That way, we can be sure none of the numbers we list will appear twice.\r\n\r\n$ 1^3\\plus{}1^3, 1^3\\plus{}2^3, 1^3\\plus{}3^3, 1^3\\plus{}4^3, 1^3\\plus{}5^3, 1^3\\plus{}6^3, 1^3\\plus{}7^3$\r\n$ 2^3\\plus{}2^3, 2^3\\plus{}3^3, 2^3\\plus{}4^3, 2^3\\plus{}5^3, 2^3\\plus{}6^3, 2^3\\plus{}7^3$\r\n$ 3^3\\plus{}3^3, 3^3\\plus{}4^3, 3^3\\plus{}5^3, 3^3\\plus{}6^3, 3^3\\plus{}7^3$\r\n$ 4^3\\plus{}4^3, 4^3\\plus{}5^3, 4^3\\plus{}6^3, 4^3\\plus{}7^3$\r\n$ 5^3\\plus{}5^3, 5^3\\plus{}6^3, 5^3\\plus{}7^3$\r\n$ 6^3\\plus{}6^3$\r\n\r\nThere are $ \\boxed{26}$ numbers.", "Solution_2": "[quote=\"RoFlLoLcOpT\"]It would be a great help if you knew the Ramanujan-Hardy number, $ 1729$ which is the first natural number that can be expressed as the sum of two distinct cubes in two different ways. That way, we can be sure none of the numbers we list will appear twice.\n\n$ 1^3\\plus{}1^3, 1^3\\plus{}2^3, 1^3\\plus{}3^3, 1^3\\plus{}4^3, 1^3\\plus{}5^3, 1^3\\plus{}6^3, 1^3\\plus{}7^3$\n$ 2^3\\plus{}2^3, 2^3\\plus{}3^3, 2^3\\plus{}4^3, 2^3\\plus{}5^3, 2^3\\plus{}6^3, 2^3\\plus{}7^3$\n$ 3^3\\plus{}3^3, 3^3\\plus{}4^3, 3^3\\plus{}5^3, 3^3\\plus{}6^3, 3^3\\plus{}7^3$\n$ 4^3\\plus{}4^3, 4^3\\plus{}5^3, 4^3\\plus{}6^3, 4^3\\plus{}7^3$\n$ 5^3\\plus{}5^3, 5^3\\plus{}6^3, 5^3\\plus{}7^3$\n$ 6^3\\plus{}6^3$\n\nThere are $ \\boxed{26}$ numbers.[/quote]\n\nWait so if the number was >1729 it might be a repeat?", "Solution_3": "[quote=\"nmani2\"][quote=\"RoFlLoLcOpT\"]It would be a great help if you knew the Ramanujan-Hardy number, $ 1729$ which is the first natural number that can be expressed as the sum of two cubes in two different ways. That way, we can be sure none of the numbers we list will appear twice.\n\n$ 1^3\\plus{}1^3, 1^3\\plus{}2^3, 1^3\\plus{}3^3, 1^3\\plus{}4^3, 1^3\\plus{}5^3, 1^3\\plus{}6^3, 1^3\\plus{}7^3$\n$ 2^3\\plus{}2^3, 2^3\\plus{}3^3, 2^3\\plus{}4^3, 2^3\\plus{}5^3, 2^3\\plus{}6^3, 2^3\\plus{}7^3$\n$ 3^3\\plus{}3^3, 3^3\\plus{}4^3, 3^3\\plus{}5^3, 3^3\\plus{}6^3, 3^3\\plus{}7^3$\n$ 4^3\\plus{}4^3, 4^3\\plus{}5^3, 4^3\\plus{}6^3, 4^3\\plus{}7^3$\n$ 5^3\\plus{}5^3, 5^3\\plus{}6^3, 5^3\\plus{}7^3$\n$ 6^3\\plus{}6^3$\n\nThere are $ \\boxed{26}$ numbers.[/quote]\n\nWait so if the number was >1729 it might be a repeat?[/quote]\n@nmani2, yes, if the number was greater than $1729$, it would be a repeat because as Rofllolcopt said, $1729$ is the first number that can be expressed as the sum of two distinct cubes in two different ways and we can assume that there are more numbers greater than $1729$ that have the same or similar property in that it can be expressed as the sum of two cubes in two or more different ways.", "Solution_4": "[quote=GameBot]How many positive integers less than 500 can be written as the sum of two positive perfect cubes?[/quote]\n\nhttps://artofproblemsolving.com/videos/counting/chapter2/186", "Solution_5": "Yuh you aint need no chart the answer's just ${7 \\choose 2} + 5$ :cool:", "Solution_6": "[quote=ihatemath123]Yuh you aint need no chart the answer's just ${7 \\choose 2} + 5$ :cool:[/quote]\n\nElaborating on this a bit, $\\binom{7}{2}$ will choose two distinct numbers from $1-7$ ($8^3$ is already too large), but we have to take out $7^3+6^3$ since it's too large, and add back the $1^3 + 1^3$ up to $6^3 + 6^3$, thus $\\binom{7}{2} - 1 + 6$.", "Solution_7": "Very nice solution, I just did the old fashioned chart" } { "Tag": [ "function", "algebra", "polynomial", "integration", "calculus", "geometry", "trapezoid" ], "Problem": "hi, \r\ni will be glad if somebody helps to solve this problem : \r\n\r\n[u]problem:[/u] \r\n\r\nlet f be a function that we can derive two times in [a;b]and f(a)=f(b)=0. \r\nlet x be a real number such that : a1 solving out c wo have the solution (a,b,(a+b-2)/(ab-1)).", "Solution_2": ":(\r\ni did this to but it's not what i want\r\nanyway thx wpolly0419 :D.", "Solution_3": "So, what are you looking for Kuba????\r\n\r\nPierre.", "Solution_4": "i don't know what i wanted when i posted this message...:(\r\nsorry", "Solution_5": "So, we may assume that the problem is solved, and move it....\r\n\r\nPierre." } { "Tag": [], "Problem": "Kto\u015b z was (polak\u00f3w :) ), opr\u00f3cz mnie :D , b\u0119dzie bra\u0142 udzia\u0142 ?? \r\nZadania wydaj\u0105 si\u0119 dosy\u0107 trudne. :huh:", "Solution_1": "Chyba wezme udzial, ... bo nie wiem czy moge dalej liczyc na OM :(\r\n(zalosnie skwasiles zadanie kombinatoryczne i jestem ogolnie zdemoralizowany)\r\nPoza tym zadania sa generalnie przepiekne na mathlinks contest :)", "Solution_2": "Zadanka rzeczywiscie bardzo fajniutkie acz trudne (co liczy sie raczej in plus). \r\n\r\nJa wystartuje (jak cos zrobie :P )" } { "Tag": [ "linear algebra", "matrix", "combinatorial inequality", "algebra", "combinatorics", "IMO", "IMO 1971" ], "Problem": "Let $ A \\equal{} (a_{ij})$, where $ i,j \\equal{} 1,2,\\ldots,n$, be a square matrix with all $ a_{ij}$ non-negative integers. For each $ i,j$ such that $ a_{ij} \\equal{} 0$, the sum of the elements in the $ i$th row and the $ j$th column is at least $ n$. Prove that the sum of all the elements in the matrix is at least $ \\frac {n^2}{2}$.", "Solution_1": "Take the the row or column (without loss of generality, a row) with the minimum possible sum $ S$ of its elements. Let $ z$ be the number of zeros in this row. We will assume $ S < \\frac{n}{2}$ once the other case is obvious. Clearly $ S \\ge n \\minus{} z \\Rightarrow z \\ge n \\minus{} S$. The total sum $ T$ of all elements of the matrix is at least the number of zeros in this row multiplicated by $ n \\minus{} S$ (because the sum of the elements of a row and a column meeting in a zero is $ \\ge n$) plus the number of nonzero elements times $ S$ (that is the minimum possible sum), it is, \\[ z(n \\minus{} S) \\plus{} (n \\minus{} z)S.\\] Note that, being $ z \\ge n \\minus{} S$ and $ n \\minus{} S \\ge S$, we can put $ z \\minus{} 1$ instead of $ z$ and $ n \\minus{} z \\plus{} 1$ instead of $ n \\minus{} z$ until we get $ z \\equal{} n \\minus{} S$, what makes the sum become smaller. So we have \\[ T \\ge (n \\minus{} S)^2 \\plus{} S^2 \\ge 2(\\frac{n \\minus{} S \\plus{} S}{2})^2 \\equal{} \\frac{n^2}{2}\\] by AM - QM inequality.", "Solution_2": "If the main diagonal contains all zeroes, we can immediately deduce from the condition that the sum $S$ of the elements is at least $n^2/2$. This motivates us to consider the largest possible subset of elements, all zero, which are pairwise in different rows of columns - let the size of this subset be $k$. (If there is no such subset, $S\\ge n^2$.)\n\nSince the condition is invariant under the operation of switching rows or columns, we may rearrange the matrix such that these $k$ entries are on the main diagonal of the matrix, i.e. $a_{11}=a_{22}=\\dots=a_{kk}=0$. Let $X=\\sum_{1\\le i,j\\le k}a_{ij}$, $Y=\\sum_{i\\le k0 such that for p >= N and all r,s,\r\n\r\nThe smallest integer greater or equal to (p-1)/3 <= f(r,s) <= the greatest integer less than or equal to 2(p-1)/3.\r\n\r\nNote: {x} : the farctional part of x.", "Solution_1": "Try http://www.mathlinks.ro/Forum/viewtopic.php?highlight=USA+TST&t=26386" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ G$ be a connected graph such that if any vertex of the graph is removed (and with it all the edges connected to it) the graph remains connected. Prove that for any two vertices, there exist two disjoint paths connecting the two vertices.", "Solution_1": "Suppose there is a pair of vertices X and Y such that any path between them contains an edge e=AB. If e is part of a cycle then there exists a path from A to B that does not include e and then we can find a new path from X to Y by traversing this new path around e - contradiction. If e is not part of a cycle then it is a cut-edge and since removing A or B removes e as well this is a contradiction. Note that if the component of A is empty after A is removed then we can instead remove B and if its component is empty as well then the graph is just a 2-vertex line and the statement doesn't hold in this case.", "Solution_2": "It's false; take $ G \\equal{} K_2$. :) \r\n\r\nThis would suggest that the above argument needs to be refined a bit.", "Solution_3": "In what sense does the argument need to be refined? dsadsa correctly noted that the statement fails for $ G \\equal{} K_2$. :maybe:", "Solution_4": "Oops! My bad.", "Solution_5": "Hmm...\r\n\r\nWhat if we define \"disjoint paths\" by \"paths that share no common vertex\"? Is the statement still true?", "Solution_6": "If all paths from X to Y go through a vertex Z with Z =/= X and Z =/= Y then removing Z will result in no paths from X to Y implying that the graph is not connected.", "Solution_7": "[quote=\"dsadsa\"]If all paths from X to Y go through a vertex Z with Z =/= X and Z =/= Y then removing Z will result in no paths from X to Y implying that the graph is not connected.[/quote]\r\n\r\nBut for the statement to be false it's not necessary that all paths from $ X$ to $ Y$ go through a vertex $ Z$...", "Solution_8": "Edit: Nevermind, I see what you mean.", "Solution_9": "If two vertices are joined by three paths and each pair of paths intersect in a vertex (but the three paths don't have any vertex in common), I believe it's possible to patch bits and pieces of the paths together to form two disjoint paths joining the vertices. (This is based on drawing a picture of what I hope is a generic situation.)", "Solution_10": "We induct on the distance between X and Y. If the distance is 1 then they are adjacent and X has at least one other vertex adjacent to it other than Y, call it Z. Then remove X and since G is still connected we can find a path from Z to Y and then add X to it to construct our second path. Suppose the statement holds for a distance of K.\r\n\r\nThen for a distance of K+1 we consider the shortest path S between X and Y. Let A be the unique vertex adjacent to X on S. We can find two disjoint paths from A to Y, call them P1 and P2. Now remove A. There exists a path L from X to Y since G-A is connected. If L is disjoint from P1 and P2 then we are done. If not then let L intersect P1 first (WLOG) at Z as one travels down the path from X to Y. Now we can easily construct two disjoint paths from X to Y. \r\n\r\nThe first path goes from X, down part of L to Z and down P1 to Y.\r\nThe second path goes from X to A and then down P2 to Y. \r\n\r\nBy POMI the statement holds for all distances between X and Y.", "Solution_11": "Nice solution, dsadsa!!" } { "Tag": [ "inequalities", "vector", "real analysis", "real analysis unsolved" ], "Problem": "Let $E_{C}$ be the set of sequences $u_{n}\\in L^{1}(\\mathbb{N})$ such that $\\|u_{n}\\|_{2}= 1$ and $\\|u_{n}\\|_{4}=C>0$. Find $\\inf \\{ \\|u\\|_{1}: u \\in E_{C}\\}$", "Solution_1": "Holder's inequality gives us the general norm inequality\r\n\\[{\\| u \\|}_{2}\\le{\\| u \\|}_{1}^{1/3}{\\| u \\|}_{4}^{2/3}\\, . \\]\r\nSo in our situation we get the lower bound ${\\| u \\|}_{1}\\ge \\frac{1}{C^{2}}$. I'm not sure whether that is the greatest lower bound.", "Solution_2": "Holder is sharp only for collinear vectors, that is, when $(u_{n})$ is a constant multiple of $(u_{n}^{4})$. This happens when all nonzero elements of $(u_{n})$ are the same. But you can't always arrange this under the stated assumptions on the norms.", "Solution_3": "Using that observation, I think we can show that my lower bound can be achieved for infinitely many values of $C$. Namely, any $C$ of the form $\\frac{1}{\\sqrt[4]{N}}$, where $N$ is a positive integer." } { "Tag": [ "vector", "trigonometry", "complex numbers" ], "Problem": "Suppose that a,b,c,d are real numbers in descending order. Given\r\na^2 + d^2 = 1 ; \r\nb^2 + c^2 =1 ;\r\nac+bd = 1/3 ; \r\n\r\nFind the value of (ab-cd).", "Solution_1": "[hide=\"brute force\"]\nwe have:\n$ a^2\\plus{}d^2\\equal{}1$ and $ b^2\\plus{}c^2\\equal{}1$\nMultiplying,we get:\n$ a^2b^2\\plus{}a^2c^2\\plus{}b^2d^2\\plus{}c^2d^2\\equal{}1$ .............(i)\n\nalso,we have:$ {(ac\\plus{}bd)}^2\\equal{}a^2c^2\\plus{}b^2d^2\\plus{}2abcd\\equal{}\\frac19$ ........(ii)\n\nSubtracting (ii) from (i),we get:\n$ a^2b^2\\plus{}c^2d^2\\minus{}2abcd\\equal{}\\frac89$\n$ {(ab\\minus{}cd)}^2\\equal{}\\frac89$\n\n :D \n\n[/hide]", "Solution_2": "OK this was the easiest method. No pun intended. \r\n\r\nThere are two more methods.The one I found really rips this apart in one line.", "Solution_3": "Well, this is very simple actually. Maybe this method is the one tht conjurer told rips the sum in one step.\r\n\r\nSolution: We can write $ (a^2 \\plus{} d^2) ( b^2 \\plus{} c^2) \\equal{} (ac \\plus{} bd)^2 \\plus{} (ab \\minus{} cd)^2 \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} > (1)$.\r\nNow , given that \r\n$ a^2 \\plus{} d^2 \\equal{} 1 \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} > (i)$\r\n$ b^2 \\plus{} c^2 \\equal{} 1 \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} > (ii)$\r\n$ ac \\plus{} bd \\equal{} 1/3 \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} \\minus{} > (iii)$\r\nSubstituting the values in the equation $ (1)$, we get that $ (ab \\minus{} cd)^2 \\equal{} 1 \\minus{} \\frac {1}{9} \\equal{} \\frac {8}{9}$ yielding us that $ (ab \\minus{} cd) \\equal{} \\pm \\frac {2 \\sqrt{2}}{3}$", "Solution_4": "Hey arvind, i thought u left Aops :rotfl: \r\n\r\nneways, nice solution :D", "Solution_5": "Well you just generalized what skand already did. No pun intended here too.\r\n\r\nMy sol:\r\n\r\nDefine x = a$ {\\hat i}$ + b$ {\\hat j}$\r\n\r\ny = c$ {\\hat i}$ + d$ {\\hat j}$\r\n\r\nWe are given x$ dot$y and we need to find x$ cross$y . Simple :)\r\n\r\nSince there is a simple vector solution, there outta be a complex no solution as well. And a trignometric too. :p", "Solution_6": "Complex: let $ x \\equal{} a \\plus{} di, y \\equal{} b \\plus{} ci$. We are given that $ |x| \\equal{} |y| \\equal{} 1$ and $ xy$ has imaginary part $ \\frac{1}{3}$...\r\n\r\nTrig: let $ a \\equal{} \\cos \\alpha, d \\equal{} \\sin \\alpha, b \\equal{} \\cos \\beta, c \\equal{} \\sin \\beta$. We are given that $ \\sin (\\alpha \\plus{} \\beta) \\equal{} \\frac{1}{3}$...\r\n\r\n(Of course, these two solutions are equivalent and pretty much equivalent to the \"algebraic\" solution as well.)", "Solution_7": "You said equivalent twice. xD" } { "Tag": [ "calculus", "integration", "vector", "trigonometry", "geometry", "parameterization", "analytic geometry" ], "Problem": "Hello all,\r\n\r\n[img]http://img244.imageshack.us/img244/218/picture8ce5.png[/img]\r\n\r\nI am completely new to this stokes theorem bussiness..what i have got so far is the nabla x F part, but i am unsure of how to find N (the unit normal field i think its called).\r\n\r\nany suggestions people?\r\n\r\ni get that nabla x F or curl(F) = [b]i[/b] + [b]j[/b] + [b]k[/b]\r\n\r\ncheers\r\n\r\n-sarah :)", "Solution_1": "Well, a way to find N is to first find any normal to the surface and then normalize it. Some lead-in work: Put the surface S into vector form, \r\n\r\n$S: \\, \\, \\, \\vec{r}(u,v)=\\left< u\\cos v,u\\sin v, v\\right> \\, \\, 0\\leq u\\leq 1,\\, 0\\leq v\\leq\\frac{\\pi}{2}$\r\n\r\nAlso, recall that the vectors $\\frac{d\\vec{r}}{du}\\mbox{ and }\\frac{d\\vec{r}}{dv}$ are tangent to S, and that the cross product of two vectors is normal to both vectors, so a normal vector to S is $\\frac{d\\vec{r}}{du}\\times\\frac{d\\vec{r}}{dv},$ but we want a unit normal to S, so we normalize the above vector (divide it by its magnitude) to get\r\n\r\n$\\vec{N}(u,v)= \\frac{\\frac{d\\vec{r}}{du}\\times\\frac{d\\vec{r}}{dv}}{\\left| \\frac{d\\vec{r}}{du}\\times\\frac{d\\vec{r}}{dv}\\right|}$", "Solution_2": "ok i do that and i get the unit normal to be $\\frac{sin(v)i-cos(v)j+u k}{\\sqrt{1+u^{2}}}$\r\n\r\nhow does that look?\r\n\r\nand then how do i use that? that is what do i do next?", "Solution_3": "Write down the Stokes' theorem. You already have the curl and normal vector; your inifinitesimal area is $du dv$. Put the boundaries of double integral and you're essentially done.", "Solution_4": "how do i work out what dS is?, i think that is my main problem....\r\n\r\nhow does this look for the integral we need to evaluate?\r\n\r\n[img]http://img151.imageshack.us/img151/5053/picture9fn3.png[/img]\r\n\r\nthat is, do we have to use the normal or the unit normal when doing this stuff?", "Solution_5": "$dS=\\left| \\frac{\\partial \\vec{r}}{\\partial u}\\times\\frac{\\partial \\vec{r}}{\\partial v}\\right|\\,du\\,dv$, this is a kind of change-of-variables formula. This term cancels the denominator in the formula for unit normal vector.", "Solution_6": "Argh, forgot the Jacobian. Sorry.", "Solution_7": "mlok, is that forumula you gave for d[b]S[/b] or dS ?", "Solution_8": "That was $dS$. The vector form of the area element ($d\\mathbf S$) is the vector of length $dS$ in the direction perpendicular to the surface. \\[d\\mathbf{S}=\\left(\\frac{\\partial \\vec{r}}{\\partial u}\\times\\frac{\\partial \\vec{r}}{\\partial v}\\right)\\,du\\,dv\\] $d\\mathbf{S}$ can be used to compute surface integrals of vector fields directly, without reducing them to a surface integral of a function. Namely, \\[\\iint_{S}\\mathbf{F}\\cdot d\\mathbf{S}=\\iint_{D}\\mathbf{F}(\\mathbf{r}(u,v))\\cdot \\left(\\frac{\\partial \\vec{r}}{\\partial u}\\times\\frac{\\partial \\vec{r}}{\\partial v}\\right)\\,du\\,dv\\] where $\\mathbf{r}$ is the parametrization of $S$ and $D$ is the parameter domain. Look, no magnitude of vectors!", "Solution_9": "ok i see. so how do you work out what dS or d[b]S[/b] is when you arnt given the parametric representation?", "Solution_10": "You have to find a parametric representation first. This may be the most difficult part of the problem. In the simplest case when your surface is a graph (say $z=g(x,y)$) you can take $x$ and $y$ to be your parameters, so that $\\mathbf{r}(x,y)=\\langle x,y, g(x,y)\\rangle$. In other cases you can try a couple of cylindrical or spherical (ugh!) coordinates as parameters. Whether or not they work will depend on your ability to solve for the third coordinate." } { "Tag": [], "Problem": "Are any of you enrolled in Kumon, a math and reading program?\r\nIf so, what level are you in? I'm in L on math and J for reading.", "Solution_1": "my bro is in it, he's in like...D, yup", "Solution_2": "I used to do the math program and got up to J. Kumon is useful for building basic skills like factoring, but I think it emphasizes rote learning far too much. I can't speak for the english part of it, though.", "Solution_3": "I finished Reading (Level L) a couple of months ago and just finished Math level Q in July.\r\n\r\nI would be lying if I said I enjoyed it. But upper levels were better. I heard in Japan math goes further to level R or S.", "Solution_4": "I think I remember my babysitter doing Kuman, and my mom wanted me to do it too, but my dad was really against it. He says that it foocusses to much on knowing how, not why. Being a math professor probably has something to do with it.", "Solution_5": "I did Kumon for maybe 5 years? My school required it, so I did it from preschool (I think) to 4th grade (I left after that year). I think I got up to level G-ish, H-ish; I can confidently say that I was faster in computations than I yam now. :cool:. It did help me boost my confidence to actually figure out that I was decent at math.", "Solution_6": "ah, man, the kids i was babysitting had to do kumon and it was HORRIBLE. i really didnt like making them do it, it seemed so busywork-esque, which i've ALWAYS ALWAYS hated. give me lots of homework, i dont care, as long as its not repetitive. its not a type of learning i necessarily agree with, but \"different strokes for different folks.\"", "Solution_7": "I did Kumon once sometime in elementary school (I don't remember what level) and couldn't stand it. They had me working on speed with basic arithmetic problems when I was well beyond that level and ready for more advanced material. Needless to say, I didn't stick with it for more than a few months, if I remember right.", "Solution_8": "Just don't go. It's a waste of your money.", "Solution_9": "My parents made me do Kumon for about 2 to 3 years in elementary school. The program was good only for practice with basic arithmetic. It might be good for a kid in elementary school who is in one of those \"new math\" curricula, but other than that, the program is essentially useless. The problems are simple enough that you don't really learn anything new by doing them (especially if you have already seen the material). I could rant and rave about Kumon for a long time, but instead I'll just say that if you are considering signing your kid up for Kumon don't. Instead, sit down with him/her and a good math book which is ahead of what he/she is covering in school. That way, by going through the book, your child will be forced to develop his/her understanding of the foundations of math in order to learn the new material. At least this is my take on the whole issue." } { "Tag": [ "floor function" ], "Problem": "Let $ \\alpha$ be the positive root of equation: $ x^2\\minus{}2008x\\minus{}1\\equal{}0$.\r\n$ x_o\\equal{}1, x_{n\\plus{}1}\\equal{}$[$ \\alpha x_n$] \r\nFind the remainder when we divide $ x_{2008}$ by 2008.\r\nEveryone help me, please!! :|", "Solution_1": "First, we know that $ a_1\\equal{}[\\alpha]$, $ a_2\\equal{}[\\alpha^2]$ ...\r\n\r\nSo that $ a_2008\\equal{}[\\alpha^{2008}]$.\r\n\r\n$ 2008<\\alpha<2009$.\r\n\r\nTherefore, $ 0<\\alpha\\minus{}2008<1$\r\n\r\nSo, $ [(\\alpha\\minus{}2008)^2008]\\equal{}1$.\r\n\r\nTherefore, the answer is $ 1$.\r\n\r\nThis may or may not be right...", "Solution_2": "[quote=\"BOGTRO\"]First, we know that $ a_1 \\equal{} [\\alpha]$, $ a_2 \\equal{} [\\alpha^2]$ ...\n\nSo that $ a_2008 \\equal{} [\\alpha^{2008}]$.[/quote] No. Consider a sequence with $ x_0 \\equal{} 1$, $ x_{n \\plus{} 1} \\equal{} \\left\\lfloor \\frac {5}{2} x_{n}\\right\\rfloor$ -- we have $ x_1 \\equal{} 2$, $ x_2 \\equal{} 5$ but $ \\left\\lfloor \\left(\\frac {5}{2}\\right)^2\\right\\rfloor \\equal{} \\left\\lfloor\\frac {25}{4}\\right\\rfloor \\equal{} 6$. The rest of your solution doesn't make much sense, either.", "Solution_3": "Anyone help me, please! :help: \r\nI think it isn't a hard problem (not to me :oops: ).\r\nSo reply as soon as possible.\r\nThank you!", "Solution_4": "[hide=\"Solution\"]a = $ \\frac{2008\\plus{}\\sqrt{2008^2 \\plus{} 4}}{2}$, which is > 2008 but < 2009. so [a] = a-2008. and $ a_2 \\equal{} [a^2 \\minus{} 2008a]$ But we know that $ a^2\\minus{}2008a \\equal{} 1$ So we get a_2 = 0. So $ a_2008 \\equal{} 0.$ [/hide]", "Solution_5": "[quote=\"pythag011\"][hide=\"Solution\"]a = $ \\frac {2008 \\plus{} \\sqrt {2008^2 \\plus{} 4}}{2}$, which is > 2008 but < 2009. so [a] = a-2008. and $ a_2 \\equal{} [a^2 \\minus{} 2008a]$ But we know that $ a^2 \\minus{} 2008a \\equal{} 1$ So we get a_2 = 0. So $ a_2008 \\equal{} 0.$ [/hide][/quote]\r\nNo, you are wrong.\r\n[a] is the greatest integer number smaller than $ a$ and {a} = a - [a]", "Solution_6": "Nobody solve this problem so I'll post my teacher's solution.\r\n[hide=\"Here\"]$ x_{n+1}=[\\alpha x_n]<\\alpha x_n<[\\alpha x_n]+1=x_{n+1}+1$\n$ \\Rightarrow \\frac{x_{n+1}}{\\alpha}O=pqr for three prime ideals is one example and

=(p^2)q is another example. I don't know these actually occur in this ring of integers, but I was hoping for a lot of help here.", "Solution_1": "I figured out the first part. Just need help with the decomposition, so you can ignore the first part of the question." } { "Tag": [], "Problem": "A train traveling from Aytown to Beetown meets with an accident after $ 1$ hr. It is stopped for $ \\frac{1}{2}$ hr., after which it proceeds at four-fifths of its usual rate, arriving at Beetown $ 2$ hr. late. If the train had covered $ 80$ miles more before the accident, it would have been just $ 1$ hr. late. The usual rate of the train is:\r\n\r\n$ \\textbf{(A)}\\ \\text{20 mph} \\qquad\r\n\\textbf{(B)}\\ \\text{30 mph} \\qquad\r\n\\textbf{(C)}\\ \\text{40 mph} \\qquad\r\n\\textbf{(D)}\\ \\text{50 mph} \\qquad\r\n\\textbf{(E)}\\ \\text{60 mph}$", "Solution_1": "[hide=\"solution\"]\ntown $ A$ and $ B$ are seperated by distance $ d$ and the trian usually travels $ v$ miles/hr then the trip usually takes $ \\frac{d}{v}$ hours. Now we can set up two equations;\n\n$ \\frac{d}{v} \\plus{} 2 \\equal{} 1 \\plus{} \\frac{1}{2} \\plus{} \\frac{d\\minus{}v}{\\frac{4v}{5}}$\n$ \\Rightarrow d\\equal{}7v \\qquad(1)$\n\n$ 1 \\plus{} \\frac{d}{v} \\equal{} \\frac{80}{v} \\plus{} \\frac{1}{2} \\plus{} \\frac{5(d\\minus{}80)}{4v}$\n$ \\Rightarrow d \\equal{} 80 \\plus{} 2v\\qquad(2)$\n\nSubbing (1) and (2)\n$ 5v \\equal{} 80 \\Longrightarrow v \\equal{} 40 \\longrightarrow \\boxed{C}$[/hide]", "Solution_2": "I think you are solving the question where the train travels 80 miles before the accident, as opposed to 80 miles MORE before the accident.\r\n\r\n[hide]\nThe train gains an hour by traveling 80 miles at full speed instead of 4/5 speed. Let r=speed. Then\n$ \\frac{80}{r}\\equal{}\\frac{80}{\\frac{4}{5}r}\\minus{}1$\n$ r\\equal{}20\\Rightarrow\\boxed{A}$\n[/hide]", "Solution_3": "[quote=\"SCHNOGGINVOGGINSCHWITZERW\"]I think you are solving the question where the train travels 80 miles before the accident, as opposed to 80 miles MORE before the accident.\n\n[hide]\nThe train gains an hour by traveling 80 miles at full speed instead of 4/5 speed. Let r=speed. Then\n$ \\frac {80}{r} \\equal{} \\frac {80}{\\frac {4}{5}r} \\minus{} 1$\n$ r \\equal{} 20\\Rightarrow\\boxed{A}$\n[/hide][/quote]SCHNOGGINVOGGINSCHWITZERW has the correct answer to this problem.", "Solution_4": "ocha, you had everything right until your second equation, at which point you just threw every thing out the window.\r\nThe left side is right, but the right side is wrong...\r\nyou should get that the LHS=$ 1 \\plus{} \\frac {80}{v} \\plus{} \\frac {1}{2} \\plus{} \\frac {5(d \\minus{} v \\minus{} 80)}{4v}$", "Solution_5": "i got what ocha got hum...", "Solution_6": "[quote=Bryanli]i got what ocha got hum...[/quote]\n\nthank you, everyone was wondering if ocha was right or wrong. In fact, research mathematicians were pondering for 8 years whether \"ocha\" or \"nittanylion\" solved the big [i]train problem[/i] correctly!", "Solution_7": "what really then why did they ask that" } { "Tag": [ "search", "puzzles" ], "Problem": "Adam, God made out of dust\r\nBut thought it best to make me first,\r\nSo I was made before man\r\nTo answer God's most Holy plan.\r\nA living being I became\r\nAnd Adam gave to me my name.\r\nI from his presence then withdrew\r\nAnd more of Adam never knew.\r\nI did my Maker's law obey\r\nNor ever went from it astray.\r\nThousands of miles I go in fear\r\nBut seldom on earth appear.\r\nFor purpose wise God did see,\r\nHe put a living soul in me.\r\nA soul from me God did claim\r\nAnd took from me the soul again.\r\nSo when from me the soul had fled\r\nI was the same as when first made.\r\nAnd without hands, or feet, or soul,\r\nI travel on from pole to pole.\r\nI labor hard by day, by night\r\nTo fallen man I give great light.\r\nThousands of people, young and old\r\nWill by my death great light behold.\r\nNo right or wrong can I conceive\r\nThe scripture I cannot believe.\r\nAlthough my name therein is found\r\nThey are to me an empty sound.\r\nNo feat of death doth trouble me\r\nReal happiness I'll never see.\r\nTo Heaven I shall never go\r\nOr to Hell below.\r\nNow when these lines you slowly read,\r\nGo search your Bible with all speed\r\nFor that my name is written there\r\nI do honestly to you declare.\r\n\r\n\r\nWho or what is the \u201cI\u201d referred to in the riddle?", "Solution_1": "[quote=\"K Sengupta\"]\n\nWho or what is the \u201cI\u201d referred to in the riddle?[/quote]\r\n\r\nI'm pretty sure that it is a star...am I right?" } { "Tag": [], "Problem": "alkene + bromine water \r\n\r\nwhat would be the products of this reaction?", "Solution_1": "two products one would be the vicinal di bromide and other the bromohydrin (major)\r\nsee [url=http://www.chemguide.co.uk/organicprops/alkenes/halogenation.html]here[/url] .\r\nalso the mechanism has bromonium ion intermediate.", "Solution_2": "That reaction is a classic test used for the identification of double bonds C=C, although for that purpose it is more commonly used a CCl4 solution of bromine.\r\n\r\nAs Pardesi said, the products are the bromohidrin and the vicinal dibromide. The first step of the mechanism involves a complexation of one bromine molecule (acting as a Lewis acid) by one molecule of alkene (the Lewis base), which is converted to a bromide ion and a cyclic bromonium ion. This step is reversible. In the next step a nucleophile (a water molecule or a bromide anion) attacks one of the carbons of the \"double bond\" in a SN2 fashion to yield the products.\r\n\r\nThe active attacking species can also be hypobromous acid (HBrO), since it is know that the halogens disproportionate in water to halide and hypohalide ions." } { "Tag": [ "algebra", "polynomial", "AMC", "AIME", "USA(J)MO", "USAMO", "inequalities" ], "Problem": "Is there a gener way of solving this kind of problem?\r\n\r\nGiven complxe numbers $a,b,c$ such that\r\n\r\n$a+b+c=\\omega$\r\n$a^2+b^2+c^2=\\beta$\r\n$a^3+b^3+c^3=\\phi$\r\nfind\r\n$a^m+b^m+c^m$\r\n\r\nwhere $\\omega,\\beta,\\phi,m$ are all real.\r\n\r\ni know it is almost impossible to solve for $a,b,c$ and i know you have to manipulate the equations in some way, but im am just not sure of a general way of doing so.\r\ncould someone please help me.", "Solution_1": "The quickest way to solve problems of the sort is to find the symmetric sum of $ab$, $a^2b$, and $abc$ by elementary manipulations of the given three identity's, then forming a polynomial with roots $a,b,c$ by relating the specific sums with Viete's formulas. From there, you just recursively apply Newton's Sums and get the desired sum.", "Solution_2": "could you please provide an example.\r\ni do not understnad how you would use Viete's Sums.", "Solution_3": "I will provide an example which was on Mildorf's Mock AIME 3, which can be found here:\r\n\r\nhttp://www.tjhsst.edu/~tmildorf/math/Adv.htm\r\n\r\nLook at problem #4, which is a problem almost identitical to the one you gave. In the solutions, Mildorf solves it exactly as I stated earlier, in much greater detail than I'm willing to write out.", "Solution_4": "this tactic is also (briefly) mentioned in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=422380&highlight=#422380]www.artofproblemsolving.com/Forum/viewtopic.php?p=422380&highlight=#422380[/url]", "Solution_5": "Try this USAMO problem from 1973: Find all solutions, real or complex, to the system of equations $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=3$.", "Solution_6": "[hide=\"here is what i got\"]\ngiven $x+y+z=3$\nwe have $(x+y+z)^2=3^2=9=x^2+y^2+z^2+2(xy+yz+zx)$\nso we find that $xy+yz+xz=3$\nthen \n$(x+y+z)(x^2+y^2+z^2)=(3)(3)=9=$\n$x^3+y^3+z^3+(x+y+z)(xy+yz+xz)-3xyz$ so $xyz=1$\nso from this info. we can use Viete's Sums and find that\nthe solutions to the system are the roots to the equation\n$x^3-3x^2+3x-1=(x-1)^3=0$\n[/hide]", "Solution_7": "i am amazed how simple that problem is for an olympiad problem", "Solution_8": "Or simply note that x^2+y^2+z^2=xy+xz+yz which is true for rearrangement only for x=y=z", "Solution_9": "[quote=\"Altheman\"]i am amazed how simple that problem is for an olympiad problem[/quote]\r\n\r\nIt is one of the first few years of the USAMO (2nd, I believe).", "Solution_10": "That's right, maokid. You could also use Newton's sums.", "Solution_11": "[quote=\"JohnConstantine\"]Or simply note that x^2+y^2+z^2=xy+xz+yz which is true for rearrangement only for x=y=z[/quote]\r\n\r\nRearrangement doesn't hold for complex numbers... A similar mistake is applying the Power-Mean Inequality to the system." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "If $\\mu(E_n)<\\infty$ for $n\\in\\mathbb{N}$ and $\\chi_{E_n}\\to f$ in $L^1$, then $f$ is a.e the characteristc function of a measurable set.", "Solution_1": "Firstly we have that $\\chi_{E_k}(x)$ converges to $f$ in measure, hence by Riesz's Theorem, there is a subsequence of $\\{\\chi_{E_k}(x)\\}$, say $\\{\\chi_{E_{k_i}}(x)\\}$, converges to $f$ almost everywhere.\r\nLet $P=\\liminf_{i\\to\\infty} E_{k_i}$. Try to prove that $f=\\chi_P$ a.e.", "Solution_2": "Thanks. I couldn't figure out whether to work with $\\limsup E_n$ or $\\liminf E_n$.", "Solution_3": "I don't really see the necessity of $ \\mu(E_n) < \\infty$.\r\nWhere does it play role?", "Solution_4": "Otherwise $ \\chi_{E_n}$ is not in $ L^1$...", "Solution_5": "Oh, I see now. That makes sense.\r\nBut it doesn't play any role whatsoever later one.\r\nYou just need to notice that $ \\chi_{liminf_{n\\to\\infty} E_n}\\equal{} liminf_{n\\to\\infty} \\chi_{E_n} \\equal{} lim_{n\\to\\infty} \\chi_{E_n} \\equal{} f$" } { "Tag": [ "geometry", "AMC", "AIME", "trigonometry", "USA(J)MO", "USAMO", "HMMT" ], "Problem": "My goal this year is to pass the AIME and make USAMO. To do this, it would help immensely to learn some geometry because that is definitely me weakest subject.\r\n\r\nI can get almost any book, and I already have Challenging Problems in Geometry (though I find this really boring), Geometry Unbound (which is pretty difficult, but does this help very much with AIME?), Art and Craft of Problem Solving, including the geometry chapters, and AoPS Volume 2.\r\n\r\nI think I might as well finish whatever sections there are on analytic geometry in Volume 2, but I am unsure what to move on to after that. Suggestions?", "Solution_1": "How far does Intro. to Geometry go? (as in how far through AMC 10, AMC 12, and AIME)", "Solution_2": "All the way to basic AIME.", "Solution_3": "Actually, I have seen several difficult AIME problems (some of them like 8+) that can be solved using things from Introduction to Geometry.", "Solution_4": "For Geometry Revisited, which chapters would you recommend the most for AIME-prep? More specifically, would chapters 4, 5, and 6 (transformations, inversive geometry, and projective geometry respectively) help any?", "Solution_5": "Yeah in AoPS Vol. 2, make sure you master the Trig chapter, the Triangles chapter, and the Cyclic Quadrilaterals chapter.\r\n\r\nAlso, along the lines of trig, it's helpful to learn the Complex Numbers chapter (highlight is De Moivre's theorem, if you're focusing on trig).", "Solution_6": "rruscyk said most of the non-trig AIME geometry can be solved using stuff from Intro to Geometry. Although you may want to learn to solve things by coordinates.", "Solution_7": "[quote=\"AwesomeToad\"]rruscyk said most of the non-trig AIME geometry can be solved using stuff from Intro to Geometry. Although you may want to learn to solve things by coordinates.[/quote]\r\nDoesn't Intro to Geometry have a chapter on analytic geometry?", "Solution_8": "[quote=\"AIME15USAMO\"][quote=\"AwesomeToad\"]rruscyk said most of the non-trig AIME geometry can be solved using stuff from Intro to Geometry. Although you may want to learn to solve things by coordinates.[/quote]\nDoesn't Intro to Geometry have a chapter on analytic geometry?[/quote]\r\n\r\nNo, I didn't see one...\r\n\r\nWish it did though...", "Solution_9": "Yes, the second edition of Intro to Geometry does have a section on Analytic Geometry.", "Solution_10": "Chapter 17 of Introduction to Geometry is Analytic Geometry. These are the sections from it.\r\n\r\n17.1 Lines\r\n17.2 Circles\r\n17.3 Basic Analytic Geometry Problems\r\n17.4 Proofs with Analytic Geometry\r\n17.5 Distance Between a Point and a Line\r\n17.6 Advanced Analytic Geometry Problems\r\n17.7 Summary\r\n\r\nI guess, from what limac says, only the second edition has Analytic Geometry. I didn't know AoPS added or cut out chapters from the books, I thought the new editions only looked nicer and they fixed the minor typos.", "Solution_11": "actually the first edition of geo. is much different from the second edition\r\nthere is an addition of chapters on [b]analytic geometry and trigonometry[/b] real important.", "Solution_12": "I'm in a similar situation (trying to make USAMO, with geometry as a weak point).\r\n\r\nI wanted to ask, do you think the geometry in Aops Vol. 2 is sufficient, accompanied with a heavy dose of past problems? I seem to be improving well so far, but I want to be able to tackle the more difficult AIME geo questions (10-15 range). And if not, would the Intro to geometry book be the best to get, assuming I could only get one more book?\r\n\r\nThanks", "Solution_13": "lol, I guess I should have mentioned...I cannot solve any AIME geometry problems. Though I can still manage to get almost every non-geo problem less than 10 right, and one of 11-15 right, this kind of sucks. The reason I am not looking for something like Intro to Geo is because I feel more mathematically mature, and able to attack harder stuff. Though, perhaps I need to humble myself...\r\n\r\nBut anyways, I will give Geometry Revisited a try. If it's too difficult, I'll figure something else out.", "Solution_14": "Well in that case, I can give you some advice since I was in that situation not too long ago.\r\n\r\nDo past problems. \r\n\r\nThis is probably the single most helpful thing to do, especially for your weak points. I learned some theorems I'd never heard of and some approaches that I'd never thought of trying. Doing lots of past AIME geo questions has gotten me to where I can get most of the easy/mid questions right. Personally, I'm wondering what to do from there.", "Solution_15": "The geo chapters in ACoPS were incredibly helpful to me. Go through those.\r\n\r\nIt really doesn't matter what you do, as long as you do a wide range of problems.", "Solution_16": "Isn't that going a little far: I guess for the really late problems but is ACoPS needed for that?", "Solution_17": "Just curious. In what way do you find challenging problems in geometry boring? I thought it was a really great geometry book. Each question teaches new methods of construction. I personally liked it and found it to be quite engaging.", "Solution_18": "[quote=\"worthawholebean\"]The geo chapters in ACoPS were incredibly helpful to me. Go through those.\n\nIt really doesn't matter what you do, as long as you do a wide range of problems.[/quote]\r\nExactly what he said. Plow through ALL the HMMT geometry problems and the easier olympiads that have geometry (canada, ireland etc.). Very helpful.", "Solution_19": "[quote=\"SnowEverywhere\"]Just curious. In what way do you find challenging problems in geometry boring? I thought it was a really great geometry book. Each question teaches new methods of construction. I personally liked it and found it to be quite engaging.[/quote]\r\n\r\nWell...maybe it's just my own motivation and interests. I've never really been fond of drawing stuff. Perhaps the whole advent of Euclidean Geometry just doesn't appeal to me.\r\n\r\nI like analytic geometry quite a bit though.", "Solution_20": "You're not fond of drawing stuff but you like analytic geometry? Isn't that kind of contradicting yourself? :P", "Solution_21": "Well, you have to draw a lot less for analytic...generally, there are fewer auxiliary lines someone has to draw and they don't have to be nearly as accurate (at least not for me)", "Solution_22": "For me, I think that mastering the use of similar triangles, power of a point, cyclic quadrilaterals, and trig identities is VERY important in the AIME. About 80% of AIME geo problems can be solved using those. Personally, I don't think that mastering analytic geo is important, since I've never seen a AIME geo problem that can be most easily solved using analytic methods. Of course, my strong point is Euclidean geo (and I've never been a fan of analytic geo), so I might be mistaken on that point.", "Solution_23": "Last year I was in a similar boat. AOPS 2 geometry and a lot of problems got me to the point where I can solve any geo problem up to 10 and about half of the ones above it. Geometry is about pattern recognition and using things that you've done in similar situations in the past, since it isn't an intuitive subject like algebra where you just do things and the answer should come out. These patterns include:\r\n-Similarity\r\n-Power of a Point\r\n-Cyclic Quadrilaterals\r\n-Angle Chasing\r\n-Basic Trig (Law of sines, cosines, tangents, & sum-to-product)\r\n-Triangle area formulas\r\n-Inradius/Circumradius relations\r\n-Basic transformations (see AOPS 1)\r\nThese are all in AOPS 2. The problems I did were all of the AIME geometry problems from 2000 up till 2009, HMMT problems from 2005 until 2009, and WOOT Problems. If you aren't in WOOT, I found ACOPS geometry really good, although you need to be patient, because they're much harder problems. However, ACOPS really lets you master and appreciate geometry at that level.", "Solution_24": "I'm not in WOOT, but could someone please post some WOOT AIME geometry problems please?", "Solution_25": "What about the trig questions on the AIME? How should you learn trig?", "Solution_26": "I was doing a couple of the HMMT Geo problems and they were helpful.\r\nAre the really hard HMMT problems around AIME level?", "Solution_27": "@mathwizarddude: I don't think we are allowed to post WOOT problems.. but there's the the olympiad resources section of the website that has a bunch of problems\r\n\r\n@Truffles: The AOPS 2 Trig & Triangles Chapters are great- they cover basic definitions, identities, graphing, properties of graphs, sum-to-product, inverse functions, law of sines, law of cosines, law of tangents, and many area formulas that use trig. You can also use 103 problems in trigonometry- it's awesome at trig, but if you have limited time, go with the AOPS route.\r\n\r\n@AlphaBetaTheta: The HMMT problems range from medium to hard AIME- they're the perfect level.", "Solution_28": "@simof: Thanks! 103 Problems in Trigonometry is a great book! I'm using it now :)" } { "Tag": [ "geometry", "3D geometry", "algorithm" ], "Problem": "Any helpful hints, tricks, or tips?", "Solution_1": "the internet is your friend", "Solution_2": "Look at other cubers pages to find the best algorithms. I personally like Bob Burton's, but thats just my opinion.", "Solution_3": "I like wrongway.org, but that's cheating :D . Good if you get stuck though", "Solution_4": "I think that this is the best tutorial for anyone who wants to take up cubing. http://lar5.com/cube/index.html", "Solution_5": "The only reason I disagree with starting by learning Petrus is that not many people do. There's a ton of people using a cross-F2L-LL method, so if you ever need advice, there's plenty of people using the same method as you. That said, Petrus has been proven fast. Anthony Hsu (who got an 11.88 today at Captains Cove) uses Petrus F2L. Others use it as well, though I cannot think of any off the top of my head (besides Lars Petrus). There is no doubt in my mind that Petrus has potential to be good, but I'm not a huge fan.\r\n\r\nmath92: It's hard to give tips when we don't know where you're at. Can you solve the cube? If so, how fast? What method? What algorithms do you know? What are your splits? What do you need advice on?", "Solution_6": "[quote=\"TimReyn\"]The only reason I disagree with starting by learning Petrus is that not many people do. There's a ton of people using a cross-F2L-LL method, so if you ever need advice, there's plenty of people using the same method as you. That said, Petrus has been proven fast. Anthony Hsu (who got an 11.88 today at Captains Cove) uses Petrus F2L. Others use it as well, though I cannot think of any off the top of my head (besides Lars Petrus). There is no doubt in my mind that Petrus has potential to be good, but I'm not a huge fan.\n\nmath92: It's hard to give tips when we don't know where you're at. Can you solve the cube? If so, how fast? What method? What algorithms do you know? What are your splits? What do you need advice on?[/quote]\r\n\r\nI can solve the cube. Average of around 3 minutes. Most of the time is taken between algorithms. I do know some finger tricks and triggers. I need advice on how to get faster. I really want my time down into the 1 min - 1:30's.", "Solution_7": "Recognition comes with practice, and after that, you could start learning new algorithms. Also, a good way to cut off a lot of time is to put the cross on the bottom. This helps decrease recognition time and you can use triggers a lot.", "Solution_8": "[quote=\"nat mc\"]Recognition comes with practice, and after that, you could start learning new algorithms. Also, a good way to cut off a lot of time is to put the cross on the bottom. This helps decrease recognition time and you can use triggers a lot.[/quote]\r\n\r\nI'll try that out next time. How long have you been cubing nat?", "Solution_9": "almost exactly one year", "Solution_10": "[quote=\"nat mc\"]almost exactly one year[/quote]\r\n\r\nI hope I can get as good as you in a year.", "Solution_11": "[quote=\"nat mc\"]Recognition comes with practice, and after that, you could start learning new algorithms. Also, a good way to cut off a lot of time is to put the cross on the bottom. This helps decrease recognition time and you can use triggers a lot.[/quote]\r\n\r\nwhat do you mean by triggers?", "Solution_12": "[quote=\"tjhance\"][quote=\"nat mc\"]Recognition comes with practice, and after that, you could start learning new algorithms. Also, a good way to cut off a lot of time is to put the cross on the bottom. This helps decrease recognition time and you can use triggers a lot.[/quote]\n\nwhat do you mean by triggers?[/quote]\r\n\r\nInstead of turning the U face with a whole hand, you can use one fast motion with your first finger. Its the motion of firing a gun with a trigger.", "Solution_13": "The most common triggers are RUR', (RUR'U')(RUR'), and (RUR'U')(RUR'U')(RUR'), for some F2Ls and also the edge inserter: (RUR'U')(L'U'L) or (L'U'LU)(RUR').\r\nThere is also (RU2R'U')(RUR') and the nice OLL F(RU'R'U')(RUR')F' and many others", "Solution_14": "My left hand is so weak. I can barely do left hand triggers. GAH.", "Solution_15": "so u guys cheated [kind of]... u didnt really figure it out on you own and get this good with your timings?", "Solution_16": "[quote=\"kstan013\"]I like wrongway.org, but that's cheating :D . Good if you get stuck though[/quote]\r\n\r\nHaha I just used that. Thank you so much. :)", "Solution_17": "I started off with Petrus because I didn't know better. Trust me, it's not good. The F2L is basically all intuitive so when you're starting out you know nothing. It's like when people start off learning the cross, they usually struggle for a bit before getting the hang of it. Starting off with Petrus is like 23094825 times that hard. My first solve was after like three days (maybe a week, I can't remember) on a virtual cube and I got 50 minutes something.", "Solution_18": "ha.. i cheated! i got an account on rubiks.com and colored the rubiks cube the way mine looked and then followed their steps! soooo easy...", "Solution_19": "[quote=\"Cyrazeno\"]ha.. i cheated! i got an account on rubiks.com and colored the rubiks cube the way mine looked and then followed their steps! soooo easy...[/quote]\r\n\r\nYeah, thats the same thing as wrongway.org.", "Solution_20": "What really is the point of the Rubic's cube if u cheat on it?\r\n(Just wondering) :huh:", "Solution_21": "for bragging :P", "Solution_22": "beating ur world record times?", "Solution_23": "it was actually made as a complex puzzle that strengthens the mind", "Solution_24": "Hey does anybody know an algorithm that keeps all pieces in place but just flips the orientation of all 12 edge pieces?", "Solution_25": "my best time is 97.04 seconds... i can do triggers but i need to learn finger tricks", "Solution_26": "LAWRENCE!\r\nyeah, i joined AoPs...\r\n\r\nmy best time is 31 seconds, but that was lucky.\r\nactual record is 38 seconds....\r\n\r\nim using the LL of the beginner method of jasmin lee...very effective....\r\n\r\nfor F2L im using my own set of moves...", "Solution_27": "[quote=\"tjhance\"]Hey does anybody know an algorithm that keeps all pieces in place but just flips the orientation of all 12 edge pieces?[/quote]\r\n\r\nR2L2U2B2(y)L2R2\r\nmakes a checkerboard pattern", "Solution_28": "[quote=\"moogra\"][quote=\"tjhance\"]Hey does anybody know an algorithm that keeps all pieces in place but just flips the orientation of all 12 edge pieces?[/quote]\n\nR2L2U2B2(y)L2R2\nmakes a checkerboard pattern[/quote]\r\n\r\nEh I mean something that will end up with every piece in the same place, but I just want the [u]orientation[/u] to be flipped.", "Solution_29": "[quote=\"tjhance\"][quote=\"moogra\"][quote=\"tjhance\"]Hey does anybody know an algorithm that keeps all pieces in place but just flips the orientation of all 12 edge pieces?[/quote]\n\nR2L2U2B2(y)L2R2\nmakes a checkerboard pattern[/quote]\n\nEh I mean something that will end up with every piece in the same place, but I just want the [u]orientation[/u] to be flipped.[/quote]\r\njust do a bunch of different commutators", "Solution_30": "If you actually wanted to learn it yourself, [url]http://www.learn2cube.com/[/url] is a good site." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra theorems" ], "Problem": "How can we prove that the set of all inner automorphisms of a finite group G is a normal subgroup of the set of all automorphisms of G? I proved 'being subgroup' part but how about the 'normality' ?", "Solution_1": "Let $G$ be a group, $\\phi$ an automorphism of, and for all $g \\in G$, $\\theta_{g}$ the inner automorphism $x \\mapsto gxg^{-1}$. Then we have that\r\n\\[\\phi \\theta_{g}\\phi^{-1}= \\theta_{\\phi(g)}\\]by a direct computation, which implies the inner automorphisms are normal in the automorphisms. Note, that I did not use the that $G$ is finite, as it is not needed for this result." } { "Tag": [], "Problem": "What is the fewest number of sundays that there can be in any one calender year?", "Solution_1": "[quote=\"mr. math\"]What is the fewest number of sundays that there can be in any one calender year?[/quote]\r\n\r\n[hide]there are a total of 54 full weeks so its 54.\n\nis this a trick question?[/hide]", "Solution_2": "No, i don't think its a trick question. Sorry, but you're answer isn't correct :( . i checked in the answer key.", "Solution_3": "[quote=\"mr. math\"]No, i don't think its a trick question. Sorry, but you're answer isn't correct :( . i checked in the answer key.[/quote]\r\n\r\n\r\nuugh crap\r\ni divided 365 by 7 wrong so 52. hope i did it rght now :rotfl: \r\nits almost 12 here\r\ni better get som rest :D", "Solution_4": "[hide=\"Sundae!\"]$\\frac{365}{7}=52.142857...$ This means there are 52 weeks and 1 day in a whole year. The most amount of sundays is 53 (if you started the year on a sunday) and the least is 52 (started the year on any other day).\n$\\boxed{52}$[/hide]", "Solution_5": "52 is correct! :lol:" } { "Tag": [ "integration", "calculus", "superior algebra", "superior algebra theorems" ], "Problem": "I know this may sound like a dumb question but I was looking through a math journal and I saw a capitol pi symbol. it looks like a series of some sort, but I have never seen it before. where would I go to study this symbol and learn about it? I say it looks like a series because it has a capitol pi with a k on top and n=4 on the bottom of it followed by something like kx (this is an example). Thanks all", "Solution_1": "See http://en.wikipedia.org/wiki/Multiplication#Capital_pi_notation .", "Solution_2": "thank you so much :)", "Solution_3": "$ \\Sigma$ = Sigma = S is the first letter in \"Sum.\"\r\n\r\n$ \\Pi$ = Pi = P is the first letter in \"Product.\"\r\n\r\nAnd $ \\int$ as in integral is really just another version of S.\r\n\r\nThere are few other symbols you could do this with, such as X (actually $ \\times$) for \"Cartesian Product\" (of sets)." } { "Tag": [], "Problem": "Sophia has 16 plants and a window. Only 5 plants can be put in the window at a time. Every plant gets the same amount of sunlight. What is the most time in the sun a plant can get in an 8 hour period?", "Solution_1": "[hide]Only five plants at any given moment so there are $5 \\cdot 8 = 40$ hours total distributed among 16 plants is $\\frac {40}{16} = 2.5$.[/hide]", "Solution_2": "okay thanks, because almost everyone in my school got that one wrong ;)" } { "Tag": [ "real analysis", "real analysis theorems" ], "Problem": "Let $A$ be connected in $\\mathbb{C}$ and $f: A\\to\\mathbb{Z}$ be a continuous function. Prove that $f$ is constant.", "Solution_1": "since f is continuous and A is connected, f(A) is connected. the only connected sets in Z are singleton sets. hence f is constant" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Find all $n$ s.t. the following polynomial is irreducible over the integers: \r\nx^n - 3 x^(n-1) + 2 x^ (n-2) + 6", "Solution_1": "people please try this problem ! it is not easy and quite nice.", "Solution_2": "The possible integer roots of $P(x)=x^n - 3 x^{n-1} + 2 x^ {n-2} + 6$ are the divisors of the constant term $\\{ \\pm{1}, \\,\\, \\pm{2}, \\pm{3}, \\,\\, \\pm{6}\\}$ .\r\n\r\nWe write $P(x)=x^{n-2}(x-2)(x-1)+6$. Then $P(2)=P(1)=6$ .Also $P(-2), \\, P(3), \\, P(6), \\, P(-3), \\, P(-6)$ are clearly not zero . \r\n\r\nWe only need to check $P(-1)=6[(-1)^{n-2}+1]$. If $n-2$ is odd $P(-1)=0$ but if $n-2$ is even $P(-1) \\neq 0$.\r\n\r\nSo for all $n$ even the polynomial is irreducible over the integers.\r\n\r\n :)", "Solution_3": "sorry for the misunderstanding. \r\nwhen i say irreducible i mean, it cannot be expressed as the product of two polynomials with integer coefficients. \r\nyou haven't proved it is irreducible for $n$ even, only that it has no linear factors, so your proof is wrong.", "Solution_4": "by generalised eisenstein it can not have factors of degree >1\r\n :D", "Solution_5": "sorry what's generalized Eisenstein's? haven't heard of it before. \r\ndoes it say that if polynomials like that one are reducible, then it must have linear factors as well? because trivially for odd n, when you divide that thing by $x+1$ you get a polynomial of degree $n-1$, which can't be reduced into linear factors.", "Solution_6": "one possible generalisation of eisenstein is:\r\n$P(x)=\\sum a_i x^i$ if $p||a_0$ $p|a_i$ for $i=1,2,...,n-k-1$ then at least one of the factors of P has degree not more than k or P is irreducible...", "Solution_7": "that's very interesting, haven't heard of it before. \r\ncould you possibly provide a proof?" } { "Tag": [ "probability", "geometry", "3D geometry", "calculus", "quadratics", "sphere", "tetrahedron" ], "Problem": "Okay, I have assembled a contest (no thanks to most of you, who didn't submit problems grrr). I will post it on this thread tomorrow probably around noon. It's quite long and contains some very hard questions, so I will accept e-mailed solutions up through noon on Sunday, December 7. My e-mail address is\r\nfuzzyjoel at yahoo dot com.\r\nPlease send all of your solutions at once, in order. Do the best you can with the geometry questions without a diagram. Also, along with your solution send me your high-school-math-class-grade-equivalent. Basically, if you are in or have already taken calc, say so. If you are not yet in calc, say how many years you would take calc in your normal system assuming you do not skip any grades. (Thus, if your are a 3rd grader taking a pre-calculus class, you would say \"I will take calculus in one year.\" If you are a 9th grader on track to take calculus senior year, you would say \"I will take calclulus in 3 years.\")\r\nIf you have any questions, I would appreciate if you would post them here as opposed to e-mailing me. Also, try to make everything as easy to understand as possible. Credit will be given according to both valid solution and correct answer, and I will make subjective estimates of how close you got.\r\nQuestions are weighted -- thus, it's worth (a little bit) more to get the harder questions correct. Also, some questions have 2 parts, in which case you get points for each part. The number of points will be written after each question.\r\n\r\nAnd, of course, don't cheat. What constitutes cheating is rather open. However, if you go to http://www.math.unl.edu, you will be cheating. (That is, don't go there. It has many of the questions I'm using, since I am not a question-producing-machine.)\r\n\r\nThere are 55 points worth of questions in 16 questions, plus possible bonuses for one question.\r\n\r\nAnyhow, if you have any questions or objections or complaints or compliments, here is a good place to put them. If you have answers, e-mail is a good place to put them. Yep. I'll post them tomorrow.", "Solution_1": "Um, two questions (ya I say that even when it's online :) ). One, would we get no credit if we accidentally send partial solutions then send the complete one later? What about if we send only the first half or so of the competition in case we don't have time to do them all but sends the rest later? Do we still get credit?\r\n\r\nTwo, can we just PM 'em to you?\r\n\r\nI would contribute problems except they are [b]way[/b] to easy for Intermediate.", "Solution_2": "It's not just intermediate -- it covers ranges from the medium end of Getting Started to the medium range of Advanced, at least in my opinion. If I do this again, submit questions.\r\n\r\nReasons not for PMing -- I don't like the format nearly as much as I like my e-mail format. And I don't know what will happen if you fill up my PM inbox.\r\n\r\nAs for e-mailing, I'm not going to be vicious on grading -- the point is to do the math after all. However, try and have courtesy -- I'm doing this on my personal account in my spare time, and I'm not paying myself to do it :).\r\n\r\nOh, also, please don't send me attached files -- just write it up in the e-mail. I don't want to have to go through the trouble that comes with attached files, which is small, but large enough to annoy me :).", "Solution_3": "Please re-read the guidelines above with respect to these questions. The questions are in no particular order. Have fun!\r\n\r\n1. Given a point O. A circle is drawn around O of radius 2 and a point A of distance 4 from O is chosen. AB is tangent to circle O at B and C is on circle O such that AOC is a straight line. Compute the areas of triangles BCO and ABO. [2]\r\n\r\n2. The quadratic equation x2 + rx + s = 0 has non-zero roots r and s. Compute the ordered pair, (r, s). [2]\r\n\r\n3. A room contains ten people wearing badges numbered 1 to 10. What is the probability that if three persons are chosen at random, the smallest of their badge numbers would be 5? [3]\r\n\r\n4. A circle is inscribed in a 2-by-2 square. Outside this circle and inside the square a second circle is drawn which is tangent to the first circle and to two sides of the square. Find the radius of the second circle. [3]\r\n\r\n5. [Note: re-worded due to original bad wording pointed out by TripleM]\r\nJoel lives with his roommate. Joel has a 70% chance of catching the flu if his roommate does not. If Joel's roommate catches the flu, Joel's probability of catching the flu will rise to 90%. Joel's roommate has a 70% chance of catching the flu. What is the total probability Joel will catch the flu? What if both Joel and his roommate get a flu shot, which will lower each of their probabilities of catching the flu to 1/4th their previous value under any circumstances? [2, 2]\r\n\r\n6. Prove: it is possible to draw 6 congruent circles on the surface of a sphere such that each intersects 4 others and any 2 intersecting circles intersect in exactly 1 point.\r\nIt is also possible to draw m congruent circles on the surface of a sphere such that each intersects exactly n others for 4 other pairs (m, n). Find these pairs. (There is no need to prove this part.) [4, 2]\r\n\r\n7. What is the probability that if 3 regular six-sided dice are rolled at least 1 shows a 6? [3]\r\n\r\n8. Four spherical oranges are stacked with three tangent oranges forming the base and the fourth centered on top. If each orange has radius 1, how high is the stack? What is the volume of the smallest tetrahedron containing all 4 oranges? [4, 4]\r\n\r\n9. If n is a positive integer and n(n + 1) is divided by 3, what are all possible values for the remainder? What if n(n + 2) is divided by 5? [2, 2]\r\n\r\n10. Given a circle whose area would be doubled if its radius were to be increased by 1. What is its current radius? [2]\r\n\r\n11. If x0.3 = 10, then x0.4 equals what? [2]\r\n\r\n12. [Note: reworded so that the question is slightly different than it used to be.]\r\nA chest consists of a rectangular box and a half-cylinder lid. If the box is 3 by 5 by 2 feet (length by width by height), what are both possibilities for its total volume (counting the lid)? [2]\r\n\r\n13. Five distinct lines are drawn in a plane. The number of distinct points where two or more lines intersect is determined. What are all possible values for this number? [3]\r\n\r\n14. Given that a line segment with 2 endpoints is the 1-D analog of a square and a single point is the 0-D analog. Define P(a shape) to be the infinite sum,\r\nP(shape) = number of vertices + sum of edge lengths + surface area + 3-D volume + 4-D volume + ...\r\nFor example:\r\nP(a point) = 1 + 0 + 0 + ... = 1\r\nP(a segment of length 3) = 2 + 3 + 0 + 0 + ... = 5\r\nP(an equilateral triangle of side s) = 3 + 3s + s2:rt3:/4 + 0 + 0 + ... = 3 + 3s + s2:rt3:/4\r\nFind a pattern in P(an n-dimensional cube of side s) by examining the smallest 4 cases (n = 0 to 4). Use this pattern to compute the 3-volume of a 5-dimensional cube.\r\nHow many of the individual measurements of a 4-dimensional cube that this method gives you can you justify without using the pattern? [3 + possible bonuses don't worry too much about the second part. It's probably not worth the work it will take you.]\r\n\r\n15. A party of 18 people went to a restaurant. The each chose a $21 meal but 4 of them forgot to bring their money. In order to pay the total bill, everyone else who brought their money had to pay how much extra? [2]\r\n\r\n16. A 6 by 6 checkerboard is covered with 18 dominoes (each covering two squares). Prove that it is possible to cut the board either horizontally or vertically into two (not necessarily equal) pieces without cutting through any dominoes. (Hint: first show that any horizontal or vertical slice cutting one domino has to cut two dominoes.) [4]\r\n\r\n17. A car traveling at 50 miles per hour covers the distance between two towns in 6 hours less than a truck which travels at 35 miles per hour. How many hours does it take the car to cover the distance? [2]", "Solution_4": "A couple of questions: I'll put them in spoiler just in case it would give anything away, it shouldn't.\n\n\n\n[hide]\n\nQuestion 5. I don't think this really makes sense.. not to me anyway. It says each have a 70% probability of catching the flu. As far as I can tell. So the total probability of Joel catching the flu is 70%. The other thing can't be taken into account.. maybe its the first sentence thats wrong, I don't know, its just confusing.\n\n\n\nQuestion 12. Which orientation is the half-cylinder lid? It could be either way, both giving different results. \n\n[/hide]", "Solution_5": "Good points, both. Although you should have been able to figure out what I meant fornumber 5, I think :).\r\nI didn't write number 12, so I can only imagine that there must have been some sort of picture that accompanied it. Well, whatever.", "Solution_6": "will the individual scores be published or kept private?", "Solution_7": "Need we show work... When are these due?", "Solution_8": "yes, work, and they are due noon, dec. 7, as it says up there.", "Solution_9": "JBL - Don't we need some sort of restrictions on number 2?", "Solution_10": "Matt, read #2 again, carefully", "Solution_11": "All of the basic rules (what to send in, when to send it in, where to send it to) are up there on that long first message. I understand that this is a math forum, and we don't all necessarily enjoy reading that much, but, I mean . . .\r\n\r\nAnd number 2 is just fine the way it is, thanks.\r\n\r\n\r\nWow, those responses were both rather mean. I'm sorry.\r\n\r\nActually, as regards the first question, I would like to reinforce that I would like *full solutions* for every question, including those that only ask for some kind of number. Now, you don't necessarily have to tell me how you figured out that solution. For example, on number 13, it is acceptable merely to list one way to achieve each of the possible values and give a brief line or two explaining why you can't reach any others. However, you will not get much (or any) credit for just saying, \"The answer is 37.\" (By the way, the answer is clearly not 37.)\r\n\r\n\r\nAnd with regards individual scores -- how would you like me to report them? Each person's score to him or her via e-mail, or should I put everyone's scores up here on the forum?\r\n\r\nIn fact, how about this -- if you don't want your score to end up on the forum, please tell me in your e-mail to me that contains your answers. Also, don't forget to write how long it will be until you take calculus. (If that doesn't make sense, go read the first message at the top of this thread again.)", "Solution_12": "could you pick out some interesting solutions and post them, or have others post their solutions? i did a good deal of these but dont plan to submit (based on time); i'd like to see if they were right.", "Solution_13": "\"based on time\"\r\n\r\nYou have, as of the time this was written, one week, 11 hours and 43 minutes. 46 minutes if I go by my watch, not my computer. Anyhow, that should be ample time to write up solutions to 17 problems.\r\n\r\nI certainly have no problem with opening this thread up to solutions after everyone has submitted theirs to me. I could also produce the \"best of\" solutions to the 17 problems. That would probably be more space-and-time efficient.", "Solution_14": "Do I have to prove? or can i just show it can be done on #6?", "Solution_15": "Showing it can be done is the same thing as proving. Just don't make any unverifiable claims, that's all. If it makes sense to you, and there's nothing you say that you then think, \"Why is that true?\" you'll probably be on safe ground.", "Solution_16": "Don't forget -- only 62 more hours left . . .", "Solution_17": "Well, I have to admit that I was rather annoyed that only 7 of you submitted any responses, and only 5 of you submitted something resembling a full set of solutions. Final scores:\r\n\r\nBugzpodder 47\r\nChinaboy 41\r\nPaladin8 39\r\nFierytycoon 24\r\nBookworm271828\r\nTopper and Confuted 6\r\n\r\n(out of a possible 55)", "Solution_18": "thanks for making this contest!", "Solution_19": "gah, i forgot all about this test, will you be posting solutions, or shall we go ahead and post for random problems.", "Solution_20": "ohhh, i missed the contest. unfortunately i had no computer until last monday, so it was difficult for me to submit solutions. Joel, if you arrange a second contest, i promise to participate, and im sure a lot more people will take part in it, especially if you warn us all a lot before the solutions are due. Also, i like that problem with the n-cubes, i thought it was really cool. where is it from?", "Solution_21": "what was the pattern to that anyways???", "Solution_22": "Would you like me to post solutions? Also, if you want me to grade it, I will, if you send it rather quickly (next couple of days).\r\n\r\nThe cube problem was my own invention, actually, although I don't claim originality, just uninfluenced discovery.", "Solution_23": "Yah, please post solutions. Thanks in advance.", "Solution_24": "yeah, i have waaaaay too much work to do to achieve a decent score, Chinaboy can vouch for me, and himself. lolo. but he did the contest earlier unlike a certain syntax error i know...\r\n\r\nso post the solutions ahoy, and thanks you too", "Solution_25": "Partial solutions:\r\nThese are solutions from the sets that various people have submitted, edited slightly. I will suppliment them with my own solutions, or possibly solutions of other solvers, in the next few days.\r\n\r\nSolutions:\r\n\r\n1. Since AB is tangent to circle O, it must make a right angle. Also, since the radius is 2 and AO is 4, it makes a 30-60-90 triangle with area 2*sqrt(3). Now, since angles BOC and AOB make a straight line and AOB is 60, COB must equal 120. You can break triangle COB into two 30-60-90 triangles with hypotenuse 2. Therefore, they have area sqrt(3)/2 and triangle COB has area sqrt(3). (Due to Paladin8)\r\n\r\n2. x^2 + rx+s = (x - r)(x - s)\r\n(x - r)(x - s) = x^2 - (r + s)x + rs\r\nrs = s\r\nr = 1 (since s is not 0)\r\n-r - s = r\r\n-s = 2r = 2\r\ns = -2\r\n(due to Confuted)\r\n\r\n3. We can choose the people corresponding to the numbers 5, a, b, where a,\r\nb are in the set {6, 7, 8, 9, 10}, since each element in the set is greater than 5. Thus our probability p = C(5, 2)/C(10, 3) = 10/120 = 1/12. (Due to Fierytycoon)\r\n\r\n5.a. 0.7 of the time your roomate gets the flu, and then you get it 0.9 of the time. 0.7*0.9=0.63. \r\n0.3 of the time he doesn't get the flu, and then you get it 0.7 of the time. 0.3*0.7 = 0.21. The sum of 0.63 and 0.21 is 0.84 = 84%\r\n(Due to topper)\r\nThe second part is identical, and we get a final answer of 18.375%.\r\n\r\n6 (part 1). Imagine a cube. Now imagine all the square faces become circles. Now there are 6 circles tangent to 4 others each. Just inflate it and fill in the missing space and you get a sphere.\r\n6 (part 2). The method mentioned above only works for regular polyhedra and there are only 5, so the other 4 are: tetrahedron (4,3) ; octahedron (8,3) ; dodecahedron (12,5) ; icosahedron (20,3). (Due to Paladin8)\r\nI thought this problem up myself -- isn't it pretty cool?\r\n\r\n7. The probability it doesn't show a 6 is 5^3/6^3 so he probability it will show at least one 6 is 1 - 125/216 = 91/216 (Due to Chinaboy)\r\n\r\n10. 2*r :^2:* :pi: = (r + 1) :^2:* :pi:, 2r :^2: = (r + 1) :^2:, r :^2: - 2r - 1 = 0, so r = :rt2: + 1 since we only want the positive root. (Due to bookworm271828)\r\n\r\n11) x^(3/10) = 10 <=> x = 10^(10/3). Then x^(4/10) = (10^(10/3))^(4/10) = 10^(4/3). (Due to Fierytycoon)\r\n\r\n\r\n13) If we move from two distinct lines into two parallel lines, we lose an intersection. If we shift a line until two intersections coincide, (colinear) we lose another. However, this transformation stops at 4.\r\n0 - all parallel\r\n1- all collinear\r\n2- not possible\r\n3- not possible\r\n4- 4 parallel, 1 distinct\r\n5- 4 collinear, 1 distinct\r\n6- 3 parallel, 2 parallel\r\n7- 3 parallel, 2 distinct\r\n8- 2 parallel, 2 parallel, 1 distinct\r\n9- 2 parallel, 4 distinct\r\n10- no 3 collinear, no 2 parallel\r\n(Due to Bugzpodder)\r\nCan you find alternate methods for any of these values?\r\n\r\n\r\n15. There are 14 paying people (18 - 4), and a bill of 378 (18*21). 378/14 = 27. So, each person pays $6 extra (27 - 21). (Due to bookworm271828)\r\n\r\n17. d = distance between two towns \r\nd/50 + 6 = d/35\r\nso d = 700... so it takes the car \r\n700/50 or 14 hours to cover the distance. (Due to Chinaboy)", "Solution_26": "thank you Mr. Joel", "Solution_27": "wow, thanks", "Solution_28": "did anyone get number 8? that was hard...", "Solution_29": "8a was gotten by Chinaboy and Paladin8. Bugspodder got partial credit on 8a and b because he did something dumb on 8a and didn't bother to actually tell me the answer in 8b so I had to spend 5 minutes checking whether he was correct or not.\r\n\r\n\r\n8. First, let us just consider a regular tetrahedron ABCD with A the top vertex. Dropping an altitude from A to plane BCD, we see in intersects the triangle exactly in its center. Let us call that point G. Then AG :^2: + GB :^2: = AB :^2: by the Pythagorean theorem. We can find (I don't want to because I'm lazy) that if the side of the tetrahedron is s, GB = s/ :rt3:. My wrists are starting to hurt, so I shall finish this later. Anyone else, feel free to carry it out if you want.\r\n\r\n14. Let's examine the polynomials.\r\nP(point) = 1\r\nP(segment length s) = s + 2\r\nP(square side s) = s :^2: + 4s + 4\r\nP(cube side s) = s :^3: + 6s :^2: + 12s + 8\r\n\r\nNow, am I the only one who notices that these correspond to (s + 2)^n for n = 0, 1, 2, and 3? Well, anyhow, they do. Thus, we can guess that for a 5D cube, we will have (s + 2) :^5: = s :^5: + 10s :^4: + 40s :^3: + 80s :^2: + 80s + 32, giving us a 3-volume of 40. Now, for the hypercube (4D), the pattern gives us s :^4: + 8s :^3: + 24s :^2: + 32s + 16. Let's see how much we can justify otherwise:\r\n\r\nWell, the first coefficient comes very easily, since it's roughly the definition of 4-volume. Now, we have 4 axes, and we need a face (a cube) for the positive and negative ends of each axis (imagine a cube or a square centered at the origin of a space or plane). That gives us 8 cubes. Faces I don't know how to explain, but this is what Bugzpodder said about it: \"There are 4 more ways to rotate the hypercube, which makes the number of faces 6*4=24.\" I don't really know if that makes sense. Anyhow, we can think of a hypercube as two cubes joined at their vertices (just like a cube is two squares joined at their vertices), so we easily get that the number of vertices is doubled and the number of edges is doubled, plus the extra edges necessary to connected the old cube to the new. (We actually can get the other measurements this way, too -- for the 24 plane faces, we have 6 in the original cube and 6 in the final cube, as well as 12 formed by dragging the edges of the cube in a 4th-dimensional direction. For the 8 cubes, we have the initial cube and final cube, as well as the cube formed by dragging each face of the original cube in a 4th dimensional direction. This process ends up being a recursive way to find the same general formula as the pattern I asked you to find.)", "Solution_30": "[quote]\nBugspodder got partial credit on 8a and b because he did something dumb on 8a and didn't bother to actually tell me the answer in 8b so I had to spend 5 minutes checking whether he was correct or not. \n[/quote]\r\nerr, sorry", "Solution_31": "Is there going to be another contest?" } { "Tag": [ "geometry", "cyclic quadrilateral", "real analysis", "real analysis unsolved" ], "Problem": "Find the area of Quadrilateral (irregular four sides shape) with sides length $ 72,23,55,59$. ?\r\n\r\n :)", "Solution_1": "It depends on the angles. Ask a different question.", "Solution_2": "Then I can't solve till I know angles .\r\n\r\nI just wanna be sure only .\r\n\r\nThank you jmerry", "Solution_3": "The area of an arbitrary quadrilateral with sides $ a,b,c,d$ is less than or equal to the area of a cyclic quadrilateral with those sides. The area of the cyclic quadrilateral is given by Brahmagupta's formula, $ \\sqrt {(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)(s \\minus{} d)},$ where $ s \\equal{} \\frac {a \\plus{} b \\plus{} c \\plus{} d}2.$", "Solution_4": "Additionally you should specify the diagonal lengths \r\n\r\nor two opposite angles of the quadrilateral.\r\n\r\nThen you can compute the area with [url=http://mathworld.wolfram.com/BretschneidersFormula.html]Bretscheider's Formula[/url]." } { "Tag": [ "inequalities", "FTW", "Vieta", "pigeonhole principle", "trigonometry", "induction", "ratio" ], "Problem": "How many problems does the topper of KSU usually get in the second level?\r\n\r\nDoes solving 3 problems ensure u are a winner?\r\n\r\nBefuddlers", "Solution_1": "Uh, I'm pretty sure you'll need more than that to beat the top people in the state. Don't bother with a certain numeric goal for KSU if you're participating; just do as best as you can and keep writing pretty proofs as the goal in mind while competing in the second round. :wink:", "Solution_2": "Well, last year I got about four problems, the year before that I got all five, and before that there was no KSU tournament. :) I don't know what the cutoff is usually like for honorable mention recognition, but it's usually around the top 30+.\r\n\r\nBy the way, I'd say getting the job done is usually more important than pretty proofs in competitions, though you should try to make all your proofs as nice as possible when you don't have severe time pressure. I think there's usually one question on this tournament that you can get if you chug away with determination for as long as it takes. Good style doesn't hurt, though, and bad style is often a symptom of an imperfectly done job.", "Solution_3": "Since all 2nd round tests must be postmarked by February 7th, please hold all discussion in this forum until Sunday, February 8th.\r\n\r\nThanks!\r\nTom", "Solution_4": "Last year I got 2.5, I think, but I was sufficiently high off my 144 last year so that only Miles moved above me.", "Solution_5": "The phrase \"sufficiently high off my 144\" sounds like 144 is some sort of illegal drug. :wink: \r\n\r\nAnyway, as to the whole stressing about how many to solve to get whatever-th place, with two hours and five problems, you are very likely to solve the same number of problems regardless of how much you stress about it, as long as you spend a reasonable amount of time on each problem. So, I would just relax, have fun, and remember not to show up 1 hour late like I did 2 years ago. :ewpu: :fail:", "Solution_6": "I think that it's ok to talk about it now. Charlie emailed me asking for a link to this forum. Feel free to talk about what you liked and didn't like.", "Solution_7": "Hmm, I thought this year's proof round was easier than last year's, but I guess I shouldn't say that cuz I got number 3 wrong...\r\n\r\n[hide=\"Number 1\"]: Mods pwn this: 24.[/hide]\n\n[hide=\"Number 2\"]: Heh, I used the product of tangents equals sum of tangents identity from last year to solve this is in a very roundabout fashion. I'm pretty sure there was an easier way to do this: 2.\n[/hide]\n\n[hide=\"Number 3\"]: Blargh... Frisbeeballer (not mentioning his real name) said you could use inequalities to solve this, but I didn't solve it, so fail...[/hide]\n\n[hide=\"Number 4\"]: Algebra-bashing with Stewart's ftw[/hide]\n\n[hide=\"Number 5\"]: Vieta's and a bit of brute-forcing for the win: can't remember the triples, but there were three, I think (correct me if I'm horribly wrong)[/hide]\r\n\r\nSo yeah, I'll probably place higher than I did last year cuz I had four legit proofs instead of one and a 144 on the first round instead of $ \\minus{} x$, where x is over 9000...\r\n\r\nThen again, this year's was kinda simple, so other people might all get 5's and 150s, so nvm...", "Solution_8": "Hm so I did #3 with pidgeonhole, possibly losing some points. Otherwise I made [decently] pretty solns.\r\n\r\n(#5: we get $ \\frac{1}{r_1}\\plus{}\\frac{1}{r_2}\\plus{}\\frac{1}{r_3}\\equal{}\\minus{}1$, methinketh?)\r\n\r\n(Edward was the only 150, was he not?)", "Solution_9": "[quote=\"Math Geek\"] Heh, I used the product of tangents equals sum of tangents identity from last year to solve this is in a very roundabout fashion. I'm pretty sure there was an easier way to do this: 2.\n\n\n Blargh... Frisbeeballer (not mentioning his real name) said you could use inequalities to solve this, but I didn't solve it, so fail...\n[/quote]\r\n\r\nYes the second problem had an easier solution i think:\r\n\r\n$ cosC*cosB$ = $ cos A$\r\n$ cosC*cosB$ = $ cos(180 \\minus{} B \\minus{} C)$\r\n$ cosC*cosB$ = $ \\minus{}cos(B\\plus{}C)$\r\n$ cosC*cosB$ = $ \\minus{}cosB*cosC \\plus{} sinB*sinC$\r\n$ 2cosC*cosB$ = $ sinB*sinC$\r\n$ tanB*tanC$ = $ 2$\r\n\r\nThe third problem could be solved by pigeon hole principle which falls into the category of combinatorics rather than inequalities. :D Yeah the third problem was just Erdos and Szekeres theorem.\r\n\r\nEDIT: Solafidefarms yes we get the expression you just stated. But by bounds you get three solutions i think\r\n\r\n$ (10,32) ; (9,27) ; (11,36)$", "Solution_10": "Are the problems available on internet?\r\n\r\nAlso, how do you register for this competition? (I know it's probably too late but I want to do it next year.)", "Solution_11": "Yeah, that's what I did for number 5, Billy. For number 2, I was kinda nervous at the time I was solving it, and cuz I looked at last year's solutions the night before, I was sorely tempted to prove the identity and then use it... In the end I used the sine of the sum of two values identity though, so I guess my proof was just a roundabout version of yours.\r\n\r\nSo I'm assuming from all of these posts that pretty much everyone aced proof round...except for me... No wonder I thought it was so easy... :) \r\n\r\nCan anyone post their proof for number 3. I hate these logic problems and need to get better at them...\r\n\r\nEDIT: I almost forgot. It turns out I did attempt number 3. Here's my proof:\r\n\r\n[hide]It is commonly known that a person's height will never remain constant throughout his lifetime. Whether he be approaching the glorious peak of adulthood or the bottomless chasm of old age and senility, his height will either increase or decrease, respectively, for the whole of his life. Ergo, if one lines up $ k^2\\plus{}1$ boys shoulder to shoulder and selects a group of $ k\\plus{}1$ boys to step forward, their heights will consistently increase or decrease, more likely the former because, after all, they're just boys! $ \\blacksquare$[/hide]\r\n\r\nI'll get points for the creativity of my proof, rest assured!", "Solution_12": "Grr, On #5, I accidentally overlooked the $ a \\neq 1$ restriction, and put $ (1 , \\minus{} t^2)$ for all integers $ t \\ge 1$, as a possibility. \r\n\r\nI couldn't get #3, but I did fine on #1, #2, and #4, so I'm going to guess that I got a 34-38 out of 50.", "Solution_13": "Did fine on #1, #2, and #4, and made a false assumption on #5 to get 2 solutions. Failed #3. \r\n\r\nWhoever from Walton who's likely to have a good chance all did 3 or 4 problems.\r\n\r\nEdward got a 150, and I think he was the only one.\r\n\r\nHere's the [url=http://math.kennesaw.edu/~ckoppelm/]KSU site[/url] from the GCTM site.", "Solution_14": "Sitan's boys in line proof FTW!!! :rotfl: \r\n\r\nI think I got 4 then...mebbe 4.1. xD", "Solution_15": "Umm, Arvind, are you sure induction works for #3? :P", "Solution_16": "Okay, here is the problem 3 solution. I think this was the hardest question on the test. It is an early result, I think, in the study of pattern-avoiding permutations.\r\n\r\n[hide]\nWe induct on $ n$. For $ n\\equal{}0$ the problem statement is trivial.\n\nNow, suppose the problem statement holds for $ n \\minus{} 1$. Then there is a sequence of $ n$ boys whose heights are monotonically increasing or decreasing. Call the boy at the end of this sequence $ a_1$. Now consider the $ n^2$ boys other than $ a_1$. By inductive hypothesis, there is another boy, $ a_2$, who is at the end of another monotonic sequence. We may continue this process to obtain $ 2n\\plus{}1$ boys who are at the end of monotonic height sequence. Without loss of generality, let at least $ n\\plus{}1$ of them be at the end of increasing height sequences; we will call these the [i]tall[/i] boys. If any one of the tall boys is taller than a tall boy who comes before him, then we may add him to that boy's sequence to obtain a sequence of $ n\\plus{}1$ boys in increasing height order. Otherwise, the tall boys form a sequence of boys in decreasing order! $ \\blacksquare$\n\nWe note also that the problem's bound is sharp. In particular, if we have $ n^2$ boys, who are (in height order) $ a_{11}, a_{12}, \\dotsc, a_{1n}, a_{21}, a_{22} , \\dotsc, a_{nn}$, then the order\n\\[ a_{n1}, a_{n2}, \\dotsc, a_{nn}, a_{(n\\minus{}1)1}, a_{(n\\minus{}1)2}, \\dotsc, a_{(n\\minus{}1)n}, \\dotsc, a_{11}, \\dotsc, a_{1n} \\]\ncontains no $ n\\plus{}1$ boys in monotonic height order. Furthermore, we may add a boy taller than any of the others between $ a_{nn}$ and $ a_{(n\\minus{}1)1}$ to obtain a sequence of $ n^2\\plus{}1$ boys in which there is only one sequence of $ n\\plus{}1$ boys in monotonic height order.[/hide]\r\n\r\nAlso, for problem 4, I think the easiest approach is to show that the cosine of the triangle's largest angle is the cosine of twice the cosine of the triangle's smallest angle. Therefore the triangle's largest angle is twice the triangle's smallest angle. The problem statement then follows from AA similarity.", "Solution_17": "[quote=\"Math Geek\"]Umm, Arvind, are you sure induction works for #3? :P[/quote]\r\n\r\ndefinitely not. but hey, at least it sounds like a solution might have happened. maybe. xDDD", "Solution_18": "Easiest for number 4 in my opinion : \r\nTriangle ABC, bisection creates point D. Let AB=4x, BC=5x, and AC=6x. So, by angle bisector theorem, AD=8/3x, DC=10/3x. Then, AB/AD=4/(8/3)=6/4, and $H_2$ + $Cl_2$ + $NaOH$))) , but not $2H_2$ + $O_2$ + $NaCl$ ?\r\n\r\n3/ While checking the migration of ions, why should we moisten the paper with sodium sulphate, but not water or others? Why?", "Solution_1": "1) Just because 2 ions have some colours in an aqueous solution doesn't mean the resulting compund will have the combination of colours, Colour of anything is related to the electonic transitions which result when a molecule absorb photons. So the color of the new cmpd. will be determined by the configuration of Molecular Orbitals(and other effect) which is very very difficult to determine exactly. I believe you can't simple determine the color!!!", "Solution_2": "2) $Na^ + $ has a great affinity for $ OH ^ - $ . You can explain this on the basis of $ \\delta G $ or any other reason(electrostataic forces) Thus, as soon as the ions are formed, this reaction takes place.\r\n\r\n\r\n3) I didn't understand this. Can you plz explain the context", "Solution_3": "1). my chem. teacher said that most copper compounds are blue. so blue would be my guess.", "Solution_4": "I would guess it would be a blend of copper's blue and permanganate's purple.\r\n\r\nIt has to do with the overvoltage of the oxidation of chloride, compared to that of hydroxide.\r\n\r\nI don't quite understand the question - does this refer to a salt bridge in a voltaic cell? Chromatography? Something else?" } { "Tag": [ "induction", "linear algebra", "matrix", "trigonometry", "linear algebra unsolved" ], "Problem": "Let $ A,B\\in M_{2}(R)$ so that $ (A\\plus{}B)^n\\equal{}A^{n}\\plus{}B^{n},\\forall n\\geq 3$. Prove that $ AB\\equal{}BA\\equal{}O_{2}$.", "Solution_1": "$ (\\mathbb{A}\\plus{}\\mathbb{B})^2 \\equal{} \\mathbb{A}^2 \\plus{} \\mathbb{B}^2$\r\n$ \\iff (\\mathbb{A}\\plus{}\\mathbb{B})^2 \\minus{} (\\mathbb{A}^2 \\plus{} \\mathbb{B}^2) \\equal{} \\mathbb{O}$\r\n$ \\iff (\\mathbb{A}^2 \\plus{} \\mathbb{AB} \\plus{} \\mathbb{BA} \\plus{} \\mathbb{B}^2) \\minus{} (\\mathbb{A}^2 \\plus{} \\mathbb{B}^2) \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{AB} \\plus{} \\mathbb{BA} \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{AB} \\equal{} \\minus{} \\mathbb{BA}$\r\n\r\n\r\n$ (\\mathbb{A}\\plus{}\\mathbb{B})^3 \\equal{} \\mathbb{A}^3 \\plus{} \\mathbb{B}^3$\r\n$ \\iff (\\mathbb{A}\\plus{}\\mathbb{B})^3 \\minus{} (\\mathbb{A}^3 \\plus{} \\mathbb{B}^3) \\equal{} \\mathbb{O}$\r\n$ \\iff (\\mathbb{A}^3 \\plus{} \\mathbb{ABA} \\plus{} \\mathbb{BAA} \\plus{} \\mathbb{BBA} \\plus{} \\mathbb{AAB} \\plus{} \\mathbb{ABB} \\plus{} \\mathbb{BAB} \\plus{} \\mathbb{B}^3) \\minus{} (\\mathbb{A}^3 \\plus{} \\mathbb{B}^3) \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{ABA} \\plus{} \\mathbb{BAA} \\plus{} \\mathbb{BBA} \\plus{} \\mathbb{AAB} \\plus{} \\mathbb{ABB} \\plus{} \\mathbb{BAB} \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{ABA} \\minus{} \\mathbb{ABA} \\plus{} \\mathbb{BBA} \\minus{} \\mathbb{ABA} \\plus{} \\mathbb{ABB} \\minus{} \\mathbb{ABB} \\equal{} \\mathbb{O} \\ \\ \\ (\\because \\mathbb{AB} \\equal{} \\minus{} \\mathbb{BA})$\r\n$ \\iff \\mathbb{BBA} \\minus{} \\mathbb{ABA} \\equal{} \\mathbb{O}$\r\n$ \\iff (\\mathbb{B}\\minus{}\\mathbb{A})\\mathbb{BA} \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{B}\\minus{}\\mathbb{A} \\equal{} \\mathbb{O} \\ \\text{or} \\ \\mathbb{BA} \\equal{} \\mathbb{O}$\r\n\r\n$ \\boxed{\\text{Case} \\ \\mathbb{B}\\minus{}\\mathbb{A} \\equal{} \\mathbb{O}}$\r\n$ \\mathbb{AB} \\equal{} \\minus{} \\mathbb{BA}$\r\n$ \\iff \\mathbb{A}^2 \\equal{} \\minus{} \\mathbb{A}^2$\r\n$ \\iff 2\\mathbb{A}^2 \\equal{} \\mathbb{O}$\r\nBut, $ \\mathbb{A}\\in M(2; \\mathbb{C})$\r\n\r\n$ \\boxed{\\text{Case} \\ \\mathbb{BA} \\equal{} \\mathbb{O}}$\r\n$ \\mathbb{AB} \\equal{} \\minus{} \\mathbb{BA}$\r\n$ \\iff \\mathbb{AB} \\equal{} \\minus{} \\mathbb{O} \\equal{} \\mathbb{O}$\r\nTherefore, $ \\mathbb{AB} \\equal{} \\mathbb{BA} \\equal{} \\mathbb{O}$\r\n\r\n\r\nSince calculate, the given statement is clearly true when $ n\\equal{}3$.\r\nNow we assume it is true for an arbitrary positive integer $ k>3$ : $ (\\mathbb{A}\\plus{}\\mathbb{B})^k \\equal{} \\mathbb{A}^k \\plus{} \\mathbb{B}^k$\r\nThen,\r\n$ (\\mathbb{A}\\plus{}\\mathbb{B})^{k\\plus{}1} \\minus{} (\\mathbb{A}^{k\\plus{}1} \\plus{} \\mathbb{B}^{k\\plus{}1})$\r\n$ \\equal{} (\\mathbb{A}\\plus{}\\mathbb{B})^k(\\mathbb{A}\\plus{}\\mathbb{B}) \\minus{} (\\mathbb{A}^{k\\plus{}1} \\plus{} \\mathbb{B}^{k\\plus{}1})$\r\n$ \\equal{} (\\mathbb{A}^k\\plus{}\\mathbb{B}^k)(\\mathbb{A}\\plus{}\\mathbb{B}) \\minus{} (\\mathbb{A}^{k\\plus{}1} \\plus{} \\mathbb{B}^{k\\plus{}1})$\r\n$ \\equal{} (\\mathbb{A}^{k\\plus{}1}\\plus{}\\mathbb{B}^{k\\plus{}1}\\plus{}\\mathbb{A}^k\\mathbb{B}\\plus{}\\mathbb{A}\\mathbb{B}^k) \\minus{} (\\mathbb{A}^{k\\plus{}1} \\plus{} \\mathbb{B}^{k\\plus{}1})$\r\n$ \\equal{} \\mathbb{A}^k\\mathbb{B}\\plus{}\\mathbb{A}\\mathbb{B}^k$\r\n$ \\equal{} (\\minus{}1)^{k\\minus{}1}\\mathbb{B}\\mathbb{A}^k\\plus{}\\mathbb{A}\\mathbb{B}^k$\r\n$ \\equal{} (\\minus{}1)^k\\mathbb{A}\\mathbb{B}\\mathbb{A}^{k\\minus{}1}\\plus{}\\mathbb{A}\\mathbb{B}\\mathbb{B}^{k\\minus{}1}$\r\n$ \\equal{} \\mathbb{A}\\mathbb{B}((\\minus{}1)^k \\mathbb{A}^{k\\minus{}1}\\plus{}\\mathbb{B}^{k\\minus{}1})$\r\n$ \\equal{} \\mathbb{O}$\r\n($ \\because \\ \\mathbb{A}\\mathbb{B} \\not\\equal{} \\mathbb{O} \\Rightarrow (\\mathbb{A}\\plus{}\\mathbb{B})^3 \\not\\equal{} \\mathbb{A}^3 \\plus{} \\mathbb{B}^3$)\r\nThus the formula works for $ n\\equal{}k\\plus{}1$.\r\nSo, by PMI, the statement is true for every integer $ n \\ge 3$.", "Solution_2": "[quote]$ (\\mathbb{A} \\plus{} \\mathbb{B})^2 \\equal{} \\mathbb{A}^2 \\plus{} \\mathbb{B}^2$[/quote]\r\nThis is not necessarily true since the property from the hypothesis is true only for $ n\\geq 3$.\r\nI also do not think that the induction part is necessary, maybe only if you are trying to prove the converse(which is easier with the binomial theorem).", "Solution_3": "The hypothesis that these are $ 2\\times 2$ is essential. A $ 3\\times 3$ counterexample: $ \\begin{bmatrix}0&1&0\\\\0&0&0\\\\0&0&0\\end{bmatrix}$ and $ \\begin{bmatrix}0&0&0\\\\0&0&1\\\\0&0&0\\end{bmatrix}$.\r\n\r\nThe field also matters: $ A\\equal{}B\\equal{}I$ is a counterexample in characteristic 2.\r\n\r\nAny argument is guaranteed to be grubby.", "Solution_4": "Since calculate, the given statement is clearly true when $ n\\equal{}3$.\r\nNow we assume it is true for an arbitrary positive integer $ k>3$ : $ (\\mathbb{A}\\plus{}\\mathbb{B})^k \\equal{} \\mathbb{A}^k \\plus{} \\mathbb{B}^k$\r\nThen,\r\n$ (\\mathbb{A}\\plus{}\\mathbb{B})^{k\\plus{}1} \\minus{} (\\mathbb{A}^{k\\plus{}1} \\plus{} \\mathbb{B}^{k\\plus{}1}) \\equal{} \\mathbb{O}$\r\n$ \\iff (\\mathbb{A}\\plus{}\\mathbb{B})^k(\\mathbb{A}\\plus{}\\mathbb{B}) \\minus{} (\\mathbb{A}^{k\\plus{}1} \\plus{} \\mathbb{B}^{k\\plus{}1}) \\equal{} \\mathbb{O}$\r\n$ \\iff (\\mathbb{A}^k\\plus{}\\mathbb{B}^k)(\\mathbb{A}\\plus{}\\mathbb{B}) \\minus{} (\\mathbb{A}^{k\\plus{}1} \\plus{} \\mathbb{B}^{k\\plus{}1}) \\equal{} \\mathbb{O}$\r\n$ \\iff (\\mathbb{A}^{k\\plus{}1}\\plus{}\\mathbb{B}^{k\\plus{}1}\\plus{}\\mathbb{A}^k\\mathbb{B}\\plus{}\\mathbb{A}\\mathbb{B}^k) \\minus{} (\\mathbb{A}^{k\\plus{}1} \\plus{} \\mathbb{B}^{k\\plus{}1}) \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{A}^k\\mathbb{B}\\plus{}\\mathbb{A}\\mathbb{B}^k \\equal{} \\mathbb{O}$\r\n$ \\iff (\\minus{}1)^{k\\minus{}1}\\mathbb{B}\\mathbb{A}^k\\plus{}\\mathbb{A}\\mathbb{B}^k \\equal{} \\mathbb{O}$\r\n$ \\iff (\\minus{}1)^k\\mathbb{A}\\mathbb{B}\\mathbb{A}^{k\\minus{}1}\\plus{}\\mathbb{A}\\mathbb{B}\\mathbb{B}^{k\\minus{}1} \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{AB}((\\minus{}1)^k \\mathbb{A}^{k\\minus{}1}\\plus{}\\mathbb{B}^{k\\minus{}1}) \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{AB} \\equal{} \\mathbb{O} \\ \\text{or} \\ (\\minus{}1)^k \\mathbb{A}^{k\\minus{}1}\\plus{}\\mathbb{B}^{k\\minus{}1} \\equal{} \\mathbb{O}$\r\n\r\n$ \\boxed{\\text{Case} \\ \\mathbb{AB} \\equal{} \\mathbb{O}}$\r\n$ \\mathbb{AB} \\equal{} \\minus{} \\mathbb{BA}$\r\n$ \\iff \\mathbb{O} \\equal{} \\minus{} \\mathbb{BA}$\r\n$ \\iff \\mathbb{BA} \\equal{} \\mathbb{O}$\r\nTherefore, $ \\mathbb{AB} \\equal{} \\mathbb{BA} \\equal{} \\mathbb{O}$\r\n\r\n$ \\boxed{\\text{Case} \\ (\\minus{}1)^k \\mathbb{A}^{k\\minus{}1}\\plus{}\\mathbb{B}^{k\\minus{}1} \\equal{} \\mathbb{O}}$\r\n$ (\\mathbb{AB})^{k\\minus{}1}$\r\n$ \\equal{} \\sqrt{2}\\sin \\left(\\frac{2k\\minus{}1}{4}\\pi\\right)\\mathbb{A}^{k\\minus{}1}\\mathbb{B}^{k\\minus{}1}$\r\n$ \\equal{} (\\minus{}1)^k\\sqrt{2}\\sin \\left(\\frac{2k\\minus{}1}{4}\\pi\\right)\\mathbb{A}^{2(k\\minus{}1)}$\r\n$ \\equal{} \\sqrt{2}\\sin \\left(\\frac{2k\\minus{}3}{4}\\pi\\right)\\mathbb{A}^{2(k\\minus{}1)}$\r\n\r\n$ (\\minus{}\\mathbb{BA})^{k\\minus{}1}$\r\n$ \\equal{} (\\minus{}1)^{k\\minus{}1}(\\mathbb{BA})^{k\\minus{}1}$\r\n$ \\equal{} (\\minus{}1)^{k\\minus{}1}(\\minus{}1)^k\\sqrt{2}\\sin \\left(\\frac{2k\\minus{}1}{4}\\pi\\right)\\mathbb{A}^{2(k\\minus{}1)}$\r\n$ \\equal{} \\sqrt{2}\\sin \\left(\\frac{2k\\minus{}1}{4}\\pi\\right)\\mathbb{A}^{2(k\\minus{}1)}$\r\n\r\nSo,\r\n$ (\\mathbb{AB})^{k\\minus{}1}\\equal{}(\\minus{}\\mathbb{BA})^{k\\minus{}1}$\r\n$ \\iff \\sqrt{2}\\sin \\left(\\frac{2k\\minus{}3}{4}\\pi\\right)\\mathbb{A}^{2(k\\minus{}1)} \\equal{} \\sqrt{2}\\sin \\left(\\frac{2k\\minus{}1}{4}\\pi\\right)\\mathbb{A}^{2(k\\minus{}1)}$\r\n$ \\iff 2\\sqrt{2}\\sin \\left(\\frac{2k\\minus{}3}{4}\\pi\\right)\\mathbb{A}^{2(k\\minus{}1)} \\equal{} \\mathbb{O}$\r\n$ \\iff \\mathbb{A}^{2(k\\minus{}1)} \\equal{} \\mathbb{O}$\r\nBut, $ \\mathbb{A}\\in M(2; \\mathbb{C})$\r\n\r\n\r\nThus the formula works for $ n\\equal{}k\\plus{}1$.\r\nSo, by PMI, the statement is true for every integer $ n \\ge 3$." } { "Tag": [ "real analysis", "number theory proposed", "number theory" ], "Problem": "Prove that $ \\frac{(2n)!}{(n)!(n\\plus{}1)!}$ is an integer for all positive integers $ n$.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?p=1280283#1280283\r\n\r\nIt equals $ \\binom{2n}{n} \\minus{} \\binom{2n}{n \\plus{} 1}$.\r\n\r\nI am tempted to give the combinatorial proof (using path counting - Catalan number definition), but I'll skip, anyone interested can look it up on wiki :)\r\n\r\nBTW, what a coincidence, it is in my school's Real Analysis homework [ok, it got nothing to do with analysis, but this is an introductory HW so it contains some NT and combinatorical problems] :P", "Solution_2": "Thank you very much Harun (Actually I got this problem from one of Huse's students:))" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ B \\equal{} (2,x)$ be the ideal generated by $ 2$ and $ x$ in the ring $ A \\equal{} \\mathbb{Z} [x]$. \r\n\r\nShow that the element $ 2\\otimes 2 \\plus{} x\\otimes x$ in $ B\\otimes _{A} B$ is not a simple tensor.\r\n\r\nIt looks like easy problem. But I can't prove it because tensor product is quite new for me. Please help me.", "Solution_1": "The trick for problems like this is constructing bilinear maps on $ B\\times B$ using the universal property of tensor products.\r\n\r\nSay $ 2\\otimes 2 \\plus{} x\\otimes x \\equal{} a(x)\\otimes b(x)$. Since $ (f,g)\\mapsto f(0)g(0)$ is a bilinear map to $ \\mathbb{Z}$, there is a map $ B\\otimes B\\to \\mathbb{Z}$ taking $ f\\otimes g$ to $ f(0)g(0)$. Applying this to our equation, we have $ a(0)b(0) \\equal{} 4$.\r\n\r\nSimilarly, $ a(0)b'(0)\\equal{}0$, and $ a'(0)b'(0)\\equal{}1$. But this is impossible." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let $a,b,c$ be positive constants. determine the maximum and minimum value of:$\\sqrt{a^2.x^2+b^2.y^2+c^2.z^2}+\\sqrt{a^2.y^2+b^2.z^2+c^2.x^2}+\\sqrt{a^2.z^2+b^2.x^2+c^2.y^2}$,subject to $x^2+y^2+z^2=1$", "Solution_1": "Oh, boy, this is so nice! The extreme values are $ a+b+c $ and $\\sqrt{3(a^2+b^2+c^2)} $. The maximum is trivial to find using Cauchy. For the minimum, by squaring it is enough to show that $\\sum_\\sqrt{a^2x^2+b^2y^+c^2z^2}\\cdot\\sqrt{a^2y^2+b^2z^2+c^2x^2}\\geq ab+bc+ca $. This follows by adding the inequalities of the form $\\sqrt{a^2x^2+b^2y^+c^2z^2}\\cdot\\sqrt{a^2y^2+b^2z^2+c^2x^2}\\geq acx^2+aby^2+bcz^2 $ found again by Cauchy. Klamkins problems are so nice!", "Solution_2": "For the minimum, it can be done by Minkowski,\r\n\r\n$\\sqrt{a^2x^2+b^2y^2+c^2z^2}+\\sqrt{a^2y^2+b^2z^2+c^2x^2}+\\sqrt{a^2z^2+b^2x^2+c^2y^2}$\r\n\r\n$\\geq \\sqrt { (ax+cx+bx)^2 + (by+ay+cy)^2 + (cz+bz+az)^2} = { a+b+c}$." } { "Tag": [ "limit", "calculus", "function", "inequalities", "real analysis", "real analysis solved" ], "Problem": "Let:\r\n \\[\\lim_{x\\to 0}f(x)=\\lim_{x\\to 0}\\frac{f(3x)-f(x)}{x}=0\\]\r\nProve that:\r\n\\[\\lim_{x\\to 0}\\frac{f(x)}{x}=0\\]", "Solution_1": "Calculus Unsolved Problems is more appropriate place for this problem ;)", "Solution_2": "Dear Myth,\r\nSince this problem was one of questions in the exam chosing the team for VMO 2005 of my friend's school ( for high school students), I post this problem to this topic.If you feel my problem being here isn't appropriate, please move it to Calculus Unsolved Problems. Thanks!", "Solution_3": "i'd have a technical question to this related to this problem - are the conditions given here enough to expand the function in Maclaurin series or not? if, so the solution is quite simple. if not, i'll have to think some more :) the thing is going on only around zero, so i guess we could do that, but i'm not sure, so please prove or disprove.", "Solution_4": "[quote=\"yrch\"]i'd have a technical question to this related to this problem - are the conditions given here enough to expand the function in Maclaurin series or not? if, so the solution is quite simple. if not, i'll have to think some more :) the thing is going on only around zero, so i guess we could do that, but i'm not sure, so please prove or disprove.[/quote]\r\nOf course, you can't use Maclaurin series, since you haven't condititon, that function is differentiable.\r\n\r\nSolution is simple enough.", "Solution_5": "right. stupid me.", "Solution_6": "Let $\\epsilon > 0$ be given. Let $\\eta > 0$ be such that for $|a| < \\eta$ we have $ | \\frac {f(3a) - f( a )} a | < \\epsilon$ (from the second hypothesis).\r\n\r\nLet $x$ such that $0 \\leq x < \\eta$ (the case $x < 0$ is similar).\r\nThen for all positive integer $k$, we have :\r\n$- \\frac {\\epsilon x } {3^k} < f(\\frac x {3^{k-1}}) - f(\\frac x {3^k}) < \\frac {\\epsilon x } {3^k}$. (1)\r\n\r\nNow let's choose a positive integer $n$ sufficientely large so that, from the first hypothesis, we have $- \\frac {\\epsilon x} 2 < f( \\frac x {3^n}) < \\frac {\\epsilon x} 2 $. (2)\r\nFor such $n$, add all the inequalities (1) for $k=1, \\cdots ,n$.\r\nUsing (2), it leads to \r\n$- \\frac {\\epsilon x} 2 - \\frac {3 \\epsilon x} 2 (1 - \\frac 1 {3^n} ) < f(x) < \\frac {\\epsilon x} 2 + \\frac {3 \\epsilon x} 2 (1 - \\frac 1 {3^n} ) $\r\nso that \r\n$- 2 \\epsilon x < f(x) < 2 \\epsilon x$\r\ntherefore $| \\frac {f(x)} x | < 2 \\epsilon.$ (3)\r\n\r\nAs claimed above, we may use a similar reasoning in the case where $x<0$.\r\nThus, from (3) the desired limit follows easily.\r\n\r\nPierre.", "Solution_7": "As I said above, solution is really simple. ;) :D", "Solution_8": "[quote=\"pigfly\"]Let:\n \\[\\lim_{x\\to 0}f(x)=\\lim_{x\\to 0}\\frac{f(3x)-f(x)}{x}=0\\]\nProve that:\n\\[\\lim_{x\\to 0}\\frac{f(x)}{x}=0\\][/quote]\r\n\r\n$\\lim_{x\\to 0}\\frac{f(3x)-f(x)}{x}=3\\lim_{x\\to 0} \\frac{f(3x)-f(0)}{3x-0}-\\lim_{x\\to 0}\\frac{f(x)-f(0)}{x-0}$\r\n\r\n\r\n$=3f'(0)-f'(0)=2f'(0)=0$, yielding $f'(0)=0$\r\n\r\nTherefore $\\lim_{x\\to 0} \\frac{f(x)}{x}=\\lim_{x\\to 0}\\frac{f(x)-f(0)}{x-0}=f'(0)=0$", "Solution_9": "In the first line, you admit that $\\lim \\frac {f(x)} x$ does exist, which is not sure.\r\nThen, you admit that $f$ is differentiable which is not given too.\r\n\r\nPierre.", "Solution_10": "You may be right. :blush:" } { "Tag": [ "probability and stats" ], "Problem": "to calculate the probablilit\u00e0 for a poker player to find itself served in hand: one brace of aces a poker a full \r\n \r\n \r\n :lol:", "Solution_1": "I'm not quite sure what you mean- I see a \"pair of aces\", but there are many variations on the basic game of poker with different rules, and I don't know what you want \"a full\" to do in that sentence.\r\n\r\nCalculating probabilities of poker hands is a standard (if long) exercise. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=136394]Here[/url] is on version." } { "Tag": [ "linear algebra", "matrix", "geometry", "analytic geometry", "projective geometry", "LaTeX" ], "Problem": "How do I write a matrix on this forum?", "Solution_1": "As an example \r\nA=\\begin{pmatrix}0&1&0&\\cdots&0&0\\\\\r\n0&0&1&\\cdots&0&0\\\\\r\n\\vdots&\\vdots&\\vdots&\\ddots&\\vdots&\\vdots\\\\\r\n0&0&0&\\cdots&0&1\\\\\r\n0&0&0&\\cdots&0&0\\end{pmatrix}\r\n\r\nyields\r\n\r\n$ A\\equal{}\\begin{pmatrix}0&1&0&\\cdots&0&0\\\\\r\n0&0&1&\\cdots&0&0\\\\\r\n\\vdots&\\vdots&\\vdots&\\ddots&\\vdots&\\vdots\\\\\r\n0&0&0&\\cdots&0&1\\\\\r\n0&0&0&\\cdots&0&0\\end{pmatrix}$\r\n\r\nOr \r\n\\left(\\begin{array}{cc}-1&5\\\\\\sqrt 2&3\\end{array}\\right)\r\n\r\nyields \r\n\r\n$ \\left(\\begin{array}{cc}\\minus{}1&5\\\\\\sqrt 2&3\\end{array}\\right)$", "Solution_2": "What about determinant?", "Solution_3": "$ \\det{A} \\equal{} \\left| \\begin{array} {cc} \\minus{}1&5\\\\ \\sqrt 2&3 \\end{array} \\right|$", "Solution_4": "[code]\\det{A} = \\begin{vmatrix} \n-1&5\\\\\n\\sqrt 2&3 \n\\end{vmatrix}[/code] is simpler\r\n$ \\det{A} \\equal{} \\begin{vmatrix} \r\n\\minus{}1&5\\\\\r\n\\sqrt 2&3 \r\n\\end{vmatrix}$", "Solution_5": "{pmatrix}\r\n\r\n$ A\\equal{}\\begin{pmatrix}1&2&3\\\\4&5&6\\\\7&8&9\\end{pmatrix}$\r\n\r\n{bmatrix}\r\n\r\n$ A\\equal{}\\begin{bmatrix}1&2&3\\\\4&5&6\\\\7&8&9\\end{bmatrix}$\r\n\r\nOn this forum, I usually (but not always) use bmatrix rather than pmatrix. The choice is a matter of cosmetic preference.", "Solution_6": "[quote=\"Kent Merryfield\"]On this forum, I usually (but not always) use bmatrix rather than pmatrix. The choice is a matter of cosmetic preference.[/quote]\r\nAs an aside, on a course I used to teach for the Open University, the bmatrix style was used for matrices in projective geometry acting on homogeneous coordinates so $ \\begin{bmatrix}1 & 2 & 3 \\\\\r\n4 & 5 & 6 \\\\\r\n7 & 8 & 9\\end{bmatrix} \\equal{} \\begin{bmatrix}2 & 4 & 6 \\\\\r\n8 & 10 & 12 \\\\\r\n14 & 16 & 18\\end{bmatrix}$, whereas pmatrix was used for the usual matrices and where $ \\begin{pmatrix}1 & 2 & 3 \\\\\r\n4 & 5 & 6 \\\\\r\n7 & 8 & 9\\end{pmatrix}\\neq\\begin{pmatrix}2 & 4 & 6 \\\\\r\n8 & 10 & 12 \\\\\r\n14 & 16 & 18\\end{pmatrix}$. Thus great care was needed.\r\n\r\nIt did cause a lot of confusion because some of the students had used bmatrix in a previous course for ordinary matrices. I had some difficulty explaining to students how mathematics notation is not always standardised and different people like using different notation. They found this an odd concept thinking that everything in mathematics is laid down and there are no personal preferences. Giving them other examples (such as soluble vs solvable groups in British vs American usage) didn't seem to help :)" } { "Tag": [ "geometry", "trapezoid", "perimeter" ], "Problem": "The measure of one of the smaller base angles of an isoceles trapezoid is $ 60^\\circ$. The shorter base is 5 inches long and the altitude is $ 2 \\sqrt{3}$ inches long. What is the number of inches in the perimeter of the trapezoid?\n[asy]import olympiad; size(150); import geometry_dev; import graph; defaultpen(linewidth(0.8));\ndraw(origin--(9,0)--(7,2*sqrt(3))--(2,2*sqrt(3))--cycle);\ndraw(Label(\"$2\\sqrt{3}$\",align=E),(2,2*sqrt(3))--(2,0),dashed);\nlabel(\"5\",(7,2*sqrt(3))--(2,2*sqrt(3)),N);\nmarkangle(Label(\"$60^{\\circ}$\"),(2,2*sqrt(3)),origin,(9,0),radius=10);[/asy]", "Solution_1": "The smallest length of the triangle is 2 (since its a 30-60-90 triangle), so the hypotenuse is 4. Thus, the perimeter is 5+4+2+2+5+4=22." } { "Tag": [], "Problem": "Having problems thinking of what the formula is for calculating the following:-\r\n\r\nSay you you have the following:-\r\n\r\nGross income 2005= \u00a38000\r\n\r\nIf this is a rise of 30% of the previous year then how would you calculate what the previous years gross was?\r\n\r\nSame applies if it was 35% , 45% etc.\r\n\r\nThanks\r\n\r\nBusby", "Solution_1": "[hide]\n8000/1.3\n(or 8000/1.35 or 8000/1.45 etc)\n[/hide]" } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "Let a,b be two positive integers that can be expressed as the sum of two squares of integer. Assume $ \\frac{a}{b}$ is an integer, prove that $ \\frac{a}{b}$ can also be expressed as the sum of two squares of integer.", "Solution_1": "[quote=\"greentreeroad\"]Let a,b be two positive integers that can be expressed as the sum of two squares of integer. Assume $ \\frac {a}{b}$ is an integer, prove that $ \\frac {a}{b}$ can also be expressed as the sum of two squares of integer.[/quote]\r\nIt is good problem . \r\nA well know lemma : \r\n[u]Lemma [/u] A number $ n$ can represent as sum of two square if and only if $ v_p(n)\\equiv 0 (\\mod 2)$ for all prime p such that $ p\\equiv 3 (\\mod 4)$\r\nNow consider problem : \r\nBecause $ a,b$ is sum of two square so $ v_p(a)\\equiv v_p(b)\\equiv 0 (\\mod 4)$ for all prime $ p\\ equiv 3 (\\mod 4)$\r\nMore than $ b|a$ so $ v_p(a)\\geq v_p(b)$ for all prime p . \r\nSo $ v_p(\\frac{a}{b})\\equiv 0 (\\mod 2 )$ for all prime $ p\\equiv 3 (\\mod 4)$\r\nIt follows that $ \\frac{a}{b}$ can represent as sum of two square .", "Solution_2": "[quote=\"TTsphn\"]\nA well know lemma : \n[u]Lemma [/u] A number $ n$ can represent as sum of two square if and only if $ v_p(n)\\equiv 0 (\\mod 2)$ for all prime p such that $ p\\equiv 3 (\\mod 4)$\n[/quote]\r\n\r\nO.K, but how to prove this lemma?", "Solution_3": "It is an well know problem . \r\nA hint : \r\n1.$ (\\frac{\\minus{}1}{p})\\equal{}1$ if $ p\\equiv 1 (\\mod 4)$ \r\n$ (\\frac{\\minus{}1}{p})\\equal{}\\minus{}1$ if $ p\\equiv 3 (\\mod 4)$\r\n2. $ (a^2\\plus{}b^2)(c^2\\plus{}d^2)\\equal{}(ac\\plus{}bd)^2\\plus{}(ad\\minus{}bc)^2$", "Solution_4": "I can see why numbers with even exponent of 4K+3 prime can be expressed as sum of two squares, but I don't know how to prove that a number of form 4K+1 which contains odd exponent of 4K+3 prime CANNOT.(for example, 21=3*7 and 165=3*11*5)", "Solution_5": "[quote=\"greentreeroad\"]I can see why numbers with even exponent of 4K+3 prime can be expressed as sum of two squares, but I don't know how to prove that a number of form 4K+1 which contains odd exponent of 4K+3 prime CANNOT.(for example, 21=3*7 and 165=3*11*5)[/quote]\r\nYes because from $ (\\frac{\\minus{}1}{p})\\equal{}\\minus{}1$ for all prime $ p\\equiv 3 (\\mod 4)$ so if $ p|x^2\\plus{}y^2$ then $ p|x,p|y$\r\nIt follow that $ v_p(x^2\\plus{}y^2)\\equal{}2.v_p (\\gcd(x,y))$", "Solution_6": "Thanks a lot. At last, how do you actually get p|x, p|y just from the fact $ p|x^2\\plus{}y^2$ for p a 4K+3 prime? That is, why t and p-t cannot both be a quadratic residue of p?", "Solution_7": "Because $ (\\frac { \\minus{} 1}{p}) \\equal{} ( \\minus{} 1)^{\\frac {p \\minus{} 1}{2}}$ (1)\r\nIf $ p\\not | x$ then $ p\\not |y$ so exist an integer k such that \r\n$ x\\equiv ky (\\mod p)$ so \r\n$ p|y^2(k^2 \\plus{} 1)$ \r\n$ \\Rightarrow p|x^2 \\plus{} 1$ \r\nIt gives contradiction from (1) .\r\nSo $ p|x$ ,follow that $ p|y$", "Solution_8": "We have\r\n$ \\frac{(pm\\minus{}qn)^2\\plus{}(pn\\plus{}qm)^2}{m^2\\plus{}n^2}\\equal{}p^2\\plus{}q^2$. \r\nCan we prove the numbers must always take this form?" } { "Tag": [ "limit", "logarithms", "integration", "calculus", "calculus computations" ], "Problem": "Evaluate:\r\n\r\n$ \\lim_{n\\to \\plus{} \\infty}\\prod_{k \\equal{} 1}^{n}(1 \\plus{} \\frac {1}{k^{3}})$", "Solution_1": "It doesn't telescope or otherwise simplify in any obvious way. It does converge - the general theory of infinite products compares its convergence to that of $ \\sum_k\\frac1{k^3},$ which is a convergent series.\r\n\r\nNumerical approximation (running this through $ n\\equal{}64$ and then using a Richardson extrapolation table) gives an approximate value of $ 2.428189792$", "Solution_2": "Mathematica gives $ \\frac{1}{\\pi} \\cosh\\left(\\frac{\\sqrt{3}}{2}\\pi\\right) \\approx 2.42819$ and that should be OK.. It matches what Kent Merryfield suggested..", "Solution_3": "$ \\lim_{n\\to \\plus{} \\infty}\\ln \\prod_{k \\equal{} 1}^{n}(1 \\plus{} \\frac {1}{k^{3}}) \\equal{} \\sum_{n \\equal{} 1}^{\\infty} \\ln \\left(1 \\plus{} \\frac {1}{n^3}\\right)\\leq \\sum_{n \\equal{} 1}^{\\infty} \\frac {1}{n^3} \\equal{} \\zeta (3)$\r\n\r\n$ \\therefore \\lim_{n\\to \\plus{} \\infty}\\prod_{k \\equal{} 1}^{n}(1 \\plus{} \\frac {1}{k^{3}})\\leq e^{\\zeta (3)}?$", "Solution_4": "Turns out that the product is equal to:\r\n\r\n$ \\exp\\left(3\\int^\\infty_0 \\frac {F(x)}{x \\left(e^x \\minus{} 1\\right)}\\,dx\\right)$\r\n\r\nwhere\r\n\r\n$ F'''(x) \\plus{} F(x) \\equal{} 1 \\, , \\qquad F(0) \\equal{} F'(0) \\equal{} F''(0) \\equal{} 0$" } { "Tag": [ "induction", "number theory proposed", "number theory" ], "Problem": "Foe each positive integer $ n$, let $ S(n)$ be the sum of the digits in the decimal expansion of $ n$. Prove that for all $ n$, $ S(2n) \\le 2S(n) \\le 10 S(2n)$ and show that there exists $ n$ such that $ S(n)\\equal{}1996 S(3n)$.", "Solution_1": "Let $ 0\\le a<10,n\\equal{}10k\\plus{}a$, then $ S(n)\\equal{}a\\plus{}S(k)$.\r\n$ S(2n)\\equal{}S(2a)\\plus{}S(2k)$ it is true for $ a<5$ and for $ a\\ge 5$.\r\nBy induktion $ S(2n)\\le 2S(n)$, generally from $ S(m\\plus{}n)\\le S(m)\\plus{}S(n)$ we get $ S(kn)\\le kS(n)$.\r\n$ S(2n)\\ge S(2k)\\plus{}1,a>0$. By induction $ 10S(2n)\\ge 2S(n)$.\r\nConsider $ n_k\\equal{}\\frac{10^k\\plus{}2}{3}$. Then $ S(3n_k)\\equal{}3, S(n_k)\\equal{}3k\\plus{}1$.\r\nTherefore if $ n\\equal{}10^{2k}n_{k\\minus{}1}\\plus{}10^kn_k\\plus{}n_k$, then $ S(3n)\\equal{}9,S(n)\\equal{}3(k\\minus{}1)\\plus{}1\\plus{}3k\\plus{}1\\plus{}3k\\plus{}1\\equal{}9k$ and $ \\frac{S(n)}{S(3n)}\\equal{}k$." } { "Tag": [ "inequalities", "induction", "limit" ], "Problem": "1. Prove for positive integers $ n$,\r\n$ \\frac{(2n)!}{2^{2n}(n!)^{2}}\\le\\frac{1}{\\sqrt{(2n)}}$\r\n\r\n2. Prove that\r\n$ \\frac{1}{31}+\\frac{1}{81}+\\frac{1}{151}+....\\frac{1}{10N^{2}+20N+1}....<\\frac{3}{40}$\r\n\r\n3. Prove by induction that for all positive integers $ n$,\r\n$ \\frac{(2n)!}{2^{2n}(n!)^{2}}\\leq\\frac{1}{\\sqrt{3n+1}}$", "Solution_1": "[quote=\"riddler\"]2. Prove that\n$ \\frac{1}{31}+\\frac{1}{81}+\\frac{1}{151}+....\\frac{1}{10N^{2}+20N+1}....<\\frac{3}{40}$[/quote]\r\n[hide=\"Solution\"]\n$ \\sum_{n=1}^{\\infty}\\frac{1}{10(n+1)^{2}-9}=\\frac{1}{10}\\sum_{n=1}^{\\infty}\\frac{1}{(n+1)^{2}-\\frac{9}{10}}<\\frac{1}{10}\\sum_{n=1}^{\\infty}\\frac{1}{(n+1)^{2}-1}=\\frac{1}{20}\\sum_{n=1}^{\\infty}\\frac{n+2-n}{n(n+2)}=$\n\n$ =\\frac{1}{20}\\sum_{n=1}^{\\infty}\\left(\\frac{1}{n}-\\frac{1}{n+2}\\right)=\\frac{1}{20}\\left(1+\\frac{1}{2}\\right)=\\frac{3}{40}$[/hide]", "Solution_2": "sorry if this is dum but how do you do from the last step? et 3/40?\r\nfrom $ \\frac{1}{n}-\\frac{1}{n+2}$", "Solution_3": "Becouse:\r\n\r\n$ \\lim_{m\\to\\infty}\\sum_{n=1}^{m}\\left(\\frac{1}{n}-\\frac{1}{n+2}\\right)=\\lim_{m\\to\\infty}\\left(1-\\frac{1}{3}+\\frac{1}{2}-\\frac{1}{4}+\\frac{1}{3}-\\frac{1}{5}+...+\\frac{1}{m-2}-\\frac{1}{m}+\\frac{1}{m-1}-\\frac{1}{m+1}+\\frac{1}{m}-\\frac{1}{m+2}\\right)=$\r\n$ =\\lim_{m\\to\\infty}\\left(1+\\frac{1}{2}-\\frac{1}{m+1}-\\frac{1}{m+2}\\right)=1+\\frac{1}{2}$\r\n\r\n[quote=\"riddler\"]3. Prove by induction that for all positive integers $ n$,\n$ \\frac{(2n)!}{2^{2n}(n!)^{2}}\\leq\\frac{1}{\\sqrt{3n+1}}$[/quote]\r\n[hide=\"solution\"]\nWe check for $ n=1$ and that's o.k., then we should prove:\n$ \\frac{(2n)!}{2^{2n}(n!)^{2}}\\leq\\frac{1}{\\sqrt{3n+1}}\\ \\Rightarrow\\ \\frac{(2n+2)!}{2^{2n+2}[(n+1)!]^{2}}\\leq\\frac{1}{\\sqrt{3n+4}}$\n\n$ \\frac{(2n+2)!}{2^{2n+2}[(n+1)!]^{2}}=\\frac{(2n)!}{2^{2n}(n!)^{2}}\\cdot\\frac{(2n+1)(2n+2)}{4(n+1)^{2}}\\leq$\n\n$ \\leq\\frac{2n+1}{2(n+2)}\\frac{1}{\\sqrt{3n+1}}$ and now we should prove:\n\n$ \\frac{2n+1}{2(n+2)}\\frac{1}{\\sqrt{3n+1}}\\leq\\frac{1}{\\sqrt{3n+4}}$\n\n$ (2n+1)\\sqrt{3n+4}\\leq 2(n+2)\\sqrt{3n+1}$\n$ (2n+1)^{2}(3n+4)\\leq 4(n+2)^{2}(3n+1)$\n...\n$ 24n^{2}+45n+12\\geq 0$\nQ.E.D.[/hide]", "Solution_4": "oh yh cheers" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "We call a real number $x$ with $0 < x < 1$ [i]interesting[/i] if $x$ is irrational and if in its decimal writing the first four decimals are equal. Determine the least positive integer $n$ with the property that every real number $t$ with $0 < t < 1$ can be written as the sum of $n$ pairwise distinct interesting numbers.", "Solution_1": "[hide]\n$n=1112$\ntake $t_{0}=0.1111$\ninteresting numbers that are smaller that $t_{0}$ begin with four zeroes so they are smaller than $0.0001$\nlet $t_{0}=\\sum_{i=1}^{n}a_{i}$ where $a_{i}$ are interesting numbers.now we have \n$t_{0}=\\sum_{i=1}^{n}a_{i}<\\sum_{i=1}^{n}0.0001=n*0.0001$ so $n>1111$ or $n\\geq 1112$\ntake any $t$ such that $00$ and $b+556\\phi<0.0001$. if $b$ is rational pick $\\phi$ irrational and vice versa. the numbers $b-556\\phi,b-555\\phi,...,b-\\phi,b+\\phi,...,b+556\\phi$ satisfy.\nif $t>t_{0}$ write it in this form $t=k*t_{0}+t^{'}$ where $k$ is an integer and $0 1$ and $ R \\neq 0$. in particular, the $ z_i$ are not left invertible, since they lie in some maximal ideal of $ R$. find $ u \\in U$ and $ r \\in R$ such that $ e_1 \\equal{} u \\plus{} r z$. then $ 1\\equal{}u \\plus{} r z_1$ shows that $ u$ is left invertible (otherwise, it lies in a maximal ideal and thus also $ u \\plus{} r z_1 \\equal{} 1$, a contradiction), and $ 0\\equal{}u \\plus{} r z_2$ shows that also $ z_2$ is left invertible, a contradiction. $ \\square$" } { "Tag": [ "algebra", "polynomial", "trigonometry", "calculus", "calculus computations" ], "Problem": "hi,\r\n\r\nI have a problem: find a polynomical funktion $f$ with $|f(x)-\\sin(x^2)|<\\frac{10^{-6}}2$ for all $x \\in [-1,1].$\r\n\r\n[Moderator edit to introduce TeX; hardly any changes needed.]", "Solution_1": "This is the particular forum where problems like this belong.\r\n\r\nIn this case, the identity $\\sin^2x=\\frac{1-\\cos 2x}2$ can be used to simplify the calculations.", "Solution_2": "thank you very much, everything figures now...", "Solution_3": "Dear Kent may be I give a stupid question,but I can't figure out.Where we have $\\sin^{2}{x}$?", "Solution_4": "Oops - I guess he did say $\\sin(x^2)$ rather than $\\sin^2x.$ In that case, we want to use the Taylor expansion for $\\sin x$ and then substitute $x^2$ for $x$." } { "Tag": [], "Problem": "okay i like james idea. if you would like to be in this election(for nothing) pplease post you would like to join. after no one joins anymore. we have a vote(every person can only vote once) and see who is teh most popular AOPser. I would like to be a nominee", "Solution_1": ":rotfl: jjx as most popular aopser? :rotfl:", "Solution_2": ":rotfl: MEWTO AS MOST POPULAR AOPSER! :huh: :rotfl:", "Solution_3": "I know I am not. That is why I did not stoop so low as to nominate myself. However, I am infinitely more popular than you.", "Solution_4": "mewto wth. i'm not saying i am teh most popular aopser. I'm probably gonna lose. butso waht im jsut signing up for fun. geez", "Solution_5": "Hmm... tinytim as the most popular AoPSer!!\r\nBut I join anyway (and tinytim) :D", "Solution_6": "okay so teh people who are in the race so far are:\r\n\r\nyongi\r\ntinytim\r\njjx1", "Solution_7": "nominate: dragon96", "Solution_8": "nominate: stevenmeow", "Solution_9": "nominate: spammer123", "Solution_10": "nominate: rrusczyk", "Solution_11": "stress and vote: rrusczyk", "Solution_12": "vote: me\r\n\r\nThe message is too small. Please make the message longer before submitting.", "Solution_13": "okay \r\n\r\nnominees: \r\njjx1:\r\nyongi: \r\ntinytim:\r\nstevenmeow: \r\ndragon \r\nspammer123 \r\n\r\nand i am not counting any admins\r\nOH AND I;M ASSUMING EVERYONE IS VOTING THEMSELVES. IF YOU'RE NOT TELL ME.OH AND ELECTION DAY ISN'T HERE", "Solution_14": "nominate: 1=2, vote: 1=2", "Solution_15": "Nominate: Randomdragoon\r\nno vote yet.", "Solution_16": "jjx1:4 \r\ntinytim:0 \r\nyongi:0 \r\nstevenmeow:0 \r\nDRAGON:0 \r\nSPAMMER:0 \r\n1=2:3 \r\nlotsofmath:0 \r\npythag:0 \r\nernie:0 \r\njames:0\r\nRandomDragon:0\r\n\r\nis he a real user?", "Solution_17": "randomdragoon is.\r\n\r\nNominate: bowei", "Solution_18": "jjx1:4\r\n1=2:3 \r\ntinytim:0 \r\nyongi:0 \r\nstevenmeow:0 \r\nDRAGON:0 \r\nSPAMMER:0 \r\nlotsofmath:0 \r\npythag:0 \r\nernie:0 \r\njames:0 \r\nRandomDragon:0\r\nbowei:0", "Solution_19": "Vote RandomDragoon.\r\n(you misspelled his name...there's 2 o's in the last part and 1 o in the first part)", "Solution_20": "jjx1:4 \r\n1=2:3\r\nRandomDragoon:1 \r\ntinytim:0 \r\nyongi:0 \r\nstevenmeow:0 \r\nDRAGON:0 \r\nSPAMMER:0 \r\nlotsofmath:0 \r\npythag:0 \r\nernie:0 \r\njames:0 \r\nbowei:0", "Solution_21": "Hmm... vote: ernie.", "Solution_22": "nominate: junggi\r\n\r\nI am \"de nominator\" [corny math joke, blah]", "Solution_23": "Nominate: MustItAlwaysBeMe", "Solution_24": "nominate: NerdOfTheAges [NotA]", "Solution_25": "jjx1:4 \r\n1=2:3 \r\nRandomDragoon:1 \r\ntinytim:0 \r\nyongi:0 \r\nstevenmeow:0 \r\nDRAGON:0 \r\nSPAMMER:0 \r\nlotsofmath:0 \r\npythag:0 \r\nernie:1\r\njames:0 \r\nbowei:0\r\nNotA:0\r\nMustItAlwaysBeMe:0\r\nJunggi:0\r\n\r\n\r\nI'm pretty sure that very few people are going to vote, but oh well.", "Solution_26": "nominatea:probabilitu1.01", "Solution_27": "nominate: MysticTerminator\r\nnominate: zeb\r\nnominate: CA Math (for doing the votecount) :lol:", "Solution_28": "jjx1:4 \r\n1=2:3 \r\nRandomDragoon:1 \r\ntinytim:0 \r\nyongi:0 \r\nstevenmeow:0 \r\nDRAGON:0 \r\nSPAMMER:0 \r\nlotsofmath:0 \r\npythag:0 \r\nernie:1 \r\njames:0 \r\nbowei:0 \r\nNotA:0 \r\nMustItAlwaysBeMe:0 \r\nJunggi:0\r\nprobabilitu1.01\r\nMysticTerminator :0\r\nzeb:0\r\nCA math:0\r\n\r\nokay from now on you can't nominate anyone. if somepone wants to be a nominee they gotta post they want to join themselves.", "Solution_29": "Note: should be probability1.01" } { "Tag": [ "trigonometry" ], "Problem": "In 1593, the Belgian mathematician Adriaan van Roomen proposed the following problem:\r\n\r\nFind the positive roots of the equation \r\n$x^{45} - 45x^{43} + 945x^{41} - 12300x^{39} + 111150x^{37} - 740459x^{35} + 3746565x^{33} - 14945040x^{31} + 469557800x^{29} - 117679100x^{27} + 236030652x^{25} - 378658800x^{23} + 483841800x^{21} - 488494125x^{19} + 384942375x^{17} - 232676280x^{15} + 105306075x^{13} - 34512074x^{11} + 7811375x^9 - 1138500x^7 + 95634x^5 - 3795x^3 + 45x = \\sqrt{\\frac{7}{4} - \\sqrt{\\frac{5}{16}} - \\sqrt{\\frac{15}{8} - \\sqrt{\\frac{45}{64}}}.}$\r\n\r\nThe French mathematician Vi\u00e8te was able to solve the equation. By hand. In just a few minutes, too, supposedly (Anecdote! One of the Bernoullis claimed to have summed the first 1000 10th powers in half of 15 minutes. My analysis professor did it in just over 8, but he was explaining it to us as he went). \r\n\r\nAnyway, anyone here want to give it a try? Or is their 16th century intellect beyond us? :P", "Solution_1": "[quote=\"Xevarion\"] Or is their 16th century intellect beyond us? :P[/quote] :huh: I think its to much for us, or at least for me.\r\n\r\nBut I might give it a try. If I were to try it I would start by simplifing the radicles, right?", "Solution_2": "The only way I could see this problem being done so quickly is with trigonometric substitution...\r\n\r\ntoo bad I don't know trigonometry well enough. :D", "Solution_3": "surge: Nice insight! That is indeed the approach that Vi\u00e8te took. Unfortunately I still can't see exactly what he did...", "Solution_4": "i think that it has something to do with\r\n\r\n$\\Im (1+i*x)^{45}$ with $x=\\tan \\theta$", "Solution_5": "Viete solved this problem by repeatedly applying trigonometric double, triple, quintic angle formulas. I read about it in Eli Maor's [u]Trigonometric Delights[/u] :)", "Solution_6": "My god! Okay, I guess those 16th century mathematicians were a lot better at brute-forcing than we are today. I'm not trying that... I do sorta wonder what the roots are, though. I bet they're not that horrible.", "Solution_7": "RHS is of course $2 \\sin 12^o$ \r\nLeft side is what you get if you expand $2 \\sin (45 y)$ in terms of $\\sin y$ and you put $2 \\sin y = x$ \r\n\r\nso you \"easily\" :) get all the roots (For example one such root is $sin (12/45)^o$ and rest you can get by adding $k\\cdot 8^o$ ($k = 0,1,2 \\cdots$ etc.( remember 360/45=8) .\r\n\r\n(To be honest, RHS was not that difficult to guess for any one who worked in the old days as us as I tell my kids .. in those days we have to do all calculations by hand, have to remember times tables up to 100, know log tables by heart and walk 10 miles uphill both ways in $40^o C$ (It looks even more terrible in Fahrenheit :) = $104^o F$ ) heat and 5 feet of snow ..:)) \r\n\r\n( and of course only thing to keep in mind for LHS was to do all middle steps of calculations on slate so not to waste too much paper) :) \r\n\r\n(Actually if you know $sin 3x = 3 sin x - 4 sin^3 x$ and $\\sin 5x = 5 \\sin x -20 \\sin^3 x + 16 \\sin^5 x$ all you have to know and you apply first formula twice and second once.)" } { "Tag": [ "geometry", "similar triangles" ], "Problem": "A neat, but somewhat tricky, problem from the 1997 Cayley:\r\n\r\nIn the diagram, ABC is equilateral, BC=2CD, AF=6, and DEF is perpendicular to AB. What is the area of the quadrilateral FBCE?\r\n\r\nHint:\r\n[hide]Extend BD to G such that AG is parallel to FD.[/hide]", "Solution_1": "Actually, muy facil way to do this.\r\n\r\n[hide=\"HINT\"] What is CE in relation to s, the side of the equilateral triangle? Then, draw up similar triangles, and you have two triangles FBE and CBE with heights which are the same. Interesting problem. In any math competition - it's very important to know all properties of equilateral, right, and isosceles triangles.[/hide]", "Solution_2": "This problem is pretty easy. If you related everything in terms of CD and numbers you can develop some nice equalities and then figure out everything about the traingles. From there the area of the quadrilateral is cake.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\"When you have no quote to post, post one about having no quote to post\"\r\n -Yours Truly", "Solution_3": "This is what I would do. \r\n\r\nConsider $\\triangle BFD$ and let $2x$ be the side length of the equilateral triangle.\r\n\r\nSince $BFD$ is $30-60-90$, it's easy to figure out that $2BF=BD$, and we find that $x=12$.\r\n\r\nIt's easy to find now that $\\triangle AEF$ has area $12\\sqrt {3}$, and we just need to subtract this from the area of the equilateral triangle." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Has anybody heard of this or learned any of it? It provides simple ways to do mental computations that would normally be done on a caclulator. Do you think this is something that should be taught in middle or elementary school in conjunction with, or instead of conventional teaching?\r\n\r\nHere's a link with a few tutorials on some of the ideas behind Vedic Math\r\n[url]http://www.vedicmaths.org/Introduction/Tutorial/Tutorial.asp[/url]", "Solution_1": "I checked it out. At first, the first few tutorials were already known, but following that, I found many of them very useful. It is excellent for competition math, especially MATHCOUNTS and AMC (I heard you can't use calculators anymore?).", "Solution_2": "I have a book at home called Vedic Mathematics. Apart from simple computation, it has ways to factor quickly and solve cubic equations, along with a variety of other things. It was the same book mentioned here.\r\nhttp://www.vedicmaths.org/Bookstore/05%20Vedic%20Maths/VM%20details.asp", "Solution_3": "My dad taught me one of those tricks when I was in elementary school and called it the \"Indian method\" for multiplication. Now I know why!", "Solution_4": "That is really great tricks for speed math! I will probably use those a lot.", "Solution_5": "I have a vedic math book. It has lots of neat tricks, but also, more importantly, it has faster ways of multiplication and long division. However, I haven't gotten into the habit of using them. vedic math is based on vinculums, like \"negative digits.\" I should try to get used to it this summer." } { "Tag": [], "Problem": "J.F. Jolly (1809-1884), a Munich teacher of physics, measured with his very sensitive self-made balance that at a depth of 5.3 m the weight of a body is greater by one and a half millionth than that on the surface. How can that be reconciled with Newton's gravitational law?", "Solution_1": "What do you mean by \"reconciled\"? By Newton's Law of Gravitation, $ F\\equal{}G\\frac{Mm}{r^{2}}$. On the surface, we can say the distance is $ r$. At a depth of 5.3m, we can say it is $ r\\minus{}5.3$. When this is plugged into the equation, you'll find that $ F$ is very slightly greater. This is because the distance between the two bodies (earth and person) is smaller, so the force between them must be greater.", "Solution_2": "What you said about the force Jrav is only true when the particle is outside of the earth, then the gravitational force on it is the same as if we put a \"point\" mass with the mass of the earth at the center. But now we put the particle inside the earth, hence the gravitational field will be differant then in the outside, so the force on the particle will not be what you said." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "I asked already very many people and anyone can find mistake in my solution of task 5 from LVII Polish Mathematical Olympiad. Good solution of this task is in attachment om57_2r.pdf. I proved in complex.pdf that thesis doesn't work in infinite number of cases. Can someone help me?", "Solution_1": "At the first glance of your note (I don't have much time right now), it seems to be correct except of one mistake: The perpendicularity of the lines $z_1z_2$ and $0z$ is not equivalent to the statement that $z_1-z_2$ is equal to $iz$ or to $-iz$, but it is equivalent to the statement that $z_1-z_2$ is equal to $rz$ for some real $r$. Your proof can be easily corrected.\r\n\r\nBy the way, it would be useful if you post the problem statement in English, as well:\r\n\r\n[color=blue][b]Problem.[/b] Let C be the midpoint of a segment AB. Some circle $o_1$ passes through the points A and C, and some circle $o_2$ passes through the points B and C. The circles $o_1$ and $o_2$ intersect at some point D (apart from C). Let P be the midpoint of the arc AD on the circle $o_1$ which doesn't contain the point C, and let Q be the midpoint of the arc BD on the circle $o_2$ which doesn't contain the point C. Prove that $PQ\\perp CD$.[/color]\r\n\r\n Darij", "Solution_2": "Sorry, I'm an idiot :blush:. But I already found this mistake and a few days ago I solved this problem. See complex.pdf." } { "Tag": [], "Problem": "For an object, with the mass $ m\\equal{}2kg$, bound at one extremity by an horizontal spring fixed on the other extremity, apply an initial horizontal speed in the sense of the stretch of the spring. The coefficient of the rub between the object and the horizontal surface is $ 0,1$. The elastic constant of the spring is $ k\\equal{}100 \\frac {N} {m}$. Determine the first three amplitudes of the spring.\r\n\r\n$ R: A_1\\equal{}0,5m,A_2\\equal{}\\frac {7} {15}m,A_3\\equal{}\\frac {13} {30}m$", "Solution_1": "I will solve the exercise, but with Romanian notations:\r\n\r\n[u][b]first amplitude[/b][/u]\r\n\r\n$ E_f\\minus{}E_i\\equal{}L \\Rightarrow \\frac {kx^2} {2}\\minus{}\\frac {mv_0^2} {2}\\equal{}\\minus{}F_fx \\Rightarrow \\frac {kx^2} {2}\\minus{}\\frac {mv_0^2} {2}\\equal{}\\minus{}\\mu mgx \\Rightarrow A_1\\equal{}x\\equal{}0,5$\r\n\r\nso $ L\\equal{}\\minus{}F_fx\\equal{}\\minus{}2 \\cdot \\frac {1} {2}\\equal{}\\minus{}1$\r\n\r\n[u][b]second amplitude[/b][/u]\r\n\r\n$ E_i\\equal{}\\frac {mv_0^2} {2}\\plus{}2L\\equal{}16\\minus{}2\\equal{}14 \\Rightarrow \\frac {ky^2} {2}\\minus{}14\\equal{}\\minus{}F_f \\cdot y \\Rightarrow A_2\\equal{}y\\equal{}\\frac {7} {15} \\Rightarrow L'\\equal{}\\minus{}F_f \\cdot y\\equal{}\\minus{}\\frac {14} {15}$\r\n\r\n[u][b]third amplitude[/b][/u]\r\n\r\n$ E_i\\equal{}\\frac {mv_0^2} {2}\\plus{}2L\\plus{}2L'\\equal{}\\frac {182} {15} \\Rightarrow \\frac {kz^2} {2}\\minus{}\\frac {182} {15}\\equal{}\\minus{}F_f \\cdot z \\Rightarrow A_3\\equal{}z\\equal{}\\frac {13} {30}$", "Solution_2": "Where is $ v_0$ ?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c,d \\ge 0$ and $a+b+c+d=1$ . Prove that\r\n$abc+bcd+cda+abd \\ge \\frac{1}{27} +\\frac{176}{27}abcd$.", "Solution_1": "Maybe not that hard.\r\nFirst, make the inequality homogenous.\r\n\\[(a+b+c+d)(abc+bcd+cda+abd) \\ge \\frac{1}{27}(a+b+c+d)^4 +\\frac{176}{27}abcd\\] \r\nIt follows that,\r\n\\[27(a+b+c+d)(abc+bcd+cda+abd) \\ge (a+b+c+d)^4 +176abcd\\]\r\nThis seems to be done if you use Muirhead.", "Solution_2": "This is incredibly obviously false. $a=1,b=c=d=0$", "Solution_3": "[quote=\"blahblahblah\"]This is incredibly obviously false. $a=1,b=c=d=0$[/quote]\r\nHe is right. Maybe it is the other way?", "Solution_4": "http://www.mathlinks.ro/Forum/viewtopic.php?t=5972&postorder=asc&highlight=so+many+strategies&start=15", "Solution_5": "Thanks [b]siuhochung[/b], I need it.\r\n\r\n[quote=\"lightrhee\"][quote=\"blahblahblah\"]This is incredibly obviously false. $a=1,b=c=d=0$[/quote]\nHe is right. Maybe it is the other way?[/quote]\r\n$a=b=c=d=\\frac{1}{4}$" } { "Tag": [], "Problem": "is there an easier way to memorize the formula for permutations and combinations? I would be happy to know some suggestions!\r\n\r\n\r\n:)", "Solution_1": "Well, it always helps to understand a formula before you begin to memorize it so here are a few explanations:\r\n[hide=\"Permutations\"] Say that there are n people and p chairs (assume that the children have names, so one child will be different from another child). We want to find out how many ways there are to seat these children. n People can be seated in the first chair. For the second chair, only n-1 people can be seated (since one boy is done). Then, n-2, and n-3, and so on. There will be n terms, thus the number of combinations is\n\\[P(n,p)=(n)(n-1)(n-2)...(n-p+1)=\\frac{n!}{p!}\\]\n[/hide]\n[hide=\"Combinations\"] Say there are n boys and m girls and you wish to seat these children. Also assume the boys are not different from one another and the girls are not different from one another. Let's seperate this into boys and girls. There is one way to organize n boys (since they are all the same), but if the boys are different, then there are n! ways. There is one way to seat the girls, but if the gilrs were different, then there would be m! ways. Thus, for every one [b]combiniation[/b] of boys, there are n! [b]permutations[/b] of boys. Similarly, for every one [b]combination[/b] for boys, there are m! [b]permutations[/b] of girls. Thus, for every one combination of a way boys and girls can be seated, there are m!n! permutations. Now, there are (m+n)! permutations in which the boys and girls can be seated (since there are m+n people total), thus,\n\\[C(m+n,m)=C(m+n,n)=\\frac{(n+m)!}{n!m!}\\]\n[/hide] Well, I hope this helped, but I am really bad at explaining.", "Solution_2": "im sure that these two formulas are worth memorizing because they appear so much. spend 5 minutes to memorize them\r\n\r\nCombinations k objects out of n objects: $\\frac{n!}{k!(n-k)!}$\r\nPermutations k objects out of n objects: $\\frac{n!}{(n-k)!}$ \r\n\r\nof course understand them too....", "Solution_3": "[quote=\"davidlizeng\"]im sure that these two formulas are worth memorizing because they appear so much. spend 5 minutes to memorize them\n\nCombinations k objects out of n objects: $\\frac{n!}{k!(n-k)!}$\nPermutations k objects out of n objects: $\\frac{n!}{(n-k)!}$ \n\nof course understand them too....[/quote]\r\nIn the first one, it that basically \"n choose k\"?", "Solution_4": "quevvy's right.", "Solution_5": "[quote=\"Quevvy\"][quote=\"davidlizeng\"]im sure that these two formulas are worth memorizing because they appear so much. spend 5 minutes to memorize them\n\nCombinations k objects out of n objects: $\\frac{n!}{k!(n-k)!}$\nPermutations k objects out of n objects: $\\frac{n!}{(n-k)!}$ \n\nof course understand them too....[/quote]\nIn the first one, it that basically \"n choose k\"?[/quote]\r\n\r\nyeah remember in your future years that \r\n\r\n$\\binom{9}{3}$ means to 9 choose 3, and you use the formula $\\frac{n!}{k!(n-k)!}$ to figure out", "Solution_6": "combinations is also\r\n\r\n(this is pretty easu)\r\n\r\n# of permutation\r\n______________\r\n(#u are picking)!\r\n\r\n\r\n-jorian", "Solution_7": "choose is important but memorizing the permutation formula is pointless\r\n\r\njust use logic\r\n\r\nthen again you really dont need to memorize choose but it pops up much more often", "Solution_8": "memorizing kills your brain\r\n\r\n\r\n\r\ndid you know that your brain slows down because your brain is too full of information as you get older? Don't stick useless stuff that you don't need to memorize exactly in your brain. :P" } { "Tag": [], "Problem": "Sea ABC un triangulo acutangulo, y sea O su circuentro. El circulo a traves de C, O y B se llama S. Las rectas AB y AC cortan al circulo S de nuevo en P y Q respectivamente. Prueba que las rectas AO y PQ son perpendiculares.", "Solution_1": "Issam: es usted de Marruecos? sie s asi porque habla espa\u00f1ol? interesante...", "Solution_2": "Si, hablo espa\u00f1ol desde que era un crio :P", "Solution_3": "sea m 1.[/hide][/hide]", "Solution_4": "[hide=\"Solution\"] The area of the above parallelogram is $ ad \\minus{} bc$ (can you prove it?). We are given that this area is $ 1$. By Pick's Theorem, and since the quadrilateral has at least $ 4$ boundary lattice points, it has exactly $ 4$ boundary lattice points and no interior lattice points.\n\nThe fact that it has no boundary lattice points implies #1. Otherwise, if $ \\frac {a}{b} \\equal{} \\frac {a'}{b'}$ is a simplification then $ (b', a')$ is on the side between $ (0, 0)$ and $ (b, a)$, and the same for the other side.\n\nThe fact that it has no interior lattice points implies #3.\n\nThe fact that the parallelogram is convex and that there are no other interior and boundary lattice points implies #2. The line from the origin to $ (b \\plus{} a, d \\plus{} c)$ is between the other two lines, so its slope is also strictly between the slopes of the other two lines. And of course the parallelogram generated by $ (b \\plus{} a, d \\plus{} c)$ and either of the other two points is still a parallelogram of area $ 1$, again by Pick's Theorem.[/hide]\r\nSee [url=http://en.wikipedia.org/wiki/Farey_sequence]Farey sequence[/url].", "Solution_5": "Thanks a lot...\r\nI wonder if there is another way of solving this", "Solution_6": "You can do everything with just algebra, but it's not quite as nice. The geometric interpretation is a much more interesting framework to view this problem.", "Solution_7": "You said: The area of the above parallelogram is (can you prove it?). Why?\r\nCould you elaborate more on your explanations, please?", "Solution_8": "[quote=\"theax\"]You said: The area of the above parallelogram is (can you prove it?). Why?\nCould you elaborate more on your explanations, please?[/quote]\r\nIf you surf the site enough, you should see a proof without words for it in that flash applet on the left side of the screen. :D", "Solution_9": "Start by proving that a triangle with coordinates (0,0),(a,b),(c,d) has area\r\n$ \\frac{|ad \\minus{} bc|}{2}$", "Solution_10": "Do you have a much simpler way?", "Solution_11": "You can prove it geometrically by drawing the rectangle with sides $ a \\plus{} b, c \\plus{} d$ around the parallelogram and subtracting the appropriate areas, but I find that proof unsatisfactory. The fun proof is with [url=http://en.wikipedia.org/wiki/Cross_product]cross products[/url].", "Solution_12": "Thanks...\r\n\r\nIs there an easier way to solve these series of problems using just algebra?", "Solution_13": "Anyone willing to help?", "Solution_14": "[hide=\"an algebraic method\"]\nSay $ \\frac {a}{b} > \\frac {c}{d}$. As they are neighbor fractions, $ ad \\minus{} bc \\equal{} 1$. Now if $ \\frac {e}{f}$ is greater than $ \\frac {c}{d}$ but less than $ \\frac {a}{b}$, then $ \\frac {e}{f} > \\frac {c}{d}\\Rightarrow ed \\minus{} cf > 0$.\n\nBut also $ ed \\minus{} cf$ is an integer, so $ ed \\minus{} cf\\geq 1$. Similarly, from $ \\frac {a}{b} > \\frac {e}{f}$, we have $ af \\minus{} be\\geq 1$.\n\nWe want to bound $ f$, so eliminate $ e$ from the above two inequalities: $ b(ed \\minus{} cf) \\plus{} d(af \\minus{} be)\\geq b \\plus{} d$. The terms with $ e$ cancel and what remains on the left hand side is $ f(ad \\minus{} bc)$, which is just $ f$. So $ f\\geq b \\plus{} d$, which is what we wanted.\n[/hide]", "Solution_15": "THANKS A LOT EVERYBODY!!!!!!!!" } { "Tag": [ "linear algebra", "matrix", "geometry", "geometric transformation" ], "Problem": "i need help in finding determinant D (nxn n>1)\r\ndeterminant D in Attachments", "Solution_1": "The matrix is:\r\n\r\n$ \\left [\\begin{array}{ccccccc}1&2&3&\\cdots&n-2&n-1&n\\\\\\noalign{\\medskip}2&3&4&\\cdots&n-1&n&n\\\\\\noalign{\\medskip}3&4&5&\\cdots&n&n&n\\\\\\noalign{\\medskip}\\vdots&\\vdots&\\vdots&\\vdots&\\vdots&\\vdots&\\vdots\\\\\\noalign{\\medskip}n-1&n&n&\\cdots&n&n&n\\\\\\noalign{\\medskip}n&n&n&\\cdots&n&n&n\\end{array}\\right ]$\r\n\r\nNow just use elementary column operations to simplify the matrix but preserve the value of the determinant.\r\n\r\n(i) Take $ -1$ times the $ n-1$st column and add it to the $ n$th column to get $ D_{1,n}=1$ and $ D_{1,i}=0$ for $ i>1$.\r\n\r\n(ii) Expand about the last column, but since you have all zeros except for the first element, you have $ |D|=(-1)^{(n+1)}|D_{1,n}|$.\r\n\r\n(iii) Now take $ -1$ times $ n-2$nd column and add it to the $ n-1$st column to get all zeros in the final column of the $ |D_{1,n}|$ matrix with the exception of $ 1$ in the upper right position.\r\n\r\n...\r\n\r\n\r\nContinuing in like fashion, I believe you get $ |D|=\\left( (-1)^{\\frac{(n+1)(n+2)}{2}-1}\\right) n$.", "Solution_2": "thank you very much\r\n\r\nbut>>>>\r\n\r\nwhat do you mean this:\r\nn-1 st ,n th,n-2 nd\r\nany column ????", "Solution_3": "Npare wrote:\r\n\r\n[quote]thank you very much\n\nbut>>>>\n\nwhat do you mean this:\nn-1 st ,n th,n-2 nd\nany column ????[/quote]\r\n\r\n\r\nThis is basically just an English construct. So for example:\r\n\r\n$ n$ th $ \\rightarrow$ column $ n$\r\n$ (n-1)$ st $ \\rightarrow$ column $ n-1$\r\n\r\nand so forth.\r\n\r\nI remember bewildering some Francophone students with the phrase $ \\text{i-i}\\grave{e}\\text{me}$ as a translation of $ i$ th. Sorry.", "Solution_4": "To add some more detail,\r\n\r\nCall the matrix with last row first column n $ A_{n}$. Notice that the last row has the same entries in each column. If $ a_{1}$ is the first column, add $ \\minus{}a_{1}$ to columns 2 through n. This does not change the determinant so you may expand along the last row. The determinant $ \\left|{A_{n}}\\right| \\equal{} (\\minus{}1)^{n\\plus{}1}n|B|$ where B is some matrix whose determinant is to be taken. Let us denote the last row of B by $ B_{(n\\minus{}1)}$. Note that the ith row of $ A_{n}\\minus{}1$, $ 1\\le i < n\\minus{}1$, is given by $ (i\\minus{}1)*B_{(n\\minus{}1)}$ + $ B_{i}$ and that $ B_{(n\\minus{}1)}$is n*$ A_{(n\\minus{}1)}$. In other words, $ |B| \\equal{}\\frac{1}{{n\\minus{}1}}|A_{n\\minus{}1}|$. Put this back in the other equation and telescope to see that\r\n\r\n$ \\left|{A_{1}}\\right|\\equal{}1$ and $ \\left|{A_{m}}\\right| \\equal{}\\minus{}m(\\minus{}1)^{\\frac{{m(m\\plus{}1)}}{2}\\minus{}1}$ for m > 1." } { "Tag": [ "algebra", "polynomial", "vector", "superior algebra", "superior algebra unsolved" ], "Problem": "Let W equal the set of all polynomials in F[x] with degree less than n such that the limit of the sum of the coefficients of the terms x_i is 0. Show that W is a subspace of V_n and find a basis of W over F.\r\n\r\n(V_n is the set of all polynomials in F[x] with degree less than n)\r\n\r\nI know that to show W is a subspace of V, I must show that under the operations of V, W is a vector space over F[x]. But I don't know how to prove that in W, distribution exists.\r\n\r\nAlso, how do I find the basis?", "Solution_1": "I have no idea what you mean by \"limit\" here.", "Solution_2": "Sorry, I just edited it.", "Solution_3": "It still doesn't make sense, but at least there's enough there to guess. Do you mean \"such that the sum of the coefficients of the $ x^i$ terms is zero\"?\r\n\r\nWhen trying to prove that a subset is a subspace, some of the axioms are automatic. You only have to worry about whether operations stay in the subspace, since the other properties are inherited from those properties in the full space.\r\n\r\n[quote=\"rawkerxxx\"]But I don't know how to prove that in W, distribution exists. [/quote]Huh?", "Solution_4": "[quote=\"jmerry\"]It still doesn't make sense, but at least there's enough there to guess. Do you mean \"such that the sum of the coefficients of the $ x^i$ terms is zero\"?[/quote]\n\nYes.\n\nWhen trying to prove that a subset is a subspace, some of the axioms are automatic. You only have to worry about whether operations stay in the subspace, since the other properties are inherited from those properties in the full space.\n\n[quote=\"rawkerxxx\"]But I don't know how to prove that in W, distribution exists. [/quote]Huh?[/quote]\r\n\r\nWell, I need to show that W is itself a vector space over F[x], so I need to show that the axioms are satisfied. So is it just automatic that since F[x] is a field then I don't need to show distribution?[/url]" } { "Tag": [ "function", "limit", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: [0; \\infty) \\to \\mathbb{R}$ a derivable function. Prove that if $ \\lim_{x \\to \\infty} f(x)$ exists and $ \\lim_{x \\to \\infty } \\int_0^x f(t) dt \\equal{}0$ then $ \\lim_{x \\to \\infty} f(x)\\equal{}0$.", "Solution_1": "If $ \\lim_{x \\to \\infty} f(x) \\equal{} L \\ne 0$ then $ \\forall \\varepsilon \\exists x_0$ s.t. $ L \\minus{} \\varepsilon < f(x) < L \\plus{} \\varepsilon$ for $ x>x_0$ and then $ \\int_{x_0}^{\\infty} (L \\minus{} \\varepsilon)dt < \\int_{x_0}^{\\infty} f(t)dt < \\int_{x_0}^{\\infty} (L \\plus{} \\varepsilon)dt$. If $ \\varepsilon$ is chosen small enough, we get a contradiction.", "Solution_2": "[quote=\"CommandoGuard\"]If $ \\lim_{x \\to \\infty} f(x) \\equal{} L \\ne 0$ then $ \\forall \\varepsilon \\exists x_0$ s.t. $ L \\minus{} \\varepsilon < f(x) < L \\plus{} \\varepsilon$ for $ x > x_0$ and then $ \\int_{x_0}^{\\infty} (L \\minus{} \\varepsilon)dt < \\int_{x_0}^{\\infty} f(t)dt < \\int_{x_0}^{\\infty} (L \\plus{} \\varepsilon)dt$. If $ \\varepsilon$ is chosen small enough, we get a contradiction.[/quote]\r\nI'm not convinced by CommandoGuard's proof : $ x_0$ is depending on $ \\epsilon$ :P \r\n$ f$ is continuous so $ \\forall_n,\\exists_{n < x_n < n \\plus{} 1},\\int_n^{n \\plus{} 1}f(t)dt \\equal{} f(x_n)$. We can write $ \\int_0^{\\infty} f(t)dt \\equal{} \\sum_{n \\equal{} 0}^{ \\plus{} \\infty}\\int_n^{n \\plus{} 1}f(t)dt \\equal{} \\sum_{n \\equal{} 0}^{ \\plus{} \\infty}f(x_n)$\r\n$ \\sum_{n \\equal{} 0}^{ \\plus{} \\infty}f(x_n)$ is convergent because $ \\int_0^{\\infty} f(t)dt$ is convergent, so $ \\lim_{n\\rightarrow \\plus{} \\infty}f(x_n) \\equal{} 0$, where $ x_n\\rightarrow \\plus{} \\infty$ . But $ \\lim_{x\\rightarrow \\plus{} \\infty}f(x)$ exists, so $ \\lim_{x\\rightarrow \\plus{} \\infty}f(x) \\equal{} 0$\r\n :cool:", "Solution_3": "I don't think there is any problem with the first solution. We are not taking anything like $ \\\\epsilon \\rightarrow 0$; just fix $ \\epsilon$ so that the two bounds are of the same sign and diverge to the infinity then that's okay.", "Solution_4": "Ok, I agree :D . Finally the proof of CommandoGuard is better, because only the convergences of $ f(\\plus{}\\infty)$ and $ \\int_0^{\\plus{}\\infty}f(t)dt$ are necessary.\r\n :cool:" } { "Tag": [], "Problem": "While traveling home from school, Alyssa fell asleep halfway through the journey. When she awoke, she still needed to travel one-fourth of the distance that she had traveled while sleeping. For what part of the journey was she awake? Express your answer as a common fraction.", "Solution_1": "Let $ x$ be the distance she traveled while asleep.\r\n\r\nLet $ y$ be the total distance.\r\n\r\n$ y \\equal{} x \\plus{} \\frac{1}{4}x \\plus{} \\frac{1}{2}y \\implies y \\equal{} \\frac{5}{4}x \\plus{} \\frac{1}{2}y$\r\n\r\n$ \\frac{1}{2}y \\equal{} x \\plus{} \\frac{1}{4}x \\implies \\frac{1}{2}y \\equal{} \\frac{5}{4}x$\r\n\r\nWhat we want to find is the simplified value of $ \\frac{1}{2}y \\plus{} \\frac{1}{4}x$.\r\n\r\nWe can find $ x$ in terms of $ y$.\r\n\r\n$ \\frac{1}{2}y \\equal{} \\frac{5}{4}x$\r\n\r\n$ \\frac{2}{5}y \\equal{} x$\r\n\r\nWe want to find $ \\frac{1}{2}y \\plus{} \\frac{1}{4}x$, which we can now simplify to:\r\n\r\n$ \\frac{1}{2}y \\plus{} (\\frac{1}{4} \\cdot \\frac{2}{5}y)$\r\n\r\nor\r\n\r\n$ \\frac{1}{2}y \\plus{} \\frac{1}{10}y$\r\n\r\nThus, our answer is $ \\frac{1}{2} \\plus{} \\frac{1}{10}$ or $ \\boxed{\\frac{3}{5}}$.", "Solution_2": "You can also imagine the trip itself. For the first 1/2 the way, Alyssa was awake. Then, she travels some distance, and when she wakes up, she needs to travel 1/4 of the the way that asleep time. Thus, 5/4 of the asleep time is equal to one half the total time, so the total time is 10/4 times the asleep time. The first half is 5/4 of the asleep time, and the woke up time is 1/4 as stated earlier, so she was awake for 6/4 of the asleep time. The fraction of time she was awake is $ \\frac{\\frac{6}{4}}{\\frac{10}{4}}\\equal{}\\frac{6}{10}\\equal{}\\frac{3}{5}$ of the total time." } { "Tag": [ "geometry", "rectangle" ], "Problem": "Question\r\n\r\nA rectangular area can be completely tiled with 200 square tiles.\r\nIf the side length of each tile was increased by 1cm, it would take only 128 tiles to tile the area.\r\nFind the side length of each tile.", "Solution_1": "If we let $ a$ and $ b$ be the lengths of the sides of the rectangle in centimeters and let $ x$ be the length od each side of the square tile in centimeters,\r\nThe number of tiles needed to cover the rectangle is $ \\frac{ab}{x^2}\\equal{}200$.\r\nIf the length of each side is increased by 1 cm then the side length is $ x\\plus{}1$.\r\nThe number of tiles that will be required now will be $ \\frac{ab}{(x\\plus{}1)^2}\\equal{}128$.\r\nSimplifying we get:\r\n $ ab\\equal{}200(x^2)$\r\n $ ab\\equal{}128(x\\plus{}1)^2$\r\n\r\nBy the Transitive property, we get:\r\n $ 200x^2\\equal{}128x^2\\plus{}256x\\plus{}128$\r\n $ 72x^2\\minus{}256x\\minus{}128\\equal{}0$\r\n $ 9x^2\\minus{}32x\\minus{}32\\equal{}0$\r\n $ (9x\\plus{}4)(x\\minus{}4)\\equal{}0$\r\n $ x\\equal{}\\minus{}\\frac{9}{4}\\text{ or }4$\r\n\r\nTherefore, the length of each side of the square tile is 4 cm." } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "Show that for y'''' = (a^4)y, y(0) = 0, y'(0) = 0, y(L) = 0, y'(L) = 0, to have a non-trivial solution cos(aL)cosh(aL) = 1. Assume a does not equal 0.\r\n\r\nI have the following hint: Show that a basis for the null space of L = x'''' - a^4 is (cosh(ax), sinh(ax), cos(ax), sin(ax)).\r\n\r\nHow would I show this?\r\n\r\nThanks for any help in advance.", "Solution_1": "I do it and I get that cos(aL) = cosh(aL), but that must be wrong. \r\n\r\nI let y(x) = Acosh(ax) + Bsinh(ax) + Ccos(ax) + Dsin(ax) for some constants A,B,C,D. \r\nFor y(0) = 0, I get that C = -A and y'(0) = 0, it seems that D = -B. \r\nI'm not sure what to do after this." } { "Tag": [ "geometry", "inradius" ], "Problem": "Square ABCD is located inside of a circle with center [i]O [/i]and radius R in such a way that verticies B and C belong to the circle and side AD contains point [i]O[/i]. Find [i]r[/i][size=59]BOC[/size]", "Solution_1": "well you only need to note that the inradi of a triangle with sides $ a,b,c$ and semiperimter $ s$ is given by:\r\n$ r\\equal{}\\sqrt{\\frac{(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}{s}}$ and use Pithagoras in triangle $ ODC$.\r\n\r\n\r\n :)" } { "Tag": [ "limit", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Can anyone help me with these two problems?\r\n\r\nlim x^2cos10x as x->0\r\n\r\nand the second:\r\n\r\nlim (1/t^2)sin^2(t/2) as t->0", "Solution_1": "$ \\lim_{x\\to 0}x^{2}\\cos(10x) \\equal{} 0\\cdot 1 \\equal{} 0$.\r\n\r\n$ \\lim_{t\\to 0}\\frac{1}{t^{2}}\\sin^{2}\\left(\\frac{t}{2}\\right) \\equal{}\\frac{1}{4}\\lim_{t\\to 0}\\left[\\frac{\\sin\\left(\\frac{t}{2}\\right)}{\\frac{t}{2}}\\right]^{2}\\equal{}\\frac{1}{4}\\cdot 1^{2}\\equal{}\\frac{1}{4}$.\r\n\r\nIn the second one I used the fact that $ \\lim_{x\\to 0}\\frac{\\sin x}{x}\\equal{} 1$." } { "Tag": [ "algebra", "polynomial", "algebra solved" ], "Problem": "Let w be a polynomial of degree two with integer coeff. Suppose that for each integer x the value w(x) is the square of an integer. Prove that w is a square of a polynomial.", "Solution_1": "This one appeared many times here.\r\nThe last and better result on that problem is Peter Scholze's proof of Harazi conjecture, namely :\r\nIf $P$ has integer coefficients and $P(2^n)$ is a square for all non-negative integers $n$, then $P$ is the square of a polynomial with integer coefficients.\r\n\r\nLook here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial+square&t=15102\r\n\r\nPierre." } { "Tag": [ "algebra", "polynomial", "linear algebra", "linear algebra unsolved" ], "Problem": "Suppose that $ P_n$ be the space of all polynomials whose degrees $ \\leq n$ with coefficients $ \\in \\mathbb{R}$. We define the inner product for given $ f(x)\\equal{}\\sum_{i\\equal{}0}^n a_ix^i , g(x)\\equal{}\\sum_{i\\equal{}0}^n b_ix^i \\in P_n$ as following : $ \\left\\langle f,g\\right\\rangle\\equal{}\\sum_{i\\equal{}0}^n a_ib_i$. Suppose that $ L \\subseteq P_n$ be the set of all polynomials like $ f$ s.t. $ f(1)\\equal{}0$ (obviously $ L$ is a subspace). For $ g \\in P_n$ we define the distance between $ g$ and $ L$ as $ \\rho (g,L)$. Prove that $ \\forall g \\in P_n , \\rho (g,L)\\equal{}\\frac{g(1)}{\\sqrt{n\\plus{}1}}$.", "Solution_1": "The codimension of $ L$ is 1. Therefore, $ \\dim L^{\\perp} \\equal{} 1$. It is easy to see that $ h(t) \\equal{} 1 \\plus{} t \\plus{}\\dots \\plus{}t^n\\in L^{\\perp}$ since $ \\langle f(t), h(t)\\rangle \\equal{} f(1)$. Therefore, $ L^{\\perp} \\equal{} \\{\\lambda h\\}$.\r\nNow the distance from $ g$ to $ L$ equals the length of the projection of $ g$ onto $ L^{\\perp}$. We have,\r\n\\[ \\rho(g, L) \\equal{} \\frac{|\\langle g, h\\rangle|}{\\|h\\|} \\equal{} \\frac{|g(1)|}{\\sqrt{n\\plus{}1}}.\\]" } { "Tag": [ "trigonometry" ], "Problem": "Find:\r\n\\[ \\frac{{\\cos 37^0 \\plus{} \\sin 37^0 }}\r\n{{\\cos 37^0 \\minus{} \\sin 37^0 }}\r\n\\]\r\n\r\n\r\n[hide]ans:$ cot8^0$[/hide]", "Solution_1": "[hide]$ {\\cos 37^\\circ\\plus{}\\sin 37^\\circ\\over\\cos 37^\\circ\\minus{}\\sin 37^\\circ}\\equal{}\\frac{\\sqrt{2}\\sin(45^\\circ\\plus{}37^\\circ)}{\\sqrt{2}\\sin(45^\\circ\\minus{}37^\\circ)}\\equal{}{\\sin 82^\\circ\\over\\sin 8^\\circ}\\equal{}\\cot 8^\\circ$[/hide]" } { "Tag": [], "Problem": "little jacob had a pie and gave 1/4 of it to his dear little mommy\r\nthen, he gave 1/3 of what was left to his dear little daddy :rotfl: \r\nlastly, he ate 3/16 of the portion left\r\nwhats left? (fraction)", "Solution_1": "[quote=\"aznmonkey1992\"]little jacob had a pie and gave 1/4 of it to his dear little mommy\nthen, he gave 1/3 of what was left to his dear little daddy :rotfl: \nlastly, he ate 3/16 of the portion left\nwhats left? (fraction)[/quote]\r\n\r\n[hide]\n$\\frac{1}{4}+\\frac{3}{4}\\cdot\\frac{1}{3} = \\frac{1}{4}+\\frac{1}{4}=\\frac{1}{2}$\n$\\frac{1}{2}\\cdot\\frac{3}{16}=\\frac{3}{32}$\n$\\frac{16}{32}+\\frac{3}{32}=\\frac{19}{32}$\n$1-\\frac{19}{32}=\\boxed{\\frac{13}{32}}$\n[/hide]", "Solution_2": "[quote=\"aznmonkey1992\"]little jacob had a pie and gave 1/4 of it to his dear little mommy\nthen, he gave 1/3 of what was left to his dear little daddy :rotfl: \nlastly, he ate 3/16 of the portion left\nwhats left? (fraction)[/quote]\r\n$1-(1/4+1/4+3/32)=1-(16/32+3/32)=13/32$", "Solution_3": "[quote=\"aznmonkey1992\"]little jacob had a pie and gave 1/4 of it to his dear little mommy\nthen, he gave 1/3 of what was left to his dear little daddy :rotfl: \nlastly, he ate 3/16 of the portion left\nwhats left? (fraction)[/quote]\r\n[hide]\n\n1-1/4=3/4\n2/3x3/4 = 6/12 = 1/2\n13/16 x 1/2 = 13/32\n[/hide]", "Solution_4": "[hide]\n\n$1 - \\frac{1}{4} = \\frac{3}{4}$\n\n$\\frac{3}{4} * \\frac{1}{3} = \\frac{1}{4}$\n\n$\\frac{1}{2} * \\frac{3}{16} = \\frac{3}{32}$\n\n$1 - \\frac{19}{32} = \\frac{13}{32}$[/hide]", "Solution_5": "[quote=\"aznmonkey1992\"]little jacob had a pie and gave 1/4 of it to his mommy\nthen, he gave 1/3 of what was left to his daddy \nlastly, he ate 3/16 of the portion left\nwhats left? (fraction)[/quote]\r\n\r\nhe started with 1. then after his mommy, he still had 3/4. He gave 1/3 of 3/4 to his daddy, or 1/4. So he had 1/2 left. Then he ate 3/16 of 1/2 or 3/32 of what was left. Therefore, there was 13/32 of the pie left." } { "Tag": [ "inequalities", "number theory solved", "number theory" ], "Problem": "Find all $x,y,z,\\in Z^+$ satisfying the equation:\r\n$\\sqrt {xyz} - \\sqrt x - \\sqrt y - \\sqrt z =2$", "Solution_1": "There's a standard way to tackle this, by using inequalities after setting $x\\le y\\le z$. However, we can do it algebraically. The number on the left is a member of the field $\\mathbb Q[\\sqrt x,\\sqrt y,\\sqrt z]$, and it can be shown that such numbers are rational iff the coefficients of each irrational among $\\sqrt x,\\sqrt y,\\sqrt z$ and the combinations formed by the multiplications of two or three of these ($\\sqrt {yz},\\sqrt{zx},\\sqrt{xy},\\sqrt{xyz}$) is $0$. If $\\sqrt x$ is irrational (for example), then $\\sqrt {yz}=1$, and this doesn't lead to a solution. This means that $x,y,z$ are, in fact, squares, so we can rewrite the equation in positive integers: $mnp-m-n-p=2$. \r\n\r\nIt's now much easier to employ the usual methods: $\\frac 2{mnp}+\\frac 1{mn}+\\frac 1{np}+\\frac 1{pm}=1$, and we may set $m\\le n\\le p$. From here we get $m\\le 2$, and if $m=2$, then $n=p=2$, so this is a solution: $(m,n,p)=(2,2,2)\\Rightarrow (x,y,z)=(4,4,4)$. If $m=1$, then $\\left(\\frac 3n+1\\right)\\left(\\frac 3p+1\\right)=4$, so $n\\le 3$. $n=3\\Rightarrow p=3$. Here's another solution: $(x,y,z)=(1,9,9)$. $n=2\\Rightarrow p=5$, so another solution is $(x,y,z)=(1,4,25)$. Finally, if $n=1$, then there's no solution. \r\n\r\nThe only solutions are thus $(x,y,z)=(4,4,4),(1,9,9)$ or $(1,4,25)$, and their permutations. \r\n\r\nI hope it's Ok." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "inequalities", "algebra", "polynomial", "function" ], "Problem": "Hello, Please comment which country conducts the hardest team selection test and national olympiad. \r\n THANKS", "Solution_1": "china, obvious. Basically CMO is harder than IMO", "Solution_2": "yeah basically the countries are china,romania ,usa.but i've never seen the vietnamese selction test nor the russian ones.duz ne1 have ne idea where they cud be found.and ,siuhochung,u have china 2002-2003 official solution? :? \r\nbest regards", "Solution_3": "[quote=\"galois\"]yeah basically the countries are china,romania ,usa.but i've never seen the vietnamese selction test nor the russian ones.duz ne1 have ne idea where they cud be found.and ,siuhochung,u have china 2002-2003 official solution? :? \nbest regards[/quote]\r\n\r\ni dun hv, but you may ask members from china, or charlie, siutsz, they have joined cmo 2003", "Solution_4": "[quote=\"galois\"]yeah basically the countries are china,romania ,usa.but i've never seen the vietnamese selction test nor the russian ones.duz ne1 have ne idea where they cud be found.and ,siuhochung,u have china 2002-2003 official solution? :? \nbest regards[/quote]\r\n\r\nthe USAMO 2003 was fairly easy (at least for four questions). if i can do them, then it is only fair to classify as fairly easy. :D :D :D \r\nI have looked at ChMO, and they are very difficult. \r\n\r\nThe CMO (canadian) from 2003 was a joke... the trival questions, two standard (aka easy) questions, and one question that is a little bit more difficult but is still easy.", "Solution_5": "For China MO official solutions they were given as a talk for leaders only, during the 2nd day of contest. I could possibly get the solution for 2002 but not 2003 because our leader in 2003 didn't attend the talk at all :( I will try to find the Macau team leader for the solution", "Solution_6": "In france, we have two olympiads :\r\n-the academic math olympiad, really easy (designed for common high school students ot the 2nd year)\r\n-the general competition, really difficult, for high school last year students, but not like the imo : there's a big problem with many many sub-questions to be solved.\r\n\r\nfor the vietnamese selction test of 1994 : (2*4h)\r\n\r\n1/ ABCD is a parallelogram. E is a point of BC and F a point of CD so that ABE and BDF have the same area. BD crosses AE at the point M and AF at N. Prove that BM, MN, ND follows the triangular inequality, and that E and F moves so that MN goes smaller, then the radius of the circle meeting the 3 summits (sorry, don't know how to say in English...) of a triangle which have for lengths BM, MN, ND also goes smaller.\r\n\r\n2/ Let the equation x+y+z+t-Nxyzt-N=0 (*) (N integer >0).\r\nProve that for an infinity of values of N, (*) have a solution in positive integers, and that for N=(4^k)(8m+7) (k, m nonnegative integers), (*) has no solutions in positive integers.\r\n\r\n3/ P is a polynomial function of degree 4, with 4 real roots.\r\nProve that the equation (1-4x)P(x)/x+(1-(1-4x)/x)P'(x)-P''(x)=0 has also 4 real roots.\r\n\r\n4/ ABC is a triangle with AB=BC=CA (dunno what's this in English...), and M is a point. D is the midpoint of AD. DM and AP crosses at point N. The images by the symetry of center M of the points A, B, C are A', B', C'. Prove that there is an unique point P so that AA=PB', PB=PC', PC=PA', and that when M moves while being different of D, the circle meeting M, N, P keeps meeting a point which is not moving.\r\n\r\n5/ Find all the functions f:R->R so that :\r\nf(x sqrt(2))+f(x (4+3sqrt(2))=2f(x (2+sqrt(2)).\r\n\r\n6/ Compute sum(1/(n1!n2!...n1994!(n2+2n3+...+1993n1994)!)), with the sum on all the 1994-uplets (n1,...,n1994) of nonnegative integers so that n1+2n2+...+1994n1994=1994.\r\n\r\nDifficult, isn't it ?", "Solution_7": "I've got a selection of past Vietnamese nat'l Tests. If anyone cares, I can post them here. General impression is that they are hard ;) ( and kind of requiring university-level knowledge, if you ask me )", "Solution_8": "it depends what you define university level by. for example in romania we do in 11th and 12th grades what people in us do in 1st and 2nd year of college. (you can see the problems in the college playground).", "Solution_9": "hi guys \r\ni just wanna say that there are selection tests are very very easy \r\nalso the preparation for the imo is very short not efficace\r\nMorrocan team is selected after just 3tests (asy tests)\r\ni hope that this problem will be solved as i was a victime of the easy tests :? \r\nany way !\r\ni just wanna ask for the mathematical program in college or high schools in romania or Russia \r\ni think our program is :blush: \r\nthank for all guys by!", "Solution_10": "[quote=\"Anonymous\"][quote=\"galois\"]yeah basically the countries are china,romania ,usa.but i've never seen the vietnamese selction test nor the russian ones.duz ne1 have ne idea where they cud be found.and ,siuhochung,u have china 2002-2003 official solution? :? \nbest regards[/quote]\n\nthe USAMO 2003 was fairly easy (at least for four questions). if i can do them, then it is only fair to classify as fairly easy. :D :D :D \nI have looked at ChMO, and they are very difficult. \n\nThe CMO (canadian) from 2003 was a joke... the trival questions, two standard (aka easy) questions, and one question that is a little bit more difficult but is still easy.[/quote]\r\n\r\nYeah, I live in Canada and I can see students in my class using a calculator for 2-3 (-1). It's sad. No offence! There are also lots of intelligent people, but hard to find.", "Solution_11": "Britain has many selection tests, but the first few ones are really easy, which is pointless to select students for the IMO team members. The first selection test is even ridiculous, which is composed of 25 IQ-like muliple choice questions.", "Solution_12": "[quote=\"boxedexe\"]Yeah, I live in Canada and I can see students in my class using a calculator for 2-3 (-1). It's sad. No offence! There are also lots of intelligent people, but hard to find.[/quote]\r\n\r\nOf course, the relationship between CMO qualifiers and the students in your class using a calculator to subtract numbers is nonexistent. So your post is pointless.\r\n\r\nBut, more importantly, don't resurrect year old threads in the future.", "Solution_13": "iran is...\r\n\r\ni think...", "Solution_14": "Do you even read threads all the way through? Don't bump old threads!", "Solution_15": "What about Argentinian TST for IMO??? (spanish version)\r\n\r\nhttp://www.oma.org.ar/enunciados/index.htm#imo", "Solution_16": "[quote=\"blahblahblah\"]Do you even read threads all the way through? Don't bump old threads![/quote]\r\n\r\nwhat do u mean? :!: \r\n\r\nif u a means every one, use Quote.", "Solution_17": "Romania have very good problems, but are and easy(some of problems) :) ! Iran and China have very big problems :( .", "Solution_18": "i dont khow a lot about other countries.but i suppose IRAN have one of most hard selections.\r\n :(", "Solution_19": "[quote=\"kelvinkklee\"]Britain has many selection tests, but the first few ones are really easy, which is pointless to select students for the IMO team members. The first selection test is even ridiculous, which is composed of 25 IQ-like muliple choice questions.[/quote]\r\n\r\nThe first few are meant to be easy. The SMC is designed to be accessible to as many people as possible.\r\n\r\nBMO 1 is also suppose to be easy so that it can interest people who've never heard of the Olympiads before and also to test a few of the better people (though not much).\r\n\r\nBMO 2 is the first test when it starts getting hard and this year was no exception. With 55 people getting 2 or less out of 180 entrants, I think its fair to say that its hardly \"too easy\".\r\n\r\n(Although compared to USAMO etc it is easy). However, there are many more TSTs once the Trinity lot have had their training and after Oundle.\r\n\r\n[b]Edit: Whoa, I didn't see how old this thread was![/b]", "Solution_20": "[quote=\"SimonM\"][b]Edit: Whoa, I didn't see how old this thread was![/b][/quote]Almost 5 years :)", "Solution_21": "Except someone else had responded to it more recently (like yesterday) which confused me.", "Solution_22": "It's still a valid topic nevertheless. Even in our modern times :lol:", "Solution_23": "What are people's opinions now? Is China still Numero Uno? \r\n\r\nI don't think India's selection tests are the hardest of the lot.. But I believe others (Soumyashant, Chappli, etc) would be better suited to comment on this than myself.", "Solution_24": "I am from China\r\nYes,I think CMO is little harder than IMO,but I should clarify that CMO is not the IMO team selection test.\r\n\r\nCMO is just used to form the national training team which has about 35 students(seems have expanded to 45 in the last two years)\r\n\r\nAnd in national training team,there are generally 6 small tests and 2 big ones.\r\nBecause all guys of national training team have the ability to get IMO medals.The problems of national training team tests are harder than those of CMO.\r\nSome problems are even much much harder which almost no one from China can solve out.", "Solution_25": "[quote=\"galoisj\"]\n... all guys of national training team have the ability to get IMO medals ...[/quote] :o Indonesia has been taking part in IMO for 20 years. But we have yet to get gold. There are only two students who got silver, in 2001 and 2007. What do you think the differences between China and Indonesia?", "Solution_26": "[quote=\"Johan Gunardi\"][quote=\"galoisj\"]\n... all guys of national training team have the ability to get IMO medals ...[/quote] :o Indonesia has been taking part in IMO for 20 years. But we have yet to get gold. There are only two students who got silver, in 2001 and 2007. What do you think the differences between China and Indonesia?[/quote]\r\n\r\n1) Math olympiad history and the structure of the olympiad in China (math camps etc)\r\n2) Number of students that participate", "Solution_27": "[quote=\"Amazigh\"][quote=\"Johan Gunardi\"][quote=\"galoisj\"]\n... all guys of national training team have the ability to get IMO medals ...[/quote] :o Indonesia has been taking part in IMO for 20 years. But we have yet to get gold. There are only two students who got silver, in 2001 and 2007. What do you think the differences between China and Indonesia?[/quote]\n\n1) Math olympiad history and the structure of the olympiad in China (math camps etc)\n2) Number of students that participate[/quote]\r\n\r\nYeah,you are right. :)", "Solution_28": "[quote=\"galoisj\"]I am from China\nYes,I think CMO is little harder than IMO,but I should clarify that CMO is not the IMO team selection test.\n\nCMO is just used to form the national training team which has about 35 students(seems have expanded to 45 in the last two years)\n\nAnd in national training team,there are generally 6 small tests and 2 big ones.\nBecause all guys of national training team have the ability to get IMO medals.The problems of national training team tests are harder than those of CMO.\nSome problems are even much much harder which almost no one from China can solve out.[/quote]\r\n\r\n[url=http://www.mathlinks.ro/resources.php?c=37&cid=47]China TST[/url] has been regarded as very challenging though not always the most beautiful problems. And the [url=http://www.mathlinks.ro/index.php?f=10]Mathlinks contest[/url] which only took place in rare cases in the most recent years. As you could see we have a few years of the China TST in the resources section. Can you tell us more about the \"national training team tests\". Do you know the original sources of those problems. Possibly you can post those problems here. \r\nRomania TST 2008 looks odd. Most problems are based on AMM problems. What is going on? Are they running out of ideas or good people who propose interesting problems. Maybe they think people are training with ISLs anyway, and why not motivate those students to read AMM more in detail?!", "Solution_29": "Well I was kind of upset on the Romanian selection tests this year. I sent in a few problems of my own, which didn't get selected (I used most of them for MLC anyway in the end.)", "Solution_30": "[quote=\"Valentin Vornicu\"]Well I was kind of upset on the Romanian selection tests this year. I sent in a few problems of my own, which didn't get selected (I used most of them for MLC anyway in the end.)[/quote]\r\n\r\nI can't seem to locate all problems from the recent MLCs. Where are they? I saw one of your problems (the 8th problem in the list which is a geometry one) as proposal for RMC 2008. For the different contests and levels 71 problems were proposed in total. I mean they have some alternatives so I don't understand they use so many known problems from AMM." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "For $ A \\equal{} \\left( \\begin{array}{cc} 1 & 2 \\\\\r\n3 & 4 \\end{array} \\right),\\ B \\equal{} \\left( \\begin{array}{cc} 0 & \\minus{} 1 \\\\\r\n\\minus{} 2 & 3 \\end{array} \\right),$ find $ (A \\plus{} B)^2, A^2 \\plus{} 2AB \\plus{} B^2.$\r\n\r\nQuestion from kunny:\r\n\r\nWhat should we learn from the problem?", "Solution_1": "[quote=\"kunny\"]Question from kunny:\n\nWhat should I learn from the problem?[/quote]\r\nOnce again, commutativity - or rather the lack of commutativity - is the issue.", "Solution_2": "Needless to say, that's right.\r\n\r\nWhat would become $ (A\\plus{}B)^2$ for matrices $ A,\\ B$ generally?\r\n\r\nI had asked a question like this before. I do know the answer.", "Solution_3": "$ A^2 \\plus{} AB \\plus{} BA \\plus{} B^2.$ For some $ A$ and $ B$, and in this case,$ AB \\neq BA,$ so you cannot simplify this using commutivity of multiplication, since it is generally NOT true given two randomly selected matrices.", "Solution_4": "[quote=\"kunny\"]\n\nWhat would become $ (A \\plus{} B)^2$ for matrices $ A,\\ B$ generally?[/quote]\n\n[quote=\"rob22\"]$ A^2 \\plus{} AB \\plus{} BA \\plus{} B^2.$ For some $ A$ and $ B$, and in this case,$ AB \\neq BA,$ so you cannot simplify this using commutivity of multiplication, since it is generally NOT true given two randomly selected matrices.[/quote]\r\n\r\nPerfect! :)" } { "Tag": [ "inequalities", "Euler", "inequalities proposed" ], "Problem": "Prove that\r\n\r\n$ 8R^2\\minus{}5r^2\\leq r_a^2\\plus{}r_b^2\\plus{}r_c^2\\leq16R^2\\minus{}24Rr\\plus{}11r^2$", "Solution_1": "[quote=\"Ligouras\"]Prove that\n\n$ 8R^2 \\minus{} 5r^2\\leq r_a^2 \\plus{} r_b^2 \\plus{} r_c^2\\leq16R^2 \\minus{} 24Rr \\plus{} 11r^2$[/quote]\r\n\r\nwe have \r\n\r\n\\[ \\sum {r_a^2 } \\equal{} (4R \\plus{} r)^2 \\minus{} 2s^2\\]\r\nand use Gerretsen Inequality $ 4R^2 \\plus{} 4Rr \\plus{} 3r^2 \\ge s^2 \\ge 16Rr \\minus{} 5r^2$ and Euler Inequality $ R \\ge 2r$ can prove it." } { "Tag": [ "trigonometry", "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ be the angles of a triangle. Prove that\r\n\r\n$ 8\\sin{\\frac{a}{2}}\\sin{\\frac{b}{2}}\\sin{\\frac{c}{2}} \\leq \\sqrt{\\cos{\\frac{a-b}{2}}\\cos{\\frac{b-c}{2}}\\cos{\\frac{c-a}{2}}}$", "Solution_1": "[quote=\"manlio\"]Let $a,b,c$ be the angles of a triangle. Prove that\n\n$ 8\\sin{\\frac{a}{2}}\\sin{\\frac{b}{2}}\\sin{\\frac{c}{2}} \\geq \\sqrt{\\cos{\\frac{a-b}{2}}\\cos{\\frac{b-c}{2}}\\cos{\\frac{c-a}{2}}}$[/quote]\r\n\r\nDear manlio, \r\nYour inequality is very nice but $ \\geq\\ $ should be changed to $ \\leq\\ $\r\n\r\nTo prove it use these indentities :\r\n$ \\sin{\\frac{a}{2}}\\sin{\\frac{b}{2}}\\sin{\\frac{c}{2}} = \\frac{r}{4R} $\r\n${ cos{\\frac{a-b}{2}}cos{\\frac{b-c}{2}}\\cos{\\frac{c-a}{2}}} = \\frac{p^2+2Rr+r^2}{8R^2} $\r\n\r\nOur inequality becomes : $ p^2+2Rr \\geq\\ 31r^2 $ which is very easy to prove.", "Solution_2": "Sorry, I will correct it at once." } { "Tag": [ "LaTeX", "integration", "calculus", "calculus computations" ], "Problem": "In my haste to learn latex I made a horrible mistake. :blush: I apologize for it. I meant to ask you all to solve \r\n$\\int\\frac{x^{5}}{(x^{3}+1)^{2}}dx$\r\n\r\nI do thank you for your efforts to solve\r\n\r\n$\\int\\frac{5x}{(x^{3}+1)^{2}}dx$\r\n\r\nActually I think the top integral is easier to solve than the bottom one.", "Solution_1": "It's far easier. Substitute $u=x^{3}+1$.", "Solution_2": "i think this is simple integration\r\nsubstitute $x^{3}= t$ then $3x^{2}= dt$\r\nnow my integration becomes $\\int{\\frac{tdt}{3(t+1)^{2}}}$\r\nwhich can be written as $\\frac{1}{6}({\\int{\\frac{(2t+2)dt}{t^{2}+2t+1}}-2\\int{\\frac{dt}{(t+1)^{2}}}})$\r\non integrating this we get\r\n$\\frac{1}{3}(ln|t+1|+\\frac{1}{t+1})+c$\r\non substituting the value of t we get\r\n$\\frac{1}{3}(ln|x^{3}+1|+\\frac{1}{x^{3}+1})+c$", "Solution_3": "my approach is to factor $\\frac{x^{5}}{(x^{3}+1)^{2}}$\r\nas\r\n\r\n\r\n$\\frac{x^{3}}{3}\\cdot\\frac{3x^{2}}{(x^{3}+1)^{2}}$\r\n\r\nThen integrate by parts\r\n\r\nI don't have the solution in front of me at present and my latex skills are not that good right now so maybe someone could show this integration by parts. BTW there is a substituition after the integration by parts but it is trivial\r\n\r\n\r\nBack to the latex to try to figure out how to write out integration by parts.\r\n\r\nIt is interesting how these integrals can have more than one solution. Is there a word for that?" } { "Tag": [ "logarithms" ], "Problem": "How do we calculate $100!$, of course without using any tools?\r\n\r\nMore specifically, what is the \"lest most significant digit\" for $100!$?", "Solution_1": "I guess you mean \"right most significant digit\" :wink: \r\n\r\nSimilar problem posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ravi&t=102715]here[/url].", "Solution_2": "[quote=\"Kurt G\u00f6del\"]I guess you mean \"right most significant digit\" :wink: \n\nSimilar problem posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ravi&t=102715]here[/url].[/quote]\r\n\r\nThe right most digit is $0$, isnt it?\r\n\r\nI meant the \"left most digit\" only... thats why I asked for a method to evaluate this monster...", "Solution_3": "[quote=\"xxxyyyy\"][quote=\"Kurt G\u00f6del\"]I guess you mean \"right most significant digit\" :wink: \n\nSimilar problem posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ravi&t=102715]here[/url].[/quote]\n\nThe right most digit is $0$, isnt it?\n\nI meant the \"left most digit\" only... thats why I asked for a method to evaluate this monster...[/quote]\r\nright most [b]significant[/b] (>0) digit", "Solution_4": "ok well 100! end in 24 zeros, so to get the rightmost non zero number we must divide by 10^24. Then, look at the leftover numbers mod 10 and simplify. It's not a hard process, but I won't actually do it as 100 is kindof large for this, I think when i saw this it was 24! instead.", "Solution_5": "and to find the leading digit you would take the logarithm of $100!$, then take the fractional part of that number, and raise $10$ to that power. the leading digit of the final number is the leading digit of $100!$." } { "Tag": [ "logarithms", "geometric series", "real analysis", "real analysis unsolved" ], "Problem": "Denote by $m(n)$ the largest prime factor of $n$,where $n$ is an integer.\r\nProve that $\\sum_{n=2}^{\\infty}\\frac{1}{n \\cdot m(n)}$ converges.", "Solution_1": "[quote=\"ffrogg\"]Denote by $m(n)$ the largest prime factor of $n$,where $n$ is an integer.\nProve that $\\sum_{n=2}^{\\infty}\\frac{1}{n \\cdot m(n)}$ converges.[/quote]\r\n\r\n$\\sum_{n=2}^{\\infty}\\frac{1}{n \\cdot m(n)}\\le \\sum_{n=2}^{\\infty}\\frac{1}{p_{n}}$ where $p_{n}$ is the n_th prime.", "Solution_2": "But $\\sum_{n=2}^{\\infty}\\frac{1}{p_{n}}$ seems to be divergent.", "Solution_3": "All terms are positive, so I can rearrange at will. Pull out all those values for which $m(n) = p$ for a fixed $p$. The sum of these terms is $\\frac{1}{p^{2}}\\sum_{n \\in S_{p}}\\frac{1}{n}$ where $S_{p}$ is the set of all integers whose prime factors are all less than or equal to $p$. This is $\\prod_{q \\leq p}\\frac{q}{q-1}$ where the product is over prime $q$. Thus our total sum is $1+\\sum_{p}\\frac{1}{p^{2}}\\prod_{q \\leq p}\\frac{q}{q-1}$. I'm not immediately sure how to show that this converges; comparison with a geometric series fails, and I don't remember any of the cleverer convergence tests.\r\n\r\nIncidentally, the limit appears to be substantially less than 3, but harmonic-y things tend to require a lot of terms to get big, so it could be substantially larger.", "Solution_4": "A simple estimate: All primes greater than 2 are odd, so $\\prod_{q\\le p}\\frac{q}{q-1}\\le \\frac21\\cdot \\frac32\\cdot \\frac54\\cdot \\frac76\\cdot \\frac98\\cdots \\frac{p}{p-1}\\le \\sqrt{(\\frac21)^{2}\\cdot(\\frac32)^{2}\\cdot\\frac43\\cdot\\frac54\\cdot \\frac65\\cdot \\frac76\\cdots \\frac{p}{p-1}}=\\sqrt{3p}$\r\n\r\nThe $n$th prime is greater than $n$, so the sum is smaller than $\\sum_{p}\\frac1{p^{2}}\\sqrt{3p}=\\sum_{p}\\sqrt{3}p^{\\frac32}< \\sqrt{3}\\sum_{n}n^{\\frac32}$. This converges.\r\n\r\nWith stronger estimates, we have something like $\\frac{1}{n^{2}\\log n}$." } { "Tag": [ "geometry open", "geometry" ], "Problem": "Let be given a circle with diameter AB and point X on AB. Let P be a arbitrary point of this circle, different from A and B.\r\nProve that \r\n\r\n [tan(APX)]/[tan(PAX)] does not depend on choice of P.", "Solution_1": "at reference frame\r\n\r\n[tan(APX)]/[tan(PAX)]=AX/XB does not depend on choice of P.\r\n[/u][/quote][/b][/i]" } { "Tag": [ "induction", "complex numbers", "algebra unsolved", "algebra" ], "Problem": "Q1 Show that $ (a_1 ^2\\plus{}b_1 ^2)(a_2 ^2 \\plus{}b_2 ^2 )...(a_n ^2 \\plus{}b_n ^2)$ can be written as sum of two squares\r\n\r\nQ2 Let z1 and z2 be two complex numbers such that $ \\left|z_1 \\right| < 1 < \\left|z_2 \\right|$ Show that $ \\left| \\frac {1 \\minus{} \\bar{z_1}z_2}{z_1 \\minus{} z_2} \\right| < 1$", "Solution_1": "[quote=\"jackie-chan01\"]Q1 Show that $ (a_1 ^2 \\plus{} b_1 ^2)(a_2 ^2 \\plus{} b_2 ^2 )...(a_n ^2 \\plus{} b_n ^2)$ can be written as sum of two squares[/quote]\r\n\r\n$ (a^2\\plus{}b^2)(c^2\\plus{}d^2)\\equal{}(ac\\plus{}bd)^2\\plus{}(ad\\minus{}bc)^2$ and an immediate induction gives the required result.", "Solution_2": "[quote=\"jackie-chan01\"]Q2 Let z1 and z2 be two complex numbers such that $ \\left|z_1 \\right| < 1 < \\left|z_2 \\right|$ Show that $ \\left| \\frac {1 \\minus{} \\bar{z_1}z_2}{z_1 \\minus{} z_2} \\right| < 1$[/quote]\r\n\r\nWe get $ (1\\minus{}|z_1|^2)(1\\minus{}|z_2|^2)<0$\r\n\r\n\r\n$ \\implies$ $ 1\\plus{}z_1\\overline{z_1}z_2\\overline{z_2}$ $ 3y - 2 < 3z - 2 => y < z.\r\nSo y < z => 3z - 2 < 3x - 2 => z < x.\r\nSo x < y < z < x, which is impossible.\r\nTherefore x, y z are equal.\r\nThey must be equal to one of the roots of the equation x - 3x + 2 = 0.\r\nSo the only solutions are (1, 1, 1) and (-2, -2, -2). :)", "Solution_8": "Hey guys, i take the time to post solutions only after i find out they are correct.\r\n\r\nmy solution:\r\n\r\n\r\nThe proof by contradiction was the basis that I used. I did not however prove it myself, I will let someone else take that. Knowing that the variables all must be equal, we have:\r\n\r\nx=y=z\r\nor\r\n\r\nx^3=3x-2\r\n\r\nor\r\n\r\n(x-1)^2(x+2)=0\r\n\r\nx=1 and -2, and they are the only solutions.... that was how I obtained my solution.\r\nk, bye.", "Solution_9": "[quote=\"Arne\"]So here goes my solution :\n\nSuppose two of x, y, z are not equal. Then WLOG x < y.\nHence x < y => 3y - 2 < 3z - 2 => y < z.\nSo y < z => 3z - 2 < 3x - 2 => z < x.\nSo x < y < z < x, which is impossible.\nTherefore x, y z are equal.\nThey must be equal to one of the roots of the equation x - 3x + 2 = 0.\nSo the only solutions are (1, 1, 1) and (-2, -2, -2). :)[/quote]\r\nyay that's what i did (i couldn't post before b/c my phone line broke AGAIN :() anways, note that if any two of the variables are equal, all three are automatically equal and you get the two solutions already found.\r\n\r\nthere was something like this on an AIME i think. it had a lot of 17s in the problem, that's all I remember...", "Solution_10": "Oh, and there are some typos in your solution Osiris ... Doesn't matter in fact.", "Solution_11": "Edited, sorry... ^^\" I guess I really shouldn't do math at 1 AM." } { "Tag": [ "number theory", "greatest common divisor", "superior algebra", "superior algebra unsolved" ], "Problem": "If gcd (a,b)=1, show that gcd (a+2b, 2a+b) = 1 or 3\r\n\r\nI've been working on this problem for so long but I can't really get anywhere.\r\n\r\nany hints?", "Solution_1": "$ \\rm{HINT}\\; (a\\plus{}2b,2a\\plus{}b) \\supset (3a,3b) \\equal{} 3(a,b) \\equal{} 3$", "Solution_2": "Wouldn't it be the opposite containment? If a is larger than b, then gcd(a+2b, 2a+b) is less than gcd(a+2b, 2a+b)...no?", "Solution_3": "My hint uses ideals. If they're not familiar then read $ (x,y)$ as $ {\\rm gcd}(x,y)$ and \"contains\" as \"divides\". Then it amounts to saying that $ d|a\\plus{}2b,2a\\plus{}b \\implies d|3a,3b \\implies d|{\\rm gcd}(3a,3b) \\equal{} 3\\; {\\rm gcd}(a,b) \\equal{} 3$." } { "Tag": [ "trigonometry", "algebra proposed", "algebra" ], "Problem": "Let $n\\in\\mathbb{N}^*$. Solve the equation $\\sum_{k=0}^n C_n^k\\cos2kx=\\cos nx$ in $\\mathbb{R}$.", "Solution_1": "note that\r\n\r\n$\\sum_{k=0}^n C_n^k\\cos2kx= \\mbox{Re}(\\sum_{k=0}^n C_n^ke^{2kxi}) = \\mbox{Re}[(e^{2xi}+1)^n]$.\r\n\r\nnow we wish to write $e^{2xi}+1 = 1+\\cos 2x+i\\sin 2x$ in trigonometric form...\r\n\r\nits norm is $\\sqrt{2+2\\cos 2x}$, and its argument, $\\gamma$, is such that \r\n\r\n$\\cos\\gamma = \\frac{1+\\cos 2x}{\\sqrt{2+2\\cos 2x}} = \\cos\\frac{ 2x}{2} = \\pm \\cos x$. so if $x$ ends in $[-\\pi/2, \\pi/2]$, we take $\\gamma = x$ and if $x$ ends in $(\\pi/2, 3\\pi/2)$ we take $\\gamma = x+\\pi$. (because in all cases $\\sqrt{2+2\\cos 2x}\\cos\\gamma = 1+\\cos 2x$ must be $\\geq 0$.)\r\n\r\nnow we have\r\n\r\n$\\mbox{Re}[(e^{2xi}+1)^n] = \\sqrt{2+2\\cos\\2x}^n\\cos n\\gamma$.\r\n\r\nif $n$ is even, then the original equation becomes $\\sqrt{2+2\\cos\\2x}^n\\cos nx = \\cos nx$.\r\n\r\nthen either $\\cos nx = 0$, in which case $x = \\frac{(2k+1)\\pi}{2n}, \\ k\\in \\mathbb Z$, or $2+2\\cos\\2x = 1$, in which case $\\cos 2x = -\\frac12$, i.e. $x = k\\pi \\pm \\frac{\\pi}{3}, \\ k\\in \\mathbb Z$. \r\n\r\nif $n$ is odd and $x$ ends in $(\\pi/2, 3\\pi/2)$, the original equation becomes $\\sqrt{2+2\\cos\\2x}^n\\cos nx = -\\cos nx$. if $\\cos nx \\neq 0$ then we get $\\sqrt{2+2\\cos\\2x}^n = -1$ which is impossible. otherwise, $\\cos nx = 0$, in which case $x = \\frac{(2k+1)\\pi}{2n}, \\ k \\in \\mathbb Z$. (all these are solutions, whether they end in the presumed interval or not...)\r\n\r\nif $n$ is odd and $x$ ends in $[-\\pi/2, \\pi/2]$, the original equation becomes $\\sqrt{2+2\\cos\\2x}^n\\cos nx = \\cos nx$, in which case either $\\cos nx = 0$, in which case we have the solutions mentioned above or $\\cos 2x = -\\frac12$, i.e. $x = 2k\\pi \\pm \\frac{\\pi}{3}$ (because we assumed $x$ ends in $[-\\pi/2, \\pi/2]$)", "Solution_2": "Nice solution!\npleurestique" } { "Tag": [ "integration", "trigonometry", "calculus", "calculus computations" ], "Problem": ":( i asked this in exam yesterday and only 3 students got it... what is going on ? :huh: \r\n\r\n\r\nshow\r\n\r\n$ \\sum_{n\\equal{}1}^\\infty\\;\\dfrac{n^2}{2^n\\cdot \\dbinom{2n}{n} }\\;\\;\\equal{}\\;\\;\\boxed{\\dfrac{8}{343}\\left(16 \\plus{} \\dfrac{37}{\\sqrt 7}\\;\\cot^{\\minus{}1} \\sqrt 7\\right)}$", "Solution_1": "They're not as smart as you? :lol:", "Solution_2": "Using \r\n\r\n$ \\bold{(i)} \\quad \\binom{2n}{n}^{ - 1} = \\frac {n}{4^n} \\beta (n, \\tfrac{1}{2} ),$\r\n$ \\bold{(ii)} \\quad \\sum_{n = 1}^{\\infty} n^3 x^n = \\frac {x(x^2 + 4x + 1)}{(1 - x)^4},$\r\n\r\nwe have\r\n\r\n\\begin{eqnarray*} & & \\sum_{n = 1}^{\\infty} \\binom{2n}{n}^{ - 1} \\frac {n^2}{2^n} \\\\\r\n& = & \\sum_{n = 1}^{\\infty} \\frac {n^3}{8^n} \\beta (n, \\tfrac{1}{2} ) \\\\\r\n& = & \\sum_{n = 1}^{\\infty} \\frac {2n^3}{8^n} \\int_{0}^{\\frac {\\pi}{2}} \\sin^{2n - 1} \\theta \\, d\\theta \\\\\r\n& = & \\int_{0}^{\\frac {\\pi}{2}} \\frac {2}{\\sin \\theta} \\sum_{n = 1}^{\\infty} n^3 \\left( \\frac {\\sin^2 \\theta}{8}\\right)^n \\, d\\theta \\\\\r\n& = & 16 \\int_{0}^{\\frac {\\pi}{2}} \\frac {\\cos^4 \\theta - 34 \\cos^2 \\theta + 97) \\sin \\theta}{(\\cos^2 \\thehta + 7)^4} \\, d\\theta \\\\\r\n& = & 16 \\int_{0}^{1} \\left( \\frac {384}{(7 + t^2)^4} - \\frac {48}{(7 + t^2)^3} + \\frac {1}{(7 + t^2)^2} \\right) \\, dt \\quad ( t = \\cos \\theta). \\end{eqnarray*}\r\n\r\nWell, we can evaluate the last integral using $ t = \\sqrt {7} \\tan \\varphi$ or something else, but I think that is not an important step of this calculation. (Frankly, my head was messed up with bunch of calculation before finishing it... [img]http://myhome.shinbiro.com/~sosQED/images/m_0503.gif[/img])" } { "Tag": [], "Problem": "$\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3}}...}}}}}$ = $3$\r\n\r\nFind x", "Solution_1": "[quote=\"bos1234\"]$\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3}}...}}}}}$ = $3$\n\nFind x[/quote]\r\n\r\n\r\n[hide]\\[\\sqrt{x+3 \\cdot 3}= 3 \\implies \\boxed{x=0}. \\][/hide]", "Solution_2": "[quote=\"bos1234\"]$\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3}}...}}}}}$ = $3$\n\nFind x[/quote]\r\n\r\n[hide]\n\n$\\left({\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3}}...\\right)^{2}}}}}}}$ = $(3)^{2}$\n\n$x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3}}...}}}}$ = $9$\n\n\n$3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3}}...}}}}$ = $9-x$\n\nNotice that dividing the sum by 3 will bring you back to the original sum since this is goes on 4ever\n\n $\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3}}...}}}}$ = $\\frac{9-x}{3}$\n\n\n$\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3\\sqrt{x+3}}...}}}}}$ = $\\frac{9-x}{3}$\n\nNotice that the LHS is equal to 3 as given in the question\n\n$3$ = $\\frac{9-x}{3}$\nsolve for that\n$x=0$[/hide]", "Solution_3": "yes both are correct" } { "Tag": [ "group theory", "abstract algebra", "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi,I am a beginner on groups.Here are two questions,thanks for help.\r\n\r\nProblem 1:Let $|G|=p^{n}$($p$ is a prime number) and let $H$ be a subgroup of $G$ such that $|H|=p^{m}$,where $m\\le n$.Show that if $m\\le k \\le n$,then there exists a subgroup $K$ of $G$ such that $|K|=p^{k}$ and $H \\subseteq K$.\r\nHint in my book\r\n[hide]Use the fact that,by the class equation,$Z(G)$ has a subgroup of order $p$.[/hide]\r\n\r\nProblem 2:Suppose $|G|=p^{n}$($p$ is a prime number),$H \\lhd G$,and $|H|=p$.Prove that $H \\subseteq Z(G)$,where $Z(G)$ is the center of $G$.", "Solution_1": "No one :(,thank you", "Solution_2": "as regarding the second question..\r\n$H$ is generated by a non-unity element of its, say $h$.\r\nnow, take $g\\in G\\setminus H$ of order $p$ (there exists one): $ghg^{-1}\\in H$, so $ghg^{-1}= h^{k}$ for some integer $0 ObOcCB are concyclic => \u2220IObOc=\u2220C/2 \u2220IOcOb=\u2220B/2\r\nAnalogously \u2220IOcOa=\u2220A/2 \u2220IOaOc=\u2220C/2\r\nIOb\u22a5OaOc IOa\u22a5ObOc IOc\u22a5OaOb => I is the orthocenter of \u25b3OaObOc\r\nWe have TcTb \u22a5 IA , so TcTb//ObOc\r\n\r\nObOc=BC*(IB/IC)* 2*(sin^2(\u2220B/2))=2*BC* sin(\u2220B/2)* sin(\u2220C/2)\r\nTcTb=2*r*cos(\u2220A/2)\r\n\t\r\nNotice in any triangle , r=4R*sin(\u2220B/2)* sin(\u2220C/2)* sin(\u2220A/2) BC=2R* sin\u2220A\r\nSo ObOc= TcTb\r\n\r\n=> TcObOcTb is a parallelogram TcOb=TbOc \r\nAnalogously, the six sides are all equality" } { "Tag": [ "induction" ], "Problem": "Prove that for $ n>1$ the $ n$-th harmonic number $ H_{n} \\equal{} 1\\plus{} \\frac{1}{2} \\plus{} \\frac{1}{3} \\plus{} \\cdots \\plus{}\\frac{1}{n}$ is not an integer.", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176128", "Solution_2": "[hide=\"Another possible approach\"]We use induction.\n\nBase cases: $ H_2\\equal{}\\frac{3}{2}$, not an integer.\n\nInductive step:\nAssume $ H_p$ is not an integer for a prime $ p$. Let $ k$ be a number such that $ p1$, $x^n+y^n>1$, the conclusion follows.[/hide]", "Solution_3": "Nice :)" } { "Tag": [ "ratio", "geometry", "circumcircle", "trigonometry", "angle bisector", "perpendicular bisector" ], "Problem": "Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively. And $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced. \r\n\r\nProve that $QO$ is perpendicular to $BC$.", "Solution_1": "Let the line PN meet the (extended) triangle side CA at R. The cross-ratio of 4 points is preserved in a central projection of the line BC to the line PR, hence,\r\n\r\n$\\frac{NB}{NC} \\cdot \\frac{MC}{MB} = \\frac{NP}{NR} \\cdot \\frac{QR}{QP}$\r\n\r\nM is the midpoint of BC, N is the midpoint of PR, hence,\r\n\r\n$\\frac{NB}{NC} = \\frac{QR}{QP}$\r\n\r\nThe angle bisector AN meets the circumcircle of the triangle $\\triangle ABC$ at the midpoint K of the arc BC opposite to the vertex A. Since $PR \\perp AN, PO \\perp AB$, it follows that $\\angle OPR = \\angle KAB = \\angle KAC = \\angle KBC$ and the isosceles triangles $\\triangle ROP \\sim \\triangle BKC$ together with the points $Q, N \\in RP$ resp. $N, M \\in BC$ on their bases are similar. Hence, the angles $\\angle NOQ = \\angle NKM$ are equal, which means that the lines $OQ \\parallel KM$ are parallel and $KM \\perp BC$ is the perpendicular bisector of BC.", "Solution_2": "[quote=\"mr.danh\"][hide=\"Another solution\"]Let $ L$ be the second intersection of $ AN$ and the circumcircle of triangle $ ABC$. Let $ D$ on $ AB$ and $ E$ on $ AC$ such that $ LD\\perp AB$, $ LE\\perp AC$. Easy to see that the Simson line $ (DMN)$ is perpendicular to $ AN$. So $ DM\\parallel PQ$. Apply Thales's theorem to pairs of parallel lines $ DM\\parallel PQ$, $ LD\\parallel OP$, we have that $ \\frac {AO}{AL} \\equal{} \\frac {AP}{AD} \\equal{} \\frac {AQ}{AM}$.\nHence $ OQ\\parallel LM$. And $ OQ\\perp BC$ (since $ LM\\perp BC$).[/hide][/quote]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=297174]Posted[/url]", "Solution_3": "Dear Mathlinkers,\nlet X be the second point of intersection of AD with the circumcircle of ABC.\nThe problem can be solved without calculation by considering the circle passing through M, N and X.\nSincerely\nJean-Louis", "Solution_4": "Let $PN$ hit $AC$ at $R$. Note that there exists a circle $\\omega$ tangent to $AP$ and $AR$. Let $Q'$ be the intersection of $PR$ and the line through $O$ perpendicular to $BC$. $Q'$ lies on the polar of $A$, so $A$ lies on the polar $\\ell$ of $Q'$. Let $E$ be the intersection of $PR$ and $\\ell$, and let $M'$ be the intersection of $AQ$ and $BC$. Because $E$ lies on the polar of $Q'$, $(P, R; Q', E) = -1$. $\\ell$ and $BC$ are both perpendicular to $OQ'$, so they are parallel. Hence, $(P, R; Q', E) = A(P, R; Q', E) = A(B, C; M', \\infty) = -1$, i.e., $M'$ and $\\infty$ are harmonic conjugates, whence $M' = M$ and $Q' = Q$.", "Solution_5": "Another solution:\n\nLet $AN$ and $AM$ cut the circumcirle of $ABC$ with centre $H$ at $F$ and $G$, respectively.\nIt is obvious that $H, M, F$ are collinear and $HF\\perp BC$. Let $HF$ cut the circumcircle again at $D, DG$ cut $BC$ at $K$.\nEasy to see that $APOE\\sim DBFC$, and since $\\angle OAQ =\\angle FDK$, line $AQ$ corresponds to $AK, OQ$ corresponds to $FK$.\nThus $\\angle NOQ = \\angle MFK$, but $\\angle MFK = \\angle MGK$ (since $MKGF$ is cyclic)$= \\angle AFD$.\n$\\angle NOQ = \\angle AFD$, this means that $OQ\\parallel FD$, thus $OQ\\perp BC$.", "Solution_6": "[hide=\"Solution\"]\nWe do not use isothermal coordinates.\n\nLet $PN \\cap AC=R$, let the perpendicular from $O$ to $BC$ meet $PR$ at $Q'$, and let $AQ'$ meet $BC$ at $M'$. It is sufficient to prove that $M'=M$.\n\nClearly $\\triangle APO \\cong \\triangle ARO$. Then by the [url=http://www.artofproblemsolving.com/blog/81708]Ratio Lemma[/url], we have $\\dfrac{PQ}{QR}=\\dfrac{PO}{OR}\\cdot\\dfrac{\\sin \\angle POQ}{\\sin \\angle ROQ}=\\dfrac{\\sin \\angle ABC}{\\sin \\angle ACB}=\\dfrac{AC}{AB}$. Then, since $AP=AR$, by the Ratio lemma we have $\\dfrac{\\sin \\angle BAQ}{\\sin \\angle CAQ}=\\dfrac{AC}{AB}$. Finally, the Ratio Lemma gives us that $\\dfrac{BM'}{M'C}=\\dfrac{AB}{AC}\\cdot \\dfrac{\\sin \\angle BAQ}{\\sin \\angle CAQ}=1$, as desired.\n\nQED [/hide]", "Solution_7": "Here's a solution for those that aren't scared of diving into a bit of algebra. It's not an efficient solution at all (NAA's above beats it with a very similar idea), but hey, it works.\n\n[hide=\"Intense Trig Bash\"]\nLet $R=AC\\cap PQ$ and $X=QO\\cap BC$. I first propose a lemma.\n\n[b]LEMMA:[/b] Suppose $a,b,c,d$ are angles such that $\\dfrac{\\sin a}{\\sin b}=\\dfrac{\\sin c}{\\sin d}$ and $a+b=c+d$. Then $a=c$ and $b=d$.\n\n[i]Proof.[/i] Let $k=a+b=c+d$. Then $b=k-a$ and $d=k-c$. Cross-multiplying and expanding gives \\begin{align*}\\sin a\\sin(k-c)&=\\sin c\\sin(k-a)\\\\\\sin a(\\sin k\\cos c-\\sin c\\cos k)&=\\sin c(\\sin k\\cos a-\\sin a\\cos k)\\\\\\sin a\\sin k\\cos c-\\sin a\\sin c\\cos k&=\\cos a\\sin c\\sin k-\\sin a\\sin c\\cos k\\\\\\sin a\\cos c&=\\sin c\\cos a\\\\\\sin(a-c)&=0\\\\a&=c,\\end{align*} as desired. $\\blacksquare$\n\nNote that it is easy to see that $\\triangle PAR$ is isosceles, which implies that $\\angle RAN=\\angle PAN=\\angle NPO\\implies AROP$ is cyclic. Therefore $\\angle PAO=\\angle PRO$ and $\\triangle POR$ is also isosceles.\n\nApplying the Ratio Lemma to $\\triangle ABC$ gives \\[\\dfrac{AB\\sin\\angle BAM}{AC\\sin\\angle MAC}=\\dfrac{BM}{CM}=1\\implies \\dfrac{AB}{AC}=\\dfrac{\\sin\\angle QAR}{\\sin\\angle PAR}.\\] Next, applying the Ratio Lemma to $\\triangle PAR$ gives $\\tfrac{PQ}{QR}=\\tfrac{AP\\sin\\angle PAQ}{AR\\sin\\angle QAR}=\\tfrac{\\sin\\angle PAQ}{\\sin\\angle QAR}$ and applying it to $\\triangle POR$ gives $\\tfrac{PQ}{QR}=\\tfrac{PO\\sin\\angle POQ}{OR\\sin\\angle QOR}=\\tfrac{\\sin\\angle POQ}{\\sin\\angle QOR}$. Therefore \\[\\dfrac{\\sin\\angle POQ}{\\sin\\angle QOR}=\\dfrac{PQ}{QR}=\\dfrac{\\sin\\angle PAR}{\\sin\\angle QAR}=\\dfrac{AC}{AB}.\\] In addition, by LoS we have $\\tfrac{AB}{\\sin\\angle ACB}=\\tfrac{AC}{\\sin\\angle ABC}$, so all in all we have $\\tfrac{\\sin\\angle POQ}{\\sin\\angle QOR}=\\tfrac{\\sin\\angle ABC}{\\sin\\angle ACB}$. But note that since $ARPO$ is cyclic we have $\\angle POQ+\\angle QOR=180^\\circ-\\angle BAC=\\angle ABC+\\angle ACB$. Therefore by our Lemma, $\\angle ABC=\\angle QOP$. It follows that $BXOP$ is cyclic, so $\\angle BXO=180^\\circ-\\angle BPO=90^\\circ$ and $BC\\perp QO$, as desired. $\\blacksquare$[/hide]", "Solution_8": "This is similar to some of the previous solutions, but uses homothety to simplify one step.\n\nLet $D$ be the other intersection point of $AN$ and the circumcircle of $ABC$. Then $DM\\perp BC$ by the South Pole Theorem. Let $E$ and $F$ be the points on $AC$ and $BC$ such that $DE\\perp AC$ and $DF\\perp BC$, respectively. Note that there is a homothety centred at $A$ sending $O$ to $D$, $P$ to $F$ and (since clearly $OR\\perp AC$) $R$ to $E$. Since $FME$ is the Simson line of point $D$, it sends the entire segment $PR$ to $FE$, and thus $Q$ to $M$. Now it is clear that $QO\\parallel MD$, so $QO\\perp BC$.", "Solution_9": "I will be using homothety. Extend PN to meet AC at J.Let $O*$ be the second intersection of AO with circumcircle of $ABC$. Consider the homothety centred at $A$ that maps $O$ to $O*$. Drop $OX\\perp AB$ and $OY\\perp AC$.$P$ maps to $X$ and $J$ maps to $Y$.So $PJ$ maps to $XY$. So, $AM\\cap PJ$ maps to $AM\\cap XY$.Invoking Simpson, easy to see that $X,M,Y$ collinear.So, $Q$ maps to $M$. $OQ\\perp BC$ iff $O*M\\perp BC$ . But this is trivial. Q.E.D. Please check my solution whether it is correct.", "Solution_10": "[quote=jayme]Dear Mathlinkers,\nlet X be the second point of intersection of AD with the circumcircle of ABC.\nThe problem can be solved without calculation by considering the circle passing through M, N and X.\nSincerely\nJean-Louis[/quote]\n\n Can you please elaborate on your method.It seems very nice.", "Solution_11": "You can also use Cartesian coordinates with, say, $N=(0,0)$, $A=(-1,0)$, $P=(0,k)$, and $O=(k^2,0)$. Since $AN$ is the angle bisector, it's actually easy to find the equation of $AC$.", "Solution_12": "Dynamic points: \nWe will vary the point $B$ on $AP$ ($A,P,O,N$ will all remain fixed). Let $X=BN\\cap QO$ and $X'=BN\\cap (ON)$. We have $B\\mapsto C\\mapsto M \\mapsto Q$ is a projective transformation, and so lines $BN$ and $OQ$ each have degree 1, therefore their intersection $X$ has degree 2. Since $(ON)$ is fixed, $X'$ has degree 1. Now to prove $X=X'$ we only must check 4 points for $B$. These four points will be $AB_{\\infty}, P, A, K$ where $K\\in AB$ with $KN\\parallel AC$. For each of these points the result is trivial, thus $X=X'$ and this implies $XO\\bot XQ$.", "Solution_13": "[quote=shobber]Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively. And $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced. \n\nProve that $QO$ is perpendicular to $BC$.[/quote]\nProof-\nLet $R$ denote intersection of $PN$ with $AC$. Let $\\omega$ denote circumcircle of triangle $APR$. \nNote by symmetry about $AN$ that $O\\in\\omega$ and is the $A-antipode$. Let $OQ\\cap\\omega=S,AQ\\cap\\omega=T$.\nNow $H(A,O,P,R)\\xRightarrow{\\text{Q}}H(T,S,R,P)\\xRightarrow{\\text{A}}H(AT,AS, AP,AR)\\implies H(AS,AM, AB,AC)$ and hence $AS\\parallel BC$ and we know $AS\\perp OS$\nSo $OQ\\perp BC$\nHence proved.", "Solution_14": "Extend $AO$ to meet $(ABC)$ at $L$. Let $D$ be the altitude from $L$ to $AB$, $E$ be the altitude from $L$ to $AC$, and $T$ be the altitude from $O$ to $AC$. Note that by Simson line wrt $\\triangle{ABC}$, we have $D, M, E$ collinear, because $LM\\perp BC$. Also, it is easy to see $APOT$ is cyclic, so by Simson line wrt $\\triangle{APT}$, we have $P, N, T$ collinear; hence, $Q\\in PT$. Finally, let $h$ be the homothety about $A$ mapping $L$ to $O$. Because $PO||LD, OT||LE$, we have that $h(D)=P$ and $h(E)=T$; hence, $h(M)$ lies on both $AM$ and $PT$, so $h(M)=Q$. Thus, by similar triangles we see $QO||ML$, meaning that $QO\\perp BC$, as desired. ", "Solution_15": "Here is my solution:\n\nLet the perpendicular from $O$ to $BC$ meet $PR$ at $Q'$. We wish to show $Q'=Q$.\nLet the parallel through $Q'$ to $BC$ meet $AB$ and $AC$ at $S$ and $T$, respectively.\nThen since $OQ' \\perp TS$, and $\\angle OPA=\\angle ORA=90$, both $PSQ'O$ and $RTOQ'$ are cyclic.\nThen $\\angle Q'OS= \\angle Q'PS = \\angle Q'PA = \\angle Q'RA = \\angle Q'OT$.\nThus, since $Q'O \\perp ST, Q'S=Q'T$, in which case $Q'$ is on the median from $A$ and $Q'=Q$.", "Solution_16": "Redefine $Q$ as $T$, let $Q = PN \\cap AC,$ note that $OQ \\perp AC$. now let $X$ be arc midpoint of $BC$, and let $E$ and $F$ be the foot of altitude from $X$ to $AB$ and $AC$. Now by simson line we know that $EMF$ is collinear and $EMF \\parallel PQ,$ so there's a homothety now at A sending $P \\to E, Q \\to F, T \\to M, O \\to X$ implying the desired result.\n", "Solution_17": "Oops it looks like I missed the homothety, and kinda rederived homothety?\n\nLet $D$ be the arc midpoint of $BC$. Let the foot from $D$ to $AB$ be $X$, and the foot to $AC$ be $Y$. Let $R=XY\\cap AN$. \n\nThen, by Simson's theorem $XMY$ are collinear, call this line $L_1$. We will attempt to show that $L_1\\parallel PNQ$. It suffices to note that\n\\[\\angle ARX = 180 - \\angle XAR - \\angle AXR = 180-\\angle DAC-\\angle BXM = 180-\\angle DBM - \\angle BDM = 90\\]\nThus, $AR\\perp XY$, but since we also have $AR\\perp PNQ$, then $XY\\parallel PNQ$.\n\nThus, we may now create a series of ratios since $PQ\\parallel XM$\n\\[\\frac{AQ}{AM}=\\frac{AP}{AX}=\\frac{AO}{AD}\\]\nThus, by SAS-ratio similarity, we have $\\triangle AQO\\sim \\triangle AMD$. Thus, $QO\\parallel MD$ and since $MD\\perp BC$, we clearly have $QO\\perp BC$ and we are done.", "Solution_18": "[hide=Solution]\n[asy]\nunitsize(0.8cm);\npair A, B, C, B1, C1, M, N, P, Q, O;\nA=(2,3sqrt(5));\nB=(0,0);\nC=(8,0);\nM=(4,0);\nN=(3.5,0);\nP=(29/4,3sqrt(5)/8);\nQ=(127/32,3sqrt(5)/64);\nO=(127/32,-15sqrt(5)/16);\nB1=(-31/16,-93sqrt(5)/32);\nC1=(221/16,-93sqrt(5)/32);\ndraw(A--B1--C1--A--(-1/4,-3sqrt(5)/8)--P--A--B--C--P--O--A--M--Q--N--P--N);\ndraw(circle(O,63sqrt(5)/32));\nlabel(\"$A$\", A, dir(90));\nlabel(\"$B$\", B, W);\nlabel(\"$C$\", C, NE);\nlabel(\"$B'$\", B1, SW);\nlabel(\"$C'$\", C1, SE);\nlabel(\"$M$\", M, S);\nlabel(\"$N$\", N, SSW);\nlabel(\"$O$\", O, S);\nlabel(\"$P$\", P, NE);\nlabel(\"$Q$\", Q, NNE);\n[/asy]\n\nConsider the circle centered at $O$ passing through $P$. Then, this circle is tangent to $AB$ and $AC$. Let $B'C'$ be the line such that $BC\\parallel B'C'$ and $B'C'$ is tangent to the circle. Then, since $PN\\perp AO$, this means that the reflection of $P$ over $N$ lies on $AB$, $PN$, and the circle. Since $AM$ is a median of $AB'C'$, this means that $PN$, $AM$, and the line through $O$ perpendicular to $B'C'$ are concurrent, so this means that $OQ\\perp B'C'$, which means $QO\\perp BC$.", "Solution_19": "Let $\\ell \\equiv \\overline{PQ}$ and $R = \\ell \\cap \\overline{AC}$. Loot at the isosceles $\\triangle APR$ (with $AP = AR$). By symmetry we have $\\overline{OR} \\perp \\overline{AC}$. Let $O' = \\overline{AO} \\cap \\odot(ABC)$ and $\\ell'$ be the Simson line of $O'$ wrt $\\triangle ABC$ ; $\\ell'$ intersect $\\overline{AB},\\overline{AC}$ at $P',R'$, respectively. As $\\overline{O'M} \\perp \\overline{BC}$ so $M \\in \\ell$. \n[asy]\nsize(200);\npair A=dir(145),B=dir(-150),C=dir(-30),M=1/2*(B+C),Op=dir(-90),Pp=foot(Op,A,B),Rp=foot(Op,A,C),N=extension(A,Op,B,C),P=extension(A,B,N,N+M-Rp),R=extension(N,P,A,C),Q=extension(N,P,A,M),O=extension(Q,foot(Q,B,C),A,N);\ndraw(circumcircle(A,B,C),red);\ndot(\"$A$\",A,dir(A));\ndot(\"$B$\",B,dir(B));\ndot(\"$C$\",C,dir(C));\ndot(\"$M$\",M,dir(-60));\ndot(\"$O'$\",Op,dir(Op));\ndot(\"$P'$\",Pp,dir(Pp));\ndot(\"$R'$\",Rp,dir(-90));\ndot(\"$N$\",N,dir(N));\ndot(\"$R$\",R,dir(90));\ndot(\"$P$\",P,dir(-170));\ndot(\"$Q$\",Q,dir(90));\ndot(\"$O$\",O,dir(-140));\ndraw(Pp--A--C--B^^Op--A--M,red);\ndraw(P--O--R^^Q--O,blue);\ndraw(Pp--Op--Rp^^Op--M,blue);\ndraw(P--R^^Pp--Rp,green);\n[/asy]\nConsider the homothety $\\mathbb H$ at $A$ sending $O \\to O'$. Note $\\mathbb H(P) = P'$ and $\\mathbb H (Q) = Q'$ as $\\overline{OP} \\parallel \\overline{O'P'}$ and $\\overline{OR} \\parallel \\overline{O'R'}$. Thus $\\mathbb H(\\ell) = \\ell'$. It follows $\\mathbb H(Q) = M$ as $$\\mathbb H (Q) = \\mathbb H ( \\ell \\cap \\overline{AM}) = \\mathbb H(\\ell) \\cap \\mathbb H(\\overline{AM}) = \\ell' \\cap \\overline{AM} = M$$\nNow since $\\overline{O'M} \\perp \\overline{BC}$ and $\\overline{OQ} \\parallel \\overline{O'M}$. Hence $\\overline{OQ} \\perp \\overline{BC}$, as desired. $\\blacksquare$", "Solution_20": "[hide = for storage]Let the circumcircle of $\\triangle{ABC}$ be $\\Omega$, let AO meet $\\Omega$ at D.\nLet E,G be the foot of the perpendiculars from D to AB and AC respectively. Now note that E,M,G are collinear by Simson's Line .Let F be the intersection of PN with AC. Note that since AN is the angle bisector of $\\angle{BAC}$, quad APOF is a cyclic kite.\nNow notice that triangles POF and EDG are similar and thus we have,\n\\\\ $\\frac{PO}{DE} = \\frac{PF}{EG} = \\frac{OF}{DG}$\n\\\\Also, since we have $OP \\parallel DE$ and $OF \\parallel DG$,\n\\\\ we have $\\frac{AF}{AG} = \\frac{OF}{DG}$ \\\\and\n\\\\ $\\frac{AP}{AE} = \\frac{OP}{DE}$\n\\\\Therefore, \\\\$\\frac{AP}{AE} = \\frac{AF}{AG} = \\frac{PF}{DG}$\n$\\Rightarrow PF \\parallel DG \\Rightarrow PQ \\parallel EM $.\n\\\\Now using Thales theorem on the triangles AEM and AMD we will get $QO \\parallel LM$. \n[/hide]", "Solution_21": "[hide=Highly recommended if you are in such a desperate situation] Very efficient coordinate bash here: https://www.math.hkust.edu.hk/excalibur/v5_n3.pdf [/hide]", "Solution_22": "one of the finest use of phantom points!\n\nconsider $PN \\cap AC=R$ and perpendicular to $BC$ from $O$ to hit $BC$ at $X$ such that $OX \\cap PR=Q'$ and $AQ' \\cap BC=M'$\\\\\nJoin $OR$ and $OC$ for later use.\\\\\n\nnow beforehand solving this problem we propose a lemma:\\\\\n\n[color=#073763][b]Lemma:-[/b][/color] In a triangle $ABC$ denote $AX$ to be a cevian, s.t. $\\angle{BAX}=\\alpha$ and $\\angle{XAC}=\\beta$ then $\\frac{BX}{XC}=\\frac{AB}{AC}\\cdot \\frac{\\sin{\\alpha}}{\\sin{\\beta}}$\\\\\n\n[color=#134F5C][b]Proof:-[/b][/color] denote $\\angle{AXB}=\\theta$ then from sine rule we get $\\frac{BX}{\\sin{\\alpha}}=\\frac{AB}{\\sin{\\theta}}$ and $\\frac{CX}{\\sin{\\beta}}=\\frac{AC}{\\sin{\\theta}}$ , dividing these we get $\\frac{BX}{CX}=\\frac{AB}{AC}\\cdot \\frac{\\sin{\\alpha}}{\\sin{\\beta}}$ $\\square$\\\\\n \nmark $\\angle{BAQ'}=x$ and $\\angle{Q'AR}=y$\\\\\n\nnow we notice that since $AN$ is angle bisector and $AN\\perp PR$ we get $\\triangle{APN}\\cong \\triangle{ARN}$ hence we get :\\\\\n\n$\\angle{APN}=\\angle{ARN}$ and hence $\\angle{OPN}=90-\\angle{APN}$ and as $N$ is mid-point of $PR$ we get $\\triangle{OPN} \\cong \\triangle{ORN}$, giving us $\\angle{ORN}=90-\\angle{APN}$\\\\\n\nhence we get $OR \\perp AC$ and therefore points $C,R,X,O$ are concyclic which gives $\\angle{XOR}=C$\\\\\n\nalso points $B,P,O,X$ are concyclic which gives us $\\angle{XOP}=B$\\\\\n\nfrom [color=#073763][b]Lemma[/b][/color] we get $\\frac{PQ'}{Q'R}=\\frac{AP}{AR}\\cdot \\frac{\\sin{x}}{\\sin{y}}$ and since $AP=AR$ we get $\\frac{PQ'}{Q'R}=\\frac{\\sin{x}}{\\sin{y}}$\\\\\n\nalso similarly we have in $\\triangle{POR}$ applying [color=#073763][b]Lemma[/b][/color] we get $\\frac{PQ'}{Q'R}=\\frac{OP}{OR}\\cdot \\frac{\\sin{B}}{\\sin{C}}$ and since $OP=OR$ we get $\\frac{\\sin{x}}{\\sin{y}}=\\frac{\\sin{B}}{\\sin{C}}$\\\\\n\nand now we again apply [color=#073763][b]Lemma[/b][/color] in $\\triangle{ABC}$ we get$ \\frac{AM'}{M'C}=\\frac{AB}{AC}\\cdot \\frac{\\sin{x}}{\\sin{y}}=\\frac{AB}{AC}\\frac{\\sin{B}}{\\sin{C}}=\\frac{2R}{2R}=1$ hence we get $AM'=M'C$ and hence $M'$ is midpoint of $BC$ which gives $AM'$ to be the median of $\\triangle{ABC}$ and hence the problem statement follows as $M=M'$ and $Q=Q'$ $\\blacksquare$", "Solution_23": "Solved with a hint to construct DE//BC\n\nThe circle centered at O through P is tangent to AB,AC, since it lies on the angle bisector. Let DE be the unique line //BC with DE tangent to this circle, with R the foot from O onto AD; since PR perp. NO (look at kite APOR, PR diagonal) along with the well known fact that AM (median in ADE), the line perp. to DE through O are concurrent, this finishes because Q is indeed the concurrence point. $\\blacksquare$", "Solution_24": "wait what\n\nLet $L$ be the midpoint of arc $BAC$ and let $K$ be the midpoint of arc $BC$.\n\nNotice that $ANML$ is cyclic so we get $APQO\\sim LCNK$.\n\nThis means \n\\[\\measuredangle(OQ,KN)=\\measuredangle(PQ,CN)\\implies \\measuredangle QOK=\\measuredangle CNP.\\]\n\nSubtract $\\measuredangle BNA$:\n\\[\\measuredangle(OQ,BC)=\\measuredangle(OQ,KN)-\\measuredangle(KN,BC)=\\measuredangle QOK-\\measuredangle BNA\\]\n\\[\\measuredangle QOK-\\measuredangle BNA=\\measuredangle CNP-\\measuredangle BNA=\\measuredangle BNP-\\measuredangle BNA=90^{\\circ}\\]\ndone!", "Solution_25": "we can also solve as follows:\n\nLet $R$ be on $(APQ)$ where $AR\\parallel BC$. Let $T=PN\\cap AC$. \n\\[-1=(\\infty_{BC},M;B,C)\\stackrel{A}{=}(R,AQ\\cap (APO);P,T)\\]\nso $\\frac{RP}{RT}=\\frac{QP}{QT}$ and $\\overline{RQO}$ bisects $\\angle PRT$.", "Solution_26": "Let $AN \\cap (ABC) = E$. By Simson, the line connecting the projections from $E$ to $AB$ and $AC$ passes through $M$. Hence the homothety at $A$ sending this line to $PN$ and $E$ to $O$ also sends $M$ to $N$. Thus\n\\[QO \\parallel ME \\perp BC. \\quad \\blacksquare\\]", "Solution_27": "[hide=Notation]By $(PQR)$, we denote the circumcircle of $PQR$.[/hide]\n[hide=Required Theorems]Simson line theorem : [url=https://artofproblemsolving.com/wiki/index.php/Simson_line]Click here to learn more[/url].[/hide]\nLet $D$ be the circumcenter of $\\Delta ABC$. Let $E$ be the concurrency point of $DM$, $AN$ and $(ABC)$. $F$ and $G$ are the perpendicular foot from $E$ to $AB$ and $AC$ respectively. $H$ is the intersection of $FG$ and $AE$. Let $\\frac{\\angle A}{2}=\\alpha$.\n[hide=Diagram][asy] \nimport graph; size(8cm); \nreal labelscalefactor = 0.5; /* changes label-to-point distance */\npen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ \npen dotstyle = black; /* point style */ \nreal xmin = -6.141728024042078, xmax = 3.174575507137483, ymin = -5.81445529676934, ymax = 3.2614274981217113; /* image dimensions */\n\n /* draw figures *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point */\ndraw((-4.88,-1.88)--(-5.211598643768283,-3.0982645825399944), linewidth(0.8)); \ndraw((-5.211598643768283,-3.0982645825399944)--(-1.494317431054321,-4.110068936296458), linewidth(0.8)); \ndraw((-2.468833572833071,-1.892791333831124)--(-5.211598643768283,-3.0982645825399944), linewidth(0.8)); \ndraw((-2.468833572833071,-1.892791333831124)--(-3.96,1.5), linewidth(0.8)); \ndraw((-3.96,1.5)--(-1.11,-1.9), linewidth(0.8)); \ndraw((-1.480249684031839,-1.4582986225585077)--(-2.468833572833071,-1.892791333831124), linewidth(0.8)); \ndraw((-2.468833572833071,-1.892791333831124)--(-1.494317431054321,-4.110068936296458), linewidth(0.8)); \ndraw((-1.480249684031839,-1.4582986225585077)--(-1.494317431054321,-4.110068936296458), linewidth(0.8)); \ndraw(circle((-1.1057141289867563,-1.0921133140036046), 3.8556451167217896), linewidth(0.8)); /* special point */\ndraw((-1.494317431054321,-4.110068936296458)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8)); \ndraw((-1.1057141289867563,-1.0921133140036046)--(-1.11,-1.9), linewidth(0.8)); \ndraw((-1.11,-1.9)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8)); \ndraw((-1.1261681919167021,-4.947704176298286)--(2.66,-1.92), linewidth(0.8)); \ndraw((-5.398473910768285,-3.784828063474779)--(-5.211598643768283,-3.0982645825399944), linewidth(0.8)); \ndraw((-5.398473910768285,-3.784828063474779)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8)); /* special point */\ndraw((0.9060966758985711,-1.0139049292406515)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8)); \ndraw((0.9060966758985711,-1.0139049292406515)--(-5.398473910768285,-3.784828063474779), linewidth(0.8)); /* special point */\ndraw((-3.96,1.5)--(-4.88,-1.88), linewidth(0.8)); \ndraw((-4.88,-1.88)--(2.66,-1.92), linewidth(0.8)); \ndraw((2.66,-1.92)--(-3.96,1.5), linewidth(0.8)); \n /* dots and labels */\ndot((-3.96,1.5),linewidth(3pt) + dotstyle); \nlabel(\"$A$\", (-4.443756574004514,1.578482344102179), NE * labelscalefactor); \nlabel(\"$B$\", (-5.345334335086407,-1.8925920360631068), NE * labelscalefactor); \nlabel(\"$C$\", (2.7087603305785044,-1.9677235161532645), NE * labelscalefactor); \nlabel(\"$M$\", (-1.0778662659654463,-2.238196844477832), NE * labelscalefactor); \nlabel(\"$N$\", (-3.1162584522915163,-1.7122764838467284), NE * labelscalefactor); \ndot((-5.211598643768283,-3.0982645825399944),linewidth(3pt) + dotstyle); \nlabel(\"$P$\", (-5.630833959429006,-3.0946957175056298), NE * labelscalefactor); \nlabel(\"$Q$\", (-2.0944703230653709,-1.5418031555221607), NE * labelscalefactor); \nlabel(\"$O$\", (-1.9193388429752132,-4.4719308790383105), NE * labelscalefactor); \ndot((-1.1057141289867563,-1.0921133140036046),linewidth(3pt) + dotstyle); \nlabel(\"$D$\", (-1.363365890308046,-0.9008564988730251), NE * labelscalefactor); \nlabel(\"$E$\", (-1.228129226145762,-5.318587528174298), NE * labelscalefactor); \ndot((-5.398473910768285,-3.784828063474779),linewidth(3pt) + dotstyle); \nlabel(\"$F$\", (-5.8562283996994795,-3.966220886551459), NE * labelscalefactor); \nlabel(\"$G$\", (0.8304733283245604,-0.8858302028549936), NE * labelscalefactor); \nlabel(\"$H$\", (-2.595522163786633,-2.9242223891810623), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */[/asy][/hide]\nBy [b]Simson line theorem[/b], we know that $F,M,G$ are collinear.\nAs, $\\angle FAE=\\angle EAG$, $\\triangle AFE\\cong\\triangle AEG\\implies AF=AG$. Thus, $AE\\perp FG\\implies PQ\\parallel FM$. Also, $OP\\parallel FE$.\nTherefore, $\\triangle OPQ$ and $\\triangle EFM$ are homothetic with center $A$, which implies $QO\\parallel ME$. But $ME\\perp BC$. So, $QO\\perp BC$ as desired. And we are done." } { "Tag": [], "Problem": "A 20-gallon container is filled halfway with a mixture that is $ 90\\%$ vinegar and $ 10\\%$ water. How many gallons of water must be added for the mixture to become $ 60\\%$ vinegar and $ 40\\%$ water?", "Solution_1": "Since only half the 20 gallon container is filled, 10% water is equal to 1 gallon and 90% vinegar is 9 gallons. In the wanted result, the 9 gallons is 60%, so the 40% water would be 6 gallons. Thus, you must 5 gallons of water to the container to get 60% vinegar and 40% water." } { "Tag": [], "Problem": "\u0398\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03c0\u03c9 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03c0\u03bf\u03c5 \u03af\u03c3\u03c9\u03c2 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b1\u03c0\u03bb\u03bf\u03ca\u03ba\u03cc, \u03b1\u03bb\u03bb\u03ac \u03bc\u03bf\u03c5 \u03ad\u03ba\u03b1\u03bd\u03b5 \u03c4\u03b5\u03c1\u03ac\u03c3\u03c4\u03b9\u03b1 \u03b5\u03bd\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7. \u0395\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03bc\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd B. C. S., \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03b9\u03bf \u03b9\u03c3\u03c7\u03c5\u03c1\u03ae \u03b3\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03bd\u03ac\u03b4\u03b1\u03c2:\r\n$ ( x1^{2} \\plus{} y1^{2} )^{2}( x2^{2} \\plus{} y2^{2} )^{2} > \\equal{} ( x1x2 \\plus{} y1y2 )^{4}$.\r\n\u0397 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c0\u03bf\u03c5 \u03b5\u03af\u03b4\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2:\r\n\u03ad\u03c3\u03c4\u03c9 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 $ a \\equal{} (x1, y1), b \\equal{} (x2,y2)$, \u03c4\u03cc\u03c4\u03b5 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03c1\u03b1\u03c6\u03b5\u03c4\u03b1\u03b9 \r\n$ |a|^{4}|b|^{4} > \\equal{} (ab)^{4} < \\equal{} > |a|^{4}|b|^{4} > \\equal{} |a|^{4}|b|^{4}|cos(a, b)|^{4}$\r\n$ < \\equal{} > 1 > \\equal{} |cos(a, b)|$, \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9.", "Solution_1": "\u03a0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 B. C. S. \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2 (\u03c3\u03c4\u03b7\u03bd \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ae \u03bc\u03bf\u03c1\u03c6\u03ae \u03c4\u03b7\u03c2):\r\n$ ( x1^{1} \\plus{} y1^{1} )^{2}( x2^{1} \\plus{} y2^{1} )^{2} > \\equal{} ( x1x2 \\plus{} y1y2 )^{2}$.\r\n\u0398\u03b1 \u03b3\u03c1\u03ac\u03c8\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b9\u03b1 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b3\u03af\u03bd\u03b5\u03b9 \u03c0\u03b9\u03bf \u03b5\u03bc\u03c6\u03b1\u03bd\u03ae\u03c2 \u03b7 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac:\r\n$ ( x1^{2} \\plus{} y1^{2} )^{2}( x2^{2} \\plus{} y2^{2} )^{2} > \\equal{} ( x1x2 \\plus{} y1y2 )^{4} < \\equal{} >$\r\n$ ( x1^{2} \\plus{} y1^{2} )^{1}( x2^{2} \\plus{} y2^{2} )^{1} > \\equal{} ( x1x2 \\plus{} y1y2 )^{2}$.\r\n\r\n\u03a3\u03c4\u03b7\u03bd \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b9\u03b1, \u03cc\u03c4\u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 $ x1, x2, y1, y2 > 0$, \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03b7 \u03c0\u03bf\u03b9\u03bf \u03b9\u03c3\u03c7\u03c5\u03c1\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1,\r\n\u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 $ ( x1^{2} \\plus{} y1^{2} )^{1}( x2^{2} \\plus{} y2^{2} )^{1} < \\equal{} ( x1^{1} \\plus{} y1^{1} )^{2}( x2^{1} \\plus{} y2^{1} )^{2}$\r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03ad\u03b3\u03b9\u03bd\u03b5 \u03b4\u03b9\u03b1\u03bd\u03c5\u03c3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac, \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03c4\u03b5\u03af \u03b3\u03b9\u03b1 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03ce\u03bd \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b5\u03c2 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2. \u0393\u03b9\u03b1 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03c0\u03b1\u03c1\u03b5\u03bd\u03b8\u03ad\u03c3\u03b5\u03c9\u03bd, \u03b1\u03c1\u03ba\u03b5\u03af \u03bc\u03b9\u03b1 \u03b1\u03c0\u03bb\u03ae \u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03ae.", "Solution_2": "[quote=\"maniopas\"]\u0398\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03c0\u03c9 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03c0\u03bf\u03c5 \u03af\u03c3\u03c9\u03c2 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b1\u03c0\u03bb\u03bf\u03ca\u03ba\u03cc, \u03b1\u03bb\u03bb\u03ac \u03bc\u03bf\u03c5 \u03ad\u03ba\u03b1\u03bd\u03b5 \u03c4\u03b5\u03c1\u03ac\u03c3\u03c4\u03b9\u03b1 \u03b5\u03bd\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7. \u0395\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03bc\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd B. C. S., \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03b9\u03bf \u03b9\u03c3\u03c7\u03c5\u03c1\u03ae \u03b3\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03bd\u03ac\u03b4\u03b1\u03c2:\n$ ( x_1^{2} \\plus{} y_1^{2} )^{2}( x_2^{2} \\plus{} y_2^{2} )^{2} > \\equal{} ( x_1x_2 \\plus{} y_1y_2 )^{4}$.\n[/quote]\r\n\r\n\u0391\u03c5\u03c4\u03ae \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b1\u03bd \u03c4\u03b7\u03bd B. C. S. ...\u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 B. C. S. \u03b1\u03c0\u03bb\u03ac \u03c5\u03c8\u03ce\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b1 2 \u03bc\u03ad\u03c1\u03b7 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03c3\u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf :!: \u0398\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b1\u03bc\u03b5 \u03b5\u03be\u03af\u03c3\u03bf\u03c5 \u03bd\u03b1 \u03c5\u03c8\u03ce\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03bf\u03c0\u03bf\u03b9\u03b1\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03bb\u03bb\u03ac\u03b6\u03b5\u03b9 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1!", "Solution_3": "\u0394\u03b9\u03ba\u03cc \u03bc\u03bf\u03c5 \u03bb\u03ac\u03b8\u03bf\u03c2: \u03bc\u03c0\u03ad\u03c1\u03b4\u03b5\u03c8\u03b1 \u03c4\u03bf\u03bd \u03c4\u03cd\u03c0\u03bf \u03c4\u03b7\u03c2 B. C. S. \u0391\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03c3\u03c4\u03b7\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1 \u03ad\u03ba\u03b1\u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c9 \u03c4\u03b7 \u03af\u03b4\u03b9\u03b1 \u03c4\u03b7\u03bd B. C. S. \u0397 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03cc\u03bc\u03c9\u03c2 \u03c0\u03b1\u03c1\u03b1\u03bc\u03ad\u03bd\u03b5\u03b9 \u03b5\u03bd\u03c4\u03c5\u03c0\u03c9\u03c3\u03b9\u03b1\u03ba\u03ae. \u0397 \u03c3\u03cd\u03b3\u03ba\u03c1\u03b9\u03c3\u03b7 \u03ad\u03c7\u03b5\u03b9 \u03b3\u03af\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1." } { "Tag": [ "Columbia", "Olimpiada de matematicas" ], "Problem": "Hola a todos.\r\n\r\nComo seguramente saben por estos d\u00edas es la XXIV Olimp\u00edada Iberoamericana de Matem\u00e1ticas en Santiago de Quer\u00e9taro (M\u00e9xico).\r\n\r\nEn este momento se est\u00e1 desarrollando la prueba del primer d\u00eda.\r\n\r\nEstar\u00eda muy agradecido si alguien puede subir los enunciados cuando \u00e9sta termine, lo mismo ma\u00f1ana, y despu\u00e9s los puntajes o dem\u00e1s novedades que vayan teniendo, as\u00ed los que no estamos all\u00e1 nos mantenemos informados :D \r\n\r\nGracias!", "Solution_1": "alguien ya puede subir los temas, solo se que fueron combinatoria, teor\u00eda de n\u00fameros y geometr\u00eda", "Solution_2": "Problema 1-Seja n um natural maior que 2.Suponhamos que n ilhas estejam localizadas ao redor de um c\u00edrculo e que entre cada duas ilhas vizinhas haja duas pontes.Partindo da ilha x_1, de quantas maneiras se podem percorrer as 2n pontes passando por cada ponte exatamente uma vez?\r\nProblema 2 - Para cada inteiro positivo n, definimos a_n=n+m, em que m \u00e9 o maior inteiro tal que 2^(2^m)<=n*2^n.Determinar quais inteiros positivos n\u00e3o aparecem na sequ\u00eancia a_n.\r\nProblema 3- Sejam C1 e C2 duas circunferencias de centros O1 e O2, com o mesmo raio, que se intersectam em A e B.Seja P um ponto sobre o arco AB de C2 que est\u00e1 dentro de C1.A reta AP intersecta C1 em C, a reta CB intersecta C2 em D e a bissetriz de = 1/(i + j) for all unequal i, j. Show that if all xi lie in the interval [0, c], then c >= 1. \r\nmy solution shows that c >= n-2/n for all n which implies c>=1.duz nebody have a different bound for c?\r\nbest regards", "Solution_1": "You can find a solution at\r\nhttp://www.kalva.demon.co.uk/short/soln/sh02a2.html." } { "Tag": [ "induction", "algebra unsolved", "algebra" ], "Problem": "Let a1=1;a(k+1)=1+ $\\frac{k}{ak}$ for all k$\\geq$1\r\nProve that:\r\n$\\sqrt{n-3/4}$ + $\\frac{1}{2}$ $\\leq$ an $\\leq$ $\\sqrt{n+1/4}$ + $\\frac{1}{2}$\r\n for all n$\\geq$1", "Solution_1": "Let $a_1=1;a(k+1)=1+ \\frac{k}{a_k}$ for all k$\\geq$1\r\nProve that:\r\n$\\sqrt{n-3/4}$ + $\\frac{1}{2}$ $\\leq$ an $\\leq$ $\\sqrt{n+1/4}$ + $\\frac{1}{2}$\r\n for all n$\\geq$1\r\n\r\nIt proved by induction. It is true for n=1. Induction is equavalent to\r\n1.$1+\\frac{n}{0.5+\\sqrt{n+0.25}}\\ge 0.5+\\sqrt{n+0.25}$ and\r\n2.$1+\\frac{n}{0.5+\\sqrt{n-0.75}}<0.5+\\sqrt{n+1.25}.$\r\nExactly $1+\\frac{n}{0.5+\\sqrt{n+0.25}}=0.5+\\sqrt{n+0.25}.$\r\n2. equavalent $\\sqrt{n^2+0.5n-1.25*0.75}-n+0.5(\\sqrt{n+1.25}-\\sqrt{n-0.75)>0.25}$." } { "Tag": [ "set theory", "real analysis", "real analysis unsolved" ], "Problem": "Hi,\r\n\r\nI can't understand how to formulate what the author describes in the following text.\r\n$ E\\equal{}\\bigcup P(E)$ is easy to prove.\r\nBut I can't formulate the \"other order\" that he talks about.\r\nHow can I formulate it ?", "Solution_1": "He's asking you to show how $ P(\\cup E)$ compares to $ E$.", "Solution_2": "yes but that is $ P(\\bigcup E)\\equal{}P(E)$.\r\nis it that simple ?", "Solution_3": "No: for example $ E \\equal{} \\{\\{1, 2\\}, \\{2, 3\\}\\}$ implies $ \\cup E \\equal{} \\{1, 2, 3\\}$ and $ P(\\cup E) \\supsetneq E$.", "Solution_4": "but JBL,I thought elements of E are just numbers not other sets.\r\nand how does $ \\bigcup{E}\\equal{}\\{1,2,3\\}$ if E contains sets.\r\nFor that matter what are the subsets of this strange set E ?\r\nwhat is P(UE) in this case ?", "Solution_5": "The elements of the $ E$ I gave you are sets: $ E$ had two elements, $ \\{1, 2\\}$ and $ \\{2, 3\\}$. (Also, in the context of set theory numbers [i]are[/i] sets, but that's irrelevant to this discussion.) More importantly, I don't think you understand the notation you're using: $ \\cup E \\overset{\\textrm{def}}{ \\equal{} } \\bigcup_{a \\in E} a$." } { "Tag": [ "projective geometry", "geometry solved", "geometry" ], "Problem": "Let ABC be arbitrary triangle and AA1, BB1 and CC1 be three concurrent lines such that A1, B1 and C1 belong to BC, CA and AB respectively. Let A1B1 and AB intersect at X, B1C1 and BC intersect at Y, A1C1 and AC intersect at Z. Prove that X,Y and Z are collinear.", "Solution_1": "Just Ceva and Menelaus we can do it.", "Solution_2": "Yes, I know that we should use Ceva and Menelaus, but I don't know the way. Can you pls post the exact solution?", "Solution_3": "By Ceva:$\\frac{BA_1}{A_1C}\\frac{CB_1}{B_1A}\\frac{AC_1}{C_1B}=1$\r\nBy Menelaus:$\\frac{BA_1}{A_1C}\\frac{CB_1}{B_1A}\\frac{AX}{XB}=1$\r\n$\\frac{BY}{YC}\\frac{CB_1}{B_1A}\\frac{AC_1}{C_1B}=1$\r\n$\\frac{BA_1}{A_1C}\\frac{CZ}{ZA}\\frac{AC_1}{C_1B}=1$\r\nSo $\\frac{BY}{YC}\\frac{CZ}{ZA}\\frac{AX}{XB}=1$\r\nThen $X,Y,Z$ is collinear.", "Solution_4": "Come on, such a problem being given on an olympiad is :spam: . It is well-known itself and it is a particular case of the even more well-known [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30323]Desargues theorem[/url].\r\n\r\n darij" } { "Tag": [ "USAMTS" ], "Problem": "Will there be a Math Jam for Round 3? If so, when will it be?", "Solution_1": "oops...we totally forgot to schedule this! :blush: \r\n\r\nWe're going to have it on Thursday, 1/11, at 7:30 Eastern (4:30 Pacific). It'll be on the official schedule shortly.", "Solution_2": "When will the round 4 problems be up?", "Solution_3": "Towards the end of the month.", "Solution_4": "By the way, we have our usual discussion embargo: please do not discuss Round 3 until after the Math Jam on Thursday.", "Solution_5": "Can we discuss it privately (pms/AIM)?", "Solution_6": "of course you can. freedom of speech, unless the Bill of Rights has a flaw", "Solution_7": "[quote=\"purple plume\"]of course you can. freedom of speech, unless the Bill of Rights has a flaw[/quote]\r\nActually, I pmed an admin and they said wait until after the math jam. About freedom of speech, the thing that prevents me from cheating on the USAMTS, is what is stopping me from discussing the round, so...", "Solution_8": "[quote=\"bpms\"][quote=\"purple plume\"]of course you can. freedom of speech, unless the Bill of Rights has a flaw[/quote]\nActually, I pmed an admin and they said wait until after the math jam. About freedom of speech, the thing that prevents me from cheating on the USAMTS, is what is stopping me from discussing the round, so...[/quote]\r\n\r\nThe USAMTS runs on an honor system, so it follows \"the thing\" you mentioned is honor. If you find it goes against said honor to discuss problems before the Math Jam, go ahead and don't. Otherwise...", "Solution_9": "[quote=\"jruzzy\"][quote=\"bpms\"][quote=\"purple plume\"]of course you can. freedom of speech, unless the Bill of Rights has a flaw[/quote]\nActually, I pmed an admin and they said wait until after the math jam. About freedom of speech, the thing that prevents me from cheating on the USAMTS, is what is stopping me from discussing the round, so...[/quote]\n\nThe USAMTS runs on an honor system, so it follows \"the thing\" you mentioned is honor. If you find it goes against said honor to discuss problems before the Math Jam, go ahead and don't. Otherwise...[/quote]\r\nDid you read my post? The admin specifically said don't discuss the third round until after the math jam.", "Solution_10": "[quote=\"bpms\"][/quote][quote=\"jruzzy\"][quote=\"bpms\"][quote=\"purple plume\"]of course you can. freedom of speech, unless the Bill of Rights has a flaw[/quote]\nActually, I pmed an admin and they said wait until after the math jam. About freedom of speech, the thing that prevents me from cheating on the USAMTS, is what is stopping me from discussing the round, so...[/quote]\n\nThe USAMTS runs on an honor system, so it follows \"the thing\" you mentioned is honor. If you find it goes against said honor to discuss problems before the Math Jam, go ahead and don't. Otherwise...[/quote][quote=\"bpms\"]\nDid you read my post? The admin specifically said don't discuss the third round until after the math jam.[/quote]\r\n\r\nJeez... no need to accuse people of failure to read posts. \r\n\r\nI did, in fact, read your post (as I do with all posts I quote) and was commenting on the second half of it. I fail to see how you can equate outright cheating on the problems to simple discussion of the problems, but I digress...\r\n\r\nThe thing is in under 24 hours anyway, so this will become a non-issue soon enough.", "Solution_11": "[quote=\"jruzzy\"][/quote][quote=\"bpms\"][/quote][quote=\"jruzzy\"][quote=\"bpms\"][quote=\"purple plume\"]of course you can. freedom of speech, unless the Bill of Rights has a flaw[/quote]\nActually, I pmed an admin and they said wait until after the math jam. About freedom of speech, the thing that prevents me from cheating on the USAMTS, is what is stopping me from discussing the round, so...[/quote]\n\nThe USAMTS runs on an honor system, so it follows \"the thing\" you mentioned is honor. If you find it goes against said honor to discuss problems before the Math Jam, go ahead and don't. Otherwise...[/quote][quote=\"bpms\"]\nDid you read my post? The admin specifically said don't discuss the third round until after the math jam.[/quote][quote=\"jruzzy\"]\n\nJeez... no need to accuse people of failure to read posts. \n\nI did, in fact, read your post (as I do with all posts I quote) and was commenting on the second half of it. I fail to see how you can equate outright cheating on the problems to simple discussion of the problems, but I digress...\n\nThe thing is in under 24 hours anyway, so this will become a non-issue soon enough.[/quote]\r\nOh sorry, my mistake. I didn't know what you were saying.", "Solution_12": "[quote=\"rrusczyk\"]Towards the end of the month.[/quote]\r\n\r\nNow that we are already reaching the end of the month, does anyone know of a more specific time that round 4 is coming out?", "Solution_13": "Probably at the end of the week -- Thurs or Fri.", "Solution_14": "How about Round 3 scores? Is it also coming at the end of the week?\r\n\r\nCan't wait... :jump:", "Solution_15": "[quote=\"vishalarul\"]How about Round 3 scores? Is it also coming at the end of the week?\n\nCan't wait... :jump:[/quote]\r\n\r\nLast time it took about five weeks, probably not for another week at least." } { "Tag": [ "vector", "analytic geometry", "induction", "linear algebra", "linear algebra unsolved" ], "Problem": "Suppose $e_1,e_2,...,e_n$ be $n+1$ vectors of an Euclidean vector space $E$ satisfied:\r\n\\[ (e_i,e_j)<0\\ \\ \\ \\forall i\\not =j \\]\r\nwhere $(.,.)$ is the inner product (definite positive scalar product) \r\nProve that any $n$ vector of them formed a basis of $E$", "Solution_1": "It's already discussed, I think.", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?t=31806", "Solution_3": "A different solution than the one harazi posted in that other thread (but also using a problem posted by alekk some time ago :)):\r\n\r\nSuppose some $n$ of those vectors are linearly dependent. Then there is a hyperplane in which they all lie, which determines two half-spaces of $E$, one of which contains all our $n+1$ vectors. It thus suffices to prove the following stronger assertion:\r\n\r\nGiven $n+1$ vectors $v_1,\\ldots,v_{n+1}$ in an $n$-dimensional uclidean space $E$ satisfying $(v_i,v_j)<0,\\ \\forall i\\ne j$, for every $v\\in E,\\ v\\ne 0$, we can find $i\\in\\overline{1,n+1}$ with $(v,v_i)<0$. \r\n\r\nSuppose the contrary, and let $v$ be such that $(v,v_i)\\ge 0,\\ \\forall i$. If we can find an $i$ for which $v_i$ and $v$ are collinear, then for all $i$ we have $(v,v_i)>0$, so we can move $v$ slightly until it's no longer collinear with any of the $v_i$'s. If we express the $v_i$'s in an orthogonal basis which $v$ is part of as the $n$'th vector, all the $v_i$ will have a non-negative $n$'th coordinate, and not all the first $n-1$ coordinates equal to zero. For each $i$, let $\\bar v_i$ be the projection of $v_i$ onto the orthogonal complement of $v$. The vectors $\\bar v_i$ are non-null and in the orthogonal complement of $v$ viewed as an $n-1$-dimensional Euclidean space with the scalar product induced from $E$ they satisfy $(\\bar v_i,\\bar v_j)<0,\\ \\forall i\\ne j$. \r\n\r\nThe above means that in an $n-1$-dimensional Euclidean space we can find $n+1=(n-1)+2$ vectors with negative mutual scalar products, but this contradicts the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=n%2B2&t=31714]posted here[/url].", "Solution_4": "This is another proof:\r\nIt is sufficient to prove that \" In an arbitrary Euclidean vector space E with finite dimention, if system of non-zero vectors $(e_1,...,e_{n+1})$ satisfied the scalar product of any two distinct vectors is negative, then every $n$ elements of them are linearly independent.\"\r\nBy induction on $n, n=1, n=2$ are trivially\r\nSuppose that the statement true for $n-1\\geq2$\r\nWith $n$, Consider the set consists of $n+1$ vector $(e_1,e_2,...,e_{n+1})$\r\nFirst, there are no couple of vectors are linear dependent. (Why?)\r\nThus, we fix $e_1$\r\nDenote $e'_i=e_i-\\frac{(e_i,e_1)}{(e_1,e_1)}e_1$\r\nThen $e'_2,...,e'_{n+1}$ are on the orthogonal complement of $L(e_1)$ (the vector space span by $e_1$)\r\nfor every $11$,\r\n\\[ (e'_i,e'_j)=(e_i,e_j)-\\frac{(e_i,e_1)}{(e_1,e_1)}(e_j,e_1)-\\frac{(e_j,e_1)}{(e_1,e_1)}(e_i,e_1)+(\\frac{(e_i,e_1)}{(e_1,e_1)}e_1,\\frac{(e_j,e_1)}{(e_1,e_1)}e_1) \\]\r\n\\[ (e'_i,e'_j)=(e_i,e_j)-\\frac{(e_i,e_1)(e_j,e_1)}{(e_1,e_1)}<0 \\]\r\nby induction hypothesis, every $n-1$ vector in $(e'_2,...,e'_{n+1})$ are linearly independent. In particular, $e'_2,...,e'_n$ are linearly independent, lie on the orthorgonal complement of $L(e_1)$. Thus, $e_1,e'_2,...,e'_n$ are linearly independent, implies, $e_1,e_2,...,e_n$ are linearly independent (why?), other subset is analogous.\r\n\r\nNice one!!" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let n be a positive integer and $X_{1}$ ,..........$X_{2n}$ a sequence of 2n integers where $X_{i}$ =$\\pm1$, $i$ =1,..........2n.\r\n\r\nDefine $S_{n}$ =$\\sum_{1}^{n}X_{2r-1}X_{2r}$\r\n\r\nFind the number of sequences with $S_{n}$ odd and the number with $S_{n}$ even.", "Solution_1": "$S_{n}\\equiv n[2]$" } { "Tag": [ "probability" ], "Problem": "A bag contains 3 tan, 2 pink, and 4 violet chips. If the 9 chips are randomly drawn from the bag, one at a time and without replacement, what the probability that the chips are drawn in such a way that the 3 tan chips are drawn consecutively, the 2 pink chips are drawn consecutively and the 4 violet chips are drawn consecutively, but not necessarily in the tan-pink-violet order? Express your answer as a common fraction.", "Solution_1": "There are 9! ways of picking any order of chips and:\r\n3! ways of arranging the tan \"bundle\"\r\n2!: pink \"bundle\"\r\n4!: violet \"bundle\"\r\n3!: rearranging the bundles.\r\nSo:\r\n\r\n$ \\frac{3*2*2*4*3*2*3*2}{9*8*7*6*5*4*3*2}\\equal{}\\frac{1}{210}$" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "let ABC be a triangle :\r\n\r\ndraw aline ,wich intersects AB and AC in points Pand Q such that: BP=PQ=QC", "Solution_1": "See [url=http://groups.google.de/group/geometry.college/browse_thread/thread/948b29d3346dd801/de28fab3d89a8f04]Hyacinthos messages #7199, #7201, #7203, #7210[/url].\r\n\r\n darij" } { "Tag": [ "rotation", "geometry", "geometric transformation", "reflection", "conics", "parabola", "function" ], "Problem": "Let's consider the cartezian system XOY and O a point light source.Fiind the ecuation of the curve y=y(x) so that if we rotate it arround Ox we obtain a mirror so that any ray going from O that reflects goes parallel with Ox to infinite.\r\nIf you know the diameter of the mirror D and the depth h fiind the focal distance of the mirror.", "Solution_1": "I got the following answers:\r\n$y^{2}=4fx$\r\n\r\n$f=\\frac{D^{2}}{16h}$", "Solution_2": "O is the focal point situated at a f-distance from the tip of the mirror so you should got an extra term in the parabola function", "Solution_3": "Is that like this?\r\n\r\n$y^{2}=4f^{2}-4fx$\r\n\r\n$f=\\frac{D^{2}}{4h}$", "Solution_4": "it's + instead of -", "Solution_5": "I think it must be minus(-).What is the value of focal length?", "Solution_6": "the focal lenght is sqrD/16h\r\nif it's minus than you get x=f>0 if y=0 and f is the distance from O(focal point) to the tip than the mirror is situated with the tip in the pozitive way of Ox but i said that the rays goes to infinite not to -infinite", "Solution_7": "Yes,you're right at that point.I didn't notice that the ray goes to +infinity :)" } { "Tag": [ "Euler", "projective geometry" ], "Problem": "An icosidodecahedron is a polyhedron with twenty triangular faces and twelve pentagonal faces. Each vertex of an icosidodecahedron is identical with two triangles and two pentagons meeting at each vertex. What is the sum of the number of vertices and the number of edges in an icosidodecahedron?", "Solution_1": "Say that name fast three time........", "Solution_2": "[hide]Well.... there's a trick:\n\nThere's a formula (maybe Pascal's theorem???)out there that says\" Verticies + Faces - edges = 2\"\nObviously, there are 20 + 12 = 32 faces.\nOver counting the edges, we see that there is 20*3 + 12*5 = 120 edges (that's a bit too much). Since every edge is shared by two figures, we should divide 120 by 2 (60). plugging this in:\n\nVerticies + 32 - 60 = 2\n There must be 30 verticies\n\n30 + 60 = [b]90[/b][/hide]", "Solution_3": "[hide]i think its eulers formula , not pascal's where if v = vertices, e = edges, f = faces, V+F=E+2[/hide]\r\n\r\nicosadodecahedron,icosadodecahedron,icosadodecahedron\r\n\r\nThere, I said it three times fast", "Solution_4": "[quote=\"philB\"][hide]Well.... there's a trick:\n\nThere's a formula (maybe Pascal's theorem???)out there that says\" Verticies + Faces - edges = 2\"\nObviously, there are 20 + 12 = 32 faces.\nOver counting the edges, we see that there is 20*3 + 12*5 = 120 edges (that's a bit too much). Since every edge is shared by two figures, we should divide 120 by 2 (60). plugging this in:\n\nVerticies + 32 - 60 = 2\n There must be 30 verticies\n\n30 + 60 = [b]90[/b][/hide][/quote]\r\nEuler's Formula, not Pascal's (I think)", "Solution_5": "[hide]first there are 20*3+12*5=120 edges without restrictions, but we overcounted in the polyhedron. so divide by 4 since each edge was counted twice to get 30. there are also 120 edges throughout the polygons, each edge is counted twice, so divide by 2 to get 60. 60+30=90[/hide]", "Solution_6": "[hide]Similiarly, each vertex is counted four times.... there are 120 verticies UNRESTICED, divided by four equals 30....[/hide]", "Solution_7": "Extension: how many space diagonals does it have?", "Solution_8": "are space diags where you connect vertices?\r\n\r\nhow do you figure out how many of the vertices lie on pentagons?\r\n\r\njorian[/hide]", "Solution_9": "[quote=\"13375P34K43V312\"]Extension: how many space diagonals does it have?[/quote]\r\nIsn't that just $\\binom{n}{2}-5k-e$ where n is the number of vertices, k is the number of pentagons, e is the number of edges?", "Solution_10": "Yeah, that would be right.\r\n\r\nAs for the solution, we would first select 2 points, which is where the ${n\\choose2}$ comes from, then subtract all of the edges, because those are generated from selecting 2 points, but don't count as space diagonals, which is where the $-e$ comes from. Finally, we have surface diagonals, the diagonals on the pentagons. There are five diagonals per pentagon, so $-5k$.", "Solution_11": "wow, the solution is pretty easy once you know how to calculate it\r\n\r\njorian" } { "Tag": [ "geometry", "3D geometry", "Duke", "college", "geometric transformation", "rotation", "probability" ], "Problem": "Welcome to the NC Cube Forum\r\nPost your cubing results.\r\nPretty Simple. :lol:\r\n\r\nMy best time is 50.45 Seconds at the First Meeting of the Raleigh Cubing Association (RCA).\r\n\r\n-Sanjay", "Solution_1": "141.7 seconds (I know, I'm terrible) :P :D", "Solution_2": "Whatever happened to North Carolina?", "Solution_3": "[quote=\"jb05\"]Whatever happened to North Carolina?[/quote]\r\n\r\nGood point. [hide=\":rotfl: \"]They lost to Duke by 11 points :lol: [/hide]\r\nI have no idea how they managed to perform so poorly.", "Solution_4": "0 seconds. Started from this position\r\nGGG\r\nGGG\r\nGGG\r\n\r\nWWWRRRYYY\r\nWWWRRRYYY\r\nWWWRRRYYY\r\n\r\nBBB\r\nBBB\r\nBBB\r\n\r\nOOO\r\nOOO\r\nOOO\r\n\r\nPS. Who goes to the RCA?? I bet the population is 1 (you.)\r\n\r\n\r\nMy best scrambled time is some 40 or 50 seconds\r\n\r\nMy worst is... 462 hours (I left it in my drawer)", "Solution_5": "Record is 46 seconds.", "Solution_6": "It takes me like 4 minutes... layer-by-layer :P \r\n\r\nNow that I'm on break, maybe I should go learn some better algorithms.", "Solution_7": "12 seconds", "Solution_8": "You know finger tricks? (most likely)", "Solution_9": "[quote=\"PiGuy\"]You know finger tricks? (most likely)[/quote]\r\n\r\nYes .", "Solution_10": "[quote=\"PI-Dimension\"]My worst is... 462 hours (I left it in my drawer)[/quote]\r\n\r\nI'm sure everyones done that before. As a matter of fact. I think the record would be a few years from the time a cube is made, to the time when some kid gets mad and throws the cube away, to when finally some hobo digs the cube up from the trash and solves it before throwing it into an incinerator. What a waste.", "Solution_11": "[quote=\"PiGuy\"][quote=\"PI-Dimension\"]My worst is... 462 hours (I left it in my drawer)[/quote]\n\nI'm sure everyones done that before. As a matter of fact. I think the record would be a few years from the time a cube is made, to the time when some kid gets mad and throws the cube away, to when finally some hobo digs the cube up from the trash and solves it before throwing it into an incinerator. What a waste.[/quote]\r\n\r\nThat's wrong Mr.Cao.\r\n\r\nthe cube is MADE finished, so the timer would start WHEN the kid messes it up", "Solution_12": "Thank you for the off-topic posts... please stop.", "Solution_13": "Are we allowed to talk about other variations of the rubik's cube because I still can't solve a rubik's revenge. The way I start is first the corners, then the center, then the layers. This proves to be very inefficient for me.\r\n\r\nBtw, does anyone know any good websites that actually teach you how to solve a professor cube without being unobtrusively confusing.", "Solution_14": "Wait people were going off-topic? Seeing as this is essentially a spam thread, I wasn't aware it was possible to go off-topic.\r\n\r\nUmm 5x5x5 cubes are not actually that difficult. It is fairly easy to get all the center blocks right, and once you get the interior edges, it becomes a 3x3x3. The hard part seems to be the interior edges. I'm sure there are plenty of good websites for that. I know there was one that helped me successfully solve mine... But by the time I solved it once, no less than three faces had broken off and resisted gluing back on, so I haven't used it since. Those things are designed horribly. Also, square-1 is kind of cool for showing off, even though it is easier than Rubik's cubes.\r\n\r\nSo yeah... you guys want to do some math now? Maybe? Hmm I have a related problem (well-known but whatever). If A is the group of cube transformations and B is the group of positions you can get by taking the cube apart and putting it back together, what is |B|/|A|? Better yet, what group is B/A isomorphic to?", "Solution_15": "Do you know? Do you even know how many combinations there are?", "Solution_16": "Well, I believe that once you have assembled the cube, you can enact any even permutation on the 24 visible faces of the edge cubies (that doesn't actually destroy the cubies) without disturbing the corner cubies, and similarly for the corner cubies. So you can look at corner cubies and edge cubies independently, and in each case, you have a group isomorphic to $ \\mathbb{Z}/2\\mathbb{Z}$. So B/A is the Klein four-group?", "Solution_17": "No. You mentioned permutation of pieces, but what about orientation of the pieces? Also, your reasoning for permutations is off.", "Solution_18": "[quote=\"jb05\"]No. You mentioned permutation of pieces, but what about orientation of the pieces? Also, your reasoning for permutations is off.[/quote]\r\nI was actually referring to the permutations of the different visible faces of the cubies which do not destroy the cubies. For instance, switching the orientation of one edge cubie permutes two visible faces, and so is an odd permutation of the visible faces. I thought this took care of orientation of the cubies (I assume that we're talking about the same things when you say \"pieces\" and I say \"cubies\"), unless you are counting the orientation of the center cubies, which I thought people usually didn't care about?\r\n\r\nIt seemed to me that the edge cubies are either solvable or not, and that any edge-solvable position can be reached from any edge-solvable position without disassembling the cube and without disturbing the corner cubies, and similarly for non-edge-solvable positions; and similarly for the corner cubies. If these conditions assumption is correct, do we not have the Klein four-group? Or are those conditions wrong?\r\n\r\nI believe you when you say I'm wrong, though; I don't have enough experience with Rubik's cubes to have confidence here.", "Solution_19": "I am fairly certain you can't talk about even-odd permutations of actual faces without missing stuff. The numerical answer is definitely 12 and I think you are missing a factor of Z/3Z.", "Solution_20": "hm I was doing some random contest problems earlier and I stumbled upon this one, hope you cubers have fun :lol: \r\n\r\nA solved Rubik's cube is a cube whose face are each painted a different color. Six fixed colors are used in constructing a Rubik's cube, though you may have noticed that they are not always painted on in the same order. How many different ways can the cube be colored? (Two \"solved\" cubes are colored the same if a rotation of one whole cube makes it look exactly like the other.)", "Solution_21": "Yay!!! I got a Rubik's Revenge over the weekend!!!", "Solution_22": "Here is an interesting problem.\r\n\r\nJack gets a scambled, solvable, Rubik's cube (the 3x3x3). He randomly turns it until it is solved. Assuming that he has unlimited time, on average, when will he solve the cube.\r\n\r\n(for the 1x1x1, its of course, 0)", "Solution_23": "Who's Jack, and if you have unlimited time, you could take as long as wanted to. What are you asking when you say when will he finish it? Also, Jack will eventually die :| ...", "Solution_24": "PI-Dimension is saying that, given that a hypothetical immortal person named Jack has a Rubik's Cube, which he turns randomly*. Given that he turns, say, once per second, how long is he expected to take in solving the cube?\r\n\r\n@PI-Dimension: Please give a better definition of \"randomly\". Depending on the interpretation, several possible answers could be correct.\r\n\r\nAnswer: Based on my definition of randomly (every possible state is equally likely at every point) I say 43,252,003,274,489,856,000 seconds.", "Solution_25": "He probably means that he has an equal chance of turning each layer.", "Solution_26": "That's how I interpreted it. It seems like it should be an extremely difficult problem.", "Solution_27": "I think it will probably need a computer... calculating expected value with like billions of possible configurations? ouch.", "Solution_28": "So what's the answer?", "Solution_29": "Err, if you've been reading the above posts, it would become apparent that it is highly unlikely that any of us know the answer.", "Solution_30": "Oh well. Here's another similar question: If Billy randomly turns a finished cube 10 times, what is the probability that after the ten moves, he is back to where he started?", "Solution_31": "I suspect that the moves need to go in (adjacent) pairs, because turning the cube isn't associative. Also, 10 is not divisible by 4, so it's gonna be like\r\n\r\n$ X_1, X_1', X_2, X_2',\\ldots, X_{10},X_{10}'$\r\n\r\nwhere $ X_1,\\ldots,X_{10}$ are moves and $ X_i'$ is the inverse of $ X_i$.\r\n\r\nWait... can a \"turn\" be a 180 degree turn? anyway, I will assume that\r\n\r\nthen there's 6 faces, each can be turned 3 ways; so 18 different moves in every situation\r\n\r\nand I am too lazy to finish but you should get the idea\r\n\r\n(however I may have left out ultra-tricky cases. hm.)", "Solution_32": "Hmm, this actually doesn't seem too difficult (the second question), merely a lot of boring computation...\r\n\r\nAssuming a turn is $ 90^\\circ$, there are 9 layers, each of which can be turned, and the total number of turns must be 0 (mod 4) or add to 0, so it shouldn't be too difficult to figure it out from there...", "Solution_33": "Oh yeah, 9 layers. Ugh. I fail to think... anyway, I need a better definition of \"turn\"", "Solution_34": "180 degrees counts as two turns.", "Solution_35": "[quote=\"Teki-Teki\"]PI-Dimension is saying that, given that a hypothetical immortal person named Jack has a Rubik's Cube, which he turns randomly*. Given that he turns, say, once per second, how long is he expected to take in solving the cube?\n\n@PI-Dimension: Please give a better definition of \"randomly\". Depending on the interpretation, several possible answers could be correct.\n\nAnswer: Based on my definition of randomly (every possible state is equally likely at every point) I say 43,252,003,274,489,856,000 seconds.[/quote]\r\n\r\nactually i was asking for how many turns, but it works too.", "Solution_36": "There seem to be error(s) in here (the final formula is wrong?), but I don't feel like looking for them. The point is that this problem requires some very powerful information that we don't have.\r\n\r\nLet $ f(k)$ be the number of ways to get from 0 back to itself in $ k$ moves (the number of ordered zero-sum sequences of the eight basic operations of length $ k$). The total number of ways to do $ k$ moves from 0 is $ 8^k$, but we need to remove possibilities that pass through 0 again, hence $ g(k)\\equal{}g^k\\minus{}\\sum_{i\\equal{}1}^{k}{f(i)g(k\\minus{}i)}$ or $ g^k\\equal{}\\sum_{i\\equal{}0}^{k}{f(i)g(k\\minus{}i)}$ with $ f(0)\\equal{}g(0)\\equal{}1$, and $ g\\equal{}g(1)$ is the number of possible \"moves\" (in this case, 8). The expected number of turns will then be $ \\frac{1}{P}\\sum_{i\\equal{}1}^{\\infty}{\\frac{g(i)i}{g^i}}\\equal{}P\\sum_{i\\equal{}0}^{\\infty}{\\frac{f(i)}{g^i}}\\equal{}43252003274489856000\\sum_{i\\equal{}0}^{\\infty}{\\frac{f(i)}{8^i}}$ (assuming that you eventually reach 0 with probability 1), where $ P$ is the number of positions of the Rubik's cube.\r\n\r\nNote that I got the last equality (between the two sums) from the generating functions $ F(gx)\\equal{}\\sum_{i\\equal{}0}^{\\infty}{f(i)x^i}$ and $ G(gx)\\equal{}\\sum_{i\\equal{}0}^{\\infty}{g(i)x^i}$. We know that $ F(gx)G(gx)\\equal{}\\frac{1}{1\\minus{}x}$, but $ G(1)$ is evidently $ P$, since we should eventually reach 0 with probability 1. Hence we can write $ G(gx)\\equal{}\\frac{1}{H(gx)}$ and $ F(gx)\\equal{}\\frac{H(gx)}{1\\minus{}x}$, with $ H(1)\\equal{}\\frac{1}{P}$, and our desired quantity $ \\frac{G'(1)g}{P}\\equal{}\\minus{}H'(1)gP$. However, $ H(gx)\\equal{}(1\\minus{}x)F(gx)$ has derivative at 1 equal to $ \\frac{1}{g}\\sum_{i\\equal{}1}^{\\infty}{i\\frac{f(i)}{g^i}\\minus{}i\\frac{f(i\\minus{}1)}{g^{i\\minus{}1}}}\\equal{}\\frac{\\minus{}1}{g}\\sum_{i\\equal{}0}^{\\infty}{\\frac{f(i)}{g^i}}$. Multiplying by $ \\minus{}Pg$ yields the equality above.\r\n\r\nUnfortunately, we can't easily finish this - it should be VERY difficult to compute $ f(i)$ in general.", "Solution_37": "[quote=\"not_trig\"]Oh yeah, 9 layers. Ugh. I fail to think... anyway, I need a better definition of \"turn\"[/quote]\r\n\r\n\r\n9 o_O I'm sorry... you guys lost me...\r\n\r\nLETS SEE\r\n\r\nFront, Left, Right, Back, Top, Bottom\r\n\r\nSince you can rotate the both back and forth (CW and CCW), that means 12...\r\n\r\nand I missing something?\r\n\r\n[hide=\"hint\"]\n\nThere are 43,252,003,274,489,856,000 possible ways of scrambling a cube\n\n[/hide]\n\n[hide=\"bigger hint\"]\n\ndivide the former number by 2 to get the average amount of time to obtain a state...\n\ndivide that by the number of ways to rearrange the centers...\n\n[/hide]\r\n\r\nEDIT: WARNING!!! THIS PROBLEM HAS A 'TRICK' TO IT!! THERE ARE MORE THAN 1 SOLVED STATE OF A RUBIK'S CUBE!!!!!!!!!!!!!!!! (read second hint)", "Solution_38": "[quote=\"PI-Dimension\"]divide the former number by 2 to get the average amount of time to obtain a state...[/quote]\r\n\r\nAverage amount of time to obtain what state?", "Solution_39": "[quote=\"jb05\"][quote=\"PI-Dimension\"]divide the former number by 2 to get the average amount of time to obtain a state...[/quote]\n\nAverage amount of time to obtain what state?[/quote]\r\n\r\nAny state............................................................\r\n\r\n\r\n=( I know... I'm a terrible teacher.", "Solution_40": "The average amount of time to reach an average state in the klein four group from 0 is 1 move, but there are 4 elements. Explanation?", "Solution_41": "ok we failed on that question...\r\n\r\nOK\r\n\r\nWhat is the minimum number of permutations of a rubik's cube, with proof (answer is already in topic)" } { "Tag": [ "geometry", "summer program", "Mathcamp" ], "Problem": "I decided to make a topic edifying people about P-adics because P-adics are cool.\r\n\r\nIn computer science terms, working with p-adics is like working with unisgned numbers. But instead of the maximum value being 65535, it is ...999999999999999999999.\r\n\r\nIn arithmetic terms, try subtracting 5 from 3. You know it's -2, but go along with me and write the numbers in a column:\r\n[code]\n 3\n- 5\n___\n\n[/code]\nWhile we're at it, write a bunch of 0s in front of the 3:\n[code]\n...00000000000000000000000000000000003\n- 5\n______________________________________\n\n[/code]\nNow subtract like you were taught in school. 5 is bigger than 3, so you change the 3 to a 13 by borrowing a 1 from the next decimal place. Subtract and you get an 8. Now you can't borrow 1 from the next decimal place because it's a 0 so you change it into a 9 and borrow a 1 from the next decimal place. But the next decimal place is a 0 so you change it into a 9 and borrow a 1 from the next decimal place. But the next decimal place...[code]\n...00000000000000000000000000000000003\n- 5\n______________________________________\n...99999999999999999999999999999999998\n[/code]\r\n\r\nBasically the point is that in p-adics positive infinite numbers are equivalent to negative finite numbers and positive finite numbers are equivalent to negative infinite numbers. Writing numbers this way is really useful because it allows you to look at infinity more closely. Here's an example.\r\n\r\nFind the sum: [tex]\\displaystyle\\sum_{k=0}^{\\infty}3^k=1+3+3^3+3^3+3^4+...[/tex].\r\n\r\nNormally you would say that this diverges, but not when you're talking about p-adics. Here are a few ways to look at this problem:\r\n\r\nLet S be the sum you're looking for. It is clear that [tex]\\frac{S-1}3=S[/tex]. Solving, you get [tex]S=-\\frac 12[/tex].\r\n\r\nAnother way: Convert everything into base 3. First off: in base 3, -1=...222222222222222, not ...99999999999999999. Writing the sum in base 3 is trivial if you know how to use other bases. It's ...1111111111111111111. You can see that [tex]...111111111111111111=\\frac{...22222222222222222222}2=-\\frac12[/tex]\r\n\r\nIf any of you has any questions or knows any cool stuff about p-adics please post. Also, does anyone know why p-adics are called p-adics?", "Solution_1": "Infinite unsigned ints... much easier to understand than the mathworld definition! What would these be important for?", "Solution_2": "I just looked at the Mathworld definition. It is indeed a lot more complicated, but it seems to be more accurate technically.\r\n\r\nP-adic numbers have been used to prove a lot of things, including the fact that it is possible to dissect a square into n triangles of equal area if and only if n is even. I took a class at Mathcamp that went over this proof, so I should know it, but unfortunately I didn't understand it and so I don't.\r\n\r\nIt would be nice if someone more knowledgeable on the subject would post something, because all I know about p-adics is the basics and that they are cool.", "Solution_3": "Did Gregory Galperin teach p-adics at MOP?", "Solution_4": "The Red group (the one I was in) wasn't taught p-adics and I don't think the other groups were either. I don't think Gregory Galperin was there at all, since I don't remember him. The class I took was at Mathcamp (USA/Canada) in 2003.", "Solution_5": "Galperin taught p-adics at mop 2002.... and we never used them on anything! how disappointing." } { "Tag": [ "number theory", "prime factorization" ], "Problem": "Hi,\r\n\r\nI just started number theory. Hopefully someone will help me cause i just can't seem to see relationships between problems, they all look so different! Heres 3 of the many im stuck on.\r\n\r\n1. Prove that if p and p+4 are consecutive primes, then 9 divides the sum of the composite numbers between them.\r\n\r\n2. Show that there are infinitely many primes of the form 6k+5.\r\n\r\n3. Suppose that n > 1 is an integer. If n is not the form 6k+3 then prove that n^2 + 2^n is a composite.\r\n\r\nThanks", "Solution_1": "I got your $1st$ problem... I think.\r\nPrime no.s end with $1,3,7,9$ so \r\n$1st$ possibility $10x+3=p$ and $p+4=10x+7$ so now there can be no prime no. in between so $sum=p+p+1+p+2+p+3$ on adding you get $3p+6$ which is multiple of $9$.\r\n$2nd$ possiblitiy $p=10x+7$ and $p+4=10x+11$ so you get $3p+6$ as the sum of the no. which is a multiple of $9$.\r\nand the $3rd$ possiblility the same way\r\nI think that is acceptable. :(", "Solution_2": "[quote=\"riddler\"]I got your $1st$ problem... I think.\nPrime no.s end with $1,3,7,9$ so \n$1st$ possibility $10x+3=p$ and $p+4=10x+7$ so now there can be no prime no. in between so $sum=p+p+1+p+2+p+3$ on adding you get $3p+6$ which is multiple of $9$.\n$2nd$ possiblitiy $p=10x+7$ and $p+4=10x+11$ so you get $3p+6$ as the sum of the no. which is a multiple of $9$.\nand the $3rd$ possiblility the same way\nI think that is acceptable. :([/quote]\r\n3p+6 isnt always divisible by 9. for example, what if p was 2?", "Solution_3": "[quote=\"kryptos\"]Hi,\n\nI just started number theory. Hopefully someone will help me cause i\njust can't seem to see relationships between problems, they all look\nso different! Heres 3 of the many im stuck on.\n\n1. Prove that if p and p+4 are consecutive primes, then 9 divides the\nsum of the composite numbers between them.\n\n2. Show that there are infinitely many primes of the form 6k+5.\n\n3. Suppose that n > 1 is an integer. If n is not the form 6k+3 then\nprove that n^2 + 2^n is a composite.\n\nThanks[/quote]\r\nnumber 2 definitely isnt a getting started problem. It might even be\r\npre-olympiad level\r\n[hide=\"hint for 2\"]The way Euclid did his argument (assuming otherwise\nand then multiplying \"all\" primes together and adding 1) can be\napplied similarly here, but it has to be implimented more\ncleverly[/hide]\r\n\r\n1.the numbers between p and p+4 are\r\np+1+p+2+p+3.\r\nwhich gives us\r\n3p+6. Now it remains to prove that p=1(mod 9) or 4(mod 9) or 7(mod\r\n9)since all of these make that divisible by 9. If p was 2(mod 9) then\r\nthe other prime would be 6(mod 9) which is divisible by 3.\r\nContradiction. p couldnt be 3(mod 9). if p was 4(mod 9), then the\r\nother would be 8(mod 9). Which could be possible, since there is no\r\nnumber guaranteed to divide it. Casework tells us nothing other than\r\nthese three work\r\n\r\nproblem 3 is a load of casework in mod 6.", "Solution_4": "scrambled, how can $p=2$ eh $p+4$ is not a prime so it is not possible. they are consecutive primes.", "Solution_5": "[quote=\"riddler\"]scrambled, how can $p=2$ eh $p+4$ is not a prime so it is not possible. they are consecutive primes.[/quote]\r\nbut you havent proved it", "Solution_6": "1. The numbers are $p, p+1, p+2, p+3, p+4$. Unless $p=3$, p isn't divisible by 3. Therefore, $p+3$ isn't divisible by 3. $p+4$ is never divisible by 3 (it's a prime larger than 3). Therefore, $p+1$ isn't divisible by 3. Since either $p+1$, $p+2$, or $p+3$ must be divisible by 3, $p+2$ is divisible by 3. The sum is equal to $3p+6=3(p+2)$, so it must be divisible by 9. If $p=3$, then $p$ and $p+4$ aren't consecutive primes. This completes the proof!\r\n\r\n2. We'll prove by contradiction. First we need the following lemma, which is very easy to prove: every prime is of the form $6k+5$ or $6k+1$. Now suppose that for a certain $m$, 6m+5 is the largest prime of that form. Consider the number $2*3*5*.......*(6m+5)-1$, which is one less than the product of all prime up to and including $6m+5$. Notice that this number is of the form $6q+5$ for some $q$. Obviously it can't be prime (or else there would be a contradiction since it would be larger than the largest prime of its form) If this number is composite, than its prime factorization consists of primes larger than those of our product. Therefore, it has to be of the form (6a+1)(6b+1)......... But a number of this form gives a remainder of 1 upon division by 6! Contradiction!\r\n\r\nHope that's correct and understandable :)", "Solution_7": "Would the fact that $2*3*5*...*(6m+5)-1$ is also of the form $6q+5$, when $6m+5$ was assumed to be the largest of its form, be enough to prove it?", "Solution_8": "cool method. :D", "Solution_9": "Thanks for all the replies! Least now i've got a few ideas :)\r\nCould someone please post a solution for Q3.", "Solution_10": "[quote=\"Holier Than Thou\"]Would the fact that $2*3*5*...*(6m+5)-1$ is also of the form $6q+5$, when $6m+5$ was assumed to be the largest of its form, be enough to prove it?[/quote]\r\n\r\nIf it's prime, then yes. If it's composite, we have to go a bit further. That's why the proof is broken up into two cases.", "Solution_11": "[quote=\"kryptos\"]Thanks for all the replies! Least now i've got a few ideas :)\nCould someone please post a solution for Q3.[/quote]\r\ni remember a recent problem discussing it but i dont think they have solution yet\r\nhere it is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41960]2^b+b^2[/url]", "Solution_12": "[quote=\"Bob\"]If it's prime, then yes. If it's composite, we have to go a bit further. That's why the proof is broken up into two cases.[/quote]\r\n\r\nAh, of course. For some reason I was thinking being in the form $6m+5$ automatically made it prime." } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "I am so rusty in linear algebra now, so I need your help guys to clarify some point.\r\n\r\nIs it possible for an infinite dimensional vector space $X$ to have a proper subspace $V$ such that $V$ is isomorphic to $X$?\r\n\r\n\r\nThanks", "Solution_1": "Sure. Every vector space has a (Hamel) basis. For an infinite-dimensional vector spact, that basis is an infinite set. Every infinite set can be placed in a one-to-one correspondence with a proper subset of itself. Use that correspondence to define your isomorphism.\r\n\r\nHere's a concrete example of this. Let $V$ be the vector space of all polynomials. Let $V'=\\{P\\in V: P(0)=0\\}.$\r\n\r\nDefine $T: V\\mapsto V'$ by $T(P)(x)=xP(x).$ That is a linear isomorphism from $V$ onto $V'.$", "Solution_2": "It is ... remember a linear map is given entirely by the values it takes on the elements of a base ..." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "For what prime $n$ does there exist solutions in integers to:\r\n\r\n$a^n+b^n=c^n+d^n$\r\n\r\nWith $a \\neq c,d$\r\n$b \\neq c,d$", "Solution_1": "Do you also exclude trivial cases like $a=-b$ etc.\u00bf", "Solution_2": "Whoops, yeah." } { "Tag": [ "LaTeX" ], "Problem": "When I use \"\", both of them are backward, $ \"\"$\r\nHow can I fix this problem?", "Solution_1": "Use `` as the first symbol.\r\n\r\n$ ``\\mathfrak{Quote.}\"$" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "probability", "AMC 10", "geometry" ], "Problem": "AMC 10 Scores\r\nHow well did YOU do? (in comparison to everyone else).\r\nEDIT: got 135 pts\r\n\r\n[color=green]Added poll. It's bugged somehow, so just vote on the first option of the two offered. -mod[/color]", "Solution_1": "BLARGH, fail, i think i got a 132 or 138. idk at this point. most likely a 132 though.", "Solution_2": "121.5 on the 10A. Though there were some easy problems I should have gotten, overall, for me, it wasn't too bad. About a 30+ improvement from last year, and 60+ from 2 years ago. I'm just happy I made AIME.", "Solution_3": "I got a 121.5 on the 10A. Okay, big improvement, but missed some easy ones. I dont really know what index score is needed to make usamo. Maybe 6-8 for a score in the 120's? I dont really know.", "Solution_4": "I'm pretty sure that this year will have a higher usamo index than last year, unless the aime is very hard.\r\n\r\nThe amc 10 was slightly easier... got a 144 since I didn't see the (r<100) and assumed it was $ r \\le 100.$\r\n\r\nOh well.\r\n\r\nAmc 12 last few problems were pretty easy... Brut3Forc3 showed the me the last five amc 12 problems....\r\n\r\nSo I'm pretty sure that you should need at least an 8 on the aime to go to usamo...", "Solution_5": "I got 144 on 10A", "Solution_6": "I got 132 on 10A.", "Solution_7": "150. Fail.\r\n\r\nAnd the first year I made AIME too... irony, anyone?", "Solution_8": "I would've gotten a perfect if I took it...", "Solution_9": "o my gosh i think the answer for #22 is incorrect, but apart from that, i got a 132.", "Solution_10": "Answer for 22 isn't wrong. After you roll the first die, there are 11 tiles left, but only 2 match, so answer is 2/11.\r\n\r\nI put 1/6...", "Solution_11": "aaaaaaaaaaa", "Solution_12": "Gaaaaah 144 by missing number *17* of all questions...I don't know anybody who missed it.", "Solution_13": "144. I happened to blank out on number 21, and write B, because it looked right. \r\n\r\nI thought the whole thing was easier than last year. I also thought that 24 and 25 were way too easy, for their numbers.\r\n\r\nAnyway, should I take the 12B or the 10B? (Btw, I'm in eighth grade, if that changes your opinion.)", "Solution_14": "I'm taking the 12B, although that's mostly because I was the school winner last year :P . Let's pretend I wasn't. If I had gotten less than a 132 on the 10A, I probably would've taken the 10B since I hopefully would be able to score higher and that could make a big difference. However, now, the only way I can score better is getting a perfect (since there's no way I'm going to leave one blank), but there's no assurance of that. So I would still take 12B, partly because it'll be better practice for the AIME and it'll be good to get used to taking it.\r\n\r\nI agree that 24 and 25 were far too easy for their numbers. I really liked #22 though, especially since it was one of those \"logical thinking\" probability problems.", "Solution_15": "I also disagree with Edison on his statement when it applies to math competitions. Success is more like 25% talent, 25% hard work, and 50% the act of not making careless mistakes. I'm serious.", "Solution_16": "[quote=\"Math Geek\"]I also disagree with Edison on his statement when it applies to math competitions. Success is more like 25% talent, 25% hard work, and 50% the act of not making careless mistakes. I'm serious.[/quote]\r\n\r\ndepends on the math contest... i guess for mathcounts, AMC 8, AMC 10/12, and MAYBE (?), AIME, this is the case, but for competitions like USAMO, its more like 40% talent, 59% hard work and 1% the act of not making careless mistakes.", "Solution_17": "I got 118.5 because I was being stupid. If I'd had another minute, I would've gotten 127.5. Oh well, there's always next year. I guess this means I disagree with Edison - not making stupid mistakes is a huge factor. But hard work still counts more than talent.", "Solution_18": "114 =(\r\n\r\nI messed up the last one after thinking abotu the 7 case. But I was too lazy to divide it!!", "Solution_19": "[quote=\"cognos599\"][quote=\"Math Geek\"]I also disagree with Edison on his statement when it applies to math competitions. Success is more like 25% talent, 25% hard work, and 50% the act of not making careless mistakes. I'm serious.[/quote]\n\ndepends on the math contest... i guess for mathcounts, AMC 8, AMC 10/12, and MAYBE (?), AIME, this is the case, but for competitions like USAMO, its more like 40% talent, 59% hard work and 1% the act of not making careless mistakes.[/quote]\r\n\r\nwhat is \"careless\" then? I mean I thought I solved a problem on USAMO last year but oops, it was wrong because my steps could've made a number negative (#5). This was an oversight, but was it \"careless\"? :/ \r\n\r\ncarelessness in the form of laziness (i.e. not justifying everything) loses points though", "Solution_20": "[quote=\"not_trig\"]\nwhat is \"careless\" then? I mean I thought I solved a problem on USAMO last year but oops, it was wrong because my steps could've made a number negative (#5). This was an oversight, but was it \"careless\"? :/ \n\ncarelessness in the form of laziness (i.e. not justifying everything) loses points though[/quote]\r\n\r\nIn my opinion, carelessness and stupid mistakes are when you know how to do something but you are stupid and skip a step or add wrong. Like if you have a system, you solve for x, and then when you go to solve for y, you substitute x into the y variable, and.... well thats what I think being careless is", "Solution_21": "if i got a 141, would a 7 on aime be mean a good chance of usamo, and an 8 like a gurantee?", "Solution_22": "[quote=\"georgiamathdude\"]if i got a 141, would a 7 on aime be mean a good chance of usamo, and an 8 like a gurantee?[/quote]\r\nwow 141 nicee, 7 is a guarantee", "Solution_23": "The act of not making careless mistakes is not an independent skill. It is 40% talent and 60% hard work.", "Solution_24": "^^ agree with above", "Solution_25": "I achieved my goal!!\r\n\r\nNow time to get ready for AIME :)", "Solution_26": "I got a 135. I made a stupid mistake and found the wrong ratio on 17", "Solution_27": "i never got to take it :( \r\n\r\n(my school fails)", "Solution_28": "I phailed. 105/150. This is my first AMC10, and I'm in 5th grade, but I still phailed.", "Solution_29": "I got a 120 on the dot. I had 123, and decided to put down two answers that I wasn't 100% sure on, but was pretty sure of. I got one of those right and it turns out one of the ones I thought was right, turned out being wrong. So 123-6-3+6=120\r\n\r\nI guess it's ok for my first try." } { "Tag": [ "probability" ], "Problem": "In the game show [i]1 vs. 100[/i], one contestant must try to eliminate a mob of 100 people in 12 questions. If successful, the contestant wins a million dollars.\r\n\r\nAssuming the following facts:\r\n[list]The contestant answers every question correctly.\nEach mob member has a constant probability $P$ (it's the same for every mob member and for every question) of answering any single question correctly.[/list]\r\n\r\nCompute $P$ so that the probability of the contestant winning a million dollars is $0.01$.", "Solution_1": "ur not including the probability of lifelines and such.........right?\r\n-jorian", "Solution_2": "More rules need to be specified.", "Solution_3": "yeah some rules might need to be applied, but this is basic i guess\r\n\r\nI know someone who likes the show, since Bob Saget is the host.", "Solution_4": "Forget about asking for help. Assume that the contestant will never ask for help. I told you, the contestant will get every question right.\r\n\r\nMaybe I should also specify that if a mob member gets a question wrong (remember, that has a constant probability $P$ for any of the 12 questions), he/she is eliminated. That should be enough information.", "Solution_5": "haha i know someone who is already obsessed with the show :D" } { "Tag": [ "modular arithmetic", "geometry", "3D geometry", "Diophantine Equations", "pen" ], "Problem": "Find all pairs $(x,y)$ of positive integers that satisfy the equation \\[y^{2}=x^{3}+16.\\]", "Solution_1": "We have $x^3 = (y-4)(y+4).$ From $\\gcd(y-4,y+4)$ divides $(y+4)-(y-4)=8$ and $y+4\\equiv y-4\\pmod{8}$ it follows that either $\\gcd(y-4,y+4)=1$ or $\\gcd(y-4,y+4)=8.$\r\nIn either case both $y-4$ and $y+4$ must be cubes. The only cubes with the difference 8 are (-8,0) and (0,8), implying that $y=\\pm 4$ and $x=0.$", "Solution_2": "$ (y\\minus{}4)(y\\plus{}4)\\equal{}x^3$\r\n$ y\\plus{}4 >y\\minus{}4$\r\nSo we take the perpuses:\r\n$ 1)$ $ y\\plus{}4\\equal{}x^3,y\\minus{}4\\equal{}1$(no integers solutions)\r\n$ 2)$ $ y\\plus{}4\\equal{}x^2,y\\minus{}4\\equal{}x$(no integers solutions)\r\nSo there are no [b]positives[/b] integers solutions...", "Solution_3": "No, dimitris, your reasoning only works for prime x. For example $ 6^3$ can also be factored as $ (2*3^2)*(2^2*3)$ or similar options.", "Solution_4": "[quote=maxal]We have $x^3 = (y-4)(y+4).$ From $\\gcd(y-4,y+4)$ divides $(y+4)-(y-4)=8$ and $y+4\\equiv y-4\\pmod{8}$ it follows that either $\\gcd(y-4,y+4)=1$ or $\\gcd(y-4,y+4)=8.$[/quote]\n\nHello, maxal!\n\nIf $y=8$, $\\gcd(y-4,y+4)=4$\n\nSo we have to consider following cases.\n\n[b]Case 1)[/b]$\\gcd(y-4,y+4)=2$\n$y-4=2k$, $y+4=2(k+4) (k:odd)$.\nSince $4k(k+4)=x^3$, $x=2m$.Then $k(k+4)=2m^3$ which is absurd.\n\n[b]Case 2)[/b]$\\gcd(y-4,y+4)=4$\n$y-4=4k$, $y+4=4(k+2) (k:odd)$.\nSince $16k(k+2)=x^3$, $x=4m$.Then $k(k+2)=4m^3$ which is absurd.\n\nThanks, Takeya.O.", "Solution_5": "[hide=Solution]Rewrite the equation as $x^3=(y-4)(y+4)$. We do casework based off the parity of $x,y$. It is easy to see they will both be the same parity.\n-------------------------------------\nCase 1: Both $x,y$ are odd.\n\nThis means both $y-4$ and $y+4$ are perfect cubes since $\\gcd(y-4,y+4)=1$ by Euclidean algorithm. We can't have two positive odd cubes that differ by $8$ so there are no solutions in this case.\n-------------------------------\nCase 2: Both $x,y$ are even.\n\nIt is easy to see that if $2|x$ then $4|y$ because otherwise $v_2(LHS)>v_2(RHS)$. Let $y=4y'$ in the original equation and we have $16y'^2=x^3+16$. If $4\\nmid x$ then we have $v_2(LHS) > v_2(RHS)$ again. Thus, $4|x$. Letting $x=4x'$ in the new equation gives $y'^2=4x'^3+1$, implying that $y'$ is odd. Letting $y'=2k+1$, $4k^2+4k+1=4x'^3+1$ which becomes $k^2+k=x^3$. This implies $k, k+1$ are both cubes by Euclidean algorithm. It follows that $k=-1,0$ are the only solutions. This results in $x=0$ which implies $y=\\pm4$.\n-----------------------------------\nThus, our solutions are $(x,y)=(0,4)$ and $(0,-4)$.[/hide]" } { "Tag": [ "ceiling function" ], "Problem": "Ten people are trying to guess a real number from 0 to 10 inclusive. You can see the other players' guesses. The eight people who are closest to the number win.\r\n\r\nProve that it is possible to guarantee a win if you go fifth or later.", "Solution_1": "$n (n\\geq 5)$ people have picked some arbitrary real numbers in [0,10]. You pick the median of those numbers. Either you choose the correct real number (and you win), or it's greater than/less than your choice. WLOG assume it was less than your choice, the worst case is that the next $9-n$ people have closer choices than you, $9-n+\\lceil\\frac{n}{2}\\rceil \\leq 7$ people have choices as good or better than yours. So in the worst case you'll be 8th and win :)", "Solution_2": "[hide=\"My Solution\"]Given the first four guesses $a_{1},a_{2},a_{3},a_{4}$, you can choose a number $x$ that is in the interval $(a_{2},a_{3})$. So there are at least 2 numbers more than yours and at least 2 numbers less than yours. Yours can be as low as the third-lowest, or as high as the third-highest.\nNow, to prove that it works when you have those conditions: testing the extremes.\nAssume that the third-lowest will lose. Then the second-lowest and lowest numbers are too far away and are losers too. But that makes at least 3 losers, and there can only be two. Contradiction. A similar proof exists for assuming you are the third-highest.\nTherefore, we can conclude that any number in the interval $(a_{2},a_{3})$ will always work.\n$\\boxed{\\text{Q.E.D.}}$[/hide]" } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "We call a positive integer an almost perfect square if it is of the form $ mn$ where the integers $ m$ and $ n$ satisfy $ m \\leq n$ and $ |\\dfrac mn \\minus{} 1 | < \\dfrac 1{2007}$ . Show that there exist infinitely many $ 6$-tuples of consecutive almost perfect squares.", "Solution_1": "$ \\left| \\frac{m}{n} \\minus{} 1 \\right| < \\frac{1}{2007} \\not \\Leftrightarrow \\left| \\frac{n}{m} \\minus{} 1 \\right| < \\frac{1}{2007}$. The problem needs to stipulate $ m > n$ or $ m < n$.", "Solution_2": "My bad, I forgot to add that in. Edited.", "Solution_3": "why do we *need* to stipulate an order on $ m$ and $ n$??", "Solution_4": "One condition is stronger than the other. The problem may intend the weak condition and be false with the strong.", "Solution_5": "I'll post some trivial remarks. First off, when $ a$ is large, then $ a^2 \\minus{} 1$ and $ a^2$ are almost perfect squares. So we get two consecutive such numbers.\r\n\r\nHere's a way to get three consecutive almost perfect squares. Set $ a \\equal{} b^2 \\plus{} b \\plus{} 1$. Then we have the factorization\r\n\\[ a^2 \\plus{} 1 \\equal{} (b^2 \\plus{} 1)(b^2 \\plus{} 2b \\plus{} 2).\r\n\\]\r\nSo $ a^2 \\minus{} 1$, $ a^2$, and $ a^2 \\plus{} 1$ are almost perfect squares when $ b$ is large.\r\n\r\nI don't know whether this approach leads anywhere, or is a deadend.", "Solution_6": "but \"...it is of the form $ mn$, where $ m,n$ satisfy $ |\\frac{m}{n}\\minus{}1|<\\frac{1}{2007}$\", without the roles of $ m,n$ distinguished, is the weaker statement. in your statement, you insist that the smaller of $ m,n$ play a particular role.", "Solution_7": "Hmm. Point taken. \r\n\r\nRavi: According to your approach, what we should be looking for is a family of six monic polynomials of degree $ 2n$, all of which factor into two monic polynomials of degree $ n$, and which differ by $ 1$. It's an interesting idea, but I can't find a good place to start.", "Solution_8": "nobody's managed to solve this problem yet?" } { "Tag": [], "Problem": "Have you ever heard of the Ku Klux Klan? Give your opinions on it.", "Solution_1": "yea, the KKK\r\n\r\nthey are very racist guys that are \"purely\" white", "Solution_2": "who would hate a group of ppl to do stuff like that? :mad: do they still exist?", "Solution_3": "There some KKK organizations in USA today, and they're legal. The members inherite everything except the actions of killing from the old KKK.\r\nThis is a good example of how free you could speak in USA, but remember NEVER talk with your friends about something from the Bible like \"burning bush\" in bars.", "Solution_4": "The \"Neo Nazi\" thread of yours has been deleted, bubka, because the nature of the post was inherently spammy.", "Solution_5": "Yes please, I think we're hear to LEARN, no to increase our post account.\r\n\r\nHere, there's a long but interesting and complete link of KKK, with general information of the \"group\": http://en.wikipedia.org/wiki/Ku_Klux_Klan", "Solution_6": "[quote=\"Jos\u00e9\"]Yes please, I think we're hear to LEARN, no to increase our post account.[/quote]\r\n\r\nOh.\r\nJust Kidding.\r\n\r\nI hate the ku klux klan because they're racist.\r\nThis topic is a bit like the one about the racist white website.", "Solution_7": "Yes I have heard of the Ku Klux Klan.\r\n\r\nThe Ku Klux Klan is infamous for its hate towards African Americans, but it should be pointed out that they also don't accept :\r\n\r\n- Jews\r\n- homosexuals\r\n- catholics (they are protestant)\r\n\r\nThe KKK is are rather \"southern\" matter. The civil war (1860-1865) between the North and South of the USA basically involved the South not willing to forbid enslavement of African Americans. Many KKK still idolize the leaders of the South like Jefferson Davis, Robert Lee, and Stonewall Jackson, and don't mind offering up a dollar or two to show that :\r\n\r\n[url]http://users.telenet.be/frederic_bel6/Atlanta/LUC3040.jpg[/url]", "Solution_8": "For that matter, they don't accept Communists as well. Or Gypsys. They are like the Neo-Nazis escept for the hair.", "Solution_9": "Look up the Aryan Nations sometime, dude. Far, far worse. Not that I'm endorsing the KKK, definitely nothing like that, it's just that I decided to look up the Aryan Nations once, because of some local news involving the group, and it scared the holy *bleep* out of me.\r\n\r\nAlso, while the history of the KKK lies in the South, please don't get it in your minds that only the South has white supremacists. The worst of them are actually in Oregon, thanks.", "Solution_10": "nevertheless, i think the KKK is a horrible group", "Solution_11": "Well they do have a point : it where white men who did all the hard work of occupying the land and fighting indians, all those Africans had to do is being brought there as slaves. So why wouldn't the USA belong to whites?", "Solution_12": "You mean why US is a multicultural country?\r\nWell in the constitution it says \"All men are created equal...\"\r\nIt means everybody are created equal. Not just whites.", "Solution_13": "[quote=\"fredbel6\"]Well they do have a point : it where white men who did all the hard work of occupying the land and fighting indians, all those Africans had to do is being brought there as slaves. So why wouldn't the USA belong to whites?[/quote]\r\n\r\nAre you kidding? Or do you seriously mean that?", "Solution_14": "[quote=\"bubka\"]\nAre you kidding? Or do you seriously mean that?[/quote]\r\n\r\n :huh: \r\n\r\nOf course I am being sarcastic.", "Solution_15": "I have always wondered: Does exist any groups that defend asian supremacy? Black supremacy? Latinoamerican supremacy?\r\nAnd -supposing there are- why we don't hear of them as we hear of neo-nazis or KKK?", "Solution_16": "I am no expert but Black Panthers seems to be one of those.", "Solution_17": "[quote=\"fredbel6\"]I am no expert but Black Panthers seems to be one of those.[/quote]\r\nActually, the Black Panthers were more of an extremist civil rights group than a black supremacy group- their mission was not to assert the superiority of African-Americans, it was to assert their equality, through whatever force they deemed necessary. The group fell apart in the early 1970's." } { "Tag": [ "abstract algebra", "algorithm" ], "Problem": "I'm intending to learn C, with a focus on USACO and kernel hacking. Do ya'll have any recommendations on good books to learn C, with a pretty good reference section too?\r\n\r\nThanks!\r\n\r\nBilly", "Solution_1": "I can't recommend any good books because I never read any but if you wish to focus on USACO, you should also consider learning C++ and the standard template library, they are priceless. However, I don't think these are available to kernel programmers. (May be wrong though.)", "Solution_2": "I must say that I think that books are absolutely useless in Computer Science nowadays. What i would do if I were you is that I would solve all of USACO training problems and the move on to other sites and eventually topcoder .\r\nhere are some links: http://infoarena.ro/links", "Solution_3": "Well, there are definitely a lot of online resources but that doesn't mean that books are useless. The good ones, that is.", "Solution_4": "What I mean is that everything is evolving so rapidly, new algoritms appear every day and books can't keep up with this. They are good for you to make a strong foundation of knowledge, but if he wants to take up USACO, online resources are more appropriate in my opinion.", "Solution_5": "The C language hardly changes every day. The C99 standard is just as valid now as it was seven years ago and the new algorithms that appear every day are most likely very specialized and of no use to ordinary sane people :) - just show me an algorithm that can be used in USACO and isn't older than 25 years.\r\n\r\nNevertheless, spix is right about one thing - there's a lot of resources online, both C / C++ language tutorials and algorithm tutorials (as well as many more advanced articles). Google for C, C++, tutorial, STL, standard template library and then try to solve the USACO training problems. That's the way I learned C++ anyway :).", "Solution_6": "Here's an update: I ended up getting K&R, and it's been great. I've kinda got the C builtins down, but I'm still working on the syntax. C really is fun though :).", "Solution_7": "You can also [url=http://tigcc.ticalc.org/]write calculator games and stuff in C[/url]. That's how I first got into it back in middle school, when I was obsessed with my TI-89. ;)" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "let $a,b,c,x,y,z$ are any real numbers then let\r\n\\[A=2(a^{2}+b^{2}+c^{2}+ab+bc+ca)\\]\r\n\\[B=43(x^{2}+y^{2}+z^{2})-42(xy+yz+zx)\\]\r\n\\[C=(a+b-c)(x+y-z)+(b+c-a)(y+z-x)+(c+a-b)(x+z-y)\\]\r\nthen prove that $AB\\ge4.C^{2}$\r\nhungktun accept this small problem gift from me.", "Solution_1": "I am not sure if I am the person Anuj Kumar sent this problem, because a bit difference in the name (oy me) he write. If that, thank you very much, Anuj Kumar. I will try it,", "Solution_2": "Today is a nice sunlight day in my hometown. I've just waken up some minute ago and think about this problem. Luckily, I found a solution instantly. I am sorry if I am not the person Anuj Kumar dedicating, here is my solution\r\n\r\nLet $b+c=m,c+a=n,a+b=p$. By some simple transformation, we get\r\n\\[2C=\\sum_{cyc}m(3y+3z-5x).\\]\r\nApplying Cauchy, we have\r\n\\[4C^{2}\\le (m^{2}+n^{2}+p^{2})\\left(\\sum (3y+z-5x)^{2}\\right).\\]\r\n\r\nJust notice that\r\n\\[m^{2}+n^{2}+p^{2}=2(a^{2}+b^{2}+c^{2}+ab+bc+ca).\\]\r\n\\[\\left(\\sum (3y+z-5x)^{2}\\right)=43(x^{2}+y^{2}+z^{2})-42(xy+yz+zx).\\]\r\n\r\nSo we have done.\r\n\r\nGood exercise for applying Cauchy :lol: .", "Solution_3": "you are the right person. sorry for spelling mistake." } { "Tag": [ "IMO", "IMO 2005" ], "Problem": "1-Is there any team who's gonna have the same stopovers \"frankfort-mexico city-merida\" and travels monday at early morning?\r\n2-few days before the competition, what do you think the most useful thing to do for a non-trained contestant could be?", "Solution_1": "[quote=\"wisemaniac\"]2-few days before the competition, what do you think the most useful thing to do for a non-trained contestant could be?[/quote]Rest :)", "Solution_2": "Albanian Team. Takes of from Tirana-Viena-Frankfort-Mexico-Merida, on Monday. Our Plane in Frankfort takes off at 1:00 PM.", "Solution_3": "i just checked ,our team too (and 7hours waiting in the airport)!! i think most of the teams are going to take one the 2 planes going from frankfurt to mexico monday at 13 :30 !! or am i wrong? :? :)", "Solution_4": "We are going Amsterdam - Houston - Merida, so there are other ways to fly to Merida. \r\n\r\nAs for what to do before the test I dare say rest is a very good advice. You might try some (simple) problems for a while to keep your mind flexible. But I strongly advice against trying to cram some new (or even old) theorems in your head, since you will not have time to learn how to use them properly and they will just confuse you.", "Solution_5": "Belgian team:\r\n11/07 PARIS CDG - MEXICO flight AM 006 departure 10.30 arrival 16.00\r\n11/07 MEXICO - MERIDA flight AM 647 departure 21.35 arrival 23.00\r\n\r\nThere's long wait in Mexico City, if any team wants to join us in our waiting, you'll probably recognize us by our big Belgian flag (black-yellow-red) and the little connect 4-game we're playing. :-)", "Solution_6": "The Finnish team goes on the route Helsinki - Frankfurt - Mexico City - Merida.\r\n\r\nWe will take the plane that leaves from Frankfurt at 13:30. Even if we don't get in the same plane, I guess we will still see each other in the airport. See you there!", "Solution_7": "I\u00b4m in Aereomexico 531 M\u00e9xico City - M\u00e9rida (12:30 AM) Someone on that plane, too?", "Solution_8": "[quote=\"Dave Vather\"]I\u00b4m in Aereomexico 531 M\u00e9xico City - M\u00e9rida (12:30 AM) Someone on that plane, too?[/quote]I\u00b4m already here :D Guys be aware it\u00b4s [b]really but really hot[/b] ...", "Solution_9": "We (Serbia&Montenegro team) are going by Belgrade-Frankfort-Mexico City-Merida line.", "Solution_10": "[quote=\"Valentin Vornicu\"][quote=\"Dave Vather\"]I\u00b4m in Aereomexico 531 M\u00e9xico City - M\u00e9rida (12:30 AM) Someone on that plane, too?[/quote]I\u00b4m already here :D Guys be aware it\u00b4s [b]really but really hot[/b] ...[/quote]\r\nIt isn\u00b4t \"The withe City\" just for its buildings, but also because many people wears withe clothes because of the hot" } { "Tag": [ "integration", "probability and stats" ], "Problem": "Form the result $ E[Y] \\equal{} \\int_{0}^{\\infty} P\\{Y > t\\} dt$, show for nonnegative random variable $ X$, that $ E[X^{n}] \\equal{} \\int_{0}^{\\infty} nx^{n\\minus{}1} P\\{X >x\\} dx$.\r\n\r\nSo I start with $ E[X^n] \\equal{} \\int_{0}^{\\infty} P\\{X^n > t\\} dt$\r\n\r\nTaking $ x^{n} \\equal{} t$ and deriving,gives $ dt \\equal{} nx^{n\\minus{}1}dx$ , thus:\r\n\r\n$ \\equal{} \\int_{0}^{\\infty} P\\{X^{n} > x^{n} \\} nx^{n\\minus{}1} dx$\r\n\r\n? $ \\equal{} \\int_{0}^{\\infty} P\\{(X^{n})^{\\frac{1}{n}} > (x^{n})^{\\frac{1}{n}}\\} nx^{n\\minus{}1} dx \\equal{} \\int_{0}^{\\infty} P\\{X>x\\} nx^{n\\minus{}1} dx$\r\n\r\nI wasn't sure if my penultimate step was correct. Anyone have any ideas?", "Solution_1": "Since $ X$ and $ x$ are nonnegative, $ P[X^n>x^n]\\equal{}P[X>x]$, so your penultimate step is justified." } { "Tag": [ "geometry", "3D geometry", "sphere", "tetrahedron" ], "Problem": "1. ABC is an isosceles triangle with AB=BC=12. P and Q are points on AB and AC such that AP=AQ=8. PC and QB intersect at X. The area of APXQ is 8. What is the area of ABC?\r\n\r\n2. ABC is an equilateral triangle. A point S is chosen in the plane of the equilateral triangle, and AS=3, BS=5, CS=8. Find BC.\r\n\r\n3. What is the side length of the smallest square that can enclose 3 non-overlapping circles of radius 1 (with proof)?\r\n\r\n4. What is the radius of the smallest sphere that can enclose 4 non-intersecting (can be tangent however) spheres of radius 1 (with proof)?\r\n\r\n5. Three circles of equal area \"cover\" a unit square (side = 1). \"Cover\" means that each point on or in the square is under at least one of the circles. What is the minimum possible radius for the circles (with proof)?", "Solution_1": "For #3 I've got the minimum when the centers of the circles form an equilateral triangle with one of the medians lies on the diagonal of the sphere, and for #4 I've got the minimum when the centers of the spheres form the vertices of a tetrahedron, but I have trouble proving they are indeed the minimum, though..." } { "Tag": [ "function", "algebra", "IMO Shortlist", "functional equation", "IMO 2008", "hojoo lee", "IMO" ], "Problem": "Find all functions $ f: (0, \\infty) \\mapsto (0, \\infty)$ (so $ f$ is a function from the positive real numbers) such that\n\\[ \\frac {\\left( f(w) \\right)^2 \\plus{} \\left( f(x) \\right)^2}{f(y^2) \\plus{} f(z^2) } \\equal{} \\frac {w^2 \\plus{} x^2}{y^2 \\plus{} z^2}\n\\]\nfor all positive real numbers $ w,x,y,z,$ satisfying $ wx \\equal{} yz.$\n\n\n[i]Author: Hojoo Lee, South Korea[/i]", "Solution_1": "Put $ w \\equal{} x \\equal{} y \\equal{} z \\equal{} t$ and observe that $ f(t^{2}) \\equal{} f(t)^{2}$. \r\nSet $ k(t) \\equal{} \\frac {f(t)}{t}$ and consider $ pq \\equal{} s ^{2}$. \r\nYou get $ \\frac {f(s^{2})}{s^{2}} \\equal{} k(s^{2}) \\equal{} \\frac {f(p^{2}) \\plus{} f(q^{2})}{p^{2} \\plus{} q^{2}}$\r\nor\r\n$ \\frac {f(s)}{s} \\equal{} \\frac { f(p) \\plus{} f(\\frac {s^{2}}{p}) }{p \\plus{} \\frac {s^{2}}{p}}$ independent of p.\r\nNow you can let $ p \\equal{} 1$ and solve the quadratic equation resulting. \r\nThe solutions are of form \r\n$ \\frac {s \\plus{} s^{ \\minus{} 1} \\plus{} / \\minus{} \\sqrt {s^{2} \\plus{} s^{ \\minus{} 2} \\plus{} 2 \\minus{} 4f(1)}}{2}$, and putting $ s \\equal{} 1$ gives\r\n$ f(1)(f(1) \\minus{} 1) \\equal{} 0$ so $ f(1) \\equal{} 1$ and you can get $ f(s)$ taking one of the values $ s$ or $ s^{ \\minus{} 1}$ at each value of $ s$. Now partition the positive reals (other than 1) into those for which $ f$ gives the identity (typical element p), and those for which $ f$ gives the inverse (typical element $ q$). The original equation contradicts the statement $ f(p^{2}q^{2}) \\in ({{p^{2}q^{2},p^{ \\minus{} 2}q^{ \\minus{} 2}}})$, so one of those partitions must be empty. Now you can go ahead and verify that either identity (everywhere) or inverse (everywhere) are satisfactory.", "Solution_2": "[quote=\"the.sceth\"]\nYou get $ \\frac {f(s^{2})}{s^{2}} \\equal{} k(s^{2}) \\equal{} \\frac {f(p^{2}) \\plus{} f(q^{2})}{p^{2} \\plus{} q^{2}}$\nor\n\\[ \\frac {f(s)}{s} \\equal{} \\frac { f(p) \\plus{} f(\\frac {s^{2}}{p}) }{p \\plus{} \\frac {s^{2}}{p}}\n\\]\nindependent of p.\n[/quote]\r\n\r\nSorry, how do you get from the first equation to the second? Did you replace $ p^2,q^2,s^2$ with $ p,q,s$?", "Solution_3": "Here is my solution .\r\nTake $ x \\equal{} y \\equal{} z \\equal{} w$ then we have\r\n\\[ f(x^2) \\equal{} f(x)^2 \\Rightarrow f(1) \\equal{} 1\r\n\\]\r\nTake $ x\\to \\sqrt {x},y\\to \\sqrt {y},z\\to sqrt{z},w\\to \\sqrt {w}$ we have :\r\n\\[ (y \\plus{} z)(f(x) \\plus{} f(w)) \\equal{} (x \\plus{} w)(f(z) \\plus{} f(y))\r\n\\]\r\nTake $ x \\equal{} w\\sqrt {yz}$ then we have :\r\n\\[ (y \\plus{} z)f(\\sqrt {yz}) \\equal{} \\sqrt {yz}(f(x) \\plus{} f(y))\r\n\\]\r\nTake $ z\\to 1$ then :\r\n\\[ \\frac {f(x)}{x} \\equal{} \\frac {f(x)^2)}{x^2 \\plus{} 1} \\Leftrightarrow (f(x) \\minus{} x)(x.f(x) \\minus{} 1) \\equal{} 0 \\Leftrightarrow f(x) \\equal{} x,f(x) \\equal{} \\frac {1}{x}\r\n\\]\r\nSuppose exist $ a,b$ is different from 1 such that $ f(a) \\equal{} a,f(b) \\equal{} \\frac {1}{b}$\r\nTake $ w \\equal{} 1,x \\equal{} yz$ then \r\n$ (y \\plus{} z)(f(yz) \\plus{} 1) \\equal{} (yz \\plus{} 1)(f(y) \\plus{} f(z))$\r\nTake $ y\\to a,z\\to b$ then\r\n\\[ (a \\plus{} b)(f(ab) \\plus{} 1) \\equal{} (ab \\plus{} 1)(a \\plus{} \\frac {1}{b})\r\n\\]\r\nIf $ f(ab) \\equal{} ab$ then \r\n$ b \\equal{} \\frac {1}{b} \\Leftrightarrow b \\equal{} 1$\r\nIf $ ab \\equal{} \\frac {1}{ab}$ then :\r\n\\[ (a \\plus{} b) \\equal{} a(ab \\plus{} 1) \\equal{} b(ab \\plus{} 1)\r\n\\]\r\nSo $ a \\equal{} b$ \r\nTherefore $ a \\equal{} b \\equal{} 1$ (contradiction ) \r\nThere are only two function satisfy condition :\r\n1) $ f(x)\\equiv x$\r\n2)$ f(x)\\equiv \\frac {1}{x}$", "Solution_4": "My solution (at home) is very similar to TTsphn's one...but quite longer and messier\r\n\r\nBy the way nice solution!\r\n\r\nDaniel", "Solution_5": "[quote=\"orl\"]Find all functions $ f: (0, \\infty) \\mapsto (0, \\infty)$ (so $ f$ is a function from the positive real numbers) such that\n\\[ \\frac {\\left( f(w) \\right)^2 \\plus{} \\left( f(x) \\right)^2}{f(y^2) \\plus{} f(z^2) } \\equal{} \\frac {w^2 \\plus{} x^2}{y^2 \\plus{} z^2}\n\\]\nfor all positive real numbes $ w,x,y,z,$ satisfying $ wx \\equal{} yz.$\n\n\n[i]Author: Hojoo Lee, South Korea[/i][/quote]\r\n\r\nhi for $ x\\equal{}y\\equal{}z\\equal{}w f(x^2)\\equal{}f(x)^2$\r\n\r\nso $ f(1)\\equal{}1 (f(x)>0)$ \r\n\r\nsupposing that $ x\\equal{}w$so $ x^2\\equal{}yz$ that mean:\r\n\r\n$ f(x^2)(y^2\\plus{}z^2) \\equal{}x^2 (f(y^2)\\plus{}f(z^2))$\r\n\r\n$ f(yz) (y^2\\plus{}z^2 )\\equal{} yz (f(y^2) \\plus{}f(z^2))$\r\n \r\nfor $ y\\equal{}1$ we have : $ f(z)(z^2\\plus{}1)\\equal{}z(f(z^2)\\plus{}1)$ wich include \r\n \r\n $ f(z)(z^2\\plus{}1)\\equal{}z(f(z)^2\\plus{}1)$ \r\n\r\n==> $ zf(z) (z\\minus{}f(z))\\plus{}(f(z)\\minus{}z) \\equal{}0$\r\n\r\n$ (z\\minus{}f(z) )(zf(z)\\minus{}1)\\equal{}0$==>$ f(z)\\equal{}z$ or $ f(z)\\equal{}(1/z)$", "Solution_6": "Note that you are not done there: there could be a horribly convoluted and nasty function which maps $ x$ to $ x$ for some real $ x$ and $ x$ to $ \\frac {1}{x}$ for others. There remains to prove that no such function satisfying the conditions of the problem exists (as TTsphn did, for example).", "Solution_7": "[quote=\"Lepuslapis\"]Note that you are not done there: there could be a horribly convoluted and nasty function which maps $ x$ to $ x$ for some real $ x$ and $ x$ to $ \\frac {1}{x}$ for others. There remains to prove that no such function satisfying the conditions of the problem exists (as TTsphn did, for example).[/quote]\r\n\r\nyes you are right we have to find contradiction for the case of f determinated us \r\n\r\n$ f$ :\r\n\r\n$ f(x)\\equal{}x$ for $ x \\in [0.a[$\r\n\r\n$ f(x)\\equal{}\\frac{1} {x}$ for $ x \\in [a.\\plus{}00[$", "Solution_8": "[quote=\"not_trig\"][quote=\"the.sceth\"]\nYou get $ \\frac {f(s^{2})}{s^{2}} \\equal{} k(s^{2}) \\equal{} \\frac {f(p^{2}) \\plus{} f(q^{2})}{p^{2} \\plus{} q^{2}}$\nor\n\\[ \\frac {f(s)}{s} \\equal{} \\frac { f(p) \\plus{} f(\\frac {s^{2}}{p}) }{p \\plus{} \\frac {s^{2}}{p}}\n\\]\nindependent of p.\n[/quote]\n\nSorry, how do you get from the first equation to the second? Did you replace $ p^2,q^2,s^2$ with $ p,q,s$?[/quote]\r\n\r\nExactly.\r\n\r\nAlso, bear in mind that the idea of the partition was equivalent to TTsphn\u00b4s method of eliminating the posibility of $ f(t)$ being somewhere $ f(t) \\equal{} t^{ \\minus{} 1} \\neq 1$ and somewhere $ f(t)\\equal{}t \\neq 1$. (I.e. as Lepuslapis pointed out.) It\u00b4s just with computation omitted.", "Solution_9": "as posted before $ f(x^2)\\equal{}f^2(x)$.. we can substitute this at the hypothesis and get $ \\dfrac{f(a)\\plus{}f(b)}{f(c)\\plus{}f(d)}\\equal{}\\dfrac{a\\plus{}b}{c\\plus{}d}$ with $ ab\\equal{}cd$...\r\n\r\nso, $ \\dfrac{f(x^3)\\plus{}f(x)}{2f(x^2)}\\equal{}\\dfrac{x^2\\plus{}1}{2x}$, and it follows that $ f(x^3)\\equal{}f(x)\\left(\\dfrac{f(x)(x^2\\plus{}1)}{x}\\minus{}1\\right)$.\r\n\r\nwe also have that \r\n$ \\dfrac{f(x^4)\\plus{}f(x)}{f(x^2)\\plus{}f(x^3)}\\equal{}\\dfrac{x^2\\minus{}x\\plus{}1}{x}$\r\n\r\nif we let $ f(x)\\equal{}a$ we have that\r\n$ f(x^4)\\equal{}(\\dfrac{x^2\\minus{}x\\plus{}1}{x})(a^2\\plus{}\\dfrac{a^2(x^2\\plus{}1)}{x}\\minus{}a)\\minus{}a\\equal{}a\\cdot \\left(\\dfrac{x^2\\minus{}x\\plus{}1}{x}\\cdot\\left(\\dfrac{a(x^2\\plus{}x\\plus{}1)}{x}\\minus{}1\\right)\\minus{}1\\right)$\r\n\r\nbut $ f(x^4)\\equal{}f^2(x^2)\\equal{}f^4(x)\\equal{}a^4$, so it follows that \r\n\r\n$ a^3\\equal{}\\dfrac{x^2\\minus{}x\\plus{}1}{x}\\cdot\\left(\\dfrac{a(x^2\\plus{}x\\plus{}1)}{x}\\minus{}1\\right)\\minus{}1$\r\n\r\nthis cubic equation has a negative solution and $ a\\equal{}x$ and $ a\\equal{}\\dfrac{1}{x}$... and this can be finished as any of the solutions above... :D", "Solution_10": "By putting $ w\\equal{}x\\equal{}y\\equal{}z$ we have $ f(x^2)\\equal{}f(x)^2$ for all $ x$.\r\nFor $ x\\equal{}1$ we have $ f(1)\\equal{}1$.\r\n\r\nSo $ f(w)^2\\plus{}f(z)^2 \\equal{} f(w^2)\\plus{}f(z^2)$, and then $ \\frac{f(x^2)\\plus{}f(y^2)}{x^2\\plus{}y^2}$ depends only on $ xy$, and so does $ \\frac{f(x)\\plus{}f(y)}{x\\plus{}y}$.\r\n\r\nSo we can say that $ \\frac{f(x)\\plus{}f(y)}{x\\plus{}y} \\equal{} \\frac{f(xy)\\plus{}f(1)}{xy\\plus{}1} \\equal{} \\frac{f(xy)\\plus{}1}{xy\\plus{}1}$.\r\nBy putting $ x\\equal{}y$ and replacing $ f(x^2)$ by $ f(x)^2$ we get the quadratic equation in $ f(x)$\r\n$ x\\cdot f(x)^2 \\minus{} (x^2\\plus{}1)\\cdot f(x) \\plus{} x\\equal{}0$\r\nthat has solutions $ f(x)\\equal{}x$, $ f(x)\\equal{}\\frac{1}{x}$.\r\n\r\nIt remains to control that there can't be $ a,b\\neq 1$ so that $ f(a)\\equal{}a$ and $ f(b)\\equal{}\\frac1b$, as TTsphn did.", "Solution_11": "Well i did the same for $ f(x^2) = (f(x))^2$. Then $ f(1) = 1$ and so $ \\displaystyle f(x) + f\\left(\\frac {1}{x}\\right) = x + \\frac 1 x$ (*). Now plug $ w = a^2, x = 1, y = a,z = a$ and we get $ \\displaystyle \\frac 1{f(x)} + f(x) = x + \\frac 1x$(**) and since $ g(x) = x + \\frac 1x$ is a strictly increasing function in $ (1,\\infty)$ we obtain that $ f(x) = x$ or $ \\displaystyle f(x) = \\frac 1x$. From (*) and (**) we get that $ \\displaystyle \\frac 1{f(x)} = f(\\frac 1x)$(***). Now to conclude use (***) suposing $ f(a) = a$, $ f(b) = \\frac 1b$ and using the fact that $ f(x^{2^k}) = (f(x))^{2^k}$ we can assume that $ a$ and $ b$ are small enough or large enough to find that $ f(ab) \\not = ab$ and $ \\displaystyle f(ab) \\not = \\frac 1{ab}$ (pluging $ w = 1, x = ab, y = a, z = b$) which is a contradiction (in fact taking a large $ b$ and and choosing $ a$ such that $ ab<1$ one gets that $ \\displaystyle f(ab)= \\frac {ab + a +\\frac 1b - b}{a+b}<0$).", "Solution_12": "[quote=\"orl\"]Find all functions $ f: (0, \\infty) \\mapsto (0, \\infty)$ (so $ f$ is a function from the positive real numbers) such that\n\\[ \\frac {( f(w) )^2 + ( f(x) )^2}{f(y^2) + f(z^2) } = \\frac {w^2 + x^2}{y^2 + z^2}\n\\]\nfor all positive real numbes $ w,x,y,z,$ satisfying $ wx = yz.$\n\n\n[i]Author: Hojoo Lee, South Korea[/i][/quote]\r\n[hide=\"1)\"]\nPut in $ (1,1,1,1)$ to find that $ f(1) = 1$\n[/hide]\n[hide=\"2)\"]\nPut in $ (1,x,\\sqrt {x},\\sqrt {x})$. Then\n\\[ \\frac {1 + f(x)^2}{2f(x)} = \\frac {1 + x^2}{2x}\n\\]\nLet $ f(x) = y$. This is clearly equivalent to\n\\[ y + \\frac {1}{y} = x + \\frac {1}{x}\\iff x - y = \\frac {x - y}{xy}\\implies x - y = 0\\text{ or }xy = 1\n\\]\nThus for all positive $ x$,\n\\[ f(x)\\in{x,\\frac {1}{x}}\n\\]\n[/hide]\n[hide=\"3)\"]\nNow suppose there were $ a,b\\not = 1$ with\n\\[ f(a) = a,f(b) = \\frac {1}{b}\n\\]\nSubstituting into the original equation shows that\n\\[ \\frac {f(\\sqrt {ab})^2 + 1}{f(a) + f(b)} = \\frac {ab + 1}{a + b}\n\\]\nFrom $ (2)$, we know that $ f(\\sqrt{ab})$ can be one of two values:\nIf $ f(\\sqrt {ab})^2 = \\frac {1}{ab}$, then\n\\[ \\frac {f(\\sqrt {ab})^2 + 1}{f(a) + f(b)} = \\frac {ab + 1}{a^2b + a} = \\frac {ab + 1}{a + b}\n\\]\nThus $ a + b = a^2b + a\\implies a = 1$, which is a contradiction.\nIf $ f(\\sqrt {ab})^2 = ab$, then\n\\[ \\frac {f(\\sqrt {ab})^2 + 1}{f(a) + f(b)} = \\frac {ab + 1}{a + \\frac {1}{b}} = \\frac {ab + 1}{a + b}\n\\]\nso $ a + \\frac {1}{b} = a + b\\implies b = 1$, contradiction.\n\nThus no such $ a,b$ exist, that is\n\\[ f(x)\\equiv x\\text{ or }f(x)\\equiv\\frac {1}{x}\n\\]\n[/hide]", "Solution_13": "This problem seems not very hard", "Solution_14": "As before, show that $ f(x)^2 \\equal{} f(x^2)$. Also similarly, substitute $ a^2, b^2, 1, ab$ to show that $ P(a, b) : f(a^2) \\plus{} f(b^2) \\equal{} f(ab) \\left( \\frac{a}{b} \\plus{} \\frac{b}{a} \\right)$. Now \r\n\r\n$ P(a, b) \\minus{} P(\\sqrt{a}, \\sqrt{b})^2 : f(a^2) f(b^2) \\equal{} f(ab)^2$.\r\n\r\nThe same quadratic-solving step (letting $ ab \\equal{} 1$) shows that $ f(a) \\equal{} a \\text{ or } \\frac{1}{a} \\forall a$, but now note that if $ f(a) \\equal{} a, f(b) \\equal{} \\frac{1}{b}$ then $ \\frac{a}{b} \\equal{} ab \\text{ or } \\frac{1}{ab}$ which is clearly only possible if $ b \\equal{} 1$ or $ a \\equal{} 1$, respectively.", "Solution_15": "We claim the answer is $f(x)=x,\\frac{1}{x}$, which both clearly work. We know prove these are the only ones.\n\nLet $P(x,y,z,w)$ be the assertion. Setting all as $t$, we get:\n\\[\\frac{2f(t)^2}{2f(t^2)}=1,\\]\n\\[f(t^2)=f(t)^2.\\]\nNote that $f(1)=1$.\n\nUsing $P(\\sqrt{t},\\sqrt{t},1,t)$, we get:\n\\[\\frac{2f(t)}{f(1)+f(t)^2}=\\frac{2t}{1+t^2},\\]\n\\[t\\cdot (f(t))^2-(1+t^2)\\cdot f(t)+t=0.\\]\nSolving for $f(t)$, we get:\n\\[f(t)=\\frac{1}{t},t.\\]\n\nNow, FTSOC, assume there exists a $f(a)$ and $f(b)$ such that $f(a)=a$ and $f(b)=\\frac{1}{b}$ with $a,b\\neq 1$. Using $P(a,b,ab,1)$, we get:\n\\[\\frac{a^2+\\frac{1}{b^2}}{f(a^2b^2)+1}=\\frac{a^2+b^2}{a^2b^2+1}.\\]\n\nIf $f(a^2b^2)=a^2b^2$, then:\n\\[b^4=1,\\]\n\\[b=1.\\]\nIf $f(a^2b^2)=\\frac{1}{a^2b^2}$, then:\n\\[a^4b^2+a^2=a^2+b^2,\\]\n\\[a^4b^2=b^2,\\]\n\\[a^4=1,\\]\n\\[a=1.\\]\n\nThus, $f(x)=x,\\frac{1}{x}$.", "Solution_16": "We claim that the only solutions are $f(x) \\equiv x$ and $f(x) \\equiv \\frac{1}{x}$.\n\nLet $P(w, x, y, z)$ denote the assertion. Then clearly $P(w, w, w, w)$ gives us that $f(w^2) = f(w)^2$. For $w = 1$ this gives $f(1) = 1$. Now we can rewrite the given as,\n\\begin{align*}\n \\frac{f(a) + f(b)}{f(c) + f(d)} = \\frac{a+b}{c+d},\n\\end{align*}\nwhere we let $a = w^2$, $b = x^2$, $c = y^2$, $d = z^2$ such that $ab = cd$. Now from $P(a^2, 1, a, a)$ we see we find after some work,\n\\begin{align*}\n (f(a) - a)(af(a) - 1) = 0.\n\\end{align*}\nThis in turn gives $f(a) = a$ or $f(a) = \\frac{1}{a}$ for all $a$ in the domain. Now to avoid the point-wise trap assume for the sake of contradiction that $f(a) = \\frac{1}{a}$ and $f(b) = b$, with $a, b \\neq 1$. Then from $P(a, b, 1, ab)$ we see that,\n\n\\begin{align*}\n ab^2 - a + b + \\frac{1}{a} = (a+b) \\cdot f(ab).\n\\end{align*}\n\nTaking cases on whether $f(ab) = \\frac{1}{ab}$ or $f(ab) = ab$ we can reduce them respectively to,\n\\begin{align*}\n (a^2 - 1)(1+ab) &= 0\\\\\n (b^2-1)(1+ab) &= 0\n\\end{align*}\nThe first case implies $a = 1$, the second case implies $b = 1$ so we are done.", "Solution_17": "The answer is $f(x)=x$ for all $x \\in \\mathbb{R^+}$ and $f(x)=\\frac{1}{x}$ for all $x \\in \\mathbb{R^+}$. It\u2019s easy to see that these functions satisfy the given equation. \\\\\nNow, we shall show that these are the only solutions. First notice that letting $w=x=y=z=a$ we have that \n$$\\frac{2f(a)^2}{2f(a^2)}=1$$\nThus, we most clearly have that $f(x)^2=f(x^2)$. Now, proceed as follows. Notice that $f(1)=f(1)^2$ gives us that $f(1)=1$. Let $w=1,x=a^2$ and $y=z=a$. Then, notice the following\n\\begin{align*}\n \\frac{f(1)+f(a^2)^2}{f(a^2)+f(a^2)} &= \\frac{1+a^4}{2a^2}\\\\\n 2a^2f(a^2)^2 - 2(a^4+1)f(a^2)+2a^2 &= 0\\\\\n a^2f(a^2)^2-(a^4+1)f(a^2)+a^2&=0\\\\\n (a^2f(a^2)-1)(f(a^2)-a^2)&=0\\\\\n\\end{align*}\nWhich means that $$f(a^2)=\\frac{1}{a^2} \\implies f(a)=\\frac{1}{a}$$ \nor $$f(a^2)=a^2\\implies f(a)=a$$\nwhich gives us that indeed the solutions to the given equation are of the claimed type.", "Solution_18": "Call the assertion $P(w,x,y,z)$. Notice $P(k,k,k,k)$ gives us\n\n\\[\\frac{2f(k)^2}{2f(k^2)}=1\\]\n\\[\\implies f(k^2)=f(k)^2\\]\n\nIn particular, $f(1)=1$.\n\nNow, $P(1,k,\\sqrt{k}, \\sqrt{k})$ gives us\n\n\\[\\frac{f(1)^2+f(k)^2}{f(k)+f(k)}=\\frac{1+k^2}{k+k}\\]\n\\[\\implies f(k) \\in \\{k, \\frac{1}{k}\\}\\]\n\nThe only other case we must check is that for some $a,b \\neq 1$, we have $f(a)=a$ and $f(b)=\\frac{1}{b}$. Plugging in $P(a,b,ab,1)$ gives\n\n\\[\\frac{a^2+\\frac{1}{b^2}}{f(a^2b^2)+1}=\\frac{a^2+b^2}{a^2b^2+1}\\]\n\nBoth cases $f(a^2b^2)=a^2b^2, \\frac{1}{a^2b^2}$ result in either $a=1$ or $b=1$, a contradiction.\n\nThus, $\\boxed{f(x)=x, \\frac{1}{x}}$.", "Solution_19": "We claim that $f(x) = x$ and $\\frac{1}{x}$ are the only solutions. It is easy to check that these satisfy the equation.\nNow, set $w = z = 1$, we get $\\frac{f(x)^2 + 1}{f(x^2)+1} = 1 \\implies f(x^2) = f(x)^2$\nTherefore the original equation is equivalent to $\\frac{f(w^2) + f(x^2)}{f(y^2)+f(z^2)} = \\frac{w^2 + x^2}{y^2 + z^2}$ or $\\frac{f(a)+f(b)}{f(c)+f(d)} = \\frac{a+b}{c+d} \\forall a, b, c, d \\in \\mathbb{R}^{+}$ satisfying $ab = cd$. \nSubstituting $(a, b, c, d) = (x, x, 1, x^2)$, we get $\\frac{2f(x)}{1+f(x)^2} = \\frac{2x}{1+x^2}$.\nSolving this, we get a quadratic in $f(x)$ with solutions $f(x) = x, \\frac{1}{x}$.\nQED", "Solution_20": "We know $f(1)=1$ from substituting $(1,1,1,1)$ and $f(x^2)=f(x)^2$ from substituting $(x,x,x,x)$ (so in reality this equation is indeed symmetric). Then substitute $(1,t,\\sqrt t, \\sqrt t)$ to get\n\\[\\frac{f(1)^2+f(t)^2}{2f(t)} = \\frac{1+t^2}{2t} \\implies (tf(t)-1)(f(t)-t)=0 \\implies f(t) \\in \\left\\{t, \\frac 1t\\right\\}.\\]\n\nNow we handle this pointwise trap. Suppose FTSOC there exist positive constants $a \\ne b$ and $a,b \\ne 1$ such that $f(a) = a$ and $f(b) = \\frac 1b$. Then substitute $(\\sqrt a, \\sqrt b, 1, \\sqrt{ab})$ to get\n\\[\\frac{a+\\frac 1b}{1+f(ab)} = \\frac{a+b}{1+ab}.\\]\n\nOur only possible choices for $f(ab)$ are $ab$ and $\\frac{1}{ab}$, but if we plug them in and try to solve the resulting equation, we find that either $a=1$ or $b=1$, contradiction.\n\nHence our solution set is $\\boxed{f(x)=\\frac 1x, \\quad f(x)=x}$, which clearly work. $\\blacksquare$", "Solution_21": "$f\\equiv x$ and $f\\equiv \\frac{1}{x}$ are the only solutions.\n\nSubstitute $z=\\dfrac{wx}{y}$ to get rid of the extra condition.\n\n$P(w,x,y)$ be the assertion.\n\n$P(1,1,1) \\implies f(1) = 1$.\n\nSo we have,\n\\begin{align*}\n \\dfrac{f(w)^2 + f(x)^2}{f(y^2) + f((wx/y)^2)} &= \\dfrac{w^2 + x^2}{y^2 + (wx/y)^2}\\\\\n &=\\dfrac{1}{\\dfrac{y^2 + (wx/y)^2}{w^2+x^2}}\\\\\n &=\\dfrac{1}{\\dfrac{f(y)^2 + f(wx/y)^2}{f(w^2)+f(x^2)}}\\\\\n &= \\dfrac{f(w^2)+f(x^2)}{f(y)^2 + f(wx/y)^2}\n.\\end{align*}\n\n$P(y,y,y) \\implies f(y^2) = f(y)^2$.\n\n$P(1,1,y) \\implies f(y)^2 + f(1/y)^2 = y^2 + 1/y^2 \\implies f(1/x) = x^2 + 1/x^2 - f(x)^2$.\n\n\\begin{align*}\n P(1,x,x^2) &\\implies\\dfrac{1+ f(x)^2}{f(x)^4 + f(1/x)^2} = \\dfrac{1+x^2}{x^4 + \\frac{1}{x^2}}\\\\\n &\\implies (1+f(x)^2)(x^6+1) = (1+x^2)(x^2f(x)^4 + x^4 + 1 - f(x)^2x^2)\\\\\n &\\implies (1+f(x)^2)(1-x^2 + x^4) = (x^2 f(x)^4 + x^4 - f(x)^2 x^2 + 1)\\\\\n &\\implies (f(x)^2 - x^2)(1-f(x)^2 x^2) = 0\\\\\n &\\implies f(x) \\in \\left\\{x,\\frac{1}{x}\\right\\}\n.\\end{align*}\n\nNow if $\\exists a,b\\in \\mathbb R^+$ such that $a,b \\neq 1$, $f(a) = a$ and $f(b) = \\frac{1}{b}$.\n\n$P(a,b,ab) \\implies \\dfrac{a^2 + \\frac{1}{b^2}}{f(ab)^2+1} = \\dfrac{a^2+b^2}{a^2b^2+1}$.\n\nIf $f(ab) = ab$, then $b=1$, contradiction.\n\nOtherwise if $f(ab) = \\frac{1}{ab} \\implies a^2b^2\\left(a^2 + \\frac{1}{b^2}\\right) = a^2 + b^2 \\implies a^4b^2 + a^2 = a^2 + b^2 \\implies a^4 = 1 \\implies a=1$, contradiction again. :yoda:", "Solution_22": "Let $P$ be the given assertion. Then $P(1,1,1,1)$ means that $f(1)^2 = f(1)$ or $f(1) = 1.$ Note that $P(w, 1, \\sqrt{w}, \\sqrt{w})$ yields,\n\\begin{align*}\n\\frac{f(w)^2 + f(1)^2}{2f(w)} &= \\frac{w^2 + 1}{2w} \\\\\n(f(w) - \\frac{1}{w})(f(w) - w) &= 0.\n\\end{align*}\nSo for any $a \\in \\mathbb{R}^+$, $f(a) \\in \\{a, \\frac{1}{a}\\}.$ To avoid the point-wise trap, assume FTSOC there exists $a,b$ such that $f(a) = a, f(b) = \\frac{1}{b}$ for $a,b \\ne 1.$ If $f(ab) = ab$ or $\\frac{1}{ab}$ we get the respective contradictions of $P(a, b, ab, 1)$ as,\n\\begin{align*}\n\t\\frac{f(a)^2 + f(b)^2}{f((ab)^2) + f(1)} &= \\frac{a^2 + b^2}{(ab)^2 + 1} \\\\\n\t\\frac{a^2 + \\frac{1}{b}^2}{(ab)^2 + 1} = \\frac{a^2 + b^2}{(ab)^2 + 1} &\\implies b^4 = 1 \\\\\n\t\\frac{a^2 + \\frac{1}{b}^2}{\\frac{1}{(ab)^2} + 1} = \\frac{a^2 + b^2}{(ab)^2 + 1} &\\implies a^4 = 1.\\\\\n\\end{align*}\nTherefore $f(x) = x$ or $f(x) = \\frac{1}{x}.$", "Solution_23": "The answers are $f(x)=x$ and $f(x)=\\frac 1x$ which both.\nLet $P(w,x,y,z)$ denote the assertion. Note that $P(1,1,1,1)$ gives $f(1)=1$ and $P(x,1,x,1)$ gives $f(x^2)=f(x)^2$.\n\n[b][color=red]Claim:[/color][/b] For each $x$, we have $f(x)=x,\\frac 1x$\n[i]Proof.[/i] $P(x,1,\\sqrt x,\\sqrt x)$ gives that \\[\\frac{f(x)^2+1}{2f(x)}=\\frac{x^2+1}{2x}.\\] This is a quadratic in $f(x)$ so it has at most $2$ solutions, but $x$ and $\\frac 1x$ work. $\\blacksquare$\nNow, if $f(a)=a$ and $f(b)=\\frac 1b$ with $a,b\\neq 1$, then $P(a,b,1,ab)$ gives \\[\\frac{a^2+\\frac{1}{b^2}}{1+f(ab)^2}=\\frac{a^2+b^2}{1+(ab)^2}.\\] Then, $f(ab)=ab$ gives $b=1$ and $f(ab)=\\frac {1}{ab}$ gives $b=1$. This finishes.", "Solution_24": "Pretty easy for p4.\nFirst of all, letting all terms equal, find that $f(x)^2$=$f(x^2)$. Then, observe that $P(x,x,1,x^2)$ yields that $f(x)(x^2+1)=x(f(x)^2+1)$ for all positive real numbers $x$.(We get $f(1)=1$) Note that,if the function $g(x)=(x^2+1)/x$ has 2 equal values for 2 inputs $x$ and $y$, then either $x=y$ or $y=1/x$. So, $f(x)$ is either $x$ or $1/x$ for all $x$. Taking $a,b$ such that neither of them is $1$, let $P(a,b,1,ab)$. Messing with those 2 nasty cases gives us a contradiction.\nTherefore, $f(x)=x$ for all $x$ or $f(x)=1/x$ for all $x$. :-D", "Solution_25": "Taking all the variables to be $1$, $f(1)=1$. Afterwards, set the variables to be equal, so $f(t)^2=f(t^2)$. Letting $(w, x, y, z)=(t, 1, \\sqrt t, \\sqrt t)$ gives\n\\[\\frac{f(t)^2+1}{2f(t)}=\\frac{t^2+1}{2t}.\\]\nIt's easy to see that $f(t)=t$ and $f(t)=\\tfrac1t$ are solutions -- since this is a quadratic in $f(t)$, there are at most $2$ solutions.\n\nNow suppose that we have $f(a)=a$ and $f(b)=\\tfrac1b$, for $a$, $b\\neq 1$. Take $(w, x, y, z)$ as $(\\sqrt a, \\sqrt b, 1, \\sqrt{ab})$. So\n\\[\\frac{f(\\sqrt{a})^2+f(\\sqrt b)^2}{f(\\sqrt{ab})^2+1}=\\frac{a+b}{ab+1}.\\]\nThe LHS is also $\\tfrac{f(a)+f(b)}{f(ab)+1}$, hence\n\\[\\frac{a+\\tfrac1b}{f(ab)+1}=\\frac{a+b}{ab+1}.\\] There are two cases for the value of $f(ab)$ -- it's either $ab$ or $\\tfrac1{ab}$.\n[list=disc]\n[*] $f(ab)=ab$: we get $b^2=1$ implying $b=1$. Contradiction.\n[*] $f(ab)=\\tfrac{1}{ab}$: we get $a^2=1$ so $a=1$. Contradiction.\n[/list]\nIt's easy to verify that $f(t)=t$ and $f(t)=\\tfrac1t$ both work. So we are done. $\\square$", "Solution_26": "[hide=\"For storage\"] \n\nObviously, $f(x)=x$ and $f(x)=\\frac1X$ are solutions to the FE and we want to prove their uniqueness. \n$P(1,1,1,1) \\Rightarrow f(1)=1$\n$P(x,1,x,1) \\Rightarrow f(x^2)=f(x)^2$\nso now we can rewrite the FE as following \n \\[ \\frac {\\left( f(w) \\right) + \\left( f(x) \\right)}{f(y) + f(z) } = \\frac {w + x}{y + z}\\] for $wx=yz$ \n\n \nNow consider $P(a^2,1,a,a)$ and solving the quadratic eqn. we have $f(a)=a$ or $1/a$\nsuppose there are reals $a,b$ none of them is $1$ satisfying $f(a)=a$ & $f(b)=1/b$ . By considering $P(a,b,ab,1)$ we have a contradiction.\n\n$$\\mathbb{Q.E.D.}$$\n\n [/hide]", "Solution_27": "We claim that $f(x) = x$ or $f(x) = \\frac{1}{x}$, which both clearly work.\n\nNow, substituting $w = x = y = z$ yields \\[f(x)^2 = f(x^2)\\text{.}\\] This implies that $f(1) = 1$. Now substitute $w = 1, x = n, y = z = \\sqrt{n}$ for some positive real number $n$. This yields\n\\begin{align*}\n \\frac{f(1)^2 + f(n)^2}{2f(n)} &= \\frac{1 + n^2}{2n} \\\\\n n(1 + f(n)^2) &= f(n)(1 + n^2) \\\\\n f(n)^2 + \\left(n + \\frac{1}{n}\\right) f(n) + 1 &= 0 \\\\\n (f(n) - n)\\left(f(n) - \\frac{1}{n}\\right) &= 0\\text{.}\n\\end{align*}\nTherefore, for each $n$, $f(n) = n$ or $\\frac{1}{n}$.\n\nNow suppose there exist $a \\neq b$, none of which are equal to $1$, such that $f(a) = a$ and $f(b) = \\frac{1}{b}$. Now\n\\begin{align*}\n \\frac{f(1)^2 + f(ab)^2}{f(a^2) + f(b^2)} &= \\frac{1 + (ab)^2}{a^2 + b^2} \\\\\n \\frac{1 + f(ab)^2}{f(a)^2 + f(b)^2} &= \\frac{1 + (ab)^2}{a^2 + b^2} \\\\\n 1 + f(ab)^2 &= \\frac{(1 + (ab)^2)(a^2 + \\frac{1}{b^2})}{a^2 + b^2} \\\\\n (1 + f(ab)^2)(a^2 + b^2) &= 2a^2 + \\frac{1}{b^2} + a^4 b^2 \\\\\n a^2 + b^2 + f(ab)^2 a^2 + f(ab)^2 b^2 &= 2a^2 + \\frac{1}{b^2} + a^4 b^2\\text{.}\n\\end{align*}\nIf $f(ab) = ab$, then we have\n\\begin{align*}\n a^2 + b^2 + a^4 b^2 + a^2 b^4 &= 2a^2 + \\frac{1}{b^2} + a^4 b^2 \\\\\n b^2 + a^2 b^4 &= a^2 + \\frac{1}{b^2} \\\\\n a^2 (b^4 - 1) + \\frac{1}{b^2} (b^4 - 1) &= 0 \\\\\n \\left(a^2 + \\left(\\frac{1}{b}\\right)^{\\!2}\\right)(b^4 - 1) &= 0\n\\end{align*}\nwhich is clearly impossible. Similarly, if $f(ab) = \\frac{1}{ab}$, then we have\n\\begin{align*}\n a^2 + b^2 + \\frac{1}{b^2} + \\frac{1}{a^2} &= 2a^2 + \\frac{1}{b^2} + a^4 b^2 \\\\\n a^2 + a^4 b^2 &= b^2 + \\frac{1}{a^2} \\\\\n b^2 (a^4 - 1) + \\frac{1}{a^2} (a^4 - 1) &= 0 \\\\\n \\left(b^2 + \\left(\\frac{1}{a}\\right)^{\\!2}\\right)(a^4 - 1) &= 0\n\\end{align*}\nwhich is also impossible.", "Solution_28": "Plug in $y=w, z=x$\n$f(x)^2+f(y)^2=f(x^2)+f(y^2)$\nWhen $x=y$, get that $f(x)^2=f(x^2)$\nSo plugging in $x=1$, get that $f(1)=1$\nIf $w=1$, then $x=yz$\n$\\frac{1+f(yz)^2}{f(y^2)+f(z^2)}=\\frac{1+y^2z^2}{y^2+z^2}$\nIf $y=z=\\sqrt{x}$, then $\\frac{1+f(x)^2}{2f(x)}=\\frac{1+x^2}{2x}$\n$2x+2xf(x)^2=2f(x)+2x^2f(x)$\n$2xf(x)^2+f(x)(-2x^2-2)+2x=0$\n$f(x)=x$ or $\\frac{1}{x}$ by the quadratic formula", "Solution_29": "The answers are $f(x) = x$ and $f(x) = \\tfrac{1}{x}$, both of which aren't too hard to verify. We now show these are the only answers.\n\nLet $P(w, x, y, z)$ denote the given assertion. From $P(1, 1, 1, 1)$, we have $f(1) = 1$. Then, from $P(1, x, \\sqrt{x}, \\sqrt{x})$, we have $\\tfrac{1 + f(x)^2}{2f(x)} = \\tfrac{1 + x^2}{2x}.$ This rearranges to \n$$xf(x)^2 - f(x) - x^2f(x) + x = 0 \\iff (xf(x) - 1)(f(x) - x) = 0.$$\nThus, for all $x$, either $f(x) = \\tfrac{1}{x}$ or $f(x) = x$. Now, suppose for the sake of contradiction there exist $a, b > 0$ and not equal to $1$ such that $f(a) = \\tfrac{1}{a}$ and $f(b) = b$. Then from $P(a, b, 1, ab)$, we see that \n$$\\frac{\\tfrac{1}{a^2} + b^2}{1 + f(a^2b^2)} = \\frac{a^2 + b^2}{1 + a^2b^2}.$$\nTaking cases on whether $f(a^2b^2) = \\tfrac{1}{a^2b^2}$ or $f(a^2b^2) = a^2b^2,$ we see that:\n[list]\n[*] If $f(a^2b^2) = \\tfrac{1}{a^2b^2}$, then $\\tfrac{1}{a^2} + b^2 + b^2 + a^2b^2 = a^2 + b^2 + \\tfrac{1}{a^2} + \\tfrac{1}{b^2}$, or $(a^2b^2 + 1)(b^4 - 1) = 0$, which implies that $b = 1$, contradiction\n[*] If $f(a^2b^2) = a^2b^2$, then $\\tfrac{1}{a^2} + b^2 + b^2 + a^2b^4 = a^2+ b^2 + a^2b^4 + a^4b^2$, or $(a^2b^2 + 1)(a^4 - 1) = 0$, so $a = 1$, contradiction\n[/list]\nSo, either $f(x) =x$ for all $x$, or $f(x) = \\tfrac{1}{x}$ for all $x$, as desired." } { "Tag": [ "Duke", "college", "inequalities", "complex numbers" ], "Problem": "dude it's a different poll\r\n\r\nDUDE GET WITH THE PROGRAM", "Solution_1": "Do all Americans use the word 'dude' that often or is that just a stereotype? :rotfl:", "Solution_2": "dude at least they don't carry mooses around with them going all like eh? we're canadian what's going on eh? do you want to sample some of our moose, eh? let's go disco in this disco bar, eh?", "Solution_3": "sure dude but no one wants a poll run by Arnav regarding basketball, dude", "Solution_4": "[quote=\"not_trig\"]sure dude but no one wants a poll run by Arnav regarding basketball, dude[/quote]\r\n\r\n\r\ndude \r\nI want a poll run by him hehe :P", "Solution_5": "[quote=\"MysticTerminator\"]dude at least they don't carry mooses around with them going all like eh? we're canadian what's going on eh? do you want to sample some of our moose, eh? let's go disco in this disco bar, eh?[/quote]\r\nI don't know anyone who says eh.... :huh: \r\nAnd we don't carry mooses around, they carry us. :rotfl:", "Solution_6": "huh you guys definitely carry mooses around, pip pip hail to the queen and all that rot, what", "Solution_7": "what i'm from canada\r\n\r\nCANADA > UNC", "Solution_8": "that . . . doesn't even make sense\r\n\r\nit's like comparing an element of the real hamiltonian quarternions cross itself to the group ring formed by the Viergruppe and the complex numbers? $(4+2i, j+4k-1) > V_{4}\\mathbb{C}$\r\n\r\nIT DOESN'T MAKE SENSE, DUDE", "Solution_9": "good point\r\n\r\nYOU CAN\"T EVEN COMPARE UNC WITH CANADA\r\n\r\nit's just TOO GOOD\r\n\r\nedit: i mean canada is too good", "Solution_10": "Well, yeah... Duke>Canada>UNC", "Solution_11": "wtf?\r\n\r\nby transitivity \r\n\r\nUNC > DUKE\r\n\r\nand\r\n\r\nCANADA > UNC\r\n\r\n=> CANADA > UNC > DUKE\r\n\r\nplus you get the same results if you use the power mean inequality", "Solution_12": "dude will you stop involving canada in this\r\n\r\nI am pretty sure the UNC team would own canada in basketball\r\n\r\nthis is mainly because canada would not be able to shoot or play any defense since countries usually just sit there\r\n\r\nwithout shooting and playing defense", "Solution_13": "It just goes to show we are peaceful. :rotfl:", "Solution_14": "this is true\r\n\r\nbut\r\n\r\nCANADA > UNC\r\n\r\nwhen it comes to ice hockey", "Solution_15": "[quote=\"not_trig\"]how bout [size=200]NO[/size]\n\npwned![/quote]\r\n\r\namusing", "Solution_16": "guys i just happened upon this thread this is like ranting about people who say \"eh\",\"dude\", and carry moose around BTW what is the plural of buffalo?", "Solution_17": "buffalae", "Solution_18": "DUDE\r\n\r\nuse correct grammar\r\n\r\n\r\nit's buffaloes\r\n\r\n\r\nbuffalae is terminator language :lol:", "Solution_19": "$latin$", "Solution_20": "snap", "Solution_21": "snap cracklpe pop\r\n\r\n\r\nPopsicles\r\n\r\nicey and popsilicious\r\n\r\ntangy and tongue-twisting\r\n\r\n bite it hard\r\n\r\nit gives fat lard,\r\n\r\n pleasing to the senses\r\n\r\nnow taste one and try!\r\n\r\nit's actually quite tasty\r\n\r\nPop....Sickle\r\n\r\nPopSnowSickle, stop the carnocentrism!", "Solution_22": "[quote=\"now a ranger\"]DUDE\n\nuse correct grammar\n\n\nit's buffaloes\n\n\nbuffalae is terminator language :lol:[/quote]\r\nI'm pretty sure it's buffalo, not buffaloes...", "Solution_23": "no.......... reallly????... i didn't know that...\r\n\r\n\r\nRagnarok is the guy!\r\n\r\n\r\nthe person with the tonka-tai! ( ask me what this is :rotfl: )\r\n\r\nhe is surreptitiously special, but not in special ed\r\n\r\nhe works hopelessly hard, but he's not Farmer Ted\r\n\r\nhe's ragnarok\r\n\r\nthe man with the plan\r\n\r\nthe gold coin can.\r\n\r\n\r\n\r\nWalk around the river is the kid with the shiver,\r\n\r\nhe eats a lot of food such as stir fried liver,\r\n\r\nwhenever he says I rock!\r\n\r\neveryone responds with a shock,\r\n\r\nhey? I Love You! You are my hero! My White Idol! \r\n\r\nThen he runs wherever his legs take him,\r\n\r\nresponding with enthusiasm and sees dark slim\r\n\r\nand TahDah! He's Michael Jackson!", "Solution_24": "[quote=\"ragnarok23\"][quote=\"now a ranger\"]DUDE\n\nuse correct grammar\n\n\nit's buffaloes\n\n\nbuffalae is terminator language :lol:[/quote]\nI'm pretty sure it's buffalo, not buffaloes...[/quote]\r\ni'll go with you seem sane", "Solution_25": "[i][/i]", "Solution_26": "[color=orange] by putting a blank post you mean \"i don't understand\" or you are watching the person but i think this post is the first 1 i have put in this forum not on testing one srry :oops: [/color] oh and sorry if i hurt ur eyes and srry mystic terminator IF DUDE YOU MEANT SOMETHING THAT I DON'T KNOW.\r\nall better, eh?", "Solution_27": ":spam: Try not to spam so much, eh?", "Solution_28": "wtf", "Solution_29": "[quote=\"Inspired By Nature\"]:spam: Try not to spam so much, eh?[/quote]\r\n\r\nDefinitely.\r\n\r\nFrom now on, I will be deleting a lot more of the spam in this forum. If people continue to spam, you may be banned from this forum." } { "Tag": [ "induction" ], "Problem": "For $ n>\\equal{}1$ show by induction that\r\n\r\na) $ 15|2^{4n} \\minus{} 1$\r\nb) $ 5|3^{3n \\plus{} 1} \\plus{} 2^{n \\plus{} 1}$ \r\nc) $ 24|2(7^n) \\plus{} 3(5^n) \\minus{} 5$", "Solution_1": "a) For $ n \\equal{} 1$, $ 2^{4} \\minus{} 1 \\equal{} 15$, which is clearly divisible by 15.\r\n\r\nAssume that for n=k, $ 16^k \\minus{} 1$ is divisible by 15. Let $ 16^k \\minus{} 1 \\equal{} 15d$\r\n\r\nFor n=k+1, $ 16^{k \\plus{} 1} \\minus{} 1 \\equal{} 16(16^k) \\minus{} 1 \\equal{} 16(15d \\plus{} 1) \\minus{} 1 \\equal{} 240 d \\plus{} 15$, which is divisible by 15. Hence proved.", "Solution_2": "b) It is obvious for n=1.\r\n\r\nFor n=k, $ 3^{3k\\plus{}1}\\plus{}2^{k\\plus{}1}\\equal{}3(27^k)\\plus{}2(2^k)$, let $ 3(27^k)\\equal{}a$ and $ 2(2^k)\\equal{}b$\r\n\r\nFor n=k+1, $ 3(27^{k\\plus{}1})\\plus{}2(2^{k\\plus{}1})\\equal{}27a\\plus{}2b\\equal{}25a\\plus{}2(a\\plus{}b)$, which is divisible by 5 since a+b is divisible by 5. And hence proved.", "Solution_3": "[hide=\"c\"]Let $ n\\equal{}1$. Then we have $ 2(7)\\plus{}3(5)\\minus{}5\\equal{}24$, which is divisible by 24. Now, assume it is true for $ n\\equal{}k$. Letting $ n\\equal{}k\\plus{}1$, we have (letting $ a\\equal{}2(7^k)$ and $ b\\equal{}3(5^k)$)\n\\[ 7\\cdot2(7^k)\\plus{}5\\cdot3(5^k)\\minus{}5\\equal{}7a\\plus{}5b\\minus{}5\\equal{}2a\\plus{}5(a\\plus{}b\\minus{}5)\\plus{}20.\\]\nSince $ (a\\plus{}b\\minus{}5)$ is divisible by 24 from our assumption, we only have to consider $ 2a\\plus{}20$, or $ 4(7^k)\\plus{}20$. Factoring, we have $ 4(7^k\\plus{}5)$. We now note that since $ 7^k$ is odd and 5 is odd, $ (7^k\\plus{}5)$ is even. Taking $ 7^k\\plus{}5\\bmod3$, we have $ 7^k\\plus{}5\\equiv1^k\\plus{}2\\bmod3$, which is clearly divisible by three. Thus, $ 4(7^k\\plus{}5)$can be factored into $ 24x$ for some positive integer $ x$, making it divisible by 24. Our induction is now complete.[/hide]" } { "Tag": [ "inequalities" ], "Problem": "Solve for $x$: $\\dfrac{a(x-1)}{x-2}>1$", "Solution_1": "Graphing $\\dfrac {x-1}{x-2}>\\dfrac1a$ will make it just a matter of moving x's and a's around.\r\n\r\nJust remember to invert signs if for negative \"a\" (clearly x cannot equal 1 or 2)", "Solution_2": "Is $a \\in \\mathbb{N}$? :D", "Solution_3": "[quote=\"chess64\"]Is $a \\in \\mathbb{N}$? :D[/quote]\r\nNo. $a \\in R$.", "Solution_4": "multiply by $(x-2)^2$ (do you see why this is helpful?), then it just becomes\r\n$a(x-1)>x^2-4x+4$\r\n$0>x^2-(4+a)x+4a$\r\n$0>(x-a)(x-4)$\r\nfrom here, it is easy, note that $x\\ne2$", "Solution_5": "[quote=\"Altheman\"]multiply by $(x-2)^2$ (do you see why this is helpful?), then it just becomes\n$a(x-1)>x^2-4x+4$\n$0>x^2-(4+a)x+4a$\n$0>(x-a)(x-4)$\nfrom here, it is easy, note that $x\\ne2$[/quote]\r\n\r\nYour work is correct... but you mean $x \\ne 4$. Also, both terms must have the same parity... so, $x > \\max(a, 4) \\text{ or } x < \\min(a, 4)$. Is there a better way to write taht?", "Solution_6": "i believe that $x\\ne 2$ because in the original expression, that would make it undefined", "Solution_7": "if you multiply by $(x-2)^2$ you have to do it to both sides.", "Solution_8": "[quote=\"Altheman\"]multiply by $(x-2)^2$ (do you see why this is helpful?), then it just becomes\n$a(x-1)>x^2-4x+4$\n$0>x^2-(4+a)x+4a$\n$0>(x-a)(x-4)$\nfrom here, it is easy, note that $x\\ne2$[/quote]\r\nYou forgot something on the left hand side... :( \r\n$\\dfrac{a(x-1)}{x-2}>1$\r\nMultiplying $(x-2)^2$ on both sides,\r\n$a(x-1)(x-2)>(x-2)^2$\r\n$a(x^2-3x+2)>x^2-4x+4$\r\n$(a-1)x^2+(4-3a)x+2a-4>0$\r\n$(x-2)[(a-1)x-a+2]>0$\r\n$\\begin{cases}x-2>0\\\\(a-1)x-a+2>0\\end{cases}$ or $\\begin{cases}x-2<0\\\\(a-1)x-a+2<0\\end{cases}$\r\n$\\begin{cases}x>2\\\\x>\\dfrac{a-2}{a-1}\\end{cases}$ or $\\begin{cases}x<2\\\\x<\\dfrac{a-2}{a-1}\\end{cases}$", "Solution_9": "I think we have to divide cases for $a\\neq 0$ to get\r\n\r\n(i)When $a=1$ , $\\{x: x>2\\}$\r\n\r\n(ii)When $a>1$ , $\\{x: x<\\dfrac{a-2}{a-1},x>2\\}$\r\n\r\n(iii)When $a<1$ , $\\{x: \\dfrac{a-2}{a-1}>r$\r\n\r\nand the formula i gave is rigth\r\n\r\n\r\nthe momentul is equal to the double of variation of energy of the ligth absorved divided by ligth speed\r\n\r\n\r\nyou can find on google this formula, or in the book i said\r\n\r\nhalliday resnick walker vomume 4\r\n\r\ni dont know the title in english\r\n\r\n\r\n\r\n\r\n\r\nhttp://www.physicsforums.com/showthread.php?t=87459", "Solution_11": "[quote=\"Thales418\"]or if e is quasi-zero\n\n\nin you case if will be parallel\n\n\nradiation pressure is not because the collision of fotons!!\n\nit's because the ligth is a elletromagnet wave\n\nfotons does not have mass. so does not have momentum!!!!\n\nyou can see it on volume 4 of halliday resnick walker\n\nand my approximation it's because i'ts non-sense the length of the mirror be too big\n\nwe can make $e$ small and $n>>r$\n\nand the formula i gave is rigth\n\n\nthe momentul is equal to the double of variation of energy of the ligth absorved divided by ligth speed\n\n\nyou can find on google this formula, or in the book i said\n\nhalliday resnick walker vomume 4\n\ni dont know the title in english\n\n\n\n\n\nhttp://www.physicsforums.com/showthread.php?t=87459[/quote]\n\nwell i feel u need to go through ur modern physics lessons again .i suppose u know there exist something called as de-broglie wavelength .and momentum of particles whose inertial mass is $0$ is given by\n\n$P$ $=$ $\\frac{h}{\\lambda}$ $=$ $\\frac{E}{c}$\n\nwhere $\\lambda$ is the de broglie wavelength of the particle .further you can't explain collisions on the basis of wave theory of light and i really doubt a book like that of halliday tells so.\n\n[quote]the momentul is equal to the double of variation of energy of the ligth absorved divided by ligth speed \n\n[/quote]\r\n\r\nthis comes by assuming elastic collisions of photons . further if the mirror is not infinitely large then you have to consider components of the force exerted by photons", "Solution_12": "you are using classic physics in quantum physics\r\n\r\n\r\nin your classic model for the collisions of the photons, the momentum is zero\r\n\r\n\r\nthe question is the ligth is tranfering energy to the mirrors and the mirrors is not absorving this energy, the reason for multiplying by 2, assuming everything is perfect\r\n\r\nthe distance of the bulb to the 2 mirrors is differente. so we have a torque\r\n\r\ni use the wave theory to explain the energy\r\n\r\nand the formula i gave is on the book\r\n\r\nwe are not able to say it is wrong", "Solution_13": "[quote=\"Thales418\"]you are using classic physics in quantum physics\n\n\nin your classic model for the collisions of the photons, the momentum is zero\n\n\nthe question is the ligth is tranfering energy to the mirrors and the mirrors is not absorving this energy, the reason for multiplying by 2, assuming everything is perfect\n\nthe distance of the bulb to the 2 mirrors is differente. so we have a torque\n\ni use the wave theory to explain the energy\n\nand the formula i gave is on the book\n\nwe are not able to say it is wrong[/quote]\n\nwho said ur formula in \"the book \" is wrong i gave that clearly in my last post\n\n\n [quote]well i feel u need to go through ur modern physics lessons again .i suppose u know there exist something called as de-broglie wavelength .and momentum of particles whose inertial mass is is given by $P$ $=$ $\\frac{h}{\\lambda}$ $=$ $\\frac{E}{c}$\n\n \n\nwhere is the de broglie wavelength of the particle .further you can't explain collisions on the basis of wave theory of light and i really doubt a book like that of halliday tells so. [/quote]\r\n\r\n\r\n\r\n\r\nfurther i suppose u have an idea what classical and what quantum m,echanics is .[i]quantum mechanics[/i] beleive on [i]\"particle nature\"[/i]\r\nof light to explain the unexplained and this is where it is applieed \r\n\r\ni suppose u have some idea about what QUANTUM MECHANICS is all about .\r\nyes energy is transferred which i already posted in previous post saying the collision was elastic.\r\nwell if think wave theory explains it then prove it here (which i am sure u won't be able to because the entire thing is derived from quantum mechanics)\r\nwell the change in momentum of the photons times the time taken is the force applied on the mirror which produces the torque\r\n\r\nnow please[i] don't tell [/i]me that[i] ur book tells energy produces torque [/i]and not force\r\n :furious:", "Solution_14": "weelll\r\n\r\n\r\nlet's start again\r\n\r\nin m physics book dont say what you are saying. i really dont understand what are you saying.\r\n\r\nthe formula is not wrong!!!!!!!!!!!\r\n\r\nwith the ligth we need to use the particular and the wave theories together\r\n\r\nthe photon have energy because it is a eletromagnetic wave etc.....\r\n\r\nthe photon collides with the mirror crashing it. justa little bit and ping pong back etc...\r\n\r\nso. all approximations i made is true. because we have everything Big and small etc. etc etc...\r\n\r\n\r\nSO. i dont understand your point saying my formula is wrong\r\n\r\n\r\ncan you explain it?", "Solution_15": "[quote=\"Thales418\"]\nwith the ligth we need to use the particular and the wave theories together\n\nthe photon have energy because it is a eletromagnetic wave etc.....\n\n\n\nSO. i dont understand your point saying my formula is wrong\n\n\ncan you explain it?[/quote]\r\n\r\nwell u need to go through the book all once again light is supposed to be considered of quantas of energy having $0$ rest mass called $photon$\r\nthey have nothing to do with [i]electromagnetic wave[/i].\r\n\r\nenergy of a photon depend on the frequency of the constituting light given by the famous $planck's$ $law$\r\n \r\n $E$ $=$ $h\\nu$\r\nunlike an EM wave whos energy depends on it's amplitude .\r\n\r\n\r\nwell i never said that ur formula is wrong . i also somewhat derived it.\r\n\r\nwhat i say is ur understanding is wrong", "Solution_16": "u are understanding me wrong man\r\n\r\n\r\ni dont know a derivation of this formula\r\n\r\nbut it is true. we just need it to solve. or not?", "Solution_17": "[quote=\"Thales418\"]u are understanding me wrong man\n\n\ni dont know a derivation of this formula\n\nbut it is true. we just need it to solve. or not?[/quote]\r\n\r\nyse u need the formulae to solve provided the mirrors are large etc.....\r\n\r\nfurther what i said is ur reasoning is completely wrontg", "Solution_18": "i assume the mirrors are not large\r\n\r\nso\r\n\r\nno problem", "Solution_19": "[quote=\"Thales418\"]i assume the mirrors are not large\n\nso\n\nno problem[/quote]\r\n\r\nif the mirror sare not large assuming the bulb as a point source of light it emanates light equally in all directions which strike the mirror at varying angles hence this becomes a case of oblique collision .hence we will have to find the angle between the applied force and the mirror or rather the incident angle.this leads to complicaton of problem.\r\n\r\nbut if the mirror is infinetly large two point on which photons collide and impart same force which are equidistant from a symmetrical line along the $y$ axis on the mirror and also equidistant from the rotational axis of the torsional scale will have their components cancelled", "Solution_20": "pardesi\r\n\r\nforget this problem\r\n\r\ni cant formulate this better\r\n\r\n\r\nso. just forget\r\n\r\n\r\nlater i will try to make the problem better etc...\r\n \r\nbut now let's close the problem\r\n\r\nthis is a hard problem. if we make assumptions we will make a lot of assumptions and approximations and the result is no torque\r\n\r\nhehe\r\n\r\n\r\nbut thanks for your post" } { "Tag": [ "email", "\\/closed" ], "Problem": "If I want to make a formal complaint about someone else in this forum posting stuff, what do I do?", "Solution_1": "PM or email me or MCrawford if ever you have a complaint about the posts on this forum." } { "Tag": [], "Problem": "Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \\leq x\\leq 8$. The sum of the largest and smallest values of $f(x)$ is\n\n$\\textbf {(A) } 1 \\qquad \\textbf {(B) } 2 \\qquad \\textbf {(C) } 4 \\qquad \\textbf {(D) } 6 \\qquad \\textbf {(E) }\\text{none of these}$", "Solution_1": "[quote=\"4everwise\"]Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \\leq x\\leq 8$. The sum of the largest and smallest values of $f(x)$ is\n\n$\\text {(A)} 1 \\qquad \\text {(B)} 2 \\qquad \\text {(C)} 4 \\qquad \\text {(D)} 6 \\qquad \\text {(E)}\\text{none of these}$[/quote]\r\n\r\n[hide]$[2,3]: f(x)=x-2+4-x-(6-2x)=2x-4$ min 0, max 2\n$[3,4]: f(x)=x-2+4-x-(2x-6)=-2x+8$ min 0, max 2\n$[4,8]: f(x)=x-2+x-4-(2x-6)=0$ max 0 min 0, \noverall, 0 min, 2 max, $\\boxed{B}$[/hide]" } { "Tag": [], "Problem": "A one-digit number is increased by 20 when multiplied by 6. What is the one-digit number?", "Solution_1": "$ 6n \\minus{} n \\equal{} 5n \\equal{} 20$. If $ 5 \\times n \\equal{} 20$, then $ n \\equal{} \\boxed{4}$." } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "[i]Find all positive integer $ n$ such that :\n\n The set $ A \\ \\equal{} \\ \\{1 ;2 ; ....; 3n \\}$ can be partitioned into $ n$ subsets $ A_1 ; A_2; ...; A_n$ with $ |A_1| \\equal{} |A_2 | \\equal{} ...\\equal{} |A_n|\\equal{}3$ and in each subset $ A_i ; i \\equal{} 1 ; 2 ; ...; n$, there is a element which is sum of two other elements in $ A_i$ .[/i]\r\n\r\n [i]Sorry for my bad English :lol: :lol:[/i]", "Solution_1": "[quote=\"nguyenvuthanhha\"][i]Find all positive integer $ n$ such that :\n\n The set $ A \\ \\equal{} \\ \\{1 ;2 ; ....; 3n \\}$ can be partitioned into $ n$ subsets $ A_1 ; A_2; ...; A_n$ with $ |A_1| \\equal{} |A_2 | \\equal{} ... \\equal{} |A_n| \\equal{} 3$ and in each subset $ A_i ; i \\equal{} 1 ; 2 ; ...; n$, there is a element which is sum of two other elements in $ A_i$ .[/i]\n\n [i]Sorry for my bad English :lol: :lol:[/i][/quote]\r\n\r\nPerhaps it's a problem in China National Olympiad (I don't remember exactly the year :mad:, but it's in 1990s)", "Solution_2": "[i] Yes , this problem was chosen for a regional contests in Vietnam in 2009 . There is no contestant have point in that problem :lol: [/i]", "Solution_3": "I don't have anything remotely close to a rigorous proof but I've checked values for n=1,4, and 5 and have come to the conclusion that the solution is all n congruent $ 0$ or $ 1\\mod 4$ the motivation for this is:\r\nthe sum of all the numbers in $ S$ is $ \\dfrac{3n(3n\\plus{}1)}{2}$ knowing this, we can conclude that the sum of the $ c$'s, where $ a_i\\plus{}b_i\\equal{}c_i$ for every $ a$, $ b$, and $ c$ in set $ A_i$, is $ \\dfrac{3n(3n\\plus{}1)}{4}$.\r\nNow, we also know that $ 3n$ must be a $ c$ in some $ A_i$, by subtracting $ 3n$ from our sum, we find the sum of the rest of the $ c$'s equals $ \\dfrac{9n(n\\minus{}1)}{4}$, must be an integer so n is either $ 0$ or $ 1$ $ \\mod 4$. I know this argument is really weak so........ :|", "Solution_4": "That's a perfectly good argument that all such $ n$ must be 0 or 1 mod 4 (though the step of subtracting $ 3n$ from the sum is completely unnecessary) -- now we just need a construction to show that it's really possible for all those values. I've tried playing around with building up from the $ n \\equal{} 4$ case, but it hasn't led anywhere yet :(", "Solution_5": "I think I have a construction that works. Notice that if we put $ 1, 2, \\ldots, n$ in different sets, we get that it's sufficient to find a string consisting of two of each of the numbers $ 1, \\ldots, n$ with $ k\\minus{}1$ numbers between the two appearances of $ k$. So for example, for $ n\\equal{}4$ we'd use the string $ 11423243$ and for $ n\\equal{}5$ we'd use the string $ 4115423253$.\r\n\r\nSo now suppose that we have the string for $ n\\equal{}4k$. We're going to exploit the fact that it naturally splits into two parts, $ A_kB_k$, with $ A_1 \\equal{} 11$ and $ B_1 \\equal{} 423243$. Now, we're going to let $ A_{k\\plus{}1} \\equal{} B_k$ and $ B_{k\\plus{}1} \\equal{} (4k\\plus{}4)(4k\\plus{}2)(4k\\plus{}3)(4k\\plus{}1)A_k(4k\\plus{}2)(4k\\plus{}1)(4k\\plus{}4)(4k\\plus{}3)$. Then we obtain the desired string for multiples of $ 4$. For example, this gives the strings $ 4232438675116587$ for $ n\\equal{}8$ and $ 8\\ 6\\ 7\\ 5\\ 1\\ 1\\ 6\\ 5\\ 8\\ 7\\ 12\\ 10\\ 11\\ 9\\ 4\\ 2\\ 3\\ 2\\ 4\\ 3\\ 10\\ 9\\ 12\\ 11$ for $ n\\equal{}12$.\r\n\r\nNow we use the fact that $ B_k$ always starts with $ 4k$ to obtain the string for $ n\\equal{}4k\\plus{}1$. To do this, replace both instances with $ 4k$ with $ 4k\\plus{}1$ and place a $ 4k$ between the first and second numbers. Then the string $ (4k)A_kB_k'$ will satisfy the desired conditions. For example, for $ k\\equal{}1$ we have $ A_1 \\equal{} 11$, $ B_1 \\equal{} 423243$, so we have $ B_1' \\equal{} 5423253$ and our string is $ 4115423253$.\r\n\r\nIt is easy to check that all of these constructions meet the necessary criteria.", "Solution_6": "[quote=\"Hamster1800\"]I think I have a construction that works. Notice that if we put $ 1, 2, \\ldots, n$ in different sets, we get that it's sufficient to find a string consisting of two of each of the numbers $ 1, \\ldots, n$ with $ k \\minus{} 1$ numbers between the two appearances of $ k$. So for example, for $ n \\equal{} 4$ we'd use the string $ 11423243$ and for $ n \\equal{} 5$ we'd use the string $ 4115423253$.\n\nSo now suppose that we have the string for $ n \\equal{} 4k$. We're going to exploit the fact that it naturally splits into two parts, $ A_kB_k$, with $ A_1 \\equal{} 11$ and $ B_1 \\equal{} 423243$. Now, we're going to let $ A_{k \\plus{} 1} \\equal{} B_k$ and $ B_{k \\plus{} 1} \\equal{} (4k \\plus{} 4)(4k \\plus{} 2)(4k \\plus{} 3)(4k \\plus{} 1)A_k(4k \\plus{} 2)(4k \\plus{} 1)(4k \\plus{} 4)(4k \\plus{} 3)$. Then we obtain the desired string for multiples of $ 4$. For example, this gives the strings $ 4232438675116587$ for $ n \\equal{} 8$ and $ 8\\ 6\\ 7\\ 5\\ 1\\ 1\\ 6\\ 5\\ 8\\ 7\\ 12\\ 10\\ 11\\ 9\\ 4\\ 2\\ 3\\ 2\\ 4\\ 3\\ 10\\ 9\\ 12\\ 11$ for $ n \\equal{} 12$.\n\nNow we use the fact that $ B_k$ always starts with $ 4k$ to obtain the string for $ n \\equal{} 4k \\plus{} 1$. To do this, replace both instances with $ 4k$ with $ 4k \\plus{} 1$ and place a $ 4k$ between the first and second numbers. Then the string $ (4k)A_kB_k'$ will satisfy the desired conditions. For example, for $ k \\equal{} 1$ we have $ A_1 \\equal{} 11$, $ B_1 \\equal{} 423243$, so we have $ B_1' \\equal{} 5423253$ and our string is $ 4115423253$.\n\nIt is easy to check that all of these constructions meet the necessary criteria.[/quote]\r\n\r\n[i]Yes , after that lemma , we have a sequence $ a_1 a_2 ....a_{2n}$ , each number $ k \\ \\in \\ \\{1;2 ;...; n \\}$ appear 2 times in that sequence with $ k \\minus{} 1$ numbers between the two appearances of $ k$\n\n Call $ \\sigma(k) ; \\sigma'(k)$ be two positions of $ k$ in that sequence . Then the partition \n$ \\{ 1,\\sigma(1) \\plus{} n,\\sigma'(1) \\plus{} n \\} , \\{ 2,\\sigma(2) \\plus{} n,\\sigma'(2) \\plus{} n \\} ; ...;\\{ n,\\sigma(n) \\plus{} n,\\sigma'(n) \\plus{} n \\}$ will satisfies all condition of the problem\n\n ( You can have $ n \\equal{} 4k ; n \\equal{} 4k \\plus{} 1$ as sansuunoousama said )\n\n Very nice problem :lol: [/i]", "Solution_7": "[quote=\"Hamster1800\"] $ 8\\ 6\\ 7\\ 5\\ 1\\ 1\\ 6\\ 5\\ 8\\ 7\\ 12\\ 10\\ 11\\ 9\\ 4\\ 2\\ 3\\ 2\\ 4\\ 3\\ 10\\ 9\\ 12\\ 11$ for $ n \\equal{} 12$\n[/quote]\r\n\r\n[i] Seems to be vey nice , but it is a wrong construction\n\nFor example , in that string , there is $ \\underline{7}$ numbers between two appearances of $ 9$[/i]", "Solution_8": "Yes, you are right. Perhaps we can modify the induction by going from $ 4k$ to $ 4k\\plus{}1$ and then from $ 4k\\plus{}1$ to $ 4k\\plus{}4$, since the problem is that the spacing doesn't work out when adding four numbers, but rather when adding three numbers (as you can see when you have just $ 423243$). I'll try that tomorrow if nobody else gets to it first.", "Solution_9": "These sequences are known to exist (they are called Skolem Sequences and http://www.research.att.com/~njas/sequences/A014552 says that the number of them is positive for these cases), but I can't figure out a correct proof. Can anyone help?", "Solution_10": "I think that besides solving in that way , there is another proof because as Hamster1800 said : the sequence has been constructed only for small $ n$\r\n\r\n and that problem was created a long time ago . :lol:", "Solution_11": "You might want to check [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1054571#1054571]this[/url] out. The things the poster proves in the solution could be helpful in the general case." } { "Tag": [ "geometry", "perimeter" ], "Problem": "An equilateral triangle has a perimeter of 24 cm. How many square centimeters are in the area of the triangle formed by connecting the midpoints of the sides of the original triangle? Express your answer in simplest radical form.", "Solution_1": "Side is 8, so midline is 4. Area: $ \\frac{4^2\\sqrt3}{4}\\equal{}4\\sqrt3$" } { "Tag": [ "inequalities", "quadratics", "algebra", "algebra unsolved" ], "Problem": "This is part of my homework exercise. I am unable to understand how it is done:\r\n\r\nFor what values of k does the system have no solution:\r\n\r\n(2x - 1)(x + 3) = y\r\n-5x^2 + 17x - k = y\r\n\r\n\r\nMostly I do not know where to start.\r\nI believe it is a question of inequalities but i'm not sure.\r\nIf anyone could point me in the right direction.", "Solution_1": "[hide=\"Hint\"]First, eliminate $y$ from both equations and set them equal.\n$2x^2+5x-3=-5x^2+17x-k$\n\n$\\Longleftrightarrow 7x^2-12x+(k-3)=0$\n\nNow, we want to find the value(s) of $k$ such that the quadratic equation $7x^2-12x+(k-3)=0$ has no solutions. So...\n[hide=\"2nd hint\"]Use discriminant ;) [/hide][/hide]", "Solution_2": "Thank you, that explanation was great.\r\n\r\nI used the Discriminant to solve for k, ended up with 57/7. \r\nI tried to check 57/7 and it fails....\r\n\r\nThe Discriminant\r\nD = 0 -- two equal, real roots\r\nD > 0 -- two real roots\r\nD < 0 -- two complex roots\r\n\r\nWhat em I missing here?\r\nI really don't think it is possible to use the discriminant to solve for k. In that case then:\r\nD = 57/7\r\nWhich means there are two real roots.\r\n\r\nStill not sure how to get the values of the k... \r\nUnless it is k > 57/7" } { "Tag": [ "MATHCOUNTS" ], "Problem": "No, this post isn't the least bit sarcastic. I've been to MC Nats, I know how extremely tough it is to do well under such extreme pressure. Especially this year, with high expectations. Beating 52 other teams is no small feat. I know you guys probably feel like you should've done better, but the intangible stuff you've taken out of Mathcounts hopefully means much more than the tangible should-haves.\r\n\r\nAnyhow, I don't know individual rankings yet except for Justin, but good job guys!", "Solution_1": "Seconded. Really good job, guys.", "Solution_2": "what??!! 5th? that's pathetic!!!!!1111one1!1!cos(0)!!1", "Solution_3": "wow lol lyndon dam i thought he was going to win seriously.\r\noh well atleast he has a chance to go to blue mop...\r\nhmm yea 5th i know fanatic probably made this so the indiana team wouldnt feel disappointed, but i dont ever remembering indiana getting worst than 3rd in like 5 years or so...", "Solution_4": "Well... stupid mistake on #1 on the team round. #9: Justin accidentally thought 8*4=12... But we can't complain because he was the only one who could do it quickly enough. 1 more team question would've gotten us 2nd place (Texas had perfect on the team round).\r\n\r\nRankings: Mark: 18th (beast)\r\nMe: 52nd (ermmm. huge deprovement.)\r\nYoukow: 76th..", "Solution_5": "is it true that you missed every other sprint question?\r\n\r\ndoes ne one know how that one kid from virginia did. the one who used to live here. i believe that his aops name is unimpossible.", "Solution_6": "who, me? :P", "Solution_7": "yes, how did you do?", "Solution_8": "Congrats to Indiana, it really was fun hanging out with you guys pretty much the whole time. Even Lyndon :lol:", "Solution_9": "[quote=\"tan90=3/0\"]yes, how did you do?[/quote]\r\n\r\n218th :(", "Solution_10": "unimpossible got 33 = 30th place. \r\n\r\nAh. Mr. Frost is making fun of you again Lyndon.", "Solution_11": "yeah, dude......mathcrazed, is David Lu better than you at math? yeah, someone told me that if i say that, you'll get mad (and no it wasn't david)\r\n\r\nbtw: tan90=3/0, who r u?", "Solution_12": "[quote=\"frost13\"]Congrats to Indiana, it really was fun hanging out with you guys pretty much the whole time. Even Lyndon :lol:[/quote]\r\n\r\n\r\nu sure?\r\n\r\n\r\ni don't know why i'm posting on indiana forum...", "Solution_13": "[quote=\"cognos599\"]yeah, dude......mathcrazed, is David Lu better than you at math? yeah, someone told me that if i say that, you'll get mad (and no it wasn't david)[/quote]\r\n\r\nUm... was that person at Nats? And did his name begin with a U?", "Solution_14": "ummmmm lyndon gets mad if anyones better at him in math.......................................right?", "Solution_15": "T.T... no I get sad xD... lol. david lu. what a beast", "Solution_16": "[quote=\"unimpossible\"][quote=\"tan90=3/0\"]yes, how did you do?[/quote]\n\n218th :([/quote]\r\nyou liar i saw ur name on top 57 students", "Solution_17": "[quote=\"mathcrazed\"]unimpossible got 33 = 30th place. [/quote]\r\n\r\nWINK WINK :wink:", "Solution_18": "YAY WE CAUGHT THE LIAR UNIMPOSSIBLE :D :D :D", "Solution_19": "alabama, massa, michigan, illinois, texas, and ohio have all invaded our forum.", "Solution_20": "[quote=\"mathcrazed\"]T.T... no I get sad xD... lol. david lu. what a beast[/quote]\r\n\r\nLol, david lu sucks at math.", "Solution_21": "[quote=\"cognos599\"]alabama, massa, michigan, illinois, texas, and ohio have all invaded our forum.[/quote]\r\n\r\nwe should try to beat missouri.\r\n\r\nEDIT: no wait wait, JAPAN HAS INVADED INDIANA", "Solution_22": "Missouri sucked this year, I blame it on my complete suckage of #10. Next year, when I inject sugar and caffeine into my veins immediately before th competition is when we will beast. Texas shall fall before us. Indiana too.", "Solution_23": "hoosiers run on corn power <3", "Solution_24": "Ohio >>>> Indiana.\r\n\r\nYou guys have black highways", "Solution_25": "[quote=\"cognos599\"]alabama, massa, michigan, illinois, texas, and ohio have all invaded our forum.[/quote]\r\n :D :D :D\r\n\r\nO_0 but how do u know im from illinois?????? hmmmmm some1's being a suspicious stalker...>_>", "Solution_26": "how did you know that i was talking to you, when i said illinois..... but now that you said it, i know you are from illinois.", "Solution_27": "[quote=\"cognos599\"]how did you know that i was talking to you, when i said illinois..... but now that you said it, i know you are from illinois.[/quote]\r\nlol :P :P :P" } { "Tag": [ "floor function", "function" ], "Problem": "define $ f: \\mathbb{N}\\rightarrow \\mathbb{N}^*$ such that $ f(m\\plus{}n)\\minus{}f(n)\\minus{}f(m)$ is either $ 0$ or $ 1$ for all $ m,n$.\r\nGiven that $ f(2)\\equal{}0, f(3)>0, f(9999)\\equal{}3333$ , find $ f(2008)$.\r\n\r\n\r\n$ \\mathbb{N}^*$ is the natural numbers joined with zero.", "Solution_1": "[hide=\"solution\"]I transform the recursion into $ f(m+n)=f(m)+f(n)+s$ for $ s\\in{0,1}$. Looking at $ f(1+1)$, we quickly see that $ f(1)=0$, and then we find $ f(3)=1$. Looking at both $ f(1+3)$ and $ f(2+2)$, we see $ f(4)=1$. Using the first three values of $ f$ and the recursion, we quickly derive the following:\n\n$ f(n+1)=f(n)$ or $ f(n+1)=f(n)+1$\nIf $ f(n-1)\\nef(n-2), f(n)=f(n-1)$.\nIf $ f(n-1)=f(n-2)=f(n-3), f(n)=f(n-1)+1$\nIf $ f(n-1)=f(n-2)\\nef(n-3), f(n)=f(n-1)$ or $ f(n)=f(n-1)+1$\n\nIn the latter case, if $ f(n)=f(n-1)$ always, then $ f(n)=\\lfloor\\frac{n}{3}\\rfloor$, $ f(2008)=669$, and $ f(9999)=3333$. Every instance where, in the latter case, $ f(n)=f(n-1)$ translates the remainder of the function one to the left. This can be done once or twice in order to maintain $ f(9999)=3333$. If done twice, then $ f(2008)=670$. So $ f(2008)$ is either $ 669$ or $ 670$.[/hide]", "Solution_2": "[hide=\"question\"]\n\nBut $ f(9999) \\ge f(9997) \\ge 4f(2008) \\plus{} 655f(3)$\n\nIf $ f(2008) \\equal{} 670$, wouldn't this lead to $ f(9999) \\ge 3335$?\n[/hide]", "Solution_3": "[quote=\"Darmani\"][hide=\"solution\"]I transform the recursion into $ f(m + n) = f(m) + f(n) + s$ for $ s\\in{0,1}$. Looking at $ f(1 + 1)$, we quickly see that $ f(1) = 0$, and then we find $ f(3) = 1$. Looking at both $ f(1 + 3)$ and $ f(2 + 2)$, we see $ f(4) = 1$. Using the first three values of $ f$ and the recursion, we quickly derive the following:\n\n$ f(n + 1) = f(n)$ or $ f(n + 1) = f(n) + 1$\nIf $ f(n - 1)\\nef(n - 2), f(n) = f(n - 1)$.\nIf $ f(n - 1) = f(n - 2) = f(n - 3), f(n) = f(n - 1) + 1$\nIf $ f(n - 1) = f(n - 2)\\nef(n - 3), f(n) = f(n - 1)$ or $ f(n) = f(n - 1) + 1$\n\nIn the latter case, if $ f(n) = f(n - 1)$ always, then $ f(n) = \\lfloor\\frac {n}{3}\\rfloor$, $ f(2008) = 669$, and $ f(9999) = 3333$. Every instance where, in the latter case, $ f(n) = f(n - 1)$ translates the remainder of the function one to the left. This can be done once or twice in order to maintain $ f(9999) = 3333$. If done twice, then $ f(2008) = 670$. So $ f(2008)$ is either $ 669$ or $ 670$.[/hide][/quote]\r\n\r\nYou wrote:\r\n$ f(n + 1) = f(n)$ or $ f(n + 1) = f(n) + 1$\r\nIf $ f(n - 1)\\nef(n - 2), f(n) = f(n - 1)$.\r\nIf $ f(n - 1) = f(n - 2) = f(n - 3), f(n) = f(n - 1) + 1$\r\nIf $ f(n - 1) = f(n - 2)\\nef(n - 3), f(n) = f(n - 1)$ or $ f(n) = f(n - 1) + 1$\r\n\r\n\r\nit is actually,\r\n$ f(n + 1) = f(n)$ or $ f(n + 1) = f(n) + 1$\r\nIf $ f(n - 1)\\not =f(n - 2), f(n) = f(n - 1)$.\r\nIf $ f(n - 1) = f(n - 2) = f(n - 3), f(n) = f(n - 1) + 1$\r\nIf $ f(n - 1) = f(n - 2)\\not =f(n - 3), f(n) = f(n - 1)$ or $ f(n) = f(n - 1) + 1$\r\nright?", "Solution_4": "Exactly, thanks. I was pretty sure the command for $ \\not \\equal{}$ was \\ne, but I was clearly wrong.\r\n\r\nThanks ffao for catching that; I figured something was wrong when I got two answers.", "Solution_5": "$ \\ne$ is \\ne. The problem with your code was that you attached the \"f\" (from f(n-3)) to it. (\\neq also works.)" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find $f: Z\\to Z$ sastifying $f(f(x))=1-x^{4}$", "Solution_1": "Nice functional equation.... :) \r\nWe have $f(f(f(x)))=1-(f(x))^{4}$ so $f(1-x^{4})=1-(f(x))^{4}$ (1)\r\nPutting x->0,1 in (1) we get \r\n$f(1)=1-(f(0))^{4}$ (2) and $f(0)=1-(f(1))^{4}$ (3)\r\n\r\n(2)-(3) => $f(1)-f(0)=((f(1))^{2}+(f(0))^{2})(f(1)-f(0)(f(1)+f(0))$\r\n\r\nIf $f(1)=f(0)$ the relation (2) gives that $(f(1))^{4}+f(1)-1=0$ which has no roots in $Z$ so $f(1)\\neq f(0)$\r\nso $((f(1))^{2}+(f(0))^{2})(f(1)+f(0))=1$ so \r\n$(f(x))^{2}+(f(0))^{2}=1$ and $f(1)+f(0)=1$ so $f(1)=1-f(0)$ (4)\r\n\r\nPutting (4) in (2) yields that $f(0)=(f(0))^{4}$ so $f(0)=0$ or \r\n$f(0)=1$ .\r\n\r\n**If $f(0)=0$ then putting x=0 in $f(f(x))=1-x^{4}$ yields $0=1$ acontradiction.\r\n\r\n**If $f(0)=1$ then by (4) $f(1)=0$ . Putting x=0 in \r\n$f(f(x))=1-x^{4}$ yields $0=1$ a contradiction again \r\n\r\nSo a function like this one does not exist .\r\nI hope I am right", "Solution_2": "yes!! your solution is the same as mine\r\nan easier problem but we can see the way to have a solution for this problem \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=110387\r\n\r\nanyone have the solution for general problem! it will be an interesting solution! and it's hard to have it :roll:" } { "Tag": [ "LaTeX" ], "Problem": "Help!!!\r\n\r\nI tried to typeset a pdf in TeXnic, and I couldn't, because at the end of the page, my sentence couldn't fit, and I don't know how to start a new page. It accounted for six errors. So, how do you start a new page?\r\n\r\nHelp!!! :huh:", "Solution_1": "LaTeX will normally insert page breaks automatically.\r\n\r\nYou're going to have to provide more specific details in order for anybody to be able to assist you.", "Solution_2": "To start a new page, insert the \\pagebreak command. So like:\r\n\r\n[code]blahblahblahblah\n\n\\pagebreak\n\nblahblahblahblah[/code]\r\n\r\n\r\n\r\nI don't know if that's what you are asking, but it's worth a shot.", "Solution_3": "The command \\newpage also works fine!" } { "Tag": [], "Problem": "Prove for every natural n (n > 2) :\r\n\\[\\begin{array}{l}\\\\ \\sum_{i = 1}^{n-1}{i\\left( \\begin{array}{l}n \\\\ i+1 \\\\ \\end{array}\\right)(n-1)^{n-i-1}}= (n-1)^{n}\\\\ \\end{array}\\]", "Solution_1": "asoone ya sakht ? :maybe:" } { "Tag": [ "AMC" ], "Problem": "Doug can paint a room in $ 5$ hours. Dave can paint the same room in $ 7$ hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let $ t$ be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by $ t$?\r\n\r\n$ \\textbf{(A)}\\ \\left(\\frac{1}{5}\\plus{}\\frac{1}{7}\\right)(t\\plus{}1)\\equal{}1 \\qquad\r\n\\textbf{(B)}\\ \\left(\\frac{1}{5}\\plus{}\\frac{1}{7}\\right)t\\plus{}1\\equal{}1 \\qquad\r\n\\textbf{(C)}\\ \\left(\\frac{1}{5}\\plus{}\\frac{1}{7}\\right)t\\equal{}1 \\\\\r\n\\textbf{(D)}\\ \\left(\\frac{1}{5}\\plus{}\\frac{1}{7}\\right)(t\\minus{}1)\\equal{}1 \\qquad\r\n\\textbf{(E)}\\ (5\\plus{}7)t\\equal{}1$", "Solution_1": "[hide]\nEach hour they work, they can paint $ \\frac{1}{5}\\plus{}\\frac{1}{7}$ of the room. They work for $ t\\minus{}1$ hours. So the correct equation is $ D$.\n[/hide]", "Solution_2": "[hide]WLOG, let the total work be 1. Then the rates of Doug and Dave are $ \\frac {1}{5}$ and $ \\frac {1}{7}$, respectively. Since they took an hour off, we have the equation\n\n$ \\left(\\frac {1}{5} \\plus{} \\frac {1}{7}\\right)(t \\minus{} 1) \\equal{} 1 \\Leftrightarrow \\boxed{D}$, as desired.[/hide]", "Solution_3": "[hide=solution]Each hour Doug paints $\\cfrac{1}{5}$ of the room and Dave $\\cfrac{1}{7}$ of the room. Combined, they paint $\\cfrac{1}{5}+\\dfrac{1}{7}$ of the room. They work for $t-1$ hours, as they do not work during lunch. Therefore the correct answer is \\[\\boxed{\\textbf{(D)} ~\\left(\\cfrac{1}{5}+\\dfrac{1}{7}\\right)\\left(t-1\\right)=1}\\]" } { "Tag": [ "modular arithmetic", "pigeonhole principle", "geometry", "rectangle" ], "Problem": "1) Show that for every n positive there exist n integers consecutives so that no one is prime or power of prime.\r\n\r\n2) Given a 4*n chessboard show that there is no n for which a knight can touch every square only one time.\r\n\r\n3) Given wichever set X of ten numbers in (10;99) it is possible to choose two sub-sets from X so that their intersection is 0 e they have the same sum.\r\n\r\n[Sorry for my terrible english, but I'll improve it, promised)", "Solution_1": "1) Is a classical consequence of the chinese remainder theorem.\r\n\r\n2) Color the $4 \\times n$ as a classical checkoard. And consider the set $E$ of the two exterior rows, and the set $I$ of the two interior rows, to prove that if a circuit does exists then the unit squares which belong to one of this set all have the same color, which is clearly absurd.\r\n\r\n3) smells like a classical pigeon-hole problem.\r\n\r\nPierre.", "Solution_2": "Is there anyone who can formalize the solutions 1 and 3, plz?", "Solution_3": "For the first problem, you can take, for example, some sequence of distinct primes $p_1,p_1',p_2,p_2',\\ldots,p_n,p_n'$. By the Chinese Remainder Theorem, thee is a number $k$ s.t. $k\\equiv -i\\pmod{p_i\\cdot p_i'},\\ \\forall i\\in\\{1,2,\\ldots,n\\}$. This simply means that $k+i$ is divisible by two distinct primes $p_i,p_i'$ (for all $i$ from $1$ to $n$), so it's not a prime power.\r\n\r\nI think Pierre was referring to something similar to this.", "Solution_4": "For 3. \r\nConsider all 1023 distinct non-empty subsets of $X$. Sum of the elements in each of these subsets is in $[10,99\\cdot 10]$ - less than 1023 different values, so due to Pigeonhole principle we conclude that two distinct subsets have equal sums. Excluding common elements from them we obtain two disjoint subsets of $X$ with the same sum.", "Solution_5": "Grobber : Not exactly ;) I tought about $n = p_i - i \\mod [p_i^2]$ for sufficientely large $n$.\r\n\r\nPierre.", "Solution_6": "Myth : That's exactly what I thought. :P \r\n\r\nPierre.", "Solution_7": "about you previous message, Pierre:\r\n\r\nIt is strange, because my idea was exactly the same as grobber's one :)\r\n :lol: :lol: :lol:", "Solution_8": "That's why I said \"something [b]similar[/b] to this\" :). I knew Pierre had something tricky up his sleeve :D.", "Solution_9": "1) Which mxn rectangles are covered by L-tetrominoes?\r\n2) Show that (1+2+3+...+n) divides (1^k + 2^k + 3^k + ... + n^k) for odd k.", "Solution_10": "1) $8\\,|\\,mn$, $m,n>1$.", "Solution_11": "[quote=\"Marzapane\"]2) Show that (1+2+3+...+n) divides (1^k + 2^k + 3^k + ... + n^k) for odd k.[/quote]\r\n\r\nwe know that (1+2+3+...+n)=n(n+1)/2. We know also that, for odd k we have:\r\nx|2*(1^k+2^k+3^k+...+(x-1)^k)=(1^k+(x-1)^k)+(2^k+(x-2)^k)...\r\n\r\nSo we have n+1|2(1^k+2^k+3^k+...+n^k) and \r\nn|2(1^k+2^k+3^k+...+(n-1)^k)+2(n^k). So, because (n,n+1)=1 we have n(n+1)/2|(1^k + 2^k + 3^k + ... + n^k) for odd k", "Solution_12": "Do you like Freddie Mercury, Simo_the_Wolf?", "Solution_13": "[quote]My soul is painted like the wings on butterflies, \nfairytales of yesterday will grow but never die, \nI can fly, my friends!! \n\n-F.Mercury-[/quote]\r\n\r\nThe Show Must Go On?", "Solution_14": "Yes, indeed!", "Solution_15": "Ok guys, you're so \"fast and furious\"! :) I like the math of domino so ...\r\n \r\nFor wich n an nxn grid can be tiled by 3x1 tiles sparing only three vertices ?", "Solution_16": "I didn't understand it :blush:", "Solution_17": "I think (not sure) he means this: you have an $n\\times n$ board. For which $n$ can you cover the entire board except for three corners (???) with $3\\times 1$ rectangles?", "Solution_18": "Ops, ya , grobber's version is right and i'm definitively a wash-out at speaking english :blush: . I beg your pardon ;)", "Solution_19": "There are no such $n$. :?", "Solution_20": "Right, but prove it :P so that everybody can learn such a beutiful math-mith branch :D .", "Solution_21": "$n$ is necessarily a multiple of $3$.\r\n\r\nAssume we want to leave the corners $(1,n),(n,1),(n,n)$ uncovered. Color the first line $1,2,3,1,2,3,\\ldots$, the second line $2,3,1,2,3,1,\\ldots$, the third line $3,1,2,3,1,2,\\ldots$, the fourth line $1,2,3,1,2,3,\\ldots$ and so on (I'm sure you can see the pattern). Each $3\\times 1$ \"domino\" covers a $1$, a $2$ and a $3$. There are equally many $1$'s, $2$'s and $3$'s on the board, so the three uncovered corners must have $1,2,3$ written in them. However, both $(1,n)$ and $(n,1)$ have a $3$, so we have a contradiction.", "Solution_22": "Yes, it is exactly my solution :)", "Solution_23": "There are N (with N>2 and odd) adjacent squares along a single line ( so Nx1), infinite 2 x 1 dimini and two players A end B. A puts down his domino for first. Who win?", "Solution_24": "Who lose in your game? The player, which cannot perform his turn. Am I right?\r\n\r\nBTW, you should make the separate topic for this problem ;)", "Solution_25": "$A$ wins for all $N\\neq 1\\pmod{4}$.", "Solution_26": "Myth, your a stud man.", "Solution_27": "mmm .... try with 13 ! :?", "Solution_28": "up, please", "Solution_29": "In your question about the 4 by n chessboard you say that it is always impossible for a knight to reach each square. I think it is possible when n=3. I think that the problem with your proof is that you can move from the interior to the interior. Before he moves from the inside to the inside he only visits squares of one color in E and the other color in I. After the knight visits the first 2n squares, he must move from the inside to the inside so that he can start visiting squares of the opposite colors." } { "Tag": [ "probability" ], "Problem": "Bob and Joe are playing a game where Bob roles three dice and Joe roles one die. If Joe's die is higher or equal to Bob's highest die, Joe wins. What is the probability of Joe winning? ;)", "Solution_1": "[quote]Bob and Joe are playing a game where Bob rolls three dice and Joe rolls one die.\nIf Joe's die is higher or equal to Bob's highest die, Joe wins.\nWhat is the probability of Joe winning?[/quote][hide]\nThere are $6^4 = 1296$ ways that the four dice can turn up.\n\nIf Joe rolls a 6, Bob can roll any number on all three dice.\nThere are $6^3 = 216$ ways this can happen.\n\nIf Joe rolls a 5, Bob must roll 5 or less on all three dice.\nThere are $5^3 = 125$ ways this can happen.\n\nIf Joe rolls a 4, Bob must roll 4 or less on all three dice.\nThere are $4^3 = 64$ ways this can happen.\n\nIf Joe rolls a 3, Bob must roll 3 or less on all three dice.\nThere are $3^3 = 27$ ways this can happen.\n\nIf Joe rolls a 2, Bob must roll 2 or less on all three dice.\nThere are $2^3 = 8$ ways this can happen.\n\nIf Joe rolls a 1, Bob must roll 1 on all three dice.\nThere is $1^3 = 1$ ways this can happen.\n\nHence, Joe will win in $216 + 125 + 64 + 27 + 8 + 1 = 441$ of the outcomes.\n\nTherefore: $P(\\text{Joe wins}) \\;= \\;\\frac{441}{1296} \\;= \\;\\frac{49}{144}$[/hide]", "Solution_2": "Bob and Joe are playing a game where Bob roles three dice and Joe roles one die. If Joe's die is higher or equal to Bob's highest die, Joe wins. After rolling, Joe and Bob notice that all 4 die have different numbers showing. What is the probability that Joe has won?", "Solution_3": "[quote]Bob and Joe are playing a game where Bob roles three dice and Joe roles one die. If Joe's die is higher or equal to Bob's highest die, Joe wins. After rolling, Joe and Bob notice that all 4 die have different numbers showing. What is the probability that Joe has won?[/quote]\r\n[hide=\"I probably got this wrong\"]\nIf Joe gets a 6, Bob can roll 5 x 4 x 3, or 60 ways\nIf Joe gets a 5, Bob can roll 4 x 3 x 2 ways, or 24 ways\nIf Joe gets a 4, Bob can roll 3 x 2 x 1 ways, or 6 ways\nIf Joe gets a 3 or lower, Bob automatically wins\n\n90/(6x5x4x3) = 90/360 = [b]1/4[/b]\n[/hide]", "Solution_4": "Good. Can you find a way to do it more quickly?", "Solution_5": "Any one played here a game called \"Risk\".. If yes, you would know such proablities by heart... :)", "Solution_6": "[quote=\"JBL\"]Treething: you're still dividing by 216. But there aren't 216 actual possibilities any more.\n\nAlso: can you find a way to do it more quickly?[/quote]\r\n\r\nNo I'm not, silly :P ." } { "Tag": [ "modular arithmetic" ], "Problem": "What is the sum of all two-digit positive integers whose squares end with the digits 01?", "Solution_1": "It must end in a 1 or a 9\r\n\r\nTesting to see if ends in 01-\r\n\r\n11-NO 121\r\n19-NO 341\r\n21-NO 441\r\n29-NO 841\r\n31-NO 961\r\n39-NO 1521\r\n41-NO 1681\r\n49-YES-2401\r\n51-YES 2601\r\n59-NO 3481\r\n61-NO 3721\r\n69-NO 4761\r\n71-NO 5041\r\n79-NO 6241\r\n81-NO 6561\r\n89-NO 7921\r\n91-NO 8281\r\n99-YES 9801\r\n\r\nWhether or not its a pattern thats there is 1,49,51,99, i dont know. But its 49+51+99=199", "Solution_2": "We must find all two digit numbers $ 10a \\plus{} b$ such that $ 100a^2 \\plus{} 20ab \\plus{} b^2\\equiv20ab \\plus{} b^2\\equiv1\\pmod{100}$.\r\nSince $ b \\equal{} 1,9$ our two cases are:\r\ni. $ b \\equal{} 1$\r\n\r\n$ 20a \\plus{} 1\\equiv1\\pmod{100}\\implies20a\\equiv0\\pmod{100}\\implies a\\equiv0\\pmod{5}$.\r\n\r\nThis means that $ 51$ is the only possible answer for this case.\r\n\r\nii. $ b \\equal{} 9$\r\n\r\n$ 180a \\plus{} 81\\equiv1\\pmod{100}\\implies80a\\equiv \\minus{} 80\\pmod{100}$\r\n\r\n$ \\implies a\\equiv \\minus{} 1\\equiv4\\pmod5$.\r\n\r\nThis means that $ 49,99$ are the only possibilities.\r\n\r\n$ 49 \\plus{} 51 \\plus{} 99 \\equal{} \\boxed{199}$." } { "Tag": [], "Problem": "Prove true for all $x\\in\\Re$ in the following equation - \r\n\\[ x^6-6x+5\\geq 0 \\]", "Solution_1": "[quote=\"eddie77\"]Prove true for all $x\\in\\Re$ in the following equation - \n\\[ x^6-6x+5\\geq 0 \\][/quote]\r\n\r\n$\\Longleftrightarrow (x-1)^2((x^2+2)(x+1)^2+3)\\geq 0$\r\n\r\nwhich is obvious", "Solution_2": "[hide]let $f(x)=x^6-6x+5$ then $f'(x)=6x^5-6=6(x^5-1)$ so $f'(x)<0$ for $x<1$ and $f'(x)>0$ for $x>1$ this means that $f(x) \\geqq f(1)=0$ for all $x$[/hide]", "Solution_3": "Just found an simple prove by AM-GM\r\n\r\n$x^6+1+1+1+1+1\\geq 6\\sqrt[6]{x^6}=6|x|\\geq 6x$ \r\n\r\nhence $x^6-6x+5\\geq 0$", "Solution_4": "that's a nice one shyong..:)", "Solution_5": "Yes, what an elegant proof!" } { "Tag": [ "MATHCOUNTS", "number theory", "least common multiple", "greatest common divisor", "prime factorization" ], "Problem": "The least common multiple of 54 and some natural number N is 432.\r\n\r\nThe greatest common divisor of 54 and N is 6.\r\n\r\nWhat is N?", "Solution_1": "48. Oops, forgot the solution :mrgreen: \r\n\r\n432/54 is 8. The prime factorization of 432 is 3 :^3: (2 :^4: ), which is equal to 9(6)(8). 9(6) is 54, so N must have 8/2 :^3: in it. If the gcf is 6, the least N can be is 48, which satisfies all the requirements.", "Solution_2": "s0mp,\r\n\r\nWhat can you say about the GCD and LCM of any two numbers?", "Solution_3": "Correct me if I'm wrong, but by the looks of it, the gcd is the product of the common prime factors of the two numbers, with the condition that you choose the lowest power possible of each factor. \r\n\r\nAnd for the lcm, its just the product of the highest power of each factor of both numbers, but if there is a 2 :^3: in one and 2 :^2: in the other, you choose the 2 :^3:.....i think. \r\n\r\nJust to clarify this as-clear-as-mud post, an example is 8(2 :^3:) and 76(19*2 :^2:) . The lcm is 19*2 :^3:, and the gcd would be 2 :^2:. :?", "Solution_4": "And since you choose the higher power for LCM and lower for GCD, what do you get when you multiply the LCM and GCD together?", "Solution_5": "Ahh.. its the product of the two numbers in question\r\n :D I see what that implies...that because gcd(lcm)=N1(N2), as this relates to your challenge question, 6(432)=54N, so all you have to do is divide by 54 and you get your answer...i get it", "Solution_6": "Bingo! Nice work.", "Solution_7": "using this formula thingie i got from my broski a few years ago the the \r\ngcd x lcm=xy\r\nso the gcd is 6 and the lcm is 432 and one of the numbers is 54 so plugging that into the equation \r\n\r\n6 x 432=(54)(y)\r\n2592=(54)(y)\r\ndivide both sides by 54 using simple algebra to get\r\n48=y\r\nvala\r\nthats a nifty little thing that one should rememorize (yes i spelled that on purpose)\r\nso remember gcd x lcm=xy\r\nokay then. bring on the next problem. nuff said", "Solution_8": "Noooooooo!\r\n\r\nDon't memorize! Understand why it's true (read above).\r\n\r\nUnderstanding is far more flexible and powerful than memorization.", "Solution_9": "that too. i figured out why and such. but yeah. oh nm. nuff said", "Solution_10": "[quote=\"MCrawford\"]Noooooooo!\n\nDon't memorize! Understand why it's true (read above).\n\nUnderstanding is far more flexible and powerful than memorization.[/quote]\r\n\r\nI think that's true, although I learned it the hard way.", "Solution_11": "of course one should always know how. sorry about that\r\n\r\n\r\nChinaboiiiiiiiiii,\r\n what did you learn the hard way?", "Solution_12": "I spent almost 2 weeks on one subject (my mom was teaching me), cause I tried to memorize the formula, but then the questions were always changing from this to that.", "Solution_13": "Yeah. Basically thats the difference between 'school math' and 'problem solving math'. At school you memorise the formulas and know when to use them, its all about whether you can use them or not. In problem solving its all about thinking about to do the problem, you told how. Once you are told how to do it the problems are (almost) always simple. Even international mathematical olympiad problems. Memorizing formulas never really helps (apart from things like double angle formulas and stuff like that.)", "Solution_14": "im sorry. if anything i hate the math school teaches cause they barely get halfway through the book ranting about some topic that has nothing to do to school. at least thats how it is in oklahoma. probably why oklahoma never does well in mathcounts. lool. its just like in geometry. if you understand the proofs, its pretty easy from there knowing the theorems and stuff.", "Solution_15": "lcm*gcd = 54*n\r\n\r\nn = 432*6/54 = 48" } { "Tag": [ "calculus", "integration", "trigonometry", "function", "limit", "calculus computations" ], "Problem": "Study the convergence of the integral\r\n\\[ \\int_1^{ \\plus{} \\infty } {\\sin \\left( {1 \\plus{} x \\plus{} x^2 \\plus{} ... \\plus{} x^{2007} } \\right)dx}.\r\n\\]\r\nI will post my solution at weekend. Have fun :roll: .", "Solution_1": "It converges. (Conditionally, of course.) Just do an integration by parts.\r\n\r\n$ \\int_1^{\\infty}\\sin(1 + x + \\cdots + x^{2007})\\,dx =$\r\n\r\n$ \\left.\\frac { - \\cos(1 + x + \\cdots + x^{2007})}{1 + 2x + \\cdots + 2007x^{2006}}\\right|_1^{\\infty} - \\int_1^{\\infty}\\frac {(2 + 3\\cdot2x + \\cdots + 2007\\cdot2006x^{2005}) \\cos(1 + x + \\cdots + x^{2007})}{(1 + 2x + \\cdots + 2007x^{2006})^2}\\,dx$\r\n\r\nThat second integral converges absolutely by comparison with a constant times $ x^{ - 2007}.$\r\n\r\nNote that these denominators have no zeros for $ x>0.$ We need that.", "Solution_2": "I suggest to use Abel test or Dirichlet test.", "Solution_3": "You can study a more general problem:\r\n\r\nLet $ f$ be a differentiable function, monotonic increasing on the inverval $ (1,\\infty)$ such that $ \\lim\\limits_{x\\rightarrow \\infty}f(x)=\\infty$. \r\n\r\nStudy the convergence of the following integral:\r\n\r\n$ \\int\\limits_{0}^{\\infty}\\sin(f(x))dx$.\r\n\r\nRemark: When $ a=0$ and $ f(x)=x^{2}$ we get the famous Fresnel integral $ \\int\\limits_{0}^{\\infty}\\sin(x^{2})dx$.", "Solution_4": "Copying the proof I gave above for the more general question will leave us with\r\n\r\n$ \\left.\\frac { \\minus{} \\cos(f(x))}{f'(x)}\\right|_0^{\\infty} \\minus{} \\int_0^{\\infty}\\frac {f''(x)\\cos(f(x))}{(f'(x))^2}$\r\n\r\nIn order to make this work, we would need to know something about the rate of growth not just of $ f$ but also of $ f'.$", "Solution_5": "You are right Kent.\r\n\r\nHere is the general case: \r\n\r\nThe integrals $ \\int\\limits_{a}^{\\infty}\\cos f(x)dx$ and $ \\int\\limits_{a}^{\\infty}\\sin f(x)dx$ converge if $ f'$ is monotonically increasing and $ \\lim\\limits_{x\\rightarrow \\infty}f'(x)\\equal{}\\infty$." } { "Tag": [ "geometry", "3D geometry", "pyramid", "ratio" ], "Problem": "A regular pyramid with a square base has edges of length 6 inches and height 4 inches. What is the ratio of the number of cubic inches in the volume of the pyramid to the number of square inches in its surface area?", "Solution_1": "Volume ($ \\frac {1}{3} B*h$)= $ \\frac {1}{3}(6^2)(4) \\equal{} 48$. \r\nSurface area ($ B \\plus{} 4 lateral sides' area$*)= $ 36 \\plus{} 4(5) \\equal{} 56$. \r\n\r\n$ \\frac {48}{56} \\equal{} \\fbox{6/7}$. \r\n\r\n*We find the slant height for the height of the triangle with the Pythagorean formula: half an edge of base ($ \\frac {1}{2}(6)$) is $ 3$. Height is $ 4$. $ 3^2 \\plus{} 4^2 \\equal{} 5^2$. \r\n\r\nEDITE: oh...whoops. :oops: Thanks for catching it!", "Solution_2": "[quote=\"r15s11z55y89w21\"]Volume ($ \\frac {1}{3} B*h$)= $ \\frac {1}{3}(6^2)(4) \\equal{} 48$. [/quote]\r\n\r\nCorrect.\r\n\r\n[hide=\"However...\"]\n\nSA = $ 36 \\plus{} 6 \\times 5 \\times 4 \\div 2$ or $ 96$\n\nThus, our answer is $ \\boxed{\\frac12}$.[/hide]" } { "Tag": [ "probability", "algebra proposed", "algebra" ], "Problem": "There are 6 doors, each one having either 1 prize behind it or 1 goat. There are a total of 4 goats and 2 prizes. By the way, a goat is different from a prize. The doors are closed and you cannot see what is behind them. You pick two of the six doors and they are fixed. I open two other doors and reveal two goats behind them. \r\n\r\nQuestion: If you are required to switch BOTH of your picked doors together in one move (and one move only), in favor of the two remaining doors to try to increase your chances of having both prizes, would you switch your two doors?", "Solution_1": "Yes :lol:", "Solution_2": "The probability you got both 0 prizes first off (4/6)(3/5)=(2/5)\r\nTherefore, after you swictch your doors, the probability you get the 2 is 3/5, much great than the 1/15 of you getting both of them off the mark..." } { "Tag": [ "calculus", "MATHCOUNTS", "USAMTS", "AMC 8", "AMC 10", "AMC" ], "Problem": "Has anyone here ever started their own math team? My school doesn't have a math team, and I have been trying to convince other students to do the AMC-12 exam, which is already extremely difficult. Furthermore, the math department leader at my school has been extremely discouraging (i.e. repeatedly telling me that no one will want to do it) and she has put the burden of paying all costs on any students who want to join. I have tried my best to ask everyone I can, but most people just don't want to. This isn't surprising since most of them associate math contest with math class -- > homework --> boring! Any tips?", "Solution_1": "Go for math teachers -- if the head of the department is being difficult, first try and line up a teacher to help you with it. You probably won't be able to get anything big, but if you can establish something with 5 or 6 people, that would be a good start. Get permission to put up posters, if your school lets you do this. Try something with an interesting problem or mathematical fact on it to try and get interested people to come to a meeting. Those would be my first suggestions. If any of that works, go talk to the department head again and show her what you've got.", "Solution_2": "Go for it! I started a math team at my school this year, and while we haven't done extremely well on, say, mandelbrot, it's a start.", "Solution_3": "And don't worry if your math team turns out to be really small. Our math team has about 25 people per grade, but there should only be about 5 people in my class. You can go faster without having to worry about the others who are behind (most of the people in my class just want to pad their GPA). Also, if you want, you could probably compete in math tournaments at local schools by yourself. Usually they aren't very expensive, and I did this one time in sixth grade when we were just starting a math team. I also agree with Joel. Show your math teacher that you are very interested in math by asking them questions or showing them interesting problems you figured out. Our math team was not student-started, but most of the people see that math is fun and cool once the \"math team season\" :D starts (which is about mid January here). \r\nGood luck!", "Solution_4": "[quote=\"mathnovice\"]Has anyone here ever started their own math team? . . . . Any tips?[/quote]\r\n\r\nI was the first coach for a new math team, beginning two years ago. A parent in our homeschooling group had a son in the UMTYMP \r\n\r\nhttp://www.itcep.umn.edu/umtymp/ \r\n\r\ncalculus class here in Minnesota, and the son had heard about math competitions from classmates. She wanted her son to have a chance to join competitions, and I was willing to give a whirl at being coach if my son (then fourth-grade age) could tag along. We first called friends and friends of friends to scratch up (barely) a six-person team for Minnesota Junior High Mathematics League \r\n\r\nhttp://members.aol.com/mathleague/ \r\n\r\nwith the families splitting the cost of registering one team. That was a pretty good experience. Some of the kids on my team were quite good, although not experienced in competition problems, but it was challenging for everybody. The MN Junior High Mathematics League founder recommends that everyone sign up for AMC 8 and for MATHCOUNTS, but we didn't that year--all this stuff was too new for us. At the end of the year, one of the parents gave me a thank-you gift: the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0425179834/learninfreed]Math Coach: A Parent's Guide to Helping Children Succeed in Math[/url] by Wayne Wickelgren, which has a lot of good tips about forming a team. I suppose this site should link to that book. \r\n\r\nLast year we did MN Junior High Math League again, with some changes in personnel, and we had five kids try the AMC 8. I was afraid our group of \"ringers\" (all recruited from a gifted student homeschooling group) would be accused of cheating, because they were in a good position to score well, but we had good proctoring from a public school in my district, and when I got the official score sheets I smiled, because each test-taker made a different pattern of mistakes, so it was pretty clear they weren't cheating off of one another. One guy in our group tied the top score in MN last year (23), and three of the five were Honor Roll, with one of the two others being Merit Roll. AGAIN we didn't try MATHCOUNTS--my daughter was just being born then, and I was too tired to take on another project. But later in the school year another parent in our group registered a team for Math Masters of Minnesota \r\n\r\nhttp://www.mathmastersmn.org/ \r\n\r\nand we entered sixth-grade and fifth-grade teams in that contest, with the fifth-grade team doing better, because it PREPARED after the sixth-grade team went in too overconfident and got creamed. (The two meets were about a month apart.) \r\n\r\nThis year we are in the MN Junior High Math League, where it all began, fielding a short-handed :cry: team that scores very well in the team part of the competition. I have just registered a team for MATHCOUNTS--but I don't know who all four of my team members will be. We just did the AMC 8, and I hope to join another homeschooling parent for her AMC 10 administration. And I HOPE we can field a Math Masters of Minnesota team (or two) again this year. \r\n\r\nNext year I'd like to get some of us into USAMTS, Mandelbrot, and other contests. \r\n\r\nHope this helps!" } { "Tag": [ "algebra", "polynomial", "algorithm", "induction", "algebra proposed" ], "Problem": "Let $P(x)$ be a polynomial with degree $ n\\in{\\mathbb{N}}$. If $ P(0),\\ P(1),\\ \\ldots ,\\ P(n)$ are integers, then prove that $ P(k)$ is integer for any integer $k$.", "Solution_1": "An algorithm for constructing the values of $ P$:\r\nLet $ \\delta P(k)\\equal{}P(k\\plus{}1)\\minus{}P(k)$, $ \\delta^2 P(k)\\equal{}\\delta P(k\\plus{}1)\\minus{}\\delta P(k)$, and so on. For $ 0\\le i\\le n$ and $ 0\\le j\\le n\\minus{}i$, $ \\delta^i P(j)$ is an integer; it comes from repeatedly taking differences of integers.\r\n\r\nNow, set $ \\delta^n P(k)\\equal{}\\delta^n P(0)$ for all $ k$; $ \\delta^n P$ must be constant since $ P$ is a polynomial of degree $ \\le n$. We can then build back down the difference table: $ \\delta^{n\\minus{}1}P(\\minus{}1)\\equal{}\\delta^{n\\minus{}1}P(0)\\minus{}\\delta^n P(\\minus{}1),\\dots,P(\\minus{}1)\\equal{}P(0)\\minus{}\\delta P(\\minus{}1)$ and similarly $ \\delta^{n\\minus{}1}P(2)\\equal{}\\delta^{n\\minus{}1}P(1)\\plus{}\\delta^n P(1),\\dots,P(n\\plus{}1)\\equal{}P(n)\\plus{}\\delta P(n)$. We repeat indefinitely to get $ P$ at all positive and negative integers. Every step of the algorithm uses only addition and subtraction, so all of the answers are integers.", "Solution_2": "My solution is almost same as yours except for my using induction, jmerry. :)", "Solution_3": "[quote=\"kunny\"]Let$ P(x)$ be polynomial with degree $ n\\in{\\mathbb{N}}$. If $ P(0),\\ P(1),\\ \\cdots P(n)$ are integers, then prove that $ P(k)$ is integer for any integers $ k$.[/quote]\r\n\r\nIf we write $ P(x)\\equal{}a_0\\plus{}\\sum_{i\\equal{}1}^na_i\\frac{\\prod_{k\\equal{}0}^{i\\minus{}1}(x\\minus{}k)}{i!}$\r\n\r\nwe have $ a_0\\equal{}P(0)$ and, for $ p\\in[1,n]$, $ P(p)\\equal{}a_0\\plus{}\\sum_{i\\equal{}1}^{p}a_i\\binom{p}{i}$, and so $ a_1\\equal{}P(1)\\minus{}P(0)$ and $ a_p\\equal{}P(p)\\minus{}a_0\\minus{}\\sum_{i\\equal{}1}^{p\\minus{}1}a_i\\binom{p}{i}$ Fpr, $ p>1$\r\n\r\nAnd so all $ a_p$ are integers.\r\n\r\nNow, for any $ m>n$ : $ P(m)\\equal{}a_0\\plus{}\\sum_{i\\equal{}1}^na_i\\frac{\\prod_{k\\equal{}0}^{i\\minus{}1}(m\\minus{}k)}{i!}$ $ \\equal{}a_0\\plus{}\\sum_{i\\equal{}1}^na_i\\binom{m}{i}$ and is an integer.\r\n\r\nNow, for any $ m<0$ : $ P(m)\\equal{}a_0\\plus{}\\sum_{i\\equal{}1}^na_i\\frac{\\prod_{k\\equal{}0}^{i\\minus{}1}(m\\minus{}k)}{i!}$ $ \\equal{}a_0\\plus{}\\sum_{i\\equal{}1}^na_i(\\minus{}1)^{i}\\binom{i\\minus{}1\\minus{}m}{i}$ and is an integer.\r\n\r\nAnd so $ P(n)$ is an integer fo any integer $ n$.", "Solution_4": "Lagrange Interpolating Polynomial." } { "Tag": [ "limit", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Let $(a_{n})_{n \\in N^{*}}$ be a sequence, strictly decreasing with $a_{n} >0$ $\\forall n \\in N^{*}$ with $\\displaystyle \\lim_{n\\to \\infty} a_{n}=0$. Let $k \\in N^{*}$, $k \\neq 1$ and $\\beta_{n,k}$ the solutions of the ecuation: $( \\log_{k+1+a_{n}} x) (\\ln (k)) =\\ln(x-1)$. Compute $L= \\displaystyle \\lim_{k\\to \\infty} \\displaystyle \\lim_{n\\to \\infty} \\beta_{n,k}$.", "Solution_1": "Do you really think there aren't other more interesting problems than those from ADMITERE ASE? I hope you won't post also those from POLITEHNICA, BAC, ...", "Solution_2": "I don't see why you always pick on me! :? :? \r\n\r\nAnyway, I can post whatever I like! I guess everybody should send you private messeges to ask you which problem to post and which not!\r\n\r\nWhatever......harazi!", "Solution_3": "I'm verry sorry that you interpreted this in this wrong way. I didn't say that you aren't allowed to post anything you want. I do not understand why consider that what I said is something bad related to your person. Please... No, I just said that some problems really aren't interesting and I don't think this site is their place. But if I'm not allowed even to say this, Ok, I apologise for having a point of view. And yes, I always pick on you, I want to see you dead :D . Come on, Lagrangia... you know, we have some ages now, we don't play anymore woth toys... And I beg everyone not to send me so many private messages, since my inbox is already full and I don't want to empty it all the time. So, everyone, please nice problems. But anyway, you can post whatever you want ( maybe this is the only good part in what you have said) . It's not my site, I'm not the judge here. I just have some observations, which seem to be missunderstood.", "Solution_4": "Well I am sorry, but this is the proper section to post this Harazi, so Lagrangia is right. \r\n\r\nMaybe a different section will be created in the future, containing all \"beautifull\" problems, but for now we just have to bear with each other. \r\n\r\nJust keep in mind that not all people have the same background/math levels, and the same preferences, so while you dislike this problems, others might enjoy it.", "Solution_5": "My dear people, but I have just expressed a simple opinion. Did I interdict to anyone posting problems? I don't remember to have done this.", "Solution_6": "Yes I know Harazi, and I understand your point of view, but I also understand Lagrangia's, that's why I was trying to create a middle path :)", "Solution_7": "Stay calm, I have nothing with Lagrangia , though he thinks the opposite. Ok, I apologise and promisse I will never say how ugly are those problems. I give my word that I will never say something about any useless problem posted. So be it!\r\n Ps: I really son't have anything with Lagrangia." } { "Tag": [ "logarithms", "trigonometry", "MATHCOUNTS", "AoPS Books" ], "Problem": "i was wondering what books are good to look at if u want to do well at mathcounts.", "Solution_1": "The AoPS books are super! Mostly the first one though; you don't usually use logarithms or advanced trigonometry at MC.\r\n\r\nAlso, Mathematical Circles is more advanced. You might want to get it after you've done AoPS Vol. 1.\r\n\r\nOld Mathcounts tests help. I think the Mathcounts Foundation publishes some of the old tests in book form.\r\n\r\nBilly", "Solution_2": "Math books, of course!!! :lol:\r\n\r\nI would also recommend the Art and Craft of Problem Solving by Paul Zeitz.", "Solution_3": "i dont think strict textbook learning is going to help because mathcounts requires some outside of the box reading", "Solution_4": "And LOTS AND LOTS OF PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE. You won't believe how many stupid mistakes I make every time. There's a Sticky in this forum about \"Mathcounts resources\".\r\n\r\nmathcounts.saab.org is what I've used." } { "Tag": [ "USAMTS" ], "Problem": "what is this?", "Solution_1": "You mean, what's USAMTS?\r\n[url]http://www.usamts.org/[/url]" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Let $f$ a degree $n$ polynomial with integer coefficients. Prove that there is $\\xi\\in I$ such that $f(\\xi)\\in\\mathbb{Z}$. ;)\r\n\r\n$I$ is the set of irational numbers.", "Solution_1": "Either I don't understand either there is something missing :| \r\n\r\nIf $f(x) = x$ such $\\xi$ can't exist." } { "Tag": [ "Euler", "geometry", "circumcircle", "perpendicular bisector", "geometry unsolved" ], "Problem": "if the euler line passes through any vertex then the triangle is either right angled or isosceles or both-prove it!", "Solution_1": "Suppose that the Euler line of $ \\triangle ABC$ passes through $ A$, and let $ O$ and $ H$ respectively denote the circumcenter and the orthocenter of $ \\triangle ABC$. If $ H\\neq A$ then we must have $ AO\\perp BC$. Hence $ A$ lies on the perpendicular bisector of $ BC$ and therefore the triangle is isosceles. If $ H\\equal{}A$ then obviously the triangle is right (also not necessarily scalene) and the conclusion follows." } { "Tag": [], "Problem": "\u0391\u03bd $a,b,c>0$ \u03bc\u03b5 $abc=1$ \u03bd\u03b4\u03bf \r\n\\[ \\frac{a}{b}+\\frac{b}{c}+\\frac{1}{a}\\geq a+b+1 \\]", "Solution_1": "\u039c\u03b9\u03b1 \u03b1\u03c0\u03bb\u03ae \u03b1\u03bb\u03bb\u03ac \u03cc\u03c7\u03b9 \u03cc\u03bc\u03bf\u03c1\u03c6\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9:\r\n\r\n\u0397 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 $\\frac{a}{b}+\\frac{b}{\\frac{1}{ab}}+\\frac{1}{a}\\geq a+b+1 \\Rightarrow (b^3-b+1) a^2 - (b^2+b) a + b \\geq 0$ \r\n\r\n\u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03c1\u03b9\u03ce\u03bd\u03c5\u03bc\u03bf \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 $a$ :D \u03bc\u03b5 \u03b4\u03b9\u03b1\u03ba\u03c1\u03af\u03bd\u03bf\u03c5\u03c3\u03b1 $D=-b(b-1)^{2}(3b+4) \\leq 0$ \r\n\r\n\u03ba\u03b1\u03b9 $(b^3-b+1)=(b+1)(b-1)^{2}+b^2 >0$ . :)\r\n\r\n\u0393\u03b9\u03b1 $D=0$ \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b3\u03b9\u03b1 $b=1$ \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $a=1$ \u03ba\u03b1\u03b9 $c=1$ \u03ac\u03c1\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03b9\u03b1 $a=b=c=1$.\r\n\r\n\u0398\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ae\u03c3\u03c9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c4\u03b9 \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf :)", "Solution_2": "\u0397 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03c0\u03b1\u03c1\u03ac\u03be\u03b5\u03bd\u03b7 ,\u03ac\u03c1\u03b1 \u03b8\u03b1 \u03b2\u03ac\u03bb\u03bb\u03c9 \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03be\u03b5\u03bd\u03b7 \u03bb\u03cd\u03c3\u03b7 . :D \r\n\r\n\u0391\u03c0\u03cc \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 Cauchy \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $\\frac{a}{b}+\\frac{b}{c}+\\frac{1}{a}\\geq\\frac{(a+b+1)^2}{ab+bc+a}$\r\n\r\n\u03ac\u03c1\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b4\u03bf $\\frac{(a+b+1)^2}{ab+bc+a}\\geq a+b+1$ \u03ae \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 \r\n\r\n$b+abc\\geq b(a+c)$ \u03ae \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 $(1-a)(1-c)\\geq 0$ \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 !!!!! \u03b1\u03c6\u03bf\u03cd \u03c4\u03bf \r\n\r\n\u03b5\u03af\u03c0\u03b5 \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03c3\u03c4\u03bf \u03c3\u03ac\u03b9\u03c4 \u03c3\u03c4\u03b9\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2 Vasc \u03b5\u03b4\u03ce\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=455275#p455275 \r\n\r\n :D", "Solution_3": "\u039c\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2 . \u03a4\u03b1 \u03c6\u03ad\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03b1 \u03c3\u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03bc\u03ad\u03bb\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03b8\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $f(a,b,c)$ \u03ba\u03b1\u03b9 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b7 \u03bc\u03ad\u03b8\u03bf\u03b4\u03bf mixing variables .\r\n\r\n\u0391\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $f(a,b,c)\\geq f(t,t,c)$ \u03cc\u03c0\u03bf\u03c5 $t=\\sqrt{ab}$\r\n\r\n\u03a3\u03c4\u03b7 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b4\u03bf $f(t,t,c)\\geq 0$ \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \r\n\r\n$\\frac{\\sqrt{ab}}{c}+\\frac{1}{\\sqrt{ab}}-2\\sqrt{ab}\\geq 0$ \r\n\r\n\u03ae \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 $\\frac{1}{c\\sqrt c}+\\sqrt{c}-\\frac{2}{\\sqrt{c}}\\geq 0$ \r\n\r\n\u03ae $(c-1)^2\\geq 0$ \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \r\n\r\n@Ramanajuan: \u03c0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03b9\u03bf \u03c6\u03c5\u03c3\u03b9\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03ae", "Solution_4": "\u03a0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b7\u03c3\u03b1 \u03bd\u03b1 \u03b2\u03c1\u03ce \u03bc\u03b9\u03b1 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03c0\u03b9\u03bf \u03c6\u03c5\u03c3\u03b9\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03bb\u03c5\u03c3\u03b7 \u03b8\u03ad\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 $a=\\frac{x}{y} ,b =\\frac{y}{z} ,c= \\frac{z}{x}$ \r\n\r\n\u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03c6\u03cd\u03b3\u03c9 \u03c4\u03b7 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \u03b1\u03bb\u03bb\u03b1 \u03b4\u03b5\u03bd \u03c4\u03b1 \u03ba\u03b1\u03c4\u03ac\u03c6\u03b5\u03c1\u03b1. \u038c\u03c4\u03b1\u03bd \u03b2\u03c1\u03ce \u03c7\u03c1\u03cc\u03bd\u03bf \u03b8\u03b1 \u03c4\u03b7\u03bd \u03be\u03b1\u03bd\u03b1\u03ba\u03bf\u03b9\u03c4\u03ac\u03be\u03c9 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03bd \u03bc\u03b5 \u03c0\u03c1\u03bf\u03bb\u03ac\u03b2\u03b5\u03b9\u03c2.\r\n\r\n :)", "Solution_5": "[quote=\"silouan\"]\n$\\frac{(a+b+1)^2}{ab+bc+a}\\geq a+b+1$\n[/quote]\r\n\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad \u03cc\u03c3\u03bf\u03bd \u03b1\u03c6\u03bf\u03c1\u03ac \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5,\u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 $a=2,b=1,c=1/2$ .\r\n\u039f Vasc \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03cc\u03c4\u03b1\u03bd \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03c3\u03b5 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ad\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2,\u03cc\u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03b4\u03c5\u03bf \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03ad\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b1 $a,b,c$ \u03ce\u03c3\u03c4\u03b5\r\n$(1-a)(1-b)\\geq 0$ \u03ae $(1-a)(1-c)\\geq 0$ \u03ae $(1-c)(1-b)\\geq 0$ ,\u03b1\u03c6\u03bf\u03cd \u03c4\u03bf \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03c4\u03bf\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd \u03c3\u03c4\u03b1 \u03b1\u03c1\u03b9\u03c3\u03c4\u03b5\u03c1\u03ac \u03bc\u03ad\u03bb\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03b8\u03b5\u03c4\u03b9\u03ba\u03cc.\r\n\u039e\u03b1\u03bd\u03b1 \u03b4\u03ad\u03c3'\u03c4\u03bf!\r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2... :!:", "Solution_6": "\u039c\u03ac\u03bb\u03bb\u03bf\u03bd \u03b4\u03b5\u03bd \u03b5\u03af\u03bc\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ac \u03c3\u03ae\u03bc\u03b5\u03c1\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf .\u0391\u03c0\u03bb\u03ac \u03c0\u03b1\u03c1\u03b1\u03c3\u03cd\u03c1\u03b8\u03b7\u03ba\u03b1", "Solution_7": "[quote=\"Ramanujan\"]\u03a0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b7\u03c3\u03b1 \u03bd\u03b1 \u03b2\u03c1\u03ce \u03bc\u03b9\u03b1 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03c0\u03b9\u03bf \u03c6\u03c5\u03c3\u03b9\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03bb\u03c5\u03c3\u03b7 \u03b8\u03ad\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 $a=\\frac{x}{y} ,b =\\frac{y}{z} ,c= \\frac{z}{x}$ \n\n\u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03c6\u03cd\u03b3\u03c9 \u03c4\u03b7 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \u03b1\u03bb\u03bb\u03b1 \u03b4\u03b5\u03bd \u03c4\u03b1 \u03ba\u03b1\u03c4\u03ac\u03c6\u03b5\u03c1\u03b1. \u038c\u03c4\u03b1\u03bd \u03b2\u03c1\u03ce \u03c7\u03c1\u03cc\u03bd\u03bf \u03b8\u03b1 \u03c4\u03b7\u03bd \u03be\u03b1\u03bd\u03b1\u03ba\u03bf\u03b9\u03c4\u03ac\u03be\u03c9 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03bd \u03bc\u03b5 \u03c0\u03c1\u03bf\u03bb\u03ac\u03b2\u03b5\u03b9\u03c2.\n\n :)[/quote]\r\n\r\n\r\n\u03c3\u03b5 \u03b1\u03c5\u03c4\u03bf \u03c3\u03c4\u03b7\u03c1\u03b9\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b7 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03bf\u03c5, \u03bc\u03b1\u03ba\u03c1\u03bf\u03c3\u03ba\u03b5\u03bb\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03be\u03b5\u03bd\u03b7..... ;)", "Solution_8": "\u03bb\u03bf\u03b9\u03c0\u03bf\u03bd, \u03b1\u03c2 \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03b1\u03c3\u03c9 \u03b5\u03bd \u03c3\u03c5\u03bd\u03c4\u03bf\u03bc\u03b9\u03b1 \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03bf\u03c5.....\r\n\r\n\u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03c5\u03bc\u03b5\u03bd\u03b1, \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \u03bf\u03b9 \u03c4\u03c1\u03b9\u03b1\u03b4\u03b5\u03c2 $(\\frac{1}{y^2},\\frac{1}{z^2},\\frac{1}{xz})$ \u03ba\u03b1\u03b9 $(xz,xy,yz)$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03b9\u03bf\u03c4\u03c1\u03bf\u03c0\u03b1 \u03b4\u03b9\u03b1\u03c4\u03b5\u03c4\u03b1\u03b3\u03bc\u03b5\u03bd\u03b5\u03c2.\r\n\u039c\u03b5\u03c4\u03b1 \u03c4\u03c3\u03b5\u03bc\u03c0\u03b9\u03c3\u03b5\u03b2 \u03ba\u03b1\u03b9 \u03bb\u03b9\u03b3\u03b5\u03c2 \u03c0\u03c1\u03b1\u03be\u03b5\u03b9\u03c2....\r\n\r\n :)", "Solution_9": "[quote=\"socrates\"]\n\u039c\u03b5\u03c4\u03b1 \u03c4\u03c3\u03b5\u03bc\u03c0\u03b9\u03c3\u03b5\u03b2 \u03ba\u03b1\u03b9 \u03bb\u03b9\u03b3\u03b5\u03c2 \u03c0\u03c1\u03b1\u03be\u03b5\u03b9\u03c2....\n :)[/quote]\r\n\r\n\u0398\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03b4\u03c9 \u03b1\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf :D :)", "Solution_10": "[quote=\"Ramanujan\"]\n\u0398\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03b4\u03c9 \u03b1\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf :D :)[/quote]\r\n\r\n\u03bf\u03ba! :) \r\n\r\n\u039c\u03b5 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03b7 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1\u03c2 \u03c4\u03c3\u03b5\u03bc\u03c0\u03b9\u03c3\u03b5\u03c6 \u03bb\u03b1\u03bc\u03b2\u03b1\u03bd\u03bf\u03c5\u03bc\u03b5 \r\n\r\n$\\text{LHS}\\geq \\frac{1}{3} \\cdot (\\frac{1}{y^2}+\\frac{1}{z^2}+\\frac{1}{xz})(xy+yz+zx)$\r\n\r\no\u03c0\u03bf\u03c4\u03b5 \u03b1\u03c1\u03ba\u03b5\u03b9 \u03bd\u03b4\u03bf $(\\frac{1}{y^2}+\\frac{1}{z^2}+\\frac{1}{xz})(xy+yz+zx)\\geq 3\\frac{x}{y}+3\\frac{y}{z}+3$\r\n\r\n\r\n\u0391\u03bd\u03b1\u03c0\u03c4\u03c5\u03c3\u03c3\u03bf\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03bf\u03bc\u03b1\u03b4\u03bf\u03c0\u03bf\u03b9\u03c9\u03bd\u03c4\u03b1\u03c2 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5\r\n\r\n$(\\frac{xz}{y^2}+\\frac{x}{y}+\\frac{x}{z})+(\\frac{xy}{z^2}+\\frac{y}{z}+\\frac{y}{x})+(\\frac{z}{y}+\\frac{y}{z}) \\geq 3\\frac{x}{y}+3\\frac{y}{z}+2$\r\n\r\n\u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03c9\u03c2 \u03b1\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03c9\u03bd\r\n\r\n$\\frac{xz}{y^2}+\\frac{x}{y}+\\frac{x}{z}\\geq 3\\frac{x}{y}$\r\n\r\n$\\frac{xy}{z^2}+\\frac{y}{z}+\\frac{y}{x}\\geq 3\\frac{y}{z}$\r\n\r\n$\\frac{z}{y}+\\frac{y}{z}\\geq 2$\r\n\r\n\u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03ba\u03c5\u03c0\u03c4\u03bf\u03c5\u03bd \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b1\u03c5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1 \u03c4\u03bf\u03c5 Cauchy (AM-GM)!\r\n\r\nE\u03bb\u03c0\u03b9\u03b6\u03c9 \u03bd\u03b1 \u03ae\u03bc\u03bf\u03c5\u03bd \u03ba\u03b1\u03c4\u03b1\u03bd\u03bf\u03b7\u03c4\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03c5\u03c1\u03b9\u03c9\u03c2 \u03c3\u03c9\u03c3\u03c4\u03bf\u03c2 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c3\u03c5\u03bb\u03bb\u03bf\u03b3\u03b9\u03c3\u03bc\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c2 \u03c0\u03c1\u03b1\u03be\u03b5\u03b9\u03c2!\r\n\u0395\u03bb\u03b5\u03b3\u03be\u03c4\u03b5 \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03b5\u03c0\u03b9\u03b2\u03b5\u03b2\u03b1\u03b9\u03c9\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03bf\u03c1\u03b8\u03bf\u03c4\u03b7\u03c4\u03b1\u03c2!\r\n\r\n\u03ba\u03b1\u03bb\u03bf \u03c0\u03b1\u03c3\u03c7\u03b1!\r\n :lol:", "Solution_11": "\u039c\u03c0\u03c1\u03ac\u03b2\u03bf Socrates \u03c0\u03bf\u03bb\u03cd \u03cc\u03bc\u03bf\u03c1\u03c6\u03b7 \u03bb\u03cd\u03c3\u03b7 :)" } { "Tag": [], "Problem": "Donna and Rod bought a new van at an \"end of the year sale\" for \"dealer cost plus $ 5\\%$.\" If they paid $ \\$17,325.00$, what was the dealer cost in dollars?", "Solution_1": "$ 1.05x\\equal{}17325 \\implies x\\equal{}\\boxed{16500}$." } { "Tag": [ "geometry", "trigonometry", "inequalities", "cyclic quadrilateral", "geometry proposed" ], "Problem": "In a cyclic quadrilateral $ ABCD, a,b,c,d$ are its side lengths, $ Q$ its area, and $ R$ its circumradius. Prove that:\r\n\r\n$ R^2\\equal{}\\frac{(ab\\plus{}cd)(ac\\plus{}bd)(ad\\plus{}bc)}{16Q^2}$.\r\n\r\nDeduce that $ R \\ge \\frac{(abcd)^{\\frac{3}{4}}}{Q\\sqrt{2}}$ with equality if and only if $ ABCD$ is a square.", "Solution_1": "$ Q\\equal{}area(ABD)\\plus{}area(BCD)\\equal{}\\frac{ad\\sin A}{2}\\plus{}\\frac{bc\\sin C}{2}\\equal{}$\r\n\r\n$ \\equal{}\\frac{(ad\\plus{}bc)\\sin A}{2}\\equal{}\\frac{(ad\\plus{}bc)BD}{4R}\\Rightarrow R\\equal{}\\frac{BD(ad\\plus{}bc)}{4Q}$\r\n\r\n$ Q\\equal{}area(ABC)\\plus{}area(ADC)\\equal{}\\frac{(ab\\plus{}cd)\\sin B}{2}\\equal{}\\frac{(ab\\plus{}cd)AC}{4R}\\Rightarrow$\r\n$ \\Rightarrow R\\equal{}\\frac{AC(ab\\plus{}cd)}{4Q}$\r\n\r\nThen $ R^2\\equal{}\\frac{AC\\cdot BD(ad\\plus{}bc)(ab\\plus{}cd)}{16Q^2}$\r\n\r\nBut ABCD is cyclic and $ AC\\cdot BD\\equal{}ac\\plus{}bd$\r\n\r\nSo $ R^2\\equal{}\\frac{(ad\\plus{}bc)(ab\\plus{}cd)(ac\\plus{}bd)}{16Q^2}$\r\n\r\nTo prove the inequality apply AM-GM inequality.\r\n\r\nThe equality holds if $ ab\\equal{}cd, \\ ac\\equal{}bd, \\ ad\\equal{}bc\\Rightarrow a\\equal{}b\\equal{}c\\equal{}d$" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "1. Let $G$ be a finite group such that $[G,G]=G$. Prove that if $K$ is a cyclic, normal subgroup of $G$ then $K$ is a subgroup of $Z(G)$.\r\n\r\n2. Prove that having a finite group $G$ and a normal subgroup $N$ of $G$ the implication is true :\r\nif there is exactly one subgroup $D$ of $G$ such that $ND=G$ and $D \\cap N =1$ then\r\n$N$ is a simple group.\r\n\r\nI'm sure it's simple but I haven't found any simple approach, any ideas ?", "Solution_1": "[quote=\"Rafal\"]1. Let $G$ be a finite group such that $[G,G]=G$. Prove that if $K$ is a cyclic, normal subgroup of $G$ then $K$ is a subgroup of $Z(G)$.\n[/quote]\r\n\r\nWrite $K=$. It is enough to show that $a \\in Z(G)$. Take $u \\in G$. Then $u = xyx^{-1}y^{-1}$ for some $x,y \\in G$. Now:\r\n\r\n$ua=au \\Leftrightarrow$\r\n $xyx^{-1}y^{-1}a=axyx^{-1}y^{-1} \\Leftrightarrow$\r\n$y^{-1}x^{-1}axy = x^{-1}y{-1}ayx$.\r\n\r\nBut both sides belong to $K$ (since $K$ is normal in $G$) and they have same order, whence they are equal, and the initial equality holds.", "Solution_2": "[quote=\"julien_santini\"]But both sides belong to $K$ (since $K$ is normal in $G$) and they have same order, whence they are equal, and the initial equality holds.[/quote]\r\nWhy are they equal then\u00bf", "Solution_3": "[quote=\"ZetaX\"][quote=\"julien_santini\"]But both sides belong to $K$ (since $K$ is normal in $G$) and they have same order, whence they are equal, and the initial equality holds.[/quote]\nWhy are they equal then\u00bf[/quote]\r\n\r\nThey are not necessarily equal, I went too fast :oops: .", "Solution_4": "Well eventually all I could get is the case where $|K|$ has a $p$-Sylow $S$ of order $|p|$, where $p$ is the smallest prime dividing $G$. In this case $G$ acts on $S$ by conjugation (since $K$ is normal, and every conjugate of an element of $K$ of order $p$ has order $p$, whence $S$ is normal in $G$ since $K$ is abelian), meaning we have an homomorphism $f: G \\rightarrow S(K)$, and since $|f(G)|=1$ or $p$, then $f(G)$ is abelian, meaning $f$ contains $G=[G: G]$ in its kernel, which gives the result.\r\nNote that I did not use the cyclicity of $K$ for this weaker result (just its commutativity).\r\n\r\nI did not find anything more in the general case; any idea ?", "Solution_5": "[quote=\"Rafal\"]2. Prove that having a finite group $G$ and a normal subgroup $N$ of $G$ the implication is true :\nif there is exactly one subgroup $D$ of $G$ such that $ND=G$ and $D \\cap N =1$ then\n$N$ is a simple group.[/quote]\r\n\r\nAs I understand the problem, it is not true. Take $n$ to be any group of odd order, and $G=N\\times\\mathbb Z_2$. Any group $D$ satisfying $ND=G,\\ N\\cap D=1$ must have order $2$, and there's only one element of order $2$ in $G$, so $D$ is unique. However, $N$ can be taken to be non-simple. WHat am I missing here? :?", "Solution_6": "your explenation seems and even is correct, I just translated it precisely from an exercise book and probably there was somewhere a typo." } { "Tag": [ "AMC" ], "Problem": "Source: IUPUI 2002 #1\r\n\r\nEvaluate\r\n[u](2*5+2)(4*7+2)(6*9+2)(8*11+2)...(1998*2001+2)[/u]\r\n(1*4+2)(3*6+2)(5*8+2)(7*10+2)...(1997*2000+2)\r\n\r\nand give your reasoning behind your answer. If you used a pattern, try to find out why this pattern is so.", "Solution_1": "The series is telescopic.\r\n\r\n(2*5+2)(4*7+2)(6*9+2)(8*11+2)...(1998*2001+2) \r\n---------------------------------------------------------------- \r\n(1*4+2)(3*6+2)(5*8+2)(7*10+2)...(1997*2000+2) \r\n\r\n 2 3 4 5...1000\r\n= ------------------\r\n 1 2 3 4...999\r\n\r\n= 1000\r\n\r\nTo prove this, we can express the kth term as\r\n\r\n[u]2k(2k+3)+2[/u]\r\n(2k-1)(2k+2)+2\r\n\r\nand the (k+1)th term as\r\n\r\n[u](2k+2)(2k+5)+2[/u]\r\n(2k+1)(2k+4)+2\r\n\r\nWe now look at the fraction f_k/f_k each kth term has to be multiplied by such that the series is telescopic:\r\n\r\n(2*5+2)/(1*4+2) = 12/6 = 12/6 *6/6 = 2\r\n(4*7+2)/(3*6+2) = 30/20 = 30/20 *10/10 = 3/2\r\n(6*9+2)/(5*8+2) = 56/42 = 56/42 * 14/14 = 4/3\r\n\r\nWe can see that the kth term of the series 6, 10, 14,... can be expressed as 4k+2. Thus, the kth term can also be expressed as\r\n\r\n(2k(2k+3)+2)/(4k+2) * (4k+2)/((2k-1)(2k+2)+2)\r\n\r\nand the (k+1)th term as\r\n\r\n((2k+2)(2k+5)+2)/(4k+6) * (4k+6)/((2k+1)(2k+4)+2)\r\n\r\nSince our original assertion was that the numerator of the kth term and the denominator of the (k+1)th term cancelled each other out,\r\n\r\n(2k(2k+3)+2)/(4k+2) * (4k+6)/((2k+1)(2k+4)+2) = 1\r\n(2k(2k+3)+2)/(4k+2) = ((2k+1)(2k+4)+2)/(4k+6)\r\n(4k+6)((2k(2k+3)+2) = (4k+2)((2k+1)(2k+4)+2)\r\n4k^3 + 12k^2 + 11k + 3 = 4k^3 + 12k^2 + 11k + 3\r\n\r\nThus, our assertion holds.", "Solution_2": "yuuupppp", "Solution_3": "mathfanatic, what does iupui stand for? sounds familiar.", "Solution_4": "Indiana University-Purdue University Indianapolis most likely", "Solution_5": "[quote=\"s0mp\"]mathfanatic, what does iupui stand for? sounds familiar.[/quote]\r\n\r\nI've heard of the name from college basketball, but yep, Syntax is right, I think.", "Solution_6": "yep, they're in the mid-continental league (meaning the crap of the crap teams). the only way a team from mid-con makes it to march madness is by winning the conference tourny, usually its valporaiso though.", "Solution_7": "I as well got 1000. No complex mathematical proof though but probably the same thinking, his was just done mathematically. First I simplified the top and bottom of the various expressions in each parenthesis. I had (12)(30)(56)(90)/(6)(20)(42)(72). I noticed several patterns emerge that proved useless. I then simplified and multiplied together to get 5. I started to think that the number of terms on the top or bottom + 1 = solution. I decided to test that by calculating the next two terms and my assertion held true. I divided 1998 by 2 and then added 1 and got 1000." } { "Tag": [ "modular arithmetic", "factorial" ], "Problem": "What is the product of the first 20 non-negative even numbers.\r\n\r\nPlease show all of your work. ;)", "Solution_1": "[hide]0 is the first non-negative even number, so the answer is 0.[/hide]", "Solution_2": "Yes....that is the correct answer....But why is 0 even? What is the definition of even? I don't know it.........", "Solution_3": "I don't know, but here is my guess. Even means not odd, and 1,3,5... and -1,-3,-5... are odd. Since 0 is not on that list, 0 is even. :?", "Solution_4": "An integer is even if it is divisible by two (in other words, an integer $n$ is even iff $n \\equiv 0 \\bmod 2$). $\\frac{0}{2}$ leaves remainder $0$ upon division by two, so it is divisible by two and it is the first non-negative even integer.\r\n\r\nHowever, if the question were modified to ask for positive integers instead, then the answer is $40!!$ (where $n!! = n(n-2)(n-4)...(4)(2)$ is the multifactorial) $= 2^{20} 20! \\approx 2.55108... \\times 10^{24}$.", "Solution_5": "Robinhe: Odd numbers are 1,3,5....etch and -1,-3,-5...etc. Even numbers are 2,4,6,8...etc and -2,-4,-6...... zero isn't any of those numbers so..............\r\n\r\nt0rajir0u: As you have said, an even number gives a remainder of 0 when divided by 2. Yet zero gives a remainder of 0 when divided by 3,1,5,7.....etc......so it might be odd......but it isn't......\r\n\r\nThis leaves us where we started..............I still am not sure why 0 is even and not odd or undefined.", "Solution_6": "[quote=\"dogseatcheese\"]Yet zero gives a remainder of 0 when divided by 3,1,5,7.....etc......so it might be odd......but it isn't......[/quote]\r\n\r\n$6$ also leaves remainder 0 when divided by 3. Is it odd? :D \r\n\r\nAn odd number is defined as one that is not divisible by two, that is, an integer $n$ is odd iff $n \\equiv 1 \\bmod 2$.", "Solution_7": "I agree with t0rajir0u:\r\n\r\nIf a number is even, then $n\\equiv0\\pmod{2}$.\r\n\r\nIf a number is odd, then $n\\equiv1\\pmod{2}$.\r\n\r\n0 clearly is not 1 (mod 2), and it is evenly divisible by 2, so it is EVEN.", "Solution_8": "Oh.......I see........thanks :lol:", "Solution_9": "hm....\r\n\r\nwhen I first learned about integers, one of my textbooks said 0 is a neutral, number still debated wheather it's even or odd...\r\n\r\nso I think you really have to watch the wordings of this problem", "Solution_10": "No, I don't think so. Any integer is either $\\equiv 0 \\bmod 2$ or $\\equiv 1 \\bmod 2$, and those definitions correspond exactly to the notions of evenness and oddness for natural numbers, so it's a perfectly logical extension to all integers, which includes zero.", "Solution_11": "[quote=\"dogseatcheese\"]Yes....that is the correct answer....But why is 0 even? What is the definition of even? I don't know it.........[/quote]\r\nA number $n$ is even iff $n\\equiv 0\\pmod 2$. \r\nA number $m$ is odd iff $m\\equiv 1\\pmod 2$.\r\n\r\nI hope that helps. :)\r\n\r\nMasoud Zargar", "Solution_12": "http://mathworld.wolfram.com/EvenNumber.html\r\n\r\nas a little extra reference.", "Solution_13": "[hide]$0$[/hide]\r\n\r\nthe link is use full...(236 factorial) ;)", "Solution_14": "What are the quotient and the rest for the following divisions:\r\n$0.\\ \\ \\boxed {0\\ : \\ 7}\\ ;\\ \\boxed {0\\ : \\ (-7)}\\ ;\\ \\boxed {3\\ : \\ 7}\\ ;\\ \\boxed {3\\ : \\ (-7)}\\ ;\\ \\boxed {(-3)\\ : \\ 7}\\ ;\\ \\boxed {(-3)\\ : \\ (-7)}\\ ;$\r\n$1.\\ \\ \\boxed {109\\ : \\ 7}\\ ;$\r\n$2.\\ \\ \\boxed {109\\ : \\ (-7)}\\ ;$\r\n$3.\\ \\ \\boxed {(-109)\\ : \\ 7}\\ ;$\r\n$4.\\ \\ \\boxed {(-109)\\ : \\ (-7)}\\ .$\r\nConclude a general rule for the division $\\boxed {\\ a\\ : \\ b\\ }$, where $a\\in Z\\,\\ b\\in Z^*.$\r\n\r\n[u]Remark.[/u] For any $a\\in Z,\\ b\\in Z^*$ there are and are unique the[b] quotient [/b]$q\\in Z$ and the [b]rest[/b] $r\\in N$ so that $0\\le r<|b|$ and $\\boxed {\\ a=q\\cdot b+r\\ }\\ .$\r\n\r\n[b]Prove that[/b] $q=(sgn\\ b)\\cdot \\left [\\frac{a}{|b|}\\right]$ and $r=|b|\\cdot \\left\\{\\frac{a}{|b|}\\right\\}$.\r\nDenote $x\\in R,\\ x=[x]+\\{x\\},\\ [x]\\le x<[x]+1,\\ 0\\le \\{x\\}<1,\\ x=|x|\\cdot sgn\\ x.$" } { "Tag": [ "geometry", "incenter", "circumcircle", "search", "geometric transformation", "homothety", "geometry unsolved" ], "Problem": "M,N,P are midpoint of BC,CA,AB respectively.F is the Feuerbach point, MF,NF,PF cut \u2299I (incircle) at R,S,T, respectively. I,G,O are incenter, centroid,circumcenter respectively.prove:\r\n(1)if NF>MF , NF>PF, then MF+PF=NF\r\n(2) AR , BS ,CT , OI , FG concurrent at Z, the interior similar center of \u2299O and \u2299I", "Solution_1": "What is Feurbach point ? Is it center of Feurbach circle ?", "Solution_2": "Dear \"thaithuan\",\r\ntha Feuerbach's point is the point of tangancy of the incircle with the nine points circle of a triangle.\r\nI found after my research the first complete synthetic proof of this result (see my website).\r\nSincerely\r\nJean-Louis", "Solution_3": "Dear plane geometry,\r\n\r\nyour first problem is quite well-known. See http://www.mathlinks.ro/viewtopic.php?search_id=1436568649&t=24959 for Darij's solution. \r\nI think the second problem would be better if you didn't give away that $ Z$ is $ X_{55}$.\r\n\r\n[b]Proof:[/b]\r\nBecause $ F_e$ is the external center of similitude of $ (I)$ and $ (W)$, we have that $ IR$ and $ WM$ are parallel. A homothecy with center $ H$ and factor $ 2$ maps $ W$ to $ O$ and $ M$ to the antipode of $ A$ in $ (O)$. Hence, $ IR \\parallel WM \\parallel OA$, which proves that $ AR$ passes through $ X_{55}$, the internal center of similitude of $ (O)$ and $ (I)$. Similarly, we prove that $ BS,CT$ pass through $ X_{55}$.\r\n\r\nNow, we only need to prove that $ F_eG$ passes through $ X_{55}$. Because $ H$ and $ G$ divide $ W$ and $ O$ harmonically, we have that $ G$ is the internal center of similitude of $ (W)$ and $ (O)$. By Monge-d'Alembert, $ FG$ passes through $ X_{55}$.", "Solution_4": "Thank you, dear Jan\uff0cIt is rather nice\uff01\r\nThis problem belongs to a friend of mine,who is very strong in Geometry" } { "Tag": [ "function", "analytic geometry", "graphing lines", "slope", "symmetry" ], "Problem": "A certain function $f$ has the properties that $f(3x)=3f(x)$ for all positive real values of $x$, and that $f(x)=1-\\mid x-2 \\mid$ for $1\\leq x \\leq 3$. Find the smallest $x$ for which $f(x)=f(2001)$.", "Solution_1": "[hide=\"Solution\"] The graph of $f(x)$ just consists of copies of the path $(1, 0) \\to (2, 1) \\to (3, 0)$ \"expanded\" by $3$ every time $x$ jumps to the next power of $3$, so you just need to figure out the distance $f(2001)$ is from the closest power of $3$ (which is $2187$), and since the graph's slope doesn't change this means that $f(2001) = f(2187)+186 = 186$.\n\nNow, between $3^{k}$ and $3^{k+1}$ the function reaches a maximum of $3^{k}$, and the smallest value of $k$ for which $3^{k}> 186$ is $k = 5$, so somewhere between $3^{k}= 243$ and $3^{k+1}= 729$ we will have another $186$. By symmetry, this occurs at $243+186 = \\boxed{ 429 }$. \n\nI think. It's late, and when it's late I mess up calculations. :([/hide]", "Solution_2": "p4: First note that $f(2001) = 3^6f\\left(\\frac{2001}{3^6}\\right) = 186$, since we can extend our given equation to $f(x)=3^k f\\left(\\frac{x}{3^k}\\right)$. Next, note that the range of $1-|x-2|$ is $[0,1]$, so we want to be sure our value of $k$ gives a value for $\\frac{186}{3^k}$ in the range $[0,1]$. The only working value is $k = 5$, giving $243 f\\left(\\frac{x}{243}\\right) = 186$. We now require $f\\left(\\frac{x}{243}\\right) = \\frac{62}{81}$, and we can solve for $x$ to get $x = \\boxed{429}, 543$.", "Solution_3": "I got 186 but forgot to look at the range of $f(x)$ :noo" } { "Tag": [], "Problem": "Ned and Nellie were playing on the elevator in their 12-story apartment complex one evening. They entered the elevator on the first floor, and decided to reach their eleventh-floor apartment by continually repeating the following sequence of steps. Ned pushed the button of the floor that was three floors above where they were currently, and upon arriving, Nellie then pressed the floor of the nearest even floor below where Ned's button led them. Assuming no one interrupted their game, how many times did they have to press a button before they could get out on the eleventh floor?", "Solution_1": "[hide]9. I did find a pattern, but there aren't like a hundred floors, so there's no point using it (btw, though, it is: After you get to the fourth floor after four button presses, every four button presses from then on will make you go higher by 4 floors). So I basically kept track of what floor I was on until I got to 11. [/hide]", "Solution_2": "[hide]9. 1->4->2->5->4->7->6->9->8->11.[/hide]", "Solution_3": "[hide]9[/hide]", "Solution_4": "[hide=\"My solution\"]\nThere are 2 ways to solve this problem:\n[list]1) [b]Just write out the numbers.[/b] There aren't a lot, making this the faster method:\n$1\\rightarrow 4\\rightarrow 2\\rightarrow 5\\rightarrow 4\\rightarrow 7\\rightarrow 6\\rightarrow 9\\rightarrow 8\\rightarrow 11$ : $10-1$ (first # doesn't count) $=9$.\n\n2) [b]Find a pattern.[/b] In this particular problem, it would take a bit longer, but it is better to use this method if you know that there are too many moves to count or if you don't know how many moves there are. Here, after reaching $4$, every two-move sequence brings you up 2 levels:\n$1\\rightarrow 4\\rightarrow 2\\rightarrow 5\\rightarrow 4$ : $4$ moves + $8-4=4$ moves (to go from $4$ to $8$) + $1$ move from $8$ to $11$ : $9$ moves total.\n[/list][/hide]\r\n\r\nEDIT: Long post, made several typos", "Solution_5": "[hide]Nine times. I didn't need to use a pattern though I did find one. 1+3=4 and 4-2=2 Every sequence of two button pushes after that was always the next even number because even+odd=odd and odd-1=even. In this case, our odd increasement is 3 so 3-1=2.[/hide]" } { "Tag": [ "inequalities", "function", "inequalities solved" ], "Problem": "a,b,c>=-3/4 and a+b+c=1 . Show that\r\na/(a 2 +1) + b/(b 2 +1) + c/(c2 +1) \\leq 9/10", "Solution_1": "Hint :\r\nconsider f(x)=x/(1+x 2 ) . Prove:\r\n If (x+y)(xy-3) \\leq 0 then f(x)+f(y) \\leq 2f((x+y)/2)", "Solution_2": "How can you use Jensen?? This function has not f''>0 or f''<0 at the interval [-3/4,+00)...", "Solution_3": "It nac't be proved by Jensen.\r\n\r\nLet's take a function f(x)=x/(x^2+1) - 18(x-1/3)/25 - 3/10.\r\n\r\nThen f'(x)=(1-x^2)/(1+x^2)^2 - 18/25, so f'(x)=0 <==> 18x^4+61x^2-7=0 <==> (9x^2-1)(2x^2+7)=0 <==> x=1/3 or -1/3.\r\n\r\nAnd we get f(x) decreases at [-3/4,-1/3], increases at [-1/3,1/3], and decreases at [1/3,+00]. \r\n\r\nf(-3/4)=0, f(1/3)=0, so f(x)<=0.\r\n\r\nHence 0>=f(a)+f(b)+f(c)=( \\sum a/(a^2+1)) - ( \\sum 18(a-1/3)/25) - 9/10\r\n\r\n=( \\sum a/(a^2+1)) - ( 18/25 * (a+b+c-1)) - 9/10\r\n\r\n=( \\sum a/(a^2+1)) - 9/10, As desired.", "Solution_4": "We have $(4a + 3)(3a - 1)^2 \\geq 0$ (and the analogous inequalities).\r\nHence $36a^3 + 3a^2 - 14a + 3 \\geq 0$ so \\[36(a^3 + a) + 3(a^2 + 1) \\geq 50a.\\]\r\nDividing by $a^2 + 1$ we get \\[36a + 3 \\geq \\frac{50a}{a^2 + 1}\\] and the two analogous inequalities.\r\nAdding these we get \\[\\sum \\frac{a}{a^2 + 1} \\leq \\frac{9 + 36 \\sum a}{50} = \\frac{9}{10}.\\]" } { "Tag": [ "algorithm" ], "Problem": "Here's a problem my uncle gave me in an email. I'm stumped on this. I'd appreciate a solution for this as good as possible. \r\n\r\n[hide=\"The problem\"]I am really stumped with this. I have researched it (online and elsewhere) and have asked for expertise from two math teacher friends, but they were equally confused. \n \nWhen dividing items among three people, what is the fairest way to take turns picking the items each person wants? This came up recently because the three kids in my family were recently dividing up the estate items of a maiden aunt who'd died. We shuffled together three pieces of paper (printed with the numbers 1, 2 and 3) and then each person blindly picked one of the pieces of paper. I thought we'd go #1, #2, #3, and then the same thing over and over again. But my brother said, no, the fairest way to do it is that the order reverses so that the \"rounds\" of picking items go like this:\n \n1, 2, 3\n3, 2, 1\n1, 2, 3\n3, 2, 1\netc.\n \nIt seems to me that while this is pretty fair to #1 and #3, it's less so to #2. But I can't find any mathematical/probability stuff on this to prove what I suspect is true. Is there some mathematically correct way to ensure fair and equal turn-taking in a situation like this?\n \nAs it turns out, I was #2 and got everything I wanted and then some--so this isn't a sour grapes issue. But I got to thinking about it purely from a math standpoint and wondering what is the simplest, fairest, most equal system. Thanks for your input.[/hide]", "Solution_1": "See http://en.wikipedia.org/wiki/Fair_division and http://mathworld.wolfram.com/CakeCutting.html .\r\n\r\nThe discrete problem is harder than the continuous problem, but incidentally a friend of mine at RSI (edit: that's the poster below!) was doing research on this subject. The three-person problem has an explicit solution, but I don't remember it.", "Solution_2": "There are a number of algorithms that can be used here to try to allocate the items fairly, but there is no \"best\" way, as long as the items are indivisible. (Imagine if all three people wanted one of the items a lot and didn't care for the other two. There is quite literally no way to resolve this issue.)\r\n\r\nOn the other hand, if you introduce a few other allowable steps, it gets more fun. For example, if all three individuals have a lot of cash, they can bid on the items to divide them among each other, by assigning a value to each item. You would want to use the Knaster procedure, which works something like this (I will use just a single item in this example--it can easily be extended to more by simply repeating the procedure):\r\n\r\nLet all three individuals \"bid\" on the item to be given out. In other words, have them each individually write down what they believe the item to be worth. Then the bids are revealed and the person who placed the highest bid receives the item. He keeps the item and pays 2/3 the value of the item. From that amount, each of the other two individuals then receives 1/3 the value of the item (as they recorded it). Since each of the other two individuals' bids were less than the bid of the person who received the highest bid, there will be money left over. This money is distributed equally among all three individuals.\r\n\r\nHere is a more concrete example. Qiaochu, Sam, and Winston have come into possession of a vintage edition of Harry Potter, a V for Vendetta DVD, and a cane. They want to divide the three items between each other.\r\n\r\nSo each person bids secretly on each item. Suppose the bids are as follows:\r\n\r\nQiaochu: \r\nHarry Potter: $210 V for Vendetta:$150\r\nCane: $60 Sam: Harry Potter:$90\r\nV for Vendetta: $60 Cane:$90\r\n\r\nWinston: \r\nHarry Potter: $240 V for Vendetta:$120\r\nCane: $75 Note that Qiaochu wins V for Vendetta, Sam wins the cane, and Winston wins Harry Potter. Now, we calculate how much money each person pays or receives (remember that if a person receives an item, they pay 2/3 of what he or she believes its value is, and if a person does not receive an item, they receive 1/3 of what he or she believes its value is): Qiaochu: 1/3($210) + V for Vendetta - 2/3($150) + 1/3($60) = V for Vendetta and pays $10 Sam: 1/3($90) + 1/3($60) + Cane - 2/3($90) = Cane and pays $10 Winston: Harry Potter - 2/3($240) + 1/3($120) + 1/3($75) = Harry Potter and pays $95 Now this means that there is 10 + 10 + 95 =$115 floating around, and it is divided equally between Qiaochu, Sam, and Winston.\r\n\r\nTo see why this algorithm works, consider the players and what they receive with respect to a single item. Either a) they receive the item and pay 2/3 of its perceived worth, or b) they do not receive the item but they receive 1/3 of its perceived worth. Either way, they should be satisfied. Also note that there is money left over that is divided equally, so each person will actually receive a bit more than 1/3 the value of the item.", "Solution_3": "[quote=\"winluo2\"]Qiaochu: \nHarry Potter: 210 dollars\nV for Vendetta: 150 dollars\nCane: 60 dollars\n\nSam: \nHarry Potter: 90 dollars\nV for Vendetta: 60 dollars\nCane: 90 dollars\n\nWinston: \nHarry Potter: 240\nV for Vendetta: 120 dollars\nCane: 75 dollars\n\nNote that Qiaochu wins V for Vendetta, Sam wins the cane, and Winston wins Harry Potter. Now, we calculate how much money each person pays or receives (remember that if a person receives an item, they pay 2/3 of what he or she believes its value is, and if a person does not receive an item, they receive 1/3 of what he or she believes its value is): \n\nQiaochu: 1/3(210) + V for Vendetta - 2/3(150) + 1/3(60) = V for Vendetta and pays 10 dollars\n\nSam: 1/3(90) + 1/3(60) + Cane - 2/3(90) = Cane and pays 10 dollars\n\nWinston: Harry Potter - 2/3(240) + 1/3(120) + 1/3(75) = Harry Potter and pays 95 dollars\n\nNow this means that there is 10 + 10 + 95 = 115 dollars floating around, and it is divided equally between Qiaochu, Sam, and Winston.[/quote]\r\n\r\nDollar signs. Tsk tsk. Also, I hate you. :P" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "1. Let G be a group. Prove that if there are 3 distinct elements x, y, z in G s.t. x^2=y^2=z^2=e, e is the identity, then G is NOT a cyclic group.\r\n\r\n2. Prove that if k>=3 then (Z_(2^k))* is NOT a cyclic group.", "Solution_1": "1.: more general, the equation $x^{n}=e$ has at most $n$ solutions in a cyclic group. We even have the converse:\r\nIf the equation $x^{n}=e$ has always at most $n$ solution in the finite group $G$, then $G$ is cyclic.\r\n\r\n2. Posted before the last days, it's not hard to find it." } { "Tag": [ "ratio", "trigonometry", "geometry proposed", "geometry" ], "Problem": "[color=blue]Let $A,C,B,D$ four collinear points (in this order) so that $\\frac{CA}{CB}=\\frac{DA}{DB}\\ .$ Denote the middlepoints $M\\ ,$ $N$ of the segments $[AB]\\ ,$ $[CD]$ and the point $P$ for which $PA=PD\\ ,$ $PA\\perp PD\\ .$ Prove that $m(\\widehat {MPN})=45^{\\circ}\\ .$\n\nSee the post http://www.mathlinks.ro/Forum/viewtopic.php?t=85445 [/color]", "Solution_1": "WLOG, A, C, B, D follow on the line in this order. Denote AB = c, BC = a, CA = b. Obviously, b > a. Point C cuts the segment AB internally in the ratio $\\frac b a$, point D cuts the segment AB externally in the ratio $\\frac b a.$\r\n\r\n$AC = \\frac{bc}{b + a},\\ \\ \\ AD = \\frac{bc}{b - a}$\r\n\r\n$CD = AD - AC = \\frac{bc}{b - a} + \\frac{bc}{b + a} = \\frac{2abc}{b^2 - c^2}$\r\n\r\nLet Q be the foot of a normal from P to the given line. Since PA = PD, Q is the midpoint of AD.\r\n\r\n$PQ = \\frac{AD}{2} = \\frac{bc}{2(b - a)}$\r\n\r\n$MQ = \\frac{AD}{2} - \\frac{AB}{2} = \\frac{bc}{2(b - a)} - \\frac c 2 = \\frac{ca}{2(b - a)}$\r\n \r\n$NQ = \\frac{AD}{2} - \\frac{CD}{2} = \\frac{bc}{2(b - a)} - \\frac{abc}{b^2 - c^2} = \\frac{bc}{2(b + a)}$\r\n\r\n$\\tan \\widehat{MPQ} = \\frac{MQ}{PQ} = \\frac a b$\r\n\r\n$\\tan \\widehat{NPQ} = \\frac{NQ}{PQ} = \\frac{b - a}{b + a}$\r\n\r\n$\\tan \\widehat{MPN} = \\frac{\\tan \\widehat{MPQ} + \\tan \\widehat{NPQ}}{1 - \\tan \\widehat{MPQ} \\tan \\widehat{NPQ}} = \\frac{\\frac a b + \\frac{b - a}{b + a}}{1 - \\frac a b \\cdot \\frac{b - a}{b + a}} =$\r\n\r\n$= \\frac{a(b + a) + (b - a)b}{b(b + a) - a(b - a)} = \\frac{a^2 + b^2}{b^2 + a^2} = 1$\r\n\r\n$\\angle MPN = 45^\\circ$" } { "Tag": [], "Problem": "http://www.youtube.com/watch?v=uSQnK5FcKas\r\n\r\nFortunately enough, there are quite some people with digital cameras or camcorders and conscience nowadays.", "Solution_1": "[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]\r\n\r\nOne still shot from a TV station does not a conspiracy make.", "Solution_2": "[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]\r\n\r\nWhat would you like to say after most of these \"innocent\" media confessed their falsifications, openly or covertly?\r\n\r\nI understand that it is very unpleasant to be proved as a liar, so the fact that some media like CNN confessed but began new falsification on the same topic is also expectable and understandable.\r\n\r\nDo not be so CNN!", "Solution_3": "Obviously, the media interprets all events through the lens that best serves its own interests (in this case, by espousing anti-Chinese rhetoric). The events in Tibet are no exception to this rule. However, there is no denying that China has ruthlessly occupied Tibet for decades and responded in a disproportionate and abhorrent manner to civil unrest in Tibet.", "Solution_4": "[quote=\"Begoner\"]However, there is no denying that China has ruthlessly occupied Tibet for decades[/quote]\n\nIf the word \"occupied\" is to be used, then you should replace \"decades\" with \"more than one thousand years\"\n\n[quote=\"Begoner\"]responded in a disproportionate and abhorrent manner to civil unrest in Tibet.[/quote]\r\n\r\nHow do you know this? Via western liars like CNN, BBC et al et al?" } { "Tag": [ "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "Consider a sequence defined with:\r\n\r\n$ \\sin(\\phi_n \\minus{} \\phi_{n \\plus{} 1}) \\equal{} \\Delta \\sin(\\phi_n)$\r\n\r\nwhere:\r\n\r\n$ n \\in \\mathbb{N}_0$\r\n\r\n$ 0 < \\Delta < 1$\r\n\r\n$ \\phi_0 \\equal{} \\sqrt {1 \\minus{} \\Delta^2}$\r\n\r\nTry to calculate:\r\n\r\n$ P \\equal{} \\frac {\\Delta}{2} \\sum^\\infty_{n \\equal{} 0} \\sin(\\phi_n)$\r\n\r\nThe problem came from a situation where you have an infinite number of sticks of height $ 1$ positioned vertically on the $ x$ axis and apart from each other by $ \\Delta$ (so the first one is at $ x \\equal{} 0$, the second at $ x \\equal{} \\Delta$, the third at $ x \\equal{} 2\\Delta$ and so on). The one at position $ x \\equal{} 0$ is fixed, while others are pushed (actually, the one at position $ x \\equal{} \\infty$ is pushed to the left, and they knock each other out - like domino) to lean one on the other. Now, the question is, what is the surface that is created under these sticks. I believe that the sum above describes it. Do you think there is another way to calculate this? An easier one, perhaps? Or can you do this one? Also, I know it's much, but what about the series. Can we get an explicit solution (my opinion - not a chance)? Or a good approximation at least? I tried with defining a new sequence by $ \\sin(\\phi_n) \\equal{} f_n$ but it didn't get me far. Any ideas are welcome!!", "Solution_1": "I realize that this does not directly address your question, but I'll throw it in anyway.\r\n\r\nSuppose we have a function $ f(x)$ such that the tangent line at any point $ (x,y)\\equal{}(t,f(t))$ intersects the $ x$ axis at the point $ (x,y)\\equal{}(t\\plus{}\\Delta, 0)$. This is fundamentally similar to your setup, except that the tangents are taken continuously and independently instead of sequentially.\r\n\r\nWe want the points $ (t,f(t))$ and $ (t\\plus{}\\Delta,0)$ to solve the equation $ y\\equal{}mx\\plus{}b$. Plugging in these two pairs gives that the tangent line is $ y \\equal{} \\minus{} \\frac{f(t)}{\\Delta}x \\plus{} \\frac{(t\\plus{}d) f(t)}{\\Delta}$. Thus, $ f'(t) \\equal{} \\minus{} \\frac{f(t)}{\\Delta} \\implies f(t) \\equal{} C e^{\\minus{} \\frac{t}{\\Delta}}$\r\n\r\nThus, the function that satisfies this continuous condition is $ f(x) \\equal{} Ce^{\\minus{} \\frac{x}{\\Delta}} \\qquad C \\in \\mathbb{R}^{\\geq 0}$\r\n\r\nWhile your condition is not the same, and the joint points on the curve are taken at at non-constant invervals, you can see that the end behavior of your function will resemble exponential decay.\r\n\r\n\r\nThat being said, I think you'd be better off allowing the sequence of points to be $ (a_n, b_n)$, where $ a_n$ and $ b_n$ are interlaced sequences, by finding the intersection between lines and circles. Then you can find a relationship between $ a$ and $ b$, so that it suffices to look at only one sequence, and each item will be defined in terms of the previous. You can then see if you can find the formula of a curve that fits those points.", "Solution_2": "More precisely, you can show that $ \\varphi_n\\equal{}A(n)(1\\minus{}\\Delta)^n$ where $ A(n)$ tends to a finite positive limit $ A$ as $ n\\to\\infty$. I doubt we can explicitly find $ A$ in terms of $ \\Delta$ though." } { "Tag": [ "function", "rate problems", "calculus", "calculus computations" ], "Problem": "Could someone please walk me through these problems including the answers! Thanks\r\n\r\nFin the average rate of change in the function -4x^2-4x-5 over the interval [-4,-4+h].\r\n\r\nFin the average rate of change in the function x^3+7x over the interval [-2,-2+h].\r\n\r\nIs this right?(f[-2+h]-f[-2])/(-2+h-(-2)\r\n=(f[-2+h]-f[-2])/h", "Solution_1": "That's the definition (in this case, for the second problem). Now calculate and simplify.", "Solution_2": "Could someone actually do the problems below for me?", "Solution_3": "[quote=\"knowltng\"]Could someone actually do the problems below for me?[/quote]\r\nWell, no. That's not what we do around here. We're willing to give you hints and suggestions, and if you'll post your own work, we'll comment on it and correct it. But you need to bring your own efforts into this.", "Solution_4": "I am very sorry and I understand I have done the work but did not want to take the time to type it all in here. Here is my work for the problems above could you tell me if they are right. If they are wrong can you correct them and walk me to the right answer?\r\nFin the average rate of change in the function x^3+7x over the interval [-2,-2+h]. \r\n\r\nf(-2+h)=(-2+h)^3-14+7h\r\n\r\nf(-2)=-22\r\n\r\nf(-2+h)-f(-2)=(-2+h)^3+8+7*h, which simplifies to\r\n19*h-6*h^2+h^3.\r\n\r\nWhen you divide this by h, you get 19-6h+h^2 is this the answer?\r\n\r\n#2\r\nFin the average rate of change in the function -4x^2-4x-5 over the interval [-4,-4+h]. \r\nF(-4+h)=(-4+h)^2-16-5\r\nF(-4)=-37\r\nF(-4+h)-F(-4)=(-4+h)^2+7*h At this point I am so lost with this problem I know I have to simplify but could someone finish the problem for me. For some reason i just can't come up with the answer", "Solution_5": "#1: You're right.\r\n#2: You haven't calculated the values at $ -4+h$ or at $ -4$ correctly, and the constant terms won't go away when you subtract unless you expand out the square first. The value at $ -4$ is actually $ -64+16-5=-53$." } { "Tag": [ "LaTeX" ], "Problem": "Is there a way to number the pages so that the first page is not numbered and the second page starts off as page 1?\r\nThanks.", "Solution_1": "On the first page put\r\n[code]\\thispagestyle{empty}\n\\setcounter{page}{0}[/code]", "Solution_2": "I would like to adjust my page numbering. What I want is a little bit more difficult:\r\n\r\nThe page numbering should look like:\r\n\r\n\\Roman{chapter}-\\arabic{page}\r\n\r\n\r\nI have tried to do use the following code:\r\n\r\n[code]\n\n\\let\\stdchapter\\chapter\n\n\n\\renewcommand{\\chapter}[1]{\\stdchapter{#1}\\addjust}\n\n\n\\newcommand{\\addjust}{\\setcounter{page}{1}}\n\n\\renewcommand{\\thepage}{\\Roman{chapter}-\\arabic{page}}\n[/code]\r\n\r\nThis isn't the way to go because you get in trouble with tableofcontents and bibliography\r\n\r\n\r\nDoes anyone know the answer?", "Solution_3": "[quote=\"Sirinial\"]This isn't the way to go because you get in trouble with tableofcontents and bibliography[/quote]\r\nI'm not sure what you mean. Provided you put the \\addjust macro before you use it, rather than after as you had it, then the page numbers of \\tableofcontents will be correct. Not sure about the relevance of the bibliography.\r\n\r\nHowever, why not use a package ready made for this [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=chappg]chappg[/url]?" } { "Tag": [], "Problem": "1. Prove that if $n$ is odd, then $\\frac{2}{n}$ can be written as the sum of two distinct unit fractions.\r\n\r\n2. Prove that any unit fraction $\\frac{1}{a}$ can be written as the sum of two distinct smaller unit fractions.", "Solution_1": "a)[hide]$\\frac{2}{2k+1}=\\frac1{k+1}+\\frac1{(k+1)(2k+1)}$[/hide]\nb)[hide]$\\frac1a = \\frac1{a+1}+\\frac1{a(a+1)}$[/hide]" } { "Tag": [ "topology", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose M and N are two orientable manifolds without boundary. Show that M x N is also orientable.", "Solution_1": "I'm not sure about this, but if we let $ M*$ an $ N*$ be the orientable double covers, we can use the fact that the product of these covering spaces is (i think) the orientable double cover for $ M\\times N$ (certainly it covers it and is 2-1, but i'm not sure if it is the cover we want to look at). it is connected if and only if $ M\\times N$ is not orientable, but because neither $ M*$ nor $ N*$ were connected to begin with, we conclude that $ M\\times N$ is orientable." } { "Tag": [ "geometry", "circumcircle", "trapezoid", "AMC", "USA(J)MO", "USAMO", "perpendicular bisector" ], "Problem": "Let $ ABC$ be a non-isosceles triangle. Let $ M,N,P$ denote the midpoints of $ BC,CA,AB$ respectively. $ D$ and $ E$ are two points on $ AM$ such that $ DA \\equal{} DB$ and $ EA \\equal{} EC$, $ BD$ meets $ CE$ at $ F$. Prove that $ A,N,F,P$ lie in a circle.", "Solution_1": "The A-median $ AM$ cuts the triangle circumcircle $ (O)$ again at $ M_o.$ $ BD, CE$ cut $ (O)$ again at $ D_o, E_o.$ Since $ PD, NE$ are perpendicular bisectors of $ AB, CA,$ $ D_oM_o \\parallel AB$ and $ E_oM_o \\parallel CA.$ In addition, $ \\angle BE_oM_o \\equal{} \\angle BAM_o$ and $ \\angle CD_oM_o \\equal{} \\angle CAM_o.$ It follows that $ BE_o \\parallel CD_o$ are both parallel to the A-symmedian $ AK.$ $ F \\equiv BD_o \\cap CE_o$ is the diagonal intersection of the isosceles trapezoid $ BE_oD_oC.$ Using central projection, project the circumcircle $ (O)$ to a circle $ (O')$ with center $ M.$ The projected $ \\triangle A'B'C'$ becomes right at $ A'$ and the A-symmedian $ AK$ of the original triangle is projected to its A'-altitude $ A'K'$. Since $ BE_o \\parallel AK \\parallel CD_o$ concur at $ Q$ at infinity, it follows that $ B'E_o', C'D_o'$ meet at $ Q' \\in A'K'.$ Since $ B'C'$ is a diameter of the projected circle $ (O'),$ the angles $ \\angle B'E_o'C', \\angle C'D_o'B'$ are right and $ F' \\in A'K'$ is the orthocenter of the $ \\triangle Q'B'C'.$ As a result, the original point $ F \\in AK.$ But the common perpendicular bisector $ OF$ of $ BE_o, CD_o$ is perpendicular to $ AK \\equiv AF,$ hence the angle $ \\angle AFO$ is right. Consequently, $ F$ is on the circle $ \\odot(APON)$ with diameter $ AO.$", "Solution_2": "[quote=\"mr.danh\"]Let $ ABC$ be a non-isosceles triangle. Let $ M,N,P$ denote the midpoints of $ BC,CA,AB$ respectively. $ D$ and $ E$ are two points on $ AM$ such that $ DA = DB$ and $ EA = EC$, $ BD$ meets $ CE$ at $ F$. Prove that $ A,N,F,P$ lie in a circle.[/quote]\r\nAn ugly solution :\r\nLet $ AM$ cut $ (ABC)$ at X .\r\nThen we have : \r\n${ (YO,YD) = (Y0,YA) (\\mod \\pi})$\r\nBut because $ A,B$ is symmetric through $ OD$ so \r\n$ (YO,YA)\\equiv (BO,BD) (\\mod \\pi)$\r\nSo $ (YO,YD)\\equiv (BO,BD) (\\mod \\pi)$\r\nIt follows that $ B,O,X,D$ is cyclic . \r\nSimilar $ C,O,E,X$ is cyclic. \r\nSo $ (CO,CF)\\equiv (BO,BF) (\\mod \\pi)$\r\n$ \\Rightarrow F \\in (BOC)$\r\n$ (FO,FD) = (FE,FO)$ (1)\r\nEasy to check that $ (FA,FO,FD,FE) = - 1$(2)\r\n(1)(2) $ \\Rightarrow FO$ and $ FA$ perpendicular . \r\nSo $ F\\in (ANP)$", "Solution_3": "[url=http://www.mathlinks.ro/viewtopic.php?p=1116181#p1116181]USAMO 2008 Problem 2[/url]", "Solution_4": "We can solve this problem by the following method:\r\nLet O be the intersection of PD and NE, hence O is the center of the circumcircle (ABC). \r\nWhen P and O is inside triangle ABC:\r\nWe have $ \\hat{BFC}\\equal{}\\hat{BDM}\\plus{}\\hat{CEM}\\equal{}2\\hat{BAM}\\plus{}2\\hat{CAM}\\equal{}2\\hat{BAC}\\equal{}\\hat{BOC}$\r\n$ \\equal{}> B,F,O,C$ are lie in a circle, hence $ \\hat{DFO}\\equal{}\\hat{OCB}\\equal{}\\hat{OBC}\\equal{}\\hat{EFO}$ => FO is bisector of angle DFE.\r\n$ \\frac{DF}{DB}.\\frac{MB}{MC}.\\frac{EC}{EF}\\equal{}1 \\equal{}> \\frac{FD}{FE}\\equal{}\\frac{BD}{CE}\\equal{}\\frac{AD}{AE}$\r\n=> FA is the external angle bisector os angle DFE. Hence angle AFO is right.\r\nSo A,N,F,P are lie a circle, whose diameter is AO.", "Solution_5": "Let\u2019s call M\u2019 the common point of AM and (ABC), different of A, B\u2019 and C\u2019 the intersection of BD and CE with (ABC). Then let\u2019s denote K, L, R and S the projections of B and C on AM, and of A on BD and CE respectively. We claim that BC=B\u2019C\u2019 and that A belongs to the angle bisector of $ \\angle B'FC'$.\r\nSince AD=BD we get $ \\angle BAM'\\equal{}\\angle ABB'$ , while AE=CE gives $ \\angle CAM'\\equal{} \\angle ACC'$, so $ \\angle BAC \\equal{} \\angle B'M'C'$, with BC=B\u2019C\u2019 and BCB\u2019C\u2019 isosceles trapezoid.\r\nThen it is easy to see $ \\triangle ABR \\equal{} \\triangle BAK$, with AR=BK, similarly getting AS=CL, but BK=CL, M being the middle of BC, hence the distances from A to FB\u2019 and FC\u2019 are equal, and AF is actually the angle bisector of $ \\angle B'FC'$, as claimed.\r\nBut from the isosceles trapezoid is follows that the angle bisector of the angle made by its diagonals is parallel to its parallel sides, BC\u2019 and CB\u2019, from here it follows that, if AF cuts again (ABC) at A\u201d, F is the middle of AA\u201d.\r\nIf AA\u2019 is a diameter of the circle (ABC), $ A'A\"\\perp AA\"$, hence $ OF\\perp AF$, and F belongs to the circle with OA diameter, which passes through N and P.\r\n\r\nBest regards,\r\nsunken rock" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "3^x+4^x=5^x is the problem. he told us that x=0,2 are answers, but there are more or something.", "Solution_1": "I think $x=2$, because equation equivalent to $f(x)=(\\frac{3}{5})^{x}+(\\frac{4}{5})^{x}=1$, note $f(2)=1$ and $f(x)$ is monotone on $\\mathbb{R}$.", "Solution_2": "@integral90: I removed your identical post from the Algebra subforum. Please, don't double post in the future." } { "Tag": [ "inequalities", "geometry", "rectangle", "perimeter" ], "Problem": "Prove that if $ ar\\minus{}2bq\\plus{}pc\\equal{}0$ and $ ac\\minus{}b^2>0,$ then, $ pr\\minus{}q^2\\leq 0.$", "Solution_1": "This question seems correct intuitively, but here is a half-proof..\r\n\r\nWe have $ ar\\plus{}pc \\equal{} bq \\plus{}bq$ and $ ac>b^2$, we want to show that $ prq^2 \\longrightarrow acpr>b^2q^2 \\longrightarrow (ar)(pr)>(bq)(bq)$\r\n\r\nSo we can reduce the equation and prove an equivalent one;\r\n If $ ab\\equal{}c^2$ proove: $ a\\plus{}b\\ge c\\plus{}c$\r\n\r\nThis is true because we know for a rectangle with given perimeter, its area maximised when it is a square.\r\n\r\nTherefore it follows that if $ pr>q^2$ then $ ar\\plus{}pc > 2bq$, but we need equality so $ pr0\\Longrightarrow ac>b^2$.[/list]\nHence,\n\n[list]$ r^2ac\\plus{}prc^2>b^2r^2\\plus{}prc^2\\Longrightarrow 2bqrc > b^2r^2\\plus{}prc^2\\geq 2brc\\sqrt{pr}\\text{ (AM\\minus{}GM)}$[/list]\nTherefore\n\n[list]$ q^2\\geq pr.$[/list]\r\nBut i think the statement should inform that $ p,r>0.$", "Solution_3": "$ 4b^2q^2\\equal{} (ar \\plus{} pc)^2 \\ge 4arpc$, from which the solution follows immediately." } { "Tag": [ "LaTeX", "inequalities", "search", "trigonometry" ], "Problem": "Let $ x,y>0$ and $ x\\plus{}y\\equal{}1$ prove that:\r\n\\[ \\frac{x}{\\sqrt{1\\minus{}x}}\\plus{}\\frac{y}{\\sqrt{1\\minus{}y}}\\geq\\sqrt2\\]\r\n\r\n[color=red][Moderator edit: LaTeX formula edited.][/color]", "Solution_1": "tabe f(x)=x/\u221a(1-x) mohadab ast pas ba jensen darim f(x)+f(y)\u22652f((x+y)/2)\r\n\r\nke masale hale :wink:", "Solution_2": "dast karie hokm:\r\n$ \\frac {x}{\\sqrt {1 \\minus{} x}} \\plus{} \\frac {y}{\\sqrt {1 \\minus{} y}}\\geq\\sqrt2$\r\n$ \\frac x{\\sqrt y} \\plus{} \\frac y{\\sqrt x}\\geq\\sqrt2$\r\n [ham makhraj va tavan 2]\r\n$ \\frac {x^3 \\plus{} y^3}{xy}\\geq2$\r\nHokme jadid:\r\n$ x^3 \\plus{} y^3\\geq2xy$\r\nVa chon $ x,y > 0$ va $ x^2 \\plus{} y^2\\geq2xy\\Rightarrow$\r\n$ x^3 \\plus{} y^3\\geq2xy$\r\n\r\nMn 3 ruz tamam say mikardam in soale mozheko ba vasete ha hal konam!mokham dasht mipokid be khoda. Fek konam ino dadashe 5om ebtedayi mam hal kone.albate man hanuz haal nakardam tabe bekhunam vaghti ketabesho baz mikonam sar dard migiram!vali 2 sal vaght daram\u2026!\r\nNokteye akhlaghi:hamishe ye rah hast ke rahat tar az unie ke behesh fekr mikonim ( :D", "Solution_3": "Saay konid LaTeX yad begirid. Az hamun link Latex balaye saretun ettelaat khubi be dast miyarid.\r\n\r\n[quote=\"chakavak\"]dast karie hokm:\nx/\u221a(1-x)+y/\u221a(1-y)\u2265\u221a2\nx/\u221ay+y/\u221ax\u2265\u221a2\n [ham makhraj va tavan 2]\n(x^3+y^3)/xy\u22652\nHokme jadid:\nx^3+y^3\u22652xy\nVa chon x,y>0 & x^2+y^2\u22652xy => \nx^3+y^3\u22652xy\n\nMn 3 ruz tamam say mikardam in soale mozheko ba vasete ha hal konam!mokham dasht mipokid be khoda. Fek konam ino dadashe 5om ebtedayi mam hal kone.albate man hanuz haal nakardam tabe bekhunam vaghti ketabesho baz mikonam sar dard migiram!vali 2 sal vaght daram\u2026!\nNokteye akhlaghi:hamishe ye rah hast ke rahat tar az unie ke behesh fekr mikonim (kalemate ghassar!jelde 2) :D[/quote]", "Solution_4": "[quote=\"chakavak\"]dast karie hokm:\n$ \\frac {x}{\\sqrt {1 \\minus{} x}} \\plus{} \\frac {y}{\\sqrt {1 \\minus{} y}}\\geq\\sqrt2$\n$ \\frac x{\\sqrt y} \\plus{} \\frac y{\\sqrt x}\\geq\\sqrt2$\n [ham makhraj va tavan 2]\n$ \\frac {x^3 \\plus{} y^3}{xy}\\geq2$\nHokme jadid:\n$ x^3 \\plus{} y^3\\geq2xy$\nVa chon $ x,y > 0$ va $ x^2 \\plus{} y^2\\geq2xy\\Rightarrow$\n$ x^3 \\plus{} y^3\\geq2xy$\n[/quote]\r\n$ \\frac {x^3 \\plus{} y^3}{xy}\\geq2$ in namosavi ghalate!mesale naghzesham $ x \\equal{} y \\equal{} \\frac {1}{2}$\r\nage makhraj moshtarak begirim va be tavane 2 beresunim in namosavi bedast nemiad!in be dast miad:\r\n$ \\frac {x^3 \\plus{} y^3 \\plus{} 2xy\\sqrt {xy}}{xy}\\geq{2}$ ke ba tavajoh be inke $ x \\plus{} y \\equal{} 1$ moadel ast ba:\r\n$ {(x \\plus{} y)((x \\plus{} y)^2 \\minus{} 3xy) \\plus{} 2xy\\sqrt {xy}\\geq{2xy}}\\Leftrightarrow{1 \\minus{} 5xy \\plus{} 2xy\\sqrt {xy}\\geq0}$ke agar gharar dahim $ t \\equal{} \\sqrt {xy}$ kafi ast sabet konim:\r\n$ 2t^3 \\minus{} 5t^2 \\plus{} 1\\geq0$\r\n$ \\Leftrightarrow{2t^3 \\minus{} t^2 \\minus{} (4t^2 \\minus{} 1)} \\equal{} (2t \\minus{} 1)(t^2 \\minus{} 2t \\minus{} 1) \\equal{} (1 \\minus{} 2t)((1 \\minus{} t)(1 \\plus{} t) \\plus{} 2t)\\geq0$\r\nke ba tavajoh be inke:$ 0\\leq\\sqrt{xy}\\equal{}t\\leq\\frac{x\\plus{}y}{2}\\equal{}\\frac {1}{2}$\r\nnamosaviye akhar niz badihi ast.", "Solution_5": "are ALI666 manham hamino bedast avordam vali hoseleye neveshtanesho inja nadashtam ke didam jensen kootahtare va oono neveshtam :) \r\n\r\nva dar rabete ba chakavack bayad goft har masaleie arzeshe khase khodesho dare va nabayad daste kam gereftesh", "Solution_6": "kheyli az hamatun mamnunam.ba m bahador movafegham kheyli zud ghezavat kardam :blush: eshtebaham maskhare bud vali ba hamunam hal mishe tozihesh bashe vase bad :!: age in dafe ham ba word type konam omid hatami davam mikone :( .pas bye ta vaghti ke latex yad begiram :lol: :P :D", "Solution_7": "rahe halaye digeie vase in soal ke taze post shode :) \r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1135885&sid=5478c9d083fd7cff746cc3ec28207242#1135885", "Solution_8": "be nazare man ke asanam soale asooni nis:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?search_id=1160077791&t=160324[/url]", "Solution_9": "Manam ye rah hal daram ke bad nist :\r\nfarz konid $ x \\leq y$ . (bedoone vared shodane eshkal be kolliat) dar in soorat agar $ x \\equal{} \\sin^{2}{a}$ va $ y \\equal{} \\cos^{2}{a}$ . mishe farz kard $ a$ too bazeye $ [0,\\frac {\\pi}{4}]$ hastesh . (ba hamoon farze $ x\\leq y$)\r\nhala hokm moadele ine ke neshoon bedim :\r\n$ \\sin^{3}{a} \\plus{} \\cos^{3}{a} \\geq \\sqrt2sin{a}cos{a}$\r\npas bayad neshoon bedim $ (\\sin{a} \\plus{} \\cos{a})(1 \\minus{} sin{a}cos{a})\\geq \\frac {\\sqrt2}{2}\\sin{2a}$\r\nama $ \\sin{a} \\plus{} cos{a} \\equal{} \\sqrt2\\sin{(a \\plus{} \\frac {\\pi}{4})}$ . vali $ a \\plus{} \\frac {\\pi}{4}\\geq 2a$ hast va darnatije :\r\n$ \\sin{(a \\plus{} \\frac {\\pi}{4})}\\geq \\sin{2a}$ \r\nham chenin $ 1\\minus{}sin{a}cos{a}\\geq \\frac {1}{2}$ ke ba estefade az ina hokm bedast miad.", "Solution_10": "[quote=\"khashi70\"]Manam ye rah hal daram ke bad nist :\nfarz konid $ x \\leq y$ . (bedoone vared shodane eshkal be kolliat) dar in soorat agar $ x \\equal{} \\sin^{2}{a}$ va $ y \\equal{} \\cos^{2}{a}$ . mishe farz kard $ a$ too bazeye $ [0,\\frac {\\pi}{4}]$ hastesh . (ba hamoon farze $ x\\leq y$)\nhala hokm moadele ine ke neshoon bedim :\n$ \\sin^{3}{a} \\plus{} \\cos^{3}{a} \\geq \\sqrt2sin{a}cos{a}$\npas bayad neshoon bedim $ (\\sin{a} \\plus{} \\cos{a})(1 \\minus{} sin{a}cos{a})\\geq \\frac {\\sqrt2}{2}\\sin{2a}$\nama $ \\sin{a} \\plus{} cos{a} \\equal{} \\sqrt2\\sin{(a \\plus{} \\frac {\\pi}{4})}$ . vali $ a \\plus{} \\frac {\\pi}{4}\\geq 2a$ hast va darnatije :\n$ \\sin{(a \\plus{} \\frac {\\pi}{4})}\\geq \\sin{2a}$ \nham chenin $ 1 \\minus{} sin{a}cos{a}\\geq \\frac {1}{2}$ ke ba estefade az ina hokm bedast miad.[/quote]\r\neyval...kheili rahe ghashangi bud...", "Solution_11": "ajab soali bude khabar nadashtim :lol: \r\n\r\nrahe khodam az hame rahattare :wink: albatte age moshkeli nadashte bashe :rotfl: \r\n\r\n$ 2xy\\sqrt{xy}$be har hal mosbate pas chon $ {x^2}\\plus{}{y^2}>2xy$\r\n\r\nbe har hal $ {x^3}\\plus{}{y^3}\\plus{}2xy\\sqrt{xy}$ ham bozorgtar mosavie $ 2xy$ e.", "Solution_12": "chakavak hamoontori ke bacheha bala goftan to nemitooni az $ 2xy\\sqrt{xy}$ sarfe nazar koni dar vaghe nemitooni begi $ x^{3}\\plus{}y^{3}\\geq 2xy$ chon in mesale naghz dare oonam $ x\\equal{}y\\equal{}\\frac{1}{2}$ hast .", "Solution_13": "bayad sabet konim\r\n\\[ \\frac x{\\sqrt y} \\plus{} \\frac y{\\sqrt x}\\geq\\sqrt2\r\n\\]\r\n\r\n\\[ \\frac x{\\sqrt y} \\plus{} \\frac y{\\sqrt x} \\equal{} \\frac {x^2}{x\\sqrt y} \\plus{} \\frac {y^2}{y\\sqrt x}\r\n\\]\r\nLemma :\r\n\\[ \\frac {a^2}{b} \\plus{} \\frac {c^2}{d}\\geq\\frac {(a \\plus{} c)^2}{b \\plus{} d}\r\n\\]\r\nis true for all\r\n\\[ a,b,c,d\\geq0\r\n\\]\r\nba tavajoh be Lem darim :\r\n\\[ \\frac {x^2}{x\\sqrt y} \\plus{} \\frac {y^2}{y\\sqrt x}\\geq\\frac {(x \\plus{} y)^2}{(x\\sqrt y \\plus{} y\\sqrt x)} \\equal{} \\frac {1}{(x\\sqrt y \\plus{} y\\sqrt x)}\r\n\\]\r\npas bayad sabet konim :\r\n\\[ \\frac {1}{(x\\sqrt y \\plus{} y\\sqrt x)}\\geq\\sqrt2\r\n\\]\r\nMidanim Min in ebarat vaghti ast ke makhraj max bashad . midanim\r\nAM-GM :\r\n\\[ {(x \\plus{} y)}\\geq{2}\\sqrt xy\r\n\\]\r\n==>\r\n\\[ \\frac {1}{2}\\geq\\sqrt xy\r\n\\]\r\ncooshy :\r\n\\[ {(x \\plus{} y)}{(1 \\plus{} 1)}\\geq{(\\sqrt x \\plus{} \\sqrt y)^2} \\equal{} \\equal{} > \\sqrt2 \\geq{(\\sqrt x \\plus{} \\sqrt y)}\r\n\\]\r\n\r\n\\[ {(x\\sqrt y \\plus{} y\\sqrt x)} \\equal{} {\\sqrt xy}{(\\sqrt x \\plus{} \\sqrt y)}\\leq \\frac {1}{2} * \\sqrt2 \\equal{} \\frac {\\sqrt2}{2}\r\n\\]\r\n==>\r\n\\[ \\frac {1}{(x\\sqrt y \\plus{} y\\sqrt x)}\\geq\\frac {1}{\\frac {\\sqrt2}{2}} \\equal{} \\sqrt2\r\n\\]\r\nsoale khoobi bood :)", "Solution_14": "age eshtebah nakonam male india 1992 bud soale rahatie rahe khodam be jensen hast vali yeki az bacheha ba miangin kheili ghasghang hal karde bud \r\n\r\nin forum ham kam kam dare khub mishe" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Given two concentric circles with radii $r$ and $R.$ Find the maximal number $n$ of circles touching both of these circles simultaneously and not intersecting each other. Show that this number $n$ lies between the numbers\r\n\r\n$\\begin{equation*}\r\n\\frac{3}{2}\\cdot\r\n\\frac{\\sqrt{R}+\\sqrt{r}}{\\sqrt{R}-\\sqrt{r}}-1\\text{\\ \\ \\ \\ \\ \\ \\ \\\r\n\\ \\ and\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ }\\frac{63}{20}\\cdot \\frac{R+r}{R-r}.\r\n\\end{equation*}$", "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url] :)" } { "Tag": [], "Problem": "Find an expression for 3/5 as a finite sum of distinct reciprocals of\r\npositive integers. (For example: 2/7 = 1/7+1/8+1/56.)\r\n\r\nProve that any positive rational number can be so expressed.", "Solution_1": "Follow these steps to get your proof:\r\n\r\na) Prove that if $ 1/(n\\plus{}1)1) prove that[img]http://www.mathlinks.ro/Forum/latexrender/pictures/ade409d2934c79aa3da282b3e87c1d7b.gif[/img] is ireducibile.(in Z[X])" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Consider a sequence $\\{a_{n}\\}_{1}^{\\infty}$ of positive reals verifying :\r\na) there exist $c \\in \\mathbb R$ such that $a_i \\leq c$ for $i=1,2,...$\r\nb) $|a_i-a_j| \\geq \\frac{1}{i+j}$ for all $i \\neq j$\r\nProve that $c \\geq 1$", "Solution_1": "Yes $Badr$ you are right, its from ISL 2002.\r\nHope we will see different solutions :?" } { "Tag": [ "inequalities", "complex numbers" ], "Problem": "Let $z_1, \\ldots, z_n$ be complex numbers. Prove that there is a set $S \\subseteq \\{ 1, \\ldots, n \\}$ such that \r\n\\[\r\n\\big|\\sum_{j \\in S} z_j\\big| \\geq \\frac{1}{\\pi} \\sum_{j=1}^n |z_j| \\, .\r\n\\]\r\n\r\nFor partial credit: Prove the result with \"$\\pi$'' replaced by a larger constant.\r\n\r\nSource: I once saw this result cited in the endnotes of Rudin's book on [i]Real and Complex Numbers[/i], in the chapter on complex measures. Unfortunately, I don't remember who to credit the result to. (If anyone knows, please let me know.)", "Solution_1": "i thought this a pretty interesting result, so i took the liberty of posting it in the inequalities forum for more people to see. it already received a response.\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=205886#p205886[/url]" } { "Tag": [ "function", "LaTeX", "calculus", "calculus computations" ], "Problem": "Suppose that f(x) tends to L as x tends to infinity, and let g(x)=f(1/x). Prove that g(x) tends to L as x tends to zero+ (tend to zero from above)\r\nIf you have any idea, your help would be valuable.", "Solution_1": "What have you tried so far? What happens when you try to directly apply the definition of limit of a function at a point?", "Solution_2": "What i cannot do is to argue about the limit as x tends to zero..I don't know how to arrive to that fact from the epsilon-delta definition of the limit of a function, which involves the limit as x tends to infinity.", "Solution_3": "There aren't any hints? I am really stuck with this problem, so any ideas would be really useful.", "Solution_4": "What does \"$ f(x)$ tends to $ L$ as $ x$ tends to infinity\" mean? Can you write this down explicitly, something like \"for every there is such that \"?\r\n\r\nIf you can, you should be well on your way - write this down in this thread and we'll take the next step.\r\n\r\nIf you can't, you have to go back to your notes or textbook and figure it out since otherwise there's no way you can get started on this problem.", "Solution_5": "it means that given any e>0 there exists a d>0 such that |f(x)-L|<=e whenever |x|>d", "Solution_6": "In $ \\text{\\LaTeX}$: \r\n\r\n1. Given any $ \\epsilon > 0$ there is a $ d$ such that $ |f(x) - L| \\leq \\epsilon$ whenever $ x > d$. \r\n\r\nSince $ x \\to \\infty$ you actually know that $ x$ is large, not that $ |x|$ is large. You don't know anything about $ f(x)$ when $ x$ is negative. \r\n\r\nThis is the data. What you [i]want [/i]to show is this:\r\n\r\n2. Given any $ \\epsilon > 0$ there is a $ \\delta$ such that $ |f(\\frac {1}{z}) - L| \\leq \\epsilon$ whenever $ |z| < \\delta$.\r\n\r\nI changed the variable to $ z$ to make it easier to see what's going on. I did not change the $ \\epsilon$ because you can, and should, rely on the $ \\epsilon$ in statement 1 to prove statement 2. \r\n\r\nYour task is to define this $ \\delta$, given $ \\epsilon$. It's unlikely you can somehow define $ \\delta$ directly from $ \\epsilon$, but it is very likely that you should make use of statement 1., which gives you a $ d$. Can you guess how $ \\delta$ should be defined as a function of $ d$? Can you make it so that if $ |x| < \\delta$, you know something about $ 1/x$ that enables you to conclude what you need about $ f(\\frac {1}{x}) - L$?" } { "Tag": [], "Problem": "can anyone tell me why the sum of the angles of a triangle is 180?", "Solution_1": "Draw a line going through a vertex parallel to the opposite side of the vertex. Then, you see that angles of the triangle can all lie on the line, thus $180^{\\circ}$.", "Solution_2": "here's a drawing. line MN is parallel to side EF. so, angle MTF is congruent to angle EFT and angle FET is congruent to NTE. since angle MTN is 180, the triangle's angles must also sum to 180.", "Solution_3": "then, how do you know that angle ftm is congruent to angle tfe? what about angle ETN and angleTEF? if you are trying to say that it's because MN is paralell to FE, can you prove that in a logical way?", "Solution_4": "[quote=\"joseph14i\"]then, how do you know that angle ftm is congruent to angle tfe? what about angle ETN and angleTEF? if you are trying to say that it's because MN is paralell to FE, can you prove that in a logical way?[/quote]\r\nSince the two lines are parallel, the two angles are congruent by alternate interior angles.", "Solution_5": "For the sake of contradiction, assume that the angles you mentioned are not congruent. If not, then draw a line (i.e. Tx) such that $\\widehat{ETN}$ and $\\widehat{TEF}$ are congruent. Then $Tx\\parallel FE$. In this case, we would have two parallels drawn from a single point to another line, but that's not possible according to Euclid's Parallel Postulate." } { "Tag": [ "integration", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "show :D\r\n\r\n$ \\sum_{m\\equal{}1}^{\\infty}\\sum_{n\\equal{}1}^{\\infty}\\;\\frac{1}{m^{2}n\\plus{}2mn\\plus{}mn^{2}}\\;\\;\\equal{}\\;\\;\\boxed{\\frac{7}{4}}$", "Solution_1": "More generally, let $ k$ be a natural number. Then \r\n\r\n$ \\sum_{m_{1},m_{2},\\cdots, m_{n}\\geq 1}\\frac{1}{m_{1}\\cdots m_{n}(m_{1}\\plus{}m_{2}\\plus{}\\cdots\\plus{}m_{n}\\plus{}k)}\\equal{}$\r\n\r\n$ \\equal{}(\\minus{}1)^{n}\\int_{0}^{1}x^{k\\minus{}1}\\ln^{n}(1\\minus{}x)dx$.\r\n\r\nWe have that\r\n\r\n$ S\\equal{}\\sum\\frac{1}{m_{1}\\cdots m_{n}}\\int_{0}^{1}x^{m_{1}\\plus{}\\cdots\\plus{}m_{n}\\plus{}k\\minus{}1}dx$\r\n\r\n$ \\equal{}\\int_{0}^{1}x^{k\\minus{}1}\\sum_{m_{1}}\\frac{x^{m_{1}}}{m_{1}}\\cdots\\sum_{m_{n}}\\frac{x^{m_{n}}}{m_{n}}dx$\r\n\r\n$ \\equal{}\\int_{0}^{1}x^{k\\minus{}1}\\left(\\ln\\frac{1}{1\\minus{}x}\\right)^{n}dx$\r\n\r\n$ \\equal{}\\int_{0}^{1}(\\minus{}1)^{n}(1\\minus{}x)^{k\\minus{}1}\\ln^{n}(x)dx$.\r\n\r\nWhen $ n\\equal{}2$ and $ k\\equal{}2$ we get that the sum equals\r\n\r\n$ \\int_{0}^{1}(1\\minus{}x)\\ln^{2}(x)dx\\equal{}\\frac{7}{4}$." } { "Tag": [ "function", "modular arithmetic", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Find all functions with real values, defined for integers and such that $ f(1)+f(ab)+f(a+b)= f(a)+f(b)+f(ab+1)$", "Solution_1": "Hi,\r\n\r\nWe can see quite quickly that :\r\n1) $f(n) = xn^{2}+yn+z$ is a family of solutions\r\n2) if $f$ and $g$ are solutions, $x f+y g$ is solution\r\n3) $f(0)$ can be any value\r\n4) the knowledge of $f$ in 1, 2, 3, 4, 5 and 6 determines totally $f$ (except $f(0)$, of course, which can take any value)\r\n\r\nWell, but, from here to the identification of all the solutions, there's still a big step...", "Solution_2": "mathmanman has stated that there's linearity (property 2), so it would be enough to find all the independent solutions (from linearity, we can get all the others).\r\nSo far, he has given some of them:\r\n$f(x)=z^{2}$,\r\n$f(x)=z$,\r\n$f(x)=1$,\r\n\r\nIf $f(0)$ is completely independent of the other values, then $f(x)=\\delta (x)$ (defined as usual, $\\delta (0)=1$, and $\\delta (x)=0$ if $x\\neq 0$) should be another independent solution.\r\nBut it's not.\r\nTaking $a+b=0$,\r\n$f(1)+f(-a^{2})+f(0)=f(a)+f(-a)+f(1-a^{2})$,\r\nwhich doesn't hold in general for $f(x)=\\delta (x)$, because\r\n$\\delta (0)=1$,\r\n$\\delta (1)=\\delta (a)=\\delta (-a)=\\delta (-a^{2})=\\delta (1-a^{2})=0$ if $a\\notin\\{-1,0,1\\}$.", "Solution_3": "Hi,\r\n\r\nFirst, in my last post, in the 4th point, the \"5\" shouldn't be here.\r\n\r\nSecond, now I have the solution, but in a very unsatisfying way. I'm interested in any direct and elegant solution.\r\n\r\n$P(a,b) : f(a+b)+f(1)+f(ab) = f(a)+f(b)+f(1+ab)$ \r\nAnd we thus have $P(a, b)$ true for all $a, b$ integers.\r\n\r\n1) the set of the solutions is a $\\mathbb{R}$-vectorial space\r\n================================\r\nobvious : $f(x)$ solution $\\implies xf$ is solution\r\n$f(x)$ and $g(x)$ solutions $\\implies f(x)+g(x)$ is solution\r\n\r\n2) the dimension of this space is at most 6\r\n==============================\r\n$P(a,2) \\implies E_{1}: f(a+2)+f(1)+f(2a) = f(a)+f(2)+f(2a+1)$\r\n$P(a,4) \\implies E_{2}: f(a+4)+f(1)+f(4a) = f(a)+f(4)+f(4a+1)$\r\n$P(2a,2) \\implies E_{3}: f(2a+2)+f(1)+f(4a) = f(2a)+f(2)+f(4a+1)$\r\n\r\n$E_{1}: f(2a+1) = f(2a)+f(a+2)-f(a)+f(1)-f(2)$\r\n$E_{3}-E_{2}\\implies E_{4}: f(2a+2) = f(2a)+f(a+4)-f(a)+f(2)-f(4)$\r\n\r\nThen let us assume $f$ fixed for 0, 1, 2, 3, 4 and 6 with the values $x_{0}, x_{1}, x_{2}, x_{3}, x_{4}$ and $x_{6}$.\r\n\r\n$E_{1}$ with $a=2$ then allows us to determine $f(5) = 2x_{4}-2x_{2}+x_{1}$, $E_{1}$ and $E_{4}$ then allow us to determine all the other values.\r\nWarning, I am not saying that $x_{0}, x_{1}, x_{2}, x_{3}, x_{4}$ and $x_{6}$ can be any values. I say that if $f$ is solution, the knowledge of these values determines entirely $f$.\r\n\r\nThis shows well that the dimension of the vectorial space is at most 6. It can possibly be less if we cannot choose $x_{0}, x_{1}, x_{2}, x_{3}, x_{4}$ and $x_{6}$ independently. I am going to show that this dimension is exactly 6 by finding a basis. \r\n\r\n3) we can find 6 solutions linearly independent.\r\n====================================\r\n$S_{1}: f_{1}(0) = 1 and f_{1}(n)=0$ for all $x > 0$\r\n$S_{2}: f_{2}(n) = 1$\r\n$S_{3}: f_{3}(n) = n$\r\n$S_{4}: f_{4}(n) = n^{2}$\r\n$S_{5}: f_{5}(n) =$\r\n36 if $n=0 \\pmod 6$\r\n11 if $n=1 \\pmod 6$\r\n20 if $n=2 \\pmod 6$\r\n27 if $n=3 \\pmod 6$\r\n20 if $n=4 \\pmod 6$\r\n11 if $n=5 \\pmod 6$\r\nChecking that $f_{5}$ is solution is done easily considering the 36 values of the couple $(x \\mod 6, y \\mod 6)$\r\n$S_{6}: f_{6}(n) =$\r\n12 if $n=0 \\pmod 6$\r\n5 if $n=1 \\pmod 6$\r\n8 if $n=2 \\pmod 6$\r\n9 if $n=3 \\pmod 6$\r\n8 if $n=4 \\pmod 6$\r\n5 if $n=5 \\pmod 6$\r\nChecking that $f_{6}$ is solution is done as for $f_{5}$.\r\n\r\nThe linear independence of these 6 solutions is quite easy to see :\r\nLet $x_{1}f_{1}(x)+x_{2}f_{2}(x)+x_{3}f_{3}(x)+x_{4}f_{4}(x)+x_{5}f_{5}(x)+x_{6}f_{6}(x) = 0$.\r\n$f_{1}, f_{5}$ and $f_{6}$ being bounded, this equality implies (let $x \\to+\\infty$) that $x_{4}$, then $x_{3}$ are zero.\r\nThen letting $a=0, 1, 2$ and 3, it comes :\r\n$x_{1}+x_{2}+36x_{5}+12x_{6}= 0$\r\n$x_{2}+11x_{5}+5x_{6}= 0$\r\n$x_{2}+20x_{5}+8x_{6}= 0$\r\n$x_{2}+27x_{5}+9x_{6}= 0$\r\nand $x_{1}= x_{2}= x_{3}= x_{4}= x_{5}= x_{6}= 0$\r\n\r\n4) the set of the solutions is thus a $\\mathbb{R}$-vectorial space of dimension 6 of which we know a basis. We thus have the set of the solutions.\r\n\r\nWe indeed find back the binomials (combinations of $f_{2}, f_{3}$ and $f_{4}$).\r\nWe find back the independence of $f(0) (f_{1})$\r\nWe add to it two strange periodical solutions : $f_{5}$ and $f_{6}$.\r\n\r\nThat's it.", "Solution_4": "Wow, mathmanman :!: Astonishing work!\r\n\r\nI still have lots of doubts about your solution, though. :( \r\n\r\nDon't worry, most of them are because of my lack of knowledge about vectorial spaces :( , so I don't quite understand the way you proved the linear independence of the solutions. But I understand it could be easily checked with advanced algebra and/or a computer, so I'm not specially worried about that.\r\n\r\nAnyway, maybe you could help me with these two easier questions:\r\n\r\nThere's some typo in your definition of $f_{1}$.\r\nDid you mean $f_{1}(0)=1$ and $f_{1}(n)=0$ for all $n\\neq 0$?\r\nI still feel that this is not a solution. Taking $a=t$, $b=-t$, where $t\\in\\mathbb{Z}\\setminus\\{-1,0,1\\}$, the functional equation doesn't hold!\r\n\r\nDid you find $f_{5}$ and $f_{6}$ by computer? They are hard to check, they must have been harder to find. Maybe some other linear combination would have been more meaningful.", "Solution_5": "I'm still stuck with this problem.\r\nIt seems that mathmanman didn't consider the negative integers.\r\nStill, if I'm not wrong, with his equations $E_{1}$ and $E_{4}$ it can indeed be proved that the dimension of the vectorial space of solutions is $6$.\r\nOn the other hand, using $E_{1}$ and $E_{4}$, $f(0)$ has an influence on the values of $f(n)$ for negative values of $n$. If $f_{1}$ isn't a solution of the functional equation, we should find a sixth independent solution.", "Solution_6": "Sorry for the late reply, but I was on a vacation last week.\r\n\r\n\r\n[quote=\"lordWings\"]Wow, mathmanman :!: Astonishing work!\n[/quote]\nThanks! :D\n\n[quote=\"lordWings\"]Did you mean $f_{1}(0)=1$ and $f_{1}(n)=0$ for all $n\\neq 0$?\nI still feel that this is not a solution. Taking $a=t$, $b=-t$, where $t\\in\\mathbb{Z}\\setminus\\{-1,0,1\\}$, the functional equation doesn't hold!\n[/quote]\nYou are right. But for me \"integers\" means \"natural integers\".\nSo I did not take into account the aspects \"relative integers\". :(\n\n[quote=\"lordWings\"]\nDid you find $f_{5}$ and $f_{6}$ by computer? They are hard to check, they must have been harder to find. Maybe some other linear combination would have been more meaningful.[/quote]\r\nYes, in my previous example, the solutions $f_{5}$ and $f_{6}$ rely on two sextuplets (36, 11, 20, 27, 20, 11) and (12, 5, 8, 9, 8, 5) that I have found experimentally.\r\n\r\nIf we replace $f_{5}$ and $f_{6}$ with :\r\n$f_{5b}= 4f_{2}+f_{5}/3-4f_{6}/3$\r\n$f_{6b}=-f_{5}/4+3f_{6}/4$\r\nWe have again a basis of the vectorial space and $f_{5b}$ and $f_{6b}$ are nicer since they rely on the following sextuplets : \r\nfor $f_{5b}$ : (0, 1, 0, 1, 0, 1)\r\nfor $f_{6b}$ : (0, 1, 1, 0, 1, 1)\r\n\r\nAnd so :\r\n$f_{5b}(x) = \\mod{(x,2)}= x-[x/2]$\r\n$f_{6b}(x) = \\mod{(x^{2}, 3)}= x^{2}-[x^{2}/3]$\r\n\r\nWe can make a last change :\r\n$f_{5c}(x) = f_{3}(x)-f_{5b}(x) = [x/2]$\r\n$f_{6c}(x) = f_{4}(x)-f_{6b}(x) = [x^{2}/3]$\r\n\r\nWhich gives a general form much more exploitable than in my previous message :\r\n\r\n$x = 0 \\implies f(0) = c_{0}$\r\n$x > 0 \\implies f(x) = c_{1}x^{2}+c_{2}x+c_{3}+c_{4}[x/2]+c_{5}[x^{2}/3]$\r\n\r\nwith $[y]$ denoting the integer part of $y$.", "Solution_7": "Will note that over all integers, the dimension of the solution set is only $5$." } { "Tag": [ "abstract algebra", "linear algebra", "matrix", "vector", "algebra", "polynomial", "modular arithmetic" ], "Problem": "Let $R$ be a ring. (This means a commutative ring with $1$ here.) Let $n$ and $m$ be two positive integers.\r\n\r\n[b]a)[/b] Show that any module homomorphism $f$ from the module $R^{n}$ to $R^{m}$ uniquely corresponds to a $n\\times m$ matrix (i. e. a matrix with $n$ columns and $m$ rows; I am not sure whether this notation is the common one) $A$ such that $Av=f\\left(v\\right)$ for every vector $v\\in R^{n}$.\r\n\r\n[b]b)[/b] Let $X_{1}$, $X_{2}$, ..., $X_{u}$, with $u=\\binom{m}{n}$, be all $n\\times n$ minors of the matrix $A$. Show that $f$ is injective if and only if there exists no nonzero element $t$ of $R$ such that $t\\cdot\\det X_{i}=0$ for every $i\\in\\left\\{1;\\;2;\\;...;\\;u\\right\\}$. [Particularly, if $n>m$, then the matrix $A$ has no $n\\times n$ minors at all, so there exists such a nonzero element $t$ of $R$ (namely, every nonzero element of $R$ can be choosen as $t$), and thus $f$ cannot be injective.]\r\n\r\n[b]c)[/b] Let $Y_{1}$, $Y_{2}$, ..., $Y_{v}$, with $v=\\binom{n}{m}$, be all $m\\times m$ minors of the matrix $A$. Show that $f$ is surjective if and only if there exists an element $\\lambda_{i}$ of $R$ for every $i\\in\\left\\{1;\\;2;\\;...;\\;v\\right\\}$ such that $\\sum_{i=1}^{v}\\lambda_{i}\\cdot\\det Y_{i}=1$. [Particularly, if $nn$ we have $ l: =k+n-i b$. Here is how the matrix $ N\\left(g\\right)$ looks like:\n\n$ N\\left(g\\right) = \\left(\\begin{array}{cccccccc}g_b & g_{b - 1} & ... & g_1 & g_0 & & & \\\\\n& g_b & g_{b - 1} & ... & g_1 & g_0 & & \\\\\n& & \\ddots & \\ddots & \\ddots & \\ddots & \\ddots & \\\\\n& & & g_b & g_{b - 1} & ... & g_1 & g_0\\end{array}\\right)$,\n\nwhere empty spaces signify zeroes. This matrix has $ a + 1$ rows and $ a + b + 1$ columns.\n\nWe know that $ M_a\\left(f\\right)$ is surjective (since $ f$ is primitive) and that $ g$ is primitive. We want to show that $ fg$ is primitive, i. e. that $ M_{a + b}\\left(fg\\right)$ is surjective. In view of $ M_{a + b}\\left(fg\\right) = M_a\\left(f\\right)N\\left(g\\right)$, this will be clear as soon as we have shown that $ N\\left(g\\right)$ is surjective. This is a simplification of the problem, because we don't need to care for $ f$ anymore (though $ a$ is still important). But here the ugly part starts: of course, I use problem [b]c)[/b] above to show the surjectivity, and I even show more (namely, I prove that for arbitrary ring elements $ g_0$, $ g_1$, ..., $ g_n$, the ideal generated by the determinants of the $ \\left(a + 1\\right)\\times\\left(a + 1\\right)$ minors of the matrix $ N\\left(g\\right)$ is the ideal $ \\left\\langle g_0,g_1,...,g_n\\right\\rangle^{a + 1}$; note that this looks like a typical result of late 19th century invariant theory). But this still requires an induction or maybe even a double induction. I can give some more details if you ask, but don't expect me to ever give a complete description of the proof, unless I succeed to find a serious simplification.\n\n darij", "Solution_12": "[quote=\"darij grinberg\"] I am interested in others's solutions, generalizations and applications.\n[/quote]it is a consequence of basic statements about Fitting ideals (I remembered them today thinking of some lecture of the Son of God [he used them to show that some coherent sheaf is a vector bundle <- this is a in the broader sense related application]). They can also be used for statements about the minimal number of generators of a module (here, this is $ 0$)." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let [tex] a,b,c [/tex] be three positive numbers. Prove that:\r\n[tex] \\frac{a^3+b^3+c^3}{4abc}+\\frac{1}{4} \\geq (\\frac{a^2+b^2+c^2}{ab+ac+bc})^2 [/tex]", "Solution_1": "It's equivalent to \r\n$(\\sum ab)^2(\\sum a^3 + abc) \\ge 4abc(\\sum a^2)^2$\r\nMinus $4abc(\\sum ab)^2$ on both sides\r\n$(\\sum ab)^2 (\\sum a^3-3abc) \\ge 4abc((\\sum a^2)^2 -(\\sum ab)^2)$\r\n$(\\sum ab)^2 (a+b+c) \\ge 4abc(\\sum a^2+\\sum ab)$ (Cancelled $a^2+b^2+c^2-ab-bc-ca$)\r\n$\\sum_{sym} (a^3b^2) + \\frac{5}{2} \\sum_{sym} (a^2b^2c) + \\sum_{sym} a^3bc \\ge 2 \\sum_{sym} a^3bc + 2 \\sum_{sym} a^2b^2c$\r\nThis is a strict inequality true by muirhead ($(3,2,0) \\succ (3,1,1)$)\r\n\r\nSo it's true with equality holds when $a=b=c$", "Solution_2": "Umm... the last inequality you wrote, it is strange, RHS is 4 and LHS is ....tatata .\r\n :roll:", "Solution_3": "[quote=\"Anh Cuong\"]Umm... the last inequality you wrote, it is strange, RHS is 4 and LHS is ....tatata .\n :roll:[/quote]\r\n\r\nbut I have canceled $a^2+b^2+c^2-ab-bc-ca$ in the previous step. So if you regroup the initial inequality you get something like $A(a^2+b^2+c^2-ab-bc-ca) \\ge 0$ where A is L.H.S. - R.H.S. in my last inequality. :lol:", "Solution_4": "$ LHS -RHS = [(a-b)^2+(b-c)^2+(c-a)^2][\\frac{a+b+c}{2abc}-\\frac{[(a+b)^2+(b+c)^2+(c+a)^2]}{(ab+ac+bc)^2}] $ \r\n It's enough to show that : $ (a+b+c)(ab+ac+bc)^2 \\geq 2abc[(a+b)^2+(b+c)^2+(c+a)^2] $, which is true by Muirhead ;)", "Solution_5": "Ahh, sorry, I get it now :)" } { "Tag": [], "Problem": "DEATH\r\n[u]+ LIFE[/u]\r\n EREFI\r\n\r\nD+E+A+T+H+L+I+F+R=IH\r\n\r\nwhat digits are represented by each letter?\r\n\r\np.s. each letter is a different positive digit, and IH is a integer, not I*H", "Solution_1": "Are all the letters different digits? or are some of them the same? Also, for $\\ D+E+A+T+H+L+I+F+R=IH$, does $\\ IH$ mean $\\ I*H$ or the integer $\\ IH$?", "Solution_2": "Just looking at the Problem...I think the Letter stand for digits so that I=I E=E etc. IH is supposed to be 1 2 digit number like um.....eg.23, 34,25, 12...etc/", "Solution_3": "Why is this starting to get ignored?\r\n\r\nI think it's the date that pulls it far back on the list", "Solution_4": "Perhaps it's too hard for the forum. I'll move it up a level." } { "Tag": [], "Problem": "Numbers of the form $ F_{n} \\equal{} 2^{2^{n}}\\plus{}1$ are called fermat numbers.Without actually calculating $ F_{5}$,show that $ F_{5}$ is divisible by $ 641$.", "Solution_1": "I'm quite sure that you mean\r\n$ 641 | F_{5}\\equal{}{2^2}^5 \\plus{} 1$ since a fermat number has the form $ F_{n}\\equal{}{2^2}^n\\plus{}1$", "Solution_2": "Yes sorry I meant $ F_{5}\\plus{}1$\r\n\r\nI have edited my post.", "Solution_3": "Since $ 5 \\cdot 2^7 \\equiv_{641} \\minus{}1$ taking the fourth powers of the congruence gives us $ 5^4 \\cdot 2^{28} \\equiv_{641} 1$ and $ 2^{32}\\plus{}1\\equal{}F_{5} \\equiv_{641} 0$ since $ 5^4 \\equiv_{641} \\minus{}16\\equal{}\\minus{}2^4$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $x,y,z$ be positive real numbers such that $x+y+z=3$.Prove that:\r\n$\\frac{\\sqrt{x+y^{2}z^{2}}}{yz}+\\frac{\\sqrt{y+z^{2}x^{2}}}{zx}+\\frac{\\sqrt{z+x^{2}y^{2}}}{xy}\\geq 3\\sqrt{2}$", "Solution_1": "We have:\r\n$\\frac{\\sqrt{x+y^{2}z^{2}}}{yz}+\\frac{\\sqrt{y+z^{2}x^{2}}}{zx}+\\frac{\\sqrt{z+x^{2}y^{2}}}{xy}$\r\n$=\\sqrt{\\frac{x}{y^{2}z^{2}}+1}+\\sqrt{\\frac{y}{z^{2}x^{2}}+1}+\\sqrt{\\frac{z}{x^{2}y^{2}}+1}$\r\nFollowwing the inequality: $\\sqrt{a_{1}^{2}+b_{1}^{2}}+\\sqrt{a_{2}^{2}+b_{2}^{2}}+\\sqrt{a_{3}^{2}+b_{3}^{2}}\\ge\\sqrt{(a_{1}+a_{2}+a_{3})^{2}+(b_{1}+b_{2}+b_{3})^{2}}$\r\nTherefore:\r\n$\\sqrt{\\frac{x}{y^{2}z^{2}}+1}+\\sqrt{\\frac{y}{z^{2}x^{2}}+1}+\\sqrt{\\frac{z}{x^{2}y^{2}}+1}$\r\n$\\ge\\sqrt{\\left(\\frac{\\sqrt{x}}{yz}+\\frac{\\sqrt{y}}{zx}+\\frac{\\sqrt{z}}{xy}\\right)^{2}+9}$\r\n$\\ge\\sqrt\\left({3\\sqrt[3]{\\frac{1}{\\sqrt{(xyz)^{3}}}}\\right)^{2}+9}\\ge\\sqrt{9+9}=3\\sqrt{2}$\r\n\r\nDone!", "Solution_2": "[quote=\"hien\"]\nFollowwing the inequality: $\\sqrt{a_{1}^{2}+b_{1}^{2}}+\\sqrt{a_{2}^{2}+b_{2}^{2}}+\\sqrt{a_{3}^{2}+b_{3}^{2}}\\ge\\sqrt{(a_{1}+a_{2}+a_{3})^{2}+(b_{1}+b_{2}+b_{3})^{2}}$\n[/quote]\r\n\r\nHow to prove it?\r\nThx..", "Solution_3": "[quote]How to prove it?[/quote]\r\ntwo times Minkowsky", "Solution_4": "[quote=\"Yosh...\"][quote=\"hien\"]\nFollowwing the inequality: $\\sqrt{a_{1}^{2}+b_{1}^{2}}+\\sqrt{a_{2}^{2}+b_{2}^{2}}+\\sqrt{a_{3}^{2}+b_{3}^{2}}\\ge\\sqrt{(a_{1}+a_{2}+a_{3})^{2}+(b_{1}+b_{2}+b_{3})^{2}}$\n[/quote]\n\nHow to prove it?\nThx..[/quote]\r\n\r\njensen on $f(x)=\\sqrt{x}\\iff 2\\sum a_{1}^{2}+2\\sum b_{1}^{2}\\geq 2\\sum a_{1}a_{2}+2\\sum b_{1}b_{2}$ done" } { "Tag": [ "calculus", "integration", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "$a \\in ]0; \\pi / 2[$, $f(a)=a \\int_0^{\\infty} \\frac{dt}{ch(t) - cos(a)}$.\r\nFind $lim_{a \\to 0} f(a)$", "Solution_1": "If $ch\\, t$ means $\\cosh t=\\frac12(e^t+e^{-t})$, the answer is $\\pi$. Indeed, it is easy to see that, for every $\\delta>0$, the integral $\\int_\\delta^\\infty$ doesn't matter and near $0$ the denominator is equivalent to $\\frac 12(t^2+a^2)$, so $a$ times the integral from $0$ to $\\delta$ is almost the same as $\\int_0^\\delta\\frac{2a\\,dt}{t^2+a^2}=2\\arctan(\\delta/a)\\to \\pi$ as $a\\to 0+$.", "Solution_2": "[quote=\"alekk\"]$\\int_0^{\\infty} \\frac{dt}{ch(t) - cos(a)}$.\n[/quote]\r\n\r\nthe integral is $\\frac{\\pi-a}{\\sin a}$\r\n\r\nthe require limit is $\\pi$" } { "Tag": [ "calculus", "integration", "limit", "real analysis", "real analysis unsolved" ], "Problem": "f:[0;1]->R , continuous such that \r\n$\\any x \\in ]0,1], f(x) \\neq 0$\r\nf(0)=0, derivable at 0 with \r\n$ f'(0)=a\\neq 0$ \r\nLet $k>1$ \r\nCompute $\\lim_{x->0}\\int_{x}^{kx}\\frac{1}{f(t)}dt$", "Solution_1": "Because of the mean-theorem, we can write : \r\nSum(x =< t =< k*x){(t/f(t))*(1/t)} = (u(x)/f(u(x)))* Sum(x =< t =< k*x){1/t} = (u(x)/f(u(x)))*ln(k) , where x < u(x) < k*x\r\nManifestly if x --> 0 then u(x)--> 0 , so (u(x)/f(u(x))) --> 1/f(0) = 1/a , and finally the initial limit is ln(k)/a \r\n :cool:" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "a,b,c are positive real,and abc=1 Prove:\r\n1/a(a+b)+1/b(b+c)+1/c(c+a)>=3/2\r\nthanks.", "Solution_1": "[quote=\"duj12358\"]a,b,c are positive real,and abc=1 Prove:\n1/a(a+b)+1/b(b+c)+1/c(c+a)>=3/2\nthanks.[/quote]\r\ndo you mean $ \\frac{1}{a(a\\plus{}b)}\\plus{}\\frac{1}{b(b\\plus{}c)}\\plus{}\\frac{1}{c(c\\plus{}a)}\\ge \\frac{3}{2}$?", "Solution_2": "See this http://www.mathlinks.ro/viewtopic.php?p=1010959#1010959", "Solution_3": "Raja Oktovin\r\nthank you very much" } { "Tag": [ "limit", "trigonometry", "function", "logarithms", "absolute value", "real analysis", "real analysis unsolved" ], "Problem": "This one is very hard, I think. You can admit here that $\\pi$ is badly approximable with rational numbers (Mahler or Mignotte or Hata's theorem). Then study the convergence of $\\sum_{n\\geq 1}{\\frac{(-1)^n\\cdot |sinn|}{n}}$.", "Solution_1": "The only thing that comes into my mind is the well-known limit:\r\n\r\n \\[ \\lim_{n\\to\\infty} \\frac{|\\sin 1|+|\\sin 2|+\\ldots+|\\sin n|}{n}=\\frac{2}{\\pi}. \\]\r\n\r\n Harazi, is this the key of solving this problem? :?", "Solution_2": "Cezar, I say I consider it very difficult, much much more difficult than that limit. Sorry, but you have to do many more steps in order to solve this problem, it's not an ordinary one. At least, I think so.", "Solution_3": "I know a proof of this one (which is not mine of course :rotfl:). Indeed this is very difficult, to say the least.\r\nI can give hints or directions if someone is interested.", "Solution_4": "Of course, you can give a hint or something.\r\n\r\nMaybe that way i'll stop looking at this problem like the calf at the new gate. (I know, it sounds horrible in English)", "Solution_5": "Please post a full solution for this serie, please. :) I tried to solve it , but no result...... :(", "Solution_6": "[quote=\"cezar lupu\"]Please post a full solution for this serie, please. :) [/quote]\r\nOK, here goes a sketch :).\r\nFirst, decompose the function $|\\sin x|$ into its Fourier series. You'll get \r\n\\[ |\\sin x|=\\sum_{m=0}^\\infty c_m\\cos mx \\]\r\nwith some $c_m$. All we need to know about $c_m$ is that they decay like $m^{-2}$.\r\nNow let $N_1M}|c_m|\\log m$ is very small and then choose $N$ such that the sums from $N_1$ to $N_2$ of every of the first $M$ series are small if only $N_2>N_1>N$. This finishes the proof of convergence.", "Solution_7": "Beautiful! In our homework, the indications were to use the Hardy-Littlewood tauberian theorem and approximations of $\\pi$, but this is much more interesting.", "Solution_8": "Harazi can you post your solution? :)", "Solution_9": "Here are the steps, I will let you complete the proof:\r\n \r\n 1) Use the fact that $\\pi$ is badly approximable with rationals to prove that $\\frac{ ln|cosn|}{lnn}$ is bounded and that for all $k>1$ the series $\\sum {ln|cos n|}{n^k}$ is convergent.\r\n\r\n 2) Put $h(x)=\\sum {\\frac{(-1)^n|sinn|x^n}{n}}$. Express $h(x)$ in terms of $\\sum_{p\\geq 1}{\\frac{1}{4p^2-1}\\cdot\\sum_{n\\geq 1}{\\frac{(-1)^{n+1}cos(2pn)x^n}{n}}=g(x)}$ where $g$ is defined on $(-1,1)$\r\n \r\n 3) Use 1) to prove that $g$ is defined in 1. \r\n\r\n 4) Prove that $h$ has a limit in 1. \r\n\r\n 5) Conclusion: use Hardy-Littlewood tauberian theorem: if $a_n=O(1/n)$ and there exists $lim_{x goes to 1}{\\sum{a_n x^n}}$ then $\\sum{a_n}$ converges. \r\n The last \"exercise\" is in fact a very serious theorem, I would be glad to see a simple approach, since the only solution I know for it is quite long and complicated.", "Solution_10": "The connection between cezar lupu's post (#2) and fedja's post (#5): cezar has essentially given the constant term, $c_0,$ of fedja's Fourier series.", "Solution_11": "Is it a oral exam or written exam of Ecole Centrale ?\r\nWich year ?", "Solution_12": "I think not even examinators at ENS can give such a problem in oral examination. It is written exam, but I don't know the year.", "Solution_13": "The convergence of the series and some other related results are also in \"Diophantine Approximations and Convergence of Alternating Series\", by N.V.Rao in Proceedings of the American Math. Soc. Vol.93, No.3 1985", "Solution_14": "Another approach that also uses the irrationality measure of pi is given here by me: https://math.stackexchange.com/questions/2202138/determine-whether-sum-n-1-infty-frac-1n-sinnn-converges/2207289#2207289\n\nMain ingredients are \n1. Koksma inequality\n2. Erdos -Turan inequality.\n", "Solution_15": "wow , just proving the convergence is that hard , finding the limit is a whole another story :maybe: there is a lemma that I think can help with both if I don't make any nonsense ! \nfor $f : \\mathbb{R} \\to \\mathbb{C}$ continuous and $2\\pi$ periodic , let $\\alpha \\in \\mathbb{R}-\\mathbb{\\pi\\mathbb{Q}}$ for a positive integer $n\\geq 1$ \n$I_{n}(f)=\\sum_{k=1}^{n} \\frac{f(k\\alpha)}{k}$ then $(\\frac{I_{n}(f)}{\\ln(n)})$ converge to $\\mu_{f}$ The average of $f$ . ", "Solution_16": "As harazi said it was problem exam Centrale 1980 MP , analysis \n\nSee also \n[url=https://math.stackexchange.com/questions/735283/convergence-of-an-alternating-series-sum-n-geq-1-frac-1n-sin-nn?noredirect=1&lq=1] see here [/url]\n\nhttps://math.stackexchange.com/questions/735283/convergence-of-an-alternating-series-sum-n-geq-1-frac-1n-sin-nn?noredirect=1&lq=1\n\n\nHere a generalization \nN.V.Rao diophantine approximation and convergence of alternating series\nProceeding of the American Mathematical Society ,volume 93, number 3,March 1985 \npage 420-422 \n\n[url=https://www.ams.org/journals/proc/1985-093-03/S0002-9939-1985-0773994-4/S0002-9939-1985-0773994-4.pdf] see here[/url]\nhttps://www.ams.org/journals/proc/1985-093-03/S0002-9939-1985-0773994-4/S0002-9939-1985-0773994-4.pdf", "Solution_17": "The problem exam Centrale 1980 M can be find in \n[url=https://concours-maths-cpge.fr/] see here [/url]\n\nhttps://concours-maths-cpge.fr/\n\nann\u00e9e\u2014>1980\nconcours\u2014> Centrale-Supelec\nfili\u00e8re \u2014> MP (ex M)\nmati\u00e8re\u2014> math\u00e9matiques \n\nand choose \"rechercher \" , you will see \n\nCentrale MP math 1 and choose \"details\" you will find \n\n\u00e9nonc\u00e9 means the problem exams \ncorrig\u00e9 means solution" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "[color=green][b] If $a,b\\in {\\mathbb R}^n \\; ,\\; a=(a_1,a_2,...,a_n)\\; ,\\; b=(b_1,b_2,...,b_n),$ then prove the inequality\n\\[ \\begin{array}{|c|} \\hline \\\\ \\left[n\\sum\\limits_{k=1}^{n}b_k^2 -\\left(\\sum\\limits_{k=1}^{n}b_k\\right)^2\\right]\\left[ \\left(\\sum\\limits_{k=1}^{n}a_k^2\\right) \\left(\\sum\\limits_{k=1}^{n}b_k^2\\right)-\\left(\\sum\\limits_{k=1}^{n}a_kb_k\\right)^2\\right]\\ge \\\\ \\\\ \\ge \\left[\\left(\\sum\\limits_{k=1}^{n}a_k\\right)\\left(\\sum\\limits_{k=1}^{n}b_k^2\\right)- \\left(\\sum\\limits_{k=1}^{n}b_k\\right) \\left(\\sum\\limits_{k=1}^{n}a_kb_k\\right)\\right]^2\\\\ \\\\ \\hline \\end{array}\\; . \\]\n[/b][/color]", "Solution_1": "Wow,I find this nice inequality is not hard:\r\njust notice:\r\n$n\\sum\\limits_{k=1}^{n}b_k^2 -(\\sum\\limits_{k=1}^{n}b_k)^2= \\sum_{i> j} (b_i-b_j)^2$\r\nand \r\n$(\\sum\\limits_{k=1}^{n}a_k^2) (\\sum\\limits_{k=1}^{n}b_k^2)-(\\sum\\limits_{k=1}^{n}a_kb_k)^2=\\sum_{i > j} (a_jb_i-a_ib_j)^2$\r\nand \r\n$(\\sum\\limits_{k=1}^{n}a_k)(\\sum\\limits_{k=1}^{n}b_k^2)- (\\sum\\limits_{k=1}^{n}b_k) (\\sum\\limits_{k=1}^{n}a_kb_k)=\\sum_{i > j} (b_i-b_j)(a_jb_i-a_ib_j)$\r\nthen use Cauchy-Schwarz.\r\nEdited a small mistake", "Solution_2": "[quote=\"zhaobin\"]Wow,I find this nice inequality is not hard: just notice ...... [/quote] \r\nOK, but please post a complete solution ! \r\nHowever, it's clear that the posted inequality implies Cauchy- Schwarz inequality. \r\nCan you post all equality cases ? Thanking for interest,", "Solution_3": "[quote=\"flip2004\"][quote=\"zhaobin\"]Wow,I find this nice inequality is not hard: just notice ...... [/quote] \nOK, but please post a complete solution ! \nHowever, it's clear that the posted inequality implies Cauchy- Schwarz inequality. \nCan you post all equality cases ? Thanking for interest,[/quote]\r\nDo you mean when does the equality holds?\r\nIt is a problem :( .Let me think about it.Now I must to go to sleep first :P", "Solution_4": "[quote=\"zhaobin\"][quote] Can you post [u][b]all equality cases[/b][/u] ? [/quote] Do you mean when does the equality holds? It is a problem ...:[/quote]\r\n[size=75][color=blue][b]I appreciate that my posted inequality is not equivalent with Cauchy-Schwarz inequality ! [/b][/color][/size]\r\nThis may be observed by investigating all equality cases (in posted inequality) , then to compare them with equality case \r\n in Cauchy-Schwarz inequality.", "Solution_5": "[quote=\"flip2004\"][quote=\"zhaobin\"][quote] Can you post [u][b]all equality cases[/b][/u] ? [/quote] Do you mean when does the equality holds? It is a problem ...:[/quote]\n[size=75][color=blue][b]I appreciate that my posted inequality is not equivalent with Cauchy-Schwarz inequality ! [/b][/color][/size]\nThis may be observed by investigating all equality cases (in posted inequality) , then to compare them with equality case \n in Cauchy-Schwarz inequality.[/quote]\r\nYes ,But I find it is not easy.Have you find all equality cases?", "Solution_6": "[quote=\"zhaobin\"]... Can you post [u][b]all equality cases[/b][/u] ? ... Have you find all equality cases?[/quote]\r\nAssuming that $n$ is a fixed positive integer, then [b]equality case holds if and only if there exist real numbers $\\alpha,\\beta,\\gamma$ \nsuch that $\\alpha a_k+\\beta b_k +\\gamma =0$ for all $k\\in \\{1,2,...,n\\}.$ [/b]\r\n...There are similar (but different) conditions for equality when $n$ is supposed to be variable in ${\\mathbb N}^* .$([size=75][color=blue][b]Try to find it ! [/b][/color][/size] )." } { "Tag": [], "Problem": "Does anyone know any \"popular\" NJ math contests? :maybe:", "Solution_1": "NJML, CJML, AMTNJ", "Solution_2": "Don't forget CML! And for high schoolers, PUMaC." } { "Tag": [ "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "Let \r\nf(x)=(a+sinx)(4+sinx)/(1+sinx)\r\nwhere a>1 , a and x are real numbers.\r\nAs x varies, find the minimum value of f(x).", "Solution_1": "$ f$ is periodic with period $ 2\\pi$, so we need only consider values of $ x$ in the interval $ [0, 2\\pi]$.\r\n\r\nSetting $ z \\equal{} 1 \\plus{} \\sin x$ may simplify the function somewhat.\r\n\r\nIn other words, consider a new function $ g(z) \\equal{} (a \\minus{} 1 \\plus{} z)(3 \\plus{} z)/z$. You could find the minimum of $ g$ on an appropriate interval using standard techniques, and then you will have found the minimum of $ f$.", "Solution_2": "Can we use high school maths to solve this problem??" } { "Tag": [ "calculus", "integration" ], "Problem": "Find the real n such that :\r\n\\[ \\frac{{{{\\left( {C_n^1} \\right)}^2}}}{1} \\plus{} \\frac{{{{\\left( {C_n^2} \\right)}^2}}}{2} \\plus{} \\frac{{{{\\left( {C_n^3} \\right)}^2}}}{3} \\plus{} ... \\plus{} \\frac{{{{\\left( {C_n^n} \\right)}^2}}}{n} \\equal{} \\frac{{475}}{{12}}\\]", "Solution_1": "hello, put $ n\\equal{}4$ in your equation.\r\nSonnhard.", "Solution_2": "Hello\r\n[hide=\"Hint (by integration)\"]\nu know $ (1\\plus{}x)^n\\equal{}C_n^0 \\plus{} xC_n^1\\plus{}x^2C_n^2\\plus{}...........\\plus{}x^{n}C_n^n$ ...........$ equation(1)$\nTransfer x by $ \\frac{1}{x}$ and then integrate the equation.\nOn RHS of ur new equation u will get $ C_n^0 (logx)\\plus{}C_n^1x\\plus{}C_n^2\\frac{x^2}{2}..........\\plus{}C_n^n\\frac{x^{n}}{n}$\nmultiply this equation with the following equation\n$ (1\\plus{}x)^n\\equal{}C_n^0x^n\\plus{}C_n^1x^{n\\minus{}1}\\plus{}............C_n^n$\ncompare the coefficient of $ x^n$ and then put x=1.\nu will get the value of n.\n[/hide]\r\nThank u.", "Solution_3": "hello, for $ n\\equal{}4$ we get $ 16\\plus{}18\\plus{}\\frac{16}{3}\\plus{}\\frac{1}{4}\\equal{}\\frac{475}{12}$, so your equation is fulfilled.\r\nSonnhard." } { "Tag": [], "Problem": "Hi\r\n\r\nProve that:\r\n\r\na/b+b/c+c/a>=(a+b)/(a+c)+(b+c)/(b+a)+(a+c)/(b+c) with (a,b,c>0)\r\n\r\n\r\nbye", "Solution_1": "This one was (1) of the thread\r\n\r\n http://www.mathlinks.ro/viewtopic.php?t=5263\r\n\r\nAnd it is really hard, see the first solution in the [url=http://www.mathlinks.ro/Forum/contest/1st%20Edition/1st%20round%20sol.pdf]proposed solutions[/url].\r\n\r\n[[b]EDIT:[/b] Links updated.]\r\n\r\n Darij", "Solution_2": "strange, both of those links refuse to open here :?" } { "Tag": [ "AMC", "AIME", "Harvard", "college", "MIT", "ARML", "HMMT" ], "Problem": "Does anyone know where I can find some AIME-level problems?\r\n\r\nThanks.", "Solution_1": "Harvard MIT guts round is good, and so is the canadian open mathematics challenge", "Solution_2": "Thanks.\r\n\r\nAnymore?", "Solution_3": "some of the last problems (probably only#25) of the Cayley competition (canadian) might help... took me a while to set up. \r\n\r\nother than that, you probably have heard of the other ones... ARML individual, HMMT, stanfordmathmeet (although in recent years ive heard that they have dropped in the quality and difficulty)." } { "Tag": [ "geometry", "analytic geometry" ], "Problem": "If x and y are each positive integers, how many distinct ordered pairs (x,y) exist such that (x,y) lies on one of the bisectors of the angles formed by the lines x-2y+5=0 and 2x-y-3=0", "Solution_1": "This is not a pre-olympiad level problem at all. It sounds like Analytic Geometry textbook.\r\n\r\nFind the locus of points equidistant from both. That's your bisector." } { "Tag": [ "probability", "symmetry", "articles" ], "Problem": "I wasn't sure where to put this but I'm going nuts that I can't figure out the logic behind this.\r\nHere's the webpage that \"explains\" it.\r\nhttp://en.wikipedia.org/wiki/Monty_hall_problem\r\n\r\nHow am I more likely to win by switching than if I stayed on the same door? To me, there's a 50/50 chance. I don't know which door the car is behind, so switching or staying has the same chance of winning me the car.\r\n\r\nThe way I understand it is you have a 66% chance of picking a goat, which means switching would give you the car.\r\nAnd you have a 33% chance of picking the car, which means switching will give you a goat.\r\n\r\nSo the chances that you picked the goat are greater than if you picked the car, which means switching is usually a better idea.", "Solution_1": "Statements about probability are in essence statements about statistics. The statement [i]\"The outcome of experiment $ E$ is $ A$ with probability $ p$\"[/i], is essentially the same as saying [i]\"If experiment $ E$ were performed $ n$ times, we should expect $ A$ to be the outcome approximately $ np$ times.\"[/i] This is a rather simplified view of probability, but 'typically' this is a good way of thinking about it.\r\n\r\nSo now let's let $ E_{\\text{switch}}$ be the Monty Hall experiment in which you always choose to switch doors, and $ A$ be the outcome that you win the car. If $ E$ is performed $ n$ times, we expect by symmetry that the door [i]originally[/i] chosen will have a car behind it about $ \\frac {1}{3}n$ times. Therefore, switching to the other door will give you a car about $ \\frac {2}{3}n$ times. Thus, if we accept the interpretation of probability as given in the first paragraph, we must conclude that the probability of event $ A$ is $ \\frac {2}{3}$.", "Solution_2": "I don't see why switching would give you a 2 in 3 chance but staying doesn't give you a 2 in 3 chance.", "Solution_3": "It can be easily seen from the picture in link you wrote, but I'll write it here:\r\n\r\n1) You firstly have chosen the right door $ \\longrightarrow$ if you change, you lose\r\n2) You firstly have chosen the goat door $ \\longrightarrow$ as you see the another goat door opened, if you change, you win\r\n\r\nProbability that 1) case will appear is $ \\frac{1}{3}$ and 2) case will apear in $ \\frac{2}{3}$ cases because there are $ 2$ goats and $ 1$ car from $ 3$ objects", "Solution_4": "[quote=\"Bugi\"]It can be easily seen from the picture in link you wrote, but I'll write it here:\n\n1) You firstly have chosen the right door $ \\longrightarrow$ if you change, you lose\n2) You firstly have chosen the goat door $ \\longrightarrow$ as you see the another goat door opened, if you change, you win\n\nProbability that 1) case will appear is $ \\frac {1}{3}$ and 2) case will apear in $ \\frac {2}{3}$ cases because there are $ 2$ goats and $ 1$ car from $ 3$ objects[/quote]\r\nI think I understand it now. It's just that there are so many explanations that make it sound a LOT more difficult than it really is.", "Solution_5": "Hmm what if the game show host did not have to open a door with a goat in it?", "Solution_6": "Then the probability is 1/2. (This is true, for example, of the game show \"Deal or No Deal.\") This restriction (that Monty is guaranteed to open a door hiding a goat) is crucial, and it's very important when you state the Monty Hall problem that you make it clear.", "Solution_7": "what if the game show host did not know what was behind each of the doors, and randomly opens one of the other two?", "Solution_8": "but then what if there was a car behind the other door?\r\nthen the game would be pointless", "Solution_9": "violin321, enough useless questions. I could have also asked some questions like: What if we can open all 3 doors? :rotfl:", "Solution_10": "[quote=\"Bugi\"]violin321, enough useless questions. I could have also asked some questions like: What if we can open all 3 doors? :rotfl:[/quote]\r\n\r\nThis is hardly a useless question. If Monty opens a door at random and reveals a goat, the answer is not the same as if Monty opens a door which he knows does not have the car behind it.", "Solution_11": "In the Wikipedia article, the figure under the Solution section clearly illustrates why its a statistically better choice to change your option when one option is removed. The key to why it can be confusing lies in that only 3 doors are considered, if you read 'Increasing the number of doors' under the section 'Aids to Understanding' this is easily seen.\r\n\r\nIncidentally, I remember this problem being mentioned on a NUMB3RS episode! Nice show... :D" } { "Tag": [ "MATHCOUNTS", "percent", "probability" ], "Problem": "Do you have any tips you can share for doing well in the Count Down round :?:", "Solution_1": "Shake hands with them as often as possible. And give good speeches.", "Solution_2": "Another one:\r\nYou have 3 seconds to answer the question, so if you're pretty fast at it, and youve just about got it, then just raise you hand, and then in those 3 seconds, finish working the question, and answer.\r\n\r\nAND THE BIG TIP:\r\n[u][b]PRACTICE!!![/b][/u] you can ask for past countdown rounds from MATHCOUNTS, and the Elisa Saabs web page has hundreds of them.", "Solution_3": "Practice also helps... lol. Lack of that is probably why I failed countdown when it counted :(.", "Solution_4": "[quote=\"dwx314\"]Another one: \nYou have 3 seconds to answer the question, so if you're pretty fast at it, and youve just about got it, then just raise you hand, and then in those 3 seconds, finish working the question, and answer. [/quote]\r\n\r\nActually, you have about 6 seconds. First they announce your name and school, then you get 3 seconds, then you can waste 1 second by starting your answer as \"umm\".\r\n\r\nAnother one~~If there is 10 seconds left and you can't get the answer, and you can see your opponent is just about to get it, buzz in, then pretend like you're thinking a lot until they ask for the answer. Then give them a huge answer that takes about 5 seconds to say. ;) Your opponent will run out of time, and no one will win.(that question)", "Solution_5": "[quote=\"ln(dx/dy)\"][quote=\"dwx314\"]Another one: \nYou have 3 seconds to answer the question, so if you're pretty fast at it, and youve just about got it, then just raise you hand, and then in those 3 seconds, finish working the question, and answer. [/quote]\n\nActually, you have about 6 seconds. First they announce your name and school, then you get 3 seconds, then you can waste 1 second by starting your answer as \"umm\".\n\nAnother one~~If there is 10 seconds left and you can't get the answer, and you can see your opponent is just about to get it, buzz in, then pretend like you're thinking a lot until they ask for the answer. Then give them a huge answer that takes about 5 seconds to say. ;) Your opponent will run out of time, and no one will win.(that question)[/quote]\r\nNow that's just cruel! But hey, whatever it takes, huh? :D My advice is to not think of how much it counts-just focus on the problem. Disregard that if you're like me and thrive under pressure, though. Adrenaline helps me think.", "Solution_6": "You definitely will not get 6 seconds at Nationals. They'll say your first name and expect an answer. \r\n\r\n3 seconds is a reasonable expectation.", "Solution_7": "I won Countdown in the Indiana State competition, but, unfortunately, it doesn't count here. I just scan the question really fast and mainly look for the numbers. Then I read the last line and see what they are asking for. It also helps that I am on the Science Bowl team. My Science Bowl team made it to Nationals in Colorado last year.", "Solution_8": "I suggest practicing without pencil and paper fairly regularly. Learn to do things like multiply, factor integers, reduce fractions, and convert between decimal/fraction in your head. Winning in the count down round has a lot to do with mental arithmetic abilities -- particularly at the state and national levels where many students know how to work nearly every problem.", "Solution_9": "I'll add just two:\r\n\r\n1 - Answer the question before your opponent does. \r\n2. Make sure that the answer is correct.", "Solution_10": "Good ones Gyan. Also, RIng the buzzer and don't raise ur hand(surprising how many people do.). Bring a pen also. Writes faster.", "Solution_11": "[quote=\"SirErnest\"]I won Countdown in the Indiana State competition, but, unfortunately, it doesn't count here. I just scan the question really fast and mainly look for the numbers. Then I read the last line and see what they are asking for. It also helps that I am on the Science Bowl team. My Science Bowl team made it to Nationals in Colorado last year.[/quote]\r\n\r\nI remember that...state CountDown I mean. Another hint that I picked up at state, when teh answer says to answer as a percent, answer as a percent not a fraction. And when the guy next to you answers as a fraction, quickly convert it to a percent and buzz in adn get the correct answer. :)", "Solution_12": "Lucky for me it doesn't count. :D", "Solution_13": "[quote=\"sponge008\"][quote=\"ln(dx/dy)\"]\n[quote=\"dwx314\"] \nAnother one: \nYou have 3 seconds to answer the question, so if you're pretty fast at it, and youve just about got it, then just raise you hand, and then in those 3 seconds, finish working the question, and answer. \n[/quote]\n\n\nActually, you have about 6 seconds. First they announce your name and school, then you get 3 seconds, then you can waste 1 second by starting your answer as \"umm\". \n\nAnother one~~If there is 10 seconds left and you can't get the answer, and you can see your opponent is just about to get it, buzz in, then pretend like you're thinking a lot until they ask for the answer. Then give them a huge answer that takes about 5 seconds to say. Your opponent will run out of time, and no one will win.(that question) \n[/quote]\n\nNow that's just cruel! But hey, whatever it takes, huh? My advice is to not think of how much it counts-just focus on the problem. Disregard that if you're like me and thrive under pressure, though. Adrenaline helps me think.[/quote]You have to think about how much it counts. If it is in sudden death, sometimes you have to break the rules a little if you want any chance of winning. Once last year in 6th grade, when we were doing a practice cd round in school, I had a coughing fit when I buzzed in, which gave me a lot of time to finish the problem. The answer was accepted, but I got it wrong anyway. :D :D :D :D :D :D :D", "Solution_14": "Countdown is unofficial in my chapter and state :D (this is NY)", "Solution_15": "[quote=Ultroid999OCPN][b][s]12.5 year bumps are lit[/s][/b]\n.........Really?\n@GrilledBrain: Please don't bump threads.\nPersonally, I feel that if you freak out, you will lose, so staying calm is important.[/quote]\n\nPlease follow your own advice\nIf you really want to tell them not to bump threads, just PM them", "Solution_16": "It really depends what level you're on. \nIf it's a school or chapter round, then countdown is probably unofficial, so you don't need to worry about it too much. Some chapters/schools have buzzers; some do not. \nIf it's state, it almost certainly has buzzers, and in some states, countdown is official, while in others, it's not. Unlike school and chapter, people will generally be much more strict about the [b]three seconds[/b] you have to answer.\nIf it's nationals, then you also have the pressure of being on live TV in front of 200+ people, besides facing some of the best mathletes in the country. Besides that, it's mostly the same as state.\n\nAll in all, just know your strengths and ability, and play accordingly.", "Solution_17": "[quote=thedoge]It really depends what level you're on. \nIf it's a school or chapter round, then countdown is probably unofficial, so you don't need to worry about it too much. Some chapters/schools have buzzers; some do not. [/quote]\nYeah lol our chapter doesn't even do countdown\nAnd why is this thread from 2005?\n", "Solution_18": "our chapter does official countdown. also, u have 45 seconds to answer questions in countdown....", "Solution_19": "[quote=MathMan5000]our chapter does official countdown. also, u have 45 seconds to answer questions in countdown....[/quote]\n\nThey mean after you hit the buzzer.", "Solution_20": "Don't do what I did this year on the Chapter Countdown Final for this year. I buzzed in and when I said the correct answer, the time for speaking ran out. And then the other person buzzed in and immediately said the same answer... Got it right. This happened twice, cuz i hesitated before answering. Somehow i still made a comeback....... :10: ", "Solution_21": "And also, it helps to be relaxed during the countdown as much as possible. Just don't stress out.", "Solution_22": "[quote=ln(dx/dy)]Do you have any tips you can share for doing well in the Count Down round :?:[/quote]\n\nwhen you're \"working out\" the problems on paper, try writing really noisily to distract and/or scare your opponent...\n\nno joke it works", "Solution_23": "Our countdowns just if you get the answer right, you move on. Make a silly mistake, well your out.", "Solution_24": "The most important thing about countdown that you need to remember is to not hesitate. Remember that it's a fast-paced round, and that if you take too long, you'll be beat to the question.\n\n\n\n[quote=e_power_pi_times_i]It is very hard to \"keep your left hand at the buzzer and right hand for writing\" when you are writing with your left hand. Thunk.[/quote]\n\n[quote=GameMaster402]then switch it lol[/quote]\n\nas a lefty individual i can confirm that this strategy is very effective", "Solution_25": "My state and chapter both use countdown officially, and this is pretty basic, but read the problem as FAST as you can and understand it quickly, as it can give you a split second advantage which could be what decides who wins that question.", "Solution_26": "um so try skipping to the end of the question to see what they want, then pick off the important info", "Solution_27": "I wouldn't recommend this strategy. Read the problem normally.", "Solution_28": "[quote=KevinTheFirst][quote=thedoge]It really depends what level you're on. \nIf it's a school or chapter round, then countdown is probably unofficial, so you don't need to worry about it too much. Some chapters/schools have buzzers; some do not. [/quote]\nYeah lol our chapter doesn't even do countdown\nAnd why is this thread from 2005?[/quote]\n\nNo KevinTheFirst, our chapter does do countdown, it's just that it's not official.", "Solution_29": "More than likely, if you read the first sentence and the last sentence, you will already know what the question will be about.\n\nEDIT:300 MSM posts!!!!!!!!!!!!!!!!!!!!!!!!!!!" } { "Tag": [ "ratio", "geometry", "rectangle" ], "Problem": "Two circles whose radii are in the ratio $4 : 1$ touch each other externally at $M$ an lie inside a rectangle $ABCD$ such that the larger circle touches sides $AD,BC$ and $CD$, and the smaller circle touches the sides $AB$ and $AD$. The common tangent at $M$ to the circles meets sides $AD$ and $AB$ at $P$ and $Q$.\r\nFind the ratios $AP/PD$ and $AQ/QB$.", "Solution_1": "Is it 1/3 and 3/5 ?\r\n\r\nMy circle in triangle geometry's kinda sketchy...", "Solution_2": "Are you sure?\r\nCan you post a complete solution?", "Solution_3": "I'm getting $\\frac{AP}{PD}=\\frac{1}{2}$ and $\\frac{AQ}{QB}=1$. Here's my solution...\r\nLet the center of the larger circle be $O$, and that of the smaller circle be $I$. Draw the radius of the larger circle, $OE$ so that $E$ is on $AD$. Now let the smaller circle touch $AB$ and $AD$ at $X$ and $Y$ respectively. Let the circles touch at $L$. Drop a perpendicular from $I$ to $OE$: call the intersection point $N$. Let $IN$ intersect $PQ$ at $M$. Let the radius of the larger circle be 4, that of the smaller 1.\r\nNow that those are all defined, the solution...\r\nClearly, $AB=8$. $OL=5, ON=4-1=3 \\implies IN=4 \\implies NX=5 \\implies AD=9$. Note the following about the following triangles. $ION~INL~QNX~QPA$. Thus, $QPA$ is a 3-4-5 right triangle. Let $AX=AY=1$, $PY=PL=n$, $QL=QX=x$. Then $(1+n) : (1+x) : (n+x)=3:4:5 \\implies n=2,x=3$.\r\n$\\therefore$ $\\frac{AP}{PD}=\\frac{n+1}{8-n}=\\frac{1}{2}$, and $\\frac{AQ}{QB}=\\frac{x+1}{7-x}=1$." } { "Tag": [ "quadratics", "number theory proposed", "number theory" ], "Problem": "Prove for all positive integer $n$: $2^n+1$ has no prime divisor in form $8k-1$.", "Solution_1": "It's from Vietnam 2004 TST. The proof is simple for those who know quadratic reciprocity and practically impossible to find otherwise." } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "Prove that:\r\n\r\n$\\displaystyle\\lim_{x\\rightarrow 0} \\frac{1-\\cos x}{x} = 0$", "Solution_1": "[quote=\"gotztahbeazn\"]Prove that:\n\n$\\displaystyle\\lim_{x\\rightarrow 0} \\frac{1-\\cos x}{x} = 0$[/quote]\r\n\r\n$= \\mathop {Lt}\\limits_{x \\to 0} \\frac{{1 - \\cos ^2 x}}{{x(1 + \\cos x)}} = \\mathop {Lt}\\limits_{x \\to 0} \\frac{{\\sin ^2 x}}{{x^2 (1 + \\cos x)}}.x = \\mathop {Lt}\\limits_{x \\to 0} \\frac{{\\sin ^2 x}}{{x^2 }}\\frac{1}{{(1 + \\cos x)}}.x = 0$", "Solution_2": "L'Hopital rule works here.\r\n\\[\\displaystyle\\lim_{x\\rightarrow 0} \\frac{1-\\cos x}{x}=\\displaystyle\\lim_{x\\rightarrow 0} \\frac{\\sin x}{1}=0\\]", "Solution_3": "Let $f(x)=\\cos x$, we have $f'(x)=-\\sin x$.\r\n\r\n$\\lim_{x\\rightarrow 0} \\frac{1-\\cos x}{x}=-\\lim_{x\\rightarrow 0}\\frac{f(x)-f(0)}{x-0}=-f'(0)=0$", "Solution_4": "$1-\\cos x\\sim \\frac{1}{2}x^2\\ (x\\to 0)$" } { "Tag": [ "ARML", "Euler", "geometry" ], "Problem": "ARML's in less than a month, so I was wondering who all from AoPS is going to ARML at Iowa in 07.\r\n\r\nI'll be there with Indiana Gold.", "Solution_1": "I'll probably be there, for MO team", "Solution_2": "hmm I'll probably go on bronze, if I'm lucky I\"ll get on silver...(from IN)", "Solution_3": "I'll be there from TX", "Solution_4": "if anyone wants to play basketball on friday, some ppl from indaina will probably be there playin so we could meet up", "Solution_5": "Well, I'll be there...\r\n\r\n...from Iowa City.\r\n\r\nI think I'll walk to the test site!", "Solution_6": "I'll be there from Chicago, hopefully Chi A.\r\n\r\nJB", "Solution_7": "[quote=\"jackb1115\"]I'll be there from Chicago, hopefully Chi A.\n\nJB[/quote]\r\n\r\nme too", "Solution_8": "[quote=\"Altheman\"][quote=\"jackb1115\"]I'll be there from Chicago, hopefully Chi A.\n\nJB[/quote]\n\nme too[/quote]ill be there with the Chicago also", "Solution_9": "[quote=\"sapphyre571\"]I'll be there from TX[/quote]\r\nsame\r\ntexas might be interested in balling against indiana", "Solution_10": "[quote=\"Klebian\"][quote=\"sapphyre571\"]I'll be there from TX[/quote]\nsame\ntexas might be interested in balling against indiana[/quote]\r\n\r\nyayuhhh thats what i'm talkin' 'bout. we'll show you how we do it up in h-town and d-town like them rockets and mavs... oh wait. :( . i meant.. san antonio...", "Solution_11": "[quote=\"minsoens\"][quote=\"Altheman\"][quote=\"jackb1115\"]I'll be there from Chicago, hopefully Chi A.\n\nJB[/quote]\n\nme too[/quote]ill be there with the Chicago also[/quote]\r\n\r\nI aswell, but I dont think I have much of a chance at A team", "Solution_12": "[quote=\"pkothari13\"][quote=\"Klebian\"][quote=\"sapphyre571\"]I'll be there from TX[/quote]\nsame\ntexas might be interested in balling against indiana[/quote]\n\nyayuhhh thats what i'm talkin' 'bout. we'll show you how we do it up in h-town and d-town like them rockets and mavs... oh wait. :( . i meant.. san antonio...[/quote]\r\n\r\nwell the pacers didn't even make the playoffs, much less win 60+ in the regular season", "Solution_13": "i'm going from IN. hopfully on Blue.\r\n\r\nNat Mc, are you \"EG\"?", "Solution_14": "I hope I'll be there with Wisconsin RED (top team)", "Solution_15": "Probably going to the lecture also, unless my team holds a Mao tournament or something :D", "Solution_16": "[quote=\"pianoforte\"]Go Missouri! :laugh: \n\nHopefully the A team.[/quote]\r\n\r\nabout sums it up", "Solution_17": "I'll be schleppin' along with the MN team, spending Saturday in the grading room, praying that there are no incorrect solutions this year :maybe: \r\n\r\nAlso, for those MN kids on here, how'd the last two practices go? (We should use the cobweb-bedecked MN forum, come to think.)", "Solution_18": "[quote=\"nat mc\"]if anyone wants to play basketball on friday, some ppl from indaina will probably be there playin so we could meet up[/quote]\r\n\r\nbasketball? where?", "Solution_19": "Well, why not add to the multiple off-topic converstions?\r\n\r\n[quote=\"generating\"]Also, for those MN kids on here, how'd the last two practices go?[/quote]\r\n\r\nGenerally speaking both practices went well - lots of 10s on team events, good relay scores, etc. In particular, the most recent practice (i.e. the one I most clearly remember) featured what was probably the best-done team event I've ever seen - everybody worked together well, all the problems were done and checked, team dynamics were great - and on our practice NYSML, the Gold team lost to NYC A by either 3 or 6 points, which is very little considering the potential total score and the fact the we were missing Nick and Rohan for a large part of the day. So, we're all excited, hoping to finally do really well again. This is a good year for us.", "Solution_20": "Wait, good relays?! That's a new one. :P\r\nGood luck, guys. I'll be looking for you in the top 7 this year.", "Solution_21": "I'll be there with the Minnesota Maroon. Fun, fun.", "Solution_22": "[quote=\"Danbert\"]Well, why not add to the multiple off-topic converstions?\n\n[quote=\"generating\"]Also, for those MN kids on here, how'd the last two practices go?[/quote]\n\nGenerally speaking both practices went well - lots of 10s on team events, good relay scores, etc. In particular, the most recent practice (i.e. the one I most clearly remember) featured what was probably the best-done team event I've ever seen - everybody worked together well, all the problems were done and checked, team dynamics were great - and on our practice NYSML, the Gold team lost to NYC A by either 3 or 6 points, which is very little considering the potential total score and the fact the we were missing Nick and Rohan for a large part of the day. So, we're all excited, hoping to finally do really well again. This is a good year for us.[/quote]\r\n\r\nI agree with this assessment of the practices, it also seems like both teams do pretty well on power comparing our scores and NYC A's. The maroon team also beat, I believe, the top teams in both divisions on the individual round questions 5 and 6.", "Solution_23": "I'll be going on either Indiana's A team or B team, not sure yet.", "Solution_24": "Yeah, the Missouri people will finalize A/B teams the Friday before the competition.\r\n\r\nProcrastinators of the world, unite! Tomorrow.", "Solution_25": "I'll be there.\r\n\r\nMichigan.", "Solution_26": "just to be clear, \r\n\r\n[b]ON FRIDAY, MEET BY THE EULER SPOILERS LECTURE[/b]\r\n\r\n[i think that is good at least]", "Solution_27": "Will be there at Penn State :)\r\n\r\nArgh, didn't see that there was a post for each location :blush:", "Solution_28": "I'm going from Indiana. Probably B team, though. What day do the actual competitions start?", "Solution_29": "The actual competitions start on Saturday, but ARML wants everyone to be there on Friday night." } { "Tag": [], "Problem": "consider :D A 4 by 8 rectangle.with 8 vertical and 4 horizontal lines if a king stays at the bottomless left in how many ways can he get to his castle just at the bottomless right in exactly n-moves(lets say n=7)if all movements are allowed(vertical,horizontal&diagonally)", "Solution_1": "Do you mean to say that the king is on a 4 square by 8 square board? And also, the word is just \"bottom\" [i](\"bottomless\" can be one of several things, none of which you probably intend to mean :P)[/i]" } { "Tag": [ "trigonometry" ], "Problem": "without calculators find the measure of acute angle x\r\n$\\cos x=\\frac{1}{1+(2+\\sqrt{3}-\\sqrt{2}-\\sqrt{6})^{2}}$", "Solution_1": "I think it must be: \r\n$\\cos x=\\frac{1}{\\sqrt{1+(2+\\sqrt{3}-\\sqrt{2}-\\sqrt{6})^{2}}}$\r\nAnd $x=\\frac{\\pi}{24}$", "Solution_2": "Could you please post how you found that?", "Solution_3": "We have: \r\n$\\cos^{2}x=\\frac{1}{1+\\left(2+\\sqrt 3-\\sqrt 2-\\sqrt 6\\right)^{2}}=\\frac{1}{1+\\left[\\left(\\sqrt 3-\\sqrt 2\\right)\\left(\\sqrt 2-1\\right)\\right]^{2}}\\\\ =\\frac{2\\sqrt 6+2\\sqrt 2+8}{16}=\\frac{\\sqrt 6+\\sqrt 2+4}{8}$\r\n$\\Longrightarrow \\cos 2x=2\\cos^{2}x-1=\\frac{\\sqrt 6+\\sqrt 2}{4}=\\cos\\frac{\\pi}{6}\\cos\\frac{\\pi}{4}+\\sin\\frac{\\pi}{6}\\sin\\frac{\\pi}{4}=\\cos\\frac{\\pi}{12}$\r\n$\\Longrightarrow x=\\pm\\frac{\\pi}{24}+k\\pi\\quad k\\in \\mathbb{Z}$\r\nOn the otherhand $x<\\frac{\\pi}{2}\\Longrightarrow x=\\frac{\\pi}{24}$" } { "Tag": [ "AMC", "AIME", "AMC 10", "AMC 12" ], "Problem": "I'm taking AMC-10 next year on Feb. 16. I hope I can make it to AIME but it's kind of nervous you know. ;) My average score is always above 100 but just little bit below than 120. :( I hope I can increase my skill better... During the winter vacation.\r\n\r\nTell which AMC you taking!", "Solution_1": "I'm taking the AMC 10B on February too. But I'm not too confident. My average practice test score is 95... :( I'm really worried. I've been studying more during this vacation, and I hope to improve. Anyone with last minute tips?", "Solution_2": "AMC 12...i didn't qualify with the AMC 10 8th grade year....with the AMC 12 9th grade year i qualified comfortably! :lol:", "Solution_3": "12A and 10B is my plan, I'm not really worried about AMC because I'm a 9th grader and all I need is a 7 or 8 on AIME(very difficult!)...", "Solution_4": "I'm taking the AMC 10A.", "Solution_5": "Probably AMC 10A only. I'll try 12B if I can handle old AMC 12 tests though. I got 112 on the one AMC 10 test I've taken so far (it was the 2002 test or something, so blank answers were 2 points instead of 2.5) so I'm not sure if that's, for me, unusually low, unusually high, or average. So for the AIME, I'm borderline at best and don't have a chance at worst. I'm glad to know qualifying isn't entirely impossible, though.", "Solution_6": "I'm doing the 10A and 12B.", "Solution_7": "12 A for sure, and, if I can arrange things with another school 12 B as well.", "Solution_8": "I'm doing 10A and 12B. I have just qualified on all the practice tests I've taken, so I think I had better take both of them.", "Solution_9": "Personally, i dont like AMC10, last year i scored higher in AMC12 than AMC10. Basically, you really have to be aware of computational mistakes in AMC10, and DON'T guess!!!! (this is the way i messed up last year) Leave them blank!!!!", "Solution_10": "I'm taking both AMC 12's. I've never taken two AMCs the same year before, is it strenuous?", "Solution_11": "AMC12, not sure which... and i think my school will only order 1 set for everybody\r\n\r\nmy school doesn't even give AMC10s, so during my 9th and 10th grade years i only had the option of taking the AMC12 =|\r\n\r\n\r\nlet's hope I qualify for the aime :?", "Solution_12": "[quote=\"Bictor717\"]I'm taking both AMC 12's. I've never taken two AMCs the same year before, is it strenuous?[/quote]\r\n\r\nI wouldn't say so; you need only study once, although you may need to review a few times.", "Solution_13": "Possibly 10A and 12B who knows..." } { "Tag": [ "search", "geometry unsolved", "geometry" ], "Problem": "In the equilateral triangle $ ABC$ is chosen a point $ O$. Its symmetric point with respect to the sides $ BC$, $ CA$ and $ AB$ are denoted respectively with $ A_1$, $ B_1$ and $ C_1$. Prove that the lines $ AA_1$, $ BB_1$ and $ CC_1$ intersects at a common point.", "Solution_1": "$\\triangle ABC$ becomes a Napoleon triangle (either internal or external) of $\\triangle A_1B_1C_1,$ so the lines $ AA_1, BB_1,CC_1$ concur at a Napoleon point of $\\triangle A_1B_1C_1.$", "Solution_2": "[quote=\"Luis Gonz\u00e1lez\"]$\\triangle ABC$ becomes a Napoleon triangle (either internal or external) of $\\triangle A_1B_1C_1,$ so the lines $ AA_1, BB_1,CC_1$ concur at a Napoleon point of $\\triangle A_1B_1C_1.$[/quote]\n1st of all, what's the proof of this? I mean $ ABC$ may be just homothetic to the Napoleon's triangle of $ A_1B_1C_1$....\n\nAnd the concurrency here is not so well know as far as I know, (and I also think that this problem is true for not for the reflected points but also for the glide reflected points)", "Solution_3": "Dear it's the celebrated Jacobi's theorem: In a $\\triangle ABC,$ points $X,Y,Z$ are constructed, all outside or all inside, such that $\\angle BCX=\\angle ACY,$ $\\angle CAY=\\angle BAZ$ and $\\angle ABZ=\\angle CBX.$ Then the lines $AX,BY,CZ$ concur.\n\nFor a proof try a search, it has been posted before." } { "Tag": [ "calculus", "derivative", "irrational number" ], "Problem": "If $ x \\plus{} y \\equal{} 1$, then the largest value of $ xy$ is:\r\n\r\n$ \\textbf{(A)}\\ 1\\qquad \r\n\\textbf{(B)}\\ 0.5\\qquad \r\n\\textbf{(C)}\\ \\text{an irrational number about }{0.4}\\qquad \r\n\\textbf{(D)}\\ 0.25\\qquad \r\n\\textbf{(E)}\\ 0$", "Solution_1": "[hide=\"Solution\"]$ x \\plus{} y \\equal{} 1\\geq 2\\sqrt {xy}$ (AM-GM)\n\n$ 1\\geq 4xy$\n\n$ xy\\leq\\frac {1}{4}$.\n\nEquality when $ x \\equal{} y \\equal{} 1/2$.\n\nAnswer: $ \\boxed{\\textbf{(D)}\\ 0.25}$.[/hide]\r\n\r\nIs this correct? It depends on $ x,y$ being positive which is not given, but if either was negative their product would be too small. (And both can't be negative).", "Solution_2": "Yeah, it's correct. You also also view this geometrically by graphing $ y \\equal{} \\minus{}x \\plus{} 1$ and finding the point at which the product $ xy$ is maximized.", "Solution_3": "or you could substitute y=1-x and find the derivative of x*(1-x)=x-x^2. The derivative is 1-2x, and the maximum is found when 1-2x=0. so x=1/2, and y=1/2, to give 1/4 or 0.25. BAM!\r\nok, i realize calculus is overkill, but it makes me feel better to actually use it once in a while." } { "Tag": [], "Problem": "Solve this equation in $ \\mathbb{Z}$ :\r\n\r\n$ x\\equal{}\\sqrt{y^2\\minus{}\\sqrt{y^2\\plus{}x}}$", "Solution_1": "Let $ k\\equal{}\\sqrt {y^2 \\plus{} x}$.\r\n\r\nThe equation \r\n$ \\Leftrightarrow \\begin{cases} x \\ge 0 , k \\ge \\sqrt {x} \\\\\r\nx^2 \\equal{} k^2 \\minus{} k \\minus{} x \\Rightarrow [k \\minus{} (x \\plus{} 1)](k \\plus{} x) \\equal{} 0 \\Rightarrow k \\equal{} x \\plus{} 1 or k \\equal{} x \\equal{} 0 \\end{cases}$\r\n\r\n$ \\Leftrightarrow (x \\plus{} 1)^2 \\equal{} y^2 \\plus{} x$ or $ x \\equal{} y \\equal{} 0$\r\n\r\nSo $ (x,y) \\equal{} (k,\\pm \\sqrt {m^2 \\plus{} m \\plus{} 1})$ for all non-negative $ m$s and $ (x,y) \\equal{} (0,0)$ are the solutions." } { "Tag": [ "AMC", "AIME" ], "Problem": "how many subsets of the set {1,2,3,4,5} contain the number 5?\r\n\r\ncan anyone tell me how to work with subsets along with this problem? :D", "Solution_1": "[hide]I don't know much about subsets, but here's what I got:\nSubsets w/5 numbers: $\\binom{4}{4}=1$\nSubsets w/ 4 numbers: $\\binom{4}{3}=4$\nSubsets w/3 numbers: $\\binom{4}{2}=6$\nSubsets w/2 numbers: $\\binom{4}{1}=4$\nSubsets w/1 numner: $\\binom{4}{0}=1$\nSo the answer is $16$.[/hide]", "Solution_2": "[hide]what about all five numbers? that makes $\\boxed{16}$[/hide]", "Solution_3": "[quote=\"SplashD\"]what about all five numbers?[/quote]\r\nI guess he thought $S\\notin\\left\\{\\text{subsets of }S\\right\\}$ :) \r\nYou could also [hide]just begin with {5} and add subsets of {1,2,3,4}. You'll get all the subsets you wanted, so we just have to count the number of subsets of {1,2,3,4}, which equals $2^{4}= 16$.[/hide]", "Solution_4": "This reminds me of a simple aime problem... we can pair up each subset with a 5 to each subset without a 5 (by ignoring the element 5, obviously) and since every subset either has a 5 or doesn't, the number of subsets with a 5 is half the number of subsets. How many subsets are there? We have 5 choices: include a 1, or not? include a 2, or not? etc. so there are 2^5 subsets. Therefore, there are 16 subsets with a five. Btw, I know this is overkill for an easy problem like this but it generalizes..." } { "Tag": [], "Problem": "Hi\r\nplease give me just a hint to solve this problem\r\n\r\nThere are $ 6$ different points in the plane. Let $ D$ be the greatest distance between them and $ d$ the smallest one. Prove that $ \\frac{D}{d}>\\sqrt{3}$.\r\n\r\n :(", "Solution_1": "[hide=\"Hint, even though I can't do it myself\"]Consider the two points that are the farthest away from each other. Where must the other 4 points lie?[/hide]", "Solution_2": "Adding on to what Brut3Forc3 said...\r\n\r\nConsider the points that are furthest apart. The distance between these points is $ D$. The other 4 points have to be in the intersection of the 2 circles with radius $ D$ with centers at the 2 points furthest points. But there's still a more restricted area...since in that intersection, the furthest possible distance of 2 points in that intersection is $ D\\sqrt {3}$. Not sure where to go from here...", "Solution_3": "Carbon57, try a little harder, you are close to the idea :) Redraw the diagrams and look closely at where those 4 points must be.", "Solution_4": "I think I have an idea. Would the maximum length of $ d$ be when the points are... [hide]in a regular hexagon?[/hide]\r\nIf so, how is it proven?", "Solution_5": "That would be the length of a smaller diagonal, the larger diagonal will have twice the length of a side of the hexagon, so a hexagon wouldn't be ideal for this problem.\r\n\r\nEDIT 10PM: Actually, equality can be achieved for $ \\frac{D}{d}\\geq\\sqrt3$. Try to find this configuration. It's easier than you think :)" } { "Tag": [ "puzzles" ], "Problem": "A small, fully loaded passenger plane tried to approach the runway during a violent snowstorm. The control tower regretfully informed the pilot that due to the inclement weather, the runways were closed to all air traffic. Furthermore, all airports within a three-hundred mile radius were also closed. Upon hearing this, the pilot immediately informed the passengers of the news while turning the plane around and heading back from where they had just come. Incredibly enough, within a half hour all the passengers were safely inside an airport terminal building.\r\nHow was this possible?", "Solution_1": "[hide=\"Because...\"]It hadn't taken off yet.[/hide]", "Solution_2": "[hide]The plane was grounded, and never took off in the first place[/hide]" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Let ABC be a triangle and P a point in its interior, not lying on any of the medians of ABC. Let $A_1$, $B_1$, $C_1$ be the intersections of PA with BC, PB with CA, PC with AB, respectively, and let $A_2$, $B_2$, $C_2$ be the intersections of $B_1C_1$ with BC, $C_1A_1$ with CA, $A_1B_1$ with AB, respectively. Prove that if some two of the circles with diameters $A_1A_2$, $B_1B_2$, $C_1C_2$ intersect, then they are coaxal.", "Solution_1": "In http://www.mathlinks.ro/Forum/viewtopic.php?t=21701 , I showed that the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ are the diagonals of a certain complete quadrilateral. Now, by the Bodenmiller theorem, the circles whose diameters are diagonals of a complete quadrilateral are coaxal.\r\n\r\nThe Bodenmiller theorem is proven at http://cut-the-knot.com/Curriculum/Geometry/CircleOnCevian.shtml#GB (\"Corollary\").\r\n\r\n Darij", "Solution_2": "What about http://cut-the-knot.com/ ? I did not see any new articles recently." } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ R$ be a commutative ring with unit, $ \\mathfrak{m}$ its maximal ideal, and $ q$ an ideal of $ R$ such that $ \\mathfrak{m}^k\\subset q$ for some $ k\\geq 1$. Prove that $ q$ is primary.", "Solution_1": "a general fact is that if the radical of an ideal is maximal ideal $ m$, then it is a $ m$-primary ideal." } { "Tag": [ "geometry", "rectangle", "similar triangles", "Pythagorean Theorem" ], "Problem": "In the diagram, the two circles are tangent, and $ PB$ and $ PD$ are common tangents to the circles. If PA=AB=6, what is the area of the larger circle?", "Solution_1": "[quote=\"hohoho11\"]In the diagram, the two circles are tangent, and $ \\line{PB}$ and $ \\line{PD}$ are common tangents to the circles. If PA=AB=6, what is the area of the larger circle?[/quote]\r\n\r\nIs what he meant to say.\r\n\r\n[hide=\"Hint\"]Draw a line from the center of the circle to the point of contact for each circles. Now, make a rectangle connecting the center of the smaller circle to the radii of the bigger circle. I believe it should be simple from there.[/hide]", "Solution_2": "yeah sorry :blush:", "Solution_3": "[hide=\"Can somebody find a simpler solution please?\"]We can create a system of equations. We can draw lines from the center of each circle to its respective points of tangency with the lines $ PD$ and $ PB$. A pair of similar right triangles will be created. Denoting the radius of the smaller circle to be $ r_1$ and that of the bigger circle to be $ r_2$, we have:\n\\[ \\frac {r_2}{12} = \\frac {r_1}{6} \\Rightarrow r_2 = 2r_1\n\\]\nFurthermore, we can find two ways to calculate the hypotenuse of the triangle with legs $ 12$ and $ r_2$. Setting them equal, we have:\n\\[ \\sqrt {12^2 + r^2_2} = r_2 + r_1 + \\sqrt {6^2 + r^2_1}\n\\]\nThen, replacing $ r_2$ with $ 2r_1$, and rearranging, we have:\n\\begin{align*} & \\sqrt {12^2 + {\\left(2r_1\\right)}^2} = 2r_1 + r_1 + \\sqrt {6^2 + r^2_1} \\\\\n& \\Rightarrow \\sqrt {144 + 4r^2_1} - \\sqrt {36 + r^2_1} = 3r_1 \\\\\n& \\Rightarrow 144 + 4r^2_1 + 36 + r^2_1 - 2\\sqrt {\\left(144 + 4r^2_1\\right)\\left(36 + r^2_1\\right)} = 9r^2_1 \\\\\n& \\Rightarrow - 2\\sqrt {\\left(144 + 4r^2_1\\right)\\left(36 + r^2_1\\right)} = 4r^2_1 - 180 \\\\\n& \\Rightarrow \\sqrt {\\left(144 + 4r^2_1\\right)\\left(36 + r^2_1\\right)} = 90 - 2r^2_1 \\\\\n& \\Rightarrow 5184 + 288r^2_1 + 4r^4_1 = 4r^4_1 - 360r^2_1 + 8100 \\\\\n& \\Rightarrow 5184 + 288r^2_1 + 4r^4_1 = r^2_1 - 45 \\\\\n& \\Rightarrow 648r^2_1 = 2916 \\\\\n& \\Rightarrow r_1 = \\frac {3\\sqrt {2}}{2} \\\\\n& \\Rightarrow r_2 = 2r_1 = 3\\sqrt {2} \\end{align*}The area is then $ \\pi\\left(3\\sqrt{2}\\right)^2=\\boxed{18\\pi}$.\n[/hide]", "Solution_4": "[hide=\"Solution.\"]Call the distance from $ P$ to the first circle $ x$. That means that the distance from the first circle to $ P$ is $ r \\plus{} x$, with $ r$ being its radius. By similar triangles, the radius of the second circle is $ 2r$ and its distance from $ P$ is $ 4r \\plus{} x$. Using the Pythagorean Theorem on both of these triangles:\n\n$ (2r)^2 \\plus{} 12^2 \\equal{} (4r \\plus{} x)^2$\n$ (r)^2 \\plus{} 6^2 \\equal{} (r \\plus{} x)^2$\n\nSimplifying and multiplying the latter by 4:\n\n$ 4r^2 \\plus{} 144 \\equal{} 16r^2 \\plus{} 8rx \\plus{} x^2$\n$ 4r^2 \\plus{} 144 \\equal{} 4r^2 \\plus{} 8rx \\plus{} 4x^2$\n\nTherefore:\n\n$ 12r^2 \\equal{} 3x^2$\n$ x \\equal{} 2r$\n\nPlugging it back into the original equation:\n\n$ r^2 \\plus{} 36 \\equal{} (3r)^2$\n$ 8r^2 \\equal{} 36$\n\nYou're interested in finding $ \\pi(2r)^2$, so divide by $ 2$ and multiply by $ \\pi$. The final answer is $ \\boxed{18\\pi}$.[/hide]", "Solution_5": "We have:\r\n+) $ IA \\equal{} 2IB$\r\n+) $ OA \\plus{} IB \\equal{} OI \\equal{} IP$\r\n+) $ IP^2 \\equal{} IB^2 \\plus{} BP^2$\r\n$ \\leftrightarrow$ $ (OA \\plus{} IB)^2 \\equal{} IB^2 \\plus{} 36$\r\n$ \\leftrightarrow$ $ (2IB \\plus{} IB)^2 \\equal{} IB^2 \\plus{} 36$\r\n$ \\leftrightarrow$ $ 9IB^2 \\equal{} IB^2 \\plus{} 36$\r\n$ \\leftrightarrow$ $ IB^2 \\equal{} \\frac {9}{2}$ $ \\leftrightarrow$ $ IB \\equal{} \\frac {3}{\\sqrt {2}}$ $ \\rightarrow$ $ OA \\equal{} 3 \\sqrt {2}$\r\n$ \\rightarrow$ $ S_{(O;OA)} \\equal{} \\pi . OA^2 \\equal{} 18.\\pi$." } { "Tag": [ "probability" ], "Problem": "[b][size=150]A box contains 3 red balls, 2 blue balls and 5 green balls. Two balls are drawn from the box, one after the other and not replaced.\n\n1.) What is the probability that they are both blue?\n\n2.) What is the probability that they are both the same colour?[/size][/b]", "Solution_1": "1)[hide]\n\nThere are 10 marbles in the box, so there is a 2/10 prob that the first marble is blue and a 1/9 prob that the second is blue, so the overall probability is $\\frac{1}{5}*\\frac{1}{9}=\\frac{1}{45}$\n\n[/hide]\n\n\n2)[hide]\n\nThere is a $\\frac{1}{45}$ prob that they are both blue, a $\\frac{1}{15}$ prob that they are both red, and a $\\frac{2}{9}$ prob that they are both green. The total is $\\frac{1}{45}+\\frac{1}{15}+\\frac{2}{9}=\\frac{14}{45}$[/hide]\r\n\r\n\r\nThis isn't quite Intermediate level. :?", "Solution_2": "Yeh...this was pretty easy. \r\n :D \r\n1. [hide]1/5*1/9 = 1/45[/hide]\n2. [hide]1/45 (see #1) + 2/9+1/15 = 14/45[/hide]" } { "Tag": [ "number theory" ], "Problem": "if:\r\n(2^n)-1 is prime \r\n[b]prove that:[/b]\r\nn is prime.", "Solution_1": "let us suppose that N is composed and d is a divisor strict of n\r\nwa can write n=dd' and:\r\n2^n-1=(2^d-1)(2^{d(d'-1)}+2^{d(d'-2)}.....+1)\r\nthe factor 2^d-1 is noncommonplace sow 2^n-1 is composed", "Solution_2": "[hide]Suppose that $ab=n$ ($a,b$ not equal to 1)\n\nIt is a well-known rule in number theory that $a-b|a^{n}-b^{n}$ for all integers $a,b$.\n\nBut $2^{n}=2^{ab}=2^{a^{b}}$\nSo $2^{a}-1|2^{a^{b}}-1^{b}=2^{n}-1$\n\nSo $2^{a}-1|2^{n}-1$\nSince $2^{a}-1$ is neither 1 nor $2^{n}-1$ it follows that $2^{n}-1$ is composite.\n[/hide]\r\n\r\nedit: I see diane got there first, but anyway...", "Solution_3": "[hide]Contrapositive- If n is not prime, $2^{n}-1$ is not prime.\nIf n is not prime, it equals 1, in which case $2^{1}-1=0$ which is not prime, or it equals ab, where a and b are some positive integers not equal to 1. If n=ab, $2^{n}-1=(2^{a})^{b}-1=(2^{a}-1)(2^{ab-1}+2^{ab-2}+...+1)$. Since both terms are >1, $2^{n}-1$, is composite, so therefore is not prime. We have proved the contrapositive, so the original statement is true.[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "I wasn't sure where to post this question but so i decided here.\r\n\r\nI am currently in an argument to prove to my friend that n/0 is not infinity. I heard there are a few paradoxes that happen if n/0 is treated as infinity. \r\n\r\nHow would you explain why n/0 is NOT infinity?", "Solution_1": "Oh geez...\r\nTry [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=203903]this[/url] if you dare...", "Solution_2": "It is undefined...\r\nbut it can be indeterminable.\r\n\r\nThe limit as x approaches 0 for $ \\frac{\\sin{x}}{x}\\equal{}1$", "Solution_3": "[quote=\"ashpotter\"]I am currently in an argument to prove to my friend that n/0 is not infinity.[/quote]\r\nHere's the proof: It's undefined.\r\n\r\nThere have been numerous topics about this; searching would have turned up some of them, like the one Brut3Forc3 pointed out.\r\n\r\nAlso, double post: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208792 (probably accidental, but still).", "Solution_4": "sorry about the double post. What happened was that my browser lagged so i refreshed the page and clicked submit twice." } { "Tag": [ "AMC", "AIME" ], "Problem": "On the AMC 12, each correct answer is worth $6$ points, each incorrect answer is worth $0$ points, and each problem left unanswered is worth $2.5$ points. If Charlyn leaves $8$ of the $25$ problems unanswered, how many of the remaining problems must she answer correctly in order to score at least $100$?\r\n\r\n$\\text {(A)} 11 \\qquad \\text {(B)} 13 \\qquad \\text {(C)} 14 \\qquad \\text {(D)} 16\\qquad \\text {(E)}17$", "Solution_1": "[hide=\"Answer\"]C.\n\nWe want the smallest integer $x$ such that $(2.5)(8)+x>100$.[/hide]", "Solution_2": "[hide]Blind luck: 17*6 = 102, so E[/hide]", "Solution_3": "[quote=\"Apprentice\"][hide]Blind luck: 17*6 = 102, so E[/hide][/quote]Read the question carefully. :)", "Solution_4": "[quote=\"Apprentice\"][hide]Blind luck: 17*6 = 102, so E[/hide][/quote]\r\nYou're forgetting that she recieves points for those 8 unanswered questions.\r\n\r\nEDIT: Oops. I posted like three seconds after you. :oops:", "Solution_5": "[hide]O...8*2.5 = 20, 100-20 = 80, 80/6 = 13.3333 so C[/hide]", "Solution_6": "[hide]Charlyn has 20 points from the unanswered questions, so she needs 80 more points. Each question is worth 6 points, so the smallest integer greater than $\\frac{80}{6}$ is $14\\implies\\boxed C$[/hide]", "Solution_7": "[hide]C[/hide]", "Solution_8": "Charlyn needs 80 points more. Then,\r\n\r\n6X\u226580\r\n\r\nX\u226580/6\r\n\r\nX\u226540/3\r\n\r\nAs X must be an integrer, X=14. So, Charlyn has to answer 14 answers correctly at least.\r\n\r\nThen, the correct answer is C.", "Solution_9": "[hide] No doubts: /boxed${14}$[/hide]", "Solution_10": "[hide]C yo[/hide]" } { "Tag": [ "number theory", "relatively prime", "combinatorics unsolved", "combinatorics" ], "Problem": "There are $n$ distinct points on a circumference. Choose one of the points. Connect this point and the $m$th point from the chosen point counterclockwise with a segment. Connect this $m$th point and the $m$th point from this $m$th point counterclockwise with a segment. Repeat such steps until no new segment is constructed. From the intersections of the segments, let the number of the intersections - which are in the circle - be $I$. Answer the following questions ($m$ and $n$ are positive integers that are relatively prime and they satisfy $6 \\leq 2m < n$).\n\n1) When the $n$ points take different positions, express the maximum value of $I$ in terms of $m$ and $n$.\n\n2) Prove that $I \\geq n$. Prove that there is a case, which is $I=n$, when $m=3$ and $n$ is arbitrary even number that satisfies the condition.", "Solution_1": "I'm pretty sure you have the additional condition that $m$ and $n$ are coprime (otherwise you can easily give a counterexample to (2)).\r\n\r\nFor the first part, it's easy to see that you can choose the points so that no three diagonals of the $n$-gon are concurrent, which means that $I\\le (m-1)n$ elements: for each of the $n$ diagonals, there are $2(m-1)$ intersection points between that diagonal and other diagonals, and each point is counted twice.\r\n\r\nFor the second part, for each diagona $A_iA_{i+m}$ consider the point of intersection between it and $A_{i+m-1}A_{i+2m-1}$. It's easy to see that this gives an injection from the set of $n$ diagonals we have drawn to $I$, so $I\\ge n$. As for the construction, start with a regular $\\frac n2-$gon $X_0X_2\\ldots X_{n-2}$, and take $X_1,X_3,\\ldots,X_{n-1}$ to be the midpoints of $X_0X_2,X_2X_4,\\ldots,X_{n-2}X_0$ respectively. After this, put $A_k=X_kX_{k+2}\\cap X_{k+3}X_{k+5},\\ \\forall k\\in\\overline{0,n-1}$, where all the indices are modulo $n$." } { "Tag": [], "Problem": "\"Question about a party?\"\r\n[hide][b]Anarchist[/b] - There should be no government; encompassing all anarchic movements even if they are at odds.\n[b]Classical Conservative[/b] - Strong government, hierarchal.\n[b]American Conservative[/b] - Domestic small-government, pro-military, relatively religious.\n[b]Classical Liberal / Libertarian[/b] - There should be fewer laws and few restrictions on people or anything else by the government.\n[b]American Liberal / New Deal Liberal /Neo-Liberal / Social Democrat[/b] - Big, economy-regulating equality-promoting socially-active program-full wealth-redistributing (usually minor) side. Somewhere between Classical Liberals and Socialists.\n[b]Democratic Socialist[/b] - Socialist as achieved by democracy (major government ownership of everything.)\n[b]Revolutionary Socialist, Communist, Marxist[/b] - Communism or Socialist as achieved by revolution.\n[b]Monarchy/Fascism/Authoritarianism[/b] - In all its forms. One guy at the top.\n\nIf you feel that you aren't represented here, are at odds with the definitions provided, or have some other beef with the poll, go ahead and comment below and I may add your choice / remedy that.[/hide]", "Solution_1": "This has been done before...check the Polls section", "Solution_2": "i like how the majority of the people say they're communists", "Solution_3": "I don't see it. Could you drop me a link, 7h3.demon.117?", "Solution_4": "I am an anarchist, but I believe in its evolution through a communist revolution. :wink:", "Solution_5": "[quote=\"Dark_Shadow\"]I don't see it. Could you drop me a link, 7h3.demon.117?[/quote]\r\n\r\nnvm it was a similar polticial question in the Round table...", "Solution_6": "Why isn't Potato on the list?\r\n\r\nIf there was a potato party, I'd totally vote for it. Potatoes are awesome.\r\n\r\nYou probably think I'm kidding, but most of my friends could tell you that I'm 100% serious. Let me illustrate some of the great things about potatoes:\r\n\r\n1. They taste awesome.\r\n2. They have most of your daily value of Vitamin C, potassium, and B6, and can act as a substitute for fiber. (Of course, wikipedia doesn't know all of the health benefits of potatoes.)\r\n3. They are much cheaper than other food, in fact, I believe they could single-handedly wipe out hunger and poverty.\r\n4. If there was a Potato party, America wouldn't be stuck with a duopoly controlling the government.\r\n5. There's more, but I have to go cook some potatoes right now. Yes, I CAN smell your envy from here.", "Solution_7": "It is interesting how there are so many communists." } { "Tag": [ "function", "inequalities solved", "inequalities" ], "Problem": "Let x, y, z be positive reals such that\r\n 1/x + 1/y + 1/z = 1, x+y+z=10\r\nFind maximal and minimal values of A = x 2 + y 2 + z 2.\r\n\r\nSource: Suggested by Namdung for School Team Selection Test, year ?", "Solution_1": "Could you please post your solution to this problem, Namdung? It is very interesting and I cannot find an elementary solution. Maybe we could study the function f(x)=(x^3-10x^2)/(1-x), but I prefer ideas rather than computations. Thanks.", "Solution_2": "1) Firstly, we show that 2 <= x, y, z <= 5. Ideed:\r\n 1-1/z = 1/x+1/y, 10-z=x+y so (1-1/z)(10-z) >= (x+y)(1/x+1/y) = 4\r\n=> z^2 - 7z + 10 <= 0 => 2 <= z <= 5. Similarly with x, y. \r\n2) Since x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx), we have to find max and min of xy+yz+zx.\r\n We have (x-2)(y-2)(z-2) >= 0 => xyz - 2(xy+yz+zx) + 4(x+y+z) - 8 >=0 => xy+yz+zx <= 32. The equality holds, for example when x=2, y=z=4.\r\n Similarly, (5-x)(5-y)(5-z)>=0 => 125 - 25(x+y+z) + 5(xy+yz+zx) - xyz >=0 => xy+yz+zx >= 125/4 (We use x+y+z=10 and xyz=xy+yz+zx).\r\nThe equality holds, for example when x=5, y=z=5/2.\r\n3) Finally, we conclude that max = 75/2, min = 36.\r\n\r\nNamdung", "Solution_3": "Super! Thank you very much, Namdung. It is absolutely splendid." } { "Tag": [ "ARML", "email", "analytic geometry", "graphing lines", "slope" ], "Problem": "So the team list is [url=http://www.nycmathteam.com/2006-nyc-math-team/arml-selections/]up[/url].\r\nI'm wondering if team members got the email too? I guess I can PM you a copy of the email (with password) if you don't have it already.", "Solution_1": "Heh, they put the two Wensi's on the same team.\r\n\r\nAlso, is Kevin really not in the top 4 at Hunter? I mean, it looks like he's fifth, since he appears to be captain of the team with the people who are almost on A, but still.\r\n\r\nI think team members did get the e-mail, although [i]I[/i] didn't.", "Solution_2": "[quote=\"JBL\"]Heh, they put the two Wensi's on the same team.[/quote]\n...again. I noticed too.\n\n[quote=\"JBL\"]Also, is Kevin really not in the top 4 at Hunter? I mean, it looks like he's fifth, since he appears to be captain of the team with the people who are almost on A, but still.[/quote]\r\nMaybe not being at NYSML counted against him...", "Solution_3": "Ah, I didn't know he missed NYSML.\r\n\r\nBreakdown of ARML members by school:\r\nStuyvesant\t29\r\nBrooklyn Tech\t9\r\nHunter\t8\r\nCardozo\t6\r\nFrancis Lewis\t3\r\nBowne\t2\r\nNewcomers\t2\r\nBayside\t2\r\nHSMSE\t2\r\nFlushing\t1\r\nSaint Ann's\t1\r\nTrinity\t 1\r\nBronx Sci\t1\r\nMidwood\t1", "Solution_4": "[quote=\"JBL\"]Heh, they put the two Wensi's on the same team.\n\nAlso, is Kevin really not in the top 4 at Hunter? I mean, it looks like he's fifth, since he appears to be captain of the team with the people who are almost on A, but still.\n\nI think team members did get the e-mail, although [i]I[/i] didn't.[/quote]\r\n\r\nhe's definitely in the top 4 at hunter...\r\n\r\nEDITED due to satisfactory explanation of situation.", "Solution_5": "As Mr. Linker BCCs all the recepients, it's hard to tell who did't get emails, though i think he might have been emailing the people who went to NYSML\r\n\r\nPS to joel: there was an email sent earlier (which you may or may not have gotten as well) about stating interest in going to ARML as coach- act upon that if you haven't gotten it, i can forward you stuff if you want (and if barry didn't already)", "Solution_6": "I'm posting a letter that I sent to the A Team Captains (slightly modified). \r\n\r\nAs of last night, several students have already expressed reservations with the published A team selections. Having befriended most of the students on the A team, I understand the disappointment that they might be feeling. \r\n\r\nThis year, we've seen an unprecedented number of A team calibre students. Unfortunately, the fact remains\r\nthat the A team ultimately comprises only 15 students and thus some very difficult choices had to be made. During the committee's deliberations, we have taken into account all available scores as well as the input of no less than a dozen indviduals (including the two captains). This process is imperfect, but necessary. Based on those judgements, the committee reached a consensus that was recently published and we continue to stand by those decisions. Our hope is that everyone can now put this behind them and focus on the preparation of ARML.", "Solution_7": "Thanks Maria, I did get that one.\r\n\r\nAnyhow, this just means that the N team is going to rock, too :lol:", "Solution_8": "Not to keep being difficult, but while you're here, Jack: Evelyn Israel from Bronx gets a 7 at NYSML, doesn't make the ARML team? Ditto for Val Zakharevich, or however he spells that, at Murrow.", "Solution_9": "The Bronx Science prom is the night before ARML [u]again[/u]. She and another Bronx Science senior said they wouldn't go to ARML at NYSML.", "Solution_10": "Prom -- feh. I figured it was probably something like that.", "Solution_11": "Val is going to some camp that weekend as well :(", "Solution_12": "Well that's odd...I think I received that e-mail, but my mind is always a slippery slope, so I'm not too sure about it. Rest assured that I do want to come, though.\r\n\r\nI was surprised a bit myself for the A team, but then again, if we are overflowing with high scorers, then some of our other teams stand a good shot of doing well as well.\r\n\r\nThe team I coached at NYSML remained relatively intact though....\r\n\r\n--KC", "Solution_13": "I suppose someone will be doing some online relay practices sometime?\r\nI'll be glad to host, or fill up a 2/3-full team, or just show up and watch.\r\nMy AIM is ManicBarry.", "Solution_14": "i ditto barry's post\r\n\r\nAIM: egyptoz", "Solution_15": "Agreed. I shall volunteer as well. I already have four sets of relays written up...\r\n\r\nAIM: ixidor1987\r\n\r\nP.S. What? I need SOMETHING to do in World Cultures: China lectures!\r\n\r\n--KC", "Solution_16": ".........Maybe i am just too weak to be on A team...... :(", "Solution_17": "It is getting more and more obvious that the nyc math team is only picking people from well-known school. It is ok to pick a lot of specialized high school students 'cause many of them are smart. (No offences to those students and schools) \"A lot\" is different from \"most.\" It seems to me that \"most\" students picked came from those schools.\r\nBut is it fair to students from other schools? \r\n\r\nI am in college right now and I still keep in touch with the math team coach in my high school. I am also helping in my school's math team. He told me that NYSML started to make changes in team selection. He told me that NYSML would pick at least one student from each school. what I was thinking right away was, \"Well, it seems that everyone can participate now.\"\r\n\r\nWhen I look at the ARML team list now, it no longer looks like a nyc math team, but a \"well-known schools math team. \" And I refer back to the NYSML individual score, I notice that 2 people receive the same score have totally different fates. The person who come from those school can stay in the original team, while another person with the same score drops to reserve team. Someone from those school stays in the team, while the other person can't even get to ARML. ...more and more examples....\r\n\r\nI understand that there are many different criteria on selection, besides NYSML score. But I personal don't think it is a good start...", "Solution_18": "Hi Simon, welcome to the forum!\r\n\r\nFirst, some numbers:\r\n\r\nAmong 68 students on the ARML team, there are 14 high schools represented. At NYSML, among 90 students, there were more than 20 schools represented. From the seven schools which lost representation, Val Zakharevich was the highest individual scorer, with a 7, and was invited back to ARML but has other obligations. Among the other students representing those schools, the highest individual score was a 3.\r\nThe Stuyvesant math team has, last I knew, more than 250 students. That is, most probably, larger than the math teams of all other 13 schools sending students to ARML combined.\r\nAs the composition of the NYSML R team shows, there was a strong preference for choosing students from schools without large representation: 7 students were the lone representative from their school. 2 other students were one of two from their school to attend. The 6 remaining students were from Tech, Hunter and Stuyvesant and accounted for the highest 4 scores on the team. There were also at least three students who were the lone representative of their schools who were placed on a different NYC team.\r\n\r\n\r\nNow, that all said, I have sympathy for the view that the NYSML and ARML team should incorporate students from as many high schools as possible, especially when the contest is downstate and so easily accessible to NYC. My conversations with those responsible for choosing the NYSML and ARML teams, as well as the actual data (especially the composition of the NYSML R team) lead me to believe that \"representing a non-Stuy/Hunter/Tech/Queens powerhouse school\" is taken seriously into account in building the NYC math team.\r\nWhen the number of students decreases by 25%, it should not be too suprising (even if it is disappointing) that the number of schools represented declines by even more than this. I'm thinking about doing a more thorough analysis of the NYSML and ARML team lists, but I strongly suspect that NYSML score will be a much stronger predictor of ARML fate than school representation.", "Solution_19": "I just want to say that the Hunter girl who replaced me has the same NYSML individual score as mine." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $p,q,r$ reals not equal $0$ and such that $p+q+r=1$and $a,b,c$ sides of a triangle. Prove that\r\n\r\n$a^2pq+b^2qr+c^2rp \\geq 0$", "Solution_1": "What if $pq<0,r=0$?", "Solution_2": "[quote=\"manlio\"]Let $p,q,r$ reals not equal $0$ and such that $p+q+r=1$and $a,b,c$ sides of a triangle. Prove that\n\n$a^2pq+b^2qr+c^2rp \\geq 0$[/quote]\r\n\r\nStill not correct; take a = b = c = 1, p = -4, q = -3, r = 8.\r\n\r\n Darij", "Solution_3": "Hey, guys, there are still some nice and difficult problems unanswered. I think it would be more interesting to solve those ones than to try to repair some trivial inequalities. My opinion... hac..." } { "Tag": [ "combinatorics solved", "combinatorics" ], "Problem": "I saw this in an incomplete paper, which only had the year:97.\r\n\r\nWe have n different primes:p_1,p_2,....p_n.and N=p_1p_2...p_n.\r\nA and B write in turn composite divisors of N.with the rule that there should never be 2 numbers such that one of them divides the other.\r\n\r\nWho can win?(A starts the game)", "Solution_1": "Could you give a little more explanation on how this game works?\r\nWhen does a player \"win\"?\r\n\r\nAnd you mean with N the product of all those primes? then I don't see how there can be 2 divisors that divide eachother...", "Solution_2": "excuse me I forgot to say that: the one who can't make a move will lose.\r\nand that N is the product of those primes.\r\n\r\nan explanation on the rule : if number p_1p_2p_3p_4 is written then no one can write p_1p_2 or p_1p_2p_3p_4p_5. because if so we may have :\r\np_1p_2 | p_1p_2p_3p_4 or p_1p_2p_3p_4 | p_1p_2p_3p_4p_5.", "Solution_3": "So, A writes N and win the game?\r\n\r\nPierre.", "Solution_4": "No...... don't put N.other wise it's not a problem.(surely not an olympiad one) by the way I found a solution:\r\n\r\nA will put p_1p_2...p_(n-1) , for the next times till the end one must use p_n and a nonempty subset of \r\nA={p_1 , ... p_(n-1)}which is not A. for the first move of B , he/ she has 2^(n-1)-2 chances. Now we calculate the number of possible positions that become impossible with each move. \r\n\r\nif one writes p_1p_2...p_kp_n we name p_1p_2...p_k the A-part of his number.in this case all the 2^k -1 nonempty subsets of the A-part of his number can not be A-part for the next numbers. and also all the \r\n2^(n-k-1)-1 nonempty subsets of A - {p_1 , p_2 ... p_n} plus his own choice that means he lessens an odd number of choices from the total of 2^(n-1)-2 possibilities.so he leaves an odd number of choices for A.and A lives an even number of choices for B and...\r\n\r\nSo ,that will be A that leaves 0 choices for B and A wins.\r\n\r\nIs there any problem? :)", "Solution_5": "Maybe I see more simple way... \r\nBTW, hossein11652 forgot to say, that players can't write 1.\r\n\r\nI think, $B$ wins.\r\nStrategy for $B$ is following: if $A$ writes $m$ then $B$ writes $N/m$.\r\nIt is clear if $A$ can make his turn, then $B$ can do it too.", "Solution_6": "actually putting N/m for B is not always a viable option-the numbers put down have to be composite,so if A chooses the product of n-1 primes,then the above move for B is violated.\r\nin fact it looks like a player A game.in terms of sets ,one has the lattice if subsets of {1,2...n} except for the empty,the singletons and the complete set.and this game is equivalent to extending an antichain in this lattice.\r\nA can start with the set {1,2...n-1},so any further choice has to be a set containing n,so A becomes player 2 in the variant of this same game on the set{1,2...n-1}(since n has to be part of every subsequent choice we can simply drop it) but with the singletons included in it.now in this poset,every element has its 'complement',so for any choice that B makes,A can choose the 'complement' choice,thus A always has a move.\r\n\r\nthus this is just a rehash of the earlier solutions!", "Solution_7": "[quote=\"seshadri\"]actually putting N/m for B is not always a viable option-the numbers put down have to be composite,so if A chooses the product of n-1 primes,then the above move for B is violated.[/quote]\r\n :blush: I didn't see that players must write composite divisors." } { "Tag": [ "function", "induction", "number theory unsolved", "number theory" ], "Problem": "Find the largest positive integer $ x$ possible which is divisible by all the positive integers less than or equal to $ \\sqrt[3]{x}$", "Solution_1": "[quote=\"creatorvn\"]Find the largest positive integer $ x$ possible which is divisible by all the positive integers less than or equal to $ \\sqrt[3]{x}$[/quote]\r\n\r\nWe are looking for pair $ x, y$ such that $ x$ is divisible by all natural numbers $ n$ with $ 1\\le n \\le y$, and such that $ (y+1)^{3}> x\\ge y^{3}$.\r\nNow look at the two functions\r\n$ y^{3}$,\r\n$ f(y) =$ the smallest integer which is divisible by all natural numbers $ n$ with $ 1\\le n \\le y$.\r\nIf we find greatest $ y$ such that $ (y+1)^{3}> f(y) \\ge y^{3}$, the solution will be $ x=f(y)$. So let's find such $ y$.\r\nWe get that $ y=7$ works, so $ x=420$ works. There are no greater solutions because for $ y\\ge 8$, $ f(y) > (y+1)^{3}$.", "Solution_2": "[quote=\"djuro\"]...There are no greater solutions because for $ y\\ge 8$, $ f(y) > (y+1)^{3}$.[/quote]\r\n\r\nThat's true (and result 420 too) but it remains to give the proof.", "Solution_3": "[quote=\"pco\"][quote=\"djuro\"]...There are no greater solutions because for $ y\\ge 8$, $ f(y) > (y+1)^{3}$.[/quote]\n\nThat's true (and result 420 too) but it remains to give the proof.[/quote]\r\n\r\nYou're right. Well, it's not hard to see that $ f(y)$ grows much faster than $ (y+1)^{3}$, but I don't have a strict proof. Do you?", "Solution_4": "[quote=\"djuro\"][quote=\"pco\"][quote=\"djuro\"]...There are no greater solutions because for $ y\\ge 8$, $ f(y) > (y+1)^{3}$.[/quote]\n\nThat's true (and result 420 too) but it remains to give the proof.[/quote]\n\nYou're right. Well, it's not hard to see that $ f(y)$ grows much faster than $ (y+1)^{3}$, but I don't have a strict proof. Do you?[/quote]\r\n\r\nI have.\r\nLet $ p_{n}$ be the $ n^{th}$ prime number ($ p_{1}=2, p_{2}=3, p_{3}=5, p_{4}=7, p_{5}=11, \\cdots$).\r\n\r\nFirst, we can show with induction that $ \\prod_{i=1}^{n}p_{i}>p_{n+1}^{3}$ $ \\forall n\\geq 5$ :\r\nIt's true for $ n=5$ : $ \\prod_{i=1}^{5}p_{i}=2*3*5*7*11=2310>p_{6}^{3}=13^{3}=2197$\r\nThen, if we have $ \\prod_{i=1}^{n}p_{i}>p_{n+1}^{3}$, then $ \\prod_{i=1}^{n+1}p_{i}>p_{n+1}^{4}$. But we have, with $ n\\geq 5$, $ p_{n+1}>8$ and then $ p_{n+1}^{4}>(2p_{n+1})^{3}$. And we know that $ 2p_{n+1}>p_{n+2}$ (Bertrand - Chebyschev) and so $ \\prod_{i=1}^{n+1}p_{i}>p_{n+2}^{3}$. Q.E.D.\r\n\r\nNow, if $ x\\geq 8^{3}$, $ x$ must be divisble by $ 2,3,4,5,6,7,8$ and so by $ 840>9^{3}$. So $ x$ must be divisble by $ 2,3,4,5,6,7,8,9$ and so by $ 2520>11^{3}=p_{5}^{3}$. But then, let $ k(x)\\geq 5$ be the natural number such that $ p_{k(x)}^{3}\\leq xp_{k(x)+1}^{3}$ which is impossible. And so $ x$ can't be $ \\geq 8^{3}$\r\n\r\nThen, if $ 7^{3}\\leq x<8^{3}$, $ x$ must be divisible by $ 1,2,3,4,5,6,7$ and so by $ 420$. So the greatest value is $ 420$" } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "$ a$,$ b$,$ c$,$ A$,$ B$,$ C$ are positive numbers such that:\r\n$ a\\plus{}A\\equal{}b\\plus{}B\\equal{}c\\plus{}C\\equal{}k$\r\nprove that :\r\n$ aB\\plus{}bC\\plus{}cAc \\geq\\frac{a\\plus{}b\\plus{}c}{2}$ and $ f(c)\\equal{}ab>0$ where $ f(x)\\equal{}x^2\\minus{}(a\\plus{}b\\plus{}c)x\\plus{}ab\\plus{}bc\\plus{}ca$ \r\nso $ f(k)>0$", "Solution_7": "It is from the USSR national olympiad, year don't remember. 8 grade, if i'm not mistaken.", "Solution_8": "We want to show $ k^2\\minus{}k(a\\plus{}b\\plus{}c)\\plus{}ab\\plus{}bc\\plus{}ca>0$.\r\nSol1.\r\n$ (k\\minus{}a)(k\\minus{}b)(k\\minus{}c)\\geq0$because$ 0$ k^3\\minus{}k^2(a\\plus{}b\\plus{}c)\\plus{}k(ab\\plus{}bc\\plus{}ca)\\minus{}abc>0$=>$ k^2\\minus{}k(a\\plus{}b\\plus{}c)\\plus{}ab\\plus{}bc\\plus{}ca\\geq0$ because $ abc>0$.\r\nSol2.\r\nLet $ f(a,b,c)\\equal{}k^2\\minus{}k(a\\plus{}b\\plus{}c)\\plus{}ab\\plus{}bc\\plus{}ca$.\r\nlet b,c is constant. so $ f(a,b,c)$ is a linear funtion. So we enough to think $ a\\equal{}0,k$. similarly we enough to think $ a,b,c\\equal{}0,k$..the next is easy.", "Solution_9": "We have\r\n$ a(k \\minus{} b) \\plus{} b(k \\minus{} c) \\plus{} c(k \\minus{} a) \\le k^2 \\implies k(a \\plus{} b \\plus{} c) \\minus{} ab \\minus{} bc \\minus{} ac \\le k^2$\r\nBut for a fixed a+b+c, ab+bc+ac is minimized when a=b=0, c equals the rest\r\nBut c must be less than k." } { "Tag": [ "AMC 12 A", "AMC" ], "Problem": "I was wondering, if I took the AMC 12 A, could I still take the 12 B? If so how would I go about doing that because I don't think my school ordered it? Thanks.", "Solution_1": "you might wanna look here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=73356\r\nit was posted just 2 hrs ago!" } { "Tag": [ "function", "number theory unsolved", "number theory" ], "Problem": "$ m,n\\in Z \\plus{} , m\\neq n$\r\nprove that $ \\exists x \\in \\Re$\r\ns.t. $ \\frac {1}{3} \\leq${$ {x(n)}$}$ \\leq \\frac {2}{3}$ and \r\n$ \\frac {1}{3} \\leq${$ x(m)$}$ \\leq \\frac {2}{3}$\r\nHere {x} $ \\equal{} x \\minus{} [x]$", "Solution_1": "Any thoughts? \r\nPlease help me if you can solve" } { "Tag": [ "function" ], "Problem": "Prove that if a relation with more than one element is transitive, it cannot be a function. (or prove it wrong)\r\n\r\nI'm not sure if this belongs here because they don't really teach this in middle school. Please PM me on where this should belong if it's in the wrong spot.", "Solution_1": "We didn't even know what elements [i]were[/i] in middle school. This definately belongs in HSB.", "Solution_2": "[quote=\"ashpotter\"]Prove that if a relation with more than one element is transitive, it cannot be a function. (or prove it wrong)\n\n[/quote]\r\nI'm not sure...\r\n[hide]If $ a$ gives $ b$ and $ b$ gives $ c$ then if it's transitive, $ a$ gives $ c$. But then $ a$ gives $ b$ and $ a$ gives $ c$. Hence it cannot be a function.[/hide]", "Solution_3": "If the relation $ R$ is defined on $ S\\times S$, and $ s$ is some fixed element of $ S$ then let $ R$ be such that $ aRb\\Leftrightarrow b \\equal{} s$. This seems to be a function to me, and the relation is also transitive. I think this is a counterexample.\r\n\r\nIn the above argument, you didn't take into account the case $ b \\equal{} c$." } { "Tag": [ "puzzles" ], "Problem": "Put the digits 1 to 9 inclusive into a grid of nine squares (3 squares to a row, 3 rows) such that the sum of the digits in every row, column, and 3-square diagonal is 15.", "Solution_1": "[quote=\"emannes\"]Put the digits 1 to 9 inclusive into a grid of nine squares (3 squares to a row, 3 rows) such that the sum of the digits in every row, column, and 3-square diagonal is 15.[/quote]\r\n[hide]Sorry for the no picture. \nTop left-4\nTop Middle-3\nTop Right-8\nMiddle Left-9\nMiddle Middle-5\nMiddle Left-1\nBottom left-2\nBottom Middle-7\nBottom Left-6[/hide]" } { "Tag": [ "function", "symmetry", "integration", "calculus", "real analysis", "calculus computations" ], "Problem": "Show that if $ g$ is not integrable on $ [0,1]$, then the sequence of averages \r\n\r\n$ \\displaystyle\\frac{1}{n}\\displaystyle\\sum_{k\\equal{}1}^{n} g (\\frac{k}{n})$\r\n\r\nmay or may not be convergent.\r\n\r\nMy question is why are we not asked to prove one way or the other.", "Solution_1": "Because you can't prove something that is false. (Well, I know some people who can, but let's not go into that.)\r\n\r\nYou need two functions $ g$: one such that the sequence of averages is convergent, the other such that the sequence is divergent. The second is quite easy: $ g(x) \\equal{} 1/x$ will work. The first one is tricky, unless one takes an easy way out and redefines $ g(x) \\equal{} 1/x$ at rational points. Let's find continuous (on $ (0,1)$) non-integrable $ g$ with this property...", "Solution_2": "if you show $ g(x)$ is not integrable with the Riemann - Stieltjes\r\n\r\nis it enough to say that for every $ \\epsilon$ and $ \\xi \\in [x_{k\\minus{}1}, x_{k}]$ that \r\n\r\n$ \\sum_{k\\equal{}1}^{n} g(\\xi)[x_{k}\\minus{}x_{k\\minus{}1}]$ is not defined particularly when $ \\xi \\equal{} 0$, thereby $ g(x)$ not integrable?\r\n\r\nThanks for the help.", "Solution_3": "Can I use $ g(x) \\equal{} \\frac{1}{x^{2}}$ to make the sequence of averages convergent(by p-series test)\r\n\r\nalso can someone answer my question from the last post(using R-S to show NOT integrable)?", "Solution_4": "Um, no. If $ g(x) \\equal{} \\frac1{x^2},$ then\r\n\r\n$ \\frac1n\\sum_{k \\equal{} 1}^ng\\left(\\frac {k}{n}\\right) \\equal{} \\frac1n\\sum_{k \\equal{} 1}^n\\frac {n^2}{k^2} \\equal{} n\\sum_{k \\equal{} 1}^n\\frac1{k^2}.$\r\n\r\nYes, $ \\sum_{k \\equal{} 1}^{\\infty}\\frac1{k^2}$ converges. All that means is that as $ n\\to\\infty,$ $ \\sum_{k \\equal{} 1}^n\\frac1{k^2} \\equal{} L \\plus{} o(1)$ for some specific positive $ L.$ (Yes, we know an exact value for $ L,$ but that's not needed for this argument.)\r\n\r\nThen $ n\\sum_{k \\equal{} 1}^n\\frac1{k^2} \\equal{} Ln \\plus{} o(n)$ and this tends to infinity as $ n\\to\\infty.$\r\n\r\nAll this gives us is a slightly more emphatic version of mlok's suggestion of $ g(x) \\equal{} \\frac1x,$ which would have given us $ \\sum_{k \\equal{} 1}^n\\frac1k,$ which is $ \\ln(n) \\plus{} O(1)$ for large $ n.$ Both examples make the same point, and neither answers the question which was the last sentence in mlok's post.\r\n\r\nFinding the answer to that last question [i]is[/i] tricky. I don't have it yet.", "Solution_5": "thanks for the explanation Kent. \r\n\r\nhow would i show that $ g(x) \\equal{}\\frac{1}{x}$ is NOT differentiable over $ [0,1]$(using Riemann -Stieltjes)?\r\n\r\n\r\nthanks", "Solution_6": "[Responding to mlok's and Kent's question]\r\n\r\nOne way out is to use symmetry, as in a principal value; try $ f(x)\\equal{}\\frac1x\\minus{}\\frac1{1\\minus{}x}\\equal{}\\frac{1\\minus{}2x}{x(1\\minus{}x)}$ on $ (0,1)$, and zero the values at the endpoints. The sum is always zero.\r\n\r\nWe could also try putting the singularity in the middle; $ f(x)\\equal{}\\begin{cases}\\frac1{2x\\minus{}1}&x\\neq\\frac12\\\\0&x\\equal{}\\frac12\\end{cases}$ works.\r\n\r\nThis is risky, of course- it doesn't work for any other point $ c$ in place of $ \\frac12$.", "Solution_7": "a big thank you jmerry, one last question:\r\n\r\ndoes anyone have an idea how to show $ g(x) \\equal{} \\frac{1}{x}$ is not integrable using Riemann-Stieltjes?\r\n\r\n\r\nthanks", "Solution_8": "Riemann-Stieljes can't be the right question without more information. To interpret $ \\int_a^b f\\,dg$ we need to know what $ g$ is.\r\n\r\nAs for the Riemann integral: any unbounded function must always fail to be Riemann integrable. Let's concentrate on $ \\frac1x$ (defined to be zero at zero) on $ [0,1].$\r\n\r\nTake any partition $ P.$ Take any $ M>0.$ The first interval of this partition has the form $ [0,a_1].$ Chose $ \\xi_1$ so that $ a_1\\cdot\\frac1{\\xi_1}>M.$ Then (since the rest of the terms are nonnegative), the Riemann sum on this partition is greater than $ M.$ Hence the Riemann sums cannot converge and the function is not integrable.", "Solution_9": "Could we have used $ g(x) \\equal{} \\begin{cases}1&x\\in\\mathbb{Q} \\\\0&x\\in\\mathbb{R\\minus{}Q}\\end{cases}$ for the case where $ g(x)$ is non integrable and the sequence of averages converges?", "Solution_10": "Yes, you could have, and mlok hinted at that. It's just that mlok decided to issue an extra challenge to get someone like jmerry to find the way." } { "Tag": [ "Putnam", "AMC", "AIME", "combinatorics proposed", "combinatorics" ], "Problem": "Find the largest two-digit prime that divides 200C100 (nCk denotes n chooses k)", "Solution_1": "[quote=\"Christine\"]Find the largest two-digit prime that divides 200C100 (nCk denotes n chooses k)[/quote]\r\n\r\nLet $ A \\equal{} \\binom{200}{100} \\equal{} \\frac {200!}{100!100!}$\r\n\r\nAssume $ 100 > p\\geq50$. Then $ p$ divides exactly twice the lower part of $ A$. If we want that $ p$ divides at least 3 times the upper part ($ 200!$), we need to have $ 3p\\leq 200$ and so $ p\\leq 66$.\r\n\r\nAnd hence the result $ p \\equal{} 61$", "Solution_2": "I've done this question before (and it's appeared on the forum before) -- what is it, an old Putnam or AIME or something?", "Solution_3": "AIME, 1983:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=51991[/url]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Given $ a,b,c > 0$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$. Prove that:\r\n$ 8(2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge (a \\plus{} bc)(b \\plus{} ca)(c \\plus{} ab)$\r\n(Tran Quoc Luat)", "Solution_1": "nice ineq\r\n\r\nmultiplying the both sides by $ \\prod{(2\\plus{}a)}$ the ineq is equivalent to :\r\n$ 8\\prod{(4\\minus{}a^2)} \\geq \\prod{(2\\plus{}a)} \\prod{(a\\plus{}bc)}$\r\n\r\nobviously $ \\prod{(2\\plus{}a)} \\leq 27$:\r\nit sufficies to show that :\r\n\r\n$ 8\\prod{(4\\minus{}a^2)} \\geq 27\\prod{(2\\plus{}a)}$\r\n\r\nbut :\\[ (\\prod(4\\minus{}a^2))^2\\equal{} (\\prod(b^2\\plus{}c^2\\plus{}1))^2 \\equal{} \\prod(a^2\\plus{}1\\plus{}b^2)(a^2\\plus{}1\\plus{}c^2) \\geq \\prod(a^2\\plus{}1\\plus{}bc)^2 \\geq \\prod(2a\\plus{}bc)^2\\] (Cauchy and am-gm)\r\nhence it's enought to prove that: $ 8\\prod(2a\\plus{}bc) \\geq 27\\prod(a\\plus{}bc)$ expanding and using $ \\sum{x^2}\\equal{}3$ it's equivalent to:\r\n$ 4abc \\plus{}5\\sum{(ab)^2} \\minus{}19(abc)^2 \\geq 0$ but it's obvious since : $ abc\\leq 1$ and $ \\sum{(ab)^2} \\geq 3(abc)^{\\frac{4}{3}}$ :wink:", "Solution_2": "[quote=\"nguyenvanhuyen\"]Given $ a,b,c > 0$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$. Prove that:\n$ 8(2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge (a \\plus{} bc)(b \\plus{} ca)(c \\plus{} ab)$\n(Tran Quoc Luat)[/quote]\r\n\r\nWe have : $ 8(2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge (a^2\\plus{}b^2)(b^2\\plus{}c^2)(c^2\\plus{}a^2)$\r\n\r\nWe have to prove that : $ (a^2\\plus{}b^2)(b^2\\plus{}c^2)(c^2\\plus{}a^2) \\geq (a \\plus{} bc)(b \\plus{} ca)(c \\plus{} ab)$\r\n\r\nIt is easily by AM-GM ;)", "Solution_3": "[quote=\"Materazzi\"][quote=\"nguyenvanhuyen\"]Given $ a,b,c > 0$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$. Prove that:\n$ 8(2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge (a \\plus{} bc)(b \\plus{} ca)(c \\plus{} ab)$\n(Tran Quoc Luat)[/quote]\n\nWe have : $ 8(2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge (a^2 \\plus{} b^2)(b^2 \\plus{} c^2)(c^2 \\plus{} a^2)$\n\n[/quote]\r\n\r\nwhy? :wink:", "Solution_4": "[quote=\"anas\"][quote=\"Materazzi\"][quote=\"nguyenvanhuyen\"]Given $ a,b,c > 0$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$. Prove that:\n$ 8(2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge (a \\plus{} bc)(b \\plus{} ca)(c \\plus{} ab)$\n(Tran Quoc Luat)[/quote]\n\nWe have : $ 8(2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge (a^2 \\plus{} b^2)(b^2 \\plus{} c^2)(c^2 \\plus{} a^2)$\n\n[/quote]\n\nwhy? :wink:[/quote]\r\nBecause $ 4 \\minus{} 2a \\equal{} b^2 \\plus{} c^2 \\plus{} (a \\minus{} 1)^2 \\geq b^2 \\plus{} c^2$." } { "Tag": [ "rotation", "conics", "ellipse", "linear algebra", "matrix", "quadratics", "linear algebra unsolved" ], "Problem": "I know that the following equation represent an ellipse:\r\n\\[ \\left( \\frac{\\minus{}2a\\plus{}2b}{3} \\right)^2 \\plus{}\r\n \\left( \\frac{a\\minus{}4b}{6} \\right)^2 \\plus{} \r\n \\left( \\frac{a\\plus{}2b}{3} \\right)^2 \\equal{} 1\r\n\\]\r\n\r\nHowever, the major axis of this ellipse is not on the $ x$-axis. So I want to use some transformation to rotate it so that the major axis coincide with the $ x$-axis. How does one proceed for problems like this? Is there a way to use linear algebra to do this?\r\n\r\nEventually, I want it look like the standard equation for an ellipse:\r\n\\[ \\frac{a^2}{m^2} \\plus{} \\frac{b^2}{n^2} \\equal{} 1.\r\n\\]", "Solution_1": "Well, you've got three terms and that's bad. Expand everything out into a form $ pa^2 \\plus{} 2qab \\plus{} rb^2 \\equal{} 1$.\r\nNow, look at the matrix $ \\begin{bmatrix}p & q \\\\\r\nq & r\\end{bmatrix}$ of this quadratic form. You want an orthogonal change of basis to diagonalize it." } { "Tag": [ "function", "integration", "limit", "Functional Analysis", "advanced fields", "advanced fields unsolved" ], "Problem": "$f: \\mathbb{R}\\to \\mathbb{R}$ a continuous and bounded function, $\\Omega$ a bounded open subset of $\\mathbb{R}^{n}$. Show that there exists at least one function $u \\in H^{1}_{0}(\\Omega)$ such that\r\n$\\Delta u = f(u)$\r\nShow that if $f$ is decreasing, this solution is unique.", "Solution_1": "We know that $\\Delta: H_{0}^{1}(\\Omega)\\to H^{-1}(\\Omega)$ is a bounded isomorphism with bounded inverse. We construct a sequence of functions by setting $u_{i+1}=\\Delta^{-1}f(u_{i})$. The $f(u_{i})$'s are bounded in $L^{2}$ and consequently they are relatively compact in $H^{-1}$. So the $u_{i}$'s are relatively compact in $H^{1}_{0}$. Now take a convergenct subsequence to obtain the desired result. (Or, you can use that other result you asked about regarding the fixed points of compact maps.)\r\n\r\nThe answer to the second question is totally independent. Given two solutions $u_{1}$ and $u_{2}$, just apply the maximum principle to $u_{1}-u_{2}$ in the region where $u_{1}\\geq u_{2}$. Actually, the proof is more technical than this, since we only have $H^{1}_{0}$ rather than $C^{2}_{0}$.", "Solution_2": "Nice :) I prefer your solution than mine (as you have guessed with fiexed point) because it is less involved, and shauder fixed point are not that easy to demonstrate :lol: Thank you", "Solution_3": "Here's another approach. Denote $U(x): =-\\int_{0}^{x}f(t)dx$ then $u$ is a solution to our problem iff it is a critical point of the functional\r\n\\[\\Phi(u)=\\frac12\\int_\\Omega|\\nabla u|^{2}dx+\\int_\\Omega U(u)dx. \\]\r\nSince $f$ is bounded $\\Phi$ is weakly lower semi-continuous coercive (i.e. $\\lim_{\\|u\\|\\to\\infty}\\Phi(u)=+\\infty$) functional on the reflexive Banach space $X: =H_{0}^{1}(\\Omega)$. It remains to make use of the following statement:\r\n\r\n[b]Proposition.[/b] Let $X$ be a reflexive Banach space, $\\Phi\\colon X\\to\\mathbf R$ a coercive weakly lower semi-continuous functional. Then $\\Phi$ attains its minimum in $X$." } { "Tag": [], "Problem": "if m,n are natural numbers prove that\r\n $\\frac{1}{m+n+1}-\\frac{1}{(m+1)(n+1)}=< \\frac{4}{45}$", "Solution_1": "I think you mean\r\n\r\n$\\frac{1}{m+n+1}-\\frac{1}{mn+m+n+1}\\leq\\frac{4}{45}$\r\n\r\n[hide=\"contradiction:\"]\n\nWe can give the fractions common denominators:\n\n$\\frac{mn+m+n+1}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}-\\frac{m+n+1}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}>\\frac{4}{45}$\n\n$\\frac{mn}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}>\\frac{4}{45}$\n\n$\\frac{45}{4}>\\frac{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}{mn}$\n\n$\\frac{45}{4}>m+n+\\frac{m}{n}+\\frac{n}{m}+3+\\frac{2}{n}+\\frac{2}{m}+\\frac{1}{mn}=m+n+3+\\frac{m+2}{n}+\\frac{n+2}{m}+\\frac{1}{nm}$\n\n45/4 is 10+1/4.\n\n$7+\\frac{1}{4}>m+n+\\frac{m+2}{n}+\\frac{n+2}{m}+\\frac{1}{nm}$\n\nWe now need to prove that, for all natural numbers m and n, that 7 and 1/4 is greater than all of that. If both m and n equal 1, then all of that equals 1+1+3+3+1, which is greater than 7.25. If m=1 and n=2: 1+2+3/2+4+1/2, which is again greater than 7.25. It doesn't matter which order we plug in values for m and n, so we don't need to find it if m=2 and n=1.\n\nIf m=3 and n=1: 3+1+5+blahblahblah, which is again greater than 7.25\n\nIt's the same thing with n=1, m=(4,5,6). We don't need to check any other pairs that have 1, because the first two terms added together will result in 8 or greater.\n\nSo let's try 2 and 2:\n\n2+2+2+2+blahblahblah, which is greater than 7.\n\n2 and 3:\n\n2+3+4/3+5/2+blahblah, which is greater than 7.25\n\n2 and 4:\n\n2+4+1+blahblah, which is greater than 7.25\n\nWith 3, it's always greater than 7.25 Therefore, all of the above statements are false. That makes\n\n$\\frac{1}{m+n+1}-\\frac{1}{mn+m+n+1}\\leq\\frac{4}{45}$ true.[/hide]\r\n\r\nI WUV algebra!!! :rotfl:", "Solution_2": "Eh, there's a problem: $\\frac{45}{4}\\neq 10+\\frac{1}{4}$, it's a 11!! \r\nNow, I don't know how to solve \"beautifully\" the problem, but I hope that your proof still holds, even if I don't really understood why you were trying just a few values of m and n. Ok, you might say that after that the sum grows too much, but it would be better to demonstrate it, don't you think? :wink:", "Solution_3": "[quote=\"nutz_for2.718281828\"]I think you mean\n\n$\\frac{1}{m+n+1}-\\frac{1}{mn+m+n+1}\\leq\\frac{4}{45}$\n\n[hide=\"contradiction:\"]\n\nWe can give the fractions common denominators:\n\n$\\frac{mn+m+n+1}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}-\\frac{m+n+1}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}>\\frac{4}{45}$\n\n$\\frac{mn}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}>\\frac{4}{45}$\n\n$\\frac{45}{4}>\\frac{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}{mn}$\n\n$\\frac{45}{4}>m+n+\\frac{m}{n}+\\frac{n}{m}+3+\\frac{2}{n}+\\frac{2}{m}+\\frac{1}{mn}=m+n+3+\\frac{m+2}{n}+\\frac{n+2}{m}+\\frac{1}{nm}$\n\n45/4 is 10+1/4.\n\n$7+\\frac{1}{4}>m+n+\\frac{m+2}{n}+\\frac{n+2}{m}+\\frac{1}{nm}$\n\nWe now need to prove that, for all natural numbers m and n, that 7 and 1/4 is greater than all of that. If both m and n equal 1, then all of that equals 1+1+3+3+1, which is greater than 7.25. If m=1 and n=2: 1+2+3/2+4+1/2, which is again greater than 7.25. It doesn't matter which order we plug in values for m and n, so we don't need to find it if m=2 and n=1.\n\nIf m=3 and n=1: 3+1+5+blahblahblah, which is again greater than 7.25\n\nIt's the same thing with n=1, m=(4,5,6). We don't need to check any other pairs that have 1, because the first two terms added together will result in 8 or greater.\n\nSo let's try 2 and 2:\n\n2+2+2+2+blahblahblah, which is greater than 7.\n\n2 and 3:\n\n2+3+4/3+5/2+blahblah, which is greater than 7.25\n\n2 and 4:\n\n2+4+1+blahblah, which is greater than 7.25\n\nWith 3, it's always greater than 7.25 Therefore, all of the above statements are false. That makes\n\n$\\frac{1}{m+n+1}-\\frac{1}{mn+m+n+1}\\leq\\frac{4}{45}$ true.[/hide]\n\nI WUV algebra!!! :rotfl:[/quote]\r\n\r\nOoh... pretty... \r\n\r\nIs that the only way to do it, or is there something... not as ugly?" } { "Tag": [ "Pythagorean Theorem", "geometry" ], "Problem": "$ \\overline{AB}$ is the hypotenuse of a right triangle $ ABC$. Median $ \\overline{AD}$ has length $ 7$ and median $ \\overline{BE}$ has length $ 4$. The length of $ \\overline{AB}$ is:\r\n\r\n$ \\textbf{(A)}\\ 10\\qquad \r\n\\textbf{(B)}\\ 5\\sqrt{3}\\qquad \r\n\\textbf{(C)}\\ 5\\sqrt{2}\\qquad \r\n\\textbf{(D)}\\ 2\\sqrt{13}\\qquad \r\n\\textbf{(E)}\\ 2\\sqrt{15}$", "Solution_1": "[hide]\n\nLet $ CE\\equal{}x$ and $ CD\\equal{}y$. By Pythagorean Theorem, $ x^2\\plus{}4y^2\\equal{}16$ and $ 4x^2\\plus{}y^2\\equal{}49$. Adding the two equations, $ 5x^2\\plus{}5y^2\\equal{}65$. Multiplying by $ \\frac{4}{5}$ gets $ 4x^2\\plus{}4y^2\\equal{}52$, so $ AB\\equal{}\\sqrt{52}\\equal{}2\\sqrt{13}$. $ \\boxed{\\textbf{D}}$\n\n[/hide]", "Solution_2": "[hide=\"Similar solution\"]Join D to E\n\nLet $ BD \\equal{} CD \\equal{} x$ and $ AE \\equal{} CE \\equal{} y$\n\nAs above we get two equations\n\n$ 4x^2 \\plus{} y^2 \\equal{} 16\\,\\, .\\, .\\, .\\, \\left(1\\right)$\n\n$ 4y^2 \\plus{} x^2 \\equal{} 49\\,\\, .\\, .\\, .\\, \\left(2\\right)$\n\nAdding these two we get\n\n$ 5\\left(x^2 \\plus{} y^2\\right) \\equal{} 65$\n\n$ \\implies x^2 \\plus{} y^2 \\equal{} 13$\n\nIn triangle $ CDE$,\n\n$ x^2 \\plus{} y^2 \\equal{} DE^2$\n\n$ \\therefore DE^2 \\equal{} 13$\n\n$ \\implies DE \\equal{} \\sqrt {13}$\n\nIt is known that the line joining two medians is both parallel to the third side and half its length. \n\n$ \\therefore AB \\equal{} 2DE \\equal{} 2\\sqrt {13} \\implies \\boxed{D}$[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "xyz=1 z,y,x>0\r\nfind the minimum value that can accept x^2+4*x*y+8*y^2+16*Z^2", "Solution_1": "Hmm...I move it into the inequalities forum...\r\n\r\nPierre.", "Solution_2": "I post my solution to it in pre-olympiad: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23742]www.mathlinks.ro/Forum/viewtopic.php?t=23742[/url]" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Quadrilateral $ ABCD$ is circumscribed, rays $ BA$ and $ CD$ intersect in point $ E$, rays $ BC$ and $ AD$ intersect in point $ F$. The incircle of the triangle formed by lines $ AB$, $ CD$ and the bisector of angle $ B$, touches $ AB$ in point $ K$, and the incircle of the triangle formed by lines $ AD$, $ BC$ and the bisector of angle $ B$, touches $ BC$ in point $ L$. Prove that lines $ KL$, $ AC$ and $ EF$ concur.\r\n\r\n[i]Author: I.Bogdanov[/i]", "Solution_1": "Let $ G \\equiv AC \\cap BD.$ Let $ (I)$ be incircle of quadrilateral $ ABCD,$ $ (X)$ incircle of triangle $ \\triangle (AB, CD, BI)$ and $ (Y)$ incircle of triangle $ \\triangle (BC, DA, BI).$ Let $ r, x, y$ be radii of $ (I), (X), (Y).$ Let $ (I)$ touch $ AB, BC, CD, DA$ at $ S, T, U, V.$ $ ST, UV, EF$ are polars WRT $ (I)$ of collinear points $ B, D, G,$ they are concurrent at the pole $ P$ of $ BD.$ By Pascal theorem for $ SSTUVV,$ $ A, G, P$ are collinear and by Pascal theorem for $ STTUUV,$ $ C, G, P$ are collinear $ \\Longrightarrow$ $ P \\in AC.$ \r\n\r\n$ \\frac {\\overline{KE}}{\\overline{KB}} \\cdot \\frac {\\overline{LB}}{\\overline{LE}} \\equal{} \\frac {\\overline{SE}}{\\overline{SB}} \\cdot \\frac {\\overline{TB}}{\\overline{TE}}$ $ \\Longleftrightarrow$ $ \\frac {\\overline{KE}}{\\overline{SE}} \\cdot \\frac {\\overline{TE}}{\\overline{LE}} \\equal{} \\frac {\\overline{KB}}{\\overline{LB}} \\cdot \\frac {\\overline{TB}}{\\overline{SB}}$ $ \\Longleftrightarrow$ $ \\frac {x}{r} \\cdot \\frac {r}{y} \\equal{} \\frac {x}{y}.$\r\n\r\nBy Menelaus theorem for $ \\triangle BEF$ cut by $ KL, ST,$ these 2 lines intersect at $ P \\equiv EF \\cap ST \\cap AC.$", "Solution_2": "Let $I$ be the incenter of quadrilateral $ABCD$, let $P,Q,R,S$ be feet from $I$ to $\\overline{AB},\\overline{BC},\\overline{CD},\\overline{DA}$, respectively. Let $I_1,I_2$ be the incenters of triangles formed by lines $AB$, $CD$ and the bisector of angle $B$ and lines $AD$, $BC$ and the bisector of angle $B$, respectively.\nObserve that by Brianchon's theorem, $\\overline{AC},\\overline{BD},\\overline{PR},\\overline{QS}$ are concurrent. Hence, by Pascal's theorem, $\\overline{EF},\\overline{AC},\\overline{PQ},\\overline{SR}$ are concurrent. Finally, it is sufficient to show $\\frac{KE}{EB}\\cdot \\frac{BF}{FL}=\\frac{KP}{PB}\\cdot \\frac{BQ}{QL}$ using Ceva/Menelaus, which is really equivalent to $\\frac{KE}{KP}\\cdot \\frac{QL}{FL}=\\frac{BE}{BF}$. Indeed, by angle bisector theorem, \\begin{align*}\n\\frac{KE}{KP}\\cdot \\frac{QL}{FL}=\\frac{I_1E}{I_1I}\\cdot \\frac{I_2I}{I_2F}=\\frac{BE}{BI}\\cdot \\frac{BI}{BF}=\\frac{BE}{BF},\n\\end{align*}as desired. $\\blacksquare$", "Solution_3": "Apologize for reviving this post, but I can't construct the diagram. Can somebody show the diagram for me? :( " } { "Tag": [], "Problem": "The result of the Indian National Astronomy Olympiad is available at [url=http://www.hbcse.tifr.res.in/olympiads/inbo06_results/inao2006_result]INAO Result 2006[/url]\r\n\r\nThe site shows that the Camp will be from [color=red]May 1 - May 20, 2006[/color]...\r\n\r\nCongrats and Best of luck! \r\n\r\nKindly continue all further discussion here! Thanks! \r\n\r\n\r\nList of students selected on merit basis from the [color=darkblue][size=150]INAO examination[/size][/color]\r\nfor the International Astronomy Olympiad Training Cum Selection Camp 2006\r\n(Note: The lists are in alphabatical order) \r\n\r\n[size=150]Junior Group[/size]\r\n\r\nRegion Name Of The Student City \r\n[color=red]Bhubneswar[/color] Soumyashant Nayak [color=red]Chandrasekharpur[/color] \r\n[color=orange]Gulbarga[/color] Sharat S Ibrahimpur [color=orange]Gulbarga[/color] \r\n[color=brown]Hyderabad A Krishna Sumanth Bagh Amberper \nHyderabad A. Mahathey Pavan Hyderabad \nHyderabad Adhokshaj Katarni Hyderabad \nHyderabad Arinjay Singh Parihar Hyderabad \nHyderabad C. V. K. Koundinya Hyderabad \nHyderabad E. Santhosh Kumar Hyderabad \nHyderabad S Mano Teja Reddy Hyderabad \nHyderabad Shruti Ranjan Satpathy Hyderabad \nHyderabad Srujan M. Hyderabad \nHyderabad T. Chenna Reddy Hyderabad \nHyderabad T. Gowtham Kumar Rajamundry [/color]\r\n[color=green]Jaipur Priyank Gupta Jaipur[/color] \r\n[color=olive]Mumbai Abhishek Padmanabhan Thane \nMumbai Adwait Vedant Mathkar Mumbai \nMumbai Kedar Shriram Tatwawadi Nashik \nMumbai Rahul Manoj Makhijani Mumbai \nMumbai Rupan Ranjan Saharoy Alibag, Raigad [/color]\r\n[color=indigo]New Delhi Adarsh Prasad New Delhi \nNew Delhi Nikhil Kumar New Delhi [/color]\r\n[color=violet]Pune Aniruddha Anand Bapat Pune \nPune Rohan Milind Khante Pune [/color]\r\n\r\n \r\n\r\n[size=150]Senior Group [/size]\r\n\r\nRegion Name Of The Student City \r\n[color=darkred]Bharuch Mohammad Fahim Bharuch [/color]\r\n[color=red]Bhubaneswar Debajyoti Manadhata Bhubaneswar \nBhubaneswar Saswat Rath Bhubaneswar [/color]\r\n[color=orange]Calicut Raziman T V Calicut[/color] \r\n[color=brown]Hyderabad Jayakiran Akurati Hyderabad [/color]\r\n[color=olive]Jaipur Gaurav Lahoti Udaipur \nJaipur Mehul Jain Jaipur \nJaipur Pranjal Bordia Udaipur \nJaipur Raman Sharma Jaipur [/color]\r\n[color=green]Lucknow Manish Singh Rathaur Lucknow \nLucknow Udbhav K. Singh Allahabad [/color]\r\n[color=olive]Mumbai Abhinav Sinha Mumbai \nMumbai Adtya Bhuvaneshwaran Mumbai \nMumbai Chiraag S Juvekar Mumbai \nMumbai Niranjan Aniruddha Viladkar Nashik \nMumbai Sivaramakrishnan Mumbai \nMumbai Tilak Sharma Mumbai [/color]\r\n[color=violet]Patna Abhay Kumar Patna \nPune Gund Ved Vishwas Pune \nPune Prashant Sachdeva Pune \nPune Shreyas Ganesh Patankar Pune \nTirupati Prathyush Manchala Secunderabad [/color]", "Solution_1": "Anyone who qualified here ?? I didn't make it but then next year's always there...", "Solution_2": "Yeah, same here. I think I was close to qualifying...lol.", "Solution_3": "[quote=\"mysmartmouth\"]Yeah, same here. I think I was close to qualifying...lol.[/quote]\r\n\r\nHmm.......I thought you said on some other post that you've never been to India though you're Indian.....so how on earth did you write the INAO ?? Do they have something for non resident Indians as well ?? :?", "Solution_4": "[quote]Anyone who qualified here ?? [/quote]\r\n\r\n\r\n\r\ni qualified :lol: \r\n\r\n\r\n\r\n\r\n\r\nudbhav", "Solution_5": "Hey udbhav, \r\n\r\nCongratulations !!!! :thumbup: \r\n\r\nAnd just by the way, do you have a copy of the senior question paper ?? Next year I guess I need to take up the senior astro-olympiad, so I'm pretty interested in getting this year's paper....", "Solution_6": "[quote=\"karthik_k\"][quote=\"mysmartmouth\"]Yeah, same here. I think I was close to qualifying...lol.[/quote]\n\nHmm.......I thought you said on some other post that you've never been to India though you're Indian.....so how on earth did you write the INAO ?? Do they have something for non resident Indians as well ?? :?[/quote]\r\n\r\nI guess I'm bad at sarcasm. LOL.", "Solution_7": "Oh............sorry I didn't understand what you meant... :oops:", "Solution_8": "the question papers aren't circulated.i'll write down the question which i remember.\r\n\r\n1) there is an aeroplane flying at the speed of 720km/h .it tries to complete a circular turn in the horizontal plane inclined at an angle of 15deg with the horizontal. find the radius of the circle in which it should turn.\r\n\r\n2) four planets are placed on each of the corners of a square of side of lengh L.find the speed which each of the planets should have in order to be in cicular orbits in the same arrangement.\r\n\r\n3) the distance of the earth from the sun is 1.5x10^11 m .the solar constant at earth is 1400 Wm^-2.s^-1.take radius of the earth as 600km.asuuming the sun to be a perfect black body find the temperature of the surface of the sun.\r\n\r\n4) by hubble's law it is known that v=hr where r is the distance of a galaxy or any such object and v is it's velocity and H is hubble's const.the wavelengh of a spectral line from a galaxy is found to be 4089 Angstrom(A) .it's laboratory wavelength is known to be 4081 A.find the distance of the galaxy from earth given that \r\n1/H=15billion seconds.\r\n\r\nNote- i have done my best to make sure that these questions resemble the original as much as possible.please point out if you find any mistakes with the questions.i can't remember the rest so i haven't written them down. \r\n\r\n\r\nUdbhav", "Solution_9": "[quote]four planets are placed on each of the corners of a square of side of lengh L.find the speed which each of the planets should have in order to be in cicular orbits in the same arrangement.\n[/quote]\r\n\r\nHey !!!!!!! We had that stuff on the junior paper too !!!!!!! Anyway, thanks a lot for those questions..... :lol:" } { "Tag": [], "Problem": "EDIT: SEE [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=189681[/url]", "Solution_1": "OKAY GUYS\r\nIGNORE THIS TOPIC PLZ\r\nALL INFO CONCERNING THIS MATCHUP CAN BE FOUND IN TEH OTHER TOPIC\r\nthank you" } { "Tag": [], "Problem": "Everyone will receive a unique piece. The game is divided up into two phases:\r\n\r\n1) The Private, \"taking\" phase (by pm)\r\nIn this phase you download all files your target has. That is, they don't lose files (but don't gain either), and you get whatever they have (barring duplicates).\r\n\r\n2) The Public, \"giving\" phase\r\nYou choose a person to \"give\" your files to (also a download). No person can be targeted more than once. For this phase you must post your target.\r\n\r\nThen it repeats until the win condition is satisfied:\r\n\r\n3) Get all the pieces to win.\r\n\r\nNote that during Phase 1 I will tell you what files you already have. Also you will only know people's starting piece, not their entire collection.\r\n\r\nSignups now :)", "Solution_1": "What do you mean by download files?", "Solution_2": "Figurative sense.\r\n\r\nSo if player 1 has Pieces 1,3,8\r\nand player 2 has Pieces 2,4,8\r\n\r\nand player 1 downloads off player 2, player 1 now has Pieces 1,2,3,4,8.", "Solution_3": "How many pieces in all?", "Solution_4": "as many as there are players, so in a game of 10 players there are 10 pieces.", "Solution_5": "\\in for this game i guess...", "Solution_6": "Sure, I'll join. This sounds fun :)", "Solution_7": "me too\r\nnow the message is long enough", "Solution_8": "i join ", "Solution_9": "in[b][/b]", "Solution_10": "I'll join.", "Solution_11": "i will join.", "Solution_12": "I'll join too.\r\n\r\nThis reminds me of torrenting, for some reason.", "Solution_13": "1) algebra2000\r\n2) AIME15\r\n3) tennis123\r\n4) yaofan\r\n5) EggyLv.999\r\n6) vahalla\r\n7) budi713\r\n8) Brut3Forc3.\r\n\r\nThis is enough, we will start now!\r\nEveryone has the piece associated with their order of joining, so vahalla has piece 6.\r\n\r\nPM your files to take, deadline on Wednesday." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "I have started learning measure theory and i need some help (all i quite easy).\r\n\r\nLet $X$ be an infinite set and let $\\mathcal{A}$ be a set of collection of all subsets $A$ of $X$ such that either $A^{c}$ or $A$ is finite.$\\mathcal{A}$ is not a $\\sigma$-algebra.\r\n\r\nFind the $\\sigma$-algebra on $\\mathbb{R}$ that is generated by the collection of all one point subsets of $\\mathbb{R}$.\r\n\r\nFind an example that the union of a collection o $\\sigma$-algebras on a set $X$ fail to be a $\\sigma$-algebra on X.\r\n\r\nIf its all very easy just give me a hint :wink: .", "Solution_1": "1) Extract a sequence $(c_{n})$ out of $X$. All the singletons $\\{c_{2n}\\}$ are in script A, but the union is not (why not?).\r\n2) At the very least, the generated set must contain all countable sets and the complements of these. This collection forms a $\\sigma$-algebra, so you get the reverse inequality. \r\n3) You want to take the union of some relatively \"small\" $\\sigma$-algebras. This can be achieved by letting $B_{n}$ be the $\\sigma$-algebra in the natural numbers generated by the first $n$ numbers." } { "Tag": [ "geometry", "trigonometry", "rectangle", "algebra proposed", "algebra" ], "Problem": "Find a geometric solution for this: \\[Arctan(\\frac{1}{2})+Arctan(\\frac{1}{3})=\\frac{\\pi}{4}\\] (Iranian Olympiad)", "Solution_1": "If you want a geometrical solution, you should have posted it in the Geometry section.\r\n\r\nLet $ACD$ be a triangle with $\\angle{D}=90^{o}$ and $\\tan{A}=\\frac{1}{2}$, and $BC'D'$ be a triangle with $\\angle{D'}=90^{o}$ and $\\tan{B}=\\frac{1}{3}$. If we let $CD=C'D'=1$, then we can use ACD and BC'D' to make a new triangle ACB such that $AC=\\sqrt{5}$, $BC=\\sqrt{10}$ and $AB=5$. By the law of cosine, we get $\\cos{\\angle{ACB}}=-\\frac{\\sqrt{2}}{2}$ which implies $\\angle{ACB}=135^{o}$. Thus $\\angle{A}+\\angle{B}=45^{o}$.", "Solution_2": "Shobber,your solution is very nice. I give an other solution. I think this is simpler.\r\n\r\nIn the following rectangle, let $AB=BC=CD=FG=GH=HE=1$ and $AF=1.$\r\nwe have $\\vartriangle BFD \\sim \\vartriangle BFC$ because $\\frac{{FB}}{{BD}}= \\frac{{BC}}{{FB}}.$Then \\[\\angle BCF = \\angle BFD \\Rightarrow \\angle BDF+\\angle BCF = \\angle BDF+\\angle BFD = \\frac{\\pi }{4}\\]" } { "Tag": [], "Problem": "!\u0627\u06cc\u0646 \u062c\u0627 \u0628\u0631\u0627\u06cc \u0645\u0633\u0627\u0644\u0647 \u0647\u0627\u06cc \u062c\u0628\u0631 \u062d\u0627\u0644\u0627 \u062f\u06cc\u06af\u0647 \u0644\u0637\u0641 \u06a9\u0646\u06cc\u062f \u0633\u0648\u0627\u0644\u0647\u0627\u062a\u0648\u0646 \u0631\u0648 \u0628\u0627 \u0647\u0645\u0648\u0646 \u0633\u06cc\u0633\u062a\u0645 \u0628\u0641\u0631\u0633\u062a\u06cc\u062f \u062f\u06cc\u06af\u0647", "Solution_1": "mish lotfan yeki ye raah hale tamiz eraae bede :huh: \r\n[url]http://www.mathlinks.ro/Forum/topic-59047.html[/url]", "Solution_2": "hatman mishe ye rahe halle tamiz erae dad . :D", "Solution_3": "0123456789", "Solution_4": "gharaar bede $a=x^3,b=y^3,c=z^3$\r\nhaalaa baayad saabet konim:\r\n$(y^3+x^3)(z^3+y^3)(x^3+z^3)\\geq2(x^3y^3z^3+(x^3+y^3+z^3)x^2y^2z^2)$\r\nyaa in ke\r\n$2x^3y^3z^3+x^6y^3+x^6z^3+y^6x^3+y^6z^3+z^6x^3+z^6y^3\\geq2(x^3y^3z^3+x^5y^2z^2+x^2y^5z^2+x^2y^2z^5)$\r\nyani dar nahaayat baayad saabet konim:\r\n$x^6y^3+x^6z^3+y^6x^3+y^6z^3+z^6x^3+z^6y^3\\geq2(x^5y^2z^2+x^2y^5z^2+x^2y^2z^5)$\r\nvali midoonim:\r\n$\\frac{x^6y^3+x^6z^3+x^6y^3+x^6z^3+y^6z^3+z^6y^3}{6}\\geq x^5y^2z^2$\r\n$\\frac{y^6x^3+y^6z^3+y^6x^3+y^6z^3+x^6z^3+z^6x^3}{6}\\geq x^2y^5z^2$\r\n$\\frac{z^6y^3+z^6x^3+z^6y^3+z^6x^3+y^6x^3+x^6y^3}{6}\\geq x^2y^2z^5$\r\nhaalaa age in 3 taaro gam konim naamosaavi saabet mishe :roll:", "Solution_5": "thanks...." } { "Tag": [ "geometry", "perimeter" ], "Problem": "If the perimeter of a square is $ 8 \\sqrt{5}$ centimeters, what is the area of the square, in square centimeters?", "Solution_1": "\\begin{align*}\r\n4s=8\\sqrt{5} &\\Rightarrow s=2\\sqrt{5} \\\\\r\n&\\Rightarrow s^2=\\boxed{20} \r\n\\end{align*}" } { "Tag": [ "function", "limit", "induction", "search", "algebra proposed", "algebra" ], "Problem": "This is a good one i believe...i find my solution approvably pretty.\r\nThis is mine of course but as you will see it's not really hard to come up with it...\r\n\r\nLet f(x) be defined in R where x is not 0. Find all functions f such that \r\nf(x-c) - f(x) = f(x-c)*f(x/c)\r\nfor all real numbers x,c such that x-c,x,x/c not 0....", "Solution_1": "I have a solution if $f$ is continuous.\r\n\r\n\\[f(x-c)-f(x)=f(x-c)\\cdot f\\left(\\frac x c\\right)\\quad(1)\\]\r\n\r\nLet $f(1)=a$. For $c=x-1$ one has\r\n\\[a-f(x)=a\\cdot f\\left(\\frac x{x-1}\\right)\\quad(2)\\]\r\n\r\nNow, let $x\\to\\infty$. From $(2)$ we have\r\n\\[a-\\lim_{t\\to\\infty}f(t)=a\\cdot f(1)\\]\r\n\\[\\lim_{t\\to\\infty}f(t)=a-a^2\\quad(3)\\]\r\n\r\nHence the above limit exists. Knowing that, we could put $c$ fixed and $x\\to\\infty$ in $(1)$ and get the following:\r\n\\[\\lim_{t\\to\\infty}f(t)-\\lim_{t\\to\\infty}f(t)=(\\lim_{t\\to\\infty}f(t))^2\\]\r\nhence\r\n\\[\\lim_{t\\to\\infty}f(t)=0\\]\r\n\r\nNow, by $(3)$:\r\n\\[a-a^2=0\\]\r\n\r\nNow we have two cases:\r\n\r\n1) $a=0$\r\nIt is obvious from $(2)$ that $f(x)=0$ for each $x$.\r\n\r\n2) $a=1$.\r\nWe will prove by induction that $f(n)=\\frac 1 n$ for each $n\\in\\mathbb{Z}$. Putting $c=1$ in $(1)$ follows:\r\n\\[f(x)=\\frac{f(x-1)}{f(x-1)+1}\\quad(4)\\]\r\nHence, by induction hypothesis,\r\n\\[f(n)=\\frac{\\frac 1{n-1}}{\\frac 1 {n-1}+1}=\\frac 1 n\\]\r\nIt is similar for negative numbers.\r\n\r\nLet $x=p,c=q$. From $(1)$ follows:\r\n\\[f\\left(\\frac p q\\right)=\\frac{f(p-q)-f(p)}{f(p-q)}=\\frac{\\frac 1{p-q}-\\frac 1 p}{\\frac 1{p-q}}=\\frac q p\\]\r\n\r\nLet $r\\neq 0$ be real number, and let $(r)_i$ be a sequence of rationals such that $\\lim_{n\\to\\infty}r_n=r$. Putting $x=r$ in $(4)$ one has:\r\n\r\n\\[\\begin{array}{rl}\r\nf(r)&=\\displaystyle\\frac{f(r-1)}{f(r-1)+1}=\\frac{f(\\lim_{n\\to\\infty}r_n-1)}{f(\\lim_{n\\to\\infty}r_n-1)+1}=\\lim_{n\\to\\infty}\\frac{f(r_n-1)}{f(r_n-1)+1}=\\\\\r\n&\\displaystyle=\\lim_{n\\to\\infty}\\frac{\\frac 1{r_n-1}}{\\frac 1{r_n-1}+1}=\\lim_{n\\to\\infty}\\frac 1{r_n}=\\frac 1 r\\end{array}\\]\r\n\r\nThus the problem is solved. There are two solutions: $f(x)\\equiv 0$ and $f(x)=\\frac 1 x$. It is easy to verify that these functions indeed satisfy the given equation.", "Solution_2": "HAHA, i just enter the forum to post this problem, (under the sourse \"DOUVROPOULOS\"...;-PPP) and i see that post.\r\n\r\nBojan Basic your solution seems ok, althgouht it can be improved.\r\nThe condition that f is continous is not needed to prove the result.\r\nI have a solution, nice but highly non-trivial, in which i use only the conditions stated in top. \r\nThe guy , who post this problem, is Douvropoulos. (giati den ebales bre fisiologiko nickname ?), And his solution is very nice and cute. (exactly as his problem). \r\n\r\nI really want to see more solutions, try it !", "Solution_3": "Well Bojab Basic your solution is good but actually not pretty because knowing a function is continuous gives you way to much info....\r\nTry simple methods....I would like to see some more trying going on....\r\nSo I'll give a hint or so ...of my solution...:\r\nSearch for a substitution that will give us another relation combining which with the first we will be able to get rid of the term f(x-c) ...and have only x,c, x/c , c/x in a new relation.....\r\nThat will be a good step i believe because doing so... having just x,c..... we can easily find symmetries......\r\n\r\n\r\nTo be even clearer:Instead of x,c using something else that will keep (x-c) as it is.Then combine the new relation with the first relation and find f(x-c)....with respect to f(x),f(c),f(x/c),f(c/x).Then replace f(x-c) in the first relation..\r\nSorry if that sounds complicated\r\n\r\nGive me what you got...The solver wins my gratitude..\r\n\r\n....\r\nAnto de nomizo na eipa tosa polla e???\r\nDikio exeis gia to nickname alla skeftika oti etsi opoios koitaei to nickname koitaei kai thn askisi....\r\nksereis an mporo na to allakso???\r\nsimeiosi:eides stergie...ego den evala simaia kai xazomares sto profile xaxaxaxaxaxaxa\r\n\r\npou na valo ekeinh thn askisi me thn anadromikh akolou8ia????", "Solution_4": "Quite nice. Consider the function $ g(x)=f(\\frac{1}{x})$. The condition of the problem can also be written $ f(y)-f(x)=f(y)f(\\frac{x}{x-y})$ for non-zero $x,y$ and different $x,y$. This becomes $ g(y)-g(x)=g(y)g(1-\\frac{x}{y})$. Take $y=1$. Then for $x$ different from 0 and 1 we have $g(1)-g(x)=g(1)g(1-x)$. Thus if $ g(1)=0$ then we obtain only the zero function. Suppose not, then we have $ g(1-\\frac{x}{y})=1-\\frac{g(\\frac{x}{y})}{ g(1)}$. Replace it in the last relation and you will find $ g(y)g(\\frac{x}{y})=g(x)g(1)$ for all $x,y$ non-zero and different. Thus $ g(y)g(1-\\frac{x}{y})= g(1) g(y-x)$ and thus we have for different $x,y$ non-zero the relations $ g(y)-g(x)=g(1) g(y-x)$ and $ g(xy)g(1)=g(x)g(y)$. Thus $ g(u+v)=g(u)+g(1) g(v)$ for $u,v$ non-zero with non-zero sum. Change them and you will find that if $g(1)$ is not 0 then $g$ is the zero function, false. Thus we must have $ g(1)=1$ and then it reduces to the classical problem of finding multiplicative and additive funtions.", "Solution_5": "[quote=\"harazi\"]...and then it reduces to the classical problem of finding multiplicative and additive funtions.[/quote]\r\n...for what, I believe, we must assume that the function is either continuous, or monotone, or bounded, or measurable (I'm not sure whether anything more is possible).\r\n\r\nReally, I was thinking some time about this problem and couldn't solve it completely without an additional assumption. Any help?", "Solution_6": "[quote=\"Bojan Basic\"][quote=\"harazi\"]...and then it reduces to the classical problem of finding multiplicative and additive funtions.[/quote]\n...for what, I believe, we must assume that the function is either continuous, or monotone, or bounded, or measurable (I'm not sure whether anything more is possible).\n\nReally, I was thinking some time about this problem and couldn't solve it completely without an additional assumption. Any help?[/quote]\r\n\r\ner, no. if you ONLY have that it's multiplicative, or ONLY that it's additive, you need additional assumptions. I believe harazi showed you have both.", "Solution_7": "You don't need any further assumption once you have shown they are both multiplicative and additive (since it follows immediately that they are increasing!), which is what I did.", "Solution_8": "Uh, sorry, I didn't know that. Thank you for clarifying.", "Solution_9": "Very nice! \r\nThe proposer's solution is similar to harazi's.\r\n\r\n :)", "Solution_10": "Well harazi.. first of all very elegant solution.. Anto is right mine is pretty similar but I find yours better\u2026\r\nWell here is mine anyway:\r\nI kind of like it because it has a lot of nice substitutions\u2026.\r\n\\[ f(x-y) - f(x) = f( x - y )\\cdot f\\left ( \\frac{x}{y} \\right ) (1) \\]\r\nNow I set x -> ( -x ) , y-> ( -y) and we have\r\n\\[ f(x - y) - f( -y) = f( x - y )\\cdot f\\left ( \\frac{y}{x} \\right ) \\]\r\nThat was what I meant as the first substitution\u2026.\r\nCombining these two we have\r\n\\[ f(x-y)= \\frac { f(x)-f(-y)}{ \\displaystyle f\\left ( \\frac{y}{x} \\right ) - f\\left ( \\frac{x}{y} \\right ) } \\]\r\nNow we replace what we found for $ f(x-y) $ in (1) and multiply out so we get:\r\n\\[ f(x)-f(-y)=f(x)\\cdot f\\left ( \\frac{y}{x} \\right ) - f(-y)\\cdot f\\left ( \\frac{x}{y} \\right ) (2)\\]\r\nThis one is very nice I think\u2026it has no $ (x-y) $ in it\u2026.\r\nAfter having y=1 in the above relation and playing a bit with it i found:\r\n\\[ f(x)\\cdot f\\left ( \\frac{1}{x} \\right ) = 1 \\]\r\nSo i used a similar technic to prove that f is multiplicative.....\r\n(2) becomes:\r\n\\[ f(x) + f(-y)\\cdot f\\left ( \\frac{x}{y} \\right ) = f(x)\\cdot f\\left ( \\frac{y}{x} \\right ) + f(-y) \\]\r\nAs we see the right part is the same if we have $ x $ or $ \\frac{y}{x} $...So the first part will be the same\r\nand we have..:\r\n\\[ f(x) + f(-y)\\cdot f\\left( \\frac{x}{y} \\right ) = f\\left ( \\frac{y}{x} \\right ) + f(-y)\\cdot f\\left ( \\frac{1}{x} \\right ) \\]\r\nUsing that and the fact that \\[ f(x)\\cdot f\\left ( \\frac{1}{x} \\right ) = 1 \\] we can easily deduce that:\r\n\\[ f(-y)=-f(x)\\cdot f\\left ( \\frac{y}{x} \\right ) \\] which will easily become \\[ f(y)=f(x)\\cdot f\\left ( \\frac{y}{x} \\right ) \\]\r\nthat of course means \\[ f(x)\\cdot f(y) = f(xy) \\]\r\nI consider \\[ g(x)=\\frac{1}{f(x)} \\]. $ g(x) $ is also multiplicative obviously but very easily it is additive..Use (1)....\r\nSo $ g(x)=x $ and thus $ f(x)=\\frac{1}{x} $\r\n\r\nThere are some cases we have to check of course but they are all alright and i am tired to write them here...." } { "Tag": [ "MATHCOUNTS", "AMC", "AIME", "factorial", "counting", "distinguishability", "number theory" ], "Problem": "I need some advice on how to write problems for mock tests, marathons, etc. All the problems I write are either ridiculously easy or have some fault in them that can't be fixed without, well, basically destroying the entire problem. \r\n\r\nSo this is why I want some advice in writing MATHCOUNTS-EASY AIME level problems :D .", "Solution_1": "Easy AIME problems (like the first 2) generally require the solver to simply follow through. They typically don't require too much ingenuity and as long as you know the concept, they're not too hard. I've found that number theory has a lot of nice little formulas and theorems, that can be turned into problems-a few years ago, there was a problem involving the number of zeros in a factorial, which is a well-known theorem. Trigonometric identities have a nice way of creating interesting problems. Maybe use something well-known but put an alternative spin on it.", "Solution_2": "I've been the head of the problem-writing committee for the high school contest my university hosts for something like 6 or 7 years. All I can really say about writing questions: [b]It's not easy![/b]\r\n\r\nI need the other committee members to bounce ideas off of, and they need to bounce their ideas off of me. Yes, we do steal ideas from other sources (no, I'm not going to reveal where), but there have to be some ideas that are original with us. My own specialty seems to be writing problems that 25%-50% of the competitors get right and that correlate pretty strongly with overall rank; I'm not so good at the questions that sort out the top handful and determine the winner.\r\n\r\nBut the process takes time. The questions have to checked and rechecked, and re-read to see if they make sense as written. Sometimes, questions are substantially edited; sometimes they're discarded. I we ever put out our first draft and called it the contest, we'd probably cause a riot.\r\n\r\n(And if a really good idea for a question occurs to me ... I'm not going to tell you. I need it for my own purposes!)", "Solution_3": "Although not completely the same (for obvious reasons, as you'll see), when you take the SAT, the questions that appear on there were not written the night before. They written several years back and field-tested to see if the question makes sense, if it's too difficult, too easy,... In fact, on every exam, there is an extra section (indistinguishable from the others) that is not scored but instead used to test questions for future years.\r\n\r\nYou can't really do that for math contests, but what you can do is find some friends who you know will not be taking your exam and have them do your questions. Get as much feedback as you can as that is the only way to get better." } { "Tag": [ "calculus", "integration", "conics", "parabola", "geometry", "analytic geometry", "calculus computations" ], "Problem": "Given a line segment $ PQ$ moving on the parabola $ y \\equal{} x^2$ with end points on the parabola. The area of the figure surrounded by $ PQ$ and the parabola is always equal to $ \\frac {4}{3}$. Find the equation of the locus of the mid point $ M$ of $ PQ$.", "Solution_1": "I get $ M\\left( t \\plus{} 1,t^2 \\plus{} t \\plus{} {\\scriptstyle\\frac {1}{2}}\\right)$ for real $ t$.", "Solution_2": "That's incorrect. :(", "Solution_3": "Let P and Q be the points $ (x_{2},y_{2})$ and $ (x_{3},y_{3})$ and let M be $ (x_{1},y_{1})$.\r\n\r\nSince we have to find equation of locus of M it means we have to find an algebraic equation which is satisfied by coordinates of every point on the path traced by M.\r\n\r\nIn other words we have to find an algebraic equation which relates $ x_{1}$ and $ y_{1}$.\r\n\r\nThus $ x_{2}+x_{3}=2x_{1}$\r\n \r\n $ y_{2}+y_{3}=2y_{1}$\r\n \r\n $ x^2_{2}+x^2_{3}=2y_{1}$\r\n\r\n $ x_{2}x_{3}=2x^2_{1}-y_{1}$\r\n\r\n $ x_{3}-x_{2}=2\\sqrt{y_{1}-x^2_{1}}$\r\n\r\nFormulating equation for areas we get\r\n\r\n $ \\frac{1}{2}(x_{3}-x_{2})(y_{2}+y_{3})-\\int_{x_{2}}^{x_{3}}x^2 dx=\\frac{4}{3}$\r\n\r\nOn simplifying,we get\r\n\r\n$ y_{1}=1+x^2_{1}$\r\n\r\nThus the required locus is $ y=1+x^2$", "Solution_4": "Is my answer correct", "Solution_5": "That's correct answer. :lol: \r\n\r\nHere is my solution:\r\n\r\nLet $ \\alpha ,\\ \\beta \\ (\\alpha < \\beta)$ be the $ x$ coordinates of the points of intersection between the parabola $ y \\equal{} x^2$ and the line $ PQ: y \\equal{} l(x)$, \r\n\r\nwe can write $ x^2 \\minus{} l(x)\\equiv (x \\minus{} \\alpha)(x \\minus{} \\beta)$, so the area is $ \\frac {4}{3} \\equal{} \\minus{} \\int_{\\alpha}^{\\beta} (x \\minus{} \\alpha)(x \\minus{} \\beta)\\ dx \\equal{} \\frac {1}{6}(\\beta \\minus{} \\alpha)^3$, yielding $ \\beta \\minus{} \\alpha \\equal{} 2$. Let $ M(X,\\ Y)$, \r\n\r\nsince $ P(\\alpha ,\\ \\alpha ^2),\\ Q(\\beta ,\\ \\beta ^2)$ without loss of generality, we have $ X \\equal{} \\frac {\\alpha \\plus{} \\beta}{2},\\ Y \\equal{} \\frac {\\alpha ^ 2 \\plus{} \\beta ^ 2}{2}\\Longleftrightarrow \\alpha \\plus{} \\beta \\equal{} 2X,\\ \\alpha ^ 2 \\plus{} \\beta ^ 2 \\equal{} 2Y$. \r\n\r\nThus use the equality : $ (\\alpha \\plus{} \\beta )^2 \\plus{} (\\alpha \\minus{} \\beta)^2 \\equal{} 2(\\alpha ^ 2 \\plus{} \\beta ^ 2)$ to get $ (2X)^2 \\plus{} 2^2 \\equal{} (2Y)^2$, yielding $ Y \\equal{} X^2 \\plus{} 1$. The desired equation is $ y \\equal{} x^2 \\plus{} 1$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "for $ a,b,c > 0$ satisfying $ a^2 \\plus{} b^2 \\plus{} c^2\\equal{}3$ prove inequality:\r\n$ (2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge \\sqrt {abc}$", "Solution_1": "[quote=\"Dumel\"]for $ a,b,c > 0$ satisfying $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$ prove inequality:\n$ (2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge \\sqrt {abc}$[/quote]\r\nLet $ a\\plus{}b\\plus{}c\\equal{}3u,$ $ ab\\plus{}ac\\plus{}bc\\equal{}3v^2$ and $ abc\\equal{}w^3.$\r\nHence, your inequality is equivalent to $ f(w^3)\\leq0,$ where $ f$ is increasing function.\r\nBut $ f$ gets a maximal value when $ w^3$ gets a maximal value, \r\nwhich happens when two numbers from $ \\{a,b,c\\}$ are equal, which very easy to check.", "Solution_2": "Hi ! \r\n\r\nYou can easily prove that : \r\n \r\n$ (2 \\minus{} a)(2 \\minus{} b)\\geq\\frac {1 \\plus{} c^{2}}{2} \\equal{}\\minus{} \\frac {a^{2} \\plus{} b^{2} \\minus{} 4}{2}$\r\n\r\n(Just expanding and SOS ! :wink: )\r\n\r\nSo we have :\r\n \r\n$ (2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} b)(2 \\minus{} c)(2 \\minus{} a)(2 \\minus{} c) \\geq\\frac {1 \\plus{} c^{2}}{2}\\frac {1 \\plus{} a^{2}}{2}\\frac {1 \\plus{} b^{2}}{2}$\r\n\r\nBut $ \\frac {1 \\plus{} c^{2}}{2} \\geq c$ (and similarly for $ a,b$) \r\n\r\nAnd we are done ! :P", "Solution_3": "nice proof! thanks a lot", "Solution_4": "[quote=\"Dumel\"]for $ a,b,c > 0$ satisfying $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$ prove inequality:\n$ (2 \\minus{} a)(2 \\minus{} b)(2 \\minus{} c) \\ge \\sqrt {abc}$[/quote]\r\nWe have \r\n$ 8(2\\minus{}a)(2\\minus{}b)(2\\minus{}c) \\ge (a\\plus{}bc)(b\\plus{}ca)(c\\plus{}ab)$" } { "Tag": [], "Problem": "Let $N_{n}$ denote the number of ordered $n$-tuples of positive integers $(a_{1},a_{2},\\ldots,a_{n})$ such that \\[1/a_{1}+1/a_{2}+\\ldots+1/a_{n}=1.\\] Determine whether $N_{10}$ is even or odd.", "Solution_1": "I'll show it is odd. Group equal $ a_i$ to obtain a sum of the form $ \\sum \\frac {n_i}{b_i}$ with $ \\sum n_i \\equal{} 10$ and $ \\{a_1,...,a_{10}\\} \\equal{} \\{b_1,...,b_r\\}$. Then the number of such solutions is $ \\frac {10!}{n_1!\\cdots n_r!}$.\r\n\r\nTo check parity, we only need to consider arrangements where this number is odd. But $ \\frac {10!}{n_1!\\cdots n_r!}$ is odd only if $ n_1 \\equal{} 2,n_2 \\equal{} 8$, or $ n_1 \\equal{} 10$. In the first we have $ 1 \\equal{} \\frac {2}{a} \\plus{} \\frac {8}{b} \\Leftrightarrow (a \\minus{} 2)(b \\minus{} 8) \\equal{} 16$, which has solutions $ (a,b)\\in\\{(3,24),(4,16),(6,12),(18,9)\\}$, each appearing $ 45$ times, $ 4\\cdot 45$ is even. In the latter case we have one solution. So the total is is odd.", "Solution_2": "So, can I ask you how many is $N_{10}$ exactly?" } { "Tag": [ "function", "inequalities unsolved", "inequalities" ], "Problem": "Find the lowest possible values from the function\r\n\\[ f(x) \\equal{} x^{2008} \\minus{} 2x^{2007} \\plus{} 3x^{2006} \\minus{} 4x^{2005} \\plus{} 5x^{2004} \\minus{} \\cdots \\minus{} 2006x^3 \\plus{} 2007x^2 \\minus{} 2008x \\plus{} 2009\\]\r\nfor any real numbers $ x$.", "Solution_1": "[hide=\"Solution\"]Notice that the function can be factored into:\n\\[ f(x) \\equal{} (x \\minus{} 1)^2(x^{2006} \\plus{} 2x^{2004} \\plus{} 3x^{2002} \\plus{} \\ldots \\plus{} 1004) \\plus{} 1005\\geq1005\\]\nHence, the minimum value is $ 1005$. It is attained when $ x \\equal{} 1$.[/hide]", "Solution_2": "[quote=\"wangsacl\"][hide=\"Solution\"]Notice that the function can be factored into:\n\\[ f(x) \\equal{} (x \\minus{} 1)^2(x^{2006} \\plus{} 2x^{2004} \\plus{} 3x^{2002} \\plus{} \\ldots \\plus{} 1004) \\plus{} 1005\\geq1005\\]\nHence, the minimum value is $ 1005$. It is attained when $ x \\equal{} 1$.[/hide][/quote]\r\nVery Nice ! :o", "Solution_3": "Ans : 2009\r\n :rotfl: \r\n\r\nfor x < 0, all terms become positive.\r\nfor 0, f(n)=2009\r\nfor 1, f(n)=1-2+3.............+2009 which >2009..\r\n\r\nconcluded!! :P", "Solution_4": "[quote=\"neverz\"]Ans : 2009\n :rotfl: \n\nfor x < 0, all terms become positive.\nfor 0, f(n)=2009\nfor 1, f(n)=1-2+3.............+2009 which >2009..\n\nconcluded!! :P[/quote]\r\n\r\n1-2+3-4+.............+2009 = 1005 ... :o", "Solution_5": "Let $\\forall n\\in {\\lbrace2,4,6,...\\rbrace} f_n(x)=\\sum_{i=1}^{n+1}(-1)^{i-1}ix^{n+1-i}$\nWe will prove that $\\forall n\\in {\\lbrace0,2,4,6,...\\rbrace} f_n(x)\\ge \\frac{n+2}{2}$ with equality iff $x=1$\nIt's clearly true for $n=2$ as $f_2(x)=(x_1)^2+2$. Now the induction step is:\n$f_{n+2}(x)=x^2f_{n}(x)-(n+2)x+(n+3)\\ge x^2\\cdot \\frac{n+2}{2}-(n+2)x+(n+3)=\\frac{n+2}{2}(x-1)^2+\\frac{n+4}{2}\\ge\\frac{(n+2)+2}{2} $\nWe see that equality is iff $((x=0\\vee x=1)\\wedge x=1)\\iff x=1$\nHence $f(x)=f_{2008}(x)\\ge \\frac{2008+2}{2}=1005$ with equality iff $x=1$", "Solution_6": "Consider $f_{2n}(x)=x^{2n}-2x^{2n+1}+\\cdots+2n+1$\\\\\n\n$\\textbf{\\textcolor{red}{Claim:-}}$ $f_{2n}(x)$ has minimum at $x=1$ which is $n+1$\\\\\n\n$\\textbf{\\textcolor{blue}{Pf:-}}$\\\\\n\n$\\textbf{Subclaim:-}$ $f_{2n}(x)=(x-1)^2(x^{2n-2}+2x^{2n-1}+\\cdots+n)+n+1$\\\\\n\n$\\textbf{\\textcolor{blue}{Pf:-}}$ $\\textbf{\\textcolor{magenta}{Base Case (n=1)}}$\\\\\n\n$f_{2}(x)=x^2-2x+3=(x-1)^2(1)+2$ which makes base case true\\\\\n\nnow we consider that $\\textbf{Subclaim}$ is true for case of $n$ , now we observe case of $n+1$\\\\\n\n we notice $f_{2n+2}(x)=x^2f_{2n}(x)-(2n+2)x+2n+3=x^2((x-1)^2(x^{2n-2}+2x^{2n-1}+\\cdots+n)+n+1)=(x-1)^2(x^{2n}+2x^{2n-2}+\\cdots+nx^2)+(n+1)(x-1)^2+n+2$\\\\\n\n$f_{2n+2}=(x-1)^2(x^{2n}+2x^{2n-2}+\\cdots+n)+n+2$ so $\\textbf{Subclaim}$ is true from induction hypothesis $\\square$\\\\\n\nso clearly we have $f_{2n}(x) \\geqslant n+1$ so minimum is at $x=n+1$ attainable at $x=1$ , so claim follows $\\square$\n\nso least possible value is $\\boxed{1005}$" } { "Tag": [], "Problem": "is it possible to boot from a bootable iso?", "Solution_1": "Hm, are you saying that you have an .iso file on your harddrive and want to boot from it? I seriously doubt that this is possible, the easiest way to do this is probaly just burn it to CD (or whatever kind of image it is)." } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Does exist function which satisfy this conditions:\r\n\r\n1)be increasing monotomicaly in [a,b] or R.\r\n2)has countable critical points .\r\n3)in any interval like [c,d] which be a subset of [a,b] or R we have at least one critical point.\r\n\r\nExpress a function if it exist.", "Solution_1": "A cheap example: $ f(x) \\equal{} \\sum_{n \\equal{} 1}^{\\infty}2^{ \\minus{} n}\\max\\{0,x \\minus{} q_n\\}$, where $ \\{q_n\\}$ is some enumeration of rational numbers in $ [0,1]$. This function is differentiable everywhere except at rational points, which are therefore critical points for $ f$. \r\n\r\nTo make things more interesting, we should insist that $ f$ be differentiable - then a countable dense set of critical points is harder to create.\r\n\r\n :idea: How about this: $ g(x)\\equal{} \\sum_{n \\equal{} 1}^{\\infty}2^{ \\minus{} n}(x \\minus{} q_n)^{1/3}$ and $ f\\equal{}g^{\\minus{}1}$?" } { "Tag": [ "geometry", "circumcircle" ], "Problem": "Suppose $ l_1$ and $ l_2$ are parralel lines and that the circle $ \\Gamma$ touches both $ l_1$ and $ l_2,$ the circle $ \\Gamma_1$ touches $ l_1$ and $ \\Gamma$ externally in $ A$ and $ B,$ respectively, and that the circle $ \\Gamma_2$ touches $ l_2$ in $ C,$ $ \\Gamma$ externally in $ D$ and $ \\Gamma_1$ externally in $ E.$ Prove that $ AD$ and $ BC$ intercect in the circumcenter of triangle $ \\Delta BDE.$", "Solution_1": "A hint:\r\n\r\n[hide]Try to show that the circumcenter of $ \\Delta BDE$ is the intersection of the three common tangents of two of the circles, respectively.[/hide]" } { "Tag": [ "number theory" ], "Problem": "Please suggest books on Number theory and inequalities. Which are the most elementary ones?\r\n\r\nI picked up \"First Steps in Number Theory-A primer on Divisibility\" by S.Shirali.Although it is a nice book,I find that it is a lot about explanations using a lot of English with little emphasis on proof writing based on mathematical language.As a result, I am sometimes forgetting what I learnt.I would like to have a book that is elementary,but uses more mathematical language and signs with good explanations.The\" Challenges and Thrill\" seemed to be a lot more advanced for a slow leaner like me.\r\n\r\nAnd I am getting stuck in Inequalities.\"Challenges\" was quite difficult for me.Can you suggest books that are very elementary?Although there are no shortcuts to labour,yet I felt the question would be best answered by the problem-solvers here.\r\n\r\nThank you in advance.", "Solution_1": "I think there is a book by Dmitri Fomin called mathematical circles or sth\r\nI found that book very entertaining and basic.\r\n\r\nbut most of the problems there are quite below olympiad level.\r\nyou can try that one :D :D", "Solution_2": "Thank you!", "Solution_3": "@muks\r\nSame with me when it comes to \"challenges\" inequalities.\r\nTry out Hall and Knight for number theory." } { "Tag": [ "inequalities", "function", "inequalities proposed" ], "Problem": "Let $ a, b, c, d$ be nonnegative real numbers such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\equal{} 1$. If $ 0intro abstract algebra\r\n\r\nI will be going to MIT CPW and Stanford Admit Weekend. I've seen Michigan (and I really liked it, but there weren't any students there). I don't have time to visit any other schools.\r\n\r\nI don't want to go to a \"party\" school (I guess all schools are party schools; I just want a group of more serious students and a reasonable amount of quiet in the dorms), but I don't think any of these schools fit into that category. (My sister went to Cal State San Diego, and it was apparently really bad.)\r\n\r\nAnybody have any advise for narrowing down this list?", "Solution_1": "UChicago.\r\n\r\nIntense academia", "Solution_2": "I have a mathematician friend who's very happy at UChicago, so I have a high opinion of it. On the other hand, it's hard to beat MIT for math. If you're curious, look at MIT's math website and look at the courses and professors to see if you're interested in the kind of stuff we do here. (But of course I'm obligated to convince you to come here :P) I should also mention that it's probably easier to do research as an undergraduate at MIT than at any of the other schools you listed.\r\n\r\nIf you want peace and quiet at MIT, you can definitely get peace and quiet. Different dorms at MIT lead quite different lives; if you live in a particular part of campus you'll learn that MIT throws surprisingly good (and loud) parties, and if you live in another part of campus you may not ever hear any dance music ever. But I think the best advertisement for MIT is the CPW experience you'll have, so see what happens after that. Feel free to PM me if you want to ask me any questions or find me during CPW :)", "Solution_3": "I went from SoCal to Berkeley, so here's what I'd make of that choice:\r\nBerkeley has all the benefits and drawbacks of being gigantic, both as a school and as a math department. There are tons of opportunities, and you can easily miss them all and fade into the bureaucracy. The dorms don't have enough space for everybody, so most students are out by the second year; one of the main perks of the Regents' scholarship is access for four years if you want it.\r\nI didn't get into research there, but I did take some introductory graduate courses, and got a lot out of them.\r\nI got plenty of peace and quiet there- too much, really. For a math major who wants it relatively quiet, the Foothill dorm is the place to start.\r\n\r\nI'll also plug the benefits of medium distance: you can easily drive up to the Bay Area to move in, but you won't have parents visiting all the time. The Chicago/MIT/Michigan options all involve flying at the beginning and end of the school year, so moving bulky stuff is a big hassle.", "Solution_4": "Back in the day, my decision was mainly between UMich and MIT. One of the main reasons I chose MIT was the people. MIT is incredibly dense with the brightest minds in the nation. Though, I have to say, it was more than just having a bunch of people that tool. The work hard, play hard principle that is the norm at MIT is what really hooked me. There's an amazing breadth of opportunities here and everyone finds a niche that they really enjoy.\r\n\r\nRegarding the party school issue...yes, you'll find a thriving party scene at any of those schools. At UMich, since you'd be in the honors dorm, it would probably be better than, say, being in a random dorm there. As for MIT, tor gave you a pretty good description. I'll add that everyone here is in the same boat in that they're extremely hosed during the week. Most of the partying is done on Friday and Saturday night. Most people head over to a fraternity to party. Though there are a few dorms and sections of dorms that party as well (Baker, EC, senior house, certain sections of other dorms). I lived in Simmons my freshman year which is definitely one of the quieter dorms, perhaps the quietest. Basically, you should have no problem finding quiet space from Sunday-Thursday anywhere and on the weekend you have to seek out parties unless you live in a place that hosts parties.\r\n\r\nThat said, the nature of social life at MIT is quiet different from that almost anywhere else. Before college I did very little socializing. I got here and lived in Simmons for a term. It was really quiet, which was nice, but lacked culture. My good friends (one of which I knew from AoPS(!)) had joined a fraternity in the fall and would be moving out of Simmons sophomore year. I really didn't understand fraternities and the value of socializing for a while (which is why I didn't rush in the fall). After a semester in Simmons, I knew I had to find something else and thus explored the fraternity my friend was in (after teasing him for a whole semester since I thought fraternities were silly...)...and it was pretty amazing. Sure they would party on Friday and/or Saturday night. But the rest of the week was spent studying and doing psets. When I got stuck, I could go ask others for help or just take my mind off of it for a while and engage in an interesting discussion.\r\n\r\nRegarding CPW, it's a really awesome experience (which I missed for a track meet my year...). You'll hopefully be able to understand the point I was trying to make in the previous paragraph by experiencing it first hand :P That said, you wont see quiet the usual MIT environment. There will be way more things going on than usual. In a few quick, sleepless days you'll be able to experience the vast diversity of things you can do at MIT. In terms of the work hard, play hard spectrum, you'll definitely see us playing hard*...so keep that in mind, i.e. it's quieter at MIT normally.\r\n\r\n* Good ol' clean fun, i.e. no alcohol. CPW is officially dry and that policy is stuck to remarkably well, especially compared to other schools that have similar policies.", "Solution_5": "As much as I hate to say \"school X is better than school Y,\" I would quickly narrow your choices down to MIT, Stanford, and Chicago, based on your stated priority of math over all else. The other schools are great, too, but the undergraduate math majors, as a whole, will not be as strong, and this has a large effect on the difficulty of your undergraduate courses.\r\n\r\nThe main advantage of Chicago for you, as I see it, is that it has a reputation for having a serious student body. However, the students you take classes with at MIT and Stanford will also be pretty serious about their studies, so this only has relevance for the overall environment (e.g. dorms). So unless that's super-important to you, I'd rule out Chicago also, since it's more expensive, and again, the math majors (as a whole) are *probably* at a *slightly* lower level than the MIT-Stanford level.\r\n\r\nThat brings us to MIT vs Stanford. At both of these schools, there are a ton of brilliant math students (and consequently a great peer group and challenging classes), so I don't think there's any point in doing a comparison there. Obviously, both are top-notch in research. Teaching quality is hard to compare, especially since the postdocs always change anyway. Luckily, the two schools are located in two very different places with very different environments, so it shouldn't be hard to choose based on the non-math factors. (I'll draw some caricatures of dubious accuracy. MIT: Cold place, urban, close to Boston, no riding in cars, sterile but cool campus, fewer women, reputation for hard work, high stress, and nerdiness, and you will be surrounded by engineers and linux. Stanford: Amazing weather, suburban but not in a crappy way, San Francisco is mostly irrelevant, beautiful sprawling Taco Bell-like campus, top-notch athletics, reputation for laid-back attitudes and quirkiness.) Honestly, if I had to make your decision (for myself, not you), based on everything I know now, I think my head might explode. \r\n\r\nThe MIT people can correct me if I'm wrong, but isn't MIT one of the few places where you can make meaningful choices about where you'll live freshman year? That may be attractive if you want quiet. Because although many students at both Stanford and MIT love quiet, there are many more who do not.\r\n\r\nP.S. For the sake of others on this board, please avoid referring to your circumstances as \"unfortunate\" in any way.", "Solution_6": "[quote=\"yenlee\"]MIT: Cold place, urban, close to Boston, no riding in cars, sterile but cool campus, [b]fewer women[/b][/quote]\nDo you know how many colleges there are in the Boston area? :P \n\nAnd yes, it's pretty cold in Boston in the winter, but I don't think no cars is a disadvantage - MIT is situated right between Boston and Cambridge and public transportation is great here, so you can get pretty much anywhere you want to go on foot or by bus and train. It's quite refreshing, at least when the weather's nice.\n\n[quote=\"yenlee\"]isn't MIT one of the few places where you can make meaningful choices about where you'll live freshman year?[/quote]\r\nQuite right. While a lot of other schools lottery you into a freshman dorm, at MIT the only restriction to where you can live freshman year is that you have to live on campus, and you can still join a fraternity or other living group without moving in. That gives each dorm a lot of power to shape its own unique culture and personality, which a lot of people wouldn't give up for anything. One of the drawbacks to this system, though, is that it makes social life at MIT somewhat segregated. It's easy to become insular. Some people prefer the lottery system because it forces you to interact with a wide variety of people, and indeed some of the newer dorms are more diverse personality-wise since they haven't had as long to establish themselves, but I think you'll find that a lot of the quirkiest and most interesting people you'll meet will inevitably find themselves drawn to the quirkiest and most interesting dorms.", "Solution_7": "[quote=\"yenlee\"]MIT: Cold place, urban, close to Boston, no riding in cars, sterile but cool campus, fewer women, reputation for hard work, high stress, and nerdiness, and you will be surrounded by engineers and linux. Stanford: Amazing weather, suburban but not in a crappy way, San Francisco is mostly irrelevant, beautiful sprawling Taco Bell-like campus, top-notch athletics, reputation for laid-back attitudes and quirkiness.) Honestly, if I had to make your decision (for myself, not you), based on everything I know now, I think my head might explode.[/quote]\r\n\r\nbased on [i]these factors alone[/i], this doesn't sound like a very difficult decision to me (i.e., Stanford and it isn't close?)", "Solution_8": "Urban vs. suburban and engineers vs. top-notch athletes are a matter of preference, which is what I think yenlee was getting at. Although I don't like the implication that Stanford somehow has a quirkier reputation than ours or that MIT people aren't laid back :P \r\n\r\nOn that subject, MIT is one of the few schools that doesn't recruit athletes, nor does it accept legacies. (We do accept \"foundational admits\" - e.g. for donating libraries - but what are you gonna do?) I have friends at other colleges who complain that it's so obvious who the people at their school are who don't belong there, in the sense that their academics just aren't up to par, and I think you'll be hard-pressed to find people like that at MIT. Our athletes are just as dedicated to their intellectual pursuits as anyone else.", "Solution_9": "[quote=\"t0rajir0u\"]Urban vs. suburban and engineers vs. top-notch athletes are a matter of preference, [/quote]\n\nOf course, in the same way that choosing a school with great weather versus a campus with cold weather, and choosing a school with a relatively even ratio of guys to girls or a school with mostly guys is a matter of preference. I mean, I can think of (good) reasons to go to MIT over Stanford, but if I were deciding on that list alone I don't think my decision would take very long.\n\n[quote=\"t0rajir0u\"] I have friends at other colleges who complain that it's so obvious who the people at their school are who don't belong there, in the sense that their academics just aren't up to par, and I think you'll be hard-pressed to find people like that at MIT. Our athletes are just as dedicated to their intellectual pursuits as anyone else.[/quote]\r\n\r\nAt one point in my life I probably would have worried about 'who belonged' at x school but it's ultimately a pretty ridiculous thing to do. As a math major you aren't going to be interacting much in classes with dumb superstar athletes, and having a wider range of people on campus is almost certainly going to create a more interesting on-campus atmosphere (not to mention that division I college sports are pretty darn exciting). I mean it's kind of hypocritical of me to say so, but nerdy math/physics/cs types aren't exactly who I want to spend most of my free time around, no matter how 'quirky' they might be.", "Solution_10": "I'll put in a few words regarding math at Stanford, although my experience is somewhat limited since I'm not a math major.\r\n\r\nAll of my math professors (four total) have been very good; if you choose the right professors to take classes with, you'll definitely get a lot out of it. I've heard that some math professors here are not great lecturers, so it's partially up to the student to try and figure out the best classes to take.\r\n\r\nIn particular, I took the 50H series last year (Leon Simon for the first 2 quarters, Simon Brendle for the 3rd). I think it is a very well-designed sequence of classes and both professors give good lectures. On the other hand, taking that sequence made me realize that I didn't really want to be a math major so make what you will of that. I will say that the classes are challenging and for the most part interesting; usually it got less interesting once I got too lost.\r\n\r\nPersonally, I chose Stanford over MIT for reasons related to what yenlee mentioned (e.g. weather, campus, distance, culture, etc). Without comparison to MIT (because I can't meaningfully make such a comparison), I think Stanford's math community is rather close-knit. There are a lot of awesome math students here, but a large number of them go into engineering (as I have), so the number of actual math majors is somewhat small.\r\n\r\nAnyway, to sum it up, Stanford's math program is really good. And the school is just awesome in general. So you should come here :).", "Solution_11": "[quote=\"blahblahblah\"]choosing a school with a relatively even ratio of guys to girls or a school with mostly guys is a matter of preference.[/quote] Both MIT and Stanford freshman classes are majority male: at MIT, the ratio is [url=http://web.mit.edu/facts/enrollment.html]54.6:45.4[/url] while at Stanford it's [url=http://www.stanford.edu/about/facts/undergraduate.html]51.3:48.8[/url] -- a difference, but not a particularly large one, especially in light of the fact that there are several tens of thousands of non-MIT undergraduates in the immediate vicinity of MIT and 0 non-Stanford undergraduates in the immediate vicinity of Stanford. Also, um, lots of people prefer to go to school in urban areas rather than suburban areas -- it's not even remotely comparable to the question of Boston vs. Palo Alto weather.", "Solution_12": "[quote=\"JBL\"]Also, um, lots of people prefer to go to school in urban areas rather than suburban areas -- it's not even remotely comparable to the question of Boston vs. Palo Alto weather.[/quote]\r\n\r\nYeah. I was being a bit sarcastic; while urban campuses versus suburban campuses definitely are a matter of personal preference, the other factors as listed by yenlee certainly seem to favor Stanford. I have no particular bias towards or against urban campuses :)", "Solution_13": "[quote=\"yenlee\"]As much as I hate to say \"school X is better than school Y,\" I would quickly narrow your choices down to MIT, Stanford, and Chicago, based on your stated priority of math over all else. The other schools are great, too, but the undergraduate math majors, as a whole, will not be as strong, and this has a large effect on the difficulty of your undergraduate courses.[/quote]\r\n\r\nJust to play devil's advocate for a minute - Does \"more difficult\" necessarily mean \"better educationally\"?\r\n\r\nAnd I think the upper level's of U. of Michigan's Honors Math program, and many of their upper level undergraduate/lower level graduate courses, will be comparable in difficulty to similar courses you'd find at MIT, Stanford, or U. of Chicago. They will also be populated by students who could have probably gone to one of those other schools.\r\n\r\nThe main difference academically may be the quality of some of the (non-honors) classes outside of the math department, and the variety of opportunities to learn mathematics outside of the classroom. Also at large Universities, the ability to get into the classes that you want can be a pain. This depends on the University though. The good news is that they usually have a large variety of classes, and they are offered frequently with a greater variety of professors to choose from.", "Solution_14": "Berkeley also has honors versions of the core upper-division classes. My experience with those was a bit uneven, but it does mean that you can find classes with other high-end students, and follow those up with graduate classes. Berkeley and Michigan come out fairly similarly on the academic options, so that choice really comes down to geography.\r\n\r\n[quote=\"gauss202\"]Also at large Universities, the ability to get into the classes that you want can sometimes be a pain.[/quote]This is not generally a problem for higher-level courses; it's the basic classes that fill, or force you into sections at odd times. Getting into Berkeley's Math H110 is easy, but Chem 1A will fill or come close. For advanced classes, having the class canceled for low enrollment is a much more relevant risk.", "Solution_15": "I don't think I care too much about the male-to-female ratio, unless it's something crazy like New Mexico Tech's. A few athletes probably won't bother me either.\r\n\r\nI think the urban-suburban thing matter quite a bit. Surely there's got to be some sort of \"college town\" with restaurants, movie theaters, grocery stores, etc. within walking distance around Stanford (I'm not expecting anything quite like Ann Arbor); it can't just be suburbs.", "Solution_16": "[quote=\"mnop\"]Surely there's got to be some sort of \"college town\" with restaurants, movie theaters, grocery stores, etc. within walking distance around Stanford (I'm not expecting anything quite like Ann Arbor); it can't just be suburbs.[/quote]\r\nClearly you haven't visited Palo Alto :P Even [url=http://www.stanforddaily.com/cgi-bin/?p=493]the Stanford Daily[/url] admits it's not a college town.", "Solution_17": "haha t0rajir0u don't insult palo alto. palo alto is awesome. there are grocery stores, several great restaurants on university avenue, and numerous movie theaters in palo alto and in neighboring cities. downtown is very un-suburbia-ish.", "Solution_18": "Funny, that description sounds... exactly like every other affluent suburb in the US.", "Solution_19": "It is not possible to dislike Palo Alto unless you are a very ignorant or confused human being.\r\n\r\nAlso, I'm currently taking the 50H series at Stanford and they're really great classes. One of the best mathematical experiences I've had. Though echoing paladin8, among everyone I know in the class, I might be the only one who hasn't been completely put off the idea of a math major because of it.", "Solution_20": "How many students are in the Math 50H course? Are they all freshman, and are there any math majors who do not go through that course?", "Solution_21": "My year, 50H had about 30, I think. I've heard it's significantly larger this year, because a lot fewer dropped out. They're mostly frosh, but not all. And there are plenty of math majors who don't go through it. If you want to get your math degree with honors, though, it's required.", "Solution_22": "Actually, after I wrote that MIT had fewer women, I did google the actual ratio (which is really not that awful) and decided to edit out that comment. I guess I forgot to do it. So I agree that the male-female ratio is not such a big distinction. But it's still true that people wear fewer clothes at Stanford. (There, against my better judgment, I said it.)\r\n\r\n@t0rajir0u, I'm sure that many people at MIT are laid back, but I think it's fair to say that, as a whole, MIT does not have a reputation for being laid back. If anything, it has the opposite reputation (fairly or not). And I'm definitely sure that MIT people can be quirky (see ZBT as Exhibit A), but Stanford does have the Stanford band, which is a significant component of Stanford culture. (And I agree that not riding in cars is a good thing. You seem to think that I came down on the side of Stanford, which is not quite true.)\r\n\r\n@mnop, Palo Alto is not a college town. It is a very wealthy and very nice suburb, with a cute little downtown area. So yes, you can easily walk or take the Marguerite bus downtown and access lots of great restaurants and whatnot, although many of them are pricey. There's also a Whole Foods and even a movie theater that plays classics. (But undergraduates don't go grocery shopping, do they?) It is definitely not the strip mall hell that you see in newer suburbs. Also of note is the Stanford Shopping Center, which is actually a very pretty outdoor mall located on Stanford property. And unlike most other suburbs, a lot of people *work* there, so it's crawling with software engineers, venture capitalists, lawyers, etc. \r\n\r\n@gauss202, I do not equate \"more difficult\" with \"better educationally.\" But if you're serious about going into math, there's nothing more important than challenging yourself and having the support you need to do so. The easiest path to challenging yourself is to take really hard classes (not that I say \"hard\" and not \"advanced\") with students of comparable ability. A harder (but often superior) way to do it is to blaze your own trail of independent study, guided by helpful professors. I guess I'm betraying my personal preference for the lazy path. I also think it's nice to find the right peer group, which is easier to do at some places than others (depending on who you are, of course). \r\n\r\nAdding to Sly Si's comment: If a student skips 50H at Stanford because he is too advanced, the requirement of taking 50H to get honors would surely be waived.", "Solution_23": "The class isn't much larger this year--I counted 36 scores in the 52H midterms, and five of those were not precisely from Stanford students.", "Solution_24": "You go there?", "Solution_25": "@t0rajir0u - Do you know any of the athletes at MIT? I'm interested in playing basketball or tennis varsity, but I'm not sure if I will be able to do a sport along with studies at MIT. Can the people there manage to have a life and learn at the same time?", "Solution_26": "Of course. MIT used to have 41 varsity sports and currently has 33 (unfortunately, we had to make some budget cuts), although you should be aware that we are a D-III school. One of my fraternity brothers is on our crew team and another fences. Generally, I think you'll find that plenty of MIT students have very active extracurricular lives.\r\n\r\nI'm not even sure what prompted this question - you must be thinking of Caltech :P" } { "Tag": [], "Problem": "Find $ x$ given $ 2^{2x\\plus{}1}\\minus{}2^{2x\\minus{}1}\\equal{}24$.", "Solution_1": "Find $ x$ given $ 2^{2x\\plus{}1}\\minus{}2^{2x\\minus{}1}\\equal{}24$. \r\n\r\n\r\nWe have $ 2^{2x \\minus{} 1} (2^{2} \\minus{} 1) \\equal{} 24$\r\n=> $ 2^{2x\\minus{}1} \\equal{} 8 \\equal{} 2^{3}$\r\n=> $ {2x \\minus{} 1} \\equal{} 3$\r\n=> $ x \\equal{} 2$ :roll:", "Solution_2": "\\begin{align*}\r\n2^{2x+1} - 2^{2x-1} &= 24 \\\\\r\n2 \\times 2^{2x} - \\frac{1}{2} \\times 2^{2x} &= 24 \\\\\r\n\\left( 2 - \\frac{1}{2} \\right) \\times 2^{2x} &= 24 \\\\\r\n\\frac{3}{2} \\times 2^{2x} &= 24 \\\\\r\n2^{2x} &= 16 \\\\\r\n2^{2x} &= 2^4 \\\\\r\n2x &= 4 \\\\\r\nx &= \\boxed{2}\r\n\\end{align*}" } { "Tag": [ "geometry", "area of a triangle", "geometry unsolved" ], "Problem": "To each side a of a convex polygon we assign the maximum area of a triangle contained\r\nin the polygon and having a as one of its sides. Show that the sum of the areas assigned to all\r\nsides of the polygon is not less than twice the area of the polygon.", "Solution_1": "Actually, IMO 2006/6. See here: [url]http://www.mathlinks.ro/viewtopic.php?p=572824#p572824[/url]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that\r\n\r\n$\\frac{1+x^2}{1+y+z^2}+\\frac{1+y^2}{1+z+x^2}+\\frac{1+z^2}{1+x+y^2} \\geq 2$\r\n\r\nwhere x,y,z are real numbers > -1", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=6150\r\n\r\n darij", "Solution_2": "[quote=\"darij grinberg\"]http://www.mathlinks.ro/Forum/viewtopic.php?t=6150\n\n darij[/quote]\r\nOh, sorry for the repost." } { "Tag": [ "summer program", "Mathcamp" ], "Problem": "Is anyone going to Mathcamp interested in playing Texas Hold'em?", "Solution_1": "I would be, but my skills/experience is kinda limited. Does anybody have any chips?", "Solution_2": "[quote=\"pilot\"]Is anyone going to Mathcamp interested in playing Texas Hold'em?[/quote]\r\n\r\nAbsolutely.\r\n\r\nAs far as chips go, we can just use spare decks of cards if nothing works out. Or pennies! Otherwise, they're probably about 10 dollars for 250. We can all chip in.", "Solution_3": "[quote=\"MithsApprentice\"]...We can all chip in.[/quote]Pun intended? :rotfl:", "Solution_4": "[quote=\"DPatrick\"][quote=\"MithsApprentice\"]...We can all chip in.[/quote]Pun intended? :rotfl:[/quote]\r\n\r\nIndeed. :lol: I didn't know how many people would get it. :)", "Solution_5": "Last year there was a bunch of people who played alot. (and I mean alot :D ). Of course, everything is differnt from year to year. I for one will play a normal game on occasion, but never anything for real money :) \r\n\r\nBut serously, your better off spening your time playing mao :D", "Solution_6": "[quote=\"salmononpi\"]Last year there was a bunch of people who played alot[/quote]\r\nI won alot, and ordered/ate pizza alot as well.", "Solution_7": "[quote=\"Geehoon\"]ordered/ate pizza alot[/quote]\r\n\r\nOh yes, good food is of the utmost importance. Someone (i.e. everyone) should acquire some phone numbers for food places as soon as we get there.", "Solution_8": "we ordered pizza (from domino's) almost every night.\r\nThere was 5-5-5 deal just on the commercials.\r\n\r\nLater, they even recognized my voice(because there arent that many people ordering pizza at 11PM there(Maine)) :rotfl:", "Solution_9": "Does anyone have chips they can bring?\r\n\r\nOtherwise we can all pitch in and buy a set as Mith suggested, but I doubt 10 dollars is enough.", "Solution_10": "Actually I just saw this and I wanted to mention that I am bringing crappy plastic chips, of 3 colors.\r\n\r\nIt is enough to play about 12 people tournament, can be stretched to maybe 16." } { "Tag": [ "geometry", "quadratics", "probability", "algebra", "polynomial", "AMC 10", "AMC" ], "Problem": "Although the 'math culture' idea did not work out as planned, I have not entirely given up in that I am talking to several non-trivial friends of mine and recommending them the AoPS books. I figured volume 1 would be more useful to a high schooler than the introductory books since volume 1 seems to be a curricular supplement whereas the introductory books are curriculum replacements, and they have already endured much of the standard curriculum. They will be taking the AMC 10 next year, and I was wondering which chapters of AoPS 1 would be most important (I could do this myself, but I don't own AoPS 1--only AoPS 2). \r\n\r\nWould it be something like:\r\n\r\nMost important: [b]25[/b]\r\nVery important: 9-18 (the geometry chapters, excepting shifts, turns, flips, whatev)\r\nImportant: 1,3-5,7, 26\r\nNot very important: everything else?\r\n\r\nI would appreciate advice on this matter from not only the administrators but anyone who has knowledge as to the content of AoPS 1 (or the new books).", "Solution_1": "I think the most important chapters are 5, 24-26 (Integers, Sequences and Series, Counting, and Probability). In my opinion, these are the subjects that are very important in math contests but are not tought much in school curriculum.\r\n\r\nChapters 9-18, 20 (Chapter 20 is just a bunch of geometry problems) are also very important, but your friends might find that they already know much of this material if they have already taken a geometry class in school (That's what happened to me :) ).\r\n\r\nChapters 1, 3-4, 6-7 are also important, but I'm sure your friends already know some, if not most of this material, especially the stuff about linear and quadratic equations.\r\n\r\nSo I would say\r\n\r\nMost Important: 5, 24-26\r\nVery Important: 9-18, 20\r\nImportant: 1, 3-4, 6-7", "Solution_2": "For me, the geometry chapters were the most important. I had learned many of the theorems and such presented in those chapters but I had no idea how to apply them to the end of the chapter problems. I don't own a copy of Volume 1 either, and that's about all I can remember.", "Solution_3": "Though it does greatly depend on the year, I would say the geometry chapters are VERY useful. Some years, they have 3-4 tougher geometry problems.\r\n\r\nCounting and probability is important, but not AS important as the algebra, try the word problems chapter.", "Solution_4": "What about volume 2?", "Solution_5": "[quote=\"Karth\"]What about volume 2?[/quote]\r\n\r\nWhat do you mean? It depends on what you're preparing for. Volume 2 wouldn't be very helpful in preparing for the AMC 10.", "Solution_6": "hi david aops is cool i am jumping through cool parts of aops2 the limits part is very nice. thx.", "Solution_7": "For volume two, definitely the stuff on Algebraic Expressions, Polynomials, Counting tricks, and Number Theory was amazing." } { "Tag": [ "inequalities" ], "Problem": "If i have to compare two equations that involve the addition of two fractions to see which is larger, how would I do that with out having to get a common denominator and putting it into 1 fraction. For example, such as 1/2 + 1/7 vs 1/3 + 6/19. Is there any way else?", "Solution_1": "Well, you could estimate. $\\frac{1}{2}= 0.5, \\frac{1}{7}\\approx 0.14$, so their sum is $\\approx 0.64$ whereas $\\frac{1}{3}\\approx 0.33, \\frac{6}{19}\\approx \\frac{6}{18}\\approx 0.33$ so their sum is $\\approx 0.66$.\r\n\r\nThis isn't totally accurate, though, so I wouldn't trust it.", "Solution_2": "Usually, the best way is to combine fractions... it is not so bad!\r\nThe only way to get around this is:\r\na. Prove an inequality about fractions or at least notice a pattern and generalize\r\nb. Use a well known inequality that happens to apply to your fractions\r\n\r\nOf course, this requires a lot of insight and doesn't work a good portion of the time, theres no issue with just combing", "Solution_3": "well, the thing I posted above was an example, but what if if the fractions were 4002/4001 + 3999/4000 vs. 3998/3999 + 3999/4000\r\nAt first I would have done the multiplying... but my bad handwriting confused me so I just guessed and got it wrong... It seemed really annoying to just multiply it out with all the 0's.. So is that the only way?... I thought that there was going to be a more elegant method.. but i dunno..", "Solution_4": "[quote=\"turtlecloud\"]well, the thing I posted above was an example, but what if if the fractions were 4002/4001 + 3999/4000 vs. 3998/3999 + 3999/4000[/quote]\r\n\r\nYou could've just posted this problem :P\r\n\r\nThis is how I would do it: Let $x = 4000$ and rewrite the two expressions as\r\n\r\n$1+\\frac{1}{x+1}+1-\\frac{1}{x}= 2-\\frac{1}{x(x+1)}$\r\n\r\nAnd\r\n\r\n$1-\\frac{1}{x-1}+1-\\frac{1}{x}= 2-\\frac{2x-1}{x(x-1)}$\r\n\r\nEdit: The problem's a little too easy this way. Did you mean, say, $\\frac{3998}{3999}+\\frac{4001}{4000}$ on the other side? That would be interesting. In the same vein, this is\r\n\r\n$1-\\frac{1}{x-1}+1-\\frac{1}{x}= 2-\\frac{1}{x(x-1)}$\r\n\r\nWhich is [b]very[/b] close to (but slightly larger than) the other expression." } { "Tag": [ "inequalities", "algebra unsolved", "algebra" ], "Problem": "Determine all triples of positive real numbers $ (x,y,z)$ such that:\r\n\r\n$ x\\plus{}y\\plus{}z\\equal{}6,$\r\n$ \\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z}\\equal{}2\\minus{}\\frac{4}{xyz}.$", "Solution_1": "$ x,y,z\\not \\equal{}0,\\frac 12$. Let $ x\\equal{}\\frac 12$, then $ y\\plus{}z\\equal{}11,5$ and $ \\frac 1y \\plus{}\\frac 1z \\equal{}\\minus{}\\frac{8}{yz}\\to y\\plus{}z\\equal{}\\minus{}8$ - contradition.\r\nLet $ xyz\\equal{}a$, then $ xy\\plus{}yz\\plus{}zx\\equal{}2a\\minus{}4$, therefore $ x,y,z$ are roots of $ t^3\\minus{}6t^2\\plus{}(2a\\minus{}4)t\\minus{}a\\equal{}0$.\r\nWe can chose one of roots $ x\\not \\equal{}0,\\frac 12$, define $ a\\equal{}\\frac{x^3\\minus{}6x^2\\minus{}4x}{1\\minus{}2x}$ and find $ y,z$ from $ y\\plus{}z\\equal{}6\\minus{}x,yz\\equal{}\\frac ax\\equal{}\\frac{x^2\\minus{}6x\\equal{}4}{1\\minus{}2x}\\equal{}b$ as roots \r\n$ t^2\\minus{}(6\\minus{}x)t\\plus{}b\\equal{}0$. For reality roots, must be $ D\\equal{}(6\\minus{}x)^2\\minus{}4b\\ge 0$.", "Solution_2": "[quote=\"moldovan\"]Determine all triples of positive real numbers $ (x,y,z)$ such that:\n\n$ x \\plus{} y \\plus{} z \\equal{} 6,$\n$ \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\equal{} 2 \\minus{} \\frac {4}{xyz}.$[/quote]\r\nApply AMGM inequality we have $ xyz\\le 8$.\r\nThe CS inequality in Engel form gives us $ \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z}\\ge \\frac {9}{6}$.\r\nThus $ \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\plus{} \\frac {4}{xyz}\\ge 2$.\r\nThe equality occurs if only if $ x \\equal{} y \\equal{} z\\equal{}2$.It is the unique solution.", "Solution_3": "Yes, I forget, that x,y,z all are positive." } { "Tag": [ "geometry", "geometric transformation", "reflection", "incenter", "symmetry", "trigonometry", "angle bisector" ], "Problem": "Hi!, How are you? ;) \r\n\r\nI ask for its aid with the following problem:\r\n\r\nIn the $\\bigtriangleup ABC$ triangle, $AB=AC$ and $ 0$\n\n$ \\minus{} a | \\minus{} a| \\equal{} \\minus{} a^2 \\geq 1$\n\n$ a^2 \\leq \\minus{} 1$\nNo real solutions.\n\nCase 2: $ a < 0$\n\n$ \\minus{} a | \\minus{} a| \\equal{} a^2 \\geq 1$\n\n$ a \\leq \\minus{} 1$ or $ a \\geq 1$ but $ a < 0$ so \n\n$ \\boxed{a \\in ( \\minus{} \\infty , \\minus{} 1] }$[/hide]\r\n\r\n\r\nEDIT: Fixed it thanks", "Solution_2": "[quote=\"modularmarc101\"][hide=\"#1\"]$ f(0) = - a | - a| \\geq 1$\n\nCase 1: $ a > 0$\n\n$ - a | - a| = - a^2 \\geq 1$\n\n$ a^2 \\leq - 1$\nNo real solutions.\n\nCase 2: $ a < 0$\n\n$ - a | - a| = a^2 \\geq 1$\n\n$ a \\leq - 1$ or $ a \\geq 1$\n\n$ \\boxed{a %Error. \"fork\" is a bad command.\n( - \\infty , - 1] \\cup [ 1 , \\infty)}$[/hide][/quote] \r\n\r\nIf $ a < 0$, then $ a \\not \\ge 1$. \r\n\r\nThus, only $ a \\in ( - \\infty , - 1]$ work." } { "Tag": [ "LaTeX" ], "Problem": "I am relatively new to LaTeX, but as a math teacher I know that there is no other way to go to get the results that I need as easily as I can in LaTeX.\r\n\r\nMy current problem is font sizes. I need a font size that is very large, something around 40pt, maybe bigger. The onyl wy I have been able to change the font size in LaTeX so far is with commands like \\Large, \\LARGE, \\huge, and \\Huge. But this still isn't big enough. Is there a way (hopefully an easy way) to set a specific font size? I will also need to be able to do this in math mode.\r\n\r\nAny suggestions or help would be appreciated.", "Solution_1": "Hi,\r\nTry the scalefnt package - it provides a command called \\scalefont{x}, which will scale your current font by a scale of x. Assuming your document has the default option of 10 pt, either:\r\n[code]\\normalsize\n\\scalefont{4}[/code]\nor\n[code]\\huge\n\\scalefont{2}[/code]\r\nshould give you a 40 pt font.\r\nSSN.", "Solution_2": "Sweet! Works like a charm. Thanks so much.\r\n\r\nCheers,\r\nBen" } { "Tag": [ "percent", "LaTeX" ], "Problem": "If I received 39/50 on an exam and the exam total was 30% of my final grade, how would I figure out what I receved out of 30. thank you.", "Solution_1": "39/50=78/100=78 %\r\n\r\n$ 78%*\\frac {3}{10} = 23.4%\n$\r\n\r\n$ 23.4% = \\frac {23.4}{100} = \\frac {2.34}{10} = \\frac {7.02}{30}\n$\r\n\r\nanswer : 7.02%\r\n\r\nEDIT: whoa my latex came out weird\r\n\r\nthanks vallon", "Solution_2": "I don't get what the question asked.\r\n\r\n39/50 is 78/100=78%\r\n\r\nYou received 30*0.78= 23.4 out of 30.\r\n\r\nHow did you get 7.02%?", "Solution_3": "[quote=\"fantasylover\"]39/50=78/100=78 %\n\n$ 78\\%\\times\\frac {3}{10} \\equal{} 23.4\\%$\n\n$ 23.4\\% \\equal{} \\frac {23.4}{100} \\equal{} \\frac {2.34}{10} \\equal{} \\frac {7.02}{30}$\n\nanswer : $ 7.02\\%$[/quote]\r\n\r\nfixed your latex\r\n\r\n\r\nedit: now that i think about it, i think that you shoulda stopped at $ 23.4$\r\nyou got $ 78\\%$ on the test, so your points out of 30 would be $ 0.78\\times30 \\equal{} 23.4$", "Solution_4": "Oh, I get it now.\r\n\r\nFantasy interpreted the total score, when I thought he meant it was just for the test.", "Solution_5": "ok...\r\ni'm not sure what you mean by that, but i think the answer is supposed to be 23.4, not 7.02%", "Solution_6": "[quote=\"captainonion\"]If I received 39/50 on an exam and the exam total was 30% of my final grade, how would I figure out [b]what I receved out of 30[/b]. thank you.[/quote]\r\n\r\nadding some text of my own", "Solution_7": "Okay, but then wouldn't the answer be the number 23.4, rather than the percent 7.02?", "Solution_8": "[quote=\"captainonion\"]If I received 39/50 on an exam and the exam total was 30% of my final grade, how would I figure out what [b]I receved out of 30[/b]. thank you.[/quote]\r\n\r\nthen what does bold part do in this problem?", "Solution_9": "i'm guessing \"what i received out of 30\" means what part of the 30% he would've gotten\r\nthat's how i understood the problem, but we should ask captainonion", "Solution_10": "I interpreted it as how many points I received out of 30.\r\n\r\nSo I thought that 23.4 was correct.\r\n\r\ncaptainonion, care to clarify :?:", "Solution_11": "hey fishy, you are right. I was looking for my total out of thirty, which would be 23.4/30=78%" } { "Tag": [ "factorial", "modular arithmetic" ], "Problem": "What three-digit integer is equal to the sum of the factorials of its digits?", "Solution_1": "Obviously, none of the digits can be $ \\ge 7$, because $ 7! > 999$.\r\n\r\nYou can deduce that this problem uses a $ 5$.\r\n\r\nThe $ 5$ probably won't be in the hundreds digit.\r\n\r\nThen, I used brute force to find $ \\boxed{145}$.", "Solution_2": "I remembered the answer is 145 AS the same question was asked in my Computer science two years ago.\r\nsuch type of nos are known by some specific term.I didn't remember the term.", "Solution_3": "These are called factorions. There are only 4 such numbers, 1,2,145, and 40585.", "Solution_4": "[quote=\"ernie\"]Obviously, none of the digits can be $ \\ge 7$, because $ 7! > 999$.\n\nYou can deduce that this problem uses a $ 5$.\n\nThe $ 5$ probably won't be in the hundreds digit.\n\nThen, I used brute force to find $ \\boxed{145}$.[/quote]\r\n\r\nOr, we can have a nice proof used by pythag011 and written by me:\r\n\r\n[hide=\":)\"]\nNo digit can be $ 7$, $ 8$, or $ 9$, because their factorials are greater than $ 999$, the largest $ 3$-digit number. Also, no digit can be $ 6$, because $ 6! \\equal{} 720$, so you need a digit in the hundreds spot that it as least $ 7$, but that is impossible. This narrows down our possibilities greatly.\n\nThe hundreds digit cannot be greater than $ 3$, because the largest possible now is $ 555$, and the sum of the factorials of their digits is $ 360$. But if the hundreds digit is $ 3$, the largest possible is $ 355$, but the sum of the factorials of the digits of $ 355$ is $ 246$. Thus the hundreds digit must be $ 1$ or $ 2$.\n\nIf the last two digits are $ 5$, the sum of the factorials of the digits is $ 240 \\plus{} \\text{first digit}$. $ 155$ obviously doesn't work, and $ 255$ doesn't either. Thus, one of the last two digits is at most $ 4$. Thus, the greatest possible sum of the factorials now is $ 5! \\plus{} 4! \\plus{} 2! \\equal{} 120 \\plus{} 24 \\plus{} 2 \\equal{} 146$. So the hundreds digit can only be $ 1$! Also, because we know that the sum of the factorials is at least $ 100$, we must have one $ 5$, because $ 4! \\cdot 3 \\equal{} 72$, which is not a $ 3$-digit number! So it must be of the form $ 1a5$ or $ 15a$. If it was in the form $ 15a$, the largest possibility would be $ 154$, but summing the factorials of the digits yields $ 120 \\plus{} 24 \\plus{} 1 \\equal{} 145$, so it can only be in the form $ 1a5$ where $ 0\\le a \\le 4$. \n\nIf we take this modulo $ 10$, we know the sum of the factorials is $ 1 \\plus{} 120 \\plus{} a! \\equiv 5 \\pmod{10}$. Thus $ a! \\equiv 4 \\pmod {10}$. Obviously, the only solution where $ 0\\le a \\le 4$ is $ 4$, so our number is $ \\boxed{145}$.[/hide]", "Solution_5": "125 works too because 1!+2!+5!=1+4+120=125 why was 125 wrong!?", "Solution_6": "$2! doesn't equal $4$, it equals $2$.", "Solution_7": "I had a question to the same problem \u2014 the solution says that \"the hundreds digit cannot be greater than 1, because the digits 5 or 6 are needed to make a 3 digit integer but will not fit with the larger hundreds digit.\" I don't understand why it can't be 2 or 3, for example.", "Solution_8": "Is there a number theory solution to this other than brute force and limiting cases? ", "Solution_9": "but 153 also works - I did it on a CALCULATOR", "Solution_10": "5 DIFFERENT calculators actually", "Solution_11": "[quote=xue1]but 153 also works - I did it on a CALCULATOR[/quote]\n\n$1! + 5! + 3! = 1+120 + 6 = 127$. $ 127 \\neq 153.$", "Solution_12": "oh, right, it's factorials, not cubes", "Solution_13": "but how do you prove it's the only answer? This one just says use brute force\n", "Solution_14": "[quote=AIME15][quote=\"ernie\"]Obviously, none of the digits can be $ \\ge 7$, because $ 7! > 999$.\n\nYou can deduce that this problem uses a $ 5$.\n\nThe $ 5$ probably won't be in the hundreds digit.\n\nThen, I used brute force to find $ \\boxed{145}$.[/quote]\n\nOr, we can have a nice proof used by pythag011 and written by me:\n\n[hide=\":)\"]\nNo digit can be $ 7$, $ 8$, or $ 9$, because their factorials are greater than $ 999$, the largest $ 3$-digit number. Also, no digit can be $ 6$, because $ 6! \\equal{} 720$, so you need a digit in the hundreds spot that it as least $ 7$, but that is impossible. This narrows down our possibilities greatly.\n\nThe hundreds digit cannot be greater than $ 3$, because the largest possible now is $ 555$, and the sum of the factorials of their digits is $ 360$. But if the hundreds digit is $ 3$, the largest possible is $ 355$, but the sum of the factorials of the digits of $ 355$ is $ 246$. Thus the hundreds digit must be $ 1$ or $ 2$.\n\nIf the last two digits are $ 5$, the sum of the factorials of the digits is $ 240 \\plus{} \\text{first digit}$. $ 155$ obviously doesn't work, and $ 255$ doesn't either. Thus, one of the last two digits is at most $ 4$. Thus, the greatest possible sum of the factorials now is $ 5! \\plus{} 4! \\plus{} 2! \\equal{} 120 \\plus{} 24 \\plus{} 2 \\equal{} 146$. So the hundreds digit can only be $ 1$! Also, because we know that the sum of the factorials is at least $ 100$, we must have one $ 5$, because $ 4! \\cdot 3 \\equal{} 72$, which is not a $ 3$-digit number! So it must be of the form $ 1a5$ or $ 15a$. If it was in the form $ 15a$, the largest possibility would be $ 154$, but summing the factorials of the digits yields $ 120 \\plus{} 24 \\plus{} 1 \\equal{} 145$, so it can only be in the form $ 1a5$ where $ 0\\le a \\le 4$. \n\nIf we take this modulo $ 10$, we know the sum of the factorials is $ 1 \\plus{} 120 \\plus{} a! \\equiv 5 \\pmod{10}$. Thus $ a! \\equiv 4 \\pmod {10}$. Obviously, the only solution where $ 0\\le a \\le 4$ is $ 4$, so our number is $ \\boxed{145}$.[/hide][/quote]\n\nyou might want to consider that 6! = 720, and since you can't use 7, you can't use 6 either\n" } { "Tag": [ "ratio" ], "Problem": "There are 8 red, 8 blue and 8 yellow marbles in a jar. What is the fewest marbles you can remove from the jar so that the ratio of red to non-red marbles is 3 to 7 and so that the ratio of yellow to non-yellow marbles is also 3 to 7?", "Solution_1": "wait, do you mean \"the fewest marbles you can remove from the jar so that the ratio of red to non-red marbles is 3 to 7 and so that the ratio of yellow to non-yellow marbles is also 3 to 7\" inside or outside the jar?\r\n\r\n\r\n[hide]well, whichever side you mean, there needs to be 10 marbles on that side because out of the 10 marbles, there can be 3 red and 3 yellow and 4 non-red-or-yellow marbles.[/hide]", "Solution_2": "[hide][color=indigo]You remove 4 (2 red and 2 yellow) :D [/color][/hide]", "Solution_3": "[hide]it's 4. 2 of both yelow and red. since there would be 20 marbles.\n6 to 14 is 3 to 7 both ways too.[/hide]", "Solution_4": "[hide]\nYou must remove 4 marbles from the jar.\n\n3:7 = 6:14\n\nSix marbles are red, and 14 are non-red.\n\nTherefore, six marbles are yellow, and 14 are non-yellow.\n\nSo there are 6 red, 6, yellow, and 8 blue marbles remaining.\n\nYou took away 4 marbles.\n\n :) \n[/hide]", "Solution_5": "[hide]\nRemove 2 red and 2 yellow. (4)\n[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Denote $ S(n)$ the sum of digits of $ n$\r\nDoes there exist an integer $ n$ such that $ S(n) \\equal{} 2009$ and $ S(n^2) \\equal{} 1996^2$?", "Solution_1": "Since the sum of the digits of $ n^2$ is $ 1996^2$, $ n^2$ has to have at least $ (1996^2)/9 \\equal{} 44134$ digits. that means that n has to have at max $ 44134/2 \\equal{} 22067$ digits. In those 22067 digits, even if the first $ 2009/9 \\equal{} 223$ digits were 9's, the square of n would not have 44134 digits, so such a number n does not exist.", "Solution_2": "Sorry,but I think it is not true.Since $ ((10^{223} \\minus{} 1)*10^{21844})^2 \\ge 10^{44133}$\r\nSo the square of $ n$ may have $ 44134$ digits\r\nAm I right? :blush:", "Solution_3": "Obviosly, if $ S(n)\\equal{}2009$, then $ n\\equal{}2\\mod 9\\to n^2\\equal{}4\\mod 9 \\to S(n^2)\\equal{}4\\mod 9 \\not \\equal{}1996\\equal{}7\\mod 9$.", "Solution_4": "[quote=\"Rust\"]Obviosly, if $ S(n) \\equal{} 2009$, then $ n \\equal{} 2\\mod 9\\to n^2 \\equal{} 4\\mod 9 \\to S(n^2) \\equal{} 4\\mod 9 \\not \\equal{} 1996 \\equal{} 7\\mod 9$.[/quote]\r\n\r\nNo,$ 1996^2 \\equal{}4 \\mod 9$ :wink:", "Solution_5": "My solution is $ 110...011...0....,,110...0999996$\r\n$ 11$ repeats $ 979$ times,and there are enough many $ 0$ between every consecutive $ 11$ \r\nHope it is right :)", "Solution_6": "[quote=\"hxy09\"]My solution is $ 110...011...0....,,110...0999996$\n$ 11$ repeats $ 979$ times,and there are enough many $ 0$ between every consecutive $ 11$ [/quote]\r\n\r\nMy check did not quite come out right. I guess the square (if there are plenty of zeros) will have\r\n979 blocks of 121\r\n478731 blocks of 242\r\none block of 99999216\r\n979 blocks of 21999912\r\n\r\nso I get for the digit sum: 3874936, while 1996^2 = 3984016\r\n\r\nBut maybe I counted wrong...", "Solution_7": "Sorry,I haven't checked it seriously :blush: \r\nAnother:$ 20..030...020...030...020...30......010...010...010...01$\r\nThere are $ 5785$ pairs of $ 2,3$ and the rest are $ 1$,$ 0$ are added between them as above." } { "Tag": [ "algebra", "polynomial", "group theory", "abstract algebra", "vector", "geometry", "geometric transformation" ], "Problem": "Let $ a,b,c$ be roots of an irreducible polynomial $ f\\in\\mathbb{Z}[X]$ such that $ a^{2}=bc$. Prove that $ a/b$ is a root of unity.", "Solution_1": "Do you have a better solution? I just crunched it. \r\n\r\nSay deg $ f$ = $ n$ and $ a = r_{1}, b = r_{2}, c = r_{3}, r_{4}, \\ldots, r_{n}$ are distinct $ n$ roots of $ f$. We can assume that $ a,b,c$ are pairwise distinct since otherwise $ \\frac{a}{b}$ is either 1 or -1, clearly a root of unity. Since $ f$ is irreducible, for each $ r_{i}$, you can find an automorphism (of say, splitting field of $ f$) sending $ a \\rightarrow r_{i}$. That is, $ r_{i}^{2}= r_{f(i)}r_{g(i)}$ for some permutations $ f,g \\in S_{n}$ with no fixed points such that $ f(i) \\neq g(i)$ for every $ i \\in \\{1, 2, \\ldots, n \\}$. To fix the idea, we have $ f(1) = 2$ and $ g(1) = 3$. \r\n\r\nTake $ G = $, a subgroup of $ S_{n}$ generated by $ f$ and $ g$. Take $ X$ to be the orbit of $ 1 \\in \\{ 1, 2, \\ldots, n \\}$ under $ G$. So $ X = \\{1, 2, 3, \\ldots \\}$. Renumbering if necessary, say $ X = \\{1, 2, 3, \\ldots, k \\}$. Then it's clear that $ \\frac{a{}^{2{}^{r}}}{b{}^{2{}^{r}}}= \\frac{r_{1}{}^{2{}^{r}}}{r_{2}{}^{2{}^{r}}}$ can be written as a quotient $ \\frac{r_{1}^{s}_{1}r_{2}^{s}_{2}\\ldots r_{k}^{s}_{k}}{r_{1}^{u}_{1}r_{2}^{u}_{2}\\ldots r_{k}{}^{u}{}^{k}}$ with $ s_{1}+\\ldots+s_{k}= u_{1}+\\ldots+u_{k}= 2^{r}$. Now if $ s_{1}+s_{2}+\\ldots+s_{k}> 2kM$ where $ M$ is the maximal of $ j$ among the lengths of products of $ f,g$ required to take $ j$ to $ 1$, then some $ s_{i}> 2M$ and applying the corresponding product of $ f,g$ required to take $ i$ to $ 1$ using the relations $ r_{i}^{2}= r_{f(i)}r_{g(i)}$, we get at the quotient with $ s_{1}> 0$ and $ u_{1}> 0$. So one can reduce the fraction $ \\frac{r_{1}^{s}_{1}\\ldots r_{k}^{s}_{k}}{r_{1}^{u}_{1}\\ldots r_{k}^{u}_{k}}$ further if $ s_{1}+\\ldots+s_{k}= u_{1}+\\ldots+u_{k}> 2kM$. This shows that $ \\frac{a{}^{2{}^{r}}}{b{}^{2{}^{r}}}$ can be written as $ \\frac{r_{1}^{s}_{1}\\ldots r_{k}^{s}_{k}}{r_{1}^{u}_{1}\\ldots r_{k}^{u}_{k}}$ with $ s_{1}+\\ldots+s_{k}= u_{1}+\\ldots+u_{k}\\leq 2kM$ for every $ r$. But there are only finitely many expressions of this form so for some $ r \\neq s$, we have $ (a/b){}^{2{}^{r}}$ = $ (a/b){}^{2{}^{s}}$. So $ (a/b){}^{2{}^{r}-2^{s}}= 1$. So $ a/b$ is a root of unity.", "Solution_2": "Here's what I did.\r\n\r\nWe assume $ \\frac{a}{b}$ is not a root of unity, and derive a contradiction.\r\n\r\nFirst, let's make the following modification: notice that for all positive integers $ n$, the conjugates of $ a^{n}$ are precisely the distinct $ n$'th powers of the conjugates of $ a$. Moreover, if $ \\frac{x}{y}$ is not a root of unity ($ x$ and $ y$ being two complex numbers), then $ x^{n}$ and $ y^{n}$ are different for all positive integers $ n$. Now, if instead of the algebraic numbers we initially had we work with their distinct $ n$'th powers for suitably chosen $ n$, then $ b^{n},a^{n},c^{n}$ will still be three distnict conjugates in arithmetic progression, and in addition, we can make sure that $ \\frac{x}{y}$ is not a root of unity for [i]any[/i] distinct conjugates $ x,y$ of $ a^{n}$. In order to keep the notation simple, I'll still name the numbers $ a,b,c$ etc., but keep in mind that now we make the extra assumption that no two conjugates of $ a$ have a root of unity quotient.\r\n\r\nNotice that [i]all[/i] roots $ a$ of our polynomial are geometric means of two other roots. This follows from the fact that the Galois group of the polynomial acts transitively on the roots. Now, if the root $ a$ has maximal absolute value, then $ b,c$ also have maximal absolute value whenever $ a^{2}=bc$. Now forget about all number-theoretic considerations. The situation is this: we have $ n\\ge 3$ complex numbers $ a_{i},\\ i=\\overline{1,n}$, all with the same absolute value, such that no quotient $ \\frac{a_{i}}{a_{j}},\\ i\\ne j$ is a root of unity. We want to reach a contradiction. Assume in addition that the set $ \\{a_{i}\\}$ is minimal with respect to these properties, and also that $ a_{n}=1$ (clearly, this does not affect the problem). \r\n\r\nThe complex numbers $ a_{i},\\ i\\le n$ lie on the unit circle; moreover, because of our assumption that $ \\frac{a_{i}}{a_{j}}$ are not roots of unity, the $ a_{i}$ are mapped in a one-to-one fashion into $ V$, the quotient of the unit circle through the group of roots of unity. Notice that $ V$ is a vector space over $ \\mathbb Q$, being a torsion-free divisible abelian group. Let $ E$ be the $ \\mathbb Q$-vector space spanned by the images $ e_{i}$ of the $ a_{i}$ in $ V$, with the numbering arranged so that $ e_{1}, \\ldots,\\ e_{t},\\ 1\\le t\\le n-1$ is a basis of $ E$. Among all the $ e_{i}$ there must be at least one ($ e_{s}$, say) which has the largest coefficient of $ e_{1}$ when written in the basis $ e_{1},\\ \\ldots,\\ e_{t}$. Then, if $ a_{s}^{2}=a_{i}a_{j}$, the coefficients of $ e_{1}$ in $ e_{i}$ and $ e_{j}$ are equal to that same maximal value. Repeat the procedure with $ a_{i},a_{j}$ instead of $ a_{s}$, and so on. The set we obtain after the finitely many distinct steps we can make has all the properties of $ \\left\\{a_{i}\\ |\\ i=\\overline{1,n}\\right\\}$ (that is, it consists of complex numbers on the unit circle, and each element is the geometric mean of two other elements), but is strictly smaller, because all $ a_{i}$ we looked at are such that the coefficient of $ e_{1}$ in $ e_{i}$ is non-zero, so, for example, the set in question does not contain $ a_{n}=1$ (which maps onto $ 0$ in $ V$). This contradicts the minimality of the set $ \\{a_{i}\\}$, and finishes the proof.", "Solution_3": "I think your proof is quite nice using | | of $ \\mathbb{C}$. I didn't think of doing this as, in the back of my mind, I replaced $ \\mathbb{Z}$ with an abstract field $ \\mathbb{F}$. \r\n\r\nSo in general, for any irreducible $ f \\in \\mathbb{F}[x]$ having some roots $ a_{0}, \\ldots, a_{k}$ in geometric progression, i.e. $ a_{0}^{k}= a_{1}a_{2}\\ldots a_{k}$ ($ k$ not necessarily the total number of roots of $ f$, i.e. $ k \\leq$ deg $ f$), $ a_{i}/a_{j}$ is a root of unity for all $ i,j$.", "Solution_4": "1) Hi sheaf.\r\nWe can have $ f(i)=f(j)$ with $ i\\not=j$ (cf. for example the polynomial $ x^{n}-\\alpha$). Perhaps you must work by contradiction as Grobber did.\r\n2) Hi Grobber.\r\nI don't understand: \"We assume $ \\frac{a}{b}$ is not a root of unity\" implies \"we can make sure that $ \\frac{x}{y}$ is not a root of unity for any distinct conjugates $ x,y$ of $ a^{n}$.\"\r\nMoreover \"assume that $ a_{n}=1$ (clearly, this does not affect the problem)\" is a detail but it isn't clear for me.\r\n3) I don't see very well the polynomial $ f$:\r\n Can somebody construct a such polynomial $ f$ which have complex roots of different modules ?\r\n Can somebody construct a such polynomial $ f$ which have all its roots of same module and s.t. $ f$ isn't a divisor of a polynomial of the form $ x^{n}-\\alpha$ ?", "Solution_5": "[quote=\"loup blanc\"]\nI don't understand: \"We assume $ \\frac{a}{b}$ is not a root of unity\" implies \"we can make sure that $ \\frac{x}{y}$ is not a root of unity for any distinct conjugates $ x,y$ of $ a^{n}$.\"\n[/quote]\n\nLet $ x_{1},\\ldots,x_{s}$ be the roots of the minimal polynomial of $ a=x_{1}$ (so $ b$ and $ c$ are also among them, of course). For all $ n\\ge 1$, the roots of the minimal polynomial of $ x_{1}^{n}$ are precisely the distinct values appearing in the multiset $ x_{1}^{n},\\ldots,x_{s}^{n}$ (why?). Now, we assumed that $ \\frac{a}{b}$ is not a root of unity. This means that $ a^{n}\\ne b^{n}$ for all $ n$ (same for $ a^{n},c^{n}$ and $ b^{n},c^{n}$), so $ a^{n},b^{n},c^{n}$ will always be distinct conjugates of one another, satisfying $ (a^{n})^{2}=b^{n}c^{n}$. Simply take $ n$ which \"eliminates\" all roots of unity of the form $ \\frac{x_{i}}{x_{j}}$, i.e. such that $ x_{i}^{n}=x_{j}^{n}$ whenever $ \\frac{x_{i}}{x_{j}}$ is a root of unity. Now, instead of working with the original algebraic numbers $ x_{1},\\ldots,x_{s}$, work with the conjugates of $ x_{1}^{n}$, which are precisely the distinct elements of the form $ x_{i}^{n}$.\n\n \n[quote=\"loup blanc\"]\nMoreover \"assume that $ a_{n}=1$ (clearly, this does not affect the problem)\" is a detail but it isn't clear for me.\n[/quote]\n\nThe problem, as stated there, no longer depends on the $ a_{i}$ being roots of an integer polynomial and all that. We're concerned only with their positions on a circle centered at the origin in the complex plane, and it's only their positions relative to one another that matter. Simply apply a homothety and a rotation around the origin to make $ a_{n}=1$. If you prefer, look at it this way: replace each $ a_{i}$ with $ \\frac{a_{i}}{a_{n}}$.\n\n \n[quote=\"loup blanc\"]\n3) I don't see very well the polynomial $ f$:\n Can somebody construct a such polynomial $ f$ which have complex roots of different modules ?\n Can somebody construct a such polynomial $ f$ which have all its roots of same module and s.t. $ f$ isn't a divisor of a polynomial of the form $ x^{n}-\\alpha$ ?[/quote]\r\n\r\nYou mean integer irreducible polynomials with three different roots $ a,b,c$ in geometric progression? I don't know if those examples exist, and right now I'm not particularly interested in the issue :)." } { "Tag": [ "quadratics", "algebra", "polynomial" ], "Problem": "Prove for every irrational real number a, there are irrational numbers b and b' such that a+b and ab' are rational while a+b' and ab are irrational.", "Solution_1": "[hide=\"Solution\"] Given our irrational real numbers $a$, let $a+b = p, ab' = q$ be two rational numbers we hope exist. Then $b = p-a, b' = \\frac{q}{a}$ and we want to show that there exist $p, q$ such that\n\n$a+\\frac{q}{a}, a(p-a)$\n\nAre irrational. \n\nIf $a+\\frac{q}{a}= x$ some irrational then $a$ satisfies the (irrational) quadratic $a^{2}-xa+q = 0$. If $a$ is not the root of a rational quadratic then any rational $q$ produces an irrational $x$. Otherwise $a$ is the root of some rational quadratic with some rational constant term, so pick $q$ not equal to that constant term and $x$ cannot then be rational.\n\nSimilarly, if $a(p-a) = y$ some rational then $a$ satisfies the (irrational) quadratic $a^{2}-pa+y = 0$. Again, if $a$ is not the root of a rational quadratic then any rational $p$ produces an irrational $y$. Otherwise $a$ is the root of some rational quadratic with some rational $a$ -coefficient, so pick $p$ not equal to that constant term and $y$ cannot be rational. [/hide]\r\nThis seems a little too easy, so I'm not sure if I'm missing subtleties.", "Solution_2": "I don't understand the part about he irrational and rational quadratics. Surely I find a constant term not equal to the term by which x is multiplied and still have it have rational roots, e.g. $x^{2}+5x+6$.", "Solution_3": "[hide=\"Some relevant background\"] [b]Definition:[/b] Let the minimal polynomial $m(x)$ over $F[x]$ (in practice, over $\\mathbb{Q}[x]$) of some $a$ be the unique monic polynomial of smallest degree such that $m(a) = 0$. \n\nIf $a$ is rational then $m(x) = x-a$. Otherwise $m(x)$ is of some higher degree. \n\n[b]Lemma:[/b] Given a polynomial $p(x)$ over $F[x]$ such that $p(a) = 0$, the minimal polynomial $m(x) | p(x)$.\n\nProof: Division algorithm. The remainder $r(x)$ when $m(x)$ divides into $p(x)$ must either be $0$ or be a polynomial of degree lower than $m(x)$ such that $r(a) = 0$ - contradiction.\n\n[b]Corollary (conjugate root theorem):[/b] If $p+\\sqrt{q}$ is a root of a rational quadratic, where $p, q$ are rationals and $q$ is not the square of a rational, then $p-\\sqrt{q}$ is also a root. [/hide]\n[hide=\"A further explanation of my logic\"] If the monic polynomial of $a$ over the rationals is quadratic, then $a$ cannot be the root of any other monic rational quadratic (or else the difference between the two quadratics, which is both linear and rational, must have $a$ as a root, contradicting $a$ irrational).\n\nHence if we fix one of the coefficients to be a rational not equal to the corresponding coefficient in the minimal polynomial of $a$ then the other coefficient cannot be rational.\n\nI am basically just using the idea of the uniqueness of the minimal polynomial, but the other results following from that idea are interesting too. [/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all natural numbers $ a,b$ such that : \r\n $ (a^2\\minus{}1)^2\\plus{}(a^2)^2\\equal{}b^2$", "Solution_1": "$ \\Box$" } { "Tag": [ "Stanford", "college" ], "Problem": "Can someone remind me what the formula is for finding the day of the week of a given date? I can't remember it...", "Solution_1": "wouldn't it depend on the month and year? :maybe:", "Solution_2": "The formula for the day of the week is called the Doomsday Formula ( or rule, like at [url]http://en.wikipedia.org/wiki/Calculating_the_day_of_the_week[/url]) \r\n\r\n\r\n[hide=\"ummm...\"]Can non-Stanford math circle participants post on this forum?[/hide]" } { "Tag": [ "AMC" ], "Problem": "How would I get involved in the AMC competitions if my school does not participate?", "Solution_1": "Find a nearby school that does participate in the AMC Contests and ask them if you could take the test along with their students. Most schools will gladly let you do so.", "Solution_2": "[quote=\"Xantos C. Guin\"]Find a nearby school that does participate in the AMC Contests and ask them if you could take the test along with their students. Most schools will gladly let you do so.[/quote]\r\n\r\nWhat if you win something? Which school will your name go under?", "Solution_3": "See the FAQ thread, which is stickied. You could also pay for the tests, in which case it would be really weird if your school didn't let you take them, especially as it gives other students at your school an opportunity to take the test.\r\n\r\nThat can get a little complicated, SplashD. It [i]should[/i] go under your real school's name, but you have to be careful to do the book-keeping right. Mistakes have occurred." } { "Tag": [ "inequalities", "rearrangement inequality", "inequalities open" ], "Problem": "let a,b,c in R .prove that\r\n$ \\sum_{sym}(\\frac {a}{b \\plus{} c})^2\\geq\\sum_{cyc}(\\frac {b}{b \\plus{} c})^2$", "Solution_1": "[hide=\"hint\"]Straightforward application of the rearrangement inequality[/hide][hide=\"solution\"]\nWLOG $ a^2 \\leq b^2 \\leq c^2 \\Rightarrow (a\\plus{}b)^2 \\leq (a\\plus{}c)^2 \\leq (b\\plus{}c)^2 \\Rightarrow (\\frac{1}{b\\plus{}c})^2 \\leq (\\frac{1}{a\\plus{}c})^2 \\leq (\\frac{1}{a\\plus{}b})^2$\n\nBecause of this, we see that the LHS of the original inequality consists of the product of similarly ordered sequences and the RHS consists of differently ordered sequences, thus it is true by the rearrangement inequality.[/hide]", "Solution_2": "$ a$, $ b$, $ c$ must be positive if you want the inequality to hold (else, $ a\\equal{}b\\equal{}1$, $ c\\equal{}\\minus{}2$ is a counterexample).\r\n\r\n darij", "Solution_3": "Yes, you're right... I just realized that if a,b,c are not nonnegative then $ a \\leq b \\leq c$ does not imply that $ (a\\plus{}b)^2 \\leq (a\\plus{}c)^2 \\leq (b\\plus{}c)^2$ but I believe my solution still works otherwise.", "Solution_4": "RHS=cyc((1-(c/b+c))^2)=3-2cyc(c/b+c)+cyc((c/b+c)^2) and our inequality is inuaqalent this cyc((a^2+c^2+2bc)/(b+c)^2)>=3 and LHS=cyc(1-(b^2-a^2)/(b+c)^2)=3-cyc((b^2-a^2)/(b+c)^2)>=3 and cyc((a^2-b^2)/(b+c)^2)>=0 and (b+c)^2=b^2+c^2+2bc<=2(b^2+c^2) and our inequality is cyc((a^2-b^2)/2(b^2+c^2))>=0 a^2=x b^2=y c^2=z and x,y,z>=0\r\nour inequality is iqualent cyc((x-y)/2(y+z))=(cyc(x^2-y^2)(z+x))/(2(y+z)(z+x)(x+y)) and its this x^3+y^3+z^3-cyc((y^2)z)>=0 and x^3+y^3+z^3>=cyc((y^2)z) its true" } { "Tag": [ "function", "combinatorics unsolved", "combinatorics" ], "Problem": "$ Problem$\r\n\r\nEnumerate the number of $ 3$-regular simple graphs.\r\n\r\n(i) Vertices are numbered.\r\n(ii) Up to Isomorphism.\r\n\r\nA graph $ G(v,e)$ with $ n$ vertices is called $ k$-regular if degree of every vertex is $ k$.", "Solution_1": "http://www.research.att.com/~njas/sequences/A005638 is unlabeled; http://www.research.att.com/~njas/sequences/A002851 is unlabeled connected; http://www.research.att.com/~njas/sequences/A004109 is labeled connected. (If you can get the connected ones you can get all of them via the exponential formula.)", "Solution_2": "The original poster asked via PM:\r\n[quote=\"I hate rating\"]I did look at the sequences. But I could not get a closed form or a gf, do you know any standard text or a paper that enumerate this ?( I dont have JSTOR access)\n\nYou said : \"If you can get the connected ones you can get all of them via the exponential formula\" . I dont understand ?[/quote]\r\nFor the first, this is unfortunately everything I know about the question. It's a natural enough question (and the sequence numbers are low enough) that I would guess not much is known about their enumeration that isn't in those sequences. (There are also several sequences for different kinds of cubic multigraphs and graphs or multigraphs with loops allowed.)\r\n\r\nFor the second question, the [url=http://en.wikipedia.org/wiki/Exponential_formula]exponential formula[/url] is just the observation that if $ A(x)$ is the exponential generating function for the sequence $ a_n$ of connected, labeled objects (e.g., connected graphs, trees, connected cubic graphs, but also many other classes) then $ B(x) \\equal{} e^{A(x)}$ is the generating function for $ b_n$, the number of not-necessarily-connected labeled objects of the same sort (e.g. graphs, forests, all cubic graphs). This becomes clear when you write down the coefficients of the latter series (as Wikipedia does in the link above)." } { "Tag": [ "algebra", "polynomial", "function", "functional equation" ], "Problem": "Find all polynomials $ f$ over $ \\mathbb{C}$ such that \r\n\\[ f(x)f(\\minus{}x) \\equal{} f(x^2)\r\n\\]", "Solution_1": "If $ f(x)$ is constant, then it is equal to $ 0$ or $ 1$.\r\n\r\nIf $ f(x)$ is nonconstant, then it has a finite number of roots. Let $ r$ be the root for which $ |r|$ is minimal. If $ r$ is a root, then $ r^2$ is a root also $ \\sqrt{r}$ is a root. Suppose for sake of contradiction that $ 1>|r|>0$. Then\r\n\\[ |r|\\minus{}|r|^2\\equal{}|r|*(1\\minus{}|r|)>0\\]\r\nwhich is a contradiction to the minimality of $ r$. For sake of contradiction, suppose that $ |r|>1$, then\r\n\\[ |r|\\minus{}|\\sqrt{r}|\\equal{}|\\sqrt{r}|*(|\\sqrt{r}|\\minus{}1)> 0\\]\r\nwhich contradicts the minimality of $ r$. \r\n\r\nConsider the cases:\r\ni) $ |r|\\equal{}0\\implies r\\equal{}0$\r\nIf $ 0$ is root, then it is a double root by considering the LHS of the functional equation. Hence $ x^2|f(x)$. At some point, $ f(x)\\equal{}x^{2n}*g(x)$ where $ g(x)$ is a polynomial that satisfies the same functional equation.\r\n\r\nii) $ |r|\\equal{}1$. Now let $ r'$ be the root where $ |r'|\\equal{}1$ and $ |\\arg (r')|>0$ is minimal. But if $ r$ is a root, then $ \\sqrt{r}$ is a root, which contradicts the minimality of $ \\arg(r')$. Hence $ \\arg(r)\\equal{}0$ so $ r\\equal{}1$. If $ f(x)\\equal{}(1\\minus{}x)*g(x)$ where $ g(x)$ is a polynomial that satisfies the same functional equation. Eventually, $ f(x)\\equal{}(1\\minus{}x)^n*g_1(x)$.\r\n\r\nIf we look at the minimal root, then divide it out, and repeat, eventually we get a constant polynomial (since $ f(x)$ has a finite degree). It is clear that any solution must be either \r\n\\[ f(x)\\equal{}(1\\minus{}x)^n*(x)^{2m} \\qquad\\text{or}\\qquad f(x)\\equal{}0\\]\r\n\r\nwhere $ m$ and $ n$ are non-negative integers. It is clear that both of these functions satisfy the equation so we are done.", "Solution_2": "This is not quite correct yet.", "Solution_3": "The function $ x^a \\minus{} x^b$ where $ a$ is an even nonnegative integer and $ b$ is an odd nonnegative integer works, but it's not in your solution space.", "Solution_4": "In addition, the polynomial $ x^2\\plus{}x\\plus{}1$ is missing.", "Solution_5": "[hide]Regard $ f(x)$ as a power series.\nLet $ f(x) \\equal{} \\sum_{k \\equal{} 1}^{\\infty} a_k x^k$.\nExpand $ f(x)f( \\minus{} x)$ and then compare coefficients with that of $ f(x^2)$.\nWe have a set of equalities:\n$ a_0^2 \\equal{} a_0$,\n$ 2 a_0 a_2 \\minus{} a_1^2 \\equal{} a_1$,\n$ 2 a_0 a_4\\minus{} a_2^2 \\equal{} a_2$,\n$ 2 a_0 a_6\\minus{} a_3^2 \\equal{} a_3$,\n$ 2 a_0 a_8 \\minus{} a_4^2 \\equal{} a_4$,\n.........\nCase 1 \n$ a_0 \\equal{} 0$\n......\nCase 2 \n$ a_0 \\equal{} 1$\n......[/hide]", "Solution_6": "Well it is pretty clear that i screwed up the $ |r| \\equal{} 1$ case...the mistake is that $ r$ is a root, implying that $ \\sqrt {r}$ is a root is not always true. $ \\pm\\sqrt {r}$ is true. \r\n\r\nBut that screws up the proof a bit because if $ \\arg(r) \\equal{} \\theta$, then $ \\arg( \\minus{} \\sqrt {r}) \\equal{} \\pi \\plus{} \\frac {\\theta}{2}$.", "Solution_7": "[hide=\"Observation\"] The important case is when $ |r| \\equal{} 1$. Let $ r \\equal{} e^{ix}$. If $ x$ is not a rational multiple of $ \\pi$, then $ e^{2^k ix}$ is also a solution for any $ x$, so there are infinitely many roots; contradiction. Hence $ r$ is an $ n^{th}$ root of unity. [/hide]", "Solution_8": "[quote=\"t0rajir0u\"][hide=\"Observation\"] The important case is when $ |r| \\equal{} 1$. Let $ r \\equal{} e^{ix}$. If $ x$ is not a rational multiple of $ \\pi$, then $ e^{2^k ix}$ is also a solution for any $ x$, so there are infinitely many roots; contradiction. Hence $ r$ is an $ n^{th}$ root of unity. [/hide][/quote]Hmm... so this means that the solution set is the set of all polynomials whose roots are roots of unity with their conjugates?\r\n\r\nEDIT: $ f(x) \\equal{} x^m(x\\minus{}1)^n\\Phi_k(x)$, cyclotomic polynomials $ \\Phi_k(x)$?", "Solution_9": "That is not true. $ \\Phi_{10}(x) \\equal{} x^4 \\minus{} x^3 \\plus{} x^2 \\minus{} x \\plus{} 1$ does not work. Neither does $ \\Phi_{12}(x)$...or for that matter, $ \\Phi_2(x) \\equal{} x \\plus{} 1$\r\n\r\nThis is because $ 2$, $ 10$ and $ 12$ are even.\r\n\r\nIf we have $ r$, then we have $ r^{2k}$. But if $ r$ is a primitive $ 2n$th root of unity, then $ r^2$ is not; it is a primitive $ n$th root of unity.\r\n\r\nI believe that cyclotomic polynomials of odd numbers all work.", "Solution_10": "[hide=\"Finishing up the solution\"] We've established that all roots of $ f(x)$ are $ k^{th}$ roots, hence primitive $ k^{th}$ roots (for appropriate $ k$). We've established that $ k$ is odd. It's also not hard to verify that $ f(x) \\equal{} x^2$ is a solution. \n\n[b]Lemma:[/b] If $ f(x), g(x)$ are solutions to the functional equation, then so is $ h(x) \\equal{} f(x) g(x)$. \n\nThe solution set is therefore closed under multiplication; it remains to consider its generators. We will do so by investigating the primitive $ k^{th}$ roots for a given $ k$ individually.\n\nLet $ n \\equal{} \\text{deg}(f)$. The leading coefficient of $ f(x)$ is either $ 1$ (when $ n$ is even) or $ \\minus{} 1$ (when $ n$ is odd). Therefore, let\n\n$ f(x) \\equal{} \\prod_{i \\equal{} 0}^{n \\minus{} 1} (r_i \\minus{} x)$. Then\n\n$ f(x) f( \\minus{} x) \\equal{} \\prod_{i \\equal{} 0}^{n \\minus{} 1} (r_i^2 \\minus{} x^2) \\equal{} f(x^2) \\equal{} \\prod_{i \\equal{} 0}^{n \\minus{} 1} (r_i \\minus{} x^2)$\n\nso the roots of $ f(x)$ have the property that when squared, they are the same set of roots. \n\nFor a given odd $ k$, let $ \\zeta_{m} \\equal{} e^{2\\pi i \\frac {m}{k} }$. Suppose that $ \\zeta_{2j}$ were a root of $ f(x)$. Then $ \\zeta_{2^{\\varphi(k) } j} \\equal{} \\zeta_j$ would also be a root. Let $ j$ be the smallest odd natural number such that $ r_j$ is a root. Then $ \\zeta_{2^l j}$ is a root for $ 0 \\le l < \\text{ord}_{k}(2)$.\n\nThis set of roots satisfies the desired property. There are $ \\frac {\\varphi(k) }{\\text{ord}_{k}(2) }$ such polynomials for every $ k$ (their product is $ \\Phi_k$), and they form the generators of our solution set. \n\nIn particular, for $ k \\equal{} 1$ we have $ f(x) \\equal{} 1 \\minus{} x$.\nFor $ k \\equal{} 3$ we have $ f(x) \\equal{} x^2 \\plus{} x \\plus{} 1$.\nFor $ k \\equal{} 5$ we have $ f(x) \\equal{} x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1$.\nFor $ k \\equal{} 7$ we have $ f(x) \\equal{} (\\zeta_1 \\minus{} x)(\\zeta_2 \\minus{} x)(\\zeta_4 \\minus{} x), (\\zeta_3 \\minus{} x)(\\zeta_6 \\minus{} x)(\\zeta_5 \\minus{} x)$ because $ \\text{ord}_7(2) \\equal{} 3$. Note that the coefficients of these polynomials is [b]not purely real.[/b]\n\nAnd so forth. [/hide]", "Solution_11": "Engel says that for p+q even we have the following solution\r\n\r\n$ f(x)\\equal{}x^p(x\\minus{}1)^q(1\\plus{}x\\plus{}x^2)^r$ for p,q,r natural numbers", "Solution_12": "Strange. Engel has missed at least the solutions $ f(x) \\equal{} 1 \\minus{} x$ and $ f(x) \\equal{} x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1$. What is the solution in the text?\r\n\r\nI have missed some solutions as well (for example, $ x(x \\minus{} 1)$). It seems that a discussion of multiplicities was more important than I had expected (particular, the multiplicity of the root $ 0$).", "Solution_13": "My Engel book gives the set of solutions $ f(x)\\equal{}(\\minus{}x)^p(1\\minus{}x)^q(x^2\\plus{}x\\plus{}1)^r,\\ p,q,r\\in\\mathbb{Z}$. For emphasis, I will say that we can even take $ p,q,r<0$. The solution is not as short as an average \"Engel\" solution size, that is, I will post the solution later (it is 2:10 a.m. now and I have to go to school tomorrow).", "Solution_14": "Is it so hard to verify that\r\n\r\n$ (x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1)(x^4 \\minus{} x^3 \\plus{} x^2 \\minus{} x \\plus{} 1) \\equal{} (x^8 \\plus{} x^6 \\plus{} x^4 \\plus{} x^2 \\plus{} 1)$?\r\n\r\nWhile Engel is correct in identifying the generator $ \\minus{}x$ (rather than the generator $ x^2$ which I identified), $ f(x) \\equal{} x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1$ is plainly in the solution set. I admit that the full solution set I gave is strange and uncharacteristically unclean; however, I don't see anything in the problem statement that suggests that I am wrong. Is there an extra condition?", "Solution_15": "Engel does not actually claim that his solution is complete. He notes that boxedexe's polynomials constitute a solution, then asks, \"Are there any others?\"", "Solution_16": "[quote=\"t0rajir0u\"]Is it so hard to verify that\n\n$ (x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1)(x^4 \\minus{} x^3 \\plus{} x^2 \\minus{} x \\plus{} 1) \\equal{} (x^8 \\plus{} x^6 \\plus{} x^4 \\plus{} x^2 \\plus{} 1)$?\n\nWhile Engel is correct in identifying the generator $ \\minus{} x$ (rather than the generator $ x^2$ which I identified), $ f(x) \\equal{} x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1$ is plainly in the solution set. I admit that the full solution set I gave is strange and uncharacteristically unclean; however, I don't see anything in the problem statement that suggests that I am wrong. Is there an extra condition?[/quote]\r\n\r\nClarification: It appears that you are frustrated, thus I must clarify that Engel does not claim that his solution set is complete, as Phelpedo stated above; I beleive you are correct.\r\n\r\nDoes anyone know the actual source of this problem?", "Solution_17": "[quote=\"Phelpedo\"]Find all polynomials $ f$ over $ \\mathbb{C}$ such that\n\\[ f(x)f( - x) = f(x^2)\n\\]\n[/quote]\r\nlet:\r\n$ A=\\{h\\in \\mathbb{C}[X]: \\ g(x)=x^2\\ or\\ g(x)=\\prod_{i=1}^{2m}(x-a^{2^{i-1}}),\\ m\\in\\mathbb{N}^*,a\\in\\mathbb{C},\\forall i\\in[|1,2m-1|],a^{2^i}\\neq a,a^{2^{2m}}=a\\}$\r\n$ B=\\{g\\in \\mathbb{C}[X]: \\ g(x)=-(x-1)\\ or\\ g(x)=-\\prod_{i=1}^{2m+1}(x-a^{2^{i-1}}),\\ m\\in\\mathbb{N},a\\in\\mathbb{C},\\forall i\\in[|1,2m|],a^{2^i}\\neq a,a^{2^{2m+1}}=a\\}$\r\nthen it's easy to prove that ${ \\{f\\in\\mathbb{C}[X]: \\ f(x^2)=f(x)f(-x)}=\\{(\\prod_{i=1}^{c}g_i)(\\prod_{j=1}^{d}h_j): \\ c,d\\in\\mathbb{N}^*,(g_i)\\in A,(h_j)\\in B\\}\\cup\\{0,1\\}$\r\nfor exemple:\r\n$ Altheman$ : $ h_i=x^2,g_i=-(x-1)$ gives $ f(x)=(1-x)^{c}x^{2d}$" } { "Tag": [ "geometry", "3D geometry", "sphere", "Gauss" ], "Problem": "Hi,\r\n\r\nI am new here. Hopefully share good things.\r\n\r\nI think this question is simple but can not understand. And don't have the tools of experiment too. Any help is very good for me. Unbelieveble confused nowadays. Please don't laugh.\r\n\r\nLet me explain the question, as you may see my attached picture, a hollowed and almost spherical conductor has (-3q) charge. Another charged conductor has (+2q). If i tie this (+2q) charged one to inside of hollowed sphere with a conductor cable what will be the charge of them? And how? I suppose the picture which i attached will explain better.\r\n\r\nMy problem is; is there a difference of directly touching to inside and tie with a cable again to inside? I suppose this question will be help to understand.", "Solution_1": "welcome fisher !!!\r\n[hide=\"hint\"]\nthe potential in both the sphere becomes same...\nsome charges get collected on the inner hollw some remain at the outer hollow.[/hide]", "Solution_2": "Thank you pardesi,\r\n\r\n[quote=\"pardesi\"]some charges get collected on the inner hollw some remain at the outer hollow.[/quote] \r\nMaybe could not understand but there will be charges inside? As i know, they must be present only outsides of conductor.\r\n\r\nDo you mean before, after, both cases? I suppose later. Then what about Z sphere? Will have charge and if it's, which kind of charge?\r\n\r\nAlso, if i would touch the Z sphere to inside directly, it would be lose all. Is this different? Couldn't understand enough.", "Solution_3": "charges can only reside on the boundary of a conductor...\r\nso some charge resides on the inner boundary and some on the outer...\r\nthe charge distribution is such that both have same potential", "Solution_4": "Thank you again,\r\n\r\nBut if; \r\n\r\n1) i touch the \"Z\" without cable(with insulated handle) to inside of \"K\"? \r\n\r\n2) \"K\" was almost closed from upside like whole sphere?\r\n\r\nMore clear, in Faraday ice pail experiment, all charges were only outside of pail. Now some of them inner, some on the outer. How?\r\n\r\n---------------\r\nFor example here:\r\nhttp://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ShellsConductors.xml\r\n\r\nWhen a charged object is placed within a metal container, an equal charge of the same sign is FORCED to the outer surface of the container.\r\n\r\n\r\nAll of the charge on any metal object resides on its outer surface if a conducting path is provided so that the charge can move there. Remember that charges will flow between two positions as long as there is a potential difference between those positions. When the voltage has been equalized, all charges will cease to flow. \r\n-----------------", "Solution_5": "[quote=\"fisher\"]\n---------------\nFor example here:\nhttp://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ShellsConductors.xml\n\nWhen a charged object is placed within a metal container, an equal charge of the same sign is FORCED to the outer surface of the container.\n\n\n-----------------[/quote]\r\n\r\nI really don't understand your question above but let me just explain the above statement. If a charged object is placed inside a metallic container, then an equal and OPPOSITE charge is forced on the inner surface of the container because even a hollow metallic container has some thickness and inside that net field must be zero. (Use gauss law).\r\n\r\nAnd so by conservation of charge equal charge of same sign is forced outside.", "Solution_6": "Excuse me Cyber-Shreyas, the link and pasted sentences are not identing my upper question. Just noticed.\r\n\r\nYesterday, pardesi and sentence from another book make me think. I suppose i will explain different now. This is;\r\n\r\nIf a hollowed cylinder(or any \"COMPLETELY\" closed) conductor charged then the charges must sit on outer surface. And everywhere has only one potential. The thing which confused me is, the figure and pictures of Faraday ice pail experiment in net, books,.... Everywhere, the pail is uncovered. Not completely closed. But still, says that if a conductor touch inside of it, all charges travels to outside of pail. So the inside of faces(facing each other) in pail has no charge, anyway everywhere has same potential at last. I mean the figure of hollow in my first post.\r\n\r\nIn that figure, the Z(hollowed) has no charge in faces at first. Now i am not sure this. After pardesi, i think, the inner faces must has charges too since, it is not completely closed sphere(without a cover). Is this true?\r\n\r\nAgain, if i understood pardesi, everytime, before touch or after touch, inner surfaces has charges. Again, if it is not completely closed with a cover, inside will has charges too. IF THIS IS TRUE, this is my question; in another experiment, an electroscope sits inside of a Faraday Cage, knob and cage tied via cable. But electroscope has no charge anytime, even cage has. We may not assume cage is COMPLETELY CLOSED. But anyhow, electroscope is never has charge. How? When inside has charges?\r\n\r\n------------\r\nBTW, does my English understandable enough?", "Solution_7": "ok u partially followed me...\r\nknow these\r\n1)charges on a conductor reside only at the boundary(the boundary may be an inner one or a outer one).\r\nThat is it is possible to have charges on the inner surface of a shell\r\nEx.U have a [b]thick[/b] shell with a point charge $ q$ at it's centre then the inner surface will have charge $ \\minus{}q$ and the outer surface has charge $ q$\r\n2)the conductor is equipotential\r\n3)in ur case the charges will redistribute on inner as well as outer surfaces\r\n4)both the surfaces will have same potential as they are connected by conducting wire", "Solution_8": "Thank you for your patience. I guess your boredom. But for long years i surmise some wrongs as true. I can not easily change my mind. Maybe for this reason, understand slow. You teach me new things, you clear my pessimism. I am thankful. But feel embarrassed too.\r\n\r\nThese are final 2 questions(certain, promise :oops: ),\r\n\r\n1) About the conductors of above figure, if i would have dip and touch the Z(ball) into the hollow(without a cover) and put out again, some charges would have left on Z(ball)? yes and no sufficient. Please.\r\n\r\n2) In my previous post, at last paragraph, mentioned about Faraday Cage experiment. The cage is not completely closed. There are plenty of holes on wire fence. But the electroscope in cage(connected to cage with a wire) don't has charges. When i think comperative, the electroscope would have had charge a bit(or more). Because the wire between cage and electroscope is contacted to outer face of cage. What did i omit now?\r\n\r\nRegards" } { "Tag": [ "Asymptote", "blogs", "\\/closed" ], "Problem": "I accidentally posted a HUGE asymptote (10000 size) in my blog, and now it doesn't load no matter what. I meant to post it as size: 1000, but I accidentally typed another zero and clicked submit. Could someone help?", "Solution_1": "I don't think it can be fixed without going into the SQL or something. Its best to kill your blog, I guess.", "Solution_2": "I've fixed the bug. It's good that you've found that hole. That's what caused the interruption in our server." } { "Tag": [ "symmetry", "combinatorics proposed", "combinatorics" ], "Problem": "Let $M$ be a finite set and $\\Omega$ be a set of subsets from $M$ such that :\r\n(i) For all $A,B \\in \\Omega$, if $|A \\cap B | \\geq 2$, then $A=B$\r\n(ii) There are $A,B,C \\in \\Omega$, pairwise distinct, such that $|A \\cap B \\cap C | = 1$\r\n(iii) For every $A \\in \\Omega$ and every $a \\in M \\setminus A$ there is a unique $B \\in \\Omega$ such that $a \\in B$ and $A \\cap B = \\emptyset.$\r\n\r\nProve that there are numbers $p$ and $s$ such that :\r\n(a) For every $a \\in M$ the number of sets which contain $a$ is $p$.\r\n(b) For every $A \\in \\Omega$, we have $|A| = s.$\r\n(c) $s+1 \\geq p.$\r\n\r\nMy solution sounds a bit strange for an olympiad problem. So what is yours? :P \r\n\r\nPierre.", "Solution_1": "Let's call the sets lines and the elements of $M$ points :). Furthermore, whenever $d,d'\\in\\Omega,d\\cap d'=\\emptyset$.\r\n\r\nThe third axiom (Euclid's $5$'th postulate, actually :)) says that $\\|$ defined by $d,d'\\in\\Omega,d\\|d'\\iff$ either $d=d'$ or $d\\cap d'=\\emptyset$ is an equivalence relation (actually, it says that the relation is transitive, the reflexivity and symmetry being obvious).\r\n\r\nWe can thus partition $\\Omega$ in $\\|$ classes. It's also clear from the third axiom that each $\\|$ class partitions $M$.\r\n\r\nIf $\\|_1,\\|_2$ are $2$ distinct $\\|$ classes (we can find at least three distinct classes according to the second axiom), then every line in $\\|_1$ has exactly $|\\|_2|$ points, because every such line intersects each line in $\\|_2$ in exactly one point, and those are all its points. This means that in each $\\|$ class $\\|_i$, all lines have the same number of points $s_i$. At the same time, all the other $\\|$ classes contain $s_i$ lines each, since each line from another $\\|$ class $\\|_j,j\\ne i$ intersects one of the lines in $\\|_i$ in exactly one point among the $s_i$. Again, this means that for all $j\\ne i$, $s_j$ is the same. This, together with the second axiom (which we fully use this time) shows that all the $s_i$ are equal, and we denote this common value by $s$.\r\n\r\nIf $p$ is the number of $\\|$ classes, then every point in $M$ belongs to $p$ lines. \r\n\r\nWe have thus shown the existence of $p,s$, and the fact that $p$ is, in fact, the number of $\\|$ classes. Take $d\\|d',a\\in d$. Each $\\|$ class different from the one to which $d,d'$ belong contains one line (and only one) passing through $a$ and some point on $d'$. Since $d'$ has $s$ points, it means that except for the class of $d,d'$, we have at most $s$ other classes. In other words, $p\\le s+1$, as desired.", "Solution_2": "Yes, that's what I did (even if I did not call the sets 'lines'). But, I thought that using an equivalence relation was a bit strange for an olympiad problem ;) \r\n\r\nPierre." } { "Tag": [ "inequalities", "absolute value" ], "Problem": "In general how do you do those problems where it gives you $ \\frac{1}{x}\\plus{}\\frac{1}{y}\\equal{}\\frac{1}{z}$ where x, y, and z are integers and it asks like what is the smallest possible value of x+y or how many total possiblilities there are.", "Solution_1": "What do you mean by how many total values? \r\n\r\nAre you sure you didn't accidentally put an \"equal\" sign instead of an \"inequalities\" sign or forget to put absolute value?\r\n\r\nI don't understand your question; make it clearer.", "Solution_2": "He probably means something like, how many ordered pairs $ (x,y)$ of positive integers satisfy $ \\frac{1}{x}\\plus{}\\frac{1}{y}\\equal{}\\frac{1}{42}$ or whatever.\r\n\r\n[url=http://www.artofproblemsolving.com/Wiki/index.php/SFFT]See here[/url]", "Solution_3": "Wait, in these types of problems, they usually give you a value of z. Multiply by xyz and factor using Simon's Favorite Factoring Trick.", "Solution_4": "what if you were asked the find the smallest possible value of x+y", "Solution_5": "$ \\frac{1}{x}\\plus{}\\frac{1}{y}\\equal{}\\frac{x\\plus{}y}{xy}$ so $ x\\plus{}y\\equal{}\\frac{xy}{z}$. The smallest sum occurs when xy is the smallest possible, so perhaps brute force and find the smallest product?", "Solution_6": "For greater numbers of unit fractions where expanding and factoring may not help as much, one can use a technique similar to that illustrated [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1336088#1336088]here[/url].", "Solution_7": "[quote=\"leewongjao\"]In general how do you do those problems where it gives you $ \\frac {1}{x} \\plus{} \\frac {1}{y} \\equal{} \\frac {1}{z}$ where x, y, and z are integers and it asks like what is the smallest possible value of x+y or how many total possiblilities there are.[/quote]\r\n\r\nGood question." } { "Tag": [ "AMC", "AIME" ], "Problem": "There are six 1 x 2 x :pi: bricks stacked one on top of another to form a tower six bricks high. Each brick can be in any orientation as long as it rests flat atop the brick (or floor) below it. How many possible heights for the tower are possible?\r\n\r\n\r\n\r\n\r\nWhy isn't the answer just 3^6?", "Solution_1": "[hide]assume that the tallest bricks are always on the bottom. Then all of the bricks on the bottom will have height pi, then all of the bricks in the middle will have height 2 and all the bricks on top will have height 1.\n\n\n\nthen we just have to divide 6 into 3 parts to get the number of possible heights, which is 28\n\n\n\nthe answer is not just 36 because if we have one arrangement of bricks and we rearrange them then we have the same height but a different tower[/hide]", "Solution_2": "Ah, silly me. Thanks.", "Solution_3": "this resembles an aime problem i did recently", "Solution_4": "For a moment I thought \"What if they were 1x2x3 bricks ? Then some of the 28 arrangements would give the same height ... but which ones, and how many ?\".\r\n\r\nThen I realised that there is a much simpler method for 1x2x3 bricks ....", "Solution_5": "Yes, 1x2x3 has a nice (simple) alternate method.", "Solution_6": "how bout 94 bricks of size 4x10x19", "Solution_7": "Not so nice, but still doable. 1x2x3 is particularly nice, though.\r\n\r\nIsn't that an old AIME or AHSME question?" } { "Tag": [], "Problem": "Hola amigos, segun veo no se ha presentado un problema en un buen tiempo, asi es que los animo a resolver este peque\u00f1o problema, y que de paso se animen a presentar sus problemas...\r\n\r\nDurante el curso de una campa\u00f1a de elecciones, p suertes de promesas diferentes han sido dadas por los diferentes partidos politicos (p>0). Mientras que varios partidos han hecho la misma promesa, toda pareja de partidos tienen la menos una promesa en comun, sin embargo ninguna pareja de partidos ha hecho lasmismas promesas. Mostrar que no existe mas de 2^(p-1) partidos.", "Solution_1": "En un conjunto $S$ de $n$ elementos no podemos encontrar mas de $2^{n-1}$ subconjuntos tal que para cualquier par $A,B \\subset S$, la interseccion de $A$ y $B$ no es nula.\r\nEsto es ya que $S$ tiene $2^n$ subconjuntos y estos forman $2^{n-1}$ parejas $X, S-X$ (donde $X \\subset S$) de las cuales solo podemos contar un miembro.", "Solution_2": "Un problema interesante (que esta un poco relacionado con lo anterior, quizas esta un poco incoherente pero me vino a la mente y esta bueno jaj) seria demostrar la siguiente generalizaci\u00f3n:\r\n\r\nDados $n$ subconjuntos convexos de $R^d$, si la interseccion de cualquiera $d+1$ subconjuntos no es nula, entonces la interseccion de todos los $n$ subconjuntos no es nula.", "Solution_3": "[quote=\"manuel\"]Un problema interesante (que esta un poco relacionado con lo anterior, quizas esta un poco incoherente pero me vino a la mente y esta bueno jaj) seria demostrar la siguiente generalizaci\u00f3n:\n\nDados $n$ subconjuntos convexos de $R^d$, si la interseccion de cualquiera $d+1$ subconjuntos no es nula, entonces la interseccion de todos los $n$ subconjuntos no es nula.[/quote]\r\n\u00bfNo eso el teorema de Helly, o al menos una generalizaci\u00f3n? Lo divertido no es ese sino verlo coloreado y toda clase de generalizaciones horribles que le han hecho a ese pobre teorema", "Solution_4": "si eso es el de helly...\r\nlo sabes demostrar dave?\r\nAh! y ense\u00f1eme la versi\u00f3n coloreada que nunca he oido de esa" } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Ya tenia rato que no posteaba nada x aki, debido a que andaba preparandome para la XX OMM, la cual fue la semana pasada y en la que sake ORO!!!!\r\njeje \r\nAki posteo el problema 1 del primer dia:\r\n\r\nSea ab un numero de dos digitos. Un entero positivo n es \"pariente\" de ab si:\r\n1)el digito de las unidades de n tambien es b\r\n2)los otros digitos de n son distintos de cero y suman a\r\nPor ejemplo, los parientes de 31 son 31, 121, 211, 1111.\r\nEncuentra todos los numeros de dos digitos que dividen a todos sus parientes.\r\n\r\nEspero sus soluciones!!!", "Solution_1": "Si $a=1$ el unico pariente de $ab$ es si mismo. \r\nSi $a=2$ el unico pariente de $ab$ es $11b$ que manualmente nos damos cuenta que no es posible llenar. \r\nSi $a\\ge 3$ entonces hay dos parientes de $ab$ que son de la forma:\r\n$n12b$ y $n21b$. La diferencia de esos dos parientes es $90$, por lo que si $ab$ los divide a los dos es porque $ab$ divide a $90$. Los divisores de 2 digitos de noventa (con $a>2$) son 30, 45, y 90. Es facil ver que los 3 funcionan.\r\nQED.", "Solution_2": "Bueno, tu solucion fue practicamente igual k la mia, solo unas cuantas diferencias:\r\n\r\nPara $a=1$ solo ab es su pariente, de manera que si se cumple.\r\nPara a$\\geq 2$ consideremos el numero $100+10(a-1)+b$ vemos que $10a+b$ debe dividir a dicho numoer, el cual es igual a $90+10a+b$. Entonces basta con que $10a+b$ divida a $90$, es decir, que sea un divisor de 90, y mayor o igual que 20 ($a\\geq 2$). Entonces se cumple para 30, 45 y 90.\r\n :lol:" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "[size=150]resolve this exercice[/size]", "Solution_1": "where is the solution \r\n\r\n :?:", "Solution_2": "[hide]Assume there is such a square, call it $x$ then for some positive integer $k$, $k\\geq 2$\n\\[\\displaystyle 2^{k+2}5^k+9=x^2\\]\n\\[\\displaystyle \\Longrightarrow 2^{k+2}5^k=(x+3)(x-3)\\]\nThen we have two equations:\n\\[2^a5^b=x+3\\]\n\\[2^{k+2-a}5^{k-b}=x-3\\]\n\\[\\Longrightarrow 2^{a-1}5^b-2^{k+1-a}5^{k-b}=3\\]\nConsider the above equation mod 2 to prove no set of integers $a,b,k$ can satisfy the equation.[/hide]", "Solution_3": "[hide]\n\nSuppose there exist a perfect square , there can be two such cases\n\n(i) When the perfect square is in the form $(10k+3)^2$ for $k\\in \\mathbb{N}$ \nThen we have $(10k+3)^2=4\\times 10^{n+1}+9$ for integer $n>0$ .\n\n$100k^2+60k=4\\times 10^{n+1}$ $\\Leftrightarrow$ $k(5k+3)=2^{n+1}5^n$\n\nSince $5k+3$ cannot be divide by $5$ , it left only $k=5^n$ . So $5^{n+1}+3=2^{n+1}$ . But $5^{n+1}>2^{n+1}$ . Hence , no solution in this case.\n\n(ii)When the perfect square is in the form $(10k+7)^2$ for $k\\in \\mathbb{N}$ similarly\n\n$100k^2+140k+40=4\\times 10^{n+1}$ $\\Leftrightarrow$ $(5k+2)(k+1)=2^{n+1}5^n$ . So this implies $k=5^n-1$ and $5^{n+1}-3=2^{n+1}$ Again we see that there are no solution here . \n\nFrom the two cases , there are no way to form a perfect square by the given number .[/hide]" } { "Tag": [], "Problem": "3 nafar miran ye goldoon mikharan 300 toman(ahde bogh masalan) pas nafari 100 toman sahmeshon mishe bad maloom mishe ke 50 toman geroon kharidan sahebe maghaze 50 toman mide be shagerdesh onam 20 toman mizare to jibesh va nafari 10 toman beheshon mide yani nafari 90 toman dadan 3*90=270 , 270+20=290 payda konid 10 tomane gom shode ra.mamnon misham rahnamaei konid", "Solution_1": "kafie yebar az dide forooshonade be ghazie negah koni yebar az dide kharidar yebaram az dide shagerde,oonvakh khodet motevajjeh mishi ke hich moshkeli nadare ghazie va hamcehi dorosteo faghat bazie ba kalamateo oon 20 tomane... :)" } { "Tag": [ "inequalities", "quadratics" ], "Problem": "I'll put here problems that i can;t solve :D, hoping that u will help me :D", "Solution_1": "a,b - two natural numbers. proof that a^2 + b^2 is the difference of two square roots only if ab is even", "Solution_2": "Assume that $ ab$ is odd then $ a,b$ are odd.\r\n\r\nSo we have $ a^2 \\plus{} b^2\\equiv 2 (\\mod 4)$, so if $ a^2 \\plus{} b^2$ is the difference of two square roots $ x$ and $ y$, we have $ x^2\\equiv y^2 \\plus{} 2 (\\mod 4)$, which is impossible", "Solution_3": "prove that\r\nx\u00b2y + z \u2265 2x\u221ayz , (v) x,y,z > 0\r\n[/code]", "Solution_4": "[quote=\"brasuceava\"]prove that\n$ x^2y \\plus{} z \\ge 2x\\sqrt{yz} \\forall x,y,z > 0$[/quote]\r\nYou know what is AM-GM inequality? - If not take time to go through this [url=http://url]link[/url]\r\nIf yes then,\r\n$ x^2y \\plus{} z \\ge 2\\sqrt{x^2yz} \\equal{} 2x\\sqrt{yz}$\r\nWhich is the same as - \r\n$ (x\\sqrt{y})^2 \\minus{} 2(x\\sqrt{y}\\cdot\\sqrt{z} \\plus{} \\sqrt{z}^2 \\ge 0 \\implies (x\\sqrt{y} \\minus{} \\sqrt{z})^2 \\ge 0$\r\nwhich is true", "Solution_5": "\\:d/ there goes another one \\:d/\r\n\r\n\r\nFind out real, positive numbers a, b, c knowing that:\r\n\r\n$ ( ab \\plus{} bc \\plus{} ca) \\minus{} 1 \\geq a^2 \\plus{} b^2 \\plus{} c^2 \\geq 3 ( a^3 \\plus{} b^3 \\plus{} c^3)$", "Solution_6": "From QM-AM we get easily that if\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} S$ then $ a \\plus{} b \\plus{} c \\le \\sqrt {3S}$ which means $ S \\plus{} 2(ab \\plus{} ac \\plus{} bc) \\le 3S$ which means $ ab \\plus{} ac \\plus{} bc \\le S$. Than, going back our inequality, we get:\r\n$ S \\minus{} 1 \\ge ab \\plus{} ac \\plus{} bc \\minus{} 1 \\ge S$ which is impossible.\r\nHence there are no such numbers.", "Solution_7": "what is qm am ?:\">", "Solution_8": "$ \\sqrt{\\frac{\\sum_{k\\equal{}1}^n a_k^2}{n}} \\le \\frac{\\sum_{k\\equal{}1}^n a_k}{n}$", "Solution_9": "correction:\r\n\r\n$ \\sqrt{\\frac{\\sum_{k\\equal{}1}^{n}a_{k}^{2}}{n}} \\geq \\frac{\\sum_{k\\equal{}1}^{n}a_{k}}{n}$\r\n\r\nThis is the Quadratic Mean-Arithmetic Mean Inequality.\r\n\r\nIf that looks alien:\r\n\r\n$ \\sqrt{\\frac{{a_1}^2 \\plus{} {a_2}^2 \\plus{} ... \\plus{} {a_n}^2}{n}} \\geq \\frac{a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n}{n}$\r\n\r\nFor real numbers $ a_1, a_2, ..., a_n$." } { "Tag": [ "superior algebra", "superior algebra theorems" ], "Problem": "I'm interested in Harari's book on Algebra in English(I don't exactly know it's name, i have Harari's book on groupes,fields,ring but they are in French)-\r\nDoes anybody have this book or books in electronic format, or maybe you know where they are available. \r\nI'd be very grateful for someone who could help me. Thanks", "Solution_1": "if it exists in electronic format, I'd also be interested.\r\n(preferrably english, but french may work too)" } { "Tag": [], "Problem": "Express three-halves of one-half as a common fraction.", "Solution_1": "naturally, we analyze the problem, break it down.....\r\n\r\n3 halves is 3/2\r\n\r\n1 half is 1/2\r\n\r\n3 halves of 1 half = 3/2*1/2 (the term \"of\"in math word problems is usually interpreted as \"multiplied\")\r\n\r\n3/2*1/2= [b]3/4[/b]\r\n\r\n[b]\nYour answer is 3/4[/b]", "Solution_2": "We have $ \\frac{3}{2}(\\frac12)\\equal{}\\boxed{\\frac34}$." } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "dear mathlinker, i have a problem \r\nlet$ x,y,z$ such that $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 1$. Prove that $ xy \\plus{} yz \\plus{} 2xz \\le \\frac {{1 \\plus{} \\sqrt 3 }}{2}$\r\n\r\nHere is my proof\r\n$ \\begin{array}{l} f(x,y,z) \\equal{} xy \\plus{} yz \\plus{} zx \\minus{} \\frac {{1 \\plus{} \\sqrt 3 }}{2} \\\\\r\nf(\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ;y,\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ) \\equal{} 2y\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} \\plus{} {x^2} \\plus{} {y^2} \\minus{} \\frac {{1 \\plus{} \\sqrt 3 }}{2} \\\\\r\nf(x,y,z) \\minus{} f(\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ;y,\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ) \\equal{} \\minus{} {(x \\minus{} z)^2}\\left( {\\frac {y}{{(x \\plus{} z) \\plus{} \\sqrt {2({x^2} \\plus{} {y^2})} }} \\plus{} 1} \\right) \\\\\r\n\\end{array}$\r\nnoitice that $ (x \\plus{} y)^2 \\le2(x^2 \\plus{} y^2)$. so we can show that \r\n\r\n\r\nif $ y > 0$ or if $ y < 0,x > 0,z > 0$ , then $ f(x,y,z) \\minus{} f(\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ;y,\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ) \\le 0$\r\nif $ y < 0$.WLOG $ x > 0,z < 0$ it follow $ xy \\plus{} yz \\plus{} 2xz < \\frac {{1 \\plus{} \\sqrt 3 }}{2}$\r\n\r\nSo, we have proved that $ f(x,y,z) \\minus{} f(\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ;y,\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ) < 0$\r\n\r\nwe have to prove that if $ x \\equal{} z \\equal{} t$ and $ y^2 \\plus{} 2t^2 \\equal{} 1$ then $ 2ty \\plus{} 2t^2 \\le \\frac {{1 \\plus{} \\sqrt 3 }}{2}$ [b] (*)[/b]\r\n\r\n[b]but i have a long proof for this, use $ f'(t)$[/b] :blush: :blush: \r\n\r\nwho can help me prove (*)\r\n\r\nThanks", "Solution_1": "[quote=\"imedeen\"]dear mathlinker, i have a problem \nlet$ x,y,z$ such that $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 1$. Prove that $ xy \\plus{} yz \\plus{} 2xz \\le \\frac {{1 \\plus{} \\sqrt 3 }}{2}$\n\nHere is my proof\n$ \\begin{array}{l} f(x,y,z) \\equal{} xy \\plus{} yz \\plus{} zx \\minus{} \\frac {{1 \\plus{} \\sqrt 3 }}{2} \\\\\nf(\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ;y,\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ) \\equal{} 2y\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} \\plus{} {x^2} \\plus{} {y^2} \\minus{} \\frac {{1 \\plus{} \\sqrt 3 }}{2} \\\\\nf(x,y,z) \\minus{} f(\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ;y,\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ) \\equal{} \\minus{} {(x \\minus{} z)^2}\\left( {\\frac {y}{{(x \\plus{} z) \\plus{} \\sqrt {2({x^2} \\plus{} {y^2})} }} \\plus{} 1} \\right) \\\\\n\\end{array}$\nnoitice that $ (x \\plus{} y)^2 \\le2(x^2 \\plus{} y^2)$. so we can show that \n\n\nif $ y > 0$ or if $ y < 0,x > 0,z > 0$ , then $ f(x,y,z) \\minus{} f(\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ;y,\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ) \\le 0$\nif $ y < 0$.WLOG $ x > 0,z < 0$ it follow $ xy \\plus{} yz \\plus{} 2xz < \\frac {{1 \\plus{} \\sqrt 3 }}{2}$\n\nSo, we have proved that $ f(x,y,z) \\minus{} f(\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ;y,\\sqrt {\\frac {{{x^2} \\plus{} {y^2}}}{2}} ) < 0$\n\nwe have to prove that if $ x \\equal{} z \\equal{} t$ and $ 2y^2 \\plus{} 2t^2 \\equal{} 1$ then $ 2ty \\plus{} 2t^2 \\le \\frac {{1 \\plus{} \\sqrt 3 }}{2}$ [b] (*)[/b]\n\n[b]but i have a long proof for this, use $ f'(t)$[/b] :blush: :blush: \n\nwho can help me prove (*)\n\nThanks[/quote]\r\nI know this is a bit off topic: Let $ \\alpha \\equal{} \\frac{1\\plus{}\\sqrt{3}}{2}$. Then $ \\frac{1}{\\alpha} \\frac{x^2 \\plus{} (\\alpha y)^2}{2} \\ge xy$, $ \\frac{1}{\\alpha} \\frac{z^2 \\plus{} (\\alpha y)^2}{2} \\ge yz$, $ x^2\\plus{}z^2 \\ge 2xz$ which gives $ xy \\plus{} yz \\plus{} 2zx \\le \\alpha(x^2\\plus{}z^2\\plus{}y^2) \\equal{} \\frac{1\\plus{}\\sqrt{3}}{2}$. Since $ 2\\alpha^2\\minus{}2\\alpha\\minus{}1 \\equal{} 0 \\iff \\frac{1}{2\\alpha}\\plus{}1 \\equal{} \\alpha$.\r\n\r\nYou have a tiny mistake: You have to prove $ 2ty \\plus{} 2t^2 \\le \\frac{1\\plus{}\\sqrt{3}}{2}$ when $ y^2 \\plus{} 2t^2 \\equal{} 1$\r\n\r\nLet $ \\alpha \\equal{} \\frac{1\\plus{}\\sqrt{3}}{2}$ again:\r\n$ 2ty \\le \\frac{1}{\\alpha} (t^2\\plus{}(\\alpha y)^2)$ Which gives $ 2ty\\plus{}2t^2 \\le \\alpha(2t^2\\plus{}y^2) \\equal{} \\frac{1\\plus{}\\sqrt{3}}{2}$", "Solution_2": "i see that your solution is not really simple :(", "Solution_3": "[quote=\"imedeen\"]i see that your solution is not really simple :([/quote]\r\nWhy don't you think so?", "Solution_4": "[quote=\"Mathias_DK\"]I know this is a bit off topic:[/quote]\r\nWhy is this off topic?", "Solution_5": "[quote=\"FelixD\"][quote=\"Mathias_DK\"]I know this is a bit off topic:[/quote]\nWhy is this off topic?[/quote]\r\nBecause he wanted a proof of (*) :P", "Solution_6": "[quote=\"Mathias_DK\"][quote=\"imedeen\"]i see that your solution is not really simple :([/quote]\nWhy don't you think so?[/quote]\r\n\r\nbecause i have a proof for (*) use $ f'$ seem nicer :) ( i think so :( )" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let AC,AB are tangents from A to (O).OA meets (O) at H ,I,D (O in segment ID,H in segment IA).Bisector of DCO meets DB,(O) at E,F;AE meets FH at K.Pover that IK perpendicular with DB", "Solution_1": "[quote=\"Herodatviet\"]OA meets (O) at H ,I,D (O in segment ID,H in segment IA)[/quote]\r\n\r\nHow on Earth does a line meet a circle in three points? :?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Show that if $ \\delta(G) \\geq 3$ then $ G$ contains a circuit with a chord.", "Solution_1": "[quote=\"ksun\"]Show that if $ \\delta(G) \\geq 3$ then $ G$ contains a circuit with a chord.[/quote]\r\nI'm sorry,but what does $ \\delta(G)$ mean?", "Solution_2": "$ \\delta (G)$ denotes, for a given graph $ G$, the minimal degree of the vertex; so if $ \\delta (G) \\geq 3$ then every vertex of the graph has a degree at least 3.", "Solution_3": "[quote=\"ksun\"]$ \\delta (G)$ denotes, for a given graph $ G$, the minimal degree of the vertex; so if $ \\delta (G) \\geq 3$ then every vertex of the graph has a degree at least 3.[/quote]\r\n[b]Proof:[/b]\r\n-Consider a path of a maximum length:$ A \\equal{} A_1A_2\\dots A_n$,\r\nObviously $ A_1$ is connected only with vertexes from $ A$(otherwise we can find a path with longer length),but degree of $ A_1$ is at least $ 3$,so there exist $ A_i,A_j\\in A$,such that $ A_1$ is connected with $ A_i$ and $ A_j$($ i,j\\neq 2$).\r\nWLOG suppose $ i > j$,but then $ A_1A_j$ is a chord in a cycle $ A_1A_2\\dots A_i$." } { "Tag": [], "Problem": "Three numbers a, b, and c, none zero, form an arithmetic progression. Increasing a by 1 or increasing c by 2 results in a geometric progression. Find b.", "Solution_1": "You posted this problem before in another forum\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=48229]http://www.mathlinks.ro/Forum/viewtopic.php?t=48229[/url]", "Solution_2": "Oups. I don't remember these things. That was a month ago though. ;) \r\n\r\nI think that this is still a good practice problem and should stay.", "Solution_3": "OK ;)\r\nSo other people, DONT click on my previous link :D", "Solution_4": "This question has not yet been answered? :huh: \r\n\r\nGuess my posts are not good... :(" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Triangle $ ABC$ with $ AB \\equal{} AC \\equal{} a$, and altitude $ AH$. Construct a circle with center $ A$, radius $ R$, $ R < a$. From points $ B$ and $ C$, draw the tangents $ BM$ and $ CN$ to this circle ($ M$ and $ N$ are the points of tangency) so that they are not symmetric with respect to the altitude $ AH$ of triangle $ ABC$. Let $ I$ be the point of intersection of $ BM$ and $ CN$.\r\n1. Find the locus of $ I$ when $ R$ varies;\r\n2. Prove that $ IB\\cdot IC \\equal{} |a^2\\minus{}d^2|$ where $ AI \\equal{} d$.", "Solution_1": "Notice BM=BM\u2019 =CN=CN\u2019 =\u3009 \u25b3BMA=\u25b3CAN =\u3009 \u2220BAC=\u2220MAN \r\n=\u3009 BACI are concyclic \r\nIB*IC=BM^2-MI^2=a^2-R^2-( d^2- R^2)= a^2- d^2\r\nWhere R is the radius of the circle A" } { "Tag": [ "search", "inequalities", "geometry", "email", "number theory", "IMO" ], "Problem": "People! \r\n\r\nI think that last year this forum ( http://www.mathlinks.ro ) was much more cooler, there were many active users which are not using this website now ... really oops!!\r\n\r\nHere is my question: \r\nwho can say what are the reasons??\r\n\r\ndavron latipov", "Solution_1": "[quote=\"Davron\"]People! \n\nI think that last year this forum ( http://www.mathlinks.ro ) was much more cooler, there were many active users which are not using this website now ... really oops!!\n\nHere is my question: \nwho can say what are the reasons??\n\ndavron latipov[/quote]\r\nHey Davron why have you disaded that now there are less active users than last year?For me I just can say that now I'm less active now than last year but I'm just have no time to spend on math,but I hope from Junuary of 2007 I will again become active in solving problems and posting solutions here:)!", "Solution_2": "That date is good for me, too! :wink:", "Solution_3": "Davron is right.\r\n\r\nWhat I really miss for example are the discussion about really difficult (but solved/solvable) nice problems (like in the beginning of MathLinks, look up thread numbers under 20000). Now most problems posted are reposts by people that don't search before. And the rest is often \"standard\" olympiad stuff, often not trivial, but still not of the type above. It got a bit better the last days (e.g. Harazi posts again), but it still is not like in earlier times (I joined more at the end of this nice period, so older users can probably judge better, but this my oppinion).\r\nThat what drove Harazi away from Inequalities is also happing in the other forums, just a bit slower.\r\nI'm not sure if one reason simply is that there are not many nice problems left in the world.\r\n\r\nI think a part of what Davron means is just the current point of a slow but constant lowering of niveau here (as all the spam shows)...\r\nBut it's difficult to express the happaning in total.", "Solution_4": "Do you believe that that there exist a way to help the forum ? :maybe:", "Solution_5": "[quote=\"silouan\"]Do you believe that that there exist a way to help the forum ? :maybe:[/quote] \r\n\r\nof course we can help the forum ...\r\n\r\ndavron latipov", "Solution_6": "Guys during the last months I wasn't so active in watching topics here and didn\u2019t really notice that this problem has grown to such big sizes, but I think the roots of this problem were seeable long ago. Saying the truth I think that one of the reasons for this is the neutral position which in some cases take moderators. What I am actually trying to say is that the moderators should sometimes be severer about persons who tries to spam here, or tries to fill themselves like they can do everything they want in this forum. This kind of behavior in some cases confusing the true funs of math and they just didn\u2019t posting here. This question is really needs to be solved and we all have to try to help moderators solving this big problem, if not in the near future our most favorite forum will notable increase in its quality and rank and soon will become like thousand other middle quality math forums in the net.\r\nThank you for attention!\r\nTigran", "Solution_7": "[quote=\"Tiks\"] What I am actually trying to say is that the moderators should sometimes be severer about persons who tries to spam here, or tries to fill themselves like they can do everything they want in this forum. [/quote]\r\nIt is true. But I see that in many forums, the moderators listed for the forum are no longer active in that forum, or even on the site at all! So perhaps it is not just because our moderators are not strict enough, but we don't have enough -- the number of moderators decreases while the number of users increases. Maybe some of us should volunteer to moderate the forums that we care about.", "Solution_8": "I generally agree with Zetax about the Old mighty Mathlinks topics(I wasnt born in mathlinks at that time , but i have seen the posts)\r\n\r\nBut Xevarion , What do you expect from the moderators , Now that the summer is over , they have their lives to cope with.\r\nAlthough, Mathlinks/AOPS site is a magnificent place to learn new facts and improve your skills, Its just an artificial world\r\nbuilt in the Memories of computers.", "Solution_9": "No, ZetaX, there are still infinitely more nice, difficult problems in the world that are still not discussed on mathlinks. This is not a reason for what is happening here. It's quite simple actually: at the beginning of mathlinks, there were much fewer users, practically all passioned about nice difficult problems and not about doing their homework or posting for 189020 times the same problem. Now, things have changed, every day there are more posts than we had in a whole week last year. Interesting enough, when you want to struggle with a hard problem, or even with an easy but nice one, you don't look through the problems posted recently (unless the topic is started by sam-n or vess or some other 2 or 3 people), but you search in the old topics (at least, this is what I'm doing). It's such a pity all the problems posted nowadays are just reformulation of old discussed problems. Why on earth do we have a search button, which works actually very well? Why on earth do we accept in the advanced area all kind of homework exercices and stuff with absolutely no value? I think I said on the forum at least 10 times that there should be created new sections, with the most valuable topics, in a few number, so that we shouldn't be obliged to search for hours for them. Of course, nothing happenned. I also asked at least 4 times to erase my name from the moderators list-name, because I have no longer the time and especially the patience to read all this boring stuff and to take actions. Nothing happened. I proposed that active users should be moderators: nothing again. Then, how can you expect the same level of discussions? Unfortunately, there are also many extremely good users who left us and this is again a pity. But how to convince them to discuss problems for homework or problems they have already seen 10 times? I really hope that mathlinks will not become just a place where each discussed topic has 4 links to previous discussions on that theme. As I said, there are still (too) many problems left!", "Solution_10": "Agreed!\r\n\r\nWhat do you think about a subforum just for threads we are sure that were posted before\u00bf\r\nOne who wants new stuff could ignore it. The difference would be that the moderators move the threads there, even when _not_ giving a link. It will be let to the proposer or friendly users to find them.\r\nAnd I would also be for the section with really good problems, but who decides\u00bf\r\n\r\n\r\nThe other is stronger actions against people reposting without any attempts of search (like even using the title gives the result).\r\n\r\nAnd well, http://www.artofproblemsolving.com/Wiki/index.php/AoPS-Mathlinks_Rules_and_Tips should finally be forced on all (new) users to be read.", "Solution_11": "wouldn't it be more practical if the moderators just started \"cleaning\" the advanced section from all kind of rubbish which flourishes among the few interesting threads?\r\nI mean taking threads as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107397]this[/url] to the preolympiad or HS basics etc. Now I don't know if that would really help solve our problem but it would at least bring some seriosity to the forum. \r\nIt is really embarracing to have such discussions in the olympiad (advanced) section. Now I understand the \"old\" users complaining for the situation at the moment :lol:", "Solution_12": "About $\\frac{2}{3}$ of the posted problems this week are ones already posted I think.\r\nWe can't simply delete them, nor can we just move them to Solved Problems without giving the author anything. Thus best is to make a subforum for this I think.\r\nAnd cause of the very good topic naming (combined with the lack of being able to search TeX-code) it's simply impossible to find anything :roll: \r\nWhen I started moderating, I regulary corrected titles, but if the user again and again post that way (some even changed them back!), you will give up or at least do it less regulary...\r\n\r\nTo the linked thread: well, at first one should delete the spam there ;)\r\nThen moving is probably best.", "Solution_13": "[quote=\"ZetaX\"]\nWe can't simply delete them, nor can we just move them to Solved Problems without giving the author anything. [/quote]\r\nZetaX I didn't get your point completely, what do you mean saying \u201cwe can\u2019t\u201d? You aren\u2019t supposed to do that or you just unable? I think if someone can delete or replace the topics (or spams) from the places that they aren\u2019t belong is the best way to get back the popularity of ML!", "Solution_14": "Well, if I want, I could, but it would not be nice to those persons who posted this (they often have no idea that it was posted before).\r\nAnother way of solving the repost-problem is deleting them and notifying the proposer that it was posted.\r\n\r\nThe problem is really caused in searching: as already said, most user don't think of it, and the way search works and how titles are named makes it hard to search.\r\nFor finding a topic I've no clue about I need more than five minutes, if I've an idea of what was posted there less (just a heuristic, I never stopped ;) ).\r\nIt would be so simple of all users would take deep searches for their stuff, it would only take them ten minutes per problem; but for the moderators, this five till ten minutes add up to hours, but if they could just move topics away without being \"forced\" to give a link, this hours will get minutes.\r\n\r\nAnd it would be nice to help moderators by always linking to older threads if possible, even at/from older topics :)\r\nA link is never spam if it has anything to do with the problem, but some people seem to see this different...\r\n\r\n\r\nWell, I will probably try a (even) stricter rule in number theory, especially deleting anything spammy without notifying. But I already see people complaining :roll:", "Solution_15": "[quote=\"harazi\"] I also asked at least 4 times to erase my name from the moderators list-name, because I have no longer the time and especially the patience to read all this boring stuff and to take actions. Nothing happened. I proposed that active users should be moderators: nothing again. Then, how can you expect the same level of discussions? Unfortunately, there are also many extremely good users who left us and this is again a pity. But how to convince them to discuss problems for homework or problems they have already seehttp://www.artofproblemsolving.com/Forum/posting.php?mode=quote&p=646797n 10 times? I really hope that mathlinks will not become just a place where each discussed topic has 4 links to previous discussions on that theme. As I said, there are still (too) many problems left![/quote]\r\n\r\nWho did you ask? Sending an email to aops@artofproblemsolving.com is the best way to get the admins' attention.", "Solution_16": "Every society,empire, forum :) etc.. has its good and bad days.\r\nLet us hope that AoPS will prosper again!!! :lol:", "Solution_17": "The reason for appearing easier and easier problems is simple: there are more people logging into forum and with growing popularity the level has to decrease, that's kind of natural to me. There are many people who after getting the problem have a strong urge to share it with the whole world.\r\n\r\nAnother subforum with extraordinary problems/solutions should be surely made. Posting new topics into it should be locked with just continuing what was already started available. Other questions is how to choose these problems as if it were only a work for moderators it would take HUGE amount of time. I'd imagine some system of notifying by users (other than PMs) would work well if made simple ,e.g.: you can just tick the problem you like if you have enough number of posts [that's the only barrier which could be made from people just doing it for fun I guess], and then moderators could see a database of votes that would contain informations on who voted and on which problem." } { "Tag": [ "LaTeX" ], "Problem": "hi\r\nI have a table like this\r\n\\begin{tabular}[hb]{c|l|c|c|c|c|}\r\n\r\nObjectifs & Sous objectifs & Paul & L-P & Kevin & P-A \\\\ \\hline\r\n\r\nD\u00e9finir le projet & Compr\u00e9hension du projet & R & R & R & R \\\\ \\cline{2-6}\r\n& D\u00e9finir les objectifs & S & R & S & S \\\\ \\cline{2-6}\r\n& Conception de la matrice des responsabilit\u00e9s & & R & & S \\\\ \\hline\r\n\r\nRecherche d'informations & Recherche(Internet, p\u00e9riodiques, etc) & R & & R & \\\\ \\cline{2-6}\r\n& Analyse de l'information trouv\u00e9e & R & & R & \\\\ \\cline{2-6}\r\n& Prise de mesures sur le moteur & A & R & & S \\\\ \\hline\r\n\r\n\\end{tabular}\r\n\r\nand i want want to enumerate the items in the first colum and have subnumber in the second colum to have something like that\r\n\r\n1. Bla 1.1 bla bla\r\n 1.2 blabla\r\n2. Buz 2.1 buzbuz\r\n 2.2 ....\r\n\r\nand so on...all in different cells\r\n\r\nalso... :blush: \r\n\r\nis there a special command to write an exponential else then \\mathrm{e}^x ?\r\n\r\nand on a title page, how can I add more info like the location or the person to who the report is given?\r\n\r\nthank you very much!!\r\n\r\nledurt", "Solution_1": "[quote]and on a title page, how can I add more info like the location or the person to who the report is given? [/quote]\r\n\r\nI mean in the same way\r\n\r\n\\title{Analyse vibratoire}\r\n\\author{Paul Doyon (04 500 800)\\and\r\nLouis-Philippe Riel (04 480 225)\\and\r\nKevin Tremblay (04 498 537)\\and\r\nPierre-Andr\u00e9 Trudel (04 515 331)}\r\n\\date{\\today}\r\n\\maketitle\r\n\r\nis there a command to add the infos I talked about?\r\n\r\nthanks again\r\nledurt" } { "Tag": [ "vector", "geometry", "geometric transformation", "reflection", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $ V$ be a finite dimensional complex vector space.\r\nLet $ T_1$, $ T_2$, ... , $ T_m$ span the dual space of $ V$. \r\n\r\nIf there exists $ x$ in $ V$ such that, for any $ y$ in $ V$,\r\n$ T_1 (x) T_1 (y) \\plus{} T_2 (x) T_2 (y) \\plus{} ... \\plus{} T_m (x) T_m (y) \\equal{} 0$\r\n\r\nShow that $ x \\equal{} 0$.", "Solution_1": "The idea is the following: consider the sum T2+...+Tm. It is easy to prove that its orthogonal is $ A_1: \\equal{} \\bigcap_{i \\equal{} 2}^m T_i^{\\perp}$, and it has to be $ \\neq 0$. Moreover, there exists $ y_1$ in $ A_1$ such that $ T_1(y_1)\\neq 0$. Using $ y \\equal{} y_1$ you show that $ T_1(x) \\equal{} 0$. So $ T_i(x) \\equal{} 0$ for all $ i$ and then x=0.", "Solution_2": "We should be able to write this entirely as matrices.\r\n\r\nWe have that $ T$ is an invertible matrix. ($ T_1,T_2,\\dots,T_n$ are the rows of that matrix.)\r\n\r\nWe are given that $ \\exists\\,x$ such that $ \\forall\\,y,\\ (Tx)^T(Ty)\\equal{}0.$ That is, $ x^TT^TTy\\equal{}0$ for all $ y.$ That means that $ x^TT^TT$ must be the zero vector, since it generates the zero functional. We can take the transpose to write this as $ T^TTx\\equal{}0.$ But since $ T$ was given to be invertible, we also have that $ T^TT$ is invertible, and hence $ x\\equal{}0.$", "Solution_3": "Yet another proof:\r\n\r\nThe condition implies that the linear functional $ \\sum_{i} T_{i}(x)T_i$ is identically zero, but since the $ T_i$ are a basis, this means that $ \\forall i$, $ T_{i}(x) \\equal{} 0$. But that means that $ x$ is mapped to zero by every single element of the dual space.\r\n\r\nThe last statement implies that $ x \\equal{} 0$, and there are a million ways to see that. Pick an inner product and consider $ \\cdot \\minus{}> < x,\\cdot >$, for example, or argue that the dual space would have dimension less than $ dim\\ V$.\r\n\r\nEdit: On further reflection, this is exactly the same as Kent's proof, though I think it still adds something to the discussion." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "[img]http://i2.photobucket.com/albums/y8/XHellsFireX/alge.png[/img]\r\n\r\nI'm not very good at Algebra. I have no idea how to go about solving both of these to figure out whether or not this is a true statement.", "Solution_1": "What is the bold equal sign in the middle supposed to be.\r\n\r\nAnyways, interpreting these equations in any of the >4ish ways possible, it seems false.\r\n\r\nThis should be moved to High School Basics. This forum is olympiad level." } { "Tag": [ "ratio" ], "Problem": "If 5 geometric means are inserted between 8 and 5832, what's the fifth term in the geometric sequence that starts with 8?", "Solution_1": "[hide=\"Solution\"] 5832/8=729, which is 3^6, conveniently. Then the common ratio is 3, and the sequence is 8,24,72,216,648,1944,5832... The fifth term is 648.[/hide]" } { "Tag": [], "Problem": "Could someone explain what an international film is, along with some examples? Too bad I don't get chance to watch movies. :(", "Solution_1": "Hmm...a film that comes from a country other than your own? I live in America and some of my favorites are Pan's Labyrinth, Delicatessen, The City of Lost Children, Amelie, Yuume Juya, Potemkin, The Bubble, The Edukators, The Seventh Seal, and Cha no Aji." } { "Tag": [ "geometry", "trigonometry", "MATHCOUNTS", "ratio" ], "Problem": "Does anybody know the area of a pentagon with all of its angles equaling 108 degrees in terms of the side length, x?", "Solution_1": "It is $\\frac{5x^2}{4\\tan{36}}$\r\n\r\nFor a regular n-gon, with side s, the area is:\r\n\r\n$\\frac{ns^2}{4\\tan{\\frac{180}{n}}}$", "Solution_2": "[quote=\"nat mc\"]It is $\\frac{5x^2}{4\\tan{36}}$[/quote]\r\n$\\tan 36^\\circ$ is a constructible number and therefore you can expressed it in terms of radicals. Challenge: find it :D", "Solution_3": "WEll, you can get\r\ntan(6*36)=tan 36", "Solution_4": "Trigonometry is definitely above the level of MATHCOUNTS. Please do not use trigonometry in this forum. :police:\r\n\r\nThere are solutions to pgpatel's question that do not require trigonometry. If nobody gives it a whirl, we'll have to move this topic up to a High School forum.", "Solution_5": "$A=\\frac{a^2}{4}\\sqrt{25+10\\sqrt{5}}$\r\n\r\nBTW, to find $\\tan{36}$ in radical form, \r\n\\[\\frac{5a^2}{4\\tan{36}}=\\frac{a^2}{4}\\sqrt{25+10\\sqrt{5}}\\]\r\n\\[\\frac{5}{\\tan{36}} \\cdot \\frac{a^2}{4}=\\frac{a^2}{4}\\sqrt{25+10\\sqrt{5}}\\]\r\n\\[\\frac{5}{\\tan{36}}=\\sqrt{25+10\\sqrt{5}}\\]\r\n\\[\\frac{5}{\\sqrt{25+10\\sqrt{5}}}=\\tan{36}\\]\r\n\\[\\tan{36}=\\frac{5\\sqrt{25+10\\sqrt{5}}}{25+10\\sqrt{5}}\\]\r\n\\[\\tan{36}=\\frac{\\sqrt{25+10\\sqrt{5}}}{5+2\\sqrt{5}}\\]\r\n\\[\\tan{36}=(\\frac{\\sqrt{25+10\\sqrt{5}}}{5+\\sqrt{20}})(\\frac{5-\\sqrt{20}}{5-\\sqrt{20}})\\]\r\n\\[\\tan{36}=(\\frac{\\sqrt{25+10\\sqrt{5}}}{5+\\sqrt{20}})(\\frac{5-\\sqrt{20}}{5-\\sqrt{20}})\\]\r\nSomeone do the simplifying for me", "Solution_6": "woah! slow down there. how did you get $A = \\frac {a^2}4 \\sqrt {25+10\\sqrt{5}}$ without knowing $\\tan 36$?\r\n\r\nalso, how would you get $\\frac{5a^2}{4\\tan{36}}=\\frac{a^2}4\\sqrt{25+10\\sqrt 5}$ without knowing $\\tan 36$ or the area formula? It seems to me that you need one or the other to get the other your way....\r\n\r\nbut anyway, I'll do the simplifying\r\n\r\n$\\tan 36 = (\\frac{\\sqrt{25+10\\sqrt 5}}{5+\\sqrt 20})(\\frac{5-\\sqrt 20}{5-\\sqrt 20})$\r\n\r\n$\\tan 36 = \\frac{5\\sqrt{25+10\\sqrt 5}-\\sqrt{500+200\\sqrt 5}}{25-20}$\r\n\r\n$\\tan 36 = \\frac{5\\sqrt{25+10\\sqrt 5}-10\\sqrt{5+2\\sqrt 5}}5$\r\n\r\n$\\tan 36 = \\sqrt{25+10\\sqrt 5}-2\\sqrt{5+2\\sqrt 5}$\r\n\r\n$\\tan 36 = \\sqrt 5\\sqrt{5+2\\sqrt 5}-2\\sqrt{5+2\\sqrt 5}$\r\n\r\n$\\tan 36 = (\\sqrt 5 - 2)(\\sqrt{5+2\\sqrt 5})$\r\n\r\n$\\tan 36 = (\\sqrt {9-4\\sqrt 5})(\\sqrt{5+2\\sqrt 5})$\r\n\r\n$\\tan 36 = \\sqrt {45 + 18\\sqrt 5 - 20\\sqrt 5 - 40}$\r\n\r\n$\\tan 36 = \\sqrt {5 - 2\\sqrt 5}$\r\n\r\nIf you can simplify that rad I don't know how...", "Solution_7": "I thought that I'd post where this came from. It comes from the equation to find the area of a regular n-gon, which is:\r\n\r\nasn, or apothem x side-length x number of sides. This can also be shown as ap or apothem x perimeter. \r\n \r\n(*the apothem is line from the center of the n-gon to one of its vertices)\r\n\r\nBy drawing out the apothems, you get several congruent triangles. The trigonometry that shouldn't be posted in this forum, :P , is used to find the area of one of these triangles, and then you multiple by the side length to get the total area.", "Solution_8": "Although the original question by pgpatel is answered, the solution required methods beyond MATHCOUNTS level.\r\n\r\nMods: pls move to appropriate level", "Solution_9": "[quote=\"Hamster1800\"]woah! slow down there. how did you get $A = \\frac {a^2}4 \\sqrt {25+10\\sqrt{5}}$ without knowing $\\tan 36$?\n[/quote]\r\n\r\n$A = \\frac {a^2}4 \\sqrt {25+10\\sqrt{5}}$ is the formula for the area of a regular pentagon. I got it from MathWorld. It has nothing to do with nat mc's formula, except for the fact that it gives the same result :lol:", "Solution_10": "but how would you derive the formula $A = \\frac{a^2}4\\sqrt{25+10\\sqrt5}$ without $\\tan 36$? I'll check mathworld but i still don't get it...\r\n\r\nalso, I don't think any regular polygons except triangles, squares, and hexagons are in mathcounts. I could be wrong though.", "Solution_11": "no, I don't think you'll need it. If they DO involve pentagons on MC problems though, chances are that they'll give you the formula, that or finding the area will be unnessesary", "Solution_12": "[color=darkblue][size=75]Moved from the Middle School MATHCOUNTS Forum by moderator rcv[/size][/color]", "Solution_13": "Definitely is. \r\n\r\nMy class at school was just introduced to it, and we're in 10th grade, pre - IB (the highest level at the school).\r\n\r\nNot saying I didn't learn it a bit before then, but that's how schools around here, and I assume most places, work.", "Solution_14": "Well, now that trig is legal, I'm still wondering how you would get the formula for area without trig. I'm watching the other post, and the golden ratio is there, but I don't quite get how you can find that that is the golden ratio (the pentagram inside the pentagon) or how you can find stuff using angle measures without using trig. I think trig is the easiest way but other ways work.", "Solution_15": "This stuff was in Mathcounts ??? Anyone telling as to how you can solve the stuff without using trignometry ??\r\nP.S : Problem is , now that I do know trignometry , that's the first thing that comes into my mind when I see such problems..... :(", "Solution_16": "well the way to get it without trig is in mathcounts now but I think trig is the easiest way :)" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "In a finite group $G$, whose order is not divisible by 3, for any $a,b\\in G$ we have $(ab)^{3}=a^{3}b^{3}$. Prove that $G$ is abelian.", "Solution_1": "${(ab)^{3}=a^{3}b^{3}}\\Rightarrow{(ba)^{2}=a^{2}b^{2}}\\Rightarrow{b^{3}a^{3}=(ba)^{3}=baa^{2}b^{2}}\\Rightarrow{b^{2}a^{3}b^{-2}=a^{3}}$. \r\nIf $|G|=3k-r,r\\in\\{1,2\\}$ then by the previous $b^{2}a^{3k}b^{-2}=a^{3k}\\Rightarrow{}b^{2}a^{r}b^{-2}=a^{r}$ i.e. $b^{2}a^{2}=a^{2}b^{2}$, whichever $r\\in\\{1,2\\}$ hence $b^{2}a^{2}=a^{2}b^{2}=(ba)^{2}\\Rightarrow{ba=ab}$", "Solution_2": "[quote=\"finister\"]... ${=(ba)^{3}=ba^{2}b^{2}a}$ ...[/quote]\r\nNo.\r\nYou can't prove it without using the second condition, there are counterexamples (even finite ones).", "Solution_3": "[quote=\"ZetaX\"]You can't prove it without using the second condition, there are counterexamples (even finite ones).[/quote]\r\n\r\nYou are right, I've corrected the proof. Sorry for not paying enough attention.", "Solution_4": "let F endmorphisme in G such that F(x)=x^3. .we have ord(G)=3k+1or3k-1 then we can prove easly that F is surjective.and we know that if a endomorphisme is surjective in G.then G is abelian. :ninja:", "Solution_5": "[quote=\"fermat3\"]and we know that if a endomorphisme is surjective in G.then G is abelian. :ninja:[/quote]\r\n\u00bf\u00bf\u00bf\r\nYou need to make some restrictions on it ;)" } { "Tag": [ "puzzles" ], "Problem": "This problem requires logical thinking, so get ready.\r\n\r\n3 Mathematicians (we call them this so that they can be considered able to figure out any logical condition) are standing in a line so that the last one can see two in front, the middle can see the one in front, and the first can see none.\r\n\r\nA hat-seller comes along with 5 hats - 3 white, 2 black. He shows them to the 3 mathematicians. He places 1 on each head randomly and hides the remaining 2. Then he says:\r\n\r\n\"I will give as much money as could possibly be wanted to the first mathematician that can name the color of the hat on his head. If you cannot justify your solution, you will not get the money.\"\r\n\r\nNo mathematician talks, turns around, or communicates in any way with anybody, looks in any mirror, or anything. [b]After a long pause[/b], one mathematician speaks up. Who is he? He states the color of the hat on his head. What color is it? He gives his justification. What is it?\r\n\r\nThe solution to this problem is under the topic \"Solution to 3 mathemeticians\" which will be posted in a few minutes. If you have a question, post it. If you have a solution, do [i]not[/i] post it. Check your answer. If you want a hint, the biggest hint is in this problem in bold.", "Solution_1": "This problem belongs in the Puzzles and Brainteasers forum.", "Solution_2": "ok suppose the first peron has a black hat, if the second person has a black hat then the third person will know right away. but if the third person doesn't know right away then the second person will know that he must have a white hat. So thus, if the second and third people don't know fairly fast then the first person knows that he must have a white hat." } { "Tag": [], "Problem": "Hey guys :) I am in year 9 and was interested in studying for the Australian mathematics competition and the best books i have seen around are the AoPS volume 1 and 2. I was wondering are these books suitable for an Australian maths comp as they are written in the USA?\r\nAn example of some practice questions are here http://www.amt.edu.au/wuamcqi04.pdf or http://www.itute.com/mathline/2006_aus_maths_comp_int_sol.pdf.(the second link is only solutions which may give an idea of the difficulty )\r\nOf what i have seen of volume 1 it seems to be all high school basics, but are there any problem solving tricks in there?\r\n :( Any help would be appreciated!", "Solution_1": "woops :blush: sorry guys i was actually asking for a friend that is in year 9, i am actually in year 12........dam it now i look like an idiot.", "Solution_2": "Please don't curse.\r\n\r\nIt should be fine for him.", "Solution_3": "sorry. Are you offended because your christian?", "Solution_4": "The content should be fine. There aren't any specifically Australian problems, but math contests have more in common with each other than with the regular curriculum anywhere.\r\nThe question I don't know the answer to- does AOPS ship them outside North America?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Show that there are no positive integers $ a,b$ such that both $ a^2\\plus{}b^2$ and $ a^2\\minus{}b^2$ are perfect squares.", "Solution_1": "$ a^2 \\minus{} b^2 \\equal{} n^2$ and$ a^2 \\plus{} b^2 \\equal{} m^2$ so we have $ m^4 \\minus{} n^4 \\equal{} (2ab)^2$\r\nnow i prove that this equation dont have solution in $ N$:$ x^4 \\minus{} y^4 \\equal{} z^2$\r\nsuppose the inverse ,take on solution such that $ x$ is minimum($ x > 0$).\r\nwe have$ (x^2 \\minus{} y^2)(x^2 \\plus{} y^2) \\equal{} z^2$ but $ (x^2 \\minus{} y^2)$ and$ (x^2 \\plus{} y^2)$ are coprime.\r\nso we have :\r\n$ (x^2 \\minus{} y^2) \\equal{} s^2$\r\n$ (x^2 \\plus{} y^2) \\equal{} r^2$\r\nthis means that:$ r^4 \\minus{} s^4 \\equal{} (2xy)^2$ but we have $ r < x$.", "Solution_2": "Nice, short and simple. Thank you for the beautiful solution. :)", "Solution_3": "[quote=\"cfheolpiixn\"][quote=\"sumita\"]$ a^2 \\minus{} b^2 \\equal{} n^2$ and$ a^2 \\plus{} b^2 \\equal{} m^2$ so we have $ m^4 \\minus{} n^4 \\equal{} (2ab)^2$\n$\\boxed{\\text{now I prove that this equation dont have solution in } \\mathbb{N}: x^4 \\minus{} y^4 \\equal{} z^2}$\nsuppose the inverse ,take on solution such that $ x$ is minimum($ x > 0$).\nwe have$ (x^2 \\minus{} y^2)(x^2 \\plus{} y^2) \\equal{} z^2$ but $ (x^2 \\minus{} y^2)$ and$ (x^2 \\plus{} y^2)$ are coprime.\nso we have :\n$ (x^2 \\minus{} y^2) \\equal{} s^2$\n$ (x^2 \\plus{} y^2) \\equal{} r^2$\nthis means that:$ r^4 \\minus{} s^4 \\equal{} (2xy)^2$ but we have $ r < x$.[/quote]\nI think the last line $ rx $, which makes the descending method fails.\n\nDoes anyone come up with other ways to prove it?[/quote]", "Solution_4": "another solution:\n[img]http://c.upanh.com/upload/6/902/3L0.11087339_1_1.bmp[/img]\n[img]http://c.upanh.com/upload/6/902/LT0.11087673_1_1.bmp[/img]\n[img]http://c.upanh.com/upload/6/902/7V0.11087736_1_1.bmp[/img]\n[img]http://c.upanh.com/upload/6/902/2L0.11087321_1_1.bmp[/img]" } { "Tag": [ "LaTeX" ], "Problem": "in $$ mode, if you want to include text into it by \\text{}, the spacing between the math stuff and the text comes out really weird. is there a way to fix that?", "Solution_1": "Have you put spaces at the beginning and end of the text box ie \\text{ this is text }? If that's not the problem could you show us a minimal example? It looks OK here: \\[\\sqrt{2}\\text{ this is text }\\frac{1}{2}\\]", "Solution_2": "thanks, that did the trick :)" } { "Tag": [ "parameterization", "probability", "geometry", "parallelogram", "integration", "calculus", "probability and stats" ], "Problem": "Let X and Y be independent random variables with an exponential distribution with parameters \u00b5 and \u03bb. Let U = min{X,Y} and V = max{X,Y}. Also, let W = V - U.\r\n\r\nShow that U and W are independent.\r\n\r\nI'm aware that in order to show this, I have to show that E(UW)= E(U)*E(W). But I'm unsure of how to go about it so any help would be great. Thanks in advance!", "Solution_1": "[quote=\"SharkBites77\"]I'm aware that in order to show this, I have to show that E(UW)= E(U)*E(W).[/quote]\r\n\r\nActually, no. Showing that $ E(UW)\\equal{}E(U)(E(W)$ would show that $ U$ and $ W$ are [i]uncorrelated[/i], but it would not show that they are [i]independent.[/i] Independent does imply uncorrelated, but uncorrelated does not imply independent.\r\n\r\nHere's a small example: Let $ (X,Y)$ be random variables distributed in such a way that the points $ (1,0),(0,1),(\\minus{}1,0),$ and $ (0,\\minus{}1)$ each have probability $ \\frac14$. Then $ E(X)\\equal{}0,E(Y)\\equal{}0,$ and $ E(XY)\\equal{}0.$ Hence $ \\text{Cov}(X,Y)\\equal{}0$ and they are uncorrelated. But they are distinctly not independent. For instance, $ P(X\\equal{}0)\\equal{}\\frac12,P(Y\\equal{}0)\\equal{}\\frac12,$ but $ P(X\\equal{}0\\text{ and }Y\\equal{}0)\\equal{}0\\ne\\frac12\\cdot\\frac12.$", "Solution_2": "A suggestion: show that $ X\\minus{}U$ and $ Y\\minus{}U$ have the same joint distribution as $ X$ and $ Y$.", "Solution_3": "[quote=\"mlok\"]A suggestion: show that $ X \\minus{} U$ and $ Y \\minus{} U$ have the same joint distribution as $ X$ and $ Y$.[/quote]\r\n\r\nThanks for the suggestion! Thank you to Kent as well for correcting my assumption.\r\n\r\nI'm guessing that to show that X-U and Y-U have the same joint distribution as X and Y, I have to perform some kind of integration. I'm unsure of what expression I am integrating, though.", "Solution_4": "I take my suggestion back: $ (X\\minus{}U,Y\\minus{}U)$ is not distributed in the same way as $ (X,Y)$. Let's use this definition of [url=http://en.wikipedia.org/wiki/Statistical_independence]independence[/url]: $ U$ and $ W$ are independent if for any real numbers $ a$ and $ b$ we have \r\n$ P[U\\le a\\, \\&\\, W\\le b]\\equal{}P[U\\le a]P[W\\le b]$. \r\n\r\nNow, $ P[U\\le a\\, \\&\\, W\\le b]\\equal{}\\mu\\lambda \\iint_{D}e^{\\minus{}\\mu x}e^{\\minus{}\\lambda y}\\,dx\\,dy$ where $ D\\equal{}\\{(x,y)\\colon \\min(x,y)\\le a, |x\\minus{}y|\\le b\\}$. Since $ D$ is the union of two parallelograms, we have \r\n$ P[U\\le a\\, \\&\\, W\\le b]\\equal{}\\mu\\lambda\\int_0^{a}e^{\\minus{}\\lambda y}dy\\int_{y}^{y\\plus{}b}e^{\\minus{}\\mu x}dx\\plus{}\\mu\\lambda\\int_0^{a}e^{\\minus{}\\mu x}dx\\int_{x}^{x\\plus{}b}e^{\\minus{}\\lambda y}dy$\r\nThe first integral is equal to $ \\frac{\\lambda}{\\mu\\plus{}\\lambda}(1\\minus{}e^{\\minus{}\\mu b})(1\\minus{}e^{\\minus{}(\\mu\\plus{}\\lambda)a})$, the second is $ \\frac{\\mu}{\\mu\\plus{}\\lambda}(1\\minus{}e^{\\minus{}\\lambda b})(1\\minus{}e^{\\minus{}(\\mu\\plus{}\\lambda)a})$. \r\nLetting $ b\\to \\infty$, we get $ P[U\\le a]\\equal{}1\\minus{}e^{\\minus{}(\\mu\\plus{}\\lambda)a}$. Keeping $ b$ fixed and letting $ a\\to\\infty$, we get a formula for $ P[W\\le b]$ which finishes the proof.", "Solution_5": "Thanks! But if you let a approach infinity while keeping b fixed, you just get \r\n\r\n1-e^(-\u00b5b), right?", "Solution_6": "No, you get $ \\frac{\\lambda}{\\mu\\plus{}\\lambda}(1\\minus{}e^{\\minus{}\\mu b})\\plus{} \\frac{\\mu}{\\mu\\plus{}\\lambda}(1\\minus{}e^{\\minus{}\\lambda b})$." } { "Tag": [], "Problem": "hameye tavabe'e $ f :\\mathbb{R}\\to\\mathbb{R}$ ra peyda konid ke baraye har $ x,y$ haghighi :\r\n$ f(yf(x)\\plus{}x)\\equal{}xf(y)\\plus{}f(x)$", "Solution_1": "soale ghashangie...\r\n\r\naval az hame gharar bedid $ y=0$ pas darim:\r\n\r\n$ f(x)=f(x)+xf(0)$ baraye har $ x$e haghighi,pas $ f(0)=0$...\r\n\r\nhamchenin age baraye $ t\\in\\mathbb{R}$ dashte bashim $ f(t)=0$ oon moghe age gharar bedim $ x=t$ natije mishe:\r\n\r\n$ f(t)=tf(y)\\Rightarrow tf(y)=0$ baraye har $ y$e haghighi,pas ya $ t=0$ ya $ f\\equiv 0$ pas tabe $ f\\equiv 0$ ye javabe...\r\n\r\nhala farz konid $ f\\not\\equiv0$ oon moghe natije mishe ke $ t=0$ yeni $ f(t)=0\\iff t=0$...\r\n\r\nkhob hala eddea mikonim ke $ f(1)=1$e...\r\n\r\n[hide=\"esbate eddea\"]\nage $ f(1)\\not= 1$ bashe oon moghe gharar bedid $ x=1$ va $ y=\\frac{1}{1-f(1)}$ ke chon $ f(1)\\not= 1$ pas $ y\\in\\mathbb{R}$ hala deghat konid ke ba in maghadire $ x,y$ darim:\n\n$ x+yf(x)=y$\n\npas moadele aslimoon baraye in $ x,y$ be shekle zir dar miad:\n\n$ f(y)=f(1)+f(y)$ pas $ f(1)=0$ amma tebghe oon chizi ke dar avvale esbatemoon goftim natije mishe ke $ 1=0$ ke inam tanaghoze pas farze $ f(1)\\not=1$ ghalat boode yeni $ f(1)=1$...\n---------------------------------------------------------\n[/hide]\n\nkhob pas ta inja natije shod ke $ \\boxed{f(t)=0\\iff t=0}$ va $ \\boxed{f(1)=1}$...\n\nkhob hala too soorate masale gharar bedid $ x=1$,pas darim:\n\n$ f(y+1)=f(y)+1$ pas $ \\boxed{f(y+1)-f(y)=1}$\n\nhamchenin be komake esteghra natije mishe ke \n\n$ \\boxed{\\forall y\\in\\mathbb{Z}: f(y)=y}$\n\nkhob hala too soorate masale age farz konim $ x\\in\\mathbb{Z}$ oon moghe khahim dasht:\n\n$ f(x+xy)=x+xf(y) (I)$ baraye har $ x$e sahih va $ y$e haghighi...\n\namma deghat konid ke ghablan dashtim $ f(y+1)-f(y)=1$ pas azin natije mishe ke $ f(x+xy)-f(xy)=x (II)$ baraye $ x$haye sahih va $ y$haye haghighi...\n\nhala az $ (I),(II)$ natije mishe ke:\n\n$ f(xy)=xf(y) (*)$\n\nkhob hala gharar bedid $ x=-1$ pas $ \\boxed{f(-y)=-f(y)}$ baraye har $ y$e haghighi...\n\nkhob hala eddea mikonim ke tabamoon jamie yani $ f(a+b)=f(a)+f(b)$...\n\n[hide=\"esbate eddea\"]\n\nbebinid darim:\n\n$ a=\\frac{a+b}{2}+\\frac{a-b}{2}=\\frac{a+b}{2}+\\frac{\\frac{a-b}{2}}{f(\\frac{a+b}{2})}.f(\\frac{a+b}{2})$\n\namma deghat konid ke too soorate masale age gharar bedim $ x=\\frac{a+b}{2}$ va $ y=\\frac{\\frac{a-b}{2}}{f(\\frac{a+b}{2})}$ natije mishe ke:\n\n$ f(a)=f(x+yf(x))=f(x)+xf(y)$\n\nbe hamin tartib darim:\n\n$ b=\\frac{a+b}{2}+\\frac{\\frac{b-a}{2}}{f(\\frac{a+b}{2})}.f(\\frac{a+b}{2})$\n\npas ba hamoon estedlale ghabli natije mishe ke:\n\n$ f(b)=f(x-yf(x))=f(x)+xf(-y)=f(x)-xf(y)$\n\npas khahim dasht:\n\n$ f(a)+f(b)=2f(x)=2f(\\frac{a+b}{2})$\n\namma too rabete $ (*)$ age gharar bedim $ x=2,y=\\frac{a+b}{2}$ natije mishe ke:\n\n$ f(a)+f(b)=2f(\\frac{a+b}{2})=f(a+b)$ yeni hamoon chizi ke mikhastim...\n\n--------------------------------------------------------\n[/hide]\r\n\r\nkhob pas didim ke $ f(a+b)=f(a)+f(b)$ baraye har $ a,b$e haghighi,pas soorate masalaro age bazesh konim mishe:\r\n\r\n$ f(x+yf(x))=f(x)+f(yf(x))=f(x)+xf(y)$\r\n\r\npas darim:\r\n\r\n$ f(yf(x))=xf(y)$hala gharar bedid $ y=1$ natije mishe:\r\n\r\n$ f(f(x))=x$ khob hala biaid too soorate masale bejaye $ x$ gharr bedid $ f(x)$ mishe:\r\n\r\n$ f(f(x)+yf(f(x)))=f(f(x))+f(x)f(y)$\r\n\r\nke age bazesh konim mishe:\r\n\r\n$ f(f(x))+f(xy)=f(f(x))+f(x)f(y)$ ke age sadash konim mishe:\r\n\r\n$ f(xy)=f(x)f(y)$ baraye har $ x,y$e haghighi,yeni tabamoon zarbi ham hast...\r\n\r\npas sabet shod ke tabamoon zarbie ghablanam sabet shod ke jamie pas tabamoon tabe hamanie yeni:\r\n\r\n$ \\forall x\\in\\mathbb{R}: f(x)=x$...\r\npas in mnasale dota javab dare:\r\n$ f\\equiv 0,f(x)=x$..." } { "Tag": [], "Problem": "This is the thread for it. Players and rules for it are on the bottom of the second page of the main thread. Remember, only posts before 5:00 PM EDT by the AoPS Clock count (as in, 4:59 counts, 5:00 doesn't).", "Solution_1": "k i posted what about ju...", "Solution_2": "Alright. I've posted... =]", "Solution_3": "I've posted as well.", "Solution_4": "This is my post.\r\n\r\nBananas are yummy.", "Solution_5": "I think this will be the last post.", "Solution_6": "I believe this will be.." } { "Tag": [ "quadratics", "Vieta", "algebra", "polynomial" ], "Problem": "This is probably not that hard though... but I need help.\r\n\r\nThis is the question:\r\n\r\n\"Let m and n be the roots of the quadratic equation 4x^2-5x+3=0 Find (m+7)(n+7)\"\r\n\r\nYou only need to do the part before \"Find (m+7)(n+7)", "Solution_1": "For these types of questions in general, you don't need to [hide=\"do this\"]solve for the actual roots.[/hide]\n\n[hide=\"Solution\"]By Vieta's Equations, $ m \\plus{} n \\equal{} 5/4$ and $ mn \\equal{} 3/4$. Therefore, $ (m \\plus{} 7)(n \\plus{} 7) \\equal{} mn \\plus{} 7(m \\plus{} n) \\plus{} 49 \\equal{} 3/4 \\plus{} 35/4 \\plus{} 49 \\equal{} \\frac {117}{2}$.[/hide]", "Solution_2": "Thanks darkdieuguerre!", "Solution_3": "[hide]\n\nBy using the quadratic equation you can find that the roots are $ \\frac{5\\plus{}\\sqrt{23}i}{8}$ and $ \\frac{5\\minus{}\\sqrt{23}i}{8}$. adding 7 to each and multiplying yields\n$ (\\frac{61\\plus{}\\sqrt{23}i}{8})(\\frac{61\\minus{}\\sqrt{23}i}{8})$\n\n$ \\frac {61^2 \\plus{} 23}{64}$\n\n$ \\frac {3744}{64}$\n\n$ \\frac {117}{2}$\n[/hide]", "Solution_4": "Yes in this case you could do both things (darkdieuguerre's and zserf's), but in other problems that give you a polynomial with a greater degree (cubic,quartic,and so on), you won't be able to find the roots at all! In those cases, you want to use Vieta's formulas (and later, Newtons Sums).\r\n\r\n[url]http://mathworld.wolfram.com/VietasFormulas.html[/url]", "Solution_5": "[hide=\"Yet another solution\"]If the roots of $ 4x^2\\minus{}5x\\plus{}3$ are $ m$ and $ n$, then $ m\\plus{}7$ and $ n\\plus{}7$ are the roots of $ 4(x\\minus{}7)^2\\minus{}5(x\\minus{}7)\\plus{}3\\equal{}4x^2\\minus{}61x\\plus{}234$.\nThen by Vieta's, the product of the roots of that thing is $ \\frac{234}{4}\\equal{}\\boxed{\\frac{117}{2}}$.[/hide]\r\n\r\ndarkdieuguerre's way is cleaner though.\r\nAnd zserf's way is the ugliest possible approach, though it still works ;)", "Solution_6": "[hide=\"Even Another Solution\"]\nWe can write $ f(x) \\equal{} 4x^2 \\minus{} 5x \\plus{} 3 \\equal{} 4(x \\minus{} m)(x \\minus{} n) \\equal{} 4(m \\minus{} x)(n \\minus{} x)$. We have $ f(\\minus{}7) \\equal{} 4(\\minus{}7)^2 \\minus{} 5(\\minus{}7) \\plus{} 3 \\equal{} 196 \\plus{} 35 \\plus{} 3 \\equal{} 234$. Thus, $ (m \\plus{} 7)(n \\plus{} 7) \\equal{} \\frac{4(m \\minus{} (\\minus{}7))(n \\minus{} (\\minus{}7))}{4} \\equal{} \\frac{f(\\minus{}7)}{2} \\equal{} \\frac{234}{4} \\equal{} \\boxed{\\frac{117}{2}}$.\n[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x_1,x_2,\\ldots,x_n$ be real numbers greater than 1. Show that \\[ \\frac{x_1x_2}{x_3}\\plus{}\\frac{x_2x_3}{x_4}\\plus{}\\cdots\\plus{}\\frac{x_nx_1}{x_2}\\ge4n\\] and determine when the equality holds.", "Solution_1": "[quote=\"Johan Gunardi\"]Let $ x_1,x_2,\\ldots,x_n$ be real numbers greater than 1. Show that\n\\[ \\frac {x_1x_2}{x_3} \\plus{} \\frac {x_2x_3}{x_4} \\plus{} \\cdots \\plus{} \\frac {x_nx_1}{x_2}\\ge4n\n\\]\nand determine when the equality holds.[/quote]\r\n\r\ntry $ x_1\\equal{}x_2\\equal{}\\cdots \\equal{}x_n\\equal{}2$, $ LHS\\equal{}2n<4n$..", "Solution_2": "The correct problem is:\r\n\r\nLet $ x_1,x_2,\\dots,x_n$ be real numbers such that $ x_i>1$ for $ i\\equal{}1,2,3,\\dots,n$. Prove that \r\n\\[ \\frac{x_1x_2}{x_3\\minus{}1}\\plus{}\\frac{x_2x_3}{x_4\\minus{}1}\\plus{}\\frac{x_3x_4}{x_5\\minus{}1}\\plus{}\\dots \\plus{} \\frac{x_nx_1}{x_2\\minus{}1} \\ge 4n\\] and determine when the equality occurs. \r\n\r\n\r\nI've ever posted it before...", "Solution_3": "I think n should be greater than 2 :blush:", "Solution_4": "Let $ y_i\\equal{}x_i\\minus{}1$ for $ i\\equal{}1,2,...,n$. Then the inequality is\r\n\r\n\\[ \\sum \\frac{y_iy_{i\\plus{}1}\\plus{}y_i\\plus{}y_{i\\plus{}1}\\plus{}1}{y_{i\\plus{}2}}\\equal{}\\sum \\frac{y_iy_{i\\plus{}1}}{y_{i\\plus{}2}}\\plus{}\\frac{1}{y_{i\\plus{}2}}\\plus{}\\sum \\frac{y_i}{y_{i\\plus{}2}}\\plus{}\\frac{y_{i\\plus{}1}}{y_{i\\plus{}2}}\\]Using the AM-GM twice, \\[ \\ge 2\\sum 2\\sqrt{ \\frac{y_iy_{i\\plus{}1}}{y_{i\\plus{}2}^2} }\\ge 4n\\sqrt[n]{\\prod\\frac{y_iy_{i\\plus{}1}}{y_{i\\plus{}2}^2}}\\equal{}4n\\] by the AM-GM on each sum.\r\n\r\n(an example of equality is when $ y_i\\equal{}1$ or $ x_i\\equal{}2$ $ i\\equal{}1,2,...,n$.", "Solution_5": "[quote=\"Altheman\"]Let $ y_i \\equal{} x_i \\minus{} 1$ for $ i \\equal{} 1,2,...,n$. Then the inequality is\n\\[ \\sum \\frac {y_iy_{i \\plus{} 1} \\plus{} y_i \\plus{} y_{i \\plus{} 1} \\plus{} 1}{y_{i \\plus{} 2}} \\equal{} \\sum \\frac {y_iy_{i \\plus{} 1}}{y_{i \\plus{} 2}} \\plus{} \\frac {1}{y_{i \\plus{} 2}} \\plus{} \\sum \\frac {y_i}{y_{i \\plus{} 2}} \\plus{} \\frac {y_{i \\plus{} 1}}{y_{i \\plus{} 2}}\n\\]\nUsing the AM-GM twice,\n\\[ \\ge 2\\sum 2\\sqrt {\\frac {y_iy_{i \\plus{} 1}}{y_{i \\plus{} 2}^2} }\\ge 4n\\sqrt [n]{\\prod\\sqrt{\\frac {y_iy_{i \\plus{} 1}}{y_{i \\plus{} 2}^2}} }\\equal{} 4n\n\\]\nby the AM-GM on each sum.\n\n(an example of equality is when $ y_i \\equal{} 1$ or $ x_i \\equal{} 2$ $ i \\equal{} 1,2,...,n$.[/quote]", "Solution_6": "See also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=240954" } { "Tag": [ "modular arithmetic", "number theory", "divisibility tests" ], "Problem": "So I have this problem:\r\nShow that n(n^2-1) is divisible by 24 when n is an odd number.\r\nAn odd number is always in the form 2k\u00b11.\r\nSo (2k\u00b11)(4k^2\u00b14k+1-1) = (2k\u00b11)(4k^2\u00b14k) = (8k^3\u00b112k^2\u00b14k).\r\nI know that's divisible by 8, 2, and 1. But how do I finally prove that it is divisible by 24?", "Solution_1": "Here is a hint:\r\n\r\n[hide]Check $n(n^2-1) \\pmod 3$[/hide]\n\nSolution:\n\n[hide]Check it mod 3. We get 0 for n=0, 1(1-1)=0 for n=1, 2(4-1)=6=0 for n=2, so its always divisible by 3.[/hide]", "Solution_2": "Considering the post by younglee1989, maybe he hasn't learned modulus yet...\r\n[hide=\"\"Here is an old-fashioned way to check divisibility by 3\"\"]\nWrite $n(n^2-1)=(n-1)n(n+1)$. So we have a product of 3 consecutive integers. But, among any 3 consecutive integers one of them is always bound to be divisible by 3!!![/hide]\r\nHopefully my observation helps out...", "Solution_3": "I see... I will try to work this one through using the second post.", "Solution_4": "here's how i did it...\r\n\r\nu can factor this into (n-1)n(n+1). Since n is an odd number, therefore (n-1) and (n+1) must both be even. We know that these 3 numbers are consecutive integers, so amongst them one must be divisible by 3. Also, with 2 consecutive even integers, one must be divisible by 4. Thus, u have one number that's divisible by 2, one by 3 and another by 4...therefore multiplying these together, the number (n-1)n(n+1) must be divisible by 2*3*4=24.\r\n\r\nHOpe this helps.\r\nCheers!\r\n\r\n\r\nP.S: how do you hide answers, like tetrahedr0n and 3X.lich did above?", "Solution_5": "[quote=\"mandy_pal\"]P.S: how do you hide answers, like tetrahedr0n and 3X.lich did above?[/quote]\r\nJust highlight your answers and click Hide (right above where you're typing).", "Solution_6": "Not the best solution, but a solution none the less.\r\n\r\nn(n 2 -1) = 0 mod 24 is what we are trying to prove for odd n.\r\n\r\nIf you plug 2k+1 = n then you can simplify n(n 2 -1) to 4k(k+1)(2k+1).\r\nIf you recognize that k(k+1)(2k+1)/6 is the sum of all squares you can see that k(k+1)(2k+1) is always divisible by 6.\r\n\r\nObviously any multiple of 6 multiplied by 4 will always be evenly divisible by 24.", "Solution_7": "Since n(n^2-1)=(n-1)n(n+1)\r\nWhen n is odd, (n-1) and (n+1) are even. One of them is divisible by 4. So it is divisible by 8.\r\nSince n-1 n and n+1 are 3 consective numbers, one of them must be divisible by 3.\r\nSo it is divisible by 24." } { "Tag": [], "Problem": "If $ (x \\plus{} y)^2 \\equal{} 105$ and $ x^2 \\plus{} y^2 \\equal{} 65$, what is the value of $ xy$?", "Solution_1": "first simplify into x^2+2xy+y^2=105. Subtract 40 from each side.\r\nx^2+2xy+y^2-40=65\r\n65=x^2+y^2\r\nSo x^2+2xy+y^2-40=x^2+y^2\r\n2xy-40=0\r\n2xy=40\r\n[b][size[/b]=18]xy=20[/size]", "Solution_2": "Alternatively, note that $ x^2 \\plus{} y^2$ factorizes as $ (x \\plus{} y)^2 \\minus{} 2xy$, so we can substitute this in our second equation to get $ 105 \\minus{} 2xy \\equal{} 65 \\implies \\minus{} 2xy \\equal{} \\minus{}40 \\implies xy \\equal{} \\boxed{20}$." } { "Tag": [ "calculus", "integration", "inequalities", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Prove the following inequality.\r\n\r\n\\[\\frac{e-1}{n+1}\\leqq\\int^e_1(\\log x)^n dx\\leqq\\frac{(n+1)e+1}{(n+1)(n+2)}\\ (n=1,2,\\cdot\\cdot\\cdot) \\]", "Solution_1": "How did you prove upper bound?\r\nI can only obtain something like $\\frac{e}{n+1}-\\frac{1}{e^n(n+1)}$. And I don't want to make further attemps too complicated.", "Solution_2": "Let $I_n=\\int^e_1( \\log x)^n dx$,by integral by parts,we have $I_{n+1}=e-(n+1)I_n$. As $I_{n+1}\\leqq I_{n }\\eqq\\cdots\\leqq I_{2}\\leqq I_{1}$,\r\n\r\nwe obtain $I_{n+1}\\leqq 1$,thus $I_{n+2}=e-(n+2)I_{n+1}=e-(n+2)\\left[e-(n+1)I_{n}\\right]\\leqq 1$,we are done.\r\n\r\nkunny" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": ":arrow: every triangle prove that: $ 9R^2$\\leq$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} (a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2$", "Solution_1": "[b]ABRORBEK wrote:[/b]\r\n$ 9.R^2\\le a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} (a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2. \\ \\ \\ (1).$ \r\nIf denote: $ p\\equal{}\\frac{a\\plus{}b\\plus{}c}{2}$ and $ p \\minus{} a \\equal{} x > 0, p \\minus{} b \\equal{} y > 0, p \\minus{} c \\equal{} z > 0$, then:\r\n$ (1) \\Leftrightarrow 9.(x \\plus{} y)^2.(x \\plus{} z)^2.(y \\plus{} z)^2\\le 64.xyz.(x \\plus{} y \\plus{} z).(x^2 \\plus{} y^2 \\plus{} z^2); (\\forall)x,y,z > 0.$ (2)\r\n[b](2) TRUE???[/b]", "Solution_2": "[quote=\"ABRORBEK\"]:arrow: every triangle prove that: $ 9R^2$\\leq$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} (a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2$[/quote]\r\nI have found that $ 9R^2\\geq a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} (a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2$ is wrong too. :wink:", "Solution_3": "every triangle prove that: $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} (a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2$\\leq$ 9R^2$" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "ABC is an isosceles right-angled triangle. K and M are given points on the hypotenuse AB. K separates A and M, and . B$ because $ A \\minus{} B \\equal{} 441 \\plus{} 126 x$ that is positive if $ x\\geq \\minus{} 3$ and let $ (2x \\plus{} 8)^3 \\equal{} 512 \\plus{} 384 x \\plus{} 96 x^2 \\plus{} 8 x^3 \\equal{} C$, $ C \\minus{} A \\equal{} \\minus{} 272 \\minus{} 36 x \\plus{} 12 x^2$. $ C \\minus{} A$ has irrational roots and $ C \\minus{} A\\geq 0$ if $ x\\geq 7$, $ \\Rightarrow$ $ A$ is between two cubes if $ x\\geq 7$ \r\n\r\nI'll continue later....\r\n\r\nConsider $ (2x\\plus{}11)^3\\equal{}D$ then $ D\\minus{}A\\equal{}547 \\plus{} 306 x \\plus{} 48 x^2$, $ D\\minus{}A>0$ Because it has imaginary roots. Consider $ (2x\\plus{}3)^3\\equal{}27 \\plus{} 54 x \\plus{} 36 x^2 \\plus{} 8 x^3\\equal{}E$, $ A\\minus{}E\\equal{}\\minus{}776 \\minus{} 396 x \\minus{} 60 x^2<0$ because it has imaginary roots. Then, if $ A$ is a cube it must be one of: $ (2x\\plus{}4)^3,(2x\\plus{}5)^3,(2x\\plus{}6)^3,(2x\\plus{}7)^3,(2x\\plus{}8)^3,(2x\\plus{}9)^3,(2x\\plus{}10)^3$. Making all equations. \r\n\r\nWe get the only solutions: $ (x,y)\\equal{}(\\minus{}5,\\minus{}6), (\\minus{}4, \\minus{}4), (\\minus{}3,4), (\\minus{}2,6)$", "Solution_2": ":D \r\n\r\nthat's nice dude." } { "Tag": [], "Problem": "[I hope to occasionally post problems from the New York State Mathematics League (NYSML). Here is my first such post.]\r\n\r\nPoints $P$ and $Q$ are on circle $O$, and chord $\\overline{PQ}$ is drawn. A second circle is drawn with diameter $\\overline{OP}$, crossing the chord at point $S$. If $OP = 7$ and $PQ = 12$, compute $PS$.", "Solution_1": "[hide]Draw a perpendicular to PQ from O. This bisects PQ, so let's call PQ's midpoint M. Right triangle OPM is inscribed in the circle whose diameter is OP. Therefore M=S, so PS=PM=6.[/hide]\r\n\r\n(Edited by moderator to hide solution.)", "Solution_2": "Nice solution." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Prove that any real matrix $A$ of trace 0 can be written as $XY-YX$ where $X$ is real symetric and Y has trace 0.", "Solution_1": "Let $tr(A) =0$ and we can use the known fact, duscussed many times on the forum that $A$ is orthogonally similar to the matrix which has zeros on the main diagonal(proof can be followed for example by induction). Let $O A O^{t}= B$ with $b_{ii}= 0$. Take any diagonal matrix $D = diag(d_{1},d_{2},...,d_{n})$ with different $d_{i}$'s and put $t_{ii}= 0, t_{ij}=\\frac{b_{ij}}{d_{i}-d_{j}}$ for $i \\neq j$. Then we have $DT-TD = B = O A O^{t}$, thus $A = O^{t}DTO-O^{t}TDO = XY-YX$, where $X = O^{t}DO$, $Y = O^{t}T O$, $X$ is symmetric and $tr(Y) = 0$.", "Solution_2": "Yep, that was also my solution." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all continuous functions $ f: R \\rightarrow R$ such that $ 9f(8x)\\minus{}9f(4x)\\plus{}2f(2x) \\equal{} 100x$ , $ \\forall x \\in R$", "Solution_1": "Let $ h(x) \\equal{} f(x) \\minus{} \\frac {5x}{2}$\r\nWe has :$ 9h(8x) \\minus{} 9h(4x) \\plus{} 2h(2x) \\equal{} 0$\r\nso $ 3(3h(8x) \\minus{} 2h(4x)) \\equal{} 3h(4x) \\minus{} 2h(2x)$\r\nLet $ m(x) \\equal{} 3h(2x) \\minus{} 2h(x)$\r\n$ 3m(2x) \\equal{} m(x)$ \r\nSo $ m(x) \\equal{} \\frac {m(\\frac {x}{2})}{3}$ so $ m(x) \\equal{} \\frac {m(\\frac {x}{2^n})}{3^n}$\r\n$ m(x)$ is continious so $ m(x) \\equal{} m(0) \\equal{} 0$\r\nSo $ 3h(2x) \\equal{} 2h(x)$ easy to check $ h(x)\\equiv 0$\r\nSO $ f(x) \\equal{} \\frac {5x}{2}$" } { "Tag": [ "geometry" ], "Problem": "There has been no geometry in orals topics. Therefore, we should do inversion.", "Solution_1": "Plane geometry just doesn't seem like it would fit with orals, and while inversion would indeed be a very interesting topic, I don't see it working very well with the orals format. If you really wanted a geometry orals topic, I think that elliptical or hyperbolic geometry would work well as an orals topic.", "Solution_2": "[url]http://www.viebach.net/ictm/HS/oral06.pdf[/url]\r\n\r\nHere are the oral topics for the 2008 season. 1A and 2A are both getting fair division, as was expected.\r\n\r\nNow, divisions 3AA and 4AA are both getting Voting Methods. Any idea what this would entail?\r\n\r\nHere's the link to the material: [url=http://bcs.whfreeman.com/fapp6/default.asp?s=&n=&i=&v=&o=&ns=0&uid=0&rau=0]Book 3AA and 4AA will be using[/url].\r\n\r\nIt looks like we might be scratching the surface of game theory, though this also seems like an extension of fair division.[/url]", "Solution_3": "[quote=\"Altheman\"]There has been no geometry in orals topics. Therefore, we should do inversion.[/quote]\r\n\r\nActually, the very first orals topic (NSML, not ICTM) I had was taxicab geometry. I think that almost qualifies.", "Solution_4": "isn't that really more combinatorics?", "Solution_5": "Kinda, but it does have the word \"geometry\" in the title at least.", "Solution_6": "This year's oral topic is Graph Theory: Networks and Circuits.\r\n\r\nWhich is actually a good topic. Who knew." } { "Tag": [ "topology" ], "Problem": "Prove that the set of all points mapped similarly from a topological space to a $ T_{1}$ space under two different continuous maps form a closed set.", "Solution_1": "This is not true, as there are $ T_1$ spaces that are not Hausdorff, and this condition would imply the opposite.\r\n\r\nFor, if $ X$ is a $ T_1$ space, we would have then that $ \\{(x,y) \\in X \\times X : y \\equal{} x\\} \\subset X \\times X$ is closed, which would imply that $ X$ is Hausdorff.\r\n\r\nThe claim is true for Hausdorff ($ T_2$), spaces, however, and is an alternate definition of being $ T_2$.", "Solution_2": "How does one prove that for a $ T_{2}$ space? Assuming i define a Hausdorff space to be a topological space where distinct points are separated by disjoint open sets", "Solution_3": "It's not too hard to show that if $ X$ is hausdorff then $ \\Delta \\equal{} \\{(x,x) \\in X \\times X\\} \\subset X \\times X$ is closed. If $ x \\ne y$ then there are disjoint open sets $ U,V$ such that $ (x,y) \\in U \\times V$. Since $ U,V$ are disjoint we have $ U \\times V \\cap \\Delta \\equal{} \\emptyset$. So $ \\Delta$ is closed.\r\n\r\nFrom there, standard argument. Let $ f,g : Y \\to X$. Then $ (f \\times g) : Y \\to X \\times X$ is induced, and $ (f \\times g)^{\\minus{}1}(\\Delta) \\equal{} \\{y \\in Y : f(y) \\equal{} g(y)\\}$ is closed.", "Solution_4": "Thanks!" } { "Tag": [ "function", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Find all monotonous functions $ f: \\mathbb{R} \\to \\mathbb{R}$ that satisfy the following functional equation:\n\\[f(f(x)) \\equal{} f( \\minus{} f(x)) \\equal{} f(x)^2.\\]", "Solution_1": "[quote=\"FelixD\"]Find all monotonous functions $ f: \\mathbb{R} \\to \\mathbb{R}$ that satisfy the following functional equation:\n$ f(f(x)) \\equal{} f( \\minus{} f(x)) \\equal{} f(x)^2.$[/quote]\r\n\r\nThis implies $ f(x)\\equal{}f(\\minus{}x)$ $ \\forall x\\in f(\\mathbb R)$. So, since $ f(x)$ is monotonous, $ f(x)\\equal{}c$ $ \\forall x\\in f(\\mathbb R)$ and, since $ f(x)\\equal{}x^2$ $ \\forall x\\in f(\\mathbb R)$, we get $ c\\equal{}c^2$\r\n\r\nAnd the only solutions are :\r\n$ f(x)\\equal{}0$\r\n$ f(x)\\equal{}1$", "Solution_2": "[quote=\"pco\"]This implies $ f(x) \\equal{} f( \\minus{} x) \\forall x\\in f(\\mathbb R)$[/quote]\r\n\r\nWhy is this true?", "Solution_3": "put x=f(x),we get f(x^2)=f^2(x)=f(-x^2) \r\nthus f(x)=f(-x), $ \\forall x\\in \\mathbb{R}$\r\nbut f(x) is monotonous,so we get f(x)=c,c=0 or 1", "Solution_4": "[quote=\"keyree10\"][quote=\"pco\"]This implies $ f(x) \\equal{} f( \\minus{} x) \\forall x\\in f(\\mathbb R)$[/quote]\n\nWhy is this true?[/quote]\r\n\r\nHello keyree10!\r\n\r\nIf $ x\\in f(\\mathbb R)$, $ \\exists y$ such that $ x\\equal{}f(y)$ and so $ f(f(y))\\equal{}f(x)$ and $ f(\\minus{}f(y))\\equal{}f(\\minus{}x)$ and, since $ f(f(y))\\equal{}f(\\minus{}f(y))$, we get $ f(x)\\equal{}\\minus{}f(x)$ $ \\forall x\\in f(\\mathbb R)$", "Solution_5": "[quote=\"pco\"][quote=\"FelixD\"]Find all monotonous functions $ f: \\mathbb{R} \\to \\mathbb{R}$ that satisfy the following functional equation:\n$ f(f(x)) \\equal{} f( \\minus{} f(x)) \\equal{} f(x)^2.$[/quote]\n\nThis implies $ f(x) \\equal{} f( \\minus{} x)$ $ \\forall x\\in f(\\mathbb R)$. So, since $ f(x)$ is monotonous, $ f(x) \\equal{} c$ $ \\forall x\\in f(\\mathbb R)$ and, since $ f(x) \\equal{} x^2$ $ \\forall x\\in f(\\mathbb R)$, we get $ c \\equal{} c^2$\n\nAnd the only solutions are :\n$ f(x) \\equal{} 0$\n$ f(x) \\equal{} 1$[/quote]\r\n\r\nIt seems to me that all you proved was that $ f(x)\\in \\{0,1\\}$ for all $ x\\in f(\\mathbb{R})$ and also since $ f(\\mathbb{R})\\subset\\{0,1\\}$ it doesn't seem like much has been done.\r\n\r\nI see that you are very good at these problems but could you give a solution with more detail?\r\n\r\nOne step:\r\nWLOG $ f$ is non decreasing.\r\n\r\nSuppose that $ x,y\\in f(\\mathbb{R})$ such that $ x\\ge y$. Then $ f(x)\\ge f(y)$, also $ \\minus{}y\\ge \\minus{}x$ so $ f(\\minus{}y)\\ge f(\\minus{}x)$ so $ f(y)\\ge f(x)$. Therefore $ f(x)\\equal{}f(y)$ for all $ x,y\\in f(\\mathbb{R})$.", "Solution_6": "[quote=\"Altheman\"] [I see that you are very good at these problems but could you give a solution with more detail?\n[/quote]\r\n\r\n :blush: Youre're right, I was a little bit too quick. Sorry for this. I forgot $ \\minus{}1$ :\r\n\r\nI think you agree with $ f(x)\\equal{}f(\\minus{}x)$ $ \\forall x\\in f(\\mathbb R)$ so we have, since $ f(x)$ is monotonous, $ f(y)\\equal{}f(x)$ $ \\forall y\\in [\\minus{}|x|,\\plus{}|x|]$ and $ f(0)\\equal{}f(x)$\r\n\r\nSo, if $ a \\in f(\\mathbb R)$ and $ b \\in f(\\mathbb R)$ we have $ f(0)\\equal{}f(a)$ and $ f(0)\\equal{}f(b)$ and so $ f(a)\\equal{}f(b)$\r\n\r\nSo $ f(x)\\equal{}c$ $ \\forall x\\in f(\\mathbb R)$\r\nAnd, since $ c\\in f(\\mathbb R)$, $ f(c)\\equal{}c$ but also $ f(c)\\equal{}c^2$ so either $ c\\equal{}0$, either $ c\\equal{}1$\r\n\r\n1) if $ c\\equal{}0$ : we know that :\r\n$ \\forall x\\in f(\\mathbb R)$, $ f(x)\\equal{}0$\r\nBut, $ \\forall x\\in f(\\mathbb R)$, $ f(x)\\equal{}x^2$\r\nso $ \\forall x\\in f(\\mathbb R)$, $ x\\equal{}0$ and so $ f(\\mathbb R)\\equal{}\\{0\\}$ and so $ f(x)\\equal{}0$ $ \\forall x$\r\n\r\n2) If $ c\\equal{}1$ : we know that :\r\n$ \\forall x\\in f(\\mathbb R)$, $ f(x)\\equal{}1$\r\nBut, $ \\forall x\\in f(\\mathbb R)$, $ f(x)\\equal{}x^2$\r\nso $ \\forall x\\in f(\\mathbb R)$, $ x^2\\equal{}1$ and so $ f(\\mathbb R)\\subseteq\\{\\minus{}1,\\plus{}1\\}$\r\n\r\nand , since $ f(\\minus{}1)\\equal{}f(1)\\equal{}1$, we have 5 solutions (remember $ f(x)$ is monotonous):\r\n\r\n$ f(x)\\equal{}1$ $ \\forall x$\r\nLet $ a<\\minus{}1$ and then $ f(x)\\equal{}\\minus{}1$ $ \\forall x\\leq a$ and $ f(x)\\equal{}1$ $ \\forall x>a$\r\nLet $ a\\leq \\minus{}1$ and then $ f(x)\\equal{}\\minus{}1$ $ \\forall x< a$ and $ f(x)\\equal{}1$ $ \\forall x\\geq a$\r\nLet $ a>1$ and then $ f(x)\\equal{}\\plus{}1$ $ \\forall xa$\r\n\r\n\r\nI think it's ok now :blush:" } { "Tag": [ "number theory", "relatively prime" ], "Problem": "This is a simple problem... yet I can't find a simple solution :D \r\nFind the hundreds digit of 4^309. any suggestions?", "Solution_1": "[quote=\"pkothari13\"]Find the hundreds digit of 4^309[/quote]\r\nDo you mean to find 3 final digit of this number?It's $ 144$", "Solution_2": "The way I found the last three digits is:\r\n[edit] My solution was wrong...I think...sry bout that.", "Solution_3": "what are the fermat theorems?\r\n\r\nthe only way i do this is:\r\ni know 2^32 = 4294967296\r\nwe only want the hundreds so just use 296 and take only the last 3 digits for the calculations\r\n2^64 = 296*296 = 616\r\n2^128 = 616*616 = 456\r\n2^256 = 456*456 = 936\r\n2^512 = 936*936 = 96\r\n2^576 = 96 * 616 = 136\r\n2^608 = 136 * 296 = 256\r\n2^618 = 256 * 24 = 144\r\n\r\nso answer is 1\r\nnot too bad", "Solution_4": "[quote=\"joml88\"]The way I found the last three digits is:\n[edit] My solution was wrong...I think...sry bout that.[/quote]\r\n\r\nturns out right though @@", "Solution_5": "Another way to solve this, with less calculation, is to invoke Euler's theorem: if $a$ and $n$ are relatively prime, $a^{\\phi(n)}\\equiv 1 \\mod n$, where $\\phi(n)$ is the number of integers between 0 and $n$ which are relatively prime to $n$.\r\n\r\nFor this problem, we need to calculate $4^{309}$ mod $1000$. We know $1000=2^35^3$, and we calculate each part separately. Obviously $4^{309}\\equiv 0 \\mod 8$.\r\nSince $\\phi(125)=100$, $4^{309}\\equiv 4^9 \\mod 125$\r\n$4^0\\equiv 1$\r\n$4^1\\equiv 4$\r\n$4^2\\equiv 16$\r\n$4^3\\equiv 64$\r\n$4^4\\equiv 256\\equiv 6$\r\n$4^5\\equiv 24$\r\n$4^6\\equiv 96$\r\n$4^7\\equiv 384\\equiv 9$\r\n$4^8\\equiv 36$\r\n$4^9\\equiv 144\\equiv 19$\r\n\r\nTherefore, $4^{309}\\equiv 144 \\mod 125$ and $4^{309}\\equiv 144 \\mod 8$, so $4^{309}\\equiv 144 \\mod 1000$.", "Solution_6": "Some comments:\r\n(The answer, if you are used to these kind of problems, could be found very fast, even without pen or paper) \r\n\r\n With the help of Little Theorem of Fermat (Or Euler's): \r\nAnd noticing 10=2*5 and \r\n\r\nphi(5) = 4\r\nphi (25) = 20 \r\nphi (125) = 100 etc...\r\n\r\nand phi(2)= 1, phi(4)=2, phi(8)=4 etc...\r\n\r\nThis means, if you want to see just the last digit the cycle will repeat after in 4, for last 2 digits it will repeat in 20, for last 3 digit it will repeat in 100 etc. (You have to be a liitle careful here the cycle could start but there could be a few \"stray terms\" in the begining)\r\n\r\n[b]This means, all you have to find is the last 3 digit of 4^9 [/b]..\r\n(As long as 4^9 >1000, you have entered the cycle)\r\nand last 3 digits of 4^109, 4^209, 4^309 etc ... are same!\r\n\r\nIf you are confused about 'stray terms' or when to start the cycle just find the value , both modulo 8 and 125 \r\nFor 4^ 9 , obviously the answer is 0 mod 8 \r\nTo find 4^9 mod 125, One can notice, all mod 125: \r\n2^7 = 128 ==3 so 2^9 == 4*3 = 12\r\nand 4^9 = 12^2 = 144.\r\n(144 happens to be 0 mod 8 so it is the correct answer, else you look for numbers which are 19 mod 125 which are divisible by 8) \r\nHope this helps.\r\nEdited later: Sorry did not see the post above, it is essentially the same method, explained above..", "Solution_7": "Yes Gyan! This 's exactly what I did." } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "Let $ f(x)\\equal{} \\frac{1}{4a^2(1\\minus{}x)}$ find all a for which there exists an integer n such that $ f(f(f(...f(0)...))) \\equal{}1$", "Solution_1": "[quote=\"kihe_freety5\"]Let $ f(x) \\equal{} \\frac {1}{4a^2(1 \\minus{} x)}$ find all a for which there exists an integer n such that $ f(f(f(...f(0)...))) \\equal{} 1$[/quote]\r\n\r\nHere is a rather ugly method. Now that I've seen the result, I'm sure there is a pretty one. Maybe somebody else will propose it. \r\n\r\nIt's rather easy to show that : $ f^{[n]}(x)^ \\equal{} \\frac {a_nx \\plus{} b_n}{c_nx \\plus{} d_n}$ with : \r\n\r\n$ b_0 \\equal{} 0$, $ b_1 \\equal{} 1$, and $ b_{n \\plus{} 2} \\equal{} 4a^2b_{n \\plus{} 1} \\minus{} 4a^2b_{n}$\r\n$ d_0 \\equal{} 1$, $ d_1 \\equal{} 4a^2$, and $ d_{n \\plus{} 2} \\equal{} 4a^2d_{n \\plus{} 1} \\minus{} 4a^2d_{n}$\r\n$ a_0 \\equal{} 1$ and $ a_{n \\plus{} 1} \\equal{} \\minus{} 4a^2b_n$\r\n$ c_0 \\equal{} 0$ and $ c_{n \\plus{} 1} \\equal{} \\minus{} 4a^2d_n$\r\n\r\nThe problem is to find $ a,n$ such that $ b_n \\equal{} d_n$. Let us now consider $ \\Delta_n \\equal{} b_n \\minus{} d_n$ as :\r\n\r\n$ \\Delta_0 \\equal{} \\minus{} 1$, $ \\Delta_1 \\equal{} 1 \\minus{} 4a^2$ and $ \\Delta_{n \\plus{} 2} \\equal{} 4a^2\\Delta_{n \\plus{} 1} \\minus{} 4a^2\\Delta_{n}$\r\n\r\nIf $ a^2 \\equal{} 1$, we get $ \\Delta_n \\equal{} \\minus{} (n \\plus{} 2)2^{n \\minus{} 1}$ and $ \\Delta_n\\ne 0$ $ \\forall n\\ge 0$\r\nIf $ a^2\\ne 1$, $ X^2 \\minus{} 4a^2X \\plus{} 4a^2 \\equal{} 0$ has two distinct roots $ r_1$ and $ r_2$ $ \\in\\mathbb C\\backslash\\{0,1\\}$ and we get :\r\n\r\n$ \\Delta_n \\equal{} \\frac {1 \\minus{} 4a^2 \\plus{} r_2}{r_1 \\minus{} r_2}r_1^n$ $ \\plus{} \\frac {1 \\minus{} 4a^2 \\plus{} r_1}{r_2 \\minus{} r_1}r_2^n$ \r\n\r\nThen $ \\Delta_n \\equal{} 0$ $ \\iff$ $ (1 \\minus{} 4a^2 \\plus{} r_2)r_1^n \\equal{} (1 \\minus{} 4a^2 \\plus{} r_1)r_2^n$\r\n\r\nUsing $ 4a^2 \\equal{} \\frac {r_1^2}{r_1 \\minus{} 1}$ (got from $ r_1^2 \\minus{} 4a^2r_1 \\plus{} 4a^2 \\equal{} 0$) and $ r_1 \\plus{} r_2 \\equal{} 4a^2$ and $ r_1r_2 \\equal{} 4a^2$, we get \r\n\r\n$ \\Delta_n \\equal{} 0$ $ \\iff$ $ (1 \\minus{} r_1)r_1^{2n} \\equal{} (1 \\minus{} \\frac {r_1^2}{r_1 \\minus{} 1} \\plus{} r_1)(\\frac {r_1^2}{r_1 \\minus{} 1})^n$\r\n\r\n$ \\Delta_n \\equal{} 0$ $ \\iff$ $ (r_1 \\minus{} 1)^{n \\plus{} 2} \\equal{} 1$ and so $ r_1 \\equal{} 1 \\plus{} e^{i\\frac {2k\\pi}{n \\plus{} 2}}$\r\n\r\nAnd so $ 4a^2 \\equal{} \\frac {r_1^2}{r_1 \\minus{} 1}$ $ \\equal{} \\frac {(1 \\plus{} e^{i\\frac {2k\\pi}{n \\plus{} 2}})^2}{e^{i\\frac {2k\\pi}{n \\plus{} 2}}}$ $ \\equal{} 2(1 \\plus{} \\cos\\frac {2k\\pi}{n \\plus{} 2})$ $ \\equal{} 4(\\cos\\frac {k\\pi}{n \\plus{} 2})^2$\r\n\r\nSo $ a \\equal{} \\pm\\cos\\frac {k\\pi}{n \\plus{} 2}$ with $ \\frac {2k}{n \\plus{} 2}$ not an integer (in order to avoid $ a \\equal{} 0$ and $ a\\equal{}\\pm 1$)\r\n\r\nAnd, with such a $ a$, $ f^{[p]}(0) \\equal{} 1$ where $ p \\equal{} \\frac {n \\plus{} 2}{\\gcd(n \\plus{} 2,k)} \\minus{} 2$ (and consider that $ f^{[n]}(0)$ is not defined for $ n > p$)\r\n\r\nThis may be simplified as :\r\n\r\n$ \\boxed{a \\equal{} \\cos(\\frac pq\\pi)}$ where $ \\gcd(p,q)\\equal{}1$, $ p,q\\in\\mathbb N$ and $ q>2$ and then $ f^{[q\\minus{}2]}(0) \\equal{} 1$" } { "Tag": [ "Alcumus" ], "Problem": "How many numbers are in the list \\[ 6,7,10,11,14,15,\\ldots,94,95,98?\\]", "Solution_1": "[hide]For every even element $ n$, the number $ 2n \\plus{} 1$ is on the list (except for $ n \\equal{} 98$).\n\nSubtracting the list of even elements by $ 2$ and dividing by $ 4$, we find that there are $ 24$ pairs of elements, which means that there are $ 24\\cdot2 \\equal{} 48$ elements total.\n\nHowever, $ 98 \\plus{} 1 \\equal{} 99$ is not on the list, so our final answer is $ 48 \\minus{} 1 \\equal{} \\boxed{47}$.\n\nEdit: Hahahaha. I was being mental.[/hide]", "Solution_2": "That's impossible. How can you have 185 terms in a sequence of integers from 6-98?", "Solution_3": "[hide]I have another solution:\nFirst, subtract five to get $1, 2, 5, 6, 9, 10 ... 89, 90, 93$. Notice that the first two numbers of every four are in the list, up to $92$. Thus, half of those numbers are in the list, giving $92/2=46$. Throwing in 93 gives $\\boxed{47}$.[/hide]", "Solution_4": "[quote=\"infinity1\"][hide=\"I have another solution:\"]\nFirst, subtract five to get $1, 2, 5, 6, 9, 10 ... 89, 90, 93$. Notice that the first two numbers of every four are in the list, up to $92$. Thus, half of those numbers are in the list, giving $92/2=46$. Throwing in 93 gives $\\boxed{47}$.[/hide][/quote]\n\nI like infinity1's solution better than the solution given by Alcumus; his (or her) approach is more direct and is easier to follow.\n\n[hide=\"Here's how I approached this problem:\"]\nFirst, $98-6+1=93$ tells us that there \"should\" be 93 terms in this sequence, but it's easy to see that a bunch of the terms are missing. Which ones are missing? Every term that is missing is a multiple of 4, and that multiple's immediate successor (i.e. 8 and 9, 12 and 13, 16 and 17, etc.) So how many multiples of 4 are there between 6 and 98? Those multiples would be $8, 12, 16, 20, ... \\text{ }, 92, 96$. Divide each term by 4 to get $2, 3, 4, 5, ... \\text{ }, 23, 24$. We then subtract $24-2+1=23$ to see that there are 23 multiples of 4 missing from the sequence.\n\nBut the sequence is not only missing its multiples of 4, it is also missing the next number after each multiple (i.e. 9, 13, 17, etc.) Since there are 23 multiples of 4, there are also 23 of their successors. Therefore, the number of terms in the sequence is $93-23-23=\\boxed{47}$.\n[/hide]\n\nTim", "Solution_5": "[quote=\"infinity1\"]I have another solution:\nFirst, subtract five to get $1, 2, 5, 6, 9, 10 ... 89, 90, 93$. Notice that the first two numbers of every four are in the list, up to $92$. Thus, half of those numbers are in the list, giving $92/2=46$. Throwing in 93 gives $\\boxed{47}$.[/quote]\n\n\nI actually took the time to start counting the pattern (thats how slow I am)...until i realized infinity1's solution...:D", "Solution_6": "[hide=This took so much bashing]The list is:\n\n6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 97, 98\n\nCount and you get 47.[/hide]", "Solution_7": "how is this level 14?\n\npair up the numbers as follows\n\n6 7\n10 11\n14 15\n...\n94 95\n98\n\nup till 98 there would be $2 \\cdot \\frac{94-2}{4} = 46$ terms. adding the $98$ term gives $46+1=\\boxed{47}$.", "Solution_8": "The Newport Crab Company marks the location of its crab pots with colored balloons. They mark every fourth pot with a red balloon, every sixth pot with a blue balloon, and every tenth pot with a yellow balloon. After placing 600 crab pots in the bay, how many pots have three different colored balloons attached?\n", "Solution_9": "Please stop posting this question everywhere. You must solve it yourself. Or you can make a new thread and ask it in there. But dont keep asking it everywhere.", "Solution_10": "[quote=PizzaPerson]The Newport Crab Company marks the location of its crab pots with colored balloons. They mark every fourth pot with a red balloon, every sixth pot with a blue balloon, and every tenth pot with a yellow balloon. After placing 600 crab pots in the bay, how many pots have three different colored balloons attached?[/quote]\n\n[quote=ibmo0907]Please stop posting this question everywhere. You must solve it yourself. Or you can make a new thread and ask it in there. But dont keep asking it everywhere.[/quote]\n\nYes, he's right. Also if you're stuck then asking questions on the message board of the course you are taking is helpful sometimes. :)", "Solution_11": "[quote=OlympusHero]how is this level 14?\n\npair up the numbers as follows\n\n6 7\n10 11\n14 15\n...\n94 95\n98\n\nup till 98 there would be $2 \\cdot \\frac{94-2}{4} = 46$ terms. adding the $98$ term gives $46+1=\\boxed{47}$.[/quote]\n\nWell maybe some people make off by one errors.\n\nHere's my solution:\nNotice that every two numbers it increases by 4, so the answer is $\\dfrac{98-6}{4} \\cdot 2 + 1 = 47.$" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "1 + 1/(1+2) + 1/(1+2+3) + .................... +1/(1+2+3+................2006) = X\r\nfind x", "Solution_1": "nth term of the series will be 1/(1+2+....+n)=2/[n(n+1)]=2[1/n - 1/(n+1)]\r\nnow adding upto 2006 we will get 2[1-1/2007]=4012/2007" } { "Tag": [ "linear algebra", "matrix", "invariant" ], "Problem": "Let $ f : R^3 \\rightarrow R^4$ be as follows:\r\n\\[ f((x_1, x_2, x_3))\\equal{}(x_1\\plus{}2x_2\\plus{}x_3, x_2\\plus{}x_3, x_1\\plus{}2x_2\\plus{}3x_3, x_1\\plus{}x_2\\plus{}2x_3)\\].\r\nDo exist bases $ A$ and $ B$ of $ R^3$ and $ R^4$ respectively such that:\r\n\\[ {M(f)^{B}_A}\\equal{}\\begin{pmatrix}1&\\minus{}1&0\\\\\r\n0&1&\\minus{}1\\\\\r\n1&0&\\minus{}1\\\\1&\\minus{}2&1\\\\\r\n\\end{pmatrix}\\] \r\n\r\n$ {M(f)^{B}_A}$ is a matrix representation of linear transformation $ f$ from base $ A$ to $ B$ I hope you understant my notation.\r\nPlease explain your reasoning. I'm interested in comprehensing the method rather than the result.", "Solution_1": "If you get to choose the bases on both sides, the only invariants are the size of the matrices and the rank of the transformation. What is the rank of $ f$, and what is the rank of that matrix $ M$?\r\n\r\n[hide=\"Proof of the claim\"]Suppose $ M$ is a $ m\\times n$ matrix of rank $ r$. There exist invertible matrices $ A$ and $ B$ ($ m\\times m$ and $ n\\times n$ respectively) such that $ A^{\\minus{}1}MB\\equal{}\\begin{bmatrix}I_{r\\times r}&0_{r\\times n\\minus{}r}\\\\0_{m\\minus{}r\\times r}& 0_{m\\minus{}r\\times n\\minus{}r}\\end{bmatrix}$; the standard rank $ r$ matrix given is the matrix of $ M$ with respect to $ A$ and $ B$.\n\nTo construct $ A^{\\minus{}1}$, row-reduce $ M$ and track the row operations. This can be done computationally by working with $ \\begin{bmatrix}M&I_{m\\times m}\\end{bmatrix}$. Then construct $ B$ by column-reducing the remaining matrix; computationally, we stick a $ n\\times n$ identity below it.\n\nSince any $ m\\times n$ matrix of rank $ r$ can be transformed into our standard matrix by a change of basis on each side, any such matrix can be transformed into any other by two such steps.[/hide]", "Solution_2": "Just to put the particular answer up here:\r\n\r\nThe rank of $ \\begin{bmatrix}1&2&1\\\\ 0&1&1\\\\ 1&2&3\\\\ 1&1&2\\end{bmatrix}$ is $ 3.$\r\n\r\nThe rank of $ \\begin{bmatrix}1&\\minus{}1&0\\\\ 0&1&\\minus{}1\\\\ 1&0&\\minus{}1\\\\ 1&\\minus{}2&1\\end{bmatrix}$ is $ 2.$\r\n\r\nBy jmerry's result, rank is the one thing that is invariant. These cannot be two matrices for the same linear transformation.", "Solution_3": "I have never taken a course in Linear Algebra (hopefully I understood things properly), but the following is my understanding and results regarding the above problem.\r\n\r\n[u]Lemma 1.[/u] If $ f: K^n\\rightarrow K^m$, $ K$ a field, is a linear mapping, then it follows that $ f$ can be represented using the matrix\r\n\\[ \\left(\\begin{array}{ccc}| & & | \\\\\r\nf(v_1) & \\hdots & f(v_n) \\\\\r\n| & & |\\end{array}\\right),\r\n\\]\r\nwhere $ \\{v_1,\\hdots,v_n\\}$ is a basis of $ K^n$.\r\n\r\n[u]Proof.[/u] Left as exercise. $ \\Box$\r\n\r\n[u]Lemma 2.[/u] If $ \\{f(a_i)\\}$ is a basis of $ \\text{im}f$. where $ f: V\\rightarrow W$ is a linear mapping, then $ V \\equal{} \\text{span}\\{a_i\\}\\oplus\\ker f$, that is, $ \\dim\\ker f \\plus{} \\dim\\text{im}f \\equal{} \\dim V$.\r\n\r\n[u]Proof.[/u] Use the definitions of direct sum. $ \\Box$\r\n\r\nSince for a matrix $ U$ that represents $ f: K^n\\rightarrow K^m$, $ \\dim\\text{im}f \\equal{} \\text{rank} U$, we have $ \\text{rank} U \\equal{} \\dim K^n \\minus{} \\dim\\ker f$. In our case, $ \\dim\\ker f \\equal{} 0$, that is, $ \\text{rank} U \\equal{} n$ for all matrices $ A$. Since $ \\dim\\text{im}f \\equal{} 3$ and $ \\text{rank} M(f)_A^B \\equal{} 2$, $ M(f)_A^B$ cannot be a representation of $ f$. Perhaps, a more experienced individual could corroborate the validity of my steps (if correct)." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Solve the equation $ 2x^3\\minus{}7x^2\\plus{}10x\\minus{}6$ knowing that one of the roots is rational.", "Solution_1": "hello, one solution is $ x\\equal{}\\frac{3}{2}$.\r\nSonnhard.", "Solution_2": "The equation is equivalent to:\r\n[url=http://www.monsterup.com][img]http://www.monsterup.com/upload/1258317411195.gif[/img][/url]\r\nso : [url=http://www.monsterup.com][img]http://www.monsterup.com/upload/125831746485.gif[/img][/url]", "Solution_3": "[quote=\"triplebig\"]Solve the equation $ 2x^3 \\minus{} 7x^2 \\plus{} 10x \\minus{} 6$ knowing that one of the roots is rational.[/quote]\ntriplebig,\n\nin your headline and in the body of your post, you did not present \nan equation (nor an inequality). It is an expression. For instance,\nthere is no equals sign as part of it. You did not give an equation\nto solve, but the polynomial you did give has roots which are also \nthe solutions to the related equation $ 2x^3 \\minus{} 7x^2 \\plus{} 10x \\minus{} 6 \\equal{} 0$.\n\n[quote=\"b.s.o\"]The equation is equivalent to:\n[url=http://www.monsterup.com][img]http://www.monsterup.com/upload/1258317411195.gif[/img][/url]\nso : [url=http://www.monsterup.com][img]http://www.monsterup.com/upload/125831746485.gif[/img][/url][/quote]\r\n\r\nb.s.o.,\r\n\r\nyou made an error twice in the factor by grouping. It appears to be a typo.\r\n\r\nThe *expression* factors into\r\n\r\n$ x^2(2x \\minus{} 3) \\minus{} 2x(2x \\minus{} 3) \\plus{} 2(2x \\minus{} 3)$ \r\n\r\nand it becomes\r\n\r\n$ (2x \\minus{} 3)(x^2 \\minus{} 2x \\plus{} 2)$,\r\n\r\nwhich factors into\r\n\r\n$ (2x \\minus{} 3)[x \\minus{} (1 \\minus{} i)][x \\minus{} (1 \\plus{} i)]$.\r\n\r\nThe roots are $ \\frac32, 1 \\minus{} i, 1 \\plus{} i$" } { "Tag": [ "articles", "percent", "LaTeX", "geometry", "Support", "blogs" ], "Problem": "Let me talk a bit about the UK academic perspective. Below I appended an article by Stephen Court who discusses the history of UK academic tenure until up the present. Especially interesting is the section on \"current trends in employment of academic staff\" starting at page 7/9. It discusses the casualization of academic staff. For a large part of the 20th century academics of all levels had job security of tenure except for certain legal cases, but this certainty to be most likely employed until retirement is gone now for most academic staff. Before 1990 86% academics were employed on permanent contracts. In 1994 60% of the academic staff had permanent contracts. And this number is actually assumed to be even lower as Higher Education Statistics Agency (HESA) does not consider staff to be contracted on less than a quarter of a full-time contract. And for the first time there were more new appointments on fixed-term than permanent contracts (49.5 percent vs 48.3 per cent). If this development continues most of staff will be employed on fixed-term contract basis and the shift from job security to insecurity will have taken place. Will these fixed-term contracts make academia less attractive in the long run?\r\n\r\nMy experience from talking to UK academic staff is that they even can dismiss whole departments if it does not make enough money. And you are most likely not appointed as academic staff if you did not prove as post-doc to be able to secure grants in academia or funded private/government projects in industry. And even at the level of senior lecturer, reader or even professor you will run the risk of getting kicked out if you do not succeed in securing enough incoming grants. A good quantity of papers and placing papers at good conferences or journals is mostly not enough. Unless you are \"exceptionally\" accomplished at doing that. Another opportunity to be able to stay on is to get involved with politics, administration, teaching etc. You may teach a wide range of courses and many of them a term, become an (under-)graduate tutor, a departmental expert on computing issues as $ \\text{\\LaTeX}$ or use of grid infrastructure for computationally expensive techniques, tutor and supervise many students etc. \r\n\r\nOf course there is also a cost of kicking out staff members for the department. It costs effort, time and money to recruit new staff though there might be plenty of people who are interested in that job. At my university teaching assistants below the lecturer level are allowed to contribute at most 10% to the students' final grade. But at other universities as the London School of Economics & Politics graduate students are asked to teach full courses due to lack of staff, e.g. a friend of mine got asked to teach a course on real analysis. But then preparing such a course requires a significant investment of time, especially if you are not \"too fluent\" in a certain course anymore you are asked to teach. And it may not coincide with your research area at all. And then again teaching in front of possibly 200+ students for the first time in your life can be quite a daunting experience. To be sufficiently assertive to deal with the students in this novel situation can be really challenging.\r\n\r\nBut this lack of loyalty may also turn out to be mutual that current staff members have less objections to move on to the next university upon arrival of a better offer. IBM experienced a similar fate. Most staff members were hired straight from university and mostly stayed on for a long time, often even until retirement. People knew they were welcome though at some point there skills may not meet the demands of the economy and they could get training to adopt to this new situation. But this also meant that many staff members were reluctant to easily move on to a different employer if a different company tried to headhunt you.\r\n\r\nI am not sure about US but I heard a considerable number of professors are tenured starting at the assistant professor level. Even with the higher-ranked universities that may often start at the associate professor level though they know there often is a long queue of highly accomplished candidates. Maybe somebody from the US can shed more light on the situation there?\r\n\r\nA related ML/AoPS thread discusses the [url=http://www.mathlinks.ro/viewtopic.php?t=242401]Economic Downturn in Academia[/url] which makes it even harder to secure an academic position these days.", "Solution_1": "In the US, the first tenure-track job for most fresh PhDs is at the Assistant Professor level. You'll get about 6 or 7 years before you must apply for tenure and if you don't make it, you leave the institution. When tenure is awarded, this Assistant Professor is generally promoted to Associate Professor. A tenured Assistant Professor would be a rare bird in the US. \r\n\r\nExperienced faculty members can be appointed at the Associate Professor or Full Professor levels - either with tenure or without. Generally, those hired without tenure are expected to apply for the permanent appointment within a few years.\r\n\r\nThe terms \"Instructor\" and \"Lecturer\" at US institutions are typically reserved for non-tenure track appointments - in many cases people without PhDs. Typically, they spend most of their time teaching and have minimal research responsibilities. The use of these titles is quite different in the US than in many other countries.\r\n\r\nSome disciplines in the US expect tenured faculty to support themselves through grant funding. Certainly medical / health sciences faculty and some Engineers need to get grants or their jobs are in peril.\r\n\r\nTenured professors in the US can also lose their jobs if the institution has financial difficulties or if the prof works in a \"low productivity\" program (not enough students to be economically viable.) Many US colleges have instituted \"post tenure review\" processes within the past 5-10 years. A few tenured faculty members have lost their positions as a result of negative evaluations.\r\n\r\nIn most previous economic downturns, academic jobs in the US have held up a bit better than those in industry - mostly because many displaced workers return to university schooling. In this downturn, some state-funded institutions are already raising admission standards so that costs may be saved through reduced enrollments (which require fewer faculty.)", "Solution_2": "Erik Ringmar has had an interesting blog entry on what PhD life is like. [url=http://ringmar.net/forgethefootnotes/?p=97]Here[/url] is the article. He is currently employed as professor by the [url=http://www.srcs.nctu.edu.tw/srcs/member/ab-11.htm]Dept of General Education, National Chiao Tung University[/url] and got well-known for this dispute on the teaching quality of the [url=http://www.lse.ac.uk/]London School of Economics and Political Science (LSE)[/url] which led him to resign from his senior lecturer position at LSE. He describes the case in the [url=http://ringmar.net/forgethefootnotes/?p=124]The Times Higher Education Supplement[/url] but it is also independently discussed in the newspaper the [url=http://www.guardian.co.uk/education/2006/may/03/highereducation.economics]Guardian.[/url]" } { "Tag": [], "Problem": "I got a 41 on r2, which means it was worth more than my 120 on r1, whee. :)\r\n\r\n[As in last year: let us not discuss rankings, even if we do know them?]", "Solution_1": "Hey Billy did ur skul teachers receive ur score as nothing is posted on the website.", "Solution_2": "Precisely.", "Solution_3": "I got a 32 on r2, which means it was [i]not[/i] worth more than my 138 on r1, rawr. :P \r\n\r\nBy the way, what's \"skul\"?", "Solution_4": "Um, I'm pretty sure \"skul\" means \"school.\" ;\\", "Solution_5": "Sitan cud u post the names of guys who placed from Northview\r\n\r\nBefuddlers", "Solution_6": "Let's not discuss rankings, as we didn't do it last year and it was more fun that way.", "Solution_7": "[quote=\"befuddlers\"]Sitan cud u post the names of guys who placed from Northview\n\nBefuddlers[/quote]\r\n\r\nNo, because I'm listening to Billy and because I don't know, actually.", "Solution_8": "Hi this is Heng, I have the results for Northview.\r\n\r\nKrishna got 39 on proof.\r\nSitan got 35 on proof.\r\nMohan got 31 on proof.\r\nBen got 29 on proof.\r\n(Krishna and Mohan are our beastly Indian twins.)\r\n\r\nThe rest of us...let's not talk about it.", "Solution_9": "Uh, for some reason, the graders took off that many points on my paper for no apparent reason, methinks. I saw my score today, and I wasn't too pleased, because I thought that all four of my proofs for #1,2,4, and 5 were completely legit. My solutions on #2 and #4 were different from the official solutions, but they were complete and correct. Are there elegance points involved, too?", "Solution_10": "40 on round 2." } { "Tag": [], "Problem": "Right folks, if you're a Harry Potter fan, here's a forum you could/should/need to go to and sign up for.\r\n\r\nDumbledore's Army: \r\n\r\nhttp://www.dumbledores-army.net\r\n\r\nOkay, when you sign up (which I know you will), your beloved *cough* moderator MithsApprentice is the referrer. Got that? That's M-i-t-h-s-A-p-p-r-e-n-t-i-c-e. Awesome :-) Hope to see some of you folks there.\r\n\r\nOh, and don't join the Death Eaters. Do yourselves a favor and don't join the Death Eaters. Please dont join the Death Eaters. It's for your own good. Really. *nods to self*", "Solution_1": "Ok. I joined cuz I'm a HP maniac. :) I'm MathFiend on there, and gave u credit miths. don't worry. :) Could you explain how the site works? What are the\"items\" in teh item shop for? I'm just a lil' but confuzzled. lol.", "Solution_2": "cool. heres sumthing else i found......ppl wrote this and stuff and its actually well written. oh, and these are separate stories and each of them starts from the first chapter so just click on the \"next chapter\" button until its done.\r\n\r\nhttp://www.schnoogle.com/authors/veela1287/JLAH01.html\r\n\r\nhttp://www.schnoogle.com/authors/kira/M01.html", "Solution_3": "[quote=\"MathFiend\"]Ok. I joined cuz I'm a HP maniac. :) I'm MathFiend on there, and gave u credit miths. don't worry. :) Could you explain how the site works? What are the\"items\" in teh item shop for? I'm just a lil' but confuzzled. lol.[/quote]\r\n\r\nAlright. Everytime you post, you get 2 casualties. And you also get other casualties for doing other stuff: contests, designs, etc. Basically you can buy items with those casualties in the itemshop and eventually you choose a side: either DA or DE (as explained) and the point is to help your team win the war.", "Solution_4": "*bump* I'm double-posting to bump this and add something.\r\n\r\nOkay, eventually they're going to add an RPG to the current board and stuff. That's what the items are actually for. When that time comes, the items will help you fight. *nods*\r\n\r\nAthena! *glares* Why haven't you signed up yet?...lolol...jkjk :P Go sign up!", "Solution_5": "okay I joined and have almost no clue what its about can someone clue me in?" } { "Tag": [ "integration", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "[color=darkred]Ascertain $ \\int\\frac {x^2}{(x^2\\minus{}4)\\sin x\\plus{}4x\\cos x}\\ \\mathrm dx$ on the interval $ (0,\\pi )\\ .$[/color]", "Solution_1": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44782]Check this[/url]" } { "Tag": [ "induction", "quadratics", "algebra", "number theory unsolved", "number theory" ], "Problem": "IF xy| x^2 + y^2 +1 , prove that 3xy = x^2 + y^2 +1\r\n\r\nI have started by saying that let there exists a number g such that \r\n\r\nxyg = x^2 + y^2 +1\r\nxyg + 2xy = x^2 + 2xy + y^2 +1\r\nxyg + 2xy -1 = (x+y)^2\r\nxy(g+2) -1 = (x+y)^2\r\n\r\nbut that has gotten me nowhere.\r\n\r\nAny pointers would be appreciated\r\n\r\nthanks", "Solution_1": "A couple tips:\r\n\r\n-Instead of working with the equation given, you may want to show that it's equivalent to working with the simultaneous equations \r\n\r\n$x|y^2+1$ and $y|x^2+1$.\r\n\r\n-You might want to try a proof by induction on whichever of $x$ and $y$ is larger (the only time they'll be equal is when $x=y=1$). Assume $x>y$, and try to construct another pair $(x', y')$ satisfying the pair of equations, but with $x'$ and $y'$ both less than $x$.", "Solution_2": "I hope it's right (made quickly ;) ):\r\nLet $t=\\frac{x^2+y^2+1}{xy}$.\r\nWe have the quadratic equation: $X^2-(ty)X+y^2+1=0$. This equation has two integer roots: $x$ and $\\frac{y^2+1}{x}$. If we assume $x>y$, then $x'=\\frac{y^2+1}{x}0$ and $S = a_2+a_3+...+a_n$ \r\n\r\nProve that \r\n\r\n$a_2^{1- \\frac{1}{2}} + a_3^{1- \\frac{1}{3}} +....... + a_n^{1- \\frac{1}{n}} < S + 2\\sqrt{S}$", "Solution_1": ":w00tb: This is very beautiful!!!!!!!!!!!! \r\nIt seems to be a very difficult problem!\r\n\r\nSilouan ,who is the constructor of this inequality?", "Solution_2": "Yes, thank you for this nice one. It seems to be hard.", "Solution_3": "it's posted before by harazi\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=cool&t=13862" } { "Tag": [ "summer program", "Mathcamp", "PROMYS", "Ross Mathematics Program", "calculus", "MathPath", "number theory" ], "Problem": "For those who didn't make into MOSP and for those who made into MOSP, what is your plan for this summer? MathCamp in Maine or PROMYS at BU, or others?", "Solution_1": "Maybe Ross, since PROMYS doesn't accept 8th graders. I felt mathcamp is too late for me to apply", "Solution_2": "If you go and look in the Other Programs forum, you can see a variety of topics related to this question.", "Solution_3": "[quote=\"beta\"]Maybe Ross, since PROMYS doesn't accept 8th graders. I felt mathcamp is too late for me to apply[/quote]\r\n\r\nIf Mathcamp is the one you are most interested in, go ahead and contact them and ask if it is too late. The director posted here about a week or so ago saying they would accept late applications.", "Solution_4": "I am taking calculus I and II at my state school and getting a job.", "Solution_5": "My son is too young to apply to MathPath (by a matter of days, because of when his birthday is), so we are saving MathPath for next year (2005). The year after that (2006) he will be old enough to get into MathCamp, and maybe will be able to get a waiver to get into Ross, if he writes a good set of application problem solutions. This year he is going to places he can go on the basis of SAT I scores, namely the Centre for Talented Youth Ireland [url=http://www.dcu.ie/ctyi/summer/academic/c_desc04.htm#Modern%20Mathematics]2004 Summer Programme[/url] and the JHU [url=http://www.jhu.edu/gifted/summer/catalogs/osmath.html]CTY summer program[/url] (second session) in Lancaster, Pennsylvania. \r\n\r\n2007 is his ninth-grade summer, so that will be a year for trying to get into MOP. I hope everyone is having an enjoyable spring and will have a summer packed with learning math.", "Solution_6": "I'm going to PROMYS as well as MOP, for a returning year.\r\n\r\nThey have good advanced lectures on top of the preliminary number theory course (which is also very good), but you have to be willing to work if you want to go at all. It does pay off, though.", "Solution_7": "I didn't apply to any any math programs. I will still try to learn more math on my own. I am signed up to take AMC/AIME Algebra and the AMC 12 Prob series on this site. I hope that I learn a lot there. Most of the summer I will be getting up at about 6 to go to cross country training. There's nothing like running in the mornings! It's optional but I love running and I want to get better just like math. Near the end of the summer I am signed up for a one week running camp in NC with my team. Back to academic. I am going to try to get myself to study some stuff over the summer(easier said than done). In particular I want to look at some science stuff(physics and chem. to be specific). Maybe some Latin and of course everyone's summer past time - reading. Mostly just ind. studying.", "Solution_8": "after MOP, I'm going to a camp in China (which is language-based rather than math-based)... :-D", "Solution_9": "[quote=\"MithsApprentice\"]after MOP, I'm going to a camp in China (which is language-based rather than math-based).[/quote]\r\n\r\nSo you're going to a language program in China? How did you sign up for that? I figured you were going over for a family visit, but I'm curious about language programs for Chinese-Americans who want to learn more Chinese.", "Solution_10": "Yeah I must admit that I am definitely not good at Chinese... which is embarrassing... but I don't have time either :-(", "Solution_11": "[quote=\"tokenadult\"]\nSo you're going to a language program in China? How did you sign up for that? I figured you were going over for a family visit, but I'm curious about language programs for Chinese-Americans who want to learn more Chinese.[/quote]\r\n\r\nHaha...it's also an excuse for my family to go back and visit relatives. But in any case, you qualify for the camp thru the HSK (HSK= [b]H[/b]an4 Yu3 [b]S[/b]hui3 Ping2 [b]K[/b]ao3 Shi4) Profiency Exam (administered annually at most Chinese embassies...we happen to have one here in Houston). The program in relatively new, in fact I believe last year was the first year they invited you to a summer vamp. In any case, they pick a certain number of top-scorers and send u off for a month to one of the universities in Beijing. \r\n\r\n(The numbers after the PingYin is the accent number)" } { "Tag": [ "logarithms", "integration", "limit", "calculus", "calculus computations" ], "Problem": "Prove that\r\n\r\n$ \\frac {\\minus{}1}{10} < \\sum_{k \\equal{} 1}^{n}{\\frac {k}{k^2 \\plus{} 1}} \\minus{} \\ln{n}\\leq\\frac {1}{2},$\r\n\r\nwith equality if and only if $ n \\equal{} 1.$", "Solution_1": "For $ n\\ge 2$, rewrite this sum as:\r\n\\[ \\sum_{k \\equal{} 1}^{n}{\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\ln{n} \\equal{} \\frac {1}{2} \\plus{} \\sum_{k \\equal{} 2}^{n}{\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\sum_{k \\equal{} 2}^{n}\\ln{\\frac {k}{k \\minus{} 1}} \\equal{} \\\\\r\n\\equal{} \\frac {1}{2} \\plus{} \\sum_{k \\equal{} 2}^{n}({\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\ln{\\frac {k}{k \\minus{} 1}})\\]\r\nBut, because $ \\frac {1}{(1 \\plus{} t)^2} < \\frac {1}{1 \\plus{} t}\\forall t > 0$, we have:\r\n\\[ \\frac{x}{1 \\plus{} x} \\equal{} \\int_{0}^{x}\\frac {1}{(1 \\plus{} t)^2}dt < \\int_{0}^{x}\\frac {1}{1 \\plus{} t}dt \\equal{} \\ln{(1 \\plus{} x)}\\]\r\nSo:\r\n\\[ {\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\ln{(1 \\plus{} \\frac {1}{k \\minus{} 1})} < {\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\frac {\\frac {1}{k \\minus{} 1}}{1 \\plus{} \\frac {1}{k \\minus{} 1}} \\\\\r\n{\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\ln{(1 \\plus{} \\frac {1}{k \\minus{} 1})} < {\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\frac {1}{k} \\\\\r\n{\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\ln{(1 \\plus{} \\frac {1}{k \\minus{} 1})} < \\minus{} \\frac {1}{k(k^2 \\plus{} 1)} < 0 \\\\\r\n\\sum_{k \\equal{} 2}^{n}{\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\ln{(1 \\plus{} \\frac {1}{k \\minus{} 1})} < \\minus{} \\sum_{k \\equal{} 2}^{n}\\frac {1}{k(k^2 \\plus{} 1)} < 0 \\\\\r\n\\frac {1}{2} \\plus{} \\sum_{k \\equal{} 2}^{n}({\\frac {k}{k^{2} \\plus{} 1}} \\minus{} \\ln{\\frac {k}{k \\minus{} 1}}) < \\frac {1}{2}\\]", "Solution_2": "[quote=\"Ji Chen\"]Prove that\n\n$ \\frac { \\minus{} 1}{10} < \\sum_{k \\equal{} 1}^{n}{\\frac {k}{k^2 \\plus{} 1}} \\minus{} \\ln{n}\\leq\\frac {1}{2},$\n\nwith equality if and only if $ n \\equal{} 1.$[/quote]\r\nfor right \r\n$ f \\equal{} \\sum_{k \\equal{} 1}^{n}(\\frac {k}{k^2 \\plus{} 1} \\minus{} \\ln{\\frac {k}{k \\minus{} 1}}) \\equal{} \\frac {1}{2} \\plus{} \\sum_{k \\equal{} 2}^{n}(\\frac {k}{k^2 \\plus{} 1} \\minus{} \\ln{\\frac {k}{k \\minus{} 1}})$\r\n$ g \\equal{} \\frac {k}{k^2 \\plus{} 1} \\minus{} \\ln{\\frac {k}{k \\minus{} 1}}$\r\n$ g' > 0$ $ \\lim_{k \\minus{} > \\plus{}\\infty}{g} \\equal{} 0$\r\nso $ f \\leq \\frac {1}{2}$", "Solution_3": "[quote=\"Ji Chen\"]Prove that\n\n$ \\frac { \\minus{} 1}{10} < \\sum_{k \\equal{} 1}^{n}{\\frac {k}{k^2 \\plus{} 1}} \\minus{} \\ln{n}\\leq\\frac {1}{2},$\n\nwith equality if and only if $ n \\equal{} 1.$[/quote]\r\nfor left,need to prove\r\n$ \\frac {1}{2} \\plus{} \\frac {2}{5} \\minus{} \\ln{2} \\plus{} \\sum_{k \\equal{} 2}^{n \\minus{} 1}(\\frac {k \\plus{} 1}{(k \\plus{} 1)^2 \\plus{} 1} \\minus{} \\ln{\\frac {k \\plus{} 1}{k}}) > \\frac { \\minus{} 1}{10}$\r\nwe just need to prove $ \\frac {1}{2} \\plus{} \\frac {2}{5} \\minus{} \\ln(2) \\minus{} \\frac {1}{2}\\sum_{k \\equal{} 2}^{n \\minus{} 1}(\\frac {1}{k \\minus{} 1} \\minus{} \\frac {1}{k}))$\r\n$ > \\frac {1}{2} \\plus{} \\frac {2}{5} \\minus{} \\ln(2) \\minus{} \\frac {1}{2}\\frac {1}{2*1} > \\frac { \\minus{} 1}{10}$ that it is.\r\n[hide=\"proof\"]\n$ f \\equal{} \\frac {n \\plus{} 1}{(n \\plus{} 1)^2 \\plus{} 1} \\minus{} \\ln(1 \\plus{} \\frac {1}{n}) \\plus{}\\frac {1}{2n(n \\minus{} 1)}$\n$ f' < 0,\\lim_{n \\to \\plus{} \\infty}{f} \\equal{} 0$\n[/hide]" } { "Tag": [ "Gauss" ], "Problem": "The sum of the digits of the page numbers of a book is 165.\r\nHow many pages are in the book?", "Solution_1": "[hide]Pages 1-9=45\nPages 10-19=55\nPages 20-29=65.\n45+55+65=165, so there are 29 pages in the book.[/hide]", "Solution_2": "[quote=\"easyas3.14159...\"][hide]Pages 1-9=45\nPages 10-19=55\nPages 20-29=65.\n45+55+65=165, so there are 29 pages in the book.[/hide][/quote]\r\n\r\nSimilar to what I did, except after finding the values of the first nine pages, I set up an equation.", "Solution_3": "[quote=\"vamathletes\"]The sum of the digits of the page numbers of a book is 165.\nHow many pages are in the book?[/quote]\r\n[hide=\"my answer\"] 1 thru 9 is 45\n10 (1) thru 19 (10) is 55\n20 (2) thru 29 (11) is 65\ntherefore there are 29 pages\ngauss method of summation is the only thing you need. or a calc[/hide]", "Solution_4": "[hide]THE number OF pages ARE 29[/hide]\r\n\r\n[color=red][size=75][Answer hidden by moderator][/size][/color]", "Solution_5": "[quote=\"solver123\"]THE number OF pages ARE ...[/quote]\r\n\r\n1. if you give an answer, give a solution, it doesnt really help people to give the answer, they want to know HOW to do it.\r\n\r\n2. hide your answers and solutions that way someone who wants to solve it themselves doesnt have an answer thrown in their face. you can hide stuff like this...\r\n\r\n[code][hide]ANSWER answer, bla bla bla, solution solution[/hide][/code]" } { "Tag": [ "geometry", "circumcircle", "perpendicular bisector", "geometry solved" ], "Problem": "1. Let ABC be an acute-angled triangle, and let M be a point on the side AC and N a point on the side BC. The circumcircles of triangles CAN and BCM intersect at the two points C and D. Prove that the line CD passes through the circumcentre of triangle ABC if and only if the right bisector of AB passes through the midpoint of MN.", "Solution_1": "First of all, let's prove this \r\n\r\nLemma: Given pts M on AC and N on BC, the perpendicular bisector of AB passes through the midpt of MN iff AMcosA=BNcosB. Here ABC is an acute triangle.\r\n\r\nProof: \r\nLet X and Y be the projections of M and N respectively on AB, and let Z be the midpt of AB. The projection of the midpt of MN on AB must be the midpoint of XY, but it also must be Z because the perpend on AB through Z passes through the midpt of MN. This means that ZX=ZY, and because ZA=ZB, we get XA=YB. But XA=AMcosA and YB=BNcosB. We got what we wanted.\r\n\r\nNow for the problem itself:\r\nIt's easy to see (good old angle-chase in the cyclic quadrilaterals MDBC and ADNC) that triangles DMB and DAN are similar (in that order of vertices). We write Ptolemy for the 2 quadrilaterals I mentioned:\r\nBM*DC=DM*BC+DB*MC (*)\r\nNA*DC=DA*NC+DN*AC; we multiply this last one by k=BM/NA=DM/DA=DB/DN to get \r\nBM*DC=DM*NC+DB*AC (**)\r\n\r\nIf we subtract (**) from (*) we get DM*BN=DB*AM, iff DM/DB=AM/BN\r\n\r\nIf the perpend. bisector of AB passes through the midpt of MN we get, from the Lemma, DM/DB=AM/BN=cosB/cosA=sin(Pi/2-B)/sin(Pi/2-A)=sinDBM/sinDMB, but we always have DBM+DMB=C=Pi/2-B+Pi/2-A, and by combining these 2 we get DBM=Pi/2-B=OCA (O is the circumcenter of ABC)=DCM=DCA (because MDBC is cyclic), so we found D, O, C collinear.\r\n\r\nOn the other hand, if we assume DCA=DCM=DBM=Pi/2-B, we get DMB=Pi/2-A, so we find AM/BN=DM/DB=cosB/cosA, and we use the Lemma again to find that the perpendicular bisector of AB passes through the midpt of MN.\r\n\r\nI hope it's Ok.\r\nSee ya! :D :D", "Solution_2": "[quote=\"grobber\"]sin(Pi/2-B)/sin(Pi/2-A)=sinDBM/sinDMB[/quote]\r\nhow do you find this, Grobber? please someone explain it to me.", "Solution_3": "Ph-An could reply my question to Grobber ?", "Solution_4": "If $C(O,R)$ is the circumcircle of the triangle $ABC$ then\r\n\r\n$D\\in [CO\\Longleftrightarrow m(\\angle DCA)=90-B\\Longleftrightarrow m(\\angle DCB)=90-A$,\r\n\r\na.s.o.", "Solution_5": "Thanks a lot." } { "Tag": [], "Problem": "A tank contains the following quantities of these gases: He = 0.6 g; dihydrogen = 0.6 g; Ar = 0.6 g. The total pressure exerted by the mixture of gases at 0 degrees Celsius is 4.6 atmospheres. \r\n\r\nPick the statements that are true in the following questions. \r\n\r\nThe partial pressure exerted by: \r\n\r\n1) each gas is 4.6 atmospheres; 2) each gas is 1.03 atm; 3) the hydrogen is 0.60 atm. 4) the argon is 10 times that of helium; 5) the hydrogen is 2 times that of helium; \r\n\r\n\r\nIncreasing the temperature of the mixture from 0 degrees Celsius to 273 degrees Celsius without change in volume will cause total pressure to: \r\n\r\n1) decrease by 1/2 to 2.3 atm; 2) increase by a factor of 2 to 9.2 atm; 3) remain unchanged; 4) increase by a factor of 273; 5) change by a factor which cannot be predicted without knowing the volume.", "Solution_1": "I solved these two questions, but don't have the answers. Could anyone check them for me? :) \r\n[hide]\nFirst question: \n\nOnly choice 5 is true. (I just used the ideal gas equation to find the independent pressure of each). \n\nSecond question: \n\nOnly choice 2 is true. [/hide]" } { "Tag": [ "floor function", "algebra unsolved", "algebra" ], "Problem": "If $ n$ runs through all the positive integers, then $ f(n) \\equal{} \\left \\lfloor n \\plus{} \\sqrt {3n} \\plus{} \\frac {1}{2} \\right \\rfloor$ runs through all positive integers skipping the terms of the sequence $ a_n \\equal{} \\left \\lfloor \\frac {n^2 \\plus{} 2n}{3} \\right \\rfloor$.", "Solution_1": "Denote increasing positive integer sequence $\\left\\{ {{b_n}} \\right\\}_{n = 1}^{ + \\infty }$ runs through all positive integers skipping the terms of the sequence of $\\left\\{ {{a_n}} \\right\\}_{n = 1}^{ + \\infty }$.\nLet ${b_n} = k$, ${a_m} < k < {a_{m + 1}}$, then there are $m$\u2019s $\\left\\{ {{a_n}} \\right\\}$ and $n$\u2019s $\\left\\{ {{b_n}} \\right\\}$ in $\\left\\{ {1,2,...,k} \\right\\}$. So $k = n + m$, and we have \n$\\left[ {\\frac{{m(m + 2)}}{3}} \\right] < n + m < \\left[ {\\frac{{(m + 1)(m + 3)}}{3}} \\right]$, and so\n$\\frac{{m(m + 2) - 2}}{3} + 1 \\leqslant \\left[ {\\frac{{m(m + 2)}}{3}} \\right] + 1 \\leqslant n + m \\leqslant \\left[ {\\frac{{(m + 1)(m + 3)}}{3}} \\right] - 1 \\leqslant \\frac{{(m + 1)(m + 3)}}{3} - 1$\nFrom this we have\n${\\left( {m - \\frac{1}{2}} \\right)^2} < {m^2} - m + 1 \\leqslant 3n \\leqslant {m^2} + m < {\\left( {m + \\frac{1}{2}} \\right)^2}$,\nHence $m - \\frac{1}{2}<\\sqrt {3n} =3.\r\nShow that n^(n^(n^n))-n^(n^n)\r\nis divisible by 1989.", "Solution_1": "Just a qustion: how do you find the questions which weren't used in the Balkan olympiad? Are there some sort of 'Balkan shortlists' :D ?", "Solution_2": "What we need to do is prove that (1) 9/E (where E is that nasty expression ); (2) 13/E; (3) 17/E because 1989=3^2*13*17. \r\n\r\n(1)\tLemma 1 3/A=n^(n^n-n)-1 if n<>0(mod 3). \r\n\r\nIf 3/n then we get 9/E so it?s enough to prove 9/E for n=1,2 (mod 3). For the proof of the lemma, let?s take n=3k+1 first, and it?s obviously true that 3/A. If n=3k+2: n^n-n is even for any n and since 2^2a=1 (mod 3) is obvious, we get 3/A. Let?s use the notation a~b if a=b(mod 9). Then if n<>0 (mod 3) we have E~n^(n^n*A)-1 (we can divide E by n^(n^n)). It?s clear that 2/n^n*A (just consider the cases n odd, n even), and because 3/n^n*A (from the lemma) we have 6/n^n*A=6a. So E~n^(6a)-1=(n^(2a)-1)(n^(4a)+n^(2a)+1). Since 3 divides each of the 2 factors we have 9/E.\r\n\r\n(2)\tIf 13/n then we are finished. Let?s use another notation: a~b if (this time) a=b (mod 13). Then E~n^(n^n*A)-1. We can easily prove that 12/n^n*A=12a, so E~(n^a)^12-1, which is divisible with 13 by Fermat?s thm.\r\n\r\n I don?t have a proof for (3). If we try the same way (Fermat's thm) we should prove that 16/n^n*A, which I?m not sure about. Maybe we have to consider some cases. I?ll try to solve it.", "Solution_3": "Yeah, it's true. 16/n^n*A for any n. If you try to prove it you'll see it's not hard. So for (3) we just repeat the proof for (2). And now: the already legendary 'hope it's correct' :D", "Solution_4": "hi grobber,\r\n\r\nI found this problem in a set of number theory problems available at \r\nhttp://www.problem-solving.be/pen/\r\nby hojoo lee. There are a lot of interesting problems extract from \r\nvarious maths olympiads. Download it ;-)\r\n\r\n[Moderator edit: Link modified, since the old one was not accessible anymore.]\r\n\r\nbye." } { "Tag": [ "inequalities" ], "Problem": "Let x,y>0. Prove the inequality is true,\r\n\r\n$ \\frac{3}{2}x^2\\plus{}\\frac{3}{2}y^2\\plus{}2xy\\minus{}x\\minus{}y\\plus{}1 \\ge (x\\plus{}y)^2$", "Solution_1": "$ \\frac {3}{2}x^2 \\plus{} \\frac {3}{2}y^2 \\plus{} 2xy \\minus{} x \\minus{} y \\plus{} 1 \\ge (x \\plus{} y)^2$\r\n\r\n$ \\Longleftrightarrow 3(x^2\\plus{}y^2)\\plus{}4xy\\minus{}2(x\\plus{}y)\\plus{}2\\geq 2(x\\plus{}y)^2$\r\n\r\n$ \\Longleftrightarrow x^2\\plus{}y^2\\minus{}2(x\\plus{}y)\\plus{}2\\geq 0$\r\n\r\n$ \\Longleftrightarrow (x\\minus{}1)^2\\plus{}(y\\minus{}1)^2\\geq 0$", "Solution_2": "one other solution is :\r\n \r\n 3/2 ( x^2+y^2) + 2xy - x - y +1>= 3/2 2xy + 2xy-x-y+1=3xy+xy-x-y+1=5xy-x-y+1=5xy-(x+y)+1>= 5xy-(2\u03c1\u03b9\u03b6\u03b1xy)+1 LET \u03c1\u03b9\u03b6\u03b1xy=u then we have 5u^2-2u+1>=0 which is true because D=4-21=-17=(a+b+c)^2/3(ab+bc+ac)>=1(since a^2+b^2+c^2>=ab+bc+ac)(Q.E.D)\n[/hide]", "Solution_2": "easy :) \r\nwith caushy:\r\n$ \\left(\\frac{c}{a\\plus{}2b}\\plus{}\\frac{a}{b\\plus{}2c}\\plus{}\\frac{b}{c\\plus{}2a}\\right)(3(ab\\plus{}bc\\plus{}ac))\\geq (a\\plus{}b\\plus{}c)^2$\r\n$ \\Leftrightarrow \\left(\\frac{c}{a\\plus{}2b}\\plus{}\\frac{a}{b\\plus{}2c}\\plus{}\\frac{b}{c\\plus{}2a}\\right) \\geq \\frac{(a\\plus{}b\\plus{}c)^2}{3(ab\\plus{}bc\\plus{}ac)}\\geq1$", "Solution_3": "Wasn't the same thing written by ghjk? Well one more method: Using Cauchy Lema \r\nWe get the given expression $ \\ge \\dfrac{(\\sum a)^2}{3\\sum ab} \\ge 1$ from the well known inequalities. Though this is the same, just posted it for those who wanted to see it done by this way.", "Solution_4": "[quote=\"Aravind Srinivas L\"]Wasn't the same thing written by ghjk? Well one more method: Using Cauchy Lema \nWe get the given expression $ \\ge \\dfrac{(\\sum a)^2}{3\\sum ab} \\ge 1$ from the well known inequalities. Though this is the same, just posted it for those who wanted to see it done by this way.[/quote]\r\nYou post was also the same as ghjk ;) I see no difference in Cauchy Schwartz and Cauchy Lemma.", "Solution_5": "[quote=\"Johan Gunardi\"]For the positive real numbers $ a,b,c$ prove the inequality\n\\[ \\frac {c}{a \\plus{} 2b} \\plus{} \\frac {a}{b \\plus{} 2c} \\plus{} \\frac {b}{c \\plus{} 2a}\\ge1.\n\\]\n[/quote]\r\nLet $ x\\equal{}a \\plus{} 2b,y\\equal{}b \\plus{} 2c,x\\equal{}c \\plus{} 2a$ then we must prove $ \\sum_{cyc} \\frac{4y\\plus{}z\\minus{}2x}{9x}\\ge 1$ or\r\n$ \\sum_{cyc}\\frac{4y\\plus{}z}{x} \\ge 15$ its obviously true by AM-GM", "Solution_6": "For $ a,\\ b,\\ c$, prove that $ \\frac{2007c}{2008a\\plus{}2009b}\\plus{}\\frac{2008a}{2009b\\plus{}2007c}\\plus{}\\frac{2009b}{2007c\\plus{}2008a}\\geq \\frac{3}{2}$.", "Solution_7": "[quote=\"kunny\"]For $ a,\\ b,\\ c$, prove that $ \\frac {2007c}{2008a \\plus{} 2009b} \\plus{} \\frac {2008a}{2009b \\plus{} 2007c} \\plus{} \\frac {2009b}{2007c \\plus{} 2008a}\\geq \\frac {3}{2}$.[/quote]\r\n\r\nLet $ 2007c\\equal{}x$, $ 2008a\\equal{}y$, and $ 2009b\\equal{}z$, then our inequality takes form \\[ \\frac{x}{y\\plus{}z}\\plus{}\\frac{y}{z\\plus{}x}\\plus{}\\frac{z}{x\\plus{}y} \\ge 1.5\\]", "Solution_8": "a,b,c are positive real numbers ???\r\n... sorry I'm not good at English", "Solution_9": "@kfc9512: it should be positive real lha.. \r\nassume that it's not positive, then let $ x\\equal{}0,y\\equal{}\\minus{}1,z\\equal{}2$ then $ LHS\\equal{}0\\minus{}\\frac{1}{2}\\minus{}2<\\frac{3}{2}$.", "Solution_10": "Sorry .... Also.. i'm not good at math.. :blush: thank you.. :blush: \r\n\r\n(and... \r\n\\frac{x}{y+z}+\\frac{y}{z+x}+\\frac{z}{x+y} \\ge 1.5by caushy)-just practice :wink:", "Solution_11": "I've just found a really beautiful proof to this problem (inspired by a proof of Nesbitt's Inequality):\r\n\r\nLet $ P\\equal{}\\sum\\frac{c}{a\\plus{}2b}$, $ Q\\equal{}\\sum\\frac{a}{a\\plus{}2b}$, $ R\\equal{}\\sum\\frac{b}{a\\plus{}2b}$. By AM-GM, we have $ P\\plus{}2Q\\equal{}\\sum\\frac{c\\plus{}2a}{a\\plus{}2b}\\ge3$ and $ 2P\\plus{}R\\ge\\frac{2c\\plus{}b}{a\\plus{}2b}\\ge3$. So $ 9P\\plus{}2Q\\plus{}4R\\equal{}4(2P\\plus{}R)\\plus{}(P\\plus{}2Q)\\ge15$, but $ 2Q\\plus{}4R\\equal{}2(Q\\plus{}2R)\\equal{}6$, so $ P\\ge1$. Q.E.D. :D", "Solution_12": "Another approach is by brute force, after computations we get \r\n$ 2\\sum_{cyc}{a^3} \\geq 3abc + \\sum_{cyc}{ a^2 (3c - 2b)} \\Leftrightarrow {\\underbrace{\\sum_{cyc}{a^3} - 3abc}}_{\\geq 0 } + {\\underbrace{\\sum_{cyc}{a^2(a + 3c -2b)}}_{\\geq 0} \\geq 0}$\r\nfor second sum we must prove somehow that is greater than zero", "Solution_13": "[quote=\"Johan Gunardi\"]For the positive real numbers $ a,b,c$ prove the inequality\n\\[ \\frac {c}{a \\plus{} 2b} \\plus{} \\frac {a}{b \\plus{} 2c} \\plus{} \\frac {b}{c \\plus{} 2a}\\ge1.\n\\]\n[/quote]\r\n$ \\sum\\frac{a}{b\\plus{}2c}\\equal{}\\frac{a^2}{ab\\plus{}2ca}\\geq \\frac{(a\\plus{}b\\plus{}c)^2}{3\\sum ab}\\geq 1$ from the trivial ineq\r\n :D", "Solution_14": "See #2 popst. You should see the former post before you post. :wink:", "Solution_15": "Hi\nUse titu\"s lemma.", "Solution_16": "it is just Cauchy Schwartz ", "Solution_17": "am i watching the similar problem too many times? I saw it above which the difference is just the number of variant became 4 instead of 3", "Solution_18": "$c^2/(ac+2bc)+a^2/(ab+2ac)+b^2/(bc+2ab)>=1$\n$(a+b+c)^2/3(ab+ac+bc)>=1$\n$a^2+b^2+c^2>=ab+ac+bc$", "Solution_19": "Titu lemma and easy problem\n" } { "Tag": [ "number theory", "prime numbers" ], "Problem": "Let $ n$ be a positive integer such that $ n^2 \\minus{} n \\plus{} 11$ is the product of 4 prime numbers, some of which may be the same.\r\nWhat is the minimum value of n? \r\n[color=white]275 but how do we get it?\nEdit : 275 is incorrect answer. Correct answer is 132[/color]", "Solution_1": "I don't think $ 275$ is right. $ 132$ gives $ 11^3 * 13$\r\n\r\nTo find this, we denote $ n \\equal{} pk \\plus{} r$. Then we have $ n^2 \\minus{} n \\plus{} 11 \\equal{} p^2k^2 \\plus{} 2pkr \\plus{} r^2 \\minus{} pk \\minus{} r \\plus{} 11$\r\nSo $ p$ divides this if and only if $ p$ divides $ r^2 \\minus{} r \\plus{} 11$, where $ r < p$. Checking this for primes up to $ 13$, we find $ p \\equal{} 11$, $ r \\equal{} 0$, $ p \\equal{} 13$, $ r \\equal{} 2$ and $ p \\equal{} 13$,$ r \\equal{} 12$ are the only times where $ p$ divides $ r^2 \\minus{} r \\plus{} 11$. So our number is of the form either $ 11k$ and of the form $ 13k \\plus{} 2$, or of the form $ 11k$ and of the form $ 13k \\plus{} 12$. Checking $ 11k$ and of the form $ 13k \\plus{} 2$, we find $ 132$ is of both these forms, and gives $ 11^3 * 13$. Checking the form $ 11k$ and of the form $ 13k \\plus{} 12$, we find $ 77$ is the only number less than $ 132$, but this does not give a product of 4 primes. so $ 132$ is the minimum.", "Solution_2": "[quote=\"malcolm\"]I don't think $ 275$ is right. $ 132$ gives $ 11^3 * 13$\n\nTo find this, we denote $ n \\equal{} pk \\plus{} r$. Then we have $ n^2 \\minus{} n \\plus{} 11 \\equal{} p^2k^2 \\plus{} 2pkr \\plus{} r^2 \\minus{} pk \\minus{} r \\plus{} 11$\nSo $ p$ divides this if and only if $ p$ divides $ r^2 \\minus{} r \\plus{} 11$, where $ r < p$. Checking this for primes up to $ 13$, we find $ p \\equal{} 11$, $ r \\equal{} 0$, $ p \\equal{} 13$, $ r \\equal{} 2$ and $ p \\equal{} 13$,$ r \\equal{} 12$ are the only times where $ p$ divides $ r^2 \\minus{} r \\plus{} 11$. So our number is of the form either $ 11k$ and of the form $ 13k \\plus{} 2$, or of the form $ 11k$ and of the form $ 13k \\plus{} 12$. Checking $ 11k$ and of the form $ 13k \\plus{} 2$, we find $ 132$ is of both these forms, and gives $ 11^3 * 13$. Checking the form $ 11k$ and of the form $ 13k \\plus{} 12$, we find $ 77$ is the only number less than $ 132$, but this does not give a product of 4 primes. so $ 132$ is the minimum.[/quote]\r\n\r\nOh 132 is less than 275. I'm sorry ! Thanks too much !", "Solution_3": "I have my method !\r\nFirst $ n^2\\minus{}n\\plus{}11$ can't divisible by $ 2,3,5,7$. We get $ n^2\\minus{}n\\plus{}11 \\ge 11^4$\r\nIt is $ n \\ge \\frac{(\\sqrt{58521}\\plus{}1)}{2}$ then $ n \\ge 122$\r\n- $ n\\equal{}122$ We get $ 122^2\\minus{}122\\plus{}11\\equal{}(11)(17)(79)$\r\nLet $ f(n)\\equal{}n^2\\minus{}n\\plus{}11$\r\nIf $ f(n)$ isn't divisible by 11. We have $ f(n) \\ge 13^4\\equal{}28561$\r\nSince $ f(i)$ isn't divisible by 11 for $ i\\equal{}123,124,...,131$.\r\nWe have $ f(i) \\ge 13^4$ for $ i\\equal{}123,124,...,131$ which give a contradiction.\r\nConsider $ f(132)$; We have $ f(132)\\equal{}(11)^3(13)$\r\nThus, $ min(n)\\equal{}132$ :)", "Solution_4": "[quote=\"HelloWorld\"]\nFirst $ n^2 \\minus{} n \\plus{} 11$ can't divisible by $ 2,3,5,7$.[/quote]\r\n\r\nhow do you know that $ n^2 \\minus{} n \\plus{} 11$ can't be divisible by $ 2,3,5,7$ :huh:", "Solution_5": "let $ n\\equal{}pk\\plus{}r$, where $ p$ is $ 2,3,5,7$ and $ 0$ is less than or equal to $ r$, and $ r$ is strictly less than $ p$. Then expanding gives several terms with $ p$ in them, and $ r^2\\plus{}r\\plus{}11$ is left over when you remove these terms, so this will only be divisible by $ p$ if $ r^2\\plus{}r\\plus{}11$ is. checking each possible value of $ r$ gives that this is never divisible by $ p$." } { "Tag": [ "trigonometry", "number theory unsolved", "number theory" ], "Problem": "Find all $ (r, q) \\in \\mathbb{Q}^2$ s.t. $ tan \\text{ } \\pi r \\equal{} q$.", "Solution_1": "It's a standard result that this is possible only when $ q \\equal{} 1, 0, \\minus{} 1$. Write $ \\sin 2 \\pi r \\equal{} \\frac {2q}{1 \\plus{} q^2}$ and $ \\cos 2 \\pi r \\equal{} \\frac {1 \\minus{} q^2}{1 \\plus{} q^2}$, which must both be rational. The only Gaussian rationals that are also roots of unity are $ \\pm 1, \\pm i$, and this is not hard to prove using unique factorization in the Gaussian integers or standard facts about cyclotomic polynomials.", "Solution_2": "Ah, I see... Thank you." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "If $ f: [a, b]\\to [c, d]$ is a monotone and bijective function, then can we say that $ f$ and $ f^{ \\minus{} 1}$ are both continuous? If so, can someone prove this?\r\nThanks.", "Solution_1": "We suppose that $ f$ is increasing.Since $ f$ is surjective we have that $ \\forall p,q \\in [a,b] , p < q$ and $ \\forall \\lambda \\in (f(p),f(q))$ there is $ c \\in [a,b]$, more precisely $ c \\in (p,q)$ such that $ f(c) \\equal{} \\lambda$.Therefore $ f$ has the Darboux property.Because $ f$ is increasing we have that there exist $ \\lim_{x\\to x_0, x < x_0}f(x) \\equal{} f(x_0 \\minus{} 0)$ and $ \\lim_{x\\to x_0,x > x_0}f(x) \\equal{} f(x_0 \\plus{} 0)$, and in addition $ f(x_0 \\minus{} 0) \\leq f(x_0) \\leq f(x_0 \\plus{} 0)$.\r\n\r\nTry to prove that if the function has $ f(x_0 \\minus{} 0)$ (or $ f(x_0 \\plus{} 0)$) and f has the Darboux property then it is necessary that $ f(x_0 \\minus{} 0) \\equal{} f(x_0)$ (respectivly $ f(x_0 \\plus{} 0) \\equal{} f(x_0)$)", "Solution_2": "If the function has the Darboux property and $ f(x_0 \\minus{} 0)$ (or $ f(x_0 \\plus{} 0)$) exists, then is it necessary that $ f(x_0 \\minus{} 0) \\equal{} f(x_0)$ (respectivly $ f(x_0 \\plus{} 0) \\equal{} f(x_0))$?\r\n\r\nMoreover how to conclude from this that $ f^{\\minus{}1}$ is also continuous?", "Solution_3": "Every discontinuous Darboux function has an essential discontinuity but a monotone function has only nonessential discontinuities, hence the function is continuous.", "Solution_4": "[quote=\"limsup\"]If the function has the Darboux property and $ f(x_0 \\minus{} 0)$ (or $ f(x_0 \\plus{} 0)$) exists, then is it necessary that $ f(x_0 \\minus{} 0) \\equal{} f(x_0)$ (respectivly $ f(x_0 \\plus{} 0) \\equal{} f(x_0))$?\n\nMoreover how to conclude from this that $ f^{ \\minus{} 1}$ is also continuous?[/quote]\r\n\r\nSince $ f$ is continuous it follows that $ f^{\\minus{}1}$ must be continuous.See the following arguments:\r\n\r\n$ \\lim_{x\\to x_0} f^{\\minus{}1}(x)\\equal{}\\lim_{x\\to y_0}f^{\\minus{}1}(f(x))\\equal{}y_0$ , where $ \\lim_{x\\to y_0}f(x)\\equal{}x_0$, because $ f$ is surjective.so we conclude that $ \\lim_{x\\to x_0}f^{\\minus{}1}(x)$ exists and is finite.\r\n\r\nNow, suppose that there is $ x_0 \\in [a,b]$ such that $ \\lim_{x\\to x_0} f^{\\minus{}1}(x) \\neq f^{\\minus{}1}(x_0)$.Because $ f$ is injective we deduce that $ f(\\lim_{x\\to x_0} f^{\\minus{}1}(x)) \\neq f(f^{\\minus{}1}(x_0))$ $ \\Leftrightarrow$ $ \\lim_{x\\to x_0} f(f^{\\minus{}1}(x)) \\neq f(f^{\\minus{}1}(x_0))$ (because f is continuous) $ \\Leftrightarrow$ $ \\lim_{x \\to x_0} x \\neq x_0$ , contradiction.Hence $ f^{\\minus{}1}$ is also continuous.", "Solution_5": "Thank you very much Svejk and Nemion for your great help. Now I ve completely understood with your arguments and by Rudin." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Here is a nice problem(actually two of them)\r\nFind $x$ for which the series defined by $f(x)$ converges\r\na). $f(x) = \\sum_{n=1}^\\infty \\frac{ \\{ nx\\}}{n}$.\r\nb). $f(x) = \\sum_{n=1}^\\infty \\frac{ \\{ nx \\}-\\frac{1}{2}}{n}$.\r\nHere ${x}= x-[x]$ is the fractional part of $x$.\r\n\r\nActually, the first one is easy using Weyl's theorem, but i think i'm not sure what to do with b).", "Solution_1": "Well as you say it is well known for (a) if $x$ is irrational then the limit is 1/2 almost surely. And obviously it is 0 for $x$ integer. \r\n\r\nI think for (b) it is also easy. Since $\\{nx\\}$ is uniformly distributed in $[0, 1]$ for $x$ irrational, then $\\{nx\\}-\\frac{1}{2}$ is uniformly distributed in $[-1/2, 1/2]$, and then the series has limit 0 almost surely. Now we don't like $x$ integer though. \r\n\r\nFor some reason it is not obvious to me what happens when $x$ is rational, not integer (in either case). Are the fractions still uniformly distributed in $\\mathbb{Z}_{q}/ q$ when $q$ is denominator? Then we get 1/2 and 0 (for (a) and (b) respectively) again for same reason.", "Solution_2": "[quote=\"Xevarion\"]Well as you say it is well known for (a) if $x$ is irrational then the limit is 1/2 almost surely. And obviously it is 0 for $x$ integer. \n[/quote]\n\nIt seems i don't understand you. What do you mean by saying that \"the limit is 1/2 almost surely\" . Did you mean the sum instead of limit ?\nIn this case, i think you are not right : if $x \\in \\mathbb{R}\\setminus \\mathbb{Z}$ the the sum is $\\infty$, if $x \\in \\mathbb{Z}$, then the sum is $0$. That's what i've got for a). \n\n[quote=\"Xevarion\"]\nFor some reason it is not obvious to me what happens when $x$ is rational, not integer (in either case). Are the fractions still uniformly distributed in $\\mathbb{Z}_{q}/ q$ when $q$ is denominator? Then we get 1/2 and 0 (for (a) and (b) respectively) again for same reason.[/quote]\r\n\r\nIn a) as i said above in case $x \\in \\mathbb{Q}$ $f(x)$ is $\\infty$ in a) and i think for b) as well.", "Solution_3": "I'm sorry, I have no idea what I was thinking. Now that I look back, none of what I said makes sense at all. :blush:" } { "Tag": [], "Problem": "$ 0