{ "Tag": [ "geometry", "function", "calculus", "derivative" ], "Problem": "There are exactly three ordered pairs which are zeroes of the function $ f(x,y) \\equal{} 2x^3 \\plus{} 12xy\\plus{}2y^3 \\minus{} 16$ and form the vertices of an equilateral triangle. Find the area of the triangle.", "Solution_1": "huh i think the problem is wrong...\r\n\r\nby quick guess and check, we can confirm that (1,1), (2,0), (0,2) are three roots.\r\n\r\nbut (1,1), (2,0), (0,2) don't form an equilateral triangle.\r\n\r\nEDIT:\r\n\r\noh...\r\n\r\noops sry :blush:", "Solution_2": "[quote=\"FantasyLover\"]huh i think the problem is wrong...\n\nby quick guess and check, we can confirm that (1,1), (2,0), (0,2) are three roots.\n\nbut (1,1), (2,0), (0,2) don't form an equilateral triangle.[/quote]\r\n\r\nYou read the problem wrong. There are actually an infinite number of roots, like $ (3,\\minus{}1)$ for instance.\r\nThe problem asks to find three roots which [i]do[/i] form an equilateral triangle, and to calculate that area.", "Solution_3": "[hide=\"Hint 1\"] Look at the three points FantasyLover has found and test the obvious conjecture. [/hide]\n[hide=\"Hint 2 (giveaway)\"] $ 2(a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc) \\equal{} (a \\plus{} b \\plus{} c)((a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2)$. [/hide]\nI don't like problems like this. It's not at all obvious how to come up with a constructive path towards the solution if you don't know the trick in advance.\n[hide=\"One possibility\"] The partial derivatives are given by $ d_x \\equal{} 6x^2 \\plus{} 12y$ and $ d_y \\equal{} 6y^2 \\plus{} 12x$. They vanish at the points $ (0, 0)$ and $ (\\minus{}2, \\minus{}2)$. The first point is not on the graph, but the second is. That tells you that $ (\\minus{}2, \\minus{}2)$ is a [url=http://mathworld.wolfram.com/SingularPoint.html]singular point[/url], which tells you that $ f(x, y) \\equal{} 0$ has genus zero and you should be looking for factors. [/hide]", "Solution_4": "[hide=\"solution\"]Factoring out a 2 gives us $ 2(x^3\\plus{}6xy\\plus{}2y^3\\minus{}8)$, which is $ 2((x)^3\\plus{}(y)^3\\plus{}(\\minus{}2)^3\\minus{}3(\\minus{}2)(x)(y))$, which can be factored into $ (x\\plus{}y\\minus{}2)((x\\minus{}y)^2\\plus{}(y\\plus{}2)^2\\plus{}(x\\plus{}2)^2)$. Gotta love that factorization. Since the squares are non-negative, they must all be 0 if their sum is 0. Thus (-2,-2) is the only solution related to that factor. The other factor says that all x and y such that x+y=2 are solutions. There is exactly one pair of points on that line such that those two and (-2,-2) is an equilateral triangle. The height of the triangle is the distance from (-2,-2) to the line x+y=2, which is $ 3\\sqrt{2}$. Therefore the base of the triangle is $ \\frac{3\\sqrt{2}}{\\sqrt{3}}*2\\equal{}2\\sqrt{6}$. The area of the triangle is therefore $ \\boxed{6\\sqrt{3}}$.[/hide]\r\n\r\nForgive any un-rigor or non-clarity.\r\n\r\nDagnabit I had a version of hint 2 but it was the wrong version." } { "Tag": [ "LaTeX", "algebra unsolved", "algebra" ], "Problem": "I need the method and solution for the following power series questions\r\n\r\nQ1\r\n\r\nsigman = at/2 [sinh at + sin at divided by cosh at - cos at]\r\n\r\nIf a = 1.08 and t = 1.3\r\n\r\ncalculate sigma correct to 5 significant figures using power series\r\n\r\n\r\nQ2\r\n\r\nEvaluate sinh (1.6), correct to 5 decimal places using power series methods.\r\n\r\nI urgently need assistance with this", "Solution_1": "Dear Johnboy 14.\r\nI don'n understand what you write. Please write it by Latex and you try to write down clearly :(" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "$\\mathrm{Prove \\;\\; that \\;\\; the \\;\\; set \\;\\; of \\;\\; all} \\;\\; n\\in\\mathbb{N} \\mathrm{\\;\\; for \\;\\; which \\;\\; the \\;\\; equation \\;\\;}$ \\[ \\frac{1}{x}+\\frac{1}{y}=\\frac{3}{n} \\]$\\mathrm{ \\;\\;doesn't \\;\\;have \\;\\; a \\;\\; solution \\;\\; (where \\;\\;} x,y\\in\\mathbb{N})$ $\\mathrm{\\;\\;can't \\;\\;be \\;\\;expressed \\;\\;as\\;\\; a \\;\\;finite \\;\\;sum\\;\\; of \\;\\;arithmetic \\;\\;progressions.}$", "Solution_1": "We could set $x=y=2n$ for all $n$, so what condition is missing\u00bf Different\u00bf Coprime\u00bf Something else\u00bf", "Solution_2": "Sorry, its $\\frac{3}{n}$. I fixed it. :oops:", "Solution_3": "some idea?", "Solution_4": "If $n\\equiv 2\\pmod {3}$ then we can choose $x=\\frac{n+1}{3}$ and $y=\\frac{n(n+1)}{3}$. If $n\\equiv 0\\pmod{3}$ we can have $x=y=\\frac{2n}{3}$" } { "Tag": [ "ARML", "AMC", "AIME", "USAMTS", "AMC 10", "AMC 12", "USA(J)MO" ], "Problem": "Its obvious that AMC 12\\sum (b\\minus{}c)^2(\\frac{a}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}\\minus{}\\frac{1}{2(ab\\plus{}bc\\plus{}ac)}) \\ge 0$\r\nI think:It's true by SOS :)", "Solution_6": "[quote=\"dduclam\"]The following is stronger\n\nLet $ a,b,c$ be nonnegative real number. Prove that\n\\[ \\left(\\frac {a}{b \\plus{} c}\\right)^2 \\plus{} \\left(\\frac {b}{c \\plus{} a}\\right)^2 \\plus{} \\left(\\frac {c}{a \\plus{} b}\\right)^2 \\plus{} \\frac1{2}\\ge\\frac {5(a^2 \\plus{} b^2 \\plus{} c^2)}{4(ab \\plus{} bc \\plus{} ca)}\n\\]\n[/quote]\r\nNice Dduclam, but the following are stronger.\r\n\r\nLet $ a,b,c$ be nonnegative real number, no two of which are zero. Prove that\r\n\r\n$ (a)\\ \\ \\ \\ \\ \\frac {a(2a\\minus{}b\\minus{}c)}{(b\\plus{}c)^2}\\plus{}\\frac {b(2b\\minus{}c\\minus{}a)}{(c\\plus{}a)^2}\\plus{}\\frac {c(2c\\minus{}a\\minus{}b)}{(a\\plus{}b)^2}\\plus{}2\\ge \\frac {2(a^2\\plus{}b^2\\plus{}c^2)}{ab\\plus{}bc\\plus{}ca}$;\r\n\r\n$ (b)\\ \\ \\ \\ \\ \\frac {a(a\\minus{}b\\minus{}c)}{(b\\plus{}c)^2}\\plus{}\\frac {b(b\\minus{}c\\minus{}a)}{(c\\plus{}a)^2}\\plus{}\\frac {c(c\\minus{}a\\minus{}b)}{(a\\plus{}b)^2}\\plus{}\\frac 3{2} \\ge \\frac {3(a^2\\plus{}b^2\\plus{}c^2)}{4(ab\\plus{}bc\\plus{}ca)}$.", "Solution_7": "[quote=\"caubetoanhoc94\"][quote=\"dduclam\"]The following is stronger\n\nLet $ a,b,c$ be nonnegative real number. Prove that\n\\[ \\left(\\frac {a}{b \\plus{} c}\\right)^2 \\plus{} \\left(\\frac {b}{c \\plus{} a}\\right)^2 \\plus{} \\left(\\frac {c}{a \\plus{} b}\\right)^2 \\plus{} \\frac1{2}\\ge\\frac {5(a^2 \\plus{} b^2 \\plus{} c^2)}{4(ab \\plus{} bc \\plus{} ca)}\n\\]\n[/quote]\n\nWe have:\n$ \\sum \\frac {a^2}{(b \\plus{} c)^2} \\ge 2 \\minus{} \\frac {10abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}$\nWe have prove that:$ \\plus{} )\\frac {5}{2} \\minus{} \\frac {10abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\ge \\frac {5}{4}\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ac}$\n$ < \\equal{} > \\sum (b \\minus{} c)^2(\\frac {a}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\minus{} \\frac {1}{2(ab \\plus{} bc \\plus{} ac)}) \\ge 0$\nI think:It's true by SOS :)[/quote]\r\nProve it ! :)", "Solution_8": "[quote=\"caubetoanhoc94\"][quote=\"dduclam\"]The following is stronger\n\nLet $ a,b,c$ be nonnegative real number. Prove that\n\\[ \\left(\\frac {a}{b \\plus{} c}\\right)^2 \\plus{} \\left(\\frac {b}{c \\plus{} a}\\right)^2 \\plus{} \\left(\\frac {c}{a \\plus{} b}\\right)^2 \\plus{} \\frac1{2}\\ge\\frac {5(a^2 \\plus{} b^2 \\plus{} c^2)}{4(ab \\plus{} bc \\plus{} ca)}\n\\]\n[/quote]\n\nWe have:\n$ \\sum \\frac {a^2}{(b \\plus{} c)^2} \\ge 2 \\minus{} \\frac {10abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}$\nWe have prove that:$ \\plus{} )\\frac {5}{2} \\minus{} \\frac {10abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\ge \\frac {5}{4}\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ac}$\n$ < \\equal{} > \\sum (b \\minus{} c)^2(\\frac {a}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\minus{} \\frac {1}{2(ab \\plus{} bc \\plus{} ac)}) \\ge 0$\nI think:It's true by SOS :)[/quote]\r\n\r\nwhat is SOS ????", "Solution_9": "[quote=\"Mr.Bikrone\"][quote=\"caubetoanhoc94\"][quote=\"dduclam\"]The following is stronger\n\nLet $ a,b,c$ be nonnegative real number. Prove that\n\\[ \\left(\\frac {a}{b \\plus{} c}\\right)^2 \\plus{} \\left(\\frac {b}{c \\plus{} a}\\right)^2 \\plus{} \\left(\\frac {c}{a \\plus{} b}\\right)^2 \\plus{} \\frac1{2}\\ge\\frac {5(a^2 \\plus{} b^2 \\plus{} c^2)}{4(ab \\plus{} bc \\plus{} ca)}\n\\]\n[/quote]\n\nWe have:\n$ \\sum \\frac {a^2}{(b \\plus{} c)^2} \\ge 2 \\minus{} \\frac {10abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}$\nWe have prove that:$ \\plus{} )\\frac {5}{2} \\minus{} \\frac {10abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\ge \\frac {5}{4}\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ac}$\n$ < \\equal{} > \\sum (b \\minus{} c)^2(\\frac {a}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\minus{} \\frac {1}{2(ab \\plus{} bc \\plus{} ac)}) \\ge 0$\nI think:It's true by SOS :)[/quote]\nProve it ! :)[/quote]\r\n\r\nI'm sory..I'm wrong! :(\r\nI will try :(" } { "Tag": [ "trigonometry" ], "Problem": "Sa se arate ca daca alegem trei puncte A,B,C pe un cerc de raza O si centru R atunci suprafata maxima a triunghiului se realizeaza $\\Longleftrightarrow$ triunghiul este echilateral.", "Solution_1": "$S=2R^{2}\\cdot \\sin{A}\\sin{B}\\sin{C}\\leq 2R^{2}\\cdot \\frac{(\\sin{A}+\\sin{B}+\\sin{C})^{3}}{27}\\leq \\frac{3\\sqrt{3}}{8}$ \r\n\r\ncu egalitate $\\Longleftrightarrow$ e echilateral.", "Solution_2": "Dap...Corect...La acelasi lucru l-am facut si eu." } { "Tag": [ "trigonometry", "ratio" ], "Problem": "Prove $\\sin 9^{\\circ}+\\cos 9^{\\circ}= \\frac{\\sqrt{3+\\sqrt{5}}}{2}$", "Solution_1": "Consider a regular pentagon $ABCDE,\\ \\triangle{ACD}$ is isosceles triangle of $36^\\circ-72^\\circ-72^\\circ.$ Let $M$ be mid point of $CD,$\r\n\r\nRemark $\\triangle{ACM},$ since $AC: CD=1: \\frac{\\sqrt{5}-1}{2}$ by golden ratio, as $\\angle{CAH}=18^\\circ,$ we have $\\sin 18^\\circ =\\frac{\\sqrt{5}-1}{4}.$ \r\n\r\nNow $(\\sin 9^\\circ+\\cos 9^\\circ)^{2}=1+2\\sin 9^\\circ \\cos 9^\\circ=1+\\sin 18^\\circ=\\frac{3+\\sqrt{5}}{4}.$\r\n\r\n$\\sin 9^\\circ+\\cos 9^\\circ>0,$ yielding $\\sin 9^\\circ+\\cos 9^\\circ=\\frac{\\sqrt{3+\\sqrt{5}}}{2}.$\r\n\r\nAlternative solution:\r\n\r\n$\\sin \\theta+\\cos \\theta =\\sqrt{2}\\left(\\cos \\theta \\cdot \\frac{1}{\\sqrt{2}}+\\sin \\theta \\cdot \\frac{1}{\\sqrt{2}}\\right)$\r\n\r\n$=\\sqrt{2}\\cos(\\theta-45^\\circ)$ \r\n\r\nThus $\\sin 9^\\circ+\\cos 9^\\circ=\\sqrt{2}\\cos 36^\\circ=\\sqrt{2}\\cdot \\frac{1+\\sqrt{5}}{4}=\\frac{\\sqrt{2}+\\sqrt{10}}{4}.$\r\n\r\nRemark: $\\sqrt{3+\\sqrt{5}}=\\sqrt{\\frac{6+2\\sqrt{5}}{2}}=\\frac{\\sqrt{5}+1}{\\sqrt{2}}=\\frac{\\sqrt{10}+\\sqrt{2}}{2}.$", "Solution_2": "we have $sin(72) = 2 sin(36) cos (36)$ by the double angle relationship. And again, \r\n$sin(72) = 4 sin(18) cos (18) (1-2sin^{2}(18)$\r\nAnd because $sin(72) = cos(18)$, we substitute to get\r\n$cos(18) = 4 sin(18) cos (18) (1-2sin^{2}(18))$ \r\nThis simplifies to \r\n$1 = 4 sin (18) (1-2sin^{2}(18))$\r\nLet x = sin(18). Then we have\r\n$1 = 4x(1-2x^{2}) \\Rightarrow (2x-1)(4x^{2}+2x-1)=0 \\Rightarrow sin (18) = \\frac{\\sqrt{5}-1}{4}$\r\n\r\nNow, expand out by double angle again to get\r\n$2sin (9)cos(9) = \\frac{\\sqrt{5}-1}{4}$\r\nAdding 1 to each side,\r\n$sin^{2}(9)+2sin (9)cos(9)+cos^{2}(9) = \\frac{\\sqrt{5}+3}{4}$\r\n\r\nNoticing that $sin^{2}(9)+2sin (9)cos(9)+cos^{2}(9) = (sin(9)+cos(9))^{2}$ and square rooting each side gives the result." } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "Find out if there exists an injective continuous function f defined on an interval [a,b] , f not constant with the property:f is not injective on every\r\nsub-interval I included in [a,b]", "Solution_1": "?????\r\nHow can a function be injective on [a,b] but not on every subset of [a,b]??\r\n\r\n\r\nPierre.", "Solution_2": "he must probably mean: find out if there is a non-injective function continous of [a,b] such that it is injective on every sub-interval of [a,b], different from [a,b].", "Solution_3": "Sorry, i wanted to write :Find out a continuous and non-constant function\r\nf defined on an interval [a,b] wich is not injective on any sub-interval I included in [a,b]", "Solution_4": "There exist functions which are differentiable on |R and monotone on none interval (thus these functions are extremely oscillating).\r\nTo construct such a function is too long and difficult to explain it here. It seems that the first construction came from Arnaud Denjoy, and uses non trivial knowledge in mesure theory.\r\n\r\nSome others may be found in the references below.\r\n\r\nNote that it answer to your question (and even more) since a continuous function is injective on an intervalle if and only if it is monotone on it.\r\n\r\nReferences :\r\n- A.Denjoy : \"Mmoire sur la drivation et son calcul inverse\"\r\nGauthier-Villard, 1954, p.210-235\r\n\r\n- Y.Katznelson, K.Stromberg \"Everywhere differentiable, nowhere monotone functions\" American mathematical monthly, avril 1974, p.349-354.\r\n\r\n- K.Stromberg \"An introduction to classical real analysis\" p.216-218\r\n\r\nPierre.", "Solution_5": "A full solution can be found at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=244776[/url]", "Solution_6": "Much simpler examples can be given for xxxxtt's question :):\r\n\r\n[url]http://planetmath.org/encyclopedia/CantorFunction.html[/url]\r\n\r\nThe function is constant on the complement of Cantor's set. Since Cantor's set has null measure, its complement has full measure, and, in particular, it's dense." } { "Tag": [ "algebra", "polynomial", "function", "trigonometry", "Rational Root Theorem" ], "Problem": "This is a polynomial that appears on an AMC:\r\n$n^3-8n^2+20n-13$ \r\nFind values n for which it is prime.\r\n\r\nThey said you need to factor it out into (n-1) and something else. My question is: how do you know (n-1) is part of the above polynomial when factored out? Is it just guess and check, or is there a technique for finding factors of the cubic polynomial? *Not the cubic formula, please*", "Solution_1": "Hi - Well, one should try some simple substitutions like n=0, +-1, +-2 to see how the function behaves, and if you are lucky, for one of these values you will get the value of the polynomial as zero. Here if you try n=1, we get:\r\n\r\n1 - 8 + 20 - 13 = 0 \r\n\r\nso (n-1) is a factor .\r\n\r\nEven more helpful is, if you are lucky enough to find the coficient of the first term 1 and last term (13 here) a prime, (of if it has only a few factors), you only have to tryout the factors of the last term (that is 1 and 13 only) \r\n\r\nSo you need to try only (+1, -1, +13,-13) \r\n\r\nIn general, if a cubic appears as a_3*x^3 + something*x^2+ otherthing*x^2 + a_0 \r\n\r\nYou try out fractions like (+-) m/n where m is a factor of a_0 and n is a factor of a_3 \r\n\r\nif if a_3 and a_0 have a few factors only ( best is if one of them - or both of them - is/are 1) \r\nThis is not only true for Cubics but all other polynomial equations..\r\n\r\nThis , of course , assumes that you are lucky and/or the person who gave you the equation has a kind heart,.. to select an equation which does have rational roots. (The theorem is knows as rational root theorem).....\r\n\r\nSo next time you see that \"clever_but_nice\" math teache gives some thing like x^4 + .......other terms + 210 = 0\r\nyou may be lucky enough to be right when you say the roots are (2,3,5,7) \r\n :)", "Solution_2": "You use the rational root theorem (I think that's what it's called....) This gives you all the POSSIBLE zeroes of any polynomial. Notice how i wrote POSSIBLE and not the ACTUAL zeroes. The theorem is:\r\n\r\n+/- p/q\r\n\r\nIn this case, p=+/- 13 and q=+/-1\r\n\r\nSo the POSSIBLE zeroes are +/- 13, +/- 1.\r\n\r\nNow you could use synthetic division to completely factor the polynomial. There are probably other ways but this is the method I know. :lol:", "Solution_3": "It gives you all possible rational zeros, thus the name rational root theorem. If a polynomial does not have rational roots, you have to resort to the more complex methods discussed in the other thread.", "Solution_4": "You can always take advantage of a graphing calculator on AMC as well.\r\n\r\n[quote=\"tetrahedr0n\"]It gives you all possible rational zeros, thus the name rational root theorem. If a polynomial does not have rational roots, you have to resort to the more complex methods discussed in the other thread.[/quote]\r\n\r\nWhat are the \"more complex methods\"? \r\n\r\nAnyway, I'm pretty sure that on AMC/AIME they wouldn't make you factor a polynomial(with degree>2) over irrational/complex numbers.", "Solution_5": "[quote=\"Beta\"]What are the \"more complex methods\"? [/quote]\nSolving cubics using trig, or perhaps the cubic formula, altough I doubt the latter will ever take place. Sometimes you can find special things about a polynomial, such as it being symmetric, having roots of unity as zeros, etc... that will allow you to find its zeros.\n\n[quote=\"beta\"]Anyway, I'm pretty sure that on AMC/AIME they wouldn't make you factor a polynomial(with degree>2) with that can't be factored over rational numbers.[/quote]\r\nNo, not if you need to find the roots. However, they might make you do a polynomial problem in which you don't need to find the zeros. Then people can waste time by trying to find the roots when there are better methods." } { "Tag": [ "arithmetic sequence" ], "Problem": "You probably know the formula for triangular numbers (n(n+1)/2) and square numbers (n :^2: )\n\n\n\nChallenge: Find a general formula for the nth pentagonal number.\n\n\n\nHint: Try to draw them first [hide]--1, 5, 12, 22...[/hide]", "Solution_1": "Give your reasoning!!", "Solution_2": "This is kind of sketchy and not at all elegant, but...\n\n[hide]n(3n-1)/2\n\nLook at the first few: 1, 5, 12, 22, 35, 51.. We see that the changes between consecutive numbers are 4, 7, 10, 13, 16..., which clearly all differ by 3. Not only does this tell us that our answer must be of degree 2, but we can use the arithmetic sequence 3n+1 of the differences; the nth term can be expressed as 1+0*3+1+1*3+1+2*3+1+3*3+...+1+(n-1)3. In this sum, we have n 1s and 1+2+3+..+(n-1) 3s, or n+3((n-1)n)/2. Multiplying this out yields (3n:^2:-3n)/2+2n/2, or (3n:^2:-n)/2. Factoring out n yields n(3n-1)/2.[/hide]" } { "Tag": [], "Problem": "\u7b1b\u5361\u5c14\u7684\u4f20\u8bb0\uff0c\u4f5c\u8005\u662fAmir D.Aczel\r\n\r\nDescartes's secret notebook", "Solution_1": "\u4f55\u4e0d\u628a\u5b83\u653e\u5230\u7f51\u4e0a\u6765\u5462? :D", "Solution_2": "\u96be\u9053\u8981\u6211\u4e00\u9875\u4e00\u9875\u626b\u63cf\u4e0a\u53bb\u5417\uff1f\r\n\u518d\u8bf4\u8fd9\u7cfb\u56fe\u4e66\u9986\u7684\u4e66\u4e5f\u3002\u3002\u3002", "Solution_3": "How secret is it??\r\nCan a secret still be a secret if it can be found in a library?\r\n\u771f\u53eb\u4eba\u8cbb\u89e3" } { "Tag": [], "Problem": "Does anyone know if Arbelos are any good? i.e. easy to read, understand, and instructive", "Solution_1": "I wouldn't say they're easy to read -- they're rather difficult. They're enjoyable, but I don't know how useful they are." } { "Tag": [ "function", "Functional Equations" ], "Problem": "Find all functions $f: \\mathbb{Q}^{+}\\to \\mathbb{Q}^{+}$ such that for all $x,y \\in \\mathbb{Q}$: \\[f \\left( x+\\frac{y}{x}\\right) =f(x)+\\frac{f(y)}{f(x)}+2y, \\; x,y \\in \\mathbb{Q}^{+}.\\]", "Solution_1": "[quote=\"Peter\"]Find all functions $f: \\mathbb{Q}^{+}\\to \\mathbb{Q}^{+}$ such that for all $x,y \\in \\mathbb{Q}$: \\[f \\left( x+\\frac{y}{x}\\right) =f(x)+\\frac{f(y)}{f(x)}+2y, \\; x,y \\in \\mathbb{Q}^{+}.\\][/quote]\r\n\r\nSetting $x=y$ in the equation we get that\r\n\\[f(x+1)=f(x)+2x+1\\]\r\n\r\ntherefore $f(x+m)=f(x+m-1)+2(x+m)-1$. Adding up those equations for \r\n\r\nall $m=1,2,...,n$ we get that $f(x+n)=f(x)+2xn+n^{2}$. Setting in the \r\n\r\nlatter $x=1$ we get that $f(n+1)=(n+1)^{2}+(f(1)-1)$.\r\n\r\nReturning to the initial equation and setting $x=1$ we get \r\n\r\n$f(y+1)=f(1)+\\frac{f(y)}{f(1)}+2y$ but we also know that \r\n\r\n$f(y+1)=f(y)+2y+1$ $\\Rightarrow$ $f(1)=1$ $\\Rightarrow$ $f(n)=n^{2}$ for all $n\\in{N}$.\r\n\r\nAgain let's go back to the initial equation and set there $x=n\\in{N}$ and \r\n\r\n$y=m\\in{N}$. We get that $f(n+\\frac{m}{n})=f(n)+\\frac{f(m)}{f(n)}+2m$. Meanwhile $f(n+\\frac{m}{n})=f(\\frac{m}{n})+2n(\\frac{m}{n})+n^{2}$ $\\Rightarrow$ \r\n\\[f(\\frac{m}{n})=\\frac{f(m)}{f(n)}=\\frac{m^{2}}{n^{2}}\\]\r\n\r\n$\\Rightarrow$\r\n\\[f(x)=x^{2}\\]\r\nfor all $x\\in{Q^{+}}$. :wink:" } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "Suppose the real valued $ g$ is defined on $ \\mathbb{R}$ and $ g'(x) < 0$ for every real $ x$. Prove there's no differentiable $ f: R \\rightarrow R$ such that $ f \\circ f \\equal{} g$.", "Solution_1": "If f(f(x)) = g then their derivatives must be equal at all points so\r\n\r\nf'( f(x) ) * f'(x) = g'(x)\r\n\r\nBecause g'(x) is negative one value must be negative, another must be positive. So d/dx (f'(f(x)) is negative at some point, positive at another. If f is differentiable f(f(x)) must also be differentiable and the curve must be continuous, so at some point d/dx f(f(x)) = 0, so at some point g'(x) must also equal 0. Contradiction.", "Solution_2": "Four lines:\r\nIf $ f\\circ f$ is injective, $ f$ is injective.\r\nIf $ f$ is injective and continuous on $ \\mathbb{R}$, it is either strictly increasing or strictly decreasing.\r\nIf $ f$ is either strictly increasing or strictly decreasing, $ f\\circ f$ is strictly increasing.\r\nCombining these, we have shown that $ f\\circ f$ is never strictly decreasing if $ f$ is continuous on $ \\mathbb{R}$.\r\n\r\nIt is that easy." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "[url=http://www.mathlinks.ro/viewtopic.php?t=111253]here[/url] it is", "Solution_1": "The guy who has given the soln for this qn has gone beating the bush beyond one point of time.. once he has proved a(a+b) and b(a+b) are rational, then it is obvious that a/b is rational for all a and b in M.. Which implies the irrationality factor in all is the same.. So we can set root(n) to be the same irrationality factor.. thats it..", "Solution_2": "Doesn't speak too highly of our problem setters! But then, many countries (smaller ones) take up problems more freely from USAMO, Russian Competitions and the like. Well, it might have been an advantage to those who have seen it, but otherwise a nice problem.\r\nSathej", "Solution_3": "[quote=\"Sathej\"]Doesn't speak too highly of our problem setters! But then, many countries (smaller ones) take up problems more freely from USAMO, Russian Competitions and the like. Well, it might have been an advantage to those who have seen it, but otherwise a nice problem.\nSathej[/quote]\r\n\r\nSadly, ppl who have already seen this problem have a much higher prob of getting through.....", "Solution_4": "Well, it happens in any exam. After all, no man-made system is immune to being defeated by man.\r\nSathej", "Solution_5": "it's ok if it is defeated by a man ....but what if 'men' where in only 30 'men'(sorry not being chauvinistic just rhyming so...)get through and from 12th only 6 :)" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Prove that for every prime number $ p$, there are infinitely many positive integers $ n$ such that $ p$ divides $ 2^n\\minus{}n$.", "Solution_1": "[quote=\"moldovan\"]Prove that for every prime number $ p$, there are infinitely many positive integers $ n$ such that $ p$ divides $ 2^n \\minus{} n$.[/quote]\r\n\r\nFor $ p\\equal{}2$ every even $ n$ satisfies the condition.\r\n\r\nFor any prime $ p>2$, $ 2^{(p\\minus{}1)^{2k}} \\equiv 1 \\equiv (p\\minus{}1)^{2k} \\pmod{p}$ for all positive integers $ k$, so $ n\\equal{}(p\\minus{}1)^{2k}$ satisfies the condition, done." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "In the book of D. Gilbarg & N. Trudinger, \"Elliptic PDE of Second Order\", there is the following piece of reasoning. For $ \\sigma\\in(0,1)$, he suggests to take a cutoff function $ \\eta \\in C_c^2(B_R)$ ($ B_R$ is the ball of redius $ R$) such that $ 0 \\le \\eta \\le 1$, $ \\eta \\equal{} 1$ on $ B_{\\sigma R}$, $ \\eta \\equal{} 0$ on $ B_R \\setminus B_{\\sigma' R}$ with $ \\sigma' \\equal{} \\frac {1 \\plus{} \\sigma}2$,\r\n\\[ \\text{(*)} \\qquad |D\\eta| \\le \\frac 4{(1 \\minus{} \\sigma)R}, \\quad |D^2\\eta| \\le \\frac {16}{(1 \\minus{} \\sigma)^2R^2}.\r\n\\]\r\nI don't see how to find such a function (I am mainly interested in (*) because the rest is easy). Can you help me? :) \r\n\r\n[b]P.S.[/b] Do you know any good book in PDE containing $ L^p$ regularity theory (don't name the book by Agmon, Douglis and Nirenberg because I have one)?", "Solution_1": "The cutoff function is going to be radial, so this is really a one-dimensional problem. \r\n\r\nSomeone who does not care about $ 4$ and $ 16$ can take any $ C^2(\\mathbb R)$ function that is $ \\equiv 1$ for $ x\\le 0$ and is $ \\equiv 0$ for $ x\\ge 1$, and shift/rescale it appropriately.\r\n\r\nIf you do care about $ 4$ and $ 16$, then you are in for some tedious interpolation with polynomials.", "Solution_2": "[quote=\"volodja\"][b]P.S.[/b] Do you know any good book in PDE containing $ L^p$ regularity theory (don't name the book by Agmon, Douglis and Nirenberg because I have one)?[/quote]\r\n\r\nTry to use a book in French by Haim Brezis" } { "Tag": [ "LaTeX", "Olimpiada de matematicas" ], "Problem": "No tengo mucho tiempo, asique lo posteo as\u00ed como lo tengo, sin pasarlo a Latex:\r\n\r\nLos tri\u00e1ngulos A1B1C1 y A2B2C2 tienen lados paralelos: A1B1||A2B2, B1C1||B2C2, C1A1||C2A2 y el tri\u00e1ngulo A2B2C2 est\u00e1 contenido en el interior del tri\u00e1ngulo A1B1C1. Un tercer tri\u00e1ngulo ABC est\u00e1 inscrito en A1B1C1 y circunscrito a A2B2C2, es decir, los puntos A, B, C pertenecen al interior de los segmentos B1C1, C1A1, A1B1 y los puntos A2, B2, C2 pertenecen al interior de los segmentos BC, CA, AB, respectivamente.\r\nSi el \u00e1rea de A1B1C1 es 63 y el \u00e1rea de A2B2C2 es 7, hallar el \u00e1rea de ABC.", "Solution_1": "[quote=\"Jos\u00e9\"]No tengo mucho tiempo, asique lo posteo as\u00ed como lo tengo, sin pasarlo a Latex:\n\nLos tri\u00e1ngulos A1B1C1 y A2B2C2 tienen lados paralelos: A1B1||A2B2, B1C1||B2C2, C1A1||C2A2 y el tri\u00e1ngulo A2B2C2 est\u00e1 contenido en el interior del tri\u00e1ngulo A1B1C1. Un tercer tri\u00e1ngulo ABC est\u00e1 inscrito en A1B1C1 y circunscrito a A2B2C2, es decir, los puntos A, B, C pertenecen al interior de los segmentos B1C1, C1A1, A1B1 y los puntos A2, B2, C2 pertenecen al interior de los segmentos BC, CA, AB, respectivamente.\nSi el \u00e1rea de A1B1C1 es 63 y el \u00e1rea de A2B2C2 es 7, hallar el \u00e1rea de ABC.[/quote]\r\n21", "Solution_2": "Bien, la respuesta es correcta... sin embargo, podr\u00edas postear tu soluci\u00f3n?" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Calculate these:\r\n1/$ \\int \\frac{sinx}{1\\plus{}sin2x} dx$\r\n2/ $ \\int \\frac{x^{2009}}{(1\\plus{}x^2)^{1006}} dx$", "Solution_1": "hello\r\n[hide=\"1\"]\n$ \\int\\frac {sinx dx}{1 \\plus{} sin2x}$\n$ \\equal{} \\frac {1}{2}\\int \\frac {2sinx}{cos^2x \\plus{} sin^2x \\plus{} 2sinxcosx}dx$\n$ \\equal{} \\frac {1}{2}\\int\\frac {cosx \\plus{} sinx}{(cosx \\plus{} sinx)^2}dx \\minus{} \\frac {1}{2}\\int\\frac {cosx \\minus{} sinx}{(cosx \\plus{} sinx)^2}dx$\n$ \\equal{} \\frac {1}{2}\\int\\frac {1}{cosx \\plus{} sinx}dx \\minus{} \\frac {1}{2}\\int\\frac {dt}{t^2}$\nfor first integration u can use\n 1. half angle formula \nor \n 2.$ \\int\\frac {1}{cosx \\plus{} sinx}dx \\equal{} \\frac {1}{\\sqrt2}\\int sec(x \\minus{} \\frac {\\pi}{4})dx \\equal{} \\frac {1}{\\sqrt2}log(tan(\\frac {x}{2} \\plus{} \\frac {\\pi}{8}) \\plus{} C$\nso \n$ \\int\\frac {sinx dx}{1 \\plus{} sin2x} \\equal{} \\frac {1}{2\\sqrt2}log(tan(\\frac {x}{2}) \\plus{} \\frac {\\pi}{8}) \\plus{} \\frac {1}{2(cosx \\plus{} sinx)} \\plus{} C$\n [/hide]\r\nthank u", "Solution_2": "[quote=\"dulldrake\"]Calculate these:\n\n2/ $ \\int \\frac {x^{2009}}{(1 \\plus{} x^2)^{1006}} dx$[/quote]\r\n\r\nHave you done this yourself by hand??? :huh: :huh: because mathematica gave a 3 page long answer(though with elementary functions :P )", "Solution_3": "Mathematica is not clever. There is an approach that makes the answer expressible in one line.", "Solution_4": "hello\r\nI m writing this post in hurry so kindly check it properly.Do let me know if i have done some mistake.\r\n[hide=\"2\"]\n $ \\int \\frac{x^{2009}}{(1\\plus{}x^2)^{1006}} dx$\n put $ (1\\plus{}x^2)\\equal{}t$\n $ \\int \\frac{(t\\minus{}1)^{1004}}{2t^{1006}}dt$\n=$ \\int (1\\minus{}\\frac{1}{t})^{1004}\\frac{dt}{2t^2}$\nput $ (1\\minus{}\\frac{1}{t})\\equal{}y$\nso that now ur integral is \n$ \\int \\frac{ y^{1004}}{2}dy$\ni think now u can proceed.[/hide]\r\nthank u", "Solution_5": "[b](2)\nvery simple put x = 1 / t. [/b]", "Solution_6": "I prefer just going straight to $ t\\equal{}1\\plus{}\\frac1{x^2}$. Doing that in steps as kabi did also works, and may be easier to see.", "Solution_7": "for 1, we can divide Nr and Dr by cos^2 x.\r\n[hide]for 2, as jagdish &jmerry pointed, i should stop use too much mathematica :wallbash: :wallbash_red: ans seems to be $ 1/(2010 (1 \\plus{} 1/x^2)^1005)$[/hide]", "Solution_8": "Yes.It's nice" } { "Tag": [], "Problem": "you know how a meter is know defined as the distance travelled by light in a certain amount of time. so now its defined in terms of time in contrast to the original standard meter. \r\nWhats the point of doing this i've been puzzled by it for ages. \r\n\r\nand doesn't time have its own disadvantages?", "Solution_1": "the definitons of fundamental; units are based on some arbitrary choices, some history and long, complicated definitions which require considerable expertise to understand. they are not exactly 'fundamental' in their present definition ---- don't worry , you're not alone!!!", "Solution_2": "I think that the point of defining one meter using time and the speed of light is that now it is exact (+/-) as opposed to defining it with standard meter - you can measure lengths with much greater accuracy. The speed of light in vacuum is constant (or so we think) and time can be measured really accurately with atomic clock. And perhaps even more importantly, you can measure the speed of light and build your own atomic clock in your basement and don't have to travel to Paris to duplicate the standard meter with 1e-15 accuracy. (Well, not really but you get the point. :))", "Solution_3": "All these Defintions have been given in the 17th General Conference on Wigths and Measures. The measurement of speed of light had become extremely precise, so it made sense to adopt the speed of light as a defined quantity and to use it to define a metre.", "Solution_4": "are their any disadvantages with the time light system" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose that $ G \\equal{} AB$, where $ A,B$ are subgroups of $ G$. Does it always follow that $ BA \\equal{} G$?\r\n\r\n[hide=\"EDIT\"]Oh yes, I don't know how I failed to see that the inclusion goes both ways... :oops: [/hide]", "Solution_1": "Yes, because $ BA\\equal{}(AB)^{\\minus{}1}$.", "Solution_2": "Can we give a simple example of a group $ G$ with sub[b]sets[/b] $ A, B$ such that $ AB \\equal{} G$ but $ BA \\neq G$?", "Solution_3": "[quote=\"JBL\"]Can we give a simple example of a group $ G$ with sub[b]sets[/b] $ A, B$ such that $ AB \\equal{} G$ but $ BA \\neq G$?[/quote]\r\nHow do you define $ AB$ if $ A$ and $ B$ are sets?", "Solution_4": "$ AB \\equal{} \\{ab: a\\in A,b\\in B\\}$ (which can be different from $ BA$ if $ G$ is nonabelian)", "Solution_5": "[quote=\"blahblahblah\"]$ AB \\equal{} \\{ab: a\\in A,b\\in B\\}$ (which can be different from $ BA$ if $ G$ is nonabelian)[/quote]\r\nUmm. I think this definition works if $ A$ and $ B$ are submagmas of $ G$. But in subsets you can't assume that one has defined the product $ ab$.", "Solution_6": "I don't really see the problem, $ ab$ corresponds to 'something' in $ G$, so under that definition $ AB$ is certainly a well defined subset of $ G$.", "Solution_7": "[quote=\"puuhikki\"]Umm. I think this definition works if $ A$ and $ B$ are submagmas of $ G$. But in subsets you can't assume that one has defined the product $ ab$.[/quote] I am also not sure what the difficulty is. $ A$ and $ B$ are sets [i]of elements of a given group[/i] $ G$, so the product $ ab$ is perfectly happy as an element of $ G$ and $ AB$ is the set of such products for $ a \\in A (\\subset G), b \\in B (\\subset G)$.", "Solution_8": "Take $ G\\equal{}S_3,A\\equal{}\\langle(12)\\rangle$ and $ B\\equal{}\\{1,(13),(123)\\}$.\r\nThen $ G\\equal{}AB\\not\\equal{}BA$." } { "Tag": [ "geometry", "incenter" ], "Problem": "Now solve this right now! I'm sure you can solve it. \r\n If there are 3 candles, there were blown out, how many candles are left?", "Solution_1": "err.. still three, just theyre not lit?", "Solution_2": "[hide] 3?\n\nI guess the trick was making you think because the candles were blown out, they were also destroyed, but that is not the case.[/hide]", "Solution_3": "lol Mikey, you and me posted on the same minute ;)", "Solution_4": "[hide]if there are 3 candles without a fire... there are still 3\n\n3 candles[/hide][/hide]", "Solution_5": "Okay. If these three candles are at the three vertices of a 3-4-5 triangle and you are at the incenter of the above triangle. . Assuming you have only one match-stick which one you will light first?", "Solution_6": "The match :D", "Solution_7": "Blown out... they were blown OUT so none are left...\r\n\r\nAnd relating to physics, a triangle with wind blowing:\r\n\r\nI would light the one on the right angle according to distance and wind resistance", "Solution_8": "[quote=\"pakagawa\"]Blown out... they were blown OUT so none are left...\n\nAnd relating to physics, a triangle with wind blowing:\n\nI would light the one on the right angle according to distance and wind resistance[/quote]\r\n\r\nNah. Definitely the match first.", "Solution_9": "The answer is 1. Only one candle is left. The others are middle, and right.\r\n\r\nYou could also say \"as many candles there are in the world.\"", "Solution_10": "[quote=\"Gyan\"]Okay. If these three candles are at the three vertices of a 3-4-5 triangle and you are at the incenter of the above triangle. . Assuming you have only one match-stick which one you will light first?[/quote]\r\n\r\n3-4-5 cm?\r\n3-4-5 km?\r\n3-4-5 Mm?\r\n3-4-5 light years?", "Solution_11": "[quote=\"pakagawa\"][quote=\"Gyan\"]Okay. If these three candles are at the three vertices of a 3-4-5 triangle and you are at the incenter of the above triangle. . Assuming you have only one match-stick which one you will light first?[/quote]\n\n3-4-5 cm?\n3-4-5 km?\n3-4-5 Mm?\n3-4-5 light years?[/quote]\r\n\r\nWhy does nobody believe me??? It definitely has to be the match which is lighted first.", "Solution_12": "[quote=\"white_horse_king88\"][quote=\"pakagawa\"][quote=\"Gyan\"]Okay. If these three candles are at the three vertices of a 3-4-5 triangle and you are at the incenter of the above triangle. . Assuming you have only one match-stick which one you will light first?[/quote]\n\n3-4-5 cm?\n3-4-5 km?\n3-4-5 Mm?\n3-4-5 light years?[/quote]\n\nWhy does nobody believe me??? It definitely has to be the match which is lighted first.[/quote]\r\n\r\nBecause I have a lighter. :P", "Solution_13": "[quote=\"Mikey\"]err.. still three, just theyre not lit?[/quote]\r\n I'm sure you're wrong. Please see my reply below.Ok?", "Solution_14": "Hi everybody! According to me, all of you are wrong. Here is my reply.\r\n There are 6 candles left. Do I have to give you an explanation?", "Solution_15": "[quote]Hi everybody! According to me, all of you are wrong. Here is my reply. There are 6 candles left. Do I have to give you an explanation?[/quote]\r\nYes, please. :?", "Solution_16": "[quote=\"yif man12\"][/quote][quote=\"white_horse_king88\"][quote=\"pakagawa\"][quote=\"Gyan\"]Okay. If these three candles are at the three vertices of a 3-4-5 triangle and you are at the incenter of the above triangle. . Assuming you have only one match-stick which one you will light first?[/quote]\n\n3-4-5 cm?\n3-4-5 km?\n3-4-5 Mm?\n3-4-5 light years?[/quote]\n\nWhy does nobody believe me??? It definitely has to be the match which is lighted first.[/quote][quote=\"yif man12\"]\n\nBecause I have a lighter. :P[/quote]\n\nThat doesn't make sense. Just use your fingers...\n\n[quote=\"bamboovn\"]Hi everybody! According to me, all of you are wrong. Here is my reply.\n There are 6 candles left. Do I have to give you an explanation?[/quote]\r\n\r\nGive me and explanation. Three candles LEFT? I don't get it.", "Solution_17": "I think that the answer is 0 when this problem is not being discussed because, when nobody is looking, they can cease to exist.", "Solution_18": "Hi all! Here is my explanation. Let's pay attention to the Verb tense of the sentence. \"There [b]are[/b] 3 candles\" \"3 [b]were[/b] blown out\" so the total is 6 candles left now. ;)", "Solution_19": "That's pretty clever, but I'm not sure that's how the question goes...", "Solution_20": "[quote=\"wereallgonnadie\"]That's pretty clever, but I'm not sure that's how the question goes...[/quote]\r\ni'm sure that my anwer is right. There is no anwer else.Anyway, let's wait emi's answer.", "Solution_21": "ooooooooooooooookay, so um... what was i gonna say? oh yeah.[size=200] THIS IS A MATH FORUM. NOT AN ENGLISH FORUM. [/size]happy?", "Solution_22": "[quote=\"gameworld7\"]ooooooooooooooookay, so um... what was i gonna say? oh yeah.[size=200] THIS IS A MATH FORUM. NOT AN ENGLISH FORUM. [/size]happy?[/quote]\r\n\r\nYeah, but emi typed the problem in a way that it's confusing and could have different meanings.", "Solution_23": "[quote=\"h_s_potter2002\"][quote=\"gameworld7\"]ooooooooooooooookay, so um... what was i gonna say? oh yeah.[size=200] THIS IS A MATH FORUM. NOT AN ENGLISH FORUM. [/size]happy?[/quote]\n\nYeah, but emi typed the problem in a way that it's confusing and could have different meanings.[/quote]\r\n\r\nI think that was gameworlds point :P The question was not a math problem, it was an english problem..", "Solution_24": "But that's the point of the problem -- it's not a mathematical question, it's a brain-teaser. That is what the objection is to (I think) -- not the discussion of the problem, but the fact that the problem is in a math forum to begin with.", "Solution_25": "Hi all, I agree with your ideas. It's not a mathematical problem especially it's an English problem. Anyway, some of us can solve it. Rethink your thought!", "Solution_26": "why dont we just answer the question for what emi meant it to be?\r\nthere are 3 left, because after the candles are blown out (as in extinguished) the candles themselves are still there.", "Solution_27": "I think since this problm lacks the exact point on where the answer is, it should be moved to Fun & Games. I won't move it yet but if discussion goes on and on about deciding whether it's english or math problem, I'll remove it since it's not more like ALL-MATH discussion.\r\n\r\nBefore it happens, please identify your question's point.", "Solution_28": "I agree with my fellow moderator Silver Falcon. If there isn't a mathematical point here, this thread should be moved to Fun and Games.", "Solution_29": "Random question:\r\n\r\nWhat exactly are you tokin', Mr. Adult?", "Solution_30": "this post is in fun and games and yes i do have a point that its an [size=200]english forum![/size] [size=59]wait i meant [/size][size=200]math forum! okay?![/size]", "Solution_31": "Um, you're contradicting yourself. Fun&Games is a place for just about anything, including English. There is just no grammar flames allowed, like being really picky over sentence structure.", "Solution_32": "ok sorry but this is a math site. u know why would they name an english learning site \"www.artofproblemsolving\" and they wouldn't name a math site \"www.englishrocks\" or something. that wouldn't make sense", "Solution_33": "So what? We discuss politics in the Round Table. Does that mean we should cut all of that because this was designed as a math site? Should we cut all the riddles and stuff because this is a math site?", "Solution_34": "interesting.....", "Solution_35": "how should i know? i don't visit the round table section.", "Solution_36": "how should i know? i don't visit the round table section.", "Solution_37": "[quote=gameworld7]how should i know? i don't visit the round table section.[/quote]\n\nLol same", "Solution_38": "*14* years later...." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Prove that $ x^4 \\minus{} 6x^3 \\plus{} 11x^2 \\minus{} 6x$, with $ x \\in N$ is a multiple of 24.", "Solution_1": "hello, factorizing the given polynomial we get $ x(x\\minus{}1)(x\\minus{}2)(x\\minus{}3)$ so it is clear that $ P(x)$ is a multiple of $ 3$ and $ 8$.\r\nSonnhard." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Does anybody know an ELEGANT way to solve $2^x = 3^y + 5$ in positive integers?", "Solution_1": "[quote=\"Arne\"]Does anybody know an ELEGANT way to solve $2^x = 3^y + 5$ in positive integers?[/quote]\r\n$x=3$, $y=1$ and $x=5$, $y=3$ and if $y>3$ then as $2^{54}=1 \\ mod \\ 3^4$ and $2^{23}=2^{77}=5\\ mod \\ 3^4$ and both are coprimes with $54=\\varphi(3^4)$ then NO more solutions.\r\n\r\nRX TCM", "Solution_2": "I didn't understand, can you explain better RobertuX, please? :D", "Solution_3": "Yes please I will be curious to see your explanations :?", "Solution_4": "How does it follow from just that, that there are no more solutions? :?\r\n\r\nI may be missing something, but it looks very tough... So far all I've got is some horrendous factorization work, such as\r\n\\[2^5(2^{2n-2}+1)(2^{n-1}+1)(2^{n-1}-1)=3^3(3^{4k}+1)(3^{2k}+1)(3^k+1)(3^k-1),\\] but I don't see how a solution should come from that, either...", "Solution_5": "I have solved this one in the forum before locking modulo $3^4$ or something like that, a very ugly brute force thing.", "Solution_6": "[quote=\"Pascual2005\"]I have solved this one in the forum [/quote]\r\nWhere, please?", "Solution_7": "[quote=\"Simo_the_Wolf\"]I didn't understand, can you explain better RobertuX, please? :D[/quote]\r\nWell :? I delete all my calculations, so maybe another can helps to you. Sorry!", "Solution_8": "I cant find it and i dont remember it properly it was lockin mod $3^t$ or $2^t$ for some $t>4$ i think", "Solution_9": "[quote=\"RobertuX\"]I delete all my calculations.Sorry![/quote]\r\n You are a joker :D", "Solution_10": "I think that I've found a solution, but it's not very elegant.\r\nAs I don't have any time to write the solution in details, I'll give an quick overview:\r\n1) we show that y = 4n +3 for some n, if x,y > 1\r\n2) we show that x = 2^k + 1 for some k\r\n3) we show that 2^x == 2 (mod 17) for k > 2 (x > 5) and find out that then 3^y == 14 (mod 17) -> y = 16k + 9 and get a contradiction with y = 4n + 3. \r\n4) k = 1 and k = 2 give us all solutions.\r\n\r\nI hope it's correct", "Solution_11": "[quote=\"wellknown\"]I think that I've found a solution, but it's not very elegant.\nAs I don't have any time to write the solution in details, I'll give an quick overview:\n1) we show that y = 4n +3 for some n, if x,y > 1\n2) we show that x = 2^k + 1 for some k\n3) we show that 2^x == 2 (mod 17) for k > 2 (x > 5) and find out that then 3^y == 14 (mod 17) -> y = 16k + 9 and get a contradiction with y = 4n + 3. \n4) k = 1 and k = 2 give us all solutions.\n\nI hope it's correct[/quote]\r\nI like your explanation. thx u!" } { "Tag": [ "percent" ], "Problem": "In a exam there are 52% fail in english and 42% fail in math. if 17% have failed in both subject then what % of student have passed in both subject?\r\n\r\nneed the solution of this math. and can you tell me where can i find this kind of math's solution.....\r\n\r\nthanks in advance", "Solution_1": "[quote=\"mayukhbd\"]In a exam there are 52% fail in english and 42% fail in math. if 17% have failed in both subject then what % of student have passed in both subject?\n\nneed the solution of this math. and can you tell me where can i find this kind of math's solution.....\n\nthanks in advance[/quote]\r\n\r\n[hide]this kind of problem can be solved with venn diagrams.\n\n$52+42 = 94$\n\nbecause 17% failed in both subjects,\n$94-17 = 77$ failed in either english or math.\n\nIf someone didn't fail in english or math, he/she must have passed both english or math, so:\n\n$100-77 = 23$ percent failed in no subjects.\n\nbtw, that's pretty bad, unless it was an extra hard test. :P [/hide]", "Solution_2": "when you get percents just assume there are 100 people...\r\nthat makes it a bit easier.", "Solution_3": "[hide]$100-(52+42-17) = 23$[/hide]\r\n\r\nBtw, good advice, davidlizeng. :)", "Solution_4": "[quote=\"davidlizeng\"]when you get percents just assume there are 100 people...\nthat makes it a bit easier.[/quote]\r\n\r\nof course. :)" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Prove that\r\n(a+b)(a+c)>=2*sqrt[abc(a+b+c)] , for all positive reals a ,b , c.\r\n23 rd brazilian mo , 2001\r\nThanks...", "Solution_1": "Construct a triangle with edges a+b, b+c, c+a.\r\nThen we have RHS=2S=(a+b)(a+c)sinA \\leq LHS.\r\nQ. E. D.", "Solution_2": "I have proved it by AM-GM\r\n\r\n(a+b)(a+c)=a(a+b+c) +bc \\geq 2 \\sqrt abc(a+b+c)\r\n\r\nDon't you think it is more simple?", "Solution_3": "Wpolly's approach is utterly wonderful but more difficult to think of.", "Solution_4": "I agree with Arne....Wpolly's solution is more interesting but more difficult" } { "Tag": [ "algebra", "linear equation" ], "Problem": "I was looking at AoPS Volume 1 Chapter 3 and got extremly confused about what a linear equation can be made up of (like subtraction, addition etc.). I know linear equations can be made up of addition, and subtraction are there any other operations that can be included in linear equations?", "Solution_1": "A linear equation can only contain terms of the form $ c_kx_k$, where $ x_k$ is a variable and $ c_k$ is an arbitrary real number, and those terms can only be added or subtracted. Basically, any given term within the linear equation can only have one variable within the term, and that variable must be expressed to the first power.", "Solution_2": "A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. In other words, no variable is raised to a power higher than 1. Most linear equations take 2 common forms:\r\n\r\n1) ax^1+by^1=c\r\n2) ax^1=c1+c2\r\n\r\nwhere a and :D c are constants\r\nand x and y are variables\r\n :roll:", "Solution_3": "But would 8*7 or 8/2 be a linear equation?", "Solution_4": "8/7 or 8*12 is not even an equation. They are expressions.\r\n\r\nBut neither are linear because linear (by definition) means that variables must be raised to the first, and at most first, power." } { "Tag": [], "Problem": "double posts :(", "Solution_1": ":)", "Solution_2": ":| What's very hard/super hard?", "Solution_3": "Did you see [url=http://www.mathlinks.ro/viewtopic.php?t=291472]this?[/url]", "Solution_4": "Wait sorry I'm not very smart :blush: and I feel stupid for asking, but what is that?????", "Solution_5": "Oh, whoops. Didn't see that thread. Thanks, kunny! :)" } { "Tag": [ "trigonometry", "logarithms", "function", "trig identities", "algebra", "log identities" ], "Problem": "Seems that everyone knows their own collection of trig identities, but does anyone know any interesting (or not-so-interesting) log identities?", "Solution_1": "Well, uh...\r\n\r\n$\\log_{b} a = \\frac{\\log_{c} a}{\\log_{c} b}$\r\n$\\ln n < \\sum_{k=1}^{n} \\frac{1}{n} < \\ln n + 1$\r\n$\\ln n! \\approx (n + \\frac{1}{2}) \\ln n - n + \\frac{1}{2} \\ln (2 \\pi)$ (Stirling's approximation in natural log form)\r\n$z = re^{ix} \\implies \\ln z = \\ln r + ix$\r\n\r\nBeyond the basic identities and applications of Euler's formula, there's not much going on.", "Solution_2": "[quote=\"t0rajir0u\"]$z = re^{ix} \\implies \\ln z = \\ln r + ix$\n[/quote]\r\n\r\nThis wouldn't be exactly accurate:\r\n\r\n$z = re^{ix} \\implies \\ln z = \\ln r + i(x+2n\\pi), n\\in\\mathbb{Z}$", "Solution_3": "could you please explain how $ln(re^{ix})=ln(r)+ix$ is not correct.\r\ni read what you posted but was not able to follow.", "Solution_4": "In other words, incomplete. He also included the mutiples of $2\\pi$. for entirety. :)", "Solution_5": "Well, here are just some logarithmic identities off the top of my head.\r\n\r\n$\\\\\\log a+\\log b+...+\\log n=\\log ab...n\\\\ \\log a-\\log b-...-\\log n=\\log \\frac{a}{b...n}\\\\ \\log a^x=x\\log a\\\\\\\\ c^{\\log_cx}=x\\\\ \\log_cx=\\loc_cx\\\\ x=x\\implies\\text{ The idenity is true}$\r\n\r\nMasoud Zargar", "Solution_6": "[quote=\"maokid7\"]could you please explain how $ln(re^{ix})=ln(r)+ix$ is not correct.\ni read what you posted but was not able to follow.[/quote]\r\n\r\nLogarithm of a complex number behaves differently than logarithm of a real number, since it's closely connected to trigonometric functions, and therefore the multiples of $2\\pi$ must be included.\r\n\r\nLook what may happen if we follow [b]t0rajir0u[/b]'s identity: We know that $e^{i\\pi}=-1$ and $e^{-i\\pi}=-1$. From what he'd written, $\\ln (-1)=\\ln e^{i\\pi}=i\\pi$ and $\\ln (-1)=\\ln e^{-i\\pi}=-i\\pi$, and from there $\\pi=-\\pi$ (!)\r\n\r\nGenerally, $\\ln e^{i\\alpha}=\\ln e^{i\\beta}\\implies \\alpha=\\beta+2n\\pi,n\\in\\mathbb{Z}.$ (And in our example it gives $\\pi=-\\pi+2n\\pi$, which obviously holds for $n=1$.)\r\n\r\nTherefore, it's not the question of entirety but of correctness :)", "Solution_7": "\\[ \\boxed {a^{\\log_b c}=c^{\\log_b a}} \\]", "Solution_8": "[b]Proof[/b]\r\n$\\\\a^{\\log_bc}=c^{\\log_ba}\\\\ \\log_bc\\log a=\\log_ba\\log c\\\\ \\frac{\\log_bc}{\\log_ba}=\\frac{\\log c}{\\log a}\\\\ \\log_ac=\\log_ac\\implies\\text{ TRUE}$\r\n\r\nmasoud zargar" } { "Tag": [ "algebra", "function", "domain", "search", "\\/closed" ], "Problem": "In my subscription in my personal profile, there are two options:\r\n\r\nUse http://www.mathlinks.ro/ domain \r\nUse http://www.artofproblemsolving.com/ domain\r\n\r\nSo what is the difference between the two? Which one is better?", "Solution_1": "Use search, it's been explained many times.", "Solution_2": "What and where is \"search\"? :huh:", "Solution_3": "http://www.mathlinks.ro/search.php" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Given $ABCD$ the tetrahedron. Let $M$ be the midpoint of $CD$. What is the terminology of the plane $(ABM)$. Many thanks!!!", "Solution_1": "Any three noncollinear points define a unique plane, so plane $ABM$ would be the plane that goes through those 3 points.", "Solution_2": "The best thing I think you should call it be the \"median plane\"!" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Why does logarithmically convex imply convex? log(f(x)) convex and f>0 => f(x) convex", "Solution_1": "Suppose $f>0$ and that $\\log(f(x))$is convex.\r\nThen $\\frac{d^{2}}{dx}\\log(f(x))=\\frac{f''(x)f(x)-(f'(x))^{2}}{(f(x))^{2}}$.\r\nSince $\\log(f(x))$ is convex, then $\\frac{d^{2}}{dx}\\log(f(x))\\geq0$ i.e. $\\frac{f''(x)f(x)-(f'(x))^{2}}{(f(x))^{2}}\\geq0$.\r\nSince $f(x)>0$ by asumption and clearly $(f'(x))^{2}>0$ we have to have $f''(x)>0$ for this relation to hold.\r\nHence $f$ is a convex function." } { "Tag": [ "algebra", "polynomial", "functional equation", "algebra unsolved" ], "Problem": "Find all polynomials $ f(x)$,such that $ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)\\plus{}3xy(x\\plus{}y)$ for all x,y\r\n\r\n\r\n\r\n\r\n\r\n\r\n__________________________________\r\n$ \\textbf{Azerbaijan Land of Fire }$ :ninja: :rotfl:", "Solution_1": "Let $ f(x) \\equal{} x^3 \\plus{} p(x)$, where $ p(x)$ is some polynomial. You get: \r\n\r\n$ p(x \\plus{} y) \\equal{} p(x) \\plus{} p(y)$\r\n\r\nThis is Cauchy's functional equation, solution of which is $ p(x) \\equal{} cx$. So, $ f(x) \\equal{} x^3 \\plus{} cx$ for some $ c\\in\\mathbb{R}$.", "Solution_2": "why should $ f(x)\\equal{}x^3\\plus{}p(x)$ with $ p$ is a polynomial? :blush:", "Solution_3": "f(x) is a polynomial so f(x)=x^3 is a polynomial as well...", "Solution_4": "[quote=\"Altheman\"]f(x) is a polynomial so f(x)=x^3 is a polynomial as well...[/quote]\r\n\r\nI think Raja means \"Why would you pick that particular $ f(x)$? Where did it come from?\" which is a good question.", "Solution_5": "[quote=\"AdrianP\"][quote=\"Altheman\"]f(x) is a polynomial so f(x)=x^3 is a polynomial as well...[/quote]\n\nI think Raja means \"Why would you pick that particular $ f(x)$? Where did it come from?\" which is a good question.[/quote]\r\n\r\nIt's not very difficult to think to the equality $ (x\\plus{}y)^3\\minus{}x^3\\minus{}y^3\\equal{}3xy(x\\plus{}y)$" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "The incircle of triangle $ ABC$ with center $ I$ touches the sides $ AB$ and $ BC$ at points $ K$ and $ P$ respectively. The bissector of angle $ C$ intersects the segment $ KP$ at point $ Q$ and the straight line $ AQ$ intersects the side $ BC$ at point $ N$. Prove that points $ A;I;N$ and $ B$ lie at a common circle.", "Solution_1": "We need to prove $ CA \\equal{} CN$.(then $ \\angle ANB \\equal{} \\angle AIB$). $ CA \\equal{} CN$ is equivalent to $ \\angle AQI \\equal{} 90$. Let incircle touch $ CA$ at $ L$. Because $ P,L$ is symmetric wrt $ CQ, \\angle ILQ \\equal{} \\angle IPQ \\equal{} \\angle IQP$. Hence $ I,L,K,Q$ are cyclic. But obviously, $ A,K,I,L$ are cyclic. Hence $ A,K,Q,I,L$ are cyclic. Hence $ \\angle AQI \\equal{} \\angle AKI \\equal{} 90$.", "Solution_2": "[quote=\"tdl\"]The incircle of triangle $ ABC$ with center $ I$ touches the sides $ AB$ and $ BC$ at points $ K$ and $ P$ respectively. The bissector of angle $ C$ intersects the segment $ KP$ at point $ Q$ and the straight line $ AQ$ intersects the side $ BC$ at point $ N$. Prove that points $ A;I;N$ and $ B$ lie at a common circle.[/quote]\r\n[hide=\"Solution\"]\nLet $ \\angle BAC \\equal{} 2\\alpha$ and $ \\angle ACB \\equal{} 2\\beta$. Thus, we have that $ \\angle ABC \\equal{} 180 \\minus{} 2\\alpha \\minus{} 2\\beta$, so $ \\angle KIP \\equal{} 2\\alpha \\plus{} 2\\beta$. This implies that $ \\angle IPK \\equal{} \\frac {180 \\minus{} \\angle KIP}{2} \\equal{} 90 \\minus{} \\alpha \\minus{} \\beta$. Also,\n\\[ \\angle ICP \\equal{} \\beta\\implies \\angle PIQ \\equal{} 180 \\minus{} \\angle PIC \\equal{} 90 \\plus{} \\beta\\implies \\angle IQK \\equal{} \\angle PIQ \\plus{} \\angle QPI \\equal{} 90 \\plus{} \\beta \\plus{} 90 \\minus{} \\alpha \\minus{} \\beta \\equal{} 180 \\minus{} \\alpha\\implies \\angle KAI \\plus{} \\angle KQI \\equal{} 180\n\\]This gives us that $ AIQK$ is cyclic, so $ \\angle AQI \\equal{} \\angle AKI \\equal{} 90\\implies \\angle IQN \\equal{} 90\\implies \\angle IQN \\plus{} \\angle IPN \\equal{} 180$, so $ QIPN$ is cyclic. Hence, $ \\angle INP \\equal{} \\angle IQP \\equal{} \\alpha \\equal{} \\angle IAB$, so $ IABN$ is cyclic, as desired. [/hide]", "Solution_3": "Dear Mathlinkers,\r\n1. A, K, I, L, Q are cocyclic (Lascases's result; see http://perso.orange.fr/jl.ayme vol. 4, An unlikely concurrence...)\r\n2. I, P, N, Q are cocyclic by the hypothesis\r\n3. Applied a converse of the pivot theorem to triangle PBK and we are done.\r\nSincerely\r\nJean-Louis" } { "Tag": [ "real analysis", "algebra", "function", "domain", "real analysis theorems" ], "Problem": "when we have $ \\mu^* (\\mathfrak{b}) \\equal{} 0$ (for [i]any[/i] $ \\mathfrak{b}\\in \\mathbb{R}^n$), why is $ \\mathfrak{b}$ also lebesgue measurable? Does this have something to do with the fact that the domain $ X \\equal{} \\mathfrak{b}$ is restricted enough so that that it is a $ \\sigma$-algebra??\r\n\r\nI am making use of the theorem (perhaps incorrectly) that: If $ \\mu^*$ is an outer measure on $ X$ and $ \\mathcal{M}$ is the collection of all $ \\mu^*$ measurable substes of $ X$, then $ \\mathcal{M}$ is a $ \\sigma$-algebra, and the restriction of all $ \\mu^*$ to $ \\mathcal{M}$ is a measure on $ X$\r\n\r\nThanks!", "Solution_1": "Lebesgue measurable sets are (by one of the definitions) exactly those sets $ E$ that enjoy the well-separation property\r\n\\[ \\mu^*(A\\cup E)\\plus{}\\mu^*(A\\setminus E)\\equal{}\\mu^*(A)\\]\r\nfor every $ A\\subset\\mathbb R^n$ where $ \\mu^*$ is the Lebesgue outer measure. This property is easy to check when $ \\mu^*(E)\\equal{}0$. Hence the conclusion. So, you are right, this is directly related to the Caratheodory theorem you quoted." } { "Tag": [], "Problem": "The average value of nine consecutive integers is 13. What is\nthe sum of the smallest and largest of these integers?", "Solution_1": "the average of consecutive integers would be the middle integer, or in this case, the 5th\r\n\r\nso let's list them out:\r\n9, 10, 11, 12, 13, 14, 15, 16, 17\r\n\r\nthe sum is obviously $ 9\\plus{}17\\equal{}26$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all positive integers $ (x,y)$ such that $ \\frac{y^2x}{x\\plus{}y}$ is a prime number", "Solution_1": "$ p(x \\plus{} y) \\equal{} y^2x$.it follows that $ p|y$ or $ p|x$. Suppose that both of them are divisable by $ p$. so there exist $ a,b$ such that \r\n$ p^{a}\\parallel{}y,p^{b}\\parallel{}x$. there are 2 cases:\r\n1-$ a\\neq{b}$ then from the above equation we'll have $ 1 \\plus{} \\min{a,b} \\equal{} 2a \\plus{} b$ which is absurd.\r\n2-$ a \\equal{} b$ then we'll have $ (\\frac {x}{p^{a}} \\plus{} \\frac {y}{p^{a}}) \\equal{} p^{a \\minus{} 1}(\\frac {y^2}{p^{2a}})(\\frac {x}{p^{a}})$\r\nso if $ \\frac {y}{p^{a}} \\equal{} m,\\frac {x}{p^{a}} \\equal{} n$ we'll have $ m^2n\\leq{m \\plus{} n}$.we can easily find all the solutions.\r\nso assume that $ p$ divides only one of them. u can see that in this case we must have $ p|x$. if $ px' \\equal{} x$ u can see that \r\n$ y |x'$ it follows $ yt \\equal{} x'$. then we'll have:$ pt \\plus{} 1 \\equal{} y^2t$ the rest follows. i wish i didn't make any mistake I MUST GO TO BED!!", "Solution_2": "$ p\\equal{}\\frac {y^2x}{x \\plus{} y} \\iff px\\plus{}py\\equal{}y^2x$\r\n\r\nTaking mod $ x$ then $ x|py$ and taking mod $ y$ then $ y|px$\r\n(1) $ py\\equal{}ax$ \r\n(2) $ px\\equal{}by$\r\n\r\nTaking mod $ p$ then $ p|y^2x$ then either $ p|x$ or $ p|y$\r\nIf $ p|x$ then $ x\\equal{}kp$ then in (1) $ py\\equal{}akp$ then $ y\\equal{}ak$ and in (2) $ pkp\\equal{}bak$ then $ p^2\\equal{}ab$ then three cases:\r\n$ (a,b)\\equal{}(1,p^2)$ or $ (p^2,1)$ or $ (p,p)$\r\n\r\nIf $ p|y$ then $ y\\equal{}kp$ then in (2) $ px\\equal{}bkp$ then $ x\\equal{}bk$ and in (1) $ pkp\\equal{}abk$ then $ p^2\\equal{}ab$ then same three cases.\r\n\r\nCase 1) $ py\\equal{}x$\r\n$ p\\equal{}\\frac{y^3p}{y(p\\plus{}1)} \\iff p\\plus{}1\\equal{}y^2 \\iff p\\equal{}(y\\plus{}1)(y\\minus{}1) \\Rightarrow y\\equal{}2$ \r\nthen $ \\boxed{(x,y,p)\\equal{}(6,2,3)}$\r\n\r\nCase 2) $ y\\equal{}px$\r\n$ p\\equal{}\\frac{p^2x^3}{x(p\\plus{}1)} \\iff p(p\\plus{}1)\\equal{}x^2 \\Rightarrow p|x \\Rightarrow x\\equal{}cp \\Rightarrow p\\plus{}1\\equal{}pc^2 \\Rightarrow 1\\equal{}p(c\\plus{}1)(c\\minus{}1)$ \r\nno solution\r\n\r\nCase 3) $ y\\equal{}x$ then $ p\\equal{}\\frac{x^3}{2x}\\equal{}\\frac{x^2}{2}$ then $ \\boxed{(x,y,p)\\equal{}(2,2,2)}$", "Solution_3": "[quote=lambruscokid]Find all positive integers $ (x,y)$ such that $ \\frac{y^2x}{x\\plus{}y}$ is a prime number[/quote]\n\n$(x,y)=d$ $x=ad , y=bd$\n$\\frac{d^{2}\u00d7b^{2}\u00d7a}{a+b}$ it must be prime \n$(a,a+b)=1$ and $(b,a+b)=1$ then $\\frac{d^2}{a+b}$$=p$ it must be integer.$p\u00d7a\u00d7b^{2}$ is prime then $b=1$\n$\\frac{d^2}{a+1}$$=p$ then $p\u00d7a$ is prime $a=1$ or $a$ is prime and $p=1$\n$p=1$ then $d^2=a+1$ $a=d^2-1$ and $a$ is prime then $a=3$ $d=2$ \n$\\boxed{(x,y,q)=(2,2,2),(6,2,3)}$" } { "Tag": [ "geometry", "inequalities", "trigonometry", "geometry unsolved" ], "Problem": "Take $\\ A_{1}$ on the side $\\ BC$ of a triangle $\\ ABC$ so that the inciclers of triangle $\\ ABA_{1}, ACA_{1}$ have equal radii $\\ r_{a}$. One defines $\\ r_{b}, r_{c}$ analogously. Prove that:\r\n$\\ 2(r_{a}+r_{b}+r_{c})+p\\leq h_{a}+h_{b}+h_{c}$", "Solution_1": "[quote=\"April\"]Take $\\ A_{1}$ on the side $\\ BC$ of a triangle $\\ ABC$ so that the inciclers of triangle $\\ ABA_{1}, ACA_{1}$ have equal radii $\\ r_{a}$. One defines $\\ r_{b}, r_{c}$ analogously. Prove that:\n$\\ 2(r_{a}+r_{b}+r_{c})+p\\leq h_{a}+h_{b}+h_{c}$[/quote]What is $\\ p, h_{a},h_{b},h_{c}$ here?", "Solution_2": "Normally $h_{a}, h_{b}, h_{c}$ are altitudes dropped on $a,b,c$ respectively. $p$ stands for semiperimeter.(also known as $s$ in some places like USA and Canada :lol: )", "Solution_3": "I think it is false,it must be like this :\r\n$\\ 2(r_{a}+r_{b}+r_{c})+p\\geq h_{a}+h_{b}+h_{c}$\r\nLet $S$ be the area of triangle.\r\nWe can easyly prove that $2r_{a}=h_{a}(1-\\sqrt{tan\\beta.tan\\gamma})$ where $2\\alpha,2\\beta$ and $2\\gamma$ are the angles of triangle.\r\nWe know that for any triangle $\\frac{h_{a}-2r}{h_{a}}=tan\\gamma.tan\\beta$\r\nWriting it for triangles $AA_{1}B$ and $AA_{1}C$ and multiplying side by side we get $(\\frac{h_{a}-2r_{a}}{h_{a}})^{2}= tan\\gamma.tan\\beta$\r\nHense $2r_{a}=h_{a}(1-\\sqrt{tan\\beta.tan\\gamma})$ \r\nSo we must prove that \r\n$p \\geq h_{a}\\sqrt{tan\\beta.tan\\gamma}+h_{b}\\sqrt{tan\\alpha.tan\\gamma}+h_{c}\\sqrt{tan\\beta.tan\\alpha}$\r\nReplacing $tan\\alpha=\\frac{r}{p-a},tan\\beta=\\frac{r}{p-b}$ and $tan\\gamma=\\frac{r}{p-c}$ we get\r\n$p \\geq \\frac{h_{a}.r}{\\sqrt{(p-b)(p-c)}}+\\frac{h_{b}.r}{\\sqrt{(p-a)(p-c)}}+\\frac{h_{c}.r}{\\sqrt{(p-b)(p-a)}}$\r\n$p \\geq \\frac{\\frac{2S}{a}.\\frac{S}{p}}{\\sqrt{(p-b)(p-c)}}+\\frac{\\frac{2S}{a}.\\frac{S}{p}}{\\sqrt{(p-a)(p-c)}}+\\frac{\\frac{2S}{a}.\\frac{S}{p}}{\\sqrt{(p-b)(p-a)}}$\r\n$p^{2}\\geq \\frac{2S^{2}}{a\\sqrt{(p-b)(p-c)}}+\\frac{2S^{2}}{b\\sqrt{(p-a)(p-c)}}+\\frac{2S^{2}}{c\\sqrt{(p-b)(p-a)}}$\r\n$p \\geq (p-a)\\frac{2\\sqrt{(p-b)(p-c}}{a}+(p-b)\\frac{2\\sqrt{(p-a)(p-c}}{b}+(p-c)\\frac{2\\sqrt{(p-b)(p-a}}{c}$\r\nBy $AM-GM$ inequality $\\frac{2\\sqrt{(p-b)(p-c}}{a}\\leq \\frac{p-c+p-b}{a}=1$.So\r\n$p=p-a+p-b+p-c \\geq (p-a)\\frac{2\\sqrt{(p-b)(p-c}}{a}+(p-b)\\frac{2\\sqrt{(p-a)(p-c}}{b}+(p-c)\\frac{2\\sqrt{(p-b)(p-a}}{c}$\r\nThat's all. :) I think it is true.", "Solution_4": "[color=darkblue]See the my topic http://www.mathlinks.ro/Forum/viewtopic.php?t=50559 $\\Longrightarrow r_{a}=\\frac{r}{a}\\cdot[p-\\sqrt{p(p-a)}]$ a.s.o.\n\n[b]Remark.[/b] $\\boxed{\\ p+2\\cdot\\sum r_{a}\\ge \\sum h_{a}\\ }$ $\\Longleftrightarrow$ $\\|\\begin{array}{c}p+2pr\\cdot\\sum\\frac{1}{a}-2r\\sqrt p\\cdot\\sum\\frac{\\sqrt{p-a}}{a}\\ge 2S\\cdot\\sum\\frac{1}{a}\\\\\\\\ 2S=ah_{a}=2pr\\end{array}$ $\\Longleftrightarrow$ $\\boxed{\\ \\sum\\frac{\\sqrt{p-a}}{a}\\le \\frac{\\sqrt p}{2r}\\ }\\ \\ (*)\\ .$ \n\n[b]Proof of the inequality[/b] $(*)\\ .$\n\n$1.\\blacktriangleright\\ \\{x,y,z\\}\\subset (0,\\infty )\\Longrightarrow$ $\\frac{4yz}{(y+z)^{2}}\\le 1$ a.s.o. $\\Longrightarrow$ $\\frac{4xyz}{(y+z)^{2}}\\le x$ a.s.o. $\\Longrightarrow$ $\\sum_{\\mathrm{cyc}}\\frac{4xyz}{(y+z)^{2}}\\le x+y+z$ $\\Longrightarrow$ $\\boxed{\\ \\sum\\frac{1}{(y+z)^{2}}\\le \\frac{x+y+z}{4xyz}\\ }\\ \\ (1)\\ .$\n\n$2.\\blacktriangleright\\ x: =p-a$ a.s.o. in the inequality $(1)$ $\\Longrightarrow$ $\\|\\begin{array}{c}\\sum\\frac{1}{a^{2}}\\le \\frac{p}{4(p-a)(p-b)(p-c)}\\\\\\\\ (p-a)(p-b)(p-c)=pr^{2}\\end{array}$ $\\Longrightarrow$ $\\boxed{\\ \\sum\\frac{1}{a^{2}}\\le \\frac{1}{4r^{2}}\\ }\\ \\ (2)\\ .$\n\n$3.\\blacktriangleright$ Apply the $C.B.S.$- inequality : $(\\sum\\sqrt{p-a}\\cdot\\frac{1}{a})^{2}\\le \\sum (p-a)\\cdot\\sum\\frac{1}{a^{2}}=p\\cdot\\sum\\frac{1}{a^{2}}\\ .$ Using the inequality $(2)$ obtain the required inequality $(*)\\ .$\n\n[b]Remark.[/b] $\\|\\begin{array}{c}xy+yz+zx\\le x^{2}+y^{2}+z^{2}\\\\\\\\ x: =\\cot A\\mathrm{\\ a.s.o.}\\\\\\\\ \\sum\\cot B\\cot C=1\\end{array}$ $\\Longrightarrow$ $\\|\\begin{array}{c}\\cot^{2}A+\\cot^{2}B+\\cot^{2}C\\ge 1\\\\\\\\ \\frac{1}{cot^{2}x}=\\frac{1}{\\sin^{2}x}-1\\end{array}$ $\\Longrightarrow$ $\\|\\begin{array}{c}\\sum\\frac{1}{\\sin^{2}A}\\ge 4\\\\\\\\ a=2R\\sin A\\mathrm{\\ a.s.o.}\\end{array}$ $\\Longrightarrow$ $\\boxed{\\ \\frac{1}{R^{2}}\\le \\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\le \\frac{1}{4r^{2}}\\ }\\ .$[/color]\r\n\r\n[color=red][b]Open problem.[/b] $\\boxed{\\ \\frac{\\sqrt p}{R}\\le \\sum\\frac{\\sqrt{p-a}}{a}\\ }\\le\\frac{\\sqrt p}{2r}\\ .$[/color]" } { "Tag": [ "probability" ], "Problem": "I came up with this problem, but I'm not sure if it can be easily solved or not. Here it is: What is the probability that the sum of the digits of a 10-digit number is divisible by 10?", "Solution_1": "Actually came out to be a lot easier than I thought. For any possibility of the first 9 digits, one out of the 10 possible last digits will make the sum a multiple of 10. So the probability is 1/10.\r\n\r\nThe problem I thought of when I saw this one is slightly harder, but fun nevertheless.. I saw a real nice solution for it in one of our camps.\r\n\r\nWhats the probability (for a 10 digit number) that the digits are increasing (ie each digit is equal or greater than the previous one)?", "Solution_2": "Very nice solution from TripleM.\r\n\r\n\r\nIf I understand correctly, his new problem if to find the number\r\nof numbers built yhis way :\r\na1a2...a10 with 1<= a1 <= a2 ... <= a10\r\n\r\n\r\nI mean the sum of the digits do not have to be divisible by 10 any more ?\r\n\r\n\r\n\r\nLet's try to generalize to n-digits numbers. I don't have a quick and astute solution as the one of TripleM, but I checked it by hand for small values of n and it seems to be correct.\r\n\r\n\r\n\r\nI claim that a number with increasing digits is of the form (in a unique way) :\r\nb1*ONE(n) + b2*ONE(n-1) + ... bn*ONE(1) \r\nwith b1 + b2 + ... + bn <= 9 and b1 >= 1\r\n\r\nwhere ONE(k) stands for the number with k '1' (ONE(3) = 111 to be clear)\r\n\r\nFor instance 378 = 3*111 + 4*11 + 3*1,\r\n\r\nThe condition b1 + b2 + ... + bn <= 9 is because carries would cancel the increase.\r\n\r\n\r\nSo we have to solve :\r\n\r\nb1 + .. + bn <= 9 with b1 >= 1.\r\n\r\nIt's the same as (1+b1') + b2 + ... + bn + B = 9\r\n\r\nBy De Moivre theorem there are binomial (8+n, n) solutions.\r\n\r\n\r\nSo the probability to have an increasing number is : \r\n\r\nbinomial (8+n, n) / 9*10^(n-1)\r\n\r\n\r\nfor n = 2 .. 10 : \r\n\r\n2 : .050\r\n3 : .0183\r\n4 : .0055\r\n5 : .00143\r\n6 : .0003336\r\n7 : .0000715\r\n8 : .0000143\r\n9 : .270111 10^-5\r\n10 : .4862 10^-6 ~ 1/2000000 (1 chance out of 2 millions)\r\n\r\n\r\n\r\n\r\nP.S. Forgive poor English", "Solution_3": "Triple M \r\n\r\nSo does that mean if all the digits are the same, it is considered increasing?\r\n ie 1111111111", "Solution_4": "If d(2) can be greater than or equal to d(1), then i get the answer as \r\n387420489/1000000000 (that should be 9 zero's). I get this because for the first digit, you have 9 possibilities. After that, you have another 9 possibilities, etc. So you have 9^10 possibilities. But then you have to divide that by the total number which is 9 for the first digit, and then 10 for all after that, i.e. 9 x 10^9.", "Solution_5": "TheSouthpaw: Yes, I never understood it really but I think the standard thing is increasing means they can be equal and strictly increasing means they can't.\r\n\r\nJelyman: The problem is that we have 9 possibilities for each number, as for example if we start with '15..' then there are only 5 possibilities for the next number.\r\n\r\ni/3s answer matches up with mine. Heres the 'nice' way I was taught:\r\nWe have 9 boxes for each of the 9 digits 1 thru 9, and we place a ball in a box for each time that digit occurs. So for example 1145577779 would be\r\n[00][][][0][00][][0000][][0]. After writing this on the whiteboard he sneakily rubbed out a couple of lines , replacing each ][ with a 1 and getting 001110100110000110. And so we all realised the answer was the number of ways of choosing 10 zeroes out of 18 places, ie C(18,10) or C(18,8).\r\n\r\nThought it was real nice..", "Solution_6": "I have no idea how that works but it does...can anyone explain triplem's method for me?", "Solution_7": "Sorry I didn't explain that very well. Every such number can be represented as a combination of 10 zeroes for the 10 digits, and 8 ones where each one implies the next digit goes up by one... and all we have to do is find all ways we can find such a combination. Nope I still don't make sense. Someone else help me here :)\r\n\r\nMaybe a smaller case: consider 2 digit numbers only using the digits 1 thru 3.\r\n[][][] three boxes, first represents number of 1s, next number of 2s, last one number of 3s. All we have to do is put two balls in them in some way.\r\nnumber 11 = [00][][] = 0011\r\nnumber 12 = [0][0][] = 0101\r\nnumber 13 = [0][][0] = 0110\r\nnumber 22 = [][00][] = 1001\r\nnumber 23 = [][0][0] = 1010\r\nnumber 33 = [][][00] = 1100\r\n\r\neach is made up of 2 ones and 2 zeroes..\r\n\r\nHope this helps, someone else can try to explain hopefully..", "Solution_8": "Meanwhile, that was only part of the problem, probably the hardest bit but we'll see. \r\nThe actual problem (difficulty pretty high):\r\nFind the of all numbers between 0 and (to make the numbers bearable) 1000000 that have either decreasing OR increasing digits. \r\nFor example some of the numbers are: 11689, 55555, 921000.", "Solution_9": "hmmm...which is easier\r\n\r\nfind all numbers that dont increase or decrease\r\n\r\nfind all numbers that do increase or decrease...\r\n\r\nwow...this is really hard...", "Solution_10": "[quote=\"akatookey\"]hmmm...which is easier\n\nfind all numbers that dont increase or decrease\n\nfind all numbers that do increase or decrease...\n\n[/quote]\r\n\r\nBasically... neither. :)", "Solution_11": "[quote=\"TripleM\"]Find the of all numbers between 0 and (to make the numbers bearable) 1000000 that have either decreasing OR increasing digits. \nFor example some of the numbers are: 11689, 55555, 921000.[/quote]\r\n\r\n\r\n\r\nI don't have time today to work on this nice problem but I would like to know if I am on the right way to the solution.\r\n\r\n\r\nLet BIG(n) = 1....10 (with n '1' and one '0') = 10*(10^n-1)/9\r\n\r\n\r\nLemma :\r\nIf In is an increasing-digits number with n digits then (BIG(n) - In) is a decreasing-digits number.\r\n\r\n\r\nThe converse is almost true : \r\nif Dn is a decreasing-digits number with no '0' inside (because of carries :-() then BIG(n) - Dn is an increasing-digits number.\r\n\r\n\r\nSo there is a one to one relationship between decreasing-digits numbers with no '0' inside and increasing-digits numbers. Numbers like 1......1 are both increasing-digits and decreasing-digits\r\n\r\n\r\nThe sum of {in|de}creasing-digits numbers with no zero inside is then : \r\n\r\nC(8+n,n) * BIG(n) - (1...1 + 2....2 + 9...9)\r\n\r\n\r\nThe problem is then to sum decreasing-digits numbers with '0' at the end.\r\nAm i far from it :?: :?:", "Solution_12": "On the right track. If you consider all the numbers as 6 digit numbers which can allow leading zeroes, then in fact you don't need a BIG(n), just a single BIG...", "Solution_13": "OK heres the solution.. (forget about 1000000 being decreasing at the moment I really meant less than a million but you can add on a million at the end.)\r\n\r\nWe can pair up every number with increasing digits with one with decreasing digits by subtracting it from 999999, eg 001256 -> 998743. Every one can be paired up in this way, and as we know from before there are 14C8 = 3003 increasing digit numbers and thus 3003 pairs each summing to 999999. So 3003 x 999999 is the sum of all pairs. Have we counted any numbers twice? Yes, all those which are both increasing AND decreasing, ie 111111,222222,...,999999. So the answer is 3003x999999 - the sum of those 9 numbers.\r\n(which is 2997997002 but thats not the point. Oh and add on 1000000 if you want :P)." } { "Tag": [ "Princeton", "college", "Columbia", "email", "ARML" ], "Problem": "The cost of the trip has been set at $\\$$130 per person. Money should be turned in to Mr. R on the day of the trip, unless you see him in person and can give him the money sooner. The seven SC students selected to participate in this trip are:\r\n\r\nSean\r\nShobhit\r\nJim \r\nAsif \r\nHarry \r\nNoah \r\nBrian \r\n\r\nThe 6 Columbia students will be driven to the airport in Charlotte by Mr. R after signing field trip forms/legal liability waivers. The plane will take off from Charlotte at x: xx on Friday, December 15th, and arrive in Princeton at x: xx. The return flight from Princeton will be on Sunday the 17th. We depart at x: xx, and arrive in Charlotte at x: xx. Mr. R will then drive the 6 Columbia students back to Columbia. \r\n\r\nPlease bring money for transportation in Princeton. This should be at least $\\$$xx.xx. Also, bring an ample amount of cash for food. You can always bring back leftover money.\r\n\r\nIf we want t-shirts, somebody send ideas to me via email ASAP. My parents can pay for (sponsor) these.\r\n\r\nAny questions, email me at the address below. Suggestions requested as well. Everything seems to finally be on track.\r\n\r\n[b]NC people reading this[/b], everything is set for you. You don't owe us any money, just bring enough for the food and travel expenses outlined 3 paragraphs above.\r\n\r\nEverything is finally falling in place, and this seems like it will be a great trip. Many thanks to Mr. R, Shobhit's mom, my parents, Mr. Lightsey, BellSouth, and everyone else who made this trip possible.\r\n\r\nSean \r\nseansoni@gmail.com", "Solution_1": "we want t-shirts, how bout top half be garnett and bottom half be unc blue or something, with some kind of expensive looking logo", "Solution_2": "OK, if the people can do t-shirts in time, we have a design. If not, people wear your pink (NC) or blue (SC) ARML shirts from last year.", "Solution_3": "Great success!\r\n\r\nResults [url=http://cgi.math.princeton.edu/mathclub/index.php/PUMaC_2006_Results]here[/url].", "Solution_4": "whoa arnav got beaten by an 8th grader, lol\r\n\r\nBerman pwned though... YEAH GO BERMAN\r\n\r\nMan, I want to go next year", "Solution_5": "the highlight of the trip was when mr. r kept showing off the trophy and every time someone asked about it, he would reply:\r\n\r\n\"why thank you\r\n\r\ni was hoping you'd ask about that\"", "Solution_6": "[quote=\"Iversonfan2007\"]the highlight of the trip was when mr. r kept showing off the trophy and every time someone asked about it, he would reply:\n\n\"why thank you\n\ni was hoping you'd ask about that\"[/quote]\r\n\r\nWhat was with that guy who was drunk or something at the train station? He was like \"My wife's birthday is today. Sing her a song!\"", "Solution_7": "guy: \"so which teams were competition\"\r\n\r\nmr. r: \"um, South and North Carolina--\"\r\n\r\nguy: \"no, tell me some real teams\"", "Solution_8": "Does anyone have a link to the contest or the Contest Questions itself?", "Solution_9": "Yea, go to the Princeton forum, and there is a linked PDF of the problems in the Solutions thread.", "Solution_10": "Thank You Very Much\r\n\r\nPS\r\nThe Problems seem to be good prep for AIME." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "algorithm" ], "Problem": "Does anyone know why r was chosen as the letter for the correlation coefficient in statistics. Just wondering.", "Solution_1": "I have no idea, maybe r for relationship?", "Solution_2": "This is sortof one of those questions like \"Why do we use m to represent slope?\" I don't know the answer but here are some suggestions:\r\n\r\n(1) Its an arbitrarily assigned letter.\r\n(2) There is something in the algorithim for finding R that involves division, hence RATIO.\r\n(3) It was invented by a guy w/a name that started w/R.\r\n(4) Maybe it means something like \"ratings\" or \"ranking,\" as in how close the data points rank in terms of being the least s.d.s away from the regression line.\r\n\r\n\r\nAll I can think of...", "Solution_3": "also, it may have been first based on a word in another language... as was much of our current english math nominclature.", "Solution_4": "Yes, like supposedly m means \"monter\" which is french for \"to climb.\"", "Solution_5": "I'm going to guess that it's because [tex]r^2[/tex] is [url=http://mathworld.wolfram.com/CorrelationCoefficient.html]based on[/url] the squared [i]residuals[/i] of the data points to the least squares fit." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "[color=blue][b]Problem.[/b]\n Let $a, b, c$ three nonnegative reals. Prove the following inequality\n\n \\[{ (a^{2}+ab+b^{2})(b^{2}+bc+c^{2})})(c^{2}+ca+a^{2})\\geq\\frac{27}{64}(a+b)^{2}(b+c)^{2}(c+a)^{2}. \\]\n\n [/color]", "Solution_1": "[quote=\"Cezar Lupu\"][color=blue][b]Problem.[/b]\n Let $a, b, c$ three nonnegative reals. Prove the following inequality\n\n \\[{ (a^{2}+ab+b^{2})(b^{2}+bc+c^{2})})(c^{2}+ca+a^{2})\\geq\\frac{27}{64}(a+b)^{2}(b+c)^{2}(c+a)^{2}. \\]\n\n [/color][/quote]\r\nToo easy\r\nJust use $a^{2}+ab+b^{2}\\geq\\frac{3}{4}(a+b)^{2}$\r\nAnd we don't need $a,b,c\\geq 0$ just $a,b,c\\in R$" } { "Tag": [ "geometry", "geometric transformation", "calculus", "integration", "inequalities", "trigonometry", "LaTeX" ], "Problem": "The 1st Southeast China Mathematical Olympiad was held in Wenzhou, Zhejiang province, July 10-11, 2004.\r\n106 students of class 10, from Zhejiang, Jiangxi and Fujian provinces, take part in this contest.\r\nMaybe there are lots of mistakes in my translation, because my English is not very good. I'll be appreciative if someone can tell me the mistakes for my improvement.\r\nHere are the problems!\r\nHave a good day!\r\n :)\r\n\r\n[b]Edited by Myth[/b]: added problems and links to the separate topics.\r\n\r\n[b]Problem 1.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26221\r\nLet $a, b, c$ be reals such that $a^2+2b^2+3c^2=\\frac{3}{2}$. Prove that $3^{-a}+9^{-b}+27^{-c}\\geq 1$.\r\n\r\n[b]Problem 2.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26225\r\nLet $D$ be a point on side $BC$ of triangle $ABC$ and point $P$ on segment $AD$. A line through $D$ intersects segments $AB$, $PB$ at $M$, $E$ respectively, also intersects the extensions of segments $AC$, $PC$ at $F$, $N$ respectively. \r\nIf $DE = DF$, prove that $DM =DN$.\r\n\r\n[b]Problem 3.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26216\r\n[b]1.[/b] Does there exists positive integral infinite sequences $\\{a_n\\}$ such that $a_{n+1}^2\\geq 2a_{n+2}a_{n}$ for any positive integer $n$?\r\n\r\n[b]2.[/b] Does there exists positive irrational infinite sequences $\\{a_n\\}$ such that $a_{n+1}^2\\geq 2a_{n+2}a_{n}$ for any positive integer $n$?\r\n\r\n[b]Problem 4.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26222\r\nGiven positive integer $n > 2004$. Numbers 1, 2, 3, \u2026, $n^2$ were filled in the squares of an $n\\times n$ checkerboard such that every square had just one number. Square $A$ was called \"good\" if\r\n\r\n[b](i)[/b] The number filled in $A$ is larger than at least 2004 numbers in the squares at the same row of $A$.\r\n[b](ii)[/b] The number written in $A$ is larger than at least 2004 numbers in the squares at the same column of $A$.\r\n\r\nFind the greatest number of \"good\" squares in $n\\times n$ checkerboard.\r\n\r\n[b]Problem 5.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26219\r\nIf the inequality $\\sqrt{2}(2a+3)\\cos\\left(x-\\frac{\\pi}{4}\\right)+\\frac{6}{\\sin x+\\cos x}-2\\sin 2x<3a+6$ holds for any real $x\\in[0,\\pi/2]$, find the value range of real number $a$ can be taken.\r\n\r\n[b]Problem 6.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26226\r\nLet $D$ be a point on base side $BC$ of isosceles triangle $ABC$. Point $F$ in triangle $ABC$ is situated on the arc of the circle passing through points $A$, $D$ and $C$. The circle passing through points $B$, $D$ and $F$ intersects side $AB$ at point $E$. Prove that \\[CD\\cdot EF + DF\\cdot AE = BD\\cdot AF.\\]\r\n\r\n[b]Problem 7.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26224\r\n$n$ teams take part in a tournament. For any two teams $A$ and $B$, if the game is played at $A$'s town, then call this game as $A$'s \"home game\", as well as $B$'s \"away game\". Any two teams play against each other just twice, each having a \"home game\" respectively. Knowing that every team can be arranged more than one \"away game\" in one week (from Sunday to Saturday); but if one team has \"home game\" in certain week, then it cannot be arranged \"away game\" in that week.\r\n\r\nIf all the games can be arranged to finish in four weeks, find the maximum possible value of $n$.\r\n\r\n[b]Problem 8.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26220\r\nFind the number of the ordered integral number groups $(x, y, z, u)$ such that $1\\leq x,y,z,u\\leq 10$ and\r\n\\[\\frac{x-y}{x+y}+\\frac{y-z}{y+z}+\\frac{z-u}{z+u}+\\frac{u-x}{u+x}>0.\\]", "Solution_1": "Obviously your English is good I think though:\r\n\r\n- in 3. (1) should be \"integer\" instead of \"integral\"\r\n- I'm not native so I won't point out mistakes in English, cuz I might be wrong also :D \r\n\r\nBut what i wanted to say: thanks for good job. :)", "Solution_2": "It is very interesting!\r\n\r\nThank you, zhaoli!\r\n\r\nBut where are separate topics for each problem? ;)", "Solution_3": "I do not know how to use LATEX, so it is difficult for me to set up separate topics in corresponding catlogs. Would you please help me to do the job, Myth?", "Solution_4": "Rrrr... [b]M[/b]yth", "Solution_5": "billzhao has just informed me that in problem 7, we may say \"home game\" for \"\u4e3b\u573a\" and \"away game\" for \"\u5ba2\u573a\", which I translated them to \"host game\" and \"guest game\", respectively. \r\nHope that it will not affect your understanding the meaning of the problem.", "Solution_6": "is it possible for hong kong to join it next yr? :?", "Solution_7": "Myth:\r\n\r\nI have also translated the 18 problems of 2004 China TST one hour ago, which posted in\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=164145#p164145\r\n\r\nAre you interested?", "Solution_8": "I have done all slave's work! See first post!", "Solution_9": "[quote=\"Myth\"]I have done all slave's work! See first post![/quote]Thank you Mikhail! :)", "Solution_10": "Myth:\r\nThanks a lot and God bless you!" } { "Tag": [ "integration", "real analysis", "real analysis solved" ], "Problem": "Find $inf \\sum_{n=0}^{\\infty}u_nS_n$ where $S_n=\\sum_{k=0}^n u_k$ and $\\sum u_n$ is a serie of positive terms with sum equal to 1.(Search the inf after all series which satisfy the hypothesis.)", "Solution_1": "Unless I'm overlooking something, it seems obvious that the infimum is $\\frac 12$. \r\n\r\nLet $S$ be the sum of the products $u_iu_j,\\ i b > c > d$ be positive integers and suppose that \\[ ac + bd = (b+d+a-c)(b+d-a+c). \\] Prove that $ab + cd$ is not prime.", "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url] :)", "Solution_2": "[b]Solution by Iura:[/b]\n\n[b]Lemma[/b]. If $ab=cd$ then there exist $m,n,p,q$ with $a=mn, b=pq, c=mp, d=nq$. [i]Proof[/i] is obvious. \n\nOpening the brackets we get $a^2+c^2-ac=b^2+bd+d^2$\nMultiplying by $4$ we get $(2a-c)^2+3c^2=(2b+d)^2+3d^2$ thus $3(c+d)(c-d)=(2b+d-2a+c)(2b+d+2a+c)$.\nNow using this Lemma we get either \n\\[c+d=mn, c-d=pq, 2a+c-2b+d=3mp, 2b+d+2a+c=nq,\\] \nor \n\\[c+d=mn, c-d=pq, 2a+c-2b+d=mp, 2b+d+2a+c=3nq.\\]\nFor the first case, notice that $16(ab+cd)=(3p^2+n^2)(3m-q)(m+q)$, and it's easy to handle it now.\nThe second case is analogous.", "Solution_3": "this is the solution i found.it is in the same spirit as the above one but a bit different.let$a=x+y+z+d ,b=x+y+d,c=d+x$.using this it is easy to see that the given relation is equivalent with $xy=(z-d)(2d+2y+x+z)$and $ab+cd=(2d+2y+x+z)(d+x)+y(y+z)$.but it is clear that $gcd(2d+2y+x+z,y)=t>1$and it is also clear that $t|ab+cd$so our number is not prime and the problem is over. :)", "Solution_4": "I'm aware that there exists a solution using the Eisenstein integers. Could someone post it?", "Solution_5": "a>b>c>d>0\r\n ac+bd=(b+d+a-c)(b+d+c-a) (*)\r\nsuppose that ab+cd -prime \r\n (*) hence (b+d+a-c)l(ac+bd) we have\r\n (b+d+a-c) l (a)(b+d+a-c)+(ac+bd)=ab+ad+a^2+bd=(a+b)(a+d) (**)\r\n\r\n (*) hence (b+d+c-a) l (ac+bd) we have\r\n (b+d+c-a) l (b+d+c-a)(c)+(ac+bd)=cb+cd+c^2+bd=(c+b)(c+d) (***)\r\n(**) (***) hence (b+d+c-a)(b+d+a-c) l [(a+b)(c+d)][(a+d)(b+c)] at this we have\r\n (ac+bd) l ((ac+bd)+(ad+bc))((ac+bd)+(ab+cd)) (****)\r\n\r\n(a-d)(b-c)>0 therefore ab+cd>ac+bd and (ab+cd)-is prime \r\nhence ((ac+bd);((ac+bd)+(ad+bc)))=1 (*****)\r\n\r\n(****) and (*****) we can say (ac+bd) l (ac+bd)+(ad+bc) at this we have (ac+bd) l (ad+bc)\r\n (ac+bd)<=(ad+bc) it is equalent (a-b)(d-c)>0 contradiction!\r\n ab+cd - is not prime", "Solution_6": "O very nice solutions ALISHER", "Solution_7": "I was trying the problem in a different approach but could not complete it. Here is my mode of approach. :maybe: \r\n\r\nGiven that a>b>c>d and (ac+bd)=(b+d+a-c)(b+d-a+c)\r\nor, ac+bd=(b+d)^2-(a-c)^2\r\nor, ac+bd=b^2+d^2+2bd-a^2-c^2+2ac\r\nor, ac+bd=a^2+c^2-b^2-d^2=(a+b)(a-b)+(c+d)(c-d)=(a+d)(a-d)-(b+c)(b-c)\r\n\r\nNow, I was trying to use the identity, (ab+cd)+(ac+bd)=(b+c)(a+d)\r\nand (ab+cd)-(ac+bd)=(a-d)(b-c)\r\nwhich gives 2[(a+b)(a-b)+(c+d)(c-d)]=(b+c)(a+d)+(a-d)(b-c)........", "Solution_8": "weird solution: :maybe: \r\nwe have:\r\n\r\n$ ac+bd=(b+d+a-c)(b+d-a+c)\\iff\\boxed{a^2-ac+c^2=b^2+bd+d^2}$ (*)\r\n\r\nnow let $ ABCD$ be a quadrilateral with $ AB=a,BC=d,CD=b,AD=c$ and $ \\angle BAD=60^\\circ$ and $ \\angle BCD=120^\\circ$.now note that according to (*) and $ \\cos$ law there exists such quadrilateral because each side of (*) is equal to $ BD^2$.\r\n\r\nnow let $ \\angle ABC=\\alpha$ hence $ \\angle CDA=180^\\circ-\\alpha$.now according to $ \\cos$ law in $ \\triangle ABC$ and $ \\triangle ACD$ we have:\r\n\r\n$ a^2+d^2-2ad\\cos\\alpha=AC^2=b^2+c^2+2bc\\cos\\alpha$\r\n\r\nhence:\r\n\r\n$ 2\\cos\\alpha=\\frac{a^2+d^2-b^2-c^2}{ad+bc}$\r\n\r\n\\begin{eqnarray*}\\Rightarrow AC^2=a^2+d^2-ad.\\frac{a^2+d^2-b^2-c^2}{ad+bc}=\\frac{(ab+cd)(ac+bd)}{ad+bc}\\end{eqnarray*}\r\n\r\nnow note that $ ABCD$ is cyclic,hence according to ptolemy's theorem we get that:\r\n\r\n$ (AC.BD)^2=(ab+cd)^2$\r\n\r\nhence:\r\n\r\n$ (ac+bd)(a^2-ac+c^2)=(ab+cd)(ad+bc)$ (**)\r\n\r\nnow note that:\r\n\r\n$ ab+cd>ac+bd>ad+bc$ (***)\r\n\r\n(the first one follows from $ (a-d)(b-c)>0$ and the second one from $ (a-b)(c-d)$)\r\n\r\nnow assume that $ ab+cd$ is a prime number,first of all from (***) we get that $ ab+cd$ and $ ac+bd$ are relatively prime.so by (**) $ ac+bd$ must devide $ ad+bc$ which is a contradiction according to (***)...", "Solution_9": "Here is my solution.\r\n\r\nLet $ P(x)\\equal{}(x\\plus{}b)(x\\plus{}d)\\minus{}(x\\minus{}a)(x\\plus{}c)$\r\n$ \\textcolor{white}{Let P(x)}\\equal{}(b\\plus{}d\\plus{}a\\minus{}c)x\\plus{}ac\\plus{}bd$\r\n$ \\textcolor{white}{Let P(x)}\\equal{}(b\\plus{}d\\plus{}a\\minus{}c)(x\\plus{}b\\plus{}d\\minus{}a\\plus{}c)$.\r\n$ P(a)\\equal{}(a\\plus{}b)(a\\plus{}d)\\equal{}(b\\plus{}d\\plus{}a\\minus{}c)(b\\plus{}c\\plus{}d)$.\r\n$ P(a)\\neq 0$, so $ b\\plus{}d\\plus{}a\\minus{}c \\mid (a\\plus{}b)(a\\plus{}d)$\r\nbut $ a\\plus{}b>b\\plus{}d\\plus{}a\\minus{}c>\\dfrac{a\\plus{}b}{2}$, therefore a part of $ b\\plus{}d\\plus{}a\\minus{}c$ must divide $ a\\plus{}d$.\r\nThus, we have $ K\\equal{}(b\\plus{}d\\plus{}a\\minus{}c,a\\plus{}d)\\equal{}(b\\minus{}c,a\\plus{}d)\\equal{}(b\\minus{}c,b\\plus{}d\\plus{}a\\minus{}c)\\neq 1$.\r\nSince $ K\\mid a\\plus{}d$, we have $ K\\le a\\plus{}d 0$ with $ a, b, c, d > 0$. This means that $ \\frac {a^2 \\plus{} c^2 \\minus{} x^2}{2ac} \\equal{} \\frac {1}{2} \\equal{} \\cos 60$. Similarly, $ \\frac {b^2 \\plus{} d^2 \\minus{} x^2}{2bd} \\equal{} \\frac { \\minus{} 1}{2} \\equal{} \\cos 120$. This means that we can consider cyclic quadrilateral $ ABCD$ with $ AB \\equal{} a$, $ BC \\equal{} c$, $ CD \\equal{} b$, and $ AD \\equal{} d$ with $ \\angle ABC \\equal{} 60$ and $ \\angle CDA \\equal{} 120$. Ptolemy's Theorem here gives that $ ab \\plus{} cd \\equal{} AC\\cdot BC$. Notice that $ BD \\equal{} x \\equal{} \\sqrt {(b \\plus{} d \\minus{} a \\plus{} c)(b \\plus{} d \\plus{} a \\minus{} c)}$. Also, $ AC > AD,CD$ by the angle condition. Furthermore, $ BD$ is greater than both of $ CB$ and $ DC$ or $ AD$ and $ AB$ since either $ \\angle BAC$ or $ \\angle BDC$ is right or obtuse. This means that $ AC, BD > 1$. Notice that $ AC$ is the squareroot of a natural number and since $ ab \\plus{} cd \\equal{} AC\\cdot BD$ is a natural number, we see that $ BD$ is in the same form as $ AC$. Yet, if $ x^2$ composite, then $ BD^2\\cdot AC^2$ must be composite, so $ BD\\cdot AC$ is composite. Then, $ x^2$ is prime. So must be $ BD^2$. If they are different primes, then $ BD\\cdot AC$ is not an integer. Then, $ BD \\equal{} AC$, so $ ABCD$ is an isoceles trapezoid. This means that either $ a \\equal{} b$ or $ c \\equal{} d$, which contradicts the problem statement. Thus, we have a contradiction, so $ ab \\plus{} cd$ is composite, as desired. \n\n[b]Remark:[/b] I believe that this is quite an interesting problem! I rarely see such geometry combined with number theory :) By the way, just to make the motivation for the geometric construction clear, it is not only the observation of the Law of Cosines that should make one think of geometry (that is rarely enough.) It is also the \"Ptolemyness\" of the problem; furthermore, notice that $ b \\plus{} d \\plus{} c \\minus{} a \\equal{} 2(s \\minus{} a)$ when $ s$ is the semiperimeter of $ ABCD$. This seems to represent Bhramagupta's Theorem in a way, since\\[ \\frac {\\sqrt {3}}{2}\\cdot 2(s \\minus{} a)(s \\minus{} c) \\equal{} \\frac {\\sqrt {3}}{2}\\cdot \\frac {(ac \\plus{} bd)}{2} \\equal{} \\frac {ac\\sin 60}{2} \\plus{} \\frac {bd\\sin 120}{2} \\equal{} [ABCD] \\equal{} \\sqrt {(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)(s \\minus{} d)}\\]This yields that $ 3(s \\minus{} a)(s \\minus{} c) \\equal{} (s \\minus{} c)(s \\minus{} d)$, which is quite nice (and quite close to what [b]Iura[/b] got. I am not sure if a proof with the Four Number Theorem follows from there; it seems quite possible...) [/hide]\n\nEDIT: Sorry, this solution is incorrect. $ BD$ can be in the form of $ \\frac {q^2}{p}$ for some $ p$, $ q$ primes.\n\nEDIT 2: I hope this works:\n[hide=\"Solution\"]\nFrom the remark before, we have that $ 3(a\\plus{}b\\plus{}d\\minus{}c)(b\\plus{}d\\plus{}c\\minus{}a)\\equal{}(a\\plus{}b\\plus{}c\\minus{}d)(a\\plus{}c\\plus{}d\\minus{}b)$. Then, for some $ w$, $ x$, $ y$, $ z$, $ a\\plus{}b\\plus{}c\\minus{}d\\equal{}3xy$, $ a\\plus{}c\\plus{}d\\minus{}b\\equal{}wz$, $ a\\plus{}b\\plus{}d\\minus{}c\\equal{}wx$, and $ b\\plus{}d\\plus{}c\\minus{}a\\equal{}yz$ (the roles of $ a\\plus{}b\\plus{}c\\minus{}d$ and $ a\\plus{}c\\plus{}d\\minus{}b$ can be switched.) This gives us that $ \\frac{3xy\\plus{}wz\\plus{}xw\\plus{}yz}{2}\\equal{}a\\plus{}b\\plus{}c\\plus{}d$. Then, we can solve for $ a$, $ b$, $ c$, and $ d$ and get that $ a\\equal{}\\frac{3xy\\plus{}wz\\plus{}xw\\minus{}yz}{4}$, $ b\\equal{}\\frac{3xy\\minus{}wz\\plus{}xw\\plus{}yz}{4}$, $ c\\equal{}\\frac{3xy\\plus{}wz\\minus{}xw\\plus{}yz}{4}$, $ d\\equal{}\\frac{wz\\plus{}xw\\plus{}yz\\minus{}3xy}{4}$. Then, \\[ ab\\plus{}cd\\equal{}\\frac{9x^2y^2\\plus{}w^2x^2\\plus{}6x^2yw\\minus{}y^2z^2\\minus{}w^2z^2\\plus{}2yz^2w\\plus{}w^2z^2\\plus{}y^2z^2\\plus{}2yz^2w\\minus{}9x^2y^2\\minus{}w^2x^2\\plus{}6x^2yw}{16}\\] Yet, this means that $ ab\\plus{}cd\\equal{}\\frac{yw(3x^2\\plus{}z^2)}{4}$, which is composite, as desired. [/hide]", "Solution_11": "I think it is an easy problem for imo6 \r\nI think it is good for imo1 :lol:", "Solution_12": "[quote=\"ufuk\"]I think it is an easy problem for imo6 \nI think it is good for imo1 :lol:[/quote]\r\n\r\nagreed. The BaBak's solution is quite predictable, upon expanding, the next step gets pretty clear, though that's too tough to be given as imo1.", "Solution_13": "[quote=\"orl\"]Let $ a > b > c > d$ be positive integers and suppose that\n\\[ ac \\plus{} bd \\equal{} (b \\plus{} d \\plus{} a \\minus{} c)(b \\plus{} d \\minus{} a \\plus{} c).\n\\]\nProve that $ ab \\plus{} cd$ is not prime.[/quote]\r\n$ ac \\plus{} bd \\equal{} (b \\plus{} d)^2 \\minus{} (a \\minus{} c)^2 \\iff$\r\n$ a^2 \\minus{} ac \\plus{} c^2 \\equal{} b^2 \\plus{} bd \\plus{} d^2$\r\nLet $ \\omega \\equal{} e^{i\\frac {2\\pi}{3}}$. Hence $ \\omega^3 \\equal{} 1, \\omega^2 \\plus{} \\omega \\plus{} 1 \\equal{} 0$. Let's work in the eisenstein integers, ie. $ \\mathbb{Z}[\\omega]$.\r\n$ (a \\plus{} \\omega c)(a \\plus{} \\omega^2 c) \\equal{} (b \\minus{} \\omega d)(b \\minus{} \\omega^2 d)$\r\n$ (a \\plus{} \\omega c)\\overline{(a \\plus{} \\omega c)} \\equal{} (b \\minus{} \\omega d)\\overline{(b \\minus{} \\omega d)} \\iff$\r\n$ x \\equal{} a \\plus{} \\omega c, y \\equal{} c \\plus{} \\omega a, z \\equal{} b \\minus{} \\omega d, t \\equal{} d \\minus{} \\omega b$\r\nLet $ x \\equal{} a \\plus{} \\omega c, \\overline{x} \\equal{} a \\plus{} \\ c \\plus{} \\omega a, y \\equal{} b \\minus{} \\omega d, \\overline{y} \\equal{} b \\minus{} \\omega^2 d$\r\nThen $ x \\overline{x} \\equal{} y \\overline{y}$\r\nLet $ n \\equal{} \\gcd(x,y)$ and let $ m \\equal{} \\frac {x}{n}$, $ n,m \\in \\mathbb{Z}$. Obviously $ m \\mid \\overline{y}$ and hence $ \\gcd(x,\\overline{y}) \\equal{} m$. Then $ x \\equal{} nm, y \\equal{} n\\overline{m}$.\r\n\r\n$ n \\equal{} p \\plus{} \\omega q, m \\equal{} r \\plus{} \\omega s, p,q,r,s \\in \\mathbb{Z}$\r\n\r\n$ x \\equal{} a \\plus{} \\omega c \\equal{} (p \\plus{} \\omega q)(r \\plus{} \\omega s) \\equal{} pr \\minus{} qs \\plus{} \\omega(qr \\plus{} ps \\minus{} qs)$\r\n\r\nHence $ a \\equal{} pr \\minus{} qs$ and $ c \\equal{} qr \\plus{} ps \\minus{} qs$.\r\n\r\n$ y \\equal{} b \\minus{} \\omega d \\equal{} (p \\plus{} \\omega q)(r \\plus{} \\omega^2 s) \\equal{} pr \\plus{} qs \\minus{} ps \\plus{} \\omega (qr \\minus{} ps)$\r\n\r\nHence $ b \\equal{} pr \\plus{} qs \\minus{} ps$ and $ d \\equal{} ps \\minus{} qr$\r\n\r\nNow $ ab \\plus{} cd \\equal{} (pr \\minus{} qs)(pr \\plus{} qs \\minus{} ps) \\plus{} (qr \\plus{} ps \\minus{} qs)(ps \\minus{} qr)$\r\n\r\nExpanding $ ab \\plus{} cd \\equal{} p^2r^2 \\plus{} pqrs \\minus{} p^2rs \\minus{} pqrs \\minus{} q^2s^2 \\plus{} s^2pq \\plus{} pqrs \\plus{} p^2s^2 \\minus{} pqs^2 \\minus{} q^2r^2 \\minus{} pqrs \\plus{} q^2rs$\r\n\r\n$ ab \\plus{} cd \\equal{} p^2r^2 \\minus{} p^2rs \\minus{} q^2s^2 \\plus{} p^2s^2 \\minus{} q^2r^2 \\plus{} q^2rs \\equal{} (p^2 \\minus{} q^2)(r^2 \\plus{} s^2) \\minus{} (p^2 \\minus{} q^2)rs$\r\n\r\n$ ab \\plus{} cd \\equal{} (p^2 \\minus{} q^2)(r^2 \\minus{} rs \\plus{} s^2)$\r\n\r\nAssume that $ ab \\plus{} cd$ is not a prime. Then either $ p^2 \\minus{} q^2 \\equal{} \\pm 1$ or $ r^2 \\minus{} rs \\plus{} s^2 \\equal{} \\pm 1$. If $ p^2 \\minus{} q^2 \\equal{} \\pm1$ then $ p \\equal{} 0,q \\equal{} \\pm 1$ or $ p \\equal{} \\pm 1, q \\equal{} 0$. If $ p \\equal{} 0$ then $ a \\plus{} b \\equal{} 0$. If $ q \\equal{} 0$ then $ c \\equal{} d$. So $ p,q\\neq 0 \\Rightarrow p^2 \\minus{} q^2 \\neq \\pm 1$.\r\n$ r^2 \\minus{} rs \\plus{} s^2 \\equal{} \\frac {1}{2}(r^2 \\plus{} (r \\minus{} s)^2 \\plus{} s^2) < \\minus{} 1$ so $ r^2 \\minus{} rs \\plus{} s^2 \\equal{} \\pm 1 \\iff r^2 \\minus{} rs \\plus{} s^2 \\equal{} 1$. This has the solutions $ (r,s) \\in \\{(0,1);(0, \\minus{} 1);(1,0);( \\minus{} 1,0);(1,1);( \\minus{} 1, \\minus{} 1)\\}$\r\nIf $ r \\equal{} 0$ then $ b \\plus{} c \\equal{} 0$. If $ s \\equal{} 0$ then $ a \\equal{} b$. If $ r \\equal{} s$ then $ a \\equal{} d$. Hence $ r^2 \\minus{} rs \\plus{} s^2 \\neq \\pm 1$. So $ ab \\plus{} cd$ are a product of to integers with numerical value at least $ 2$. Hence $ ab \\plus{} cd$ is composite (Therefore not prime).", "Solution_14": "[quote=\"orl\"]Let $a >0, b > c > d\\geq0$ be integers and suppose that \\[ ac + bd = (b+d+a-c)(b+d-a+c). \\] Prove that $ab + cd$ is not prime.[/quote]$ab+cd=\\frac{(b^2-c^2)(b^2+bd+d^2)}{ab-cd-bc},$\n\nwhere $(b^2+bd+d^2)-(b^2-c^2)=c^2+d^2+bd>0,$\n\n$(b^2-c^2)-(ab-cd-bc)=\\frac{(b-c)(c-d)(b^2-c^2+d^2+ad+bd)}{a(a+b-c+d)}>0,$\n\n$ab-cd-bc=\\frac{(b-c)[a^2+b^2+c^2+d^2+bc+2da+(b-c)(a+d)]}{2a+b-c+2d}>0.$\n\n[hide=\"Remark\"]$ab+cd-\\frac{(b^2-c^2)(b^2+bd+d^2)}{ab-cd-bc}$\n\n$=\\frac{b^2[ac+bd-(b+d+a-c)(b+d-a+c)]}{ab-cd-bc}=0,$\n\n$(b^2-c^2)-(ab-cd-bc)-\\frac{(b-c)(c-d)(b^2-c^2+d^2+ad+bd)}{a(a+b-c+d)}$\n\n$=\\frac{(bc+cd-c^2-ab-bd)[ac+bd-(b+d+a-c)(b+d-a+c)]}{a(a+b-c+d)}=0,$\n\n$ab-cd-bc-\\frac{(b-c)[a^2+b^2+c^2+d^2+bc+2da+(b-c)(a+d)]}{2a+b-c+2d}$\n\n$=\\frac{(b+c)[ac+bd-(b+d+a-c)(b+d-a+c)]}{2a+b-c+2d}=0.$[/hide]\n\nSee also : http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150525", "Solution_15": "[hide=solution][quote=ISL 2001 N5]Let $a > b > c > d$ be positive integers and suppose that \\[ ac + bd = (b+d+a-c)(b+d-a+c). \\] Prove that $ab + cd$ is not prime.[/quote]\nBruh N5?clearly N4 is harder.\n$b+d+a-c|ac+bd \\implies b+d+a-c|ac+bd+ab+ad+a^2-ac=(a+b)(a+d)$\nSimilarly $b+d-a+c|(c+b)(c+d)$\nMultiplying the equations give\n\\begin{align*} (b+d+a-c)(b+d-a+c)=ac+bd|(a+b)(a+d)(c+b)(c+d) \\\\\nac+bd|(ac+bd+ad+bc)(ac+bd+ab+cd) \\\\\n\\implies ac+bd|(ad+bc)(ab+cd) \\end{align*}\nAgain,\n$a(c-d)>b(c-d)(a>b,c>d) \\implies ac+bd>ad+bc$\nand $a(b-c)>d(b-c)(a>d,b>c) \\implies ab+cd>ac+bd>ad+bc$.\nThis implies as $ac+bd \\not | ad+bc \\implies gcd(ac+bd,ab+cd)>1$,\nhence done $\\blacksquare$[/hide]", "Solution_16": "Delete this post", "Solution_17": "We all love geometry\n\nLet $p=ac+bd,q=ab+cd,r=ad+bc.$ Define cycilc quadrilateral $ABCD.$ Note that we have the following relations:\n\n\\[AC^{2}=(AB\\cdot CD+BC\\cdot DA){\\frac {AB\\cdot DA+BC\\cdot CD}{AB\\cdot BC+DA\\cdot CD}}\\]\\[BD^{2}=(AB\\cdot CD+BC\\cdot DA){\\frac {AB\\cdot BC+DA\\cdot CD}{AB\\cdot DA+BC\\cdot CD}}\\]\nLet $AB=a,BC=d,CD=b,DA=c.$ We get \\[AC^2=(ab+cd)\\frac{ac+bd}{ad+bc}=\\frac{pq}{r}\\]and \\[BD^2=(ab+cd)\\frac{ad+bc}{ac+bd}=\\frac{qr}{p}\\]\nNote that \\[ ac + bd = (b+d+a-c)(b+d-a+c). \\]simplifies to \\[b^2+d^2+bd=a^2+c^2-ac\\]\nObserve that we can make $\\angle BCD=120^\\circ$ that wouldn't change our expressions for $AC$ and $BD$ at all. This implies that \\[BD^2=b^2+d^2+bd=\\frac{qr}{p}\\]\nFinally, we see that $p(b^2+bd+d^2)=qr.$ If $q$ is a prime then $p(b^2+bd+d^2)$ implies either $q|p$ or $q|(b^2+bd+d^2).$\n\nBut since $q$ is prime, $q=p$ or $q=b^2+bd+d^2.$ By rearrangement lemma $ab+cd>ac+bd$ so $q\\neq p.$ Thus, $q=b^2+bd+d^2$ which implies $p=r$ which is also false by the rearrangement lemma. Thus, $q$ isn't prime.", "Solution_18": "\\begin{align*} ac+bd=(b+d+a-c)(b+d-a+c) \\\\ \\implies a^2-ac+c^2=b^2+bd+d^2 \\\\ \\implies (ab+cd)(ad+bc)=ac(b^2+bd+d^2)+bd(a^2-ac+c^2) \\\\ \\implies (ab+cd)(ad+bc)=(ac+bd)(a^2-ac+c^2)~~~~~(*)\\end{align*}\nSuppose the contrary. Then, $$a>b>c>d \\implies ab+cd>ac+bd>ad+bc.$$ We have that $ac+bd$ is relatively prime to $ab+cd.$ But, $$(*)\\implies ac+bd|ad+bc,$$ contradiction.", "Solution_19": "Let $ABCD$ be a quadrilateral such that $AB=a,BC=c,CD=b,DA=d$ and $\\angle B=60,\\angle D=120.$ Notice $ABCD$ exists as the conditions is equivalent to $$AC^2=a^2+c^2-ac=b^2+d^2+bd$$ by LoC. By Strong Ptolemy, $$(a^2+c^2-ac)(ac+bd)=(ab+cd)(ad+bc),$$ or $ac+bd\\mid (ab+cd)(ad+bc).$ Evidently, $ab+cd>ac+bd>ad+bc>1.$ Hence, if $ab+cd$ is prime, then $ac+bd\\mid ab+cd$ or $ab+bd\\mid ad+bc.$ Both are absurd, so $ab+cd$ is not prime. $\\square$", "Solution_20": "[hide=Storage]\n\nExpand and simplify to get $a^2 + c^2 - ac = b^2 + d^2 + bd$. Constuct a cyclic quadrilateral $ABCD$ with $AB = a, BC = c, CD = b, DA = d$ and $\\angle B = 30, \\angle D = 150$. Then, $AC^2 = (ab + cd)\\frac{ad + bc}{ac + bd}$ is an integer. However, $ab + cd < ad + bc$ and $ac + bd < ab + cd$, so $ab + cd$ can not be prime.\n\n[/hide]", "Solution_21": "== Solution ==\nEquality is equivalent to \n$ a^2 - ac + c^2 = b^2 + bd + d^2 (1) $.\n\nLet $ABCD$ be the quadrilateral with $AB = a$, $BC = d$, $CD = b$, $AD = c$, $ \\angle BAD =\n60^\\circ$, and $ \\angle BCD = 120^\\circ$. Such a quadrilateral exists by $(1)$ and the Law of Cosines.\n\nBy Strong Form of Ptolemy's Theorem, we find that;\n\n$BD^2 = \\frac{(ab+cd)(ad+bc)}{ac+bd}$\n\nand by rearrangement inequality;\n\n$ab+cd > ac+bd > ad+bc$.\n\nAssume $ab+cd = p$ is a prime, since $a^2 - ac + c^2 = BD^2$ is an integer $p \\times \\frac{ad+bc}{ac+bd}$ must be an integer but this is false since $(p,ac+bd) = 1$ and $ac+bd > ad+bc$. Thus $ab+cd$ can not be a prime.\n", "Solution_22": "The solutions that used Four number Lemma posted here are all incomplete(or incorrect). When u get $3(ab+cd)=F\\times G$, how could u state that it is a prime? ", "Solution_23": "We can rewrite the equation as\n\\[ac+bd=(b+d)^2-(a-c)^2\\implies a^2-ac+c^2=b^2+bd+d^2.\\]\nInterpret this as\n[quote=rewrite]\t\nWe have a cyclic quadrilateral, two of whose opposite angles are $60$ and $120$. The sides adjacent to the $60$ are $a$ and $c$, and the sides adjacent to the $120$ are $b$ and $d$. Prove that the product of the diagonals of this cyclic quadrilateral must be composite, given $a>b>c>d$ and all four are integers.\n[/quote]\t\nLet $e$ and $f$ be the length of those diagonals. (Doesn't quite matter what we denote them by.)\n\\[e^2=\\frac{(ab+cd)(ad+bc)}{ac+bd}\\text{ and } f^2=\\frac{(ab+cd)(ac+bd)}{ad+bc}.\\]\nIf $e\\ne f$, then say $e^2=1$ and $f^2=p^2$. Notice that $e^2=1$ is impossible as $(ab+cd)(ad+bc)>ac+bd$ and similarly if $f^2=1$. Therefore, we must have them equal and their value $p$. But it's impossible for a square to be a prime, since it's an integer by assumption. So we are done. $\\blacksquare$", "Solution_24": "Note that $b+d+a-c \\mid ac+bd + a(b+d+a-c) = (a+b)(a+d)$ and similarly $b+d-a+c \\mid (c+b)(c+d)$. \n\nLet $(a+b)(a+d) = x(b+d+a-c)$ and $(c+b)(c+d) = y(b+d-a+c)$ for integers $x$ and $y$. Multiplying the two equations yields $$(a+b)(c+d)(a+d)(c+b) = xy(ac+bd)$$ $$(ac+bd+ad+bc)(ac+bd+ab+cd) = xy(ac+bd)$$ Taking modulo $ac+bd$ both sides yields $$ac+bd \\mid (ad+bc)(ab+cd)$$ Since $a>b>c>d$ we have $ab+cd > ac+bd > ad+bc$ so if $ab+cd$ is prime then $ac+bd \\mid ad+bc$, contradiction. Thus $ab+cd$ is not prime.\n", "Solution_25": "[b]CLAIM:[/b] We have \\[ ac + bd | (ab+cd)(ad+bc). \\]\n [u]PROOF:[/u]\n b+d+a-c | (ac+bd) + a(b+d+a-c) = (a+d)(a+b)\n b+d+a-c | (ac+bd) + c(b+d-a+c) = (c+b)(c+b)\n So, (b+d+a-c)(b+d-a+c) | (a+d)(a+b)(c+b)(c+b).\n ac+bd | ((ac+bd) + (ad+bc))((ac+bd)+(ab+dc))\n Proved....\n NOW, as a>b>c>d, \n ad+cd > ac+bd > ad+bc (By rearrangement inequality)\n If ab+cd is prime , then ac+bd | ad+bc which implies ac+bd ac+bd>ad+bc$ by Rearrangement inequality; so if $ab+cd$ is prime then $ac+bd\\mid ad+bc\\Rightarrow ad+bc\\geq ac+bd$, a contradiction", "Solution_28": "The expansion is equivalent to $a^2-ac+c^2 = b^2+bd+d^2$. Equivalently, $a, c, d, b$ are the side lengths of a cyclic quadrilateral with opposite angles $60^\\circ$ and $120^\\circ$. This implies that $b^2+bd+d^2 = \\frac{(ab+cd)(ad+bc)}{ac+bd}$ is an integer.\n\nAssume for the sake of contradiction that $ab+cd$ is prime; then as $ac+bd > ad+bc$, follows that $ab+cd \\mid ac+bd$, but $ac+bd < ab+cd$, contradiction.", "Solution_29": "Note that $$ac + bd = (b+d-a+c)(b+d+a-c) \\iff (a+b)(a+d) = (b+d+c)(b+d+a-c)$$ So let's use the Four Number Lemma to conclude: $$\\begin{array}{c} a+b =pq \\\\ a+d =rs \\\\ b+d+c = ps \\\\ b+d+a-c =qr \\end{array}$$ from where we get that $$\\begin{array}{c} a=\\frac{1}{3}(2pq+2rs-ps-qr) \\\\ \\\\ b =\\frac{1}{3}(ps+qr+pq-2rs) \\\\ \\\\ c=\\frac{1}{3}(ps-2qr+rs+pq) \\\\ \\\\ d=\\frac{1}{3}(ps+qr+rs-2pq) \\end{array}$$ So we have $$ab+cd = \\frac{(2pq+2rs-ps-qr)(ps+qr+pq-2rs)+(ps-2qr+rs+pq)(ps+qr+rs-2pq)}{9} = \\frac{r(2p-r)(q^2-qs+s^2)}{3}$$ and from $a>b>c>d$ we can easily deduce $p>r$ and $q>s$ so $2p-r\\ge 3$ and $q^2-qs+s^2=(q-s)^2+qs \\ge 3$. Lastly, we can easily check that if the equality holds for either of them then $r\\neq 1$." } { "Tag": [ "AMC" ], "Problem": "has anyone tried that book\r\n\r\nis it good for the amc 10 \r\n\r\nthanks\r\n\r\n :starwars: :pilot:", "Solution_1": "I bought this book and I'm not gonna lie, it was totally not worth it (especially given the cost), it didn't help much, it just had a bunch of solutions to freely existing AMC problems, it'd be easier, more tuned to you, and cheaper to simply post problems you have trouble with on the forums and then read the responses.\r\n\r\nThat being said, if the money is of little concern to you, and if you absolutely need to be able to work on your AMC game without having a computer near you, you should strongly consider buying it.", "Solution_2": "so you are saying it just has problems and stuff\r\n\r\nbut does it have strategies\r\n\r\nwould aops book 1 or 2 be better?\r\n\r\n\r\nthanks" } { "Tag": [], "Problem": "How many ways can the positive integer $ n$ can be written as an ordered sum of at least one positive integer? For example,\r\n\r\n\\[ 4\\equal{}1\\plus{}3\\equal{}3\\plus{}1\\equal{}2\\plus{}2\\equal{}1\\plus{}1\\plus{}2\\equal{}1\\plus{}2\\plus{}1\\equal{}2\\plus{}1\\plus{}1\\equal{}1\\plus{}1\\plus{}1\\plus{}1\r\n\\]\r\n\r\nso when $ n\\equal{}4$, there are eight such ordered partitions.", "Solution_1": "[quote=\"#H34N1\"]How many ways can the positive integer $ n$ can be written as an ordered sum of at least one positive integer? For example,\n\\[ 4 \\equal{} 1 \\plus{} 3 \\equal{} 3 \\plus{} 1 \\equal{} 2 \\plus{} 2 \\equal{} 1 \\plus{} 1 \\plus{} 2 \\equal{} 1 \\plus{} 2 \\plus{} 1 \\equal{} 2 \\plus{} 1 \\plus{} 1 \\equal{} 1 \\plus{} 1 \\plus{} 1 \\plus{} 1\n\\]\nso when $ n \\equal{} 4$, there are eight such ordered partitions.[/quote]\r\n[hide=\"Solution\"]\nConsider a string of $ n$ ones with small gaps between them. Between each gap, we will choose to or to not put a bar between the two adjacent numbers. If we put a bar in front of the $ i$th one, for example, and then the next bar is in front of the $ i\\plus{}j\\minus{}1$th, then the number corresponding to these bars is $ j$. Here is an example for $ n\\equal{}8$: $ 1, 1, 1, | 1, 1, |1,| 1, 1$ turns into $ 3\\plus{}2\\plus{}1\\plus{}2$. As there are $ n\\minus{}1$ gaps and $ 2$ possibilities per gap, there are $ 2^{n\\minus{}1}$ such partitions. [/hide]", "Solution_2": "well, for example when you evaluate for n=5, you have the equations\r\na=5,a+b=5,a+b+c=5,a+b+c+d=5,a+b+C+d+e=5 all as possible cases, with a,b,c,d,e>1\r\nthe first equation has 4c4=1 solution, and the second equation becomes a+b=3 with a,b>0, so by balls and urns this becomes4c1 solutions, and then 4c2 and 4c3, and 4c4. in general, you get a sum of nCk's where k ranges from 1 to n, so you could probably turn this into a proof. \r\n\r\ni propose that the number of ordered partitions of a number n is 2^[n-1]\r\n\r\nedit:beaten", "Solution_3": "It's easy. $ f(n)\\equal{}f(n\\minus{}1)\\plus{}f(n\\minus{}2)\\plus{}...\\plus{}f(1)\\plus{}1$ We can get this by considering the first number in the ordered pair. The last 1 is for when the first number = n. Then do strong induction.", "Solution_4": "The identity $ \\sum_{k \\equal{} 0}^{n\\minus{}1} \\binom{n\\minus{}1}{k} \\equal{} 2^{n \\minus{} 1}$ makes [b]mathking123[/b]'s solution incorrect.\r\n\r\nUsing that identity correctly with this problem yields what [b]The QuattoMaster 6000[/b] said.\r\n\r\nEDIT Thanks for the clarification, [b]t0rajir0u[/b].", "Solution_5": "The identity you want is $ \\sum_{k\\equal{}0}^{n\\minus{}1} {n\\minus{}1 \\choose k} \\equal{} 2^{n\\minus{}1}$, and the combinatorial interpretation of this identity is very familiar. (You may have gotten confused with the identity from [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]this post[/url].)", "Solution_6": "Makes much more sense now. :) Thanks and edited." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "the vertices of triangle $ ABC$ are lattice points.two of its sides have lengths which belong to set$ /{{sqr root 17,sqr root 1999,sqr root 2000}}/.$what is the maximum possible area of $ ABC$.....\r\nfacing difficulty to write second bracket :(", "Solution_1": "can anyone help....i am facing with difficulties to find the lattice points....help :maybe:", "Solution_2": "noone interested????????? :maybe: :(" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "a+b+c+d+e+f+g = 1 and a, b, c, d, e, f, g are non-negative. Find the minimum value of max(a+b+c, b+c+d, c+d+e, d+e+f, e+f+g).", "Solution_1": "We have\r\n\\begin{eqnarray*}\r\n\\lefteqn{\\max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g)} \\\\ \r\n & \\geq & \\max(a + b + c, c + d + e, e + f + g) \\\\\r\n & \\geq & \\frac{(a + b + c) + (c + d + e) + (e + f + g)}{3} \\\\\r\n & = & \\frac{a + b + c + d + e + f + g}{3} + \\frac{c + e}{3} \\\\\r\n & = & \\frac{1}{3} + \\frac{c + e}{3} \\\\\r\n & \\geq & \\frac{1}{3}.\r\n\\end{eqnarray*}\r\nThus the minimum of the expression is at least $\\frac{1}{3}$.\r\n\r\nTo see that the minimum is equal to $\\frac{1}{3}$, consider the following assignment of variables: $a = d = g = \\frac{1}{3}$, and the rest of the variables are 0." } { "Tag": [ "articles", "LaTeX" ], "Problem": "Hello,\r\n\r\nI want to use the \\chapter level so that it displays, for example, 4.2.3. But it is not really appropriate for my paper to display \"Chapter 4\". How do I get rid of that line? \r\n\r\nThank you.", "Solution_1": "That doesn't make much sense since chapter is the highest level in the book and report classes whereas 4.2.3 would be the number for a subsection ie chapter 4, section 2, subsection 3, so you how can a chapter be numbered 4.2.3?.\r\nYou don't have to use \\chapter if you don't need it but you can redefine it so the name chapter doesn't appear\r\n[code]\\renewcommand{\\chaptername}{}[/code] You can also redefine the chapter counter eg\n[code]\\renewcommand{\\thechapter}{\\arabic{chapter}.\\arabic{section}.\\arabic{subsection}}[/code]but, as I say, this doesn't make sense.", "Solution_2": "I think the idea is to not have \\chapter make \"Chapter x\" appear, but still have the automatic numbering of sections etc.\r\nI have a similar question: How can I change the formatting (I'm in an article class) of the text that appears from \\section{Blah}, and have autonumbering of theorems, etc work?", "Solution_3": "[quote=\"caffeineboy\"]I have a similar question: How can I change the formatting (I'm in an article class) of the text that appears from \\section{Blah}[/quote]Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=titlesec]titlesec package[/url] \n[quote=\"caffeineboy\"]and have autonumbering of theorems, etc work?[/quote]Theorems are automatically numbered by default. So\r\n[code]\\newtheorem{theorem}{Theorem}\n...\n\\begin{theorem}\nThis is a theorem\n\\end{theorem}[/code]gives\r\n[b]Theorem 1.[/b] [i]This is a theorem[/i]\r\nSee Section 3.7 of [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LATEX2e[/url].", "Solution_4": "Sorry, I left out that I wanted the theorems to be numbered including the section counter, ie \\newtheorem{thm}{Theorem}[section], but titlesec handles this fine, thanks.", "Solution_5": "Sorry if I didn't explain myself well. As Caffeine said, \"...the idea is to not have \\chapter make \"Chapter x\" appear, but still have the automatic numbering of sections etc.\"\r\n\r\nWould I still use \\renewcommand{\\chaptername}{} ?", "Solution_6": "If you want chapter numbering but no chapter heading then don't put one in with \\chapter{...}. Instead use \\stepcounter{chapter} to increment the chapter counter. But it all seems very strange to have chapters with no titles - why bother with chapters at all?", "Solution_7": "Thinking about it, what might make sense is to have chapter titles but without the word \"Chapter\" or the numbering that follows. That can be done with\r\n[code]\\renewcommand{\\chaptername}{}\n\\renewcommand{\\thechapter}{}\n\\renewcommand{\\thesection}{\\arabic{chapter}.\\arabic{section}}[/code]The first two lines get rid of the chapter name and number and the third line puts the chapter back into the section numbering.", "Solution_8": "Yeah that last one is what I was looking for. Thanks." } { "Tag": [ "Mafia", "FTW", "\\/closed" ], "Problem": "Read the ? above.", "Solution_1": "Otherwise, someone would send some large amount of spam to everyone.", "Solution_2": "No ability to mass PM makes sending Mafia Role PMs a serious headache :stretcher:. Maybe if the mods or authorized people could do it, sending them would be a lot easier.\r\nSame with big games, FTW tournaments, etc...", "Solution_3": "There was an issue with this a while back during the AoPS Team Chess Tournament. If you're going to mass message someone, do it on email. The problems with mass PMing outweigh the benefits." } { "Tag": [ "Diophantine Equations" ], "Problem": "Is there a positive integer $m$ such that the equation \\[\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{abc}= \\frac{m}{a+b+c}\\] has infinitely many solutions in positive integers $a, b, c \\;$?", "Solution_1": "For $ m\\equal{}12$ then it has many solution.\r\nProve with the Pell's equation and has discuss .", "Solution_2": "Can you please prove it has infinitely many for 12?", "Solution_3": "$ x\\equal{}a\\minus{}k,y\\equal{}a,z\\equal{}a\\plus{}k$\r\nThen the equation equivalent:\r\n$ a^2\\minus{}3k^2\\equal{}1$\r\nIt is the Pell equation.\r\nThe root of equation is :\r\n$ a_0\\equal{}1,k_0\\equal{}0$\r\n$ a_1\\equal{}2,k_1\\equal{}1$\r\n$ a_{n\\plus{}2}\\equal{}4a_{n\\plus{}1}\\minus{}a_n$\r\n$ k_{n\\plus{}2}\\equal{}4k_{n\\plus{}1}\\minus{}k_n$\r\nEasy to check $ a>k$ so the equation has positive solution.\r\nMore consider the equation:\r\n$ \\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z}\\plus{}\\frac{c}{xyz}\\equal{}\\frac{3c\\plus{}9}{x\\plus{}y\\plus{}z}$\r\nWith the same method we can prove this equation has infinite natural solution.", "Solution_4": "http://www.kalva.demon.co.uk/short/soln/sh02n4.html" } { "Tag": [], "Problem": "Here are the rules:\r\nKeep naming songs as long as you can!\r\nNote: name 1 song per post!\r\n\r\nI'll start the chain:\r\nHoliday (green day)", "Solution_1": "Ass like that (Eminem)", "Solution_2": "yesterday-beatles", "Solution_3": "WELCOME HOMe-Coheed and cambria", "Solution_4": "I like Gorrilaz :D :rotfl:", "Solution_5": "First Date(Blink 182)", "Solution_6": "\"Epitaph, including March For No Reason and Tomorrow And Tomorrow\" - King Crimson.", "Solution_7": "Miss Murder- AFI" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Points $ A_1,B_1,C_1$ are chosen on the sides $ BC,CA,AB$ of and isosceles triangle $ ABC(AB\\equal{}BC)$.It is known that $ \\widehat{BC_1A_1}\\equal{}\\widehat{CA_1B_1}\\equal{}\\widehat{A}$ Let $ BB_1$ and $ CC_1$ meet at $ P$.Prove that $ AB_1PC_1$ is cyclic.", "Solution_1": "It's easy. $ \\angle{A}\\equal{}\\angle{C}$ the conditions imply that]. $ C_1A_1CA$ and $ BA_1B_1A$ are cyclic. (Directly from the given angles) Then, $ \\angle{B_1BA_1}\\equal{}\\angle{A_1AB_1}$ and $ \\angle{C_1CB}\\equal{}\\angle{C_1AA_1}$ this means $ \\angle{BPC}\\equal{}180^o\\minus{}\\angle{A}$ and hence the result.", "Solution_2": "There are another solution:\r\nConstruct P' is the intersection of CC1 and cicumcircle of BA1C1, then simple to relize that C1P'B1A is cyclic. So $ 27abc \\minus{} (a \\plus{} b \\plus{} c)^3 \\plus{} \\frac {12(ab \\plus{} bc \\plus{} ca) \\minus{} 36abc}{abc} \\plus{} \\frac {8 \\minus{} 8abc}{abc} \\geq 0$\r\nWe have:\r\n$ \\frac {12(ab \\plus{} bc \\plus{} ca) \\minus{} 36abc}{abc} \\equal{} \\frac {8c(a \\minus{} b)^2 \\plus{} (4a \\plus{} 4b)(a \\minus{} c)(b \\minus{} c)}{abc}$\r\n$ 27abc \\minus{} (a \\plus{} b \\plus{} c)^3 \\equal{} \\minus{} (a \\plus{} b \\plus{} 7c)(a \\minus{} b)^2 \\minus{} (a \\minus{} c)(b \\minus{} c)(4a \\plus{} 4b \\plus{} c)$\r\n$ \\frac {8 \\minus{} 8abc}{abc} \\equal{} \\frac {8(a \\plus{} b \\plus{} 7c)(a \\minus{} b)^2 \\plus{} 8(4a \\plus{} 4b \\plus{} c)(a \\minus{} c)(b \\minus{} c)}{27abc}$\r\nSo this inequality becomes:\r\n$ M(a \\minus{} b)^2 \\plus{} N(a \\minus{} c)(b \\minus{} c) \\geq 0$\r\nwith $ M \\equal{} \\frac {8c}{abc} \\plus{} \\frac {8(a \\plus{} b \\plus{} 7c)}{27abc} \\minus{} a \\minus{} b \\minus{} 7c$\r\nand $ N\\equal{}\\frac{4a\\plus{}4b}{abc}\\plus{}\\frac{8(4a\\plus{}4b\\plus{}c)}{27abc}\\minus{}4a\\minus{}4b\\minus{}c$\r\nWe will prove that $ M \\geq 0;N \\geq 0$\r\n$ M \\geq 0$<=>$ \\frac {8c}{abc} \\plus{} \\frac {8(3 \\plus{} 6c)}{27abc} \\geq 6c \\plus{} 3$\r\n<=>$ \\frac {88}{9ab} \\plus{} \\frac {24}{27abc} \\geq 6c \\plus{} 3$\r\nWe need prove that $ \\frac {88}{9ab} \\geq 6c \\plus{} \\frac {19}{9}$\r\n<=>$ 88 \\geq 54abc \\plus{} 19ab$\r\nThis inequality true because $ ab \\leq abc \\leq 1$\r\n We need prove that $ N \\geq 0$\r\nor $ \\frac {24}{27abc} \\plus{} \\frac {100(a \\plus{} b)}{9abc} \\geq 3a \\plus{} 3b \\plus{} 3$\r\nBut we have:\r\n$ \\frac {24}{27abc} \\plus{} \\frac {100(a \\plus{} b)}{9abc} \\geq \\frac {24}{27} \\plus{} \\frac {200}{9c}$\r\nWe will prove $ \\frac {200}{9c} \\geq 3a \\plus{} 3b \\plus{} \\frac {19}{9}$\r\n<=>$ 200 \\geq 27c(a \\plus{} b) \\plus{} 19c$\r\nBut by AM-GM $ 27c(a \\plus{} b) \\leq \\frac {243}{4}$\r\nAnd we need prove $ \\frac {557}{4} \\geq 19c$\r\n<=>$ \\frac {557}{76} \\geq c$\r\nDone!", "Solution_3": "Nice, Chien Than.\r\nBut, since $ 8(abc\\plus{}\\frac 1{abc}) \\ge 16$, it suffices to show that\r\n$ 19abc\\plus{}12(\\frac 1{a}\\plus{}\\frac 1{b}\\plus{}\\frac 1{c}) \\ge 55$.", "Solution_4": "Other problem:\r\nLet $ a;b;c$ be positive real numbers.Find min of:\r\n$ S\\equal{}(1\\plus{}\\frac{a}{3b})(1\\plus{}\\frac{b}{3c})(1\\plus{}\\frac{c}{3a})$" } { "Tag": [ "limit", "algebra", "polynomial", "algebra unsolved" ], "Problem": "1. Prove that exists some number n, such that in decimal representation of $ 5^n$ there is block of at least 100 zeros.\r\n\r\n2. Find limit\r\n\r\n$ \\mathop {\\lim }\\limits_{n \\to infinity} \\{ { (5 \\plus{} \\sqrt {24} )^n } \\}$", "Solution_1": "$ n\\to \\plus{}\\infty\\Rightarrow L\\to \\plus{}\\infty\r\n n\\to \\minus{}\\infty\\Rightarrow L\\to 0$:wink:", "Solution_2": "[quote=\"karis\"]$ n\\to \\plus{} \\infty\\Rightarrow L\\to \\plus{} \\infty n\\to \\minus{} \\infty\\Rightarrow L\\to 0$:wink:[/quote]\r\n\r\nThe mistake is corrected.\r\n\r\nI am new here, so I am already not good in Latex...", "Solution_3": "$ (5 \\plus{} \\sqrt {24})^n \\plus{} (5 \\minus{} \\sqrt {24})^n \\in \\mathbb{N}$, so the fractional part is $ 1 \\minus{} (5 \\minus{} \\sqrt {24})^n$ and approaches $ 1$.", "Solution_4": "[quote=\"t0rajir0u\"]$ (5 \\plus{} \\sqrt {24})^n \\plus{} (5 \\minus{} \\sqrt {24})^n \\in \\mathbb{N}$, so the fractional part is $ 1 \\minus{} (5 \\minus{} \\sqrt {24})^n$ and approaches $ 1$.[/quote]\r\n\r\nThat's beautiful! Thank you very much. And could somebody say me the main idea of the second problem? I seen many problems of that type, but I cannot imagine how to start with it.", "Solution_5": "One of the ideas is that $ s_n \\equal{} (5 \\plus{} \\sqrt {24})^n \\plus{} (5 \\minus{} \\sqrt {24})^n$ satisfies the second-order linear recurrence\r\n$ s_{n \\plus{} 2} \\equal{} 10s_{n \\plus{} 1} \\minus{} s_n$\r\n\r\nso every term in the sequence is an integer. There is a very general theory about how to handle sequences of this type; see [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=215833]here[/url] and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url]. There are two other ways to look at this:\r\n\r\n- When you add $ (5 \\minus{} \\sqrt {24})^n$, the terms in the binomial expansion containing the odd powers of $ \\sqrt {24}$ cancel, so you are left with a sum of integers. This is a good elementary perspective but does not readily generalize without more work.\r\n\r\n- The sum of the powers of the roots of $ x^2 \\minus{} 10x \\plus{} 1$ are symmetric polynomials in the roots, hence polynomials in the coefficients. This is closely related to the linear recurrence idea; see [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums]Newton sums[/url] and the [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_symmetric_polynomials#The_fundamental_theorem_of_symmetric_polynomials]fundamental theorem of symmetric polynomials[/url]." } { "Tag": [ "function", "arithmetic sequence", "number theory proposed", "number theory" ], "Problem": "$(F_n)$ denote Fibonacci numbers and $(L_n)$ Lucas numbers.\r\nProve that $F_{2005}$ and $L_{2005}$ are coprime.", "Solution_1": "I only have two things to say :):\r\n\r\n$F_n+L_n=2F_{n+1}$\r\n\r\n$(F_n,F_{n+1})=1$.", "Solution_2": "[quote=\"Collins\"]$(F_n)$ denote Fibonacci numbers and $(L_n)$ Lucas numbers.\nProve that $F_{2005}$ and $L_{2005}$ are coprime.[/quote]\r\n\r\nUse [i]soustraction formula[/i]: $2F_{m-n}=(-1)^n(F_mL_n-L_mF_n)$ with $(m,n)=(2006,2005)$.\r\nIn fact $\\gcd (F_n,L_n)=1$ if $n\\equiv 1 or 2[3]$ and $\\gcd (F_n,L_n)=2$ if $n\\equiv 0[3]$.\r\n\r\n :( You are too fast for me, Grobber!!!", "Solution_3": "Actually, there's still the possibility that $2|F_n,L_n$, so what I wrote wasn't complete. We still have to show that $F_{2005},L_{2005}$ are not both even, but that's not hard. :)", "Solution_4": "loocking both sequences modulo 2 and finding a period will do the job.", "Solution_5": "Here's a nice fact: if $k$ is odd, $\\gcd(F_k,L_m)|2$.", "Solution_6": "I bet some nice and slick forumla solves this one, but here's a solution which doesn't use any slick formulae :). \r\n\r\nLet $u=\\frac{1+\\sqrt 5}2$ and $\\bar u=\\frac{1-\\sqrt 5}2$. Assume an odd prime $p|F_k=\\frac 1{\\sqrt 5}\\cdot(u^k-\\bar u^k)$ and $p|L_m=u^m+\\bar u^m$. If $p$ is $5$, we're done: $5$ never divides an $L_m$. Otherwise, work in $\\mathbb Z_p[\\sqrt 5]$. We have $u^{2k}+1=0$ and $u^{2m}+(-1)^m=0$. \r\n\r\nIf $m$ is even, this is impossible, because the order of $u$ in the multiplicative group of our field divides $4k$, and, on the other hand, we find that it must divide $4m$ and it must contain all $2$'s in the decomposition of $4m$, and there are more than $2$ (because $m$ is even). \r\n\r\nIf $m$ is odd, the order of $u$ divides $2m$, but it also divides $4k$ and has all the $2$'s in the decomposition of $4k$, again a contradiction.", "Solution_7": "jaja, thats what i had in mind!!!!!!!!!!! Nice Grobber!", "Solution_8": "[quote=\"grobber\"]I bet some nice and slick formula solves this one[/quote]\r\nWhy bother with anything so crude as a formula?\r\n\r\n(Let $p$ be a prime in the following)\r\nIn any sequence obeying the Fibonacci recurrence (or any second-order recurrence which is reversible mod $p$), the zeros mod $p$, if they exist, must form a single arithmetic sequence with difference independent of the initial conditions; if the series have zeros mod $p$, they differ mod $p$ only by a constant multiple and a horizontal translation.\r\nThe absolute values of the Lucas sequence are an even function, so an element divisible by $p$ must be mirrored by another with the negative of its index. For $p>2$, this means that if the Lucas sequence is ever zero mod $p$, the difference between zeros mod $p$ is even. This difference must carry over to the Fibonacci sequence as well; since $p|F_0$, $p$ must not divide any Fibonacci number with odd index.\r\n\r\nFor powers of 2, we directly show that no Lucas numbers are divisible by 4." } { "Tag": [ "trigonometry", "Euler", "algebra unsolved", "algebra" ], "Problem": "Following equation i've to reform it like \"a+bi\" :\r\n\r\ncos(2t-1)+i sin(2t+1)\r\n\r\nCan someone help me solving this prob?\r\n\r\nTHX", "Solution_1": "It's already in the form $ a\\plus{}bi$ with $ a\\equal{}\\cos (2t\\minus{}1)$, $ b\\equal{}\\sin (2t\\plus{}1)$. Can you make your question more specific?", "Solution_2": "[quote=\"JoeBlow\"]It's already in the form $ a \\plus{} bi$ with $ a \\equal{} \\cos (2t \\minus{} 1)$, $ b \\equal{} \\sin (2t \\plus{} 1)$. [/quote]\r\n\r\nThis was my first thought also but i was also thinking that it was too easy, so i started using Euler, but i got stuck.\r\n\r\nThx JoeBlow for the reply" } { "Tag": [ "puzzles" ], "Problem": "1. How would you rearange the letters in the word \"new door\" to make one word?\r\n\r\n2. Yes, I am heavy. but backwards I am not. Who am I?", "Solution_1": "[hide=\"1\"]new door$\\rightarrow$one word[/hide]\n[hide=\"2\"]ton[/hide]", "Solution_2": "[hide]1. New door is one word rearanged\n\n2.ton not[/hide]" } { "Tag": [ "number theory", "least common multiple" ], "Problem": "I am SO FRUSTRATED with this problem. I have spent hours and hours on it checking it and re-checking it, reworking it, and I cannot get the correct solution. I am willing to bet it is some minor oversight, but it completely eludes me. Someone please show me the error of my ways before I blow a gasket.\r\n\r\nThis is my feeble attempt to work the problem. What is really bizzarre is that this problem consistently gives me the same wrong answer!\r\n\r\nSolve the equation. Check the result.\r\n\r\n[(3 + p) / 3] - 4p = 1 - [(p + 7) / 2]\r\n\r\n----------------------------------------------------------------------------------------------------------\r\n\r\nThe first thing we need to do is eliminate the fractions by multiplying both sides of the equation by the LCD. In this case it is 6.\r\n\r\n(6){[(3 + p) / 3] - 4p} = (6){1 - [(p + 7) / 2]}\r\n\r\nUse the distributive property:\r\n\r\n(6)[(3 + p) / 3] - (6)4p = (6)1 - (6)[(p + 7) / 2]\r\n\r\nSimplify non-fractional terms:\r\n\r\n(6)[(3 + p) / 3] - 24p = 6 - (6)[(p + 7) / 2]\r\n\r\nUse the commutative property:\r\n\r\n(6/3)(3 + p) - 24p = 6 - (6/2)(p + 7)\r\n\r\nSimplify:\r\n\r\n(2)(3 + p) - 24p = 6 - (3)(p + 7)\r\n\r\nDistribute:\r\n\r\n(2)3 + (2)p - 24p = 6 - (3)p + (3)7\r\n\r\n6 + 2p - 24p = 6 - 3p + 21\r\n\r\nCombine like terms and place in conventional order:\r\n\r\n(-22p) + 6 = (-3p) + 27\r\n\r\nAdd 3p to both sides:\r\n\r\n(-22p) + 6 + 3p = (-3p) + 27 + 3p\r\n\r\n(-19p) + 6 = 27\r\n\r\nSubtract 6 from both sides:\r\n\r\n(-19p) + 6 - 6 = 27 - 6\r\n\r\n(-19p) = 21\r\n\r\nUse the associative property and divide both sides by -19:\r\n\r\n(-19)(p) / -19 = 21 / -19\r\n\r\np = -21/19\r\n\r\n----------------------------------------------------------------\r\n\r\nTo check our solution, we plug it back into the original equation:\r\n\r\n[(3 + p) / 3] - 4p = 1 - [(p + 7) / 2]\r\n\r\n[(3 + (-21/19)) / 3] - 4(-21/19) = 1 - [((-21/19) + 7) / 2]\r\n\r\n----------------------------------------------------------------\r\n\r\nI will begin by solving the FIRST expression:\r\n\r\n[(3 + (-21/19)) / 3] - 4(-21/19)\r\n\r\nSolve the numerator first:\r\n\r\n[((57/19) + (-21/19)) / 3] - 4(-21/19)\r\n\r\n[(36/19) / 3] - 4(-21/19)\r\n\r\nThen the fraction itself:\r\n\r\n(12/19) - 4(-21/19)\r\n\r\nMultiply and simplify:\r\n\r\n(12/19) - (-84/19)\r\n\r\n(12/19) + (84/19) = 96/19\r\n\r\nThe first expression simplifies to 96/19.\r\n\r\n----------------------------------------------------------------\r\n\r\nNow we check the SECOND expression:\r\n\r\n1 - [((-21/19) + 7) / 2]\r\n\r\nDo the numerator first:\r\n\r\n1 - [((-21/19) + (133/19)) / 2]\r\n\r\n1 - [(112/19) / 2]\r\n\r\nThen the / 2 in the fraction:\r\n\r\n1 - (56/19)\r\n\r\n(19/19) - (56/19) = -37/19\r\n\r\n----------------------------------------------------------------\r\n\r\n96/19 =? -37/19 FALSE!!!\r\n\r\nAnd false again, and again", "Solution_1": "[quote=\"kamikaze762\"]I am SO FRUSTRATED with this problem. I have spent hours and hours on it checking it and re-checking it, reworking it, and I cannot get the correct solution. I am willing to bet it is some minor oversight, but it completely eludes me. Someone please show me the error of my ways before I blow a gasket.\n\nThis is my feeble attempt to work the problem. What is really bizzarre is that this problem consistently gives me the same wrong answer!\n\nSolve the equation. Check the result.\n\n[(3 + p) / 3] - 4p = 1 - [(p + 7) / 2]\n\n----------------------------------------------------------------------------------------------------------\n\nThe first thing we need to do is eliminate the fractions by multiplying both sides of the equation by the LCD. In this case it is 6.\n\n(6){[(3 + p) / 3] - 4p} = (6){1 - [(p + 7) / 2]}\n\nUse the distributive property:\n\n(6)[(3 + p) / 3] - (6)4p = (6)1 - (6)[(p + 7) / 2]\n\nSimplify non-fractional terms:\n\n(6)[(3 + p) / 3] - 24p = 6 - (6)[(p + 7) / 2]\n\nUse the commutative property:\n\n(6/3)(3 + p) - 24p = 6 - (6/2)(p + 7)\n\nSimplify:\n\n(2)(3 + p) - 24p = 6 - (3)(p + 7)\n\nDistribute:\n\n(2)3 + (2)p - 24p = 6 - (3)p + (3)7\n[/quote]\r\n\r\nYour error is right here. You distributed incorrectly, there should be a minus in front of $ (3)7$. If you repeat the calculation from here with the minus there, you should get the correct answer $ p\\equal{}\\frac{21}{19}$", "Solution_2": "He's right. Here's what should have happened:\r\n\r\n$ \\frac{3\\plus{}p}{3}\\minus{}4p\\equal{}1\\minus{}\\frac{p\\plus{}7}{2}$\r\n\r\n$ 6\\left(\\frac{3\\plus{}p}{3}\\minus{}4p\\right)\\equal{}6\\left(1\\minus{}\\frac{p\\plus{}7}{2}\\right)$\r\n\r\n$ 2(3\\plus{}p)\\minus{}6(4p)\\equal{}6(1)\\minus{}3(p\\plus{}7)$\r\n\r\n$ 6\\plus{}2p\\minus{}24p\\equal{}6\\minus{}3p\\minus{}21$ (Your mistake was here. When you distributed, you didn't distribute the minus sign to the 21.)\r\n\r\n$ 21\\equal{}19p$\r\n\r\n$ \\frac{21}{19}\\equal{}p$\r\n\r\n[u][b]Check:[/b][/u]\r\n\r\n$ \\frac{3\\plus{}\\frac{21}{19}}{3}\\minus{}4\\left(\\frac{21}{19}\\right)\\equal{}1\\minus{}\\frac{\\frac{21}{19}\\plus{}7}{2}$\r\n\r\n$ \\frac{\\frac{57}{19}\\plus{}\\frac{21}{19}}{3}\\minus{}\\frac{84}{19}\\equal{}1\\minus{}\\frac{\\frac{21}{19}\\plus{}\\frac{133}{19}}{2}$\r\n\r\n$ \\frac{\\frac{78}{19}}{3}\\minus{}\\frac{84}{19}\\equal{}1\\minus{}\\frac{\\frac{154}{19}}{2}$\r\n\r\n$ \\frac{26}{19}\\minus{}\\frac{84}{19}\\equal{}\\frac{19}{19}\\minus{}\\frac{77}{19}$\r\n\r\n$ \\minus{}\\frac{58}{19}\\equal{}\\minus{}\\frac{58}{19}$", "Solution_3": "First off, I really appreciate your responses, and I see where I messed up, but I am still having trouble grasping the concept of negative distibution and/or distribution after a minus sign.\r\n\r\nLet me make certain I understand this rule.\r\n\r\n6 - (3)(p + 7)\r\n\r\nOk, we have a minus just hanging around there ready to perform subtraction.\r\n\r\nSince (3) comes after it, (3) becomes negative once it is distributed?\r\n\r\nIf that were the case, why does the problem not end up like this since p and 7 are both positive?\r\n\r\n6 - (-3p) - 7\r\n= 6 + 3p - 7\r\n\r\nMy question is, why does the sign of the first term remain the same while the sign of the second term changes? After all, both terms receive the same treatment here, so it seems counter-intuitive to change only the last term.\r\n\r\nFor additonal clarity:\r\nThis is the correct distribution, but how?\r\n\r\n6 - (3)(p + 7)\r\n\r\n6 - (3)p + (-3)7\r\n\r\nThat blows my mind.\r\n\r\nCan someone lay this out for me in a logical way so I can see how this works? Any help is greatly appreciated. :)", "Solution_4": "Wait a sec. I think I understand.\r\n\r\nAs a matter of logic, it would actually be done in this way?\r\n\r\n6 - (3)(p + 7) \r\n\r\n6 - (3)p + - (3)7 \r\n\r\nSimply saying that since the (p + 7) is grouped and appears after a minus sign, it must be interpreted that that operation to be performed on the group must be performed on all terms within the group. In this case, we are multiplying all terms within the group, and also subtracting all terms within the group. \r\n\r\nIs that the correct logical analysis of this rule?", "Solution_5": "yes, that is correct.\r\n\r\n$ 6\\minus{}3(p\\plus{}7)\\equal{}6\\plus{}(\\minus{}3(p\\plus{}7))$, so we must distibute the negative.\r\n\r\n$ 6\\minus{}3(p\\plus{}7)\\equal{}6\\minus{}(3p\\plus{}21)\\equal{}6\\plus{}(\\minus{}1(3p\\plus{}21))$. We must distibute the $ \\minus{}1$.", "Solution_6": "Well done and thank you.", "Solution_7": "btw I got something totally different i think\r\n\r\n((3+p)/3)-4p=1-((p+7)/2)\r\n [color=orange] (+4p)[/color]\r\n((3+p)/3) = 4p+1-((p+7)/2)\r\n [color=orange] (+((p+1)/2)[/color]\r\n((3+p)/3)+((p+7)/2) = 4p+1\r\n\r\n(2(3+p)+3(p+7))/6 = 4p+1\r\n\r\n6+2p+3p+21 = 24p+6\r\n\r\n5p+27 = 24p+6\r\n [color=orange](-5p)[/color]\r\n27 = 19p+6\r\n [color=orange](-6)[/color]\r\n21 = 19p\r\n [color=orange] (/19)[/color]\r\n[size=150][color=blue]p = 21/19 \n (or 1 2/21)[/color][/size]\r\n\r\nI checked it and am quite sure it works", "Solution_8": "That solution has already been stated. If you saw the 58/19, then you were probably looking at the checking portion.\r\n\r\nBTW: Your second gold line should probably say \"(+((p+[b]7[/b])/2))", "Solution_9": "[quote=\"fishythefish\"]That solution has already been stated. If you saw the 58/19, then you were probably looking at the checking portion.\n\nBTW: Your second gold line should probably say \"(+((p+[b]7[/b])/2))[/quote]\r\n\r\nthanks. Yes it should say (+((p+7)/2)))", "Solution_10": "You dont need to use all the terms like commutative , distributive...\r\n\r\n\r\nits a simple format question and you do not need divide by 19 and all.\r\nJust take the LCM both the sides treating p + 7 as a whole.\r\n\r\npretty easy.", "Solution_11": "I don't quite understand what you're saying. I understand that there might be an alternate method of solving, I was just showing this one step by step.\r\n\r\nFeel free to post your solution if this bothers you so much." } { "Tag": [ "function", "integration", "calculus", "complex analysis", "complex analysis unsolved" ], "Problem": "Hello,\r\n\r\nI was trying to do a problem of the forum and I am stuck on the following: does the fact that $\\forall z \\in \\mathbb{C}, \\int_{\\gamma_{z}}{\\frac{Re(f(\\omega))}{w-z}}=0$ imply that $f$ is constant ? Here $\\gamma_{z}$ denotes any simple piecewise smooth path such that $Ind(\\gamma,z)=1$, and $f$ is an entire function.\r\n\r\nNote that it would be enough to show that $Re(f)$ is entire ( I say that in case we could apply Morera's theorem ). Anyway, I'm note even sure the result holds.", "Solution_1": "This is a bit strange question: take $\\gamma_z$ to be a very small circle around $z$. Then the integral is almost $Re(f(z))$ and we conclude that $Re(f(z))=0$ for all $z$, which, of course, implies that $f$ is a (purely imaginary) constant. Are you sure that you stated the question correctly? :?", "Solution_2": "No, my question actually does not make any sense at all. Sorry for that." } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "projective geometry", "cyclic quadrilateral", "geometry proposed" ], "Problem": "Let K, L, M be the feet of the internal bisectors of the angles $\\angle A,\\ \\angle B,\\ \\angle C$ of a triangle $\\triangle ABC.$ Let D, E, F be the tangency points of the triangle incircle (I) with the sides BC, CA, AB and D', E', F' reflections of D, E, F in the angle bisectors AK, BL, CM. Show that the points $X \\equiv LM \\cap EF',\\ Y \\equiv MK \\cap FD',\\ Z \\equiv KL \\cap DE'$ lie on the line IO, where O is the triangle circumcenter.", "Solution_1": "[quote=\"yetti\"]Let K, L, M be the feet of the internal bisectors of the angles $\\angle A,\\ \\angle B,\\ \\angle C$ of a triangle $\\triangle ABC.$ Let D, E, F be the tangency points of the triangle incircle (I) with the sides BC, CA, AB and D', E', F' reflections of D, E, F in the angle bisectors AK, BL, CM. Show that the points $X \\equiv LM \\cap EF',\\ Y \\equiv MK \\cap FD',\\ Z \\equiv KL \\cap DE'$ lie on the line IO, where O is the triangle circumcenter.[/quote]\r\n\r\nFirst we will prove that the lines $FD'$,$F'D$ and $MK$ are collinear.\r\n\r\nLet's use this well know fact.\r\n\r\n[color=blue]Theorem 1.Let $ABCD$ is an inscribed quadrilateral.Let $X\\in{AB\\cap{CD}}$,$Z\\in{AC\\cap{BD}}$ and the tangents lines to the circumscribed circle at the point $A$ and $B$ intesecting at $Y$,then $X$,$Y$ and $Z$ are collinear.[/color]\r\n\r\nReally this theorem is special case of the Pascal's theorem about hexagon!\r\n \r\nIt's easy to see that the point's $F$,$D$,$D'$ and $F'$ are on the \r\n\r\nincircle,and lines $MF$,$MF'$,$KD$ and $KD'$ are touching the incircle.\r\n\r\n\r\nSo if $FD$ intersecting $F'D'$ at $T$ and $FD'$ intersecting $DF'$ at $R$,then from the theorem we have that the points $T$,$R$,$M$ and $T$,$R$,$K$ are collinear=>the points $M$,$R$,$K$ are collinear too,=>$Y\\equiv{R}$=> the lines $FD'$,$DF'$ and $MK$ are collinear.\r\n\r\n\r\nNow let's continiu the lines $FF'$,$DD'$ and $EE'$,they will construct a triangle $A'B'C'$,but us we have that $FF'\\parallel DL$ and $F'\\in(I)$,we can easely see that (I) is the Euler's circle for the triangle $A'B'C'$;$F'$,$D'$ and $E'$ are feet of altituds and $D,E,F$ are middelpoint of it's sides.\r\nFrom another well-know fact we know that the line $IO$ is same with Euler's line of the triangle $DEF$,and Euler's line of the triangle $DEF$ is same with the Euler's line of the triangle $A'B'C'$.\r\n\r\n\r\nSo finaly we have to prove that the points $X,Y,Z$ are collinear with Euler's line of the triangle $A'B'C'$,let's write this as another problem.\r\n\r\n\r\n[color=blue]Problem 2.In the triangle $ABC$,$A_{1}$ and $B_{1}$ are fetes of altitudes from $A$ and $B$,and $A_{2}$ and $B_{2}$ are middelpoints of the sides $BC$ and $AC$ respectively.Let $H$ is the orthocentr and $M$ median's intersection point of $ABC$,also $X\\in{A_{1}B_{2}\\cap{A_{2}B_{1}}}$.Prove that $H$ ,$M$ and $X$ are collinear. [/color] \r\n\r\n\r\nBut if we remember the Papus' theorem we can see that the [color=blue]Problem 2 [/color] is the special case of that theorem! :wink:", "Solution_2": "Very nice, thank you :!:\r\n\r\nLet KD', LE' meet at P. KPLC is a (concave) tangential quadrilateral, with diagonals KL, PC and DEE'D' its contact cyclic quadrilateral with diagonals DE', E'D. These 2 quadrilaterals have the same diagonal intersection $Z \\equiv KL \\cap PC \\equiv DE' \\cap ED'.$ Similarly, $X \\equiv LM \\cap EF' \\cap FE'$ and $Y \\equiv MK \\cap FD' \\cap DF'.$\r\n\r\nLet $U \\equiv KD' \\cap CA,\\ V \\equiv LE' \\cap BC.$ From the kites ABKU and BVLA, $AU = AB = BV$ According to [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=105048[/url] etc., $UV \\perp IO.$ ED' is a polar of U and DE' is a polar of V with respect to the incircle (I). The poles U, V of ED', DE' are collinear with the direction UV perpendicular to IO, i.e., with the pole W of IO at infinity, hence the polars ED', DE', IO of the collinear points U, V, W (the last one at infinity) are concurrent at Z. Similarly, X, Y also lie on IO.", "Solution_3": "Thank you for corrections and for very nice problem [b]yetti[/b] :!:" } { "Tag": [ "integration" ], "Problem": "\u03a3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03b7\u03c1\u03b9\u03b1 \u03c3\u03b5 \u03bf\u03c3\u03bf\u03c5\u03c2 \u03be\u03b5\u03ba\u03b9\u03bd\u03b7\u03c3\u03b1\u03bd \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03c4\u03b7\u03c1\u03bf\u03c5\u03bd \u03b1\u03c5\u03c4\u03b7 \u03b5\u03b4\u03c9 \u03c4\u03b7\u03bd \u03ba\u03bf\u03b9\u03bd\u03bf\u03c4\u03b7\u03c4\u03b1 \u03b1\u03bb\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b9\u03c0\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b9\u03bf\u03c5\u03bd\u03c4\u03b1\u03b9 \u03b5\u03bd\u03b5\u03c1\u03b3\u03b1 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03b7! \r\n\u0391\u03c2 \u03b2\u03b1\u03bb\u03c9 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03c9 \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03bc\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\r\n\r\n \u039d\u03b1 \u03b5\u03be\u03b5\u03c4\u03b1\u03c3\u03c4\u03b5\u03b9 \u03b1\u03bd \u03c3\u03c5\u03b3\u03ba\u03bb\u03b9\u03bd\u03b5\u03b9 \u03b7 \u03c3\u03b5\u03b9\u03c1\u03b1\r\n\r\n\r\n\r\n$ \\sum_{n=1}^{\\infty}{\\int_0^1\\cos(nt^2)dt}$", "Solution_1": "[quote=\"mathsforfun!\"]\u039d\u03b1 \u03b5\u03be\u03b5\u03c4\u03b1\u03c3\u03c4\u03b5\u03b9 \u03b1\u03bd \u03c3\u03c5\u03b3\u03ba\u03bb\u03b9\u03bd\u03b5\u03b9 \u03b7 \u03c3\u03b5\u03b9\u03c1\u03b1\n\n$ \\sum_{n = 1}^{\\infty}{\\int_0^1\\cos(nt^2)dt}$[/quote]\r\n\r\nFirst off, welcome :welcomeani:\r\n\r\n\u0391\u03bd \u03b4\u03b5\u03bd \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2, \u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03b1\u03c0\u03bf\u03ba\u03bb\u03af\u03bd\u03b5\u03b9... \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 $ u = t\\sqrt {n}$ \u03bf \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc\u03c2 \u03cc\u03c1\u03bf\u03c2 \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 $ \\frac {1}{\\sqrt {n}} \\int_0^{\\sqrt {n}} \\cos(u^2) du$ \u03ba\u03b1\u03b9 \u03c4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03cc \u03cc\u03c1\u03b9\u03bf ($ \\sqrt {\\pi}/4$? \u03ba\u03ac\u03c4\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf) \u03ba\u03b1\u03b8\u03ce\u03c2 $ n\\to\\infty$, \u03bf\u03c0\u03cc\u03c4\u03b5...\r\n\r\n\u0392\u03ad\u03b2\u03b1\u03b9\u03b1, \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c3\u03c5\u03b3\u03ba\u03bb\u03af\u03bd\u03b5\u03b9 \u03c3\u03b5 \u03b8\u03b5\u03c4\u03b9\u03ba\u03ae \u03c4\u03b9\u03bc\u03ae \u03b8\u03ad\u03bb\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b4\u03bf\u03c5\u03bb\u03af\u03c4\u03c3\u03b1, \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bc\u03c9\u03c2 \u03b3\u03b5\u03bd\u03b9\u03ba\u03ac \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae (\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03be\u03ad\u03c1\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ae \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03c3\u03b7). \u0391\u03bd \u03b8\u03ad\u03bb\u03b5\u03c4\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b5\u03c2 \u03bb\u03b5\u03c0\u03c4\u03bf\u03bc\u03ad\u03c1\u03b5\u03b9\u03b5\u03c2, let me know.\r\n\r\nCheerio,\r\n\r\nDurandal 1707\r\n\r\nPS: T\u03b1 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03bc\u03b1\u03c4\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c6\u03c4\u03b7\u03bd\u03cc lower bound, \u03b1\u03bb\u03bb\u03ac \u03c4\u03ce\u03c1\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b3\u03ac \u03ba\u03b1\u03b9 \u03b2\u03b1\u03c1\u03af\u03b5\u03bc\u03b1\u03b9 \u03bd\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03ce... \u03b1\u03c0\u03cc \u03b1\u03cd\u03c1\u03b9\u03bf :)", "Solution_2": "[quote=\"Durandal\"][quote=\"mathsforfun!\"]\u039d\u03b1 \u03b5\u03be\u03b5\u03c4\u03b1\u03c3\u03c4\u03b5\u03b9 \u03b1\u03bd \u03c3\u03c5\u03b3\u03ba\u03bb\u03b9\u03bd\u03b5\u03b9 \u03b7 \u03c3\u03b5\u03b9\u03c1\u03b1\n\n$ \\sum_{n = 1}^{\\infty}{\\int_0^1\\cos(nt^2)dt}$[/quote]\n\nFirst off, welcome :welcomeani:\n\n\u0391\u03bd \u03b4\u03b5\u03bd \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2, \u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03b1\u03c0\u03bf\u03ba\u03bb\u03af\u03bd\u03b5\u03b9... \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 $ u = t\\sqrt {n}$ \u03bf \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc\u03c2 \u03cc\u03c1\u03bf\u03c2 \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 $ \\frac {1}{\\sqrt {n}} \\int_0^{\\sqrt {n}} \\cos(u^2) du$ \u03ba\u03b1\u03b9 \u03c4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03cc \u03cc\u03c1\u03b9\u03bf ($ \\sqrt {\\pi}/4$? \u03ba\u03ac\u03c4\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf) \u03ba\u03b1\u03b8\u03ce\u03c2 $ n\\to\\infty$, \u03bf\u03c0\u03cc\u03c4\u03b5...\n\n\u0392\u03ad\u03b2\u03b1\u03b9\u03b1, \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c3\u03c5\u03b3\u03ba\u03bb\u03af\u03bd\u03b5\u03b9 \u03c3\u03b5 \u03b8\u03b5\u03c4\u03b9\u03ba\u03ae \u03c4\u03b9\u03bc\u03ae \u03b8\u03ad\u03bb\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b4\u03bf\u03c5\u03bb\u03af\u03c4\u03c3\u03b1, \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bc\u03c9\u03c2 \u03b3\u03b5\u03bd\u03b9\u03ba\u03ac \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae (\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03be\u03ad\u03c1\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ae \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03c3\u03b7). \u0391\u03bd \u03b8\u03ad\u03bb\u03b5\u03c4\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b5\u03c2 \u03bb\u03b5\u03c0\u03c4\u03bf\u03bc\u03ad\u03c1\u03b5\u03b9\u03b5\u03c2, let me know.\n\nCheerio,\n\nDurandal 1707\nPS: T\u03b1 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03bc\u03b1\u03c4\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c6\u03c4\u03b7\u03bd\u03cc lower bound, \u03b1\u03bb\u03bb\u03ac \u03c4\u03ce\u03c1\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b3\u03ac \u03ba\u03b1\u03b9 \u03b2\u03b1\u03c1\u03af\u03b5\u03bc\u03b1\u03b9 \u03bd\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03ce... \u03b1\u03c0\u03cc \u03b1\u03cd\u03c1\u03b9\u03bf :)[/quote]\r\n\u0394\u03b5\u03c2 \u03c3\u03c4\u03bf http://en.wikipedia.org/wiki/Cornu_spiral#Cornu_spiral\r\n\u039f\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03bc\u03b1 Fresnel \u03c7\u03c1\u03ae\u03c3\u03b9\u03bc\u03bf \u03c3\u03c4\u03b7\u03bd \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03bf\u03c0\u03c4\u03b9\u03ba\u03ae" } { "Tag": [ "IMO", "IMO Shortlist" ], "Problem": "Here is the IMO shortlist problems 1992, I got it from internet.", "Solution_1": "IMO 1992 LongList and ShortList:", "Solution_2": "I notice that problem 45 of the [b]1992[/b] short list appears on the [b]1993[/b] IMO. Does this strike anyone else as strange?", "Solution_3": "[b][size=160]IMO LongList 1992 Problems (Posted In The Forums)[/size][/b]\n\nProblem 1 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004282\n\nProblem 2 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004289\n\nProblem 3 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004807\n\nProblem 4 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004811\n\nProblem 5 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004306\n\nProblem 6 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004818\n\nProblem 7 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005940\n\nProblem 8 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004820\n\nProblem 9 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004823\n\nProblem 10 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366404\n\nProblem 11 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004308\n\nProblem 12 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004829\n\nProblem 13 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004837\n\nProblem 14 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004841\n\nProblem 15 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004846\n\nProblem 16 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004310\n\nProblem 17 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366410\n\nProblem 18 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004855\n\nProblem 19 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004863\n\nProblem 20 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004866\n\nProblem 21 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004869\n\nProblem 22 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366420\n\nProblem 23 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004314\n\nProblem 24 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004925\n\nProblem 25 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004935\n\nProblem 26 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366399\n\nProblem 27 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004965\n\nProblem 28 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=126017\n\nProblem 29 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222516\n\nProblem 30 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2002366\n\nProblem 31 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222819\n\nProblem 32 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004995\n\nProblem 33 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2002365\n\nProblem 34 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005037\n\nProblem 35 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222519\n\nProblem 36 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004327\n\nProblem 37 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005065\n\nProblem 38 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366415\n\nProblem 39 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005071\n\nProblem 40 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005073\n\nProblem 41 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005088\n\nProblem 42 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222521\n\nProblem 43 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005096\n\nProblem 44 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849468\n\nProblem 45 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005112\n\nProblem 46 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005114\n\nProblem 47 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004331\n\nProblem 48 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005214\n\nProblem 49 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005480\n\nProblem 50 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005699\n\nProblem 51 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222525\n\nProblem 52 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005715\n\nProblem 53 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=154338\n\nProblem 54 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005719\n\nProblem 55 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222823\n\nProblem 56 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005726\n\nProblem 57 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005730\n\nProblem 58 : False\n\nProblem 59 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005737\n\nProblem 60 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222532\n\nProblem 61 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005739\n\nProblem 62 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005745\n\nProblem 63 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005749\n\nProblem 64 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005753\n\nProblem 65 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005754\n\nProblem 66 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005759\n\nProblem 67 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005765\n\nProblem 68 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004333\n\nProblem 69 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222814\n\nProblem 70 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005771\n\nProblem 71 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005889\n\nProblem 72 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005893\n\nProblem 73 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005906\n\nProblem 74 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005912\n\nProblem 75 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005916\n\nProblem 76 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005919\n\nProblem 77 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005922\n\nProblem 78 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005925\n\nProblem 79 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222822\n\nProblem 80 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005926\n\nProblem 81 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005928\n\nProblem 82 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005932", "Solution_4": "Here is typed form of IMO 1992 ShortList." } { "Tag": [ "Euler", "logarithms" ], "Problem": "For $ n$ a positive integer, if $ \\frac {1}{(4n^2 \\minus{} 2n)^2}$ is the general term,\r\nfind the exact value of the sum of the following infinite series:\r\n\r\n$ \\frac{1}{2^2} \\plus{} \\frac{1}{12^2} \\plus{} \\frac{1}{30^2} \\plus{} ...$", "Solution_1": "[quote=\"Arrange your tan\"]For $ n$ a positive integer, if $ \\frac {1}{(4n^2 \\minus{} 2n)^2}$ is the general term,\nfind the exact value of the sum of the following infinite series:\n\n$ \\frac {1}{2^2} \\plus{} \\frac {1}{12^2} \\plus{} \\frac {1}{30^2} \\plus{} ...$[/quote]\r\n[hide]We want to find $ \\sum_{n\\equal{}1}^{\\infty} \\frac{1}{(4n^2\\minus{}2n)^2}$. This splits into $ \\sum_{n\\equal{}1}^{\\infty} (\\frac{1}{n}\\plus{}\\frac{1}{4n^2}\\minus{}\\frac{2}{2n\\minus{}1}\\plus{}\\frac{1}{(2n\\minus{}1)^{2}})$[/hide]\r\nThis is what I have so far...unfortunately I have to go...:| (I'll work on this later)", "Solution_2": "[quote=\"Arrange your tan\"]For $ n$ a positive integer, if $ \\frac {1}{(4n^2 \\minus{} 2n)^2}$ is the general term,\nfind the exact value of the sum of the following infinite series:\n\n$ \\frac {1}{2^2} \\plus{} \\frac {1}{12^2} \\plus{} \\frac {1}{30^2} \\plus{} ...$[/quote]\r\n\r\nSuggested hint for starting:\r\n[hide]\nThe general term can be written as $ \\frac {1}{[(2n \\minus{} 1)(2n)]^2}$.\n\nAnd the first few terms can be written as:\n\n$ \\frac {1}{(1 \\cdot 2)^2} \\plus{} \\frac {1}{(3 \\cdot 4)^2} \\plus{} \\frac {1}{(5 \\cdot 6)^2} \\plus{} ...$ \n[/hide]\n\nAdditional hint:\n[hide]\nThe general term can be rewritten as $ (\\frac {1}{2n \\minus{} 1} \\minus{} \\frac{1}{2n})^2$.\n\nAnd the first few terms can be rewritten as:\n\n$ ( \\frac 11 \\minus{} \\frac12)^2 \\plus{} (\\frac13 \\minus{} \\frac14)^2 \\plus{} (\\frac15 \\minus{} \\frac16)^2 \\plus{} ...$\n\n[/hide]", "Solution_3": "[hide=\"Solution\"]\nAs mentioned in the tip, the general term can be rewritten as $ (\\frac{1}{2n\\minus{}1}\\minus{}\\frac{1}{2n})^{2}$. \nThus, we have:\n$ (\\frac{1}{1}\\minus{}\\frac{1}2)^{2}\\plus{}(\\frac{1}3\\minus{}\\frac{1}4)^{2}\\plus{}(\\frac{1}5\\minus{}\\frac{1}6)^{2}\\plus{}...$.\nIf we expand the terms and reorganize, we get:\n$ \\left ( \\frac{1}{1^{2}} \\plus{}\\frac{1}{2^{2}}\\plus{}\\frac{1}{3^{2}}\\plus{}...\\right )\\minus{}2\\left ( \\frac{1}{1*2}\\plus{}\\frac{1}{3*4}\\plus{}\\frac{1}{5*6}\\plus{}...\\right )$\n\nWe have now expressed the series as the sum of two other series.\nThe first series is the $ \\frac{1}{n^{2}}$ series, whose sum Euler proved to be $ \\frac{\\pi ^{2}}{6}$.\nThe second series is not as obvious. Disregarding the factor of $ 2$, the general term can be expressed as $ \\frac{1}{2n(2n\\minus{}1)}$, which can be rewritten as $ \\frac{1}{2n\\minus{}1}\\minus{}\\frac{1}{2n}$. Thus, we have $ \\left (\\frac{1}{1}\\minus{}\\frac{1}{2} \\right )\\plus{}\\left (\\frac{1}{3}\\minus{}\\frac{1}{4} \\right )\\plus{}\\left (\\frac{1}{5}\\minus{}\\frac{1}{6} \\right )\\plus{}...$. This is merely the alternating harmonic series, whose sum is $ \\ln 2$.\n\nSo, the answer is $ \\boxed{\\frac{\\pi ^{2}}{6}\\minus{}2\\ln 2}$.\n[/hide]\r\n\r\nI liked the problem, may I ask how you came up with it?" } { "Tag": [ "function", "algebra", "polynomial", "trigonometry", "linear algebra", "matrix", "calculus" ], "Problem": "Of course the solutions of $y''+y=0$ differential equation are well known.I need a particular solution for the differential equation $y''+y=4cosx$.", "Solution_1": "One of the method of finding the particilar solution is using [b]Differential Operator.[/b]\r\n\r\nLet $D=\\frac{d}{dx},$ we can rewrite the given differential equation as $(D^{2}+1)y=Re 4e^{ix}.$ Thus $y=\\frac{4}{D^{2}+1}e^{ix}$\r\n\r\n$\\therefore y=\\frac{4}{(D+i)(D-i)}e^{ix}.$", "Solution_2": "This is a constant-coefficient linear differential equation for which the nonhomogeneity lies within the class of functions (linear combinations of exponentials times polynomials) that are themselves annihilated by some such operator.\r\n\r\nNotation: define the operator $L(y)=y''+y.$ The characteristic polynomial associated to this operator is $\\lambda^{2}+1,$ which has roots $\\pm i.$ What expontial $e^{\\lambda x}$ is $4\\cos x$ associated with? That's also $\\pm i,$ and that's a complication, since it means that $L(4\\cos x)=0.$\r\n\r\nThe situation is essentially the same as having a double root of the characteristic polynomial; we get ourselves out of it by multiplying by $x.$\r\n\r\nSo: Assume $y_{p}(x)=Ax\\cos x+Bx\\sin x.$\r\n\r\nDifferentiate that at put it into the differential equation - can you now solve for $A$ and $B?$", "Solution_3": "Thank u very much. I found $y_{p}(x)=2x\\sin x$ is a particular solution. Thanks again.", "Solution_4": "What I showed you above was the method of undetermined coefficients. One could also use the method of variation of parameters.\r\n\r\nStart by assuming that $y=u\\cos x+v\\sin x.$ We're going to take advantage of the fact that the solution space of the homogeneous equation is spanned by $\\{\\cos x,\\sin x\\}.$ Since we've introduced two degrees of freedom, we will impose one extra condition when it is convenient to do so.\r\n\r\n$y'=u'\\cos x+v'\\sin x-u\\sin x+v\\cos x.$\r\n\r\nWe impose the condition $u'\\cos x+v'\\sin x=0.$ That gives us\r\n\r\n$y'=-u\\sin x+v\\cos x$\r\n\r\n$y''=-u'\\sin x+v'\\cos x-u\\cos x-v\\sin x$\r\n\r\n$y''+y=-u'\\sin x+v'\\cos x=4\\cos x$ by the differential equation.\r\n\r\nThat gives us two linear equations for the two unknown functions $u'$ and $v'.$ In matrix form:\r\n\r\n$\\begin{bmatrix}\\cos x&\\sin x\\\\-\\sin x&\\cos x\\end{bmatrix}\\begin{bmatrix}u'\\\\ v'\\end{bmatrix}=\\begin{bmatrix}0\\\\ 4\\cos x\\end{bmatrix}.$\r\n\r\nThis has the solutions $u'=-4\\sin x\\cos x$ and $v'=4\\cos^{2}x.$ There's still quite a bit of integration and algebraic manipulation to go, but with patience, you will arrive at the same answer as above. \r\n\r\nIn this particular instance, the technique of undetermined coefficients proved far simpler; but the technique of variation of parameters can reach cases that are beyond the reach of undetermined coefficients." } { "Tag": [ "induction" ], "Problem": "Consider the integers $2$, $3$, and $6$. They are distinct, and their reciprocals add up to $1$.\r\n\r\nCan you think of four distinct integers whose reciprocals add up to $1$?\r\n\r\nHow about five such integers?\r\n\r\nAnd six?\r\n\r\nAnd $n$ such integers?", "Solution_1": "This is related to perfect numbers.\r\n\r\nIf n = sum-of-proper-divisors-of-n, then dividing both sides by n gives $1 = 1/d_{1}+1/d_{2}+...$.\r\n\r\n28 = 1 + 2 + 4 + 7 + 14\r\n1 = 1/28 + 1/14 + 1/7 + 1/4 + 1/2", "Solution_2": "You know that this works for $n=3$, then you can show that it works for $n=4$:\n[hide]for example\n$1=2^{-1}+4^{-1}+6^{-1}+12^{-1}$[/hide]\n and then do the easy induction step from $n$ to $n+2$\n[hide]simply replace the reciprocal of the largest integer $n_{max}^{-1}$ by the sum\n$n_{max}^{-1}(2^{-1}+3^{-1}+6^{-1}).$[/hide]", "Solution_3": "And then do the even easier step from $n$ to $n+1$ - halve it!" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:\r\n$ \\frac{a^{3}+b^{3}+c^{3}}{a+b+c}\\geq\\frac{a^{3}b^{3}+a^{3}c^{3}+b^{3}c^{3}}{a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}}.$", "Solution_1": "(a,b,c)=(1,1,1/3),then \r\n\r\n(a^3+b^3+c^3)/(a+b+c)\r\n\r\n-(b^3*a^3+b^3*c^3+c^3*a^3)/(a^2*b^2+b^2*c^2+c^2*a^2)=-4/693.", "Solution_2": "[quote=\"fjwxcsl\"](a,b,c)=(1,1,1/3),then \n\n(a^3+b^3+c^3)/(a+b+c)\n\n-(b^3*a^3+b^3*c^3+c^3*a^3)/(a^2*b^2+b^2*c^2+c^2*a^2)=-4/693.[/quote]\r\nYes, you are right. My mistake. Thank you, fjwxcsl. :(", "Solution_3": "oh my god, the first time, argady have made mistake :P I hope you will correct it and you will have another nice inequality :D If it is symetric, maybe Schur will kill it ..." } { "Tag": [ "trigonometry", "quadratics", "inequalities" ], "Problem": "Find all nagetive $a$ which makes the quadratic inequality $\\sin^2{x}+a\\cos{x}+a^2 \\geq 1+ \\cos{x}$ true for all $x \\in R$.", "Solution_1": "[quote=\"shobber\"]Find all nagetive $a$ which makes the quadratic inequality $\\sin^2{x}+a\\cos{x}+a^2 \\geq 1+ \\cos{x}$ true for all $x \\in R$.[/quote]\r\n\r\nIs it this way ? \r\nWe have $cos^2 x+(1-a)cos x - a^2\\leq 0$ after simplying . So taking discriminant , \r\n\r\n$(1-a)^2+4a^2=5(a-\\frac{1}{5})^2+\\frac{4}{5}>0$ which is always true for all $a$ . Since $a$ is non-negative,so $a\\geq 0$ . :?", "Solution_2": "Be more careful.\r\n(1) $a$ is negative.\r\n(2) $|\\cos{x}| \\leq 1$", "Solution_3": "[quote=\"shobber\"]Be more careful.\n(1) $a$ is negative.\n(2) $|\\cos{x}| \\leq 1$[/quote]\r\n\r\n$\\cos ^2 x \\le \\cos x\\,\\,\\,\\,\\forall x$ such that cosx positive\r\n\r\nso, $\\cos ^2 x + (1 - a)\\cos x\\, - a^2 \\le 0\\,\\,\\,$\r\n\r\n$ \\Rightarrow cos x + (1 - a)\\cos x\\, - a^2 \\, \\le 0$\r\n\r\n$ \\Rightarrow (2 - a)\\cos x\\, - a^2 \\, \\le 0$\r\n\r\nfor this to be true for all x, \r\n\r\n$\\frac{{a^2 }}{{2 - a}} \\ge 1$ and as a is given negative, this inequality gives $a \\le - 2$", "Solution_4": "[quote=\"shobber\"]Be more careful.\n(1) $a$ is negative.\n(2) $|\\cos{x}| \\leq 1$[/quote]\r\n\r\nOpps .... misread as $a$ is non-negative :blush: But I cant figure out how $|\\cos{x}| \\leq 1$ would have affected the roots . :o", "Solution_5": "[quote=\"vidyamanohar\"][quote=\"shobber\"]Be more careful.\n(1) $a$ is negative.\n(2) $|\\cos{x}| \\leq 1$[/quote]\n\n$\\cos ^2 x \\le \\cos x\\,\\,\\,\\,\\forall x$\n\nso, $\\cos ^2 x + (1 - a)\\cos x\\, - a^2 \\le 0\\,\\,\\,$\n\n$ \\Rightarrow cos x + (1 - a)\\cos x\\, - a^2 \\, \\le 0$\n\n$ \\Rightarrow (2 - a)\\cos x\\, - a^2 \\, \\le 0$\n\nfor this to be true for all x, \n\n$\\frac{{a^2 }}{{2 - a}} \\ge 1$ and as a is given negative, this inequality gives $a \\le - 2$[/quote]\r\n\r\nHow this $ \\Rightarrow cos x + (1 - a)\\cos x\\, - a^2 \\, \\le 0$ can happen ?\r\nAnd why this is true ? $\\cos ^2 x \\le \\cos x\\,\\,\\,\\,\\forall x$\r\nIf $x$ is $120^0$ i.e , that cant be true ...\r\n\r\nIm abit confused for what you have write . Can you explain in word ?", "Solution_6": "[quote=\"shyong\"][quote=\"vidyamanohar\"][quote=\"shobber\"]Be more careful.\n(1) $a$ is negative.\n(2) $|\\cos{x}| \\leq 1$[/quote]\n\n$\\cos ^2 x \\le \\cos x\\,\\,\\,\\,\\forall x$\n\nso, $\\cos ^2 x + (1 - a)\\cos x\\, - a^2 \\le 0\\,\\,\\,$\n\n$ \\Rightarrow cos x + (1 - a)\\cos x\\, - a^2 \\, \\le 0$\n\n$ \\Rightarrow (2 - a)\\cos x\\, - a^2 \\, \\le 0$\n\nfor this to be true for all x, \n\n$\\frac{{a^2 }}{{2 - a}} \\ge 1$ and as a is given negative, this inequality gives $a \\le - 2$[/quote]\n\nHow this $ \\Rightarrow cos x + (1 - a)\\cos x\\, - a^2 \\, \\le 0$ can happen ?\n\nIm abit confused for what you have write . Can you explain in word ?[/quote]\r\n\r\nwe are looking at the smallest value that can be bigger than $\\cos ^2 x + (1 - a)\\cos x\\, - a^2$.\r\n\r\nWhen positive, cosx is bigger than $\\cos ^2 x$. \r\n\r\nI have compared the $cos x + (1 - a)\\cos x\\, - a^2$ with zero, in place of $\\cos ^2 x + (1 - a)\\cos x\\, - a^2$.\r\nand $|\\cos{x}| \\leq 1$ is reflected in taking $\\frac{{a^2 }}{{2 - a}} \\ge 1$\r\n\r\nhope this clears your doubt", "Solution_7": "[quote=\"vidyamanohar\"][/quote][quote=\"shyong\"][quote=\"vidyamanohar\"][quote=\"shobber\"]Be more careful.\n(1) $a$ is negative.\n(2) $|\\cos{x}| \\leq 1$[/quote]\n\n$\\cos ^2 x \\le \\cos x\\,\\,\\,\\,\\forall x$\n\nso, $\\cos ^2 x + (1 - a)\\cos x\\, - a^2 \\le 0\\,\\,\\,$\n\n$ \\Rightarrow cos x + (1 - a)\\cos x\\, - a^2 \\, \\le 0$\n\n$ \\Rightarrow (2 - a)\\cos x\\, - a^2 \\, \\le 0$\n\nfor this to be true for all x, \n\n$\\frac{{a^2 }}{{2 - a}} \\ge 1$ and as a is given negative, this inequality gives $a \\le - 2$[/quote]\n\nHow this $ \\Rightarrow cos x + (1 - a)\\cos x\\, - a^2 \\, \\le 0$ can happen ?\n\nIm abit confused for what you have write . Can you explain in word ?[/quote][quote=\"vidyamanohar\"]\n\nwe are looking for the smallest value that can be bigger than $\\cos ^2 x + (1 - a)\\cos x\\, - a^2$.\n\nWhen positive, cosx is bigger than $\\cos ^2 x$. \n\nI have compared the $cos x + (1 - a)\\cos x\\, - a^2$ with zero, in place of $\\cos ^2 x + (1 - a)\\cos x\\, - a^2$.\nand $|\\cos{x}| \\leq 1$ is reflected in taking $\\frac{{a^2 }}{{2 - a}} \\ge 1$\n\nhope this clears your doubt[/quote]\r\n\r\nWhy can we compare $cos x + (1 - a)\\cos x\\, - a^2$ with zero in place of $\\cos ^2 x + (1 - a)\\cos x\\, - a^2$ ? I cant get it :(" } { "Tag": [ "vector", "geometry", "geometric transformation", "reflection", "advanced fields", "advanced fields unsolved" ], "Problem": "Sketch the vector field on the real line, find all the fixed points, classify their stability, and sketch the graph of x(t) for different initial conditions.\r\n\r\n(a) x* = 1 - (x^14)\r\n(b) x* = sinx/(e^x)\r\n(c) x* = x - (x^3)\r\n\r\nThe roots for (a) and (b) are x=+/- 1 but I don't know how to draw these or where to take it from here.\r\nI'm not sure about (c). I think sinx=0 so that corresponds to x=0, 2pi, pi.....???", "Solution_1": "For drawing, all you need to do is to figure out when the vector points to the right and when it points to the left (i.e., when the expression is positive and when it is negative). Once you figured that out, imagine (or try to draw) an arrow at each point of the real line in the direction corresponding to that point. Now just follow the arrows and see whether they lead you toward a fixed point (stability) or away from it (instability). Your typical drawing will look like this:\r\n\r\n[asy]size(250);\nreal f(real x){return 0.14*sgn(x);}\nD((0,0.2),rgb(0.95,0.99,0.99));\nD((-2,0)--(2,0));\nD(\"-1\",(-1,0));\nD(\"0\",(0,0));\nD(\"1\",(1,0));\nfor(real x=-1.8;x<1.9;x+=0.2)\nif(abs(x-x^3)>0.0001) \n{\nreal y=x+f(x-x^3);\nD((x,0)--(y,0),black+linewidth(1), EndArrow);\n}[/asy]\r\n\r\nIf you want to be more accurate, you should make the lengths of the arrows reflect the size of the right hand side (the larger the size, the longer the arrow) but it is not so important.", "Solution_2": "But don't you need to know what the graph looks like in order to know what the vector field is? How do you graph x-x^3? Or sinx/e^x, or x^14?" } { "Tag": [ "function", "integration", "analytic geometry", "topology", "complex analysis", "complex analysis unsolved" ], "Problem": "Let $ \\Omega$ be an open set of the complex plane $ F$ be a family of analytic functions $ f: \\Omega \\rightarrow \\mathbb{C}$ .\r\nSuppose that there is an $ M>0$ such that\r\n$ \\int\\int_{\\Omega} |f(x,y)| dxdy=0) x^2^n \\geq \\sum (n>=0) \\int (n-n+1) x^2^t dt and also \\sum (n>=0) x^2^n=x+ \\sum (n>=1) x^2^n<=x+ \\sum (n>=1) \\int (n-1-n) x^2^t dt. Thus, \\sum (n>=0) x 2 2 2 n - \\int (0 to infinity) x^2^t dt . Make the change of variable in the integral and find that it is 1/ln2* \\int (1 to infinity) x^u/u du and make the same trick: write the last integral as an infinite sum of integrals and use the same inequality. I think you've got the idea.", "Solution_4": "Thanks for showing your solution to me. There was a tiny mistake in mine (a typo), so I edited the post.\r\n\r\nRegarding (b), to be honest, I don't think the limit exists. I tried to use Mathematica to make a plot of f'-g', but I only did this for some values of n, not the whole series, and it's really crazy-looking. \r\n\r\nAnyway, we definitely can't say somthing like \"f-g is increasing\" (or \"decreasing\").", "Solution_5": "I have ploted f-g in MathCad. It looks like decreasing continuos concave (maybe, approximately) function.\r\n\r\nMy solution of a) is similar to grobber's one.", "Solution_6": "Well, I didn't know how to plot x+x2+.. in Mathematica and I only worked with approximations of f'-g', but I think there's a region \"realy close\" to 1 where f-g becomes \"very decreasing\" :).", "Solution_7": "I was trying to make some progress on this problem, when I realized that my solution for (a) doesn't show squat :). The first inequality shows that $(f-g)(x)\\le 0$, but the second inequality shows that $(f-g)(x)\\ge (1-\\frac 1{\\ln 2})(x+x^2+x^4+\\ldots)$, and this isn't bounded from below on $(0,1)$. It tends to $-\\infty$ when $x\\to 1,\\ x<1$.\r\n\r\nAny progress anyone? (I mean for point (b))", "Solution_8": "For a) I used $1-x\\leq \\frac{2}{3}\\cdot(1-x^2)$ for $x\\in [1/2,1]$ and $0\\leq\\log_2(1+x)-x\\leq (1-\\frac{1}{2\\ln 2})(1-x)$.", "Solution_9": "I can't manage to show that when you sum members like the last one for $x^{2^n}$ instead of $x$ you get a bounded sum, even with your hints. Anyway, I think Harazi's solution gives the best idea about what's going on. I think we get a tight bound by using that method: $(f-g)(x)\\ge\\frac {-x}{\\ln 2}$. \r\n\r\nYours might be just as good, but I'm sort of tired right now, and this problem is giving me a really hard time. I can't try to solve other problems because I'm obsessed, but, on the other hand, I have no idea how to continue :(.", "Solution_10": "2grobber. For a).\r\n\\[\r\n\\sum_{n=1}^{\\infty}\\left(\\log_2(1+x^{2^n})-x^{2^n}\\right)=\\sum_{x^{2^n}>1/2}+\\sum_{x^{2^n}\\leq 1/2}.\r\n\\]\r\nSecond sum is obviously bounded. For the first sum we have:\r\n\\[\r\n\\sum_{x^{2^n}>1/2} = \\sum_{n=1}^{k}\\leq \\sum_{n=1}^{k} C\\cdot(1-x^{2^{k+1}})\\left(\\frac{2}{3}\\right)^n\\leq C',\r\n\\]\r\nwhere $x^{2^k}>1/2$ and $x^{2^{k+1}}\\leq 1/2$.", "Solution_11": "Thanks! Pretty stupid of me not to see that :).", "Solution_12": "I think that I have finally managed to prove that the limit exists. Anyway, there are so many computations that I am sure it is wrong. \r\n Let us take a number $0\\frac12$, period. And it is hard to believe that $2$ is special here. I understand that this is not a solution, but I hope the argument is fairly convincing even if we agree to outrule using computers for this problem...\n[/hide]", "Solution_16": "The equality from http://www.mathlinks.ro/Forum/viewtopic.php?t=28814 seems unbelievable at first sight too. :)", "Solution_17": "OK. I figured the one I need out. If you apply $x\\frac d{dx}$ to it many times and choose $x$ appropriately, one term will dominate the sum of all others. :lol:", "Solution_18": "Sorry. What are talking about? Do you speak russian?", "Solution_19": "[quote=\"Myth\"]Sorry. What are talking about? Do you speak russian?[/quote]\r\nThe identity I declared \"unbelievable\" in my previous post. Da, ja govorju po russki luchshe chem po anglijski... ;)", "Solution_20": "[quote=\"fedja\"][quote=\"Myth\"]Sorry. What are talking about? Do you speak russian?[/quote]\nThe identity I declared \"unbelievable\" in my previous post. Da, ja govorju po russki luchshe chem po anglijski... ;)[/quote]\r\n\u042f \u043f\u0440\u043e?\u0442\u043e \u043d\u0435 \u043f\u043e\u043d?\u043b, \u043a \u0447\u0435\u043c\u0443 \u0442\u044b ?\u043a\u0430\u0437\u0430\u043b \u043f\u0440\u043e \u0434\u0438\u0444\u0444\u0435\u0440\u0435\u043d\u0446\u0438\u0440\u043e\u0432\u0430\u043d\u0438\u0435. \u0422\u044b \u0443\u0447\u0438\u0448\u044c??/\u0440\u0430\u0431\u043e\u0442\u0430\u0435\u0448\u044c \u0432 \u0421\u0428??\r\n\r\nTo Valentin. I have own opinion when I am allowed to write in Russian ;)", "Solution_21": "Why don't we switch to PM's not to spam the forum? ;)", "Solution_22": "Aww well guys I didnt get what you are trying to say & the link is not opening in my browser.\r\nSo that leaves with the million dollar question whether the limit exists or not.According to my \"wrong\"calculations it doesnot... :(", "Solution_23": "[quote=\"the game\"]Aww well guys I didnt get what you are trying to say & the link is not opening in my browser.[/quote]\r\nSorry, the game! :blush: Do you understand what was written in my posts #16 and #18? Together they are assumed to prove that there is no limit...", "Solution_24": "[quote=\"fedja\"][quote=\"the game\"]Aww well guys I didnt get what you are trying to say & the link is not opening in my browser.[/quote]\nSorry, the game! :blush: Do you understand what was written in my posts #16 and #18? Together they are assumed to prove that there is no limit...[/quote]\r\n\r\nSorry fedja the link didnt open .Maybe some kind of browser problem.Thats what I meant :blush:" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Let p(x) be a polynomial with integer coefficients. Prove that if k is an even integer, we have p(k)=p(0)(mod 2) and if k is odd, then we have p(k)=p(1)(mod 2)", "Solution_1": "[hide=\"First Problem\"]\nWhen k is a even integer(2b):\n$ P(k) \\equal{} (a_n)(2b)^n \\plus{} (a_{n \\minus{} 1})(2b)^{n \\minus{} 1} \\plus{} ... \\plus{} a_0.$\nAnd\n$ p(0) \\equal{} a_0$\n\nWe know that $ a_0\\equiv a_0(mod 2)$ and that if $ a\\equiv b(mod z)$ and $ c\\equiv d(mod z)$ then $ a \\plus{} c\\equiv b \\plus{} d(mod z)$.\n\nSince every term that contains (2b) is even and congruent to 0 mod 2, when we add the terms in $ p(k)$ in (mod 2) we are left with $ a_0$ and that equals the value of $ p(0)$.\n[/hide]\n[hide=\"Second Problem\"]\nTo Prove that when k is odd, $ p(k)\\equiv p(1)(mod 2)$ we in other words have to prove that $ p(k)\\minus{}p(1)\\equiv 0(mod 2)$\n\nLet's say that $ k\\equal{}2b\\plus{}1$\nSo we have\n$ p(k)\\equal{}(a_n)(2b\\plus{}1)^n\\plus{}(a_{n\\minus{}1})(2b\\plus{}1)^{n\\minus{}1}\\plus{}\\cdots\\plus{}a_0$\n$ p(1)\\equal{}(a_n)(1)^n\\plus{}(a_{n\\minus{}1})(1)^{n\\minus{}1}\\plus{}\\cdots\\plus{}a_0$\n\n$ (a_n)(2b\\plus{}1)^n\\minus{}(a_n)(1)^n\\equiv (a_n)((2b)^n\\plus{}(2b)^{n\\minus{}1}\\plus{}\\cdots\\plus{}1)\\minus{}(a_n)(1)^n(mod 2)$\n\nSince we know that every term that contains $ 2b$ is congruent to 0 (mod 2) we are left with:\n\n$ (a_n)(2b\\plus{}1)^n\\minus{}(a_n)(1)^n\\equiv 0(mod 2)$\nWe repeat this process for every n:\n$ (a_{n\\minus{}1})(2b\\plus{}1)^{n\\minus{}1}\\minus{}(a_{n\\minus{}1})(1)^{n\\minus{}1}\\equiv 0(mod 2)$\n$ (a_{n\\minus{}2})(2b\\plus{}1)^{n\\minus{}2}\\minus{}(a_{n\\minus{}2})(1)^{n\\minus{}2}\\equiv 0(mod 2)$\n$ \\vdots$\n$ (a_{0})(2b\\plus{}1)^{0}\\minus{}(a_{0})(1)^{0}\\equiv 0(mod 2)$\nSo this proves that $ p(k)\\minus{}p(1)\\equiv 0(mod 2)$ or $ p(k)\\equiv p(1)(mod 2)$\n[/hide]" } { "Tag": [ "analytic geometry" ], "Problem": "The path of a ball is given by y=(-1/20)x^2 + 3x + 5 , where y is the height in feet and x is the horizontal distance in feet.\r\n\r\n(a) Find the maximum height of the ball\r\n\r\n(b) Which term determines the height at which the ball was thrown? Does changing this term change the coordinates of the maximym height of the ball? Explain. \r\n\r\nhave fun", "Solution_1": "Complete the square for (a). You know squares cannot be less than 0, so replacing the square with 0 gives the maximum height.\r\n\r\nThe height at which the ball is thrown is the height at x=0.", "Solution_2": "haha\r\nyeah\r\ni know that i have to complete the square\r\nbut the prob is that wen i do\r\n\r\ni get a really weird fraction\r\n\r\ni dont know what im getting wrong\r\n\r\nand the initial height should be 5\r\nnot 0\r\nbecause of the value of c", "Solution_3": "Zuton Force said it occurs [i]at[/i] x=0, so the initial height is 5. :wink:", "Solution_4": "For future reference, this problem would be more appropriate in High School Basics.", "Solution_5": "Ya i was wondering y it was in the pre-olympiad section when it can be easily solved.." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "$P(x)=ax^{3}+bx^{2}+cx+d$ is a polynomial with integer coefficients and $a$ is non-zero .We have $xP(x)=yP(y)$ for many pairs $(x,y)$ of unequal integers. Show that $P(x)$ has an integer root.", "Solution_1": "This problem is A3 in IMO2002 Shortlist. \r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118702#p118702]here[/url] is the respective thread.\r\nWhile [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2002]here[/url] you can see all problems of IMO2002 Shortlist.", "Solution_2": "yeah, thanks" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $a, b, c$ be positive real numbers such that\r\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2$.\r\nProve that\r\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1$", "Solution_1": "[quote=\"Beat\"]Let $a, b, c$ be positive real numbers such that\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2$.\nProve that\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1$[/quote]\r\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2\\Leftrightarrow ab+ac+bc+2abc=1.$\r\nLet $\\sqrt[3]{abc}=x.$ Then $1=ab+ac+bc+2abc\\geq3x^{2}+2x^{3}.$\r\nBut $2x^{3}+3x^{2}-1=(x+1)^{2}(2x-1).$ Hence, $x\\leq\\frac{1}{2}.$\r\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1\\Leftrightarrow1+2(a+b+c)\\geq32abc.$\r\n$1+2(a+b+c)\\geq1+6x.$\r\nRemain to prove that $1+6x\\geq32x^{3},$ which true since,\r\n $32x^{3}-6x-1=(2x-1)(16x^{2}+8x+1)\\leq0.$", "Solution_2": "[quote=\"Beat\"]Let $a, b, c$ be positive real numbers such that\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2$.\nProve that\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1$[/quote]\r\nNicer:\r\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1\\Leftrightarrow\\sum_{cyc}\\left(\\frac{1}{4a+1}-\\frac{1}{a+1}+\\frac{1}{3}\\right)\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{(2a-1)^{2}}{(4a+1)(a+1)}\\geq0.$ :)", "Solution_3": "I found even more nicer:\r\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2$, equivalently\r\n$ab+bc+ca+2abc=1$.\r\nBy the Cauchy-Schwarz inequality it follows that:\r\n$((a+b+c)+\\frac{1}{2})(bc+ca+ab+2abc)\\ge (4\\sqrt{abc})^{2}=16abc$\r\nOr equivalently: $2(a+b+c)+1\\ge 32abc$, $\\frac{1}{1+4a}+\\frac{1}{1+4b}+\\frac{1}{1+4c}\\ge1$", "Solution_4": "[quote=\"Beat\"]\nBy the Cauchy-Schwarz inequality it follows that:\n$((a+b+c)+\\frac{1}{2})(bc+ca+ab+2abc)\\ge (4\\sqrt{abc})^{2}=16abc$\n[/quote]\r\nNice! :lol:", "Solution_5": "See also here .\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=%2A32abc%2A&t=79554&sid=7a3f16552285148fbb15b3b510b1c3d9\r\n\r\nCampos posted a very nice solution using Cauchy . :)", "Solution_6": "[quote=Beat]Let $a, b, c$ be positive real numbers such that\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2$.\nProve that\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1$[/quote]\n[b]Easy.[/b]\n\n $\\sum_{cyc}\\frac{1}{4a+1}=\\sum_{cyc}(\\frac{1}{4a+1}+\\frac{1}{3})-1\\geq\\sum_{cyc}\\frac{4}{(4a+1)+3}-1=1.$", "Solution_7": "[quote=sqing][quote=Beat]Let $a, b, c$ be positive real numbers such that\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2$.\nProve that\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1$[/quote]\n[b]Easy.[/b]\n\n$\\sum_{cyc}\\frac{1}{4a+1}=\\sum_{cyc}(\\frac{1}{4a+1}+\\frac{1}{3})-1\\geq\\frac{4}{(4a+1)+3}-1=1.$[/quote]\n\nIs it?\n$\\sum_{cyc}\\frac{1}{4a+1}=\\sum_{cyc}(\\frac{1}{4a+1}+\\frac{1}{3})-1\\geq\\sum_{cyc}\\frac{4}{(4a+1)+3}-1=1.$", "Solution_8": "Thanks.\n$\\sum_{cyc}\\frac{1}{4a+1}=\\sum_{cyc}(\\frac{1}{4a+1}+\\frac{1}{3})-1\\geq\\sum_{cyc}\\frac{4}{(4a+1)+3}-1=1.$", "Solution_9": "[quote=Beat]Let $a, b, c$ be positive real numbers such that\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2$.\nProve that\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1$[/quote]\n\n$\\frac{1}{{4a + 1}} \\ge \\frac{1}{{a + 1}} - \\frac{1}{3} \\Leftrightarrow \\left( {1 - 2a} \\right)^2 \\ge 0$\n$ \\Rightarrow \\sum {\\frac{1}{{4a + 1}}} \\ge \\sum {\\frac{1}{{a + 1}}} - 1 = 1$", "Solution_10": "[quote=scpajmb][quote=Beat]Let $a, b, c$ be positive real numbers such that\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=2$.\nProve that\n$\\frac{1}{4a+1}+\\frac{1}{4b+1}+\\frac{1}{4c+1}\\ge 1$[/quote]\n\n$\\frac{1}{{4a + 1}} \\ge \\frac{1}{{a + 2}} - \\frac{1}{3} \\Leftrightarrow \\left( {1 - 2a} \\right)^2 \\ge 0$\n$ \\Rightarrow \\sum {\\frac{1}{{4a + 1}}} \\ge \\sum {\\frac{1}{{a + 2}}} - 1 = 1$[/quote]\n$\\sum {\\frac{1}{{4a + 1}}} \\ge \\sum \\left({\\frac{1}{{a + 1}}- \\frac{1}{3}}\\right)= 1$" } { "Tag": [ "floor function", "fractional part", "algebra", "IMO Shortlist" ], "Problem": "Let $ a, b, c$ be positive integers satisfying the conditions $ b > 2a$ and $ c > 2b.$ Show that there exists a real number $ \\lambda$ with the property that all the three numbers $ \\lambda a, \\lambda b, \\lambda c$ have their fractional parts lying in the interval $ \\left(\\frac {1}{3}, \\frac {2}{3} \\right].$", "Solution_1": "Why are some other files written that the interval is $ \\left(\\frac{1}{3},\\frac{2}{3}\\right]$ not $ \\left(\\frac{1}{2},\\frac{2}{3}\\right]$?\r\nAnd what is the correct statement?", "Solution_2": "I checked the IMO Compendium, and it says that (1/3, 2/3] is the correct interval.", "Solution_3": "If $\\lambda\\in\\left(\\frac{3k+1}{3a}, \\frac{3k+2}{3a}\\right]$, then $\\frac{1}{3}<\\{\\lambda a\\}\\leq \\frac{2}{3}$. Define the intervals similarly for $b$ and $c$. Call the interval for $a$ - $L_k$, the interval for $b$ - $M_k$ and the interval for $c$ - $N_k$. Now the length of each $M_k$ is $\\frac{1}{3b}$ which are spaced $\\frac{2}{3b}$ apart from each other. The interval for each $N_k$ is $\\frac{1}{3c}$ and they are spaced $\\frac{2}{3c}<\\frac{1}{3b}$ apart. Thus no $M_k$ can not contain an interval $N_k$ because the intervals not containing $N_k$ are all of length less than $\\frac{1}{3b}$. Thus each interval $M_k$ must intersect some interval $N_k$.\n\nNow what remains to prove find is some interval $M_k$ that is contained entirely within some interval $L_k$. If that is true, then we want some $x$ and $y$ such that\n\\[\\frac{2x+1}{3a}\\leq \\frac{3y+1}{3b}<\\frac{3y+2}{3b}\\leq \\frac{3x+2}{3a}.\\]\nThat is equivalent to finding some integer $x$ and $y$ such that\n\\[\\frac{3x+1}{3y+1}\\leq \\frac{a}{b}\\leq \\frac{3x+2}{3y+2}.\\]\nNow if $\\frac{a}{b}\\leq \\frac14$, then let $y=2^{e}$ where $e$ is the maximal $e$ such that $\\frac{1}{3\\cdot 2^e+1}\\leq \\frac{a}{b}$. The minimal $e$ is $0$ and thus the minimum $\\frac{a}{b}$ is $\\frac{1}{4}$ which explains the bound. Now then $\\frac{1}{3\\cdot 2^{e-1}+1}\\geq \\frac{a}{b}$ by definition so $\\frac{2}{3\\cdot 2^e+2}\\geq \\frac{a}{b}$. Therefore, if we let $x=0$ and $y={2^e}$, we have\n\\[\\frac{2x+1}{3y+1}=\\frac{1}{3\\cdot 2^e+1}\\leq \\frac{a}{b}\\leq \\frac{2}{3\\cdot 2^e+2}=\\frac{3x+2}{3y+2}.\\]\nNow if $\\frac{a}{b}\\geq \\frac{1}{4}$. Note that $\\frac{a}{b}<\\frac12$ because $b>2a$. Now the sequence $\\frac{3\\cdot2^e-2}{3\\cdot2^{e+1}-2}$ approaches $\\frac12$ as $e$ approaches infinity. Now choose the maximal $e$ such that $\\frac{3\\cdot2^e-2}{3\\cdot2^{e+1}-2}\\leq \\frac{a}{b}$. Then $\\frac{3\\cdot 2^{e+1}-2}{3\\cdot 2^{e+2}-2}=\\frac{3\\cdot 2^e-1}{3\\cdot 2^{e+1}-1}\\geq \\frac{a}{b}$. Thus let $x=2^{e}-1$ and $y=2^{e+1}-1$ and thus that satisfies\n\\[\\frac{3x+1}{3y+1}\\leq \\frac{a}{b}\\leq \\frac{3x+2}{3y+2}.\\]\nTherefore it is always possible to find some integer $x,y$ such that $M_k$ is contained entirely within some $L_k$, and the $M_k$ will always intersect some $N_k$ which proves that there exists some real number $\\lambda$ that is in all three intervals so the fractional parts of $\\lambda a, \\lambda b,$ and $\\lambda c$ are all in the interval $\\left(\\frac13, \\frac23\\right]$. $\\blacksquare$", "Solution_4": "[quote=\"JSGandora\"]That is equivalent to finding some integer $x$ and $y$ such that\n\\[\\frac{3x+1}{3y+1}\\leq \\frac{a}{b}\\leq \\frac{3x+2}{3y+2}.\\][/quote]\nAlternatively, note that this is equivalent to $\\frac{1}{3} \\le \\frac{ya - xb}{b-a} \\le \\frac{2}{3}$. If $a=gu$, $b=gv$ with $(u,v)=1$, we then just need to find $k\\ge1$ with $\\frac{1}{3} \\le \\frac{kg}{b-a} = \\frac{k}{v-u} \\le \\frac{2}{3}$ (since $g = na-mb$ for some positive integers $n,m$). But $b>2a\\implies v>2u$, so $v-u \\ge u+1 \\ge 2$, and we can simply take $k = \\lfloor{(v-u)/2}\\rfloor$ to finish (the \"hardest\" case is when $v-u=3$).\n\nI'm not sure if we can use rational approximation when $a/b$ is irrational (the problem still holds in this case by the previous solution), but at least the two-dimensional version of Kronecker's theorem doesn't seem to work directly.", "Solution_5": "Consider a \"good $x$ interval\" a interval of continuous $\\lambda$s that work for $x$. It measures $1/3x$, and by inequality, any good-b interval intersects some good-c interval, so is enough to prove some good-b interval is completely inside a good-a interval. Is trivial to see that the $(\\lambda_1+1)/2$'th good-b interval and the $(\\lambda_2+1)/2$'th good-a interval satisfy this iff\n\n$ \\frac{ | \\lambda_1b - \\lambda_2a | }{2} \\le \\frac{b-a}{3}$.\n\nBy bezout lemma, is easy to find odd $\\lambda$s that satisfy this (one needs to divide in cases $a|b$ and $a$ doesnt divide $b$), so we finish.", "Solution_6": "Solved with [b]Smileyklaws[/b]\n\nConsider all intervals of the form $\\left(\\frac{3k+1}{3b}, \\frac{3k+2}{3b}\\right]$ for $k \\ge 0$. Each of these intervals has length $\\frac{1}{3b}$. Now consider all intervals of the form $\\left(\\frac{3k+1}{3c}, \\frac{3k+2}{3c}\\right]$ for $k \\ge 0$. Each of these intervals has length $\\frac{1}{3c}$, and the space between two consecutive intervals is $\\frac{2}{3c}$. But $c > 2b \\implies \\frac{2}{3c} < \\frac{1}{3b}$, so it follows that each $c$-interval overlaps with a $b$-interval. We wish to show that there exists some $\\lambda$ in an $a$-interval, a $b$-interval, and a $c$-interval.\n\nNow we claim that there exists a $c$-interval that is contained within an $a$-interval. Note that each $c$-interval has length $\\frac{1}{3c}$, while the space between two consecutive intervals is $\\frac{2}{3c}$. Then $$c>4a \\implies \\frac{4}{3c}<\\frac{1}{3a} \\implies \\frac{1}{3c}+\\frac{2}{3c}+\\frac{1}{3c} < \\frac{1}{3a}$$so with some effort it follows that there exists a $c$-interval contained within an $a$-interval.\n\nSince each $c$-interval overlaps with a $b$-interval, by our work above, there exists an $a$-interval, a $b$-interval, and a $c$-interval that overlap. Consider some real $d$ in the union. Then for some positive integer $m$, $$\\frac{3m+1}{3a} < \\lambda \\le \\frac{3m+2}{3a} \\implies m+\\frac{1}{3} < \\lambda a \\le m+\\frac{2}{3} \\implies \\frac{1}{3} < \\{\\lambda a\\} \\le \\frac{2}{3}$$Similarly, $\\frac{1}{3} < \\{\\lambda b\\} \\le \\frac{2}{3}$, $\\frac{1}{3} < \\{\\lambda c\\} \\le \\frac{2}{3}$. This is what we needed, so we are done. $\\blacksquare$", "Solution_7": "[quote=JSGandora]If $\\lambda\\in\\left(\\frac{3k+1}{3a}, \\frac{3k+2}{3a}\\right]$, then $\\frac{1}{3}<\\{\\lambda a\\}\\leq \\frac{2}{3}$. Define the intervals similarly for $b$ and $c$. Call the interval for $a$ - $L_k$, the interval for $b$ - $M_k$ and the interval for $c$ - $N_k$. Now the length of each $M_k$ is $\\frac{1}{3b}$ which are spaced $\\frac{2}{3b}$ apart from each other. The interval for each $N_k$ is $\\frac{1}{3c}$ and they are spaced $\\frac{2}{3c}<\\frac{1}{3b}$ apart. Thus no $M_k$ can not contain an interval $N_k$ because the intervals not containing $N_k$ are all of length less than $\\frac{1}{3b}$. Thus each interval $M_k$ must intersect some interval $N_k$.\n\nNow what remains to prove find is some interval $M_k$ that is contained entirely within some interval $L_k$. If that is true, then we want some $x$ and $y$ such that\n\\[\\frac{2x+1}{3a}\\leq \\frac{3y+1}{3b}<\\frac{3y+2}{3b}\\leq \\frac{3x+2}{3a}.\\]\nThat is equivalent to finding some integer $x$ and $y$ such that\n\\[\\frac{3x+1}{3y+1}\\leq \\frac{a}{b}\\leq \\frac{3x+2}{3y+2}.\\]\nNow if $\\frac{a}{b}\\leq \\frac14$, then let $y=2^{e}$ where $e$ is the maximal $e$ such that $\\frac{1}{3\\cdot 2^e+1}\\leq \\frac{a}{b}$. The minimal $e$ is $0$ and thus the minimum $\\frac{a}{b}$ is $\\frac{1}{4}$ which explains the bound. Now then $\\frac{1}{3\\cdot 2^{e-1}+1}\\geq \\frac{a}{b}$ by definition so $\\frac{2}{3\\cdot 2^e+2}\\geq \\frac{a}{b}$. Therefore, if we let $x=0$ and $y={2^e}$, we have\n\\[\\frac{2x+1}{3y+1}=\\frac{1}{3\\cdot 2^e+1}\\leq \\frac{a}{b}\\leq \\frac{2}{3\\cdot 2^e+2}=\\frac{3x+2}{3y+2}.\\]\nNow if $\\frac{a}{b}\\geq \\frac{1}{4}$. Note that $\\frac{a}{b}<\\frac12$ because $b>2a$. Now the sequence $\\frac{3\\cdot2^e-2}{3\\cdot2^{e+1}-2}$ approaches $\\frac12$ as $e$ approaches infinity. Now choose the maximal $e$ such that $\\frac{3\\cdot2^e-2}{3\\cdot2^{e+1}-2}\\leq \\frac{a}{b}$. Then $\\frac{3\\cdot 2^{e+1}-2}{3\\cdot 2^{e+2}-2}=\\frac{3\\cdot 2^e-1}{3\\cdot 2^{e+1}-1}\\geq \\frac{a}{b}$. Thus let $x=2^{e}-1$ and $y=2^{e+1}-1$ and thus that satisfies\n\\[\\frac{3x+1}{3y+1}\\leq \\frac{a}{b}\\leq \\frac{3x+2}{3y+2}.\\]\nTherefore it is always possible to find some integer $x,y$ such that $M_k$ is contained entirely within some $L_k$, and the $M_k$ will always intersect some $N_k$ which proves that there exists some real number $\\lambda$ that is in all three intervals so the fractional parts of $\\lambda a, \\lambda b,$ and $\\lambda c$ are all in the interval $\\left(\\frac13, \\frac23\\right]$. $\\blacksquare$[/quote]\n\nI think you meant $3x+1$ in LHS of the inequality instead of $2x+1$.", "Solution_8": "Note that $\\{\\lambda x\\}\\in \\left(\\tfrac {1}{3}, \\tfrac {2}{3} \\right]$ is equivalent to $\\lambda \\in \\left(\\tfrac{3k+1}{3x},\\tfrac{3k+2}{3x}\\right]$ for some integer $k$. Call these the $x$-intervals, for $x=a,b,c$. Each $x$-interval has a starting point of $\\tfrac{3k+1}{3x}$, a length of $\\tfrac{1}{3x}$ and are $\\tfrac{2}{3x}$ apart. Since $\\tfrac{2}{3c}<\\tfrac{1}{3b}$, each $b$-interval intersects some $c$-interval at least somewhere.\n\\\\$~$\\\\\nIt remains to show that some $b$-interval is contained in another $a$-interval. To do this, some $b$-interval's starting point must be in \\[\\left(\\frac{k}{a}+\\frac{1}{3a},\\frac{k}{a}+\\frac{2}{3a}-\\frac{1}{3b}\\right] = \\left(\\frac{3bk+b}{3ab},\\frac{3bk+2b-a}{3ab}\\right]\\]\nMeanwhile, the starting points of the $b$-intervals are\n\\[\\frac{1}{3b}+\\frac{l}{b}=\\frac{3la+a}{3ab}\\]\nClearly, if $b>4a$ then $2b-a-b>3a$ so there must exist some $l$ that makes $3la+a$ fall into the interval $(3bk+b,3bk+2b-a]$. If $b=4a$ then $7a$ and $2b-a$ coincide.\n\\\\$~$\\\\\nNow, we claim that no matter what $r=\\tfrac{b}{a}$ is, we will be able to find a value $k$ and $l$. We have already done $k\\ge 4$. Consider $l=2k+1$. We have $3la+a=(6k+4)a$. As long as \n\\[b(3k+1) < a(6k+4) \\le b(3k+2)-a \\implies 2+\\frac{1}{3k+2}\\le r < 2+\\frac{2}{3k+1}\\]\nThus, it is true for $r\\in[2+\\tfrac1{3k+2},2+\\tfrac2{3k+1})$. Note that the lower bound can get arbitrarily close to $2$. It remains to show that the upper bound is at least equal to the lower bound of the previous, that for all $k\\ge 1$, \\[\\frac2{3k+1}\\ge \\frac1{3k-1}\\]\nWhich is true. Note that when $k=0$ the upper bound is $4$. Therefore, if $r\\in (2,4)$ we can find a value of $k,l$ that fit. We are done. ", "Solution_9": "onk\n\nb-interval intersects c-interval\n\nprove b-interval is contained by a-interval\n\nnow do spam stuff\n\n(oops i actually goofed and got to @above for $k=0$ and didn't generalize L me)\n\nwait just do what gandora did trust\n\nokay yeah cool my brain just farted but this works. hopefully i don't forget this but the moral of the story is to just do it smh my head" } { "Tag": [ "geometry", "geometric transformation", "reflection", "modular arithmetic", "national olympiad" ], "Problem": "16th Hungary-Israel olympiad\r\nIsrael, 28.3.2006\r\n\r\nday 1\r\n\r\n1. let x,y,p,n,k be naturals satisfying\r\nx^n + y^n = p^k\r\nif n>1 is odd and p is odd prime prove that n is power of p.\r\n\r\n2. Box with dimensions aXbXc is paved by 3D domino blocks 1X1X2\r\nthat have sides parallel to the box sides. any side of the box have\r\nequal number of blocks paralleling. what is the possible dimensions of the box?\r\n(domino block is paralleling to side of the box if it's side of length 2 is paralleling to\r\nthe box side).\r\n\r\n3. Let H be convex polygon with n sides. let A(1), A(2), ..., A(n) be his vertexes and\r\nA'(1), A'(2), ..., A'(n) be the reflection by the middle section of the neighboring diagonal\r\nA(i-1)A(i+1) (here A(n+1)=A(1)). The vertex A(i) is good if A'(i) is inside of H.\r\nprove that H have at least n-3 good vertexes.\r\n\r\nday 2\r\n\r\n4. P is point inside a given circle. there are three cords passing through P\r\nwith equal length. prove that P is the center of the circle.\r\n\r\n5. For any non-negative x, y, z let:\r\nS= x^2y + y^2z + z^2x + xy^2 + yz^2 + zx^2\r\nC= x^2y + y^2z + z^2x\r\nfind the maximal values of S and C when x + y + z = 1.\r\n\r\n6. Group of 100 contestants numbered 1, 2, 3, ..., 100 playing the following game:\r\nthe judge writing the numbers 1, 2, 3, ..., 100 on 100 cards, turn them over\r\nand place them on a table in arbitrary order.\r\nfor any j, contestant number j entering the room and flip 50 cards.\r\nhe gain one point if the number j is on one of the flipped cards.\r\nis there a strategy giving more than 1% to win?\r\n(contestants can't talk to each other when the game begin).", "Solution_1": "1. assume that $ p^m|x$, and $ p^m$ is the highest power doing so. write $ x \\equal{} p^mt$, $ t\\in \\mathbf{N}$. clearly $ mn < k$, and $ y^n \\equal{} p^k \\minus{} p^{mn}t \\equal{} p^{mn}(p^{k \\minus{} mn} \\minus{} t)$, so the highest power of $ p$ dividing $ y$ is $ p^m$. dividing out the highest power of $ p$ we reduce the problem to the case where $ p$ doesn't divide $ x$ or $ y$. now since $ n$ is odd, we may write the equation as\r\n\r\n$ p^k \\equal{} (x \\plus{} y)(x^{n \\minus{} 1} \\minus{} x^{n \\minus{} 2}y \\plus{} x^{n \\minus{} 3}y^2 \\minus{} \\cdots \\plus{} y^{n \\minus{} 1})$ (*)\r\n\r\nclearly $ (x \\plus{} y) > 1$, so $ p|x \\plus{} y$, i.e. $ y\\equiv \\minus{} x \\pmod{p}$.\r\n\r\ndenote by $ E$ the second factor on the RHS of (*). we then have \r\n\r\n$ E \\equiv x^{n \\minus{} 1} \\plus{} x^{n \\minus{} 1} \\plus{} \\cdots \\plus{} x^{n \\minus{} 1} \\equal{} nx^{n \\minus{} 1} \\pmod{p}$ (**)\r\n\r\nnow if both $ x$ and $ y$ were 1, we would have $ p^k \\equal{} 2$, but $ p$ is odd. so one of $ x,y$ is $ > 1$. thus, since $ n > 1$, $ x \\plus{} y < x^n \\plus{} y^n \\equal{} p^k$, so $ E > 1$. so $ p|E$, and in light of (**) it follows that $ p|n$.\r\n\r\nnow we have $ (x^{p})^{n/p} \\plus{} (y^{p})^{n/p} \\equal{} p^k$. if $ n/p \\equal{} 1$, we're done. otherwise, $ n/p > 1$, so we may apply the above argument. continuing like this, we must eventually reach $ n/p^i \\equal{} 1$ for some $ i$, and so we end.\r\n\r\n4. if $ P$ weren't the center, and the point $ O$ instead were the center, we would get three congruent right triangles all with the same hypotenuse, $ OP$, so that the leg ending at point $ P$ is the same length for all three of them. but there are at most two such triangles. \r\n\r\n6. i've heard this problem somewhere else. you forgot to say that the goal of the game is that at least one of the contestants turn over the correct card.", "Solution_2": "Problem 6 is a very classic problem.... (I have heard of a version with prisoners and boxes... same idea...)\r\n\r\nthe strategy is that the nth contestant first goes to the nth card on the table, and then if the card said k, he would go to the kth card, and so on until they either find their card, or exhaust all 50 tries. This would increase the chances of them winning by a great deal, as the card numbers will cycle (i.e. they will lead to each other through a set repetition of values). Thus, each competitor will eventually find the winning card through the maximum cycle length (in the worst case scenario).\r\n\r\nSo, for every single competitor to win, all cycles must have a length that is less than 50... Thus, since there are 100! permutations, and of these, there are 100!/N that are bad for every N>50... we can see that the number of good permutations is: 100! - 100! / 51 - 100! / 52 - ... - 100!/100\r\n\r\nThis yields an approximate winning rate of: 31.1%... not bad! :P :P", "Solution_3": "In 6 i forgot to mention that the aim is to get 100 points.\r\nNawahd Werdna can you expalin how you get the method to win?\r\nCan someone give a link to similar problems?", "Solution_4": "Well... to find this method, you just need to play around with the problem for a while... and you'll get it...\r\n\r\nTo give you an example why this solution works, lets take an example with 10 cards...\r\nLets say the cards are in the order\r\n10 2 4 5 7 1 9 6 3 8\r\n\r\nNow, contestanat 1 goes to the table, and he goes to card position 1, which is 10... then, he goes to card position 10, with value 8... he then goes to card position 8, which is 6, and to the 6th position, which has value 1.... and so on \r\n\r\nWe can write this as a cycle, namely: \r\n\r\n10, 8, 6, 1, 10 , ...\r\nand so, since this cycle has length that less than 5, we can see that every one of the numbers 10, 8, 6, and 1 will be guaranteed to find their card using this method... :D" } { "Tag": [ "calculus", "Princeton", "college", "Stanford", "articles", "topology" ], "Problem": "This might not be appropriate for this category but those of you taking single variable Calculus and algebra based physics do any of you use the Osteebe, Zorn Calculus book and Giancoli for physics. If yes, do you think the books are good?", "Solution_1": "I use a book written by Paul A. Foerster for Calculus, and it's pretty good.\r\n\r\nAs far as physics goes... I used a Princeton Review AP Physics book a bunch, and I've read about half of Richard Feynman's Lectures on Physics Volume 1. That's a tough read; though. You need to pretty much know the stuff before hand.", "Solution_2": "Feynman's books are fantastic, but confuted is exactly right. According to legend, when Feynman actually gave the course (a freshman physics course), he quickly ran off all the students - but all the later year students and grad students (and some profs) started coming to solidify their knowledge. \r\n\r\nThe Feynman lectures are excellent for exactly that purpose. If you find yourself really into physics, get them.", "Solution_3": "[quote=\"hello\"]This might not be appropriate for this category but those of you taking single variable Calculus and algebra based physics do any of you use the Osteebe, Zorn Calculus book and Giancoli for physics. If yes, do you think the books are good?[/quote]\r\n\r\nHmmm, I think below I'll put my VERY rough draft of a FAQ on studying calculus. To answer your basic question, I have the preliminary edition of the Ostebee, Zorn textbook and it looks all right as a high school textbook for an AP calculus class. It has a very strong \"reform\" flavor. I have several other calculus textbooks at home. By far my favorite, among all the calculus books I hope to slog through in the next few years (I've never had a calculus class) is Michael Spivak's textbook Calculus. \r\n\r\nWe haven't had an algebra-based physics textbook around the house yet. I imagine next year we will buy whatever EPGY specifies for its computer-based beginning physics course. I recommend LOTS of reading in adult popular books about physics to supplement what you find in your textbook, as I do for studying any science course. \r\n\r\nHope this helps! \r\n\r\nVERY rough draft of calculus study FAQ follows, and I'd love to hear comments from participants here with suggestions on how to improve the FAQ. [FAQ begins here:] The best calculus textbook, at least in English, is Michael Spivak's book Calculus (Publish or Perish Press, third edition). This textbook is not specifically written to the AP Calculus syllabus and thus is unknown to many AP calculus teachers, but it is unanimously recommended by mathematicians. \r\n\r\nIn answer to a question in the Usenet sci.math newsgroup, I wrote \r\n\r\n> What comes after calculus? \r\n\r\nWhat comes after a first course in calculus is a second try at learning calculus to really UNDERSTAND calculus. Hardly any universities use Michael Spivak's book Calculus (third edition, Publish or Perish Press) as a textbook for a first calculus course, but it is positively SCARY how unanimous mathematicians are in recommending that book for people who really want to understand calculus. Spivak's book also has good humor, as most good mathematical books do. \r\n\r\nA little while earlier, someone else on the same newsgroup asked a series of detailed questions about studying calculus. \r\n\r\n> Anyone got any idea as to good book on calculus? \r\n\r\nThe calculus book that I have seen most consistently get glowing reviews from professional mathematicians, with many calling it the \"best\" calculus book, is Calculus by Michael Spivak (3rd edition, Publish or Perish Press). I own Spivak's book, and know several people locally who think it is wonderful. However, Spivak's Calculus is not an EASY introduction to calculus--it's very useful to have for clear explanations and working through it would probably give you a deep understanding of calculus that would be difficult to forget, but it wouldn't be the fastest way to learn, for example, the calculus tested on the AP calculus exams. \r\n\r\nI second-source (actually, quintuple-source or more) calculus books, as I am also trying to learn calculus by self-study. In the end, the only way to learn calculus is to do A LOT of problems that build progressively on earlier understanding--on that, all teachers of calculus agree, whatever else they disagree on. And calculus is, indeed, one of the math courses for which there is the most varied assortment of textbooks, sometimes taking wildly differing approaches. \r\n\r\n> Can I learn calculus in the internet? \r\n\r\nThere is a distance learning course on calculus for which you can get college credit (if you PAY for the college credit and keep up with your required homework) offered by the University of Illinois Urbana-Champaign \r\n\r\nhttp://www-cm.math.uiuc.edu/ \r\n\r\ncalled Calculus and Mathematica which some math professors like a lot, and others decry. I MAY just buy the book for that course through Amazon.com and work through it myself with a student copy of Mathematica, forgoing the college credit, but I'm still trying to decide. \r\n\r\nThere is another distance learning course about calculus offered by the Education Program for Gifted Youth (EPGY) at Stanford University \r\n\r\nhttp://epgy.standford.edu/ \r\n\r\nbut that course, as its name implies, is for gifted young people who haven't yet graduated from high school. That course uses a combination of the Anton textbook and innovative course software developed by Stanford. My oldest son is taking EPGY courses that are prerequisites for the calculus course right now, and may be taking calculus within two years. \r\n\r\n> Is there a web site to learn calculus? \r\n\r\nThere are a lot of great Web sites with articles ABOUT calculus, including what amounts to a complete, FREE textbook by a famous mathematician in at least one case. Those include \r\n\r\nhttp://www.math.gatech.edu/~morley/1507/MeaninglessSymbolicManipulation.html \r\n\r\n(approaches to studying Calculus, including a recommendation of some of the books I just mentioned) \r\n\r\nhttp://www.math.washington.edu/~duchamp/124notes/k124-main.pdf \r\n\r\nhttp://www.math.washington.edu/~duchamp/125notes/k125-main.pdf \r\n\r\n(Two lengthy Adobe Acrobat .PDF files with a college calculus course, free for the downloading, by Professor Neal Koblitz of the University of Washington) \r\n\r\nhttp://www.math.hawaii.edu/~lee/calculus/ \r\n\r\n(Web page with links to many articles by Professor E. Lee Lady of the University of Hawaii) \r\n\r\nThere is a lot more where these links came from; you could find good stuff about calculus on the Web for hours. Try Google \r\n\r\nhttp://www.google.com/ \r\n\r\nto run more searches of your own. \r\n\r\nThat's as far as I wrote for the Usenet inquirer. Most parents of homeschooled kids are looking for something like a high school calculus book. The first place to look is in your friendly local public library, where you may find several different calculus books aimed at high school students. Stewart's book is used here in Minnesota by the University of Minnesota Talented Youth Mathematics Program (UMTYMP) and the Anton book is used by EPGY; I can get both in my local library systems. How to Learn Calculus: A Streetwise Guide looks pretty good to me, although I would have your local library request it rather than buying it yourself. Calculus Made Easy is NOT a complete calculus course, but it is good for motivation and keeping the learner from being psyched out. Spivak's Hitchhiker's Guide to Calculus is a good introductory text; there are lots of high school cram books on calculus in good public libraries. There are many good calculus preparation sites for university students, for example, \r\n\r\nhttp://www.math.mun.ca/~apics/calculus/ \r\n\r\nThe person I replied to on Usenet asked another useful question: \r\n\r\n> What about other books that would serve as backgrounders\r\n> along the same vein as the above point? \r\n\r\nThere are LOTS of books for \"transition courses\" that bridge the gap between undergraduate calculus and math courses beyond calculus for math and science majors. Those books often have titles like \"Introduction to Proof,\" or \"Problem Solving\" or \"Introduction to Advanced Mathematics\" or \"Mathematical Thinking\" or phrases more or less like that. It is a good idea to start reading a book like that, and working through its exercises, BEFORE finishing a calculus course. Probably your local university has several book like this in the campus bookstore, and the math department could recommend whatever title is used where you go to school. \r\n\r\n[End of calculus FAQ] \r\n\r\nWhat do the rest of you have to say about studying calculus?", "Solution_4": "Is Spivak's [i]Calculus on Manifolds[/i] the \"Volume II\" of his [i]Calculus[/i]?", "Solution_5": "Please don't revive old threads. Also, I'm sending this off to another forum, where it belongs.", "Solution_6": "[quote=\"confuted\"]I use a book written by Paul A. Foerster for Calculus, and it's pretty good.\n[/quote]\r\nWe use the same book. It has good problems, and clearly outlines topics. However, the author's sense of humor (i.e. the name puns) I could do without." } { "Tag": [ "modular arithmetic" ], "Problem": "1. Denote by $p_{k}$ the $k$th prime number. Show that $p_{1}p_{2}\\cdots p_{n}+1$ cannot be the perfect square of an integer.\r\n\r\n2. Prove that it is impossible for three consecutive squares to sum to another perfect square.", "Solution_1": "[hide=\"Hint for 1\"] $\\bmod 4$. [/hide] [hide=\"Hint for 2\"] $\\bmod 3$. [/hide]", "Solution_2": "[hide=\"Solution to #2\"]\nLet the squares be $(x-1)^{2}$, $x^{2}$, and $(x+1)^{2}$. Their sum is $3x^{2}+2\\equiv 2\\pmod 3$.\n\nSince $0^{2}\\equiv 0\\pmod 3$, $1^{2}\\equiv 1\\pmod 3$, and $2^{2}\\equiv 1\\pmod 3$, it follows that a perfect square cannot be congruent to 2 mod 3. [/hide]", "Solution_3": "[hide=\"1\"]Since all primes except 2 is in the form $4n\\pm1,$\n$p_{1}p_{2}\\cdots p_{n}+1=2\\cdot\\pm1+1=2+1=3\\mod 4$\nbut a square cannot be $3\\mod4$[/hide]", "Solution_4": "[hide=\"Solution to #1\"]\nAssume $p_{1}p_{2}\\cdots p_{n}+1=m^{2}\\implies p_{1}p_{2}\\cdots p_{n}=(m+1)(m-1)$.\n\nSince $p_{1}=2$, the LHS is even, so $m$ must be odd. Then $m+1$ and $m-1$ are both even and the RHS is divisible by 4. Contradiction.[/hide]", "Solution_5": "[quote=\"junggi\"][hide=\"1\"]Since all primes except 2 is in the form $4n\\pm1,$\n$p_{1}p_{2}\\cdots p_{n}+1=2\\cdot\\pm1+1=2+1=3\\mod 4$\nbut a square cannot be $3\\mod4$[/hide][/quote]\r\n\r\nHow do you know that it is 2+1 = 3 instead of 2-1 = 1?\r\n\r\n(because $p_{1}p_{2}\\cdots p_{n}+1=2\\cdot\\pm1+1=2+1=3\\mod 4$ )", "Solution_6": "[quote=\"computer_nuke\"][quote=\"junggi\"][hide=\"1\"]Since all primes except 2 is in the form $4n\\pm1,$\n$p_{1}p_{2}\\cdots p_{n}+1=2\\cdot\\pm1+1=2+1=3\\mod 4$\nbut a square cannot be $3\\mod4$[/hide][/quote]\n\nHow do you know that it is 2+1 = 3 instead of 2-1 = 1?\n\n(because $p_{1}p_{2}\\cdots p_{n}+1=2\\cdot\\pm1+1=2+1=3\\mod 4$ )[/quote]\r\nSorry, I texed it weird...I meant it to be like this:\r\n$p_{1}p_{2}\\cdots p_{n}+1=2\\cdot1+1=2+1=3\\mod 4$\r\nor\r\n$p_{1}p_{2}\\cdots p_{n}+1=2\\cdot(-1)+1=-2+1=-1=3\\mod 4$", "Solution_7": "[quote=\"junggi\"][quote=\"computer_nuke\"][quote=\"junggi\"][hide=\"1\"]Since all primes except 2 is in the form $4n\\pm1,$\n$p_{1}p_{2}\\cdots p_{n}+1=2\\cdot\\pm1+1=2+1=3\\mod 4$\nbut a square cannot be $3\\mod4$[/hide][/quote]\n\nHow do you know that it is 2+1 = 3 instead of 2-1 = 1?\n\n(because $p_{1}p_{2}\\cdots p_{n}+1=2\\cdot\\pm1+1=2+1=3\\mod 4$ )[/quote]\nSorry, I texed it weird...I meant it to be like this:\n$p_{1}p_{2}\\cdots p_{n}+1=2\\cdot1+1=2+1=3\\mod 4$\nor\n$p_{1}p_{2}\\cdots p_{n}+1=2\\cdot(-1)+1=-2+1=-1=3\\mod 4$[/quote]\r\n\r\nYeah that makes more sense, thanks.", "Solution_8": "[hide=\"1\"]Since 2 is a prime, $p_{1}p_{2}\\cdots p_{n}$ is always 2mod4. You add 1 to get 3mod4, but a square can only be 0 or 1 mod 4.[/hide]\n\n[hide=\"2\"]$n^{2}+n^{2}+2n+1+n^{2}+4n+4=3n^{2}+6n+5$\n\nSince that will be congruent to 2 mod 3, and squares can only be 0 or 1 mod 3, that equation can't be another square.[/hide]" } { "Tag": [ "calculus", "integration", "algebra unsolved", "algebra" ], "Problem": "find all integeral solution pf the equation \r\n $[2x -y -2 )^2 =7 * ( x-2y-y^2-1)$", "Solution_1": "hello, solving for $x$ we get\n$x=1/2\\,y+{\\frac {15}{8}}+1/8\\,\\sqrt {-112\\,{y}^{2}-168\\,y+49}$\nor\n$x=1/2\\,y+{\\frac {15}{8}}-1/8\\,\\sqrt {-112\\,{y}^{2}-168\\,y+49}$\nand now it must be\n$-112y^2-168y+49\\geq 0$\nand we get\n$-\\frac{7}{4}\\le y\\le \\frac{1}{4}$\nSonnhard." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Evaluate\r\n\\[ \\sum_{n=-\\infty}^\\infty{(x-n)}^{-k} \\] where k>1 is an integer", "Solution_1": "Does anyone know how to do this? It's related to the Huruwitz Zeta function. Can it even be done by reasonably elementary methods?" } { "Tag": [ "LaTeX" ], "Problem": "Hi all, \r\n\r\nJust wondering, is it possible to have multiple headers per page? Like for example if I had two different half-page tests with the same header, and wanted to print them on the same page to save paper. Then one header would be at the top of the page, and one would be halfway down the page. \r\n\r\nI'm using the fancyhdr package", "Solution_1": "Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=twoinone]twoinone[/url] or [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=twoupltx]twoupltx[/url] packages to see how well they work.", "Solution_2": "You know, you could just manually make a header by using the center function.", "Solution_3": "...which wouldn't show up on the ToC.", "Solution_4": "hmmmm... alright I'll check those things out\r\n\r\nthanks for the replies everyone!" } { "Tag": [ "ratio", "trigonometry", "algebra", "polynomial", "symmetry", "absolute value", "Hi" ], "Problem": "Suppose $\\alpha$ and $\\beta$ are in $(0,\\frac{\\pi}{2})$ and are rational multiples of $\\pi$. Is it possible that $\\frac{\\cos{\\alpha}}{\\cos{\\beta}}$ is a rational number? (The $(0,\\frac{\\pi}{2})$ restriction is only to avoid the obvious examples\u2014where the ratio is $1$ or $1/2$ or $2$ in magnitude.)\r\n\r\nWhat I already know: If $\\cos{\\theta}$ is rational and $\\theta$ is a rational multiple of $\\pi$, then $\\left vert|\\cos{\\theta}\\right vert|$ is either 0,1, or $\\frac{1}{2}$. This is because $2\\cos{n\\theta}$ is a monic polynomial $P_n$ in $2\\cos{\\theta}$ with integer coeffecients, and so if $\\theta$ is $\\frac{p}{q}\\pi$, then $2\\cos{\\theta}$ is a root of $P_{2q} = 2$, so it must be an integer. Of course, if those polynomials were irreducible, it would be easy to prove that the above ratio cannot be rational, but I don't know if they are irreducible or not.", "Solution_1": "For each $n$, there is a monic polynomial $p_n$ of degree $n$ with integer coefficients such that $p_n(2\\cos\\theta)=2\\cos n\\theta$. Using this, we can show that if $a$ is twice the cosine of a rational multiple of $\\pi$, it is an algebraic integer between $-2$ and $2$ such that all of its conjugates are real numbers between $-2$ and $2$.\r\n\r\nIf $a$ and $b=r\\frac mn a$ are both algebraic integers, then $\\frac1n a$ is an algebraic integer, and its conjugates are $\\frac1n$ times the conjugates of $a$. Therefore, if $\\frac a2$ and $\\frac b2$ are both cosines of rational angles, $\\frac1n a$ is an algebraic integer between $-\\frac2n$ and $\\frac2n$, with all of its conjugates in that same interval. This is only possible if $n=1$ or $a=\\pm2$. Similarly, either $m=1$ or $b=\\pm2$, and the examples you have already noted are the only ones.", "Solution_2": "[quote=\"jmerry\"]For each $n$, there is a monic polynomial $p_n$ of degree $n$ with integer coefficients such that $p_n(2\\cos\\theta)=2\\cos n\\theta$. [/quote] Agreed.\n\n[quote]Using this, we can show that if $a$ is the cosine of a rational multiple of $\\pi$, it is an algebraic integer between $-2$ and $2$ such that all of its conjugates are real numbers between $-2$ and $2$.[/quote] Did you mean $a$ is [i]twice[/i] the cosine of ...? It is an algebraic integer because it satisfies a monic polynomial, but why are its conjugates in that interval, and why are they even real? Am I missing something obvious?\n\n[quote]If $a$ and $b=r\\frac mn a$ are both algebraic integers, then $\\frac1n a$ is an algebraic integer, and its conjugates are $\\frac1n$ times the conjugates of $a$. Therefore, if $a$ and $b$ are both cosines of rational angles, $\\frac1n a$ is an algebraic integer between $-\\frac2n$ and $\\frac2n$, with all of its conjugates in that same interval. [/quote] By $r\\frac mn$, do you mean \"$r+\\frac mn$, where $\\gcd(m,n)=1$ and $m\\leq n$?\" In any case, I understand the statement about $\\frac 1n a$ and its conjugates.\n[quote]This is only possible if $n=1$ or $a=\\pm2$. Similarly, either $m=1$ or $b=\\pm2$, and the examples you have already noted are the only ones.[/quote] I don't see why, sorry.", "Solution_3": "The conjugates of $a$ lie in the same intervall since they are the other zeros of the desired polynomial: $cos(nx)$ is periodically $\\mod \\frac{2\\pi}{n}$, so we know all zeros, namely $2 \\cos(\\theta), 2\\cos(\\theta+\\frac{2\\pi}{n}), 2\\cos(\\theta+\\frac{4\\pi}{n}), ..., 2\\cos(\\theta+\\frac{2\\pi (n-1)}{n})$.\r\n\r\nFor the last part: the norm of $\\frac{a}{n}$ is the product over it's conjugates and it's an integer. But $n>2$ would give that this norm has absolute value $<1$, so $a=0$.", "Solution_4": "Great, I think I understood that. Let me just write down the complete proof, and kindly tell me if something is wrong:\r\n\r\nIf $a$ is two times the cosine of a rational multiple of $\\pi$, then $a$ in an algebraic integer, all of whose conjugates are real numbers in $(-2,2)$ (assume that $a$ is neither $-2$ nor $2$ nor $0$, because those cases are easy).\r\n If $b$ is also twice the cosine of a rational multiple of $\\pi$ and $b=\\frac mn a$ with $\\gcd(m,n)=1$, then as $\\frac 1n a$ is a $\\mathbb Z$-linear combination of $a$ and $b$, it is also an algebraic integer, and has all conjugates in $(\\frac{-2}{n},\\frac{2}{n})$ (and nonzero). But the product of all these conjugates is an integer (the constant term of the polynomial for $\\frac 1n a$, upto sign), so if $n\\geq2$, then each of them is of magnitude less than $1$, so their product cannot be an integer. Hence, $n = 1$, i.e, $b$ is an integer multiple of $a$. But then, by symmetry, $a$ must also be an integer multiple of $b$, so $a$ and $b$ are equal (upto sign).", "Solution_5": "[quote=\"shreevatsa\"] Did you mean a is twice the cosine of ...?[/quote]\r\nFixed. There was another related mistake farther down." } { "Tag": [ "MATHCOUNTS", "analytic geometry", "geometry", "area of a triangle", "Heron\\u0027s formula", "absolute value" ], "Problem": "There are multiple ways to do this problem and I would like to see a number of different solutions on this one!\r\n\r\nWhat is the area of the triangle with coordinates (-1, -1), (3, 0), and (1, -9)?", "Solution_1": "it is a nice problem, and i urge you to generalize to any coordinates, and maybe even to an n-sided polygon. (you can just present a method/algorithm, not a full-fledged formula).", "Solution_2": "[hide]I think the determinant way is the easiest for this one. Finding the sides for heron's is a bother and could get messy, while for Pick's you have to draw out the diagram and count all the boundary, interior points, a method easy to mess up on when you are making a rough drawing with your hand.[/hide]", "Solution_3": "1. High school students should really sit on the sidelines and let middler school students try to attack these MATHCOUNTS problems [i]before[/i] chiming in -- even with spoilers. There is plenty of time to comment later and it's best for MATHCOUNTS students to be forced to wade blindly through the possibilities themselves first.\r\n\r\n2. Fortunately nobody has given an answer yet, so we can see how the MATHCOUNTS students can apply these methods.\r\n\r\n3. There are still yet unmentioned methods for solving this problem. :)", "Solution_4": "Hmm...as an middle schooler, the only method I understood was the Heron's formula, and even that I doubt all middle schoolers learn.\n\n\n\nHeron's formula: [hide] :sqrt: s(s-a)(s-b)(s-c) whereas s=(a+b+c)/2[/hide]\n\n\n\nYou could also draw an auxiliary line such that the line goes through both the vertex (any one of them) and form a right angle with the side opposite the vertex you chose, but that'll be even messier...\n\nYou could even go and move the triangle so the triangle's only in the first quadrant and put one of the vertex in the origin, then apply some formula to it, but sadly I do not know of such formula...", "Solution_5": "This is easy with Heron's, but I'm pretty sure your supposed to do it otherwise...\n\n\n\nHint: [hide] Note that the sides are 2 :rt: 17, :rt:17, and :rt5::rt:17[/hide]", "Solution_6": "The area can be computed with determinants using the coordinates with a generalized method for finding the area. \r\n\r\nSolution: (-1,-1)\r\n (3,0)\r\n (1,-9)\r\n (-1,-1) (list the first coordinate again)\r\nmultiply the 1st coordinate on the top with its alternate pair one row down, seperate the sides you multiply by and add them independently. Subtract your sum on the right by the sum on your left, find the absolute value of that and divide by two.\r\n\r\nNow assuming that the coordinates can be replaced with variables such as (X1,Y1)\r\n(X2,Y2)\r\n(X3,Y3)\r\n(X1,Y1) \r\n\r\nour formula would be: (x1y2+x2y3+x3y1)-(x2y1+x3y2+x1y3)/2, and finally finish by finding its absolute value as the answer. \r\nCorrect me if I am wrong but my answer was [b]12[/b][/i]", "Solution_7": "[hide=\"Solution using Shoestring\"]\nPlacing these points in clockwise order:\n-1,-1\n3,0\n1,-9\n-1,-1\nNow multiplying diagonally, \nWe get a sum of 6 and - 28 in each.\nThe absolute value of each is 22.\nNow we divide by 2.\nSo we get that the area is 11.\n[/hide]" } { "Tag": [ "function", "number theory proposed", "number theory" ], "Problem": "For a natural number $ n$ we denote $ 1\\cdot11\\cdot111\\cdot\\ldots\\cdot\\underbrace{1\\ldots1}_{n}$ with $ \\left[n\\right]!$. For example $ \\left[4\\right]!\\equal{}1\\cdot11\\cdot111\\cdot1111$. Prove that for every natural $ m$ and $ n$ $ \\left[m\\right]!\\cdot\\left[n\\right]!$ divides $ \\left[m\\plus{}n\\right]!$.", "Solution_1": "Let $ x_k\\equal{}\\frac{10^k\\minus{}1}{9}$, then $ A\\equal{}\\frac{[m\\plus{}n]!}{[m]![n]!}\\equal{}\\frac{x_{m\\plus{}1}...x_{m\\plus{}n}}{x_1...x_n}.$\r\nObviosly $ x_k|x_l$ if and only if $ k|l$. Therefore $ A\\in Z$." } { "Tag": [ "geometry", "perimeter", "inequalities", "search", "triangle inequality" ], "Problem": "How many distinct, nondegenerate triangles are there with integer side lengths such that its perimeter is 30.", "Solution_1": "[quote=\"da ban man\"]How many distinct, nondegenerate triangles are there with integer side lengths such that its perimeter is 30.[/quote]\r\n\r\n[hide]Brute-force!\nNo, but something like it should work.\nSuppose the sides are $ a,b,c$ with $ a\\le b\\le c$.\nIf one of the side lengths is $ 1$, then by the Triangle Inequality,\n$ 1+b>c$\n$ c-b<1$,\nclearly impossible since they are integers and their sum is $ 29$, which is odd.\n\nIf one of the side lengths is $ 2$, then \n$ c-b<2$, and we get that\n$ (b,c) = (14,14)$\nis the only possible solution.\n\nExperimenting more, we get\n$ (a,b,c)=(2,14,14)$\n$ (a,b,c)=(3,13,14)$\n$ (a,b,c)=(4,13,13)$\n$ (a,b,c)=(4,12,14)$\n$ (a,b,c)=(5,12,13)$\n$ (a,b,c)=(5,11,14)$\n$ (a,b,c)=(6,12,12)$\n$ (a,b,c)=(6,11,13)$\n$ (a,b,c)=(6,10,14)$\n$ (a,b,c)=(7,11,12)$\n$ (a,b,c)=(7,10,13)$\n$ (a,b,c)=(7,9,14)$\n$ (a,b,c)=(8,11,11)$\n$ (a,b,c)=(8,10,12)$\n$ (a,b,c)=(8,9,13)$\n$ (a,b,c)=(8,8,14)$\n$ (a,b,c)=(9,10,11)$\n$ (a,b,c)=(9,9,12)$\n$ (a,b,c)=(10,10,10)$\n\n$ 1+1+2+2+3+3+4+2+1 = \\boxed{19}$[/hide]\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1358695267&t=157412[/url]\r\nis almost the exact same problem, and it's posted by you..." } { "Tag": [], "Problem": "N is an odd integer such that $-596$ are a (well?) known theorems in group theory. I proved $Aut(Q_{8})~=S_{4}$ and $Aut(D_{4})~=D_{4}$ some time ago.\r\nFor the other groups I used the theorem $Inn(G)~=G/Z(G)$ and calculated the center.", "Solution_4": "The earliest known example is due to G. A. Miller, ``A non-abelian group whose of isomorphisms is abelian'' in Messenger Math. $\\pmb{ 43}$, 124--125, 1913.\r\n\r\nHis example is a group of order 64. The group has four generators $a,b,c,d$ and relations \\[a^{8}=b^{2}=c^{2}=d^{2}=[a,c]=[b,c]=[c,d]=[d,a]=1, bab=a^{5},bdb=cd.\\] The autormphism group is an elementary abelian group of order $2^{7}.$" } { "Tag": [ "vector", "trigonometry" ], "Problem": "I don't know if this is the right place to place a thread on vectors, in case not you are of course welcome to move it to the right subforum - would probably have done it anyway :) \r\n\r\nFirst of, i have to say i don't know how to make a vector sign, but until i find out how to do it i will just write $a>$ which here is a vector. How would you solve this problem?\r\n\r\n$|a>|=3$, $|a>+b>|=5$ and $a\\cdot b=2$\r\n\r\nFind $|b>|$ and the angle between $a>$ and $b>$.", "Solution_1": "$|\\vec{a}+\\vec{b}|=5$\r\n\r\nsquare both side you get $|\\vec{a}|^2+|\\vec{b}|^2+2|\\vec{a}\\cdot\\vec{b}|=25$\r\n\r\nSo $|\\vec{b}|=\\pm 2\\sqrt{3}$\r\n\r\nAnd for the angle , just use\r\n\r\n$\\cos\\theta =\\frac{\\vec{a}\\cdot\\vec{b}}{|\\vec{a}||\\vec{b}|}$", "Solution_2": "OK, i see how you get from $|\\vec a+\\vec b|=5$ to $|\\vec a|^2+|\\vec b|^2+2|\\vec a\\cdot \\vec b|=25$, but i don't see how you get from here to $|\\vec b|=\\pm2\\sqrt3$. BTW: What does $|\\vec a \\cdot \\vec b|$ actually mean? I know that $\\vec a\\cdot \\vec b=|\\vec a|\\cdot |\\vec b|\\cdot \\cos \\theta$, but i am not sure that is the same as $|\\vec a \\cdot \\vec b|$, is it?", "Solution_3": "the question give us $|\\vec{a}|=3$ and $\\vec{a}\\cdot\\vec{b}=2$ .So just substitute inside and get the answer . \r\n\r\nwell i guess it shud be $2\\vec{a}\\cdot\\vec{b}$ rather than a putting a modulus :P . just some typo", "Solution_4": "I don't see how you get $|\\vec b|=\\pm2\\sqrt3$, but nvm with that, because i get $|\\vec b|=\\pm\\sqrt{12}$ which obviously is the same.\r\n\r\nSo the end result is:\r\n\r\n$\\theta=\\pm78.90$\r\n\r\nThx for your help ;)" } { "Tag": [ "number theory open", "number theory" ], "Problem": "Find all the triplets $ (s,t,x), s,t,x \\in N, t>s$ such that:\r\n$ (1) ((t^{2}\\minus{}s^{2})y)^{2}\\plus{}64sty$ is a perfect square and $ 2|t\\minus{}s$\r\n$ (2) ((t^{2}\\minus{}s^{2})y)^{2}\\plus{}8sty$ is a perfect square\r\n\r\n([u]not[/u] necessarily both)\r\n\r\nI think there are no solutions. Can anyone prove it?", "Solution_1": "There exist solutions. \r\nIn particular, $ t\\equal{}2$, $ s\\equal{}1$, $ x\\equal{}1$ imply\r\n\\[ ((t^2\\minus{}s^2)x^2)^2\\plus{}8stx \\equal{} 5^2.\\]", "Solution_2": "You're right, my claim is too strong. Here's a (hopefully) correct one-\r\n\r\nThere are no triplets $ (s,t,x)$ of natural numbers, $ t > s$, such that:\r\n$ (1) ((t^{2} \\minus{} s^{2})x)^{2} \\plus{} 4stx$ is a perfect square and $ 2|t \\minus{} s$.\r\n$ (2) ((t^{2} \\minus{} s^{2})x)^{2} \\plus{} 8stx$ is a perfect square and $ 2|t \\minus{} s$.\r\n\r\nThese problems are equivalent to a problem from a recent math olympiad, so they're suppose to be true. But I might have done something wrong.", "Solution_3": "[quote=\"bambaman\"]There are no triplets $ (s,t,x)$ of natural numbers, $ t > s$, such that:\n$ (1) ((t^{2} \\minus{} s^{2})x)^{2} \\plus{} 4stx$ is a perfect square and $ 2|t \\minus{} s$.[/quote]\r\n\\[ ((18^2\\minus{}16^2)\\cdot 1)^2 \\plus{} 4\\cdot 16\\cdot 18\\cdot 1 \\equal{} 76^2\\]" } { "Tag": [], "Problem": "A set of five positive integers has a median of 3 and a mean of 11. What is the maximum possible value of the set's largest element?", "Solution_1": "We know the numbers are like this:\r\n\r\n_ _ 3 _ x\r\n\r\nand they all add up to 55. To make the lasrgest number as large as possible, we try to make the other numbers as small as possible. For the first two blanks, we fill them with 1's. The 4th blank may not be lower than 3, so the smallest possible value of that would be 3. Thus we get:\r\n\r\n1+1+3+3+x=55\r\nSo the largest value of x is 47.", "Solution_2": "Maximize largest element by making the rest smaller. 1,1,3,3,47, max is 47." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Say ABCD is a quadrilateral so that a circle can be inscribed in this which touches the quadrilateral at E, F, G and H on AB, BC, CD, DA.\r\n\r\nProve that, DA, BC and GE are concurrent.\r\nProve that, AC, BD, EG, FH are concurrent.", "Solution_1": "Hmm... i think the first one can't be true. because if we take a triangle ABC and its one excircle (opposite to A say), then we can make another side of quadrilateral arbitrarily so that A, BC and circles tangent point and the other point can't be collinear.", "Solution_2": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=77663 for part #2 and more. For part #2, see also http://www.mathlinks.ro/Forum/viewtopic.php?t=19549 .\r\n\r\n darij" } { "Tag": [ "inequalities", "number theory" ], "Problem": "For every positive a, b, c and d, show that \r\n\r\n\r\n(a+b)(b+c)(c+d)(a+d) is at least equal to 16abcd.", "Solution_1": "First of all, this isn't number theory (mods: this could go to Intermediate, probably).\r\n\r\nSecond of all, \r\n\r\n[hide=\"Hint\"]Multiplying AM-GM for all four pairs on the left hand side will give right hand side[/hide]" } { "Tag": [ "IMO", "IMO 2008" ], "Problem": "Okay, this is our team :\r\n\r\n1. Beka Ergemlidze\r\n2. Lasha lakirbaia\r\n3. Nika Machavariani\r\n4. Lasha Peradze\r\n5. Tsotne Tabidze \r\n6. Levan Varamashvili\r\n\r\n~~~~~~~~~~~~~~\r\n\r\nGood luck, boys!!!", "Solution_1": "good luck ;)\r\n\r\n6 medals approaching :D" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve the equation $\\ [x]+[x^2]=[x^3]$.", "Solution_1": "The transitions will happen at:\r\n$0, \\pm 1, \\pm 2, \\pm 3, \\dots$\r\n$0, \\pm 1, \\pm\\sqrt{2}, \\pm\\sqrt{3}, \\dots$\r\n$0, \\pm 1, \\pm\\sqrt[3]{2}, \\pm\\sqrt[3]{3}, \\dots$.\r\n\r\nLooking at the values of $\\ [x]$, $[x^2]$, $[x^3]$, it's easy to see that there won't be any solution with $x<-1$ or $x>\\sqrt{3}$.\r\n\r\nThe hard part is to order the intervals. I must confess I used a calculator :blush:\r\n$0<1<\\sqrt[3]{2}<\\sqrt{2}<\\sqrt[3]{3}<\\sqrt[3]{4}<\\sqrt[3]{5}<\\sqrt{3}<\\dots$.\r\n\r\nLooking at the values at every interval, the only solutions are:\r\n$x \\in [-1,0) \\cup [0,1) \\cup [\\sqrt[3]{2},\\sqrt{2}) \\cup [\\sqrt[3]{3},\\sqrt[3]{4})$.", "Solution_2": "[quote=\"lordWings\"]The hard part is to order the intervals. I must confess I used a calculator :blush:\n$0<1<\\sqrt[3]{2}<\\sqrt{2}<\\sqrt[3]{3}<\\sqrt[3]{4}<\\sqrt[3]{5}<\\sqrt{3}<\\dots$..[/quote]\r\n\r\nI don't understand are you joking ;) ?\r\n\r\nTo order that you just need to rise all the terms at the power 6 (all the value are positiv so you can do that) and it becomes easy that $0<1<2^2=4<2^3=8<3^2=9<4^2=16<5^2=25<3^3=27$" } { "Tag": [ "probability", "probability and stats" ], "Problem": "There are 2 boxes with $ n$ pencils each one.\r\n\r\nRandomly we remove a pencil of these boxes, and is not import of what box we remove the pencil.\r\n\r\nIn a inesperd moment, a box of them this empty one...\r\n\r\nCalculate the probability that in the other box (nonempty) they are $ k$ pencils.", "Solution_1": "[quote=\"jorgeston\"]In a inesperd moment, a box of them this empty one... [/quote]\r\n\r\nEr, what?\r\n\r\n :?:\r\n\r\nEDIT: Googling \"inesperd\", I found that your question has been answered elsewhere: http://www.mathhelpforum.com/math-help/advanced-probability-statistics/33980-probability-problem.html" } { "Tag": [ "articles", "algebra", "polynomial" ], "Problem": "How do you find the equation of a sequence 7, 23, 55, 109, 191,307, 463, ...?", "Solution_1": "[hide=\"hint\"]\nLook at those 3rd differences...\n\n[/hide]", "Solution_2": "i have read one article regarding this, i just don't what next. yup, its a third degree polynomial. i have already got the equation but i don't know what to do next.\r\n\r\nI tried to pattern it to what I had read and the equation is \r\n\r\n$ f(x) \\equal{} 7_xC_0 \\plus{} 16_xC_1 \\plus{} 16_xC_2 \\plus{} 16_xC_4$\r\n\r\nWhat's next? How do I simplify this into a 3rd degree polyonmial.", "Solution_3": "That's a fourth-degree polynomial (did you mean $ _x C_3$ for the last one?) Now, $ {x \\choose 0} \\equal{} 1, {x \\choose 1} \\equal{} x, {x \\choose 2} \\equal{} \\frac {x(x \\minus{} 1)}{2}, {x \\choose 3} \\equal{} \\frac {x(x \\minus{} 1)(x \\minus{} 2)}{6}$. Remember the definition of a combination!", "Solution_4": "an equation of the sequence is a third degree polynomial (cubic).\r\n\r\n$ S_n \\equal{} n^3 \\plus{} 2n^2 \\plus{} 3n \\plus{} 1$" } { "Tag": [ "trigonometry", "algebra", "polynomial", "geometry", "trig identities", "Law of Sines", "geometry proposed" ], "Problem": "$ ABC$ a triangle such that $ AB\\equal{}AC$,the bissector of $ \\angle B$ intersect $ (AC)$ at $ D$,we suppose that:$ BC\\equal{}BD\\plus{}AD$.\r\nFind $ \\angle A.$", "Solution_1": "Here is what I have so far:\r\nLet $ \\angle DBC \\equal{} x$, AD=n, BD=m. By Law of Sines on $ \\triangle BDC$, \r\n$ DC \\equal{} m/2cosx$. \r\nSo by LOS on $ \\triangle ABC$, we have \r\n$ sin4x/(m \\plus{} n) \\equal{} sin2x/(n \\plus{} (m/2cosx))$ \r\nAlso, by LOS on $ \\triangle ABD$, \r\n$ sin4x/m \\equal{} sinx/n$, hence $ n \\equal{} msinx/sin4x$. \r\nSubstituting, we have \r\n$ sin 4x/(m \\plus{} msinx/sin4x) \\equal{} sin2x/(msinx/sin4x \\plus{} m/2cosx)$. \r\nM cancels, this simplifies to \r\n$ 1 \\plus{} 2cos2x \\equal{} 2cosx \\plus{} 1/(2cos2x)$. \r\nIf $ cosx \\equal{} p$, \r\n$ 16p^4 \\minus{} 8p^3 \\minus{} 12p^2 \\plus{} 4p \\plus{} 1 \\equal{} 0$.\r\nMy calculator says this polynomial is 8(2p-1)(p-cos140)(p-cos100)(p-cos20).\r\nSo x=20, and $ \\angle A \\equal{} 100$, I can't prove it without a calculator.", "Solution_2": "Here is a better method:\r\nExtend BD to E so DE=AD and form quadrilateral AEBC. If $ \\angle ABE \\equal{} x$, by angle chase, we can find all angles in the figure in terms of x.\r\nVirgil Nicula showed me a property that in any quadrilateral (let's use this one as an example), $ sinABEsinCAEsinBECsinACB \\equal{} sinCBEsinBACsinAEBsinECA$.\r\nSo we have:\r\n$ sinx \\cdot sin(3x/2) \\cdot sin(90 \\minus{} x/2) \\cdot sin2x \\equal{} sinx \\cdot sin4x \\cdot sin(3x/2) \\cdot sin(90 \\minus{} 5x/2)$\r\n$ cos(x/2) \\cdot sin2x \\equal{} sin4x \\cdot cos(5x/2)$\r\n$ cos(x/2) \\equal{} 2cos2x \\cdot cos(5x/2)$\r\n$ cos(x/2) \\equal{} 2cos2x \\cdot (cos2x \\cdot cos(x/2) \\minus{} sin2x \\cdot sin(x/2))$\r\n$ (2cos^2(2x) \\minus{} 1) \\cdot cos(x/2) \\equal{} 2sin2xcos2x \\cdot sin(x/2)$\r\n$ cos 4x \\cdot cos(x/2) \\equal{} sin4x \\cdot sin(x/2)$\r\n$ tan4x \\cdot tan(x/2) \\equal{} 1$\r\n$ 4x \\plus{} x/2 \\equal{} 90m$ (m is odd)\r\n$ x \\equal{} 20m$\r\nSo x is 20, $ \\angle A$=100\r\nx can't be higher than that (like 60) or it wouldn't make a triangle.", "Solution_3": "$ a \\equal{} BC, b \\equal{} CA \\equal{} AB.$\r\n$ BC \\equal{} BD \\plus{} AD \\Longleftrightarrow$ $ a \\equal{} \\frac {\\sqrt {ab(a \\plus{} 2b)a}}{a \\plus{} b} \\plus{} \\frac {b^2}{a \\plus{} b}.$\r\nChoosing $ b \\equal{} 1,$ this becomes $ a(a \\plus{} 1) \\equal{} a \\sqrt {a \\plus{} 2} \\plus{} 1.$ Squaring to get rid of $ \\sqrt {\\ \\ },$ $ a^4 \\plus{} a^3 \\minus{} 3a^2 \\minus{} 2a \\plus{} 1 \\equal{} 0.$\r\nThis has obvious root $ a \\equal{} \\minus{} 1,$ not acceptable. (The only rational roots can be $ \\pm 1,$ this is the thing to look for.) Factoring it out by long division, $ a^3 \\minus{} 3a \\plus{} 1 \\equal{} 0.$ We need $ \\cos B \\equal{} \\frac {a}{2b} \\equal{} \\frac {a}{2}$ $ \\Longrightarrow$ $ \\cos 3B \\equal{} 4 \\cos^3 B \\minus{} 3 \\cos B \\equal{} \\minus{} \\frac {_1}{^2},$ $ 3B \\equal{} k \\cdot 120^\\circ,$ where $ k \\mod 3 \\neq 0.$ $ B \\equal{} k \\cdot 40^\\circ < 90^\\circ$ $ \\Longrightarrow$ $ B \\equal{} 40^\\circ$ or $ 80^\\circ$ for $ k \\equal{} 1, 2.$ Suppose $ B \\equal{} 80^\\circ > 60^\\circ,$ then $ BD \\plus{} AD > AB > BC,$ not acceptable. (In fact, then $ AD \\minus{} BD \\equal{} BC.$) The only possible result is $ B \\equal{} 40^\\circ,$ $ A \\equal{} 180^\\circ \\minus{} 2B \\equal{} 100^\\circ.$" } { "Tag": [ "linear algebra", "matrix", "real analysis", "real analysis unsolved" ], "Problem": "Find equation of all such possible curves which pass through (1,0) and are given by the determinant eqn::", "Solution_1": "Solve the determinant and then the differential equation.", "Solution_2": "[quote=\"eureka-123\"]Find equation of all such possible curves which pass through (1,0) and are given by the determinant eqn::[/quote]\r\nthe second matrix is not square. what do you mean by determinant????? :wink: :wink:" } { "Tag": [], "Problem": ":) Where would u most likely Stop, Drop, and Roll?\r\n\r\nAny discussion?\r\n\r\nI like stop drop and roll when the surface is not hard and when the surface does not stink. If there was a pile of dollar bills on the ground (I mean MOUNTAINS) I would.", "Solution_1": "i put other. u see me everyday and i'm pretty sure you know by now you know what i'd do with stop, drop and roll. if you don't, just ask me at school." } { "Tag": [ "articles", "geometry" ], "Problem": "Below is a piece of news I just read. Maybe some of you already know this but I found it interested to share it to others:\r\n\r\n[quote][b]Did Jesus Ask Judas to Betray Him?[/b] \n\n[i]Updated 3:48 PM ET April 6, 2006[/i]\n\nIt is a mystery 2,000 years in the making, buried in the desert and fueled by centuries of debate and doubt, theft and deceit. The question: Was there ever a Gospel according to Judas? And if there was, what did it reveal? \n\nThe mystery began to unravel almost 30 years ago, according to a new National Geographic Channel documentary. \n\n[b]Watch the full story on ABC's \"Primetime\" Thursday at 10 p.m. ET and \"Nightline\" at 11:35 p.m. ET and get more information from National Geographic by clicking here. Watch the full documentary \"The Gospel of Judas\" at 8 p.m. ET Sunday, April 9, on the National Geographic Channel.[/b] \n\nA farmer looking for treasure in a cave in Egypt instead found a decaying leather-bound book, a codex. \n\n\nBecause the text was written in ancient Coptic, the farmer did not know what it was, but he figured he could sell it. He did, to an antiquities dealer, but still no one knew its secret text -- and no one ever would if it continued to disintegrate before it could be translated. \n \n\nFive years after it was found, the first glimpse of the book's meaning was gleaned when scholar Stephen Emmel was asked to look at the document. \n\n\"We were told we were not allowed to make any photographs, we were not allowed to make any notes,\" Emmel told \"Primetime.\" \"I leafed through and by chance spotted a dialogue between Jesus and Judas and his disciples and in particular the name Judas came up again and again. Judas said ... the Lord said, blah, blah, blah.\" \n\nCould it be a Gospel according to Judas? Scholars had long believed such a Gospel existed but that it had been banned by the early church, called blasphemous and ordered destroyed. Had a copy somehow survived, telling the story of the most reviled man in history? \n\n\n[b]A Great Betrayal[/b]\n\"[Judas is] the one who handed over his friend,\" said Marvin Meyer, co-chair of the religious studies department at Chapman University. \"He's the one who brought about the crucifixion, and he's the one who is damned for all time.\" \n\nThe Bible says Judas betrayed Jesus for 30 pieces of silver. The Bible also tells how the other disciples let Jesus down: Peter denied him three times but is nevertheless honored with the basilica in Rome. It is Judas alone who is not forgiven, condemned to the seventh level of hell in Dante's \"Inferno,\" eaten head-first by Lucifer. \n\n\"Often they think of him as somebody who was greedy, avaricious, who was more interested in making money than in being faithful to his master,\" said Bart Ehrman, chair of the religious studies department at the University of North Carolina. \n\nAnd Judas over the centuries also became a symbol of anti-Semitism. \n\n\"Traditionally in Christian circles, Judas in fact has been associated with Jews,\" Ehrman said. \"Of being traitors, avaricious, who in fact, betray Jesus, who are Christ-killers. And this portrayal of Judas of course also leads then to horrendous acts of anti-Semitism through the centuries.\" \n\nBut what if there were more to the story? The first task was to see if the document was genuine. For 16 years, the Codex sat crumbling in the most unlikely of places -- a safe-deposit vault in a Citibank on Long Island, N.Y., until the year 2000, when it was purchased by a former antiquities dealer, Frieda Nussberger-Tchacos. \n\n\"I think the circumstances of this manuscript coming to me were predestined,\" she said. \"Judas was asking me to do something for him.\" \n\n\n[b]Searching for Answers[/b]\nFinally, with the funding of the National Geographic Society and two foundations, a dream team of scientists and scholars was assembled to determine if the document was really an ancient text. \n\nVerification was an enormous challenge. The 13 pages of papyrus, with writing on the back and front, were in a thousand pieces. The box with what might be the lost Gospel was shipped to Swiss restorer Florence Darbre. \n\n\"Your heart is beating a little bit faster the first time you take it in your hand ever,\" she said, \"in your hand, yes, because you can see it but you don't know how he is, how brittle he is, how delicate it is.\" \n\nPainstakingly, Darbre and her partner fit the tiny pieces together. But just because it looked authentic, it still had to be proved whether it was written 1,800 years ago. For that, Emmel, the scholar, and Rodolphe Kasser, another expert, were brought in. \n\n\"I've looked at hundreds of papyri, Coptic papyri, in my career, and this is absolutely typical of ancient Coptic manuscripts,\" Emmel said. \"I'm completely convinced. The number of people who could create such a text is very small,\" he said, perhaps just 25. Kasser estimated there were just four or five people who could create a convincing imitation of an ancient Coptic document. \n\n\"We're two of them,\" Emmel said. \"Now, who would create such a thing? Someone would have to know Coptic better than we do. And there isn't anybody who knows Coptic better than we do. I'm sorry.\" \n\nBut the final test of authenticity meant taking a radical step -- destroying tiny pieces of the document to carbon date it. \n\n\"I will have to burn the Gospel of Judas in order to be able to date it,\" said Timothy Jull, a carbon-dating expert at the University of Arizona's physics center. \n\nFifteen hours later, he had a final answer. The text was real. The Gospel according to Judas, written between the third and fourth century, was believed to be a copy of a much older document written in Greek in the second century. \n\n\"Radio carbon dating of the papyrus from the Gospel of Judas confirms that it's from the third to the fourth century A.D.,\" Jull said, \"and this supports the authenticity of the Gospel of Judas.\" \n\n\n[b]Did Jesus Ask to Be Betrayed?[/b]\n\"You can't fake a codex like this,\" said Elaine Pagels, a professor at Princeton University and one of the world's foremost experts on ancient religious texts, especially the so-called Gnostic or secret Gospels, which include the Gospel of Judas, all written in the first and second centuries and banned by the early church. \n\n\"It's crumbling; it's a particular kind of papyrus; it's a particular kind of script,\" Pagels said. \"It would be absolutely not worth anyone's while, and far too difficult, to try to fake this kind of text. This is a genuine ancient text.\" \n\nSo what is in the Gospel of Judas? It is a dialogue that claims to be a conversation between Jesus and Judas in which Jesus asks Judas to betray him. \n\n\"Judas has the terrible task of taking it upon himself to turn him over to the authorities for this reason,\" Pagels said. \"Now, the Gospel of Judas also has Judas say to Jesus in fear and terror that he has a dream that the other disciples will hate him and will stone him to death, will attack him. \n\n\"And Jesus says, 'Yes, in fact, they will think that you are a terrible person because of what you did. This is part of the burden that you bear. But they will be wrong about that.' So it is an extraordinary transformation of the ordinary understanding of Judas Iscariot.\" \n\nPagels said the text shows that Christ, in fact, asked Judas to betray him for an undisclosed reason. \"The Gospel of Judas does suggest that the betrayal of Jesus is not a reprehensible act, not the act of a traitor, you know, the worst villain in the history of the world, but that it's a secret mystery between him and Jesus,\" she said. \n\nHerbert Krosney, author of \"The Lost Gospel: The Quest for the Gospel of Judas Iscariot,\" one of two new books and a National Geographic Channel documentary about the Gospel of Judas, said the finding is significant. \n\n\"Judas is not the betrayer,\" Krosney said. \"Judas is, rather, the favored disciple of Jesus. He is the one whose star shines in the heavens and in the skies, and Judas, therefore, becomes unique. He is Jesus' best friend rather than his betrayer.\" \n\n\n[b]A New Perspective[/b]\nToday the Gospel of Judas got its first public outing at a news conference, and it is on display at the National Geographic Society in Washington, D.C. It will eventually return to Egypt to be housed in Cairo's Coptic Museum. It is also available online, in Coptic and English, and is the cover story of the new National Geographic magazine. \n\nBut while the document is a real one, is what it claims also true? Did the New Testament Gospels of Matthew, Mark, Luke and John get it wrong? Did Jesus askJudas to betray him? \n\n\"I don't think that we have in this Gospel what I would call historical proof,\" Pagels said. \"We also don't have that in the other Gospels.\" \n\nShe said there may never be an answer. \"I don't see how we would, although you see we could always find next week or in 50 years other evidence that we don't have now.\" \n\nIn the end, science may have its answers, but questions of spirit and soul cannot be analyzed like a piece of papyrus. If Judas did not betray Jesus and was part of a grand plan, does that change anything for Christian theology? \n\nThe discovery does not alter the belief of evangelical scholar Ben Witherington that Judas did indeed betray Jesus. \"Well, it would mean among other things that Jesus had some kind of death wish, for a start,\" Witherington said. \"And it would raise some questions about his character. \n\n\"I would think there are some questions of integrity that would be raised about that, for him to sort of script it in such a way that he's using his disciples to go and set this up would suggest a sort of level of hands-on intention to it, where he's not just submitting to the will of God in his life,\" Witherington said. \"He's actually got his hands on the wheel, and he's driving the wheel of history in a particular direction. And some would find that troubling.\" \n\nThe scholars interviewed by \"Primetime\" agreed that the real importance of the Gospel of Judas is the window it provides into what some early Christians were thinking. But they acknowledged that some in the organized church will not like the discovery of this Gospel. \n\n\"Absolutely, they won't,\" Pagels said. \"Some will be very offended, and they'll say this proves that all of those texts are rubbish, because it's an utterly preposterous idea that Judas could have been involved in a secret mystery with Jesus.\" \n\nBut that would miss the point, she added. \"The Christian message is a message about faith and hope and, you know, the relationship between God and human beings. It's not a matter of historical fact.\" \n\nPagels said she hoped the find will have a big impact. \n\n\"I would hope that people appreciate the excitement of this discovery and recognize that it's all right to ask the kinds of questions that sometimes they're afraid to ask, and say, 'What else didn't we know about the early Christian movement? Could, for example, Judas be forgiven?'\" she said. \n\n\"And when people start asking that question, they'll realize that it doesn't destroy faith, it actually can strengthen it. But it's a different kind of faith; it's informed by what we understand about our past.\" [/quote]", "Solution_1": "This article misses a major point when it talks about how hard this gospel would be to fake.\r\n\r\nWhat is really meant is that it would be really hard to fake it today. If only 20 people could fake it today, how many could fake it 100, 300, 800 years ago? It is illogical to assume that such a document is a copy of a prior document. We have the history of copying documents for the church documents. We also have many copies showing up across the world for most of the church documents that are included in the Bible. This doesn't apply to this paper. It is also illogical to assume that if it is a fake it could only have been faked by those living today.\r\n\r\nThe dating, if correct, says that it was created in the second or third century. There would be many people who would have a reason to come up with something discrediting the Christian story, or redeeming the name of Judas. (Remember the descendent of Dr. Mudd (who treated Mr. Booth, Abe Lincon's killer) who spent his entire life trying to get a legal clearing of Dr. Mudd? There may have been many who could have made such a document as a fake, joke, or fiction; which lay undiscovered until recently, where the context of the document not available.\r\n\r\nJust because something is old doesn't make it necessarly true. Those creating it didn't say, \"Now let's put this here where it will be found thousands of years later.\r\n\r\nI don't know, but let's not jump to illogical conclusions in looking at on document.", "Solution_2": "on a comical side note : where there lots of people called judas before that?\r\n\r\n\r\nIt's the same thing with Adolf : the name practically vanished the last half century :oops:", "Solution_3": "I think no........", "Solution_4": "i thought judas was a pretty common name back then", "Solution_5": "It was common. Read on if you want a full answer, rather than just the conclusion. \r\n\r\nThanks to computer science (or a concordance) one can easily determine that the name did not appear once in the old testament, but had the following appearances in the new testament.\r\n\r\nOne of Jesus' half brothers (different father) was named Judas (Mark 6:3).\r\n\r\nWhat were the names of the twelve apostles?\r\n\r\nI suspect not one in ten thousand Christians can name them all, even though they are listed in the Gospels. See Luke 6:13 (written by Luke a physician who accompanied Paul, neither of who were among the original 12 Apostles, but together wrote far over half of the new testament), when the 12 are first called Apostles by Jesus.\r\n\r\nSimon (Peter)\r\nAndrew\r\nJames\r\nJohn (the Gospel writer and youngest Apostle)\r\nPhilip\r\nBartholomew \r\nMatthew (the Gospel writer and former tax collector)\r\nThomas (the doubter)\r\nJames (not Jesus' half brother)\r\nSimon (the Zealot - this was a sect of religion at the time, not a character statement)\r\nJudas (son of James, yet another James)\r\nJudas Iscariot (the Betrayer)\r\n\r\nThe list of the twelve original Apostles in Matthew 10:3 is the same except Judas (son of James) is replaced by Thaddeus.\r\n\r\nThe Gospel of John (14:22) reports a discussion between \"Judas (not Judas Iscariot)\" and Jesus, after Jesus rose from the dead.\r\n\r\nActs 1:13 reports Judas (son of James) as one of the original Apostles present when Matthias was selected by lot from those with Jesus from the beginning. (This is not surprising, as Acts is written by Luke, who wrote the Gospel with his name.)\r\n\r\nIn Acts 5:37, a Pharisee named Gamaliel argued in the Sanhedrin (Jewish high court) for the lives of the 12 disciples, after Jesus' death. Part of his argument was that others' movements died out after their leader died, and this Jesus sect should be treated the same way. One of the prior revolts was lead by a person designated as \"Judas the Galilean.\" (Gamaliel was one of Paul's teachers.)\r\n\r\nIn Acts 9:11, after Paul is struck blind on the road to Damascus, he is taken to the house of a man called Judas on Straight Street, where Jesus tells a Christian named Ananias to go find and bless Paul, who has been beating, killing and imprisoning Christians and who has a legal warrant to collect Christians from Damascus and take them back to Jerusalem for trial. (Ananias was initially reluctant to go and find Paul.)\r\n\r\nIn Acts 15:22, the church in Jerusalem, led by Peter, sent two of its members to go with Paul back to Antioch to tell them that God did not require that the Jewish law applied to Christians. One of these is \"Judas (called Barsabbas).\" The accompanying letter designated him only as \"Judas.\"\r\n\r\nMaybe too much Bible for some, but this exhaustive description will allow all the gentle readers determine for themselves if they are convinced that the name Judas was common at the time and in the area where Jesus lived." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given are $\\triangle ABC$ prove that\r\n$sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2}\\le \\frac{3}{2}+\\frac{r}{2R}(4cos\\frac{A}{2}.cos\\frac{B}{2}.cos\\frac{C}{2}-(cos\\frac{A}{2}+cos\\frac{B}{2}+cos\\frac{C}{2}))$ :lol:", "Solution_1": "Nobody? :(", "Solution_2": "Maybe it is \r\n$sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2}\\le \\frac{3}{2}+\\frac{1}{3}(4cos\\frac{A}{2}.cos\\frac{B}{2}.cos\\frac{C}{2}-(cos\\frac{A}{2}+cos\\frac{B}{2}+cos\\frac{C}{2}))$ \r\nAre anybody interested in? :)", "Solution_3": "Are you sure about $\\frac{r}{2R}\\geq \\frac{1}{3}$ ??\r\n\r\nIt's well known that $OI^{2}= R^{2}-2Rr \\geq 0$\r\n\r\nso $R\\geq 2r$. Therefore, $\\frac{r}{2R}\\geq \\frac{1}{3}$ is wrong.", "Solution_4": "ok you are right Chang Woo-JIn, because my second ineq is true, too and it is stronger than the first, but maybe it is easier than the first! :) :)", "Solution_5": "[quote=\"gemath\"]Maybe it is \n$sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2}\\le \\frac{3}{2}+\\frac{1}{3}(4cos\\frac{A}{2}.cos\\frac{B}{2}.cos\\frac{C}{2}-(cos\\frac{A}{2}+cos\\frac{B}{2}+cos\\frac{C}{2}))$ \nAre anybody interested in? :)[/quote]\r\nI'm very sorry it is flase :blush:,therefore I'm not sure the first is true but the second must be $sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2}\\le \\frac{3}{2}+\\frac{1}{6}(4cos\\frac{A}{2}.cos\\frac{B}{2}.cos\\frac{C}{2}-(cos\\frac{A}{2}+cos\\frac{B}{2}+cos\\frac{C}{2}))$\r\nI also want to find the inequality stronger than the basic inequality I made some mistake therefore hope anybody are interested in it excuse for me :blush:" } { "Tag": [ "limit", "integration", "calculus", "calculus computations" ], "Problem": "show :D\r\n\r\n$ \\lim_{k\\to\\infty}\\;\\left\\{\\dfrac{\\sqrt{\\;k\\plus{}1\\;}\\plus{}\\sqrt{\\;k\\plus{}2\\;}\\plus{}\\sqrt{\\;k\\plus{}3\\;}\\plus{}\\;\\;\\ldots\\;\\;\\plus{}\\sqrt{\\;2k\\minus{}1\\;}}{k^{\\frac{3}{2}}}\\right\\}\\;\\equal{}\\;\\boxed{\\frac{2}{3}\\left(\\sqrt 8\\minus{}1\\right)}$", "Solution_1": "That limit is equal to $ \\lim_{k\\to\\infty}\\sum_{i\\equal{}0}^{k\\minus{}1}\\frac{1}{k}\\cdot \\sqrt{1\\plus{}\\frac{i}{k}}\\equal{}\\int_{0}^1\\sqrt{1\\plus{}x}\\,\\text{dx}$." } { "Tag": [ "articles", "Support", "search" ], "Problem": "We know that since Al Gore dropped out. (slander), there are alot of democrats running. Lets see who you like the most. EVEN REPUBLICANS. Currently the order of the poll (dean, gephardt, ....) is the rank they are in. \r\nI personally like Kerry, a decorated Vietnam veteran.", "Solution_1": "Hmm, if only we could vote for more than 1 ... like, all of them. Honestly, in terms of fundamental beliefs, I'd take any of the 9. And, actually, just about anybody. Heck, I could name Republicans who I'd feel glad to have as president instead of W. \r\n\r\nBy the way, did any of you hear the results from D.C.? They held an unofficial primary, and all the main candidates except Dean asked not to be included, so it was only Dean (42%), Sharpton (35%), Braun and Kucinich.", "Solution_2": "wasn't that DC primary mostly for protesting the citizens' civil rights because their votes don't count in the real thing etc.", "Solution_3": "Yes. They do get one or two electoral votes, though. (That is, they help elect the president, even though they aren't represented by a voting member of congress.)", "Solution_4": "I'd say anyone but John Kerry, I've seen two of his ads and they were peeuuu! He clearly said he only \"approved\" of this ad and he didn't write it himself and he filled it with scenarios in which poor or the sad are shown and not specific political acts he said he'll do.\r\n\r\nStill I'd rather have him then our current president...then again everyone knows he'll still win so *sigh*", "Solution_5": "Lieberman?! Lieberman has the most votes so far? That's like voting for Zell Miller for Democrat, or Jim Jeffords for Republican ...", "Solution_6": "[quote=\"JBL\"]Lieberman?! Lieberman has the most votes so far? That's like voting for Zell Miller for Democrat, or Jim Jeffords for Republican ...[/quote]\r\nAh, but it's not like the 2004 Democratic Candidate even matters. I think we can all admit that.", "Solution_7": "That's interesting, given that 50% of the country is Democratic and as of now, at most 55% of Americans on polls would like to see him re-elected. That's not too good for a sitting president.", "Solution_8": "All I know is I don\u2019t like Sharpton. Liebermans ok I guess. Id like never to see a Clinton in office again even if it was Hillary, who I like, I just don\u2019t want to see them in office ever again. I know they are good for the U.S. but when it comes down to it they are awful with foreign affairs (that\u2019s just my belief.) I don\u2019t know if Id like to see Bush get reelected or not because I have a list of his pros and cons; one of the pros is his secretary of state. I like Powell he is a smart man, one of the smartest secretary\u2019s we\u2019ve ever had. I like him a lot I just wish he \u2018d have chosen a different party to run with. I like some of the things about democrats but then again I like some of the things about republicans.\r\n\r\nDo yall know who has more power than the president?", "Solution_9": "Please try making your \"sentences\" easier to read; I had to skip your whole post because it was giving me a headache.", "Solution_10": "Ok its fixed. I can use proper grammer when I want to. I tire of it real fast because I use it so much at school. I guess thats what you would call lazy. I have the TAKS test as proof of being grammatically correct. For whoever doesnt know what that is, it is a state mandatory test that we have to take in Texas its horrible. Thats all the teachers teach about. Oh you have to do this and you have to that in order to pass this test. You know what, Ive never once used one of their stratagies and I make academic recognition on it every time they are like the easiest tests in the world. I made a perfect score on my writing and grammer exam last year and in 4th grade (those are the only two years you take the writing exam.) I just am lazy I guess. Oh well all I have to worry about this year is passing the American history exam thats like gonna be hard for me because all I have memorized is my 50 states and their capitals, the first presidential cabinent, and the preamble to the Declaration of Independence. I dont even have my presidents memorized; how many of them were there again?", "Solution_11": "Clinton was the 42nd, I believe. Depending on whether you count Cleavland once or twice :).", "Solution_12": "news update\r\ni hear it is a four guy race now that kerry and edwards have catched up (the two others are gehpart, den) . LETS GO FOR KERRY.\r\ndean has lost a few points. oh and never hear about clak and lieberman in iowa its because they aren't capiangin there.", "Solution_13": "I personally would like to see Bush reelected. \r\n\r\naalindsey-there have been 44 presidents counting Bush in office now. I know you should be able to name them because I can. Washington, Adams, Jefferson,Madison, and Monroe in order just to name a few.", "Solution_14": "[quote=\"JBL\"]That's interesting, given that 50% of the country is Democratic and as of now, at most 55% of Americans on polls would like to see him re-elected. That's not too good for a sitting president.[/quote]\r\nAh who cares, I bet on Bush.\r\n\r\nHilary is smart by waiting unlike Gore :lol:\r\n\r\nugh I can now clearly see Bush is making U.S. education even worse than it is now, and I thought it can't get any worse...memorization can get you no where but they have to make you memorize or else you'll realize the horrible truth about the U.S. and it's dirty dirty past...\r\n\r\nI'd say Bush is the strongest though I'll beat him soon enough...:twisted:", "Solution_15": "HUH?!?\r\n\r\nBHorse, you're making us Republicans look, well, not good.", "Solution_16": "Huh?\r\n\r\nWe republicans? I am not either!\r\n\r\nI am independent! besides, I can't vote yet!\r\n\r\nI am not making republicans look bad.", "Solution_17": "Clinton? Sure he's a pressy boy but Bush is just downright dirty when it comes to recruiting, not only did he avoid getting recruited during Vietnam or Korea (I think it's Vietnam but I'm not sure; I read it in Time like an year ago so maybe it's online somewhere) using his dad's political power he also made it [b]look[/b] like he actually was in that war by going to Texas aviation school or something and he took photos of him in a plane which never went to the war field at all. Besides he cares more about his own life than any other so don't be surprised if he decides to sacrifice you guys for money or oil or whatever he want.", "Solution_18": "tare Bush never ran away from those wars I know that at least he may not have participated but then again I don't know if there was major recruting for those war. I just know he didn't run away my parents were army.", "Solution_19": "BHorseMath, there was no draft during Desert Storm...plus, wasn't he governor of Arkansas at that time? It would have been completely unnecessary for a man his age to have been worried about being drafted. On Vietnam, he was a Rhodes Scholar, which would have made him exempt (as long as he was a college student), so he never draft-dodged.", "Solution_20": "Um, it was in [b]Time magazine[/b] and I know your parents are in the Texan army but I'm still going to take Time's word for it since I know Time pretty well but I don't know your parents at all :lol:", "Solution_21": "Tare, all writers have their own bias, so you can't always take what Time says by face value. Although it's my favorite news magazine too.", "Solution_22": "I understand that but it wasn't in the editorial section it was like the second biggest topic there and I'm sure that it is a fact rather than an opinion sort of a thing.", "Solution_23": "First of all: Clinton was far too old to be in Desert Storm. You're thinking of Vietnam. Neither Clinton nor Bush, nor, for that matter, most politicians, went to Vietnam. Mostly, this is due to the political influence of their families. (Kerry is a notable exception.) I wasn't exactly able to figure out in a quick search of the internet how Clinton avoided the draft. Partly it was due to a student deferment while he was at college, but beyond that it's a little unclear. Roughly, I think he was able to put it off until Nixon suspended the draft and thus was never drafted. I do know that Bush's family pulled political strings to get him a spot in the Texas National Guard that he ended up not showing up for for 15 months or so.", "Solution_24": "[quote=\"Tare\"]I understand that but it wasn't in the editorial section it was like the second biggest topic there and I'm sure that it is a fact rather than an opinion sort of a thing.[/quote]\r\n\r\nhmm...don't think you bother checking the \"setting the record straight\" where TIME corrects its errors, which there are quite a few of...", "Solution_25": "BBOO YAH I WAS THE ONLY ONE WHO VOTED FOR KERRY AND I WAS THE ONE WHO CREATED THE POLL. BOY AM I PHYSIC. Kerry wins the Big Tuesday and Edwards drops out. Kerry wins all except for one: Vermont.", "Solution_26": "Hehe, but he got beat nearly 2-to-1 by a man no longer running in the race. (On the other hand, he did get nearly 75% of the vote in MA.)\r\n\r\nAnyhow, Kerry 2004 -- yeehaw!\r\n\r\n(By the way, I think you meant \"psychic,\" not \"physic.\"", "Solution_27": "On the news the analyst people were saying that Vermont was not anti-Kerry; they were just pro-Dean.b", "Solution_28": "I am voting for either Poppin' Fresh or the Jolly Green Giant.\r\n\r\nWould it be any worse to have a fictional president than a real one?\r\n\r\nFYI -- Poppin' Fresh has mad math skills and would likely cause an international economic boom.", "Solution_29": "For a moment there I thought you were churchill :lol:\r\n\r\nI have no idea who they are, but I'm still rooting for Kelly...I'm choosing the red pill :lol:" } { "Tag": [ "inequalities unsolved", "inequalities", "algebra" ], "Problem": "Let $a;b;c>0$. Prove that:\r\n\r\n$A=\\frac{1}{\\sqrt{a}}+\\frac{3}{\\sqrt{b}}+\\frac{8}{\\sqrt{3c+2a}}\\geq\\frac{16\\sqrt{2}}{\\sqrt{3(a+b+c)}}$", "Solution_1": "Let $f(a,b,c)=\\frac{1}{\\sqrt{a}}+\\frac{3}{\\sqrt{b}}+\\frac{8}{\\sqrt{3c+2a}}-\\frac{16\\sqrt{2}}{\\sqrt{3(a+b+c)}}$.\r\nIt is easy to show that $f(a,b,0) \\ge 0$, $f(a,b,c)\\ge 0$ when $c\\to \\infty$.\r\n$\\frac{\\partial f}{\\partial c}= 0 \\iff a+3c=3b$.\r\nSo it is enough to show that $f(3(b-c),b,c) \\ge 0$.\r\nLet $b-c=x, b=y$.\r\n$f(3(b-c),b,c)\\ge 0 \\iff \\frac{1}{\\sqrt{x}}+\\frac{3\\sqrt{3}}{\\sqrt{y}}\\ge \\frac{8}{\\sqrt{x+y}}$, it is obvious by jensen.", "Solution_2": "Let $a,b>0$. Prove that\n$$\\frac{1}{\\sqrt{a}}+\\frac{4}{\\sqrt{3a+2b}}\\geq\\frac{3\\sqrt{6}}{\\sqrt{5a+2b}}$$\n$$\\frac{1}{\\sqrt{a}}+\\frac{4}{\\sqrt{a+2b}}\\geq\\frac{3\\sqrt{6}}{\\sqrt{3a+2b}}$$\n$$\\frac{1}{\\sqrt{a}}+\\frac{4}{\\sqrt{2a+b}}\\geq\\frac{3\\sqrt{6}}{\\sqrt{4a+b}}$$\n$$\\frac{1}{\\sqrt{a}}+\\frac{2}{\\sqrt{a+2b}}\\geq\\frac{\\sqrt{6(1+\\sqrt[3]{2}+\\sqrt[3]{4})}}{\\sqrt{3a+2b}}$$\n", "Solution_3": "[quote=sqing]Let $a,b>0$. Prove that\n$$\\frac{1}{\\sqrt{a}}+\\frac{4}{\\sqrt{3a+2b}}\\geq\\frac{3\\sqrt{6}}{\\sqrt{5a+2b}}$$\n$$\\frac{1}{\\sqrt{a}}+\\frac{4}{\\sqrt{a+2b}}\\geq\\frac{3\\sqrt{6}}{\\sqrt{3a+2b}}$$\n$$\\frac{1}{\\sqrt{a}}+\\frac{4}{\\sqrt{2a+b}}\\geq\\frac{3\\sqrt{6}}{\\sqrt{4a+b}}$$\n$$\\frac{1}{\\sqrt{a}}+\\frac{2}{\\sqrt{a+2b}}\\geq\\frac{\\sqrt{6(1+\\sqrt[3]{2}+\\sqrt[3]{4})}}{\\sqrt{3a+2b}}$$[/quote]" } { "Tag": [ "limit", "inequalities", "calculus", "search", "quadratics", "calculus computations" ], "Problem": "Hello there,\r\n\r\nI am trying to prove the following limit using the delta-epsilon definition. I have worked backwards to find a suitable epsilon value but am stuck on how to write the actual proof using the minimum of this value and my assumed delta value.\r\n\r\nIn class, my instructor simply ends the proof with a new statement under (10) below saying that if $ \\delta \\text{=min}\\left\\{ \\text{1, }\\frac{57\\varepsilon }{20} \\right\\}$, then QED. However, I want to go beyond this and actually arrive at $ |f(x) - L | = \\abs{\\frac{x + 3}{x^2 + x + 4} - \\frac{2}{3}} < \\varepsilon$. \r\n\r\nThank you very much.\r\n\r\n---\r\n\r\n\\begin{align}\r\n & \\text{Prove the following statements directly using the formal }\\varepsilon \\text{,}\\delta \\text{ definition}\\text{.} \\\\ \r\n & \\text{ii) }\\underset{x\\to 1}{\\mathop{\\lim }}\\,\\frac{x+3}{{{x}^{2}}+x+4}=\\frac{2}{3} \\\\ \r\n & ---- \\\\ \r\n & \\text{My work is below}\\text{. I am working backwards from the epsilon inequality} \\\\ \r\n & \\text{to find a suitable delta}\\text{.} \\\\ \r\n & \\text{As question states: }\\text{ }0<\\left| x-1 \\right|<\\delta \\Rightarrow \\left| \\frac{x+3}{{{x}^{2}}+x+4}-\\frac{2}{3} \\right|<\\varepsilon \\\\ \r\n & \\left| \\frac{3(x+3)-2({{x}^{2}}+x+4)}{3({{x}^{2}}+x+4)} \\right|<\\varepsilon \\Rightarrow \\left| \\frac{(2x+1)(x-1)}{3({{x}^{2}}+x+4)} \\right|<\\varepsilon \\\\ \r\n & \\text{Let }\\left| x-1 \\right|<\\frac{1}{2}. \\\\ \r\n & \\text{Then }\\frac{1}{2}45$ and $\\angle{B}>45$.\r\nIn the interior of the triangle we construct a right-angled and isosceles triangle $ABD$ with $\\widehat{ADB}=90$ .\r\nNow in the exterior of the triangle we construct the right-angled and isosceles triangles $BCE$ and $ACZ$ with $\\widehat{BEC}=90$ and $\\widehat{AZC}=90$ respectively .Prove that \r\n$DECZ$ is a paralellogram .", "Solution_1": "Assume $A, B, C$ are positively oriented. Let $M$ be the midpoint of $AB$. Then $MD$ is the rotated image of $MB$ for $90^\\circ$ CCW - expressing that with complex numbers, we get $d-m=i(b-m)\\iff d={a+b\\over 2}+i{b-a\\over 2}$, since $m={a+b\\over 2}$\r\n\r\nSimilarly, $e={b+c\\over 2}+i{b-c\\over 2}, z={a+c\\over 2}+i{c-a\\over 2}$\r\n\r\nNow $e-d+z={b+c-a-b+a+c\\over 2}+i{b-c-b+a+c-a\\over 2}=c$, which yields $e-d=c-z$. The conclusion follows.", "Solution_2": "Thank you , nice solution . :)" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "a tricky problem (one of those with short enounces :D) let R be a ring such that xn = x. Prove that for each x,y we have xyn-1=yn-1x.", "Solution_1": "solved it :D", "Solution_2": "[quote=\"Valentin Vornicu\"]let R be a ring such that xn = x. Prove that for each x,y we have xyn-1=yn-1x.[/quote]\r\n\r\nFor n = 2\r\n\r\nx+y=(x+y)= x+xy+yx+y => xy=-yx \r\na+1= (a+1) = a+2a+1 => 2a=0 => -a=a\r\n\r\nxy=yx", "Solution_3": "[quote=\"Valentin Vornicu\"] let R be a ring such that xn = x. Prove that for each x,y we have xyn-1=yn-1x.[/quote]\r\n\r\nRemind me the theorem:Let R a ring with a unit element such that for any x\\inR, there exist a positive integer nx such that x nx=x, then R is commutative\r\n\r\nBiography of Jacobson\r\nhttp://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Jacobson.html", "Solution_4": "Warning ! It seems, in the V.V. problem the ring hasnt unit element for the multiplicative operation ... Am I right ?", "Solution_5": "[quote=\"Diogene\"]Warning ! It seems, in the V.V. problem the ring hasnt unit element for the multiplicative operation ... Am I right ?[/quote]\r\n\r\nUnit or not unit element that's the question, V.V will tell us", "Solution_6": "one does not need an unit in the ring to prove the property.", "Solution_7": "Well, clearly n>=2 ==> x^(2n-2) = x^(n-1) \r\nAlso we have that x^2 = 0 ==> x^n = 0 ==> x = 0\r\n\r\nLet z = x^(n-1) and y be in the ring\r\n\r\n(zyz-yz)^2 = (zyz-zy)^2 = 0 ==> yz = zy ==> x^(n-1) y = y x^(n-1)" } { "Tag": [], "Problem": "Would taking this class only look frivolous on a college application? Would it look the same if I took a class like this somewhere else? This is one of the subjects I want to study over the summer, and I wouldn't mind getting credit for it at my high school.", "Solution_1": "You are talking about [url=http://epgy.stanford.edu/courses/english/EG20/]EG20[/url], right? That's a good course. It wouldn't look silly on a high school transcript, and the knowledge is useful.", "Solution_2": "[quote=\"tokenadult\"]You are talking about [url=http://epgy.stanford.edu/courses/english/EG20/]EG20[/url], right?[/quote]\r\nYep, that's it. Thanks." } { "Tag": [ "calculus", "Euler", "logarithms", "limit" ], "Problem": "Hello, I have found an approximation to $\\sum_{k=1}^n\\frac{1}{k}$. Would you say that this is something good. If it is, does it worth being reviewed by some people? I'm not sure about such things. By the way, I also have an algebraic proof for it, including some calculus. Woudl you please assist me? Thanks in advance.\r\n\r\nMasoud Zargar", "Solution_1": "How good\u00bf\r\nSomething near to best possible was given in http://www.mathlinks.ro/Forum/viewtopic.php?t=46075 ...", "Solution_2": "well, I do not know much calculus; I have never studied it in depth and neither taken a course, but my formula gives it to an accuracy of approximately $\\pm 0.06$, and as $n\\rightarrow\\infty$, the accuracy increases. The formula I have found is obvious that it can be improved. I'll post it here later.", "Solution_3": "[url=http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html]Euler-Maclaurin[/url] might be helpful.\r\n$\\sum_{i=1}^{n}\\frac{1}{i}\\sim \\ln n+\\gamma+\\frac{1}{2n}-\\sum_{k=1}^{\\infty}\\frac{B_{2k}}{2k}\\frac{1}{n^{2k}}$. Please notice that this is not an equality. What it really means is $\\lim_{n \\to \\infty}n^{2m}\\left[\\sum_{i=1}^{n}\\frac{1}{i}-\\left(\\ln n+\\gamma+\\frac{1}{2n}-\\sum_{k=1}^{m-1}\\frac{B_{2k}}{2k}\\frac{1}{n^{2k}}\\right)\\right]=-\\frac{B_{2m}}{2m}$" } { "Tag": [ "geometry", "MIT", "college", "complex analysis" ], "Problem": "Suppose your college does not offer a particular math course that interests you. Are there some good sources for correspondence or online courses in advanced math?\r\n\r\nAs an example, where might one find such a course in Game Theory? Or Analysis of Time Series?\r\n\r\nThank you.", "Solution_1": "Just because a course isn't in the catalog doesn't mean that you can't take it---many college faculty have areas of expertise not reflected in any course offered in their departments. So, I have two suggestions. (1) Ask around in the department to see if anyone is able and willing to offer an independent study in the topics you want to learn. (2) Check out MIT's Open Courseware and see if the topics you want are covered there.", "Solution_2": "you may also want to consider some sort of study abroad (one example is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68721]BSM[/url])", "Solution_3": "Hi,\r\n\r\nThank you for the responses to my original question, but none of them is quite what I was hoping for.\r\n\r\nI am looking for a place that gives correspondence or online courses in advanced math.\r\n\r\nI did check the MIT site, and that is close, but not really an active course with an instructor who will answer questions.\r\n\r\nAny more ideas?\r\n\r\nThank you.", "Solution_4": "There might be a very small number of colleges that offer math classes beyond calculus/linear algebra on-line. The only program that I know of is Stanford's EPGY [url]http://www-epgy.stanford.edu/courses/math/[/url]. They offer several courses that are common to many college-level math programs. Nothing real esoteric though - partial differential equations and complex analysis are the most advanced courses. High-school graduates (older than 18) might not be able to take the EPGY university-level courses - I'm not that familiar with the program's admission requirements.\r\n\r\nThere's probably not much demand for on-line advanced math courses among students already attending colleges. There's not much demand for the traditional-delivery of these courses on most campuses either. I think it would be tough for most colleges to find a critical mass of quality students to take these kinds of classes on-line." } { "Tag": [], "Problem": "Identify the units on the variable x \r\n\r\nx=1/2 at^2 (at^2 is atfter the whole 1/2 part, not on bottom or top) \r\n\r\na is measured in meters/sec^2 and t is measured in seconds", "Solution_1": "$x=\\frac12 a t^2$\r\nNewton's second law with initial velocity $v_0=0$\r\nThe unit $=(m)(s)^2\\cdot (s)^2=m$", "Solution_2": "[quote=\"shobber\"]$x=\\frac12 a t^2$\nNewton's second law with initial velocity $v_0=0$\nThe unit $=(m)(s)^2\\cdot (s)^2=m$[/quote]\r\nit's $\\frac{m}{s^2}*s^2$ :lol:" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Suppose $ f$ is absolutely continuous on $ (a,b)$ and that $ f'(x) \\equal{} 0$ almost everywhere (in the sense of Lebesgue). Then $ f$ is constant.\r\none proof is in Royden's book.\r\nbut\r\nOne approach is to define a signed measure on $ (a,b)$ by\r\n$ \\mu(c,d) \\equal{} f(d) \\minus{} f(c)$ , and apply the Radon-Nikodym theorem.\r\nbut how is this done?\r\ncan anyone write something in more detalis?\r\nThanks.", "Solution_1": "Try page 8 in [url=http://www.emis.de/journals/DM/v81/art9.pdf]this[/url] link. Is this what you're hoping for?\r\n\r\nThe proof I'm familiar with uses Vitali coverings." } { "Tag": [ "logarithms", "calculus", "derivative", "calculus computations" ], "Problem": "Find an expression for $\\frac{dy}{dx}$ if:\r\n$y^{2}=x^{x}$", "Solution_1": "Taking natural logarithm of both sides and use the differentiation of implicit function.", "Solution_2": "Implicit differentiation: \\[y^{2}=x^{x}\\Longrightarrow 2y\\frac{dy}{dx}= \\frac{d}{dx}\\exp(x\\log x)\\\\ \\Longrightarrow \\frac{dy}{dx}= 1/2 x^{ x/2}\\left( \\ln \\left( x \\right)+1 \\right)\\]", "Solution_3": "[hide=\"My Solution\"]\nTake logs of both sides:\n$2lny=xlnx$\n\n$\\frac{2}{y}\\frac{dy}{dx}=1+lnx$\n\n$\\frac{dy}{dx}=\\frac{y(1+lnx)}{2}$\n[/hide]", "Solution_4": "Ottem, your answer is correct.\r\ncoffeym, $\\ln y^{2}=\\ln |y|^{2}=2\\ln |y|.$" } { "Tag": [ "geometry", "3D geometry", "invariant" ], "Problem": "Let $n \\geq 3$ be a positive integer. Prove that the sum of the cubes of all natural numbers, coprime and less than $n$, is divisible by $n$.", "Solution_1": "[hide=\"Solution\"] We wish to show that the sum of the cubes of the members of $U_{n}$ is zero. \n\nThis is rather simple. The additive inverse of the cube of a member of $U_{n}$ is the cube of its additive inverse (edit: in other words, $(a+b) | (a^{3}+b^{3})$), so we add each number to its additive inverse (invariant under cubing) for a sum of zero. \n\nFor $n \\ge 3$ we must have $2 | \\varphi(n)$ and so such a pairing always exists. (Or to be more explicit: For $n \\ge 3$ we have $\\frac{n}{2}\\not \\in U_{n}$, so every member of $U_{n}$ has an additive inverse different from itself.) [/hide]" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "This is from the Dummit and Foote section on conjugacy and the class equation:\r\nLet $ G$ be a finite group and choose a system of representatives $ g_1, \\cdots, g_r$ for the conjugacy classes of $ G$. Show that if the $ g_i$ pairwise commute then $ G$ is abelian.", "Solution_1": "[quote=\"JustSee12\"]This is from the Dummit and Foote section on conjugacy and the class equation:\nLet $ G$ be a finite group and choose a system of representatives $ g_1, \\cdots, g_r$ for the conjugacy classes of $ G$. Show that if the $ g_i$ pairwise commute then $ G$ is abelian.[/quote]\r\n\r\nIf $ G$ is non-abelian, then $ |G|>r$ and, without loss of generality, we may assume that $ Z(G)\\equal{}\\{g_1, \\cdots, g_s \\},$ for some $ 1 \\leq s < r.$ Since the $ g_i$ pairwise commute, we have $ [G: C(g_j)] \\leq \\frac{|G|}{r},$ \r\n\r\nfor all $ j.$ But then $ |G|\\equal{}s\\plus{}\\sum_{j\\equal{}s\\plus{}1}^r[G: C(g_j)] \\leq s\\plus{}\\frac{r\\minus{}s}{r}|G|,$ which gives us the contradiction $ |G| \\leq r.$" } { "Tag": [ "vector", "geometry", "circumcircle", "analytic geometry" ], "Problem": "[i]Let $z_{1},z_{2},z_{3}\\in\\mathbb{C}$ such that $|z_{1}|=|z_{2}|=|z_{3}|=R$ and $z_{2}\\neq z_{3}$. Prove that \\[\\min_{a\\in\\mathbb{R}}|az_{2}+(1-a)z_{3}-z_{1}|=\\frac{1}{2R}|z_{1}-z_{2}|\\cdot |z_{1}-z_{3}|.\\] [/i]", "Solution_1": "[hide]\nLet $z_{1}, z_{2}, z_{3}$ be three vectors in complex plane that start at the origin. Connect the three endpoints we get a triangle with circumcenter at the origin. Let the end point of $z_{1}$ be $A$, $z_{2}$ be $B$, and $z_{3}$ be $C$, then by simple analytic geometry (and simple algebra arrangement) the given equation is the same as:\n\n$\\min|k(z_{2}-z_{3})+z_{3}-z_{1}|=\\frac{bc}{2R}$\n\n(for notation sake, I change the $a$ in the original equation to $k$, because $a$ will stand for a side of my triangle).\n\nNow look at the expression in the min. We see that $z_{3}-z_{1}$ is a vector parallel to and the same length as the side $b$, and $z_{2}-z_{3}$ is parallel and the same length as the side $a$. using tip to tail vector addition, we see that the minimum expression becomes the minimum length from $A$ to $BC$, which is the length of the altitude from $A$ to $BC$ by simple geometry.\n\nSo, we wish to prove that $h_{A}= \\frac{bc}{2R}$. Recall that $\\frac{ah_{A}}{2}= \\triangle$, and also $\\triangle = \\frac{abc}{4R}$, equate and solve we see that the equation indeed is true.\n\nIf $z_{1}$ equals to either $z_{2}$ or $z_{3}$, plug in $k=1$ we get both sides to be zero, so the equation holds all the time.\n[/hide]", "Solution_2": "[quote=\"pkerichang\"][hide]\nLet $z_{1}, z_{2}, z_{3}$ be three vectors in complex plane that start at the origin. Connect the three endpoints we get a triangle with circumcenter at the origin. Let the end point of $z_{1}$ be $A$, $z_{2}$ be $B$, and $z_{3}$ be $C$, then by simple analytic geometry (and simple algebra arrangement) the given equation is the same as:\n\n$\\min|k(z_{2}-z_{3})+z_{3}-z_{1}|=\\frac{bc}{2R}$\n\n(for notation sake, I change the $a$ in the original equation to $k$, because $a$ will stand for a side of my triangle).\n\nNow look at the expression in the min. We see that $z_{3}-z_{1}$ is a vector parallel to and the same length as the side $b$, and $z_{2}-z_{3}$ is parallel and the same length as the side $a$. using tip to tail vector addition, we see that the minimum expression becomes the minimum length from $A$ to $BC$, which is the length of the altitude from $A$ to $BC$ by simple geometry.\n\nSo, we wish to prove that $h_{A}= \\frac{bc}{2R}$. Recall that $\\frac{ah_{A}}{2}= \\triangle$, and also $\\triangle = \\frac{abc}{4R}$, equate and solve we see that the equation indeed is true.\n\nIf $z_{1}$ equals to either $z_{2}$ or $z_{3}$, plug in $k=1$ we get both sides to be zero, so the equation holds all the time.\n[/hide][/quote]\r\nCongratulations on writing something about vectors that I understand. Your proof was very clear and intuitive.", "Solution_3": "[quote=\"PenguinIntegral\"][quote=\"pkerichang\"][hide]\nLet $z_{1}, z_{2}, z_{3}$ be three vectors in complex plane that start at the origin. Connect the three endpoints we get a triangle with circumcenter at the origin. Let the end point of $z_{1}$ be $A$, $z_{2}$ be $B$, and $z_{3}$ be $C$, then by simple analytic geometry (and simple algebra arrangement) the given equation is the same as:\n\n$\\min|k(z_{2}-z_{3})+z_{3}-z_{1}|=\\frac{bc}{2R}$\n\n(for notation sake, I change the $a$ in the original equation to $k$, because $a$ will stand for a side of my triangle).\n\nNow look at the expression in the min. We see that $z_{3}-z_{1}$ is a vector parallel to and the same length as the side $b$, and $z_{2}-z_{3}$ is parallel and the same length as the side $a$. using tip to tail vector addition, we see that the minimum expression becomes the minimum length from $A$ to $BC$, which is the length of the altitude from $A$ to $BC$ by simple geometry.\n\nSo, we wish to prove that $h_{A}= \\frac{bc}{2R}$. Recall that $\\frac{ah_{A}}{2}= \\triangle$, and also $\\triangle = \\frac{abc}{4R}$, equate and solve we see that the equation indeed is true.\n\nIf $z_{1}$ equals to either $z_{2}$ or $z_{3}$, plug in $k=1$ we get both sides to be zero, so the equation holds all the time.\n[/hide][/quote]\nCongratulations on writing something about vectors that I understand. Your proof was very clear and intuitive.[/quote]\r\n\r\nThanks. The first time I saw this one I feel \"There must be some geometry in this or else I'm not going to bother with all the nasty algebra stuff\", and it comes out nice :)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let \\[a, b, c, d\\] be positive numbers satisfying equations: \\[a^{3}+b^{3}+c^{3}= 3d^{3}\\] \\[b^{4}+c^{4}+d^{4}= 3a^{4}\\] \\[c^{5}+d^{5}+a^{5}= 3b^{5}\\] Proof that \\[a = b = c = d\\]", "Solution_1": "Using Am-Gm:\r\n$b\\geq a\\geq c\\geq d$\r\n(add $d^{3}, a^{4}, b^{5}$ to each side of each equation repectively then powermean)\r\nwe now need to prove $d\\geq b$ to maintain equality\r\nbut from first eq. $3d^{3}\\geq 3c^{3}, c\\geq d \\implies c=d \\implies 2d^{3}\\geq 2a^{3}\\implies d=a \\implies d=b=a=c$", "Solution_2": "Does anyone look at these problems at all? I think my solution might be wrong but right idea.\r\n\r\nWhat do you think, gollywog?" } { "Tag": [ "Euler", "geometry", "incenter" ], "Problem": "Daca in triunghiul $\\ ABC$ avem $\\ m(\\angle ACB)=120$, demonstrati ca centrul cercului inscris triunghiului $\\ ABC$ este situat pe [i]Dreapta lui Euler[/i] $\\triangle ABC$.", "Solution_1": "Cine poate da o solutie?", "Solution_2": "Problema e gre\u015fit\u0103, deci e un pic cam greu de rezolvat. Vezi urmatoarele doua linkuri pentru varianta corecta: [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=277944&highlight=incenter+line+euler#277944]Link 1[/url] \u015fi [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=445562&highlight=incenter+line+euler#445562]Link 2[/url].", "Solution_3": "perfect_radio, eu asa am vazut-o scrisa intr-o cartea a domnului Cristinel Mortici.\r\n[hide=\"post scriptum\"]Am inteles ce ziceau oamenii aia...in linkurile date de tine.Cat de cat.. :) [/hide]" } { "Tag": [ "function", "LaTeX", "number theory unsolved", "number theory" ], "Problem": "Find the minimum of natural number $ n$ such that there exist a function which is a nonconstant function $ f: Z\\mapsto R_{+}\\cap\\{0\\}$ for $ x,y\\in{Z}$,we have:\r\n a) $ f(xy)=f(x)f(y)$\r\n b)$ 2f(x^{2}+y^{2})-f(x)-f(y)\\in\\{0,1,2,...,n\\}$ \r\nAnd for this above value $ n$, find all which function $ f(x).$\r\n\r\nEdited by N.T.TUAN\r\n[code]Find the minimum of natural number $n$ such that there exist a function which is a nonconstant function $f: Z\\mapsto R_{+}\\cap\\{0\\} for $x,y\\in{Z}$,we have:\n a) $f(xy)=f(x)f(y)$\n b)$2f(x^{2}+y^{2})-f(x)-f(y)\\in\\{0,1,2,...,n\\}$ \nAnd for this above value $n$, find all which function $f(x).$[/code]", "Solution_1": "[quote=\"Nbach\"]Find the minimum of natural number $ n$ such that there exist a function which is a nonconstant function $ f: Z\\mapsto R_{+}\\cap{0}$ for $ x,y\\in{Z}$,we have: \n\na) $ f(xy)=f(x)f(y)$ \nb) $ 2f(x^{2}+y^{2})-f(x)-f(y)\\in{0,1,2,...,n}$\n\nAnd for this above value $ n$,find all which function $ f(x)$.[/quote]\r\n\r\nTip: Regular text looks horrible in $ \\text{\\LaTeX}$.", "Solution_2": "[quote=\"Nbach\"]Find the minimum of natural number $ n$ such that there exist a function which is a nonconstant function $ f: Z\\mapsto R_{+}\\cap\\{0\\}$ for $ x,y\\in{Z}$ [...][/quote]\r\n$ \\mathbb{R}_{+}\\cap \\{0\\}\\subseteq \\{0\\}$, so what!? :huh: And what's then $ \\mathbb{R}_{+}$? I would guess the set of all positive real numbers, but I feel somewhere there's something absolutely wrong...", "Solution_3": "suppose that $ f: \\ Z\\to B$ :D \r\n\r\n$ f$ nonconstant gives $ \\exists a: \\ f(a)\\neq f(0)$\r\nso $ f(0)f(a)=f(0)=f(0)f(0)$ gives $ f(0)=0$\r\nand $ y=0$ gives $ 2f(x^{2})-f(x)-f(0)=f(x)(2f(x)-1)\\in\\{0,1,2,...,n\\}$\r\nthen $ \\forall x\\in Z: \\ f(x)\\in H=\\{\\frac{1\\pm \\sqrt{8a+1}}{4}|\\ a\\in \\{0,1,2,...,n\\}\\}$\r\nwe take $ n_{min}=a$\r\nthen for some $ x\\in Z$: $ f(x)=\\frac{1\\pm \\sqrt{8a+1}}{4}$ and we have $ (f(x))^{2}=f(x^{2})\\in H$ so we must have $ \\frac{(1\\pm \\sqrt{8a+1})^{2}}{16}=\\frac{1\\pm \\sqrt{8b+1}}{4}$ for some $ b\\in \\{0,1,2,...,a\\}$\r\n :arrow: $ 8a-2\\pm 2\\sqrt{8a+1}=\\pm \\sqrt{8b+1}$ \r\n\r\nit's nothing, but it's can gives an idea for some one :wink:", "Solution_4": "Trust me, I'm still wondering how to interpret the problem. Maybe it should be $ \\mathbb{R}_{+}\\!\\cup \\{0\\}$, and not $ \\mathbb{R}_{+}\\!\\cap \\{0\\}$, I don't know... :maybe: Anyway, let me say your $ B$ is fantastic! :D", "Solution_5": "OK,I'm so sorry.It's the first time I use LaTeX...\r\nIt's $ R_{+}\\cup{0}$.", "Solution_6": "[quote=\"Nbach\"]Find the minimum of natural number $ n$ such that there exist a function which is a nonconstant function $ f: \\mathbb{Z}\\mapsto \\mathbb{R}_{+}\\cup\\{0\\}$ for $ x,y \\in \\mathbb{Z}$, we have:\n a) $ f(xy)=f(x)f(y)$\n b) $ 2f(x^{2}+y^{2})-f(x)-f(y)\\in\\{0,1,2,...,n\\}$[/quote]\r\nWell, I'm proving the minimum is attained for $ n = 1$. Indeed, since $ f$ is multiplicative, we need $ f(0) = f(0)^{2}$ and $ f(1) = f(1)^{2}$. Hence $ f(0), f(1) \\in \\{0, 1\\}$. Suppose $ f(0) = 1$. Then $ 1 = f(x \\cdot 0) = f(x) \\cdot f(0) = f(x)$, for any $ x \\in \\mathbb{Z}$, i.e. $ f$ is constant. So necessarily $ f(0) = 0$. In a similar way, admit $ f(1) = 0$. Then $ f(x) = f(1 \\cdot x) = f(1) \\cdot f(x) = 0$, for any $ x \\in \\mathbb{Z}$, viz $ f$ is constant again. So a fortiori $ f(1) = 1$. Then $ 2 f(1^{2}+0^{2})-f(1)-f(0) = 1$, yielding $ n \\ge 1$. Now let\r\n\\[ g(x) = \\left\\{ \\begin{array}{ll}0 & \\mbox{ if }x = 0 \\\\ 1 & \\mbox{ if }x \\in \\mathbb{Z}, x \\ne 0 \\end{array}\\right.. \\]\r\nThen clearly $ g$ is a multiplicative function $ \\mathbb{Z}\\to \\mathbb{R}_{+}\\! \\cup \\{0\\}$ and $ 0 \\le 2g(x^{2}+y^{2})-g(x)-g(y) \\le 1$, for any $ x, y \\in \\mathbb{Z}$. Thus $ n = 1$ and the claim is proved. :wink: So we are left with sieving for any other function $ f: \\mathbb{Z}\\mapsto \\mathbb{R}_{+}\\!\\cup\\{0\\}$, apart from $ g$, such that $ f(xy) = f(x) f(y)$ and $ 0 \\le 2f(x^{2}+y^{2})-f(x)-f(y) \\le 1$, whenever $ x, y \\in \\mathbb{Z}$. First, note that $ f$ is uniquely determined by its values on the set $ \\mathbb{P}= \\{p \\in \\mathbb{N}: p \\mbox{ is prime}\\}$, because of its multiplicativity (here $ \\mathbb{N}= \\{1, 2, \\ldots\\}$).\r\n\r\nSecond, by way of contradiction, take for granted there is $ x_{0}\\in \\mathbb{Z}$ such that $ f(x_{0}) \\not \\in \\{0, 1\\}$. Since $ f(x_{0}^{n}) = f(x_{0})^{n}$, for any $ n \\in \\mathbb{N}= \\{1, 2, \\ldots\\}$, then $ 0 \\le f(x_{0})^{2n}-f(x_{0})^{n}= f(x_{0}^{2n}+0^{2})-f(x_{0}^{n})-f(0) \\le 1$, i.e. $ 0 \\le f(x_{0})^{n}\\cdot (2f(x_{0})^{n}-1)) \\le 1$. Anyway this is absurd, as either $ f(x_{0})^{n}\\to 0^{+}$ or $ f(x_{0})^{n}\\to+\\infty$, as $ n \\to \\infty$ (recall that $ f(\\mathbb{Z}) \\subseteq \\mathbb{R}_{+}\\!\\cup \\{0\\}$ and hence $ f(x_{0}) > 0$). Thus in fact $ f(\\mathbb{Z}) = \\{0, 1\\}$. Now, $ 1 = f(1) = f(-1)^{2}$. Therefore $ f(-1) = 1$ and $ f(x) = f(-1) \\cdot f(-x) = f(-x)$, for any $ x \\in \\mathbb{Z}$, i.e. $ f$ has to be an even function.\r\n\r\nAt this point, admit $ f(2) = 0$. Then $ 2f(1^{2}+1^{2})-f(1)-f(1) = 2f(2)-2 =-2$. But actually from (b): $ 0 \\le 2f(1^{2}+1^{2})-f(1)-f(1) \\le 1$. Hence necessarily $ f(2) = 1$. Now consider that\r\n\\[ \\forall x, y \\in \\mathbb{Z}: \\begin{array}{ll}\\left\\{ \\begin{array}{l}2f(x^{2}+y^{2})-f(x)-f(y) \\ge 0 \\\\ f(x^{2}+y^{2}) = 0 \\end{array}\\right.\\quad\\Longrightarrow\\quad f(x) = f(y) = 0 & \\quad\\qquad (1) \\\\ & \\\\ \\left\\{\\begin{array}{l}2f(x^{2}+y^{2})-f(x)-f(y) \\le 1 \\\\ f(x) = f(y) = 0 \\end{array}\\right.\\quad\\Longrightarrow\\quad f(x^{2}+y^{2}) = 0 & \\quad\\qquad (2) \\end{array}\\]\r\nand set $ \\mathcal{S}= \\{n \\in 2\\mathbb{N}+1: \\exists^{no}a, b \\in \\mathbb{N}_{\\;\\!0}\\mbox{ s.t. }\\!\\gcd(a,b) = 1\\;\\wedge\\;n = a^{2}+b^{2}\\}$ and $ \\mathcal{D}= \\{n \\in \\mathcal{S}: f(n) = 0\\}$ (here as usual, $ \\mathbb{N}_{\\;\\!0}= \\mathbb{N}\\cup \\{0\\}$). Note $ 1 \\not\\in \\mathcal{D}$, though $ 1 \\in \\mathcal{S}$. By way of contradiction, suppose $ \\mathcal{D}\\ne \\emptyset$ and let $ n_{0}= \\min(\\mathcal{D})$ (recall that $ \\mathbb{N}$ is well-ordered). Then there must exist $ a, b \\in \\mathbb{N}$ such that $ n_{0}= a^{2}+b^{2}$ and $ \\gcd(a,b) = 1$. Thus $ f(a) = f(b) = 0$ from (1). Moreover, as $ n_{0}$ is odd, we lose no generality assuming $ a = 2 \\alpha$, for some $ \\alpha \\in \\mathbb{N}$. \r\n\r\nThen $ f(\\alpha) = f(2) f(\\alpha) = f(a) = 0$ and $ f(\\alpha^{2}+b^{2}) = 0$ from (2). Letting $ \\alpha^{2}+b^{2}= 2^{k}u$, where $ u \\in 2\\mathbb{N}+1$ and $ k \\in \\mathbb{N}_{0}$, it follows $ f(u) = 0$. Besides, $ u$ is representable by the form $ c^{2}+d^{\\;\\!2}$, for suitable $ c, d \\in \\mathbb{N}$. In fact, $ u \\mid (\\alpha^{2}+b^{2})$ and $ \\gcd(\\alpha, b) = 1$. Hence there exists no prime $ p \\in \\mathbb{N}$ such that $ p \\equiv 3 \\bmod 4$ and $ p \\mid u$, which is enough to grant that $ u$ and any of its divisors $ w \\in \\mathbb{N}$ can be expressed as a sum of squares of positive integers. In particular, this is the case when\r\n\\[ w = \\left(\\frac{c}{\\gcd(c,d)}\\right)^{\\!2}+\\left(\\frac{d}{\\gcd(c,d)}\\right)^{\\!2}, \\]\r\nwhether $ c, d \\in \\mathbb{N}$ and $ u = c^{2}+d^{\\;\\!2}$. But actually we got an absurd, since then $ w \\in \\mathcal{D}$ and $ w < n_{0}= \\min(\\mathcal{D})$ in the same time. So necessarily $ f(n) = 1$, whenever $ n \\in \\mathcal{S}$. In particular, $ f(p) = 1$, if $ p \\in \\mathbb{N}$ is a prime $ \\equiv 1 \\bmod 4$.\r\n\r\nIndeed, we're near to the end, don't give up! :rotfl: Well, note there can't exist two distinct primes $ p, q \\in \\mathbb{N}$ such that $ p \\equiv q \\equiv 3 \\bmod 4$ and $ f(p) = f(q) = 0$. Otherwise, it would be $ f(p^{2}+q^{2}) = 0$ on the one hand [according to (1)] and $ f(p^{2}+q^{2}) = 1$ on the other [since apparently $ (p^{2}+q^{2})$ has a divisor $ w \\in \\mathcal{S}$ - guess what? :wink: ], absurd! So either $ f \\equiv g$ or there is one only prime $ q \\in \\mathbb{N}$, with $ q \\equiv 3 \\bmod 4$, such that $ f(q) = 0$.\r\n\r\nIndeed, we want to prove the last posit is consistent. In fact, it's enough to prove it doesn't clash [btw, is \"clash\" the right verb to use in this context? :maybe: Thanks!] with b). So fix $ x, y \\in \\mathbb{Z}$ and let $ h(x,y) = 2f(x^{2}+y^{2})-f(x)-f(y)$ for notational convenience. If $ q \\mid (x^{2}+y^{2})$, then necessarily $ q \\mid \\gcd(x,y)$ (it's a well-known fact!) and trivially $ t(x,y) = 0$. Hence suppose $ q \\nmid (x^{2}+y^{2})$. Then again $ t(x,y) = 0$, whether $ q \\nmid xy$, and $ t(x,y) = 1$, if $ q \\mid x$ or $ q \\mid y$ (all is quite easy to check, so I'm leaving out details). Thus everything is right and we're done. Nice problem! :)", "Solution_7": "Is it possible to know who's the author of the problem? Thanks! :)", "Solution_8": "Yes, it is possible. :wink:", "Solution_9": "OK.Gabriel Dospinescu is author of this above problem.", "Solution_10": "O.K., thank you so much. :) Btw, note I've just edited my first post to fix a minor detail." } { "Tag": [], "Problem": "The set A contains 10 distinct integers. The smallest of these integers is -6 and the largest is 5. The sum of the integers in set A is 1. How many possible set As are there?", "Solution_1": "well, the range for the remaining 8 is -5,-4,-3,-2,-1,0,1,2,3,4 or 10 numbers\r\n\r\nso, 10*9*8*7*6*5*4*3 /me is too lazy to do calculations.", "Solution_2": "Since they're distinct integers and two of the spots are already taken, we have 8 integers left to choose from between $ \\minus{}5$ and $ 4$ inclusive. There are $ 10$ integers there, and assuming that permutations of a set are the same as the set, then there are obviously $ \\binom{10}{8}\\equal{}45$ possible sets." } { "Tag": [ "trigonometry", "logarithms", "function", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Prove that\r\n\\[ \\sum_{i \\equal{} 1}^{n}\\frac {Cos(ix)}{i} \\geq{ \\minus{} 1}\r\n\\]", "Solution_1": "The sum can be computed exactly:\r\n\r\n$ \\sum_{k \\equal{} 1}^{\\infty}\\frac {\\cos kx}{k} \\equal{} \\minus{} \\text{Re}(\\log|1 \\minus{} e^{ix}|) \\equal{} \\minus{} \\ln\\left|2\\sin\\frac {x}2\\right|.$\r\n\r\nThe minimum occurs when $ x \\equal{} \\pi$ and the minimum value is $ \\minus{} \\ln2,$ which is greater than $ \\minus{} 1.$\r\n\r\nThis computation comes from a combination of the power series identity $ \\sum_{k \\equal{} 1}^{\\infty}\\frac {z^k}{k} \\equal{} \\minus{} \\log(1 \\minus{} z)$ for $ |z| < 1$ together with Abel's Theorem for extending the function to the boundary of the disk.", "Solution_2": "Erm, the question was about partial sums, not only about the full series ;)", "Solution_3": "Oops, sorry, misread that. Looking at it numerically, I certainly believe it is true, but I don't have the proof yet.", "Solution_4": "One simple way to approach the problem is to notice that $ \\frac 34\\plus{}\\cos x\\plus{}\\frac{\\cos2x}2\\plus{}\\dots\\plus{}\\frac{\\cos nx}n$ has almost convex Fourier coefficients, so if we move everything down by $ \\frac 1n(\\cos x\\plus{}\\cos 2x\\plus{}\\dots\\plus{}\\cos nx)$, we'll get at least $ \\minus{}\\frac 34$ for the resulting sum. Now if we could bound the sum of pure cosines by $ \\minus{}\\frac n4$ from below, we would be done. Unfortunately, it is not the case and we get only $ \\minus{}7/4$ this way if we use the trivial estimate $ \\minus{}n$. On the other hand, we are fine with this approach if $ \\pi>|x|>10/n$, say, so the problem really occurs only at a small neighborhood of the origin, where it is easy to deal with: the first $ cn$ terms give you something of order $ \\log n$ and the remaining terms sum to a constant at worst, so for large $ n$ we get what we need. All estimates are effective, so now one needs just to play with a computer a bit to finish the problem, but I hate this way. There should be much neater approach.", "Solution_5": "This inequality was proved by Young in 1913. I can give the exact reference if you want me to.", "Solution_6": "Please, give the reference...", "Solution_7": "You can read about this and similer inequalities in Mitrinivic, Pekaric, Fink, Classical and new inequalities in analysis, Kluwer, 1993 on page 615. The original reference is W.H.Young, On a certain series of Fourier, Proc.London Math.Soc. 2(11), 1913, 357-366." } { "Tag": [ "probability", "number theory", "relatively prime" ], "Problem": "Ten points in the plane are given, with no three collinear. Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$", "Solution_1": "[hide=\"Solution\"]\nThere are $\\binom{10}{2}=45$ segments.\n\nThere are $\\binom{45}{4}$ ways to choose four segments.\n\nThere are $\\binom{10}{3}$ ways to choose three points for a triangle. After we choose this triangle we have $45-3=42$ segments left to choose from.\n\nThus our answer is $\\frac{42\\binom{10}{3}}{\\binom{45}{4}}=\\frac{16}{473}.$ Our answer is thus $16+473=\\boxed{489}.$[/hide]", "Solution_2": "[hide]In order for a triangle to be formed, two of the segments must have endpoints on the same point, let it be $A$. Also, the two other endpoints, $B$ and $C$, must be connected. The fourth segment can be any other one.\n\nThe probability that a second segment $BC$ shares the same endpoint as the first one, $AB$, is $8/44=2/11$, since there are $\\binom{10}{2}=45$ total segments. The probability that a third segment has endpoints $B$ and $C$ is $1/43$. We don't care about the fourth segment; there are $42$ possibilities for it. So our answer is $42(2/11)(1/43)=84/473$.[/hide]\r\n\r\nWhat's wrong with this solution?", "Solution_3": "At a quick glance, you are overcounting (obviously). I think this is occuring because it doesn't seem that you are taking into account the overcount produced since you can start with AB, BC, or CA and still form triangle ABC.", "Solution_4": "So, divide by $3!$? It's still off, though... :(", "Solution_5": "atm, i cant see the problem with chess's solution, but joml, i believe you are over counting as well.\r\nyou took into account the fact that order doesnt matter for the first 3 segments, but what about the last one?\r\ni believe, you have to divide your answer by 4, so you end up with [hide=\"this\"]$\\frac{4}{473}$ so $m + n = \\boxed{477}$[/hide] am i wrong?", "Solution_6": "[quote=\"chess64\"][hide]In order for a triangle to be formed, two of the segments must have endpoints on the same point, let it be $A$. Also, the two other endpoints, $B$ and $C$, must be connected. The fourth segment can be any other one.\n\nThe probability that a second segment $BC$ shares the same endpoint as the first one, $AB$, is $8/44=2/11$, since there are $\\binom{10}{2}=45$ total segments. The probability that a third segment has endpoints $B$ and $C$ is $1/43$. We don't care about the fourth segment; there are $42$ possibilities for it. So our answer is $42(2/11)(1/43)=84/473$.[/hide]\n\nWhat's wrong with this solution?[/quote]\r\n\r\nThough there are 42 possibilities for the last part, the probability of getting a last part is 1 ($42/42$ since it doesn't matter). Also the $8/44$ should $16/44=4/11$ since it it can stem from either A or B. So your answer should be $4/11\\cdot1/43=4/473$, which is $477$ and still $4$ times off of what joml got... Hmmm...?", "Solution_7": "[quote=\"kops723\"]atm, i cant see the problem with chess's solution, but joml, i believe you are over counting as well.\nyou took into account the fact that order doesnt matter for the first 3 segments, but what about the last one?\ni believe, you have to divide your answer by 4, so you end up with [hide=\"this\"]$\\frac{4}{473}$ so $m + n = \\boxed{477}$[/hide] am i wrong?[/quote]\r\n\r\nYou are wrong :) Look at my solution again. I am not taking order into consideration when I chose the first three segments. I choose the whole triangle as a first step. I don't break the triangle down into three separate components. So my answer is valid...kalva agrees http://www.kalva.demon.co.uk/aime/aime99.html", "Solution_8": "[quote=\"bob123\"][quote=\"chess64\"][hide]In order for a triangle to be formed, two of the segments must have endpoints on the same point, let it be $A$. Also, the two other endpoints, $B$ and $C$, must be connected. The fourth segment can be any other one.\n\nThe probability that a second segment $BC$ shares the same endpoint as the first one, $AB$, is $8/44=2/11$, since there are $\\binom{10}{2}=45$ total segments. The probability that a third segment has endpoints $B$ and $C$ is $1/43$. We don't care about the fourth segment; there are $42$ possibilities for it. So our answer is $42(2/11)(1/43)=84/473$.[/hide]\n\nWhat's wrong with this solution?[/quote]\n\nThough there are 42 possibilities for the last part, the probability of getting a last part is 1 ($42/42$ since it doesn't matter). Also the $8/44$ should $16/44=4/11$ since it it can stem from either A or B. So your answer should be $4/11\\cdot1/43=4/473$, which is $477$ and still $4$ times off of what joml got... Hmmm...?[/quote]\r\n\r\nYeah, that's what I got before but then I changed it because it was wrong...but it's still wrong :stink:", "Solution_9": "Not so sure...\r\n\r\n[hide]\nCase 1: We form the triangle with the first 3 segments we choose.\nThe probability for this was already calculated by bob123: $\\frac{4}{473}$\n\nCase 2: We form the triangle only after the $4th$ segment is chosen.\n\nSubcase 1: First segment can be any segment, second segment we choose shares an endpoint with the first, third segment shares no endpoint with the other two, fourth segment completes triangle.\n\nProbability = $1\\times\\frac{16}{44}\\times\\frac{42}{43}\\times\\frac{1}{42}=\\frac{4}{473}$\n\nSubcase 2: First segment can be any segment, second segment does not share an endpoint with the first, third segment shares an endpoint with at least one of the two previous segments, fourth segment completes the triangle.\n\nThis is one of the parts i'm not sure about. We have another two subcases to deal with: the third segment shares an endpoint with exactly one of the previous two segments, or with both. In the case of the former (of which there are four possible segments), we then have 2 choices for the fourth segment whereas we only have 1 choice for the latter.\n\nProbability for the first case: $\\frac{44-16}{44}\\times\\frac{4}{43}\\times\\frac{2}{42}=\\frac{4}{11\\times43\\times3}$\n\nProbabilitiy for the second case: $\\frac{44-16}{44}\\times\\frac{24}{43}\\times\\frac{1}{42}=\\frac{4}{11\\times43}$\n\nBut when I add all these probabilities together, I get a different answer...[/hide]", "Solution_10": "[quote=\"joml88\"][quote=\"kops723\"]atm, i cant see the problem with chess's solution, but joml, i believe you are over counting as well.\nyou took into account the fact that order doesnt matter for the first 3 segments, but what about the last one?\ni believe, you have to divide your answer by 4, so you end up with [hide=\"this\"]$\\frac{4}{473}$ so $m + n = \\boxed{477}$[/hide] am i wrong?[/quote]\n\nYou are wrong :) Look at my solution again. I am not taking order into consideration when I chose the first three segments. I choose the whole triangle as a first step. I don't break the triangle down into three separate components. So my answer is valid...kalva agrees http://www.kalva.demon.co.uk/aime/aime99.html[/quote]\r\n\r\ni think you misunderstood my argument, but if you have a reference youre probably right, but i understand that you didnt take into account how you chose the segments of the triangle, but how come the way you did doesnt imply that you [i]have[/i] to choose the 3 segments of the triangle first and only [i]then[/i] can you choose the 4th segment, instead of being able to choose the 4th segment as any of the 4 picks", "Solution_11": "[quote=\"kops723\"][quote=\"joml88\"][quote=\"kops723\"]atm, i cant see the problem with chess's solution, but joml, i believe you are over counting as well.\nyou took into account the fact that order doesnt matter for the first 3 segments, but what about the last one?\ni believe, you have to divide your answer by 4, so you end up with [hide=\"this\"]$\\frac{4}{473}$ so $m + n = \\boxed{477}$[/hide] am i wrong?[/quote]\n\nYou are wrong :) Look at my solution again. I am not taking order into consideration when I chose the first three segments. I choose the whole triangle as a first step. I don't break the triangle down into three separate components. So my answer is valid...kalva agrees http://www.kalva.demon.co.uk/aime/aime99.html[/quote]\n\ni think you misunderstood my argument, but if you have a reference youre probably right, but i understand that you didnt take into account how you chose the segments of the triangle, but how come the way you did doesnt imply that you [i]have[/i] to choose the 3 segments of the triangle first and only [i]then[/i] can you choose the 4th segment, instead of being able to choose the 4th segment as any of the 4 picks[/quote]\r\n\r\nhuh? I think it does imply that. There are 42 choices for the last segment...does not that imply that the other 3 sides were already decided ;)", "Solution_12": "no, you still misunderstand me. let me give an example:\r\nby your method, clearly we can pick the 4 segments like this(i'll call the points p1, p2... p10):\r\np2p4, then p7p4, then p2p7, then p1p8\r\n\r\nbut it doesnt seem to me that your allowing for the fact that that is the same as:\r\np2p4, p1p8, p7p4, lastly p2p7... in which you are placing the abitrary segment pick in between two other picks", "Solution_13": "Order doesnt matter in this case.", "Solution_14": "Can someone provide a solution that calculates the probability directly, choosing one segment at a time?", "Solution_15": "[quote=\"kops723\"]no, you still misunderstand me. let me give an example:\nby your method, clearly we can pick the 4 segments like this(i'll call the points p1, p2... p10):\np2p4, then p7p4, then p2p7, then p1p8\n\nbut it doesnt seem to me that your allowing for the fact that that is the same as:\np2p4, p1p8, p7p4, lastly p2p7... in which you are placing the abitrary segment pick in between two other picks[/quote]\r\n\r\nPlease reread my solution carefully.\r\n\r\nWe want to find how many ways we can choose 4 segments such that 3 of them make a triangle. What I am doing is first choosing the triangle. Then I choose the fourth segment. By choosing the whole triangle I am in no way, shape, or form considering the order that the segments are chosen. Neither am I considering whether the triangle or extra line segment was chosen first (I would have to multiply my answer by 2 if that were the case).", "Solution_16": "Let's first pick the three points of the triangle. There are $\\binom{10}{3}$ ways to do this. Then, out of the $\\frac{10 \\cdot 9}{2} = 45$ segments there are $42$ of them left, so we multiply by $42$. The total amount of ways is $\\binom{45}{4}$. Dividing, we get a result of $\\frac{16}{473} \\implies \\boxed{489}$." } { "Tag": [ "function", "real analysis", "real analysis theorems" ], "Problem": "Does anyone know why harmonic functions are called \"harmonic\"? There should be some reason for this name but I couldn't think of any when asked this question two days ago... :oops:", "Solution_1": "[url=http://www.mathpages.com/home/kmath214/kmath214.htm]This webpage[/url] explains the name as follows: solving Laplace's equation by separating variables, one obtains the equation of harmonic motion. (The latter is probably named after harmonics in music.)\r\n\r\nWhy is the harmonic series called harmonic? I don't see how it connects either to harmonic functions or to harmonic oscillator.", "Solution_2": "Well, the harmonic series most certainly has to do with the music concept of harmonics. The overtones produced by a vibrating string (or any sound wave) are given by $1,\\frac{1}{2}, \\frac{1}{3}, ...$ of the fundamental wavelength. Thus producing the harmonic series (both mathematically and musically)." } { "Tag": [ "search", "combinatorics proposed", "combinatorics" ], "Problem": "Prove that an equilateral triangle could not decomposed to more than on equilateral triangles with different size.", "Solution_1": "it was discussed on forum do the search to find omid", "Solution_2": "Dear sam-n, I searched but I didn't find anything. (Maybe it because of my bad search). Still I suggest try this challenging problem." } { "Tag": [ "LaTeX" ], "Problem": "HI!\r\n\r\nFor the titlepage of my work I need one word to be very large.\r\nOne letter should be about 4cm high. I tried to get that done by using \\fontsize but\r\nit's not getting taller than aprox. 1.5 cm.\r\n\r\nWaht can I do?\r\n\r\nThanks\r\nBye", "Solution_1": "\\usepackage{scalefnt}\r\n\r\n...\r\n\r\n\r\n\\begingroup \\scalefont{1000} word\\endgroup will make the word as big as it can be. (I think) I've had the same problem..." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Prove that $\\sum_{n=1}^{\\infty}\\int_{\\frac{1}{n+1}}^{\\frac{1}{n}}{\\left|\\frac{1}{x}\\sin \\frac{\\pi}{x}\\right| dx}$ diverge for $x>0.$", "Solution_1": "Let $S$ be your sum. Then\r\n\r\n$S=\\sum_{n+1}^{\\infty}\\int_{n}^{n+1}\\frac{|\\sin(\\pi y)|}{y}dy=\\sum_{n+1}^{\\infty}\\int_{0}^{1}\\frac{|\\sin(\\pi z)|}{n+z}dz\\geq \\sum_{n=1}^{\\infty}\\int_{0}^{1}\\frac{|\\sin(\\pi z)|}{n+1}dz$\r\n\r\n$=\\sum_{n+1}^{\\infty}\\frac{1}{n+1}\\int_{0}^{1}\\sin \\pi z dz=\\sum_{n=1}^{\\infty}\\frac{2}{\\pi(n+1)}=\\infty$." } { "Tag": [ "MATHCOUNTS" ], "Problem": "what state is going to win Nationals this year at National Mathcounts?", "Solution_1": "north carolina?\r\n\r\nhow are you assuming that california will not win? or florida? or new york?\r\n\r\non the bright side a non-massachusetts resident spelled massachusetts right.\r\n\r\nbut we all know it's too early to predict these things well, although deep deep down we know massachusetts will obviously win.", "Solution_2": "Indiana isn't that good this year. People like me might make Nats", "Solution_3": "uhh...\r\n\r\nI think Indiana isn't that bad. Next year I think we have a legitimate shot at winning it all. This year it looks like I'm the only eighth grader [not a guarantee for people that will go OMG you assumed you'd make nats and bug me about it -_-], with an outside shot of jeffery shen and a more likely but still not >50% shot of joseph botros making it.\r\n\r\nI think our top 3 is very solid in terms of winnability, just our fourth might be our achilles heel. I'm very interested in finding out who he/she will be, or if one of the \"solid 3\" will make a slipup at state, which is possible. I'll probably screw up :rotfl:. More likely though is me getting 2nd -- I have a knack of always getting 1 worse than my goals -- mathcounts 2 years in a row [9th, 5th when i was aiming for 8th, 4th], history day[runner up, aiming for 2nd place who is the other person that makes nationals], and science fair [honorable mention aka 4th, goal was 3rd] -_-.\r\n\r\nand indianamath, don't turn into ubemaya 2.0 :P", "Solution_4": "well, i just said i MIGHT make it. at least i have a better shot than jeffrey.", "Solution_5": "By \"ubemaya 2.0\" i meant \"don't put yourself down too much\"... ubemaya is king of putting himself down. You said \"even people like me\".", "Solution_6": "MA has four people who were at nats last year, all eighth graders now. We have a decent shot.", "Solution_7": "[quote=\"perfect628\"]MA has four people who were at nats last year, all eighth graders now. We have a decent shot.[/quote]\r\n\r\nYou do not know if they can all get into the team or not.\r\n\r\nIt is too early to have this poll. You are able to do a good prediction ONLY after each team has been formed.\r\n\r\nMy rough prediction:\r\n\r\nCA, WA, MI, and NC have the best shots.", "Solution_8": "[quote=\"perfect628\"]MA has four people who were at nats last year, all eighth graders now. We have a decent shot.[/quote]\r\n\r\nUh, I thought I had a roommate from Massachusetts that was an 8th grader...I am not so sure that I was ignorant of the fact that he was a 7th grader the whole time.", "Solution_9": "I'm pretty sure it was 8-7-7-7.", "Solution_10": "Yes, but a kid who made it to nats last year from HI now lives in MA", "Solution_11": "owned\r\n\r\nseriously what is it with this north carolina crap? BFG what are you basing your predictions on?", "Solution_12": "[quote=\"BFG\"][quote=\"perfect628\"]MA has four people who were at nats last year, all eighth graders now. We have a decent shot.[/quote]\n\nYou do not know if they can all get into the team or not.\n\nIt is too early to have this poll. You are able to do a good prediction ONLY after each team has been formed.\n\nMy rough prediction:\n\nCA, WA, MI, and NC have the best shots.[/quote]\r\n\r\nWell, I know that, but we'll probably have at least 2-3.", "Solution_13": "What state is HI? :blush:\r\n\r\nHawaii???", "Solution_14": "Yes. His name is Dong something.", "Solution_15": "[quote=\"13375P34K43V312\"]\n\nBFG what are you basing your predictions on?\n\n[/quote]\r\n\r\nBased on your signature. I am truly a member of the Arnav appriciation club. You are the fake one. ARML.", "Solution_16": "[quote=\"BFG\"][quote=\"13375P34K43V312\"]\n\nBFG what are you basing your predictions on?\n\n[/quote]\n\nBased on your signature. I am truly a member of the Arnav appriciation club. You are the fake one. ARML.[/quote]\r\nJust because they are good at high school math doesn't mean they can do anything at all in MC.\r\nLook at NH (though that's mostly because of exeter I suppose)", "Solution_17": "im all with the whole\r\n\r\nits too early to know. i mean, we dont even know who will be on nats teams for sure.", "Solution_18": "I think it will be California because so many middle schoolers are there. I too got 5th in district mathcounts last year (4th to 5th by countdown) but I'm going to Florida state this year...any idea what 4th place cutoff is? (The thread for this topic only shows top 10 cutoff)", "Solution_19": "[quote=\"LawOfSigns\"]I think it will be California because so many middle schoolers are there. I too got 5th in district mathcounts last year (4th to 5th by countdown) but I'm going to Florida state this year...any idea what 4th place cutoff is? (The thread for this topic only shows top 10 cutoff)[/quote]\r\n\r\nwell, last year it was around 35. not too high. but this year you get to play vs bpms and ragnarok and myself.", "Solution_20": "no offense, but I think this topic is pointless.\r\n\r\n\r\nStates aren't even over. :D", "Solution_21": "States aren't even started.", "Solution_22": "[quote=\"perfect628\"]States aren't even started.[/quote]\r\n\r\ni thought states started in the beginning of march...so like a week ago?", "Solution_23": "[quote=\"Walk Around The River\"][quote=\"perfect628\"]States aren't even started.[/quote]\n\ni thought states started in the beginning of march...so like a week ago?[/quote]\r\nSo, a week from now?", "Solution_24": "[quote=\"bpms\"][quote=\"Walk Around The River\"][quote=\"perfect628\"]States aren't even started.[/quote]\n\ni thought states started in the beginning of march...so like a week ago?[/quote]\nSo, a week from now?[/quote]\r\n\r\nwhat do you mean? wasnt today like march 12th...?\r\n\r\nEDIT: okay, apparently its still february. i dont really pay attention to that stuff.", "Solution_25": "But how is someone 2 weeks off?", "Solution_26": "lmfao......march 12th :rotfl: :rotfl: :rotfl:", "Solution_27": "Dude, why is Illinois up there? Take it down, we suck. I only voted for it because I WISH we'll win...", "Solution_28": "[quote=\"Fanatic\"]\nMore likely though is me getting 2nd -- I have a knack of always getting 1 worse than my goals -- mathcounts 2 years in a row [9th, 5th when i was aiming for 8th, 4th], history day[runner up, aiming for 2nd place who is the other person that makes nationals], and science fair [honorable mention aka 4th, goal was 3rd] -_-.\n\n[/quote]\r\n\r\nAim for 0th.\r\n :lol:", "Solution_29": "Please don't spam :) \r\nThank you!\r\n\r\n[hide]\nSomeone's gonna say that I'm one to talk :D[/hide]\r\n-Jorian", "Solution_30": "Really?\r\nI always get my goal times about $\\frac{1}{2}$ rankwise.\r\n50-22\r\n10-3\r\n7-4\r\n57-27\r\n1-1\r\n1-2 (the only time I place lower than my goal)", "Solution_31": ":o u didn't put west virginia on ur poll list!! i feel insulted...", "Solution_32": "[quote=\"espark52\"]:o u didn't put west virginia on ur poll list!! i feel insulted...[/quote]\r\nLets look at who VA has.\r\nThey're only returnee got $\\frac{2}{3}$ of my score at nats.\r\nInyoung is a beast, but she can't carry the team alone.\r\n\r\nMaybe I'm missing someone, but I doubt it.\r\n\r\nFL should be on the poll, as well as NJ.\r\nMD and NC aren't going to win, I don't think.\r\nIL doesn't have any returnees, I doubt they're going to win.\r\nCA should definitely be on the poll, they have Albert Gu and Alexander Lin contending for the team. They're actually the favorites since they have no returnees this year.", "Solution_33": "YOU DIDN'T PUT NY ON THE LIST! I FEEL INSULTED!", "Solution_34": "neither did they with Connecticut, but that's not the point\r\n\r\nand by the way, they did, it's under the \"other\" section\r\n\r\nhas a team not from the states ever done particulary well?\r\n\r\n-jorian", "Solution_35": "VI is like so gonna win", "Solution_36": "This thread has degenerated to a venue for SPAM and dissing your competitors.\r\n\r\nThread locked." } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Evaluate $ \\int_{0}^{\\infty }\\frac {dx}{ ( x \\plus{} \\sqrt {1 \\plus{} x^2 } ) ^ n }$ .\r\n\r\n please help me", "Solution_1": "Note that we need $ n>1$ in order for the integral to converge, as the integrand looks like a constant times $ x^{\\minus{}n}$ for large $ x.$\r\n\r\nMake the substitution $ x\\equal{}\\sinh t,$ from which $ dx\\equal{}\\cosh t\\,dt.$\r\n\r\n$ \\int_0^{\\infty}\\frac1{(x\\plus{}\\sqrt{x^2\\plus{}1})^n}\\,dx\\equal{}\\int_0^{\\infty}\\frac{\\cosh t}{(\\sinh t\\plus{}\\cosh t)^n}\\,dt\\equal{}\\int_0^{\\infty}\\frac{\\frac12(e^t\\plus{}e^{\\minus{}t})}{e^{nt}}\\,dt$\r\n\r\n$ \\equal{}\\frac12\\int_0^{\\infty}e^{\\minus{}(n\\minus{}1)t}\\plus{}e^{\\minus{}(n\\plus{}1)t}\\,dt\\equal{}\\frac12\\left(\\frac1{n\\minus{}1}\\plus{}\\frac1{n\\plus{}1}\\right)\\equal{}\\frac{n}{n^2\\minus{}1}.$\r\n\r\nWe don't even need for $ n$ to be an integer for this to make sense - just that $ n>1.$", "Solution_2": "Put $ t \\equal{} x \\plus{} \\sqrt {x^2 \\plus{} 1}$ and solving for $ x$ we get $ x \\equal{} \\frac {1}{2}\\left( t \\minus{} \\frac {1}{t} \\right)$ and then $ \\frac {1}{2}\\int_{1}^{\\infty }{\\left( t^{ \\minus{} n} \\plus{} t^{ \\minus{} (n \\plus{} 2)} \\right)\\,dt} \\equal{} \\frac {n}{n^{2} \\minus{} 1}.$\r\n\r\nEssentially the same solution.", "Solution_3": "thankyou very much" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "last year i was given nice problem but i failed to solve now i remember it from dusty part of of my memory :) \r\nsuppose that we have $f:\\mathbb{Z}^2\\rightarrow \\mathbb{Z}^2$.such that $\\frac{|f(x)-f(y)|}{|x-y|}$ is bounded prove that there exists $r$ such that each circle of radius $r$ in the plane contain at least one point of $f(\\mathbb{Z}^2)$.", "Solution_1": "sam-n,first congratulation for the iran team in imo $4$th of the world.\r\nNow $f:\\mathbb{Z}^2 \\longrightarrow \\mathbb{Z}^2$\r\nSo what is the meaning of $f(x)-f(y)$, \r\nshouldnet it be $f(x,y)-f(y,x)$ :?: \r\n\r\nP.S well done to all of the iran contestantes\r\nMr.Hatami(By idea Omid Hatami) can you say who did the full mark? :D", "Solution_2": "yes that's the best result since 1999 and about your question $x$ and $y$ are points in $\\mathbb{Z}^2$.", "Solution_3": "How do you divide two elements of $ \\mathbb{Z}^2 $?", "Solution_4": "Sam_n , If by $f(x)-f(y)$ you means the distances between the $f(x)$ and $f(y)$ then i thing it is wrong :? \r\n\r\nFor every points in the plane like $A=(x,y)$ let the $f(A)=B=(0,0)$ and we have that $\\frac{f(x)-f(y)}{x-y}$is bounded (it is actually zero ) so if there exist a \r\n$r$ then get away from the $(0,0)$ more than $r$ so the circle wont have any point of $f(\\mathbb{Z}^2)$.\r\n\r\nAm i saying something stupid? :?", "Solution_5": "He means bounded both from above and away from zero. In other words, there are two positive constants $c_1,c_2>0$ s.t. $c_1<\\frac{|f(x)-f(y)|}{|x-y|}R satisfying f[x-f(x)]=x/2, for real x.", "Solution_1": "Here it is:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=absurd&t=29819[/url]" } { "Tag": [ "geometry", "geometric transformation", "reflection", "rectangle", "Pythagorean Theorem", "AMC" ], "Problem": "A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?\r\n\r\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=611[/img]\r\n\r\n$ \\textbf{(A) } 20\\pi \\qquad \\textbf{(B) } 25\\pi \\qquad \\textbf{(C) } 30\\pi \\qquad \\textbf{(D) } 40\\pi \\qquad \\textbf{(E) } 50\\pi$", "Solution_1": "[hide]Each side of the square has length $\\sqrt{40}=2\\sqrt{10}$. Draw a radius from the midpoint of the side along the diameter (the center of the circle) to one of the points where the square touches the circle. By Pythag. Theorem we find the square of the radius is 50, so the circle has area 50*pi, so the semicircle has area 25*pi. B.[/hide]", "Solution_2": "[hide]\nIf the area of the square is $40$, then the length of a side is $\\sqrt{40}$. We draw an auxiliary line segment from the center of the semicircle to a vertex of the square, forming a right triangle. The center of the semicircle bisects the side of the square on the diameter, so one leg of this triangle has length $\\frac{\\sqrt{40}}{2}=\\sqrt{10}$. The other leg, obviously, has length $\\sqrt{40}$. By the Pythagorean Theorem, the length of the hypotenuse is $\\sqrt{10+40}=\\sqrt{50}$ so the area of the semicircle is $\\frac{50 \\pi}{2}=25 \\pi$.[/hide]", "Solution_3": "I just reflect the picture in the diameter. Then I have a rectangle with sides $a$ and $2a$ its diagonal is a diameter of a circle. By Pitagorean Theorem:\r\n$d^2=a^2+(2a)^2$. From now the answer is obvious", "Solution_4": "Wait why was this topic revisited after one month?", "Solution_5": "I have no idea. But it was the Problem of the Day on March 17, even though it was posted on February 17. :?" } { "Tag": [ "search", "LaTeX", "algebra", "polynomial", "linear algebra", "matrix", "combinatorics solved" ], "Problem": "Let $ G^{\\prime}$ be a $ 4$-regular graph. Arbitrarily add one edge to $ G^{\\prime}$, and denote the resulting graph by $ G$. Prove that the graph $ G$ has a $ 3$-regular subgraph.\n\n[i]Note.[/i] A graph is called $ k$-regular if each vertex of the graph has exactly $ k$ neighbours.", "Solution_1": "I think it is based on the number of edges & its parity.\r\nLet there be n vertex in the graph.\r\nThen initially there are 4*n / 2 = 2*n edges,in G\", and in G 2n + 1.(odd)\r\nAlso, in the new G, let A & B be the 2 vertexes that have the new edge between them. Counting the number of vertices & knowing that we don't have any triangles, for any 2 that know each other, except A & B, we have\r\n8 neighbours. Counting very carefully I think should give us the result :)))", "Solution_2": "hello DusT :) the problem is mor harder than you explain if you are sure that your idea works post it completely then i post the sol :D", "Solution_3": "Hi!\r\nSorry, I missunderstood the problem. I considered that we have to prove there is a triangle, not a 3regular graph. Sorry again :). I'll read more carefull next time :D", "Solution_4": "Hy!\r\nI give up. \r\nDoes anybody else think about it ?\r\nI have no starting idea.\r\nCan you at least give a hint?\r\nBtw, where can the Iranian Math Olympiad Problems be found?\r\nNowhere on the INET i think :)", "Solution_5": "well Dust, it's simple: search of the forum, after the source of problem only, and type Iran* in the search field. You will find a lot of problems from Iran right here on the forum.", "Solution_6": "I listened to your advice, Valentin.\r\nMost of the Iranian Problems are at the UNSOLVED section :) :) :D\r\nI was searching for somekind of training :)\r\nAnyway, what about the solution for this problem ?", "Solution_7": "So, when will we see the solution for this problem?\r\nSam-n ?", "Solution_8": "I 'm trying to use Latex it's boring so if there is any eror plz excuse me", "Solution_9": "now you can see the sol ;) there is useful theorem which its name is shouvale(I am not sure that its dictation is right so if it's wrong don't laugh at me :blush: :D )\n\nsho... theorem: consider a system of $n$ equations $f_j(x_1,...,x_m)\\equiv 0 (modp)$ s.t $j=1,2,...,n $ and $f_j$ is a polynomial with degree $r_j$ and $p$ is prime number and also $r_1+r_2+...+r_n 0$ and suppose $I=\\left\\{ e_i\\mid i \\in J \\right\\}$. now we have for any $i$ in $J$ the relation $x_i^2\\equiv 1$ (mod $3$) so $\\sum_{e_i\\in I} a_{i,j} \\equiv \\sum a_{i,j}x_i^2 \\equiv 0$.\n\nuh now we are done\nisn't it interesting?", "Solution_10": "Great solution, but about this \"Theorem\", could you please prove it or post the correct name, so I could seach it on the internet.\r\nThnx", "Solution_11": "Here is a little clarification about the \"Theorem\", DusT. Sam-n is talking about the famous Chevalley-Warning Theorem (now you can Google it :) .)\r\n\r\nBy the way, this problem is a special case with $p=3$ of a theorem of Alon:\r\n\r\nTheorem (Alon, 1999). Let $p$ be a prime and $G$ be a graph (without multiple edges) of average degree $> 2p-2$ and maximal degree $\\leq 2p-1$. Then, $G$ has a $p$-regular subgraph.\r\n\r\nIt has recently been proved that this assertion also holds for prime powers $p$, although it is an open problem whether this holds for all positive integers $p$.\r\n\r\nAlso, I cannot resist mentioning the following relevant (or is it? :lol: ) conjecture of Berge and Sauer proved by Taskinov:\r\n\r\nTheorem. (Taskinov). Every *simple* 4-regular graph has a 3-regular subgraph.\r\n\r\nThis is not true if we remove the assumption that the graph is simple (check it!), but one additional edge entails the existence of a 3-regular subgraph even in this more general case. \r\n\r\n--Vesselin", "Solution_12": "You may find a proof here :\r\nhttp://www.kenjioba.net/images/RegSubgraphs.pdf\r\n\r\nPierre.", "Solution_13": "Thanks a lot.\r\nThere is also another problem with a solution like that:\r\nIt is another post of another person from Iran. I think they know a lot of theorems. I wonder if they use this theorems in the olympiad.\r\nhttp://www.mathlinks.ro/viewtopic.php?t=5656\r\nHere is the link to the topic that uses another theorem I don't know.\r\nI would be glad if somebody would explain it to me.\r\nThanks.", "Solution_14": "I think the theorem you're referring to was discussed here:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=5976[/url]\r\n\r\nIt's equivalent to the problem Darij posted in the first message, and you can find at least $2$ proofs for that ($3$ if you consider mine a proof :))." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "n,a,b are natural number and $(a,b)=1$ prove that:\r\n$1/a + 1/a+b + .... +1/a+nb$ isn't natural.", "Solution_1": "I believe what you mean is $\\frac{1}{a} + \\frac{1}{a+b} + \\ldots + \\frac{1}{1+nb}$\r\n\r\nLet $\\prod_{i=0}^{n} (a+ib) = M$. Note that $\\frac{1}{a} + \\frac{1}{a+b} + \\ldots + \\frac{1}{1+nb} = \\frac {\\sum_{i=0}^n \\frac{M}{a+ib}}{M}$\r\n\r\nNote that the integers $a+b, a+2b, \\ldots, a + nb$ are each coprime to $a$, thus if $a>1$, then taking mod $p$ for some prime divisor $p$ of $a$, the numerator is $\\not\\equiv 0 \\pmod{p}$ (since for $i = 1, 2, \\ldots, n$, the integer $\\frac{M}{a+ib}$, being divisible by $a$, is $\\equiv 0 \\pmod{p}$, while $\\frac{M}{a}$ is $\\not\\equiv 0 \\pmod{p}$), while the denominator is $\\equiv 0 \\pmod{p}$.\r\n\r\nNow for the case $a=1$, if $b=1$, it then reduces to proving tt $1+ \\frac{1}{2} + \\ldots + \\frac{1}{n+1}$ is never an integer $\\forall$ positive integers $n$, which is a well known result (there are many methods of proving this, a simple way would be to consider powers of $2$ that divide the numerator & denominator). So assuming $b>1$, we note that $a+b = b+1$ is coprime to $a+ib = 1+ib$ for $i = 2, 3, 4, \\ldots, n$. By a similar process, we note that for some prime divisor $q$ of $b+1$, the numerator is $\\not\\equiv 0 \\pmod{q}$, while the denominator is $\\equiv 0 \\pmod{q}$.", "Solution_2": "[quote]Note that the integers $a+b, a+2b, \\ldots, a + nb$ are each coprime to $a$,[/quote]\r\nHow about $a+ab, a+2ab,\\dots$? :?", "Solution_3": "For the case of $\\frac{1}{a} + \\frac{1}{a+ab} + \\ldots + \\frac{1}{a+nab}$, by taking out the factor of $\\frac{1}{a}$, it reduces to the similar case of $1+ \\frac{1}{1+b} + \\frac{1}{1+2b} + \\ldots + \\frac{1}{1+nb}$.", "Solution_4": "[quote=\"dblues\"]Note that the integers $a+b, a+2b, \\ldots, a + nb$ are each coprime to $a$, thus if $a>1$, then taking mod $p$ for some prime divisor $p$ of $a$, the numerator is $\\not\\equiv 0 \\pmod{p}$ (since for $i = 1, 2, \\ldots, n$, the integer $\\frac{M}{a+ib}$, being divisible by $a$, is $\\equiv 0 \\pmod{p}$, while $\\frac{M}{a}$ is $\\not\\equiv 0 \\pmod{p}$), while the denominator is $\\equiv 0 \\pmod{p}$.[/quote]\r\nHe-he. It is completey wrong... ;)", "Solution_5": "[quote=\"grodij\"]n,a,b are natural number and $(a,b)=1$ prove that:\n$1/a + 1/a+b + .... +1/a+nb$ isn't natural.[/quote]\r\n\r\nI think this it's too easy: if $p$ is the greatest prime less than $n+1$, then exists one or two integers \r\n$1\\le k_1\\le n$, $k_2=k_1+p$ uch that $k_1n+1=0 \\ mod \\ p$ so in the case of one it's clear that the numerator is not divisible by $p$. And in the case of two is harder but an analogous idea shows that numerator is not divisible by $p^2$ (If someone has a nice idea to solve this second case, please write it).\r\n\r\nBest Regards.\r\nRX the Crazy Math", "Solution_6": "[quote=\"RobertuX\"]\n\nI think this it's too easy: if $p$ is the greatest prime less than $n+1$, then exists one or two integers \n$1\\le k_1\\le n$, $k_2=k_1+p$ uch that $k_1n+1=0 \\ mod \\ p$ so in the case of one it's clear that the numerator is not divisible by $p$. And in the case of two is harder but an analogous idea shows that numerator is not divisible by $p^2$ (If someone has a nice idea to solve this second case, please write it).\n\nBest Regards.\nRX the Crazy Math[/quote]\r\n\r\nPlease explain again. I don't get it, our denominators are of the form $ib+a$, u used $in+1$ up there, why?", "Solution_7": "[quote=\"ikap\"]\nPlease explain again. I don't get it, our denominators are of the form $ib+a$, u used $in+1$ up there, why?[/quote]\r\nBecause I simplify the expression original to $\\frac 1 a\\sum_{k=0}^n \\frac 1 {1+kb}$ so I don't care the value of $a$.", "Solution_8": "what problem are you talking about?\r\nu don't keep $b$ natural this way :?\r\n\r\nAnyway, I got the idea. Still hard to solve the second case.\r\nIn fact, the fact that $p^2$ doesn't divide the numerator reduces to $a+ib+a+(i+p)b\\neq 0 (\\mod p^2)$ which doesn't seem to be generally true.", "Solution_9": "[quote=\"Myth\"][quote=\"dblues\"]Note that the integers $a+b, a+2b, \\ldots, a + nb$ are each coprime to $a$, thus if $a>1$, then taking mod $p$ for some prime divisor $p$ of $a$, the numerator is $\\not\\equiv 0 \\pmod{p}$ (since for $i = 1, 2, \\ldots, n$, the integer $\\frac{M}{a+ib}$, being divisible by $a$, is $\\equiv 0 \\pmod{p}$, while $\\frac{M}{a}$ is $\\not\\equiv 0 \\pmod{p}$), while the denominator is $\\equiv 0 \\pmod{p}$.[/quote]\nHe-he. It is completey wrong... ;)[/quote] I don't get it..\r\nWhy is it [u]completely[/u] wrong? We can modify the solution as such: We now let $p^k \\mid \\mid a$, so taking $\\mod p^k$, each of the terms are divisible by $p^k$ except for $\\frac Ma$, so the numerator is $\\not \\equiv 0 \\pmod{p^k}$, while the denominator is $\\equiv 0 \\pmod{p^k}$.", "Solution_10": "Wait.. I reread my proof, and it's perfectly correct to me.. u don't even need to consider $p^k$ since $\\frac Ma \\not \\equiv 0 \\pmod{p}$.\r\n\r\nAs for the case when $a=1$, the same idea is used, just that now your $p$ is some prime divisor of the terms $1+b, 1+2b, \\ldots, 1+nb$.\r\n\r\nIt should be correct.", "Solution_11": "[quote=\"dblues\"]Wait.. I reread my proof, and it's perfectly correct to me.. u don't even need to consider $p^k$ since $\\frac Ma \\not \\equiv 0 \\pmod{p}$.\n\nAs for the case when $a=1$, the same idea is used, just that now your $p$ is some prime divisor of the terms $1+b, 1+2b, \\ldots, 1+nb$.\n\nIt should be correct.[/quote]You haven't understood fedja's remark yet... ;) It is possible $\\frac Ma \\equiv 0 \\pmod{p}$", "Solution_12": "Oh gosh.. I finally see why my solution is wrong.. I had been too hasty. To [i]fedja[/i], sorry I didn't give a second thought to your post. I originally thought you were referring to a new $b^{\\prime} = ab$, which was obviously not what you meant. :blush: \r\nTo [i]Myth[/i], do you have a solution?" } { "Tag": [ "ratio", "geometry", "incenter", "circumcircle", "inradius" ], "Problem": "Find the ratio of the legs of a right triangle if the half of the hypotenuse (from the vertex to the midopoint) is seen from the incenter at the right angle.", "Solution_1": "$ \\triangle ABC$ has right $ \\angle C,$ I is the incenter, O midpoint of AB (the circumcenter) and M the midpoint of AO. By definition, $ \\angle AIO$ is right $ \\Longrightarrow$ IM = AM = OM, the $ \\triangle AMI$ is isosceles with $ \\angle MIA \\equal{} \\angle MAI \\equal{} \\frac{\\angle A}{2} \\Longrightarrow$ $ \\angle AMI \\equal{} \\pi \\minus{} \\angle A,$ $ MI \\parallel AC.$ F is foot of a normal from M to AC, MF = r (inradius) and at the same time, $ \\frac{MF}{BC} \\equal{} \\frac{MA}{AB} \\equal{} \\frac{1}{4},$ $ MF \\equal{} \\frac{a}{4} \\equal{} r \\equal{} s \\minus{} c,$ where $ s \\equal{} \\frac{a \\plus{} b \\plus{} c}{2}$ is the triangle semiperimeter. $ a \\plus{} 2b \\equal{} 2c,$ $ a^2 \\plus{} 4ab \\plus{} 4b^2 \\equal{} 4c^2 \\equal{} 4a^2 \\plus{} 4b^2,$ $ 4ab \\equal{} 3a^2,$ $ \\frac{BC}{CA} \\equal{} \\frac{a}{b} \\equal{} \\frac{4}{3}.$" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra solved" ], "Problem": "Find all matrices with complex coefficients that have the class of similitude closed.", "Solution_1": "The diagonalizable matrices?", "Solution_2": "yes :)", "Solution_3": "How did you guys find them?\r\n\r\nI figured we could do something like this:\r\n\r\nTo show that non-diagonalizable matrices don't work, we could turn the matrix we're loking at into its Jordan form, and then, by similarity through matrices of the type $diag(1,\\ldots,1,\\alpha,1,\\ldots,1)$, we could make the $1$'s below the diagonal tend to $0$. This means that the closure of the similairty class of every non-diagonalizable matrix contains a diagonal matrix, meaning that the class is not closed.\r\n\r\nTo show that diagonalizable matrices work, take such a matrix, and observe that matrices from the closure of the similarity class satisfy the same minimal polynomial as the matrix we chose, and this minimal polynomial has distinct roots, meaning that all matrices in the closure are diagonalizable as well. Since they also have the same eigenvalues with the same multiplicities, the conclusion follows." } { "Tag": [ "induction", "inequalities proposed", "inequalities", "222" ], "Problem": "If $ x_n\\geq 0,x_0\\equal{}0,x_n^2\\plus{}x_n\\minus{}1\\equal{}x_{n\\minus{}1}^2,$ then $ 1\\minus{}\\frac{2^n}{3^n}\\leq x_n \\leq 1\\minus{}\\frac{1}{3^n}.$", "Solution_1": "[quote=\"Ji Chen\"]If $ x_n\\geq 0,x_0 \\equal{} 0,x_n^2 \\plus{} x_n \\minus{} 1 \\equal{} x_{n \\minus{} 1}^2,$ then $ 1 \\minus{} \\frac {2^n}{3^n}\\leq x_n \\leq 1 \\minus{} \\frac {1}{3^n}.$[/quote]\r\n\r\nThe induction works well. :wink: :)", "Solution_2": "Here is a more sophisticated proof, but it can tell us where $ 1/3^n$ and $ 2^n/3^n$ comes from.\r\nDenote $ y_n \\equal{} 1 \\minus{} x_n$, by induction method we have $ y_n$ decrease and $ y_n \\in (0,1)$. We have\r\n$ y^2_n \\minus{} 3y_n \\equal{} y_{n \\minus{} 1}^2 \\minus{} 2y_{n \\minus{} 1}$ so\r\n1) $ 3y_n \\equal{} y_n^2 \\minus{} y_{n \\minus{} 1}^2 \\plus{} 2y_{n \\minus{} 1} < 2y_{n \\minus{} 1} \\Rightarrow y_n < y_02^n/3^n,$\r\n2) $ 3y_n \\equal{} y_n^2 \\minus{} y_{n \\minus{} 1}^2 \\plus{} 2y_{n \\minus{} 1} > y_{n \\minus{} 1} \\Rightarrow y_n > y_0/3^n.$" } { "Tag": [ "AMC", "AIME", "inequalities", "triangle inequality", "complex numbers" ], "Problem": "given thath $a_1+a_2=17$ \r\nfind the minimum value of \r\n$\\displaystyle\\sqrt{1^2+a_1}+\\sqrt{3^2+a_2}$\r\n\r\nthe real question was \r\ngiven that ${a_1,a_2,\\dots,a_n\\in\\{R}$\r\nand $S_n$ is the minimum value of\r\n$\\displaystyle\\sum_{k=1}^{n}\\sqrt{(2k-1)^2+a_k}$\r\nwhile \r\n$a_1+a_2+a_3+\\dots+a_n=17$\r\nfind $S_{10}$", "Solution_1": "this is a fun problem...i believe it is from aime?\r\n\r\nanyway, there are a few solutions...here is a complex number solution\r\n\r\nsay we have\r\n\r\n$|a_1+i|+|a_2+3i|+...+|a_n+(2n-1)i|\\ge |\\sum a_i+n^2i|$\r\n\r\n(note that this is equivalent to the sum)\r\n\r\nby the complex number triangle inequality, with equality when the complex numbers are scalar multiples to each other, then the min is:\r\n\r\n$\\sqrt{17^2+n^4}$\r\n\r\ni think that there is more to the problem though...", "Solution_2": "The question was to find the unique $n$ such that $S_n$ was integer, I think. I've solved it before.\r\n\r\nEdit: This is AIME 1991, Problem 15. A solution is here:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=431997#p431997" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $x_i>0, i=\\overline{1,n}$ and $\\gamma=(1+\\frac{1}{n})^n$, then:\r\n$\\sum_{i=1}^n\\sqrt[k]{x_1 x_2 ... x_k} < \\gamma.\\sum_{i=1}^{n}a_k$\r\n... :)\r\nOf course, $n\\in\\mathbb{N}$ :)", "Solution_1": "[quote=\"Lovasz\"]Let $x_i, i=\\overline{1,n}$ and $\\gamma=(1+\\frac{1}{n})^n$, then:\n$\\sum_{i=1}^n\\sqrt[k]{x_1 x_2 ... x_k} < \\gamma.\\sum_{i=1}^{n}a_k$\n... :)[/quote]\r\nMaybe $x_i>0$ and $\\gamma=(1+\\frac{1}{n})^n, n\\in\\mathbb N$ then:\r\n$\\sum_{i=1}^n\\sqrt[k]{x_1 x_2 ... x_k} < \\gamma\\cdot\\sum_{i=1}^{n}x_k$ ?\r\n\r\nG.Polya in 1925 found beautiful proof for this inequality.", "Solution_2": "Arqady, can you post Polya's proof? :)", "Solution_3": "[quote=\"Cezar Lupu\"]Arqady, can you post Polya's proof? :)[/quote]\r\nPolya's hint: \r\nLet $c_{1}\\cdot c_{2}\\cdot...\\cdot c_{k}=(k+1)^k.$ ;)", "Solution_4": "[quote=\"arqady\"][quote=\"Cezar Lupu\"]Arqady, can you post Polya's proof? :)[/quote]\nPolya's hint: \nLet $c_{1}\\cdot c_{2}\\cdot...\\cdot c_{k}=(k+1)^k.$ ;)[/quote]\r\nExcuse me, I don't understand what $c_i$ is, arqady.", "Solution_5": "That's probably $x_i$, and wow, this is a good hint, I would never have thought about that :blush: !", "Solution_6": "$\\sum_{k=1}^n\\sqrt[k]{x_1 x_2 ... x_k}=\\sum_{k=1}^n\\frac{\\sqrt[k]{x_1c_1 x_2c_2 ... x_kc_k}}{k+1}...$ ;)", "Solution_7": "Maybe the book had the same idea... but It's not quite natural...", "Solution_8": "Let $c_k=\\frac{(k+1)^k}{k^{k-1}}.$ Then $c_1c_2...c_k=(k+1)^k.$ Hence,\r\n$\\sum_{k=1}^n\\sqrt[k]{x_1 x_2 ... x_k}=\\sum_{k=1}^n\\frac{\\sqrt[k]{x_1c_1 x_2c_2 ... x_kc_k}}{k+1}\\leq\\sum_{k=1}^n\\frac{x_1c_1+x_2c_2+...+x_kc_k}{k(k+1)}=$\r\n$=\\sum_{k=1}^n x_kc_k\\sum_{i=k}^n\\frac{1}{i(i+1)}=$\r\n$=\\sum_{k=1}^n x_kc_k(\\frac{1}{k}-\\frac{1}{k+1}+\\frac{1}{k+1}-\\frac{1}{k+2}+...+\\frac{1}{n}-\\frac{1}{n+1})<\\sum_{k=1}^nx_kc_k\\cdot\\frac{1}{k}=$\r\n$=\\sum_{k=1}^nx_k\\frac{(k+1)^k}{k^{k-1}}\\cdot\\frac{1}{k}=\\sum_{k=1}^nx_k(1+\\frac{1}{k})^k<(1+\\frac{1}{n})^n\\sum_{k=1}^nx_k.$", "Solution_9": "i was just wandering:can we find the smallest constant $\\gamma_0$ so that \r\nthe first inequality holds with this $\\gamma_0$ instead of $\\gamma$ for all natural n\r\nthe proof that arqady presented above doesnt help much (at least at first sight) eventhough it is really brilliant \r\n :)" } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "i hope i have not asked this one before... :maybe: \r\n\r\n\r\nshow :D\r\n\r\n$ \\dfrac{1^2}{1 \\plus{} 1^3} \\minus{} \\dfrac{2^2}{1 \\plus{} 2^3} \\plus{} \\dfrac{3^2}{1 \\plus{} 3^3} \\minus{} \\dfrac{4^2}{1 \\plus{} 4^3} \\plus{} \\dfrac{5^2}{1 \\plus{} 5^3} \\minus{} \\dfrac{6^2}{1 \\plus{} 6^3}\\; \\plus{} \\; \\minus{} \\;\\ldots\\; \\equal{} \\;\\boxed{\\frac {1}{3}\\left(1 \\minus{} \\ln\\,2 \\plus{} \\pi\\cdot\\text{sech}\\left(\\frac {\\sqrt 3\\,\\pi}{2}\\right)\\right)}$", "Solution_1": "You asked this one [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=156669]here[/url]. :gleam:" } { "Tag": [ "LaTeX" ], "Problem": "hello all:\r\n\r\nduring the installation of MikTeX, the MikTex wizard asks me to select a path to the local package repository! Could anyone tell me what that means and what i need to do. Thanks in advance and appreciate the help.\r\n\r\ncheers,\r\nnatasha", "Solution_1": "Hi,\r\nThe directory you specify as the local repository during installation will hold all the packages that MikTeX downloads and loads at every startup. If, in the future, you need any packages that MikTeX did not download (ie you did not specify), you will have to download them to this directory and then update MikTeX's file name database. \r\nSSN." } { "Tag": [ "inequalities", "calculus" ], "Problem": "My computer is currently not working properly. So I don't have my drawing and PDF answers with me. So, for #2, I hope contestant themselves can demonstrate it.\r\n\r\nOr, I'll add the links. :D \r\n\r\nProblem is in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40244]here[/url]\r\n\r\nI'll write up the inequality part very soon.", "Solution_1": "For #1, I reduced it to\r\n\r\n\\[2(a^3+b^3+c^3) \\leq a^2(b+c)+b^2(a+c)+c^2(a+b)\\]\r\n\r\n :blush: I'm inexperienced with Olympiad inequalities.", "Solution_2": "Could I ask which ones I got right? Because if I got number 2 right then I'll post it.", "Solution_3": "I think both of you (you and BornForMath) got number one right..\r\n\r\nI will be greatly appreciate if you two post #2 as well because I haven't got time to grade them yet.", "Solution_4": "Can you post for me please, silverfalcon? I lost my solution for number 2. :blush: and i am too lazy to type it up again.", "Solution_5": "I'm going to discribe my general method and hopefully that will be good enough...\r\n\r\nOkay. Start with a circle.\r\nGot it?\r\nYay. Now choose any two points on the circumferance of that circle such that the internal angle is less then 180 but larger then 90. Do that for the other point, with the requirement that the three angles add up to 360. \r\n\r\nYay. Now, the equality holds, because the line segments in question are tangent to the circle, and the angles are acute because the internal angles on the circle are less then 180 but greater then 90.\r\n\r\nAll lengths are represented because we can change the radius of the circle.\r\nAll angles are represented because we can pick our three points to make all combinations.\r\n\r\nDone! \r\nYay!\r\nQED", "Solution_6": "Can someone please post the solution to Problem 1? :)", "Solution_7": "Okay, here's number 1. \r\n[hide]\n1. We break the inequality into three parts. \\vskip .4in\n\nPart 1: $2(pr + ps) \\geq 2(pr + ps)$ \\vskip .2in\n\nThis part is relativly simple. $2(pr + ps) = 2(pr + ps)$, for all $p,s$. Therefore $2(pr + ps) \\geq 2(pr + ps)$ for all $p,s$. \\blacksquare \\vskip .3in\n\nPart 2: $4a^3 + 4b^3 + 4c^3 \\geq 2a^2b + 2a^2c + 2ab^2 + 2b^2c + 2ac^2 + 2bc^2$ \\vskip .2in\n\nConsider the expression $(a - b)^2(a + b)$ Because $a,b,c$ are all greater then or equal to zero, $a + b$ must be greater then or equal to zero. Also, $(a - b)^2$ is greater then zero by the trivial inequality, which states that a square of real number is greater then or equal to zero. Thus, $(a-b)^2(a+b) \\geq 0$. Because $a,b,c \\geq 0$, $(a-b)^2(a+b) \\geq 0$, $(a-c)^2(a+c) \\geq 0$, $(b-c)^2(b+c) \\geq 0$. Because all three of these expressions are greater then or equal to zero, the sum of these expressions must be greater then or equal to zero. So we have $(a-b)^2(a+b)+ (a-c)^2(a+c) + (b-c)^2(b+c)\\geq 0$\nExpanding, we have $2a^3 - a^2b - a^2c + 2b^2 - b^2a - b^2c + 2c^3 - c^2a - c^2b \\geq 0$\nOr equivalently, $2a^3 + 2b^3 + 2c^3 \\geq a^2b + a^2c + b^2a + b^2c + c^2a + c^2b$\nMultiplying each side by 2, we have $4a^3 + 4b^3 + 4c^3 \\geq 2a^2b + 2a^2c + 2ab^2 + 2b^2c + 2ac^2 + 2bc^2$, as desired. \\blacksquare \\vskip .3in\n\nPart 3: $2(1 + i)^j \\geq 2ij + 2$ \\vskip .2in\n\nI'm going to use some calculus here because it's the easiest way to prove it. Okay. \nLet $f(x) = x^j$. Then $f'(x) = jx^{j-1}$ and $f''(x) = j(j-1)x^{j-2}$. $f(1) = 1$, $f'(1) = j$, and $f''(x)>0$ if $x > 0$ and $j \\geq 1$, meaning the graph is concave for $x \\geq 0$. Therefore the graph of $y = f(x)$ is always above the tangent line as long as $x \\geq 0$ Making the substitution $i = x - 1$, we know that $(1 + i)^j \\geq 1 + ij$ for $i \\geq -1$ and $j \\geq 0$. Because $i,j \\geq 0$ in our problem, the inequality holds. Multiplying by two gives $2(1 + i)^j \\geq 2ij + 2$, as desired. \\blacksquare \\vskip .2in\n\nIf you add the previous three inequalities, you get:\n $\\displaystyle 2(1+i)^j + 4a^3+4b^3+4c^3+2(pr+qs) \\geq 2+2a^2b+2a^2c+2ab^2+2b^2c+2ac^2+2bc^2 + 2(ps+qr) + 2ij$, which we wanted to prove in the first place. QED \\vskip .3in\n[/hide]\r\nThe only problem with this solution is that I had to use calculus for the third part." } { "Tag": [ "trigonometry", "function" ], "Problem": "$\\sec^{4}(\\pi/7)+\\sec^{4}(2\\pi/7)+sec^{4}(3\\pi/7)= 416$", "Solution_1": "Please use a \\ before the trig function when typing them in LaTeX. For example, \\sec x gives $\\sec x$.", "Solution_2": "let $z=e^{\\frac{2\\pi i}{14}}$, and make the appropriate substitutions and mass expand...\r\n\r\nalternately, we could develop a linear recurrence that describes the sequence $s_{k}=\\sec^{k}\\frac{\\pi}{7}+\\sec^{k}\\frac{2\\pi}{7}+\\sec^{k}\\frac{3\\pi}{7}$...this leads to a nice generalization, but it is still a lot of work." } { "Tag": [ "vector", "abstract algebra", "linear algebra", "linear algebra unsolved" ], "Problem": "Suppose that U and V are finite dimensional vector spaces and \r\nthat S belong to L(V,W) , T belongs L(U,V) Prove that \r\n\r\nDim null ST <= dim null S +Dim null T\r\n\r\nTHANK U IN ADVANCE", "Solution_1": "Pick a basis (x1,...,xp) of Ker(T) and complete it into a basis (x1,...,xp,...,xn) of U. Then x is in Ker (ST) <=> x is in Ker (T) or T*(x) in Ker(S). Since the restriction T* of T to vect(x_(p+1),...,xn) is an isomorphism onto Im(T), Im(T) Intersection Ker(S) will be of dimension at most Ker(S). Now E={x; T*(x) in Ker(S)} is of dimension at most Ker(S), and since E and Ker(T) are disjoint, the result follows.", "Solution_2": "[quote=\"julien_santini\"]Pick a basis (x1,...,xp) of Ker(T) and complete it into a basis (x1,...,xp,...,xn) of U. Then x is in Ker (ST) <=> x is in Ker (T) or T*(x) in Ker(S). Since the restriction T* of T to vect(x_(p+1),...,xn) is an isomorphism onto Im(T), Im(T) Intersection Ker(S) will be of dimension at most Ker(S). Now E={x; T*(x) in Ker(S)} is of dimension at most Ker(S), and since E and Ker(T) are disjoint, the result follows.[/quote]\r\n\r\nSince the restriction T* of T to vect(x_(p+1),...,xn) is an isomorphism onto Im(T),\r\nCan u explain why it that", "Solution_3": "If $E$ and $F$ are two vector spaces with finite dimension, and $f: E \\rightarrow F$ is linear, then:\r\n$rank(f)+dim(Ker(f))=dim(E)$. In particular if you restrict $f$ to a supplementary $S$ of its kernel, then $dim(S)=dim(E)-dim(Ker(f))=rank(f)$, i.e $f$ isomorphism." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "find all positive integers $a,b$such that:$5a^{b}-b=2004$", "Solution_1": "[hide=\"(Ugly) Solution\"]\n$5a^{b}=2004+b$\nExamining mod 5, we see that $b \\equiv 1 \\mod 5$.\nLet's assume that $b \\neq 1$.\nHence $b \\geq 6$.\nHence $a < 3$.\nAlso, if $a=2$, we can see that $b\\leq 11$, and by trying the $b=1,6$ we find that there are no solutions with $a=2$.\nSo $a=1$ if $b \\neq 1$. Hence $b=-1999$, which isn't possible. So, $b = 1$.\nPlugging this in, we find that $a=401$.\n\nSo, the only solution is $a=401, b=1$.\n\n[/hide]" } { "Tag": [ "ratio", "number theory unsolved", "number theory" ], "Problem": "Let $ S(n)$ denote the sum of all digits of positive integer $ n$.\r\n\r\nProve that there exist infinitely many $ n$ such that $ S(n)>2007S(3n)$", "Solution_1": "The problem is too easy:\r\nIt is sufficient to take \r\n$ 3n\\equal{}100\\dots011$(we take $ 0$-$ k$ times),then $ S(3n)\\equal{}3$,and $ S(n)\\equal{}3(k\\plus{}1)\\plus{}7$.Now by taken any $ k>2004$ we obtain result." } { "Tag": [ "quadratics" ], "Problem": "Suppose we are given a quadratic equation with integer coefficients. Can we have a discriminant of 23?\r\n :D :D :D", "Solution_1": "mod 4 again !", "Solution_2": "indeed. Fascinating! :)" } { "Tag": [ "algebra", "polynomial", "function", "search", "geometry", "analytic geometry" ], "Problem": "A sequence consists of the numbers $1$, $3$, $140$, $142$, $144$, $131$...\r\n\r\nWhat is the $17^{th}$ term in the sequence?", "Solution_1": "No takers yet?\r\n\r\nJust in case anyone's wondering, this sequence definitely does work. I've checked it over multiple times to make sure I didn't make any mistakes. ;)", "Solution_2": "Hmm... I'm getting no results from the Online Encyclopedia of Integer Sequences.\r\n\r\nThis looks like a tricky one - the terms go up and down. Better get a piece of paper out...", "Solution_3": "[hide=\"Hint 1\"]It's not found by some super-complex equation. It's rather straightforward once you get the concept.[/hide]\n\n[hide=\"Hint 2\"]It requires lateral thinking to find the answer, but it's purely mathematical. It's just something that you're not very likely to think of. ;)[/hide]", "Solution_4": "Well there are a lot of solutions, I really don't like these kinds of problems at all.", "Solution_5": "There may be a lot of solutions, but there's only one correct solution, as I made this problem up...", "Solution_6": "[quote=\"JesusFreak197\"]There may be a lot of solutions, but there's only one correct solution...[/quote]\r\n\r\nExactly--how are we supposed to know which one it is?\r\n\r\nWhatever, somebody else can probably figure it out.", "Solution_7": "Just tell me if I'm on the right track...\r\nSums of digits? Or products of digits?", "Solution_8": "[quote=\"JesusFreak197\"]There may be a lot of solutions, but there's only one correct solution, as I made this problem up...[/quote]\r\n\r\nDisagree, these type of problem consist of infinite number of patterns, as any polynomial with degree > 4 satisfy \r\n\r\nP(1)=1st number, \r\nP(2)=2nd number, \r\nP(3)=3rd number,\r\nP(4)=4th number\r\nP(5)=5th number\r\n\r\nwill work. (and there are infinite number of such polynomials)\r\n\r\nSo yeah don't post this kind of problem in future.", "Solution_9": "If you don't believe me, here is a polynomial that works\r\n\r\n$P(x)=-\\frac{37}{8}(x-1)(x-2)(x-3)(x-4)(x-5)+\\frac{135}{8}(x-1)(x-2)(x-3)(x-4) -45(x-1)(x-2)(x-3) + \\frac{135}{2}(x-1)(x-2)+2x-1$\r\n\r\nAnd you can have fun calculating $P(17)$.\r\n\r\nOh if you don't like that how about\r\n\r\n$P(x)=2^{120!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)-\\frac{37}{8}(x-1)(x-2)(x-3)(x-4)(x-5)+\\frac{135}{8}(x-1)(x-2)(x-3)(x-4) -45(x-1)(x-2)(x-3) + \\frac{135}{2}(x-1)(x-2)+2x-1$\r\n\r\nMaybe this looks even better?\r\n\r\n$P(x)=2^{120!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-2006^{2006})-\\frac{37}{8}(x-1)(x-2)(x-3)(x-4)(x-5)+\\frac{135}{8}(x-1)(x-2)(x-3)(x-4) -45(x-1)(x-2)(x-3) + \\frac{135}{2}(x-1)(x-2)+2x-1$\r\n\r\nHopefully you get my point.", "Solution_10": "As I said in my first hint, and I will make clear now for any who did not want to read it, the answer is not found by some kind of super-complex function. The method is actually quite simple once you figure out what it is.\r\n\r\nAnd I can still say that there is only one correct solution, even though many other solutions will give answers, since there is one solution that I have in mind.", "Solution_11": "[quote=\"JesusFreak197\"]As I said in my first hint, and I will make clear now for any who did not want to read it, the answer is not found by some kind of super-complex function. The method is actually quite simple once you figure out what it is.\n\nAnd I can still say that there is only one correct solution, even though many other solutions will give answers, since there is one solution that I have in mind.[/quote]\r\n\r\nOH YEAH MY ANSWER \r\n$P(x)=-\\frac{37}{8}(x-1)(x-2)(x-3)(x-4)(x-5)+\\frac{135}{8}(x-1)(x-2)(x-3)(x-4) -45(x-1)(x-2)(x-3) + \\frac{135}{2}(x-1)(x-2)+2x-1$\r\n\r\nIS RIGHT(PROVE ME WRONG) AND YEAH I AM ALWAYS RIGHT \r\n\r\nbut seriously you can't claim that the particular function you are thinking of is correct when there are infinite number of them .\r\n\r\nOkay here's my problem: find $x+y$.\r\n\r\n [hide=\"answer\"]2 because the value of x and y I have in mind is 1 and 1. ANY OTHER ANSWER IS WRONG WRONG WRONG [/hide]", "Solution_12": "[quote=\"beta\"][quote=\"JesusFreak197\"]As I said in my first hint, and I will make clear now for any who did not want to read it, the answer is not found by some kind of super-complex function. The method is actually quite simple once you figure out what it is.\n\nAnd I can still say that there is only one correct solution, even though many other solutions will give answers, since there is one solution that I have in mind.[/quote]\n\nOH YEAH MY ANSWER \n$P(x)=-\\frac{37}{8}(x-1)(x-2)(x-3)(x-4)(x-5)+\\frac{135}{8}(x-1)(x-2)(x-3)(x-4) -45(x-1)(x-2)(x-3) + \\frac{135}{2}(x-1)(x-2)+2x-1$\n\nIS RIGHT(PROVE ME WRONG) AND YEAH I AM ALWAYS RIGHT \n\nbut seriously you can't claim that the particular function you are thinking of is correct when there are infinite number of them .\n\nOkay here's my problem: find $x+y$.\n\n [hide=\"answer\"]2 because the value of x and y I have in mind is 1 and 1. ANY OTHER ANSWER IS WRONG WRONG WRONG [/hide][/quote]\r\n\r\nI understand what you're saying. But the solution he has in mind does not use a polynomial. The idea is not to use math skills really, but to use lateral thinking. For example, take the series\r\n\r\n$1, 11, 21, 1211, 111221, ...$\r\n\r\nOf course, we could find a polynomial to describe this sequence. However, there is a way which is not associated with a polynomial, a simple one with little math (which I'm sure most of you know). The idea is to get something like that. So, in truth, JesusFreak's problem is NOT a math problem, having no one defined answer. But then, does any problem in life have a defined answer? There is always some fuzziness.", "Solution_13": "Yeah exactly, that's why this type of pattern problems should not be posted, at least not in this forum, as you can find infinite number of functions/patterns describing this sequence. And it's a matter of opinion which sequence is the \"best\" one (you can be really cheap and say\r\n f(x) =0 for x>6, which is a perfectly okay function to describe this sequence).", "Solution_14": "I'd actually have to agree. These types of problems are always very annoying because, mathematically speaking, while it's true you want the simplest pattern, there are literally an infinity of patterns to search through, so it's just a matter of whether you manage to get the one the problem-giver asks for.\r\n\r\nOr another argument: There isn't any way to attack these types of problems generally other than guess-and-check, so they don't contribute anything significant mathematically.", "Solution_15": "[quote=\"K81o7\"]For example, take the series\n$1, 11, 21, 1211, 111221, ...$[/quote]Next term is 312211. Read the sequences out loud. one, then describe it. one one. Then describe that, two ones. Describe: one two one one. one one one two two ones. Fun problem.", "Solution_16": "[quote=\"Hokkage\"][quote=\"K81o7\"]For example, take the series\n$1, 11, 21, 1211, 111221, ...$[/quote]Next term is 312211. Read the sequences out loud. one, then describe it. one one. Then describe that, two ones. Describe: one two one one. one one one two two ones. Fun problem.[/quote]\r\n\r\nExactly. Even if I had some other way of describing the sequence than you did, your answer wouldn't be wrong. The idea is of course to find a non-polynomial solution.", "Solution_17": "[quote=\"K81o7\"][quote=\"Hokkage\"][quote=\"K81o7\"]For example, take the series\n$1, 11, 21, 1211, 111221, ...$[/quote]Next term is 312211. Read the sequences out loud. one, then describe it. one one. Then describe that, two ones. Describe: one two one one. one one one two two ones. Fun problem.[/quote]\n\nExactly. Even if I had some other way of describing the sequence than you did, your answer wouldn't be wrong. The idea is of course to find a non-polynomial solution.[/quote]\r\nHmm what's wrong with a polynomial solution that works lol :D \r\nWell the ability to recognize patterns is important in math, as you can make conjectures based on your pattern. But then I don't like the idea of making contrived pattern problems.", "Solution_18": "I'm sorry if I posted this in the wrong forum. I thought the answer would be more interesting for this one than the Puzzles/Brainteasers Forum once someone found it.\r\n\r\nAlso, another hint: it's even more mathematical than the sequence that K81o7 gave. ;)", "Solution_19": "[quote=\"beta\"]\nHmm what's wrong with a polynomial solution that works lol :D \nWell the ability to recognize patterns is important in math, as you can make conjectures based on your pattern. But then I don't like the idea of making contrived pattern problems.[/quote]\r\n\r\nThere IS nothing wrong with a polynomial solution. A polynomial solution usually doesn't take much thinking or creativity. I'm sure JesusFreak's solution isn't the only non-polynomial one.", "Solution_20": "[quote=\"K81o7\"]\n\nThere IS nothing wrong with a polynomial solution. A polynomial solution usually doesn't take much thinking or creativity. I'm sure JesusFreak's solution isn't the only non-polynomial one.[/quote]\r\n\r\nFinding a polynomial(without a calculator) is not that easy, at least when you have like 4+ terms, although I figured out an algorithm(see my polynomial for this problem) that will work for any sequence", "Solution_21": "[url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange interpolation[/url] is another way to fit a polynomial function to a given set of points. It looks a lot like what beta wrote (similar reasoning I would believe).\r\n\r\nAnyway, problems of this type should almost always go in the Puzzles & Brainteasers forum. I don't know how his sequence is constructed though so I can't really say if it OK to be in this forum (I'll trust him though). Although we know that this type of question should be avoided at all costs on competitions (*ahem*iTest*ahem) and standardized tests, it should be OK to pose it as a challenge on a math forum.\r\n\r\nOh, and JesusFreak, it would be better to say something like \"that answer works but it's not the intended answer\" or \"look for a simpler pattern\" instead of saying that there is only one correct answer because you said so. It's like giving someone a geometry problem and they prove it using a really messy coordinate approach when there is a nice synthetic approach. You wouldn't tell them they're wrong just that there is a more elegant approach they should try to find.", "Solution_22": "JF197, can you just tell us what it is, because apparently nobody feels like figuring it out. :D", "Solution_23": "One non-polynomial possibility: Maybe it has something to do with the digits themselves, like if we concatenate them, we get $13140142144131$, and this seems like being periodical, with the period $13140142144$. So, if we decide that all the following terms are at most three-digit, we'll get $1, 3, 140, 142, 144, 131, 401, 421, 441, 314, 14, 214, 413, 140, 142, 144, 131$, and the 17th is $131$.", "Solution_24": "Heh, I didn't even notice that similarity. Nice try, but not the answer I have in mind.\r\n\r\nTry something with bases. That should help someone get it. ;)", "Solution_25": "[hide=\"Possibility\"]I think it's $243$.[/hide]", "Solution_26": "Yep, that's correct. ;) Want to say how you got that?", "Solution_27": "[hide=\"Explanation\"]Well, I've played with bases before this, but never thought of a negative one :)\n\nIn base $(-5)$ (but with digits $0, 1, 2, 3, 4$ - just imagine! :) ) we have $1=1_{10}, 3=3_{10}, 140=5_{10}, 142=7_{10}, 142=9_{10}, 131=11_{10}$. Hence, 17th member should be $33_{10}$ and that would be $243$ in the base $(-5)$ (Calculation: If $\\overline{abc}_{(-5)}=33_{10}$, then $25a-5b+c=33$. Hence $5|c-3\\Rightarrow c=3$. Now $5a-b=6\\Rightarrow 5|b+1\\Rightarrow b=4$. Now $a=2$.)[/hide]" } { "Tag": [ "LaTeX", "Olimpiada de matematicas" ], "Problem": "[color=blue][size=150]PERU TST IMO - 2006[/size]\nSabado, 20 de mayo.[/color]\r\n\r\n[b]Pregunta 04[/b]\r\nEn un triangulo acutangulo $ABC$ se trazan: su circunferencia\r\ncircunscrita $w$ con centro $O$, la circunferencia circunscrita\r\n$w_1$ del triangulo $AOC$ y el diametro $OQ$ de $w_1$. Se eligen\r\nlos puntos $M$ y $N$ sobre las rectas $AQ$ y $AC$,\r\nrespectivamente, de tal manera que el cuadrilatero $AMBN$ sea un\r\nparalelogramo. Probar que el punto de interseccion de las rectas\r\n$MN$ y $BQ$ pertenece a la circunferencia $w_1.$\r\n\r\n$\\text{\\LaTeX}{}$ed by Carlos Bravo 2006 - MathLinks AoPS", "Solution_1": "Alguna idea ..??\r\n\r\n\r\nCarlos Bravo ;)\r\nLIMA - PERU", "Solution_2": "yo lo acabo de hacer pero son las 12:15 am en mexico y tengo el suficiente sue\u00f1o para resistirme y escribirlo ma\u00f1ana\r\n\r\npero me gustaria ver otras respuestas, tu como lo resolviste carlos bravo" } { "Tag": [], "Problem": "Let $ f(x)\\equal{}\\frac{6*(6*6)^x}{6\\plus{}(6*6)^x}$. Evaluate $ \\displaystyle\\sum_{i\\equal{}1}^{666} f(\\frac{i}{666})$", "Solution_1": "$ f(x)\\plus{}f(1\\minus{}x)\\equal{}6$\r\nAns:$ \\frac{14001}{7}$", "Solution_2": "thanks. \r\n\r\nthat was very easy. :)" } { "Tag": [ "limit", "Olimpiada de matematicas" ], "Problem": "[color=blue]No logro hacer el c\u00e1lculo siguiente:\n$\\mathop{\\lim }_{n \\to \\infty }\\frac{{\\left({1+\\sqrt 5 }\\right)^{n+1}-\\left({1-\\sqrt 5 }\\right)^{n+1}}}{{\\left({1+\\sqrt 5 }\\right)^{n}-\\left({1-\\sqrt 5 }\\right)^{n}}}$\nSeg\u00fan mi parecer, deber\u00eda dar el siguiente resultado:\n$\\frac{{\\left({1+\\sqrt 5 }\\right)}}{2}$\n\u00bfPodr\u00eda alguien ayudarme?\nGracias por adelantado.[/color]", "Solution_1": "[quote=\"R.G.A.M.\"][color=blue]bueno creo que seria lo mismo que:$\\mathop{\\lim }_{n \\to \\infty }\\frac{F_{n+1}}{F_{n}}$\n.[/color][/quote]", "Solution_2": "Sea $L=\\mathop{\\lim}_{n \\to \\infty}\\frac{F_{n+1}}{F_{n}}$ entonces, tambi\u00e9n tenemos que $L=\\mathop{\\lim}_{n \\to \\infty}\\frac{F_{n+2}}{F_{n+1}}$, luego (multiplicando) tenemos que $L^{2}=\\mathop{\\lim}_{n \\to \\infty}\\frac{F_{n+2}}{F_{n}}$, como $F_{n+2}=F_{n+1}+F_{n}$ : \\[L^{2}=\\mathop{\\lim}_{n \\to \\infty}\\frac{F_{n+2}}{F_{n}}=\\mathop{\\lim}_{n \\to \\infty}( 1+\\frac{F_{n+1}}{F_{n}})=1+L\\] Ahora, como $L\\geq 1$, ya se puede obtener el valor de $L$.\r\n\r\n$Tipe$", "Solution_3": "[color=blue]Una idea similar hab\u00eda tenido.\n\nHall\u00e9 la ecuaci\u00f3n caracter\u00edstica de la sucesi\u00f3n de Fibonacci para dar con la fracci\u00f3n escrita en el primer post.\n\nAgradezco mucho por la soluci\u00f3n. No sab\u00eda c\u00f3mo trabajar con l\u00edmites cuando el valor del exponente tiende a infinito. \u00bfExiste alg\u00fan m\u00e9todo general para ello?\n\nMuchas gracias, nuevamente.[/color]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ \\{a_n\\}$ be a sequence given by the formulas:\r\n$ a_1 \\equal{} 1$;\r\n$ a_{n \\plus{} 1} \\equal{} a_n \\plus{} \\frac {1}{a_n}$; $ n \\equal{} 1,2,...$\r\nProve that $ 14 < a_{100} < 18$.", "Solution_1": "square the relation and sum it over $ n \\equal{} 2, ..., 100$, to get $ a_{100}^2 \\equal{} 1 \\plus{} 2\\cdot 99 \\plus{} \\frac {1}{a_1^2} \\plus{} \\cdots \\frac {1}{a_{99}^2} \\equal{} 199 \\plus{} \\frac {1}{a_1^2} \\plus{} \\cdots \\frac {1}{a_{99}^2}$. since $ 0 < \\frac {1}{a_1^2} \\plus{} \\cdots \\frac {1}{a_{99}^2} < 99$ (because $ a_i\\geq 1$ for all $ i$, the result follows." } { "Tag": [ "function", "topology" ], "Problem": "Let X be a complete Hausdoff space and f:X -> R is continuous. Prove that the function P: x -> (x, f(x)) is continuous.", "Solution_1": "If $ f: A\\to B$ is continuous, the map $ A\\to A\\times B, x\\mapsto (x,f(x))$ is always continuous. One need not to assume hausdorff or completeness\r\n(by the way: what should that mean? There is no notion of completeness for topological spaces. Did you forget to mention that $ X$ is metric or uniform?)\r\n\r\nSo why is that? It's an easy consequence of the definition of product. Just check it yourself. There is no trick or hidden catch, everythings straight foreward. You can do it by yourself.", "Solution_2": "I'm sorry. X is metric space. I'm still quite confused on how to prove it to be continuous. Can you help me a little bit more? Thanks", "Solution_3": "Take the definition of \"continuous\" and of the product. That's all you need and as I said: There is no hidden catch, everthing's straight foreward. Just try it...", "Solution_4": "Although it's going to be a bit more of a kludge if you are chasing distances around using that $ X$ is a metric space. I assure you, it's not necessary.", "Solution_5": "I think you do not need the completeness condition in here. And I do not see the need to use $ T_{2}$. The proof is like, suppose $ L \\equal{} U\\times V$ is an open set in $ A\\times B$, we need to prove $ P^{ \\minus{} 1}L$ is open. However if $ y\\in V \\minus{} f(U)$, then $ P^{\\minus{}1}(U,y)$ is empty(which is open), and the same is $ (x,y)$ with $ x\\in U \\minus{} f^{ \\minus{} 1}(V)$. So only the points as $ (W,f(W)$ is there, $ W$ is $ U\\cap f^{ \\minus{} 1}(V)$. Since $ W$ is open, $ L \\equal{} W$ is also open. I think we only need to deal with basis of product topology, then the desired open set can come out of inverse mapping and union of open sets quite easily.", "Solution_6": "The whole point of the product topology is that a map from $ X$ to $ A \\times B$ is continuous iff it is continuous in each coordinate. :)" } { "Tag": [ "function", "limit", "algebra", "polynomial", "trigonometry", "calculus", "calculus computations" ], "Problem": "If the function f is continuous for all real numbers and if f(x) = [x^2 - 7x + 12] / [x-4] when x does not = 4, then f(4) = ?\r\n\r\n\r\na) 8/7\r\nb) -1\r\nc) 0\r\nd) undefined\r\ne) 1", "Solution_1": "The key is that the function is continuous. For a function to be continuous, it must meet three criterion, it is defined at the point, the limit exists to that point, and the limit and actual value agree with each other. Our function is\r\n$ f(x)\\equal{}\\frac{x^{2}\\minus{}7x\\plus{}12}{x\\minus{}4}$\r\nThis can be broken down into\r\n$ f(x)\\equal{}\\frac{(x\\minus{}4)(x\\minus{}3)}{(x\\minus{}4)}\\equal{}x\\minus{}3$\r\nSo now, we need only take the limit as $ x$ approaches 4.\r\n$ \\lim_{x\\rightarrow 4}x\\minus{}3\\equal{}1$\r\nSo in order for the function to be continuous,\r\n$ \\boxed{f(4)\\equal{}1}$", "Solution_2": "let $ f(x)\\equal{}x^2\\minus{}7x\\plus{}12$ & $ g(x)\\equal{}x\\minus{}4$, note that $ \\frac{f(4)}{g(4)}$ is indeterminate, then:\r\n$ \\displaystyle\\lim_{x\\to4}\\frac{f(x)}{g(x)}\\equal{}\\lim_{x\\to4}\\frac{f\\prime(x)}{g\\prime(x)}\\equal{}2(4)\\minus{}7\\equal{}1$", "Solution_3": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176952]This topic[/url] has the same kind of language - a continuous function given by a certain formula except at one point - but I pushed a little harder by asking for not just the value but also the third derivative.\r\n\r\nAs for igiul's solution - why use that technique? There's nothing wrong with using straightforward polynomial algebra when it's appropriate.", "Solution_4": "igiul's solution is just an application of L'Hopital's rule and I like such solutions (I just used the one with less calculus). If I recall Kent, didn't you give a L'Hopital's Rule problem on your final that most people answered wrong?", "Solution_5": "I actually love the rule myself. Used it to beat a nasty limit on my AB semester final just a few days ago.", "Solution_6": "From my L'H\u00f4pital is not everything campaign:\r\n\r\nFind $ \\lim_{x\\to0}\\frac{1\\minus{}\\cos(x^3)}{\\sin^3 (x^2)}.$\r\n\r\nBear in mind that there are a significant number of regular readers here who knew the answer in 10 seconds or less, without writing anything down - and they're not doing L'H\u00f4pital in their heads. There are other ways of thinking.", "Solution_7": "[quote=\"Kent Merryfield\"]From my L'H\u00f4pital is not everything campaign:\n\nFind $ \\lim_{x\\to0}\\frac {1 \\minus{} \\cos(x^3)}{\\sin^3 (x^2)}.$\n\nBear in mind that there are a significant number of regular readers here who knew the answer in 10 seconds or less, without writing anything down - and they're not doing L'H\u00f4pital in their heads. There are other ways of thinking.[/quote]\r\n\r\nAh!! Yes! We used to get those in our final exams(mostly involving a 1/sin x term that always used to stick around if we used l'hospitals rule). We could almost always solve the problem by using power series more elegantly :D", "Solution_8": "[quote=\"Kent Merryfield\"]From my L'H\u00f4pital is not everything campaign:\n\nFind $ \\lim_{x\\to0}\\frac {1 \\minus{} \\cos(x^3)}{\\sin^3 (x^2)}.$\n\nBear in mind that there are a significant number of regular readers here who knew the answer in 10 seconds or less, without writing anything down - and they're not doing L'H\u00f4pital in their heads. There are other ways of thinking.[/quote]\r\n\r\nlol kent, I'm in high school!" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Suppose S\u2282R^n is compact, f : S -> R^n is continuous, and f(x) > 0 for every x\u2208S . Show that there is a number \r\nc > 0 such that f(x) >= 1/c for every x\u2208S. I tried to use the definition of compactness but i'm not sure i solved the problem.", "Solution_1": "The big theorem on compactness: The continuous image of a compact set is compact.\r\n\r\nNow, the set $\\{f(x): x\\in S\\}$ is compact as a set of real numbers. A compact subset of $\\mathbb{R}$ has a minimum (and it has a maximum, and it's closed). This minimum is $\\frac1c$.", "Solution_2": "well we have $Imf=[\\alpha, \\beta]$ with $\\alpha >0$ thus we can choose $c$ such as ${\\frac{1}{c}=\\alpha}$ and that's it :thumbup:" } { "Tag": [ "algebra", "polynomial", "inequalities", "algebra unsolved" ], "Problem": "prove that the equation $x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e=0$ have $5$ reel roots then \r\n$2a^{5}<5b$.", "Solution_1": "I believe this follows from Maclaurin's inequalities. \r\n\r\nhttp://en.wikipedia.org/wiki/MacLaurin's_inequality", "Solution_2": "But MacLaurin's inequality only works for positive reals!", "Solution_3": "Derive the polynomial $3$ times, and know this has $2$ distinct real roots, using the mean value theorem.", "Solution_4": "yes Olorin i have the same idea but i do not find the desirid condition :|", "Solution_5": "[quote=\"arkhammedos\"]prove that the equation $x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e=0$ have $5$ reel roots then \n$2a^{5}<5b$.[/quote]\r\nIt is tupo. Consider $(x+t)(x+2t)(x+3t)(x+4t)(x+5t)=x^{5}+15tx^{4}+85t^{2}x^{3}+...$\r\nBut $2a^{5}-5b=2*15^{5}t^{5}-425t^{2}>0$ for $t>0$.\r\nCondition must be unique deg, were $deg(a)=1, deg(b)=2,deg(c)=3,deg(d)=4,deg(e)=5$, for example $a^{5}-bc+e$.", "Solution_6": ":lol: The correct condition is $2a^{2}>5b$. Didn't see that.", "Solution_7": "[quote=\"olorin\"]:lol: The correct condition is $2a^{2}>5b$. Didn't see that.[/quote]\r\nNo.\r\nLet $x\\to x+t$, then \r\n$a\\to a+5t,b\\tob+4at+10t^{2},c\\to c+3bt+6at^{2}+10t^{3},d\\to d+2ct+3bt^{2}+4at^{3}+5t^{4},e\\toe+dt+ct^{2}+bt^{3}+at^{4}+t^{5}$.\r\nCondition must be invariant.", "Solution_8": "And it is!? :huh: \r\n$2a^{2}>5b\\Leftrightarrow 2(a+5t)^{2}>5(b+4at+10t^{2})$", "Solution_9": "$2a^{2}>5b$ needed, because $f'''(x)=6(10x^{2}+4ax+b)$ must had real roots. But it is not sufficiently condition." } { "Tag": [ "inequalities", "geometry", "perimeter", "induction", "algorithm", "combinatorics unsolved", "combinatorics" ], "Problem": "In a small town, there are $n \\times n$ houses indexed by $(i, j)$ for $1 \\leq i, j \\leq n$ with $(1, 1)$ being the house at the top left corner, where $i$ and $j$ are the row and column indices, respectively. At time 0, a fire breaks out at the house indexed by $(1, c)$, where $c \\leq \\frac{n}{2}$. During each subsequent time interval $[t, t+1]$, the fire fighters defend a house which is not yet on fire while the fire spreads to all undefended [i]neighbors[/i] of each house which was on fire at time t. Once a house is defended, it remains so all the time. The process ends when the fire can no longer spread. At most how many houses can be saved by the fire fighters?\r\nA house indexed by $(i, j)$ is a [i]neighbor[/i] of a house indexed by $(k, l)$ if $|i - k| + |j - l|=1$.", "Solution_1": "Finally we can discuss it :)", "Solution_2": "I personally think that it takes 20 minutes to read and understand the question completely. :(", "Solution_3": "to be frank, i still dun really understand this question till now............. :(\r\ndoes it mean that, only ONE random house broke out of fire at first?", "Solution_4": "[quote=\"siuhochung\"]to be frank, i still dun really understand this question till now............. :(\ndoes it mean that, only ONE random house broke out of fire at first?[/quote]\r\n\r\nYes, I believe that $c$ is a fixed number. Only house is on fire initially. And we seek for an answer in terms of $n$ and $c$.", "Solution_5": "Answer: $n^2 - n$ \r\n\r\n\r\n[hide]\nThere are a maximum of 4 neighbouring houses for any given house: \none just above and one just below , one to the right and one to the left. (Similar to all the positions in which a king can move on a chess board at any given time)\n\nAssume that at time t=0 the fire started on house (1,1).\n\nIn time [0,1] \ndefend the house (1,2) while the fire spreads to house (2,1)\n\nIn time [1,2] \ndefend the house (2,2) while the fire spreads to house (3,1)\n\nIn time [2,3] \ndefend the house (3,2) while the fire spreads to house (4,1)\n\n .\n .\n .\ncontinue as above till time [n-1,n] during which you defend the house (n,2).\n\nThus you have defended all the houses in the 2nd column, and the fire cannot spread to 3rd ,4th , etc columns. \n\nSo you have defended $n^2 - n$ houses.[/hide]", "Solution_6": "Answer: $(n-c)*n$\r\n\r\nThe trick is to first defend the house (1,c+1) then (2,c+1) , then (3,c+1) and so on ... till (n,c+1).\r\n\r\nThus you would have defended all the houses in the coloumn $c+1$\r\n\r\nhence total number of houses defended would be $(n-c)*n$", "Solution_7": "Well, symmetricmean, of course you could defend some more houses (namely $c-2$ ) once you've saved $(1, c+1), ..., (n, c+1)$, so it would be $n^2-cn+c-2$, but that isn't optimal. I didn't get this problem at the APMO but other contestants told me that saving $(2, c)$ first, one would get the fire spreading in 2 directions and could save houses on each side alternately giving a V-like pattern (this is easier to see in a diagram), which is more efficient. However, no one I know gave a rigorous proof that this was optimal, either.", "Solution_8": "I agree with Severius; this v-shaped pattern yields the formula $n(n-c)+c^2-c$, which is larger than $n(n-c)$. However, I do not have a rigorous proof that this is optimal either.", "Solution_9": "It isn't hard at all to show that the fighters can save at least $n(n-c)+c(c-1)$ houses... But then, why can't they save mor houses? Here are some ideas i thought, although i don't know if they work because it is very dificult to formally prove them. In the test i did a solution, but asumming this two lemmas were true :lol: \r\n (1) First name the houses with numbers, acording to their minimal distance (how many houses they have to walk) to the house $(1,c)$. Then it would help a lot if we could prove that \"the more convenient\" thing to do is to save houses with different numbers.. i.e. not to save a house with number 5 in the fourth turn, because saving a house with number 4 would \"help\" also to protect that house with number 5. \r\n (2) In an optimal pattern, it is \"not convenient\" to save two houses in the same column. This is intuitively right, because how could it be convenient to save two houses from the same column, if we want to make an \"horizontal barrier\"?? (well, we can make a vertical barrier with $c=1$, but in the other cases is almost evident that the best thing to do is a conected barrier that covers from row 1 to $n$). \r\n\r\nIt can be easily proved that we must have at least one saved house per column. So here we can also think about the number of turns the fire can spread. \r\n\r\n If anyone has solved the problem i would like him/her to post some hints to formalize things here...", "Solution_10": "I have not solved the problem, but it seems like the following logic is workable:\r\n\r\n1) Show that after the kth rescue (defending a house at time k), there are at least k dangerous houses (a dangerous house is a house on fire that can still spread to at least one of its neighbours)\r\n\r\n2) Using 1), show that after the kth rescue, at least k+1 more houses will be lit up on fire.\r\n\r\n3) Using 2), we can effectively calculate the rest of the problem.\r\n\r\nNow of course, 1) wouldn't be true for big k (namely k>c) but for the time being, we can suppose that c is sufficiently large for us to recognize and formally prove the above patterns.\r\n\r\nDoes this help anyone?", "Solution_11": "I triedo some inequalities with the perimeter of the endangered zone and the longest walk of house on fire....It could work but i reallly screwed everything in this exam... :(", "Solution_12": "Can someone post a valid solution please?", "Solution_13": "This isn't something I came up with myself, it's the official solution.\r\n\r\nLabel the houses according to their \"distance\" from (1,c) as suggested by RaMlaF: say that house (i, j) is at level t if $\\mid i - 1\\mid + \\mid j - c\\mid = t $. Let d(t) be the number of houses at level t defended by time t, p(t) the number of houses at level t + 1 or greater defended by time t. Clearly $p(t + 1) + d(t + 1) \\leq p(t) + 1. $ (*)\r\n\r\nLet s(t) be the number of houses at level t which are not burning at time t. We prove by induction that $s(t) \\leq t - p(t)$ for $1\\leq t \\leq n-1$. The base case t = 1 is clearly true, so suppose it is true when t = k. At time k + 1, consider the set M(k+1) of the s(k+1) - d(k+1) houses at level k+1 that do not burn but are not themselves defended. These houses survive because all their neighbours at level k do not burn. There are at least s(k+1) - d(k+1) houses at level k that are neighbours of at least one house in M(k+1), so there must be at least s(k+1) - d(k+1) houses at level k that do not burn; that is, $s(k) \\geq s(k+1) - d(k+1).$ But by the inductive hypothesis $s(k) \\leq k - p(k)$, so $s(k+1) - d(k+1) \\leq k - p(k)$. Hence, using (*), $s(k+1) \\leq k - p(k) + d(k+1) \\leq k + 1 - p(k+1)$ also.\r\n\r\n$p(t) \\geq 0$ for all t, so from that it follow that $s(t) \\leq t$ for all $1\\leq t \\leq n-1$. Summing, we find that at most $\\frac 12 n(n-1)$ houses at levels n-1 or below can be defended. It is easy to see that the v-shaped algorithm described by Severius and others saves exactly this many houses at levels n-1 and below, and every house at level n and above, so this must be the optimal algorithm, and at most $n(n-c) + c(c-1)$ houses can be saved.", "Solution_14": "the Solution : n^2+c^2-nc-c\r\n\r\nI have a hard copy of the APMO solutions\r\n\r\nbut I can't post to here ,I can't use the system", "Solution_15": "Can we extend the problem with arbitary (i, j). Anyone has some idea for it ?", "Solution_16": "Can we extend the problem with arbitary (i, j). Anyone has some idea for it ?", "Solution_17": "There is the solution" } { "Tag": [ "geometry", "Asymptote", "FTW", "similar triangles" ], "Problem": "In the diagram, RECT and LONG are rectangles. How many square units are in the area of LONG?\n\n[asy]pair L,O,N,G,R,E,C,T; unitsize(.5cm);\nT = (0,7); C = (13,7); E = (13,0);\nL = (0,4); O = (1,7); N = (13,3); G= (12,0);\ndraw(R--E--C--T--R); draw(L--O--N--G--L);\nlabel(\"R\",R,W);label(\"E\",E,dir(0));label(\"C\",C,NE);label(\"T\",T,NW);\nlabel(\"L\",L,W);label(\"O\",O-(3.5,0),N);label(\"N\",N-(3,0),E);label(\"G\",G,S);\nlabel(\"3\",(T+L)/2,W); label(\"1\",(T+O)/2-(3.5,0),N);[/asy]", "Solution_1": "Triangle OCN ~ Triangle LTO\r\nSo OC/CN=3\r\n\r\nOC=3x\r\nCN=x\r\n\r\nTC=3x+1, TR=3+x\r\nOL=sqrt(10)\r\nON=sqrt(10)*x\r\n\r\nSo area of LONG=10x\r\narea of RECT=3x^2+10x+3\r\n\r\nI do not think this can be solved.\r\n\r\n[geogebra]ceeb7d8f9a711d6da8aaf24d771deb7d21074510[/geogebra] \r\nSlide C. We cannot determine the area given what we currently know.\r\n\r\nACCORDING to the asymptote file, x=4.\r\nTherefore area=40.", "Solution_2": "[quote=\"james4l\"]Triangle OCN ~ Triangle LTO\nSo OC/CN=3\n\nOC=3x\nCN=x\n\nTC=3x+1, TR=3+x\nOL=sqrt(10)\nON=sqrt(10)*x\n\nSo area of LONG=10x\narea of RECT=3x^2+10x+3\n\nI do not think this can be solved.\n\n[geogebra]ceeb7d8f9a711d6da8aaf24d771deb7d21074510[/geogebra] \nSlide C. We cannot determine the area given what we currently know.\n\nACCORDING to the asymptote file, x=4.\nTherefore area=40.[/quote]\nI did this problem on FTW and it said the answer was 70.\nI dont know how to do it. can someone show me?", "Solution_3": "In the original problem, CN is given as 7.\nSide OC is then 21. Use similar triangles to calculate.\nSegment ON is 7sqrt 10\nSegment LO is sqrt 10\nThe product is 70" } { "Tag": [ "convex polygon", "geometry", "pentagon", "angles", "IMO Shortlist" ], "Problem": "In the convex pentagon $ ABCDE,$ the sides $ BC, CD, DE$ are equal. Moreover each diagonal of the pentagon is parallel to a side ($ AC$ is parallel to $ DE$, $ BD$ is parallel to $ AE$ etc.). Prove that $ ABCDE$ is a regular pentagon.", "Solution_1": "Let $DE$ meet $AB$ at $M$ and $DC$ meet $AB$ at$N$.\r\n$EC$ parallel to $AB$=>$EC$ parallel to$MN$,but $ED=DC$=>$DM=DN$=> $EM=CN$.\r\n$AE$paralel to $BD$ =>$MA/AB=ME/ED$and $BC$paralel $AD$=>$NB/AB=NC/CD$,but\r\n$ME/ED= NC/CD$=>$MA/AB=NB/AB$=>$MA=NB$\r\n$MA=NB$,$EM=CN$and$$AE=BC$\r\nRespectiveli $BC=AB$=>$BA=ED$=>$MA=ME$=>$y^2$. Put $x=y^2+k$ where $k$ is positive integer. We have\r\n$(y^2+k)^2+(y^2+k+1)^2=y^4+(y+1)^4$ so then\r\n$k(2y^2+k)=(2y-k)(2y^2+2y+k+2)$ so $k<2y$ and $(2y-k)(k+2)-2yk-k^2$ is divisible by $2y^2$. This implies $4y-2k^2-2k$ is divisible by $2y^2$ and then $2y=k^2+k$ or $4y-2(k^2+k)\\geq 2y^2$ or $-4y+2(k^2+k)\\geq 2y^2$\r\n\r\nThe second is fail! So \r\n1) If $2y=k^2+k$ then $2y^2+k=k(2y^2+2y+k+2)$ which is fail cause the LHy$ which gives $x\\geq y^2+y+1$ then\r\n$x^2+(x+1)^2\\geq (y^2+y+1)^2+(y^2+y+2)^2>y^4+(y+1)^4$\r\nThe left is $t=1$ and then \r\n$y=k^2+k$ and then $2y^2+k=(2k^2+k)(2y^2+2y+k+2)>2y^2+k$ , i think :D \r\nSo we are done!", "Solution_2": "You could also write the equation as:\r\n\r\n((y+1)^2 - x)*((y+1)^2 + x) = (x+1 - y^2)*(x+1 + y^2)\r\n\r\nFrom here we see that y^2 <= x < (y+1)^2. So write x = y^2 + r, where 0 <= r < 2y + 1. Then the equation becomes\r\n\r\n(2y + 1 - r)*(2y^2 + 2y + r + 1) = (r + 1)*(2y^2 + r + 1)\r\n\r\nFrom here one can see that when r > y then the RHS > LHS and when r <= y then RHS < LHS, so the equation can never hold.", "Solution_3": "$x^2+(x+1)^2=y^4+(y+1)^4$\r\n$2x^2+2x+1=2y^4+4y^3+6y^2+4y+1$\r\n$x^2+x+1=y^4+2y^3+3y^2+2y+1$\r\n$x^2+x+1=(y^2+y+1)^2$\r\n\r\nBut $x^2document properties->fonts. Don't forget that fonts sometimes have to be paid for.\r\n\r\nAlso have a look at the examples at [url=http://www.tug.dk/FontCatalogue/]The LaTeX Font Catalogue[/url]" } { "Tag": [ "function", "probability", "complex analysis", "complex analysis unsolved" ], "Problem": "I'd like to see different solutions for this problem:\r\nsuppose $ f_{1}(t)$ and $ f_{2}(t)$ is coresponding characteristic function of independent random variables $ \\xi_{1},\\xi_{2}$\r\nif $ f_{1}(t)f_{2}(t) \\equal{} e^{\\minus{}t^{2}}$\r\nthen $ f_{1}$ and $ f_{2}$ is analytical on $ \\mathbb{C}$.", "Solution_1": "The shortest one I know is the following: $ P(\\xi_1\\plus{}\\xi_2>t\\plus{}C)\\ge P(\\xi_1\\ge C)P(\\xi_2>t)$. Thus, since $ \\xi_1\\plus{}\\xi_2$ is Gaussian, the probability that $ \\xi_2>t$ decays like $ e^{\\minus{}c t^2}$ as $ t\\to\\plus{}\\infty$ (just choose $ C\\in\\mathbb R$ such that $ P(\\xi_1\\ge C)>0$). Similarly, the probability that $ \\xi_2\\lambda\\}$ with appropriately chosen balls, and apply the basic covering lemma." } { "Tag": [], "Problem": "Determine all the natural number pairs of $ a$ and $ b$ such that $ \\frac{a\\plus{}1}{b}$ and $ \\frac{b\\plus{}1}{a}$ are also natural numbers.", "Solution_1": "Sorry, wrong question", "Solution_2": "The only solutions are $ (1,1),(1,2),(2,1),(2,3),(3,2)$.\r\n\r\nLet $ b \\plus{} 1 \\equal{} ma$ and $ a \\plus{} 1 \\equal{} nb$ for positive integers $ m$ and $ n$. Then $ b \\equal{} ma \\minus{} 1$, so we have $ a \\plus{} 1 \\equal{} n(ma \\minus{} 1) \\equal{} mna \\minus{} n \\implies a(1 \\minus{} mn) \\equal{} \\minus{} n \\minus{} 1$. Therefore, $ a \\equal{} \\frac {n \\plus{} 1}{mn \\minus{} 1}$. Similarly, $ b \\equal{} \\frac {m \\plus{} 1}{mn \\minus{} 1}$,\r\n\r\nLooking at $ a$, if $ m \\equal{} 1$, then we have $ n \\equal{} 2,3$. If $ m \\equal{} 2$, we have $ n \\equal{} 1$. If $ m \\equal{} 3$, we also have $ n \\equal{} 1$, and for $ m\\geq 4$, we have no solution. The same can be said of $ b$.\r\n\r\nSo our only solutions are the ones stated above.", "Solution_3": "[quote=\"modularmarc101\"]Determine all the natural number pairs of $ a$ and $ b$ such that $ \\frac {a \\plus{} 1}{b}$ and $ \\frac {b \\plus{} 1}{a}$ are also natural numbers.[/quote]\r\n\r\nClearly $ a\\plus{}1 \\ge b$ and $ b\\plus{}1 \\ge a$\r\n\r\nSo $ a\\plus{}1 \\ge b \\ge a\\minus{}1$\r\n\r\nWhen $ b\\equal{}a\\plus{}1$ we get $ a|[(a\\plus{}1)\\plus{}1]$, gives $ a\\equal{}1,2 \\Rightarrow (a,b)\\equal{}(1,2),(2,3)$\r\n\r\nWhen $ b\\equal{}a$ we get $ a|(a\\plus{}1)$, gives $ a\\equal{}1 \\Rightarrow (a,b)\\equal{}(1,1)$\r\n\r\nWhen $ b\\equal{}a\\minus{}1$, we get $ (a\\minus{}1)|(a\\plus{}1)$, gives $ a\\equal{}2,3 \\Rightarrow (a,b)\\equal{}(2,1),(3,2)$", "Solution_4": "I like that solution. :)" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Do they exist positive number solution of\r\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4 \\plus{} e^4 \\equal{} 3333^{3333}$.\r\nIf there are exist give some example of solution.", "Solution_1": "$ LHS\\equiv 5(8)$ so $ a,b,c,d,e$ are all odd.\r\nBecause of $ LHS\\equiv 9(16)$, and $ odd^4\\equiv 1,9(16)$, we get that there are not solutions.", "Solution_2": "I don't understand what pelao wrote.\r\nFirst, it seems there's confusion between the $ LHS$ and the $ RHS$.\r\nSecond, $ RHS\\equiv 5(\\text{ mod }16)$ and not $ 9$ which gives no contradiction.", "Solution_3": "Anyone? I can't solve this problem!", "Solution_4": "Aaaaargh. $ 3333^{3333}$ is a sum of [b]six [/b]fourth powers, but I'm still not sure about 5 :maybe: \r\n\r\nTo make 6, notice that \r\n\r\n$ 3333 \\equal{} 3^4\\plus{}3^4\\plus{}5^4\\plus{}5^4\\plus{}5^4\\plus{}6^4$\r\n\r\nso \r\n\r\n$ 3333^{3333} \\equal{} 3333^{3332} \\times 3333 \\equal{} (3333^{833})^4(3^4\\plus{}3^4\\plus{}5^4\\plus{}5^4\\plus{}5^4\\plus{}6^4) \\equal{}$\r\n\r\n$ \\equal{} (3m)^4\\plus{}(3m)^4\\plus{}(5m)^4\\plus{}(5m)^4\\plus{}(5m)^4\\plus{}(6m)^4$ where $ m \\equal{} 3333^{833}$.\r\n\r\nI don't think there's any simple argument that shows that 5 fourth-powers are impossible by reducing modulo some $ m$. By mod 16 you can see that $ a,b,c,d,e$ are all odd, and by mod 5 you can see that two of them are divisible by 5 and three aren't. But that doesn't get you very far." } { "Tag": [ "algebra open", "algebra" ], "Problem": "Color all points in plane with RED or GREEN .\r\nLet $ABC$ be a triangle .\r\nProve that There exists a triangle $MNP$~$ABC$ such that $M,N,P$ are the same color.", "Solution_1": "Hi,Minh! :) \r\nI think it's a combinatoric's problem,not algebra.Look: http://www.mathlinks.ro/Forum/topic-26773.html" } { "Tag": [ "calculus", "integration", "geometry", "calculus computations" ], "Problem": "Evaluate\r\n\r\n\\[\\frac{2005\\displaystyle \\int_0^{1002}\\frac{dx}{\\sqrt{1002^2-x^2}+\\sqrt{1003^2-x^2}}+\\int_{1002}^{1003}\\sqrt{1003^2-x^2}dx}{\\displaystyle \\int_0^1\\sqrt{1-x^2}dx}\\]", "Solution_1": "that would be a nice new year problem =P\r\n\r\nafter simplifying, you get the (area of quartercircle with radius 1003 - area of quartercircle with radius 1002)/(area of quartercircle with radius 1)\r\nanswer: 2005", "Solution_2": "Yes, you are right, isaacchao :) \r\n\r\nkunny" } { "Tag": [ "geometry", "incenter", "circumcircle", "trapezoid", "inradius", "algebra", "function" ], "Problem": "Dear Mathlinkers, \r\nlet ABCD a square, M a point on segment CD, I, X, Y the incenters wrt MAB, MAD, MBC, \r\nP, Q the meetpoints of the parallel to AD passing through X with BI, AC, \r\nS, R the meetpoints of the parallel to AD passing through Y with AI, BD. \r\nWe know, from another problem that M, X, P, S and Y are concyclic. \r\nLet be (1) this circle, U, V the second meetpoints of AC, BD with (1) and W the meetpoint of PU and SV.\r\nProve (perhaps with polar) that QW and PS are parallel.\r\nSincerely \r\nJean-Louis", "Solution_1": "$ (2)$ is circumcircle of isosceles trapezoid $ XQRY$ with center $ O.$ $ \\triangle PSW, \\triangle QRO$ are both isosceles right with $ PW \\equal{} SW, QO \\equal{} RO.$ From another problem, we also know that $ PS \\parallel QR$ and $ O \\in PS$ $ \\Longrightarrow$ the isosceles right $ \\triangle PSW \\cong QRO$ are congruent and $ W \\in QR.$", "Solution_2": "Dear Yetti and Mathlinkers,\r\nI was waiting your answer and was a little bit certain that you will used that O is on PS...\r\nMy question is : can we prove first that W is on QR, then it follows that O is on PS.\r\nThank you and sincerely\r\nJean-Louis", "Solution_3": "Dear Mathlinkers,\r\nsorry, I come back withe the same question:\r\n\r\ncan we prove first that W is on QR, then it follows that O is on PS.\r\n \r\nSincerely \r\nJean-Louis", "Solution_4": "$ K, L$ are midpoints of $ XQ, YR.$ $ a$ is the square side, $ x, y, r$ are radii of incircles $ (X), (Y), (I).$ $ XQ \\equal{} a \\minus{} 2x,$ $ YR \\equal{} a \\minus{} 2y,$ $ KL \\equal{} \\frac {_1}{^2}(XQ \\plus{} YR) \\equal{} a \\minus{} (x \\plus{} y).$ By the infamous inradii problem (see [url]http://www.mathlinks.ro/viewtopic.php?t=123243[/url], [url]http://www.mathlinks.ro/viewtopic.php?t=270926[/url], etc.), \r\n\r\n$ \\left(1 \\minus{} \\frac {2x}{a}\\right)\\left(1 \\minus{} \\frac {2y}{a}\\right) \\equal{} 1 \\minus{} \\frac {2r}{a}\\ \\Longrightarrow$ $ \\color{blue}\\boxed{r \\equal{} x \\plus{} y \\minus{} \\frac {2xy}{a}}$\r\n\r\nOn the other hand, for a square (see [url]http://www.mathlinks.ro/viewtopic.php?t=264276[/url]),\r\n\r\n$ \\frac {1}{r} \\equal{} \\frac {1}{a \\minus{} 2x} \\plus{} \\frac {1}{a \\minus{} 2y} \\equal{} \\frac {1}{XQ} \\plus{} \\frac {1}{YR} \\equal{} \\frac {2\\ KL}{XQ \\cdot YR}.$\r\n\r\nFrom similar triangles $ \\triangle ASY \\sim \\triangle AXD,$ $ \\triangle BPX \\sim \\triangle BYC$ (see [url]http://www.mathlinks.ro/viewtopic.php?t=273356[/url]), \r\n\r\n$ \\color{blue}\\boxed{PQ \\equal{} RS \\equal{} a \\minus{} \\left(2x \\plus{} 2y \\minus{} \\frac {2xy}{a}\\right) \\equal{} a \\minus{} (x \\plus{} y \\plus{} r)}$\r\n\r\n$ 2\\ PQ \\cdot KL \\equal{} 2\\ KL^2 \\minus{} 2r \\cdot KL \\equal{} KL^2 \\plus{} \\frac {_1}{^4} (XQ \\plus{} YR)^2 \\minus{} XQ \\cdot YR \\equal{}$ $ KL^2 \\plus{} \\frac {_1}{^4} (XQ \\minus{} YR)^2 \\equal{} QR^2$\r\n\r\nSince the triangles $ \\triangle QRO, \\triangle PSW$ are both isosceles right, it follows that\r\n\r\n$ \\color{blue}\\boxed{[PQRS] \\equal{} PQ \\cdot KL \\equal{} \\frac {_1}{^2}QR^2 \\Longleftrightarrow O \\in PS \\Longleftrightarrow W \\in QR}$\r\n\r\nAs synthetic as it can be. I do not see anything simpler than this calculation. Using inradius $ r$ simplifies it a lot, but is not really necessary.", "Solution_5": "Dear Vladimir,\r\nthank you very much for your help and your new idea with area...\r\nNow, I begin to think that we are in a limiting situation concerning two domains : calculation which appears to be necessary and \"shape\" which apparently doesn't work. \r\nI will continue to investigate this intersting parallelism...\r\nSincerely\r\nJean-Louis", "Solution_6": "Well, I think the two domains here are:\r\n\r\n1. The first equation for inradius $ r$ holds, when $ ABCD$ is just a parallelogram, with $ a$ as the distance between $ AB, CD,$ [color=red]but $ \\color{red}M$ must be on $ \\color{red}CD.$[/color]\r\n\r\n2. The second equation for inradius $ r$ holds even for $ M \\not\\in CD,$ (when $ M_1 \\equiv MA \\cap CD, M_2 \\equiv MB \\cap CD$ and $ x, y$ are inradii of $ \\triangle M_1AD, \\triangle M_2BC$), [color=red]but $ \\color{red}ABCD$ must be a square.[/color]\r\n\r\nProperties the 2 configurations cannot lead to a conclusion, which is beyond the capabilities of either of them, hence the shape does not work. But the full constraint for $ a, x, y$ with $ ABCD$ a square and $ M \\in CD$ is factorable into the above two simple equations for $ r,$ hence calculation works:\r\n\r\n$ a \\equal{} MD \\plus{} MC \\equal{} \\frac {2x(a \\minus{} x)}{a \\minus{} 2x} \\plus{} \\frac {2y(a \\minus{} y)}{a \\minus{} 2y} \\equal{}$\r\n\r\n$ \\equal{} \\frac {2a \\left(x \\plus{} y \\minus{} \\frac {2xy}{a}\\right)(a \\minus{} x \\minus{} y)}{(a \\minus{} 2x)(a \\minus{} 2y)} \\equal{} a \\left(x \\plus{} y \\minus{} \\frac {2xy}{a}\\right) \\left(\\frac {1}{a \\minus{} 2x} \\plus{} \\frac {1}{a \\minus{} 2y}\\right)$\r\n\r\n[hide=\"geogebra test\"] [geogebra]72ddc082d9c2aba9317c6a9a54cd93fd26f61dfa[/geogebra] [/hide]", "Solution_7": "I think you make it much more complicated than it is.\r\n In fact, the trapezoid XQRY is a pseudosquare (it's diagonals \r\n are congruent and orthogonal). The midpoints of the sides are \r\n vertices of a square, PWSZ are square vertices as well.\r\n Thus there are plenty of squares inscribed into XQRY.\r\n One of the diagonals of these squares envelopes a parabola, \r\n while the other passes thru a fixed point - it's focus.\r\n Miquel theorem must be right there.\r\n\r\n Hope this removes the last stumbling block on the path to a nice\r\n article in French on your wonderful site, jayme !\r\n\r\n Do not forget to include a historical note on Jules Mathot's point,\r\n or Anti-center as we know it presently.\r\n\r\n\r\n M.T.", "Solution_8": "[quote=\"armpist\"]I think you make it much more complicated than it is.\n In fact, the trapezoid XQRY is a pseudosquare (it's diagonals \n are congruent and orthogonal). The midpoints of the sides are \n vertices of a square, PWSZ are square vertices as well.\n Thus there are plenty of squares inscribed into XQRY.\n One of the diagonals of these squares envelopes a parabola, \n while the other passes thru a fixed point - it's focus.\n Miquel theorem must be right there.\n\n Hope this removes the last stumbling block on the path to a nice\n article in French on your wonderful site, jayme !\n\n Do not forget to include a historical note on Jules Mathot's point,\n or Anti-center as we know it presently.\n\n\n M.T.[/quote]\n\nThe parabola with focus O, in this case for an isosceles right pedal triangle rotated about its pedal point O, it exists in any triangle for any pedal point. It proves nothing about W, unless you already know that O is on PS, and then you do not need the parabola. After[quote=\"jayme\"]\n... can we prove first that W is on QR, then it follows that O is on PS...[/quote] then from the definition of P, S, you are supposed to prove either \"W is on QR\" or \"O is on PS\", not their trivial equivalence, which is all you did. Maybe you should cut such pretense of insight.", "Solution_9": "Many thanks for an educated advice, sorry for the trivial stuff.\r\n\r\n Hope the following will be a bit more complicated.\r\n\r\n\r\n From Yaglom we know that incircle (I) of AMB and the incircles (X) and (Y) have \r\n a common tangent. When a variable circle tangent to side AB and this common tangent \r\n moves along, the lines from A and B thru it's center intersect the verticals XQ and YR\r\n in 4 points. Lines connecting these points envelope two ellipses with some tangent lines easily \r\n available. Have to show that when this circle is in the position of AMB incircle, this tangent to one\r\n ellipse is in position PR. \r\n\r\n Still immaterial?\r\n\r\n\r\n M.T.", "Solution_10": "$ frac 1 (2)$", "Solution_11": "Dear Ruby,\r\nwhat is the sense of your message?\r\nSincerely\r\nJean-Louis" } { "Tag": [ "inequalities", "geometry", "circumcircle", "trigonometry", "complex numbers", "inequalities solved" ], "Problem": "In any triagle ABC prove that\r\nR/2r >=ma/ha>=(b/c+c/b)/2\r\nThanks!!", "Solution_1": "I've only worked out the first part:\r\n\r\nR/2r>=ma/ha iff pR/2S>=ma/ha=a*ma/a*ha=a*ma/2S iff p*R>=a*ma. Let z1, z2, z3 be the complex numbers corresponding to A, B, C, respectively. Let O (the circumcenter) be the origin. We thus have |z1|=|z2|=|z3|=R. We must prove R(|z1-z2|+|z2-z3|+|z3-z1|)>=|z2-z3|*|z2+z3-2z1| (do the calculations and you'll see) iff R(|z1-z2|+|z2-z3|+|z3-z1|)>=|z2^2-z3^2-2z1*z2+2z1*z3|. \r\n\r\nWe have |z2^2-z3^2-2z1*z2+2z1*z3|=|(z2^2-z1*z2)+(z1*z3-z3^2)+(z1*z3-z1*z2)|<= |z2||z1-z2|+|z3||z3-z1|+|z1||z2-z3| = R(|z1-z2|+|z2-z3|+|z3-z1|) Q.E.D.", "Solution_2": "For the second part:\r\n\r\nIt's easy (LOTS of computations) to see that 4a^2*ma^2=(b^2-c^2)^2+16S^2. We know that 4a^2*ha^2=16S^2, so ma^2/ha^2=(b^2-c^2)^2/16S^2 + 1 = (b^2-c^2)^2/(2bc)^2*sinA^2 + 1 >= (b^2-c^2)^2/(2bc)^2 + 1 = (b^2+c^2)^2/(2bc)^2 iff ma/ha>=(b^2+c^2)/2bc = (b/c+c/b)/2, Q.E.D.\r\n\r\nThe nasty thing about inequalities like these is that one should know things like 4a^2*ma^2=(b^2-c^2)^2+16S^2 practically by heart, since it's VERY boring to deduce stuff like this. I saw this particular formula in a book and realized I could use it here, but that's not a nice way to solve a problem...", "Solution_3": "I have nice approach for the second part:\r\n\r\nLet take triangle ABC with M is midpoint of BC. The line AM meets circumcircle again at D. We have\r\n AM.MD = MB.MC = a^2/4\r\nBut MD = AD - AM <= 2R - AM and we have\r\n m_a(2R - m_a) >= a^2/4\r\n=> 2Rm_a >= m_a^2 + a^2/4 = (b^2+c^2)/2 (1)\r\nBut h_a = 2S/a = bcsinA/a = bc/2R and (1) is exactly the required ineq.\r\n\r\nThis approach can be use for other special lines in triangle to get the interesting ineq. \r\n\r\nNamdung", "Solution_4": "[quote=\"grobber\"]It's easy (LOTS of computations) to see that 4a^2*ma^2=(b^2-c^2)^2+16S^2.[/quote]\r\n\r\nWell, the idenitity $4a^2m_a^2=\\left(b^2-c^2\\right)^2+16S^2$, where S is the area of triangle ABC and $m_a$ is the median from the vertex A, can be proven geometrically:\r\n\r\nLet $h_a$ be the altitude of triangle ABC from the vertex A. Then, the area of triangle ABC equals $S=\\frac12\\cdot a\\cdot h_a$. Hence, we can equivalently transform the identity in question as follows:\r\n\r\n$4a^2m_a^2=\\left(b^2-c^2\\right)^2+16S^2$ is equivalent to\r\n$4a^2m_a^2=\\left(b^2-c^2\\right)^2+16\\cdot\\left(\\frac12\\cdot a\\cdot h_a\\right)^2$ is equivalent to\r\n$4a^2m_a^2=\\left(b^2-c^2\\right)^2+4a^2h_a^2$ is equivalent to\r\n$4a^2m_a^2-4a^2h_a^2=\\left(b^2-c^2\\right)^2$ is equivalent to\r\n$\\left(2a\\right)^2\\left(m_a^2-h_a^2\\right)=\\left(b^2-c^2\\right)^2$ is equivalent to\r\n$m_a^2-h_a^2=\\left(\\frac{b^2-c^2}{2a}\\right)^2$.\r\n\r\nNow, in order to prove this equality, we consider the midpoint M of the side BC of triangle ABC, and the foot H of the altitude from the vertex A. Then, $AM=m_a$ and $AH=h_a$, and the triangle AHM is right-angled, so that the Pythagoras theorem yields $AM^2=AH^2+HM^2$. Consequently,\r\n\r\n$HM^2=AM^2-AH^2=m_a^2-h_a^2$.\r\n\r\nOn the other hand, since M is the midpoint of the side BC of triangle ABC, we have $BM=\\frac{BC}{2}=\\frac{a}{2}$, and since the point H is the foot of the altitude from the vertex A, the triangle BHA is right-angled, and thus $BH=AB\\cdot\\cos\\measuredangle ABH=c\\cdot\\cos B$. Thus,\r\n\r\n$HM=\\left|BM-BH\\right|=\\left|\\frac{a}{2}-c\\cdot\\cos B\\right|=\\left|\\frac{a}{2}-c\\cdot\\frac{c^2+a^2-b^2}{2ca}\\right|=\\left|\\frac{a}{2}-\\frac{c^2+a^2-b^2}{2a}\\right|$\r\n$=\\left|\\frac{a^2-\\left(c^2+a^2-b^2\\right)}{2a}\\right|=\\left|\\frac{b^2-c^2}{2a}\\right|$,\r\n\r\nand thus $HM^2=\\left|\\frac{b^2-c^2}{2a}\\right|^2=\\left(\\frac{b^2-c^2}{2a}\\right)^2$. Comparing this with $HM^2=m_a^2-h_a^2$, we get $m_a^2-h_a^2=\\left(\\frac{b^2-c^2}{2a}\\right)^2$, and the proof is complete.\r\n\r\n darij" } { "Tag": [ "function" ], "Problem": "Let $f(x)$ be a function which satisfies \\[f(x+29)=f(29-x)\\] for all real values of $x$. If $f(x)$ has exactly three real roots $\\alpha$, $\\beta$, $\\gamma$, determine the value of $\\alpha + \\beta + \\gamma$", "Solution_1": "$f(x)$ is symmetric about $x=29$.\r\nhence one root must be $29$, and the sum of the other two is $29\\cdot 2=58$.\r\nhence $\\alpha+\\beta+\\gamma = 29+58=87$." } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "Solve the equation \r\n\r\n$ \\cos^{\\minus{}1} x\\sqrt{3} \\plus{} \\cos x \\equal{} \\dfrac {\\pi}{2}$", "Solution_1": "[quote=\"corps0507\"]Solve the equation \n\n$ \\cos^{ \\minus{} 1} x\\sqrt {3} \\plus{} \\cos x \\equal{} \\dfrac {\\pi}{2}$[/quote]\r\n\r\n[color=darkblue]Can you give us exact of your problem $ \\cos^{ \\minus{} 1} (x\\sqrt {3}) \\plus{} \\cos x \\equal{} \\dfrac {\\pi}{2}$ OR $ (\\cos^{ \\minus{} 1}x). \\sqrt {3} \\plus{} \\cos x \\equal{} \\dfrac {\\pi}{2}$[/color]", "Solution_2": "actually that is the exact question ^^\r\n\r\ni'm still struggling to solve it, but i find it hard, and i think that the second term is really $ \\cos^{\\minus{}1} x$.", "Solution_3": "hello, let $ f(x)=arccos(x\\sqrt{3})+\\cos(x)-\\frac{\\pi}{2}$, then is $ f^'(x)=-\\frac{\\sqrt{3}}{\\sqrt{1-3x^2}}-\\sin(x)$ and $ f^'(x)$ is negative for all $ x$ with $ -\\frac{\\sqrt{3}}{3}0$. It follows that $ f(x)$ is strictly monotonic decreasing and we have only one intersection point with the $ x$-axis it is nearly $ 0.4520877966566117605650938111977306975578144064999181181478659087703923167869282588990606060510949004$.\r\nSonnhard." } { "Tag": [ "calculus", "integration", "geometry", "parameterization", "function", "abstract algebra", "algebra" ], "Problem": "I also think it seems too late for a 23-year-old man to start university life from 2005, eventhough his major is math!\r\nPart 2 of the 2-paged paper is achived on 2007-02-02, and part 3 is achived 2008-04-24. I submitted it to JAMS online.I am not sure whether it is suitable or not. It's my wishes to make everyone believe Fermat and to be a mathman accepted by Wiles.Would you please do me a favour to help me check it and send me your review please?", "Solution_1": "The first proof is absurd. You appear to be abusing the Chinese Remainder Theorem and conflating equality $ \\bmod uvw$ with equality. The fact that there are an infinite number of solutions to a modular system doesn't mean there are an infinite number of solutions over $ \\mathbb{Q}$ (which I assume is your base field) of an algebraic curve.\r\n\r\nAs for the second proof, $ \\gcd(z \\minus{} y, x) \\equal{} 1$ does not imply $ z \\minus{} y \\equal{} 1$. \r\n\r\n[b]Check your work.[/b] Very good mathematicians have been working on Fermat's Last Theorem for 350 years, and if you think a proof that simple hasn't been found by now, you don't understand any of the work that's been done in the 1800s alone. And you are just one in a [b]long line[/b] of mathematicians who think they've found proofs that turn out to be flawed. And the flaws here are very obvious.", "Solution_2": "about part 3 ,z-y is the factor of x^n.\r\nabout part 2,there are an infinite number of solutions over a straight line . so it's my argument.\r\n\r\nThank you. I have no idea how to express these details.", "Solution_3": "[quote=\"Taoli\"]about part 3 ,z-y is the factor of x^n.[/quote]\n\nSorry, I pointed out the wrong error. (I believe there are several, one of which is that your proof that $ (z \\minus{} y, x) \\equal{} 1$ is wrong.) Here is another: you assume there exists $ b$ such that $ z^b \\plus{} y^b \\equiv 0 \\bmod x$, but all you can really deduce is that $ z^n \\minus{} y^n \\equiv 0 \\bmod x$. You have done nothing to prove that $ b$ exists.\n\n[quote=\"Taoli\"]about part 2,there are an infinite number of solutions over a straight line .[/quote]\r\n\r\nThat's what I'm saying. There aren't. You've abused the Chinese Remainder Theorem. When you write $ (ua \\plus{} t)^n \\plus{} (vb \\plus{} t)^n \\equal{} (wc \\plus{} t)^n$, you are applying CRT to a particular solution $ x, y, z$. If you change the value of $ t$, you also change the value of $ a, b, c$ (assuming $ u, v, w$ is fixed) - in other words, the infinite number of solutions by CRT $ \\bmod uvw$ give [b]the same solution $ x, y, z$ over the integers.[/b]", "Solution_4": "My argument is that:\r\n\r\nno (x,y,z) over N whose b doesnot exist is the solution of Fermat equation.\r\n\r\nthe infinite number of solutions by CRT are given by the same solutionover the integers whose existance is supposed.\r\n\r\nI think that is the point.", "Solution_5": "[quote=\"Taoli\"]the infinite number of solutions by CRT are given by the same solutionover the integers whose existance is supposed.[/quote]\r\n\r\nThose solutions do not constitute an infinite number of solutions in the sense of Siegel's theorem. The infinite number of solutions in $ \\mathbb{Z}_{uvw}$ (which is not a field) translate to a finite number of solutions in $ \\mathbb{Z}$. \r\n\r\nMost mathematicians believe that it is highly unlikely that Fermat's original proof was correct. One common proposal is that Fermat incorrectly supposed that the ring of integers of a [url=http://mathworld.wolfram.com/CyclotomicField.html]cyclotomic field[/url] has unique prime factorization (which would imply FLT, but it isn't true); work in this direction led to the notion of [url=http://mathworld.wolfram.com/ClassNumber.html]class number[/url].", "Solution_6": "[quote=\"t0rajir0u\"]\nMost mathematicians believe that it is highly unlikely that Fermat's original proof was correct. One common proposal is that Fermat incorrectly supposed that the ring of integers of a [url=http://mathworld.wolfram.com/CyclotomicField.html]cyclotomic field[/url] has unique prime factorization (which would imply FLT, but it isn't true); work in this direction led to the notion of [url=http://mathworld.wolfram.com/ClassNumber.html]class number[/url].[/quote]\r\n\r\nI have learned some historical details. You are my first kindhearted friend here! Glad to see you! Many details are passed over since I am not able to express them exactly. I try to verify that there are an infinite number integral points both over an algebraic curve and a straight line. I refer to the geometry the sense of Siegel's theorem.\r\n\r\nI am a green hand. There is a long way,and I need a teacher to speed my study up.", "Solution_7": "[quote=\"Taoli\"]I try to verify that there are an infinite number integral points both over an algebraic curve and a straight line. [/quote]\r\n\r\nTry to write the equation of the line.", "Solution_8": "One of its forms is ua-vb-x+y=0.", "Solution_9": "The only way I open this address is to go to the netbar outside my campus, so I usually log here once a week.\r\nI am eager to learn further math.Thank you for your help.", "Solution_10": "[quote=\"Taoli\"]One of its forms is ua-vb-x+y=0.[/quote]\r\n\r\nThink about what your variables are. $ ua, vb$ are expressions that came from your use of CRT, so the only way this can be interpreted as a line is if $ x, y$ are your variables and $ ua, vb$ are fixed. But if $ ua, vb$ are fixed then $ t$ is fixed. This is not the equation of an algebraic curve at all; it is a trivial consequence of the way you defined $ t$. Do you see your error now?", "Solution_11": "[quote=\"t0rajir0u\"]When you write $ (ua \\plus{} t)^n \\plus{} (vb \\plus{} t)^n \\equal{} (wc \\plus{} t)^n$, you are applying CRT to a particular solution $ x, y, z$. If you change the value of $ t$, you also change the value of $ a, b, c$ (assuming $ u, v, w$ is fixed) - in other words, the infinite number of solutions by CRT $ \\bmod uvw$ give [b]the same solution $ x, y, z$ over the integers.[/b][/quote]\r\n\r\nWe can treat u,v,x,y as constant and a,b as variable. I use x,y,z as the solution of Fermat equation traditionally.", "Solution_12": "Yes, but when you change $ a, b$ you also change $ t$. I have yet to figure out how you got rid of $ t$ in your algebra but there is clearly something wrong in that pile of combinatorial expansions.", "Solution_13": "Through a parameter transformation from one space to another space.", "Solution_14": "Nonsense. The point is that you cannot legitimately use CRT to generate an infinite number of distinct integer solutions to a plane algebraic curve. What you'll end up doing (and this has to have happened in your proof - don't ask me to point out where because I'm not willing to sift through that mess of algebra) is generating an infinite number of distinct integer solutions to an algebraic variety of degree greater than $ 1$, which is no contradiction.", "Solution_15": "I am enjoying the feelings of thinking. But I am such an anxious fool that I have made one mistake after one. Ha, who are my mates? Let's learn more.\r\n\r\nThank you! I think I understand you now.", "Solution_16": "\\section{an elementary proof}\r\nLemma: If $ x^{n}\\plus{}y^{n}\\equal{}z^{n}(n>2), x, y, z\\in \\mathbb{Q}(w), w^{n\\plus{}1}\\equal{}\\minus{}1, n$ \r\nis odd, there is the solution of the following equations: $ z^b\\plus{}y^b\\equiv0, (modx)$.\r\n\\newline Proof: Since $ z^{n}\\plus{}w(wy)^{n}\\equal{}x^{n}$, hence $ z^{n(n\\plus{}1)}\\plus{}y^{n(n\\plus{}1)}\\equiv0, (modx)$.$ \\Box$\r\n\\newline There is an elementary proof of the following form of Fermat's last\r\ntheorem: there is no positive integer solution over $ \\mathbb{Q}(w), w^{n\\plus{}1}\\equal{}\\minus{}1$ meeting Fermat\r\nequation $ x^{n}\\plus{}y^{n}\\equal{}z^{n}(n>2), n$ is odd.\r\n\\newline Proof: Suppose $ (x,y,z)$ is one such solution of Fermat\r\nequation and $ (x,y)\\equal{}1, (y,z)\\equal{}1, (z,x)\\equal{}1, x$ is odd. Considering $ n$ is odd we let $ z>y, x\\equiv r, 12), x\\in \\mathbb{N}, y, z\\in \\mathbb{Q}(r,\r\nw), r\\equal{}(n\\plus{}1)^{\\frac{1}{n}}, w^{n\\plus{}1}\\equal{}\\minus{}1, n$ is odd.\r\n\\newline Proof: Suppose $ (x,y,z)$ is one such solution of Fermat\r\nequation and $ (x, y)\\equal{}1, (y, z)\\equal{}1, (z, x)\\equal{}1$. Since\r\n$ z^n\\plus{}w_i(w_iy)^n\\equal{}x^n$, hence $ z^{n(n\\plus{}1)}\\plus{}(w_iy)^{n(n\\plus{}1)}\\equiv0,\r\n(modx^n)$. Then we get $ w_i^nz^{n(n\\plus{}1)}\\plus{}y^{n(n\\plus{}1)}\\equiv0,\r\n(modx^n)$. It follows that\r\n$ \\Sigma_0^nw_i^nz^{n(n\\plus{}1)}\\plus{}(n\\plus{}1)y^{n(n\\plus{}1)}\\equiv0, (modx^n)$. Since\r\n$ w^{n\\plus{}1}\\plus{}1\\equal{}0$, hence $ \\Sigma_0^nw_i^n\\equal{}0$. It follows that\r\n$ (n\\plus{}1)y^{n(n\\plus{}1)}\\equiv0, (modx^n)$. Since $ (x, y)\\equal{}1$, hence\r\n$ x|(n\\plus{}1)^{\\frac{1}{n}}$. This is contradictory.$ \\Box$", "Solution_21": "[quote=\"Taoli\"]\n$ z^n \\plus{} w_i(w_iy)^n \\equal{} x^n$, hence $ z^{n(n \\plus{} 1)} \\plus{} (w_iy)^{n(n \\plus{} 1)}\\equiv0, (modx^n)$[/quote]\r\nIt seems to me that this is true only when $ n$ is even, not odd. And you're still using the greatest common divisor.", "Solution_22": "[quote=\"t0rajir0u\"][quote=\"Taoli\"]\n$ z^n \\plus{} w_i(w_iy)^n \\equal{} x^n$, hence $ z^{n(n \\plus{} 1)} \\plus{} (w_iy)^{n(n \\plus{} 1)}\\equiv0, (modx^n)$[/quote]\nIt seems to me that this is true only when $ n$ is even, not odd. And you're still using the greatest common divisor.[/quote]\r\n\r\n\r\nConsidering $ n$ is odd, we get $ n\\plus{}1$ is even.\r\nSince $ z^n \\plus{} w_i(w_iy)^n \\equal{} x^n$, hence $ z^{n(n \\plus{} 1)} \\plus{} (w_iy)^{n(n \\plus{} 1)}\\equiv z^{n(n \\plus{} 1)} \\minus{}(w_i)^{n \\plus{} 1} (w_iy)^{n(n \\plus{} 1)}\\equiv0, (modx^n)$.\r\n\r\nCould coprime be defined with any common divisor equal to 1?", "Solution_23": "One way to define a \"greatest common divisor\" is a divisor such that every other common divisor divides it. Over a ring with prime factorization, this divisor exists and is unique up to multiplication by units. Over a ring without prime factorization, this is no longer true.\r\n\r\nThe problem remains that you're assuming things like \"if $ \\gcd(a, b) \\equal{} 1$ and $ a | bc$ then $ a | c$,\" which don't hold when the class number is greater than $ 1$. You would probably benefit significantly from a text on algebraic number theory, although I'm afraid I don't have the experience to recommend you one.", "Solution_24": "For deeper studies, I can recommend J. Neukirch: Algebraic Number Theory (especially the first chapter). But it requires a good knowledge in superior algebra (rings, ideals, modules, fields, Galois theory). For superior algebra itself (including unique factorisation domains, euclidean domains), maybe S. Lang: Algebra is a good choice (up to what I've heard, I don't have that book).", "Solution_25": "Thank you! I need some chance to enter a good graduate school.\r\nI will work hard for it!", "Solution_26": "[quote=\"t0rajir0u\"]One way to define a \"greatest common divisor\" is a divisor such that every other common divisor divides it. Over a ring with prime factorization, this divisor exists and is unique up to multiplication by units. Over a ring without prime factorization, this is no longer true.\n\nThe problem remains that you're assuming things like \"if $ \\gcd(a, b) \\equal{} 1$ and $ a | bc$ then $ a | c$,\" which don't hold when the class number is greater than $ 1$. You would probably benefit significantly from a text on algebraic number theory, although I'm afraid I don't have the experience to recommend you one.[/quote]\r\n\r\nI haven't looked up the following lemmas:\r\nLemma1:If $ a^n|b^n$, then $ a|b$.\r\nLemma2:$ ye$ has the same factor $ y_i(\\neq e)$ as $ y$, here $ e$ is unity.\r\n\r\nPerhaps they works.", "Solution_27": "I will go for summer vocation.\r\n\r\nI will not be back until this September.\r\n\r\nGood luck, everyone!", "Solution_28": "[quote=\"Taoli\"]I see the former mistake in the lemma. help me check the following, I need some time to master the background knowledge.\n\n\n\\section{an elementary proof}\nThere is an elementary proof of the following form of Fermat's last\ntheorem: there is no positive integer solution meeting Fermat\nequation $ x^n \\plus{} y^n \\equal{} z^n(n > 2), x\\in \\mathbb{N}, y, z\\in \\mathbb{Q}(r, w), r \\equal{} (n \\plus{} 1)^{\\frac {1}{n}}, w^{n \\plus{} 1} \\equal{} \\minus{} 1, n$ is odd.\n\\newline Proof: Suppose $ (x,y,z)$ is one such solution of Fermat\nequation and $ (x, y) \\equal{} 1, (y, z) \\equal{} 1, (z, x) \\equal{} 1$. Since\n$ z^n \\plus{} w_i(w_iy)^n \\equal{} x^n$, hence $ z^{n(n \\plus{} 1)} \\plus{} (w_iy)^{n(n \\plus{} 1)}\\equiv0, (modx^n)$.\n\n\n\n\n Then we get $ w_i^nz^{n(n \\plus{} 1)} \\plus{} y^{n(n \\plus{} 1)}\\equiv0, (modx^n)$. \n\n[/quote]This is wrong.", "Solution_29": "Fermat's last theorem: there is no solution of Fermat equation $ x^n \\plus{} y^n \\equal{} z^n, x,y,z\\in \\mathbb{Z},xyz\\neq0,n$ is odd prime.\r\nProof: Suppose $ (x,y,z)$ is one such solution of Fermat equation and $ (x, y) \\equal{} 1, (y, z) \\equal{} 1, (z, x) \\equal{} 1$. \r\nLet $ z^n\\equal{} w^p,y^n\\equal{} v^p,x^n\\equal{} u^p,p\\equal{}\\phi(n), n\\equiv r \\bmod p$. \r\nSince $ n$ is odd prime, we can suppose $ (n,xy)\\equal{}1$. \r\nOver $ \\mathbb{Q}(d^{r/p}),d\\neq 0$, we get $ w^p\\equal{}u^p \\plus{} v^p \\equiv 1\\plus{}1 \\bmod n$.\r\nThis is contradictory.$ \\Box$" } { "Tag": [ "function", "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ A$ be a finite set and $ f: A\\to A$ a function.\r\nProve that there exists $ A_0,A_1,A_2,A_3$ pairwise disjoint subsets of $ A$ so that $ A\\equal{}A_0\\cup A_1\\cup A_2\\cup A_3$ with the property that $ f(x)\\equal{}x,\\forall x\\in A_0$ and $ f(A_i)\\cap A_i\\equal{}\\emptyset,i\\equal{}1,2,3$.\r\nIs this true if $ A$ is infinite?", "Solution_1": "Use induction on the number $ n$ of elements of $ A$.\r\n\r\nConsider the directed graph whose vertices are the element of $ A$, and the edge $ xy$ is drawn iff $ f(x)=y$.\r\n\r\nWlog, we may assume that the undirected graph with same vertices and edges is connected, otherwise just follow the reasoning on each connected component and form the desired sets from those found on each of these components (with respect that $ A_0 = \\{x\\in A/ f(x)=x\\}.$)\r\n- If there is some $ x\\in A$ which is the endpoint of at least two edges of the directed graph, then there is at least one vertex, say $ y$, which is the endpoint of none of the edges (and the converse is true). Note that $ f(y) \\neq y.$\r\nUse the induction hypothesis on $ A-\\{y\\}$. Now according to the set which contains $ f(y)$, it leaves at least two of the sets for $ y$.\r\n\r\n- If each vertex is the endpoint of exactly one edge, then the graph is a cycle, which can always be $ 3$-coloured (use two colours alternatively until you reach the last vertex and give it the third colour). this colouration leads to the desired partition.\r\n\r\nNote that in the case of some isolated vertices ($ f(a)=a$) and triangles ($ f(a)=b,f(b)=c,f(c)=a$), the partition cannot be realised into less than four sets.\r\n\r\nPierre." } { "Tag": [ "algebra", "function", "domain", "logarithms", "inequalities" ], "Problem": "Let $ n$ be a positive integer. If the number of integers in the domain of $ y=\\log ((1-x)(x-n))$ is $ 2n-6$, the what is $ n$?", "Solution_1": "[quote=\"ordinary boy\"]Let $ n$ be a positive integer. If the number of integers in the domain of $ y=\\log ((1-x)(x-n))$ is $ 2n-6$, the what is $ n$?[/quote]\r\n\r\nI'm assuming base 10.\r\n\r\n[hide]The domain of that is all $ x\\in \\mathbb{R}$ such that\n\n$ (1-x)(x-n) > 0.$\n\nSolving this inequality for $ x$ yields:\n$ 1 0.$\n\nSolving this inequality for $ x$ yields:\n$ 10, not equal to 1)" } { "Tag": [ "function", "integration", "algebra", "domain", "Functional Analysis", "complex analysis" ], "Problem": "Hi! I have two problems on approximating a function $ 1/z$ on a unit circle $ \\mathbb{T} : \\equal{} \\{z \\in \\mathbb{C} | |z| \\equal{} 1\\}$ with polynomials.\r\n\r\n1. Let $ \\ A \\subsetneq \\mathbb{T}$ be a compact subset. Prove that for $ \\epsilon > 0$ there exist $ \\ a_0, \\ldots, a_n \\in \\mathbb{C}$ such that\r\n\\[ \\sup_{z \\in A} \\left|\\frac {1}{z} \\minus{} \\sum_{j \\equal{} 0}^{n}a_jz^j\\right| < \\epsilon.\r\n\\]\r\n2. Prove that for any $ \\ a_0, \\ldots, a_n \\in \\mathbb{C}$\r\n\\[ \\sup_{z \\in \\mathbb{T}} \\left|\\frac {1}{z} \\minus{} \\sum_{j \\equal{} 0}^{n}a_jz^j\\right| \\geq 1.\r\n\\]", "Solution_1": "#2:\r\n$ 2\\pi\\equal{}\\left|\\int_{\\mathbb{T}}p(z)\\minus{}\\frac1z\\,dz\\right|\\le \\int_{\\mathbb{T}}\\left|p(z)\\minus{}\\frac1z\\right|\\,|dz|\\le 2\\pi\\max_{z\\in \\mathbb{T}}\\left|p(z)\\minus{}\\frac1z\\right|$\r\nActually, this applies to any approximation with a function analytic on the closed disk.", "Solution_2": "I can do #1 with a standard functional analysis argument (something like a functional analysis oriented proof of runge's theorem), but maybe there is a much simpler argument for the particular domain in question.\r\n\r\nedit: Ohps, #1 is in fact a direct application of Runge's theorem. I forgot exactly what the theorem said in more precise versions :P" } { "Tag": [ "geometry" ], "Problem": "$\\overline{AB}$ and $\\overline{ED}$ are parallel tangents of the circle $O$ at points $A$ and $E$, respectively, cutting a third tangent $\\overline{BCD}$ at points $B$ and $D$, respectively. If $AB = 8$ and $ED = 18$, what is the radius of the circle?", "Solution_1": "[hide=\"Answer\"]$r=\\sqrt{AB\\cdot ED}$.[/hide]" } { "Tag": [ "logarithms" ], "Problem": "Could Someone proof the NERNST EQUATION?", "Solution_1": "consider the equation:\r\n$aA+bB\\leftrightharpoons cC+dD$\r\nwe know:\r\n$Q=\\frac{[C] ^c[D]^d}{[A]^a[B]^b}$\r\n$\\Delta G =\\Delta G^{\\circ} +RT\\ln{Q}$\r\nand\r\n$\\Delta G=-nF\\Delta E$\r\n$\\Rightarrow -nF\\Delta E=\\Delta G^{\\circ} +RT\\ln{Q}\\Rightarrow \\Delta E=\\Delta E^{\\circ} -\\frac{RT}{nF}\\ln{\\frac{[C] ^c[D]^d}{[A]^a[B]^b}}$\r\n$\\mathbb{QED}$ :)\r\n\r\nnote that in equilibrium $\\Delta E^{\\circ}=0, Q_{eq}=K$" } { "Tag": [ "probability", "conditional probability" ], "Problem": "Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?\r\n\r\nThanks for the help!", "Solution_1": "[hide=\"Counting\"]Team B only won 1 game: well, only 1 case - ABAA\nTeam B wins twice, then there's 3 places to put the second game: BBAAA, ABBAA, ABABA\n\nSo, it's a $ \\frac{1}{4}$ chance.[/hide]", "Solution_2": "pascal, you missed the fact that they aren't equally likely. You have to weight them appropriately.\r\n\r\n[hide=\"Solution\"] We have one of the following two cases, where _ denotes an unknown:\n\n_ B _ _ A\n_ B _ A\n\nIn the first case, we have to fill in the blanks with an A, another A, and a B. The probability that the first is a B is $ \\frac {1}{3}$.\n\nIn this case, A must win all the games, so the only possibility is ABAA. So the probability for this case is $ 0$.\n\nWe add to get $ 0 \\plus{} \\frac13 \\equal{} \\boxed{\\frac {1}{3}}$.[/hide]\r\n\r\nEDIT: I realized my mistake. Ignore this solution.", "Solution_3": "@PowerOfPi: You're right that the cases have to be weighted, but I don't think you did it exactly right.\r\n[hide=\"My Solution\"]By the rule for conditional probability, $ P(A|B) \\equal{} \\frac {P(A\\&B)}{P(B)}$ ($ P(A|B)$ is the probability of event $ A$ given event $ B$ occurs). In this case, $ A$ is the event that team B wins the first game, and $ B$ is the event that team B wins the first game and team A wins the series.\n\n$ P(A\\&B)$\nThe only possibility if team B wins the first and second games but team A wins the series is BBAAA. This occurs with probability $ \\left(\\frac {1}{2}\\right)^5 \\equal{} \\frac {1}{32}$, since there is a 1/2 chance that a given game will match this pattern. Thus, $ P(A\\&B) \\equal{} \\frac {1}{32}$.\n\n$ P(B)$.\nAs mentioned in the previous solutions, the only possibilities are ABAA, ABABA, ABBAA, and BBAAA. The probability that ABAA will occur is $ \\left(\\frac {1}{2}\\right)^4$, and the probability for each of the remaining possibilities is $ \\left(\\frac {1}{2}\\right)^5$. Thus, $ P(B) \\equal{} \\frac {1}{16} \\plus{} \\frac {3}{32} \\equal{} \\frac {5}{32}$.\n\nNow, we just divide:\n$ P(A|B) \\equal{} \\frac {P(A\\&B)}{P(B)} \\equal{} \\frac {1/32}{5/32} \\equal{} \\boxed{\\frac {1}{5}}$.[/hide]\r\n\r\nAnother solution:\r\n[url]http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10A_Problems/Problem_18[/url]\r\n\r\nEDIT: Changed answer to 1/5 instead of 5/32, as grn_trtle pointed out.", "Solution_4": "Ugh, great, we now have 4 differing solutions?\r\nMine's probably wrong, since I didn't take into account that they're weighted differently.\r\n\r\nMathWise:\r\nIt's already a given that B won the 2nd game, so wouldn't $ P(A\\&B) \\equal{} \\left(\\frac{1}{2}\\right)^{4} \\equal{} \\frac{1}{16}$?\r\nSo then ABAA would be $ \\left(\\frac{1}{2}\\right)^{3}$ and so the total would then be $ \\frac{5}{16}$", "Solution_5": "Intriguingly, none of your answers is right :lol:\r\n\r\n[hide=\"Solution\"]\nThe only possibilities are: ABAA, BBAAA, ABBAA, ABABA\nABAA will happen 1/16 of the time, the rest 1/32 of the time.\n\nIn only one of these does B win the first game, so $ \\frac {1/32}{3/32 \\plus{} 1/16} \\equal{} \\boxed{\\frac {1}{5}}$.\n[/hide]\r\n\r\nEDIT: Actually, MathWise was right but the very last computation is wrong (try multiplying by $ \\frac{32}{32}$). However, conditional probability is really not necessary.", "Solution_6": "[hide=\"@pascal12\"][quote=\"pascal12\"]It's already a given that B won the 2nd game, so wouldn't $ P(A\\&B) \\equal{} \\left(\\frac {1}{2}\\right)^{4} \\equal{} \\frac {1}{16}$?\nSo then ABAA would be $ \\left(\\frac {1}{2}\\right)^{3}$ and so the total would then be $ \\frac {5}{16}$[/quote]\n\nWe are trying to calculate $ P(A\\&B)$, not $ P(A\\&B|B)$. Basically, instead of trying to find $ P(A|B)$ we are trying to find $ \\frac{P(A\\&B)}{P(B)}$, as this is often easier to find. However, when we calculate $ P(A\\&B)$ and $ P(B)$, we do not assume that B is a given. If you do assume that B is a given, then $ P(A\\&B)\\equal{}P(A|B)$, and $ P(B)\\equal{}1$, so it's the same as just trying to calculate $ P(A|B)$.\n\nThus, if we assume B is given, then $ P(A\\&B)\\equal{}\\frac{5}{16}$ and $ P(B)\\equal{}1$. (In order to calculate $ P(A\\&B)$ when B is given, we have to solve the problem, as this is what the problem is asking for.)[/hide]" } { "Tag": [ "pigeonhole principle", "MIT", "college", "Putnam", "arithmetic sequence" ], "Problem": "Let $A$ be any set of $19$ distinct integers chosen from the arithmetic progression $1, 4, 7, \\cdots 100$. Prove that there must be two distinct integers in $A$ whose sum is $104$.\r\n\r\nOops, sorry about the typo :blush:, should be $104$.\r\n\r\nI got this from the website for a problem-solving course at MIT, and the worksheet where I saw it stated that the $20$ could be replaced by a $19$.", "Solution_1": "[quote=\"Fraxia\"]Let $A$ be any set of $19$ distinct integers chosen from the arithmetic progression $1, 4, 7, \\cdots 100$. Prove that there must be two distinct integers in $A$ whose sum is $10$.[/quote]\r\n\r\nI don't think this is possible, since you can have the set of\r\n\r\n${13,16,...,67}$ and none of them add up to 10.", "Solution_2": "The problem that is of this type on Putnam 1978 is\r\n[i]Let S = {1, 4, 7, 10, 13, 16, ... , 100}. Let T be a subset of 20 elements of S. Show that we can find two distinct elements of T with sum 104.[/i]", "Solution_3": "[quote=\"boxedexe\"]The problem that is of this type on Putnam 1978 is\n[i]Let S = {1, 4, 7, 10, 13, 16, ... , 100}. Let T be a subset of 20 elements of S. Show that we can find two distinct elements of T with sum 104.[/i][/quote]\r\n\r\nWell, that explains it\r\n\r\nLet us break the subsets in the following manner.\r\n\r\n{1}, {4,100}, {7,97},...{49,55}, {52}\r\n\r\nThere are 18 sets here, so let us assume that after 18 pulls one element from each set is pulled to enforce maximization. Then this means that we have two draws remaining. It can't be from {1} or {52} since they are already drawn. So the two remaining elements have to be drawn from the 16 sets remaining, and hence there will be a set with two elements pulled by the pigeonhole principle.\r\n\r\nWhich implies that the sum is going to be 104 for any drawing of 20 integers.\r\n\r\nQED" } { "Tag": [], "Problem": "A square $ PQRS$ has sides of length $ x$. $ T$ is the midpoint of $ QR$ and $ U$ is the foot of the perpendicular from $ T$ to $ QS$. What is the length of $ TU$?", "Solution_1": "[quote=\"AndrewTom\"]A square $ PQRS$ has sides of length $ x$. $ T$ is the midpoint of $ QR$ and $ U$ is the foot of the perpendicular from $ T$ to $ QS$. What is the length of $ TU$?[/quote]\r\n[hide]Triangle QTU is a 45-45-90 with hypotenuse length being $ \\frac{x}{2}$. Thus the length of $ TU$ is $ \\frac{x\\sqrt{2}}{4}$.[/hide]" } { "Tag": [ "inequalities", "triangle inequality" ], "Problem": "Suppose that for any number $a$ there is a point on the graph of $y=f(x)$ closest to the point $(a,0)$ (this is guaranteed when $f$ is continuous, but that's not important). Define $g(a)$ as the distance from $(a,0)$ to that point, and prove that for all $c$ and $d$, $g(c)-g(d)\\le|c-d|$.", "Solution_1": "We will call $C$ and $D$ the points of the graph of $f$ achieving the required minimal distances.\r\nAssume the result to be false. Then $\\exists c,d, g(c) > |c-d| + g(d)$. That means $C$ must lie outside the circle with center $(c,0)$ and radius $|c-d|+g(d)$. But every point $D$ must lie inside this circle (by the triangle inequality); in particular, $d((c,0),D) \\leq |c-d|+g(d)$, which is absurd.", "Solution_2": "Right. It's easier to give a direct proof though.\r\n\r\nLet $s$ be the distance from $(c,0)$ to $D$. By the triangle inequality, $s\\le{g}(d)+|c-d|$. But $g(c)\\le{s}$ by the definition of $g(c)$." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Evaluate\r\n\\[ 1^2+4^2+7^2+\\cdots+97^2+100^2 \\]", "Solution_1": "[hide]Using finite differences, you find the constant at cubic ($18$).\n\n$a+b+c+d=1$\n$8a+4b+2c+d=17$\n$27a+9b+3c+d=66$\n$64a+16b+4c+d=166$\n\nSolve for a,b,c,d.\n$a=3$\n$b=-1.5$\n$c=-.5$\n$d=0$\n\nFactored, your equation is: $\\frac{1}{2}(6x^3-3x^2-x)$\n\nEvaluate with $x=34$ (34 groups until 100)\n\n$\\frac{1}{2}[(6)(34)^3-3(34)^2-34]=\\boxed{116161}$\n\nI hope I didn't make any algebraic errors (in solving the equations).[/hide]", "Solution_2": "[quote=surge]Using finite differences, you find the constant at cubic ($18$).[/quote]\r\n\r\nWhat does this mean?", "Solution_3": "[quote=\"DanK\"][quote=\"surge\"]Using finite differences, you find the constant at cubic ($18$).[/quote]\n\nWhat does this mean?[/quote]\r\n\r\nIt's this ugly process that you learn in Algebra II/Trig that I promptly forgot.", "Solution_4": "it's some really ugly stuff, but is actually quite useful when determining which degree the polynomial is.\r\nfind the difference between two consecutive numbers\r\nif not all of the differences are the same, repeat procedure for the first differences\r\nrepeat until all the numbers in the row are constant. that is your finite difference", "Solution_5": "[hide=\"Rather screwy method\"]\n$\\sum_{k=0}^{n} (1+3k)^2 = \\sum_{k=0}^{n} (1 + 6k + 9k^2) = \\left( \\sum_{k=0}^{n} 1 \\right) + 6 \\left( \\sum_{k=0}^{n} k \\right) + 9 \\left( \\sum_{k=0}^{n} k^2 \\right)$\n\nWe know how to evaluate all of these. We find that the expression becomes\n\n$(n+1) + 6 \\frac{n(n+1)}{2} + 9 \\frac{n(n+1)(2n+1)}{6}$\n\nFor $n = 33$, this becomes\n\n$\\boxed{ 116161 }$.\n[/hide]" } { "Tag": [ "puzzles" ], "Problem": "......*. . *\r\n*. . *. . * . . *\r\n*. . *. . * . . *\r\n......*. . *\r\n\r\n\r\nPicture the 12 asterisks above as dots making a square-shaped\r\ncross, and the periods are to be ignored.\r\n\r\nIf you wanted to place them as points in the xy-plane,\r\nthey could be listed (from lowest row to highest row) as: \r\nA(1,0), B(2,0), C(0,1), D(1,1), E(2,1), F(3,1),\r\nG(0,2), H(1,2), I(2,2), J(3,2), K(1,3), L(2,3)\r\n\r\nTwo or more dots (all to act only as vertices of squares)\r\non part of the 12 dots (all of which at a time are to be \r\nconnected to form squares are to form different squares\r\nfor a total count. If a set of dots can make more than \r\none square, then you count all of them.\r\n\r\nSummary\r\n------------\r\n\r\nYou are choosing all ways of 2, 3, and 4 dots at a time in the \r\n12-dot diagram to be vertices only of squares and add\r\nthe total. \r\n\r\nNote: If one choice of dots gets the same square (same size\r\nand same location) as another, then it does NOT add to \r\nthe total number.\r\n\r\nExample) # of squares picking dots B,E,F --> 1\r\nExample) # of squares picking dots B,I --> 2 (one on each side)", "Solution_1": "Is it 75?\r\n\r\nBreakdown:\r\nFor choosing 4 dots: 11 ways\r\nFor choosing 3 dots: 8 ways\r\nFor choosing 2 dots: 56 ways:\r\nAt least one of those is probably wrong though :maybe:", "Solution_2": "There will be a delay in getting back with specifics about this.\r\nI expect about one day.\r\n\r\nIt is good that you showed some partial work (breakdown).\r\nThat is, it makes it more helpful about checking parts of \r\nwhere subtotals came from.\r\n\r\n(Others might post in the meantime too.)", "Solution_3": "cyberspace,\r\n\r\nI got the same subtotals for 3 dots and for 4 dots,\r\n\r\nas you, respectively.\r\n\r\nHowever, I got a larger number for 2 dots than you \r\n\r\ndid, but I want to study the case for 2 dots again." } { "Tag": [ "integration", "advanced fields", "advanced fields unsolved" ], "Problem": "$(\\Omega, \\mu)$ a mesurable space, $\\mu( \\Omega )=1$. For $1 \\leq p < \\infty$, suppose that $S$ is a closed subspace of $L^p(\\Omega, \\mathbb{C})$, included in $L^{\\infty}(\\Omega, \\mathbb{C})$. Show that $S$ is a finite dimensional subspace.", "Solution_1": "I don't have the time to write down a full proof now but the way to proceed seems clear: take any bounded sequence $(f_{n})$ in $L_{p}$, and extract a subsequence $g_{n}$ such that $\\int {g_{n}}$ converges on every interval $I_{k,n}=[k.2^{-n},(k+1).2^{-n}]$, and such that $|\\int {(g_{n},I_{k,n})}-lim \\ \\int {(g_{n},I_{k,n})}| \\leq 2^{-n!}$ for every $k$. Now I guess we can prove that $(g_{n})$ is Cauchy for the $L_{p}$ norm, and hence convergent to an element of $S$ (since $S$ is closed). But this proves that $S$ is finite dimensional (by Riesz's theorem)." } { "Tag": [], "Problem": "Find the sum of the digits of the number: $ 10^{2008}\\minus{}2008$\r\n\r\nI'll post the answer on thursday.", "Solution_1": "I think I know the answer but I don't know how to \"hidden content\" something. Can you help???!!!\r\n\r\nThis is my first post! YAY :D\r\n\r\nIt ends in 045 right?", "Solution_2": "To post a hidden answer just type \"[;hide=Whatevertext]what's hidden[/hide]\" But with out the quote marks and take out the semi colon.", "Solution_3": "Ok thanks\r\n\r\n[hide=\"Answer\"]You do 10000-2008=7992 7+9+9+2=27 \nNow all the other digits besides the last 4 digits are 9s and there are 2003 other digits \n2003*9=18027\n18027+27=18054 ->[b]18054[/b] [/hide]\r\n\r\nI meant 054 in the first one :P", "Solution_4": "Yeah, I agree. I got 18063, but I'm pretty sure I'm wrong, since subtracting 2008 takes off one digit.", "Solution_5": "well I guess ill post the answer now.\r\n\r\nok micrim you almost had it, you were missing one 9.\r\ndarkmattter got it right.\r\n\r\nAnswer:\r\n$ 10^5 \\minus{} 2008 \\equal{} 97992$\r\n$ 10^6 \\minus{} 2008 \\equal{} 997992$\r\n\r\nnotice that the last four digits are always 7992 wich is 27, \r\nand that the number of nines is always the power -4\r\n\r\nso here is a equation for the sum of the digits of any 10^x -2008; where x>4:\r\n$ 27 \\plus{} 9(x \\minus{} 4)$\r\n\r\nso just plug in 2008 for x and you get:\r\n\r\n$ 10^{2008} \\minus{} 2008$:\r\n$ 27 \\plus{} 9(2008 \\minus{} 4)$\r\n$ \\equal{} \\boxed{18063}$", "Solution_6": "Oh I know what I did wrong. 10^2008 has 2008 0's not digits......\r\n Righttttttttttt... :wallbash_red:", "Solution_7": "Sweet! :lol: I made the same mistake as micrim, but a typing error on the calculator accidentally gave me the answer! :rotfl:", "Solution_8": "$ 10^{2008} \\minus{} 2008 \\equal{} 10000000000.......0000 \\minus{} 2008 \\equal{} 999999999999999.......7992$\r\n\r\n\r\nThe sum of the digits is equal to $ n$\r\n\r\n$ n \\equal{} (2008 \\minus{} 4)*9 \\plus{} 7 \\plus{} 9 \\plus{} 9 \\plus{} 2 \\equal{} (2008 \\minus{} 4)*9 \\plus{} 9 \\plus{} 9 \\equal{} (2008 \\minus{} 4 \\plus{} 3)*9 \\equal{} (2008 \\minus{} 1)9$\r\n\r\n$ n \\equal{} (2007)9$\r\n\r\nWhat do you think :D", "Solution_9": "@elio(n)\r\n\r\nYep that works,Although you missed a $ \\plus{}9$ when you wrote $ (2008\\minus{}4)*9\\plus{}9\\plus{}9$", "Solution_10": "[hide] first I did 10 to a low power (to the fifth) minus 2008\n\nI did 100000-2008\n\nI got 97992\n\nso I realized that it always ended in 7992 and the rest were 9s.\n\nI knew that when 10 is to the x power, it has one more digit than x\n\nso 10^2008 has 2009 digits. Taking away 2008 from that number, we get 2008 digits. Then adding the 7 and the 2 together, it becomes 9, so I did 9 x 2007\n\nI got 18063[/hide]", "Solution_11": "i calculated the answer multiple times and i have concluded that by cheating off of ya'll's answer that the real answer is 10863. yeah :w00t:" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "let $a,b \\in \\mathbb{N}$. Prove that $(a,b)=(3,5)$ is only solution for equation $a^3=b^2+2$", "Solution_1": "Factor it as $a^3 = (b+ \\sqrt{-2})(b-\\sqrt{-2})$.\r\nNow $\\gcd(b+ \\sqrt{-2} , b- \\sqrt{-2}) = \\gcd(b+ \\sqrt{-2}, 2 \\sqrt{-2} ) =1$ since $\\sqrt{2} \\nmid b$ because $2 \\nmid b$ (the last one can be seen $\\mod 8$).\r\nThus by unique factorisation in $\\mathbb{Z}[\\sqrt{-2}]$ we get that $b+\\sqrt{-2} = (x+y\\sqrt{-2})^3$ is a perfect cube.\r\nThis gives us:\r\n$x^3-6xy^2=b$\r\n$y(3x^2-2y^2)=3x^2y-2y^3=1$\r\nthus $y = \\pm 1$, so $3x^2-2 = y= \\pm 1$ giving finally $y=1 , x= \\pm 1$.\r\nSo $b=x(x^2-6y^2) = \\pm 5$, done.", "Solution_2": "i know solution using ring $\\mathbb{Z}[\\sqrt{2}]$. I'm intersted in lementary solution if it's possible...\r\n\r\n[mod.: never, really never, take anything like numbers into \\mathbb{...}]", "Solution_3": "There is no elementary method.\r\n\r\nBomb", "Solution_4": "[quote=\"bomb\"]There is no elementary method.\n\nBomb[/quote]\r\nI would formulate it in another way: there seems to be no method not being the same as the above one up to elementarisation.", "Solution_5": "why there is no elementary solution" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "let $ k>0$ be an integer and define $ f(x)\\equal{}\\sum_{n\\equal{}0}^\\infty c_n x^n,$ where $ c_{n\\plus{}k}\\equal{}c_n$ for each $ n\\ge0.$ Find the interval of convergence of the series and compute $ f(x)$ explicitly.\r\n\r\ndon't know how to find $ f(x)$ explicitly.", "Solution_1": "If not all terms of $ \\{ c_n \\}$ vanish, then root test is applicable, yielding 1 as the radius of convergence for this series. Also,\r\n\r\n\\begin{eqnarray*}\r\nf(x) & = & \\sum_{n=0}^{\\infty} c_n x^n \\\\\r\n& = & \\sum_{q=0}^{\\infty} \\sum_{r=0}^{k-1} c_{qk+r} x^{qk+r} \\qquad (n = qk+r)\\\\\r\n& = & \\sum_{q=0}^{\\infty} \\sum_{r=0}^{k-1} c_{r} x^r \\cdot x^{qk} \\\\\r\n& = & \\left( \\sum_{r=0}^{k-1} c_{r} x^r \\right) \\left( \\sum_{q=0}^{\\infty} x^{qk} \\right) \\\\\r\n& = & \\frac{1}{1-x^q} \\sum_{r=0}^{k-1} c_{r} x^r\r\n\\end{eqnarray*}" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "You are given the sequence $a_1 = 1, a_2 = 0, a_3 = 1, a_4 = 0, a_5 = 1, a_6 = 0$ and a prime number $p$.\r\n\r\nThe rest of the sequence is defined as $a_{n+6} = \\sum_{i=0}^{5}a_{n+i} \\mod p$ \r\n\r\nFind all primes $p$ for which the sequence contains $0,1,0,1,0,1$ as some six consecutive terms.", "Solution_1": "I have been able to show that p must be one of 2,3,7,11 for that to happen. I was able to verify, using a computer that only for 2,3 and 11 does the sequence contain 0,1,0,1,0,1.\r\n\r\nSo, try the modified problem (which does not involve a computer):\r\n\r\nShow that the set of primes p for which the sequence contains 0,1,0,1,0,1 is finite. \r\n\r\nIn fact, it is a subset of {2,3,7,11}." } { "Tag": [ "Online Math Open", "search" ], "Problem": "A\u03bd $ a,b,c$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03af \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03b9 \u03ce\u03c3\u03c4\u03b5 $ ab\\plus{}bc\\plus{}ca\\plus{}2abc\\equal{}1$ \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 $ 2(a\\plus{}b\\plus{}c)\\plus{}1\\geq 32abc$", "Solution_1": "Profanos $ abc$ diaforo tu midenos,opote $ ab\\plus{}bc\\plus{}ca\\equal{}1\\minus{}2abc$ \r\ndld\r\n$ \\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\equal{}\\frac{1\\minus{}2abc}{abc}$\r\n\r\nomos\r\n$ (a\\plus{}b\\plus{}c)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c})>\\equal{}9$\r\nopote\r\n$ a\\plus{}b\\plus{}c>\\equal{}\\frac{9abc}{1\\minus{}2abc}$\r\n\r\n$ 2(a\\plus{}b\\plus{}c)>\\equal{}\\frac{18abc}{1\\minus{}2abc}$\r\n\r\ndiladi $ 2(a\\plus{}b\\plus{}c)\\plus{}1>\\equal{}\\frac{16abc\\plus{}1}{1\\minus{}2abc}$\r\nopote arki ndo\r\n$ \\frac{16abc\\plus{}1}{1\\minus{}2abc}>\\equal{}32abc$\r\n\r\nisodinama afu $ 2abc<1$ ($ >\\equal{}1$ apokliete afu $ a,b,c$ 8etiki pragmatiki..)\r\n\r\n$ 16abc\\plus{}1>\\equal{}32abc\\minus{}64(abc)^2$\r\n\r\ndld $ (8abc\\minus{}1)^2>\\equal{}0$ pu isxii kai i isotita isxii profanos otan $ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{2}$", "Solution_2": "\u0394\u03b5\u03af\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03b5\u03b4\u03ce . \u00a8\u03b5\u03c7\u03c9 \u03c0\u03bf\u03c3\u03c4\u03ac\u03c1\u03b5\u03b9 \u03bc\u03b9\u03b1 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03ba\u03b1\u03bb\u03ae \u03bb\u03cd\u03c3\u03b7 only Cauchy :) \r\nhttp://www.mathlinks.ro/viewtopic.php?search_id=1564802248&t=80027", "Solution_3": "\u0391\u03c1\u03ba\u03b5\u03c4\u03ac \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1. :huh: \r\n\r\n\u03a7\u03ac\u03c1\u03b7\u03ba\u03b1 \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03c0\u03bf\u03bb\u03cd \u03ce\u03c1\u03b1 \u03b2\u03c1\u03ae\u03ba\u03b1 \u03bc\u03b9\u03b1 \u03c0\u03bf\u03bb\u03cd original \u03bb\u03cd\u03c3\u03b7. :idea: \r\n\r\n\u0398\u03ad\u03c4\u03bf\u03c5\u03bc\u03b5 $ a\\equal{}(1\\minus{}x)/x$ \u03ba\u03c4\u03bb. \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b7 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7 \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9 \u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03ae $ x\\plus{}y\\plus{}z\\equal{}2$ (!) \u03b5\u03bd\u03ce \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03c1\u03bf\u03c2 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03ae:\r\n\r\n$ x/(4\\minus{}3x)\\plus{}y/(4\\minus{}3y)\\plus{}z/(4\\minus{}3z)>\\equal{}1$\r\n\r\n(\u0391\u03bd \u03ba\u03ac\u03c4\u03b9 \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b1\u03b9\u03c4\u03ad\u03c1\u03c9 \u03b5\u03be\u03ae\u03b3\u03b7\u03c3\u03b7 \u03c0\u03b5\u03af\u03c4\u03b5 \u03bc\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03c4\u03bf \u03b2\u03ac\u03bb\u03c9 \u03b1\u03cd\u03c1\u03b9\u03bf \u03b3\u03b9\u03b1\u03c4\u03af \u03bd\u03c5\u03c3\u03c4\u03ac\u03b6\u03c9 \u03bb\u03af\u03b3\u03bf)\r\n\r\n\u03a0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03ce\u03c1\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03ba\u03bb\u03ac\u03c3\u03bc\u03b1 \u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03c9 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03b7\u03c4\u03ae \u03c4\u03bf\u03c5 \u03ba\u03b1\u03b9 Andreescu \u03ba\u03b1\u03b9 \u03c4\u03ad\u03bb\u03bf\u03c2.\r\n\r\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03b4\u03b9\u03ba\u03b1\u03b9\u03bf\u03cd\u03bc\u03b1\u03b9 \u03ad\u03bd\u03b1 Q.E.D. :D", "Solution_4": "\u03a7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 $ AM \\minus{} HM$ \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5\r\n$ (a \\plus{} b \\plus{} c \\plus{} 1/2)/4 > \\equal{} 4/(1/a \\plus{} 1/b \\plus{} 1/c \\plus{} 2)$\r\n\u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5\r\n$ (a \\plus{} b \\plus{} c \\plus{} 1/2) > \\equal{} 16abc$\r\n\u03a0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03bc\u03b5 2 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03b7" } { "Tag": [ "calculus", "integration", "algebra", "partial fractions", "calculus computations" ], "Problem": "A sequence $\\{a_n\\}$ is defined by $a_n=\\int_0^1 x^3(1-x)^n dx\\ (n=1,2,3.\\cdots)$\r\n\r\nFind the constant number $c$ such that $\\sum_{n=1}^{\\infty} (n+c)(a_n-a_{n+1})=\\frac{1}{3}$", "Solution_1": "This is not so hard by conventional methods, although there may be a cleverer way. I'm leaving out most of the calculations because I am lazy.\r\n\r\nWe can simply calculate $a_n - a_{n-1}$ very straight-forwardly. Adding the integrals adds the integrands, giving us $a_n - a_{n-1} = \\int_0^1x^4(1-x)^n dx$. Doing the RHS by parts and combining like terms, $a_{n+1} = \\frac{1+n}{5+n}a_n$ from which it follows that $a_n = \\frac{5!n!}{(n+4)!}a_1$. Direct calculation gives $a_1 = \\frac{1}{20}$, so $a_n = \\frac{6}{(n+1)(n+2)(n+3)(n+4)}$. (We also could have done integration by parts 3 times to get this result.) Now, if we take this and stick it back into the original summation, we find $\\sum_{n=1}^{\\infty} (n+c)(a_n-a_{n+1})=\\frac{1}{3}$, \r\nand we can decompose the expression in that sum in order to write it as\r\n$(c - 1)\\sum_{n=1}^{\\infty}\\frac{24}{(n+1)(n+2)(n+3)(n+4)(n+5)} + \\sum_{n=1}^{\\infty}\\frac{24}{(n+2)(n+3)(n+4)(n+5)} = \\frac{1}{3}$\r\n\r\nEach of those sums can be gotten by standard partial fractions, although they're a bit messy (especially the first one), and then we have a linear system whose solution is $c = 5$.", "Solution_2": "Thank your for your reply, JBL.\r\n\r\nYour answer is correct.\r\n\r\nkunny" } { "Tag": [ "geometry" ], "Problem": "Given that BDEF is a square and AB = BC = 1, find the number of square units in the area of the regular octagon.\n\n[asy]pair A,B,C,D,E,F;\nA = (-3,1); B = (-3,3); C = (-1,3); D = (3,3); E = (3,-3); F = (-3,-3);\nfor(int i = 0; i < 4; ++i){\ndraw(rotate(90*i)*(B--D));\ndraw(rotate(90*i)*(A--C));\n}\nlabel(\"A\",A,W); label(\"B\",B,NW);label(\"C\",C,N);\nlabel(\"D\",D,NE);label(\"E\",E,SE);label(\"F\",F,SW);[/asy]", "Solution_1": "The side length of the octagon is $ \\sqrt{2}$. The side length of the square, therefore, is $ 2 \\plus{} \\sqrt{2}$. Simplify $ (2 \\plus{} \\sqrt{2})^2$ for the area of the square. This is equal to $ 4 \\plus{} 4\\sqrt{2} \\plus{}2$, or $ 6 \\plus{} 4\\sqrt{2}$. Then, subtract the areas of the $ 4$ triangles. Each triangle's area is $ \\frac{1}{2} \\times 1 \\times 1$, or $ \\frac{1}{2}$. Subtract $ 2$ from $ 6 \\plus{} 4\\sqrt{2}$, to receive the final answer of $ \\boxed{4 \\plus{} 4\\sqrt2}$." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Evaluate $ \\int_0^{2\\pi } {\\frac{{\\cos 2x}}{{5 \\minus{} 4\\sin x}}\\,dx} .$", "Solution_1": "[hide]If $ I$ is the integral, then substituting for $ 2\\pi \\minus{}x$ gives\n\\[ I\\equal{}\\int_0^{2\\pi} \\frac{\\cos 2x}{5\\plus{}4\\sin x}\\ dx\\]\nAdd to the original integral and divide by 2 to get\n\\[ I\\equal{}5\\int_0^{2\\pi} \\frac{\\cos 2x}{25\\minus{}16\\sin^2 x}\\ dx\\equal{}5\\int_0^{2\\pi} \\frac{\\cos 2x}{17\\plus{}8\\cos 2x}\\ dx\\]\nNote that this can be expressed as\n\\[ 10\\int_0^\\pi \\frac{\\cos 2x}{17\\plus{}8\\cos 2x}\\ dx\\equal{}10\\int_{\\minus{}\\frac{\\pi}{2}}^\\frac{\\pi}{2} \\frac{\\cos 2x}{8\\cos 2x \\minus{}17}\\ dx\\]\nLet $ u\\equal{}\\tan x$, so that $ \\ du\\equal{}(1\\plus{}u^2)\\ dx$, giving the improper integral\n\\[ 10\\int_{\\minus{}\\infty}^\\infty \\frac{u^2\\minus{}1}{(1\\plus{}u^2)(9\\plus{}25u^2)}\\ du\\equal{}10\\int_{\\minus{}\\infty}^\\infty \\frac{1}{8(1\\plus{}u^2)}\\minus{}\\frac{17}{8(9\\plus{}25u^2)}\\ du\\]\n\\[ \\equal{}10\\left[ \\frac{1}{8}\\arctan u\\minus{}\\frac{17}{120}\\arctan \\frac{5}{3}u\\right]_{\\minus{}\\infty}^\\infty \\equal{}\\boxed{\\minus{}\\frac{\\pi}{6}}\\][/hide]", "Solution_2": "Generally, $ \\int_{0}^{2\\pi} \\frac{e^{i n x}}{5\\minus{}4\\sin x} \\, dx \\equal{} \\frac{2 \\pi}{3} \\left( \\frac{i}{2} \\right)^n$." } { "Tag": [], "Problem": "\"The integer $ 999,919$ is a product of a three digit prime and a four digit prime. Let A be the sum of the digits of the three digit prime, and B be the sum of the digits of the four digit prime. Compute the sum $ A \\plus{} B$.\"\r\n\r\nSince $ 999,919$ is a product of two primes, then that must mean that it's only factors are $ 1$, $ 999919$, three-digit prime, four-digit prime. That's as far as I have gotten. Calculators are allowed, but not TI-89 or higher (or the HP equivalents). Note that I'm only supposed to spend about 3 minutes on this problem, so sitting around and brute forcing isn't a good idea.", "Solution_1": "It's one of those notice random things about the number problems:\r\n[hide]\n$ 999919 \\equal{} 1000000 \\minus{} 81 \\equal{} 1000^2 \\minus{} 9^2 \\equal{} (1000 \\plus{} 9)(1000 \\minus{} 9) \\equal{} 1009*991$\nI'm going to take the problem's word that those numbers are prime.\nAnyway, the desired answer should be $ 1 \\plus{} 9 \\plus{} 9 \\plus{} 9 \\plus{} 1 \\equal{} \\boxed{29}$\n[/hide]", "Solution_2": "You are correct! Thanks for the help!" } { "Tag": [ "integration", "inequalities", "function", "logarithms", "real analysis", "complex analysis", "real analysis unsolved" ], "Problem": "I understand how to use them in proofs... but I really wish I knew where the equation:\r\n\r\n$ \\frac{1}{p}\\plus{} \\frac{1}{q}\\equal{}1$ came from! \r\n\r\nFor example, I can prove that for a bounded linear functional one $ L^p$, $ F(f)\\equal{} \\int fg$ we have $ \\parallel{}F\\parallel{} \\equal{} \\parallel{}g\\parallel{}_q$ and the proof is all algebra. But, it just seems like magic. I don't know WHY it works. Can someone give me a hint as to the bigger picture here?", "Solution_1": "I'm not sure it is the ultimate reason, but Young's inequality $ \\frac {x^p}p\\plus{}\\frac {y^{q}}q\\ge xy$, which is behind most of the \"magic\" concerning the $ L^p$ spaces is just a partial case of $ \\Phi(x)\\plus{}\\Psi(y)\\ge xy$ where $ \\Phi$ and $ \\Psi$ are any convex conjugate functions vanishing at $ 0$ (i.e., for some positive increasing continuous function $ f: [0,\\plus{}\\infty)\\to[0,\\plus{}\\infty)$ with $ f(0)\\equal{}0$, we have $ \\Phi(x)\\equal{}\\int_0^x f$, $ \\Psi(x)\\equal{}\\int_0^x f^{\\minus{}1}$ where $ f^{\\minus{}1}$ is the inverse function to $ f$. The proof is the same picture with the graph of $ f$ and two curvilinear triangles covering a rectangle. Now, given any $ \\Phi$, you can compute $ \\Psi$ and it just happens that if $ \\Phi(x)\\equal{}\\frac{x^p}{p}$, then $ \\Psi(y)\\equal{}\\frac{y^q}q$ with $ q\\equal{}\\frac{p}{p\\minus{}1}$.\r\n\r\nAnother good reason is that if you want to have the inequality $ \\int_{\\mathbb R} fg\\le \\|f\\|_p\\|g\\|_q$, both sides should scale the same way when you change $ f(x)$ and $ g(x)$ to $ f(x/t)$ and $ g(x/t)$. Now, the left hand side scales as $ t$ and the right hand side as $ t^{\\frac 1p\\plus{}\\frac 1q}$. This doesn't explain why everything works when $ \\frac 1p\\plus{}\\frac 1q\\equal{}1$ but, at least, explains why it doesn't work otherwise.", "Solution_2": "The answer in my opinion is just one word: Convexity. That is certainly what I think of when I see the equation $ 1/p\\plus{}1/q\\equal{}1$. If you look at Rudin's Real and Complex Analysis, you'll find his chapter on $ L^p$ spaces starts with a section on convex functions.\r\n\r\nFedja mentioned the inequality $ xy\\leq \\frac{x^p}{p}\\plus{}\\frac{y^q}{q}$. That inequality, and some clever manipulations, are really all that is behind the basic results for $ L^p$ spaces, conjugate indices, duality, etc. And that inequality comes directly from the convexity of the exponential function: $ e^{(1/p)\\log{x^p}\\plus{}(1/q)\\log{y^q}}\\leq (1/p)e^{\\log{x^p}}\\plus{} (1/q)e^{\\log{y^q}}$.", "Solution_3": "Folland remarks in his Real Analysis:\r\n[quote] In some respects it is unfortunate that $ L^p$ spaces were not named $ L^{1/p}$ spaces[/quote]\r\nbecause of the role of linear relations in $ 1/p$.", "Solution_4": "[quote=\"mlok\"]because of the role of linear relations in $ 1/p$.[/quote]\r\nFor instance, Young's inequality for convolutions: $ \\|f*g\\|_r\\le\\|f\\|_p\\|g\\|_q,$ whenever $ \\frac1r\\equal{}\\frac1p\\plus{}\\frac1q\\minus{}1$ and $ p,q,r\\in[1,\\infty].$", "Solution_5": "Thanks-- I'll look at Rudin, we skipped convexity... so maybe that's why I'm confused." } { "Tag": [ "calculus", "integration", "algebra proposed", "algebra" ], "Problem": "Assume that the cubiic equation $4x^3-(a-2)x-(a+4)=0\\ (a\\in{\\mathbb{Z}})$ has a positive rational root not integer.\r\nFind the root satisfying this property.", "Solution_1": "You said quartic and gave a cubic. The roots are stated to all be positive rationals but sum to zero (no $x^2$ term). Some mistake?", "Solution_2": "He didn't say that [i]all[/i] roots are positive rationals.\r\nI think what he meant was that the equation has at least one positive rational root :roll:", "Solution_3": "let the non-integral rational root be p/q. then q must divide 4 => q is 2 or 4.\r\n\r\ndivide the above integral by (qx-p) for each case of p. factorise and then u'll get an eqn in a and p. but values of p are limited.", "Solution_4": "To everyone:Sorry for that I made a typo.I have just edited.\r\n\r\nkunny" } { "Tag": [ "function", "trigonometry", "calculus", "derivative", "inequalities proposed", "inequalities" ], "Problem": "Let $\\alpha ,\\ \\beta ,\\ \\gamma$ be real numbers. Find the maximum value of $k$ satisying $\\alpha^{2}+\\beta^{2}+\\gamma^{2}+\\cos^{2}\\alpha+\\cos^{2}\\beta+\\cos^{2}\\gamma\\geq k$.", "Solution_1": "For any real $x$, we have\r\n\r\n$\\left|x\\right|\\geq\\left|\\sin x\\right|\\ \\Longleftrightarrow\\ x^{2}\\geq\\sin^{2}x\\ \\Longleftrightarrow x^{2}+\\cos^{2}x\\geq \\sin^{2}x+\\cos^{2}x=1$.\r\n\r\nNeedless to say, equality holds if and only if $x=0$.\r\n\r\n[i]Very[/i] meaningful problem...\r\n\r\n darij", "Solution_2": "Thank you! :) I have no comment more than this. That's very background of this problem.", "Solution_3": "Just minimize the function x^2+(cosx)^2\r\nAs the function is continuous and bounded at the bottom (the function is always grater than 0), the minimal point will be achieved when derivative is 0. \r\nThe derivative is (x^2+(cosx)^2)'=2x-2cosxsenx, and is equal with 0 when x=cosxsenx, but this can only happen into the [-1,1] bound. If we plug x=0 into the formula, then we have 0=cos0sen0 which is true. Function f=cosxsenx has derivative (cosx)^2-(senx)^2. As f passes through (0,0), the derivative for this point is 1, so, the y=x line is tangent to senxcosx on (0,0), (cosx)^2-(senx)^2 decreases when x goes from 0 to 1 or -1, so the function cosxsenx is between the x axis and the funcition y=x, this means the only x for which x=cosxsenx, is 0. So, the minimum of x^2+(cosx)^2 is 1 (x=0). Then k=3." } { "Tag": [ "analytic geometry", "geometry", "geometric transformation", "reflection", "inequalities" ], "Problem": "In Cartesian coordinate, let be given two points $A(0,6),B(2,5)$ and line $(d): x-2y+2=0$. Find point $M \\in (d)$ such that $|MA-MB|$ takes the greatest value.", "Solution_1": "Reflect A through d to A', then we get A': (4,-2). Hence by triangular inequality, $MA-MB =MA'-MB \\leq AB = \\sqrt{17}$, thus $M= A'B \\cap d$. So $M: \\left( \\frac{38}{7}, \\frac{26}{7}\\right)$." } { "Tag": [ "calculus", "function", "integration", "geometry", "trapezoid", "algebra", "polynomial" ], "Problem": "HI\r\n\r\nI need help with this.\r\nWork out the extended newton cotes formula (with error term) for the case d = 3.\r\n\r\nthanks", "Solution_1": "Do you mean Newton-Cotes quadrature formulae? :)", "Solution_2": "[quote=\"Cezar Lupu\"]Do you mean Newton-Cotes quadrature formulae? :)[/quote]\r\n\r\nyup\r\nthanks", "Solution_3": "I'll post it later with the proof. Now I have to go! ;)", "Solution_4": "numero1: Numerical analysis is not number theory; your post does not belong here. Numerical analysis is also not complex analysis. It might be calculus - real analysis, but I think the best place to collect together numerical analysis questions would be in College Playground - Advanced Fields.\r\n\r\nA moderator should move this topic.", "Solution_5": "[quote=\"Cezar Lupu\"]I'll post it later with the proof. Now I have to go! ;)[/quote]\r\n\r\nBro, can you post it. I need as soon as possible. :lol: \r\nThanks a lot :)", "Solution_6": "I think googling might help :) :\r\nhttp://www.google.com/search?q=newton+cotes+quadrature+formula", "Solution_7": "Well here it is in six knots. If $f: [a,b]\\to\\mathbb{R}$ is a $C^{6}[a, b]$ function then there is $c\\in (a, b)$ such that\r\n\r\n \\[ \\int_{a}^{b} f(x)dx=\\frac{(b-a))}{90}[7f(a)+32f(\\frac{3a+b}{4})+12f(\\frac{a+b}{2})+32f(\\frac{a+3b}{4})+7f(b)]-\\frac{(b-a)^{7}}{1935360}\\cdot f^{(6)}(c). \\]\r\n\r\n This is a quite strong quadrature formulae. :)", "Solution_8": "[quote=\"Cezar Lupu\"]Well here it is in six knots. If $f: [a,b]\\to\\mathbb{R}$ is a $C^{6}[a, b]$ function then there is $c\\in (a, b)$ such that\n\n \\[ \\int_{a}^{b} f(x)dx=\\frac{(b-a))}{90}[7f(a)+32f(\\frac{3a+b}{4})+12f(\\frac{a+b}{2})+32f(\\frac{a+3b}{4})+7f(b)]-\\frac{(b-a)^{7}}{1935360}\\cdot f^{(6)}(c). \\]\n\n This is a quite strong quadrature formulae. :)[/quote]\r\n\r\nWas that for the case d = 3?\r\nCould you explain what all that means...I am still very confused.\r\nthanks", "Solution_9": "A Newton-Cotes formula involves equally spaced points within an interval. A \"closed\" Newton-Cotes formula would use the endpoints; an \"open\" formula would not.\r\n\r\nThe closed Newton-Cotes formula using two points is the trapezoid rule, and is a second order method - it integrates first degree polynomials exactly and has an error that depends on the second derivative. The 3-point closed Newton-Cotes formula is what is usually taught in calculus textbooks as Simpson's rule. It's a 4th order method. THe 4-point closed Newton-Cotes forumla is also 4th order. Cezar Lupu has given the formula for the 5-point closed Newton-Cotes formula, which is (as you can see from the error formula provided), 6th order.\r\n\r\nOne can devise these formulas for all orders, although there are some diminishing returns aspects of pushing them too far." } { "Tag": [ "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Does anyone know how to solve $\\int_{-\\infty}^\\infty e^{-x^{2}}dx$?", "Solution_1": "Yes. It's [i]very[/i] well-known. Here's the most recent [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=150137]appearance[/url].", "Solution_2": "You may also want to look [url=http://en.wikipedia.org/wiki/Gaussian_integral]here[/url]", "Solution_3": "Or [url=http://mathworld.wolfram.com/Erf.html]here[/url]", "Solution_4": "Just as a side note, when I read this post it came to my mind that, according to a book I read some time ago, this was Lord Kelvin's favourite integral. It is said, that, during one of his lectures to engineering students he uses the word \"mathematician\". He then stops, turns to his students and ask them \"Do you know what a mathematician is?\". Then Lord Kelvin goes to the blackboard and write in it the above integral. Then he says to the class: \"A mathematician is someone for whom this integral is as obvious as that twice two makes four is to you.\"" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "If f:R\u2192R, solve this functional equation.\r\n\r\nf(x-f(y))=f(f(y))+xf(y)+f(x)-1 for all x, y\u2208R.", "Solution_1": "This is IMO 99/6th problem.\r\n[url]http://www.mathlinks.ro/viewtopic.php?p=131856#131856[/url]" } { "Tag": [ "trigonometry", "factorial", "real analysis", "real analysis unsolved" ], "Problem": "Calculate\r\n\\[ \\frac{1\\plus{}\\frac{{\\pi}^4}{2^4\\cdot 4!}\\plus{}\\frac{{\\pi}^8}{2^8\\cdot 8!}\\plus{}\\frac{{\\pi}^{12}}{2^{12}\\cdot 12!}\\plus{}\\cdots}{\\frac 1{2!}\\plus{}\\frac{{\\pi}^4}{2^4\\cdot 6!}\\plus{}\\frac{{\\pi}^8}{2^8\\cdot 10!}\\plus{}\\frac{{\\pi}^{12}}{2^{12}\\cdot 14!}\\plus{}\\cdots}\\]", "Solution_1": "First, define:\r\n\r\n$ f(x) \\equal{} \\sum ^\\infty_{n \\equal{} 0} \\frac {x^{4n \\plus{} 2}}{(4n \\plus{} 2)!}$\r\n\r\nSuccessive differentiating gives:\r\n\r\n$ f'(x) \\equal{} \\sum ^\\infty_{n \\equal{} 0} \\frac {(4n \\plus{} 2) x^{4n \\plus{} 1}}{(4n \\plus{} 2)!} \\equal{} \\sum ^\\infty_{n \\equal{} 0} \\frac {x^{4n \\plus{} 1}}{(4n \\plus{} 1)!}$\r\n\r\n$ f''(x) \\equal{} \\sum ^\\infty_{n \\equal{} 0} \\frac {(4n \\plus{} 1) x^{4n}}{(4n \\plus{} 1)!} \\equal{} \\sum ^\\infty_{n \\equal{} 0} \\frac {x^{4n}}{(4n)!}$\r\n\r\n$ f'''(x) \\equal{} \\sum ^\\infty_{n \\equal{} 0} \\frac {(4n) x^{4n \\minus{} 1}}{(4n)!} \\equal{} \\sum ^\\infty_{n \\equal{} 1} \\frac {x^{4n \\minus{} 1}}{(4n \\minus{} 1)!}$\r\n\r\n$ f^{(4)}(x) \\equal{} \\sum ^\\infty_{n \\equal{} 0} \\frac {(4n \\minus{} 1) x^{4n \\minus{} 2}}{(4n \\minus{} 1)!} \\equal{} \\sum ^\\infty_{n \\equal{} 1} \\frac {x^{4n \\minus{} 2}}{(4n \\minus{} 2)!} \\equal{} \\sum ^\\infty_{n \\equal{} 0} \\frac {x^{4n \\plus{} 2}}{(4n \\plus{} 2)!} \\equal{} f(x)$\r\n\r\nFrom this we can find $ f$:\r\n\r\n$ f(0) \\equal{} 0$\r\n\r\n$ f'(0) \\equal{} 0$\r\n\r\n$ f''(0) \\equal{} 1$\r\n\r\n$ f'''(0) \\equal{} 0$\r\n\r\n$ f^{(4)}(x) \\equal{} f(x)$\r\n\r\nThis is a linear differential equation so we can observe the characteristic equation (define $ f(x) \\equal{} e^{rx}$):\r\n\r\n$ r^4 \\equal{} 1$\r\n\r\n$ r_1 \\equal{} 1\\quad r_2 \\equal{} \\minus{} 1\\quad r_3 \\equal{} i \\quad r_4 \\equal{} \\minus{} i$\r\n\r\nAnd this boils down to:\r\n\r\n$ f(x) \\equal{} C_1 e^x \\plus{} C_2 e^{ \\minus{} x} \\plus{} C_3 \\sin x \\plus{} C_4 \\cos x$\r\n\r\nNow, from the initial conditions we have:\r\n\r\n$ (1) \\qquad f(x) \\equal{} C_1 e^x \\plus{} C_2 e^{ \\minus{} x} \\plus{} C_3 \\sin x \\plus{} C_4 \\cos x \\Rightarrow f(0) \\equal{} C_1 \\plus{} C_2 \\plus{} C_4 \\equal{} 0$\r\n\r\n$ (2) \\qquad f'(x) \\equal{} C_1 e^x \\minus{} C_2 e^{ \\minus{} x} \\plus{} C_3 \\cos x \\minus{} C_4 \\sin x \\Rightarrow f'(0) \\equal{} C_1 \\minus{} C_2 \\plus{} C_3 \\equal{} 0$\r\n\r\n$ (3) \\qquad f''(x) \\equal{} C_1 e^x \\plus{} C_2 e^{ \\minus{} x} \\minus{} C_3 \\sin x \\minus{} C_4 \\cos x \\Rightarrow f''(0) \\equal{} C_1 \\plus{} C_2 \\minus{} C_4 \\equal{} 1$\r\n\r\n$ (4) \\qquad f'''(x) \\equal{} C_1 e^x \\minus{} C_2 e^{ \\minus{} x} \\minus{} C_3 \\cos x \\plus{} C_4 \\sin x \\Rightarrow f'''(0) \\equal{} C_1 \\minus{} C_2 \\minus{} C_3 \\equal{} 0$\r\n\r\nSumming up $ (2)$ and $ (4)$ we get $ C_1 \\equal{} C_2$. Also, $ (1)$ and $ (3)$ give us $ C_1 \\plus{} C_2 \\equal{} \\frac {1}{2} \\Rightarrow C_1 \\equal{} C_2 \\equal{} \\frac {1}{4}$. The rest is easy - $ C_3 \\equal{} 0$ and $ C_4 \\equal{} \\minus{} \\frac {1}{2}$.\r\n\r\nThis gives us:\r\n\r\n$ f(x) \\equal{} \\frac {1}{4}\\left(e^x \\plus{} e^{ \\minus{} x}\\right) \\minus{} \\frac {1}{2} \\cos x \\equal{} \\frac {1}{2}\\left(\\cosh x \\minus{} \\cos x\\right)$\r\n\r\nNow, all that we have to see is that your quotient can be written as:\r\n\r\n$ Q \\equal{} \\frac {\\sum ^\\infty_{n \\equal{} 0} \\frac {1}{(4n)!} \\left(\\frac {\\pi}{2}\\right)^{4n}}{\\sum ^\\infty_{n \\equal{} 0} \\frac {1}{(4n \\plus{} 2)!} \\left(\\frac {\\pi}{2}\\right)^{4n}} \\equal{} \\frac {f''\\left(\\frac {\\pi}{2}\\right)}{\\left(\\frac {\\pi}{2}\\right)^{ \\minus{} 2} f\\left(\\frac {\\pi}{2}\\right)} \\equal{} \\left(\\frac {\\pi}{2}\\right)^2 \\frac {f''\\left(\\frac {\\pi}{2}\\right)}{f\\left(\\frac {\\pi}{2}\\right)}$\r\n\r\nTo proceed, we have to calculate $ f''(x)$:\r\n\r\n$ f'(x) \\equal{} \\frac {1}{2}\\left(\\sinh x \\plus{} \\sin x\\right)$\r\n\r\n$ f''(x) \\equal{} \\frac {1}{2}\\left(\\cosh x \\plus{} \\cos x\\right)$\r\n\r\nAnd the result is:\r\n\r\n$ Q \\equal{} \\left(\\frac {\\pi}{2}\\right)^2 \\frac {\\cosh \\frac {\\pi}{2} \\minus{} \\cos \\frac {\\pi}{2}}{\\cosh \\frac {\\pi}{2} \\plus{} \\cos \\frac {\\pi}{2}} \\equal{} \\left(\\frac {\\pi}{2}\\right)^2$\r\n\r\nTo sum up:\r\n\r\n$ \\boxed{\\boxed{\\frac {1 \\plus{} \\frac {{\\pi}^{4}}{2^{4}\\cdot 4!} \\plus{} \\frac {{\\pi}^{8}}{2^{8}\\cdot 8!} \\plus{} \\frac {{\\pi}^{12}}{2^{12}\\cdot 12!} \\plus{} \\cdots}{\\frac {1}{2!} \\plus{} \\frac {{\\pi}^{4}}{2^{4}\\cdot 6!} \\plus{} \\frac {{\\pi}^{8}}{2^{8}\\cdot 10!} \\plus{} \\frac {{\\pi}^{12}}{2^{12}\\cdot 14!} \\plus{} \\cdots} \\equal{} \\frac {\\pi^2}{4}}}$", "Solution_2": "Though, in retrospect I suppose, one might have also immediately written the numerator as\r\n\\[ \\sum_{k = 0}^\\infty \\left(\\frac {\\pi}{2}\\right)^{4k}\\frac {1}{(4k)!} = \\frac {1}{2}\\left(\\cosh \\frac {\\pi}{2} + \\cos \\frac {\\pi}{2}\\right)\\]\r\nand the denominator as\r\n\\[{ \\sum_{k = 0}^\\infty \\left(\\frac {\\pi}{2}\\right)^{4k}\\frac {1}{(4k + 2)!} = \\left(\\frac {2}{\\pi}\\right)^2\\sum_{k = 0}^\\infty \\left(\\frac {\\pi}{2}\\right)^{4k + 2}\\frac {1}{(4k + 2)!} = \\left(\\frac {2}{\\pi}\\right)^2\\left(\\frac {1}{2}\\right)\\left(\\cosh \\frac {\\pi}{2} - \\cos \\frac {\\pi}{2}\\right)}\\]\r\nThis is of course in retrospect after glancing at milin's solution, but it's not too unnatural an approach; any time you see a sum of powers of a constant over factorials it's worth eyeballing it to see if you can manipulate it into some combination of trigonometric and exponential series by canceling out unwanted terms." } { "Tag": [ "MATHCOUNTS", "AMC", "AIME", "USA(J)MO", "USAMO", "AMC 8" ], "Problem": "im in 5th grade, so how do i join the actual competition? :o :?:", "Solution_1": "You'll have to talk to your principal about participating.", "Solution_2": "When does the actual competition take place? Is there even a winner?", "Solution_3": "There is an award for perfect scorers.", "Solution_4": "Oh!!! Is it MONEY!!! :whistling:", "Solution_5": "No, just a plaque or something, I believe.", "Solution_6": "Oh, well that is still rewarding.... ( :first: )\r\n\r\nIs MOEMS even that hard, or is it really easy?", "Solution_7": "Presuming that this test includes 4th graders, its pretty easy.", "Solution_8": "There are different tests for different grade levels.\r\n\r\nIn comparison to MathCounts, it's fairly easy. Probably around the same level as the AMC 8.", "Solution_9": "I've heard that the AMC 8 is around the same level as MC...\r\n\r\nI'd say that Chapter MC Sprint is about as hard as AMC 8, but when you get to the Target and Team Rounds, and higher levels, it becomes harder.\r\n\r\nThe time restriction on MOEMS is a lot more generous than those on AMC 8 and MathCounts, yet the problems aren't much harder, if at all.", "Solution_10": "[quote=\"AIME15\"]I've heard that the AMC 8 is around the same level as MC...\n\nI'd say that Chapter MC Sprint is about as hard as AMC 8, but when you get to the Target and Team Rounds, and higher levels, it becomes harder.\n\nThe time restriction on MOEMS is a lot more generous than those on AMC 8 and MathCounts, yet the problems aren't much harder, if at all.[/quote]\r\n\r\n1.) AMC 8 is extremely easy. The only problems that are MC Chapter level are like the last 7 of them. I was actually just working on a copy of AMC 8 from 2001. It wasn't as hard as MC. It was about the same level as school. \r\n\r\n2.) I think that you get a pretty good amount of time for AMC 8.", "Solution_11": "You get 40 minutes for 25 problems, which is much less than 3-5 minutes per problem on MOEMS.", "Solution_12": "Oh!!! Well I didn't expect MOEMS to have so much time. Usually the AMC competitions aren't based on time. At the more difficult rounds, like AIME, USAMO, and IMO you get a lot of time." } { "Tag": [], "Problem": "If $ a, b$ and $ c$ are positive integers, $ 0 < a < b < c < 10$ and\n$ \\frac{1}{a} \\plus{} \\frac{1}{b} \\plus{} \\frac{1}{c} \\equal{} 1$, what is the value of $ c$?", "Solution_1": "If $ a \\geq 3$ clearly the maximum sum is $ \\frac13\\plus{}\\frac14\\plus{}\\frac15<1$\r\n\r\nTherefore a=2.\r\n\r\nWE now have $ 2= MG + MA pt n variabile pozitive, unde \r\n\r\nMP=media patratica\r\nMH=media armonica\r\nMG=media geometrica\r\nMA=media aritmetica", "Solution_1": "Ar fi utila, daca ar fi adevarata.\r\n[u]Contraexemplu[/u]: $n=4$, $a_{1}=1$, $a_{2}=a_{3}=a_{4}=2$ (se verifica).\r\nDeci nu mai e nevoie sa incerci sa o demonstrezi :)", "Solution_2": "pt n=4 , a1 = 1, a2=a3=a4=2 inegalitatea este adevarata...\r\n\r\nsi este adevarata in general...verifica-ti calculele.", "Solution_3": "Tu sustii ca $\\frac{8}{5}+\\frac{\\sqrt{13}}2 \\geq \\frac74+\\sqrt[4] 8$? Ce e in stanga e $\\approx 3.40$, iar ce e in dreapta e $\\approx 3.43$...", "Solution_4": "...:| am calculat eu prost inseamna...eu am gasit problema pt 2 variabile a si b ...si era data si generalizarea....asta in mai multe cartzi...deci nu e inventata sau ceva:)", "Solution_5": "Cred ca ce ai g\u00e5sit tu era [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sierpinski&t=52180]Inegalitatea lui Sierpinski[/url]. O inegalitate e fals\u00e5 atunci cand nu e adevarat\u00e5 \u00een fiecare caz, deci nu avea cum s\u00e5 fie asta. :!: C\u00e2nd postezi o ipotez\u00e5, poti face si c\u00e2teva verific\u00e5ri." } { "Tag": [ "geometry", "circumcircle", "inradius", "trigonometry", "inequalities", "perimeter", "algebra" ], "Problem": "Prove that \r\n$h_a+h_b+h_c\\le 3(R+r)$\r\nwith $h_a,h_b,h_c$ are the altitudes, $R$ is the circumradius and $r$ is the inradius", "Solution_1": "[quote=\"Stun\"]Prove that \n$h_a+h_b+h_c\\le 3(R+r)$\nwith $h_a,h_b,h_c$ are the altitudes, $R$ is the circumradius and $r$ is the inradius[/quote]\r\n\r\nOne solution I immediately have is the following one:\r\n\r\nIt is well-known that $h_a=2R\\sin B\\sin C$, $h_b=2R\\sin C\\sin A$, $h_c=2R\\sin A\\sin B$ and $R+r=R\\left(\\cos A+\\cos B+\\cos C\\right)$, so that your inequality\r\n\r\n$h_a+h_b+h_c\\leq 3\\left(R+r\\right)$\r\n\r\ncan be rewritten as\r\n\r\n$2R\\sin B\\sin C+2R\\sin C\\sin A+2R\\sin A\\sin B\\leq 3R\\left(\\cos A+\\cos B+\\cos C\\right)$,\r\n\r\nwhat simplifies to\r\n\r\n$2\\left(\\sin B\\sin C+\\sin C\\sin A+\\sin A\\sin B\\right)\\leq 3\\left(\\cos A+\\cos B+\\cos C\\right)$,\r\n\r\nor, equivalently,\r\n\r\n$\\sin B\\sin C+\\sin C\\sin A+\\sin A\\sin B\\leq \\frac32\\left(\\cos A+\\cos B+\\cos C\\right)$.\r\n\r\nBut this is easy: Since we have\r\n\r\n$\\sin B\\sin C+\\sin C\\sin A+\\sin A\\sin B\\leq \\left(\\cos A+\\cos B+\\cos C\\right)^2$\r\n\r\n(see http://www.mathlinks.ro/Forum/viewtopic.php?t=5716 ) and\r\n\r\n$\\cos A+\\cos B+\\cos C\\leq \\frac32$,\r\n\r\nwe have\r\n\r\n$\\sin B\\sin C+\\sin C\\sin A+\\sin A\\sin B\\leq \\left(\\cos A+\\cos B+\\cos C\\right)^2$\r\n$=\\left(\\cos A+\\cos B+\\cos C\\right)\\cdot\\left(\\cos A+\\cos B+\\cos C\\right)\\leq \\frac32\\cdot\\left(\\cos A+\\cos B+\\cos C\\right)$,\r\n\r\nand we are done.\r\n\r\nBut I think there is also a more elegant solution: Consider the case when the triangle ABC is acute-angled. Then, a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24870]well-known formula[/url] states that AH + BH + CH = 2 (R + r), where H is the orthocenter of triangle ABC; thus, $R+r=\\frac{AH+BH+CH}{2}$, and your inequality $h_a+h_b+h_c\\leq 3\\left(R+r\\right)$ rewrites as $h_a+h_b+h_c\\leq 3\\frac{AH+BH+CH}{2}$, or, equivalently, $2\\left(h_a+h_b+h_c\\right) \\leq 3\\left(AH+BH+CH\\right)$. But if A', B', C' are the feet of the altitudes of triangle ABC, then we have $h_a=AA^{\\prime}=AH+HA^{\\prime}$, and similarly $h_b=BH+HB^{\\prime}$ and $h_c=CH+HC^{\\prime}$, so that the inequality $2\\left(h_a+h_b+h_c\\right) \\leq 3\\left(AH+BH+CH\\right)$ rewrites as\r\n\r\n$2\\left(\\left(AH+HA^{\\prime}\\right)+\\left(BH+HB^{\\prime}\\right)+\\left(CH+HC^{\\prime}\\right)\\right) \\leq 3\\left(AH+BH+CH\\right)$.\r\n\r\nThis simplifies to $2\\left(HA^{\\prime}+HB^{\\prime}+HC^{\\prime}\\right)\\leq AH+BH+CH$, what immediately follows from the Erd\u00f6s-Mordell inequality for the triangle ABC and its interior point H (with HA', HB', HC' being the distances from this point H to the sides BC, CA, AB of triangle ABC). But I don't succeed to extend this solution to the case of obtuse-angled triangles ABC. Can someone help?\r\n\r\n Darij", "Solution_2": "Here is a stronger inequality: Let [tex] w_a , w_b , w_c [/tex] are bisectors of the angles A , B , C . Then [tex] w_a+w_c+w_c\\leq3(R+r) [/tex]\r\n\r\nProve :\r\n[tex] (\\Sigma w_a)^2=(\\Sigma\\frac{2bc}{b+c}cos\\frac{A}{2})^2\\leq(\\Sigma\\sqrt{bc}cos\\frac{A}{2})^2\\leq\\Sigma{ab}\\Sigma{cos^2\\frac{A}{2}\\leq4(R+r)^2\\frac{9}{4}[/tex] , so we have the result.\r\nI hope you know how to prove the well know ineqs:\r\n1) [tex] \\Sigma ab\\leq4(R+r)^2[/tex]\r\n2)[tex]\\Sigma cos^2\\frac{A}{2}\\leq\\frac{9}{4}[/tex]", "Solution_3": "[quote=\"Stun\"]If $ h_a,h_b,h_c$ are the altitudes of a triangle $ ABC$, $ R$ is the circumradius and $ r$ is the inradius, prove that\n\n$ h_a \\plus{} h_b \\plus{} h_c\\le 3(R \\plus{} r)$.[/quote]Actually, the following stronger inequality holds \n\n$ h_a \\plus{} h_b \\plus{} h_c\\le 2R \\plus{} 5r$,\n\nwith equality if and only if $ \\triangle ABC$ is equilateral.\n\n[b]Proof[/b] A formula states that \n\n$ h_a \\plus{} h_b \\plus{} h_c \\equal{} \\frac {s^2 \\plus{} r^2 \\plus{} 4Rr}{2R}$,\n\nwhere $ s$ is the semi-perimeter of $ \\triangle ABC$, and the inequality rewrites as\n\n$ s^2\\leq 4R^2 \\plus{} 6Rr \\minus{} r^2$,\n\nwhich immediately follows from the well-known inequalities $ s^2 \\leq 4R^2 \\plus{} 4Rr \\plus{} 3r^2$ and $ R\\geq 2r$.\n\nOn the other hand, the Horizon Mood's inequality holds\n\n$ h_a \\plus{} w_b \\plus{} m_c\\le 3(R \\plus{} r)$,\n\nwhere $ w_b$ is the angle-bisector and $ m_c$ is the median.\n\n[b]Proof[/b] It is well-known that\n\n$ h_a \\equal{} \\frac {2\\sqrt {s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}}{a}, w_b \\equal{} \\frac {2\\sqrt {cas(s \\minus{} b)}}{c \\plus{} a},m_c \\equal{} \\frac {\\sqrt {2a^2 \\plus{} 2b^2 \\minus{} c^2}}{2}$,\n\n$ R \\equal{} \\frac {4abc}{\\sqrt {s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}},r \\equal{} \\sqrt {\\frac {(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}{s}}$,\n\nwhere $ a,b,c$ are sides of $ \\triangle ABC$. \n\nSubstitute $ a \\equal{} y \\plus{} z,b \\equal{} z \\plus{} x,c \\equal{} x \\plus{} y$, the geometric inequality is equivalent to the following algebraic inequality:\n\n$ 8xyz(x \\plus{} y \\plus{} z)(2y \\plus{} z \\plus{} x) \\plus{} 8y(y \\plus{} z)(x \\plus{} y \\plus{} z)\\sqrt {zx(x \\plus{} y)(y \\plus{} z)}$\n\n$ \\plus{} 2(y \\plus{} z)(2y \\plus{} z \\plus{} x)\\sqrt {xyz(x \\plus{} y \\plus{} z)[4z(x \\plus{} y \\plus{} z) \\plus{} (x \\minus{} y)^2]}$\n\n$ \\leq 3(y \\plus{} z)^2(z \\plus{} x)(x \\plus{} y)(2y \\plus{} z \\plus{} x) \\plus{} 12xyz(y \\plus{} z)(2y \\plus{} z \\plus{} x)$\n\nfor positive real numbers $ x,y,z$. \n\nSince $ [y(z \\plus{} x) \\plus{} 2zx]^2 \\minus{} 4zx(x \\plus{} y)(y \\plus{} z) \\equal{} y^2(z \\minus{} x)^2\\geq 0$, we have to prove\n\n$ 2(y \\plus{} z)(2y \\plus{} z \\plus{} x)\\sqrt {xyz(x \\plus{} y \\plus{} z)[4z(x \\plus{} y \\plus{} z) \\plus{} (x \\minus{} y)^2]}$\n\n$ \\leq 3(y \\plus{} z)^2(z \\plus{} x)(x \\plus{} y)(2y \\plus{} z \\plus{} x) \\plus{} 12xyz(y \\plus{} z)(2y \\plus{} z \\plus{} x)$\n\n$ \\minus{} 8xyz(x \\plus{} y \\plus{} z)(2y \\plus{} z \\plus{} x) \\minus{} 4y(y \\plus{} z)(x \\plus{} y \\plus{} z)[y(z \\plus{} x) \\plus{} 2zx]$\n\n$ \\equal{} (3y^2 \\minus{} 2yz \\plus{} 3z^2)x^3 \\plus{} (5y^3 \\plus{} 9yz^2 \\plus{} 6z^3)x^2 \\plus{} (y \\plus{} z)^2(2y^2 \\plus{} 8yz \\plus{} 3z^2)x$\n\n$ \\plus{} yz(y \\plus{} z)^2(2y \\plus{} 3z)$,\n\nor, equivalently, \n\n$ F(x,y,z)\\equiv [(3y^2 \\minus{} 2yz \\plus{} 3z^2)x^3 \\plus{} (5y^3 \\plus{} 9yz^2 \\plus{} 6z^3)x^2 \\plus{} (y \\plus{} z)^2(2y^2 \\plus{} 8yz \\plus{} 3z^2)x$\n\n$ \\plus{} yz(y \\plus{} z)^2(2y \\plus{} 3z)]^2 \\minus{} 4xyz(y \\plus{} z)^2(x \\plus{} y \\plus{} z)(2y \\plus{} z \\plus{} x)^2[4z(x \\plus{} y \\plus{} z) \\plus{} (x \\minus{} y)^2]$\n\n$ \\geq 0$,\n\nwith equality if and only if $ x \\equal{} y \\equal{} z$.\n\n[i]Case 1[/i]: $ x \\equal{} \\min\\{x,y,z\\}.F(x,x \\plus{} s,x \\plus{} t)$\n\n$ \\equal{} 768(4s^2 \\minus{} 8st \\plus{} 5t^2)x^8 \\plus{} 64(112s^3 \\minus{} 100s^2t \\minus{} 84st^2 \\plus{} 161t^3)x^7$\n\n$ \\plus{} 16(420s^4 \\plus{} 152s^3t \\minus{} 664s^2t^2 \\plus{} 546st^3 \\plus{} 699t^4)x^6$\n\n$ \\plus{} 16(196s^5 \\plus{} 382s^4t \\minus{} 100s^3t^2 \\plus{} 299s^2t^3 \\plus{} 963st^4 \\plus{} 390t^5)x^5$\n\n$ \\plus{} 4(180s^6 \\plus{} 796s^5t \\plus{} 973s^4t^2 \\plus{} 1806s^3t^3 \\plus{} 3305s^2t^4 \\plus{} 2284st^5 \\plus{} 476t^6)x^4$\n\n$ \\plus{} 16(s \\plus{} t)(4s^6 \\plus{} 38s^5t \\plus{} 108s^4t^2 \\plus{} 264s^3t^3 \\plus{} 326s^2t^4 \\plus{} 145st^5 \\plus{} 19t^6)x^3$\n\n$ \\plus{} 4t(s \\plus{} t)^2(12s^5 \\plus{} 112s^4t \\plus{} 313s^3t^2 \\plus{} 307s^2t^3 \\plus{} 82st^4 \\plus{} 5t^5)x^2$\n\n$ \\plus{} 4st^2(s \\plus{} t)^3(16s^3 \\plus{} 48s^2t \\plus{} 40st^2 \\plus{} 5t^3)x \\plus{} s^2t^2(2s \\plus{} 3t)^2(s \\plus{} t)^4\\geq 0$,\n\nwhich is clearly true for\n\n$ 112s^3 \\minus{} 100s^2t \\minus{} 84st^2 \\plus{} 161t^3 \\equal{} 4(28s \\plus{} 31t)(s \\minus{} t)^2 \\plus{} t^2(52s \\plus{} 37t)$,\n\n$ 420s^4 \\plus{} 152s^3t \\minus{} 664s^2t^2 \\plus{} 546st^3 \\equal{} 2s(210s \\plus{} 271t)(s \\minus{} t)^2 \\plus{} 2st(225s^2 \\plus{} 2t^2)$.\n\n[i]Case 2[/i]: $ y \\equal{} \\min\\{x,y,z\\}. F(y \\plus{} t,y,y \\plus{} s)$\n\n$ \\equal{} 768(5s^2 \\minus{} 2st \\plus{} t^2)y^8 \\plus{} 64(161s^3 \\plus{} 81s^2t \\plus{} 23st^2 \\plus{} 7t^3)y^7$\n\n$ \\plus{} 16(699s^4 \\plus{} 1166s^3t \\plus{} 716s^2t^2 \\plus{} 74st^3 \\plus{} 5t^4)y^6$\n\n$ \\plus{} 16s(390s^4 \\plus{} 1281s^3t \\plus{} 1523s^2t^2 \\plus{} 457st^3 \\plus{} 17t^4)y^5$\n\n$ \\plus{} 4s(476s^5 \\plus{} 2660s^4t \\plus{} 5545s^3t^2 \\plus{} 3274s^2t^3 \\plus{} 541st^4 \\plus{} 4t^5)y^4$\n\n$ \\plus{} 16s^2(s \\plus{} t)(19s^4 \\plus{} 160s^3t \\plus{} 454s^2t^2 \\plus{} 191st^3 \\plus{} 20t^4)y^3$\n\n$ \\plus{} 4s^2(s \\plus{} t)^2(5s^4 \\plus{} 86s^3t \\plus{} 375s^2t^2 \\plus{} 95st^3 \\plus{} 5t^4)y^2$\n\n$ \\plus{} 4s^3t(s \\plus{} t)^3(5s^2 \\plus{} 44st \\plus{} 5t^2)y \\plus{} 9s^4t^2(s \\plus{} t)^4\\geq0$.\n\n[i]Case 3[/i]: $ z \\equal{} \\min\\{x,y,z\\}.F(z \\plus{} s,z \\plus{} t,z)$\n\n$ \\equal{} 768(s^2 \\plus{} 4t^2)z^8 \\plus{} 64(7s^3 \\plus{} 52s^2t \\plus{} 148st^2 \\plus{} 112t^3)z^7$\n\n$ \\plus{} 16(5s^4 \\plus{} 102s^3t \\plus{} 1080s^2t^2 \\plus{} 1304st^3 \\plus{} 420t^4)z^6$\n\n$ \\plus{} 16t(13s^4 \\plus{} 549s^3t \\plus{} 1804s^2t^2 \\plus{} 1158st^3 \\plus{} 196t^4)z^5$\n\n$ \\plus{} 4t( \\minus{} 4s^5 \\plus{} 521s^4t \\plus{} 3302s^3t^2 \\plus{} 5613s^2t^3 \\plus{} 2044st^4 \\plus{} 180t^5)z^4$\n\n$ \\plus{} 16t^2(16s^5 \\plus{} 180s^4t \\plus{} 561s^3t^2 \\plus{} 566s^2t^3 \\plus{} 110st^4 \\plus{} 4t^5)z^3$\n\n$ \\plus{} 4st^2(5s^5 \\plus{} 81s^4t \\plus{} 424s^3t^2 \\plus{} 751s^2t^3 \\plus{} 460st^4 \\plus{} 36t^5)z^2$\n\n$ \\plus{} 4s^2t^3(s \\plus{} t)(5s^3 \\plus{} 38s^2t \\plus{} 72st^2 \\plus{} 40t^3)z \\plus{} s^2t^4(s \\plus{} t)^2(3s \\plus{} 2t)^2$\n\n$ \\equal{} 768(s^2 \\plus{} 4t^2)z^8 \\plus{} 64(7s^3 \\plus{} 52s^2t \\plus{} 148st^2 \\plus{} 112t^3)z^7$\n\n$ \\plus{} 8(9s^4 \\plus{} 204s^3t \\plus{} 2160s^2t^2 \\plus{} 2608st^3 \\plus{} 840t^4)z^6$\n\n$ \\plus{} 16t(13s^4 \\plus{} 549s^3t \\plus{} 1804s^2t^2 \\plus{} 1158st^3 \\plus{} 196t^4)z^5$\n\n$ \\plus{} 4t^2(521s^4 \\plus{} 3302s^3t \\plus{} 5613s^2t^2 \\plus{} 2044st^3 \\plus{} 180t^4)z^4$\n\n$ \\plus{} 16t^2(16s^5 \\plus{} 180s^4t \\plus{} 561s^3t^2 \\plus{} 566s^2t^3 \\plus{} 110st^4 \\plus{} 4t^5)z^3$\n\n$ \\plus{} 4st^2(3s^5 \\plus{} 81s^4t \\plus{} 424s^3t^2 \\plus{} 751s^2t^3 \\plus{} 460st^4 \\plus{} 36t^5)z^2$\n\n$ \\plus{} 4s^2t^3(s \\plus{} t)(5s^3 \\plus{} 38s^2t \\plus{} 72st^2 \\plus{} 40t^3)z \\plus{} s^2t^4(s \\plus{} t)^2(3s \\plus{} 2t)^2$\n\n$ \\plus{} 8z^2s^4(z^2 \\minus{} st)^2\\geq0$.\n[quote=\"nickolas\"]Let $ w_a , w_b , w_c$ are bisectors of the angles $ A , B , C$, then $ w_a \\plus{} w_c \\plus{} w_c\\leq3(R \\plus{} r)$.[/quote]Here is a stronger inequality: \r\n\r\n$ w_a \\plus{} w_c \\plus{} w_c\\leq 9r \\plus{} k(R \\minus{} 2r)$,\r\n\r\nwhere $ k \\equal{} 2.1707\\cdots$ be a root of the following irreducible polynomial \r\n\r\n$ 72k^5 \\minus{} 1497k^4 \\plus{} 12450k^3 \\minus{} 51109k^2 \\plus{} 102227k \\minus{} 78656$;\r\n\r\nwith equality if $ b \\equal{} c \\equal{} 1,a \\equal{} 0.62892\\cdots$ be a root of the following irreducible polynomial \r\n\r\n$ a^5 \\plus{} 12a^4 \\plus{} 39a^3 \\plus{} 17a^2 \\minus{} 7a \\minus{} 14$." } { "Tag": [], "Problem": "Whats $ \\frac{f_{1}}{5^{1}}+\\frac{f_{2}}{5^{2}}+...+\\frac{f_{n}}{5^{n}}+.......$ where $ f_{1}=1$,$ f_{2}=2$, and $ f_{n}=f_{n-1}+f_{n-2}$\r\n\r\n\r\nTheres a simple common fractional answer to this. Its just a simple trick..", "Solution_1": "call that sum S. You can find an equation soley for S if you multiply that by 5 and subtract the originial S from it." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "1. Prove that there exists an infinite sequence of pairwise co-prime positive $>1$, none of which divides a number of the form $2^{m}+2^{n}+1$, where $m,n \\in \\mathbb{N}$. \r\n\r\n2*. [Open to me. ] Are there infinitely many primes which do not divide any number of the form $2^{m}+2^{n}+1$?", "Solution_1": "1. Consider $a=2^{p}-1$. If $ak=2^{m}+2^{n}+1,m\\ge n$, then $k=-1+2^{l}a$ a - odd and $a2^{l+p}-2^{l}a-2^{p}=2^{m}+2^{n}.$\r\nIt give $l=p$ and $a2^{2p}+(a+1)=2^{m-p}+2^{n-p}$ contradition for prime p$\\ge 3$.\r\nObviosly $(2^{p}-1,2^{q}-1)=1$ for different primes p and q.\r\n2. Let $P|2^{p}-1$ prime divisor of Mersen's number (p prime). Then $r_{m}=2^{m}(mod \\ P), r_{n}=2^{n}.$\r\n$P|2^{m}+2^{n}+1\\to r_{m}+r_{n}=P-1$.\r\nObviosly if P is mersen's prime it give contradition. If there are infinetely many Mersenn's primes, then 2. is true. \r\nBut it is true for many (I think for infinetely many) prime divisisors of Mersen's numbers.\r\nFor example $2^{11}-1=23*89$ and $2^{n}(mod \\ 89)=1,2,4,8,16,32,64,39,78,67,45.$ Obviosly $r_{n}+r_{m}\\not = 88$." } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "If $ f(0) \\equal{} g(0) \\equal{} 0$ and $ g'$ and $ f''$ are continuous, show that: \r\n$ \\int_{0}^{a} f(x)g''(x) dx \\equal{} f(a)g'(a)\\minus{}f'(a)g(a)\\plus{}\\int_{0}^{a} f''(x)g(x) dx$", "Solution_1": "That is straightfor using successively integration by parts." } { "Tag": [ "puzzles" ], "Problem": "I just got this puzzule from my friend. I want to share it with everyone:\r\n\r\n To walk one mile in desert, an elephant must eat one pound of bananas. The maximum load of bananas the elephant can walk with is 1000lb. A desert is 2000 miles wide and there is no banana in it. On both side of the desert an unlimited supply of bananas is stored.\r\n1. Can the elephant cross the desert? [Hint: the elephant can make cashes.]\r\n2. If the answer is possible, what is the shortest distance it should cover?\r\n3. Can the elephant cross the desert if it is N miles wide? [N is a finite but maybe is very big number]\r\n \r\n Have fun! :D", "Solution_1": "i guess that the elephant could take trips back and forth and put a supply of bananas in the middle of the desert, and go back and repeat, i.e. go 500 mi, deposit 500 bananas, and do that 2 times, then go 500 mi past, and make sure to have enough bananas at each place, and in this way, the elephant could go infinite distance, since it could make check points with infinite bananas, and repeat", "Solution_2": "Moved to Puzzles.", "Solution_3": "you mean, caches, like it can drop bananas in a part of the desert and return to them later(assuming it can find the bananas and they don't dessicate and all that jazz)", "Solution_4": "Mine is probably redudant, but:\r\n\r\n1. Go 250 mi., drop 500, go back. (500 mi.)\r\n2. Go 250 mi., pick up 250, go 250 more, drop 500, go back 250, pick up 250, go back. (1000 mi.)\r\n3. Same as 1. (500 mi.)\r\n4. Same as 2, but go out 250 more miles and drop 500 at the 750 mi. point. (1500 mi.)\r\n5. Same as 1. (500 mi.)\r\n6. Same as 2. (1000 mi.)\r\n7. Same as 1. (500 mi.)\r\n8. Same as 4, but go to the 1000 mi. point. (2000 mi.)\r\n9. Repeat steps 1 - 8. (7500 mi.)\r\n10. Go all the way across (2000 mi.)\r\n\r\nTotal: 17000 mi.\r\n\r\nAs I said, probably redundant.", "Solution_5": "Yeah, it could do something like\r\nTake 1000, go some distance, drop some, use the rest of the bananas to come back. Repeat. Essentially that point in the desert now has an infinite number of bananas. Hence the elephant has moved the starting point forward, and shortened the distance. So the elephant can travel any distance required.", "Solution_6": "Width of the desert = 2000 miles\r\nConsumption rate of the elephant = 1 lb/mile\r\nMaximum carriage capacity of the elephane = 1000 lb \r\nLet the initial quantity of bananas = Q lb (Here Q is unlimited!!!)\r\n[code] C\n.-----------------------.----------------------------.\nA <------------- 2000 miles -----------------------> B [/code]\r\n\r\n1. Yes, the elephant can cross the desert. \r\n\r\n2. The shortest distance would be covered by the elephant when it reaches a point C with 1000 pounds of bananas where the distance CB = 1000 miles.\r\n\r\nThen, it would require 2000 lb of bananas at a point that is 1000/3 = 333.33 miles prior to C. (You will get the logic if you just think for a while :lol: )\r\n\r\nThe elephant would require 3000 lb of bananas at a point that is 1000/3 + 1000/5 = 533.33.... miles prior to C. \r\n\r\nProceeding in this way, the elephant would require 7000 lb of bananas at a point that is 1000/3 + 1000/5 + 1000/7 + 1000/9 + 1000/11 + 1000/13 = 8604800/9009 miles = 955.1338 miles prior to C. \r\n\r\n1000 - (8604800/9009) = 404200/9009 = 44.8662 miles. \r\n\r\nShortest distance covered by the elephant = (404200/9009)*15 + (1000/13)*13 + (1000/11)*11 + (1000/9)*9 + (1000/7)*7 + (1000/5)*5 + (1000/3)*3 + 1000 = [size=150][b]7672.993673[/b][/size] miles.\r\n\r\n3. Since, there is unlimited supply of bananas on the sides of the desert, the maximum width of the desert that the elephant can cross = 1000 + 1000/3 + 1000/5 + 1000/7 + 1000/9 + 1000/11 + 1000/13 + ....... \r\n= 1000 * (1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 + 1/17 + ...) \r\n= 1000 *{ (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + ...) - (1/2 + 1/4 + 1/6 + 1/8 + 1/10 + .....)} \r\n= 1000 *{ (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + ...) - (1/2)*(1 + 1/2 + 1/3 + 1/4 + 1/5 + .....)} \r\n= 1000*(1/2)*(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + ...)\r\n= 500*(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + ...)\r\n= 500*S where S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + ...\r\n\r\nNow, the series S diverges as S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + ... = S = 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ... = 1 + 1/2 + 1/2 + 1/2 + 1/2 + ...........\r\n\r\n[i][b]The elephant can cross any desert N miles wide[/b][/i] as ahown above. \r\n\r\n[b][i][/i][/b]" } { "Tag": [ "function", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Suppose that $: \\mathbb R^{m}\\to \\mathbb R$ satisfies two conditions i)For each compact set $K,f(K)$ is compact. ii)For any nested decreasing sequence of compacts$(K_{n}),\\ f(\\bigcap K_{n})=\\bigcap f(K_{n})$.\r\nProve that $f$ is continuous.", "Solution_1": "You have a confusing typo. State the range of f & where is K contained, just in case. It may be useful to know whether the range of f is open, closed, or whatnot.", "Solution_2": "Consider a sequence $x_{n}\\to x$ in $\\mathbb R^{n}$. We shall prove that $f(x_{n}) \\to f(x)$.\r\n\r\nLet $K_{n}= \\{x_{i}: i\\geq n\\}\\cup \\{x\\}$. Each set $K_{n}$ is compact. So, by the condition, each set $f(K_{n})$ is compact. Let $y$ be a limit point of the sequence $f(x_{i})$. Since all except a finite number of $f(x_{i})$ belong to $f(K_{n})$ and $f(K_{n})$ is closed, $y\\in f(K_{n})$. We have\r\n\\[y\\in \\cap_{n=1}^\\infty f(X_{n}) = f(\\cap_{n=1}^\\infty X_{n}) = f(\\{x\\}). \\]\r\nWe proved that $y= f(x)$. So $f(x_{i})$ is a sequence in a compact set, whose only limit point is $f(x)$. Therefore, $f(x_{n}) \\to f(x)$. This concludes the proof.", "Solution_3": "Here's my solution. Let $R^{p}$be the range of f and take a nonempty closed set $C \\subset R^{p}$and let b be a limit point of $f^{-1}(C)$. Take a decreasing sequence of closed balls $(B(b;{\\textstyle{1 \\over n}})) = (K_{n})$ so that $\\cap K_{n}= b$.\r\n\r\n$\\cap (f(K_{n}) \\cap C) = ( \\cap f(K_{n})) \\cap C = \\{ f(b)\\}= f( \\cap K_{n}) \\cap C)$\r\n\r\nTherefore $b \\in f^{-1}(C)$. $f^{-1}(C)$ is closed.\r\n\r\nI hope this is right. I did not use the first property so it might not be. :oops:" } { "Tag": [], "Problem": "In a coding system, a single bit is encoded as a triple, consisting of a short binary string S and 2 integers, K and N. A single \"processing step\" takes a binary string and replaces each 1 by 0, and each 0 by the input string S. The bit that was encoded can be recovered by applying this \"processing step\" successively K times, beginning with the string 1, and then extracting the N_th bit from the resulting string. Your job is to decode, that is, given S, K and N, extract the said bit. \r\n\r\nINPUT\r\nThe input file will contain several test cases. The first line contains an integer t, the number of test-cases. The succeeding lines contain the test-cases. \r\n\r\nInput Format:\r\nEach test-case comprises the following:\r\n\r\nLine 1: K \r\nLine 2: N \r\nLine 3: S \r\n\r\n\r\nOUTPUT\r\n The output should comprise the solutions to all the test cases, in the order of the test cases in the input. There should NOT be any blank line in the output.\r\n\r\nOutput Format:\r\nLine 1: 1 or 0, the N_th bit that is extracted after K processing steps. \r\n\r\nPOINTS and CONSTRAINTS\r\nTotal points: 200 \r\n\r\nEasy Set (50 points): 1<=t<=20, 0<=K<=2000, 1<=N<=1000, 1<=Length of S<=10.\r\nHard set: 1<=t<=20, 0<=K<=1000000, 1<=N<=100000000000000, 1<=Length of S<=10.\r\n\r\n\r\nTime Limit: 5 seconds\r\n\r\n\r\nSAMPLE INPUT\r\n2\r\n3\r\n5\r\n001\r\n2\r\n1\r\n101\r\n\r\nSAMPLE OUTPUT\r\n0\r\n1\r\n\r\nExplanation: In the first test-case, the 3 processing steps are: 1 -> 0 -> 001 -> 0010010. The 5th bit of 0010010 is 0. \r\n\r\ncan sumone explain the solution for this", "Solution_1": "OK, here's a hint: suppose you start with \"1\" and do $n$ steps. Can you calculate the length of the resulting string? Can you calculate the length of the resulting string if you start with \"0\"? Can you calculate this for every possible $n$ from 1 to 1000000 and for both starting strings (\"0\" and \"1\") in linear time?" } { "Tag": [ "function", "analytic geometry", "graphing lines", "slope", "linear algebra" ], "Problem": "According to the linear function represented in this table, what is the value of $ x$ when\n$ y \\equal{} 8$?\n\\begin{tabular}{ c | c }\n x&y\\\\ \\hline\n -4&23\\\\\n 1&20\\\\\n 6&17\\\\\n\\end{tabular}", "Solution_1": "[quote=\"GameBot\"]According to the linear function represented in this table, what is the value of $ x$ when $ y \\equal{} 8$?\n\\[ \\begin{tabular}{ c | c } x & y \\\\\n\\hline \\minus{} 4 & 23 \\\\\n1 & 20 \\\\\n6 & 17 \\\\\n\\end{tabular}\n\\]\n[/quote]\r\n\r\n$ \\Delta x \\equal{} 5 \\to \\Delta y \\equal{} \\minus{} 3$.\r\nSo,\r\n$ 8 \\minus{} 17 \\equal{} \\minus{} 9 \\equal{} 3\\times( \\minus{} 3) \\equal{} 3 \\times \\Delta y$.\r\nTherefore, $ 3\\times\\Delta x \\equal{} 3 \\times 5 \\equal{} 15$.\r\nThus, $ 6 \\plus{} 15 \\equal{} \\boxed{21}$.", "Solution_2": "[quote=\"GameBot\"]According to the linear function represented in this table, what is the value of $ x$ when\n$ y \\equal{} 8$?\n\\begin{tabular}{ c | c }\n x&y\\\\ \\hline\n -4&23\\\\\n 1&20\\\\\n 6&17\\\\\n\\end{tabular}[/quote]\n\nWhat is up with this graph", "Solution_3": "[hide=\"Solution\"]Since the function is linear, we can find the slope, then the equation, then substitute in $y=8$ to find $x$.\nFirst, let's find the slope.\nThese are the given points:\n$(-4, 23)$\n$(1, 20)$\n$(6, 17)$\n\nLet's take the bottom two points and use the slope formula:\n$\\frac{17-20}{6-1}\n=-\\frac{3}{5}$\n\nTherefore the slope is $-\\frac{3}{5}$.\n\nWe use the point-slope formula to find the equation using the point $(1, 20)$.\n$y-20 = -\\frac{3}{5}(x-1)$\n\n$y-20 =-\\frac{3}{5}x + \\frac{3}{5}$\n\n$y = -\\frac{3}{5}x + 20\\frac{3}{5}$\n\nNow we plug in $y=8$ into the equation:\n$8 = -\\frac{3}{5}x + 20\\frac{3}{5}$\n\n$\\frac{3}{5}x = \\frac{63}{5}$\n\n$x = \\boxed{21}$[/hide]", "Solution_4": "[quote=\"minimario\"][quote=\"GameBot\"]According to the linear function represented in this table, what is the value of $ x$ when\n$ y \\equal{} 8$?\n\\begin{tabular}{ c | c }\n x&y\\\\ \\hline\n -4&23\\\\\n 1&20\\\\\n 6&17\\\\\n\\end{tabular}[/quote]\n\nWhat is up with this graph[/quote]\nMaybe he didn't put the $$.", "Solution_5": "[quote=\"GameBot\"]\n\\begin{tabular}{ c | c }\n x&y\\\\ \\hline\n -4&23\\\\\n 1&20\\\\\n 6&17\\\\\n\\end{tabular}[/quote]\n\n[quote=\"Kouichi Nakagawa\"]\\[ \\begin{tabular}{ c | c } x & y \\\\\n\\hline \\minus{} 4 & 23 \\\\\n1 & 20 \\\\\n6 & 17 \\\\\n\\end{tabular}\n\\][/quote]\n\nNotice the differences. Simply a coding error." } { "Tag": [ "limit", "conics", "parabola" ], "Problem": "Can someone explain to me where my reasoning went wrong, becuase obviously the proof is wrong.\r\n\r\n---------------------------------------------------------------------------------------------------\r\n1*1 = 12 \r\n12 = 1\r\n02 = 0\r\n\r\nx2 = x\r\nx = 1, 0\r\n\r\ninfinity2 = infinty\r\nso if we substituted infinity into the equation x2 = x, we could say that infinity = 1 or infinty = 0\r\n\r\nBut obviously that's not true.", "Solution_1": "You can't treat infinity like a real number. If you do a lot of things will be screwed up. Infinity means growing without bound. Since you can't define infinity as one exact number you can't use it in equations like it is one exact number. The main point is that you have to be careful when working with infinity.", "Solution_2": "You really can't do infinity2 = infinity. It doesn't make sense. The closest you can get is\r\n\r\n[tex]\\displaystyle \\lim_{x\\rightarrow\\infty}x^2 = \\infty[/tex].", "Solution_3": "I get it now, infinity isn't a real number like 23 or something.\r\n\r\nBut what is [tex]\\displaystyle \\lim_{x\\rightarrow\\infty}x^2 = \\infty[/tex]?", "Solution_4": "That is read the limit as x grows w/out bound of x squared goes to inifinity. It just means that if you picked values of x to put into x^2 and each time you increased the value you chose that the result would keep getting bigger and bigger but it would not approach any value. Limits aren't getting started but I'll leave this topic here since it was informational.", "Solution_5": "Does $\\displaystyle \\lim_{x\\rightarrow\\infty}x^2 = \\infty$ mean that $x^{2}$ approaches $\\infty$?", "Solution_6": "Yes as x approach infinity so does x^2. This should be pretty easy to see. If you look at a graph of it you will see a parabola. The parabola keeps increasing without bound as the x value increases.", "Solution_7": "Right so as the value of x gets bigger and bigger (approaches infinity), the value of x^2 gets bigger and bigger (infinity), so the limit (what happens to f(x) as x approaches infinity) is equal to infinity." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "The expressions $ a \\plus{} b \\plus{} c, ab \\plus{} ac \\plus{} bc,$ and $ abc$ are called the elementary symmetric expressions on the three letters $ a, b, c;$ symmetric because if we interchange any two letters, say $ a$ and $ c,$ the expressions remain algebraically the same. The common degree of its terms is called the order of the expression. Let $ S_k(n)$ denote the elementary expression on $ k$ different letters of order $ n;$ for example $ S_4(3) \\equal{} abc \\plus{} abd \\plus{} acd \\plus{} bcd.$ There are four terms in $ S_4(3).$ How many terms are there in $ S_{9891}(1989)?$ (Assume that we have $ 9891$ different letters.)", "Solution_1": "Am I missing something or is it:\r\n\r\n$ \\displaystyle {9891 \\choose 1989}$", "Solution_2": "[quote=\"SimonM\"]Am I missing something or is it:\n\n$ \\displaystyle 9891 \\choose 1989$[/quote]\r\n\r\nYou are missing the fact that the \\displaystyle command, the way you use it, only applies to the $ 9891$ and not the whole binomial coefficient. Brackets help:\r\n\r\n$ \\displaystyle {9891 \\choose 1989}$.\r\n\r\nOtherwise, the answer is right. Orl, what have I told you about the quality of ILL problems? ;)\r\n\r\n darij" } { "Tag": [ "trigonometry", "complex numbers" ], "Problem": "The complex number $ w$ equals $ \\minus{} 128 \\minus{} 128\\cdot\\sqrt {3}\\cdot i$, where $ i^2 \\equal{} \\minus{} 1$, and $ a$ and $ b$ are real numbers. There are four distinct complex numbers $ v$ such that $ v^4 \\equal{} w$. Find these four numbers, and express them in $ a \\plus{} bi$ form.", "Solution_1": "This is from the CJML website, right?\r\n\r\n[hide=\"Solution\"]\nLet $ \\text{\\cis}\\ \\theta = \\cos \\theta + i \\sin \\theta$. Then $ -128-128\\cdot 3 \\cdot i = 256\\ \\text{cis}\\ 240^{\\circ}$, and so by De Moivre's, $ v = (256\\ \\text{cis}\\ 240^{\\circ})^{1/4} = 4 \\text{cis}\\ 60+90k$ where $ k \\in \\{0,1,2,3\\}$. The answers are \n\\begin{align*}\n4\\ \\text{cis}\\ 60 &= 2 + 2\\sqrt{3} i\\\\\n4\\ \\text{cis}\\ 150 &= -2\\sqrt{3} + 2i\\\\\n4\\ \\text{cis}\\ 240 &= -2 - 2\\sqrt{3} i\\\\\n4\\ \\text{cis}\\ 330 &= 2\\sqrt{3} - 2i\n\\end{align*}\n[/hide]", "Solution_2": "Yes it is." } { "Tag": [ "linear algebra", "matrix", "vector", "advanced fields", "advanced fields solved" ], "Problem": "What is a spectrum??????????? (Not sure if it belongs to this forum, if not please move it to the correct one).......... thanks.....", "Solution_1": "Spectrum of what?", "Solution_2": "[quote=\"amfulger\"]Spectrum of what?[/quote]\r\n\r\nI don't know, any example.... I just heard this name and I wanted to know something about it.....", "Solution_3": "The set of the eigenvalues of a matrix (in linear algebra).", "Solution_4": "For example if it is spectrum of a linear operator $A$ in a vector space $V$, then the spectrum is the set of the eigenvalues of $A$, i.e. the set of $\\lambda$ for which there exists $x \\neq 0$ such that $Ax= \\lambda x$.", "Solution_5": "[quote=\"eugene\"]For example if it is spectrum of a linear operator $A$ in a vector space $V$, then the spectrum is the set of the eigenvalues of $A$, i.e. the set of $\\lambda$ for which there exists $x \\neq 0$ such that $Ax= \\lambda x$.[/quote]\r\n\r\nThanks eugene. I find this: http://mathworld.wolfram.com/Spectrum.html" } { "Tag": [ "geometry", "3D geometry", "quadratics", "trigonometry", "algebra", "quadratic formula" ], "Problem": "How do I solve for x\r\n\r\nx^3=216\r\n\r\nI know its 6, but what are the steps to get the answer?\r\n\r\n\r\nThanks for any help. :)", "Solution_1": "[quote=\"The Will\"]How do I solve for x\n\nx^3=216\n\nI know its 6, but what are the steps to get the answer?\n\n\nThanks for any help. :)[/quote]\r\n\r\n\r\nYou just cube root it to get the answer...\r\n\r\nBut actually techinically, there are 3 answers, except the other 2 are imaginary.\r\nYou can use DeMoivre's Therom and find the answers are actually\r\n\r\n6(cos0+i*sin0)\r\n6(cos 120+i*sin120)\r\n6(cos 240+i*sin240) where i=sqrt{-1}\r\n\r\nand the things are in degrees..", "Solution_2": "cube root of $216$ is it's only real answer.but to find the conjugate none-real roots you have to use the cubic formula.\r\nit has 3 answer which are.\r\n$x=6$\r\n$x=-3-5.19615i$\r\n$x=-3+5.19615i$", "Solution_3": "[quote=\"The Will\"]How do I solve for x\n\nx^3=216\n\nI know its 6, but what are the steps to get the answer?\n\n\nThanks for any help. :)[/quote]\r\n[hide=\"All solutions\"]$x^{3}= 216$\n$x^{3}-216 = 0$\n$(x-6)(x^{2}+6x+36) = 0$\n\nHence, $x=6$ or a root of $x^{2}+6x+36$. By the quadratic formula,\n$x = 6,-3\\pm 3i\\sqrt{3}$[/hide]\r\n\r\nEDIT: Grr.... I always forget the negative sign...", "Solution_4": "how did u change the roots from decimal to $x=3\\pm3i\\sqrt{3}$ ? isnt it $x=-3\\pm3i\\sqrt{3}$?\r\ncan you explain please how u changed from decimal to that form? is there a general rule?", "Solution_5": "it is -3, I assume it is a typo.", "Solution_6": "[quote=\"The Will\"]How do I solve for x\n\nx^3=216\n\nI know its 6, but what are the steps to get the answer?[/quote]\n\nThe first thing you might try is assuming $x$ is an integer. That assumption works out nicely in this case, since you don't have to test many values of $x$ until that works, but if that assumption doesn't pan out, then $x$ is not going to be very nice at all, and all you can do is take a cube root with a calculator or what have you.\n\n[quote=\"binomial_4eva\"]is there a general rule?[/quote]\r\n\r\nYes. We will be making use of Euler's formula,\r\n\r\n$e^{ix}= \\cos x+i \\sin x$\r\n\r\nAnd the corollary, De Moivre's Formula,\r\n\r\n$(\\cos x+i \\sin x)^{n}= \\cos nx+i \\sin nx$\r\n\r\nFirst we will find the cube roots of $1$. To get the cube roots of $216$ we merely multiply by $6$. Now, when we have\r\n\r\n$(\\cos x+i \\sin x)^{3}= \\cos 3x+i \\sin 3x = 1$\r\n\r\nWe know that $3x = 0, 2\\pi, 4\\pi \\implies x = 0, \\frac{2\\pi}{3}, \\frac{4\\pi}{3}$. Hence the three cube roots of $1$ are given by\r\n\r\n$e^{0}= 1$\r\n$e^{i \\frac{2\\pi}{3}}= \\cos \\frac{2\\pi}{3}+i \\sin \\frac{2\\pi}{3}=-\\frac{1}{2}+i \\frac{ \\sqrt{3}}{2}$\r\n$e^{i \\frac{4\\pi}{3}}= \\cos \\frac{4\\pi}{3}+i \\sin \\frac{4\\pi}{3}=-\\frac{1}{2}-i \\frac{\\sqrt{3}}{2}$", "Solution_7": "Thank you all very much for your answers. :lol: \r\n\r\nNow I need to go figure out how to get the cube root \"without a calculator\" \r\n\r\n\r\nHow do you know that its x^3 is the cube root of 216? It is, but what there tells you that x is the cube root of 216? :huh:", "Solution_8": "[quote=\"The Will\"]but what there tells you that x is the cube root of 216? :huh:[/quote]\r\n\r\nThat's the definition of a cube root. If $x^{3}= 216$, then we say that $x$ is a cube root of $216$, just the same way if $x^{2}= 36$, then we say that $x$ is a square root of $36$.", "Solution_9": "[quote=\"t0rajir0u\"][quote=\"The Will\"]but what there tells you that x is the cube root of 216? :huh:[/quote]\n\nThat's the definition of a cube root. If $x^{3}= 216$, then we say that $x$ is a cube root of $216$, just the same way if $x^{2}= 36$, then we say that $x$ is a square root of $36$.[/quote]\r\n\r\nlol ok \r\n\r\nThanks makes sense.", "Solution_10": "why are people still posting roots of unity when like 50 people have already posted the same exact solution and also it doesnt belong here" } { "Tag": [ "algorithm", "number theory", "greatest common divisor", "calculus", "integration", "induction", "modular arithmetic" ], "Problem": "My friend showed me this problem. What does $ (a_1,...a_n)|c$ mean?\r\n\r\nShow that given $ a_1,...a_n$, the equation $ a_1x_1\\plus{}..a_nx_n\\equal{}c$ has integer solutions iff $ (a_1,...a_n)|c$.\r\nHow can solutions be found using the Euclidean Algorithm and substitution?", "Solution_1": "$ (a_1, \\ldots, a_n)$ is the greatest common divisor of the numbers $ a_1$, $ a_2$, ..., $ a_n$.\r\n\r\nThe vertical bar $ |$ denotes \"divides\". Thus the condition you quote says that $ c$ must be divisible by the greatest common divisor of the $ a_i$ in order for the linear equation to have a solution. \r\n\r\nThe key to this is to see how the Euclidean algorithm can be used to write the gcd of a set of numbers as a linear combination in those numbers. Start with two numbers, say 30 and 105. Apply the Euclidean algorithm to find their gcd. Now trace back and try to use the equations you got along the way to write the gcd $ d$ as a combination $ 30a \\plus{} 105b \\equal{} d$. Generalize.\r\n\r\n- R", "Solution_2": "[quote=\"randomgraph\"]\n\nNow trace back and try to use the equations you got along the way to write the gcd $ d$ as a combination $ 30a \\plus{} 105b \\equal{} d$. Generalize.\n\n- R[/quote]\r\n\r\nWhat equations did we get along the way? How can the gcd of two numbers be a combination of the two numbers? Like 3 is the gcd of 3 and 15... What would be the point of writing $ 3\\equal{}3*1\\plus{}0*15$?", "Solution_3": "[hide=\"If GCD does not divide c then no solution\"]If the GCD of the set of $ a_i$ does not divide $ c$, then note that any integral linear combination of all $ a_i$ will contain a factor that does not divide $ c$, and we have no solution.[/hide]\n\n[hide=\"Simplification of other direction\"]If the GCD of the set $ a_i$ does divide $ c$, then let $ a_i/GCD[\\{ a_i\\}] \\equal{} b_i$, and let $ c/GCD[\\{ a_i \\}] \\equal{} d$. Then, $ GCD[\\{ b_i\\}]\\equal{}1$ and $ a_1 x_1 \\plus{} \\ldots \\plus{} a_n x_n \\equal{} c \\iff b_1 x_1 \\plus{} \\ldots \\plus{} b_n x_n \\equal{} d$. It suffices to prove the latter, in which the GCD is 1.[/hide]\n\n[hide=\"Induction\"]\n[b]Base case (n=2):[/b] \n\nSee [url]http://en.wikipedia.org/wiki/Euclidean_algorithm#Proof[/url]\n\n[b]Inductive Step:[/b] Suppose it works for the case of $ n\\equal{}m$. We want to show there is a solution to the case of $ n\\equal{}m\\plus{}1$, which takes the form:\n\n$ (b_1 x_1 \\plus{} \\ldots \\plus{} b_m x_m) \\plus{} b_{m\\plus{}1} x_{m\\plus{}1} \\equal{} d$\n\nLet $ k$ be the GCD of the first $ m$ values of $ b_i$. Let $ f_i \\equal{} b_i / k, \\; \\forall 1 \\leq i \\leq m$. We then have:\n\n$ (f_1 x_1 \\plus{} \\ldots f_m x_m)k \\plus{} b_{m\\plus{}1} x_{m\\plus{}1} \\equal{} d$\n\nWe know that $ GCD[b_1 , \\ldots b_{m\\plus{}1}]\\equal{}1 \\implies GCD[k,b_{m\\plus{}1}]\\equal{}1$\nWe also know that $ GCD[f_1 \\ldots f_m]\\equal{}1$.\n\nFrom the case of $ n\\equal{}2$, we know there is a solution $ (x_*,x_{m\\plus{}1})\\equal{}(u,v)$ to $ kx_*\\plus{}b_{m\\plus{}1} x_{m\\plus{}1} \\equal{} d$.\nFrom the case of $ n\\equal{}m$, we know there is a solution to $ f_1 x_1 \\plus{} \\ldots f_m x_m\\equal{}u$. Combining the two, we find that there is a solution to $ (b_1 x_1 \\plus{} \\ldots \\plus{} b_m x_m) \\plus{} b_{m\\plus{}1} x_{m\\plus{}1} \\equal{} d$; that is, the case of $ n\\equal{}m\\plus{}1$\n\nThe induction is complete.[/hide]", "Solution_4": "Let's take those one by one...\r\n\r\n[quote=\"aufha\"]\nWhat equations did we get along the way? \n[/quote]\nLet's say you're calculating the gcd of 63 and 13 using the Euclidean Algorithm. It looks something like this:\n\n$ 63 \\equal{} 4*13 \\plus{} 11$\n$ 13 \\equal{} 1*11 \\plus{} 2$\n$ 11 \\equal{} 5*2 \\plus{} 1$\n\nso the gcd is 1, and tracing back we see that\n\n$ 1 \\equal{} 11 \\minus{} 5*2 \\equal{} 11 \\minus{} 5*(13 \\minus{} 1*11) \\equal{} \\minus{} 5*13 \\plus{} 6*11 \\equal{} \\minus{} 5*13 \\plus{} 6*(63 \\minus{} 4*13) \\equal{} \\minus{} 29*13 \\plus{} 6*63$\n\nand here you go: the gcd is a linear combination with integer coefficients of the original numbers.\n\n[quote=\"aufha\"]\nHow can the gcd of two numbers be a combination of the two numbers? \n[/quote]\nI'm not sure I understand what you're asking exactly. How is it possible? Well, we just saw an example, and you mentioned one yourself.\n\n[quote=\"aufha\"]\nLike 3 is the gcd of 3 and 15... What would be the point of writing $ 3 \\equal{} 3*1 \\plus{} 0*15$?\n[/quote]\r\n\r\nI would write this as $ 3 \\equal{} 1*3 \\plus{} 0*15$ just to be consistent, but that is correct. What's the point? Well, the gcd of two integers is:\r\n\r\n1. The largest number which divides them both, and\r\n2. The smallest positive number which can be written as a linear combination (with integer coefficients) of the two of them.\r\n\r\nThis is actually a very important observation, useful in many number-theoretic and ring-theoretic problems and proofs. It is true that if one of the two numbers happens to [b]be[/b] the gcd, this expression doesn't seem very illuminating. That doesn't mean it's not helpful or useful in other cases.\r\n\r\nThe point of my previous post was to try and show the way for you to see how to solve the problem. If that wasn't clear enough, I hope that now it is - if not, please ask again.\r\n\r\n- R", "Solution_5": "This is also known as the [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity#Generalizations]generalized form[/url] of [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity]Bezout's Identity[/url].\r\n\r\nThis not only holds for the integers $ \\mathbb{Z}$, but also for any [url=http://mathworld.wolfram.com/EuclideanRing.html]Euclidean ring[/url].\r\n\r\n[hide=\"An example of how to solve such a system\"]\nTake a look at the equation $ 7x \\plus{} 5y \\equal{} 11$. Remember that if $ (a,b) \\nmid c$ (does not divide c), there would be no solutions. However, since $ (7,5) \\equal{} 1 \\mid 11$, we can proceed. To solve it, first eliminate the $ y$ term by writing the equation in $ \\mod 5$. We have $ 7x \\equal{} 11 \\equiv 1 \\pmod 5$. We add multiples of 5 until we can divide out the $ 7$: $ 7x \\equiv 21 \\pmod 5$, so $ x \\equiv 3 \\pmod 5$. If we let $ x \\equal{} 5k \\plus{} 3$ for $ k \\in \\mathbb{Z}$, we can solve for $ y$. A simple solution would be $ (3, \\minus{} 2)$. \nIn the more general case, solutions to $ ax \\plus{} by \\equal{} c, (a,b) \\equal{} 1$ are of the form $ \\left(k,\\displaystyle\\frac {c \\minus{} ak}{b}\\right)$, with $ k$ being any solution to $ ak \\equiv c \\pmod b$.\n[/hide]" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Find $ f(10)$, if $ f(n) \\equal{} n \\minus{} 10$ when $ n > 100$ and $ f(n) \\equal{} f(f(n \\plus{} 11))$ in all other cases.", "Solution_1": "Please show me the smart way to do this problem? I think mine is just dumb, and probably wrong too.\r\n\r\n[hide]We see that $ f(10) \\equal{} f(f(f(f(f(f(f(f(f(f(109))))))))))$.\n\nSimplyfing the $ f(f(f(109)))$, we see that it becomes $ f(f(99))$, becoming $ f(f(f(110))) \\equal{} f(f(100)) \\equal{} f(f(f(111))) \\equal{} f(f(101)) \\equal{} f(91) \\equal{} f(f(102)) \\equal{} f(92) \\equal{} f(f(103)) \\equal{} \\ldots \\equal{} f(98) \\equal{} f(f(109))$. As we keep going, we see there is a pattern. Every pair of operations on the numbers increases the number by one, until attains a value above a $ 100$, at which time the number dips below $ 100$. Each time this cycle operation gets rid of one left parentheses.\n\nI believe the answer is $ 91$? Please correct me if I'm wrong.[/hide]", "Solution_2": "No my friend, the answer is indeed $ 91$ :) The official solution is also along these lines, although a bit more rigorous.", "Solution_3": "can someone guide me to where i can find about these kinda maths i mean the topic is it polynomials or something else,, a link or a site, or a book. 123456789 which theorem or what you used to solve these with?.please", "Solution_4": "@binomial_4eva, I didn't use a \"theorem.\" I used logic.", "Solution_5": "binomial_4eva, if you notice that $ f(f(f(109))) \\equal{} f(f(99)) \\equal{} f(f(f(110))) \\equal{} f(f(100)) \\equal{} f(f(f(111))) \\equal{} f(f(101)) \\equal{} f(91) \\equal{} f(f(102)) \\equal{} f(92) \\equal{} f(f(103)) \\equal{} f(93) \\equal{} f(f(104)) \\equal{} f(94) \\equal{} f(f(105)) \\equal{} f(95) \\equal{} f(f(106)) \\equal{} f(96) \\equal{} f(f(107)) \\equal{} f(f(97)) \\equal{} f(f(108)) \\equal{} f(98) \\equal{} f(f(109))$, (as 123456789 pointed out) then we basically have $ f(f(109)) \\equal{} f(f(f(109)))$ and repeatedly applying this to $ f(10) \\equal{} f(f(f(f(f(f(f(f(f(f(109))))))))))$ we have $ f(10) \\equal{} f(f(109))$. Similar to what we have above, $ f(f(109))) \\equal{} f(99) \\equal{} f(f(110)) \\equal{} f(100) \\equal{} f(f(111)) \\equal{} f(101) \\equal{} 91$, so our answer is indeed $ 91$.\r\n\r\nNo theorem needed, just lots of algebra." } { "Tag": [ "geometry", "trigonometry", "rectangle", "number theory", "relatively prime", "trig identities", "Law of Cosines" ], "Problem": "I just drew pentagon $LINGO$, where $\\overline{LI} \\cong \\overline{IN} \\cong \\overline{NG} \\cong \\overline{GO} \\cong \\overline{OL}$. Also, $\\overline{LI}\\parallel\\overline{NG}$ and $\\overline{LI}\\perp\\overline{IN}$. If $LN=\\sqrt2$:\r\n\r\nA) Compute the product of all the possible areas of pentagon $LINGO$.\r\nB) Assume that $LINGO$ is a convex polygon. If I drew $\\overline{IO}$, its length could be expressed in the form $\\frac{\\sqrt{a+b\\sqrt{c}}}d$. Compute the ordered quadruple $(a,b,c,d)$ if $a,b,c,d\\in\\mathbb{N}$, $a$ and $b$ are relatively prime, and $c$ is prime.", "Solution_1": "[hide=\"A\"]So basically you have square $LING$ and equilateral triangle $LGO$. $LN=\\sqrt{2} \\implies LI=IN=NG=GO=OL=LG=1$. Therefore the area is $1^2+\\frac{1^2 \\sqrt{3}}{4}= \\boxed{1+\\frac{\\sqrt{3}}{4}}$.[/hide]\n\n[hide=\"B\"] By Law of cosines $IO^2 = 2-2\\cos(\\frac{5\\pi}{6})=2+\\sqrt{3}$\n$IO=\\sqrt{2+\\sqrt{3}} \\therefore (a,b,c,d) = \\boxed{(2,1,3,1)}$[/hide]", "Solution_2": "Cincodemayo:\r\n[hide]\nPart A wants the product of all possible areas... There is one more :)\n[/hide]", "Solution_3": "Hmm... Law of Cosines... didn't think of that. I guess that works too. What I had in mind was this:\r\n[hide]Extend $\\overline{LI}$ to point $X$ so that $\\overline{LX}\\perp\\overline{OX}$. $LX=1+\\frac{\\sqrt3}2$ and $OX=\\frac12$, so apply the Pythagorean Theorem to get $\\sqrt{2+\\sqrt3}$. So the answer would be $\\boxed{2,1,3,1}$.[/hide]\n\nAnd A is wrong:\n[hide]Point $O$ could be also positioned inside square $LING$, making the area [hide]$1-\\frac{\\sqrt{3}}4$. So the answer is [hide]$\\frac{13}{16}$.[/hide][/hide][/hide]\n\nI'll give out more parts to do. If $LINGO$ is convex:\nC) Extend $\\overline{GO}$ to point $A$ so that $L$, $I$, and $A$ are collinear. Draw point $B$ so that $N$, $G$, and $B$ are collinear. Now draw $C$, the midpoint of $\\overline{AB}$. If the length of $LC$ can be expressed as $\\frac{a+b\\sqrt{c}}d$, compute the ordered quadruple $(a,b,c,d)$ if $a,b,c,d\\in\\mathbb{N}$, $a$ and $b$ share no common factor, and $c$ is prime.\nD) Prove that $\\bigtriangleup LGO \\cong \\bigtriangleup ABO$.\nE) Name all line segments with a length of $\\sqrt3$.\nF) If you drew lines connecting $A$ to $I$ and $B$ to $N$, you'd form a quadrilateral, $ABNI$. What is this quadrilateral's area?\n\nThese build upon each other.\n\n[hide=\"Hint for C\"]Pythagorean Theorem[/hide]\n[hide=\"Second hint for C\"]$\\bigtriangleup LAC$[/hide]\n[hide=\"Another method for C\"]Law of Cosines[/hide]\n[hide=\"Hint for E\"]There are two.[/hide]", "Solution_4": "Assuming $\\overline{LO}$ is extended to $B$...\r\n[hide=\"C\"]\nIf you draw it out, you have that $\\triangle OAB$ is an equilateral triangle w/sides 1. $OC$ would then be $\\frac{\\sqrt3}{2}$. Since $m\\angle LOC=150$. Then by law of cosines, $LC^2=1+\\frac34-\\frac{\\sqrt3}{2}\\cos150=\\frac{13}{4}$ so $LC=\\frac{\\sqrt{13}}{2}$ so $(a,b,c,d)=(0,1,13,2)$?\n[/hide]", "Solution_5": "An error by saying $\\frac{\\sqrt3}2$ in your calculations, but you got the right answer in the end.\r\n\r\nAgain, assume that $LINGO$ is convex.\r\nG) The length of $\\overline{NC}$ can be expressed in the format $\\frac{\\sqrt{a+b\\sqrt{c}}}d$, where $a,b,c,d\\in\\mathbb{N}$, $a$ and $b$ are relatively prime, and $c$ is prime. Compute the ordered quadruple $(a,b,c,d)$.", "Solution_6": "Hints for D:\r\n[hide]$\\overline{LI}\\parallel\\overline{NG}$. $L$, $I$, and $A$ are collinear, and $B$, $G$, and $N$ are collinear. What can you conclude about $\\overline{ILA}$ and $\\overline{NGB}$?[/hide]\n[hide]Alternate Interior Angles[/hide]\n[hide]$\\overline{LG}\\parallel\\overline{IN}$[/hide]", "Solution_7": "[hide=\"D\"]\nASA: \nboth triangles are equilateral with side length 1: GBA and LAB are right. GAB and LBA are $60^\\circ$ b/c LAG and LBG are $30^\\circ$. $AB \\cong IN =1$, and the same with $\\triangle LOG$. I don't feel like typing out the formalities but you get the gist of it. [/hide]\n\n[hide=\"E\"] $\\boxed{AL \\text{ and }BG}$.$\\angle ALB$ and $\\angle BGA$ are $30^\\circ$ and $AB$ is $1$ so $LB=AG=\\sqrt{3}$.[/hide]\n\n[hide=\"F\"] $AINB$ is a rectangle (right angles), and $BN=BG-NG=\\sqrt{3}-1$. So the area is $1(\\sqrt{3}-1)=\\boxed{\\sqrt{3}-1}$[/hide]\n\n[hide=\"G\"] $NC^2 = NB^2 + BC^2 \\implies NC=\\sqrt{(\\sqrt{3}-1)^2+(\\frac{1}{2})^2}=\\frac{\\sqrt{17-8\\sqrt{3}}}{2}$. So $(a,b,c,d)=\\boxed{17,-8,3,4}$[/hide]", "Solution_8": "E, F, and G are incorrect. D is good." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "homothety", "ratio", "power of a point", "radical axis" ], "Problem": "Given a triangle $ ABC$. The angle bisectors of the angles $ ABC$ and $ BCA$ intersect the sides $ CA$ and $ AB$ at the points $ B_1$ and $ C_1$, and intersect each other at the point $ I$. The line $ B_1C_1$ intersects the circumcircle of triangle $ ABC$ at the points $ M$ and $ N$. Prove that the circumradius of triangle $ MIN$ is twice as long as the circumradius of triangle $ ABC$.", "Solution_1": "[color=darkblue][b]Remark[/b] ([u]Bulgaria, 1997[/u]). If the side $AB$ separates the points $M$ and $C$ then $\\frac{1}{BM}=\\frac{1}{AM}+\\frac{1}{CM}$ and $\\frac{1}{CN}{=\\frac{1}{AN}+\\frac{1}{BN}\\ .}$[/color]\r\n\r\n[color=darkred][b]Generalization[/b] ([u]own[/u]). Let $ABC$ be a triangle inscribed in the circle $w$ and for the points $X\\in (AC)\\ ,$ $Y\\in (AB)$ define the points $\\{M,N\\}=XY\\cap w\\ .$ Then for any point $D\\in \\{M,N\\}$ there is the following relation : \\[ \\boxed {\\left|b\\cdot \\frac{YB}{YA}\\cdot \\frac{1}{DB}-c\\cdot \\frac{XC}{XA}\\cdot \\frac{1}{DC}\\right|=a\\cdot \\frac{1}{AD}}\\ . \\] [b]Remark.[/b] If $I\\in BX\\cap CY$ then we obtain the above problem from Bulgaria, 1997.[/color]", "Solution_2": "Man.. You have to wonder how these people manage to solve the problems in contest :). I have a pretty short and simple proof, but it took me a while to see what was going on, and I used dynamic geometry software.\r\n\r\nLet $I_b,I_c$ denote the $B$ and, respectively, $C$-excenters of $ABC$. If we prove that the circle $(MIN)$ passes through these two points we're done, because then $(MIN)$ would be the circumcircle of $II_bI_c$, while the circumcircle of $ABC$ is its nine-point circle. $ABC$ is the orthic triangle of $II_bI_c$, so $B_1C_1$ is the orthic axis of $II_bI_c$, which coincides with the radical axis of its circumcircle and nine-point circle. These two circles do intersect, because $\\angle I_bII_c>\\frac\\pi 2$, so the two points of intersection must be precisely the points of intersection between $B_1C_1$ and $(ABC)$, i.e. $M$ and $N$.", "Solution_3": "Here is my solution to the problem.\r\n\r\nI base my proof on the following lemma (I predicted this lemma, and fortunately it is true)\r\n\r\n[i]Lemma[/i] Suppose $(C,h)$ and $(O, r)$ are two given distinct circles. A point $P$ moves in the circle $(C)$ such that the tangent at $P$ intersects $(O)$ at two points $A$ and $B$. Then the circumcenter $J$ of $ABC$ moves on a circle with center $O$.\r\n\r\n[hide=\"Proof\"] Up till now, I just have a brute force proof. Maybe I 'm not good enough today to find a synthetic one. But anyway, let $OC = m, AB =c, CA=b, BC = a, S_{ABC} = S, R_{ABC} = R$ and $M$ be the midpoint of $AB$. Then by Leibnitz theorem, we obtain\n$a^2(b^2+c^2-a^2)OA^2 + b^2(c^2+a^2-b^2)OB^2 + c^2(a^2+b^2-c^2)OC^2$ \n\n$= \\sum (2b^2c^2 - a^4) KJ^2+a^2(b^2+c^2-a^2)JA^2 + b^2(c^2+a^2-b^2)JB^2 + c^2(a^2+b^2-c^2)JC^2$ \n\nThus $OJ^2 = r^2 - R^2 + \\frac{(m^2-r^2)c^2(a^2+b^2-c^2)}{16S^2}$\n\n$= r^2 - R^2 + \\frac{(m^2-r^2)(2h^2+PA^2+PB^2-AB^2)}{4h^2}$ \n\n$= r^2 - R^2 + \\frac{PA^2.PB^2}{4h^2} + \\frac{(m^2-r^2)(2h^2+PA^2+PB^2-AB^2)}{4h^2}$\n\n$= \\frac{2r^2+2m^2-h^2}{4} + \\frac{2(m^2-r^2)PA.PB - PA^2.PB^2 - h^2(PA^2+PB^2)}{4h^2}$\n\n$= \\frac{2r^2+2m^2-h^2}{4} + \\frac{2(m^2-r^2)(PO^2- r^2) - (PO^2-r^2)^2 - h^2(2PO^2 +2r^2 - 4MO^2)}{4h^2}$\n\n$= u + \\frac{2m^2PO^2 - PO^4 - 2h^2PO^2 + 4h^2MO^2}{4h^2}$, here $u$ is a constant in term of $m, h, r$.\n\nIt is now sufficient to show that the numerator is a constant in term of $m, h, r$. We have $PO^2 = MO^2 + m^2 - (h-MO)^2 = m^2-h^2 + 2hMO$. Replacing this into the numerator we have \n\n$4h^2MO^2 + 2m^2(m^2+2hMO - h^2) - (m^2+2hMO - h^2)^2 - 2h^2(m^2+2hMO - h^2)$ \n\n$= (m^2-h^2)^2$. \n\nHence the lemma is proved. [/hide]\r\n\r\nConsider the $A$-excircle is $(C)$. You may notice that if we let the chord $AB$ in the lemma be $BC, AC, AB$ respectively, then the circumcenter $J$ will be the midpoints of $I_aI, I_aI_c, I_aI_b$ which lie on a circle center $O$ (however, we know that this circle is in fact the circumcircle $(O)$). Let the chord $AB$ in the lemma be the two chords of circumcircle $(O)$ that are tangent with $(I_a)$ at the two circle's intersections. Then without difficulty, you can see that the midpoints of $I_aM$ and $I_aN$ are the circumcenter $J$ in this time which lie on the circumcircle $(O)$ too. Hence, the homothety center $I_a$ with ratio $2$ maps these midpoints to $I, I_b, I_c, M, N$. Therefore they both lie on a circle the circumradius of which is twice as long as the circumradius of $ABC$.", "Solution_4": "@ Virgil: I was aware of the Bulgarian problem; it was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=3807 . The generalization can basicly be proven in the same way.\r\n\r\n@ Grobber: It is not that hard to come up with the idea; actually, what can a circle passing through I and having its radius equal to twice the circumradius of triangle ABC be? The circumcircle of triangle $II_bI_c$ is the only such circle known, so one tries to show that it coincides with the circumcircle of triangle MIN, and this is not particularly hard.\r\n\r\n@ Treegoner: Your Lemma was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19000]a Tournament of Towns 2004 problem[/url] ;) .\r\n\r\n Darij", "Solution_5": "Thank you, Darij! Your proof to the lemma is wonderful. :)", "Solution_6": "wait i dont understand. as i know I_aI_bI_c has ABC as its nine point circle. why is ABC the nine point circle of II_bI_c?", "Solution_7": "[quote=\"marko avila\"]wait i dont understand. as i know I_aI_bI_c has ABC as its nine point circle. why is ABC the nine point circle of II_bI_c?[/quote]\r\n\r\nBecause $A$ is the foot of altitude from $I$ to $I_{b}I_{c}$\r\n$C$ is the foot of altitude from $I_{b}$ to $II_{c}$\r\n$B$ is the foot of altitude from $I_{c}$ to $II_{b}$ \r\n\r\nThe conclusion follows :)", "Solution_8": "[quote=grobber]Man.. You have to wonder how these people manage to solve the problems in contest :). I have a pretty short and simple proof, but it took me a while to see what was going on, and I used dynamic geometry software.\n\nLet $I_b,I_c$ denote the $B$ and, respectively, $C$-excenters of $ABC$. If we prove that the circle $(MIN)$ passes through these two points we're done, because then $(MIN)$ would be the circumcircle of $II_bI_c$, while the circumcircle of $ABC$ is its nine-point circle. $ABC$ is the orthic triangle of $II_bI_c$, so $B_1C_1$ is the orthic axis of $II_bI_c$, which coincides with the radical axis of its circumcircle and nine-point circle. These two circles do intersect, because $\\angle I_bII_c>\\frac\\pi 2$, so the two points of intersection must be precisely the points of intersection between $B_1C_1$ and $(ABC)$, i.e. $M$ and $N$.[/quote]\n\nSweet!! \n\n", "Solution_9": "Call the circumcircle $\\omega$ and call the circle $I I_b I_c$ as $\\Gamma$. it\u2019s quite well known that $AIBI_c$ and $AICI_b$ are cyclic. Call them $\\omega_1$ and $\\omega_2$. Now we get that $C_1$ is the radical center of $\\omega\\Gamma\\omega_1$ and $B_1$ is the radical center of $\\omega\\Gamma\\omega_2$ so $B_1,C_1$ are on the radical axis of $\\omega$ and $\\Gamma$. Thus $M,N$ are the intersections of $\\omega$ and $\\Gamma$ so $IMNI_b I_c$ is cyclic. Now because of the homothety taking $\\omega$ to $\\Gamma$ we conclude the problem.", "Solution_10": "Let $P,Q$ be the midpoints of minor arcs $AC$ and $AB$ and let $X,Y$ be the $B,C$ excenters. By PoP, we have that $(NC_1)(MC_1) = (AC_1)(C_1B) = (IC_1)(C_1Y)$ so $Y$ lies on $(MIN)$ and similarly so does $X$. Since $P,Q$ are midpoints of $IX, IY$, $(MIN)$ has double the radius of $(PIQ)$. \n\nBut since $QA = QI$ and $PI =PA$, we have $\\triangle PAQ$ and $\\triangle PIQ$ are congruent so the circumradii are equal too, so $(MIN)$ has double the radius of $(ABC)$, as desired. ", "Solution_11": "Denote by $I_A,I_B,I_C$ respective excenters of $ABC.$ Since $AICI_B$ is cyclic, point $B_1$ (and analogously $C_1$) lies on radical axis of $\\odot (ABC),\\odot (II_BI_C),$ therefore $M,N\\in \\odot (II_BI_C).$ We are done, because in triangle $I_AI_BI_C$\n\n$\\bullet \\text{ } I$ is the orthocenter;\n$\\bullet \\text{ } \\odot (ABC)$ is the nine-point circle.", "Solution_12": "Well, I have pretty complicated proof.. Also special thanks to L567!\n[asy]/* File unicodetex not found. */\n\n /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */\nimport graph; size(14.2cm); \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 = -37.7, xmax = 12.5, ymin = -8.9, ymax = 21.5; /* image dimensions */\npen wrwrwr = rgb(0.4,0.4,0.4); \n\ndraw((-7.2,6)--(-7.7,-2.2)--(2.5,-2.2)--cycle, linewidth(0.4) + wrwrwr); \n /* draw figures */\ndraw((-7.2,6)--(-7.7,-2.2), linewidth(0.4) + wrwrwr); \ndraw((-7.7,-2.2)--(2.5,-2.2), linewidth(0.4) + wrwrwr); \ndraw((2.5,-2.2)--(-7.2,6), linewidth(0.4) + wrwrwr); \ndraw(circle((-2.5,1.7), 6.4), linewidth(0.4) + wrwrwr); \ndraw(circle((-4.8,13.3), 12.7), linewidth(0.4) + wrwrwr); \ndraw((-8.9,1.2)--(-4.8,0.5), linewidth(0.4) + wrwrwr); \ndraw((-4.8,0.5)--(3.5,3.6), linewidth(0.4) + wrwrwr); \ndraw((-8.9,2)--(1.6,6.5), linewidth(0.4) + wrwrwr); \ndraw((-13,3.5)--(2.5,-2.2), linewidth(0.4) + wrwrwr); \ndraw((-7.7,-2.2)--(1.6,6.5), linewidth(0.4) + wrwrwr); \ndraw((1.6,6.5)--(8,12.5), linewidth(0.4) + wrwrwr); \ndraw((-8.9,1.2)--(3.5,3.6), linewidth(0.4) + wrwrwr); \ndraw((-13,3.5)--(8,12.5), linewidth(0.4) + wrwrwr); \ndraw((-2.6,-4.7)--(-7.2,6), linewidth(0.4) + wrwrwr); \ndraw((-26.2,-2.1)--(-7.7,-2.2), linewidth(0.4) + wrwrwr); \ndraw((-26.2,-2.1)--(-13,3.5), linewidth(0.4) + wrwrwr); \ndraw((-26.2,-2.1)--(-8.9,1.2), linewidth(0.4) + wrwrwr); \n /* dots and labels */\ndot((-7.2,6),linewidth(5pt) + dotstyle); \nlabel(\"$A$\", (-7,6.4), NE * labelscalefactor); \ndot((-7.7,-2.2),linewidth(5pt) + dotstyle); \nlabel(\"$B$\", (-7.5,-1.8), NE * labelscalefactor); \ndot((2.5,-2.2),linewidth(5pt) + dotstyle); \nlabel(\"$C$\", (2.7,-1.8), NE * labelscalefactor); \ndot((-2.8,2.4),linewidth(4pt) + dotstyle); \nlabel(\"$B_1$\", (-2.7,2.6), NE * labelscalefactor); \ndot((-7.4,1.5),linewidth(4pt) + dotstyle); \nlabel(\"$C_1$\", (-7.3,1.8), NE * labelscalefactor); \ndot((-4.8,0.5),linewidth(4pt) + dotstyle); \nlabel(\"$I$\", (-4.7,0.8), NE * labelscalefactor); \ndot((-8.9,1.2),linewidth(4pt) + dotstyle); \nlabel(\"$M$\", (-8.8,1.5), NE * labelscalefactor); \ndot((3.5,3.6),linewidth(4pt) + dotstyle); \nlabel(\"$N$\", (3.6,3.9), NE * labelscalefactor); \ndot((-8.9,2),linewidth(4pt) + dotstyle); \nlabel(\"$M_C$\", (-8.8,2.3), NE * labelscalefactor); \ndot((1.6,6.5),linewidth(4pt) + dotstyle); \nlabel(\"$M_B$\", (1.7,6.8), NE * labelscalefactor); \ndot((-13,3.5),linewidth(4pt) + dotstyle); \nlabel(\"$I'$\", (-12.9,3.8), NE * labelscalefactor); \ndot((8,12.5),linewidth(4pt) + dotstyle); \nlabel(\"$I'_{1}$\", (8.1,12.8), NE * labelscalefactor); \ndot((-2.6,-4.7),linewidth(4pt) + dotstyle); \nlabel(\"$K$\", (-2.4,-4.4), NE * labelscalefactor); \ndot((-26.2,-2.1),linewidth(4pt) + dotstyle); \nlabel(\"$T$\", (-26,-1.8), NE * labelscalefactor); \ndot((-3.7,-2.2),linewidth(4pt) + dotstyle); \nlabel(\"$A_1$\", (-3.5,-1.9), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */[/asy]\nIntroduce the following points\n\n $\\bullet$ $I_C,I_B$ as the $C,B$ excenters\n $\\bullet$ $A_1$ as $AI\\cap BC$\n $\\bullet$ $M_C, M_B$ as midpoint of arc $AB,AC$ respectively. \n\n[color=#6FA8DC][b]Claim[/b][/color]: The circumradius of $II_CI_B$ is twice the circumradius of $ABC$.\n\n[b]Proof[/b]: By incenter-excenter lemma, note that $M_C$ is the midpoint of $II_C$ and $M_B$ is the midpoint of $II_B$.\nSo the circumradius of $II_CI_B$ is twice the circumradius of $IM_CM_B$. \n\nSo we will show that the circumradius of $IM_CM_B=$ the circumradius of $ABC$. For that, note that by sine law,\n$$ \\frac{\\sin \\angle IM_CM_B}{ IM_B}=\\frac{\\sin \\angle IM_CM_B}{M_BC}=\\frac{\\sin\\angle M_BBC}{M_BC}.$$\n\n[color=#6FA8DC][b]Claim[/b][/color]: $ I_CI_B,C_1B_1,BC$ are concurrent\n\n[b]Proof[/b]: Let $C_1B_1\\cap BC=T.$ Note that $(T,A_1;B,C)=-1.$ But note that $\\angle I_CAA_1=90, \\angle BAA_1=\\angle CAA_1$. So $T-I_C-A-I_B$. \n\nNow, it is well known that $I_CI_BCB$ is cyclic. So by POP, we have $$TM\\cdot TN=TB\\cdot TC=TI_C\\cdot TI_B\\implies II_CI_BMN\\text{ is cyclic }.$$\nAnd we are done as The circumradius of $II_CI_BMN$ is twice the circumradius of $ABC$.\n\n", "Solution_13": "Quite nice problem.\n\nLet $I_B$ and $I_C$ be the $B$ and $C$-excenters respectively. Notice that $B_1$ is the radical center of $(IACI_B)$, $(ABC)$, and $(I_BII_C)$, and similar for $C_1$, which implies $M, N$ both lie on $(I_BII_C)$. On the other hand by incenter-excenter lemma the radius of $(I_BII_C)$ is equal to the radius of $(I_CI_AI_B)$, which is twice the circumradius of $(ABC)$, as needed.", "Solution_14": "Absolutely brutal\n[hide = solution]\nLet $I_A$ denote the $A$-excentre of $ABC$. Define $I_B, I_C$ analogously. Let line $BB_1$ meet $\\odot(MIN)$ at point $X$. Observe that $B_1I \\cdot B_1X = B_1M \\cdot B_1N = B_1A \\cdot B_1C = B_1A \\cdot B_1I_B$ where the last equality follows due to the existence of $\\odot(AIBI_B)$. It follows that $X = I_B$, in particular $I_B \\in \\odot(MIN)$. Similarily one deduces $I_C \\in \\odot(MIN)$. As $I$ is the orthocentre of $\\triangle I_AI_BI_C$ it follows that $\\odot(MIN)$ and $\\odot(I_AI_BI_C)$ have equal radius. Finally, remark that $\\odot(ABC)$ is the nine point circle of $\\triangle I_AI_BI_C$ and so it has half the radius as $\\odot(I_AI_BI_C)$, as required. $\\square$\n[/hide]" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "$A$ is n by n, invertible matrix on field $K$\r\n\r\n$AB$=$BA$ for every n by n invertible matrix $B$ in $K$\r\n\r\nFind $A$", "Solution_1": "I would say if and only if $A = \\lambda I_n$ ?\r\n\r\nTaking $B = I_n + \\Delta_i^j$ with $\\Delta_i^j$ is the matrix with all null entries except for $(i,j)=1$\r\n$B$ is invertible, $B^{-1}$ is easy to compute, it is $I_n - \\Delta_i^j$\r\n$AB = BA$ for all $B$ gives that $A$ is diagonal." } { "Tag": [], "Problem": "ok as MOST of you know, my first problem i mean I'M BACK!!!!!!!!!!!!!!!!! (the title of the problem) will not be so ... difficult :twisted: ok lets start:\n\n(mods...don't delete this please!)\n\nthere's a person named... bob...he wants to destroy the world. there are 7 continents in the world(everybody knows that) and lets just say that there are 4 on one side and 3 on the other. for one continent, it took him 23 nuclear bombs and it took 14 hours. for another it took 35 nuclear bombs and that took 21 hours. for another it took 53 nuclear bombs and that one took 1 day and 8 hours and so on (look for a pattern). that is the side with 4 continents so you need to figure out how many nuclear bombs and how many hours it took to destroy the last side. oh yeah the ones in italics are the questions. also figure out how many nuclear bombs and how many hours in total it took to destroy the first half. now for the other side. sadly, it was easier to destroy it so...it took him 76 nuclear bombs to destroy one continent and 59 hours. then another continent took 61 nuclear bombs and it took 47 hrs. (look for a different pattern) find the number of bombs and hrs it took to destroy the last continent (round to the nearest one). second to last but not least, find the number of bombs and hrs that destroyed the side with 3 continents. last but not least, add up all of the total # of bombs and hrs it took to destroy earth. and PLEASE DON'T DELETE THIS, IT TOOK ME 20 MINS TO CREATE! i triple checked my work so i know my answer is right. i'll post my answer on a different topic but don't look at it unless you know your answer. [hide]1st half 80 bombs, 48.5 hrs and the second half 40 bombs 37.4 hrs 1st half 191 bombs in all, 115.5 hrs 2nd half 177 bombs and 143.4 hrs[/hide]", "Solution_1": "oh yeah i forgot to tell you guys... [size=200][u][i][b]don't look at the spoilers!!!!!!!!!!![/b][/i][/u][/size]", "Solution_2": "[quote=\"gameworld7\"][i][size=100]figure out how many nuclear bombs and how many hours it took to destroy the last side.[/size][/i][/quote]\r\n\r\nI think that should be continent instead of side for the last word.", "Solution_3": "i never even knew there were spoilers! THANK YOU for telling me!\r\n\r\nHOW could ANYONE not tell that you are the same, geeky, author of I'm Back! (j/k)", "Solution_4": "Don't post the answer in another topic. That just clogs up the forum. \r\n\r\nAlso, am I the only one who sees no pattern whatsoever there? How can you see a pattern with only 2 terms? It could be a million things.\r\n\r\nExample- Find the next term in this pattern.\r\n\r\n3,8\r\n\r\nIt could be 13,24,3/8,63, or a ton of stuff.", "Solution_5": "[quote=\"yif man12\"]Don't post the answer in another topic. That just clogs up the forum. \n\nAlso, am I the only one who sees no pattern whatsoever there? How can you see a pattern with only 2 terms? It could be a million things.\n\nExample- Find the next term in this pattern.\n\n3,8\n\nIt could be 13,24,3/8,63, or a ton of stuff.[/quote]\r\n\r\nSince you asked, no, you're not.", "Solution_6": "do i really really really really really need to make myself clear ok i'll say it 1 more time. no 5 more times. no 7 more times is good.\r\n[size=200][u][i][b]don't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!![/b][/i][/u][/size] :evil: :evil: :evil: :evil: :evil: :evil:", "Solution_7": "replying to the person that said that 2 examples is not enough, i agree but lets just say that the pattern is the same example:\r\n2,5,? well yeah it could be *5/2 or +3 but lets also say in my problem it is dividing and another operation", "Solution_8": "That's definitely not specific enough. It could be a million different things. That would make the question totally invalid. I'll give you another example.\r\n\r\n3,8\r\n\r\nThe pattern is dividing and then another operation. Is it divide by 3 and times 8? Or divide by 6 and plus 7.5? There's an infinite number of possible combinations.\r\n\r\nAlso, stop being so paranoid about people looking at your spoiler. It's getting to be really pathetic.", "Solution_9": "[quote=\"gameworld7\"]do i really really really really really need to make myself clear ok i'll say it 1 more time. no 5 more times. no 7 more times is good.\n[size=200][u][i][b]don't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!!\ndon't look at the spoilers!!!!!!!!!!![/b][/i][/u][/size] :evil: :evil: :evil: :evil: :evil: :evil:[/quote]\r\n\r\nHoly crackers. We have cracker alert. And no I don't see the pattern anyways. These problems are so vague. Since you asked a \"nice\" question I'll give you a \"nice\" answer. Read the Bible. I think it predicts the Armageddon in there, though I may be wrong.", "Solution_10": "Hmm, I'm surprised no one has inserted a \"certain president's\" name in this topic yet.", "Solution_11": "[hide]I looked at the spoiler[/hide] Don't look at the spoiler! Don't look at the spoiler!!!", "Solution_12": "A+MATH wrote:[hide]I looked at the spoiler[/hide] Don't look at the spoiler! Don't look at the spoiler!!!\n\n\n\nOMG! You DID NOT just write that!", "Solution_13": "that gameworld7 or whoever you are,\r\n\r\nplease stop posting your tedious problems. They don't make sense and are poorly worded. and everyone will just get annoyed..... and NO ONE cares you are back.....", "Solution_14": "That \"certain president\" - was that Carter or Clinton.\r\n\r\nBesides, it is silly to put the answer right afterwards and expect no one to look. I looked first thing :P , as soon as I was told not to.\r\n\r\nBilly", "Solution_15": "yif man12 wrote:A+MATH wrote:[hide]I looked at the spoiler[/hide] Don't look at the spoiler! Don't look at the spoiler!!!\n\nOMG! You DID NOT just write that!\n\n\n\n[hide]LOL[/hide]DON'T LOOK AT THE SPOILER!!!!!!!!!!!", "Solution_16": "[u][i][b][size=200] :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: :twisted: [/size][/b][/i][/u] don't look at the spoilers! and you make me so sad :( :( :(", "Solution_17": "Why don't you stop whining about the spoilers and address the incompetency of your problem? If you didn't want people to know the answers, DONT POST THEM!", "Solution_18": "beta wrote:that gameworld7 or whoever you are,\n\nplease stop posting your tedious problems. They don't make sense and are poorly worded. and everyone will just get annoyed..... and NO ONE cares you are back.....\n\nTHANK YOU!\n\nyif wrote:Why don't you stop whining about the spoilers and address the incompetency of your problem? If you didn't want people to know the answers, DONT POST THEM!\n\n\n\nthank you again!\n\n[hide](not that im any better)[/hide]", "Solution_19": "[hide]you will die in seven days... (wonder why no one has done this yet. oh well.)[/hide]OMGOMGOMG DON'T LOOK AT THE SPOILER OR ELSE!!!", "Solution_20": "[quote=\"gameworld7\"][b]PLEASE DON'T DELETE THIS, IT TOOK ME 20 MINS TO CREATE![/b][/quote]\r\n\r\nThen I kindly advise you to spend your valuable time working on something more worthwhile (and coherent).", "Solution_21": "yif man12 wrote:[hide]you will die in seven days... (wonder why no one has done this yet. oh well.)[/hide]OMGOMGOMG DON'T LOOK AT THE SPOILER OR ELSE!!!\n\n\n\nuh.... WHAT was THAT?... ur as disturbing as ken...", "Solution_22": "but the spoilers.....ITS SO TEMPTING!!! i cant live without the spoilers!! dont tell me not 2 look at them i NEED them!!! WOE IS ME!!! um...i didnt read the question...but i saw something about nuclear bombs.....and i just read the replies. \r\n\r\n[quote]uh.... WHAT was THAT?... ur as disturbing as ken...[/quote]\r\n\r\nwell gee, SUNNYRAY. maybe that was SARCASM or AN IMITATION of some randomly disturbing person that is randomly called gameworld7 who does not exist because the topic here is just our imaginations....ooooooh aahhhhh.", "Solution_23": "[quote=\"Athena\"]but the spoilers.....ITS SO TEMPTING!!! i cant live without the spoilers!! dont tell me not 2 look at them i NEED them!!! WOE IS ME!!! um...i didnt read the question...but i saw something about nuclear bombs.....and i just read the replies. \n\n[quote]uh.... WHAT was THAT?... ur as disturbing as ken...[/quote]\n\nwell gee, SUNNYRAY. maybe that was SARCASM or AN IMITATION of some randomly disturbing person that is randomly called gameworld7 who does not exist because the topic here is just our imaginations....ooooooh aahhhhh.[/quote]\r\n\r\nGee, how'd you know?", "Solution_24": "I'm starting to feel sorry for theone, and anyone else who goes to Miller.", "Solution_25": "I think we went a little out of control there. I don't think we should be posting things as blunt as some of the previous comments on here. But hey I'm not a moderater so there is nothing I can do. ;)", "Solution_26": "[quote] I think we went a little out of control there. I don't think we should be posting things as blunt as some of the previous comments on here.[/quote]\r\n\r\nYeah, I agree, that was kind of mean to world7. I'm sure that he/she tried to make a good problem and just worded it a little wrong. I think the problem is funny, I mean who would have that many nuclear bombs? :D :D :D", "Solution_27": "oh yeah i think i'm a he and i think that i don't like to be called \"world7\" and theone doesn't go to miller anymore and lets say that the fed gov. gave all the nukes for hmmm... 5 trillion bucks?", "Solution_28": "JUST edit your post and delete the spoilers lol", "Solution_29": "Ken- helpful hint- double check ur problem with a partner before u post it, quantity= spam, please also stop spamming- post more thought out problems\r\n\r\nOther people- please stop spamming this topic!\r\n\r\nNoTe: I am also back... start looking for problems...", "Solution_30": "first i'm too lazy to edit my post and second, umm, is there a second? um i DON'T want to double check or whatever" } { "Tag": [], "Problem": "so i decided to start a book club to discuss books. ive been reading the heir books by cinda williams chima from where my battle of the roses game came from. they are pretty good books with all the suspense and turns of events", "Solution_1": "girls in my school say Twilight is good.", "Solution_2": "First rule of Book Club: Don't talk about Book Club", "Solution_3": "[quote=\"mathguy999\"]girls in my school say Twilight is good.[/quote]\r\n\r\nhaha it is really good. but not just girls say that i know a couple of guys that like that series =]", "Solution_4": "Anyone ever read Dr. Seuss?", "Solution_5": "yes but.. wait just how old are you?", "Solution_6": "[quote=\"guineapiggies\"][quote=\"mathguy999\"]girls in my school say Twilight is good.[/quote]\n\nhaha it is really good. but not just girls say that i know a couple of guys that like that series =][/quote]\r\nThat doesn't sound right...\r\nAt my school just about every girl loves the series. No boys are talking about it.", "Solution_7": "[quote=\"evilhamster\"][quote=\"guineapiggies\"][quote=\"mathguy999\"]girls in my school say Twilight is good.[/quote]\n\nhaha it is really good. but not just girls say that i know a couple of guys that like that series =][/quote]\nThat doesn't sound right...\nAt my school just about every girl loves the series. No boys are talking about it.[/quote]\r\n\r\n... me too... (twilight is about girls and vampires falling in love)\r\nWe have a twilight lover who might go to watch the twilight movie on the first day.", "Solution_8": "guys like me in my school don't talk about twilight, but some of the you-know-what guys do. it sounds kind of a weird book to me, but they say its good.", "Solution_9": "In my school, the girls are crazy over twilight( I despise twilight).", "Solution_10": "So do I. At a recent math competition, one of the rounds' problems were all about Twilight. *groan*", "Solution_11": "hm i just finished reading The Death Collector...", "Solution_12": "i luv twilight!! the movie is coming out this friday!!! and yes. i do know about 2 or 3 guys that are going to see it. and no they are not you-know-what. they are just awesome like that :D you should read it!! oh another good book is... umm... let's see... kite runner i guess. (random book i randomly saw on my shelf just now that happens to be good) read it for a class over the summer... i'd say for 8th grade and above...but it's good =]", "Solution_13": "I've read Kite Runner before. It's really sad.", "Solution_14": "At my school some of the girls are like \"Have you read Twilight?\" and if you say no then their eyes bug out of their sockets and they practically scream \"YOU HAVEN'T READ TWILIGHT?\" At least not all the girls in my school do that. But it's annoying.", "Solution_15": "Exactly why aren't people talking about Brisingr and the Iliad?", "Solution_16": "hm\r\n\r\ni think i'll read brisingr soon\r\n\r\ni've read eragon and eldest and they are pretty good", "Solution_17": "I put a hold on Brisingr at our library system, but I am about hold #200, so it probably won't arrive for another month or so.\r\nAnybody read Pendragon? I haven't read it, but a lot of my classmates like the book, boys and girls alike. I'm wondering if it's really that good.", "Solution_18": "I've read Pendragon. I definately recommend it.\r\n\r\nMacHale's last book in this series should be coming out. I'm really excited for it since the last one had hell breaking loose - I'm not sure if the Travelers can actually score a win at the end. :|", "Solution_19": "Well, I mean, being \"children's books\" (I loosely apply that term) the good side will win. Even in adult books the good side frequently wins. \r\nSome of my fav books/authors:\r\nFootprints of God (All other books by Greg Iles r crazy)\r\nBlasphemy (similar to above, about AI)\r\nJoseph Finder (workplace crime)\r\nMichael Crichton (Fiction about controversial topics)\r\nLincoln Child and Douglas Preston (crime with a twist?)\r\netc", "Solution_20": "The Physics of Superheroes by James Kakalios is hilarious and very entertaining not only to the geek side but to everyone :)" } { "Tag": [], "Problem": "\u039d\u03b1 \u03c0\u03c1\u03bf\u03c1\u03b4\u03b9\u03bf\u03c1\u03b9\u03c3\u03c4\u03bf\u03cd\u03bd \u03cc\u03bb\u03b5\u03c2 \u03bf\u03b9 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ad\u03c2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \u03bc\u03b5 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03cc\u03c1\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b7\u03bd \u03b5\u03be\u03ae\u03c2 \u03b9\u03b4\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1: \u03b1\u03bd $ S$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03c3\u03b5\u03b9\u03c1\u03ac\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc $ a$ \u03bc\u03b5 $ 0 2$ because there must exist an odd prime $ p$ to a certain power $ a$ that divides $ n$ since $ n$ is odd and composite, and $ \\phi(p^a) \\equal{} p^{a\\minus{}1}(p\\minus{}1)$, which is even since $ p\\minus{}1$ is even, so since the totient function is multiplicative, $ \\phi(n)$ is even for $ n > 2$, so $ \\phi(n)$ cannot divide $ n$ because that would imply that $ n$ is even, which is a contradiction.\n\n[/hide]", "Solution_3": "[quote=\"Ihatepie\"][hide=\"Solution\"]\nBy Fermat's Little Theorem, $ 2^{n}\\equiv 2 \\mod n$ if $ n$ is odd\n[/hide][/quote]\n\nFLT requires that $ n$ is prime\n\nEDIT: okay its been proven, here is another [hide=\"solution\"] \n\nClearly $ n$ cannot be even, so let $ n$ be odd. and let $ S$ be the set of integers, $ n$, such that $ 2^n \\equiv 1 \\mod n$. We will show that this set is empty by showing there does not exist a least element, $ n_0$.\n\nNow, there must exist $ i, j \\in \\mathbb{N}$ such that $ 2^i \\equiv 2^j \\mod n \\rightarrow 2^j(2^{i \\minus{} j} \\minus{} 1) \\equiv 0 \\mod n$\n\nSince $ \\gcd(2,n) \\equal{} 1$ we have $ 2^{i \\minus{} j} \\equiv 1 \\mod n$. So we know there must exist $ i \\minus{} j \\equal{} a$ such that $ 2^a \\equiv 1 \\mod n$. So assume $ a$ is the smallest such integer. \n\nWe know $ a < n$ because $ 2^k$ cycles through integers in $ \\mod n$ in periods of $ a$, and $ 2^k \\in \\{1,2,3...,n \\minus{} 1\\}$ which has only $ n \\minus{} 1$ elements. therefore if $ 2^n \\equiv 1 \\mod n$ then $ n$ is a mltiple of $ a$, or $ a|n$\n\nNow, assume $ n_0$ is the least element in $ S$, we have $ 2^{n_0} \\equiv 1 \\mod n$. However $ a | n_0$ and $ 2^a \\equiv 1 \\mod n$, Therefore $ 2^a \\equiv 1 \\mod a$ and $ a \\in S$\nBut $ a < n$ so this contradicts the assumption that $ n_0$ is the least element.\n\nHence $ S$ is empty and there are no $ n \\in \\mathbb{N}$ such that $ 2^n \\equiv 1 \\mod n$\n[/hide]", "Solution_4": "Oops :blush: \r\n\r\nFor some reason I just thought 2 had to be relatively prime with n.", "Solution_5": "Euler's Theorem/FLT holds true only if the base and modulus are relatively prime.", "Solution_6": "yup, you are right. the base and the moduli should be relatively prime." } { "Tag": [ "counting", "derangement", "combinatorics unsolved", "combinatorics" ], "Problem": "Here is the problem:\r\n Find the number of all permutations of n elements , such that:\r\n a) exactly m (0<=m<=n) elements are on their starting places.\r\n b) there is not an element that is on its starting place\r\n\r\nI hope i have written it clear", "Solution_1": "You're asking for the number of permutations of an $ n$-set with $ m$ fixed points. It's easy to see that if we can solve this question for $ m = 0$ then we can solve it for all $ m$. A permutation with 0 fixed points is a \"derangement,\" and the problem of counting derangements has been discussed here several times in the past -- try a search." } { "Tag": [], "Problem": "find the ordered pair (x,y) that is the solution to the system:\r\n\r\n(x+2y)/xy=11/12\r\n\r\n(2x-3y)/xy=2/3", "Solution_1": "[hide=\"Hint\"]Split the fractions and put some substitutions.[/hide]", "Solution_2": "[hide=\"solution\"]We cross multiply each fraction:\n\n11xy=12x+24y\n2xy=6x-9y\n4xy=12x-18y\n42y=7xy\nx=6\ny=12/7[/hide]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f: R\\to R$ such that\r\n$ f(x\\plus{}y)\\plus{}f(x)(f(y)\\equal{}f(xy)\\plus{}f(x)\\plus{}f(y)$", "Solution_1": "[quote=\"sylvect\"]Find all functions $ f: R\\to R$ such that\n$ f(f(x) \\plus{} y^2) \\equal{} f^2(x) \\minus{} f(x)f(y) \\plus{} xy \\plus{} x$[/quote]\r\nSorry,but it from our test,Lock it mod http://www.mathlinks.ro/Forum/viewtopic.php?t=168836" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Find a particular solution to the given equation:\r\n\r\ny\"+y = sinx + x cosx", "Solution_1": "[quote=\"gambs\"]Find a particular solution to the given equation:\n\ny\"+y = sinx + x cosx[/quote]\r\n\r\nIf I did'nt misscalculated:\r\n$y=\\frac{x^{2}-Asin(2x+B)+C}{4}sinx-\\frac{x+Acos(2x+B)}{4}cosx$\r\n\r\nWhere A,B,C are arbitrary.", "Solution_2": "How did you arrive at that answer? What were your initial steps?\r\nIt isn't the answer given in the book but is similar. Also, the problem is to solve for the coefficients.\r\n\r\nyp(x) = 1/4 (x^2 sinx - x cosx)", "Solution_3": "[quote=\"gambs\"]How did you arrive at that answer? What were your initial steps?\nIt isn't the answer given in the book but is similar. Also, the problem is to solve for the coefficients.\n\nyp(x) = 1/4 (x^2 sinx - x cosx)[/quote]\r\n\r\nIf A=0 and C=0 u got the answer from the book I presume:))\r\nOfcourse the homogen part of equation is the wellknown equation $~y''+y=0~$ wich has its solutions: $~y=C_{1}sinx+C_{2}cosx$, $C_{1},C_{2}\\in R$\r\nSo I searched your eqation's particular solutions in $~f(x)sinx+g(x)cosx$ form." } { "Tag": [], "Problem": "I got this from another forum. \r\n\r\nHow to play: \r\nEach person makes a wish (Ex. I wish for a cookie) then the next person crushes that persons wish (Ex. But the cookie was made of pure bleach) and makes their own wish (Ex. I wish for a bleach free cookie).\r\nI'll start.\r\n\r\nI wish for a soda.", "Solution_1": "the soda was a carryon bomb\r\n\r\ni wish for a element skateboard.", "Solution_2": "btw there was another topic exactly like this(and 15+ pages too i think)", "Solution_3": "The skateboard was broken in half\r\n\r\nI wish for 100 wishes\r\n\r\n[quote=\"junggi\"]btw there was another topic exactly like this(and 15+ pages too i think)[/quote]\r\n\r\nLink Please?", "Solution_4": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=wish+corrupt&t=109351[/url]\r\nhaha nvm it's 36 pages", "Solution_5": "You get 100 wishes, but they expire in 1 nanosecond.\r\n\r\nI wish I was able to memorize 100 digits of pi (I only know 57 digits).", "Solution_6": "You memorize 100 digits but they're the wrong ones.\r\n\r\nI wish this was the original thread.", "Solution_7": "i pm the mod to delete this thread and you revive the original thread.\r\n\r\ni wish the world exploded", "Solution_8": "your bomb is a dud. i wish i had read a question on the nationwide test today correctly. (no mention of ___ whatsoever :P )", "Solution_9": "[quote=\"junggi\"]i pm the mod to delete this thread and you revive the original thread.\n\ni wish the world exploded[/quote]\r\n\r\nI revived the original. Thanks.", "Solution_10": "[quote=\"footballrocks41237\"][quote=\"junggi\"]i pm the mod to delete this thread and you revive the original thread.\n\ni wish the world exploded[/quote]\n\nI revived the original. Thanks.[/quote]\r\nlollll i was jk in that post\r\nsorry", "Solution_11": "LOL. :blush: Still, the original is back and running...", "Solution_12": "you read the question correctly, but all the other questions you read wrong :lol: \r\n\r\ni wish for a non-exploding-non-leaking-non-dud-and-100%-drinkable-with-nothing-happening-to-me soda\r\n\r\n :rotfl:", "Solution_13": "Soda is another name for sodium bicarbonate, or baking soda. Your soda is dissolved in water. You drink it. It tastes terrible. Also, you just drank lemon juice. You vomit and burp a lot. You feel miserable.\r\n\r\nI wish money grew on trees (a classic).", "Solution_14": "wg, but there was a typhoon and your money tree got washed away before you got to pick the money\niwf, a friend forever", "Solution_15": "There is no reason to revive a 6-year-old thread, especially in the Fun Factory, where you are not posting something useful." } { "Tag": [], "Problem": "[b]5050(nr.1/2009):[/b] Demonstrati ca daca numerele $ a; b; c>0$ satisfac inegalitatea: $ a^3\\plus{}b^3\\plus{}c^3<5.abc$, \r\natunci numerele a, b si c sunt lungimile laturilor unui triunghi.([i]V.I. Kaskevici[/i]-Minsk)\r\n[b]5051(nr.1/2009):[/b] Construiti un patrulater care este atat inscriptibil cat si circumscriptibil, cunoscand cazele celor doua cercuri\r\nR si r si unghiul dintre diagonalele sale. ([i]A. Zaslavsky[/i])\r\n[b]5055(nr.2/2009):[/b] In triunghiul ABC ducem mediana AM si inaltimea AH. Demonstrati ca: $ MH\\ge\\frac {|AB \\minus{} AC|}{2}.$ ([i]S.I. Tokarev[/i])\r\n[b]5063(nr.3/2009):[/b] Aratati ca in orice triunghi are loc inegalitatea: $ a^2\\plus{}b^2\\plus{}c^2\\plus{}\\frac{1}{16}.(|a\\minus{}b|\\plus{}|b\\minus{}c|\\plus{}|c\\minus{}a|)\\le 9.R^2.$([i]S.G. Abdullaev[/i])", "Solution_1": "Can anyone please translate the problem and give me the link to the website of this journal? I'm really curious to find out more about Romanian Math Journal, especially I only know the \"Gazetta Matematica\" journal. Does anyone know when will the contests hosted by this journal be in this year?" } { "Tag": [ "logarithms", "inequalities", "inequalities proposed" ], "Problem": "$x$ is a positive real number and $n \\in Z$ and $n<0$.prove that:\r\n$\\frac{(n+1)^{n+1}}{n^{n}}>\\frac{n^{n}}{(n-1)^{n-1}}$ :idea:", "Solution_1": "$[x\\log x]''=\\frac{1}{x},$ thus $x\\log x$ is strictly convex. Now if we apply Jensen we get \\[e^{x\\log x+y\\log y}> e^{(x+y)\\log (x+y)/2}\\] thus \\[\\frac{x^{x}}{\\left(\\frac{x+y}{2}\\right)^{(x+y)/2}}>\\frac{\\left(\\frac{x+y}{2}\\right)^{(x+y)/2}}{y^{y}}\\;\\;\\;\\;\\;(1)\\] for positive reals $x,y.$ About negatives, $x\\log x$ is concave, so we have \\[(-x)^{-x}(-y)^{-y}< \\left[\\left(-\\frac{x+y}{2}\\right)^{-(x+y)/2}\\right]^{2}\\;\\;\\;\\;\\;\\Leftrightarrow\\;\\;\\;\\;\\;\\frac{1}{x^{x}y^{y}}<\\frac{1}{\\left(\\frac{x+y}{2}\\right)^{x+y}}\\] which is exactly (1). Thus the inequality (1) is true for all reals $x,y$ such that $xy>0.$", "Solution_2": "thank you.it seems that your solution is very hard,but nice.\r\nis there any easy one?", "Solution_3": "$1(1+\\frac{1}{n})^{n}\\leq (\\frac{1+n(1+\\frac{1}{n})}{n+1})^{n+1}$(amgm)", "Solution_4": "can you explain your solution.plz.because may be it will be more usefull. :D" } { "Tag": [ "logarithms", "number theory", "prime numbers", "number theory unsolved" ], "Problem": "Let $\\sigma(n)$ be sum of all positive diviosrs of natural number $n$. Prove that if $\\sigma(n)=5n$ than $n$ has more than 5 distinct prime divisors.", "Solution_1": "Let $\\omega(n)$ be the number of distinct primes $p_1, p_2, \\ldots, p_{\\omega(n)}$ dividing $n$. Wlog, suppose $p_1 < p_2 < \\ldots < p_{\\omega(n)}$. You get $\\frac{\\sigma(n)}{n} < \\prod_{k=1}^{\\omega(n)} \\frac{p_k}{p_k - 1} \\leq 3\\cdot \\prod_{k=3}^{\\omega(n)} \\frac{p_k}{p_k - 1} \\leq 3 \\cdot (\\frac{5}{4})^{\\omega(n) - 2}$, as the map $]1,+\\infty[\\; \\mapsto \\mathbb{R}: x \\mapsto \\frac{x}{x-1}$ is strictly decreasing on its domain. Now consider (taking logarithms and making some calculations!) that $(\\frac{5}{4})^{\\omega(n) - 2} \\leq \\frac{5}{3}$, if $\\omega(n) < 5$, and deduce that $5 = \\frac{\\sigma(n)}{n}$ only if $\\omega(n) \\ge 5$, q.e.d.", "Solution_2": "Nice, and by the way, good to see you back here.", "Solution_3": "Oh, thanks! :blush:", "Solution_4": "Can somebody explain to me why\r\n$\\frac{\\sigma(n)}{n} < \\prod_{k=1}^{\\omega(n)} \\frac{p_k}{p_k - 1}$ ?", "Solution_5": "Note that $\\sigma (p^k) = \\frac{p^{k+1}-1}{p-1}<\\frac{p^{k+1}}{p-1}$ which gives by multiplicativity the desired inequality.", "Solution_6": "Thanks, I understand. ;)", "Solution_7": "[quote=\"hitlEULER\"] Now consider (taking logarithms and making some calculations!) that $(\\frac{5}{4})^{\\omega(n) - 2} \\leq \\frac{5}{3}$, if $\\omega(n) < 5$, and deduce that $5 = \\frac{\\sigma(n)}{n}$ only if $\\omega(n) \\ge 5$, q.e.d.[/quote]\r\nYou have a mistake in your solution!\r\nSuppose $\\omega(n)=4$ then we get from your solution $\\frac{25}{4} \\leq \\frac{5}{3}$ - contradiction\r\nYou can solve it like this:\r\nbecause as you said $\\frac{x}{x-1}$ is strictly dicreasing so we need to show that with 5 smallest prime numbers except 1 that $\\prod_{k=1}^{5} \\frac{p_k}{p_k - 1} \\leq 5$ so:\r\n$\\frac{2}{1}\\cdot \\frac{3}{2} \\cdot \\frac{5}{4} \\cdot \\frac{7}{6} \\cdot \\frac{11}{10}=4.8125$\r\nSo if $\\sigma(n)=5n$ then there are more than 5 distinct prime divisors." } { "Tag": [], "Problem": "Last year, I became interested in: how many arrangements of an unrooted tree up to graph isomorphism are there for n vertices?\r\n\r\nThe sequence begins: 1, 1, 1, 1, 2, 3, 6, 11, 23, 47...\r\n\r\n[url]http://www.research.att.com/~njas/sequences/A000055[/url]\r\n\r\nI am really rather interested in this, but so far I have not been able to find much information on it. Does a definitive recurrence or an explicit formula exist for this?", "Solution_1": "There's a formula on that page you linked to :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Determine all pairs $(x, y)$ of positive integers with $y \\vert x^{2}+1$ and $x \\vert y^{3}+1$.", "Solution_1": "I think it may help to you :[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=25716[/url]", "Solution_2": "It may not help me :wink: Because there isn't a solution ,either." } { "Tag": [ "ARML", "email", "Support", "MATHCOUNTS", "HCSSiM", "geometry", "probability" ], "Problem": "AoPS hosts a number of local communities including those for Alabama, Florida, Texas, Upstate New York, and Southern California. These communities are a great way to create social networks between high caliber students and also facilitate ARML teams and other local events. These communities are hidden until you are added as a member by a moderator. If there are active members of our forum who would like to take on the responsibility of moderating other such local communities, please contact me at crawford@artofproblemsolving.com.\r\n\r\nIf you think you can round up enough AoPSers for a local community not listed above, PM me and we will create a community for your city or state area.\r\n\r\n[b]Edit: Much of this post no longer applies as we have revamped the entire local/national community forum area.[/b]", "Solution_1": "The Southern California group (one of the ones MCrawford refers to above) is run as a \"hidden forum\", meaning that you would only see it as an option if the moderator (me) has admitted you to the group. I would assume that the other groups referred to are run the same way.\r\n\r\nIf you are a high school or middle school student in Southern California and are interested in joining an ARML team or otherwise in finding out what this is about, send me a private message.", "Solution_2": "Hmm... sounds very cool.\r\n\r\nI would love to have one for South Carolina but honestly I don't really think there are too many South Carolinians on the forum....I don't know of any active ones right now. I'm sure that Marcus (gauss202) would be probably be able to get in touch with more schools and such to get people on here. But the last time he was on here was a little after ARML.\r\n\r\n", "Solution_3": "[quote]Alabama, Florida, Texas, Upstate New York, and Southern California.[/quote]\r\n\r\nIs there one for the midwest? I find that a lot of communities revolve around bigger cities where interest is great. For those of us stuck in states that don't have numerous programs, it is a little bit more difficult. If AoPS decides to create a community for the midwest - or specifically for my home state, MI, I volunteer to moderate it.", "Solution_4": "Likewise, students, teachers, and math enthusiasts in Alabama who wish to join the Alabama community can email me at crawford@artofproblemsolving.com.\r\n\r\nWe may soon have a way to see which groups are available and have an automated email sent to these community moderators when students wish to join these hidden communities.", "Solution_5": "[quote=\"tetrahedr0n\"]\nIs there one for the midwest? I find that a lot of communities revolve around bigger cities where interest is great. For those of us stuck in states that don't have numerous programs, it is a little bit more difficult. If AoPS decides to create a community for the midwest - or specifically for my home state, MI, I volunteer to moderate it.[/quote]\r\n\r\nI think Michigan is certainly large enough to have its own community. If you would be interested in moderating it or know the ARML coach (to see if they would be interested), I could set that up soon. Let me know.", "Solution_6": "[quote=\"joml88\"]Hmm... sounds very cool.\n\nI would love to have one for South Carolina but honestly I don't really think there are too many South Carolinians on the forum....I don't know of any active ones right now. I'm sure that Marcus (gauss202) would be probably be able to get in touch with more schools and such to get people on here. But the last time he was on here was a little after ARML.\n\n[/quote]\r\n\r\nPerhaps the Carolinas together would make a good sized community? Marcus would certainly make a good moderator if he is interested. Otherwise, either you or someone else could moderate? Let me know.", "Solution_7": "[quote]I think Michigan is certainly large enough to have its own community. If you would be interested in moderating it or know the ARML coach (to see if they would be interested), I could set that up soon. Let me know.[/quote]\r\n\r\nI am interested in moderating. The issue with ARML is that it is hard to say what is going to happen this year. Up to last year, every year MI has had an ARML team known as the \"MI All Star Team\" determined by finish in the MMPC competition. MMPC director David Redman and former director Robert Messer were coaches. Last year, there was a lack of funding/interest, so the All Star Team system stopped. ICAE - a local math club/circle offered to send a team, altough it was difficult to go due to disorganization in travel plans. This year might be different - either Mr. Messer will use the old system, or ICAE might continue the trend started last year. I'll try to find some more information.", "Solution_8": "Hi, I've just emailed to indicate an interest in forming and maintaining a Minnesota AoPS community too. I just met someone this weekend who would be good to recruit into the group.", "Solution_9": "Hey Tetrahedr0n, did you get my PM?", "Solution_10": "[quote]Hey Tetrahedr0n, did you get my PM?[/quote]\r\n\r\nI did. I'm sorry I didn't respond a little earlier, but I've been very busy with finals the last two days. I'm going to post here soon to attract more members as well.", "Solution_11": "Is this only for the usa? I wouldn't mind to do Belgium/Flanders if you're Europe is okay too :)", "Solution_12": "[quote=\"Peter VDD\"]Is this only for the usa? I wouldn't mind to do Belgium/Flanders if you're Europe is okay too :)[/quote]\r\n\r\nWe plan to build some national communities in the relatively near future. They may even have some different features, though we are in discussion of the ideas. We will keep you posted.", "Solution_13": "ok, if you ever need me just drop me a pm or a mail. I'll unsubscribe from this topic for now. :)", "Solution_14": "Do you suppose the Massachusetts group would be interested in expanding to a New England group? I don't suppose Vermont could support a forum of its own ;-)", "Solution_15": "ondrob, yes we could do that. But please remember local and national communities are just for accomodation and for other (local) discussions you may want to have. \r\n\r\nwe expect all members to use the main forums just as before, and even more. \r\n\r\nplease present to me on email a possible structure of the forums the Czech-Slowack community should posses.", "Solution_16": "I'd moderate or just join Connecticut if there were one. It it's too small, we can probably combine with Rhode Island to make Rhodenecticut. (Or just southern new england, to be bland. :) )", "Solution_17": "[quote=\"SnowStorm\"]I could do Colorado although I doubt there is anyone from here on these forums.[/quote]\r\n\r\nThere is already a Colorado community. The moderator is rcv. It is one of the new ones and is just getting going.", "Solution_18": "[quote=\"SnowStorm\"]I could do Colorado although I doubt there is anyone from here on these forums.[/quote]\r\nHi, SnowStorm! I have added you to the Colorado group. Send me a PM or introduce yourself over on the Colorado Forum. See you there!", "Solution_19": "Everyone can request to join the usergroup by going to usergroups -> selecting the group and pressing the join group button (on the second row).", "Solution_20": "I don't any of the admin to feel rushed by this, but I'm just curious. Is there going to be one made for Connecticut? If so, there's no rush. I'm just curious about it.", "Solution_21": "[quote=\"yif man12\"]I don't any of the admin to feel rushed by this, but I'm just curious. Is there going to be one made for Connecticut? If so, there's no rush. I'm just curious about it.[/quote]\r\n\r\nAre you volunteering to moderate?", "Solution_22": "Sure, I'd be glad to!", "Solution_23": "[quote=\"yif man12\"]Sure, I'd be glad to![/quote]Done.", "Solution_24": "just wondering... if even hungarian is in now, why not dutch? :) I know some people of this board are from belgium/holland..", "Solution_25": "Is there one for Northern California, specifically the SFBA?", "Solution_26": "[quote=\"aznphatso\"]Is there one for Northern California, specifically the SFBA?[/quote]\r\n\r\nYes, and you're a member. Do you not see it in the North American communities area?", "Solution_27": "So there it is, staring at me right in the face. Thanks!", "Solution_28": "[quote=\"Peter VDD\"]Is this only for the usa? I wouldn't mind to do Belgium/Flanders if you're Europe is okay too :)[/quote]Well, now you have your own Dutch/Flanders community :)", "Solution_29": "[quote=\"Valentin Vornicu\"][quote=\"Peter VDD\"]Is this only for the usa? I wouldn't mind to do Belgium/Flanders if you're Europe is okay too :)[/quote]Well, now you have your own Dutch/Flanders community :)[/quote] :clap: :clap2: :omighty:" } { "Tag": [ "Putnam", "linear algebra", "linear algebra unsolved" ], "Problem": "Suppose $P$ and $Q$ are $n\\times n$ matrices over some field. Suppose that $P^2=P,\\,Q^2=Q,$ and $I-P-Q$ is invertible.\r\n\r\nProve that $P$ and $Q$ have the same rank.", "Solution_1": "$P$ has the same rank as $P(I-P-Q)=-PQ$, and $Q$ has the same rank as $(I-P-Q)Q=-PQ$, so the two ranks are equal.", "Solution_2": "I gave that problem to my \"Putnam preparation\" class, and one of my students came up with precisely that proof. Once you've found the argument that grobber just gave, there's no point looking for any alternatives." } { "Tag": [], "Problem": "I have no idea on how to approach this problem...\r\nA square array of dots with $10$ rows and $10$ columns is given. Each dot is coloured either blue or red. Whenever two dots of the same colour are adjacent in the same row or column, they are joined by a line segment of the same color as the dots. If they are adjacent but of different colors, they are joined by a green line segment. In total, there are $52$ red dots. There are $2$ red dots at the corners with an additional $16$ red dots on the edges of the array. The remainder of the red dots are inside the array. There are $98$ green line segments. The number of blue line segments is:\r\n\r\n$(A) 36$ $(B) 37$ $(C) 38$ $(D) 39$ $(E) 40$\r\n\r\nSource: CMC-Fermat 2001- Question #25\r\n\r\n[hide=\"some really obvious stuff I came up with\"]\nThere are a total of $48$ blue dots.\n$16$ blue dots are on the edges. \n$2$ blue dots are on corners.\nThe maximum number of green line segments is $180$[/hide]", "Solution_1": "Hi Hellstar. Below is hint but reply for solution if you want it :)\r\n\r\n[hide=\"hint\"]Idea is simple: Look at each dot. Each segment contains 2 dots. Count the number of segments incident from blue dots (note that some dots have different number of segments incident from it) But there are two types of segments... oops did I say too much?[/hide]", "Solution_2": "[hide=\"so basically...\"]\nIf you focus on segments the total number of segments will be $180$. So the total number of blue and red segments will be $82$ since the number of green segments is $98$. So following your hint, since there are a total of $48$ blue dots ($2$ on corners, $16$ on edges, and $30$ inner), there will be a total of $2*2+16*3+30*4=172$ segments issuing from blue dots. These segments can only be blue or green, because at least one dot is blue. Subtracting the $98$ green segments, we are left with $74$ blue segments. These are segments that connect blue to blue so we overcounted by a factor of $2$ and we divide $2$ to get $37$. Is that right?[/hide]", "Solution_3": "Yup :thumbup: That's exactly what I had in mind" } { "Tag": [ "geometry", "circumcircle", "perpendicular bisector", "geometry proposed" ], "Problem": "In triangle ABC, bisectors of angles B and C intersect ar point E. The area of the circumcirle of triangle BCE equals [i]q[/i]. Find the area of the circumcircle of triangle ABC if BC=[i]d[/i]", "Solution_1": "Perpendicular bisector of BC meets BC at its midpoint M and the circumcircle (O) at P, Q. The intersection on the opposite side of BC than A is circumcenter of the $ \\triangle EBC.$ Let $ R,\\ \\rho$ be radii of the circles (O), (P), OB = OC = R, PQ = 2R. The line BC is inversion of (O) in (P), $ PM \\cdot PQ \\equal{} \\rho^2,$\r\n$ OM \\equal{} OP \\minus{} PM \\equal{} R \\minus{} \\frac {\\rho^2}{2R},\\ \\ \\ OM^2 \\equal{} R^2 \\minus{} \\rho^2 \\plus{} \\frac {\\rho^4}{4R^2}$\r\n$ \\frac {d^2}{4} \\equal{} BM^2 \\equal{} OB^2 \\minus{} OM^2 \\equal{} \\rho^2 \\minus{} \\frac {\\rho^4}{4R^2}$\r\n$ R^2 \\equal{} \\frac {\\rho^4}{4\\rho^2 \\minus{} d^2},\\ \\ \\ \\pi R^2 \\equal{} \\frac {(\\pi \\rho^2)^2}{4 (\\pi \\rho^2) \\minus{} \\pi d^2}$" } { "Tag": [ "algebra", "polynomial", "email" ], "Problem": "Darn, a double post...", "Solution_1": "Oh, yeah, nemo. I may not be able to understand much about your polynomial problems, but ateast I had the satisfaction of seeing you needing help on a concept. that's almost unheard of for you :rotfl: btw, out of curiosity, can you give me a brief explanation of what you're learning so far?", "Solution_2": "Umm... dude this is a forum, not a discussion page. If you really need to talk email or pm me. Anyways, can someone look at my other post." } { "Tag": [], "Problem": "A circle in the 1st quadrant touches the x-axis at point A and touchest the y-axis at point B. It touches the line $x+2y-3-\\sqrt5 = 0$ at point C(a,b). Find a+b", "Solution_1": "Be more specific...By touches, do you mean that it is tangent to???", "Solution_2": "Yes.. I mean tangent.\r\nSorry about that.", "Solution_3": "From my pics. :D \r\n[hide=\"My answer\"]\n$OB=\\frac{3+\\sqrt5}2\\\\\nOA=3+\\sqrt5\\\\\nAB=\\frac{5+3\\sqrt5}2\\ \\ (from\\ Pythagoras)$\n\\begin{eqnarray*}\n&p+r&=\\frac{3+\\sqrt5}2&...(1)\\\\\n&r+q&=3+\\sqrt5&...(2)\\\\\n&p+q&=\\frac{5+3\\sqrt5}2&...(3)\\\\\n\\frac{(1)+(2)+(3)}2;&p+q+r&=\\frac{7+3\\sqrt5}2&...(4)\\\\\n(4)-(3);&r&=1\n\\end{eqnarray*}\nradias of this circle is 1 and center is (1,1)\nand equation of this circle is $(x-1)^2+(y-1)^2=1$\nfor (a,b) is common point in 2 equation\n$(a-1)^2+(b-1)^2=1\\ \\ and\\ \\ a+2b-3-\\sqrt5=0$\nI got $a=\\frac{5+\\sqrt5}5\\ \\ and\\ \\ b=\\frac{5+2\\sqrt5}5$\nThen $a+b=\\frac{10+3\\sqrt5}5$ :blush: [/hide]", "Solution_4": "That works. :)" } { "Tag": [ "trigonometry", "limit", "inequalities", "algebra unsolved", "algebra" ], "Problem": "Determine the number of real solutions of the system \\[\\left\\{ \\begin{aligned}\\cos x_{1}&= x_{2}\\\\ &\\cdots \\\\ \\cos x_{n-1}&= x_{n}\\\\ \\cos x_{n}&= x_{1}\\\\ \\end{aligned}\\right.\\]", "Solution_1": "Only one solution $x_{i}=x_{0}$, were $cos(x_{0})=x_{0}$, because $|cos'x_{0}|=sinx_{0}<1$.", "Solution_2": "Note $ x_{0}=\\cos x_{0},$ $\\cos^{(n)}x= \\underbrace{\\cos\\dots\\cos}_n x $\n$ x_{i} \\in (0,1),$ $ i =1, \\dots,n $\nfor $ n=1 $,it\u2019s obvious.\n$ n\\geq2,x_{n}=\\cos x_{n-1} \\in (-1,1),$ $ x_{1}=\\cos x_{n} \\in (\\cos1,1) \\subset (0,1),$ $\\dots,$ $ x_{n}= \\cos x_{n-1} \\in (0,1).$ so $ x_{i} \\in (0,1),i =1,\\dots,n $\nWhen $ x_{1}=x_{2},x_{1}= \\cos x_{1},$ $ x_{1}=x_{0}. $ Hence $ x_{i}= x_{0},i =1,\\dots,n $ is the solution.\nWhen $ x_{1} > x_{2},n\\neq2.n $ must be an even number.\nIf $ n=2k+1,k \\in N^+, x_{1} > x_{2} \\Rightarrow \\cos x_{1} > \\cos x_{2}\\Rightarrow x_{2} > x_{3}\\dots$\nWe get $ x_{1} > x_{2}, x_{2} < x_{3},\\dots, x_{2k} > x_{2k+1}, x_{2k+1} > x_{1}, x_{1} < x_{2} $.contradiction. \nWhen $ n=2 $,$ \\cos^{(2)} x_{1} = x_{1}. $ let $ g(x)=\\cos(\\cos x) - x ,g $'$ (x)=\\sin(\\cos x)\\sin x -1<0 $,and $ g(x_{0})=0, x_{0} $ is the only solution. but $ x_{1} > x_{2} $ contradiction.\nFor $ n\\geq4,$ let $ n=2k,$ suppose $ x_{1} > x_{3} $ or $ x_{1} < x_{3},( x_{1}\\neq x_{3}, $ see $ n=2) $,$ x_{1} > x_{3} \\Rightarrow \\cos^{(2)} x_{1} > \\cos^{(2)} x_{3} \\Rightarrow x_{3} > x_{5},\\dots,$ have $ x_{1} > x_{3}>$ $\\dots $ $ > x_{2k-1} > x_{1} $contradiction. when $ x_{1} < x_{3}\\dots $\nIn similar way,thereis no solution when $ x_{1} < x_{2} $\nas Rust said $ x_{i}=x_{0}$ is the only solution.\n[hide]For $00.\\]\r\nDefine $ f(y)$ to be the smallest $ n$ such that \r\n\\[ X_1\\plus{}X_2\\plus{}\\cdots\\plus{}X_n\\geq y.\\]\r\nFind the limit\r\n\\[ \\lim_{y\\rightarrow \\infty}\\frac{f(y)}{y}.\\]", "Solution_1": "Compare [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20848]this old topic.[/url]\r\n\r\nOne difference: in that other topic, we were assuming that $ X_i$ was nonnegative. bravado does not seem to be making that assumption here. Does that make a difference? Maybe not, but I haven't checked to be sure.", "Solution_2": "Fix $ y > 0$. We have $ S_{f(y) \\minus{} 1} < y \\leq S_{f(y)}$ by definition. Hence\r\n\r\nDivide through by $ f(y)$, so that\r\n\\[ \\frac {S_{f(y) \\minus{} 1}}{f(y) \\minus{} 1}\\cdot \\frac {f(y) \\minus{} 1}{f(y)} < \\frac {y}{f(y)} \\leq \\frac {S_{f(y)}}{f(y)}\r\n\\]\r\nSince $ f(y)\\to\\infty$ as $ y\\to\\infty$ (why?), using the SLLN on both sides of the above shows that a.s. $ \\frac {y}{f(y)}\\to v$, or $ \\frac {f(y)}{y}\\to \\frac {1}{v}$." } { "Tag": [ "geometry", "trigonometry", "quadratics" ], "Problem": "Here are some problems in Math Zoom Year-round classes. In the following exercises, find real solutions unless indicated otherwise:\r\n\r\n1. Solve for $ x,y,z$:\r\n\\[ \\left\\{\r\n\\begin{array}{rcl}\r\nx^2 \\plus{} xy \\plus{} xz \\minus{} x &\\equal{}& 2 \\\\\r\ny^2 \\plus{} yz \\plus{} yx \\minus{} y &\\equal{}& 4 \\\\\r\nz^2 \\plus{} zx \\plus{} zy \\minus{} z &\\equal{}& 6\r\n\\end{array}\r\n\\right.\\]\r\n\r\n2. Solve for $ x$ and $ y$:\r\n\\[ x^2 \\minus{} 2x\\sin(xy) \\plus{}1 \\equal{} 0\\]\r\n\r\n3. Let $ x$, $ y$, and $ z$ be positive real numbers, and\r\n\\[ (x^2\\plus{}1)(y^2\\plus{}2)(z^2\\plus{}8) \\equal{} 32xyz\\]\r\nFind $ x$, $ y$, and $ z$.\r\n\r\n4. Given that $ a^2\\plus{}2a\\minus{}5\\equal{}0$, and $ b^2\\plus{}2b\\minus{}5\\equal{}0$, find all possible values of $ \\frac{b^2}{a}\\plus{}\\frac{a^2}{b}$.\r\n\r\n5. Given that $ a$ and $ b$ are distinct positive integers, and the two equations $ (a\\minus{}1)x^2 \\minus{}(a^2\\plus{}2)x \\plus{} (a^2\\plus{}2a)\\equal{}0$ and $ (b\\minus{}1)x^2 \\minus{}(b^2\\plus{}2)x \\plus{} (b^2\\plus{}2b)\\equal{}0$ have a common root, find all possible values of $ \\frac{1}{2a^2}\\plus{}\\frac{4}{3b^2}$.\r\n\r\n6. Solve for $ x$, $ y$, and $ z$: \r\n\\[ \\left\\{\r\n\\begin{array}{ccl}\r\nx\\plus{}y &\\equal{}& 4 \\\\\r\nxy\\minus{}z^2 &\\equal{}& 4\r\n\\end{array}\r\n\\right.\\]\r\n\r\n7. In $ \\triangle ABC$, $ \\angle C\\equal{}90^\\circ$, and the area of the triangle is $ 30$. What is the minimum possible value for the hypotenuse $ c$?\r\n\r\n8. Solve for $ x$:\r\n\\[ \\frac{1}{x^2\\plus{}2x\\minus{}3} \\plus{} \\frac{18}{x^2\\plus{}2x\\plus{}2} \\minus{} \\frac{18}{x^2\\plus{}2x\\plus{}1} \\equal{} 0\\]\r\n\r\n9. Solve for $ x$:\r\n\\[ \\frac{4x}{x^2\\plus{}x\\plus{}3} \\plus{} \\frac{5x}{x^2\\minus{}5x\\plus{}3} \\equal{} \\minus{}\\frac{3}{2}\\]\r\n\r\n10. Solve for $ x$:\r\n\\[ \\sqrt{4x\\minus{}1} \\plus{} \\sqrt{4x\\plus{}1} \\minus{} \\sqrt{16x^2\\minus{}1} \\equal{} 4x\\]", "Solution_1": "[hide]1. Let $ s \\equal{} x \\plus{} y \\plus{} z$, Add the equations to get $ s^2 \\minus{} s \\equal{} 12\\implies s \\equal{} 4, \\minus{} 3$ Each equation is $ x(s \\minus{} 1) \\equal{} 2$, etc answer follows.\n\n2. Rearrange the equation $ (x \\minus{} \\sin xy)^2 \\equal{} \\minus{} cos^2xy$. So $ x \\equal{} \\sin xy$ and $ \\cos xy \\equal{} 0$. Clearly, $ \\sin xy \\equal{} \\pm 1 \\equal{} x$, and we easily solve for $ y$.\n\n3. $ x^2 \\plus{} 1\\ge 2x$, $ y^2 \\plus{} 2\\ge 2\\sqrt {2} x$, $ z^2 \\plus{} 8\\ge 4\\sqrt {2}z$ by AM-GM, multiply them to get that $ (x^2 \\plus{} 1)(y^2 \\plus{} 2)(z^2 \\plus{} 8)\\ge 32xyz$ with equality iff $ x \\equal{} 1$, $ y \\equal{} \\sqrt {2}$, $ z \\equal{} 2\\sqrt {2}$.\n\n4. If $ a \\equal{} b$, then we get $ 2( \\minus{} 1\\pm\\sqrt {6})$. If $ a\\ne b$, then they are the distinct roots of the quadratic $ y^2 \\plus{} 2y \\minus{} 5 \\equal{} 0$, we want $ \\frac {a^3 \\plus{} b^3}{ab} \\equal{} \\frac {(a \\plus{} b)^3 \\minus{} 3ab(a \\plus{} b)}{ab}$ and the answer follows.\n\n5. Factor the equations to get $ (x \\minus{} a) ( \\minus{} 2 \\minus{} a \\minus{} x \\plus{} a x) \\equal{} 0$. Let $ r$ be the common root, then we have that $ a \\equal{} r$, $ b \\equal{} \\frac {r \\plus{} 2}{r \\minus{} 1} \\equal{} 1 \\plus{} \\frac {3}{r \\minus{} 1}$. Clearly $ r > 0$ is an integer so $ r \\minus{} 1 \\equal{} 1,3$. Check if $ a,b$ are distinct, etc and compute. \n\n6. $ (t \\minus{} x)(t \\minus{} y) \\equal{} t^2 \\minus{} t(x \\plus{} y) \\plus{} xy \\equal{} t^2 \\minus{} 4t \\plus{} 4 \\plus{} z^2 \\equal{} 0\\implies (t \\minus{} 2)^2 \\equal{} \\minus{} z^2$ so $ z \\equal{} 0$, $ x,y \\equal{} 2$.\n\n7. $ ab \\equal{} 60$, $ c \\equal{} \\sqrt {a^2 \\plus{} b^2}\\ge \\sqrt {2}ab \\equal{} 60\\sqrt {2}$; equality occurs when $ a \\equal{} b$.\n\n8. Let $ u \\equal{} x^2 \\plus{} 2x \\plus{} 1$. The equation is then $ \\frac {1}{u \\minus{} 4} \\plus{} \\frac {18}{u \\plus{} 1} \\minus{} \\frac {18}{u} \\equal{} 0\\implies u \\equal{} 8,9$; the solutions for $ x$ follow.\n\n9. Let $ u \\equal{} x^2 \\plus{} x \\plus{} 3$, then we have $ \\frac {4x}{u} \\plus{} \\frac {5x}{u \\minus{} 6x} \\plus{} \\frac {3}{2}$ factors as $ (u \\minus{} 4x)(u \\plus{} 4x)$ and the answer follows.\n\n10. Let $ u \\equal{} \\sqrt {4x \\minus{} 1}$ and $ v \\equal{} \\sqrt {4x \\plus{} 1}$. Then $ u \\plus{} v \\minus{} uv \\equal{} \\frac {1}{2}\\cdot (u^2 \\plus{} v^2)\\iff 2(u \\plus{} v) \\equal{} (u \\plus{} v)^2$. So either $ u \\plus{} v \\equal{} 0$ or $ u \\plus{} v \\equal{} 2$. In the first case, $ u,v\\ge 0$, so $ u \\equal{} 0;v \\equal{} 0$ is inconsistent. In the second case, $ v^2 \\minus{} u^2 \\equal{} 2$, so $ v \\minus{} u \\equal{} 1$, $ (u,v) \\equal{} (1.5,0.5)$ which gives $ x \\equal{} \\frac {5}{16}$ which works.[/hide]" } { "Tag": [ "calculus", "factorial", "algebra", "polynomial", "logarithms", "limit" ], "Problem": "sorry if this is dumb but what is $0^0$? and $0/0$? practically I even know it will be $0$ but...\r\nand one more.... you have this equation....\r\n$x+2=0$\r\nMultiply by $x-2$ both sides.... you get a different solution to the question?huh :wallbash_red: this is weird or what? \r\nI will appreciate if you explain your answer. This question is really annoying :mad:", "Solution_1": "$\\frac{0}{0}=\\frac{a}{b}$\r\nMultiplying by $b$ to clear the denomenator, we have\r\n$\\frac{0 \\cdot b}{0}=a$\r\n$\\frac{0}{0}=a$\r\nMultiply by $0$ to get\r\n$0=0$\r\nSo it doesn't matter what $a$ or $b$ are; $\\frac{0}{0}$ can equal anything, so it is undefined.", "Solution_2": "Depending on the context, $0^0$ is considered to be of interdeterminant form, 1, or undefined.", "Solution_3": "removed", "Solution_4": "hey, chess64 how can it be indetermined?eh? :rotfl: because $0^0$ should be either $0$ or $1$.\r\nBut what????????????????????????? :( :? \r\nI need the exact answer. :(", "Solution_5": "[quote=\"riddler\"]sorry if this is dumb but what is $0^0$? and $0/0$? practically I even know it will be $0$ but...\nand one more.... you have this equation....\n$x+2=0$\nMultiply by $x-2$ both sides.... you get a different solution to the question?huh :wallbash_red: this is weird or what? \nI will appreciate if you explain your answer. This question is really annoying :mad:[/quote]\r\n\r\nIt's strange because anything to the 0th power is supposed to be 1, but 0 to any power is 0. These two contradict, and I think that it is undefined. \r\n\r\n0/0 also is strange because it has three possible solutions. \r\n-anything divided by itself is 1\r\n-0 divided by anything is 0\r\n-anything divided by 0 is undefined. \r\n\r\nI don't know what to tell you about that. I think it's undefined. \r\n\r\nWhen you multiply both sides by x-2, the left side turns into $(x+2)(x-2)$ which turns into $x^2 - 4$. The right side is 0, and multiplying it by x-2 still makes it 0. (do you see why?)\r\n\r\nSolving, $x^2 - 4 = 0$\r\n$x^2 = 4$\r\nso, $x = 2$, [i]or[/i] $x=-2$\r\n\r\nAnd the solution to x + 2 = 0 is $x = -2$.", "Solution_6": "i understand but this should happen in math??!!!! :mad: its really annoying!", "Solution_7": "Oh things can be worse than this.\r\nA moderator by the name of jili posted this one: \r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38781[/url]", "Solution_8": "[quote=\"riddler\"]hey, chess64 how can it be indetermined?eh? :rotfl: because $0^0$ should be either $0$ or $1$.\nBut what????????????????????????? :( :? \nI need the exact answer. :([/quote]\r\n\r\nIndeterminant form. Do you really want me to get into calculus? :P", "Solution_9": "[quote=\"236factorial\"]Oh things can be worse than this.\nA moderator by the name of jili posted this one: \n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38781[/url][/quote]\r\n\r\nthat ones easier than mine besides mines impossible. :P", "Solution_10": "Of course its impossible. Its hard to find the value of an indeterminant number :D", "Solution_11": "[b]What does \"0/0\" mean?[/b]0/0 means the the number c such that 0\u00d7c = 0. What value of c will make this equation true? How about 1? or 2, or -26/31? Yes! c can be any number and still satisfy 0\u00d7c = 0. Therefore, 0/0 does not mean any particular number - or even anything until we give it some new meaning. \r\n\r\n[b]What does \"00\" mean?[/b]\r\nIn trying to answer this, we could first ask what \"0n\" means for n \u2260 0, n is an integer. \r\nFor instance, \"03\" = 0\u00d70\u00d70 = 0. \r\n01 = 0.\r\nSo 0n is the value of the string where 0 occurs n times. Then, 00 is the value of the string where 0 occurs 0 times. But if 0 occurs 0 times, the string has no value and meaning\r\n\r\nAll of this was gotten from [url=http://www.mathpath.org/concepts/division.by.zero.htm]http://www.mathpath.org/concepts/division.by.zero.htm[/url]", "Solution_12": "I just found this out: :blush: \r\n\r\n[hide]0^0= (0^1)/(0^1)\n0^0= 0/0\n\n\nSo, the values of 0^0 and 0/0, whatever they may be, are the same.[/hide]", "Solution_13": "Why does 0^0 = 0^1 ????", "Solution_14": "It doesn't. [b]0^1/0^1[/b] should be 0^(1-1)=0^0.\r\n\r\n0^1/0^1 = 0/0.\r\n\r\n[hide]The weird part is...\n\nIf you cancel out 0^1/0^1, you get 1. But 0/0 can be any number.\n\n :arrow: Nothing's perfect, even math. There are little aspects in the subject that simply can't be solved. [/hide]", "Solution_15": "Well, it has been established that : 0^0 = 1\r\nIt's juste like : 0! = 1 (where \"!\" = \"factorial\")", "Solution_16": "0^0 is NOT equal to 1. By the quotient of powers, we can consider\r\n\r\n$\\frac{0^n}{0^n}=0^0$. This would work for any base other than 0. But the problem with $\\frac{0^n}{0^n}$ is because you have 0's in both the numerator and the denominator and that's $\\frac{0}{0}$. Which of course is undefined. Therefore 0^0 is undefined.\r\n\r\nFor x+2=0, when you multiply both sides by x-2, you get $x^2-4=0, x^2=4, x=2 or -2$. But one of them is an extraneous solution.\r\n\r\nAll right? 0^0 is NOT equal to 1 or 0 because it's undefined, 0/0 is NOT equal to 0 because it's undefined too, and for the equation, when you introduce a square term you also introduce an extraneous solution most of the time.\r\n\r\nHope that helps.", "Solution_17": "OK, my mistake!\r\n0^0 is undefined, but a^0 = 1 (except when a = 0, duh)", "Solution_18": "[quote=\"riddler\"]sorry if this is dumb but what is $0^0$? and $0/0$? practically I even know it will be $0$ but...\nand one more.... you have this equation....\n$x+2=0$\nMultiply by $x-2$ both sides.... you get a different solution to the question?huh :wallbash_red: this is weird or what? \nI will appreciate if you explain your answer. This question is really annoying :mad:[/quote]\r\n\r\n$\\frac{0}{0}$ is $0$.\r\n\r\n$0^0$ is 1\r\n\r\n$x+2=0$'s answer is -2. x=-2", "Solution_19": "$0^0$ is considered undefined because you can't tell if it is 0 or 1. It can never be defined, and if someone says it can, they are wrong.", "Solution_20": "Same goes with $\\frac{0}{0}$. It has three answers: 1, 0, and undefined.", "Solution_21": "But it can be proved that it is not equal to 0 or 1, so it can only be undefined.", "Solution_22": "hey. this was moved. :D", "Solution_23": "[quote=\"chess64\"]Depending on the context, $0^0$ is considered to be of interdeterminant form, 1, or undefined.[/quote]\r\n\r\nis there a word called interdeterminant?", "Solution_24": "I believe he means [url=http://mathworld.wolfram.com/Indeterminate.html]Indeterminate[/url].", "Solution_25": "thanks. also, i didn't mean to double post. :)", "Solution_26": "Riddler, just to try to answer the question about $x+2=0$...\r\nSuppose you take a polynomial $(x-a_1)(x-a_2)(x-a_3)...(x-a_n)=0$. That's a polynomial of degree n. Now multiply both sides by $(x-2)$. Notice that when $x=2$, $(x-2)=0$. Thus, you've just increased the degree to $n+1$, and you've added the root $2$ by putting that in! Multiplying both sides by something doesn't always give the same equation: you just won't remove any roots. Realize that when you multiply by $(x-2)$, you're implicitly assuming that $x\\neq2$, because if x=2, then you're just multiplying both sides of the equation by 0, and getting no information. Same way, if you multiply both sides by $\\frac{1}{x-2}$, you're implicitly assuming that $x\\neq2$.\r\nAnother way to look at it: whenever you get roots which you had to dramatically change the original equation to get, you MUST check them, because many may not be valid, some may yield identitities orthings that aren't possible.", "Solution_27": "$0^0$ and $\\frac{0}{0}$ are both undefined!!!\r\n\r\nand for x+2=0\r\nthe ans. is -2 ,\r\n\r\nsince you times (x-2) to both sides\r\nthe (x+2)(x-2)=0,\r\nx= 2 or -2\r\n\r\nBut the reality is that x+2=0\r\nso the ans is -2 only. :P", "Solution_28": "Indeterminate forms can only be evaluated through the process of limit.", "Solution_29": "Well, $x^0=1$ according to exponential rules. I have come up with the follwoing:\r\n\r\n$x^0=x^{a-a}=\\frac{x^a}{x^a}$, and since anything divided by itself equals $1$, $x^0=1$.\r\n\r\nSomethings in life are meant to stay undefined for the intelligibility of humans.\r\n\r\nMasoud Zargar", "Solution_30": "[quote=\"boxedexe\"]Well, $x^0=1$ according to exponential rules. I have come up with the follwoing:\n\n$x^0=x^{a-a}=\\frac{x^a}{x^a}$, and since anything divided by itself equals $1$, $x^0=1$.\n\nSomethings in life are meant to stay undefined for the intelligibility of humans.\n\nMasoud Zargar[/quote]\r\n\r\nObjection.\r\n\r\nEverything you've said is true EXCEPT when x=0. 0 divided by 0 is not 1. Then you'd find yourself trying to calculate $\\frac{0^a}{0^a}=\\frac{0}{0}$. On a side note, we'd better hope that $a \\neq 0$.\r\n\r\n0 divided by 0 is not 1. You can make it equal anything with calculus. Or by these means:\r\n\r\nWe see that\r\n0*0=0\r\n0*1=0\r\n0*2=0\r\n0*3=0\r\n0*4=0\r\n0*5=0\r\n0*6=0\r\n..........\r\n\r\nDividing both sides by zero (with a fire extinguisher ready in case your paper bursts into flames), we get\r\n\r\n$\\frac{0}{0}=0$\r\n$\\frac{0}{0}=1$\r\n$\\frac{0}{0}=2$\r\n$\\frac{0}{0}=3$\r\n$\\frac{0}{0}=4$\r\n$\\frac{0}{0}=5$\r\n$\\frac{0}{0}=6$\r\n.........\r\n\r\nSo $\\frac{0}{0}$ is indeterminate, and so is $0^0$ because of the arguments earlier in this thread.", "Solution_31": "0/0 is not undefined, it's indeterminate form. I'm not sure about 0^0 though, I've never run across it before. If 0/0 = 0^0, then I guess it's indeterminate form too, but I'm not sure if they're equal.", "Solution_32": "Ah, didn't see that last post. It seems this was already cleared up, sorry.", "Solution_33": "Tip: In the future if you find that your post is bad, use the delete button if nobody's replied to it. If you find there's something wrong with it, use the edit button. Don't double post. I realized you didn't know, but now you do.", "Solution_34": "[quote=\"mathnerd314\"][quote=\"boxedexe\"]Well, $x^0=1$ according to exponential rules. I have come up with the follwoing:\n\n$x^0=x^{a-a}=\\frac{x^a}{x^a}$, and since anything divided by itself equals $1$, $x^0=1$.\n\nSomethings in life are meant to stay undefined for the intelligibility of humans.\n\nMasoud Zargar[/quote]\n\nObjection.\n\nEverything you've said is true EXCEPT when x=0. 0 divided by 0 is not 1. Then you'd find yourself trying to calculate $\\frac{0^a}{0^a}=\\frac{0}{0}$. On a side note, we'd better hope that $a \\neq 0$.\n\n0 divided by 0 is not 1. You can make it equal anything with calculus. Or by these means:\n\nWe see that\n0*0=0\n0*1=0\n0*2=0\n0*3=0\n0*4=0\n0*5=0\n0*6=0\n..........\n\nDividing both sides by zero (with a fire extinguisher ready in case your paper bursts into flames), we get\n\n$\\frac{0}{0}=0$\n$\\frac{0}{0}=1$\n$\\frac{0}{0}=2$\n$\\frac{0}{0}=3$\n$\\frac{0}{0}=4$\n$\\frac{0}{0}=5$\n$\\frac{0}{0}=6$\n.........\n\nSo $\\frac{0}{0}$ is indeterminate, and so is $0^0$ because of the arguments earlier in this thread.[/quote]\r\n\r\nYes, you're correct. :oops: I forgot to give the conditions.\r\n\r\nMasoud Zargar", "Solution_35": "With all due respect, boxedexe, a great deal of the posts I have seen you make end in something along the lines of \"...hence, the result is incomprehensible to the human mind\". Althought this is true in many cases, just because an idea dosent fit our human intuition dosen't mean it is impossible or undefinable or that it will remained unsolved among future generations. Instead of pulling out the phrase like a rifle give it some thought, as in this case it back-fired. :D", "Solution_36": "[quote=\"G-UNIT\"]With all due respect, boxedexe, a great deal of the posts I have seen you make end in something along the lines of \"...hence, the result is incomprehensible to the human mind\". Althought this is true in many cases, just because an idea dosent fit our human intuition dosen't mean it is impossible or undefinable or that it will remained unsolved among future generations. Instead of pulling out the phrase like a rifle give it some thought, as in this case it back-fired. :D[/quote]\r\n\r\nWell, most of my posts are not among those lines. However, I still agree with you. :) I actually believe that even if something is not the way the human mind understands it, it could still be intelligible. An example would be Quantum physics and its myteries. Next time, I'll try to word my ideas in a way that it is both clear and that my ideas are not misunderstood.:) In this case, I don't believe it backfired; it was just a simple mistake of not giving the conditions. \r\n\r\nMasoud Zargar", "Solution_37": "hey riddler , as for the x-2=0 , let me tell you another way similar to k81o7\r\n\r\nlook, you know $ x-2=0 $\r\n\r\nwhen you multiply by x+2 you get $ x^2-4 = 0 $ \r\n\r\nnote that these 2 expressions are different !when you have multiplied by x+2 , you have actually CHANGED the equation . its no longer the same equation as x-2! hence the answers will obviusly be different \r\n\r\ni think you mean to say that on multiplying both sides of the equation by the same ummmm thing , (here x+2) the equation shouldnt change because the 2 things get cancelled . WHICH IS WRONG because they do change. It will ONLY not change if the 'thing ' is a CONSTANT like 2,3 ,4 , ok ??\r\n\r\nlike if you have n=m+1 doesnt mean that this is same as n^2 = (m+1)n !!", "Solution_38": "[quote=\"eminem\"]like if you have n=m+1 doesnt mean that this is same as n^2 = (m+1)n !![/quote]\r\nYou can do this, you just have to make sure you eliminate the extraneous solution $n=0$ that you gain by multiplying both sides by $n$. (Of course, it is possible that $n=0$ really is a solution, so you have to test for this before eliminating it.)", "Solution_39": "Since you guys have spent two pages on this, I think I'll step in.\r\n\r\nIf by $\\dfrac00$ you mean literally zero divided by literally zero divided by literally zero, then the fraction must remain undefined. There's no way to give it a unique meaning. As someone pointed out, the equation $0\\cdot x=0$ has many solutions, so you can't single out just one.\r\n\r\nOn the other hand, if you mean $\\dfrac00$ as a shorthand for \"the limit of an expression in which both the numerator and denominator tend to zero,\" or less formally, as \"very small divided by very small\", then you have what is referred to in calculus as an indeterminate case - anything can still happen, and we have to do more work to decide the case. We have to ask HOW small over HOW small. \r\n\r\nAll indeterminate cases result from the conflict of tendencies that pull in opposite directions. A numerator that tends to zero would tend to make the fraction go to zero; a denominator that tends to zero would tend to make the fraction go to infinity. Those are conflicting tendencies, and we must do more work to gauge their relative strengths.\r\n\r\nIt's much the same for $0^0.$\r\n\r\nIf we mean literally zero raised to the literally zero power, than we must give up and leave the quantity undefined. If we're faced with a troublesome question of how to define $a^b$ we use $a^b=e^{b\\ln a}.$ Since $\\ln 0$ is not defined, that doesn't help us.\r\n\r\nHOWEVER, there is one exception to the above paragraph, which has to do with representing a polynomial or power series in summation notation. By agreed upon convention, the expression $P(x)=\\sum\\limits_{k=0}^na_kx^k$ has the same identical meaning as $P(x)=a_0+a_1x+a_2x^2+\\cdots a_nx^n.$ Included in this is the notion that $P(0)=a_0.$ But this isn't a definition of $0^0$; it's a notational convention having to do with summation notation.\r\n\r\nIf by $0^0$, we mean a shorthand for a limit in which a base tends to zero, raised to a power that tends to zero, then we have another indeterminate case. Anything can still happen, and we have work to do. Again, there are conflicting tendencies. The power tending to zero is trying to pull this limit in to 1. The base tending to zero is trying to push the limit outwards to either zero or infinity. A good first step for the work needed to sort out the conflicting tendencies is usually to take the logarithm of it.\r\n\r\nSome examples:\r\n\r\n$\\lim\\limits_{x\\to0^+}x^x=1.$\r\n\r\n$\\lim\\limits_{x\\to0^+}x^{\\left(\\dfrac1{\\ln x}\\right)}=e.$\r\n\r\n$\\lim\\limits_{x\\to0^+}\\left(e^{-\\dfrac1x}\\right)^{\\dfrac1{\\ln x}}=\\infty$\r\n\r\nDiscussion of examples like this would belong in \"Calculus Computations and Tutorials.\"" } { "Tag": [ "geometry", "3D geometry", "modular arithmetic", "number theory", "relatively prime", "number theory unsolved" ], "Problem": "find all primes $ p$ and $ q$ such that $ p \\plus{} q \\equal{} (p \\minus{} q)^3$\r\n :D", "Solution_1": "It posted before .You can solve the general \r\n$ m\\plus{}n\\equal{}(m\\minus{}n)^3$\r\nWith your problem solution is \r\n$ (p,q)\\equal{}(5,3)$", "Solution_2": "[hide]\n$ p\\plus{}q$ divides $ (p\\minus{}q)^3$ and $ (p\\plus{}q)^3$, so it divides\n\\[ (p\\plus{}q)^3\\minus{}(p\\minus{}q)^3\\equal{}2q(3p^2\\plus{}q^2)\\]\nSo $ p\\plus{}q$ divides $ 2(3p^2\\plus{}q^2)$ as $ p\\plus{}q$ and $ q$ are relatively prime ($ p\\equal{}q$ clearly doesn't work).\n\nSimilarly, $ p\\plus{}q$ divides $ (p\\plus{}q)^3\\plus{}(p\\minus{}q)^3$ from which we can conclude $ p\\plus{}q$ divides $ 2(p^2\\plus{}3q^2)$.\n\nNow we get that $ p\\plus{}q$ divides $ 2(p^2\\plus{}3q^2)\\plus{}2(3p^2\\plus{}q^2)\\equal{}8(p^2\\plus{}q^2)$ and $ 8((p\\plus{}q)^2\\minus{}p^2\\minus{}q^2)\\equal{}16pq$. Because $ p\\plus{}q$ is relatively prime to $ p$ and $ q$, $ p\\plus{}q$ divides 16. But also $ p\\plus{}q$ is a cube greater than 1. We get $ p\\plus{}q\\equal{}8$. For $ p>q$ this occurs only when $ p\\equal{}5,q\\equal{}3$.\n[/hide]", "Solution_3": "[u]Solution.[/u] Take the equation modulo $ p\\plus{}q$ to obtain $ 0\\equiv (p\\minus{}q)^3\\equiv 8p^3\\pmod{p\\plus{}q}$. Clearly, $ \\gcd(p,q)\\equal{}1\\iff \\gcd(p,p\\plus{}q)\\equal{}1$. Hence, $ p\\plus{}q|8$. Take the equation modulo $ p\\minus{}q$ to obtain $ 0\\equiv p\\plus{}q\\equiv 2p\\pmod{p\\minus{}q}$. Since $ \\gcd(p,p\\minus{}q)\\equal{}1$, $ p\\minus{}q|2$. It is obvious that $ (p,q)\\equal{}(5,3)$. $ \\Box$" } { "Tag": [], "Problem": "\u0397 \u03b9\u03b4\u03b5\u03b1 \u03c4\u03b7\u03c2 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03b7\u03c2-\u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b1\u03b3\u03b7\u03c2 \u03c0\u03c1\u03bf\u03b5\u03ba\u03c5\u03c8\u03b5 \u03b1\u03c0\u03bf \u03bb\u03c5\u03ba\u03b5\u03b9\u03b1\u03ba\u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \r\n\u0395\u03c3\u03c4\u03c9 $ C_1,C_2,C_3$ \u03c4\u03c1\u03b5\u03af\u03c2 \u03bf\u03bc\u03cc\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 \u03ba\u03b5\u03bd\u03c4\u03c1\u03bf\u03c5 $ O$ \u03ba\u03b1\u03b9 $ AA',BB',CC'$ \u03c4\u03c1\u03b5\u03af\u03c2 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c4\u03c1\u03bf\u03b9 \u03c4\u03c9\u03bd $ C_1,C_2,C_3$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03bf\u03b3\u03c1\u03b1\u03bc\u03bc\u03b1 $ OAKB,OALC,OBMC$.\r\n\u0391\u03bd $ OMHA',OLIB',OKUC'$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03cc\u03b3\u03c1\u03b1\u03bc\u03bc\u03b1 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03bf\u03c4\u03b9\r\n$ (OH)+(OU)+(OI)\\geq{(OA)+(OB)+(OC)}$\r\n\r\n(G.Bas. easy)", "Solution_1": "\u0394\u03b5\u03bd \u03c3\u03c5\u03b3\u03ba\u03b9\u03bd\u03b5\u03af \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1\u03bd \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd\r\n\u0391\u03bd\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 \u03c3\u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03c9\u03c3\u03b7 $ |a\\plus{}b\\minus{}c|\\plus{}|a\\minus{}b\\plus{}c|\\plus{}|\\minus{}a\\plus{}b\\plus{}c|\\geq{|a|\\plus{}|b|\\plus{}|c|}$ \u03bf\u03c0\u03bf\u03c5 a,b,c \u03c4\u03b1 \u03b4\u03b9\u03b1\u03bd\u03c5\u03c3\u03bc\u03b1\u03c4\u03b1 \u039f\u0391,\u039f\u0392,\u039fC \u03ae \u03b5\u03b9\u03ba\u03cc\u03bd\u03b5\u03c2 \u03c4\u03c9\u03bd \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03c9\u03bd.........\r\n\r\n\u0393\u03b5\u03b9\u03ac\u03c2 \u03c3\u03b1\u03c2" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Give $a,b \\in Z^{+}$ such as $ab(a+b)\\vdots a^{2}+ab+b^{2}$\r\nProve that :\r\n$\\mid a-b\\mid >\\sqrt[3]{ab}$", "Solution_1": "[quote=\"Redriver\"] $ab(a+b)\\vdots a^{2}+ab+b^{2}$[/quote]\r\n\r\nIs that a \"divides\" sign?", "Solution_2": "The problem is from Russia 2001 and is $a^{2}+b^{2}+ab|ab(a+b)$. Let $d=gcd(a,b)$ and $a=a_{1}d,b=b_{1}d$, then $a_{1}^{2}+a_{1}b_{1}+b_{1}^{2}|da_{1}b_{1}(a_{1}+b_{1})$. Simply we can prove that $gcd(a_{1}b_{1}(a_{1}+b_{1}),a_{1}^{2}+a_{1}b_{1}+b_{1}^{2})=1$. Thus $a^{2}+ab+b^{2}|d^{3}$. So $|a-b|\\geq gcd(a,b)=d\\geq (a^{2}+b^{2}+ab)^{\\frac13}\\geq\\sqrt[3]{3ab}>\\sqrt[3]{ab}$" } { "Tag": [], "Problem": "1--Sydsaeter, Hammond - Mathematical Methods of Economics\r\n2-- RGD Allen, Mathematical Analysis for Economics\r\n\r\n3--R.Dornbusch, S.Fischer - Macroeconomics\r\n4--5Macroeconomics by Makiw\r\n5--Mankiw - Macro + Micro - Study Guide\r\n6--Macroeconomic Theory and Policy by William H. Branson\r\n\r\n\r\n7--microeconomics by Mankiw\r\n8--Intermediate Microeconomics by Hal Varian\r\n9--workouts in intermediate microeconomics by hal varian\r\n10-- Microeconomics Analysis by varian\r\n11--Microeconomics by Pindyck and Rubinfeld\r\n\r\n12--Basic Econometrics by Damodar N Gujarati\r\n13--Essentials of Econometrics by Damodar N Gujarati\r\n14--applied statistics for business and economics: an essential version by allen L.webster\r\n15--Statistics for Business and Economics (with CD-ROM) by David R. Anderson, Dennis J. Sweeney, and Thomas A. Williams\r\n\r\n16--Krugman and Obstfeld (KO), 8th edition (Pearson Low Price Edition)\r\n\r\n\r\n\r\nI would be thankful if someone can provide me with the ebooks and solution manuals of books mentioned above.I am really in need of these books for my exams and therefore one of my friend referred me to this site...i am really hopeful.\r\nmost of solution manual are available in authors or publisher website..but it wont let me download as they require some code or something.\r\nany torrents or link to these books would be highly appreciated.\r\nthank you", "Solution_1": "please someone help me...i urgently need solution manuals..and ebooks..please\r\n\r\nthanks" } { "Tag": [ "modular arithmetic" ], "Problem": "i know how to compute $ \\pmod{n}$, where $ n$ is a large number, but how does one compute $ 4^{10}\\plus{}5^9\\plus{}6^8\\plus{}7^7\\pmod{13}$?\r\n\r\nThanks in advance", "Solution_1": "[u]Strategy:[/u]\r\nTake it term by term. Each term:\r\n- convert smaller bases into larger bases while reducing the exponent.\r\n- convert the larger bases into smaller bases via mod13, which does not raise the exponent\r\nThus, in effect, repeating the above pair of steps over and over will just reduce your exponents until they are very reasonable.\r\n\r\n[hide=\"Specific Solution\"]\n\n$ 4^2 \\equal{} 16 \\equiv 3 \\pmod{13}$\n$ \\implies 4^{10} \\equiv 3^5 \\equiv 3 \\cdot 9^2 \\equiv 3 \\cdot 16 \\equiv 3 \\cdot 3 \\equiv 9 \\pmod{13}$\n\n$ 5^9 \\equiv 5 \\cdot (25)^4 \\equiv 4 \\cdot ( \\minus{} 1)^4 \\equiv 5 \\pmod{13}$\n\n$ 6^8 \\equal{} 36^4 \\equiv ( \\minus{} 3)^4 \\equiv 3^4 \\equiv ( \\minus{} 4)^2 \\equiv 16 \\equiv 3 \\pmod{13}$\n\n$ 7^7 \\equiv ( \\minus{} 6)^7 \\equiv \\minus{} 6 \\cdot (36)^3 \\equiv \\minus{} 6 \\cdot ( \\minus{} 3)^3 \\equiv 6 \\cdot (27) \\equiv 6 \\pmod{13}$\n\nAdding the above, we get $ 4^{10} \\plus{} 5^9 \\plus{} 6^8 \\plus{} 7^7 \\equiv 9 \\plus{} 5 \\plus{} 3 \\plus{} 6 \\equiv 10 \\pmod{13}$[/hide]" } { "Tag": [ "probability" ], "Problem": "A boy has four red marbles and eight blue marbles. He arranges his twelve marbles randomly, in a ring. What is the probability that no two red marbles are adjacent?", "Solution_1": "[hide]Suppose he first puts the 8 blue marbles in a ring. Then, he puts the four red marbles one at a time. The probabilities that the first red marble is not adjacent to any other red marbles in the ring is $1$. The probability for the second is $\\frac{11}{13}$, for the third, $\\frac{10}{14}$, and for the fourth, $\\frac{9}{15}$. So then, the total probability in question is $\\frac{11*10*9}{13*14*15}=\\frac\\boxed{{33}{91}}$[/hide]" } { "Tag": [ "modular arithmetic", "geometry", "3D geometry", "Diophantine Equations" ], "Problem": "Prove that if $n$ is a positive integer such that the equation \\[x^{3}-3xy^{2}+y^{3}=n.\\] has a solution in integers $(x,y),$ then it has at least three such solutions. Show that the equation has no solutions in integers when $n=2891$.", "Solution_1": "Note that if $ (x,y)$ is a solution in inetegers to $ x^3 \\minus{} 3xy^2 \\plus{} y^3 \\equal{} n$ then $ (y \\minus{} x, \\minus{} x)$ and $ ( \\minus{} y,x \\minus{} y)$ are also solutions in integers to $ 3 \\minus{} 3xy^2 \\plus{} y^3 \\equal{} n$, because:\r\n$ (y \\minus{} x)^3 \\minus{} 3(y \\minus{} x)( \\minus{} x)^2 \\plus{} ( \\minus{} x)^3 \\equal{} y^3 \\minus{} 3y^2x \\plus{} 3x^y \\minus{} x^3 \\minus{} 3(x^2y \\plus{} x^3) \\minus{} x^3 \\equal{} x^3 \\minus{} 3xy^2 \\plus{} y^3 \\equal{} n$;\r\n$ ( \\minus{} y)^3 \\minus{} 3( \\minus{} y)(x \\minus{} y)^2 \\plus{} (x \\minus{} y)^3 \\equal{} \\minus{} y^3 \\plus{} 3(x^2y \\minus{} 6y^2x \\plus{} y^3) \\plus{} x^3 \\minus{} 3x^2y \\plus{} 3y^2x \\minus{} y^3 \\equal{} x^3 \\minus{} 3xy^2 \\plus{} y^3 \\equal{} n$.\r\nNow, we can never have $ (x,y) \\equal{} (y \\minus{} x, \\minus{} x)$ because this yields $ x \\equal{} y \\equal{} 0$ and then $ n \\equal{} 0$ which is impossible. The same happens when $ (x,y) \\equal{} (y,x \\minus{} y)$ and $ (y \\minus{} x, \\minus{} x) \\equal{} (y,x \\minus{} y)$. Therefore, if $ (x,y)$ is an integer solution to $ x^3 \\minus{} 3xy^2 \\plus{} y^3 \\equal{} n$ then $ (y \\minus{} x, \\minus{} x)$ and $ (\\minus{}y,x \\minus{} y)$ are two other different solutions in integers to $ 3 \\minus{} 3xy^2 \\plus{} y^3 \\equal{} n$, this proves part one.\r\n\r\nAssume there do exist integers $ x,y$ such that $ x^3 \\minus{} 3xy^2 \\plus{} y^3 \\equal{} 2891$. Now, $ 2891\\equiv 2\\pmod 9$; because any cube of an integer is $ \\minus{} 1,0,$ or $ 1\\pmod 9$ then if $ x\\equiv 0\\pmod 3$ or $ y\\equiv 0\\pmod 3$ then $ x^3 \\minus{} 3xy^2\\equiv 0\\pmod 9$ or $ \\minus{} 3xy^2 \\plus{} y^3\\equiv 0\\pmod 9$, respectively, then $ y^3\\equiv 2\\pmod 9$ or $ x^3\\equiv 2\\pmod 9$, respectively which is impossible. Therefore, we cannot have $ x$ or $ y$ congruent to $ 0\\pmod 3$. Therefore, $ x^3,y^3$ can be conruent to $ \\minus{} 1$ or $ 1\\pmod 9$.\r\nNow, if $ x^3\\equiv \\minus{} 1\\pmod 9, y^3\\equiv 1\\pmod 9$ or $ y^3\\equiv \\minus{} 1\\pmod 9, x^3\\equiv 1\\pmod 9$ then we must have $ \\minus{} 3xy^2\\equiv 2\\pmod 9$ which is impossible as $ 3|( \\minus{} 3xy^2)$.\r\nTherefore, we must have $ x^3\\equiv y^3\\pmod 9$ then $ x^3 \\plus{} y^3\\equiv 2$ or $ \\minus{} 2\\pmod 9$ then $ \\minus{} 3xy^2\\equiv 0$ or $ 4\\pmod 9$. However, $ \\minus{} 3xy^2$ cannot be $ 4\\pmod 9$ as $ 3|( \\minus{} 3xy^2)$. On the other hand, $ \\minus{} 3xy^2$ cannot be $ 0\\pmod 9$ because then $ 3|(xy^2)$ so $ x$ or $ y$ is congruent $ 0\\pmod 3$ which we proved before cannot be possible. Therefore, we get a contradiction and the equation $ x^3 \\minus{} 3xy^2 \\plus{} y^3 \\equal{} 2891$ has no integer solutions QED.", "Solution_2": "Let $\\omega$ be a ninth root of unity. Note that $x^3-3xy^2+y^3=(x-y(\\omega+\\frac{1}{\\omega}))(x-y(\\omega^2+\\frac{1}{\\omega^2}))(x-y(\\omega^4+\\frac{1}{\\omega^4}))$.\nNow note that $(x-y(\\omega^2+\\frac{1}{\\omega^2}))=(x+2y-(\\omega+\\frac{1}{\\omega})^2y)$. Now note that $(\\omega+\\frac{1}{\\omega})(\\omega^2+\\frac{1}{\\omega^2})(\\omega^4+\\frac{1}{\\omega^4})=-1$ and $(\\omega+\\frac{1}{\\omega})(x-y(\\omega^2+\\frac{1}{\\omega^2}))=(-y+(\\omega+\\frac{1}{\\omega})(x-y))$. So using $x^3-3xy^2+y^3=(x-y(\\omega+\\frac{1}{\\omega}))(x-y(\\omega^2+\\frac{1}{\\omega^2}))(x-y(\\omega^4+\\frac{1}{\\omega^4}))$ we get that if $f(x,y)=x^3-3xy^2+y^3$ then $f(x,y)=f(-y,x-y)$ and also $f(x,y)=f(-y,x-y)=f(y-x,-x)$. Hence one solution induces at least three solutions. Now $2891=7^2*59$. Using the $x=t+\\frac{1}{t}$ substitution which makes $x^3-3x+1=t^3+1+\\frac{1}{t^3}$, then cyclotomic polynomials and field theory implies that for any prime divisor $p$ of $x^3-3x+1$, we have $p$ is $9k \\pm 1$ for some $k$. Thus if $7|x^3-3xy^2+y^3$, we see that $7|x,y$. Thus, a solution to $x^3-3xy^2+y^3=2891$ does not exist.\n\n" } { "Tag": [ "function", "LaTeX", "quadratics", "algebra", "quadratic formula" ], "Problem": "Let $ f(x) \\equal{} x^3 \\plus{} 3x \\plus{} 1$, where $ x$ is a real number. Given that the inverse function of $ f$ exists and is given by\r\n\\[ f^{\\minus{}1}(x) \\equal{} \\left( \\frac {x \\minus{} a \\plus{} \\sqrt {x^2 \\minus{} bx \\plus{} c}}{2} \\right)^{1/3} \\plus{} \\left( \\frac {x \\minus{} a \\minus{} \\sqrt {x^2 \\minus{} bx \\plus{} c}}{2} \\right)^{1/3}\r\n\\] where $ a$, $ b$ and $ c$ are positive constants, find the value of $ a \\plus{} 10b \\plus{} 100c$.", "Solution_1": "Eh, I used a really lame brute-force method that is probably wrong...\r\n\r\n[hide]\nSince $ f(f^{ \\minus{} 1}(x)) \\equal{} x$ for all x, let's substitute some convenient values!\nFirst, $ x \\equal{} 0$ gives $ f^{ \\minus{} 1}(0) \\equal{} (\\frac { \\minus{} a \\plus{} \\sqrt {c}}{2})^{1/3} \\plus{} (\\frac { \\minus{} a \\minus{} \\sqrt {c}}{2})^{1/3}$.\nAlso, solving for $ x$ in $ x^3 \\plus{} 3x \\plus{} 1 \\equal{} 0$ using the cubic formula, the above value is also equal to $ (\\frac {( \\minus{} 1) \\plus{} \\sqrt {5}}{2})^{1/3} \\plus{} ...$(messy LaTeX)\nWhich gives $ a \\equal{} 1$ and $ c \\equal{} 5$ since they are both positive.\n\nSimilarly, I found $ b \\equal{} 2$ using the cubic formula with $ f(f^{ \\minus{} 1}( \\minus{} 1)) \\equal{} \\minus{} 1$.\nSo the answer is $ \\boxed{521}$.[/hide]", "Solution_2": "[hide]\nFor the sake of convenience, let $ A \\equal{} x \\minus{} a$, $ B \\equal{} x^2 \\minus{} bx \\plus{} c$, then $ f^{ \\minus{} 1}(x) \\equal{} \\left(\\dfrac{A \\plus{} \\sqrt {B}}{2}\\right)^{1/3} \\plus{} \\left(\\dfrac{A \\minus{} \\sqrt {B}}{2}\\right)^{1/3}$.\n\nLet $ y \\equal{} f(x) \\equal{} x^3 \\plus{} 3x \\plus{} 1$, then $ f^{ \\minus{} 1}(y) \\equal{} x$, so\n\\[ \\left(\\dfrac{A \\plus{} \\sqrt {B}}{2}\\right)^{1/3} \\plus{} \\left(\\dfrac{A \\minus{} \\sqrt {B}}{2}\\right)^{1/3} \\minus{} x \\equal{} 0\n\\]\nTherefore\n\\[ \\dfrac{A \\plus{} \\sqrt {B}}{2} \\plus{} \\dfrac{A \\minus{} \\sqrt {B}}{2} \\minus{} x^3 \\equal{} \\minus{} 3x\\cdot \\left(\\dfrac{(A \\plus{} \\sqrt {B})(A \\minus{} \\sqrt {B})}{4}\\right)^{1/3}\n\\]\nSimplifying\n\\[ A \\minus{} x^3 \\equal{} \\minus{}3x\\cdot \\left(\\dfrac{A^2\\minus{}B}{4}\\right)^{1/3}\n\\]\nSubstituting $ A \\equal{} y \\minus{} a, B \\equal{} y^2 \\minus{} by \\plus{} c, y \\equal{} x^3 \\plus{} x \\plus{} 1$\n\\[ 3x \\plus{} 1 \\minus{} a \\equal{} \\minus{} 3x\\cdot \\left(\\dfrac{(b \\minus{} 2a)(x^3 \\plus{} x \\plus{} 1) \\plus{} a^2 \\minus{} c}{4}\\right)^{1/3}\n\\]\nPlugging in $ x \\equal{} 0$, we get $ a \\equal{} 1$, so\n\\[ x \\equal{} \\minus{} x \\cdot \\left(\\dfrac{(b \\minus{} 2)(x^3 \\plus{} x \\plus{} 1) \\plus{} 1 \\minus{} c}{4}\\right)^{1/3}\n\\]\nDividing both sides by $ x$\n\\[ \\dfrac{(b \\minus{} 2)(x^3 \\plus{} x \\plus{} 1) \\plus{} 1 \\minus{} c}{4} \\equal{} \\minus{} 1\n\\]\nRearranging\n\\[ (b \\minus{} 2)x^3 \\plus{} 3(b \\minus{} 2)x \\plus{} b \\minus{} c \\plus{} 3 \\equal{} 0\n\\]\nHence, $ b \\equal{} 2, c \\equal{} 5$, and our answer is $ 521$.\n[/hide]", "Solution_3": "Solving the equation isn't really brute force.\r\n\r\nSubsitute $ x \\equal{} y \\minus{} \\frac{1}{y}$\r\n\r\nto get $ f(y) \\equal{} y^{6} \\plus{}y^{3} \\minus{} 1$, which can easily be solved with the quadratic formula", "Solution_4": "[quote=\"iniquitus\"]Solving the equation isn't really brute force.\n\nSubsitute $ x \\equal{} y \\minus{} \\frac {1}{y}$\n\nto get $ f(y) \\equal{} y^{6} \\plus{} y^{3} \\minus{} 1$, which can easily be solved with the quadratic formula[/quote]\r\n\r\nIs there any motivation behind this substitution? It clearly gets the job done, but I don't see how you would come up with that directly from the given cubic.", "Solution_5": "Apparently it is the same idea used in the derivation of cubic formula:\r\nhttp://mathworld.wolfram.com/VietasSubstitution.html", "Solution_6": "yes. The cubic equation is actually two simple substitutions.\r\n\r\nFor\r\n\r\n$ x^3 \\plus{} ax^2 \\plus{} bx \\plus{} c \\equal{} 0$ (divide by the leading coefficient),\r\n\r\nsubstitute $ x \\equal{} y \\minus{} \\frac {a}{3}$ to obtain\r\n\r\n$ y^3 \\plus{} dy \\plus{} e \\equal{} 0$ (the depressed cubic), and then substitute $ y \\equal{} z \\minus{} \\frac {d}{3z}$ to obtain a quadratic.", "Solution_7": "The substitution is apparently a common technique when seling with cubics, as it gets rid of the $ x^2$ term. \r\nFor example, \r\nLet $ P(x) \\equal{} x^3\\minus{}3x^2\\plus{}5x$. If $ P(a)\\equal{}1$ and $ P(b)\\equal{}5$, then find $ a\\plus{}b$. \r\nTry a substitution here." } { "Tag": [ "Ross Mathematics Program", "geometry", "geometric transformation", "reflection", "email" ], "Problem": "I have been thinking about what to do this summer, and based on my interests, I have narrowed down my choices to these:\r\n\r\n1. Attend Ross Mathematic Program for the second year as a junior counselor\r\n2. Internship related to mathematics (working with professors, etc.)\r\n3. Work (tutoring, teaching, etc.)\r\n4. Any other suggestions?\r\n\r\nI am also applying to RSI, but I probably won't get in... :)\r\nThe reason this is in the college forum is because I want to know what colleges will think of each of these options. I know I should put my interests and my passion in front of what colleges think, but at this point, all of my options reflect my passion.", "Solution_1": "My opinion is that you should do what you think is the most unique and fruitful opportunity (the best combination of the two).\r\n\r\n- For example, you can always teach or tutor -- however, if it's a particularly special teaching opportunity, then you might consider it.\r\n- If you've already been to Ross, it might not be as fruitful to do that again as it would be to do an internship under a professor, especially if the latter involves research.\r\n- At the same time, if interning for the professor actually just means making his/her research paper look presentable or stapling copies, then it's not so hot.\r\n\r\nMy suggestion is that, to judge the fruitfulness aspect, write down a list of all the skills and experience you hope to gain from each one, and then rank them the set of skills/experience that seems most attractive. Then, take into account which ones you will opportunities you will be able to put off (most likely tutoring/teaching) and decide which one of them seems most attractive in the short run.\r\n\r\nNote that it is [i]always[/i] a good idea to think about the skills and experience you hope to gain from something you plan to engage yourself in, and also to reflect on them seriously when you're done. This way, you'll internalize their value, and be able to express yourself most effectively when college applications come around, and again later when interviews for jobs/internships come around.", "Solution_2": "I'd say, do whatever you think you'll enjoy most... likely, that's the thing you'll be able to talk about most truthfully when it comes time for college apps.", "Solution_3": "So I guess working with a professor would be better than going to Ross again. Yeah, I'll put in plenty of work if I work with a professor, it won't be mainly the professor's work.", "Solution_4": "My question is how did u find a professor to work with? inside connections?", "Solution_5": "Just send emails to professors in the field you are interested in... But a LOT of them say no. Maybe about 90% haha.", "Solution_6": "[quote=\"buzzer11\"]Just send emails to professors in the field you are interested in... But a LOT of them say no. Maybe about 90% haha.[/quote]\r\n\r\nI had emailed about 10 professors, and only 2 had even responded. One rejected me and one welcomed me, which was pretty awesome. I mean, I had expected a lot of rejections, but not a lot of ignorals.", "Solution_7": "Yeah, same with me. About half of my contacts didn't respond.", "Solution_8": "That's nothing to get surprised or demoralized about -- professors are busy people! Some of them are particularly bad with e-mail, and occasionally forget to respond to e-mails from their own students.\r\n\r\nKeep in mind that there's absolutely nothing rude in shooting a follow-up e-mail if you haven't heard back in a week (of course, keep it polite, respectful, and sensitive to their busy schedule)." } { "Tag": [ "algebra", "polynomial", "complex numbers", "calculus", "calculus computations" ], "Problem": "prove :D \r\n\r\n$ \\sum_{k=1}^{6}\\;\\sec^{6}\\;\\frac{k\\pi}{13}\\;=\\;\\boxed{326592}$", "Solution_1": "If we define $ P_m(n) \\equal{} \\sum_{k \\equal{} 1}^n \\sec^{2m} \\left(\\frac {k\\pi}{2n \\plus{} 1}\\right)$ then\r\n\r\n$ P_1(n) \\equal{} 2\\, n(n \\plus{} 1)$\r\n\r\n$ P_2(n) \\equal{} \\frac {2^3}{3} \\, n(n \\plus{} 1)(n^2 \\plus{} n \\plus{} 1)$\r\n\r\n$ P_3(n) \\equal{} \\frac {2^3}{3\\cdot 5} \\, n(n \\plus{} 1)(8n^4 \\plus{} 16n^3 \\plus{} 19n^2 \\plus{} 11n \\plus{} 6)$\r\n\r\n$ P_4(n) \\equal{} \\frac {2^7}{3^2\\cdot 5\\cdot 7}\\, n(n \\plus{} 1)(17n^6 \\plus{} 51n^5 \\plus{} 82n^4 \\plus{} 79n^3 \\plus{} 54n^2 \\plus{} 23n \\plus{} 9)$\r\n\r\nIt looks like $ P_m$ is always a polynomial with $ \\deg(P_m) \\equal{} 2m$ and $ n(n \\plus{} 1)|P_m(n)$ .\r\n\r\nIn particular $ P_3(6) \\equal{} 326592$ .\r\n\r\n\r\nSimilarly we can define $ Q_m(n) \\equal{} \\sum_{k \\equal{} 1}^n \\cot^{2m} \\left(\\frac {k\\pi}{2n \\plus{} 1}\\right)$ .\r\n\r\n$ Q_1(n) \\equal{} \\frac13 \\, n(2n \\minus{} 1)$\r\n\r\n$ Q_2(n) \\equal{} \\frac {1}{3^2\\cdot 5}\\, n(2n \\minus{} 1)(4n^2 \\plus{} 10n \\minus{} 9)$\r\n\r\n$ Q_3(n) \\equal{} \\frac {1}{3^3\\cdot 5\\cdot 7}\\, n(2n \\minus{} 1)(32n^4 \\plus{} 112n^3 \\plus{} 8n^2 \\minus{} 252n \\plus{} 135)$\r\n\r\n$ Q_4(n) \\equal{} \\frac {1}{3^4\\cdot 5^2\\cdot 7}\\, n$ $ (2n \\minus{} 1)(192n^6 \\plus{} 864n^5 \\plus{} 496n^4 \\minus{} 2248n^3 \\minus{} 1388n^2 \\plus{} 3834n \\minus{} 1575)$ .\r\n\r\nHere $ Q_m$ seems to be always a polynomial with $ \\deg(Q_m) \\equal{} 2m$ and $ n(2n \\minus{} 1)|Q_m(n)$ .\r\n\r\n\r\nDoes somebody see a general formula for $ P_m$ and $ Q_m$ ?", "Solution_2": "I also have been looking for the proving of\r\n(k=1..n), SUM[cosec(k*pi/(2n+1))^2)] = 2/3*n*(n+1) \r\n(it can be deduced from Q1 (n) mentioned above)\r\nI have found a proving for this special case [Q1(n)] in the Wikipedia at Basel-Problem ( http://en.wikipedia.org/wiki/Basel_problem )\r\n but I think it's too complicated and hope there exists another way of proving, in which there is no need for using Complex-numbers, Moivre's-, Newton's- (binomial)-, and Viete's- formulas, so there can be a proving which uses only equivalent trigonometric and algebric transformation." } { "Tag": [], "Problem": "What is the sum of the first six positive multiples of 3?", "Solution_1": "The first six positive multiples of 3 are 3,6,9,12,15,18\r\nNoticing that 3+18=6+15=9+12=21, the answer is 3*21=$ \\boxed {63}$" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Given three non-collinear points $A,B,C$, construct a circle with centre $C$ such that the tangents from $A$ and $B$ are parallel.", "Solution_1": "[quote=\"chess64\"]Given three non-collinear points $A,B,C$, construct a circle with centre $C$ such that the tangents from $A$ and $B$ are parallel.[/quote]\r\n\r\nLet $D$ and $E$ be the tangency points, since the tangents are parallel $D,C,E$ would be collinear, $C$ is the centre so $C$ is the middlepoint of $DE$, also $DE$ is perpendicular to both tangents.\r\n\r\n$DC=CE$ would be the circle's radium, now we have the radium, so by compass we can construct the circle.\r\n\r\n :D", "Solution_2": "[hide]Find $M$ the midpoint of $AB$, draw $l_1$ tru $A$, $l_1 \\|l$. Find the distance from $M$ to $l_1=d$. Draw a circle with center $C$ and radius $d$. \nif $C=M$ draw circle with radius $AM$\n [/hide]\r\nAre u interested in constuction problems? wud u join such a thread? Pls reply!", "Solution_3": "[quote=\"delta\"][hide]Find $M$ the midpoint of $AB$, draw $l_{1}$ tru $A$, $l_{1}\\|l$. Find the distance from $M$ to $l_{1}=d$. Draw a circle with center $C$ and radius $d$. \nif $C=M$ draw circle with radius $AM$\n [/hide][/quote]\nOf course you mean that $l$ is the line that joins $M$ and $C$\n\n[quote]Are u interested in constuction problems? wud u join such a thread? Pls reply![/quote]\r\nI love geometry! \r\nMost of all, I love construction problems!! :D :)", "Solution_4": "I have a solution, could someone please tell me if it works pls\n\n[hide = Solution]\nLet $M$ be the midpoint of segment $AB$. Let the line $MC$ be $\\ell$. Now let $\\ell_A$ and $\\ell_B$ be the lines parallel to $\\ell$ passing through points $A$ and $B$, respectively. Let points $P$ and $Q$ be the projections of $C$ in the direction perpendicular to $\\ell$ onto $\\ell_A$ and $\\ell_B$, respectively. The the circle passing through points $P$ and $Q$ with centre $C$ works, as $CP \\perp \\ell_A$ and $CQ \\perp \\ell_B$.\n[/hide]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let $n$ points be given on a circle, and let $nk + 1$ chords between these points be drawn, where $2k+1 < n$. Show that it is possible to select $k+1$ of the chords so that no two of them intersect.", "Solution_1": "Number the points clockwise from $1$ to $n$. Then each chord is a pair $(a, b)$ with $1 \\leq a < b \\leq n$.\n\nWe'll show the problem for $k=1$ first: in this case, we want to prove that there are no $n+1$ chords such that any two of them intersect. Let $(a, b), (c, d)$ be any two chords with $a \\leq c$. Then they intersect iff $c \\leq b \\leq d$. Now, take a $2 \\times n+1$ board; write a chord on each column with the smaller number on top. Order the chords by smaller first element, from left to right (if $a \\leq c$, then $(a, b)$ goes first), and order those with equal first element by smaller second element. Now note that no number on the second row can be smaller than a number on the first row (as otherwise the corresponding chords would be disjoint). Also, for two chords with different 1st element, the one on the right must have greater or equal 2nd element (as otherwise the second would be contained in the first, and they wouldn't intersect). So the second row is ordered as well. Let the first row be $a_1, \\ldots, a_{n+1}$ and the second be $b_1, \\ldots, b_{n+1}$. We know that $a_1 \\leq \\ldots \\leq a_{n+1} \\leq b_1 \\leq \\ldots \\leq b_{n+1}$. Consider $b_{n+1}-a_1$. Clearly this number is $>0$ and $\\leq n-1$. On the other hand, each chord has greater sum of elements than the one to the left, so that $a_{i+1}+b_{i+1}-a_i-b_i \\geq 1$ for $i=1, \\ldots, n$.\n\nNow consider the sum of the terms $a_{i+1}+b_{i+1}-a_i-b_i$ for $i=1, \\ldots, n$. It is at least $n$, but it telescopes to $a_{n+1}+b_{n+1}-a_1-b_1 \\leq b_{n+1}-a_1 2$ be given, and starting $ a_0 \\equal{} 1, a_1 \\equal{} a$ define recursively:\r\n\r\n\\[ a_{n\\plus{}1} \\equal{} \\left(\\frac{a^2_n}{a^2_{n\\minus{}1}} \\minus{} 2 \\right) \\cdot a_n.\\]\r\n\r\nShow that for all integers $ k > 0,$ we have: $ \\sum^k_{i \\equal{} 0} \\frac{1}{a_i} < \\frac12 \\cdot (2 \\plus{} a \\minus{} \\sqrt{a^2\\minus{}4}).$", "Solution_1": "[quote=\"\"ISL questions and solutions from 1995-2001\" by A.Ajorloo\"]\nnote that $ a > 2$,so there exists a positive real $ b$ such that $ a \\equal{} b \\plus{} \\frac 1b$ hence:\n\n$ a^2 \\equal{} b^2 \\plus{} \\frac 1{b^2} \\plus{} 2\\Rightarrow a^2 \\minus{} 2 \\equal{} b^2 \\plus{} \\frac 1{b^2}$\n\nso we have:\n\n$ a_2 \\equal{} a(a^2 \\minus{} 2) \\equal{} \\left(b^2 \\plus{} \\frac 1{b^2}\\right)\\left(b \\plus{} \\frac 1b\\right)$\n\n$ a_3 \\equal{} \\left((\\frac {a_2}{a_1})^2 \\minus{} 2\\right)a_2 \\equal{} \\left((b^2 \\plus{} \\frac 1{b^2})^2 \\minus{} 2\\right)a_2 \\equal{} \\left(b^4 \\plus{} \\frac 1{b^4}\\right)\\left(b^2 \\plus{} \\frac 1{b^2}\\right)\\left(b \\plus{} \\frac 1b\\right)$\n\ncontinuing this,we get that:\n\n$ a_n \\equal{} \\left(b^{2^n \\minus{} 1} \\plus{} \\frac 1{b^{2^n \\minus{} 1}}\\right)\\cdots \\left(b^2 \\plus{} \\frac 1{b^2}\\right)\\left(b \\plus{} \\frac 1b\\right)$\n\nso we have:\n\n$ \\sum_{i \\equal{} 0}^n\\frac 1{a_i} \\equal{} 1 \\plus{} \\frac {b}{b^2 \\plus{} 1} \\plus{} \\frac {b^3}{(b^2 \\plus{} 1)(b^4 \\plus{} 1)} \\plus{} \\ldots \\plus{} \\frac {b^{2^n \\minus{} 1}}{(b^2 \\plus{} 1)(b^4 \\plus{} 1)\\cdots (b^{2^n} \\plus{} 1)}$\n\non the other hand we have:\n\n$ \\frac 12\\left(a \\plus{} 2 \\minus{} \\sqrt {a^2 \\minus{} 4}\\right) \\equal{} \\frac 12\\left(b \\plus{} \\frac 1b \\plus{} 2 \\minus{} (b \\minus{} \\frac 1b)\\right) \\equal{} 1 \\plus{} \\frac 1b$\n\nso we have to show that:\n\n$ \\frac {b^2}{1 \\plus{} b^2} \\plus{} \\frac {b^4}{(1 \\plus{} b^2)(1 \\plus{} b^4)} \\plus{} \\ldots \\plus{} \\frac {b^{2^n}}{(1 \\plus{} b^2)(1 \\plus{} b^4)\\cdots (1 \\plus{} b^{2^n})} < 1$\n\nnow note that:\n\n$ \\frac {b^{2^k}}{(1 \\plus{} b^2)\\cdots (1 \\plus{} b^{2^k})} \\equal{} \\frac 1{(1 \\plus{} b^2)\\cdots (1 \\plus{} b^{2^k \\minus{} 1})} \\minus{} \\frac 1{(1 \\plus{} b^2)\\cdots (1 \\plus{} b^{2^k})}$\n\nnow summing up this equation for $ 1\\leq k\\leq n$ we get that:\n\n$ \\frac {b^2}{1 \\plus{} b^2} \\plus{} \\ldots \\plus{} \\frac {b^{2^n}}{(1 \\plus{} b^2)\\cdots (1 \\plus{} b^{2^n})} \\equal{} 1 \\minus{} \\frac 1{(1 \\plus{} b^2)\\cdots (1 \\plus{} b^{2^n})} < 1$\n\nQED[/quote]", "Solution_2": "Let $S(a)$ be the sum of the LHS of the inequaltiy when $a_1=a$ when $k$ goes to infinity.\n\nThus, $S(a)=\\frac{1}{1}+\\frac{1}{a}+\\frac{1}{ (a^2-2)a }+\\frac{1}{((a^2-2)^2-2)(a^2-2)a}+\\cdots$. We have that $a(S(a)-1)=S(a^2-2)$.\n\nLet $S(a)=f(a)+\\frac{2+a-\\sqrt{a^2-4}}{2}$. Subsituting and simplifying: $af(a)=f(a^2-2)$.\n\nIt is clear to see that $1\\le S(a)\\le 2$ (since $a>2$). Thus, if $f(a)=\\epsilon \\neq 0$, then there exists $x$ such that $|f(x)|>2$, but that would imply that $S(a)=f(a)+\\frac{2+a-\\sqrt{a^2-4}}{2}$ is outside of the range of $[1,2]$, contradiction. Thus, $f(a)\\equiv 0$ and $S(a)=\\frac{2+a-\\sqrt{a^2-4}}{2}$.\n\nThus, all the partial sums are less than $S(a)=\\frac{2+a-\\sqrt{a^2-4}}{2}$, as desired.", "Solution_3": "The solution as given in Bhabak Ghalebi's post is hard to realize(probably too hard....its not an usual transormation).Are we to have no easier solutions???" } { "Tag": [ "MIT", "college", "Princeton", "Harvard" ], "Problem": "How can an international applicant applying for MIT, Princeton, Harvard etc. stand out? Australia doesn't really have a 'good' high school qualification to be honest. Will things like an Olympiad medal/certificate help? I have read about the requirements of some University but don't quite get them.", "Solution_1": "International applicants stand out the same way domestic applicants do. \r\n\r\nThat having been said, international admissions at selective colleges is [b]very[/b] competitive; at MIT, for example, the admission rate is about 5%. Even if you are extremely well-qualified, it is not easy to get accepted.", "Solution_2": "In my mind, instead of applying to anywhere he weakly hopes to be accepted one is at the state to choose applying to more logical place to be able to enter.\r\nFor example, Harvard is the exact strongest university all over the world for its intellectual capacities and international qualities, but i think number of who choose MIT is bigger than number of who choose Harvard.", "Solution_3": "What does the [url=http://www.amt.canberra.edu.au/]Australian Mathematics Trust[/url] recommend to strong high school students in Australia for getting ready for challenging university courses?", "Solution_4": "What do you mean Tokenadult?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "A graph has 120 points. A weak quartet is a set of four points with just one edge. What is the maximum possible number of weak quartets?", "Solution_1": "Are you finding a solution? It is here http://www.mathlinks.ro/Forum/viewtopic.php?t=15588 .", "Solution_2": "I had that document but the solution is too long, I don't understand because my mothertongue is English. So, who can solve it in a shorter solution? Thanks a lot :)", "Solution_3": "It seems that your mothertongue is [i]not[/i] English, but before you tell us what exactly you don't understand we cannot help you. Please ask any questions in the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=17341 .\r\n\r\nNote that this is one of the hardest combinatorics problems of IMO Shortlists, so there shouldn't be a shorter solution.\r\n\r\n darij" } { "Tag": [ "geometry", "geometric transformation", "homothety", "geometry unsolved" ], "Problem": "1.In an acute angled $ \\triangle ABC$ points $ D,E,F$ are located on the sides $ BC,CA,AB$ respectively,such that \r\n$ \\frac{CD}{CE}\\equal{}\\frac{CA}{CB}$,\r\n\r\n$ \\frac{AE}{AF}\\equal{}\\frac{AB}{AC}$,\r\n\r\n$ \\frac{BF}{BD}\\equal{}\\frac{BC}{BA}$.\r\n\r\nProve that $ AD,BE,CF$ are altitudes of $ \\triangle ABC$.\r\n\r\n\r\n2.The circumference of a circle is divided into eight areas by a convex quadrilateral $ ABCD$ with four arcs lying inside the quadrilateral and the remaining four lying outside it.The lengths of the arcs lying inside the quadrilateral are denoted by $ p,q,r,s$ in anticlockwise direction starting from some arc.\r\nSuppose $ p\\plus{}r\\equal{}q\\plus{}s$.\r\nProve that $ ABCD$ is a cyclic quadrilateral.\r\n\r\n\r\n3.Let $ BE$ and $ CF$ be the altitudes of an acute angled $ \\triangle ABC$,with $ E$ on $ AC$ and $ F$ on $ AB$.Let $ O$ be the point of intersection of $ BE$ and $ CF$.Take any line $ KL$ through $ O$ with $ K$ on $ AB$ and $ L$ on $ AC$.Suppose $ M$ and $ N$ are located on $ BE$ and $ CF$ respectively such that $ KM$ is perpendicular $ BE$ and $ LN$ is perpendicular to $ CF$.Prove that $ FM$ is parallel to $ EN$.", "Solution_1": "1. Let $ AD',BE',CF'$ be the altitudes of $ \\triangle ABC$. Then $ D'E'\\parallel DE,E'F'\\parallel EF,F'D'\\parallel FD$. This is only possible if $ D'\\equal{}D,E'\\equal{}E,F'\\equal{}F$. Q.E.D.", "Solution_2": "[b]2[/b] [geogebra]92cc838cc92b6cc6fbe776f238cae2f6e53965a3[/geogebra] \r\nFor the denomination of points see the picture. Then, by the assumption $ 2K_DK_B\\plus{}2L_BL_D\\equal{}L_CK_A\\plus{}K_DK_B\\plus{}L_AK_C\\plus{}L_BL_D$ \r\n(by $ XY$ we denote the half of the degree mesaure of arc $ \\smallsmile XY$). So $ \\angle ABC\\plus{}\\angle CDA\\equal{}(K_DK_B\\plus{}K_BL_A\\plus{}L_AK_C\\plus{}K_CL_B\\plus{}L_BL_D\\minus{}K_AL_C)\\plus{}(L_BL_D\\plus{}L_DL_C\\plus{}L_CK_A\\plus{}K_AK_D\\plus{}K_DK_B\\minus{}L_AK_C)\\equal{}2K_DK_B\\plus{}2L_BL_D\\plus{}K_BL_A\\plus{}K_CL_B\\plus{}L_DL_C\\plus{}L_CK_A\\plus{}K_AK_D\\equal{}K_AK_D\\plus{}K_DK_B\\plus{}K_BL_A\\plus{}L_AK_C\\plus{}K_CL_B\\plus{}L_BL_D\\plus{}L_DL_C\\plus{}L_CK_A\\equal{}\\pi.$ Thus $ ABCD$ is cyclic.", "Solution_3": "[b]3[/b]\r\n$ KM \\bot BE, BE \\bot AC$ so $ KM \\| EL$. Thus $ \\triangle KMO \\sim \\triangle LEO$ with the coefficient $ \\frac{KO}{OL}$. Analogiously $ \\triangle OFK \\sim \\triangle ONL$ with the same coefficient. So there exists a homothety with the center $ O$, which transforms $ \\triangle MKF$ into $ \\triangle NLE$. Thus $ FM \\| EN$ Q.E.D.", "Solution_4": "1\nThe first given equality is equivalent to $CD\\cdot BC=CE\\cdot AC$ $\\implies$ $\\triangle ACD \\sim \\triangle BCE$, hence $\\widehat {ADC} = \\widehat {BEC}$ $(1)$.\nSimilarly, from $BD\\cdot BC=BF\\cdot BA$ we get $\\widehat{ADB}=\\widehat{BFC}$ $(2)$.\nAs $\\widehat{ADB}+\\widehat{ADC}=180^\\circ$ , we get $\\widehat{BFC}+\\widehat{BEC}=180^\\circ$ $(3)$.\nFrom $AF\\cdot AB=AE\\cdot AC$ the quad $BCEF $ is cyclic, hence $\\widehat{BFC}=\\widehat{BEC}$ $(4)$, from the relations $(3)$ and $(4)$ getting $BE\\perp AC$ and $CF\\perp AB$, done\n\nBest regards,\nsunken rock", "Solution_5": "3\nFrom $AFOE$, $FKMO$ and $ONLE$ cyclic, we get $\\widehat{BFM}+\\widehat{CEN}=\\hat {A}$, done.\n\nBest regards,\nsunken rock" } { "Tag": [ "geometry", "cyclic quadrilateral", "geometry unsolved" ], "Problem": "Let $ABCD$ be a cyclic quadrilateral of area 8. If there exists a point $O$ in the plane of the quadrilateral such that $OA+OB+OC+OD = 8$, prove that $ABCD$ is an isosceles trapezoid.", "Solution_1": "We have $OA+OB+OC+OD \\geq AC+BD \\geq 2 \\sqrt{AC.BD} \\geq 2 \\sqrt{2S} = 8$.\r\n\r\nThe equality happens when $AC = BD$ and $AC$ is perpendicular to $BD$. Hence $ABCD$ is an isosceles trapezoid.", "Solution_2": "Quite a contrived problem - any such convex quadrilateral will have equal-length, perpendicular diagonals, with only possible positioning for point $O$ at their meeting point ... whence the trivial conclusion - fortunately the other problems in the set were slightly more challenging :)", "Solution_3": "OA+OB+OC+OD>=AC+BD>=2\u221a(AC*BD)>=2*\u221a(2*S)=2*\u221a(2*8)=2*4=8 equality for AC=BD and AC\u27d8BD and how ABCD is quadrilateral resulting conclusion." } { "Tag": [ "AoPSwiki", "number theory", "modular arithmetic" ], "Problem": "Apparently, most schools do not teach modular arithmetic, but it is required for most math competitions.\r\n\r\nCan somebody please post a link to a website that explains (in-depth) modular arithmetic, from the basics to the more advanced?\r\n\r\nThanks.", "Solution_1": "Beginning and intermediate number theory from AoPSWiki:\r\n[url]http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction[/url]\r\n[url]http://www.artofproblemsolving.com/Wiki/index.php/Intermediate_modular_arithmetic[/url]\r\nThis pdf by Naoki Sato has a section (#4) on modular arithmetic:\r\n[url]http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf[/url]" } { "Tag": [ "factorial", "number theory", "least common multiple", "prime numbers", "prime factorization" ], "Problem": "well im new here and im kind of lost. :? . well anyway someone have told me that i can ask a question throught here so here it is:\r\n\r\nFind a number that can be devided by 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 :D \r\n\r\nIt was really hard so......i thought you could help me. thank you. ;)", "Solution_1": "[hide]umm all u really have to do is find a number with factors of 1,2,3,4,5,6,7,8,9, and 10. So you can just multiply them all together and you get 3628800. :D [/hide]", "Solution_2": "If instead, you wish to find the [i]smallest[/i] such number, you would first prime factorize all the numbers:\r\n\r\n[hide](don't need 1 because everything's divisible by 1)\n$2=2$\n$3=3$\n$4=2^2$\n$5=5$\n$6=2\\cdot3$\n$7=7$\n$8=2^3$\n$9=3^2$\n$10=2\\cdot5$\n\nThen, we see that the only prime numbers used was 2, 3, 5, and 7.\n\nWe find the number that has the greatest exponent for each of them, namely, $2^3,3^2,5,7$\n\nMultiply them together to get $2520$ as the smallest possible such number.[/hide]", "Solution_3": "Well...... danfvh i think im finding for the LEAST NUMBER that can be.......u know. Well anyway i think Zyloch gave the best information for me to do the problem. :lol: But i still dont much get it. :( Sorry im just a 7th grader. Could somebody explain it better??\r\n :D", "Solution_4": "well i think i get it now my father explained it to me a while ago ( after i read Zyloch's message) :lol: He's good in math you know :first:", "Solution_5": "It's a matter of finding the least common multiple. So, I'll give a simple example.\r\n\r\nWhat is the smallest integer that can be divided evenly by 6 and 8?\r\n\r\nThe answer is 24. That shouldn't be too hard, just test it yourself.\r\n\r\nNow, notice 24 = 2 * 2 * 2 * 3. And 6 = 2 * 3. And 8 = 2 * 2 * 2. If we multiple 6 * 8, we are multiplying (2 * 3) * (2 * 2 * 2), by substitution. Follow?\r\n\r\nBut we don't need that 2 in the prime factorization of 6 - factoring a number into the primes which when multiplied together gives that number - since the 2's in the 8 supply plenty of 2 factors. So instead, we have 3 * 2 * 2 * 2 as the smallest integer that evenly divides 6 and 8.\r\n\r\nSame with 4 and 6. 4 = 2 * 2. 6 = 2 * 3. The number only needs to be divisible by 2 * 3 and 2 * 2, so how about 2 * 2 * 3? That works. Your answer is 2 * 2 * 3 = 12.\r\n\r\nSo with 10!, which is 10 factorial, equal to 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, we factor each of the numbers involved.\r\n\r\nNow follow Zyloch's reasoning about the prime factorization of the LCM, least common multiple of those numbers, and you get 2520.", "Solution_6": "um.......uh........i think i get it know. :wacko: thanks! in anyway! :)", "Solution_7": "If you're a 7th grader, you might want to go take a look in the middle school forums.", "Solution_8": "actually JBL here in the philippines im just 1st years highschool! begginer in highschool stuffs like that! in our country 1st year highschools are BEGGINER!!!!! :mad:", "Solution_9": "No offense was intended, certainly. I think there must be confusion about our grade numbering systems. In the United States, 7th grade is the next-to-last year of middle school, typically serving students 11 to 13 years old. High school is from grades 9 to 12 (ages approximately 14 to 18). Is the numbering system different in the Phillipines?", "Solution_10": "yes!!!! very much different highschool levels starts with first year approximately 12-14 years old and ends at forth year 15-17 years old\r\n\r\nim just 13 years old so im just a 1st year highschool here in the philippines and a 7th grader in the united states\r\n\r\nohhhhhhhhhhh!!!\r\n\r\nSo confusing :wacko:" } { "Tag": [ "geometry", "circumcircle", "incenter", "cyclic quadrilateral", "geometry proposed" ], "Problem": "Given cyclic quadrilateral $ ABCD$. Four circles each touching its diagonals and the circumcircle internally are equal. Is $ ABCD$ a square?\r\n\r\n[i]Author: C.Pohoata, A.Zaslavsky[/i]", "Solution_1": "Let $ I_a, I_b, I_c, I_d$ be the incenters of triangles $ BCD, ACD, ABD, ABC$. Applying [b]Sawayama-Thebault's[/b] theorem we have $ I_a$ lies on $ O_2O_3$, i.e $ I_b, I_c, I_d$. But $ R_{O_1}\\equal{}R_{O_2}\\equal{}R_{O_3}\\equal{}R_{O_4}$ then $ R_{I_a} \\equal{} R_{I_b} \\equal{} R_{I_c} \\equal{} R_{I_d}$\r\nThis result is well-known: $ I_aI_bI_cI_d$ is a rectangle. But $ I_aI_b//CD$, i.e then $ ABCD$ is a rectangle. Moreover, 4 radius of $ (O_1), (O_2), (O_3), (O_4)$ are equal thus we can easy show that $ ABCD$ is a square." } { "Tag": [ "geometry", "trapezoid" ], "Problem": "Let $ ABCD$ be a trapezoid inscribed in a circle C of diameter AB, with AB and CD parallel, AB=2CD. Let Q be the point where AD meets BC and P the intersection of the tangents to C at B,D. Calculate the area of ABPQ in terms of the area of PDQ.", "Solution_1": "You've got 2002 and 2006 in Resources. http://www.mathlinks.ro/viewtopic.php?p=1313573#1313573" } { "Tag": [ "articles" ], "Problem": "Please post your opinion:\r\n\r\n(Translated by AltaVista)\r\n\r\nThe kidnapped young person identifies itself with his detector Natascha Kampusch, that now is 18 years old, remained captive from the 10 in the cellar of a house in the outskirts of Vienna. His kidnapper, an electrician of 44 years killed itself throwing itself underneath a train. According to the experts, the young sample symptoms of the \"syndrome of Stockholm\". Austria does not leave the astonishment and the commotion by the case of Natascha Kampusch, a young person who had been kidnapped with 10 years of age and reappeared yesterday, eight years later, with 18. According to one inquired, after the appearance of young the his detector was killed throwing itself under the routes of the train, to the north of Vienna. On the other hand, the psychologists who attend the Natascha detailed that he shows symptoms of the \"syndrome of Stockholm\", a phenomenon that usually is observed in victims of kidnappings and that imply the development of affection and attachment with the detector. Natascha had disappeared in March of 1998. A companion hers related then that boarded and was put in a light truck to the force by a stranger when it went to the school. She never knew herself more of her. Until now. According to one inquired, the young person she escaped yesterday and she managed to hide in the garden of a house in Strasshof, to the north of Vienna, near the house that became its jail during eight years. She was found by a woman who alerted to the police, after which the pale Natascha told him who was and that its adolescence catched in a cellar had lived all. \"I am Natascha Kampusch\", said to him to the authorities; and in addition, it revealed that his kidnapper, identified like Wolfgang Priklopil, a technician of electricity of 44 years, had left towards Vienna in his BMW 850. Shortly after one knew that the man had killed itself throwing itself under the routes of a train. Last night the police registered the house and found the hiding place. It had three meters in length, 1.6 meters wide and 2 of depth, constructed from the grave of an accessible garage and by a hollow of 50 by 50 centimeters that were closed with an electronic system. The hiding place had been transformed into a species of dormitory equipped for the young person. There was a bed and a small bookcase with infantile books, one radio and a television set. In that reduced space, Natascha lived at least eight years, although lately it could have left some times. The first declarations of the young person threw that it was not always locked up, but that in the last years his kidnapper allowed him to accompany it to the supermarket, and even, in some trip of vacations or strolls by the garden of the house. Nevertheless, Natascha seems to have undergone strong pressures or threats so that it is not contacted with strangers. Everything aims at that his kidnapper was practically the only person with whom he maintained relation in these eight years. The parents of the young person recognized it in a meeting that made cry to the victim. In a letter directed to the press and spread today by average Austrians, the parents requested \"patience\", and three days of tranquillity. For that reason, one hopes that the details of the case begin to know themselves slowly. In as much, the victim was under the supervision of psychologists of the police, who have adopted a strategy of extreme caution in their interrogations. The drama of Natascha would have arrived at its aim, although parecec\u00eda to be to the doors of another one.", "Solution_1": "interesting... bad grammar by AltaVista though", "Solution_2": "just to let you know, AltaVista isn't a user on some random forum, it's an online translating device :p", "Solution_3": "[url=http://www.cnn.com/2006/WORLD/europe/08/25/austria.kidnap/index.html]CNN[/url] and BBC (which I apparently can't get access to right now) offer articles in English.", "Solution_4": "The story is amazing. What I would like to know is \"WHY\"... why did he, the kidnapper, do what he did. Maybe he wanted to have someone that can love him (that what he thought)", "Solution_5": "[url=http://news.bbc.co.uk/2/hi/europe/5285290.stm]BBC article in English[/url]\r\n\r\nWell, I think it's the same as Belgium's Dutroux : paedophilia.\r\n[url=http://news.bbc.co.uk/2/hi/europe/3829075.stm]BBC article on Dutroux[/url]\r\n\r\nThis girl in Austria caused quite a stir because Europe (and lots of countries in the world) were still shocked about the Dutroux case.", "Solution_6": "well jose I don't kwow what kind of love is this.pure Natasha", "Solution_7": "[quote=\"kostas\"]well jose I don't kwow what kind of love is this.pure Natasha[/quote]\r\n :huh: pure Natasha??", "Solution_8": "[quote=\"fredbel6\"][quote=\"kostas\"]well jose I don't kwow what kind of love is this.pure Natasha[/quote]\n :huh: pure Natasha??[/quote]\r\n\r\nMaybe \"poor Natascha\"", "Solution_9": "[quote=\"fredbel6\"]Well, I think it's the same as Belgium's Dutroux : paedophilia.[/quote]\r\nIs this a good comparision? Of course, this was a kidnapping, but Dutroux kidnapped 6 kids and killed 4 of them, no? And this one's also different because of the Stockholm-syndrom-stuff. Really, I think the Dutroux-story was on a completely different level.", "Solution_10": "[quote=\"Kurt G\u00f6del\"][quote=\"fredbel6\"]Well, I think it's the same as Belgium's Dutroux : paedophilia.[/quote]\nIs this a good comparision? Of course, this was a kidnapping, but Dutroux kidnapped 6 kids and killed 4 of them, no? And this one's also different because of the Stockholm-syndrom-stuff. Really, I think the Dutroux-story was on a completely different level.[/quote]\r\nOf course!\r\n\r\nBut we can say that now, but at that time in Austria in 1998, those people didn't know that.", "Solution_11": "Is it just me, or is crime in the USA much more horrific than in other countries?", "Solution_12": "[quote=\"Xevarion\"]Is it just me, or is crime in the USA much more horrific than in other countries?[/quote]\r\nYou mean perception or the crime itself?", "Solution_13": "[color=white]No, fred, he means that the kidnapping of an American person is much more horrifying than the kidnapping of some Polish lady. The Holocaust should've wiped them all out, he's cleaning up the stragglers. That's what he's trying to say, fred, that's what he's trying to say.[/color]\r\n\r\n\r\nForesight tells me I should either delete the above or post a big [color=red]I'M BEING SARCASTIC, THIS IS WHITE HUMOR, HE MEANS THAT CRIME IN AMERICA IS THOUGHT OF MUCH MORE HORRIFIC-LY THAN IT IS IN OTHER COUNTRIES[/color] notice below my post. That or white it out.", "Solution_14": "I can't tell if this is crude spam or some kind of subtle jab. \r\n\r\nYou're frick'in awesome when you do that.", "Solution_15": "Actually, what I meant was that there are crimes that were so much more violent, cruel, disturbing, or whatever in America that I am mildly surprised by the idea of there being a large amount of national interest in something as \"tame\" as kidnapping a girl for 8 years. (I recall at least two crimes like that in America, one of which was resolved relatively recently. The other was a girl called Elizabeth Smart, I think.)\r\n\r\nedit: I should specify that I may be much more jaded than the average US citizen, and I have very little capacity for empathy. So it might just be me. On the other hand, the USA has way higher rate of murders per capita than most other countries in the world....", "Solution_16": "[quote=\"PenguinIntegral\"]I can't tell if this is crude spam or some kind of subtle jab. \n\nYou're frick'in awesome when you do that.[/quote]\r\n\r\ncrude spam\r\n\r\nman you're subtle too", "Solution_17": "i'm not sure if high-profile crimes tend to be worse in the u.s.\r\nthere is some pretty cruel and weird stuff that goes on in europe and other parts of the world.\r\nconsider this example. the movie hostel was based off of some underground hotel in thailand (i think) that advertised torturing people for an hourly rate.\r\nof course thats just one example, but i do not think that high-profile crimes in the u.s. are necessarilly worse than those in other parts of the world.", "Solution_18": "yeah... they're just covered more\r\n\r\nlike take that terry schiavo thing, in most any other country they would've let her go and made no big deal about it... here? people get upset about letting a vegetable die...", "Solution_19": "[quote=\"Treething\"]yeah... they're just covered more\n\nlike take that terry schiavo thing, in most any other country they would've let her go and made no big deal about it... here? people get upset about letting a vegetable die...[/quote]\r\nWell, at least the USA is not forcing abortions on people with too many kids. But yes, Terry Schiavo would have received euthanasia. I never got the moral ethics. It's not okay to let someone fall asleep forever peacefully, but it's okay to let them die of thirst? Do it and do the job well or don't do it.\r\n\r\nThe worst things happen in Africa, there you have children who are abducted when they're 10, and force to serve as sex slaves for rebel leaders and fight (even kill their own parents). Very notorious for this is the Lord's Resistance Army in Uganda. It is often said that nobody cares about this due to Missing White Woman Syndrome.", "Solution_20": "I am thinking of places where the government is stable enough that these kinds of crimes are due to a few individuals, not to the army or to large organizations like drug cartels. \r\n\r\nI recall some incident in Germany where two guys met in an online chat room or something and met so taht one of them could eat the other... the Rammstein song Mein Teil is based on that event. \r\n\r\nBut does Europe have many people like Ed Gein, Jeffrey Dahmer, the Boston Strangler, the BTK dude (Suicide Commando has an album with that title; some of the songs are rather disturbing), Ted Bundy, Gacy, ...\r\nMaybe it is just b/c in the USA, only USA crimes make big news most of the time, but the only one outside of US that I can think of is Jack the Ripper.", "Solution_21": "[quote=\"Xevarion\"]I am thinking of places where the government is stable enough that these kinds of crimes are due to a few individuals, not to the army or to large organizations like drug cartels. \n\nI recall some incident in Germany where two guys met in an online chat room or something and met so taht one of them could eat the other... the Rammstein song Mein Teil is based on that event. \n\nBut does Europe have many people like Ed Gein, Jeffrey Dahmer, the Boston Strangler, the BTK dude (Suicide Commando has an album with that title; some of the songs are rather disturbing), Ted Bundy, Gacy, ...\nMaybe it is just b/c in the USA, only USA crimes make big news most of the time, but the only one outside of US that I can think of is Jack the Ripper.[/quote]\r\nYou've been brought up in the US, of course you're going to remember mainly famous US crimes. \r\n\r\n\r\nThere's definately a lot of criminals in every country.", "Solution_22": "Actually I usually read BBC news when I read news at all, but that's not going to help me learn about crimes in the past.", "Solution_23": "And what about those boys that went into their school and killed lots of people (in Germany)??? that was terrible too", "Solution_24": "USA does it better, USA does it first. \r\nhttp://en.wikipedia.org/wiki/School_massacre#Infamous_school_massacres\r\n\r\nActually, the most deadly one on that list is the Russia incident from 2 yrs ago, but that wasn't done by students." } { "Tag": [ "calculus", "real analysis", "real analysis theorems" ], "Problem": "Has anybody used [i]Multivariable Calculus [/i]by Theodore Shifrin. If so, what did you think about the book?\r\n\r\nNot sure if this is in the right forum (the calculus forum is centered more on analysis, which this book does not cover)", "Solution_1": "Well, I think it does belong in calculus/analysis, as an \"other question.\" Personally, I'm not familiar with the book." } { "Tag": [ "floor function", "combinatorial geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "how many convex 4-gons can be formed if we are given n non-colinear points in 2D?", "Solution_1": "Well, if the points are in convex position (e.g., along a circle) then the answer is just $ \\binom{n}{4}$, but of course the answer can be smaller. With five points in general position, there is always at least one convex quadrilateral (this is the [url=http://en.wikipedia.org/wiki/Happy_Ending_problem]Happy Ending problem[/url]), and I don't have any idea what the lower bound should be in general.", "Solution_2": "The minimum number is $ f(n)\\cdot\\binom{n}{4}$, where $ f$ has the following properties: it's nondecreasing, positive for $ n\\ge 5$, and bounded above by some $ c<1$. There is a limit, and I have no idea how to calculate it.", "Solution_3": "[quote=\"JBL\"]Well, if the points are in convex position (e.g., along a circle) then the answer is just $ \\binom{n}{4}$, but of course the answer can be smaller. With five points in general position, there is always at least one convex quadrilateral (this is the [url=http://en.wikipedia.org/wiki/Happy_Ending_problem]Happy Ending problem[/url]), and I don't have any idea what the lower bound should be in general.[/quote]\r\n\r\nif you put one point inside the convex polygonal region formed by n points, then the number of convex 4-gons formed is $ C_{n \\minus{} 2}/C_{n \\minus{} 3}*(n \\minus{} 3)$+$ \\binom{n}{3}*(n\\minus{}3)/4$", "Solution_4": "if you put two points inside the convex hull formed by the aforementioned n points, that is, total n+2 points, then the total number of convex 4-gons formed by all those points is at least $ \\lfloor_{n/2}\\rfloor*n$ $ \\minus{}$ $ \\displaystyle\\sum_{t \\equal{} 1}^{\\lfloor_{n/2}\\rfloor}{2t}$ $ \\plus{}$ $ 2*C_{n \\minus{} 2}/C_{n \\minus{} 3}*(n \\minus{} 3)$ + $ \\binom{n}{3}*(n \\minus{} 3)/4$ and at most $ \\lfloor_{n/2}\\rfloor*n$ $ \\minus{}$ $ \\displaystyle\\sum_{t \\equal{} 1}^{\\lfloor_{n/2}\\rfloor}{2t}$ $ \\plus{}$ $ 2*C_{n \\minus{} 2}/C_{n \\minus{} 3}*(n \\minus{} 3)$+$ \\binom{n}{3}*(n \\minus{} 3)/4$ + $ \\displaystyle\\sum_{k \\equal{} 2}^{\\lfloor_{n/2}\\rfloor}{t}$.", "Solution_5": "dear petronius, how on earth did you derive those eequations? Can you expalin to me, please? And what does $ C_{n\\minus{}2}$ mean? What does C stand for? Thank you so much if you could explain to me. *Big kiss*", "Solution_6": "Petronius has pseudo-responded here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=317887" } { "Tag": [ "number theory", "prime numbers" ], "Problem": "The digits 1, 2, 3, 4, 5, 6, 7, and 9 are used to form four two-digit prime numbers, with each digit used exactly once. What is the sum of these four primes? \\[ \\text {(A) } 150 \\qquad \\text {(B) } 160 \\qquad \\text {(C) } 170 \\qquad \\text {(D) } 180 \\qquad \\text {(E) } 190 \\]", "Solution_1": "[hide]2, 4, 5, and 6 must be the tens digits, since the numbers would not be prime if they were the units digits. Thus, 1, 3, 7, and 9 are the units digits. $20+40+50+60+1+3+7+9=190\\implies\\boxed E$[/hide]", "Solution_2": "[hide=\"Answer\"]Since they are all two-digit, the units digits must be the number 1, 3, 7, 9. Therefore, we have $1+3+7+9+20+40+50+60=190\\Rightarrow \\boxed{E}$.[/hide]", "Solution_3": "Ahh... the above methods were excellent, but I did it the dumb way (which was a lot more fun). \r\n\r\n[hide]We know that the tens digits have to be 2, 4, 5, and 6, since if they were units digits, the resulting number would be composite. Playing with resrictions on units digits and placing digits, we get four numbers: 23, 41, 67, and 59. This sum is 190. [/hide]\r\n\r\nAnd catcurio, you had the right answer, but the wrong answer choice...", "Solution_4": "[quote=\"236factorial\"]...we get four numbers: 23, 41, 67, and 59...[/quote]\r\n\r\nActually, there are four possibilites: (23, 41, 59, 67), (23, 47, 59, 61), (29, 41, 53, 67), (29, 47, 53, 61). So the above methods cover for all of them.", "Solution_5": "[quote=\"Farenhajt\"][quote=\"236factorial\"]...we get four numbers: 23, 41, 67, and 59...[/quote]\n\nActually, there are four possibilites: (23, 41, 59, 67), (23, 47, 59, 61), (29, 41, 53, 67), (29, 47, 53, 61). So the above methods cover for all of them.[/quote]\r\n\r\nYeah I was suspecting that but I didn't really want to go through all that... I'm really tired. :sleeping:", "Solution_6": "[quote=\"236factorial\"]And catcurio, you had the right answer, but the wrong answer choice...[/quote]\r\nOops... edited. :oops:" } { "Tag": [ "LaTeX" ], "Problem": "I recently attempted to typeset a document in LaTeX involving arcs of circles. However, I was unable to find any means of creating the arc symbol. I know that it must be possible since it is done in the [i]Art of Problem Solving[/i] books, but I have no idea how. Can anybody help?", "Solution_1": "You can do it crudely with $\\overset{\\frown}{ABC}$, but you'd want to make the frown wider. I'll look into that. If you find it, please do tell me." } { "Tag": [ "factorial", "number theory" ], "Problem": "Find the last nonzero digit of $100!$", "Solution_1": "[hide=\"A bit inelegant, but works\"]Clearly, only the last non-zero digit of each factor in the factorial is significant. Thus, if $n$ is the last non-zero digit of $9!$, then we want the residue of $n^{11}$ mod 10 (10 $n$s for the non-multiples of 10; 1 $n$ for the multiples of 10, as 100 obviously has no effect whatsoever). We can easily see that $n=8$, so by Fermat's Little Theorem, the answer must be congruent to 2 mod 5 and 0 mod 2, i.e., it must be $\\boxed{2}$, Q.E.D.[/hide]" } { "Tag": [], "Problem": "let ABC be a triangle and (C)be a circle with the diameter AB.(C)intersects CB and CA in points Pand Q,respectivly.\r\n\r\nand let the tangents to (C),from Pand Q ,intersect each other in point X.\r\n\r\nprove:CX is prependicualr to AB.", "Solution_1": "$\\angle{C}=\\frac{1}{2}(180^2-\\angle{POQ})=\\frac12 \\angle{PXQ}$ and $PX=QX$, So $X$ is the circumcentre of $\\triangle{CPQ}$. \r\n\r\n$CX \\perp AB \\Longleftrightarrow \\angle{PCX}+\\angle{A}=90^o$\r\nSince $\\angle{PCX}=\\angle{CPX}$, $\\angle{A}=\\angle{APO}$, so:\r\n\\[\\angle{PCX}+\\angle{A}=\\angle{CPX}+\\angle{A}=90^o\\]", "Solution_2": "thanks shobber ;)" } { "Tag": [], "Problem": "How many prime numbers less than 100 have a units digit of 3?", "Solution_1": "You can quickly see that the only values that aren't prime that have 3 as a unit's digit are the ones with multiples of 3 as the ten's digit. Thus there are $ 10\\minus{}3\\equal{}\\boxed{7}$ such primes.", "Solution_2": "Hello!\n\n[hide=My answer]There are 10 integers that have a units digit of 3 that are less than 100.\nThe only integers that have a units digit 3 that are NOT prime are $\\boxed{33}$$\\boxed{63}$$\\boxed{93}$\nNotice that the tens digit is a multiple of 3.\n$10-3=7$\nAnswer: $\\boxed{7}$[/hide]\n\nTHANKS!!! {:)+<:" } { "Tag": [ "function", "trigonometry", "algebra", "polynomial", "logarithms", "domain", "complex analysis" ], "Problem": "Is there some sort of general way to show that a function is analytic? For example, how would you show that $ \\cos^2(x)$ is analytic?", "Solution_1": "I'm assuming you mean \"analytic\" as \"locally given by a power series.\"\r\n\r\nThe simplest way to do that is to explicitly find a power series, although that's not always available. It is here:\r\n\r\n$ \\cos^2x\\equal{}\\frac12(1\\plus{}\\cos 2x)$\r\n\r\n$ \\equal{}\\frac12\\left(1\\plus{}1\\minus{}\\frac{2^2x^2}2\\plus{}\\frac{2^4x^4}{4!}\\plus{}\\cdots\\right)$\r\n\r\nwhich you can then simplify. This is, in fact, an entire function (given by a power series with an infinite radius of convergence.)\r\n\r\nIn general, we need some theorems:\r\n\r\n1. The sum of analytic functions is analytic (where both defined - this is understood below). \r\n\r\n2. The product of analytic functions is analytic.\r\n\r\n3. The quotient of analytic functions is analytic as long as we stay away from the zeros of the denominator.\r\n\r\n4. The composition of analytic functions is analytic.\r\n\r\n5. Many familiar functions are analytic: polynomials; the exponential (and sine and cosine), the logarithm (in the interior of its domain), the inverse trig functions (in the interiors of their domains).\r\n\r\nIn this case, we can see that the square of a cosine is the composition of two analytic functions, hence analytic." } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "quadratics", "algebra" ], "Problem": "What do you guys think about the CNMLs this year, compared to last?\r\nI think they're quite a bit easier, last year I was getting 2's and 3's, but this year I've got two 5's and 6.\r\n\r\nAlso, what did you think of today's CNML? I messed up on the one with the irregular numbers, put 60 instead of 36. :(", "Solution_1": "I think you should wait about a week next time before discussing answers.\r\n\r\nBut yeah, I made the same mistake you did.\r\n\r\nI don't really see a change in difficulty though, since last year I got nothing but 5's and 6's and this year I have 6/5/6.\r\n\r\nMaybe you're just getting better.", "Solution_2": "This round was really really really easy, relative to the previous rounds. I got a 6. :)", "Solution_3": "Does the school co-ordinator give out free (or at a cost) solutions after the math contest session?", "Solution_4": "I'm doing the cnml as well. my score so far is 5/4/5", "Solution_5": "The CNML is the worst contest that has ever existed in the history of human-kind :furious: \r\nI do each contest in under 10 min.\r\nBasically all of the problems IN HARDER FORM :rotfl: have appeared somewhere before. \r\nIt is sooo easy to cheat (though i don't do that), but at least in my school, some people write the same thing on different days. :!: \r\nI know some incidents when the wrong score was recorded for some people. :mad: \r\nAbout 80 % of the problems can be done win nearly completely trial and error. :mad: \r\nThe wording of the problems is horrible.:stretcher: \r\nThere many other bad things about this contest. :mad:", "Solution_6": "What cnml? :S", "Solution_7": "yeah CNML does suck, I made a mistake in the first one, with negative signs, no more perfect. I don't even think you win anything anyway, it's not challenging enough.", "Solution_8": "it's just a series of fun contests... nothing more...", "Solution_9": "They are not fun at all :furious:", "Solution_10": "Well, it's just kinda a funny tool to get little kids involved in the mathematics. :D", "Solution_11": "It is, but i think it makes high school students who take it seriously a bit more stupid. :(", "Solution_12": "Well...I usually use them as a practice of avoiding \"2+5=8\" type mistakes.\r\nBut yeah...it is sort of stupid to take it too seriously.", "Solution_13": "Did anyone get a certificate for being the top student in their school?", "Solution_14": "[quote=\"liudaveliu\"]Did anyone get a certificate for being the top student in their school?[/quote]\r\nMy school hands out a certificate and book prize to the top student in each grade at the end of the school year. :)", "Solution_15": "[quote=\"Jeff219\"][quote=\"liudaveliu\"]Did anyone get a certificate for being the top student in their school?[/quote]\nMy school hands out a certificate and book prize to the top student in each grade at the end of the school year. :)[/quote]\r\n\r\nMy school gives book prizes to the top two of all grades...\r\nBTW, I got perfect this year! lol... :rotfl:", "Solution_16": "PERFECT? :ninja:", "Solution_17": "I got 35. btw cnml 6 was such a joke. i want my school to give out prizes too but it doesn't :(", "Solution_18": "i agree with alex, cnml is such a trashy contest, the only difficult in it is a) understanding what they want you to do (the questions are poorly worded) and b) making sure that your TI89 has batteries (and c) making sure you dont make adding mistakes, but if you take care of b) you should be ok with that). If you go to their website, you will see that people frequently \"challenge\" the questions (they try to agrue that their incorrect answers actually deserve the marks) in a comicaly hillarius fasion.", "Solution_19": "[quote=\"mnica\"]i agree with alex, cnml is such a trashy contest, the only difficult in it is a) understanding what they want you to do (the questions are poorly worded) and b) making sure that your TI89 has batteries (and c) making sure you dont make adding mistakes, but if you take care of b) you should be ok with that). If you go to their website, you will see that people frequently \"challenge\" the questions (they try to agrue that their incorrect answers actually deserve the marks) in a comicaly hillarius fasion.[/quote]\r\nYay. You joined :)", "Solution_20": "is it for high school students?\r\nwhich level is it?", "Solution_21": "It's intended for grades 11 and 12, but it's a very easy contest compared to AIME, USAMO, IMO, etc...very easy. Grades 9 and 10 may also write it if their teachers allow them.", "Solution_22": "oh really?\r\nsounds lik euclid.\r\ni wanna have a try hehe", "Solution_23": "CNML is to Hypatia as Hypatia is to Part B of the Canadian Open as Part B of the Canadian Open is to the .. IMO?", "Solution_24": "[quote=\"Effervescence\"]CNML is to Hypatia as Hypatia is to Part B of the Canadian Open as Part B of the Canadian Open is to the .. IMO?[/quote]\r\n\r\nToo true.\r\nBut still, CNML is as fun and as stupid as you want it to be... :roll:", "Solution_25": "Just looking over this message board here and decided to sign up (after years of viewing the AoPS forums). CNML really is a joke contest and frankly it disgusts me how some schools (cough Pushkin) use it as a means of making a reputation for themselves. I was one of four students to score perfect in 2003-04, but 2 of the students were from Pushkin which is now not allowed to participate in the CNML. I am a member of the Euclid marking committee and remember hearing numerous teachers and other members of the math community discuss how Pushkin ruined the contest by cheating. Anyone could cheat on the contest because the moderator has a copy of the solutions and could tell any of the students them, further it is marked by the moderator who could submit any grades he wishes. It is amusing however how pushkin was able to have like 10 perfect students last year, but yet still unable to have one student score over 70 on Euclid!\r\n\r\nCNML questions can be worded weirdly, and they leave a lot of room for ambiguity with solutions because they say \"all equivalent values are acceptable\". I remember hearing of a question that asked you to find the roots of this one quadratic, and one student wrote the quadratic equation and did not simplify but received credit because it was equivalent to the solution (i think the solution was a and b, but his enormous values received credit).\r\n\r\nAs for prizes, if you finish first overall, you receive a shiny plaque with your name on it! lol. That's about that though.", "Solution_26": "[quote=\"jonnyg\"]\nAs for prizes, if you finish first overall, you receive a shiny plaque with your name on it! lol. That's about that though.[/quote]\r\n\r\nlol...wonder where those book prizes came from then..." } { "Tag": [ "function", "integration", "algebra", "polynomial", "number theory", "relatively prime", "real analysis" ], "Problem": "Let $f$ be a continuous function defined on $[0,1]$ and let for all $n$ the sequence $r_1, r_2,..., r_{\\phi(n)}$ be those numbers smaller than $n$ and relatively prime to it. Find the limit of $\\frac{ f(\\frac{r_1}{n})+...+f(\\frac{ r_{\\phi(n)}}{n})}{\\phi(n)}$", "Solution_1": "I don't get it :?. What if $f$ is constant with value $1$, for example? For every $n$, that number is $\\frac{\\varphi(n)}n$, and this sequence certainly doesn't have a limit as $n\\to\\infty$. In fact, the closure of its set of values is the whole interval $[0,1]$. What am I missing?", "Solution_2": "Terribly sorry, I forgot that at denominator we have $\\phi(n)$. Sorry again.", "Solution_3": "limit=$\\int_0^1f(x)dx$", "Solution_4": "It's clear that it cannot be anything else (just consider prime $n$). But how about the proof? ;)", "Solution_5": "is it obvious? :?", "Solution_6": "I think that a solution would be more indicated than 19 posts, don't you think? I really don't think the problem is obvious.", "Solution_7": "It isn't obvious (as Myth told me once, nothing is :) ) but it is still quite manageable. The solution I found runs like this. Let $S(n,f)$ be our sum. First of all, since $a$ is relatively prime to $n$ if and only if $n-a$ is, we see that the $S(n,f)=0$ for any function $f$ satisfying $f(1-x)=-f(x)$. This allows to reduce the problem to the case when $f(0)=f(1)$. Now, using the Weierstrass theorem on density of trigonometric polynomials, we see that it is enough to consider the case $f(x)=e_k(x)=e^{2\\pi i kx}$. It is clear that for $k=0$, we have $S(n,e_0)=1=\\int_0^1 e_0$ for all $n$. Now, it is a straightforward exercise in elementary number theory to show that $S(n,e_k)$ is a multiplicative function of $n$ for fixed $k$. Now observe that, for any prime number $p$, $S(p^m,e_k)=0$ whenever $p^{m-1}>k$. Also, $S(p,e_k)=-\\frac1{p-1}$ for $p>k$. From here it follows that, for any fixed $k\\ne 0$, we have $S(n,e_k)\\to 0=\\int_0^1e_k$ as $n\\to\\infty$ and we are done. What is your solution, harazi?", "Solution_8": "Very nice. I also used Weierstrass theorem, but for polynomials. In order to prove the result for $x^k$ I used the inclusion-exclusion principle. The computations are however quite messy." } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "derivative", "calculus computations" ], "Problem": "$ \\int \\frac{1}{\\sin x \\sqrt{\\sin^2 x \\plus{} a^2}} dx$.\r\n\r\nI am looking for the most elegant and compact answer, so different methods will be appreciated.", "Solution_1": "i don't know about elegant, but i know the following method and another one using DuIs.\r\n\r\n$ \\int \\dfrac { \\sin \\, x \\; dx } { \\sin^2 \\, x \\cdot \\sqrt { a^2 \\plus{} \\sin^2 \\, x } } \\;\\equal{}\\; \\int \\dfrac { \\sin \\, x \\; dx } { ( 1 \\minus{} \\cos^2 \\, x ) \\cdot \\sqrt { (a^2 \\plus{} 1) \\minus{} \\cos^2 \\, x } }$\r\n\r\n$ \\equal{}\\; \\int \\dfrac { \\sec^2 \\, x \\;\\tan\\, x \\; dx } { ( \\sec^2 \\, x \\minus{} 1 ) \\cdot \\sqrt { (a^2 \\plus{} 1 ) \\sec^2 \\, x \\minus{} 1} }$\r\n\r\nnow put $ u^2 \\;\\equal{}\\;(a^2 \\plus{} 1 ) \\sec^2 \\, x \\minus{} 1$ \r\n\r\n$ I\\; \\equal{}\\;\\dfrac { 1 } { a^2 \\plus{} 1} \\; \\int \\dfrac { u\\; du } { ( \\dfrac { u^2 \\plus{} 1} { a^2 \\plus{} 1} \\minus{} 1 ) \\cdot u } \\;\\equal{}\\; \\int \\dfrac { du } { u^2\\minus{}a^2}$\r\n\r\n$ \\;\\equal{}\\; \\dfrac {1} {2a} \\ln \\bigg| \\dfrac {u\\minus{}a} {u\\plus{}a } \\bigg | \\qquad \\text{or} \\qquad \\minus{} \\frac {1}{a} \\text{coth}^ {\\minus{}1} \\; \\dfrac { u} { a}$\r\n\r\n$ \\;\\equal{}\\; \\dfrac {1} {2a} \\ln \\bigg| \\dfrac { \\sqrt { a^2 \\plus{} \\sin^2 \\, x } \\minus{}a\\cos\\, x } {\\sqrt { a^2 \\plus{} \\sin^2 \\, x } \\plus{}a\\cos\\, x } \\bigg | \\qquad \\text{or}$\r\n\r\n$ \\minus{} \\frac {1}{a} \\text{coth}^ {\\minus{}1} \\left( \\dfrac { \\sqrt { a^2 \\plus{} \\sin^2 \\, x } } { a \\cos\\, x } \\right)$", "Solution_2": "Thanks, that method is indeed nice. Could I see the Differentiation under the integral sign method?", "Solution_3": "[quote=\"Gib Z\"]$ I \\equal{} \\int \\frac {1}{\\sin x \\sqrt {\\sin^2 x \\plus{} a^2}}\\ \\mathrm {dx}\\ ,\\ a > 0$ .[/quote]\r\n[color=darkblue][b][u]Proof.[/u][/b] I\"ll use the substitution $ \\left\\|\\begin{array}{c} \\boxed {t \\equal{} \\phi (x) \\equal{} \\cos x} \\\\\n \\\\\n\\phi'(x) \\equal{} \\minus{} \\sin x \\\\\n \\\\\n\\left(\\ b \\equal{} \\sqrt {a^2 \\plus{} 1}\\ \\right)\\end{array}\\right\\|$ $ \\Longrightarrow$\n\n$ I \\equal{} \\int \\frac {(\\cos x)'}{\\left(\\cos^2 x \\minus{} 1\\right) \\sqrt {b^2 \\minus{} \\cos^2x}}\\ \\mathrm {dx} \\equal{}$ $ F(\\cos x) \\plus{} \\mathcal C$ , where \n\n$ F \\equal{} \\int\\frac {1}{\\left(t^2 \\minus{} 1\\right)\\sqrt {b^2 \\minus{} t^2}}\\ \\mathrm {dt}$ . Now I\"ll use the [b]Abel[/b]'s substitution \n\n$ \\left\\|\\begin{array}{c} \\boxed {y \\equal{} \\psi (t) \\equal{} \\frac {t}{\\sqrt {b^2 \\minus{} t^2}}}\\ ,\\ t^2 \\equal{} \\frac {b^2y^2}{y^2 \\plus{} 1} \\\\\n \\\\\n\\psi'(t) \\equal{} \\frac {b^2}{\\left(b^2 \\minus{} t^2\\right)\\sqrt {b^2 \\minus{} t^2}} \\\\\n \\\\\n\\left(\\ t^2 \\minus{} 1 \\equal{} \\frac {\\left(b^2 \\minus{} 1\\right)y^2 \\minus{} 1}{y^2 \\plus{} 1}\\ ,\\ b^2 \\minus{} t^2 \\equal{} \\frac {b^2}{y^2 \\plus{} 1}\\ \\right)\\end{array}\\right\\|$ $ \\Longrightarrow$\n\n$ F \\equal{} \\int\\frac {1}{\\left(t^2 \\minus{} 1\\right)\\sqrt {b^2 \\minus{} t^2}}\\cdot\\frac {\\left(b^2 \\minus{} t^2\\right)\\sqrt {b^2 \\minus{} t^2}}{b^2}\\cdot \\psi '(t)\\ \\mathrm {dt} \\equal{}$\n\n$ \\int\\frac {b^2 \\minus{} t^2}{b^2\\left(t^2 \\minus{} 1\\right)}\\cdot \\psi '(t)\\ \\mathrm {dt} \\equal{}$ $ G(\\frac {t}{\\sqrt {b^2 \\minus{} t^2}}) \\plus{} \\mathcal C$ ,\n\nwhere $ G \\equal{} \\int\\frac {1}{\\left(b^2 \\minus{} 1\\right)y^2 \\minus{} 1}\\ \\mathrm {dy} \\equal{}$ $ \\frac {1}{2a}\\cdot\\ln\\left|\\frac {ay \\minus{} 1}{ay \\plus{} 1}\\right| \\plus{} \\mathcal C$ a.s.o.\n\n[b]Remark.[/b] You can use directly the substitution $ t\\equal{}\\frac {\\cos x}{\\sqrt {b^2\\minus{}\\cos x}}\\ .$[/color]" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "LaTeX" ], "Problem": "A 10-foot plank is used to brace a basement wall during construction of a home. The plank is nailed to the wall 6 feet above the floor. The slope of the plank is ________", "Solution_1": "[quote=\"grizzland\"]A 10-foot plank is used to brace a basement wall during construction of a home. The plank is nailed to the wall 6 feet above the floor. The slope of the plank is ________[/quote]\r\n\r\n[hide]Its a 6-8-10 triangle so $\\frac{3}{4}$[/hide]", "Solution_2": "[hide][latex]\\frac{3}{4}[/latex][/hide]\r\n\r\nI don't know why everyone hides their answers, but I guess I'll do it too.", "Solution_3": "[quote=\"Temperal\"][hide][latex]\\frac{3}{4}[/latex][/hide]\n\nI don't know why everyone hides their answers, but I guess I'll do it too.[/quote]\r\nthe [latex] tags are for the wiki I think, use dollar signs for latex on the forums.", "Solution_4": "you need to hide it so other people dont accidentally look at your solution. technically, only the first couple posts should be hidden, but still hide all your solutions anyways. also, if they werent hidden, the page would get so long and it would take longer to load the page", "Solution_5": "[hide]Depending on how the coordinate system is placed, the slope could be either $\\frac34$ or $-\\frac34$. The sides of the triangle are 6, 8, and 10, so the slope is rise over run = $\\frac68$.[/hide]", "Solution_6": "no, for the wiki you have to use and \r\n\r\nby the way, when i was putting up an old IMO problem, i got errors when i did x^2 + y^2 + z^2 = b^2\r\nbut not when i did x^2+ y^2 + z^2 = b^2. any idea why?", "Solution_7": "[quote=\"Teki-Teki\"]no, for the wiki you have to use and \n\nby the way, when i was putting up an old IMO problem, i got errors when i did x^2 + y^2 + z^2 = b^2\nbut not when i did x^2+ y^2 + z^2 = b^2. any idea why?[/quote]\r\nProbably because of missing brackets {}, but still, that's very odd :huh:", "Solution_8": "Oh, yeah. Another math forum I go to uses [latex][/latex], so...", "Solution_9": "[quote=\"Temperal\"]Oh, yeah. Another math forum I go to uses [latex][/latex], so...[/quote]\r\n\r\nReally? Which forum (I'm interested just to see)?", "Solution_10": "[quote=\"b-flat\"][quote=\"Temperal\"]Oh, yeah. Another math forum I go to uses [latex][/latex], so...[/quote]\n\nReally? Which forum (I'm interested just to see)?[/quote]\r\nhttp://www.xpmath.com/forums\r\nIt's a small forum run off VBulletin.\r\nBut this is off topic.", "Solution_11": "I don't use Latex :P \r\n\r\n[hide]3/4 because the sides are 6,8, and 10, so 6/8 which is 3/4.[/hide]\r\nYeah http://www.xpmath.com is awesome!", "Solution_12": "You should. Otherwise you can't write stuff like summation symbols." } { "Tag": [], "Problem": "What is the value of \\\\\n$ \\frac{1}{2}\\times4\\times\\frac{1}{8}\\times16\\times\\frac{1}{32}\\times64\\times\\frac{1}{128}\\times256\\times\\frac{1}{512}\\times1024$?", "Solution_1": "You basically group the numbers by 2s, and you should notice that the product of every two is always 2. Since there are 5 pairs of numbers/fractions, the answer is $ 2^5\\equal{}32$." } { "Tag": [ "geometry", "3D geometry", "pyramid" ], "Problem": "A grocer stacks oranges in a pyramid-like stack whose rectangular base is $ 5$ oranges by $ 8$ oranges. Each orange above the first level rests in a pocket formed by four oranges in the level below. The stack is completed by a single row of oranges. How many oranges are in the stack?\r\n\r\n$ \\textbf{(A)}\\ 96 \\qquad\r\n\\textbf{(B)}\\ 98 \\qquad\r\n\\textbf{(C)}\\ 100 \\qquad\r\n\\textbf{(D)}\\ 101 \\qquad\r\n\\textbf{(E)}\\ 134$", "Solution_1": "[hide]\nyou notice that each new stack decreases in both dimensions by 1.\nthat means: \n5*8 + 4*7 + 3*6 + 2*5 + 1*4 =100, C\n [/hide]", "Solution_2": "[quote=\"#H34N1\"]A grocer stacks oranges in a pyramid-like stack whose rectangular base is 5 oranges by 8 oranges. Each orange above the first level rests in a pocket formed by four oranges in the level below. The stack is completed by a single row of oranges. How many oranges are in the stack?\n\n(A) 96 \n(B) 98 \n(C) 100 \n(D) 101 \n(E) 134[/quote]\r\n\r\n[hide=\"solution\"]$5\\cdot 8+4\\cdot 7+3\\cdot 6+2\\cdot 5+1\\cdot 4=\\boxed{100}$[/hide]" } { "Tag": [ "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "This is probably easy, but I can't find the subject in my calculus book. Here's an example from an assignment of mine...\r\n\r\nC(X) = 15 + 7 log8 X, where X>1\r\n\r\nx is number of years on market\r\nrate of sales for years 4,8,10", "Solution_1": "alright, I figured the derivitive as\r\n\r\n$\\ c'(x)=\\frac8{xln(5)}$\r\n\r\nnow I need for find the derivitive when X is 4,8, and 10\r\n\r\nyet I don't know how to find derivitives for the fraction\r\nI've seen fraction examples but the answers make no sense to me, help!", "Solution_2": "you just have to put in your values for x, if f(x) = x^2, at x=3 it equals 9.\r\n\r\n$c'(4)= \\frac{7}{4ln(8)}$\r\n$c'(8) = \\frac{7}{8ln(8)}$\r\n$c'(10) = \\frac{7}{10ln(8)}$" } { "Tag": [], "Problem": "How many 9-step paths are there from $ E$ to $ G$?[asy]size(4cm,4cm);int w=6;int h=5;int i;for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label(\"$G$\", (w-1,0), SE);label(\"$E$\", (0,h-1), NW);[/asy]", "Solution_1": "R : advance to the right\r\nD : advance to the down\r\nThe route supports that I arrange five R and fore D to one line.\r\nTherefore,\r\n$ \\dfrac{(5 \\plus{} 4)!}{5! \\ 4!} \\equal{} \\binom{9}{5} \\equal{} \\boxed{126}$", "Solution_2": "[hide=\"My Solution\"]\nThe number of steps it takes to go from $E$ to $F$ is 4 and the number of steps to $G$ from $F$ is 5.\n\nThere are $\\dbinom{4}{1}=\\dfrac{4\\cdot3\\cdot2\\cdot1}{1}=4$ ways to go from $E$ to $F$.\n\nThere are $\\dbinom{5}{2}=\\dfrac{5\\cdot4\\cdot3\\cdot2\\cdot1}{2\\cdot1}=10$ ways to get from $F$ to $G$.\n\nMultiplying these gives: $4\\cdot10=40$.\n\n[hide=\"answer\"]There are $\\boxed{40}$ 9-step paths.[/hide][/hide]", "Solution_3": "Why is it in the official solution $\\binom{9}{4}$? Why is it selecting four steps from a group of 9 steps? Don't we need all 9 steps?", "Solution_4": "I think kangaroos is right because doesn't the combo rule only apply when you can only go downwards and right for this problem? But this problem let's you go up down right left, right?", "Solution_5": "you can't go left or up because then you take too many steps to get to G. Also, $\\binom{4}{1}=\\frac{4!}{3!1!}=\\frac{4}{1}=4$\nand\n$\\binom{5}{2}=\\frac{5!}{3!2!}=\\frac{5\\cdot4}{2}=10$.", "Solution_6": "Nvm about my questions got them answered", "Solution_7": "I believe Kouichi Nakagawa is correct. It would be $\\binom{9}{5}=\\binom{9}{4}$ because we have to move 9 steps in all, 5 rights and 4 downs. If we choose where to put the rights, the placement of the downs is decided. Similarly, if we choose the downs first, then the rights are decided.", "Solution_8": "[hide=Sol]Ans = (5+4)c4 or (5+4)c5 = 126[/hide]", "Solution_9": "I used adding (like Pascal's triangle). There is one way to stay at E, one way to go east, and one way to go south. There are two ways to go to the southeast square from the start, which comes from 1+1. (This is a proven, yet not stated method. Aops didn't explain this solution because it didn't want you cheating like this.)", "Solution_10": "just some basic addition.", "Solution_11": "$\\binom{9}{5}=126$", "Solution_12": "or pascal's triangle", "Solution_13": "$\\binom{9}{4}\\times\\binom{9-4}{5}=\\binom{9}{4}=126$", "Solution_14": "[hide] Since there are 9 steps and 4 of them have to be downwards, the answer would be $C(9,4)=\\boxed{126}$ ways.[/hide]" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ A \\equal{} \\times \\cdots \\times $ be a finite abelian group with $ |x_i| \\equal{} n_i$ for all $ 1 \\leq i \\leq r$. Prove that if $ G$ is any group containing commuting elements $ g_1, \\cdots, g_r$ such that $ g_i^{n_i} \\equal{} 1$ for all $ 1 \\leq i \\leq r$, then there is a unique homomorphism from $ A$ to $ G$ which sends $ x_i$ to $ g_i$ for all $ i$.\r\n\r\nI believe the assumption that the order of $ g_i$ divide $ n_i$ is unnecessary. Can someone come up with a counterexample? If so, I'll post my proof, since I can't find an error in it.", "Solution_1": "Without that order condition, you don't get a homomorphism. We have $ e\\equal{}f(e)\\equal{}f(x_1^{n_1})\\equal{}f(x_1)^{n_1}\\equal{}g_1^{n_1}$ for any homomorphism $ f$." } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "Hi, I have a question about inequalities.\r\n\r\nIf an inequality is symmetric for variables $ a_1, a_2, \\cdots a_n$, we may assume that $ a_1\\ge a_2\\ge \\cdots \\ge a_n$.\r\n\r\nIs there such assumption we may make for cyclic inqualities?\r\n\r\nThanks. :)", "Solution_1": "[quote=\"FantasyLover\"]Hi, I have a question about inequalities.\n\nIf an inequality is symmetric for variables $ a_1, a_2, \\cdots a_n$, we may assume that $ a_1\\ge a_2\\ge \\cdots \\ge a_n$.\n\nIs there such assumption we may make for cyclic inqualities?\n\nThanks. :)[/quote]\r\n\r\nIf an inequality is cyclic for variables $ a_1, a_2, \\cdots a_n$, we may assume that $ a_1\\equal{}\\max \\{a_1, a_2, \\cdots ,a_n \\}$ or $ a_1\\equal{}\\min \\{a_1, a_2, \\cdots ,a_n \\}$.", "Solution_2": "Oh, okay.\r\n\r\nThank you! :)" } { "Tag": [], "Problem": "#2. [color=white]The answer is 0 because it's like matching. You can't get only one wrong.[/color]", "Solution_1": "May I ask the purpose of this poll?\r\n\r\nThe reason is we have enough new getting started stuffs and we don't want anything that's not \"math.\"\r\n\r\nIf not replied in next two days, it will be removed.\r\n\r\nRegards,\r\n\r\nSilverfalcon", "Solution_2": "You're answering number 2 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18791]this thread[/url], right? When you have the answer to a question, you post it directly in the question thread. There's no need to start a new topic. Or if it's really necessary, which I don't think it is, at the very least post the original question so people know what you're talking about.", "Solution_3": "Sorry, you can delete the poll please, or tell me how.\r\n\r\nI didn't know the rule about having polls and I was just interested in the poll feature (and also wondered if most people thought the problem was easy). I also accidentally pressed the wrong button to create a new topic.\r\n\r\nI will repost my answer to the question in the proper place." } { "Tag": [ "inequalities", "function", "inequalities unsolved" ], "Problem": "[tex] a;b;c [/tex] are positive real numbers.Prove that\r\n[tex]\\frac{a^3}{(a+b)^3}+\\frac{b^3}{(b+c)^3}+\\frac{c^3}{(c+a)^3}\\ge \\frac{3}{8}[/tex]", "Solution_1": "Put \r\n $\\displaystyle x=\\frac{b}{a}$\r\n $\\displaystyle y=\\frac{c}{b}$\r\n $\\displaystyle z=\\frac{a}{c}$\r\nWe have $xyz=1$\r\nNow must prove\r\n$\\displaystyle \\frac{1}{(x+1)^3}+\\frac{1}{(y+1)^3}+\\frac{1}{(z+1)^3} \\ge \\frac{3}{8}$\r\n\r\nIt quite easy because This one is more strong:\r\n\r\n$\\displaystyle \\frac{1}{x+1}+\\frac{1}{y+1}+\\frac{1}{z+1}\\ge \\frac{3}{2}$\r\n\r\n\r\n :D :D :D", "Solution_2": "[quote=\"Viet Math\"]\nIt quite easy because This one is more strong:\n\n$\\displaystyle \\frac{1}{x+1}+\\frac{1}{y+1}+\\frac{1}{z+1}\\ge \\frac{3}{2}$\n\n\n :D :D :D[/quote]\r\n\r\nWell your inequality is false for $x\\longrightarrow0,y=z\\longrightarrow\\infty$.\r\nBut the inequality is true for the power 2( http://www.mathlinks.ro/Forum/viewtopic.php?t=32031 and http://www.mathlinks.ro/Forum/viewtopic.php?t=30724)\r\nThe cubic power can be reached by using Power Mean Inequality.", "Solution_3": "[quote=\"hardsoul\"][quote=\"Viet Math\"]\nIt quite easy because This one is more strong:\n\n$\\displaystyle \\frac{1}{x+1}+\\frac{1}{y+1}+\\frac{1}{z+1}\\ge \\frac{3}{2}$\n\n\n :D :D :D[/quote]\n\nWell your inequality is false for $x\\longrightarrow0,y=z\\longrightarrow\\infty$.\nBut the inequality is true for the power 2( http://www.mathlinks.ro/Forum/viewtopic.php?t=32031 and http://www.mathlinks.ro/Forum/viewtopic.php?t=30724)\nThe cubic power can be reached by using Power Mean Inequality.[/quote]\r\n\r\nOh,i'm sorry :blush: :blush:", "Solution_4": "It's easy to prove : \r\n $ \\frac{1}{(1+x)^2} + \\frac{1}{(1+y)^2} \\geq \\frac{1}{1+xy} = \\frac{a}{c+a} $ \r\n $ \\frac{a}{c+a} + \\frac{c^2}{(c+a)^2} \\geq \\frac{3}{4} $ \r\n After that , applying Power mean inequality ==> Q.E.D", "Solution_5": "Does anyone have the full contest?\r\nPlease post it on th forum\r\nThank you", "Solution_6": "I think the stronger ineq. holds\r\n$\\frac{a^3}{(a+b)(a^2+b^2)}+\\frac{b^3}{(b+c)(b^2+c^2)}+\\frac{c^3}{(c+a)(c^2+a^2)}\\ge \\frac{3}{4}$", "Solution_7": "technically, no proof has been shown, so i will quickly show one\r\n\r\nas before, we reduce to xyz = 1, need to show:\r\n\r\n$\\sum \\frac{1}{(x+1)^3} \\geq \\frac{3}{8}$\r\n\r\nBy jensen, $LHS \\geq 3f(\\frac{x+y+z}{3}) \\geq 3f(\\sqrt[3]{xyz}) = RHS$, where $f(x) = \\frac{1}{(x+1)^3}$ is concave.\r\n\r\n---", "Solution_8": "[quote=\"Singular\"]$3f(\\frac{x+y+z}{3}) \\geq 3f(\\sqrt[3]{xyz}) = RHS$, ---[/quote]\r\nI don't think it true ;)", "Solution_9": "Yes I agree, $f$ is a decreasing function, hence $f(\\frac{x+y+z}{3}) \\leq f(\\sqrt[3]{xyz})$ Which means that your proof is wrong :lol: \r\nDoes anyone has a right Proof ?? :?", "Solution_10": "I think you maybe see the problem following:\r\n$a,b,c,d$ are four positive numbers such that $abcd=1$\r\nthen we have $\\frac{1}{(1+a)^2}+\\frac{1}{(1+b)^2}+\\frac{1}{(1+c)^2}+\\frac{1}{(1+d)^2} \\geq 1$\r\nthen let $d=1$,we have $\\frac{1}{(1+a)^2}+\\frac{1}{(1+b)^2}+\\frac{1}{(1+c)^2} \\geq \\frac{3}{4}$\r\nuse power mean inequality. :)", "Solution_11": "[quote=\"minhkhoa\"]I think the stronger ineq. holds\n$\\frac{a^3}{(a+b)(a^2+b^2)}+\\frac{b^3}{(b+c)(b^2+c^2)}+\\frac{c^3}{(c+a)(c^2+a^2)}\\ge \\frac{3}{4}$[/quote]\r\ncan you post your solution for this?\r\nthanks", "Solution_12": "I don't want to say this but I don't have the solution yet :blush: :P", "Solution_13": "oops, yea, jensen too strong", "Solution_14": "This inequality has been posted before:\r\nhttp://www.mathlinks.ro/Forum/topic-17246-20.html" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c>0$. Prove that:\r\n$ \\sqrt{\\frac{a^3}{a^3\\plus{}(b\\plus{}c)^3}}\\plus{}\\sqrt{\\frac{b^3}{b^3\\plus{}(c\\plus{}a)^3}}\\plus{}\\sqrt{\\frac{c^3}{c^3\\plus{}(a\\plus{}b)^3}}\\geq 1$", "Solution_1": "Hint\r\n\\[ \\sqrt {\\frac{{{a^3}}}{{{a^3} \\plus{} {{\\left( {b \\plus{} c} \\right)}^3}}}} \\ge \\frac{{{a^2}}}{{{a^2} \\plus{} {b^2} \\plus{} {c^2}}}\\]", "Solution_2": "[quote=\"dvtvd\"]Hint\n\\[ \\sqrt {\\frac {{{a^3}}}{{{a^3} \\plus{} {{\\left( {b \\plus{} c} \\right)}^3}}}} \\ge \\frac {{{a^2}}}{{{a^2} \\plus{} {b^2} \\plus{} {c^2}}}\n\\]\n[/quote]\r\n\r\nWhy is it? I had started expanding...it won't work.", "Solution_3": "[quote=\"Stephen\"][quote=\"dvtvd\"]Hint\n\\[ \\sqrt {\\frac {{{a^3}}}{{{a^3} \\plus{} {{\\left( {b \\plus{} c} \\right)}^3}}}} \\ge \\frac {{{a^2}}}{{{a^2} \\plus{} {b^2} \\plus{} {c^2}}}\n\\]\n[/quote]\n\nWhy is it? I had started expanding...it won't work.[/quote]\r\nBecause it 's equivalent to:$ 2a^2(b^2\\plus{}c^2)\\plus{}(b^2\\plus{}c^2)^2 \\ge a(b\\plus{}c)^3$\r\nWhich is obvious true by AM-GM :wink:", "Solution_4": "[quote=tdl]Let $ a,b,c>0$. Prove that:\n$$ \\sqrt{\\frac{a^3}{a^3\\plus{}(b\\plus{}c)^3}}\\plus{}\\sqrt{\\frac{b^3}{b^3\\plus{}(c\\plus{}a)^3}}\\plus{}\\sqrt{\\frac{c^3}{c^3\\plus{}(a\\plus{}b)^3}}\\geq 1$$[/quote]\nLet $a,b,c\\in R^{+}$. [url=https://artofproblemsolving.com/community/c6h1998437p13957733]Prove the following inequality:[/url]\n$$\\sqrt{\\frac{a^{3}}{b^{3}+(c+a)^{3}}}+\\sqrt{\\frac{b^{3}}{c^{3}+(a+b)^{3}}}+\\sqrt{\\frac{c^{3}}{a^{3}+(b+c)^{3}}}\\geq1$$\n", "Solution_5": "So we have:\n $\\sqrt{\\frac{a^{3}}{b^{3} + (a + c)^{3}}} = \\sqrt{\\frac{1}{1 + (\\frac{b}{a} + \\frac{c}{a})^{3}}}$\n\nHave: \n$\\sqrt{1 + k^{3}} = \\sqrt{(1 + k)(k^{2} - k + 1)} \\leq \\frac{1+k + k^{2} - k + 1}{2} = 1 + \\frac{k^{2}}{2}$\n\n $\\sqrt{\\frac{1}{1 + (\\frac{b}{a} + \\frac{c}{a})^{3}}} \\geq \\frac{1}{1 + \\frac{1}{2}(\\frac{b}{a} + \\frac{c}{a})^{2}} \\geq \\frac{1}{1 + \\frac{b^{2} + c^{2}}{a^{2}}} = \\frac{a^{2}}{a^{2} + b^{2} + c^{2}}$\n\nDo the same => Done", "Solution_6": "[quote=tdl]Let $ a,b,c>0$. Prove that:\n$$ \\sqrt{\\frac{a^3}{a^3\\plus{}(b\\plus{}c)^3}}\\plus{}\\sqrt{\\frac{b^3}{b^3\\plus{}(c\\plus{}a)^3}}\\plus{}\\sqrt{\\frac{c^3}{c^3\\plus{}(a\\plus{}b)^3}}\\geq 1$$[/quote]\nfor $a,b,c>0.$ [url=https://artofproblemsolving.com/community/c6h62256p372570]Prove that[/url]\n$$1 \\leq \\sqrt{\\frac{a^3}{a^3+(b+c)^3}}+ \\sqrt{\\frac{b^3}{b^3+(c+a)^3}}+\\sqrt{\\frac{c^3}{c^3+(a+b)^3}} < \\sqrt 2$$\n", "Solution_7": "Does anybody have solution for this using Holder's Inequality? :icecream: " } { "Tag": [], "Problem": "There is a now a [url=http://www.artofproblemsolving.com/Forum/index.php?f=320]Canadian sub-forum[/url]! :D \n\nYou also access the forum from the index page under\nNational & Local Communities > North America > Canada\n\nI've moved all the past Canadian topics over there. And please continue our Canada-related discussion in the new location.", "Solution_1": "I was the one who asked Valentin for it :D :D :D", "Solution_2": "Enjoy it, but keep the spam at low levels, ok? :)" } { "Tag": [ "calculus" ], "Problem": "How can I find how fast this sum diverges in terms of x?\r\n\r\n$ \\sum_{n \\equal{} 0}^\\infty \\sum_{k \\equal{} 0}^{n} [{2n\\choose2k} {(2k \\plus{} 1)}^{ \\minus{} x}] / 2^{2n \\plus{} 1}$\r\n\r\n0>3222<<[/hide]\r\nMight be more...", "Solution_2": "dwx, both 4-digit numbers have to be even, if I am understanding the problem correctly.", "Solution_3": "SOURCE: from workout 6, 2004-2005 (jli was hiding it with small size fonts)\r\n\r\nDWX: you almost solved it. Consider mathmom's comments", "Solution_4": "dwx314, you are very close.\r\n\r\nhint: [hide]rearrange the digits[/hide]\n\nmy answer: [hide]2232, 1116[/hide]", "Solution_5": "[quote=\"jli\"][size=59]from workout 6, 2004-2005[/size]\n\nA positive, even four-digit integer contains only two different digits, and the sum of its digits is 9. If this integer is exactly twce another number also satisfying these conditions, what is the integer(s)?[/quote]\r\n\r\n[hide]2232-1116=1116[/hide]", "Solution_6": "yeah, there was only one answer in the answer book. it should be 2232", "Solution_7": "*laughs nervously*\r\n*Hits himself for his dub mistake*\r\n[hide=\"-\"]*Adds that to his endless list of dumb mistake made*\n*Sighs, and resolves to start reading the problem*\n*(In vain)*[/hide]", "Solution_8": "[quote=\"dwx314\"]*laughs nervously*\n*Hits himself for his dub mistake*\n[hide=\"-\"]*Adds that to his endless list of dumb mistake made*\n*Sighs, and resolves to start reading the problem*\n*(In vain)*[/hide][/quote]\r\n\r\nDub?\r\n\r\nBetter hit yourself again", "Solution_9": "[hide]I got $2232$ and therefore $1116$ as the second number (logic; the first number has to be a multiple of four)[/hide]" } { "Tag": [ "geometry" ], "Problem": "A square is circumscribed about a circle of radius $ r$. What is the number of square units in the area of the square? Express your answer in terms of $ r$.", "Solution_1": "draw a picture.\r\n\r\na length of the diagonal is $ 2r^2$.\r\n\r\nso the area of the square is $ 2r*2r/2\\equal{}2r^2$\r\n\r\nanswer : $ 2r^2$", "Solution_2": "That's what I got, 2r^2, but the system said it was incorrect....\r\nit said that it was $ 4R^2$", "Solution_3": "[quote=\"dwei1019\"]That's what I got, 2r^2, but the system said it was incorrect....\nit said that it was $ 4R^2$[/quote]\r\n\r\nReport an error.", "Solution_4": "No, it is not an error. Because the square is circumscribed about the circle, its sidelength is twice the radius of the circle. Thus, the side of the square is $ 2r$. Therefore, the area of the square is $ 2r\\cdot 2r \\equal{} \\boxed{4r^2}$. The picture looks like this:\r\n[asy]pair A=(0,0), B=(2,0), C=(-2,-2), D=(-2,2), E=(2,2), F=(2,-2);\ndraw(C--D--E--F--cycle);\ndraw(Circle(A,2));\ndot(A);\ndraw(A--B);\nlabel(\"$r$\", (1,0), S);\nlabel(\"$2r$\", (0,-2), S);[/asy]", "Solution_5": "FAIL.\r\nIt's circumscribed, not inscribed.\r\n[quote=\"FantasyLover\"]\ndraw a picture.\n\na length of the diagonal is $ 2r^2$.\n\nso the area of the square is $ 2r*2r/2\\equal{}2r^2$\n\nanswer : $ 2r^2$\n[/quote]\r\nand that just barely makes any sense :)", "Solution_6": "[b]this is how i did it \n\nwe know that the radius is $r$ \n\nthe radius is tangent to 1 side of the square so the length of the square of 1 side will be $2r$\n\nand the equation to find the area for a square is \n\n$2r$*$2r$=4r^2 \n\nso the answer is $\\boxed{4r^2}$ [/b]" } { "Tag": [ "geometry", "perimeter", "rhombus", "rectangle", "circumcircle" ], "Problem": "Take a polygon with the perimeter of 1. Prove that the more sides the polygon has, the more area it has.\n\n\n\nHint: [hide]Use the fact a rhombus is less efficient than a rectangle.[/hide]", "Solution_1": "eum... perhaps you could tell us what you're talking about? :?\r\n\r\n\"efficience\"? :?", "Solution_2": "anyone?", "Solution_3": "Are you talking about regular polygons, becuase a triangle with sides all 0.3333... has a bigger area than a rectangle with two sides 0.49 and two sides 0.01.\r\n\r\nAnd what does this problem have to do with circles?", "Solution_4": "yes.\r\na circle has infinitly many sides so thats why the topic is so titled", "Solution_5": "Considering the case of a regular polygon,\r\n\r\none can arrrive at the following formula for area\r\n\r\nA = (nR^2/2)* sin(2*pi/n)\r\n\r\nwhere R = radius of the circumcircle of the polygon\r\n\r\nand n = no. of sides of the polygon.\r\n\r\nAgain , the expression for perimeter of a polygon is \r\n\r\nP = 2nR* sin(pi/n)\r\n\r\nSubstituting this in the expression for Area, we get \r\n\r\nA = (PR/2)*cos(pi/n)\r\n\r\nHere P =1\r\nSo \r\nA(n) = (R/2)*cos(pi/n)\r\n\r\nNow cos(x) is decreasing in [0,pi/2]\r\n\r\nand pi/n belongs to [0,pi/2] for n>=3\r\n\r\nSo cos(pi/(n+1)) > cos(pi/n)\r\n\r\n=> A(n+1) > A(n)\r\n\r\nThat is the Area sequence is an increasing one" } { "Tag": [], "Problem": "Xinran walks 3 miles per hour (mph) uphill, 4 mph on a flat surface and 5 mph downhill. If he hikes one mile uphill, then one mile on the level and then returns by the same route to his starting point, how many minutes does he walk?", "Solution_1": "He walks 1 mile uphill at a rate of 3 mph so it took him 1/3 of an hour which is 20 minutes. \r\n\r\nHe then walks 1 mile on the level at a rate of 4 mph so it would be 1/4 of an hour which is 15 minutes. \r\n\r\nHe returns the same way so he walks 1 mile on a flat surface for another 15 minutes and walks downhill at a rate of 5 mph so he walked for 1/5 of an hour which is 12 minutes.\r\n\r\nAdd the minutes up and you get 62 minutes.", "Solution_2": "Alternatively, his average speed is the harmonic mean of 3,4,4,5.\r\n\r\n$ \\frac{4}{\\frac13\\plus{}\\frac14\\plus{}\\frac14\\plus{}\\frac15}\\equal{}\\frac{4}{\\frac{20}{60}\\plus{}\\frac{15}{60}\\plus{}\\frac{15}{60}\\plus{}\\frac{12}{60}}\\equal{}\\frac{4}{\\frac{62}{60}}\\equal{}\\frac{4\\cdot60}{62}$\r\n\r\nThus he moves on average 240 miles in 62 hours. Dividing top and bottom by 60, he moves 4 miles in $ \\boxed{62}$ minutes.", "Solution_3": "Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1281660#1281660]this[/url] for the reason that the harmonic mean works here." } { "Tag": [], "Problem": "Elizabeth visits her friend Andrew and then returns home by the same route. She always walks 2km/h when going uphill, 6km/h when going downhill and 3km/h when on level ground. If her total walking time is 6 hours, then what is the total distance she walks in km?", "Solution_1": "hi math92:\r\n\r\nAre you sure you gave us all the information?\r\n\r\nWhat we don't know is for how long she walked going uphill, going downhill or on level ground.\r\n\r\nYou have three unknowns. How are these unknowns related to each other?", "Solution_2": "[quote=\"ritchjp\"]hi math92:\n\nAre you sure you gave us all the information?\n\nWhat we don't know is for how long she walked going uphill, going downhill or on level ground.\n\nYou have three unknowns. How are these unknowns related to each other?[/quote]\r\ni think ritchjp is right. u hav at least 3 unknowns that r necessary 4 this prob.", "Solution_3": "No, you can solve this.", "Solution_4": "[hide]We're not given how much of the different elevation changes there are, so let's assume it doesn't matter. If it doesn't matter, let's say that he's walking on all flat ground. Then, he would have to go for 6*3 = $\\boxed{18\\text{ km}}$\n\n[/hide]", "Solution_5": "[hide=\"Hint\"]The amount of uphill, downhill or level ground doesn't matter[/hide]", "Solution_6": "oops, overlooked some info, yeah it is solvable.", "Solution_7": "the level ground amount only matters, so it is 3km*6hours to get [b]18[/b]", "Solution_8": "[hide=\"Answer\"]\n\n$a$=distance uphill , $b$=distance downhill , $c$=distance level ground\n\n In an eqation her total walking time looks like:\n\n$6=(\\frac{a}{2} + \\frac{b}{6} + \\frac{c}{3}) + (\\frac{a}{6} + \\frac{b}{2} + \\frac{c}{3})$\n\n However on her trip back, uphill turns to downhill and downhill to uphill. So we get:\n\n$6(\\frac{a}{2} + \\frac{b}{6} + \\frac{c}{3} + \\frac{b}{2} + \\frac{a}{6} + \\frac{c}{3})=36$\n\n Simplifying we get:\n\n$4(a+b+c)=36$\n\n And see see $x+y+z$ equals 9. \n\n Plugging this in we see that she walks a total of $\\boxed{18}$ kilometres. [/hide]", "Solution_9": "I guess the question is how much of that distance is level and how much of that distance is not level. On the round trip, for the up and down distance, you would be averaging 4 km/hr and on the level ground you would be walking 3 km/hr. So how much of that time would you be averaging 4 km/hr and how much of that time would you be walking 3 km/hr?", "Solution_10": "[hide]\nSince the problem doesn't give the time that Elizebeth was walking uphill or downhill, or on flat ground, assume that it doesn't matter. So the answer is 3*6=18[/hide]" } { "Tag": [], "Problem": "Mr. Jones has eight different children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. \"Look, daddy!\" she exclaims. \"That number is evenly divisible by the age of each of us kids!\" \"That's right,\" replied Mr. Jones, \"and the last two digits just happen to be my age.\" Which of the following is not the age of one of Mr. Jones's children?\r\nA) 4\r\nB) 5\r\nC) 6\r\nD) 7\r\nE) 8", "Solution_1": "[quote=\"mathdog15\"]Mr. Jones has eight different children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. \"Look, daddy!\" she exclaims. \"That number is evenly divisible by the age of each of us kids!\" \"That's right,\" replied Mr. Jones, \"and the last two digits just happen to be my age.\" Which of the following is not the age of one of Mr. Jones's children?\nA) 4\nB) 5\nC) 6\nD) 7\nE) 8[/quote]\r\n\r\n[hide]It is B) 5. I used divisibility rules and guess and check basically, but the license plate is 5544. This is divisible by 9, 8, 7, 6, 4, 3, 2, and 1. [/hide]", "Solution_2": "All of the AMC problems have already been solved:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=433616#433616\r\nhttp://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2006" } { "Tag": [ "mathleague.org", "ARML", "ratio", "geometry", "3D geometry", "octahedron", "calculus" ], "Problem": "Who all knows about this contest? The web site (mathleague.org) is fairly limited. Looks like it's just Kansas, Missouri, and Iowa. Tim Sanders (ARML president) ran this contest for a while, I'm not sure if he started it. I just won the Missouri state contest on Saturday, with scores of Target 80/80 (I protested number 8, it was thrown out) and Sprint 110/120 (used 3x/8 instead of 5x/8 for the ratio in the infinite series on 14, and they rejected my protest on number 16, the instructions don't say anything about what to do if there's not enough information), so total 190/200, second place was 170. Does anyone know the criterion for being invited to Regional Championship? It's at KU on May 14, I'm just curious who all else (and about how many people) will be there, and winning scores in Kansas and Iowa.", "Solution_1": "hehe, u did pretty well...\r\n\r\n95 on sprint (missed 16 as well; stupid mistakes on #2 and #4, the border problem and the octahedron problem)\r\n\r\n60 on target (missed 45 degrees, and the geometric series)...yea, number 8 on that test was sketchy...almost half my team, including me, put down \"no positive integral solution\"...i think they should have just counted that as the correct answer, instead of tossing out the question entirely...\r\n\r\nanyways, cya at regionals...7 from my school qualified, ~5 or 6 ppl will be there", "Solution_2": "oh yea, criteria: \r\n1.) top 3 in grade level in each indiv. event\r\n2.) top 1 in each \"team\" event (i.e. team, relay, power)\r\n3.) i think, but not sure, top sweepstakes?\r\n\r\ni got 2nd in target, 3rd in sprint, 1st team, 2nd relay, 3rd power, team got 1st overall sweeps...\r\n\r\nbtw, the test (every round, except maybe the team round) this year was WAAAY too easy...far easier than it had been in previous years...i didnt like it very much..." } { "Tag": [ "search", "function", "geometry", "3D geometry", "octahedron", "\\/closed" ], "Problem": "The search function brings up \"No topics or posts met your search criteria\" no matter what I search for (I tried generic things that are very common).", "Solution_1": "I just tried now a few different searches and all worked. What exactly did you search for?\r\n\r\nEdit: You are right, the search doesn't look into 2 or 3 letter words. I will try to fix that this weekend.", "Solution_2": "And it doesn't look into some longer words like \"value\", too...", "Solution_3": "Here's an interesting one- wildcards at the beginning of words don't work. For example, a search for *hedron generates only six results, all of which have \"hedron\" somewhere in them. Words like \"octahedron\" and \"polyhedron\" are highlighted, but don't count as a match to generate results in the first place.", "Solution_4": "Yes the index has to be rebuilt. I will try to do that this weekend sometime. It seems that it got corrupted during the move." } { "Tag": [ "factorial" ], "Problem": "Find the sum...\r\n$ \\frac{220!}{219!\\plus{}218!}\\plus{}\\frac{221!}{220!\\plus{}219!}\\plus{}\\dots\\plus{}\\frac{228!}{227!\\plus{}226!}$", "Solution_1": "[hide]\n$ \\frac{218!}{218!}\\left(\\frac{220\\cdot 219}{219\\plus{}1}\\right)\\plus{}\\frac{219!}{219!}\\left(\\frac{221\\cdot 220}{220\\plus{}1}\\right)\\plus{}...\\plus{}\\frac{226!}{226!}\\left(\\frac{228\\cdot 227}{227\\plus{}1}\\right)$\n$ \\frac{9\\left( 219\\plus{}227\\right)}{2}\\equal{} 2007$.\n[/hide]" } { "Tag": [ "articles" ], "Problem": "I have been doing many problems in AoPS recently. The USA uses imperial units as their primary measurement system. However, I live in Singapore where we only use SI units. Thus I know almost nothing about the imperial units.\r\n\r\n[quote]The rails of a railroad are 30 feet long. As a train passess over the point where the rails are joined, there is an audible click, The speed of the train in mph is approximately the number of clicks heard in how many seconds?[/quote]\n\nThat was from the 1955 AHSME. The maths is very easy, but the terminology, no.\n\nAccording to Wikipedia, 1 international mile is 5280 feet. Do I use nautical or the international mile?[/quote]", "Solution_1": "\"Mile\" without qualification pretty much always refers to the international mile in a general context.", "Solution_2": "Even in the USA, the nautical mile is never used, unless specified.\r\n\r\nYou can always interpret \"mile\" as the international mile, if nothing is there to tell you otherwise.", "Solution_3": "I have been having similar problems as i generally use metric measurements.\r\nAlso, I think I have it sorted now but I was having trouble with some of the money related problems.", "Solution_4": "Watch out for tons (metric ton = 1000 Kg; ton = 2000 lb in the US) and billions (1 billion = 1000 millions in the US).\r\n\r\nI don\"t know of any other country where 1 billion = 1000 millions, do you?", "Solution_5": "In my country, 1 billion is equal to a thousand millions. ($ 10^9$)", "Solution_6": "I'm pretty sure a billion is defined as $ 10^9$ for everyone.", "Solution_7": "[quote=\"herefishyfishy1\"]I'm pretty sure a billion is defined as $ 10^9$ for everyone.[/quote]\r\nUntil I came to the US, 1 billion was the 12th power of 10, one trillion the 18th power, and so on. One n-illion would be the 6n power of ten (simpler than the 3n+3 power).\r\nI remember reading that to the French, 1000 millions is a \"milliard\" and one million millions is a \"billion\".\r\nI also read (recently) an article titled \"when a billion is not a billion\" or some close rewording of that. I'll try to google my way back to it.", "Solution_8": "Here are the references I could google:\r\n\r\nhttp://faculty.kutztown.edu/schaeffe/Trivia/trivia.html\r\nhttp://en.wikipedia.org/wiki/Long_and_short_scales#Short_scale_countries" } { "Tag": [ "modular arithmetic" ], "Problem": "For what units digit $ d$ is the five-digit number $ 23,45d$ a multiple of 9?", "Solution_1": "We must have \\[ 2\\plus{}3\\plus{}4\\plus{}5\\plus{}d\\equiv 0\\pmod{9} \\Rightarrow d\\equal{}\\boxed{4}\\]" } { "Tag": [ "calculus", "derivative", "number theory proposed", "number theory" ], "Problem": "If $a$ is a real root of $x^5 - x^3 + x - 2 = 0$, show that $[a^6] = 3$. \r\n\r\n\r\nIndian 2004!", "Solution_1": "$P(x)=x^5-x^3+x-2$ is increasing because its derivative is positive $P'(x)=5x^4-3x^2+1 >0 ,\\forall x\\in R$\r\n(One) $P(1.201) < 0 \\Longrightarrow a > 1.201 \\Longrightarrow a^6 > 3 $ (make the calculation :D )\r\n(Deux) $P(1.25) > 0 \\Longrightarrow a < 1.25 \\Longrightarrow a^6 < 4$ (make the calculation :D )\r\n(Trei) Trivial conclusion : $[a^6]=3$\r\n\r\n :cool: Do you dislike this kind of proof :cool:", "Solution_2": "It isn't too ugly! :) :) \r\n\r\nI was trying to prove it by proving that $43 + 1). You can find a definition at [url]http://mathworld.wolfram.com/EllipticIntegral.html[/url]\r\n\r\nThere is some information about quintic polynomials at\r\n[url]http://mathworld.wolfram.com/QuinticEquation.html[/url]", "Solution_25": "I think I can remember reading that it can be proven that no formula exists for n>=5, but what abt n=4?", "Solution_26": "There's a solution for the quartic in terms of radicals.", "Solution_27": "That won't work in general, of course, since not all cubics or quartics have rational roots." } { "Tag": [ "quadratics", "algebra", "polynomial" ], "Problem": "In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and obtains $ 8$ and $ 2$ for the roots. Another student makes a mistake only in the coefficient of the first degree term and find $ \\minus{}9$ and $ \\minus{}1$ for the roots. The correct equation was:\r\n$ \\textbf{(A)}\\ x^2\\minus{}10x\\plus{}9\\equal{}0 \\qquad\\textbf{(B)}\\ x^2\\plus{}10x\\plus{}9\\equal{}0 \\qquad\\textbf{(C)}\\ x^2\\minus{}10x\\plus{}16\\equal{}0\\\\\r\n\\textbf{(D)}\\ x^2\\minus{}8x\\minus{}9\\equal{}0 \\qquad\\textbf{(E)}\\ \\text{none of these}$", "Solution_1": "[hide]\n\nThe first statement says that the sum of the roots is 10, since the coefficient of x is correct.\nThe second statements says that the product of the roots is 9, since the constant term is correct.\n\nTherefore, the polynomial is $ x^2 \\minus{} 10x \\plus{} 9 \\equal{} 0\\implies\\boxed{\\text{A}}$[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that $ 2^n \\minus{} 1$ cannot divide $ 3^n \\minus{} 1$ for any $ n > 1$", "Solution_1": "If $ n$ is even then $ 2^n\\minus{}1$ is a multiple of $ 3$ and $ 3^n\\minus{}1$ is not; so $ n$ is odd\r\nThen you can see a more general problem here:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=1148144#1148144\r\n\r\nDaniel" } { "Tag": [ "trigonometry", "calculus", "integration", "function", "real analysis", "real analysis unsolved" ], "Problem": "Show that the series $ \\sum_{n\\equal{}1}^{\\infty} \\frac{\\sin n}{n}$ is conditionally convergent.", "Solution_1": "This follows from the Abel-Dirichlet criterion ( http://en.wikipedia.org/wiki/Dirichlet%27s_test ) since\r\nI. $ 1/n$ is a decreasing sequence that tends to $ 0$.\r\nII.\r\n\\[ \\left|\\sum_{n \\equal{} 1}^k \\sin n \\right| \\equal{} \\left|\\frac {\\cos\\frac {1}{2} \\minus{} \\cos\\frac {k \\plus{} 1}{2}}{\\sin\\frac {1}{2}}\\right| \\text{ is bounded.}\r\n\\]", "Solution_2": "OK, that's a proof of convergence (summation by parts; it has appeared plenty of times before).\r\n\r\nNow, can you find the exact sum?", "Solution_3": "I wonder whether we've a simpler proof for it's convergence or not :?:", "Solution_4": "The proof of Abel-Dirichlet is about as simple as it gets; it's not a deep proof at all.", "Solution_5": "[quote=\"Timestopper_STG\"]I wonder whether we've a simpler proof for it's convergence or not :?:[/quote]\r\n\r\nI'd guess that my proof is the simplest :D \r\n\r\nIf you are willing to use complex or Fourier analysis you can get a [i]shorter[/i] proof, and even compute the sum (which is $ (\\pi \\minus{} 1)/2$). Whether this proof is [i]simpler[/i] depends on your taste. :)", "Solution_6": "Since I said [b]conditional[/b] convergence you also have to prove that $ \\sum_{n\\equal{}1}^{\\infty} \\frac{|\\sin n|}{n}$ is divergent. Or am I missing something?", "Solution_7": "That is trivial.", "Solution_8": "[quote=\"Timestopper_STG\"]I wonder whether we've a simpler proof for it's convergence or not :?:[/quote]\r\n\r\nWell, there is another option, though maybe not simpler.\r\nFirst, we use integral test. So we want to prove the integral converges. But just split the integral at each 0 of the function, so we have an alternating sequence whose terms go to 0 which converges.\r\n\r\nEDIT: As said below, this argument is wrong. Sorry.", "Solution_9": "The integral test does not apply. One of its hypotheses is that the function is nonnegative and decreasing.", "Solution_10": "The integral test only works for decreasing, nonnegative sequences. The given sequence is neither ;)\r\n\r\nEdit: too slow.", "Solution_11": "[quote=\"jmerry\"]\n\nNow, can you find the exact sum?[/quote]\r\n\r\nUse Fourier serie", "Solution_12": "[hide]$ \\sum_{n\\equal{}1}^{\\plus{} \\infty} \\frac{\\sin(n \\theta)}{n} \\equal{} \\frac{\\pi \\minus{} \\theta}{2}$\n\nSee, for example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110277]here[/url].[/hide]", "Solution_13": "[quote=\"Yury\"]That is trivial.[/quote]\r\nCan you please explain?", "Solution_14": "One way is to show that for every real number $ n$, either $ |\\sin n| \\ge 0.1$ or $ |\\sin(n \\plus{} 1)| \\ge 0.1$.", "Solution_15": "[b]Hint 1[/b]: Prove that for every $ x$, either $ |\\sin x| \\geq \\sin (1/2)|$ or $ |\\sin (x \\plus{} 1)| \\geq \\sin(1/2)$ (this is what Ravi B suggested);\r\n\r\nor\r\n\r\n[b]Hint 2[/b]: Prove that $ |\\sin x| \\plus{} |\\sin (x \\plus{} 1)| \\geq \\sin 1$;\r\n\r\nor \r\n\r\n[b]Hint 3[/b]: Note that $ |\\sin x| \\plus{} |\\sin (x \\plus{} 1)|$ is a continuous function, and therefore attains its minimum on $ [0, 2\\pi]$ and thus (why?) on $ \\mathbb R$. On the other hand, $ |\\sin x| \\plus{} |\\sin (x \\plus{} 1)| \\neq 0$. Therefore, $ |\\sin x| \\plus{} |\\sin (x \\plus{} 1)| \\geq \\min_x (|\\sin x| \\plus{} |\\sin (x \\plus{} 1)|) \\equiv c > 0$.", "Solution_16": "Hint 4: use the equidistribution theorem." } { "Tag": [], "Problem": "If 1 ounce is equivalent to approximately 28.35 grams, how many ounces are in 541 grams? (Answer to the nearest tenth of an ounce.)", "Solution_1": "$ \\frac{28.35}{1}\\equal{}\\frac{541}{x}$\r\n$ x\\equal{}\\frac{541}{28.35}\\equal{}19.08$", "Solution_2": "If $ \\frac{a}{b} \\equal{} \\frac{c}{d}$, then $ ad \\equal{} bc$.\r\n\r\nAnd also, the answer would be $ \\boxed{19.1}$, because it says to the nearest [b]tenth[/b]." } { "Tag": [], "Problem": "Three pies and four cakes sell for $ \\$35$ while four pies and five cakes sell for $ \\$44.50$. What is the cost to purchase one pie and one cake?", "Solution_1": "hello\r\n$ 4x\\plus{}3y\\equal{}35$\r\n$ 5x\\plus{}4y\\equal{}44.50$\r\n$ x\\plus{}y\\equal{}9.5$\r\nthank u" } { "Tag": [ "function", "calculus", "integration", "algebra proposed", "algebra" ], "Problem": "Let $f$ be a function defined on the positive integers, taking positive integral values, such that\r\n\r\n$f(a)f(b) = f(ab)$ for all positive integers $a$ and $b$,\r\n$f(a) < f(b)$ if $a < b$,\r\n$f(3) \\geq 7$.\r\n\r\nFind the smallest possible value of $f(3)$.", "Solution_1": "ok ill post my solution:\r\n\r\nNow clearly $f(x)=x^2$ is a function that satisfies all the conditions, yielding $f(3)=9$\r\n\r\nBut with this conjecture we need to prove that $f(3)\\neq7, 8$\r\n\r\nSeparate this into two cases.\r\n\r\nCASE 1 $f(3)=7$\r\n\r\n$\\implies 2\\leq f(2) \\leq 6$ \r\n\r\nWe need to check that for each of these possible values of $f(2)$, we arrive at a contradiction.\r\n\r\nIf $f(2) = 2$\r\n\r\n$\\implies f(4)=4, f(3)=7$. contradiction because $f(a)3$\r\n\r\nthen $f(8)>64$ but $f(9) = 49$ contradiction.\r\n\r\nCASE 2 is similar to CASE 1. you know how to do it :P im too lazyyy\r\n\r\nand after proving all this we realise that the minimum value of $f(3)=9$ :D", "Solution_2": "I remember that it have been in a Chinese competition.\r\nwe claim that all the function satisfied the given condition is $f(x)=x^a$\r\nwhere a is a real number.", "Solution_3": "But this is a function from the positive integers to the positive integers.", "Solution_4": "This gives that $a$ is a positive integer, which solves the problem!", "Solution_5": "Does it? I mean, we know that multiplicative functions defined on the reals are of the form $f(x) = x^a$, but does this also hold for multiplicative functions defined on the positive integers?", "Solution_6": "No, the original problem (the one from China) was also from $\\mathbb{N}$ to some set ($\\mathbb{R}$ or so), and had also the condition that the function is increasing.", "Solution_7": "Oh, I see, didn't know that :) the fact that $f$ is increasing changes a few things indeed.", "Solution_8": "Btw, here it is: http://www.mathlinks.ro/Forum/viewtopic.php?t=55870", "Solution_9": "So observe that $f(x)=x^2$ is a solution.We will prove that $f(3)\\neq7,8$\nFor that we see that f(1)=1 and $f(2)\\ge 3$.Also $f(16)0$ \u03bd\u03b4\u03bf :\r\n\r\n$ \\frac {a}{a^2 \\plus{} 2} \\plus{} \\frac {b}{b^2 \\plus{} 2} \\plus{} \\frac {c}{c^2 \\plus{} 2} \\leq 1$", "Solution_1": "\u039c\u03b9\u03c7\u03ac\u03bb\u03b7 a,b,c \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af????", "Solution_2": "\u0391\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03cc Baltic Way. \u0395\u03cd\u03ba\u03bf\u03bb\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9...\r\n\r\n\u0388\u03bd\u03b1 \u03bc\u03b9\u03ba\u03c1\u03cc hint:[hide]$ a/(a^2\\plus{}2)<\\equal{}a/(2a\\plus{}1)$[/hide]", "Solution_3": "\u039d\u03b1\u03b9 \u03b5\u03c5\u03c4\u03c5\u03c7\u03ce\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03bf \u03b5\u03af\u03c0\u03b5\u03c2 \u039c\u03ae\u03c4\u03c3\u03bf... \u03a4\u03bf \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b1 :wink:", "Solution_4": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c6\u03b9\u03ad\u03c1\u03c9\u03c3\u03b7 ... \u038c\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b5 \u03bf \u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \r\n\r\n[hide]$ \\frac{a}{a^2\\plus{}2}\\leq\\frac{a}{2a\\plus{}1}$ \u03ac\u03c1\u03b1 \u03bc\u03ad\u03bd\u03b5\u03b9 \u03bd\u03b4\u03bf $ \\sum\\frac{a}{2a\\plus{}1}\\leq 1$ \u03c0\u03bf\u03c5 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b5\u03af \u03bc\u03b5 \n\n$ a\\plus{}b\\plus{}c\\geq 3$ \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 qed [/hide]", "Solution_5": "\u03ae\u03c4\u03b1\u03bd \u03bb\u03af\u03b3\u03bf \u03b5\u03c5\u03ba\u03bf\u03bb\u03ac\u03ba\u03b9.... \u03b4\u03b5\u03c2 \u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03b2\u03b1\u03bb\u03b1 \u03c4\u03ce\u03c1\u03b1 :)" } { "Tag": [ "geometry", "rectangle", "analytic geometry", "graphing lines", "slope", "geometric transformation", "reflection" ], "Problem": "Let $ n$ be a positive integer. Consider a rectangle $ (90n\\plus{}1)\\times(90n\\plus{}5)$ consisting of unit squares. Let $ S$ be the set of the vertices of these squares. Prove that the number of distinct lines passing through at least two points of $ S$ is divisible by $ 4$.", "Solution_1": "Hope I didn't make some mistake again...\r\n\r\nFirst of all, put origin of cartesian coordinate system in lower left corner.\r\nThere are four types of these lines: horizontal, vertical, with positive slope, and with negative slope.\r\nThere are $ 180n + 8$ horizontal and vertical lines, so its divisible with 4. Number of lines with positive slope is the same as the number of lines with negative slope, by simetry, so the number of all lines is divisible with 4 iff number of lines with positive slope is even. We are going to prove that:\r\n\r\nThe slopes of these lines are rational numbers of the form: $ \\frac ba, a\\in {1,...,90n + 1\\}, b\\in \\{1,...,90n + 5\\}, gcd(a,b)=1}$, and let $ A$ be the set of all the slopes of these lines (there is no line with slope smaller then $ \\frac 1{90n + 1}$ or bigger then $ 90n + 5$). It's not hard to se that number of lines with slope $ \\frac ba$ is $ b\\cdot (90n + 1 + 1 -a) + a\\cdot (90n + 5 + 1 - b) - ab = b(90n + 2 ) + a(90n + 6) - 3ab\\equiv_2 ab$. If $ \\frac ba \\in A$ and $ (a,b)\\not = (1,1)$ and $ b\\leq 90n + 1$ then $ \\frac ab$ is also in $ A$, so the number of lines with slope $ \\frac ba$ + the number of lines with slope $ \\frac ab$ is $ 2ab \\mod 2$ so it's even. \r\n\r\nNow we have to count lines with slopes $ 1, \\frac {90n + 2}a, \\frac {90n + 3}a, \\frac {90n + 4}a, \\frac {90n + 5}a$. There are $ 180n + 5$ lines with slope 1, and there are even number of lines with slopes of the form $ \\frac {90n + 2}a$ and $ \\frac {90n + 4}a$, so we have to deal with slopes $ \\frac {90n + 3}a$ and $ \\frac {90n + 5}a$.\r\n\r\nWe saw that if $ a$ is even, then number of lines with slope $ \\frac {90n + 3}a$ is even, but when $ a$ is odd, then there is odd number of those lines. But there are even number of $ a$'s which are relatively prime with $ 90n + 3$ and odd ($ 4|\\varphi(90n + 3)$), so the number of all lines with slope $ \\frac {90n + 3}a$ is even. \r\nOn the other hand, there is odd number of odd $ a$ which are relatively prime with $ 90n + 5$ and not bigger then $ 90n + 1$ (again $ 4|\\varphi (90n + 5)$ but $ a$ can't be $ 90n+3$) so there is odd number of lines with slopes of the form $ \\frac {90n + 5}a$, but the number of lines with slope $ 1$ is also odd. So, the number of all lines with positive slope is even. QED", "Solution_2": "hmmm.\r\n\r\ncan't you just take the two perpendicular bisectors of the sides of the rectangle as your x and y axis? then for each such line except the lines passing through the origin and the lines parallel to the axis there will be 4 images (reflection on x axis, y axis 180 degrees rotation by the origin and the line itslef) that are distinct and this gives a partition into four-tuples for these lines. together with the 180n+8 horizontal or vertical lines we only need to show that the number of lines passing through the origin is divisible by 4.\r\nnow this number corresponds to twice the number of rational numbers $ \\frac{a}{b}$ with $ a\\in \\{1,3,5,\\dots, 90n\\plus{}1\\},b\\in \\{1,3,\\dots,90n\\plus{}5\\}$ which we have to show is even.\r\nas we can pair $ \\frac{a}{b}$ with $ \\frac{b}{a}$ then we only need to consider $ b\\in \\{90n\\plus{}3,90n\\plus{}5\\}$ which gives us\r\n$ \\varphi (90n\\plus{}5)\\plus{}\\varphi (90n\\plus{}3)$\r\nthe first term is divisible by 4 as $ 5|90n\\plus{}5$ and the second term is divisible by four as $ 90n\\plus{}3$ contains at least two odd prime divisors", "Solution_3": "Well, yeah, yours is a bit smarter :) When I solved it, i didn't try to improve it. \r\nI am just not sure, how do you exclude cases $ b\\in \\{90n \\plus{} 2, 90n \\plus{} 4\\}$?", "Solution_4": "note that in my coordinate system the points can not have integer coordinates, instead they are of the form $ (\\frac{2p\\plus{}1}{2},\\frac{2q\\plus{}1}{2})$\r\ndo you see it now?", "Solution_5": "yuup :)\r\nbye", "Solution_6": "dude no guys\r\n\r\nthis method fails. \r\n\r\nhere's a hint: are you sure (Albanian Eagles) that what you want to consider is phi(90n+5)+phi(90n+3) ?", "Solution_7": "Yes pretty sure.", "Solution_8": "no dude listen\r\n\r\nyou want to consider gamma(93)+gamma(95), where the gamma function is the number of ODD numbers relatively prime to whatever. \r\n\r\nbtw if you use your method you end up with 2 mod 4, since the line y=x is paired with y=x when you pair reciprocal slopes. \r\n\r\nthat's why you actually need to show gamma(93)+gamma(95) is odd\r\nrather than phi(93)+phi(95) is even", "Solution_9": "another way:(Hope i don't mistake .please check it)\nyou can take perpendicular bisector of the sides of rectangle the $X,Y$ axes.the number of vertical and horizontal lines are $180n+8$ that is divisible by $4$.the number of lines that pass the origin is $2(45n+3)+2(45n+1)$ that is divisible by $4$.rest of the lines can be partitioned into groups such that the number of each group is $4$.because we can consider of reflections of each line through the axes$X,Y$. and that is exactly $4$." } { "Tag": [ "AMC" ], "Problem": "This is the thread for the posting of the American Mathematics Competitions 10. Please read over the directions and submit answers by Private Message to [b]anirudh[/b]. Please include all answers in a readable answer with solutions.\r\n\r\nThank you. :)\r\n\r\nSorry for the Mis-typed file name. Some of the problems are probably too hard though.\r\n\r\nSign up in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029.\r\n\r\nDownload the problems from http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=6273 if you cannot access the attachment.\r\n\r\nAlternate Download: [url=http://www5.upload2.net/download/2fed2efdeec894596fb24c9f6c0848e0/4588b97e/X09B1MPZEX7P25e/AMC10.pdf]here[/url]\r\n\r\nPlease do not post in this thread. Do all posting in the other post.", "Solution_1": "Update: ERRATA HAS BEEN FIXED!!! DOWNLOAD THE NEW VERSION!!!", "Solution_2": "[size=200]As posted in the other thread, as not to cause confusion, no more solutions are required.[/size]\r\n\r\nMy original plan was to take people's best solutions and put them together in a solution booklet, but I'm just going to create the solutions myself.\r\n\r\nUnfortunately, I can't edit the above posts, so please just ignore the part about solutions." } { "Tag": [ "trigonometry", "trig identities", "tangent half angle" ], "Problem": "could someone please give me some trig problems for practice? i just started to really learn about it yesterday. i know a bit about graphs and some identities. thanks", "Solution_1": "Show that $ \\tan{\\frac{x}{2}}\\equal{}\\frac{\\sin{x}}{1\\plus{}\\cos{x}}$.\r\nThere are many ways to do this problem so see if you can find them. Oh, and you can't use any formula that you don't derive (except sum and difference formulas).", "Solution_2": "[hide=\"One Proof\"]First, we use the double angle formulas for cosine, which are easily derived from the general cosine sum formula and the identity $ \\sin^2 \\theta + \\cos^2 \\theta = 1$:\n[b]Sine:[/b]\n\\begin{align*} \\cos 2\\theta = 1 - 2\\sin^2 \\theta & \\Rightarrow 2\\sin^2 \\theta = 1 - \\cos 2\\theta \\\\\n& \\Rightarrow \\sin^2 \\theta = \\frac {1 - \\cos 2\\theta}{2} \\\\\n& \\Rightarrow \\sin \\theta = \\pm \\sqrt {\\frac {1 - \\cos 2\\theta}{2}} \\end{align*}\n[b]Cosine:[/b]\n\\begin{align*} \\cos 2\\theta = 2\\cos^2 \\theta - 1 \\\\\n& \\Rightarrow 2\\cos^2\\theta = 1 + \\cos 2\\theta \\\\\n& \\Rightarrow \\cos^2 \\theta = \\frac {1 + \\cos 2\\theta}{2} \\\\\n& \\Rightarrow \\cos \\theta = \\pm \\sqrt {\\frac {1 + \\cos 2\\theta}{2}} \\end{align*}\n[b]Tangent Half-Angle:[/b]\nIf we let $ x = \\frac {\\theta}{2}$, and use $ \\tan \\theta = \\frac {\\sin \\theta}{\\cos \\theta}$, we have (the plus-or-minus sign drops out because the the numerator and denominator will be positive for any angle $ x$):\n\\[ \\tan \\frac {x}{2} \\& = \\sqrt {\\frac {1 - \\cos x}{1 + \\cos x}} \\\\\n\\& = \\frac {\\sqrt {1 - \\cos x}}{\\sqrt {1 + \\cos x}} \\\\\n\\& = \\frac {\\sqrt {1 - \\cos^2 x}}{1 + \\cos x} \\\\\n\\& = \\frac {\\sqrt {\\sin^2 x}}{1 + \\cos x} \\\\\n\\& = \\boxed{\\frac {\\sin x}{1 + \\cos x}}\n\\]\n[/hide]", "Solution_3": "[hide=\"Another simpler one\"]\nWrite $ \\sinh = 2 \\cdot \\sin {\\frac {x}{2}} \\cdot \\cos {\\frac {x}{2}}$\nAnd $ 1 + \\cos x = 2 \\cdot cos^2 {\\frac {x}{2}}$ :) \n[/hide]", "Solution_4": "1. Given $ \\sin ^2 \\plus{} \\cos^2 \\equal{} 1$, derive the other Pythagorean Identities (yes this is a simple one)." } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities proposed" ], "Problem": "For $a,b,c$ positive real numbers, prove that\r\n \r\n$\\frac{a^2+2bc}{(b+c)^3}+\\frac{b^2+2ca}{(c+a)^3}+\\frac{c^2+2ab}{(a+b)^3}\\ge \\frac{27}{8(a+b+c)}$", "Solution_1": "I dedicate the idea of this proof to darij grinberg.\r\n\r\nBy generalized schur inequality (Vornicu-schur :D),\r\n\\[\r\n\\sum {\\frac{{(a - b)(a - c)}}{{(b + c)^3 }}} \\ge 0\r\n\\]\r\nSo we have to prove,\r\n\\[\r\n\\left( {a + b + c} \\right)\\left( {ab + bc + ca} \\right)\\sum {\\frac{1}{{(a + b)^3 }}} \\ge \\frac{{27}}{8}\r\n\\]\r\nNow,\r\n\\[\r\n\\begin{array}{l}\r\n \\left( {a + b + c} \\right)\\left( {ab + bc + ca} \\right)\\sum {\\frac{1}{{(a + b)^3 }}} \\\\ \r\n \\ge \\frac{1}{3}\\left( {a + b + c} \\right)\\left( {ab + bc + ca} \\right)\\sum {\\frac{1}{{(a + b)^2 }}} \\sum {\\frac{1}{{(a + b)}}} \\\\ \r\n = \\frac{1}{3}\\left( {a + b + c} \\right)\\sum {\\frac{1}{{(a + b)}}} \\left( {ab + bc + ca} \\right)\\sum {\\frac{1}{{(a + b)^2 }}} \\\\ \r\n \\ge \\frac{1}{3}\\left( {\\frac{9}{2}} \\right)\\left( {\\frac{9}{4}} \\right) \\\\ \r\n = \\frac{{27}}{8} \\\\ \r\n \\end{array}\r\n\\]\r\nby chebyshev, am-hm and iran 1996.", "Solution_2": "I don't remember this one\r\n\r\n \r\n$\\sum {\\frac{{(a - b)(a - c)}}{{(b + c)^3 }}} \\ge 0$\r\n\r\n\r\nWhere did you find it?", "Solution_3": "[quote=\"manlio\"]I don't remember this one\n\n \n$\\sum {\\frac{{(a - b)(a - c)}}{{(b + c)^3 }}} \\ge 0$\n\n\nWhere did you find it?[/quote]\r\n\r\nit is a special case of vornicu-schur inequality.", "Solution_4": "Where did you find it?", "Solution_5": "[quote=\"manlio\"]Where did you find it?[/quote]\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212 :D", "Solution_6": "Sorry, I didn't remember it :blush: :blush:", "Solution_7": "With the same conditions, the generalized inequality \r\n\r\n$ \\frac {a^2 \\plus{} kbc}{(b \\plus{} c)^{3}} \\plus{} \\frac {b^2 \\plus{} kca}{(c \\plus{} a)^{3}} \\plus{} \\frac {c^2 \\plus{} kab}{(a \\plus{} b)^{3}}\\geq\\frac {9(1 \\plus{} k)}{8(a \\plus{} b \\plus{} c)}$\r\n\r\nholds if and only if $ \\minus{} 219.06\\cdots \\equal{} k_3\\leq k\\leq \\frac {23}{7} \\equal{} 3.2857\\cdots,$\r\n\r\nwhere $ k_3$ is the least real root of the following irreducible polynomial over $ \\mathbb{Q}:$\r\n\r\n$ 4k^5 \\plus{} 875k^4 \\minus{} 280k^3 \\minus{} 274k^2 \\plus{} 28k \\minus{} 49;$\r\n\r\nwith equality if $ k \\equal{} k_3,b \\equal{} c \\equal{} 1,a \\equal{} 6.7506\\cdots$ is the greatest real root of the following irreducible polynomial\r\n\r\n$ 2a^5 \\plus{} a^4 \\minus{} 56a^3 \\minus{} 226a^2 \\minus{} 354a \\minus{} 199.$\r\n\r\nI'm expecting nice solutions for $ k \\equal{} \\frac {23}{7}:$\r\n\r\n$ \\frac {7a^2 \\plus{} 23bc}{(b \\plus{} c)^{3}} \\plus{} \\frac {7b^2 \\plus{} 23ca}{(c \\plus{} a)^{3}} \\plus{} \\frac {7c^2 \\plus{} 23ab}{(a \\plus{} b)^{3}}\\geq\\frac {135}{4(a \\plus{} b \\plus{} c)}$ \r\n\r\nand $ k \\equal{} \\minus{} 219:$\r\n\r\n$ \\frac {a^2 \\minus{} 219bc}{(b \\plus{} c)^{3}} \\plus{} \\frac {b^2 \\minus{} 219ca}{(c \\plus{} a)^{3}} \\plus{} \\frac {c^2 \\minus{} 219ab}{(a \\plus{} b)^{3}}\\geq\\frac { \\minus{} 981}{4(a \\plus{} b \\plus{} c)}.$" } { "Tag": [ "modular arithmetic", "induction", "function", "number theory", "totient function" ], "Problem": "Prove that $ 2^{3^n} \\plus{}1$ is divisible by $ 3^{n\\plus{}1}$ for all $ n$.", "Solution_1": "so you need to show that\r\n\r\n$ 2^{3^n} \\equiv \\minus{} 1 \\pmod{3^{n \\plus{} 1}}$\r\n\r\n\r\n\r\nFirst, by euler's totient theorem\r\n\r\n$ a^{3^n(3 \\minus{} 1)} \\equiv 1 \\pmod{3^{n \\plus{} 1}}$\r\n\r\nfor all $ a$ not divisible by $ 3$, and therefore also for $ a \\equal{} 2$. Now let $ 2^{3^n} \\equal{} x$. Our relation is then\r\n\r\n$ x^2 \\equiv 1 \\pmod{3^{n \\plus{} 1}}$\r\n\r\nand hence $ 3^{n \\plus{} 1} | x^2 \\minus{} 1 \\equal{} (x \\minus{} 1)(x \\plus{} 1)$. Therefore at least one of these factors must be divisible by $ 3$.\r\n\r\nCase 1: $ 3|x \\minus{} 1$ which implies $ x \\equiv 1 \\pmod{3}$, $ x \\plus{} 1 \\equiv 2 \\pmod{3}$ and therefore $ 3$ does not divide $ x \\plus{} 1$. Thus, $ 3^{n \\plus{} 1} | x \\minus{} 1$ implies $ x \\equiv 1 \\pmod{3^{n \\plus{} 1}}$.\r\n\r\nCase 2: $ 3|x \\plus{} 1$ which implies $ x \\equiv \\minus{} 1 \\pmod{3}$, $ x \\minus{} 1 \\equiv 1 \\pmod{3}$ and therefore $ 3$ does not divide $ x \\minus{} 1$. Thus, $ 3^{n \\plus{} 1} | x \\plus{} 1$ implies $ x \\equiv \\minus{} 1 \\pmod{3^{n \\plus{} 1}}$, as desired.\r\n\r\nOur proof will be complete if we can show the impossibility of case 1. That is, we only need to show that $ x \\equal{} 2^{3^n} \\equiv 2 \\pmod{3}$ for any $ n$. This is simple: Since $ 3^n$ is always odd, there exists an integer $ k$ such that $ 2k \\plus{} 1 \\equal{} 3^n$. Therefore, $ 2^{3^n} \\equiv 2^{2k \\plus{} 1} \\equiv 4^k \\cdot 2 \\equiv 2 \\pmod{3}$, exactly as desired.", "Solution_2": "How would we prove the Euler's Totient Theorem and the congruence $ 2\\cdot 4^k\\equiv 2\\mod 3$?\r\n\r\nEDITED TYPO", "Solution_3": "Induction: $ 2^{3^n} \\plus{} 1 \\equal{} \\left(2^{3^{n \\minus{} 1}} \\plus{} 1\\right) \\left[\\underbrace{\\left(2^{3^{n \\minus{} 1}}\\right)^2}_{\\equiv 1 \\pmod{3}} \\minus{} \\underbrace{\\left(2^{3^{n \\minus{} 1}}\\right)}_{\\equiv \\minus{} 1 \\pmod {3}} \\plus{} 1\\right]$.\r\n\r\n\r\n\r\n[quote=\"mathwizarddude\"]Could would we prove the Euler's Totient Theorem and the congruence $ 2\\cdot 4^k\\equiv 2\\mod 3$?[/quote]\r\n\r\n'Prove [[Euler's Totient Theorem]]' as in a proof of the theorem?\r\n\r\n$ 2 \\cdot 4^k \\equiv 2 \\cdot 1^k \\equiv 2 \\pmod{3}$.", "Solution_4": "On [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_function#Formulas]this page[/url] where it says\r\n\"$ p_1^{e_1\\minus{}1}p_2^{e_2}\\cdots p_m^{e_m} \\plus{} p_1^{e_1}p_2^{e_2\\minus{}1}\\cdots p_m^{e_m} \\plus{} \\cdots \\plus{} p_1^{e_1}p_2^{e_2}\\cdots p_m^{e_m \\minus{} 1}.$\r\n\r\nWe can factor out, though:\r\n\r\n$ p_1^{e_1\\minus{}1}p_2^{e_2\\minus{}1}\\cdots p_m^{e_m\\minus{}1}(p_1\\plus{}p_2\\plus{}\\cdots \\plus{} p_m).$\"\r\n\r\nI don't think the factorization is correct.", "Solution_5": "azips: Is there an elementary proof of Euler's Totient Theorem?", "Solution_6": "[quote=\"mathwizarddude\"]azips: Is there an elementary proof of Euler's Totient Theorem?[/quote]\r\n\r\nYes there is;\r\nEuler's Totient theorem is a generalisation of Fermat's little theorem, so i suggest you proove that one fist.\r\n[url=http://www.cut-the-knot.org/blue/Fermat.shtml//url]Fermat's Little Theorem[/url]\r\n[url=http://www.cut-the-knot.org/blue/Euler.shtml//url]Euler's Totient Function[/url]\r\n\r\nEDIT: sorry the links don't work very well, they will only take you to the website index, but the proofs are easy enough to find from there.", "Solution_7": "[quote=\"mathwizarddude\"]azips: Is there an elementary proof of Euler's Totient Theorem?[/quote]\r\n\r\nAlso added a quick proof to wiki - [[Fermat's Little Theorem]] which can be generalized to Euler's." } { "Tag": [ "USAMTS", "email" ], "Problem": "when i first registered for usamts last year, i was under 13. i did not participate, however. this year i am over 13 but it says that i need to submit the under 13 form on the usamts website. i have changed by grade under my profile. do i have to send the under 13 permission form?", "Solution_1": "Send an email to usamts@usamts.org about your age; include your USAMTS Id #. Don't worry about the form.", "Solution_2": "thank you, I have sent the email." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $x,y,z$ be nonegative real numbers, prove that \\[(x^{2}+y^{2})(y^{2}+z^{2})(z^{2}+x^{2})(x+y-z)(y+z-x)(z+x-y)\\leq 8(xyz)^{3}.\\]", "Solution_1": "a counter-example\r\n\r\n$[x=1,y=1/2,z=1]$" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AIME I", "AIME II" ], "Problem": "Hi, \r\n\r\nI'm a Junior in High School and I just scored a 6 on the AIME (I made some retarded mistakes). This is my last year to report my scores to colleges and makeing USAMO would be so incredibly helpful. Does anyone know If I can take the test on the alternate date, or if that is only for people who didn't take the first one?", "Solution_1": "You can only take one or the other, not both.", "Solution_2": "Thanks,\r\n\r\nI guess I'm just screwed then. There's always next year, but then it wont count for college.", "Solution_3": "Don't worry, a 6 is still pretty good.", "Solution_4": "Qualifying for USAMO senior year, while it won't help with college admissions, is still something to be proud of, and something you might want to attain if you enjoy math :D", "Solution_5": "I was sick on AIME day, sick enough that i couldnt go. I was hoping to have a shot at usamo for college admissions. Now im not sure if my teacher will let me take the AIME II, guess i have to look around more. Next year i will still try to get usamo, but not for college admissions, for my own sake, and because i like math.", "Solution_6": "It won't help for admissions, but if you get waitlisted, it definitely improves your chances. Just make sure to call in or write them a letter stating your outstanding achievement, and who knows, you might just get in.", "Solution_7": "Thats all true,\r\n\r\nI'll prepare and make it last year. I only wish it had dawned on me to prepare for these thigns earlier. You hear that kids, start NOW.", "Solution_8": "Well actually, amazingly enough, sometimes you can randomly take the AIME II too.", "Solution_9": "Reporting next year's AIME score may be too late for college admissions, but a 6 is noteworthy. Contrary to what the poll on this forum may have lead you to believe (crazy nerds -.- *mutters* ) that is quite an achievement. I'd put it down when you're applying to college next year." } { "Tag": [ "probability", "AMC 10", "AoPS Books" ], "Problem": "hi, im new to problem solving and just took the AMC 10 A. Just wondering how u do the following problems:\r\n\r\n10. Coin A is flipped three times and coin B is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?\r\n\r\n13. At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?\r\n\r\n17. Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?\r\n\r\nAlso I just started reading the AoPs books, im wondering what chapters or topics should I read in order to be able to do these problems?", "Solution_1": "[quote=\"02\"]hi, im new to problem solving and just took the AMC 10 A. Just wondering how u do the following problems:\n\n10. Coin A is flipped three times and coin B is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?\n[/quote]\r\n\r\nYou break it up into cases. \r\n\r\nCase 1:\r\n3 heads from coin A and from B\r\n\r\nfor coin A it would just be 1/2^3 x 3C3=1/8\r\ncoin B=1/2^4 x 4C3=1/4\r\n\r\nmultiply the probabilities (independent events) to get 1/32.\r\n\r\nCase 2:\r\n2 heads from coin A and from B\r\n\r\np(A)=probability of A in this case (same for p(B))\r\n\r\np(A)=1/2^3 x 3C2=3/8\r\np(B)=1/2^4 x 4C2=3/8\r\n\r\np(A) x p(B)=9/64\r\n\r\nCase 3:\r\n1 head from coin A and coin B\r\n\r\np(A)=1/8 x 3C1=3/8\r\np(B)=1/16 x 4C1=1/4\r\n\r\np(A) x p(B)=3/32\r\n\r\nCase 4:\r\n0 heads from coin A and coin B\r\n\r\np(a)=1/8\r\np(b)=1/16\r\n\r\np(A) x p(B)=1/128\r\n\r\nnow you add up all the probabilites\r\n\r\n1/32+9/64+3/32+1/128=35/128", "Solution_2": "[quote=\"02\"]\n13. At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?\n[/quote]\r\n\r\nThis was basically an inverse propotion. \r\n\r\nm=men\r\nw=women\r\n\r\nmw=12x3=36\r\n\r\nYou're given that each woman danced with 2 men, so 2w=36\r\n\r\nThus, there were 18 women.", "Solution_3": "[quote=\"02\"]\n\n17. Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in [/quote]\r\n\r\nI used logic on this problem. When Brenda and Sally meet, they are obviously at the same point. So the next time they meet is when Brenda's Distance (B) and Sally's Distance (S) covers the whole track.\r\n\r\nSo if in the whole track, Sally ran 150 meters, then in half the track she would run 75 meters.\r\n\r\nSince Brenda ran 100 meters in half of the track, the distance of half of the track is 100+75=175. Thus, the length of the whole track is 2(175)=350.", "Solution_4": "Quote:13. At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party? \n\n\n\nTo further explain why Neal multiplied 12 and 3 together: (highlight the text below to view it)\n\n[hide]\n\nEach man danced with three women, and there were 12 men, so there were a total of 36 man-woman pairs. Since each woman participated in exactly two of those pairs, there were 36/2 = 18 women at the party.\n\n\n\nAlternatively, since each man had 3/2 as many partners as each woman did, only 2/3 as many men as women were necessary. Thus, if w is the number of women, we have 2/3*w = 12, so w = 18.[/hide]", "Solution_5": "I missed #17 on 10A...:-(\r\n\r\n17. Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters? \r\n\r\nI used algebra.\r\n\r\nBrenda's rate --- b \r\nSally's rate --- s\r\nlength of the track --- l\r\nl/2 = bt + st = 100 + st\r\nl = 200 + 2st\r\nl = 2bt + 150\r\n2l = 2(bt+st) + 350\r\n2l = l + 350\r\nl = 350", "Solution_6": "I missed 17 on the test also - along with 8 others...", "Solution_7": "What if some men danced with the same woman? Would the answer change?" } { "Tag": [ "logarithms", "function", "complex numbers", "absolute value" ], "Problem": "What is a^z defined as, where a is a real number and z is a complex number. And better yet, what is z_1^z_2, where they're both complex numbers?", "Solution_1": "Well, those definitions are slightly problematic, because you end up with multiple-valued things. If we take a positive real number to an imaginary power, we need the following definition of complex numbers: z = r*e^(i:theta:), where r is the absolute value of z and :theta: is the angle it makes with the positive x axis. (This is another way of writing the polar form of complex numbers.) If we just look at numbers z in the upper-half-plane, we can talk about a^(x+iy), where y is positive, we see that it can be written as a^x*e^(i*ln(y)) (where ln is the logarithm base e for real numbers). So that's nicely defined. I'm going to let Simon say anything else about this, because he knows it better than I do and I'd just be guessing.", "Solution_2": "We start by defining e^z. I feel somewhat bad saying this since it turns some of Euler's most amazing work into mere definition, but I'll do it anyway. If z=x+iy, then e^z=e^x*e^(iy), and e^(iy)=cos(y)+i*sin(y).\r\n\r\nNow we can turn a^z into a number of the form e^z'. Write a=e^ln(a). (I should note that ln is a real-valued real function only, taking positive reals to reals. ln(-1) and ln(i) are not defined.) a^z is therefore equal to e^(z*ln(a)), or e^(x*ln(a))*e^(iy*ln(a)), or a^x*(cos(y*ln(a))+i*sin(y*ln(a))).\r\n\r\nNow a bit more about logarithms: we should write log(z) for the logarithm of a complex number, where it is understood that the base is e. Unfortunately, the complex logarithm function is not uniquely defined. In fact, it has infinitely many values. (To see this, recall that e^(2*k*i*pi)=1 for any integer k.) Thus log(z) is referring the set of values whose antilog is z. We can make matters a bit better defining Log(z) (note the capitalization) to be the value of log(z) with -pi 4 Xe -+ 2 XeO3 + 24 HF + 3 O2.\r\n\r\nXeF6 + 3 H2O ------------------> XeO3 + 6 HF", "Solution_1": "Well what do you want explained?", "Solution_2": "Of course the reason why these products are formes :wink:", "Solution_3": "In $ \\ce{XeF_4}$ Xe is in +4 state so the hyrolysis leads to a disproportionation reaction.\r\n\r\nBut in $ \\ce{XeF6}$ Xe is already in the +6 state and hence can't disproportionate and hence undergoes normal hydrolysis to give $ \\ce{XeO3}$. In some cases partial hydrolysis can take place to give $ \\ce{XeOF_4}$ and $ \\ce{XeO_2 F_2}$.", "Solution_4": "Side questions:\r\n\r\n1) What are the geometries of the Xenon species $ \\ce{XeF2}, \\ce{XeF4}$, $ \\ce{XeF5 \\minus{} }$, $ \\ce{XeF6}$, and $ \\ce{XeO6^{4\\minus{}}}$?\r\n\r\n2) Can $ \\ce{XeF6}$ be stored in a glass container?", "Solution_5": "1. $ {XeF_2}$ ----Linear\r\n $ {XeF_4}$-----Square planar\r\n $ {XeF_6}$-----Distorted Octahedron\r\n $ {XeO_6^4 \\minus{} }$--square bipyramidal\r\n\r\n i think $ {XeF_5^ \\minus{} }$ has square pyramidal.. structure...\r\n\r\n2.I don think so cause $ {XeF_6}$ is a powerful flourinating agent and so the flourine will etch the glass...", "Solution_6": "1)\tXeF2-Linear, XeF4-square planar, XeF5-probably planar( pentagonal with two lone pairs occupying the axial positions of the pentagonal bipyramid) ,XeF6-distorted octahedron ,XeO6(4-)octahedral\r\n2)\tNope, it is stored in nickel or monel containers, as XeF6 reacts with SiO2 to give SiF4 and XeOF4", "Solution_7": "sorry for double posting\r\n[quote=\"Euclidean Geometer\"]Could some one explain the following two reactions:\n 6XeF4 + 12 H2O -----------> 4 Xe -+ 2 XeO3 + 24 HF + 3 O2.\n\nXeF6 + 3 H2O ------------------> XeO3 + 6 HF[/quote]\r\nProbably all compounds of Xe in the lower oxidation are unstable. And this makes them better oxidizing agents, than Xef6, and in almost all of their reactions, XeF4 and XeF2 are either reduced or disproportionate, whereas almost all reactions of xe in VI state( reactions of XeF6) are not redox reactions.\r\nP.S: only guessing\r\n :D", "Solution_8": "chemrock's answer was more appropriate Varun....", "Solution_9": "oh, first of all i hadnt seen his post. but his post does not explain why XeF6 doesnt get reduced, or why XeF2 doesnt disproportionate and only gets reduced. :D", "Solution_10": "Could smone temme partial equations if possible for these reactions?" } { "Tag": [], "Problem": "[color=darkred]Let $ ABC$ be a triangle for which $ AB\\perp AC$ and $ B < 60^{\\circ}\\ .$ Prove that exists $ D\\in (AB)$ so that $ 2\\cdot AD = BC\\ .$\n\nConsider the point $ E\\in (AB)$ for which $ DB = DE\\ .$ Prove that $ EB = EC\\Longleftrightarrow B = 36^{\\circ}\\ .$\n\n\n[b][u]Another equivalent enunciation.[/u][/b] Let $ ABC$ be a triangle for which $ A=90^{\\circ}$ and $ B<60^{\\circ}$. \n\nFor a point $ D\\in (AB)$ consider the point $ E\\in (AD)$ for which $ DB=DE$. Prove that if two from \n\nthe following three sentencies are truly, than and the another sentence is truly : $ \\{\\begin{array}{cc}\n1\\blacktriangleright & 2\\cdot AD=BC\\\\\\\\\n2\\blacktriangleright & EB=EC\\\\\\\\\n3\\blacktriangleright & B=36^{\\circ}\\end{array}$[/color]", "Solution_1": "because B < 60 so $ cosB\\equal{}\\frac{AB}{AC} > cos60 \\equal{}1/2$\r\nthen 2AB > BC so there exist point D satisfying 2AD=BC\r\n\r\nI is the midpoint of EC we have ID = BC/2 = AD so angle AID = angle IAD , furthermore angle IAD = angle IEA = 2B , then angle AID = 2 B , however we already have angle EID = angle ECB = B , so angle AIE=B\r\nnow look at triangle IAE , angle IAE + angle AIE +angle IEA =180 , in other words , 5 B =180 ; B=36", "Solution_2": "Let's note $m(\\hat B)=x$; take $M$ - the midpoint of $BC$. Then $AM=AD=MB$. This implies $MD\\parallel CE$ and $MD=BD$. Consequently $m(\\widehat{AMC})=2x, m(\\widehat{BMD})=x$ and $\\widehat{AMD}=\\widehat{ADM}$. But $\\widehat{AMD}=180^\\circ-3x$ and $\\widehat{ADM}=2x$, so $180^\\circ-3x=2x\\iff x=36^\\circ$.\n\nBest regards,\nsunken rock" } { "Tag": [ "Putnam", "linear algebra", "matrix", "integration" ], "Problem": "Problem $ B_5$) Let $ A$ be the $ n$x$ n$ matrix with $ a_{ij} = cos((i + j)\\phi)$ ,$ \\phi = \\frac {2\\pi}{n}$, $ n \\in \\mathbb{N},n > 2$\r\nFind the determinant of $ I + A$\r\n - I learn method from this solution .It is very nice , It can find eigenvalues and eigenvectors but I try general with $ \\forall \\phi \\in \\mathbb{R}$ do not work ! :( But can for solution with $ B$ be the $ n$x$ n$ matrix with $ b_{ij} = sin((i + j)\\phi)$ then can proof that : $ \\det(I + A) = \\det(I + B) = 1 - \\frac {n^2}{4}$\r\n General It to see more nice problem !!!\r\nLemma 1) Let $ x_1,x_2,..,x_n,y_1,y_2,...,y_n$ are real numbers . Put :\r\n $ A_1 = \\begin{bmatrix}sin(x_1 + y_1) & sin(x_1 + y_2) & .... & sin(x_1 + y_n) \\\\\r\nsin(x_2 + y_1) & sin(x_2 + y_2) & .... & sin(x_2 + y_n) \\\\\r\n... & ... & ... & ... & ... \\\\\r\nsin(x_n + y_1) & sin(x_n + y_2) & .... & sin(x_n + y_n)\\end{bmatrix}$\r\n$ A_2 = \\begin{bmatrix}cos(x_1 + y_1) & cos(x_1 + y_2) & .... & cos(x_1 + y_n) \\\\\r\ncos(x_2 + y_1) & cos(x_2 + y_2) & .... & cos(x_2 + y_n) \\\\\r\n... & ... & ... & ... & ... \\\\\r\ncos(x_n + y_1) & cos(x_n + y_2) & .... & cos(x_n + y_n)\\end{bmatrix}$\r\nThen $ rank(A_1) = rank(A_2) \\leq 2$\r\n Proof lemma 1)For $ A_1$ are similary for $ A_2$\r\nPut $ C_{j} = \\begin{bmatrix}sin(x_1 + y_j) \\\\\r\nsin(x_2 + y_j) \\\\\r\n... \\\\\r\nsin(x_n + y_j)\\end{bmatrix} = cos(y_j)\\begin{bmatrix}sin(x_1) \\\\\r\nsin(x_2) \\\\\r\n... \\\\\r\nsin(x_n)\\end{bmatrix} + sin(y_j)\\begin{bmatrix}cos(x_1) \\\\\r\ncos(x_2) \\\\\r\n... \\\\\r\ncos(x_n)\\end{bmatrix} = cos(y_j)\\alpha + sin(y_j)\\beta \\in Vect(\\alpha,\\beta)$\r\n easy see $ Vect(C_1,C_2,..,C_n) \\subset Vect(\\alpha,\\beta)$\r\n Then $ rank(A_1) = \\dim(Vect(C_1,C_2,..,C_n)) \\leq \\dim(Vect(\\alpha,\\beta)) \\leq 2$ \r\nSimilary we have $ rank(A_2) \\leq 2$\r\n Lemma 2) (It is very interesting with calculate $ \\int$ ,....)\r\n $ \\forall x \\neq k\\pi ,k \\in \\mathbb{Z}$ then \r\n $ (\\frac {sin(nx)}{sinx})^2 = n^2 - 4\\sum_{1 \\leq i < j \\leq n}{sin^{2}(j - i)x}$ !\r\n Proof lemma 2)\r\n Put $ M = \\sum_{k = 1}^{n}{sin(2kx)}$,$ N = \\sum_{k = 1}^{n}{cos(2kx)}$\r\nWe have :\r\n $ 2sin(x)M = \\sum_{k = 1}^{n}{2sin(2kx)sin(x)} = \\sum_{k = 1}^{n}{(cos(2k - 1)x - cos(2k + 1)x)} = cos(x) - cos(2n + 1)x = 2sin(nx)sin(n + 1)x$\r\n Hence $ M = \\frac {2sin(nx)sin(n + 1)x}{sinx}$\r\nthe same :\r\n $ 2sin(x)N = \\sum_{k = 1}^{n}{2cos(2kx)sin(x)} = \\sum_{k = 1}^{n}{(sin(2k + 1)x - sin(2k - 1)x)} = sin(2n + 1)x - sin(x) = 2sin(nx)cos(n + 1)x$\r\n Hence : $ N = \\frac {2sin(nx)cos(n + 1)x}{sinx}$\r\nThen : \r\n$ (\\frac {sin(nx)}{sinx})^2 = (\\frac {2sin(nx)sin(n + 1)x}{sinx})^2 + (\\frac {2sin(nx)cos(n + 1)x}{sinx})^2 = M^2 + N^2 = (\\sum_{k = 1}^{n}{sin(2kx)})^2 + (\\sum_{k = 1}^{n}{cos(2kx)})^2 = n + \\sum_{1 \\leq i < j \\leq n}{(2sin(ix)sin(jx) + 2cos(ix)cos(jx)) = n + 2\\sum_{1 \\leq i < j \\leq n}{cos2(j - i)x} = n^2 - 4\\sum_{1 \\leq i < j \\leq n}{sin^{2}(j - i)x}}$\r\n Proof complete lemma2 )\r\nNow become problem for lemma 1 ) $ x_i = y_i = i \\phi$ , $ \\phi = \\frac {2p\\pi}{n},\\frac {2p}{n} \\notin \\mathbb{Z},p \\in \\mathbb{Z}$ (putnam 1999 $ p = 1$)\r\nthen by $ rank(A_1) = rank(A_2) \\leq 2$ Hence example $ A_1$\r\n easy $ M = N = 0$ and $ \\sum_{1 \\leq i < j \\leq n}{sin^{2}(j - i)\\phi} = \\frac {n^2}{4}$ \r\nWe have :\r\n $ \\det(I + A_1) = 1 + Tr(A_1) + \\sum_{1 \\leq i < j \\leq n}{det(\\begin{bmatrix}a_{ii} & a_{ij} \\\\\r\na_{ji} & a_{jj}\\end{bmatrix})}$\r\n We have $ Tr(A_1) = M = 0$ and \r\n$ \\sum_{1 \\leq i < j \\leq n}{det(\\begin{bmatrix}a_{ii} & a_{ij} \\\\\r\na_{ji} & a_{jj}\\end{bmatrix})} = \\sum_{1 \\leq i < j \\leq n}{(sin(2i \\phi})sin(2j\\phi) - sin^{2}((i + j)\\phi)) = \\frac {1}{2}\\sum_{1 \\leq i < j \\leq n}{cos(2(j - i)\\phi) - cos(2(i + j)\\phi) - 1 + cos(2(i + j)\\phi)} = - \\sum_{1 \\leq i < j \\leq n}{sin^{2}(j - i)\\phi} = - \\frac {n^2}{4}$\r\n easy with $ A_2$ (similary a result)\r\n and $ \\det(I + A_1) = \\det(I + A_2) = 1 - \\frac {n^2}{4}$\r\nNow for $ \\phi = \\frac {\\pi}{n + 1}$, ..,.... Very interesting !\r\n More : To more interesting You can calculate determinant $ I + K$ Here $ K = [k_{ij}]_{1 \\leq i,j \\leq n}$\r\n Here $ k_{ij} = C_{n}^{i}cos((i + j)\\phi)$ , to similary $ k_{ij} = C_{n}^{i}sin((i + j)\\phi)$ , $ \\phi = \\frac {2\\pi}{n},n > 2 , n \\in \\mathbb{N}$\r\n You can proof that :\r\n1) $ rank(K) \\leq 2$\r\n2) $ 4\\sum_{1\\leq i < j\\leq n} {C_{n}^{i}C_{n}^{j}sin^{2}((i - j)x)} = (\\sum_{k = 1}^{n}{C_{n}^{k}})^2 - 4^ncos^{2n}{(x)} + 2^{n + 1}cos^{n}{(x)}cos(nx) - 1$\r\n You can check It It is don't dificult :P ......", "Solution_1": "What was the point of this rambling? Please ask clear questions.", "Solution_2": "[quote=\"jmerry\"]What was the point of this rambling? [/quote]\r\n :) I will thinks !\r\nBut I find a some different point can be constant .....\r\n Example : Here $ (\\frac {sin(nx)}{sinx})^{2} \\equal{} n^{2} \\minus{} 4\\sum_{1\\leq i < j\\leq n}{sin^{2}(j \\minus{} i)x}$ \r\n We have $ cos(ix) \\equal{} ch(x)$ and $ sh(x) \\equal{} \\minus{} isin(ix)$ I find :\r\n$ M_1 \\equal{} \\sum_{k \\equal{} 1}^{n}{sh(2kx)} \\equal{} \\frac {2sh(nx)sh(n \\plus{} 1)x}{shx}$,$ N_1 \\equal{} \\sum_{k \\equal{} 1}^{n}{ch(2kx)} \\equal{} \\frac {2sh(nx)ch(n \\plus{} 1)x}{shx}$\r\n $ (\\frac {sh(nx)}{shx})^{2} \\equal{} n^{2} \\plus{} 4\\sum_{1\\leq i < j\\leq n}{sh^{2}(j \\minus{} i)x}$ ,$ x \\neq 0$\r\nPoint $ x$ to $ \\frac {sh(nx)}{shx} \\equal{} ?$ ....." } { "Tag": [], "Problem": "A sequence of letters is formed by writing 1 A, 2 B's, 3 C's, and so forth, increasing the number of letters written by one each time the next letter of the alphabet is written. What is the 200th letter in the sequence?", "Solution_1": "200 is between the 19th and 20th triangular numbers (190 and 200)\r\n\r\nso our answer is the 20th letter of the alphabet, T." } { "Tag": [ "blogs", "articles" ], "Problem": "the day before my english essay was due, i had already finished it and brought it to school. during lunch, i let my friend read it, and i made him promise me that he wouldnt plagiarize it. but a few days after the essay was due, i asked him to see his essay and the whole intro was almost exactly like mine, and so was the rest of the essay. is this considered plagiarism on my part?", "Solution_1": "If I understand your story correctly, it would be plagiarism on his part. Also, why are you friends with people that would do that?", "Solution_2": "Plagiarism (by you)? Probably not, although your school (or even your English department) might have a formal definition you can read. Ethically questionable? Absolutely, especially if this was even remotely predictable (which it sounds like it was). Grounds for finding a new friend? That's up to you, but I might consider it.", "Solution_3": "Well i don't think it's plagiarism because you didn't \"copyright\" it? or something along those lines...for example if two people accross the world wrote a similar essay they cannot call it plagarism because they \r\n\r\na) did not see the other's essay (not online or in a book, etc.) so they could not copy it because it was not available to them\r\nb) they do not even know each other \r\n\r\n\r\nhowever i believe your friend was simply out of ideas...", "Solution_4": "Unless he is in New Zealand.", "Solution_5": "Hopefully, your english teacher won't read it and will just give you credit for doing them.\r\n\r\n\r\nI wouldn't let ANY of my friends copy my paper almost exactly if I had one. :blush:", "Solution_6": "No, not plagiarism because it wasn't really published work (even if \"published work\" now means Wikipedia). However, it was certainly cheating. If he really needed help, you should have helped him, not let him read your paper. Unless he edited it for you, there was little to no benefit to you, and in return you gave him your paper. It was cheating because you said that he blatantly copied it. I would talk to him about it, because you could both get in serious trouble if your teacher notices.", "Solution_7": "Let's be clear here: plagiarism is copying without attribution. It doesn't matter whether the work is published, copywrited, or spray-painted on a subway car: if you copy work without attribution, that's plagiarism. Your friend is unquestionably guilty of plagiarism. In fact, some kinds of plagiarism are worse than others, and your friend is unquestionably guilty of a \"bad\" kind.\r\n\r\nIf your friend needed help, he should have talked to the teacher, done his own research and written his own paper. Regardless of whether you benefited from it, whether your teachers caught it, whether your friend was \"out of ideas,\" he acted in a dishonest and ethically unsound way, and you helped him.\r\n\r\nfuncia, if you can explain the meaning of that comment, I won't delete it.", "Solution_8": "I think funcia means there is a different rule in New Zealand about plagiarism maybe?\r\n\r\n\r\n\r\n\r\nI know several classmates who got a zero for plagiarism on a historical fiction critical analysis book report. \r\n\r\nThey copied, and even though it was not published, just a rough draft, still a big fat plump 0.", "Solution_9": "[quote=\"now a ranger\"]Hopefully, your english teacher won't read it and will just give you credit for doing them.\n\n\nI wouldn't let ANY of my friends copy my paper almost exactly if I had one. :blush:[/quote]\r\nwait do u mean if u had one friend or if u had one piece of paper?", "Solution_10": "JBL is correct about the defintition of plagiarism. Copying someone else's work is plagiarism if you don't give the first author credit. When doing a school paper you should always do some of your own thinking and own writing. \r\n\r\nCopyright belongs in a work as soon as it is created, under modern copyright law. You don't need to register the copyright nor do you need to put a copyright notice on the work (although both of those steps give you additional legal protections in court). If someone else wrote something, and you copy it, you may also be infrininging a copyright in addition to committing plagiarism.", "Solution_11": "[quote=\"ProtestanT\"][quote=\"now a ranger\"]Hopefully, your english teacher won't read it and will just give you credit for doing them.\n\n\nI wouldn't let ANY of my friends copy my paper almost exactly if I had one. :blush:[/quote]\nwait do u mean if u had one friend or if u had one piece of paper?[/quote]\r\n\r\n\r\nIf I had a paper to write, i meant.\r\n\r\n\r\nsorry for being ambiguous", "Solution_12": "[quote=\"tokenadult\"]Copyright belongs in a work as soon as it is created, under modern copyright law. You don't need to register the copyright nor do you need to put a copyright notice on the work (although both of those steps give you additional legal protections in court). If someone else wrote something, and you copy it, you may also be infringing a copyright in addition to committing plagiarism.[/quote]\r\n\r\nEven in the US? That's something I truly didn't know if so.\r\n\r\nOh and yes that is what I meant with NZ.", "Solution_13": "Hmmm.\r\n\r\nWhy does it matter so much?", "Solution_14": "[quote=\"Treething\"]Hmmm.\n\nWhy does it matter so much?[/quote]\r\n\r\nBecause you can get it REALLY big trouble.", "Solution_15": "[quote=\"7h3.D3m0n.117\"][quote=\"Treething\"]Hmmm.\n\nWhy does it matter so much?[/quote]\n\nBecause you can get it REALLY big trouble.[/quote]\r\nMissed the sarcasm tag? (I hope)", "Solution_16": "Because it's intellectually lazy and dishonest.\r\n\r\n[url=http://timpanogos.wordpress.com/2007/02/19/a-little-plagiarism-a-little-book/]Here[/url] is a blog post I just recently ran across about plagiarism and what's wrong with it.", "Solution_17": "[quote=\"JBL\"]Because it's intellectually lazy and dishonest.\n\n[url=http://timpanogos.wordpress.com/2007/02/19/a-little-plagiarism-a-little-book/]Here[/url] is a blog post I just recently ran across about plagiarism and what's wrong with it.[/quote]\r\n\r\n...reading over this article, I'd say some are simply too lazy to care. If the sin of plagiarism lies in morals and ethics, where does the stereotypical lazy, amoral, Asian honors student lie?" } { "Tag": [ "geometry", "3D geometry", "sphere", "Locus", "Locus problems", "IMO", "IMO 1963" ], "Problem": "Point $A$ and segment $BC$ are given. Determine the locus of points in space which are vertices of right angles with one side passing through $A$, and the other side intersecting segment $BC$.", "Solution_1": "Let F be the foot of a normal from the point A to the line BC and P an arbitrary point on the segment BC. If the point A does not lie on the line BC, the locus of the vertices V of the right angles $\\angle AVP$ is a sphere with a diameter AP and passing through the point F. This sphere is centered on the midline M, N of the triangle $\\triangle ABC$ parallel to the side BC, where M, N are the midpoints of the sides AB, AC. If the point A lies on the line BC, then the points A, F and the locus of the vertices V of the right angles $\\angle AVP$ is a sphere centered on segment M, N, where M, N are the midpoints of the segments AB, AC. As the point P moves on the segment BC, the locus of the vertices V becomes a portion of the pencil of spheres centered on the segment MN and passing through the points A, F. If the point A does not lie on the line BC, the pencil of spheres is elliptic (the spheres intersect in a fixed circle with the diameter AF and perpendicular to the line BC). If the point A lies on the line BC, the pencil of spheres is parabolic (the spheres touch the plane perpendicular to the line BC at the point A).", "Solution_2": "Let the intersection of the right angle with $BC$ be $P.$ Therefore, given $A$ and $P,$ all possible vertices of the right angle lie on the circle with diameter $AP.$ Thus, when all possible circles are drawn, it encloses a region as shown in the diagram 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} { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a,b,c,d>0 a\u00b2+b\u00b2+c\u00b2+c\u00b2+d\u00b2=1\r\nprove that \r\n$ \\ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} d^3$+ 8(1-a)(1-b)(1-c)(1-d) \u2265 1", "Solution_1": "We will use Lagrange multipliers to prove this ineq $ 01 [b]prove that:[/b]\r\n[b]((a^m) -1,(a^n) -1)=a^(m,n) -1[/b]", "Solution_1": "Ah...have done this before, but too lazy to redo it. If you let $m=x(m, n)$ and $n=y(m, n)$, then you just have to prove that $a^{(x-1)(m, n)}+a^{(x-2)(m, n)}+...+a^{(m, n)}+1$ and $a^{(y-1)(m, n)}+a^{(y-2)(m, n)}+...+a^{(m, n)}+1$ are relatively prime, which follows from repeatedly using the division algorithm, because x and y are relatively prime...", "Solution_2": "What? Come on, Krishanu! Just WOP it! :D\r\n\r\nLet $d = (a^{m}-1, a^{n}-1)$. Then\r\n\r\n$a^{m}\\equiv a^{n}\\equiv 1 \\bmod d$\r\n$\\implies a^{(m, n)}\\equiv 1 \\bmod d$\r\n$\\implies d | a^{(m, n)}-1$\r\n\r\nAnd of course we can show $a^{(m, n)}-1 | d$. Hence the two are equal :D", "Solution_3": "Too lazy...that was the point" } { "Tag": [ "linear algebra", "matrix", "combinatorics solved", "combinatorics" ], "Problem": "Let $x_1$, $x_2$, ..., $x_n$ and $y_1$, $y_2$, ..., $y_n$ be real numbers. Let $A$ be a $n \\times n$ matrix such that $a_{ij}=1$ if $x_i \\geq y_j$, and $a_{ij}=0$ otherwise. Let $B$ be a $n \\times n$ matrix made of $0$'s and $1$'s such that the sum of any row of $A$ equals the sum of the corresponding row of $B$, and the same for the columns. Prove that $A=B$.", "Solution_1": "Is there a crucial difference with that one?\r\nhttp://www.mathlinks.ro/viewtopic.php?t=5758\r\n\r\nPierre.", "Solution_2": "It's the same, just take $-y_i$ instead of $y_i$. ;)", "Solution_3": "It is proposed by Iran in IMO shortlist 2003" } { "Tag": [ "geometry", "AMC", "AIME", "analytic geometry", "calculus", "integration", "trigonometry" ], "Problem": "I'm in this online course that helps you prepare for the AIME, and even our teacher had problems solving this one... :wallbash_red: \r\nI don't get this that well, either, but you guys are geniuses (improper grammar, i know :rotfl: )\r\nso good luck!! :thumbup: \r\n\r\n[img]http://i17.tinypic.com/62sh4zk.jpg[/img]", "Solution_1": "lol i thought about coordinates :rotfl: \r\n\r\n[hide]\nBy Power of a Point, $ DA^{2}= 3*1 \\Rightarrow DA = \\sqrt{3}$ (Because DA = A(extension of AD onto circle). Similarly, $ CE=\\sqrt{3}$. Hence, $ \\angle DOA = 60$, so $ \\angle DOC = 30$. Now, the area of the sector is the area of sector $ ODE$ - 2*$ [DBO]$ = $ \\frac{\\pi}{3}-(\\sqrt{3}-1)$ so yeah \n[/hide]", "Solution_2": "Umm >_>\r\n\r\n[hide=\"Daesunbash\"]\n...so $ BD = \\sqrt3-1$, meaning that the shaded area is slightly more than $ \\frac{(\\sqrt3-1)^{2}}{2}\\approx 0.268$. Choice A is the closest.\n[/hide]", "Solution_3": "I think this is the reason that calculators are now banned from the AMC.", "Solution_4": "mkay...This takes like 5 seconds with calculus if you know your basic integrals.\r\n\r\n$ \\int_{1}^{\\sqrt{3}}\\sqrt{4-x^{2}}-1 \\, dx=2\\sin^{-1}\\frac{x}{2}+\\frac{x}{2}*\\sqrt{4-x^{2}}-x |_{1}^{\\sqrt{3}}=\\frac{\\pi}{3}+1-\\sqrt{3}$", "Solution_5": "that's not exactly a basic integral", "Solution_6": "I am sure that trig substituions are part of the AP calculus BC curriculum. I guess that is what I mean by basic.", "Solution_7": "yea, it's definitely not a \"hard\" integral to solve, but it's not exactly a quick integral", "Solution_8": "here's a simple way to do it, pretty sure it works\r\n\r\n[hide]draw a line from O to D and O to E, now OE and OD are both 2 since they are the radii, We know CO and AB is 1, so we can now find the lenth of CE and DA, which is $ \\sqrt{3}$, so you can see the triangle is a 30-60-90 triangle, so angle DOA is 60 degrees, that means COD is 30 degrees, and from the other triangle, we can see that COE is 60 degrees, that means EOA is 30, so COA-COD-EOA=30\ndegrees, now find the area of region DOE, it is $ \\frac{30}{360}*2^{2}\\pi$ now you can find the lenth of BE and BD by subtracting CB and AB from CE and AD, so BE and BD is $ \\sqrt{3}-1$ now draw a line from BO, and we can prove\ntriangle DOB and OBE is equal by sss, to find the area of OBE, we know the height which is CO=1, and we know the base, BE=$ \\sqrt{3}-1$, and we can find the area of OBE, just remember to multiply by two to get the area of triangle ODB also, now subtract the area of the two triangles from the region DOE\n$ \\frac{\\pi}{3}-(\\sqrt{3}-1)$ it equals answer choice A[/hide]", "Solution_9": "thanks!\r\nbut the integral part confused me, cuz i didn't take calculus.. i know just a teeny bit of derivatives, and that's it", "Solution_10": "[quote=\"Elemennop\"]yea, it's definitely not a \"hard\" integral to solve, but it's not exactly a quick integral[/quote]\r\n\r\nWhen you look at an integral, first you go throught the obvious set of techniques\r\n\r\ni) direct use of an integral rule \r\nii) u substition \r\niii) integration by parts \r\niv) trig substitution\r\n\r\nit is obvious that iv) is what you do here...the problem is solved by the 'AP calculus' algorithm, hence I would consider it to be a mechanical, easy problem when you use the calculus approach.", "Solution_11": "[hide]Split the region bound by C, O, E, and the arc. It is formed by a sector of area $ r\\theta = \\pi/3$ and a triangle of area $ \\sqrt{3}/2.$ (Note: you know these are 30,60,90 triangles because the hypotenuse/radius 2, is twice the leg 1) The total area is $ \\pi/3+\\sqrt{3}/2.$\n\nThe area of the region bound by C, O, A and D has same area. Hence, the total area of everything in the quarter circle excluding the desired region is\n\\[ 2\\left(\\pi/3+\\sqrt{3}/2\\right)-1, \\]\nwhere 1 is the area of the square.\n\nSubtract from $ \\pi$ to get A as answer.[/hide]" } { "Tag": [ "geometry", "3D geometry", "algorithm", "rotation" ], "Problem": "a little bit hard for me\r\nhow about you?\r\nshare your experience^^ :lol:\r\n\r\nyeh....\r\nsorry...I mean Rubik's cube :P", "Solution_1": "Try the $3 \\times 3 \\times 3 \\times 3$ one ;)", "Solution_2": "By 3x3x3 magic cube, do you mean a 3x3x3 cube filled with the numbers 1-27 such that the sum of the numbers in any row, column, or diagonal is constant?\r\n\r\nOr do you mean some sort of Rubik-based puzzle?", "Solution_3": "I think she means the rubix cube. In chinese, it literally translates into magic cube. I don't know about other places though.", "Solution_4": "In Germany, it's called both \"Rubix' Cube\" and \"Zauberw\u00fcrfel\" (=\"magic cube\").", "Solution_5": "I used to know how to do it, its not too complicated but you have to learn how. Its nearly impossible without knowledge of exactly how to do it in my opinion.", "Solution_6": "my best time is like 15 minutes, :blush:", "Solution_7": "[quote=\"math92\"]my best time is like 15 minutes, :blush:[/quote]\r\n\r\nwell, if you did it without memorizing the algorithms, then thats pretty good...\r\ni used to be impressed until i learned how, and found that it was just applying these methods in various cases... \r\n\r\nanyways, try doing it in the 27 turns or so which is calculatedto be the minimum number of turns in order to solve the cube in any position... o.0", "Solution_8": "Hm...the first time i did it took like a week of on and off thinking.\r\nAfter that, I structured my method more, made it more methodical. Then I sped it up, adopted things from other methods. It turned out my method was very similar to Lars Petrus's...at the start at least.\r\nRecently, I took an average around 19 seconds.", "Solution_9": "I know someone who takes about 8 or 9", "Solution_10": "[quote=\"kyyuanmathcount\"]I know someone who takes about 8 or 9[/quote]\r\n\r\nReally? Average? ...\r\nIf that were so, that person would hold the world record by a bit.", "Solution_11": "[quote=\"K81o7\"][quote=\"kyyuanmathcount\"]I know someone who takes about 8 or 9[/quote]\n\nReally? Average? ...\nIf that were so, that person would hold the world record by a bit.[/quote]\r\n\r\nagreed.\r\nim not sure its even possible to solve a well mixed cube in 8 seconds...", "Solution_12": "[quote=\"kyyuanmathcount\"]I know someone who takes about 8 or 9[/quote]\r\n\r\nThat's not true - the official world record is just over 12.", "Solution_13": "Maybe he means 8 or 9 minutes? :D \r\n\r\nI do the layer by layer method because I just got a cube for Christmas.\r\n\r\nI got my time down to one minute.\r\n\r\nBut I think I need to use a better method if I want to break 30 seconds.\r\n\r\nBy the way, I think some algorithm proved that it takes at most 20 moves to solve any given cube, given that you make the right moves of course!", "Solution_14": "ok, 2X moves to solve then =D\r\n\r\nhow would you prove such a thing anyways? :? \r\n\r\nmy own algorithm gets me about 30 sec, which is not bad. i guess it also depends on the \"turnability\" of the cube, and the speed of your turns...", "Solution_15": "Wow, you people are so good...my record irs 51.18 seconds.\r\n\r\nDoes anyone have suggestions for lubricating a cube with common household materials (I don't have silicon/silicone (spelling)).", "Solution_16": "Petroleum jelly?", "Solution_17": "You can prove it by developing a very complex set of algorithms.\r\n\r\nThe reason people don't use such an available algorithm is because there are too many cases to consider while cubing.", "Solution_18": "sometimes some lubricants make the cube messy...i could also suggest taking your cube apart and just wiping everything, if that makes sense...\r\n\r\nof course, using the jelly would also require taking the cube apart:\r\nfor those of you who dont know, taking the cube apart is quite simple. all you have to do is turn one face to a 45 degree angle and pop out an edge piece with your finger if your cube is extremely loose, or a pen. then the rest of the cube sort of falls apart.", "Solution_19": "I use Vaseline and it seems to work. Yeah, a 3x3 is easy to put back together but a 5x5 is not :lol:", "Solution_20": "About the 8 or 9 seconds, the world record is 7 isnt it?", "Solution_21": "erm, no, i think its around 12, 11.75 to be exact...", "Solution_22": "Best time : 55 seconds after gotten lucky", "Solution_23": "The current world record for 3x3x3 two-handed is 11.75 for a single solve, held by Jean Pons from France. Average of 5 is 14.59, held by Shotaro Makisumi ([url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=2889]litoxe2718281828[/url]). One-handed single-solve is 22.05, held by Ryan Patricio, and one-handed average of 3 is 26.73, also held by Ryan Patricio. Blindfolded is 1:46.47, by Leyan Lo.\r\n\r\nAs for my personal records, I have ~27 seconds single solve (not lucky), average is somewhere in the low 30's, and my one-handed is ~1:06. As you can see, I haven't accurately timed myself very much. :P Also, I'm pretty sure I've done better than these at times, but never timed them, so I don't know exactly how fast they were.\r\n\r\nAlso, the unofficial WR for fewest moves to solve a cube is 21, and I think the theoretical minimum is 20, although I could be wrong. Official WR is 28. My best is 48 (that I counted for).\r\n\r\nThe fastest lucky solve is difficult to determine. Some have done it in under 5 seconds, but often because the cube wasn't scrambled enough. The fastest I could find that I was sure was scrambled properly and simply lucky (unofficial, though) was Macky's 7.40 second solve, where he basically skipped half of the last layer. :P The fastest unofficial non-lucky solve is 9.54 seconds.", "Solution_24": "Fastest solve: 33.95 seconds\r\nFastest average of 10 (12, but with the fastest and slowest taken out): 39.18 seconds\r\n\r\nCubing is my current addiction. :)", "Solution_25": "My best average is 18.76, best solve without luck is 14.40 and best solve with luck is 13.90. It's been a while though so I'm probably out of shape.\r\n\r\n[b]Don't[/b] lubricate with vaseline. It contains petroleum and will wear down your cube eventually. Use a silicon based lubricant instead. Preferably some kind of spray. (Any professional cuber will tell you this :)).\r\n\r\nHave any of you solved blindfolded? I can do it in about 15 minutes. It's really frustrating until you do it right for the first time. When you do it wrong, of course, you have little clue of _where_ you went wrong, so getting it right for the first time is difficult. But when it happens - ohh man. :D", "Solution_26": "For lubrication, I first used Snap silicone spray. Works quite well. However, I haven't bothered to reapply it, and that was something like 10 months ago. While I'm sure my cube would spin better if I did reapply, at the moment it works just fine through wearing down inside and being lubricated by cube dust. :D", "Solution_27": "I really hate to be an Admin. but if you didn't notice there is a whole forum on rubik's cube, and similar math based games! ;) ;) ;).\r\n\r\nFor those who still have not found it: :D\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/index.php?f=360]www.artofproblemsolving.com/Forum/index.php?f=360[/url]\r\n\r\nBut, personally, i have seen people use almost any kind of lubriant, everything from Baby oil to olive oil (don't ask). :huh:", "Solution_28": "my record is like somewhere around 40 or 50 seconds. and btw... its all about silicon lube. thats teh best stuff out there for cubes. some ppl i kno r super fast meaning less than 7 seconds as a record.", "Solution_29": "Fire, it's not your duty to be an administrator, and the capital font could be misinterpreted as shouting, which by all means isn't the message you [i]want[/i] to be communicating. :P", "Solution_30": "Speed depends on\r\n\r\n1. How easily the springs and sides turn (some people call it the smoothness)\r\n2. How well the springs stretch (the margin of error)\r\nAnd of course, the cuber. But really, one needs a balance between weak springs and strong springs. Too strong means harder to turn, almost no margin of error. Too weak means it'll literally fall apart in your hands. I had a cube which was extremely easy to turn and the margin of error was about 60% (literally). As long as you didn't go crazy on it, it didn't pop (however, you could rotate corners in place with easy, and pop out pieces very easily as well. You could also put corners directly in). However, I lost that...yet it also has to do with wearing the cube in, so my other cube has some potential. It is a bit harder to turn and has less margin of error, but then again, it's sturdier, and I haven't used it as long.", "Solution_31": "it takes me about 1 second to rotate once.....\r\n\r\nlol..", "Solution_32": "Do you use your fingers rather than your whole hand to turn a side? If you learn to just flick it with your finger, it'll be much faster. However, your cube needs to be fairly loose to do that. Just lubricate it with something good, preferably silicone spray, and work it around a little, and you'll be good to go. Make sure to work with it for about an hour after lubricating to get the excess out, washing your hands every 5 to 10 minutes, since you don't want to keep that stuff on your hands. ;)", "Solution_33": "Yeah, you ideally combine flicking and turning to get really really fast (you want smooth transitions) but ultimately you really only touch the cube with your fingertips (pretty much, and you mostly use thumb and index fingers, some ring finger).", "Solution_34": "I forget my best time, but I haven't solved a Rubik's Cube since MP2005, my best time then was... under a minute I think...\r\n\r\nI remember I solved one while submerged underwater, so it couldn't have been over one and a half minutes.", "Solution_35": "Once you know how, you can scramble and unscramble the cube in less than 5 min. I can!", "Solution_36": "i can generally solve the cube in 2-3 minutes, i did recently time myself but i forgot what the time was (that's effective :lol: ). the 5x5 takes me about 25 minutes (is this extremely long?) and the 4x4 somewhere in between there.\r\n\r\nas for lubricating them, my old 3x3 turned horribly, but my new one (25th anniversary cube) i got at christmas, it turns really well. the 5x5 is sometimes a pain, though. i actually haven't ever lubricated/taken apart any of my cubes, strangely enough, but i assume it will happen inevitably...", "Solution_37": "I can do it using the internet. :D Does that count??", "Solution_38": "hm... my record was 2 min (im pretty bad yes)", "Solution_39": "WHAT???!!!\r\n\r\nThat's the record I'll never break!\r\n\r\n*fanfare*", "Solution_40": "I think my best time was 1:16, but I had a sucky cube..\r\nI don't speedcube anymore, I find it a waste of time." } { "Tag": [ "number theory open", "number theory" ], "Problem": "Find all the numbers n such that n!+1 is a perfect square.\r\nCheers!", "Solution_1": "This seems to be a very serious conflict. I have just one question: do you have solutions for these problems? I may be stupid, but they seem very hard. If they really are, they I do not understand your point. But I'm not interested, so I let Valentin to think about them.", "Solution_2": "As far I as know this problem is not only difficult but this also is an open problem.\r\nThus, I do not see any interest to your way of asking problems even if they are not given to me....\r\n\r\nYou will learn more about the problem at :\r\nhttp://mathworld.wolfram.com/BrocardsProblem.html\r\n\r\nPierre." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "It is given the sequence $y_1=y_2=1, y_{n+2}=(4k-5)y_{n+1-y_n}+4-2k, n\\ge 1$.\r\nFind all integer numbers numbers $k$, for which all the terms of the sequence are perfect squares. (Enjoy)", "Solution_1": "Just to make sure: $y_n$ on RHS [i]should[/i] be in the index?", "Solution_2": "[quote=\"Beat\"]It is given the sequence $y_1=y_2=1, y_{n+2}=(4k-5)y_{n+1-y_n}+4-2k, n\\ge 1$.\nFind all integer numbers numbers $k$, for which all the terms of the sequence are perfect squares. (Enjoy)[/quote]\nIt would be:\n[quote]It is given the sequence $y_1=y_2=1, y_{n+2}=(4k-5)y_{n+1}-y_n+4-2k, n\\ge 1$.\nFind all integer numbers numbers $k$, for which all the terms of the sequence are perfect squares.(Enjoy)[/quote]", "Solution_3": "Yes, now it`s correct", "Solution_4": "My answer is: there's no integer value of k satisties the problem's condition.\r\nIs it true?, or false? :)", "Solution_5": "False :( k=1 and k=3 :)", "Solution_6": "Maybe I had a mistake in calculating. Can you post your solution? :(", "Solution_7": "I solved this very beautiful problem .yes the only solution for $k$ is $1$ & $3$ .you can solved it ,only take sequence mod $k-2$ ,and prove \r\n$k-2$ must be $1$ or $-1$ so this very very very good problem can solved. :D :D", "Solution_8": "[quote=\"dna\"]only take sequence mod $k-2$ ,and prove \n$k-2$ must be $1$ or $-1$ so this very very very good problem can solved. :D :D[/quote]\r\n\r\n\r\nHow :?: \r\n\r\nAnd further..." } { "Tag": [ "percent" ], "Problem": "Kyle's science project grade has two parts. The oral presentation is worth 30 points, and the written report is worth 70 points. Kyle earned $ 84\\%$ of the possible points for his written report. To earn at least $ 87\\%$ of the possible points for the entire project, what percent of the possible points for the oral presentation must he earn?", "Solution_1": "If he earned 84 percent on the oral presentation, then he earned $ 0.84 \\times 70 \\equal{} 58.8$ points. He wants a total of 87 points, so he needs $ 87 \\minus{} 58.8 \\equal{} 28.2 \\div 30 \\equal{} \\boxed{94}$ percent." } { "Tag": [ "topology", "function", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $A$ and $B$ be the [b]topological space [/b]that both have fix point property and $A\\cap B=\\bar{A}\\cap B=A\\cap\\bar{B}=\\{c\\}$ then $A\\cup B$ also have [b]fix point property[/b].\r\nThe set have fix point property respect to its topology if any function from this set to itself have fix point.\r\n\r\nSaman [b]Gharib[/b]", "Solution_1": "it is challenging problem so it's nice to thinking.\r\n\r\nSaman Gharib", "Solution_2": "Does $\\bar{A}$ mean the closure of A?", "Solution_3": "Of course, \"any function\" must mean any continuous function. And I'm not sure what topological separation axioms we might need - I think we need points to be closed.\r\n\r\nDefine the function $p_A: A\\cup B\\mapsto A$ by $p_A(x)=\\{\\begin{array}{cc}x&\\text{ if }x\\in A\\\\c&\\text{ if }x\\in B\\end{array}.$\r\n\r\n$p_A$ is a continuous function. We see that by taking a closed $F\\subset A\\cup B.$ We have that either $p_A^{-1}(F)=F\\cap A$ or $p_A^{-1}(F)=(F\\cap A)\\cup \\{c\\}.$ Since $A$ and $B$ are both closed in $A\\cup B,$, this is closed.\r\n\r\nSimilarly, define $p_B(x)=\\{\\begin{array}{cc}x&\\text{ if }x\\in B\\\\c&\\text{ if }x\\in A\\end{array}.$\r\n\r\nNow, suppose $f: A\\cup B\\mapsto A\\cup B$ is continuous. Define a new function $g$ by $g=p_A\\circ f,$ restricted to $A.$ $g: A\\mapsto A$ is continuous, hence by hypothesis has a fixed point. If this fixed point is in $A\\setminus\\{c\\},$ then we have a fixed point of $f$ and are done. The alternative is that $f(c)\\in B.$\r\n\r\nRepeat this argument with $h=p_B\\circ f,$ restricted to $B.$ As before, either we find a fixed point of $f$ in $B,$ or $f(c)\\in A.$ But if both $f(c)\\in B$ and $f(c)\\in A,$ then $f(c)=c$ and $c$ is a fixed point." } { "Tag": [ "geometry", "probability", "MATHCOUNTS" ], "Problem": "It's one month till state for me, I and I got these books because they were the top 3 on the mathcounts list and I didn't think I would have time for 4 books. [b]Anyone who hasn't done the 2007 state yet[/b], what chapters should I read? (I'm in geometry honors, but I got the geometry book because its challenge problems never come near to being covered at school) I'm worst at # theory, poor at prob, and ok at geometry; (by mathcounts standards anyways) which chapters are most helpful, from your experience with 2006 & earlier competition scores?", "Solution_1": "[quote=\"LawOfSigns\"]It's one month till state for me, I and I got these books because they were the top 3 on the mathcounts list and I didn't think I would have time for 4 books. [b]Anyone who hasn't done the 2007 state yet[/b], what chapters should I read? (I'm in geometry honors, but I got the geometry book because its challenge problems never come near to being covered at school) I'm worst at # theory, poor at prob, and ok at geometry; (by mathcounts standards anyways) which chapters are most helpful, from your experience with 2006 & earlier competition scores?[/quote]Because of the boldface in the original posters' question, I don't feel comfortable leaving this discussion open. (Basically, there's no way to police this.) So although this is a good question, I feel like I need to lock this thread.\r\n\r\nPlease refrain from asking or suggesting particular topics to study until the conclusion of all the state contests.\r\n\r\nI'm sorry I have to do this, but I feel it is necessary to protect the integrity of the contest." } { "Tag": [ "LaTeX" ], "Problem": "Hi. I have a friend who started a new forum for school. But he's currently planning to put a math section into the forum. He was wondering how you guys put LaTeX into the forum so we could do stuff like $\\frac{x}{y}$, $\\sqrt[z]{x}$, or $x^{2}-4=0$ and stuff like that. His forum is php or IPB(I forget which). Please help my friend, that would mean a lot to him.\r\nThanks.", "Solution_1": "[url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url]", "Solution_2": "I installed latexrender and set up the paths ..etc\r\n\r\nBUT it doesn't render any .gif files.\r\n\r\nWhen I hit the render button I get a dos window that appears and disappears very quickly, so that I can't read it.\r\n\r\n where does latexrender place its output files??\r\n\r\nAnd what is the problem/fix?\r\n\r\nTIA", "Solution_3": "Please don't post in two threads. I can answer your problem with the detail you give in the other thread.", "Solution_4": "My friend used the LaTeX renderer on his forum, but the forum required didn't enable the html as well as one of the other incompatible choices. How do you enable html well without an ipb forum?", "Solution_5": "I would need a lot more detail before being able to make any suggestions. How about posting a link to the forum so I could see what you are referring to?", "Solution_6": "I try to download LATEX but I always end up downloading RIBBIT. Does anybody know why this happens and what's RIBBIT's difference with LATEX.", "Solution_7": "Ribbit is a piece of software that uses LaTeX to add equations into Word. So it isn't LaTeX at all.\r\n\r\nHave you followed the instructions and links at [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php]AoPS Downloads for LaTeX[/url] which gets you to download MiKTeX - a LaTeX distribution for Windows?", "Solution_8": "Well, I think so :blush: Anyway, thanks for the information. I'll try again... :wink:" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all functions $ f:\\mathbb{R}\\to\\mathbb{R}$ such that \n\\[ f(x\\plus{}y\\plus{}z)\\equal{}f(x)f(1\\minus{}y)\\plus{}f(y)f(1\\minus{}z)\\plus{}f(z)f(1\\minus{}x)\\] for all $ x,y,z\\in\\mathbb{R}$.", "Solution_1": "[quote=\"Pirkuliyev Rovsen\"]Find all functions $ f\\colon \\mathbb{R}\\to\\mathbb{R}$ such that $ f(x \\plus{} y \\plus{} z) \\equal{} f(x)f(1 \\minus{} y) \\plus{} f(y)f(1 \\minus{} z) \\plus{} f(z)f(1 \\minus{} x)$ for all $ x,y,z\\in\\mathbb{R}$[/quote]\r\n\r\nLet $ P(x,y,z)$ be the assertion $ f(x \\plus{} y \\plus{} z) \\equal{} f(x)f(1 \\minus{} y) \\plus{} f(y)f(1 \\minus{} z) \\plus{} f(z)f(1 \\minus{} x)$\r\n\r\n\r\n$ P(x,1 \\minus{} y,y)$ $ \\implies$ $ f(x \\plus{} 1) \\equal{} (f(x) \\plus{} f(1 \\minus{} x))f(y) \\plus{} f(1 \\minus{} y)^2$\r\n$ P(x,1 \\minus{} z,z)$ $ \\implies$ $ f(x \\plus{} 1) \\equal{} (f(x) \\plus{} f(1 \\minus{} x))f(z) \\plus{} f(1 \\minus{} z)^2$\r\n\r\nSubtracting these two lines, we get : $ (f(x) \\plus{} f(1 \\minus{} x))(f(y) \\minus{} f(z)) \\equal{} f(1 \\minus{} z)^2 \\minus{} f(1 \\minus{} y)^2$ and so :\r\n\r\nIf $ f(x)$ is not constant ($ \\exists f(y)\\ne f(z)$), then $ f(x) \\plus{} f(1 \\minus{} x) \\equal{} a$ but then $ P(x,1 \\minus{} y,y)$ becomes $ f(x \\plus{} 1) \\equal{} af(y) \\plus{} f(1 \\minus{} y)^2$ constant hence contradiction.\r\nSo $ f(x) \\equal{} c$ constant. Plugging back in the original equation, we get $ 3c^2 \\equal{} c$ and $ c\\in\\{0,\\frac 13\\}$\r\n\r\nHence the only two solutions :\r\n$ f(x) \\equal{} 0$ $ \\forall x$\r\n\r\n$ f(x) \\equal{} \\frac 13$ $ \\forall x$" } { "Tag": [ "modular arithmetic", "quadratics", "number theory" ], "Problem": "Um can someone prove this question?\r\n\"Prove that when you take an odd number and square it followed by subtracting 1, it can always be divided by 8.\" Ex: 5 squared is 25, then minus one is 24 which is divisible by 8. :lol: :D :P", "Solution_1": "proofs don't belong here :wink:", "Solution_2": "Thats not a proof. You just showed for one case. You have to prove [b]ALL[/b] cases.", "Solution_3": "$ 2n\\plus{}1$ will always be an odd number because $ Even\\plus{}1\\equal{}Odd$\r\nTherefore,\r\n$ (2n \\plus{} 1)^{2} \\minus{} 1 \\equal{} 4n^2 \\plus{} 4n$\r\nNow we will assume that that is divisible by 8\r\n$ 4n^2 \\plus{} 4n \\equal{} 8k$\r\n$ n^2 \\plus{} n \\equal{} 2k$\r\nNow this means that as long as any number squared, added to the original number, is divisible by 2,then any odd number squared subtracted by 1 will be divisible by 8.\r\nAnd since any number squared, and then added to the original number will be even, because $ Even \\plus{} Even \\equal{} Even$ and $ Odd \\plus{} Odd \\equal{} Even$, then any off number squared subtracted by 1 will be divisible by 8.\r\nI suck at proofs.\r\n :wink:", "Solution_4": "Make the last statement stronger. Other than that it is ok.", "Solution_5": "[quote=\"ZhangPeijin\"]$ 2n \\plus{} 1$ will always be an odd number because $ Even \\plus{} 1 \\equal{} Odd$\nTherefore,\n$ (2n \\plus{} 1)^{2} \\minus{} 1 \\equal{} 4n^2 \\plus{} 4n$\nNow we will assume that that is divisible by 8\n$ 4n^2 \\plus{} 4n \\equal{} 8k$\n$ n^2 \\plus{} n \\equal{} 2k$\nNow this means that as long as any number squared, added to the original number, is divisible by 2,then any odd number squared subtracted by 1 will be divisible by 8.\nAnd since any number squared, and then added to the original number will be even, because $ Even \\plus{} Even \\equal{} Even$ and $ Odd \\plus{} Odd \\equal{} Even$, then any off number squared subtracted by 1 will be divisible by 8.\nI suck at proofs.\n :wink:[/quote]The ideas contained in here are sufficient enough to solve the problem but the way this proof is written, it is not logically correct. First, you do not assume the conclusion, ie 'Now we will assume that that is divisible by 8' and then prove it??? Brainstorming $ \\neq$ proving. Here is a typical way to improve your proof and shorten it:\r\n\r\n$ 2n \\plus{} 1$ will always be an odd number because $ Even \\plus{} 1 \\equal{} Odd$\r\n\r\nTherefore,\r\n$ (2n \\plus{} 1)^{2} \\minus{} 1 \\equal{} 4n^2 \\plus{} 4n\\equal{}8\\frac{n(n\\plus{}1)}2$.\r\n\r\n$ n(n\\plus{}1)$ is an even integer because either $ n$ or $ n\\plus{}1$ is even.\r\n\r\nSo, $ (2n \\plus{} 1)^{2} \\minus{} 1 \\equal{}8\\frac{n(n\\plus{}1)}2\\equal{}8k$, where $ k\\equal{}\\frac{n(n\\plus{}1)}2$ is some integer.", "Solution_6": "[quote=\"hunter34\"]Thats not a proof. You just showed for one case. You have to prove [b]ALL[/b] cases.[/quote]\r\nI know. I just wanted to show an example", "Solution_7": "Sorry. Did not see the EX in front of it.", "Solution_8": "[hide=\"if n is any odd integer, then\"] \n$ n\\pmod{8}\\equiv \\{1,3,5,7\\}\\equiv \\{1,3, \\minus{} 3, \\minus{} 1\\}$ which leads to: $ n^2\\pmod{8}\\equiv \\{1\\}\\Rightarrow (n^2 \\minus{} 1)\\equiv 0\\pmod{8}$.\nThe latter statement verifies the equation $ n^2 \\minus{} 1 \\equal{} 8k$. An exception to LHS divisibility by eight is $ k \\equal{} 0$ which implies $ n\\neq\\pm1$.\n\nTherefore $ 8\\mid{(n^2 \\minus{} 1)}$ if and only if $ |n|\\neq1$.[/hide]\r\nI liked that. :)\r\n\r\nEDIT: Modular arithmetic is clearly a powerful tool here, but only because we so easily defined n as any odd integer. This problem is related to what is known as [url=http://mathworld.wolfram.com/QuadraticCongruenceEquation.html]quadratic congruency[/url], and that can become messy in a quick way if you don't know what you're doing. AOPS volume II has a nice section on it, as does Introduction to Number Theory by Mathew Crawford (I think).", "Solution_9": "[quote]\nTherefore $ 8\\mid{(n^2 \\minus{} 1)}$ if and only if $ |n|\\neq1$.[/quote]\r\n\r\nare you sure?\r\n\r\nif $ n\\equal{}2$ \r\n$ 8\\not|(2^2\\minus{}1)$ \r\nis it ?", "Solution_10": "[quote=\"igiul\"]An exception to LHS divisibility by eight is $ k \\equal{} 0$ which implies $ n\\neq\\pm1$.\\[/quote]\r\n\r\nZero divides nothing but everything (integers except zero) divides zero. That is, $ z|0$ for all integers $ z\\neq0$ (where we throw out zero since $ \\frac {0}{0}$ is indeterminate), so there is no exception to the result you got. Nice proof, though!", "Solution_11": "[quote=\"binomial_4eva\"][quote]\nTherefore $ 8\\mid{(n^2 \\minus{} 1)}$ if and only if $ |n|\\neq1$.[/quote]\n\nare you sure?[/quote]\r\n\r\nFor odd $ n$, yes. What is the problem?", "Solution_12": "oh sorry i didnt read the odd bit", "Solution_13": "An alternate solution:\r\n\r\nWe want to prove that $ 8\\mid (n^2\\minus{}1)$ for all odd $ n$. But, $ n^2\\minus{}1\\equal{}(n\\plus{}1)(n\\minus{}1)$ For odd $ n$, both of $ n\\plus{}1$,$ n\\minus{}1$ are even, and since of every two consecutive evens one is a multiple of 4, we have that their product is a multiple of $ 2\\cdot 4\\equal{}8$.\r\n\r\nOr this could be slightly changed by letting $ n\\equal{}2k\\plus{}1$ so that $ (2k\\plus{}1)^2\\minus{}1\\equal{}(2k\\plus{}2)(2k)\\equal{}4(k\\plus{}1)k$ and exactly one of $ k$, $ k\\plus{}1$ is even." } { "Tag": [ "inequalities" ], "Problem": "Suppose you need to use at least 1 dollar worth of stamps to mail a package. You have as many 3 cent stamps as you need, but only four 32 cent stamps. How many of each stamp can you use?\r\n\r\nWhat systems of inequalities would describe this situation?", "Solution_1": "[hide=\"Solution?\"]Let $ x$ be the number of $ 3$-cent stamps, and $ y$ be the number of $ 32$-cent stamps. We have the following three inequalities: $ 3x \\plus{} 32y \\ge 100$ and $ x \\ge 0$ and $ 0 \\le y \\le 4$.[/hide]", "Solution_2": "If you're just looking for systems of inequality, then I think darkdieuguerre has the right idea. As for how many you can use, that would be infinite, I assume, cause you have an unlimited number of 3 cent stamps....." } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Call an integer $m$ [b]good[/b] if $|m|$ is not a square. Determine all integers $m$ with the following property :::\r\n\r\n$m$ can be represented , in infinitely many ways, as a sum of three distinct [b]good[/b] integers whose product is the square of an odd integer.", "Solution_1": "Posted before:\r\n\r\nIMO Shortlist 2003, problem N5 (or N6)..." } { "Tag": [ "integration", "function", "limit", "algebra unsolved", "algebra" ], "Problem": "Let f and g be two continious, distinct funtions from $[0, 1]\\rightarrow R$ such that:\r\n$\\int\\limits_{0}^{1}f(x)d(x)=\\int\\limits_{0}^{1}g(x)d(x)$.\r\nLet ${y_n=\\int\\limits_{0}^{1}\\frac{f^{n+1}(x)dx}{g^n(x)}}$, for ${n\\geq0}$, natural.\r\nProve that $y_n$ is an increasing and divergent sequence", "Solution_1": "$f^{n+1}$ is $f\\circ f\\circ f\\circ \\ldots \\circ f$ ?", "Solution_2": "No, it isn't. i remember that i posted a solution to this nice problem before. Anyway, here it goes again :D : \r\n\r\n [color=blue] [b] Problem. [/b]\n\n Let $f, g: [0,1]\\to (0,\\infty)$ two continous functions such that \\[ \\int_{0}^{1} f(x)dx=\\int_{0}^{1} g(x)dx. \\]\n Let us define the sequence \\[ x_{n}=\\int_{0}^{1} \\frac{f^{n+1} (x)}{g^{n} (x)} dx. \\]\n Prove that $(x_{n})_{n\\in\\mathbb{N^{*}}}$ is increasing and divergent.\n [/color]\r\n\r\n [i] Solution. [/i]\r\n\r\n It is easy to see that there is $c\\in[0,1]$ such that $f(c) >g(c)$ if $f\\neq g$. From here we deduce that there is\r\n a compact interval $I$ such that $f(x)> g(x)$. So we have\r\n\r\n \\[ \\int_{0}^{1} \\frac{f^{n+1} (x)}{g^{n} (x)}dx=\\int_{0}^{1} f(x)\\cdot\\frac{f^{n} (x)}{g^{n} (x)}dx\\geq (\\inf_{x\\in I} \\frac{f(x)}{g(x)})^{n}\\cdot\\int_{I} f(x)dx. \\]\r\n\r\n Since $\\lim_{n\\to\\infty}(\\inf_{x\\in I} \\frac{f(x)}{g(x)})^{n}=\\infty$ by the sandwich theorem the conclusion follows.", "Solution_3": "And by the way, this isn't an algebra problem for sure. I think that this should be moved to College Playground-Analisys.... ;)", "Solution_4": "Wow, I forgot to prove that the sequence is increasing. Here it goes:\r\n\r\n \\[ x_{n}-x_{n-1}=\\int_{0}^{1} \\frac{(f^{n}(x)-g^{n}(x))(f(x)-g(x))}{g^{n}(x)} dx\\geq 0. \\]" } { "Tag": [ "LaTeX" ], "Problem": "I want to change the font of the page number that is automatically generated on the first page of a new chapter by the \\chapter{} function. \r\n\r\nAlso is it possible to change the alignment of this page number? i.e. from center to outer?\r\n\r\nmany thanks in advance\r\n\r\n....ok, so now I have an identical problem with the Index page.... i.e. different font and alignment for the page number compared to teh rest on the doc.", "Solution_1": "The only reason that I can see for a different font for the page numbers is that you have changed the font for parts of the document but without a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] it is impossible to tell. \r\n\r\nThe best way to deal with footer (and header) layout is to use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancyhdr]fancyhdr package[/url]; \\fancyfoot[LE,RO]{\\thepage} for example puts the page number on the left on even pages and on the right on odd pages.", "Solution_2": "Ok, sorry for not including any code. I know about fancyhdr and am using it but when you invoke a new chapter it produces a new page format of its own automatically; i.e. it suppresses the header and footer rules and text that I have setup in fancyhdr and produces a blank header and footer without rules and with a centered page number in a serif font.\r\n\r\nI don't mind the blank headers and footers but would like the page number font to match and aslo possibly to align it at 'outer' like the rest of my two-side book. The page number revets back to my settings after the first page of the chapter without me having to do anything. As I say, I have the same issue with my index pages which are also producing a centered serif font page number.\r\n\r\nAny ideas?\r\n\r\nAt the start of the doc I use;\r\n\r\n\\fancyfoot[LE,RO]{\\sffamily{\\thepage}}\r\n\r\nto setup my page numbers.\r\n\r\nA typical chapter start looks like;\r\n\r\n\\cleardoublepage\r\n\\mainmatter\r\n\\chapter{Introduction}\r\n\r\n\\fancyhead[LE,RO]{\\sffamily\\textbf{Introduction}}\r\n\\renewcommand{\\headrulewidth}{1.0pt} %Underlines the header. (Set to 0pt if not required).\r\n\\renewcommand{\\footrulewidth}{1.0pt} %Underlines the footer. (Set to 0pt if not required).\r\n\r\nmany thanks", "Solution_3": "Unfortunately it's impossible to tell exactly what you are doing as you haven't provided a [b]complete[/b] minimal example with the preamble etc, so I am left with guessing what is happening.\r\n\\chapter redefines the page style to plain, so I suspect that you haven't redefined the plain pagestyle. Section 7 of the fancyhdr manual shows you how to do this.", "Solution_4": "Ok, that was all I needed. I have redefined the plain pagestyle and that has solved the problem.\r\n\r\nThanks very much - your help is really appreciated" } { "Tag": [ "probability", "geometry", "geometric transformation", "reflection", "analytic geometry", "graphing lines", "slope" ], "Problem": "here is my question:\r\nThree points are chosen randomly and independently on a circle. What is the probability that all 3 pairwise distances between the points are less than the radius of the circle?", "Solution_1": "is it even possible?\r\n0?", "Solution_2": "Of course it's possible. Set it so that the two outer points have less than the distance of the radius between them.\r\n\r\nI remember seeing this problem before, but I'm not completely sure how to solve it. Obviously, the placement of the first point is irrelevant. After that, the second point has a $\\frac{1}{3}$ chance of being within the possible area. However, I get stuck after this. Essentially, if the second point is placed (an arbitrarily small distance under) one radial length from the first point, then the third point must be placed between them, so we have a $\\frac{1}{6}$ chance of meeting the conditions. However, the closer the second point is placed to the first, the larger the chance is that the third point will be placed so that the conditions are met. How am I supposed to approach this?", "Solution_3": "Woo... awesome problem...\r\n\r\n[hide][img]http://img127.imageshack.us/img127/7399/amc2003b254qe.png[/img]\n\nLook at the picture (circle has circumference $1$). Call the smallest distance between two points $x$. Call the left point the \"center,\" that is, both the other points are closer to the \"center\" than each other. The third point is located on the arc of length $1-x$.\n\n(1) Then the distance between the left point and the third point is smaller than $\\frac{1-x}{2}$ (to maintain the \"center\" status) but greater than $x$ (so that $x$ is still the smallest distance). However, we also note that $x$ has to be less than $\\frac{1-x}{2}$ or the third point will become the center, so $x < \\frac{1-x}{2} \\Rightarrow x < \\frac{1}{3}$. \n\n(2) Similarly apply this argument for the third point (reflect the graph as shown below), and we get a graph that looks like this ($x$-axis is the distance from the center to the right point, $y$-axis is the distance from the third point to the center).\n\n[img]http://img423.imageshack.us/img423/9702/amc2003b2514ci.png[/img]\n\nThe red indicates the area described by (1) and the blue is (2). The yellow is the area such that the sum of the arcs is less than $\\frac{1}{6}$, or equivalently the distance is less than one radius (note that by picking a \"center\" we guarantee this is the total area, because the sum of the distances is always greater than the independent distances).\n\nHence our answer is $\\frac{\\frac{1}{72}}{\\frac{1}{9}+2\\left(\\frac{1}{36}\\right)} = \\frac{\\frac{1}{72}}{\\frac{1}{6}} = \\frac{1}{12}$. Whew.[/hide]", "Solution_4": "paladin8, how'd you get from x<1/3 to your graphs which show x<1/2?\r\n\r\nSorry I'm confused.", "Solution_5": "[quote=\"WarpedKlown1335\"]paladin8, how'd you get from x<1/3 to your graphs which show x<1/2?\n\nSorry I'm confused.[/quote]\r\n\r\nSo for part (1), you get the red portion, where $x < \\frac{1}{3}$ and (call the other arc length $y$ for now) $y < \\frac{1-x}{2}$, $y > x$ (explanation in previous post).\r\n\r\nThen in part (2), flip the graph over the line $y=x$ and apply the same argument, to fill in the blue portion.\r\n\r\nEssentially, it is necessary to create the $x < \\frac{1}{3}$ condition because as you can see the slope of the graph changes as it gets to $\\frac{1}{3}$ (also to maintain the center condition).", "Solution_6": "For a revamped solution, see [url=http://wangsblog.com/jeffrey/?p=16]here[/url]. The same idea, hopefully better explained." } { "Tag": [ "geometry", "circumcircle", "ratio", "function", "trigonometry", "rectangle", "cyclic quadrilateral" ], "Problem": "Let $K$ is the point of intersection the diagonals of cyclic quadrilateral $ABCD$. In the triangle\r\n$AKD$ exists the point $P$, such that $\\angle APC=\\angle ADC+90^\\circ$ and $\\angle BPD=\\angle BAD+90^\\circ$. Prove that the diagonals of\r\nthe convex quadrilateral, formed the foots of perpendiculars from $P$ on sides of $ABCD$, are perpendicular.\r\n(Sorry for my English).", "Solution_1": "Let the tangents to the quadrilateral circumcircle (O) at the vertices A, C meet at a point U and let the tangents at the vertices B, D meet at a point V. The circle (U) with radius UA = UC meets the circle (V) with radius VB = VD at points P, Q, the point P lying inside the quadrilateral ABCD. As long as $\\angle D < 90^\\circ$ and $\\angle A < 90^\\circ$, the point P lies inside the triangle $\\triangle AKD$ and obviously, $\\angle APC = \\angle D + 90^\\circ$ and $\\angle BPD = \\angle A + 90^\\circ$. PQ is the radical axis of the circles (U), (V) and the circumcircle (O) is perpendicular to both of them, hence, it is centered on their radical axis, $O \\in PQ$. The diagonal AC is the radical axis of the circles (O), (U) and the diagonal BD is the radical axis of the circles (O), (V). The diagonal intersection K is then the radical center of the circles (O), (U), (V) lying on the radical axis PQ of (U), (V).\r\n\r\nLet P' be an arbitrary point on PQ and E', F', G', H' the feet of normals from P' to the quadrilateral sides AB, BC, CD, DA. When the point P' coincides with the circumcenter O, then E', F', G', H' are the midpoints of AB, BC, CD, DA, so that $\\frac{AE'}{AB} = \\frac{DG'}{DC}$ and $\\frac{BF'}{BC} = \\frac{AH'}{AD}$. When the point P' coincides with the diagonal intersection K, then E', F', G', H' are the feet of the K-altitudes of the triangles $\\triangle ABK, \\triangle BCK, \\triangle DCK, \\triangle ADK$. Since the triangle pairs $\\triangle ABK \\sim \\triangle DCK, \\triangle BCK \\sim \\triangle ADK$ are similar, again, $\\frac{AE'}{AB} = \\frac{DG'}{DC}$ and $\\frac{BF'}{BC} = \\frac{AH'}{AD}$. Since these ratios are linar functions of the position of the point $P' \\in PQ$ and since they are equal for 2 positions of this point, they are equal for all positions of $P' \\in PQ$. In particular, they are equal when P' coincides with the point P. Thus if we denote E, F, G, H the feet of normals to AB, BC, CD, DA from P, we have $\\frac{AE}{AB} = \\frac{DG}{DC}$ and $\\frac{BF}{BC} = \\frac{AH}{AD}$.\r\n\r\nLet the lines AB, CD meet at X and the lines BC, DA at Y. Since $\\angle A < 90^\\circ, \\angle D < 90^\\circ$, the point X lies on the opposite side of the line BC than the quadrilateral vertices A, D. Point U is the pole of the diagonal AC and point V is the pole of the diagonal BD with respect to the circumcircle (O). Hence, UV is a polar of the diagonal intersection K, which is identical with the line XY, i.e., the points U, V, X, Y are collinear. Let (X) be a circle centered at the point X and with radius XP. Using the sine theorem for the triangles $\\triangle AUX, \\triangle BVX$,\r\n\r\n$\\frac{UA}{UX} = \\frac{\\sin \\widehat{UXA}}{\\sin \\widehat{UAX}},\\ \\frac{VB}{VX} = \\frac{\\sin \\widehat{VXB}}{\\sin \\widehat{VBX}}$\r\n\r\n$\\frac{UX}{VX} = \\frac{UA}{VB} \\cdot \\frac{\\sin \\widehat{UAX}}{\\sin \\widehat{VBX}}$\r\n\r\nbecause $\\sin \\widehat{UXA} = \\sin \\widehat{VXB}.$ But\r\n\r\n$\\angle UAX = \\angle BDA = \\angle BCA = 180^\\circ - (\\angle B + \\angle CAB) =$\r\n\r\n$= 180^\\circ - (\\angle B + \\angle VBC) = \\angle VBX$\r\n\r\nand consequently,\r\n\r\n$\\frac{UX}{VX} = \\frac{UA}{VB} = \\frac{UP}{VP}$\r\n\r\nThus PX bisects the angle $\\angle UPV$ in the triangle $\\triangle UVP$. But since $(U) \\perp UP, (V) \\perp VP, (X) \\perp XP$, the circle (X) bisects the angle formed by the circles (U), (V). \r\n\r\nInvert the quadrilateral ABCD with the circumcircle (O) and the circles (U), (V), (X) in a circle (P) with radius PX (same as radius of the circle (X)). The pencil of circles (U), (V), (X) (with collinear centers and all passing through the point P) is carried into a pencil of concurrent lines u, v, x, such that the line x bisects the angle formed by the lines u, v, because inversion preserves angles between curves. The circle (O) centered on the radical axis PQ of this pencil and perpendicular to 2 circles (U), (V) of this pencil (i.e., perpendicular to all circles of this pencil) is carried into a circle (O') centered at the concurrency point O' of the lines u, v, x. The quadrilateral vertices A, B, C, D are carried into the intersections A', B', C', D' of the lines u, v with the circle (O'), i.e., the quadrilateral ABCD is carried into a rectangle A'B'C'D'. Since the circle (X) is congruent to the inversion circle (P), the line x (midline of the rectangle A'B'C'D') is the perpendicular bisector of the segment PX. Therefore, the triangles $\\triangle D'PA' \\cong \\triangle C'XB'$ are congruent. But the triangles $\\triangle APD \\sim \\triangle D'PA'$ with the common angle at the vertex P are similar by the basic properties of inversion $PA \\cdot PA' = PD \\cdot PD' = PX^2$. It follows that the triangles $\\triangle APD \\sim \\triangle D'PA' \\cong \\triangle C'XB'$ are similar. The triangle $\\triangle C'XB'$ is the inverted triangle $\\triangle CXB$ in the circle (P) and the triangle $\\triangle EFG$ is the pedal triangle of the triangle $\\triangle CXB$ with respect to the point P. This means that the triangles $\\triangle C'XB' \\sim \\triangle EFG$ are also similar and consequently, the triangles $\\triangle APD \\sim \\triangle D'PA' \\cong \\triangle C'XB' \\sim \\triangle EFG$ are all similar. In entirely the same way, we can show that the triangles $\\triangle BPC \\sim \\triangle EHG$ are also similar.\r\n\r\nIf the triangles $\\triangle APD, \\triangle BPC$ are carried into the triangles $\\triangle EFG, \\triangle EHG$ with the common side EG by similarity transformations, then using the relation $\\frac{BF}{BC} = \\frac{AH}{AD}$ proved previously, this means that the points $H \\in AD, F \\in BC$ are both carried into the same point $L \\in EG$. Since $FL \\perp EG, HL \\perp EG$, the points F, L, H are collinear, i.e., L is the diagonal intersection of the quadrilateral EFGH and $FH \\perp EG$. :cool:", "Solution_2": "[hide=\"Here are some interesting ideas.\"][color=darkred][b]Lemma 1.[/b] Let $ABCD$ be a quadrilateral $(AB=a\\ ,$ $BC=b\\ ,$ $CD=d\\ ,$ $DA=d\\ ,$ $AC=e\\ ,$ $BD=f)\\ .$ Then $A+C\\in \\{90^{\\circ},270^{\\circ}\\}$ if and only if $e^2f^2=a^2c^2+b^2d^2\\ .$[/color]\n\n[color=darkblue][b]Lemma 2.[/b] Let $ABCD$ be a cyclic quadrilateral and let $I$ be a point of the its interior. Denote the projections $M$, $N$, $P$, $R$ of the point $I$ on the sides $[AB]$, $[BC]$, $[CD]$, $[DA]$ respectively. Then the quadrilateral $MNPR$ is orthodiagonally $(MP\\perp NR)$ if and only if $\\frac{IA^2+IC^2}{IB^2+ID^2}=\\left(\\frac ef\\right)^2\\ .$[/color][/hide]I will come back soon. Here time is 2.51 AM !" } { "Tag": [], "Problem": "$ \\overline{BC}$ is parallel to the segment through $ A$, and $ AB \\equal{} BC$. What is the number of degrees represented by $ x$?\n\n[asy]draw((0,0)--(10,0));\ndraw((0,3)--(10,3));\ndraw((2,3)--(8,0));\ndraw((2,3)--(4,0));\nlabel(\"$A$\",(2,3),N);\nlabel(\"$B$\",(4,0),S);\nlabel(\"$C$\",(8,0),S);\nlabel(\"$124^{\\circ}$\",(2,3),SW);\nlabel(\"$x^{\\circ}$\",(4.5,3),S);[/asy]", "Solution_1": "Since we have parallel lines, we know $ \\angle ABC \\equal{} 124^\\circ$. $ \\triangle ABC$ is isosceles, so that means $ \\angle BAC \\equal{} \\angle BCA$, so they are $ \\frac{180\\minus{}124}{2} \\equal{} 28$. Therefore, $ x \\equal{} 124 \\minus{} 28 \\equal{} 96$.", "Solution_2": "Actually, $ x\\equal{}180\\minus{}124\\minus{}28\\equal{}28$, or alternatively we have $ \\angle BCA \\equal{} x$ because the lines are parallel." } { "Tag": [ "MATHCOUNTS", "LaTeX", "ARML", "USAMTS", "search", "AMC", "AIME" ], "Problem": "I need really a lot practice.. So, here 's what great community memebers can do to me:\r\n1. Post the question.....\r\n2. Answer the question..... (with explanation)\r\nI hope questions will in middle level becaus I'm attending MATHCOUNTS for state and I think those questions are pretty high level... So, here' s my question.. \r\n\r\nThe sum of three numbers a, b, and c is 77. If a decreased by 5 is equal to b increased by 5 and also equal to c multiplied by 5, what is the value of b?\r\n\r\n[hide]\nThere are several points that I want to add in my application.\n\n1. In 8th grade, I was the one who received Jay Ernest Award for Science.\n2. Under School Extracurricular Activities, Gifted club includes activities like Model UN, Scholastic Scrimmage, Stock Market Game, and several math competitions like AMC (American Mathematics Competition) and Bucknell University Gold Exam. I had to take an exam in junior high with psychologist and passed it to become part of gifted. \n3. In addition for School Extracurricular Activities: Scholastic J 9-12 and Varsity J 10-12.\n4. Under Community Activities, I brought up the idea of volunteering as an assistant coach for junior high math competition, Mathcounts, which I entered in my junior high years and became the only person from our school to qualify to states. I had to receive permissions from both junior high and high school, and tried to go there almost once in a week to teach the kids in junior high gifted.\n5. Art of Problem Solving is an internationally respected forum devoted for learning math, both for classroom and competition. This forum is run by very talented mathematicians who have all qualified to national or international level in math competition. I was asked personally by one of these respected administrators of the forum. I have been a moderator more than four years, and in the process, I learned a mathematical language for computer called LaTeX to help more effectively.\n6. I participated in numerous math competitions, including ARML (American Regional Math League - a national competition for highly exceptional high school students from each state), USAMTS (USA Mathematical Talent Search - A national competition that requires students to write full-written proofs for five problems on four rounds), Lehigh University Math Contest for High School Students, Bucknell Gold Exam (first place in my school all participating years; second place in the competition in sophomore year), AMC/AIME (American Mathematics Competition/American Invitational Mathematics Examination - national level math competition for high school students; AIME only picked certain high scorers from AMC).\n7. With the work activities, I want to add that although I do not have any work experience, it was because I wanted to devote my time fully into academics and community services. Academics has been my number one priority during high school, and after that is community service. Because of these two factors, I did not have much time to find a job because I'm either volunteering in one place or working on my academics. [/hide]", "Solution_1": "[quote=\"Silverfalcon\"]I need really a lot practice.. So, here 's what great community memebers can do to me:\n1. Post the question.....\n2. Answer the question..... (with explanation)\nI hope questions will in middle level becaus I'm attending MATHCOUNTS for state and I think those questions are pretty high level... So, here' s my question.. \n\nThe sum of three numbers a, b, and c is 77. If a decreased by 5 is equal to b increased by 5 and also equal to c multiplied by 5, what is the value of b?[/quote]\r\n\r\nwe'll write this in terms of b. \r\nb+(b+10)+(b+5)/5=77.\r\n2b+10+b/5+1=77\r\n2b+b/5=66\r\n10b/5+b/5=66\r\n11b/5=66\r\n11b=330.\r\nb=30\r\n\r\nheres my question:\r\nWhat is the remainder when 5^1993 is divided by 7?", "Solution_2": "5^1 mod 7 = 5\r\n5^2 mod 7 = 4\r\n5^3 mod 7 = 6\r\n5^4 mod 7 = 2\r\n5^5 mod 7 = 3\r\n5^6 mod 7 = 1\r\n5^7 mod 7 = 5 repeats from here, so 5^1993 mod 7= 5^(1993 mod 6) mod 7=5 mod 7 = 5\r\n\r\nwhat is the greatest possible value of g in p(x)=x^8-6x^7+ax^6+bx^5+cx^4+dx^3+ex^2+fx+g\r\nif all roots of the polynomial are positive", "Solution_3": "I hope I don't mean anything by this but I meant that I want the question to be in middle level, not like this 8th power polynomial questions.. I know some of polynomial questoins but this is more like to be from Math Olympiad.. So, here's my question from state competition in MATHCOUNTS...\r\n\r\nThe surface area of a cube is numerically equal to twice its volume. Find the length of a diagonal of the cube.", "Solution_4": "Scrambled's question was already fairly hard compared to the original. Demon's is way harder. He's not getting any help if you just keep on feeding him really hard questions that he doesn't want.\n\n\n\nAnyway, to answer yours:\n\n[hide]\n\n6 sides of a cube.\n\n\n\n6d^2 = 2d^3\n\nd^3 - 3d^2 = 0\n\nd^2(d-3) = 0\n\nd=3, 0\n\n\n\n0 doesn't work, so d=3. If you want the diagonal, that would be sqrt(3^2 + (3 :rt2: )^2) which is :sqrt: 27 or 3 :rt3: .\n\n[/hide]\n\n\n\nHere's a geometry problem to mix it up:\n\n\n\nYou have a bunch of identical circles. How many circles can you place around a central circle so that all circles are tangent to one another without overlapping? Justify.", "Solution_5": "Mine wasn't really a polynomial question, if you know the formula sum and product of the roots, you get sum=6 and product=g. the maximum value of the product is achieved when all the roots are equal 6/8=3/4 so all the roots are 3/4 the product in this case is just (3/4)^8", "Solution_6": "Thanks to you, rlee for understanding that I don't truly get the hard questoins.. But what's those signs you put? Are they the answer for my question? If you are saying root 3 or root 2 are answer, it is incorrect.. And what does your question mean? Can you post the question with drawing or more explanation?", "Solution_7": "[quote=\"Demon\"]Mine wasn't really a polynomial question, if you know the formula sum and product of the roots, you get sum=6 and product=g. the maximum value of the product is achieved when all the roots are equal 6/8=3/4 so all the roots are 3/4 the product in this case is just (3/4)^8[/quote]\r\nArg not Viete's...I [b]still[/b] don't get how to use that thingy, can someone explain that in plain english?", "Solution_8": "Oh.. I didn't understand the picture..\r\nYes, it is 3 :sqrt: 3...\r\nThen what does his question mean?", "Solution_9": "I don't get it...\r\nI have algebra II but I don't get what you meant by that...\r\nWhat's that formula for?", "Solution_10": "The formula is the sum of the roots of a polynomial to the n degree.\n\n\n\nFor the circle question you don't really need a picture. The problem is just that you don't understand the question. \n\n\n\nThe question says that you have a circle. Around that circle you draw identical circles that are tangent to it( tangent means that they touch it at one and only one point). How many of these tangent circles can you draw around it so that none of them overlap each other?\n\n\n\nSolution:[hide]\n\nThis question is a lot easier to explain with a picture but I don't have any software for that so I will try my best to do it in words.\n\n\n\nI started with a circle of radius r ( you can pick a value for the radius but ultimately it doesn't matter because they all have the same radius). Then I drew a circle tangent to this one and connected the centers with a line. Tangent to both of these circles I drew another circle which means that now I have the original and two circles tangent to it. Connecting the centers of all the circles I was left with a triangle. This triangle was equilateral because each side was equal to 2r. This means that the angles were all 60. Since a circle has 360 degrees and every circle I draw tangent will make a 60 degree angle I can conclude that there will be 6 of these angles and thus 6 circles can be drawn around the first circle.[/hide]", "Solution_11": "Sorry. I accidently posted the same thing twice. The first time I did it my computer froze while I was looking over it so I thought it didn't get posted but it did.", "Solution_12": "[quote=\"Demon\"]Mine wasn't really a polynomial question, if you know the formula sum and product of the roots, you get sum=6 and product=g. the maximum value of the product is achieved when all the roots are equal 6/8=3/4 so all the roots are 3/4 the product in this case is just (3/4)^8[/quote]\r\n\r\nIt's an interesting problem, but are you sure that it responds to the request of the person who originally posted the thread, and that it really fits on the Getting Started board? I rather think that a problem solution stated in those terms is into the intermediate level. \r\n\r\nBTW, joml88 has a great verbal explanation, written in spoiler :) to the question posed by rlee. That's great stuff to have in this forum. \r\n\r\n(After edit:) I will be deleting posts liberally in this forum, particularly in this thread, so that the original poster can readily find the answers to the very on-topic and on-level questions he is posing.", "Solution_13": "[quote=\"Tare\"][quote=\"Demon\"]Mine wasn't really a polynomial question, if you know the formula sum and product of the roots, you get sum=6 and product=g. the maximum value of the product is achieved when all the roots are equal 6/8=3/4 so all the roots are 3/4 the product in this case is just (3/4)^8[/quote]\nArg not Viete's...I [b]still[/b] don't get how to use that thingy, can someone explain that in plain english?[/quote]\r\n\r\nLets say you have a polynomial such as x^3 - 34x^2 + 12x - 9. I just made that up so it probably doesn't have 3 real roots but forget that. Lets say its roots are a, b, and c. Then we can write it as (x-a)(x-b)(x-c). If you expand that, you get x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc. \r\nSo we must have a+b+c = 34, ab+bc+ca = 12, and abc=9. \r\nAnd of course you can do that for any polynomial you want.", "Solution_14": "Marathon is not ended...\r\nSince we spent a lot time in really hard question, let's go with little bit nerves-off questions....\r\n\r\nThere is this octahedron. When two faces met, the edge is colored orange. How many orange edges are there?", "Solution_15": "[quote=\"demji\"][quote=\"beta\"]\nLet f be a function for which f(x/3) = x :^2: + x + 1. Find the sum of all values of z for which f(3z) = 7.[/quote]\n[/quote]\r\n\r\nMy answer seems to disagree with Silverfalcon's answer.\r\n\r\nBut, then again, maybe someone can tell me where I went wrong.\r\n\r\nYou have f(x/3) = x^2 + x + 1, and you want z for f(3z) = 7.\r\n\r\nSo let f(x/3) = g(y). \r\n\r\ng(y) = (3y)^2 + 3y + 1.\r\n\r\nSo, we have f(3z), which by our ealier function, is (3(3z))^2 + 3(3z) + 1 = 81z^2 + 9z + 1 = 7.\r\n\r\nSo, 81z^2 + 9z - 6 = 0.\r\n\r\nAnd 3(27z^2 + 3z - 2) = 0.\r\n\r\nSo, 3(9z-2)(3z+1) = 0.\r\n\r\nSo, the solutions are 2/9 and -1/3, which summed equals-1/9.\r\n\r\nDiscrepency... yikes.", "Solution_16": "[quote=\"white_horse_king88\"]Question: How many positive integers less than 2002 are divisible by either 3, 4, and/or 5?[/quote]\r\n\r\nI'm not sure if this is right, but...\r\n\r\n[hide]Number divisible by 3 - 667\ndivisible by 4, but not 12 - 333\ndivisible by 5, but not 15 or 20 - 240\n\ntotal - 1240[/hide]\r\n\r\nIf ${f(x) = \\frac{x!}{f(x-1)}}$ and $f(1) = 1$, what is $f(5)$?", "Solution_17": "i have a nice question but don't know if it belongs here so i'll put it down.\r\nyou have 2 parallel chords in a circle. I line drawn perpendicular to one of the chords intersects the other at a distance of 1. The outer chord is 6 and the one closer to the center is 8. What is the diameter of the circle. \r\n\r\nHint[hide] use the Pythagorean Theorem[/hide]\r\nthis problem might have been on one of the mock amc's", "Solution_18": "[quote=\"h_s_potter2002\"]\nIf ${f(x) = \\frac{x!}{f(x-1)}}$ and $f(1) = 1$, what is $f(5)$?[/quote]\r\n\r\n[hide]\n$f(1)=1$\n$f(2)=\\frac{2!}{1}=2$\n$f(3)=\\frac{3!}{2}=3$\n$f(4)=\\frac{4!}{3}=8$\n$f(5)=\\frac{5!}{8}=15$\n[/hide]\r\n\r\nHow many zeroes are at the end of $\\frac{200!}{5^3}$?", "Solution_19": "[quote=\"sam3.14\"]How many zeroes are at the end of $\\frac{200!}{5^3}$?[/quote]\n[hide=\"Answer\"]\n$[200/125]=1$\n$[200/25]=8$\n$[200/5]=40$\n$40+8+1=49$\n$49-3=46$[/hide]\n\n[quote=\"demji\"][quote=\"Rep123Max\"]For what value of n does (1^3+2^3+3^3+...n^3)/(1+2+3+...n)=36[/quote][/quote]\n\n[hide=\"Solution\"]$\\frac{\\left(\\frac{n(n+1)}{2}\\right)^2}{\\frac{n(n+1)}{2}}=\\frac{n(n+1)}{2}$\n\n$n(n+1)=72$\n\n$n=8 \\text{ or } n=-9$[/hide]\r\n\r\nWhat are the number of paths from (0,0) to (6,6) with x >= y at all points. Or if you'd like, y >= x at all points.", "Solution_20": "RC-7th, you might want to look at your reply with the Latex on the second one.\r\n\r\nIt doesn't look so nice.", "Solution_21": "(it needs a space between \\\\ and [, or it considers \\[ to be starting displaymath).", "Solution_22": "isn't anyone going to answer the circle problem\r\nextra hint: use 2 triangles.", "Solution_23": "[quote=\"RC-7th\"][quote=\"sam3.14\"]How many zeroes are at the end of $\\frac{200!}{5^3}$?[/quote]\n[hide=\"Answer\"]\n$[200/125]=1$\n$[200/25]=8$\n$[200/5]=40$\n$40+8+1=49$\n$49-3=46$[/hide]\n\n[quote=\"demji\"][quote=\"Rep123Max\"]For what value of n does (1^3+2^3+3^3+...n^3)/(1+2+3+...n)=36[/quote][/quote]\n\n[hide=\"Solution\"]$\\frac{\\left(\\frac{n(n+1)}{2}\\right)^2}{\\frac{n(n+1)}{2}}=\\frac{n(n+1)}{2}$\n\nThere's only one solution. Ignore the negative.\n$n(n+1)=72$\n\n$n=8 \\text{ or } n=-9$[/hide]\n\nWhat are the number of paths from (0,0) to (6,6) with x >= y at all points. Or if you'd like, y >= x at all points.[/quote]", "Solution_24": "ok i'll post the circle solution since no one is trying it.\r\n[hide]\nlet r be the radius of the circle. Let s be the distance from the midpoint of the chord of length 8 to the center of the circle.\nWe are then able to make 2 triangles with 1 leg equal to half each chord.\nr 2 =4 2 +s 2 \nr 2 =3 2 +(s+1) 2\nthen we can set them equal\n16+s 2 =9+s 2 +2s+1\n2s=6\ns=3\nthen we use the pythagorean theorem:\n3 2 +4 2 =r 2 \nr 2 =25\nr=5\n\nbut the question asks for the diameter. \nd=2r\nd=10\n[/hide]", "Solution_25": "Just in case anyone forgot: \r\n\r\nWhat are the number of paths from (0,0) to (6,6) with x geq y at all points. Or if you'd like, y geq x at all points.", "Solution_26": "When u say paths do u mean shortest paths as in no overlaps? If so\r\n\r\nSolution \r\n[hide]\nHere is some basic theory. Suppose that we live in a town whose streets run in two perpendicular directions (as in this case), where all the streets of the town are represented in the form of horizontal and vertical lines. We can number the horizontal lines with number $0, 1, 2, 3, ...$ and do the same for the vertical lines. Then we can denote their interesctions by pairs of coordinates $(m,n)$, where $m$ is the number of the \"vertical\" street which passes through the intersection and $n$ is the number of the \"horizontal street\". Suppose that we have to go from a house located at the intersection $(0,0)$ (as in this case) to a house located at the intersection $(m,n)$. There will then be $\\displaystyle{m+n\\choose n}$ shortest paths joining the two houses, for each of these shortest paths is the $m+n$ blocks long--$m$ blocks in the horizontal direction and $n$ blocks in the vertical direction. Applying this knowledge to our problem at hand we see that the number of paths is\n\n\\[\\displaystyle{6+6\\choose 6}=\\displaystyle{12 \\choose 6}=924\\]\n\n\n[/hide]\r\n\r\nI believe that this is the correct solution for any integer points $(x,y)$ where $0 \\leq x,y \\leq 6$.", "Solution_27": "Is that right? It seems a bit big but I think I did it right unless I did something really stupid ;)", "Solution_28": "AHHH...crap....\"$x\\ge{y}$\" woops....", "Solution_29": "Answer\r\n\r\n[hide]\n$132$\n[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c,d$ are real numbers \r\n\r\n$ (4,0,0,0) \\plus{} 3(2,2,0,0) \\geq 4(3,1,0,0)$ or \r\n\r\n$ 3(a^{4} \\plus{} b^{4} \\plus{} c^{4} \\plus{} d^{4}) \\plus{} 6(a^{2}b^{2} \\plus{} a^{2}c^{2} \\plus{} a^{2}d^{2} \\plus{} b^{2}c^{2} \\plus{} b^{2}d^{2} \\plus{} c^{2}d^{2}) \\geq 4(a^{3}b\\plus{}b^{3}a\\plus{}a^{3}c\\plus{}c^{3}a\\plus{}a^{3}d\\plus{}d^{3}a\\plus{}b^{3}c\\plus{}c^{3}b\\plus{}b^{3}d\\plus{}d^{3}b\\plus{}c^{3}d\\plus{}d^{3}c)$", "Solution_1": "what do you mean by the first inequality?", "Solution_2": "[quote=\"aadil\"]what do you mean by the first inequality?[/quote]\r\n\r\n\r\nthe first inequality is equivalent to the second inequality.\r\n\r\nin general $ (x_{1},x_{2},...,x_{n})\\equal{}\\sum_{cyc} a_{1}^{x_{1}} a_{2}^{x_{2}}... a_{n}^{x_{n}}$. :D", "Solution_3": "yes but not cyclic it's symmetric", "Solution_4": "[quote=\"mestav\"]Let $ a,b,c,d$ are real numbers \n\n$ (4,0,0,0) \\plus{} 3(2,2,0,0) \\geq 4(3,1,0,0)$ or \n\n$ 3(a^{4} \\plus{} b^{4} \\plus{} c^{4} \\plus{} d^{4}) \\plus{} 6(a^{2}b^{2} \\plus{} a^{2}c^{2} \\plus{} a^{2}d^{2} \\plus{} b^{2}c^{2} \\plus{} b^{2}d^{2} \\plus{} c^{2}d^{2}) \\geq 8(a^{3}b \\plus{} a^{3}c \\plus{} a^{3}d \\plus{} b^{3}c \\plus{} b^{3}d \\plus{} c^{3}d)$[/quote]\r\nThis is not true, for counter-example, $ \\, a\\equal{}9,\\, b\\equal{}8,\\, c\\equal{}7,\\, d\\equal{}6.$", "Solution_5": "[quote=\"mestav\"]yes but not cyclic it's symmetric[/quote]\r\n\r\noh yeah, sorry :oops:", "Solution_6": "original inequality is $ a^{3}(b \\plus{} c \\plus{} d) \\plus{} b^{3}(a \\plus{} c \\plus{} d) \\plus{} c^3(a \\plus{} b \\plus{} d) \\plus{} d^3(a \\plus{} b \\plus{} c)\\le\\frac {3}{4}(a^{2} \\plus{} b^{2} \\plus{} c^{2} \\plus{} d^{2})^2$\r\n\r\nI multiplied and I have this inequality\r\n\r\nis original inequality false? or my operations false?", "Solution_7": "It's equivalent to $ (a \\minus{} b)^4 \\plus{} (a \\minus{} c)^4 \\plus{} (a \\minus{} d)^4 \\plus{} (b \\minus{} c)^4 \\plus{} (b \\minus{} d)^4 \\plus{} (c \\minus{} d)^4 \\ge 0$, but I think the right side of the inequality should be $ 4(a^3b \\plus{} b^3a \\plus{} a^3c \\plus{} c^3a \\plus{} a^3d \\plus{} d^3a \\plus{} b^3c \\plus{} c^3b \\plus{} b^3d \\plus{} d^3b \\plus{} c^3d \\plus{} d^3c)$ instead of what you have.", "Solution_8": "I am sorry I fixed it and thank you dgreenb801", "Solution_9": "The convention in this forum is to use square brackets for symmetrical sums, like $ [3,1,0,0]$. Note that this sum has always $ n!$ terms, e.g. $ [4,0,0,0] \\equal{} 6\\sum a^4$, as the symmetrical sum counts [i]all [/i]the permutations. \r\n\r\nIf one writes $ \\sum_{sym}$, it is understood to count each different term only once, e.g. if we deal with 4 variables, $ \\sum_{sym}a^3b$ has 12 terms, not 24. And $ [3,1,0,0]\\equal{}2\\sum_{sym}a^3b$.\r\n\r\nThis sometimes leads to confusion, let's just be coherent in each line. :)" } { "Tag": [ "ratio", "calculus", "integration", "function", "logarithms", "analytic geometry", "graphing lines" ], "Problem": "Thermodynamics is an important branch of Classical Physics, and its application to chemical systems plays a decisive role in the understanding of their behaviour. This thread contains a selection of problems collected or adapted from several sources: Thermodynamics books, university Physical Chemistry exams, several Internet resources (including this forum), and a few elaborated by me. The subjects covered by these problems are the following: basic definitions in Thermodynamics; Zeroth, First, Second, and Third Laws of Thermodynamics; Phase Equilibrium; Reaction Equilibrium; and Solutions. The material here can be used both by students preparing for Chemistry Olympiads and by university students. Note that the problems do not follow the order of the subjects covered as presented above: this because it is my opinion that a given problem is more challenging, and there is a better chance for the student to develop its understanding, if the problem is not explicitly linked to a given subject. I hope you find this material useful. Also, feel free to make any comments or suggestions.\r\n\r\n[b]Important note:[/b] in order to prevent this thread to become a mess, please [i]do not post here discussions or solutions relative to these problems[/i] \u2013 use instead [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185861]this thread[/url] to do that. However, feel free to post here any interesting problems you might have from other sources.\r\n-------------------------------------------------------------------------------\r\n\r\n[size=150][b]Problem 1[/b][/size]. Consider the reversible expansion of an ideal gas from $ V_1$ to $ V_2$.\r\n\r\n(a) Deduce the expression for the work done by the system on the surroundings if the process is isothermic.\r\n\r\n(b) If the process is performed in an adiabatic way, will the final pressure of the system be greater, equal, or lower compared with (a)? Justify your answer.\r\n\r\n[size=150][b]Problem 2[/b][/size]. For a fixed amount of a perfect gas, which of the following statements must be true?\r\n\r\n(a) U and H each depend only on T.\r\n(b) $ C_P$ is a constant.\r\n(c) $ P\\,dV \\equal{} nR\\,dT$ for every infinitesimal process.\r\n(d) $ C_P \\minus{} C_V \\equal{} nR$\r\n(e) $ dU \\equal{} C_V\\,dT$ for a reversible process.", "Solution_1": "[size=150][b]Problem 3[/b][/size]. For each of the following processes deduce whether each of the quantities $ q$, $ w$, $ \\Delta U$, $ \\Delta H$, $ \\Delta S$, and $ \\Delta S_{univ}$ is positive, zero, or negative:\r\n\r\n(a) Reversible melting of solid benzene at 1 atm and the normal melting point.\r\n(b) Reversible melting of ice at 1 atm and 0\u00baC.\r\n(c) Reversible adiabatic expansion of a perfect gas.\r\n(d) Reversible isothermal expansion of a perfect gas.\r\n(e) Adiabatic expansion of a perfect gas into a vacuum.\r\n(f) Reversible heating of a perfect gas at constant P.\r\n(g) Reversible cooling of a perfect gas at constant V.\r\n(h) Joule-Thomson adiabatic throttling of a perfect gas.\r\n(i) Combustion of benzene in a sealed container with rigid, adiabatic walls.\r\n(j) Adiabatic expansion of a nonideal gas into vacuum.\r\n\r\n[size=150][b]Problem 4[/b][/size]. A perfect gas undergoes an expansion process at constant pressure. Does its internal energy increase or decrease? Justify your answer.\r\n\r\n[size=150][b]Problem 5[/b][/size]. A perfect gas with $ C_{V,m} \\equal{} 3R$ independent of T expands adiabatically into a vacuum, thereby doubling its volume. Two students present the following conflicting analyses. John uses the expression $ TV^{\\gamma \\minus{} 1} \\equal{} constant$ to write $ T_2/T_1 \\equal{} (V_1/2V_1)^{1/3} \\equal{}> T_2 \\equal{} T_1/\\sqrt[3]{2}$. Wendy writes $ \\Delta U \\equal{} q \\plus{} w \\equal{} 0 \\plus{} 0 \\equal{} 0$ and $ \\Delta U \\equal{} C_V \\Delta T$, so $ \\Delta T \\equal{} 0$ and $ T_2 \\equal{} T_1$. Which student is correct? What error did the other student make?", "Solution_2": "[size=150][b]Problem 6[/b][/size]. Classify each of the following sentences as true or false. Always justify your choice.\r\n\r\n(1)\tA closed system cannot interact with its surroundings.\r\n\r\n(2)\tThe Avogadro constant has no units.\r\n\r\n(3)\tFor every cyclic process, the final state of the system is the same as the initial state.\r\n\r\n(4)\t$ \\Delta T \\equal{} 0$ for every isothermal process.\r\n\r\n(5)\tEvery closed system is isolated.\r\n\r\n(6)\tDoubling the absolute temperature of an ideal gas at fixed volume and amount of gas will double its pressure.\r\n\r\n(7)\tDensity is an intensive property.\r\n\r\n(8)\tA thermodynamic process is defined by the final state and the initial state.\r\n\r\n(9)\t$ \\Delta U \\equal{} 0$ for every cyclic process.\r\n\r\n(10)\tThe Atlantic Ocean is an open system.\r\n\r\n(11)\tFor a fixed amount of an ideal gas, the product PV remains constant during any process.\r\n\r\n(12)\tThe P-V work in a mechanically reversible process in a closed system always equals $ \\minus{}P \\Delta V$.\r\n\r\n(13)\tThe ratio PV/mT is the same for all gases in the limit of zero pressure.\r\n\r\n(14)\tFor every process, $ \\Delta E_{syst} \\equal{} \\minus{} \\Delta E_{surr}$.\r\n\r\n(15)\tA homogeneous system must be a pure substance.\r\n\r\n(16)\t$ \\Delta U \\equal{} q \\plus{} w$ for every thermodynamic system at rest in the absence of external fields.\r\n\r\n(17)\tA system containing only one substance must be homogeneous.\r\n\r\n(18)\tThe pressure of a nonideal gas mixture is equal to the sum of the partial pressures defined by $ P_i \\equal{} x_i P$.\r\n\r\n(19)\tA change of state from state 1 to state 2 produces a greater increase in entropy when carried out irreversibly than when done reversibly.\r\n\r\n(20)\tMole fractions are intensive properties.", "Solution_3": "(21)\t$ \\Delta U \\equal{} 0$ for a reversible phase change at constant T and P.\r\n\r\n(22)\tFor every cyclic process, the final state of the surroundings is the same as the initial state of the surroundings.\r\n\r\n(23)\tOn the Celsius scale, the boiling point of water is slightly less than 100.00\u00baC.\r\n\r\n(24)\tThe value of the work in a reversible process in a closed system can be found so long as we know the initial state and the final state of the system.\r\n\r\n(25)\tEvery isolated system is closed.\r\n\r\n(26)\tIdeal gas isotherms further away from the axes of a PV plot correspond to higher temperatures.\r\n\r\n(27)\tAll ideal gases have the same density at 25\u00baC and 1 bar.\r\n\r\n(28)\tIncreasing the temperature of the hot reservoir of a Carnot cycle engine must increase the efficiency of the engine. \r\n\r\n(29)\tThe infinitesimal P-V work in a mechanically reversible process in a closed system always equals $ \\minus{}P dV$.\r\n\r\n(30)\tFor a closed system at rest with no fields present, the sum $ q \\plus{} w$ has the same value for every process that goes from a given state 1 to a given state 2.\r\n\r\n(31)\tThe ratio PV/nT is the same for all gases in the limit of zero pressure.\r\n\r\n(32)\tThe value of the integral $ \\int_1^2 P\\,dV$ is fixed once the initial and final states 1 and 2 and the equation of state P = P(T,V) are known.\r\n\r\n(33)\tEvery process for which $ \\Delta T \\equal{} 0$ is an isothermal process.\r\n\r\n(34)\t$ C_P$ is an extensive property and a state function.\r\n\r\n(35)\tWe have $ \\int_1^2 P\\,dV \\equal{} \\int_1^2 nR\\,dT$ for every reversible process in an ideal gas.\r\n\r\n(36)\tIf systems A and B each consist of pure liquid water at 1 bar pressure and if $ T_A > T_B$, then the internal energy of system A must be greater than that of B.\r\n\r\n(37)\tThe equation $ w_{rev} \\equal{} \\minus{} \\int_1^2 P\\,dV$ applies only to constant-pressure processes.\r\n\r\n(38)\tThe quantities $ H$, $ U$, $ PV$, $ \\Delta H$, and $ P \\Delta V$ all have the same dimensions.\r\n\r\n(39)\tThe heat $ q$ for an irreversible change of state from state 1 to state 2 might differ from the heat for the same change of state carried out reversibly. \r\n\r\n(40)\t$ q$ must be zero for an isothermal process.", "Solution_4": "(41)\t$ \\Delta H$ is a state function.\r\n\r\n(42)\tFor a closed system at rest in the absence of external fields, $ U \\equal{} q \\plus{} w$.\r\n\r\n(43)\tThe value of the work done by a system in a given process can only be calculated if the process is reversible.\r\n\r\n(44)\t$ \\Delta T \\equal{} 0$ for every adiabatic process in a closed system.\r\n\r\n(45)\tDecreasing the temperature of the cold reservoir of a Carnot cycle engine must increase the efficiency of the engine.\r\n\r\n(46)\t$ \\Delta G \\equal{} 0$ always for a reversible process in a closed system capable of P-V work only.\r\n\r\n(47)\t$ C_V$ is independent of T for every perfect gas.\r\n\r\n(48)\tThe higher the absolute temperature of a system, the smaller the increase in its entropy produced by a given positive amount $ \\delta q_{rev}$ of reversible heat flow.\r\n\r\n(49)\tIf a closed system undergoes a reversible process for which $ \\Delta V \\equal{} 0$, then the P-V work done on the system in this process must be zero.\r\n\r\n(50)\tA closed system process that has $ \\Delta T \\equal{} 0$, must have $ \\Delta U \\equal{} 0$.\r\n\r\n(51)\t$ \\Delta H$ is defined only for a constant-pressure process.\r\n\r\n(52)\tFor an adiabatic process in a closed system, $ \\Delta S$ must be zero.\r\n\r\n(53)\tThe entropy of 20 g of $ H_2O(l)$ at 300 K and 1 bar is twice the entropy of 10 g of $ H_2O(l)$ at the same T and P.\r\n\r\n(54)\tU remains constant in every isothermal process in a closed system.\r\n\r\n(55)\tThe relation $ \\Delta G \\equal{} \\Delta H \\minus{} T \\Delta S$ is valid for all processes.\r\n\r\n(56)\tA Carnot cycle is, by definition, a reversible cycle.\r\n\r\n(57)\tFor a constant-volume process in a closed system, $ \\Delta U \\equal{} \\Delta H$.\r\n\r\n(58)\tAn adiabatic process cannot decrease the entropy of a closed system.\r\n\r\n(59)\t$ q \\equal{} 0$ for every cyclic process.\r\n\r\n(60)\tThe molar entropy of 20 g of $ H_2O(l)$ at 300 K and 1 bar is equal to the molar entropy of 10 g of $ H_2O(l)$ at the same T and P.", "Solution_5": "(61)\tFor every closed system in thermal and mechanical equilibrium and capable of P-V work only, the state function G is minimized when material equilibrium is reached.\r\n\r\n(62)\tIf a closed system at rest in the absence of external fields undergoes an adiabatic process that has $ w \\equal{} 0$, then the system\u2019s temperature must remain constant. \r\n\r\n(63)\tThe quantities SdT, TdS, VdP, and $ \\int_1^2 V\\,dP$ all have dimensions of energy.\r\n\r\n(64)\tThe chemical potential $ \\mu_i$ is a state function and an intensive property.\r\n\r\n(65)\tFor a reversible isothermal process in a closed system, $ \\Delta S$ must be zero.\r\n\r\n(66)\t$ \\Delta G$ is undefined for a process in which T changes.\r\n\r\n(67)\tP-V work is usually negligible for solids and liquids.\r\n\r\n(68)\tSince a Carnot cycle is a cyclic process, the work done in a Carnot cycle is zero.\r\n\r\n(69)\tThe Gibbs energy of 12 g of ice at 0\u00baC and 1 atm is less than the Gibbs energy of 12 g of liquid water at 0\u00baC and 1 atm.\r\n\r\n(70)\tIf neither heat nor matter can enter or leave a system, that system must be isolated.\r\n\r\n(71)\t$ \\Delta G \\equal{} 0$ for a reversible phase change at constant T and P.\r\n\r\n(72)\tWe always have $ \\int_1^2 \\frac{C_V}{T}\\,dT \\equal{} C_V \\ln \\frac{T_2}{T_1}$.\r\n\r\n(73)\t$ C_P \\minus{} C_V \\equal{} nR$ for all gases.\r\n\r\n(74)\tFor a closed system with P-V work only, a constant-pressure process that has $ q > 0$ must have $ \\Delta T > 0$.\r\n\r\n(75)\tFor an adiabatic process in a closed system, $ \\Delta S_{syst}$ must be zero.\r\n\r\n(76)\tFor a closed system, $ \\Delta S$ can never be negative.\r\n\r\n(77)\tFor a reversible process in a closed system, $ \\Delta S$ must be zero.\r\n\r\n(78)\tFor a closed system, equilibrium has been reached when S has been maximized.\r\n\r\n(79)\tThermodynamics cannot calculate $ \\Delta S$ for an irreversible process.\r\n\r\n(80)\tFor a reversible process in a closed system, $ \\delta q \\equal{} T\\,dS$.", "Solution_6": "(81)\tFor every process in an isolated system, $ \\Delta T \\equal{} 0$.\r\n\r\n(82)\tFor every process in an isolated system that has no macroscopic kinetic or potential energy, $ \\Delta U \\equal{} 0$.\r\n\r\n(83)\tFor an adiabatic process in a closed system, $ \\Delta S$ cannot be negative.\r\n\r\n(84)\tFor every process in an isolated system, $ \\Delta S \\equal{} 0$.\r\n\r\n(85)\t$ \\Delta S \\equal{} 0$ for every adiabatic process in a closed system.\r\n\r\n(86)\tFor every reversible process in a closed system, $ \\Delta S \\equal{} \\Delta H/T$.\r\n\r\n(87)\tFor a process in an isolated system, $ \\Delta S$ cannot be negative.\r\n\r\n(88)\tFor every isothermal process in a closed system, $ \\Delta S \\equal{} \\Delta H/T$.\r\n\r\n(89)\t$ q \\equal{} 0$ for every isothermal process in a closed system.\r\n\r\n(90)\tFor the conversion of 1 mol of $ N_2(g)$ from 25\u00baC and 10 L to 25\u00baC and 20 L, $ \\Delta S$ is the same independently of the conversion being reversible or irreversible.\r\n\r\n(91)\tFor a reversible process in a closed system, $ \\Delta S_{univ}$ must be zero.\r\n\r\n(92)\t$ C_P \\minus{} C_V \\equal{} TV \\alpha ^2/ \\kappa$ for any substance.\r\n\r\n(93)\tFor an irreversible, isothermal, isobaric process in a closed system with P-V work only, $ \\Delta G$ must be negative.\r\n\r\n(94)\t$ \\Delta S$ is positive for every irreversible process.\r\n\r\n(95)\tThe work done by a closed system can exceed the decrease in the system\u2019s internal energy.\r\n\r\n(96)\tThe Gibbs energy of a closed system with P-V work only is always minimized at equilibrium.\r\n\r\n(97)\tIf a system remains in thermal and mechanical equilibrium during a process, then its T and P are constant during the process.\r\n\r\n(98)\tThe entropy S of a closed system with P-V work only is always maximized at equilibrium.\r\n\r\n(99)\t$ \\Delta G \\equal{} 0$ for the freezing of supercooled water at -10\u00baC and 1 atm.\r\n\r\n(100)\t$ \\Delta S_{syst} \\plus{} \\Delta S_{surr} > 0$ for every irreversible process.", "Solution_7": "(101)\t$ G_{syst} \\plus{} G_{surr}$ is constant for any process.\r\n\r\n(102)\tIf a system absorbs heat, then its temperature will always increase.\r\n\r\n(103)\tIf the temperature of a system increases, then it must have absorbed heat.\r\n\r\n(104)\tBy no means can a given quantity of heat be completely converted into work, according to the second law.\r\n\r\n(105)\tFor every spontaneous process $ \\Delta S > 0$.\r\n\r\n(106)\tFor every isochoric process in a closed system $ \\Delta U \\equal{} q_V$, where $ q_V$ is the heat exchanged at constant volume.\r\n\r\n(107)\tIf the temperature of a system increases then its volume must also increase.\r\n\r\n(108)\t$ \\Delta G < 0$ for every spontaneous process.\r\n\r\n(109)\tFor every isobaric process in a closed system $ \\Delta H \\equal{} q_P$, where $ q_P$ is the heat exchanged at constant pressure.\r\n\r\n(110)\tOn isothermal conditions, if the pressure of a system increases then its volume must decrease.", "Solution_8": "[size=150][b]Problem 7[/b][/size]. Because the heat capacities per unit volume of gases are small, accurate measurement of $ C_P$ or $ C_V$ for gases is not easy. However, accurate measurement of the adiabatic coefficient $ \\gamma$ of a gas (for example, by measurement of the speed of sound in the gas) is easy. For gaseous $ CCl_4$ at 0.1 bar and 20\u00baC, experiment gives $ \\gamma \\equal{} 1.13$. Find $ C_{P,m}$ and $ C_{V,m}$ for $ CCl_4(g)$ at 20\u00baC and 1 bar.\r\n\r\n\r\n[size=150][b]Problem 8[/b][/size]. Which of the following cyclic integrals must vanish for a closed system with P-V work only?\r\n\r\n(a) $ \\oint P\\,dV$\r\n\r\n(b) $ \\oint (P\\,dV \\plus{} V\\,dP)$\r\n\r\n(c) $ \\oint V\\,dV$\r\n\r\n(d) $ \\oint \\frac{\\delta q_{rev}}{T}$\r\n\r\n(e) $ \\oint H\\,dT$\r\n\r\n(f) $ \\oint dU$\r\n\r\n(g) $ \\oint \\delta q_{rev}$\r\n\r\n(h) $ \\oint \\delta q_P$\r\n\r\n(i) $ \\oint \\delta w_{rev}$\r\n\r\n(j) $ \\oint \\frac{\\delta w_{rev}}{P}$", "Solution_9": "[size=150][b]Problem 9[/b][/size]. The molar heat capacity of oxygen at constant pressure for temperatures in the range 300 to 400 K and for low or moderate pressures can be approximated as $ C_{P,m} \\equal{} a \\plus{} bT$, where $ a \\equal{} 6.15\\,\\,cal\\,K^{\\minus{}1}\\,mol^{\\minus{}1}$ and $ b \\equal{} 0.00310\\,\\,cal\\,K^{\\minus{}2}\\,mol^{\\minus{}1}$.\r\n\r\n(a) Calculate $ q$, $ w$, $ \\Delta U$, and $ \\Delta H$ when 2.00 mol of $ O_2$ are reversibly heated from 27\u00baC to 127\u00baC with P held fixed at 1.00 atm. Assume perfect-gas behaviour.\r\n\r\n(b) Calculate $ q$, $ w$, $ \\Delta U$, and $ \\Delta H$ when 2.00 mol of $ O_2$ initially at 1.00 atm are reversibly heated from 27\u00baC to 127\u00baC with V held fixed.\r\n\r\n(c) Find $ \\Delta S$ when 2.00 mol of $ O_2$ are heated from 27\u00baC to 127\u00baC with P held fixed.\r\n\r\n\r\n[size=150][b]Problem 10[/b][/size]. The molar heat of vaporization of Ar at its normal boiling point 87.3 K is 1.56 kcal/mol.\r\n\r\n(a) Compute $ \\Delta S$ for the vaporization of 1.00 mol of Ar at 87.3 K and 1 atm.\r\n\r\n(b) Find $ \\Delta S$ when 5.00 g of Ar gas condenses to liquid at 87.3 K and 1 atm.", "Solution_10": "[size=150][b]Problem 11[/b][/size]. Explain how liquid water can go from 25\u00baC and 1 atm to 30\u00baC and 1 atm in a process for which $ q < 0$.\r\n\r\n\r\n[size=150][b]Problem 12[/b][/size]. One mole of liquid water at 30\u00baC is adiabatically compressed, with P increasing from 1.00 to 10.00 atm. Since liquids and solids are rather incompressible, it is a fairly good approximation to take V as unchanged for this process. With this approximation calculate $ q$, $ \\Delta U$, and $ \\Delta H$ for this process.\r\n\r\n\r\n[size=150][b]Problem 13[/b][/size]. Calculate $ \\Delta U$ and $ \\Delta H$ for each of the following changes in state of 2.50 mol of a perfect monoatomic gas with $ C_{V,m} \\equal{} 1.5R$ for all temperatures:\r\n\r\n(a) (1.50 atm, 400 K) -----> (3.00 atm, 600 K)\r\n\r\n(b) (2.50 atm, 20.0 L) -----> (2.00 atm, 30.0 L)\r\n\r\n(c) (28.5 L, 400 K) -----> (42.0 L, 400 K)\r\n\r\n\r\n[size=150][b]Problem 14[/b][/size]. Can $ q$ and $ w$ be calculated for the processes of the previous problem? If the answer is yes, calculate them for each process.", "Solution_11": "[size=150][b]Problem 15[/b][/size]. For each of the following sets of quantities, all the quantities except one have something in common. State what they have in common and state which quantity does not belong with to each set (in some cases, more than one answer for the property in common might be possible).\r\n\r\n(a) $ H$, $ U$, $ q$, $ S$, $ T$.\r\n\r\n(b) $ T$, $ \\Delta S$, $ q$, $ w$, $ \\Delta H$.\r\n\r\n(c) $ q$, $ w$, $ U$, $ \\Delta U$, $ V$, $ H$.\r\n\r\n(d) $ C_V$, $ C_P$, $ U$, $ T$, $ S$, $ G$, $ A$, $ V$.\r\n\r\n(e) $ \\rho$, $ S_m$, $ M$, $ V$.\r\n\r\n(f) $ \\Delta H$, $ \\Delta S$, $ dV$, $ \\Delta P$.\r\n\r\n(g) $ H$, $ U$, $ G$, $ S$, $ A$.\r\n\r\n(h) $ U$, $ V$, $ \\Delta H$, $ S$, $ T$.", "Solution_12": "[size=150][b]Problem 16[/b][/size]. For each of the following statements, tell which state function(s) is (are) being described.\r\n\r\n(a) It enables one to find the rates of change of $ H$ and of $ S$ with respect to T at constant P.\r\n\r\n(b) They determine whether substance $ i$ in phase $ \\alpha$ is in phase equilibrium with $ i$ in phase $ \\beta$.\r\n\r\n(c) It enables one to find the rates of change of $ U$ and of $ S$ with respect to T at constant V.\r\n\r\n(d) It is maximized when an isolated system reaches equilibrium.\r\n\r\n(e) It is maximized when a system reaches equilibrium.\r\n\r\n(f) It is minimized when a closed system capable of P-V work only and held at constant T and P reaches equilibrium.\r\n\r\n\r\n[size=150][b]Problem 17[/b][/size]. Show that it is impossible to attain the absolute zero of temperature.", "Solution_13": "[size=150][b]Problem 18[/b][/size]. Derive expressions for $ \\left(\\frac{\\partial S}{\\partial T}\\right)_P$, $ \\left(\\frac{\\partial S}{\\partial T}\\right)_V$, and $ \\left(\\frac{\\partial H}{\\partial P}\\right)_T$.\r\n\r\n\r\n[size=150][b]Problem 19[/b][/size]. Show that $ \\left(\\frac{\\partial U}{\\partial P}\\right)_T \\equal{} PV \\kappa \\minus{} TV \\alpha$.\r\n\r\n\r\n[size=150][b]Problem 20[/b][/size]. Show that $ \\left(\\frac{\\partial U}{\\partial T}\\right)_P \\equal{} C_P \\minus{} PV \\alpha$.", "Solution_14": "[size=150][b]Problem 21[/b][/size]. Show that $ \\left(\\frac{\\partial H}{\\partial V}\\right)_T \\equal{} \\frac{\\alpha T \\minus{} 1}{\\kappa}$.\r\n\r\n\r\n[size=150][b]Problem 22[/b][/size]. Show that $ \\mu_J \\equal{} \\frac{P \\minus{} \\alpha T / \\kappa}{C_V}$, where $ \\mu_J$ is the Joule coefficient.\r\n\r\n\r\n[size=150][b]Problem 23[/b][/size]. Show that $ \\mu_{JT} \\equal{} \\frac{V(\\alpha T \\minus{} 1)}{C_P}$, where $ \\mu_{JT}$ is the Joule-Thomson coefficient.", "Solution_15": "[size=150][b]Problem 24[/b][/size]. Prove the [i]Gibbs-Helmholtz equation[/i], $ \\left[\\frac {\\partial (G/T)}{\\partial T}\\right]_P \\equal{} \\minus{} \\frac {H}{T^2}$.\r\n\r\n\r\n[size=150][b]Problem 25[/b][/size]. Prove [i]Mayer\u2019s Relation[/i], $ C_P \\minus{} C_V \\equal{} \\frac {TV \\alpha ^2}{\\kappa}$.\r\n\r\n\r\n[size=150][b]Problem 26[/b][/size]. Show that $ \\left(\\frac {\\partial C_P}{\\partial P}\\right)_T \\equal{} \\minus{} T \\left(\\frac {\\partial ^2V}{\\partial T^2}\\right)_P$. The volumes of substances increase approximately linearly with T, so $ \\frac {\\partial ^2V}{\\partial T^2}$ is usually quite small. Therefore, the pressure dependence of $ C_P$ can usually be neglected unless one is dealing with high pressures.", "Solution_16": "[size=150][b]Problem 27[/b][/size]. Show that $ \\left(\\frac {\\partial C_V}{\\partial V}\\right)_T \\equal{} T \\left(\\frac {\\partial ^2P}{\\partial T^2}\\right)_V$.\r\n\r\n\r\n[size=150][b]Problem 28[/b][/size]. A certain gas obeys the equation of state $ PV \\equal{} nRT(1 \\plus{} bP)$, where $ b$ is a constant. For this gas, find expressions for (a) $ \\left(\\frac {\\partial U}{\\partial V}\\right)_T$, (b) $ C_P \\minus{} C_V$, and (c) $ \\mu_{JT}$.\r\n\r\n\r\n[size=150][b]Problem 29[/b][/size]. A certain author states that a given solid obeys the two equations of state shown bellow, where $ V_o$, $ a$, $ b$, and $ c$ are constants. Show that this statement is wrong, because both equations cannot describe the same substance.\r\n\r\n$ V \\equal{} V_o \\minus{} aP \\plus{} bT$ and $ U \\equal{} cT \\minus{} bPT$", "Solution_17": "[size=150][b]Problem 30[/b][/size]. Show that $ \\frac{\\left(\\frac{\\partial P}{\\partial V}\\right)_S}{\\left(\\frac{\\partial P}{\\partial V}\\right)_T} \\equal{} \\gamma$ (this is [i]Reech\u2019s Relation[/i]).\r\n\r\n\r\n[size=150][b]Problem 31[/b][/size]. Show that\r\n\r\n$ \\left(C_P \\minus{} C_V \\right) \\left( \\frac{\\partial^2 T}{\\partial P \\partial V} \\right) \\plus{} \\left( \\frac{\\partial C_P}{\\partial P} \\right)_V \\left( \\frac{\\partial T}{\\partial V} \\right)_P \\minus{} \\left( \\frac{\\partial C_V}{\\partial V} \\right)_P \\left( \\frac{\\partial T}{\\partial P} \\right)_V \\equal{} 1$.\r\n\r\n\r\n[size=150][b]Problem 32[/b][/size]. The behaviour of a given gas is described by the thermal equation of state $ PV \\equal{} kT^3$ and by the energetic equation of state $ U \\equal{} bT^3 \\ln\\left(\\frac{V}{V_o}\\right)$, where $ k$ and $ V_o$ are known constants. \r\n\r\n(a) Express the constant $ b$ as a function of $ k$ and $ V_o$.\r\n\r\n(b) Is $ C_V$ for this gas compatible with the Third Law of thermodynamics?", "Solution_18": "[size=150][b]Problem 33[/b][/size]. An ideal gas goes through a reversible adiabatic process from the initial state $ (P_1, V_1, T_1)$ to the final state $ (P_2, V_2, T_2)$. You may assume the heat capacities to be constant.\r\n\r\n(a) Show that for such a process Poisson\u2019s equation $ PV^{\\gamma} \\equal{} const.$ holds.\r\n\r\n(b) Derive an expression for the work done by (or on) the gas.\r\n\r\n(c) Show that, if a quantity of heat equivalent to the work done by the gas in the previous process is added to the gas in the final state $ (P_2, V_2, T_2)$ while keeping the volume constant, the temperature of the gas will return to $ T_1$.\r\n\r\n\r\n[size=150][b]Problem 34[/b][/size]. Show that the internal energy of a system whose equation of state has the form $ P \\equal{} T \\cdot f(V)$ (where $ f(V)$ is a function of V only) is independent of volume.", "Solution_19": "[size=150][b]Problem 35[/b][/size]. The internal energy per unit volume of a gas, $ u$, is a function of T only, and the equation of state for the gas is given by $ P \\equal{} \\frac {1}{3} u(T)$. Is the heat capacity at constant volume for this gas compatible with the Third Law of thermodynamics?\r\n\r\n\r\n[size=150][b]Problem 36[/b][/size]. Show that the Hessian determinant $ H$ for the internal energy $ U \\equal{} U(S,V)$ is given by $ H[U(S,V)] \\equal{} \\minus{} \\frac{T}{C_V} \\left(\\frac{\\partial P}{\\partial V}\\right)_T$.", "Solution_20": "[size=150][b]Problem 37[/b][/size]. (a) Provide a deduction of [i]Clapeyron equation [/i] (for a bonus, try to do this in at least two different ways).\r\n\r\n(b) Use Clapeyron equation to estimate the slope of the equilibrium curve solid-liquid of water, knowing that the enthalpy of fusion of water is $ \\Delta H_{fus} \\equal{} 6.008\\,kJ/mol$, the densities of ice and liquid water at 0\u00baC are $ \\rho_{ice} \\equal{} 0.91671\\,g/mL$ and $ \\rho_{lw} \\equal{} 0.99984\\,g/mL$, and the molar mass of water is 18 g/mol.", "Solution_21": "[size=150][b]Problem 38[/b][/size]. The expansion and pressure coefficients for a given system are given by\r\n\r\n$ \\alpha \\equal{} \\frac {b}{V_o \\minus{} aP \\plus{} bT}$ and $ \\alpha _P \\equal{} \\frac {b}{aP}$\r\n\r\nwhere $ a,\\,b, \\,V_o$ are constants. Derive the thermal state equation for this system.\r\n\r\n\r\n[size=150][b]Problem 39[/b][/size]. The heat capacity at constant volume for a system is given by\r\n\r\n$ C_V \\equal{} A \\frac {e^{ \\minus{} k/T}}{T^3}$,\r\n\r\nwhere [i]A[/i] and [i]k[/i] are constants. Is this expression compatible with the Third Law of Thermodynamics?", "Solution_22": "[size=150][b]Problem 40[/b][/size]. Is the following state equation for a gas, $ PV \\equal{} nRT \\left(1\\minus{}\\frac{c}{V}\\right)$, where $ c$ is a positive constant, acceptable for all values of pressure and volume?", "Solution_23": "[size=150][b]Problem 41[/b][/size]. Show that $ \\left(\\frac {\\partial H}{\\partial P}\\right)_T \\equal{} V \\minus{} T \\left(\\frac {\\partial V}{\\partial T}\\right)_P$.", "Solution_24": "[size=150][b]Problem 42[/b][/size]. Prove the equation $ \\left(\\frac {\\partial U}{\\partial V}\\right)_T \\equal{} T \\left(\\frac {\\partial P}{\\partial T}\\right)_V \\minus{} P$ by at least three different ways. This is a very important expression, since it relates the thermal and energetic equations of state.", "Solution_25": "The Joule-Thomson or \u201cthrottling\u201d process is an experimental procedure of great practical importance that is commonly used to cool and liquefy gases. It was originally devised in 1853 by James Joule and William Thomson (in later life, Lord Kelvin). In this process (refer to figure I below), a gas is made to pass through a porous barrier from a region maintained at the higher pressure $ P_1$ to a region maintained at a lower pressure $ P_2$. Note that all the walls of the system (including the pistons and porous plug) are adiabatic. The throttling barrier is porous but not greatly so, which allows a gas to pass slowly from one chamber to the other, therefore maintaining pressure equilibrium on both sides. The porous wall was originally made of a wad of cotton, but nowadays it can be made of glass fibbers or a porous ceramic material, or the porous plug can be replaced by a narrow opening (a needle valve). The last two options are used in industrial applications, where the pressure difference is maintained continuously by a pump.\r\nTo treat thermodynamically this process, start by noting that all the pressure drop occurs at the porous wall, therefore the gas on the left is maintained at constant pressure $ P_1$ and the gas that emerges on the right is at constant pressure $ P_2$. As there is no transfer of energy to/from the gas as heat, we can write for $ \\Delta U \\equal{} W_T$, where $ W_T$ is total amount of work done on/by the gas. If one mole of the gas occupying a volume $ V_1$ is made to pass through the barrier by the left piston, the work done [i]on[/i] the gas is $ W_{left} \\equal{} P_1V_1$. As this amount of gas emerges on the right, it will occupy a new volume $ V_2$ (since the pressure is different on the right) and it will push the right piston, therefore the work done [i]by[/i] the gas is $ W_{right} \\equal{} \\minus{} P_2V_2$. Hence,\r\n\r\n$ \\Delta U \\equal{} W_T \\equal{} W_{left} \\plus{} W_{right} \\equal{} > U_2 \\minus{} U_1 \\equal{} P_1V_1 \\minus{} P_2V_2 \\equal{} > U_1 \\plus{} P_1V_1 \\equal{} U_2 \\plus{} P_2V_2$\r\n\r\nRemembering the definition of enthalpy, $ H \\equiv U \\plus{} PV$, the equation above can be written as\r\n\r\n$ H_1 \\equal{} H_2$\r\n\r\nand so the Joule-Thomson process occurs with constant enthalpy. As we are usually interested in the change of temperature that occurs with the forced change in pressure, this process allow us to measure the partial derivative\r\n\r\n$ \\left(\\frac {\\partial T}{\\partial P}\\right)_H$\r\n\r\n(considering small changes in pressure), which is commonly called the [i]Joule-Thomson coefficient[/i] and symbolized by $ \\mu _{JT}$. Now we are to apply the mathematical formalism of thermodynamics in order to relate this partial derivative with readily measured quantities. We have\r\n\r\n$ \\left(\\frac {\\partial T}{\\partial P}\\right)_H \\left(\\frac {\\partial P}{\\partial H}\\right)_T \\left(\\frac {\\partial H}{\\partial T}\\right)_P \\equal{} \\minus{} 1 \\equal{} > \\left(\\frac {\\partial T}{\\partial P}\\right)_H \\equal{} \\minus{} \\frac {1}{\\left(\\frac {\\partial P}{\\partial H}\\right)_T \\left(\\frac {\\partial H}{\\partial T}\\right)_P }$\r\n\r\n$ \\equal{} > \\left(\\frac {\\partial T}{\\partial P}\\right)_H \\equal{} \\minus{} \\frac {\\left(\\frac {\\partial H}{\\partial P}\\right)_T}{\\left(\\frac {\\partial H}{\\partial T}\\right)_P}$\r\n\r\nBut $ \\left(\\frac {\\partial H}{\\partial T}\\right)_P \\equal{} C_P$, and from the definition of enthalpy, differentiating, and using $ dU \\equal{} TdS \\minus{} PdV$ we have\r\n\r\n$ dH \\equal{} TdS \\plus{} VdP$\r\n\r\nNow considering a constant T process and dividing the last equation by $ dP$ we get\r\n\r\n$ \\left(\\frac {\\partial H}{\\partial P}\\right)_T \\equal{} T \\left(\\frac {\\partial S}{\\partial P}\\right)_T \\plus{} V$\r\n\r\nBut from $ dG \\equal{} \\minus{} SdT \\plus{} VdP$ we can write \r\n\r\n$ \\minus{} \\left(\\frac {\\partial S}{\\partial P}\\right)_T \\equal{} \\left(\\frac {\\partial V}{\\partial T}\\right)_P \\equal{} > \\left(\\frac {\\partial S}{\\partial P}\\right)_T \\equal{} \\minus{} \\alpha V$, since by definition $ \\alpha \\equal{} \\frac {1}{V} \\left(\\frac {\\partial V}{\\partial T}\\right)_P$.\r\n\r\nTherefore, \r\n\r\n$ \\left(\\frac {\\partial T}{\\partial P}\\right)_H \\equal{} \\minus{} \\frac { \\minus{} \\alpha VT \\plus{} V}{C_P} \\equal{} > \\left(\\frac {\\partial T}{\\partial P}\\right)_H \\equal{} \\frac {V(\\alpha T \\minus{} 1)}{C_P}$.\r\n\r\nIn the throttling experiment we always have $ dP < 0$. Also, the stability of thermodynamic systems requires that $ C_P > 0$. Therefore, from the last expression it is easy to see that $ \\alpha T > 1$ (where T is the initial temperature of the gas) implies $ dT < 0$, and the throttling process cools the gas. If $ \\alpha T < 1$ then $ dT > 0$ and the gas is heated. For $ \\alpha T \\equal{} 1$, the partial derivative vanishes, and an infinitesimal pressure change neither heats nor cools the gas. Since the condition $ \\alpha T \\equal{} 1$ defines the turn point from an heating process to a cooling process, we therefore define the [i]inversion temperature[/i] $ T_{inv}$ of a gas by\r\n\r\n$ T_{inv} \\equal{} \\frac {1}{\\alpha}$\r\n\r\nNote that the inversion temperature of a gas [i]is not[/i] a constant, since the coefficient of thermal expansion $ \\alpha$ is an intensive state function and is, therefore, a function of temperature and pressure. \r\nExperimentally, it is found that the T versus P curve for a series of throttling experiments with the same initial state (defined by its T and P) in common has the form shown in figure II: note that the extremum corresponding to the inversion condition $ \\mu _{JT} \\equal{} 0$ is a maximum. As we have shown above, to the right of point [i]a[/i] $ \\mu _{JT} < 0$ and a gas is heated in that region; to left of [i]a[/i], $ \\mu _{JT} > 0$ and a gas is cooled. Note also that the curve shown connects points with the same enthalpy, and is therefore an [i]isenthalpic[/i] curve. From the considerations given until now, one may conclude that a gas must be first pre-cooled below its inversion temperature (therefore assuring that $ \\mu _{JT} > 0$) in order for it to be cooled through a Joule-Thomson experiment. Otherwise, heating occur instead.\r\nFor most gases at atmospheric pressure the inversion temperature is quite higher than room temperature - for example, 621 K (348\u00baC) for $ \\ce{N2}$ and 764 K (491\u00baC) for $ \\ce{O2}$ -, and so these gases can be cooled at room temperature by isenthalpic expansion. On the other hand, at atmospheric pressure the gases $ \\ce{He}$ and $ \\ce{H2}$ have very low inversion temperatures (-222\u00baC for Helium), and so these two gases will warm by a throttling process at room temperature.", "Solution_26": "[size=150][b]Problem 43[/b][/size]. In the text above it was shown that the Joule-Thomson process occurs at constant enthalpy. However, this statement is not completely correct. Why? What better statement can we make?\r\n\r\n\r\n[size=150][b]Problem 44[/b][/size]. Is $ \\mu _{JT}$ a state function? If yes, then is it extensive or intensive?\r\n\r\n\r\n[size=150][b]Problem 45[/b][/size]. Compute $ \\mu _{JT}$ for an ideal gas, and comment your result.\r\n\r\n\r\n[size=150][b]Problem 46[/b][/size]. For air at temperatures near 25\u00baC and pressures in the range 0 to 50 bar, the $ \\mu _{JT}$ values are all reasonably close to 0.2\u00baC/bar. Estimate the final temperature of the gas if 58 g of air at 25\u00baC and 50 bar undergoes a Joule-Thomson throttling to a final pressure of 1 bar.\r\n\r\n\r\n[size=150][b]Problem 47[/b][/size]. Find an expression for the inversion temperature of real gases, assuming them to be described by the Van der Waals equation of state.\r\n\r\n\r\n[size=150][b]Problem 48[/b][/size]. A gas has the following equations of state, $ U \\equal{} PV$ and $ T \\equal{} 3B \\sqrt[3]{\\frac{U^2}{NV}}$, where B is a positive constant, and the system obeys the Nernst postulate. The gas, at an initial temperature $ T_1$ and initial pressure $ P_1$, goes through a Joule-Thomson experiment, with the final pressure being $ P_2$. Find the final temperature $ T_2$ of the gas.", "Solution_27": "[size=150][b]Problem 49[/b][/size]. For a particular gaseous system it has been determined that the energy is given by $ U \\equal{} \\frac{5}{2}PV \\plus{} B$, where B is a constant. Find the equation of the adiabats in the P-V plane for this gas.\r\n\r\n\r\n[size=150][b]Problem 50[/b][/size]. The energy of a particular system, of one mole, is given by $ U \\equal{} AP^2V$, where A is a positive constant. Find the equation of the adiabats in the P-V plane.", "Solution_28": "[size=150][b]Problem 51[/b][/size]. Let [i]n[/i] moles of a perfect gas undergo an adiabatic free expansion into a vacuum (the Joule experiment). Express $ \\Delta S$ in terms of the initial and final temperatures and volumes.\r\n\r\n\r\n[size=150][b]Problem 52[/b][/size]. Calculate $ \\Delta S$ for the conversion of 1 mole of supercooled liquid water at -10\u00baC and 1 atm to 1 mole of ice at -10\u00baC and 1 atm.", "Solution_29": "[size=150][b]Problem 53[/b][/size]. The specific heat capacity $ c_P$ of water is nearly constant in the temperature range 25\u00baC to 75\u00baC at 1 atm, having the value of 1.00 cal/g/\u00baC.\r\n\r\n(a) Find $ \\Delta S$ when 100 g of water are reversibly heated from 25\u00baC to 50\u00baC at 1 atm.\r\n\r\n(b) Without doing the calculation, state whether $ \\Delta S$ for heating 100 g of water from 50\u00baC to 75\u00baC at 1 atm will be greater than, equal to, or less than $ \\Delta S$ for the previous 25\u00baC to 50\u00baC heating. Justify your answer.\r\n\r\n(c) Find $ \\Delta S$ when 100 g of water are reversibly heated from 50\u00baC to 75\u00baC at 1 atm. Does your result confirm the answer given to part (b)?" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Yes I'm an idiot...I would know this stuff, but I haven't been able to attend class lately.\r\n\r\nHelp and/or answers would be GREEEEEEEEEEATLY appreciated w/ factoring the following problems:\r\n\r\n1.) X^2-X-12 (X^2 means squared just in case nobody knew that)\r\n\r\n2.) X^2-4X-12\r\n\r\n3.) X^2-X-6\r\n\r\n4.) X^2-5X+6\r\n\r\n5.) X^2-7X+12\r\n\r\nAgain Thanks a lot if you can help me.", "Solution_1": "1:\r\n$x^2-x-12=(x-4)(x+3)$\r\n\r\n2:\r\n$x^2-4x-12=(x-6)(x+2)$\r\n\r\n3:\r\n$x^2-x-6=(x-3)(x+2)$\r\n\r\n4:\r\n$x^2-5x+6=(x-2)(x-3)$\r\n\r\n5:\r\n$x^2-7x+12=(x-3)(x-4)$\r\n\r\n6:\r\nThis is not suitable for Algebra Unsolved Problems\r\n\r\n7:\r\nThis thread should be posted at the Getting Started forum.", "Solution_2": "My bad...Thanks for your help though.\r\n\r\nI'm gonna go post it over there but is there any way you can show me how you arrived at your conclusions? Thanks again.", "Solution_3": "$x^2-(p+q)x+pq$\r\n$=x^2-px-qx+pq$\r\n$=x(x-p)-q(x-p)$\r\n$=(x-p)(x-q)$", "Solution_4": "or u can do it with the hard way and efficient way :? \r\n :ninja: Discriminant :ninja:" } { "Tag": [ "integration", "LaTeX" ], "Problem": "[b][u]Question #1[/u][/b]\r\nHow can i make an expression appear in the middle of the screen ?\r\nAlso is there a way to make it appear anywhere we want ??\r\n\r\n$\\int f(x) dx$\r\n\r\n[b][u]Question #2[/u][/b]\r\nWhy there is a space in the first line of the post below and is there is a way to avoid it ???\r\n\r\n$(a+b)^2 =(a+b)(a+b) = \\\\ a(a+b)+b(a+b)=a^2+ab+ba+b^2 =\\\\ a^2+2ab+b^2$ .\r\n\r\n[b][u]Question #3[/u][/b]\r\n\r\nFor the expression above : Why the [b]dot[/b] at the end does not appear correctly ?\r\nI dont use correctly this -> \\\\ ?\r\n\r\n[b][u]Question #4[/u][/b]\r\n\r\nWhen i write a lengthy expression i have to change lines because it appears \r\nsomething like this... \r\n\r\n$\\frac{a}{b+2c+d}+\\frac{b}{c+2d+a}+\\frac{c}{d+2a+b}+\\frac{d}{a+2b+c}= \\frac{a}{S-(a-c)}+\\frac{b}{S-(b-d)}+\\frac{c}{S+(a-c)}+\\frac{d}{S+(b-d)}$\r\n\r\n\r\nThat is for now :D . Thanks a lot :)", "Solution_1": "Q1: [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#spacing]Try here[/url] You type \\begin{center} [i] text [/i] \\end{center}\r\nQ2: [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#eqnarray]This is really useful too[/url] Use the \\begin{eqnarray*} tool" } { "Tag": [ "function", "limit", "integration", "logarithms", "calculus", "inequalities", "real analysis" ], "Problem": "Given that $f:[0,1]\\rightarrow\\mathbb{R}$ is a continuous function, compute the following limit:\r\n\r\n\\[ \\lim_{n\\rightarrow\\infty} \\int_0^1 \\frac{x^{\\frac{1}{n}}-1}{\\frac{1}{n}} \\cdot f(x)\\ dx \\]", "Solution_1": "The pointwise limit of $\\frac{x^{\\frac1n}-1}{\\frac1n} = \\ln x$. Furthermore (as can be shown by using Taylor's Theorem on $x^{\\frac1n}=e^{\\frac{\\ln x}n}$), each term of this sequence is bounded by $|\\ln x|$. Since $f$ is continuous on [0, 1], there is some $M$ such that $|f(x)|\\le M$, so the integrands are dominated by the integrable function $M|\\ln x|$. By the Dominated Convergence Theorem, the limit of the integral is the integral of the limit, namely $\\int_0^1 f(x)\\ln x dx$. Of course, this could have been done by more elementary means without quoting the DCT, but the result is the same.", "Solution_2": "A somewhat related question, but with a different answer. Again, assume that $f$ is continuous on [0, 1]. Compute\r\n$\\lim_{n\\to\\infty} n\\int_0^1 f(x)(1-x)^{n-1}dx$.", "Solution_3": "I'm sorry, can you post a more rigorous and elementary proof? This was intented for students aged 18-19 who do not posess superior knowledge in the integration theory area. Thank you very much!", "Solution_4": "OK, I'll give it a try. Interchanging limits of sequences of functions with integrals isn't normally dealt with in the first calculus sequence.\r\n\r\nThe key inequality is that for $y\\le 0, 1+y\\le e^y\\le 1+y+\\frac{y^2}2$.\r\nSo, $1+\\frac{\\ln x}n\\le e^{\\ln x/n}\\le 1 + \\frac{\\ln x}n+\\frac{(\\ln x)^2}{2n^2}$\r\n$\\frac{\\ln x}n\\le x^{1/n}-1\\le \\frac{\\ln x}n+\\frac{(\\ln x)^2}{2n^2}$\r\n$\\ln x\\le \\frac{x^{1/n}-1}{1/n}\\le \\ln x + \\frac{(\\ln x)^2}{2n}$\r\n$0\\le \\frac{x^{1/n}-1}{1/n}-\\ln x\\le \\frac{(\\ln x)^2}{2n}$\r\n$\\left|\\frac{x^{1/n}-1}{1/n}-\\ln x\\right|\\le\\frac{(\\ln x)^2}{2n}$\r\nSince $f$ is continuous, there exists $M$ such that $|f(x)|\\le M$.\r\nThus, $\\left|\\int_0^1\\frac{x^{1/n}-1}{1/n}\\cdot f(x)dx - \\int_0^1\\ln x\\cdot f(x)dx\\right|\\le\\int_0^1\\left|\\frac{x^{1/n}-1}{1/n}-\\ln x\\right||f(x|dx$\r\n$\\le \\int_0^1 \\frac{M(\\ln x)^2}{2n}dx = \\frac{M}n \\to 0$ as $n\\to\\infty$.", "Solution_5": "[quote=\"Kent Merryfield\"]A somewhat related question, but with a different answer. Again, assume that $f$ is continuous on [0, 1]. Compute\n$\\lim_{n\\to\\infty} n\\int_0^1 f(x)(1-x)^{n-1}dx$.[/quote]\r\n\r\nAs for this one, I'll assume $f(x)\\ge 0$, because otherwise we should employ some boring $|\\cdot|$ :). Moreover, we can assume $f(x)=0$, because, since $\\int^1_0n(1-x)^{n-1}dx=1$, we can replace $f$ with $g(x)=f(x)-f(0)$. Now we write the integral as $\\int^{a_k}_0+\\int^1_{a_k}$. The second one is $\\le M(1-a_k)^n$, while the first one is $\\le M_k(1-(1-a_k)^n)$, where $M,M_k$ are bounds for $f$ on [$0,1$] and [$0,a_k$] respectively. When $n\\to \\infty$, the second integral goes to $0$, while the first integral becomes $\\le M_k+\\varepsilon$ with $\\varepsilon$ as small as we want. If we make $a_k\\to 0$, using $f(0)=0$, we get the result $n\\int^1_0f(x)(1-x)^{n-1}\\to 0$. \r\n\r\nFor general $f$ the limit is thus $f(0)$.", "Solution_6": "That's it, grobber, clean and simple enough. An old Fourier analyst like me can't resist a good \"approximation to the identity\"." } { "Tag": [ "function", "calculus", "derivative", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Suppose that $f(x)=\\sum_{k\\geq 1}{a_k e^{ikx}}$ is a continuous function (attention, the convergence is only simple). Then the $a_k$ are its Fourier coefficients. I have a very beautiful proof for this one, but I'd like to see many approaches.", "Solution_1": "\"simple\" meaning \"pointwise\", i.e., the assumption being that the series converges to $f(x)$ for every $x$? :?", "Solution_2": "Yes, sorry, I was too influenced by the french vocabulary. :blush:", "Solution_3": "please can you give a hint to your proof ...it's very strange , my teacher had never talk about this result even if it's very interesting", "Solution_4": "the trick is to introduce $F(x) = \\sum_{k \\geq 1}\\frac{a_{k}}{k^{2}}e^{ikx}$, hopping that it will looks like a function such that $F''=-f$. This time everything converges normally because the $(a_{k})$ are bounded, so one knows the fourier coefficient of $F$, and when one computes $DF(x)$, the pseudo second derivative, one can see that $DF(x)=-f(x)$, where $Dg(x) = \\lim_{h \\to 0}\\frac{g(x+h)+g(x-h)-2g(x)}{h^{2}}$. So if $G \\in C^{2}(\\mathbb{R})$ satisfies $G''(x)=f(x)$, one can see that $D(F+G)=0$, which means that $F+G$ is an affine function. From here, the conclusion follows." } { "Tag": [ "Gauss", "limit", "trigonometry", "floor function", "real analysis", "real analysis solved" ], "Problem": "Congratulations on Mathlink's recovery! :) :) \r\n\r\nI am very glad to see this forum again.Thank you for your taking troubles to whom it may concern.\r\n\r\nEvaluate \\[\\lim_{n\\to\\infty}\\sin \\left( 2\\pi\\sqrt{n^2+\\left[\\frac{n}{3}\\right]}\\right )\\]", "Solution_1": "$\\lfloor\\frac{n}3\\rfloor=\\frac{n}3+O(1)$\r\n\r\n$1+\\frac1{n^2}\\lfloor\\frac{n}3\\rfloor= 1+\\frac1{3n}+O\\left(\\frac1{n^2}\\right)$\r\n\r\n$\\sqrt{n^2+\\lfloor\\frac{n}3\\rfloor}= n\\sqrt{1+\\frac1{n^2}\\lfloor\\frac{n}3\\rfloor}= n\\left(1+\\frac1{6n}+O\\left(\\frac1{n^2}\\right)\\right)$\r\n\r\n$2\\pi\\sqrt{n^2+\\lfloor\\frac{n}3\\rfloor}= 2\\pi n+\\frac{2\\pi}6+O\\left(\\frac1n\\right)$\r\n\r\nSo the limit is $\\sin\\left(\\frac{\\pi}3\\right)=\\frac{\\sqrt{3}}2.$\r\n\r\nNewton's binomial series is a powerful thing.", "Solution_2": "[quote=\"kunny\"]Congratulations on Mathlink's recovery! :) :) \n\nI am very glad to see this forum.Thank you for your taking troubles to whom it may concern.\n[/quote] Altough it's a little bit off-topic here, you're welcome. ;)", "Solution_3": "@kunny. For whom this problem was proposed? I mean what level of students was being assumed?", "Solution_4": "Here is my solution.\r\n\r\n$\\frac{n}{3}-1<[\\frac{n}{3}]\\leqq\\frac{n}{3}$ ,so we have $\\frac{1}{3}-\\frac{1}{n}<\\frac{1}{n}[\\frac{n}{3}]\\leqq\\frac{1}{3}$ by Archimedes'axiom we have $\\lim_{n\\to\\infty}\\frac{1}{n}[\\frac{n}{3}]=\\frac{1}{3}$.\r\n Analogously $\\lim_{n\\to\\infty}\\frac{1}{n^2}[\\frac{n}{3}]=0$\r\n\r\nThen $\\lim_{n\\to\\infty}(\\sqrt{n^2+[\\frac{n}{3}]}-n)=\\lim_{n\\to\\infty}\\frac{[\\frac{n}{3}]}{\\sqrt{n^2+[\\frac{n}{3}}]+n}=\\lim_{n\\to\\infty}\\frac{\\frac{1}{n}[\\frac{n}{3}]}{\\sqrt{1+\\frac{1}{n^2}[\\frac{n}{3}]}+1}=\\frac{1}{6}$\r\n\r\nThus $\\lim_{n\\to\\infty}\\sin (2\\pi\\sqrt{n^2+[\\frac{n}{3}]})=\\lim_{n\\to\\infty}\\sin (2\\pi(\\sqrt{n^2+[\\frac{n}{3}]}-n))$.\r\n\r\nBy The continuity of $\\sin x$, the desired limit is $\\sin \\frac{2\\pi}{6}=\\frac{\\sqrt{3}}{2}$" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "The convex hexagon $ABCDEF$ satisfies $\\angle A + \\angle C + \\angle E = 360^{\\cdot}$ and $AB \\cdot CD \\cdot EF = BC \\cdot DE \\cdot FA$. Show that $AB \\cdot FD \\cdot EC = BF \\cdot DE \\cdot CA$.", "Solution_1": "At least posted before:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=48100\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=1117\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=19479\r\n\r\n darij" } { "Tag": [ "graph theory", "combinatorics unsolved", "combinatorics" ], "Problem": "For what positive integers $ n$ is it possible to divide the complete graph on $ n$ points into triangles so that each edge is used exactly once?", "Solution_1": "See http://mathworld.wolfram.com/SteinerTripleSystem.html" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c\\geq\\ 0$ such that $ a \\plus{} b \\plus{} c \\equal{} 3$.Find $ max$\r\n[size=150]$ D \\equal{} (a^\\frac {7}{5} b \\plus{} a^\\frac {3}{2} b \\plus{} 1)( b^\\frac {7}{5} c \\plus{} b^\\frac {3}{2} c \\plus{} 1)(c^\\frac {7}{5} a \\plus{} c^\\frac {3}{2} a \\plus{} 1)$[/size]\r\nIt is my sorry to can_hang2007", "Solution_1": "The maximum is $ 27$, use the following\r\n\\[ a^{3/2}b \\plus{}b^{3/2}c\\plus{}c^{3/2}a \\le 3\\]\r\n:)", "Solution_2": "[quote=\"can_hang2007\"]The maximum is $ 27$, use the following\n\\[ a^{3/2}b \\plus{} b^{3/2}c \\plus{} c^{3/2}a \\leq 3\n\\]\n:)[/quote]\r\nit is not enough,how about $ a^{7/5}b \\plus{} b^{7/5}c \\plus{} c^{7/5}a$,can you detail it,i think it is not easy", "Solution_3": "[quote=\"Allnames\"][quote=\"can_hang2007\"]The maximum is $ 27$, use the following\n\\[ a^{3/2}b \\plus{} b^{3/2}c \\plus{} c^{3/2}a \\leq 3\n\\]\n:)[/quote]\nit is not enough,how about $ a^{7/5}b \\plus{} b^{7/5}c \\plus{} c^{7/5}a$,can you detail it,i think it is not easy[/quote]\r\nWe also have the inequality\r\n\\[ a^{7/5}b\\plus{}b^{7/5}c\\plus{}c^{7/5}a \\le 3\\]\r\nand it follows from the inequality\r\n\\[ a^{3/2}b\\plus{}b^{3/2}c\\plus{}c^{3/2}a \\le 3\\]\r\n;)", "Solution_4": "how about 2 ineq,who can post the proof" } { "Tag": [ "category theory", "abstract algebra", "invariant", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello\r\n\r\nConsider R-modules\r\n\r\nlet F be a covariant functor to the abelian group category such that each short exact sequence\r\n\r\n\r\n$0\\rightarrow A\\buildrel{f}\\over{ \\rightarrow}B\\buildrel{g}\\over{\\rightarrow}C\\rightarrow 0$\r\n\r\nis transformed into a new short exact sequence\r\n\r\n\r\n$0\\rightarrow FA\\buildrel{F(f)}\\over{\\rightarrow}FB\\buildrel{F(g)}\\over{\\rightarrow}FC\\rightarrow 0$\r\n\r\nThat is our definition of exact functor.\r\n\r\nProve now that each long exact sequence is also transformed into a long exact sequence.\r\nIt's not as easy as it looks! I got stuck... :(", "Solution_1": "It's obvious after showing the following important equivalence:\r\n\r\n$F$ is exact $\\Longleftrightarrow$ for all exact sequences $\\bullet \\buildrel{f}\\over{\\rightarrow} \\bullet \\buildrel{g}\\over{\\rightarrow} \\bullet$ the sequence $\\bullet \\buildrel{F(f)}\\over{\\rightarrow} \\bullet\\buildrel{F(g)}\\over{\\rightarrow} \\bullet$ is also exact.\r\n \r\nwhere $F$ is an additive functors between abelian categories. The important stuff are the arrows, so I don't denote the objects.\r\n \r\n$\\Longleftarrow$ :\r\n\r\nLet $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{f}\\over{ \\rightarrow} \\bullet \\buildrel{g}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$ be a short exact sequence. Then $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{f}\\over{ \\rightarrow} \\bullet$ is exact, hence $F(0) \\buildrel{0} \\over{ \\rightarrow} \\bullet \\buildrel{F(f)}\\over{ \\rightarrow} \\bullet$ by assumption. It follows $0 = im(0) = ker F(f)$, so that $F(f)$ is a monomorphism. Dually, we see that $F(g)$ is an epimorphism. So, $0 \\buildrel{0} \\over{ \\rightarrow} \\bullet \\buildrel{F(f)}\\over{ \\rightarrow} \\bullet$ and $\\bullet \\buildrel{F(g)}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$ are exact. Besides, the exactness of $\\bullet \\buildrel{f}\\over{ \\rightarrow} \\bullet \\buildrel{g}\\over{ \\rightarrow} \\bullet$ implies the exactness of $\\bullet \\buildrel{F(f)}\\over{ \\rightarrow} \\bullet \\buildrel{F(g)}\\over{ \\rightarrow} \\bullet$. Composing gives, that $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{F(f)}\\over{ \\rightarrow} \\bullet \\buildrel{F(g)}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$ is exact.\r\n \r\n$\\Longrightarrow$:\r\n \r\nLet $f$ be an arbitrary epimorphism. Then $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{ker(f)}\\over{ \\rightarrow} \\bullet \\buildrel{f}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$ is exact, hence $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{F(ker f)}\\over{ \\rightarrow} \\bullet \\buildrel{F(f)}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$. Therefore, $F(f)$ is an epimorphism. Thus $F$ preserves epimorphisms, and dually we derive that monomorphisms are preserved.\r\n\r\nNot, let $f$ be an arbitrary morphism. It can be uniquely (up to isomorphism) factorized by an epimorphism $coim(f)$ and a monomorphism $im(f)$. Then $F(f) = F(im(f) \\circ coim(f))=F(im(f)) \\circ F(coim(f))$. We've seen that $F(im(f))$ is a monomorphism and that $F(coim(f))$ is an epimorphism. So, $F(im(f))=im(F(f))$ and $F(coim(f))=coim(F(f))$. Besides, the exactness of $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{ker(f)}\\over{ \\rightarrow} \\bullet \\buildrel{coim(f)}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$ yields the exactness of $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{F(ker(f))}\\over{ \\rightarrow} \\bullet \\buildrel{F(coim(f))}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$. Thus, $F(Ker(f))=ker(F(coim(f))=ker(coim(F(f))=ker(F(f))$. Dually it follows $F(coker(f))=coker(F(f))$.\r\n\r\nNow, we have established enough invariants to proof the assertion:\r\n\r\nLet $\\bullet \\buildrel{f}\\over{ \\rightarrow} \\bullet \\buildrel{g}\\over{ \\rightarrow} \\bullet$ be exact. This means, that $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{ker(g)}\\over{ \\rightarrow} \\bullet \\buildrel{coker(f)}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$ is exact, hence also $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{F(ker(g))}\\over{ \\rightarrow} \\bullet \\buildrel{F(coker(f))}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$ by assumption. This sequence is isomorphic to $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{ker(F(g))}\\over{ \\rightarrow} \\bullet \\buildrel{coker(F(f))}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$. So it is also exact, hence also $\\bullet \\buildrel{F(f)}\\over{ \\rightarrow} \\bullet \\buildrel{F(g)}\\over{ \\rightarrow} \\bullet$.", "Solution_2": "Danke, du bist wirklich sehr sehr gut! :D \r\nAber :\r\n\r\n\r\nI have some questions. First of all, I 'should' be able to deduce that the decomposition in an abelian category of a random morphism into an epic and then a monic morphism is 'essentially unique', right? I had a go at it, and I needed to consider a lot of things, is it possible you explicitly need for instance the assumption that every monic morphism is a kernel of its own cokernel (as far as I know that is an explicit axiom of abelian category)\r\n\r\nYou said : this sequence is isomorphic to, what exactly do you mean by isomorphic sequence?\r\n\r\nThen finally, what I initially asked for was : prove that a covariant exact functor from $R-mod$ to $Ab$ maps lojng exact sequence to long exact sequence.\r\nNow I think I could at least change $Ab$ into $S-mod$\r\n\r\nYou however, worked with abelian categories (which is nicer as it is more general) but you assumed additivity of your functor F. That is 'bad' for me as it in conflict with my assumptions in the original problem, but .. where do you use it in your proof?", "Solution_3": "[quote=\"fredbel6\"]Danke, du bist wirklich sehr sehr gut! :D [/quote]\nUnd du kannst sehr sehr gut Deutsch, lass uns doch so weiterschreiben? ;-)\n\n[quote=\"fredbel6\"]is it possible you explicitly need for instance the assumption that every monic morphism is a kernel of its own cokernel [/quote]\nYes - this is quite important.\n\n[quote=\"fredbel6\"]You said : this sequence is isomorphic to, what exactly do you mean by isomorphic sequence?[/quote]\nSequences of the same \"type\" are isomorphic, when there are isomorphisms between the occuring objects (of the same position) so that the resulting diagram commutes. This should remind you of the category of chain complexes ;)\n\n[quote=\"fredbel6\"]Now I think I could at least change $Ab$ into $S-mod$[/quote]\nYes, of course. As I already said, the whole think works for arbitrary abelian categories.\n\n[quote=\"fredbel6\"]... you assumed additivity of your functor F ... where do you use it in your proof?[/quote]\r\nGood question. :-) At the beginning, I use that the null morphism is mapped to the null morphism. I think that's it.", "Solution_4": "Nein, es tut mir leid aber ich glaube nicht das wir k\u00f6nnen weiterschreiben in Deutsch :oops: . Even if I would call it a language that I speak, it would be very low on the list, number four actually, and I cannot say difficult sentences in German, like this one!\r\n\r\nBut I must admit that is wrong ;) the functor had to be additive and covariant and from $R-mod$ to $Ab$\r\nYou used only abelian categories and additive covariant functor\r\nSo I guess nothing is lost!", "Solution_5": "Actually, I am quite ashamed that I only think of it now, but I have been working in$R-mod$ for so long, how exactly do you define exact sequence in a general abelian category.\r\n\r\nMy guess would be : \r\n\r\n\r\n$\\bullet \\buildrel{f}\\over{ \\rightarrow} \\bullet \\buildrel{g}\\over{ \\rightarrow} \\bullet$ is said to be exact if :\r\n\r\n1. $g\\circ f=0$\r\n\r\n2. $coker(f)\\circ ker(g)=0$\r\n\r\nOr am I leaving out a crucial assumption?\r\n\r\nThe confusing this is that you use exactness of the last sequence I wrote to prove exactness of $\\bullet \\buildrel{f}\\over{ \\rightarrow} \\bullet \\buildrel{g}\\over{ \\rightarrow} \\bullet$, while I use that sequence to define exactness! :ninja:", "Solution_6": "The definition of exactness is correct. But I don't understand your confusion ...", "Solution_7": "This is all so nice and interesting but stuff like that should be included in my syllabus in my opinion then :| \r\n\r\n[quote]Let $\\bullet \\buildrel{f}\\over{ \\rightarrow} \\bullet \\buildrel{g}\\over{ \\rightarrow} \\bullet$ be exact. This means, that $0 \\buildrel{0}\\over{ \\rightarrow} \\bullet \\buildrel{ker(g)}\\over{ \\rightarrow} \\bullet \\buildrel{coker(f)}\\over{ \\rightarrow} \\bullet \\buildrel{0}\\over{ \\rightarrow} 0$ is exact[/quote]\r\n\r\nSo how would you go about proving this then?\r\nThis is confusing at it involves kernels of cokernels", "Solution_8": "The definition of exactness can be replaced by im f = ker g, and he have Im ker g = ker coker ker g = ker g, ker coker f = im f ...." } { "Tag": [ "Stanford", "college", "Harvard", "MIT" ], "Problem": "For getting into a top tier college (Stanford, Harvard, MIT, Princeton...), how important is it to be well-rounded. I'm a high school sophomore right now, and I swim, but that takes up a lot of time every day and my junior year is going to be packed. Is it worth it to keep swimming through high school? Or should I devoted more time to studying?", "Solution_1": "I'd say you shouldn't base your decision about swimming just on what would look good on your college applications. Do you like swimming? If not, do something else that you like more. That will probably make you look well-rounded still, and you'll also have more fun. If you do like swimming, then keep doing it! It's not hurting you.", "Solution_2": "In high school I was under a lot of pressure to do a sport, because if I didn't, I wouldn't be \"well-rounded\" and no top-tier college would want me. I resisted, and ultimately proved the prevailing wisdom wrong. So, I would take any advice along the lines of \"you must do X in order to be a well-rounded person and get into a good school\" with a grain of salt. \r\n\r\nIf you truly enjoy swimming, by all means, do it! If you don't, find something else you like. Don't worry about what other people think. Feeling empowered to do activities that you enjoy, and that are meaningful to you-- to [b]you[/b], not to parents, teachers, or colleges-- is absolutely priceless, and will serve you well for the rest of your life." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "I know it's well-known but I couldn't find a proof:\r\n\r\nLet $ (a_n)_{n\\ge 1}$ be a sequence of real numbers with $ \\sum \\frac{a_n}{n}$ convergent. Prove that $ \\lim_{n\\to \\infty}\\frac{a_1\\plus{}a_2\\plus{}\\dots \\plus{}a_n}{n}\\equal{} 0$", "Solution_1": "If $ b_n \\geq 0$ and $ \\sum b_n$ converges, then:\r\n\r\n1) $ \\lim b_n \\equal{} 0$;\r\n\r\n2) $ \\lim n b_n \\equal{} 0$.\r\n\r\nIn your case, we can conclude that $ \\lim n \\cdot \\frac {a_n}{n} \\equal{} \\lim a_n \\equal{} 0$. Now we need a theorem about sequences:\r\n\r\n(3) If $ \\lim a_n$ exists then $ \\lim \\frac {a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n}{n}$ also exists and we have\r\n\r\n$ \\lim \\frac {a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n}{n} \\equal{} \\lim a_n$.\r\n\r\nTherefore, regarding your problem, we can then conclude that $ \\lim \\frac {a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n}{n} \\equal{} 0$. I will leave the proofs of (1), (2), and (3) as exercises for you.", "Solution_2": "Don't underestimate this problem.\r\n\r\n[quote=\"Carcul\"]\n\n2) $ \\lim n b_n \\equal{} 0$.\n\nIn your case, we can conclude that $ \\lim n \\cdot \\frac {a_n}{n} \\equal{} \\lim a_n \\equal{} 0$. \n[/quote]\r\n\r\nNot true. Consider $ a_n\\equal{}(\\minus{}1)^n$ so $ b_n\\equal{}\\frac{(\\minus{}1)^n}{n}$ and the series $ \\sum b_n$ converges conditionally. So it can not be solved with a simple application of the Cezaro-Stolz lemma.", "Solution_3": "You are right: I was considering sequences of non-negative real numbers in my post.", "Solution_4": "Let $ b_n \\equal{} a_n / n$ so that $ a_n \\equal{} n b_n$.\r\nThen $ \\sum b_n$ converges.\r\nLet $ f(n)\\equal{}\\frac{b_1\\plus{}2b_2\\plus{}\\ldots\\plus{}nb_n}{n}$\r\nWe want to show that $ \\lim_{n\\to\\infty}f(n)\\equal{}0$\r\nLet $ s_n\\equal{}b_1\\plus{}\\ldots\\plus{}b_n$.\r\nThen $ f(n)\\equal{}s_n\\minus{}\\frac{s_1\\plus{}s_2\\plus{}\\ldots\\plus{}s_{n\\minus{}1}}{n}$ (just consider the number of times each $ b_i$ appears).\r\n\r\nBut $ \\lim_{n\\to\\infty}s_n$ does exist (since the sum converges), so we can use Stolz-Cesaro on $ \\frac{s_1\\plus{}s_2\\plus{}\\ldots\\plus{}s_{n\\minus{}1}}{n}$, and the desired conclusion follows.\r\n\r\nI think this works." } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "$P(x)\\in \\mathbb{Z}[x],P(x)=a_{0}+\\dots+a_{p-1}x^{p-1}$ and if we know that for prime $p>2$ following holds: if $p$ does not divide $a-b$ then $p$ does not divide $P(a)-P(b)$.prove that $p|a_{p-1}$", "Solution_1": "In $\\mathbb F_{p}$ (the field with $p$ elements), we have relations of the form $P(x_{i})=y_{i},\\ i=\\overline{0,p-1}$, where both $\\{x_{i}\\}$ and $\\{y_{i}\\}$ are permutations of $\\overline{0,p-1}$. \r\n\r\nSince our polynomial has degree $\\le p-1$, we can find it in terms of $a_{i}$ and $b_{i}$ through Lagrange interpolation, and writing down the expression for the interpolation polynomial shows the result (the main coefficient $a_{p-1}$ will turn out to be $0+1+\\ldots+p-1=0$)." } { "Tag": [ "function", "geometry", "search", "Support", "vector" ], "Problem": "[size=167][b][color=red]Opera[/color] for Windows [ Win 95/98/ME/NT/2000/XP ][/b][/size]\r\n[b]Surf the Internet in a safer, faster, and easier way[/b]\r\n\r\nVisit [url=http://www.opera.com]Opera web site[/url].\r\n\r\nThe most full-featured Internet power tool on the market, Opera includes pop-up blocking, tabbed browsing, integrated searches, and advanced functions like Opera's groundbreaking E-mail program, RSS Newsfeeds and IRC chat. And because we know that our users have different needs, you can customize the look and content of your Opera browser with a few clicks of the mouse.\r\n\r\n[size=125]Key features in Opera 8[/size]\r\n\r\n:arrow: [b]Fresh new look[/b]\r\nWe've cleaned up our front yard. The Opera 8 interface is designed to make the advanced functions easy and effective to use. Menus, toolbars and other elements have undergone our \"slim and clean\"-routine. The licensed version has the largest browsing area in the industry. \r\n\r\n:arrow: [b]Security[/b]\r\nStay away from spy-ware, viruses, and other malicious applications that silently attack your computer while you are surfing the Web. For example, Opera displays security information inside the address bar, located next to the padlock icon that indicates the level of security present on a site. Opera 8 also provides protection against phishing attacks and automatically checks for security updates.\r\n\r\n:arrow: [b]Speed[/b]\r\nWe're holding on to our claim: \"The Fastest Browser on Earth\". In Opera 8, further improvements have been made in the way the browser reads pages to allow you to fly the Web. \r\n\r\nFor further information on just how good our rewritten scripting engine is, go to [url=http://www.howtocreate.co.uk/browserSpeed.html](Browser speed comparison)[/url]\r\n\r\n:arrow: [b]Tabbed browsing and pop-up blocking[/b]\r\nOpera pioneered tabs and pop-up blocking ages ago. But they're still key features that more and more people start enjoying. With tabs you can open multiple pages within the same application window. Opera lets you control whether to block all pop-ups, or open only the ones that you have requested. \r\n\r\n:arrow: [b]Integrated search[/b]\r\nSearch your favorite sites, for example Google, eBay, or Amazon without having to go to their Web pages. Use the integrated search window or search directly in the address field using shortcuts (e.g. writing \"g Opera\" in the address field will search for \"Opera\" using Google). \r\n\r\n:arrow: [b]Customization[/b]\r\nUsing the 'appearance' dialog you can make Opera look almost any way you want. Move buttons and search fields, add and remove toolbars, and so forth. Opera skins can change the look of your browser by giving it the icons and buttons of your choice. \r\n\r\n:arrow: [b]Password manager[/b]\r\nOpera's password manager, the Wand, remembers your usernames and passwords so you will not have to. The password manager can store more than one username and password combination for each page, and will let you choose which one to use if multiple combinations are stored. \r\n\r\n:arrow: [b]Fit to window width[/b]\r\nOpera's \"fit to window width\"-feature means no more horizontal scrolling and improved ability to print entire Web pages. In combination with Opera's Zoom function, you can magnify Web pages dramatically and still view them without having to scroll sideways. \r\n\r\n:arrow: [b]Voice[/b]\r\nOpera is the first browser to prepare for a future of Web sites offering interactive, voice-enabled shopping and booking systems. You can browse the Web using spoken commands, such as \"Opera next link\", \"Opera back\", or \"Opera speak\". The latter command will make Opera read Web page content and e-mail messages to you aloud ,adding usability as either a screen reader or advanced dictionary. Voice is currently offered in English and works on Windows 2000 and XP. A headset with a microphone is required to use Voice. \r\n\r\n:arrow: [b]Additional features[/b]\r\n :!: Full support for Gmail\r\nSupport for Atom newsfeeds\r\nFirst Web browser to natively support Scalable Vector Graphics (SVG)", "Solution_1": "I have used Opera for ages and I recommend it as the best ever tool for intenet surfing.", "Solution_2": "This is interesting, the site gives me permanent refused connection :what?:", "Solution_3": "Sorry for that awful typo.", "Solution_4": "Looks a lot nicer, and seems to be a bit faster. However I still don't like the ads too much :(", "Solution_5": "Don't listen to him!\r\n\r\nOpera Software has special [b]FREE OF CHARGE[/b] offers for educational institutions.", "Solution_6": "Is there any way to configure the fonts in Opera? The posts on AoPS are not displaying properly ;)", "Solution_7": "All questions to administrators.", "Solution_8": "[quote=\"fanaticsm\"]Is there any way to configure the fonts in Opera? The posts on AoPS are not displaying properly ;)[/quote]As in any browser there must be ways to configure font sizes. Also there is a small switch in the upper right corner on the index page (if you are using the AoPS skin) which you might try tweaking at the same time with the browser font settings.\r\n\r\nMikhail, but how to obtain the add-free version?! :?", "Solution_9": "The ad-free version can be obtained by purchasing the browser. More information can be found at [url=https://secure.bmtmicro.com/servlets/Orders.ShoppingCart?CID=310&PRODUCTID=3100024&AID=812930]this[/url] website.\r\n\r\nI believe the cost is 39 USD..", "Solution_10": "Is there possible a way to obtain the ad-free version free and [i]legally[/i]?", "Solution_11": "[quote=\"Myself\"]Is there possible a way to obtain the ad-free version free and [i]legally[/i]?[/quote]\r\nI doubt that you can, but you can always use Mozilla Firefox, which is free and also very good. :D", "Solution_12": "[quote=\"fanaticsm\"]The ad-free version can be obtained by purchasing the browser. More information can be found at [url=https://secure.bmtmicro.com/servlets/Orders.ShoppingCart?CID=310&PRODUCTID=3100024&AID=812930]this[/url] website.\nI believe the cost is 39 USD..[/quote]Please, don't give a partial information.\n\nStudent license costs 20 USD.\n\nAlso visit [b]Opera Higher Education Program[/b] at http://www.opera.com/education/\n\n[quote]Opera offers free site licenses to higher education institutions.[/quote]\n[quote=\"stupidkid\"]...but you can always use Mozilla Firefox, which is free and also very good.[/quote] This opinion comes only from lack of knowledges. Opera's features cover everything what you have with IE or Firefox, moreover it gives many unique great things, which help with internet surfing.\r\nOpera contains built-in mail client M2, but I don't use it (here I think that Thunderbird is very good).", "Solution_13": "[quote=\"stupidkid\"][quote=\"Myself\"]Is there possible a way to obtain the ad-free version free and [i]legally[/i]?[/quote]\nI doubt that you can, but you can always use Mozilla Firefox, which is free and also very good. :D[/quote]Not very good when it comes to java unfortunately :(", "Solution_14": "[quote=\"Valentin Vornicu\"][quote=\"stupidkid\"][quote=\"Myself\"]Is there possible a way to obtain the ad-free version free and [i]legally[/i]?[/quote]\nI doubt that you can, but you can always use Mozilla Firefox, which is free and also very good. :D[/quote]Not very good when it comes to java unfortunately :([/quote]\r\nReally? I don't know, I use it all the time, both in Linux and Windows and usually Java is not a problem. Sometimes Flash is.", "Solution_15": "My Firefox (1.0.3) crashes about 2-3 times a day, especially when loading pages with JavaScript. Opera, IE and Netscape don't have such problems. \r\n\r\nIf only IE had Tabs :(", "Solution_16": "Well, I don't know, there are things that I don't like about the Explorer, but mainly is because I like Linux and open source stuff. And yes, I love the tabs in FireFox. If my wife wouldn't hate Linux, it would be the only operative system in my computer (unfortunatelly she does :( ).\r\n\r\nAs I told, it is interesting, I have never had FireFox crashing down with Java Applets or JavaScript. But it never shows the Flash stuff, even when I have tried to configure the pluggin. My version is 1.0, released on November 7, 2004!\r\n\r\nI know I have ages without using Opera (probably at least three or four years) but when I used long time ago, it was not that good as I remember. I suppose that they have corrected almost all the thousands of bugs that it had. Nevertheless, I prefer FireFox", "Solution_17": "It just crashed 1 minute ago when I was trying to post a reply. This was the 21st crash in the past week! :mad:", "Solution_18": "BTW, this topic is devoted to the best browser on the Earth! So I don't understand your discussion here. Indeed, Firefox and IE are not so good.\r\nSorry, but I even can't understand who seriously can use Firefox and IE for work. :? \r\n\r\n[size=125][color=red][b]OPERA[/b][/color] forever! [/size]:first:", "Solution_19": "They're giving out free registration codes today only! Now's a great time to try it if you haven't already. \r\nLink:\r\n[url=http://my.opera.com/community/party/reg.dml]my.opera.com/community/party/reg.dml[/url]\r\n\r\nDisclaimer: I'm still using Firefox because of certain invaluable extensions. :)", "Solution_20": "I use FF too. David, you've GOT to get a newer version. (Flash works fine for me on Deer Park Alpha 1/2, but Java takes forever (if not a crash too))\r\n\r\nAnd yes, although I don't use it much, I've gotten my free registration, and also :) posted it in F&G.", "Solution_21": "[quote=\"Myth\"]\n[quote=\"stupidkid\"]...but you can always use Mozilla Firefox, which is free and also very good.[/quote] This opinion comes only from lack of knowledges. Opera's features cover everything what you have with IE or Firefox, moreover it gives many unique great things, which help with internet surfing.[/quote]\r\nFirefox still has some features unavailable in Opera, most importantly extensions. (Does Opera support extensions? I don't think so...)", "Solution_22": "Does it support Greasemonkey?! Platypus?! Undo Close Tab for any tab up to one closed 25 tabs ago? Hm????", "Solution_23": "[quote=\"solafidefarms\"]Does it support Greasemonkey?! Platypus?! Undo Close Tab for any tab up to one closed 25 tabs ago? Hm????[/quote]\r\n\r\nLet me guess... No." } { "Tag": [ "geometry", "circumcircle", "parallelogram", "geometric transformation", "homothety", "ratio", "geometry unsolved" ], "Problem": "Consider a triangle ABC and select a point M on the circumcircle of the given triangle. The feet of the perpendiculars from M to the sides of the triangle all lie on the same line, Simson's line (I can prove this). Prove that Simson's line bisects the segment MH, where H is the orthocenter of the triangle ABC. Also prove that the midpoint of the segment MH lie on the Euler's circle! \r\n\r\nPlease help me to prove this! Thanks", "Solution_1": "[hide=\"solution\"]before proving let's have some terminologies\n$ ABC$ is the triangle\n$ P$ is a point on the circumcircle($ \\Gamma$)\n$ U$ is the other point of intersection of simson's line with $ \\Gamma$\n$ AH$ extended meets $ BC$ at $ D$ and $ \\Gamma$ at $ D^{'}$\n$ PD^{'}$ meets $ BC$ at $ Q$ .\n$ HQ$ meets $ PU$ at $ V$\n$ A_{1},B_{1},C_{1}$ are the feet of perpendiculars from $ P$ onto $ BC,AC,CB$,\n[u][b][color=red]Lemma[/color][/b][/u]:\n$ AU$ is parallel to $ A_{1}B_{1}$\n[u][b]Proof of Lemma[/b][/u]\nconsider the cyclic quadrilaterals $ PAUC$ and $ PB_{1}A_{1}C$\nangle $ PUA \\equal{} PCA \\equal{} PA_{1}B_{1}$\n\n[u][b]proof[/b][/u]\nsince $ HD \\equal{} DD^{'}$ we and that $ PV$ and $ AD$ are both parallel since they are perpendicular to the same line.\nwe have $ \\Delta HQD^{'}$ and $ \\Delta PQV$ are isosceles.\nnow angles $ D^{'}HV \\equal{} PVH \\equal{} D^{'}PU \\equal{} DAU$\nso $ HV$ is parallel to $ AU$ is parallel to simson line of $ P$(by lemma)\nalso note that in $ \\Delta PHV$ $ A_{1}B_{1}$ is parallel to $ HV$ and bisects $ PV$ hence it bisects $ PH$\nthus we are done\n[/hide]", "Solution_2": "[quote=\"pardesi\"]\n$ U$ is the other point of intersection of simson's line with $ \\Gamma$\n$ PD$ meets $ BC$ at $ Q$ .\n[/quote]\r\nI think you meant $ U$ is the other intersection of $ PA_{1}$ with $ \\Gamma$, and $ PD'\\cap BC \\equal{}\\{Q\\}$.\r\n\r\nA somewhat similar solution is the following:\r\nLet $ AH$ meet the Simson line at $ A'$. Since $ A'H\\|PA_{1}$, we have to prove $ PA'HA_{1}$ is a parallelogram, or $ PA' \\equal{} A_{1}H$. Let $ D$ and $ D'$ be points of intersection of $ AH$ with $ BC$ and $ \\Gamma$. Using similar angle chasing as in pardesi's solution, we obtain that $ PA'D'A_{1}$ is an isosceles trapezoid. Then $ PA' \\equal{} A_{1}D' \\equal{} A_{1}H$.\r\n\r\nNow since a homothety of center $ H$ and ratio $ \\frac{1}2$ sends $ \\Gamma$ to Euler's circle $ \\omega$, we see that the midpoint of $ MH$ lies on $ \\omega$.", "Solution_3": "yes sorry for the mistake" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "(a) Let $a,b>1$ be real numbers such that $a+b \\leq 4$ find the minimum value of the\r\n expression:\r\n\r\n $F = \\frac{a^4}{(b-1)^3} + \\frac{b^4}{(a-1)^3}$\r\n\r\n(b) Find the maximum value of the expression:\r\n\r\n $\\frac{x+y}{1+z} + \\frac{y+z}{1+x} + \\frac{z+x}{1+y}$\r\n\r\n where x,y,z are real numbers in the interval $[\\frac{1}{2} ; 1]$", "Solution_1": "1. a=b=2,F=32,\r\n2. x=y=z=1, max=3.", "Solution_2": "[quote=\"Rust\"]1. a=b=2,F=32,\n2. x=y=z=1, max=3.[/quote]\r\n\r\nhow do you get that? :lol:" } { "Tag": [ "calculus", "integration", "logarithms", "trigonometry", "calculus computations" ], "Problem": "Evaluate $ \\int_0^2 \\frac{2x\\plus{}1}{\\sqrt{x^2\\plus{}4}}dx$\r\n\r\nNote that you aren't allowed to use the formula $ \\int \\frac{1}{\\sqrt{x^2\\plus{}a}}dx\\equal{}...$", "Solution_1": "[hide=\"Solution\"]Lets calculate $ I\\equal{}\\int\\frac{2x\\plus{}1}{\\sqrt{x^2\\plus{}4}}dx\\equal{}\\int\\frac{2x}{\\sqrt{x^2\\plus{}4}}dx\\plus{}\\frac{1}{2}\\int\\frac{dx}{\\sqrt{(\\frac{x}{2})^2\\plus{}1}}$\n\n1st integral is ofc $ 2\\sqrt{x^2\\plus{}4}$, and now the 2nd:\n$ \\frac{1}{2}\\int\\frac{dx}{\\sqrt{(\\frac{x}{2})^2\\plus{}1}}\\equal{}\\left\\|\\begin{array}{l} x\\equal{}2\\sinh t \\\\ dx\\equal{}2\\cosh t \\end{array}\\right\\|\\equal{}\\int\\frac{\\cosh t}{\\sqrt{\\sinh^2t\\plus{}1}}dt\\equal{}\\int\\frac{\\cosh t}{\\cosh t}dt\\equal{}\\int dt\\equal{}t\\plus{}C\\equal{}\\mbox{arsinh}\\frac{x}{2}\\plus{}C\\equal{}\\ln\\left(\\frac{x}{2}\\plus{}\\sqrt{\\frac{x^2}{4}\\plus{}1}\\right)\\plus{}C$, hence:\n\n$ \\int_0^2\\frac{2x\\plus{}1}{\\sqrt{x^2\\plus{}4}}dx\\equal{}\\left[2\\sqrt{x^2\\plus{}4}\\plus{}\\ln\\left(\\frac{x}{2}\\plus{}\\sqrt{\\frac{x^2}{4}\\plus{}1}\\right)\\right]^2_0\\equal{}4(\\sqrt{2}\\minus{}1)\\plus{}\\ln (1\\plus{}\\sqrt{2})$[/hide]", "Solution_2": "That's right.\r\n\r\nTo Japanese high school students.\r\n\r\nLet $ x\\equal{}2\\tan \\theta$, we have $ \\int_0^2 \\frac{1}{\\sqrt{x^2\\plus{}4}}dx\\equal{}\\int_0^{\\frac{\\pi}{4}} \\frac{\\cos \\theta}{1\\minus{}\\sin ^ 2 \\theta}d\\theta$\r\n\r\n$ \\sin ht\\equal{}\\frac{e^t\\minus{}e^{\\minus{}t}}{2},\\ \\cos ht\\equal{}\\frac{e^t\\plus{}e^{\\minus{}t}}{2}$", "Solution_3": "For the last one you can also set $ u\\equal{}x\\plus{}\\sqrt{x^2\\plus{}4}.$" } { "Tag": [ "search", "real analysis", "real analysis unsolved" ], "Problem": "Riemann proved that for convergent series $ \\sum x_{n}, x_{n}\\in \\mathbb{R}$, such that $ \\sum |x_{n}|$ ,doesn't converges and for each $ \\sigma \\in \\mathbb{\\overline{R}}$ there exists bijection $ s: \\mathbb{N}\\rightarrow \\mathbb{N}$ such that $ \\sum_{n=1}^{\\infty}x_{s(n)}=\\sigma$.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?search_id=1503399436&t=67043", "Solution_2": "Sorry didn't know it was posted already. Grobber's last post gives the solution to this problem, and it's more rigorous then my solution, so I wont post it. \r\nNow try to solve the other problem that I gave :P ." } { "Tag": [ "LaTeX" ], "Problem": "I 'm trying t insert a table in my page headers, but latex seems to put the header always wrong.\r\n\r\nThis is what i do to make the table:\r\n\r\n\\newpagestyle{teststyle}{\r\n \\sethead{\r\n{\\footnotesize\r\n\\begin{tabular}{|p{2.6cm}|p{3.6cm}|p{3.6cm}|p{3.6cm}|}\r\n\\hline\r\n\\multicolumn{1}{|l|}{Kwaliteitshandboek} &\r\n\\multicolumn{3}{l|}{\\thechapter.~\\chaptertitle}\\\\\r\n\\multicolumn{1}{|l|}{} &\r\n\\multicolumn{3}{l|}{\\hspace{3mm}\\thesection ~\\sectiontitle} \\\\\r\n\\hline\r\nOpgesteld door: & Goedgekeurd door: & Paraaf: &\r\nDatum Goedkeuring:\\\\\r\n& & & 01-01-2005\\\\\r\nRes. Avondrust & \\large Deswert Luk & . . . . . . . . . . . . . . .\r\n. &\r\nDatum toepassing:\\\\\r\n& & & 01-01-2005\\\\\r\n\\hline\r\n\\end{tabular}\r\n}\r\n }\r\n {}\r\n {}\r\n}\r\n\r\nThe table is perfect, but it always apears on top of the text of my document. When i increase the headersize, the table just goos down as well...\r\n\r\nThese are my page settings: \r\n\\documentclass[a4paper,oneside,11pt]{book}\r\n\r\n\\usepackage[dvips]{graphicx}\r\n\\usepackage[dutch]{babel}\r\n\\usepackage{makeidx}\r\n\\usepackage{wrapfig}\r\n\\usepackage{url}\r\n\\usepackage{picins}\r\n\\usepackage[ansinew]{inputenc}\r\n\\usepackage[big, pagestyles]{titlesec}\r\n\\usepackage{eso-pic,graphicx}\r\n\r\n\r\n\r\n\\setlength{\\topmargin}{1cm}\r\n\\setlength{\\headheight}{10cm}\r\n\r\n\\setlength{\\parskip}{1ex plus 0.5ex minus 0.2ex}\r\n\\setlength{\\parindent}{0pt} \\sloppy \\setlength{\\textwidth}{15cm}\r\n\\setlength{\\oddsidemargin}{0.5in}\r\n\\addtolength{\\textheight}{54,4pt}\r\n\\renewcommand{\\baselinestretch}{1.2}\r\n\r\nPlease help me out here :)\r\ntnx a lot!", "Solution_1": "Got it!\r\nJust need to put [b] next to the table definition... :) GREAT" } { "Tag": [ "Euler", "geometry proposed", "geometry" ], "Problem": "Triangle ABC is acute. P is satisfying $ \\angle APB\\equal{}\\angle BPC\\equal{}\\angle CPA$. Show that three Euler lines of triangles $ APB,BPC,CPA$are concurrent.", "Solution_1": "Who can solve it?does anyone have some idea?", "Solution_2": "[quote=\"mr.danh\"]Triangle ABC is acute. P is satisfying $ \\angle APB \\equal{} \\angle BPC \\equal{} \\angle CPA$. Show that three Euler lines of triangles $ APB,BPC,CPA$are concurrent.[/quote]\r\nIt is very nice, but not hard. Here is my solution:\r\nConstruct the equilateral triangle BCE.\r\nEasy to show that A,P,E are collinear and P,B,C,E lie on a circle, whose center is denoted by I.\r\nK,G are the centroids of triangles BPC,ABC respectively. Hence, IK is the Euler line of triangle BPC. \r\nLet's prove that I,K,G are collinear. Similarly, three Euler lines of triangles APB,BPC,CPA are concurrent at the centroid of triangle ABC.\r\nLet M be the midpoint of side BC. we have $ \\frac{MI}{ME}\\equal{}\\frac{MK}{MP}\\equal{}\\frac{MG}{MA}\\equal{}\\frac{1}{3}$, hence I,K,G lie on a line parallel to the line (AKE).\r\nDone.", "Solution_3": "http://www.mathlinks.ro/viewtopic.php?t=6188" } { "Tag": [ "AMC", "AMC 10" ], "Problem": "I found some AHSME problems on resource sections, but most of them are incomplete and I can't find solutions. Anyone knows where to find ASHME problems and solutions (1970s and 1980s), maybe other websites?", "Solution_1": "You can purchase them from the AMC's website.", "Solution_2": "Purchase the AHSME VOl II CD from the AMC office, it has the complete set of contests, problems and solutions from 1975-2000, exactly as they appeared. The Volume I CD has all the contests 1950-1974. Details are at:\r\nhttp://www.unl.edu/amc/d-publication/d1-pubarchive/20090515-PubRev.pdf\r\n\r\nAlternatively, the Contest Problem Book Series from the MAA Bookstore has the contest problems and detailed solutions in book form. You probably want Problem Books IV, V, VI for the 70's and 80's.\r\n\r\n-- Steve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_3": "Are there any free ones available of the web, like amc10 and ahsme/amc12??", "Solution_4": "AoPS/MathLinks has thousands of problems from various old competitions. Go to the \"Contests\" or \"Resources\" button at the top of the page.", "Solution_5": "Does anyone know where to find the ahsme's from like 1990-1995 for free?", "Solution_6": "[quote=\"mathemagician1729\"]Does anyone know where to find the ahsme's from like 1990-1995 for free?[/quote]\r\n\r\nYou have to buy them.\r\nLong time ago, I found them through a friend, but if there's something I can tell you, it's that most of \"good\" problems in those years are found in AoPS book, especially the 2nd one." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ x,y,z > 0$\r\n$ xyz \\geq xy \\plus{} yz \\plus{} zx$\r\nProve\r\n$ xyz\\geq3(x \\plus{} y \\plus{} z)$\r\n\r\nIndia 2001", "Solution_1": "Please edit it :wink:", "Solution_2": "Its ok now :) ?", "Solution_3": "let $ \\frac {1}{x} \\equal{} a,\\frac {1}{y} \\equal{} b,\\frac {1}{z} \\equal{} c$\r\nthus the given condition is $ \\sum a \\leq 1$\r\n $ \\Rightarrow 3(\\sum ab) \\leq (\\sum a)^{2} \\leq 1$\r\ndone", "Solution_4": "[quote=\"pardesi\"]let $ \\frac {1}{x} \\equal{} a,\\frac {1}{y} \\equal{} b,\\frac {1}{z} \\equal{} c$\nthus the given condition is $ \\sum a \\leq 1$\n $ \\Rightarrow 3(\\sum ab) \\leq (\\sum a)^{2} \\leq 1$\ndone[/quote]\r\n\r\nIts the solution I had.\r\nIs there any other solution :D ?", "Solution_5": "$ xyz(zy\\plus{}yz\\plus{}xz)\\ge (xy\\plus{}yz\\plus{}zx)^2\\ge 3xyz(x\\plus{}y\\plus{}z)$ then $ xyz\\ge xy\\plus{}yz\\plus{}zx\\ge 3(x\\plus{}y\\plus{}z)$ :)", "Solution_6": "[quote=quangpbc]$ (xy\\plus{}yz\\plus{}zx)^2\\ge 3xyz(x\\plus{}y\\plus{}z)$ [/quote]\ncan anyone pls tell how we got this?? :help:" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Matrix A = \r\n0 0 0 1 2 \r\n0 0 4 2 0 \r\n0 3 5 -1 1 \r\n2 1 0 2 -1 \r\n\r\n\r\nFind square matrix F such that: \r\n\r\n\r\nAF = 3 -1 0 5 0 \r\n6 2 8 6 0 \r\n-3 6 16 -7 -1 \r\n6 -2 2 11 0 \r\nand \r\n\r\nFA= 2 -2 -1 4 -4 \r\n2 -2 -9 2 0 \r\n-2 -4 -1 2 2 \r\n2 4 9 4 2 \r\n\r\n\r\nThank you", "Solution_1": "Thank you. I hope someone understands the question." } { "Tag": [ "analytic geometry" ], "Problem": "On a rectangular coordinate system, what is the number of units in the distance between $ (4,1)$ and $ (16,\\minus{}4)$?", "Solution_1": "Using the distance formula gives: $ \\sqrt{12^2\\plus{}5^2}\\equal{}\\sqrt{169}\\equal{}\\boxed{13}$." } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $A \\in M_n(R)$ such that: $A=A^t$ and ${a_{ii}>\\sum_{j=1;j \\neq i}^n|a_ia_j}|$.\r\n Prove that $det(A- xI_n)$ has all positive roots", "Solution_1": "Do you mean $a_{ii}>\\sum_{j=1;j \\neq i}^n|a_{ij}|$?\r\n\r\nIt is a classical problem.", "Solution_2": "Let $\\lambda$ be an eigenvalue of $A$. (Since $A$ is symmetric, all eigenvalues are real)\r\nSuppose $A\\alpha=\\lambda\\alpha$, where $\\alpha = (x_1,\\dots,x_n)^T\\neq 0$\r\nLet $|x_i| = \\max \\{|x_1|,\\dots,|x_n|\\}$, then $x_i\\neq 0$\r\nWe have\r\n$a_{i1}x_1+a_{i2}x_2+ \\cdots + a_{in}x_n = \\lambda x_i$\r\nthen\r\n$a_{i1}\\frac{x_1}{x_i} + \\cdots + a_{ii} + \\cdots + a_{in}\\frac{x_n}{x_i} = \\lambda$\r\n\r\nNotice that $a_{ii} > \\sum_{j\\neq i} |a_{ij}|$ and $|x_j/x_i| \\leq 1$\r\nwe have\r\n$\\lambda = a_{i1}\\frac{x_1}{x_i} + \\cdots + a_{ii} + \\cdots + a_{in}\\frac{x_n}{x_i} \\geq a_{ii} - \\sum \\left|a_{ij}\\frac{x_j}{x_i}\\right| \\geq a_{ii} - \\sum |a_{ij}| > 0 $", "Solution_3": "There is also another approach with Gershgorin Circle Theorem:\r\nas liyi said our matrix $A$ has only real eigenvalues, since $\\displaystyle a_{ii}>\\sum_{j \\neq i}^n|a_{ij}|$, $a_{ii}>0$ and all the Gershgorin Circles have their centers at $a_{ii}$ on positive real axis and their radiuses $0$ such that for any $\\varepsilon >0$ we have $|d(x,x_0)-d(y,x_0)|<\\varepsilon$ whenever $d(x,y)<\\delta$. \r\n\r\nFrom the triangle inequality we have \r\n\\[ d(x,x_0)\\le d(x,y)+d(y,x_0), \\]\r\nand similarly $d(y,x_0)\\le d(x,y)+d(x,x_0)$, so that we draw the conclusion that\r\n\\[ |d(x,x_0)-d(y,x_0)|\\le d(x,y)<\\delta : =\\varepsilon , \\]\r\nwhenever $d(x,y)<\\delta$. This also implies that $f_{x_0}$ is uniformly continuous. \r\n\r\nAm I correct? If not, should I generalize to some more arbitrary metric on $\\mathbb{R}$?", "Solution_1": "that is completely correct in my opinion\r\nI don't think that it will work for every metric on the reals : consider the discrete metric\r\nIt will of course work for every equivalent metric, as it gives the same topology\r\n\r\nIf you want something else to work on, take a random subset A, nonempty in your metric space\r\n\r\nand define $d(x,A)=inf\\{d(x,a)|a\\in A\\}$\r\n\r\nshow that this function is also uniformly continuous. When is it zero?", "Solution_2": "[quote=\"fredbel6\"]that is completely correct in my opinion\nI don't think that it will work for every metric on the reals : consider the discrete metric\nIt will of course work for every equivalent metric, as it gives the same topology\n\nIf you want something else to work on, take a random subset A, nonempty in your metric space\n\nand define $d(x,A)=inf\\{d(x,a)|a\\in A\\}$\n\nshow that this function is also uniformly continuous. When is it zero?[/quote]\r\n\r\nRight, accepting your challange I get\r\n\\[ \\inf _{a\\in A}d(x,a)\\le \\inf_{a\\in A}(d(x,y)+d(y,a))=d(x,y)+d(y,A) \\]\r\nand as above we have \r\n\\[ |d(x,A)-d(y,A)|\\le d(x,y), \\]\r\nwhich proves uniform continuity. Also, it is zero for any $x\\in \\overline{A}$ (closure), I guess.\r\n\r\nThanks for your response!" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Does anyone know of any quick ways of working out the determinant of a matrix of the form D-I, where I is the identity and D has determinant one?", "Solution_1": "There's really no shortcut, unless you know more about $D$ than that. Of course, finding determinants isn't very hard.", "Solution_2": "For $D$ diagonal, there's a simple rule saying that just like $det(D) = \\prod(\\lambda_i)$, we have that $det(D-I)=\\prod(\\lambda_i-1)$. Note that the diagonal elements on a diagonal matrix are the eigenvalues of that matrix, and that the product of the eigenvalues is always the determinant of a matrix :)" } { "Tag": [ "modular arithmetic" ], "Problem": "Before saying anything else, I want people not to get angry that I made both this poll and the one on invented vs discovered. I am just curious as to the difference in what a person says in regards to this question as opposed to math being invented or discovered. I want to see if I am understanding people views correctly and see people's opinions on this topic.\r\n\r\nI hope this vote does not turn out negatively, I have no bad intentions. I will only say that I believe that 1+1 is always 2 because of the universal meaning of 1 and +. \r\n\r\nWhat do you guys think? I encourage all opinions and the same to continue in the other thread, I apologize for saying everything is obvious.", "Solution_1": "It depends how you define, 1, +, =, and 2. If we use our modern definitions, then yep. \r\n\r\nBut of course we can always define 1 to be 2 and then we have 2+2=2.", "Solution_2": "[quote=\"WindSlicer\"]It depends how you define, 1, +, =, and 2. If we use our modern definitions, then yep. \n\nBut of course we can always define 1 to be 2 and then we have 2+2=2.[/quote]\r\nThat's not correct because if I replace '2' by $x$, I get $x+x=x$ and this is not the same as $1+1=2$. Maybe you mean something like $2+2=5$, with '2' replacing 1 and '5' replacing 2.", "Solution_3": "[quote=\"10000th User\"][quote=\"WindSlicer\"]It depends how you define, 1, +, =, and 2. If we use our modern definitions, then yep. \n\nBut of course we can always define 1 to be 2 and then we have 2+2=2.[/quote]\nThat's not correct because if I replace '2' by $x$, I get $x+x=x$ and this is not the same as $1+1=2$. Maybe you mean something like $2+2=5$, with '2' replacing 1 and '5' replacing 2.[/quote]\r\n\r\n Thats assuming that \"+\" is defined using our \"modern definitions\". As windlsicer said, \"=\" \"+\" and numbers can be defined differently, which means algebra which only holds for our modern definitons, will not hold for our new cases.", "Solution_4": "Well to be technical, 1+1=2 in base 10. If it were in base 2, it would be 1+1=10", "Solution_5": "[quote=\"G-UNIT\"][quote=\"10000th User\"][quote=\"WindSlicer\"]It depends how you define, 1, +, =, and 2. If we use our modern definitions, then yep. \n\nBut of course we can always define 1 to be 2 and then we have 2+2=2.[/quote]\nThat's not correct because if I replace '2' by $x$, I get $x+x=x$ and this is not the same as $1+1=2$. Maybe you mean something like $2+2=5$, with '2' replacing 1 and '5' replacing 2.[/quote]\n\n Thats assuming that \"+\" is defined using our \"modern definitions\". As windlsicer said, \"=\" \"+\" and numbers can be defined differently, which means algebra which only holds for our modern definitons, will not hold for our new cases.[/quote]\r\nI'm sorry but I don't understand what you mean \"modern definitions\". I understand WindSlicer 3rd sentence: \"we can always define 1 to be 2 and then we have 2+2=2\". He define \"1 to be 2\" and everything same but he put \"2+2=2\", and this is not correct proof.", "Solution_6": "i'm a little suprised that 75% of people vote \"no\".\r\nis it bacause of the different bases?", "Solution_7": "[quote=\"forever\"]i'm a little suprised that 75% of people vote \"no\".\nis it bacause of the different bases?[/quote]\r\nI know it is only different in base 2 because $1_2+1_2=10_2$ but $1_b+1_b=2_b$ in $b>2$\r\n\r\nI know there is a $1+1=0$ or $1+1=1$ in engineering??? I don't know where its from but I know there are formulas like that :blush:", "Solution_8": "[quote=\"10000th User\"]I know there is a $1+1=0$ or $1+1=1$ in engineering??? I don't know where its from but I know there are formulas like that :blush:[/quote]\r\nBoolean ring?\r\nBoolean Algebra?", "Solution_9": "I was thinking about Chemistry when I voted for no. \r\n$50ml ( \\text{spirit})+50ml(\\text{water}) < 100ml$.", "Solution_10": "Isn't 1 + 1 = 3 for very large values of 1? ;)", "Solution_11": "[quote=\"10000th User\"][/quote][quote=\"G-UNIT\"][quote=\"10000th User\"][quote=\"WindSlicer\"]It depends how you define, 1, +, =, and 2. If we use our modern definitions, then yep. \n\nBut of course we can always define 1 to be 2 and then we have 2+2=2.[/quote]\nThat's not correct because if I replace '2' by $x$, I get $x+x=x$ and this is not the same as $1+1=2$. Maybe you mean something like $2+2=5$, with '2' replacing 1 and '5' replacing 2.[/quote]\n\n Thats assuming that \"+\" is defined using our \"modern definitions\". As windlsicer said, \"=\" \"+\" and numbers can be defined differently, which means algebra which only holds for our modern definitons, will not hold for our new cases.[/quote][quote=\"10000th User\"]\nI'm sorry but I don't understand what you mean \"modern definitions\". I understand WindSlicer 3rd sentence: \"we can always define 1 to be 2 and then we have 2+2=2\". He define \"1 to be 2\" and everything same but he put \"2+2=2\", and this is not correct proof.[/quote]\r\n\r\nYeah, I wasn't really paying attention to the end product, my focus was on the fact that you can define anything to be something else, therefore 1+1 may not always equal 2 depending how how you define the characters.", "Solution_12": "Yeah, technically you [i]can[/i] define the symbols to be anything. You could redefine the '+' operator such that it can spit out different values for the same inputs. But reasonably speaking, it is a common consensus that '+' means addition and that by using the most commonly held axioms 1+1=2...ALL THE TIME :D (well, mathematically speaking...as shobber points out it is a little different when dealing with chemistry and one could make other arguments in other fields I presume).", "Solution_13": "As posted earlier, if we do addition and subtraction on the set 0,1 and take it (mod 2), all of our \"modern\" definitions are satisifed (it is a field) and we get\r\n\r\n1+1=0.\r\n\r\n\r\nNo real problem here.", "Solution_14": "[quote=\"Arne\"]Isn't 1 + 1 = 3 for very large values of 1? ;)[/quote]\r\nSorry, but what did you mean by \"very large values of 1\"?", "Solution_15": "I think he's trying to round down somehow.", "Solution_16": "Well, if you read the nutrition labels on food items, you will find that not only does $1+1\\not=2$, but $1+0+0+0=3$, $4\\cdot 2=0$, and $9\\cdot k\\equiv 0\\pmod{10}$, for all $k\\in\\mathbb{Z}$.", "Solution_17": "In the children's world, 1+1=11.", "Solution_18": "[quote=\"frt\"][quote=\"Arne\"]Isn't 1 + 1 = 3 for very large values of 1? ;)[/quote]\nSorry, but what did you mean by \"very large values of 1\"?[/quote]\r\n\r\nI was actually joking :) Don't worry about it.", "Solution_19": "from 1984:\r\n-Two and two are four.\r\n-Sometimes, Winston. Sometimes they are five. Sometimes they are three. Sometimes they are all of them at once. You must try harder.\r\n\r\nso, 1+1 doesn't have to equal two, just like 2+2 doesn't have to equal 4. :lol:", "Solution_20": "People, sorry I didnt mention in the original post, i mean the meaning of 1+1 always gives the same asnwer, knowing what \"1\" and \"+\" means, that is what I was refering to.", "Solution_21": "I am surprised that no accountant has yet attempted to answer the question. If (s)he (the accountant) is anyting like the one who does tax returns would make sure that all the windows are closed and would wispher:\r\n[quote] How much you want it to be? [/quote]\r\n\r\n(Yeah I know, an old joke...! :)\r\n(Not to mention, if the calculator you use is MS windows 3.1 or have an old Pentium chip - your answer may vary)", "Solution_22": "if it was in base 2, it would be 10, not 2 :lol:", "Solution_23": "Suppose it was $1_{\\infty}+1_{\\infty}=2_{\\infty}=2\\cdot \\infty=\\infty$ or indetermined.", "Solution_24": "Sorry I did not say in my first post, but I meant that the meaning of 1+1 in our basic arithmetical answer produces the answer 2, u cant changed the meaning of 1 or + in what I was asking about in my question. How would the answer then change?" } { "Tag": [], "Problem": "Peter and Darij play the game of the three knights. It is played on a chess board, initially three knights are placed on positions $h1,g2$ and $f3$. See the left image below. Alternately the players move one of the knights to an empty field whereas contrary to chess the knight may just perform the four moves shown in the right image. The goal is to move the knight to its home. A knight is at home when it is one of the hatched fields $a8,a7,b8$ or $b7$. The person who gets the third knight home will win the game. If there might be a situation before that no movement is possible anymore there will be a tie. Peter starts.\r\n\r\nIs there a winning strategy for any of the players ? If so, what is the strategy and why does it force the victory ?", "Solution_1": "Hmm. Nobody there who wants to help Darij and Peter ? :D", "Solution_2": "can the knights take each other..lol? and also are the hatching fields for each knight respectively or for all the knights?", "Solution_3": "[quote=\"DPopov\"]can the knights take each other..lol? and also are the hatching fields for each knight respectively or for all the knights?[/quote]\r\n\r\nObviously the first point is not true, otherwise you cannot move all knights in the upper left four squares. For the second point: Finally all the three knights will be in the previously mentioned squares whereas one of fours squares remains empty. :)" } { "Tag": [ "function", "analytic geometry", "graphing lines", "slope" ], "Problem": "A botanist measures the growth of a plant, in centimeters, every day. Connecting the dots placed by him on a graph, we get the picture beside. If you always keep this relationship between time and height, the plant will, in the 30 (thirtieth) day, a height equal to:\r\n\r\n[img]http://img4.imageshack.us/img4/9803/cals.jpg[/img]\r\n\r\na) 5 cm\r\nb) 6 cm \r\nc) 3 cm\r\nd) 15 cm\r\ne) 30 cm\r\n\r\nI'm very novice in this subject (Functions), can anyone explain me more clearly ?\r\nThanks. :lol:", "Solution_1": "The key is to determine the equation of the graph. It's pretty clear that it's a line so what is its equation? We know that $ (10,2)$ and $ (5,1)$ are both points, so the slope of the line is $ \\frac{2\\minus{}1}{10\\minus{}5}\\equal{}\\frac{1}{5}$. So the line has equation $ y\\equal{}\\frac{x}{5}$. So when $ x\\equal{}30$ (30 days), $ y\\equal{}\\frac{30}{5}\\equal{}6$. So the height is 6cm.", "Solution_2": "Thanks, but is the equation Y = Ax + B ?\r\nAnd B, what's it value ?\r\nA = x/5 ?", "Solution_3": "$ A\\equal{}\\frac{1}{5}, B\\equal{}0$.", "Solution_4": "Ok, but how do you know that B = 0 ?", "Solution_5": "[quote=\"luiseduardo\"]Ok, but how do you know that B = 0 ?[/quote]\r\nStraight line from the origin", "Solution_6": "hum ... :( \r\nThanks, but i think it's difficult to understand. It's my first time with exercises like that.", "Solution_7": "Ah ok :lol: \r\nNo problem. I understand now. Thanks.", "Solution_8": "So can say that the line has equation $ y\\equal{}\\frac{x}{5}\\plus{}b$ if you want to, but since $ (0,0)$ is a point, you get that $ 0\\equal{}0\\plus{}b$, so $ b\\equal{}0$.", "Solution_9": "Basically b is the number's y intercept, and since it goes through the origin, its y intercept is 0.", "Solution_10": "actually, there is not even the need to find the equation of ther line.you can use the similarity between the triangles. :wink: (this is much more simple, i think)", "Solution_11": "It is b\r\n\r\nIt starts from the origin, and has slope of 1/5. So you just count up 6 and move over 30 \r\n\r\n:starwars: :spam: :nhl:", "Solution_12": "[quote=\"v235711\"]actually, there is not even the need to find the equation of the line.you can use the similarity between the triangles. :wink: (this is much more simple, i think)[/quote]\r\nNot if you're trying to understand functions and equations of lines, which is what the OP is trying to do.", "Solution_13": "Thanks JRav.\r\nI understand the problem now. :lol:" } { "Tag": [ "calculus", "derivative", "trigonometry", "calculus computations" ], "Problem": "Well I have this problem here, and so far the calculus isn't the problem, its the algebra. I just want to see if anyone can solve this for me, as I am stuck and its valentine's day and the combination of not having a girlfriend and doing f--king problems like these is making me want to find a tall building.\r\n\r\nx(arc cosy) = (arc sinxy)\r\n\r\nPerhaps I am missing a simpler way to do this, or maybe this is just one ugly mother. I tried getting y on the left side, but then after taking the derivative (using multiple chain/product rules), there are some dy/dx's wrapped up in the right side. \r\n\r\nAlso, these inverses are NOT 1/sin (csc), 1/cos (sec), these are the inverse trig fxns (someone told me they are also knows as arc fxns)\r\n\r\nI have the formulas for the arc fxns:\r\nd/dx[arc sin(u)]=1/[sqr rt=1-(u^2)]\r\nd/dx[arc cos(u)]=(-)1/[sqr rt=1-(u^2)]\r\n\r\nAny help would be appreciated as it is surely not my time to leave this world, but every second i spend on this problem speeds it towards me.", "Solution_1": "I finally got dy/dx, if someone still wants to do this i would like to compare answers and methods.\r\n-- Thanks", "Solution_2": "I started writing this up before your second post:\r\n\r\nLet me just double-check: are you trying to find the derivative $\\frac{dy}{dx}$ given the equation $x \\cdot \\cos^{-1}y = \\sin^{-1}(xy)$? That equation isn't going to solve for a nice expression of $y$ in terms of $x$, which suggests that you have to use implicit differentiation. After you do that, it should be easy to solve for $\\frac{dy}{dx}$ (although the resulting expression is pretty ugly).", "Solution_3": "Yes sir, thats the equation. I got an answer that checks out, and yea youre right i had to use implicit diff. i just had to work it out in my head before i messed with in on paper to make sure... thanks! like i said if you wanna try it, ill compare notes." } { "Tag": [], "Problem": "Solve for x:\r\n\\[ y \\equal{} \\frac{2x}{t\\minus{}1}\\]", "Solution_1": "\\[ y \\equal{} \\frac{2x}{t\\minus{}1} \\]\r\n\r\n\\[ y(t\\minus{}1) \\equal{} 2x \\]\r\n\r\n\\[ \\frac{y(t\\minus{}1)}{2} \\equal{} x \\]", "Solution_2": "whoops sorry, the t is supposed to be another x. so instead its:\r\n\\[ y \\equal{} \\frac{2x}{x\\minus{}1}\\]", "Solution_3": "For other problems like those, first cross-multiply to eliminate fractions. Then solve the equation to get the answer (which isabella already gave) \\[ \\frac{y(t \\minus{} 1)}{2} \\equal{} x\\]", "Solution_4": "[quote=\"Lord Voldemort\"]whoops sorry, the t is supposed to be another x. so instead its:\n\\[ y \\equal{} \\frac {2x}{x \\minus{} 1}\n\\]\n[/quote]\r\nthat's what i meant, sorry for the confusion.", "Solution_5": "Then substitute the $ t$ in the answer for $ x$. I don't think you can get any simpler than that :) .", "Solution_6": "$ x\\equal{}\\frac{y}{y\\minus{}2}$\r\n\r\nI'll explain it better when I get home.", "Solution_7": "[hide=\"solution\"]\n$ y\\equal{}\\frac{2x}{x\\minus{}1}$ \n\n$ yx\\minus{}y\\equal{}2x$\n\n$ \\minus{}y\\equal{}2x\\minus{}yx$\n\n$ \\minus{}y\\equal{}x(2\\minus{}y)$\n\n$ \\frac{\\minus{}y}{2\\minus{}y}\\equal{}x$ \n\n$ \\boxed{\\frac{y}{y\\minus{}2}\\equal{}x}$\n\n[/hide]\r\ni guess i'll save markkram the time.", "Solution_8": "Minus one point for forgetting restrictions. Remember that y can't be 2.", "Solution_9": "Thanks you guys!!!!", "Solution_10": "Ok we have: y=(2x)/(x-1)\r\n\r\nWE need to isolate \"x\" or just get it on one side to show what it equals..\r\n\r\nFirst: we need to multiply each side by (x-1) to get rid of the denominator.\r\n\r\nNow we have: xy-y=2x <--- a lot simpler huh?\r\n\r\nNow, our final step. WE just need to divide each side by 2 to get rid of the coefficient and completely isolate x.\r\n\r\nThe answer: [(xy-y)/2]=x", "Solution_11": "[quote]\n$ x\\equal{} \\frac{xy\\minus{}y}{2}$\n[/quote]\r\n\r\nbut x isn't completely isolated. You need to bring xy to the other side, then factor out x." } { "Tag": [ "trigonometry" ], "Problem": "i have posted this exercire in the High school intermediate topic but no one answer \r\nso i post in this topic , please help me solve the equation \r\n(sin2x)^2 + sin2x + sinx + 1 = 0", "Solution_1": "hello, simplifying your equation we get\r\n$ 4\\, \\left( \\sin \\left( x \\right) \\right) ^{2} \\left( \\cos \\left( x\r\n \\right) \\right) ^{2}\\plus{}2\\,\\sin \\left( x \\right) \\cos \\left( x \\right) \r\n\\plus{}\\sin \\left( x \\right) \\plus{}1\\equal{}0$\r\nconverting this in $ \\tan(\\frac{x}{2})$ we get\r\n$ \\displaystyle{16\\,{\\frac { \\left( \\tan \\left( 1/2\\,x \\right) \\right) ^{2} \\left( 1\\minus{}\r\n \\left( \\tan \\left( 1/2\\,x \\right) \\right) ^{2} \\right) ^{2}}{\r\n \\left( 1\\plus{} \\left( \\tan \\left( 1/2\\,x \\right) \\right) ^{2} \\right) ^{4\r\n}}}\\plus{}4\\,{\\frac {\\tan \\left( 1/2\\,x \\right) \\left( 1\\minus{} \\left( \\tan\r\n \\left( 1/2\\,x \\right) \\right) ^{2} \\right) }{ \\left( 1\\plus{} \\left( \\tan\r\n \\left( 1/2\\,x \\right) \\right) ^{2} \\right) ^{2}}}\\plus{}2\\,{\\frac {\\tan\r\n \\left( 1/2\\,x \\right) }{1\\plus{} \\left( \\tan \\left( 1/2\\,x \\right) \r\n \\right) ^{2}}}\\plus{}1\\equal{}0}$\r\nsubstituting $ \\tan(\\frac{x}{2})\\equal{}t$ and factorring we have to solve\r\n$ \\left( t\\plus{}1 \\right) \\left( {t}^{7}\\minus{}3\\,{t}^{6}\\plus{}23\\,{t}^{5}\\minus{}21\\,{t}^{4}\r\n\\minus{}5\\,{t}^{3}\\plus{}15\\,{t}^{2}\\plus{}5\\,t\\plus{}1 \\right)\\equal{}0$.\r\nSonnhard." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that for $a,b,c$ are nonegative real, we have:\r\n\r\n$\\frac{2a^2}{b^2+c^2}+\\frac{2b^2}{a^2+c^2}+\\frac{2c^2}{a^2+b^2} \\ge \\frac{(a+b+c)^2}{ab+bc+ca}$\r\n\r\n\r\nPham Kim Hung", "Solution_1": "I use Vasc'c inequality $a,b,c>0$ then $\\frac{a^2}{b^2+c^2}+\\frac{b^2}{c^2+a^2}+\\frac{c^2}{a^2+b^2}\\geq\\frac{a}{b+c}+\\frac{b}{c+a}+\\frac{c}{a+b}$ and BCS\r\n$LHS\\geq\\frac{2a}{b+c}+\\frac{2b}{c+a}+\\frac{2c}{a+b}=\\frac{2a^2}{ab+ac}+\\frac{2b^2}{bc+ba}+\\frac{2c^2}{ca+cb}\\geq\\frac{2(a+b+c)^2}{2(ab+bc+ca)}=RHS$\r\n\r\nSorry , i cannot find the link with Vasc's inequality,,,,", "Solution_2": "Very nice solution and It's much simpler than I think at first. \r\nOther hand, $2$ is the best constant for this.", "Solution_3": "[quote=\"hungkhtn\"]\nOther hand, $2$ is the best constant for this.[/quote]\r\n\r\nWhat do you mean?" } { "Tag": [ "search", "algebra", "binomial theorem" ], "Problem": "Let's say we have $ (x+y)^{2}$ . We'd automatically think it equals $ x^{2}+2xy+y^{2}$. However, this presents a problem when dealing with coefficients.\r\n\r\nIf we have, for instance $ (3a+4b)$, the only way this can be factored with the above formula is to do the following:\r\n\r\n$ (3+4)^{2}= (3^{2}+2(3)(4)+4^{2}) = (9+24+16)$ \r\n\r\n$ (a+b)^{2}= (1)a^{2}+(2)ab+(1)b^{2}$\r\n\r\nIgnoring the coefficients of the second equation and replacing them with the coefficients of the first equation, we have:\r\n\r\n$ 9a^{2}+24ab+16b^{2}$ which is correct.\r\n\r\nBut instead of doing all that work, wouldn't it make more sense just to use a more thorough formula such as $ (cx+dy)^{2}= c^{2}x^{2}+2cdxy+d^{2}x^{2}$ where $ c,d$ are coefficients and $ x,y$ are variables? This way, we could get our answer instantly. Example:\r\n\r\n$ (8x+9y)^{2}= 8^{2}x^{2}+2(8)(9)xy+9^{2}y^{2}= 64x^{2}+144xy+81y^{2}$\r\n\r\nSorry if the question seems really confusing, my main issue is whether to list factoring formulas with or without variables representing coefficients.", "Solution_1": "I think when factoring formulas are listed, they usually omit the coefficients because we can assign $ a = 8x$ and $ b = 9y$, for example, in $ (a+b)^{2}$ and we get $ a^{2}+2ab+b^{2}= (8x)^{2}+2(8x)(9y)+(9y)^{2}$. So it's easy to substitute the coefficients into the formula and we don't have to explicitly represent coefficients in the formula?", "Solution_2": "If I understand your question correctly, you're starting to touch on something called the binomial theorem. Do a search on the site for it.", "Solution_3": "[quote=\"aidan\"]I think when factoring formulas are listed, they usually omit the coefficients because we can assign $ a = 8x$ and $ b = 9y$, for example, in $ (a+b)^{2}$ and we get $ a^{2}+2ab+b^{2}= (8x)^{2}+2(8x)(9y)+(9y)^{2}$. So it's easy to substitute the coefficients into the formula and we don't have to explicitly represent coefficients in the formula?[/quote]\n\nThat pretty much answered my question. Thank you!\n\n[quote=\"max_tm\"]If I understand your question correctly, you're starting to touch on something called the binomial theorem. Do a search on the site for it.[/quote]\r\n\r\nI've seen the binomial theorem before, but that's not what I'm asking about. Thanks anyway." } { "Tag": [ "calculus", "integration", "function" ], "Problem": "Prove or disprove, for any $n$\r\n\r\n$ \\displaystyle \\sum^n_{k=3} \\frac{1}{k^3}<\\frac{1}{12}$", "Solution_1": "Isn't it enough to notice that for $n=3$, $\\frac{1}{k^3}=\\frac{1}{27}$, which is less than $\\frac{1}{12}$, and that for any $n>3$, $\\displaystyle \\sum^n_{k=3}k^3$ will be greater than when $n=3$, and thus $\\displaystyle \\displaystyle \\sum^n_{k=3} \\frac{1}{k^3}$ will be less than $\\frac{1}{27}$ (and consenquently less than $\\frac{1}{2}$?", "Solution_2": "[quote=\"Elemennop\"]Isn't it enough to notice that for $n=3$, $\\frac{1}{k^3}=\\frac{1}{27}$, which is less than $\\frac{1}{12}$, and that for any $n>3$, $\\displaystyle \\sum^n_{k=3}k^3$ will be greater than when $n=3$, and thus $\\displaystyle \\displaystyle \\sum^n_{k=3} \\frac{1}{k^3}$ will be less than $\\frac{1}{27}$ (and consenquently less than $\\frac{1}{2}$?[/quote]\r\nYes, all of the individual terms will be less than $\\frac{1}{12}$, but, how do you know that $\\frac{1}{27}+\\frac{1}{64}+\\frac{1}{125}+\\frac{1}{216} + ... + \\frac{1}{n^3} < \\frac{1}{12}$?", "Solution_3": "[quote=\"Elemennop\"]Isn't it enough to notice that for $n=3$, $\\frac{1}{k^3}=\\frac{1}{27}$, which is less than $\\frac{1}{12}$, and that for any $n>3$, $\\displaystyle \\sum^n_{k=3}k^3$ will be greater than when $n=3$, and thus $\\displaystyle \\displaystyle \\sum^n_{k=3} \\frac{1}{k^3}$ will be less than $\\frac{1}{27}$ (and consenquently less than $\\frac{1}{2}$?[/quote]\r\nI think what you have proved is that as n increases each term will decrease, but this is not enough to say that the whole sum will decrease. \r\n$\\sum^n_{k=3}k^3 $is not the reciprocal of $\\sum^n_{k=3} \\frac{1}{k^3}$", "Solution_4": "So how do you do it?", "Solution_5": "Hmm.\r\n\r\nI would say the way to prove it now would be show that if $n=3$, the expression is less than $\\frac{1}{12}$, then find what it converges to as $n$ approaches $\\infty$ (and show it is less than $\\frac{1}{12}$), and then show that the expression is decreasing for each $n$ as $n$ approaches $\\infty$.", "Solution_6": "i guess we can't express zeta(3) as an exact number yet. i looked it up (http://64.233.187.104/search?q=cache:IkjY0zJIzu8J:mathworld.wolfram.com/AperysConstant.html+zeta(3)&hl=en), and all i found was an approximation ~1.2020569, so i dont know if thats the way to go. telescoping it seems more likely.", "Solution_7": "Hum diddly dum.\r\n\r\n$\\sum_3 1/k^3<\\sum_2 1/[k(k+1)(k+2)]=1/4-1/6=1/12$\r\n\r\nI think there's some minor pedagogical value for you guys to figure out why what I wrote is true without me telling you so; if you still have questions in a day, I will be glad to answer them.", "Solution_8": "Very good!", "Solution_9": "Out of curiosity, is that how you constructed the problem? Because an integral test doesn't work very well - I'm not sure but I suspect that we need to sum a large number of terms by hand because the integral gives us an estimate that is not very sharp.", "Solution_10": "That indeed is how it was derived. It was actually a problem in an IrMO, a few years ago... How does the integral solution go? Im not so sharp on that kind of thing... :(", "Solution_11": "We have\r\n\r\n$\\sum_{k=n}^{\\infty} \\frac {1}{k^3}<\\int^{\\infty}_{n-1} \\frac {dt}{t^3}=\\frac{1}{2(n-1)^2}$\r\n\r\nThis method actually works if we take $n=6$ and sum the first three terms manually.", "Solution_12": "How'd you get the first bit? Or is it something general like\r\n\r\n$\\displaystyle \\sum_{k=f}^\\infty f(k) < \\int_{f-1}^\\infty f(k) dx$", "Solution_13": "See if you can prove that it holds for any monotonically decreasing $f$", "Solution_14": "$\\sum_{k=1}^\\infty{1 \\over k^{3}}$", "Solution_15": "Woah you did some serious digging! Dspite massive attempts, no one can calculate this value! Can anyone think of a function that has $\\frac{1}{k^{3}}$ as roots for all $k \\in \\mathbb{N}$?", "Solution_16": "hwo did you compute $\\sum_{2}1/[k(k+1)(k+2)]=1/4-1/6=1/12$ ?", "Solution_17": "Do you know partial fractions at all?" } { "Tag": [ "MATHCOUNTS", "conics", "parabola", "algebra", "polynomial" ], "Problem": "What is your mathcounts index school+chapter+state=94(my score).", "Solution_1": "112 *sniff* i'm sad.", "Solution_2": "indianamath i guess you pwned me, BTW who was the 4th 7th grader who made it to aime. it was you, me , lynodon and someone else.", "Solution_3": "i'm guessing youkou homa (sp?) but i'm not sure if he made it.", "Solution_4": "I know I'm not supposed to be in an Indiana forum\r\nbut oh well\r\n\r\nSchool: $44$\r\nChapter: $39$\r\nState: $30$\r\n\r\nso\r\n\r\n$44+39+30=\\fbox{113}$", "Solution_5": "It was Youkow Homma with a 135 on the 10B\r\n\r\nMy index:\r\n\r\nSchool: 44\r\nChapter: 44\r\nState: I can't reveal... greater than 38 shouldn't be that surprising though.\r\n\r\nTotal index: 126+", "Solution_6": "43 (school) + 39 (chapter) + 32 (state) = 114 (i think)", "Solution_7": "41 + 34 + 31 = 106\r\n\r\nschool, chapter, and state, respectively.", "Solution_8": "[quote=\"indianamath\"]112 *sniff* i'm sad.[/quote]\r\n\r\nyou got 112?? i soo did not beat you! i couldn't have", "Solution_9": "i got 2 pm's from rcv(a mod) about spamming....", "Solution_10": "haha, but u weren't the only one spamming. at least i didn't spam that much (did I?). it's about time someone took control of this place.... :lol:", "Solution_11": "basically everyone was spamming...", "Solution_12": "i'm pretty sure he's checking every comment i make. i got my second pm because i said i got 112 index and didn't make it.\r\n\r\noh yeah lol, that index was about chapter+state. i added my school too.lol", "Solution_13": "what did u get on school?", "Solution_14": "School: 44\r\nChapter: 44\r\nState: -_- lol\r\nIndex: like over 88 <(^^)> lol...", "Solution_15": "[quote=\"mathcrazed\"]\nIndex: like over 88 <(^^)> lol...[/quote]\r\n\r\nthat says a lot! lol. wow, major hint to what u got at states!! :lol:", "Solution_16": "yep it could be... possibly in the range from 0-46. Wow, we narrowed it down to only 47 numbers guys (and girls XD)!", "Solution_17": "[quote=\"mathcrazed\"]yep it could be... possibly in the range from 0-46. Wow, we narrowed it down to only 47 numbers guys (and girls XD)![/quote]\r\n\r\nmore like [i]girl[/i]. lol. i am the only one right? i feel so special...(not ed. i'm one step ahead of all u ppl! :lol: )", "Solution_18": "yep message too short -_-... GOD ------", "Solution_19": "Well no we've narrowed it down to 45. You couldn't have gotten a 45, because the paper said I got a \"near perfect\" so I couldn't have gotten a 46.\r\n\r\nXD crazy logic.\r\n\r\nPlus mwp probably knows her own score and you got a higher score", "Solution_20": ":D XD aasd asd ASZX", "Solution_21": "[quote=\"Fanatic\"].\nPlus mwp probably knows her own score and you got a higher score[/quote]\n\nis it just me or did that kinda not make sense. obviously mathcrazed got a higher score than me...\n\n[quote=\"mathcrazed\"] XD aasd asd ASZX[/quote]\r\n\r\nyou ppl are crazy...", "Solution_22": "not necessarily you could have lost a tiebreaker from 13th to 4th.", "Solution_23": "true, yet not true...", "Solution_24": "In my post earlier, my point was that mwp could further narrow down lyndon's score by using her score as the minimum for his.", "Solution_25": "but i think i already know the top 4, as i mentioned earlier...who knows, i could be wrong.", "Solution_26": "[quote=\"cognos599\"]not necessarily you could have lost a tiebreaker from 13th to 4th.[/quote]\r\n\r\n :rotfl: :rotfl: :rotfl:", "Solution_27": "sorry yuqing, if that was the case i have deeply insulted you by saying you got the same score as my sister.", "Solution_28": "lol i'm not laughing at that. If 13th tied with 4th, it would REALLYREALLYREALLY suck to be 13th. lol lose all the tiebreakers lol....lol....lmao...", "Solution_29": "hey, i can see all this, just in case you didn't know! jk, but that would suck. terribly. i think i did lose to some ppl due to tiebreakers, though...", "Solution_30": "yeah, a lot of people got 32.", "Solution_31": "I know 12th place, and she got a 33.", "Solution_32": "was that stevens? it's good that i came first in tiebreaker at least...", "Solution_33": "i have a question (math-related). it may sound stupid but bear with me...\r\n\r\njust say you have a parabola $-3{x^{2}}+15x+30 = y$ . if you set $y = 0$. \r\n\r\nthen you can get $-3{x^{2}}+15x+30 = 0$.\r\n\r\nif you factor out the $-3$ and divide throughout, you get ${x^{2}}-5x-10 = 0$.\r\n\r\nnow since they are both equal to zero, you can set them equal to each other:\r\n\r\n$-3{x^{2}}+15x+30 ={x^{2}}-5x-10$\r\n\r\nbut by using common sense, you know the two sides aren't equivalent. i'm just confused how that works. could someone explain it to me, thanks", "Solution_34": "well, when you set them equal to 0, you found that the zeros of the two polynomials were equal, but the graphs actually equal-you could divide by -3 because you are finding the zeros. To begin with,their \"a\" values are opposite-one is posite the other negative-so they arent even going in the same direction. And when you graph parabolas like that, the equation isnt like ax^2+bx+c=0, it is just ax^2+bx+c, so in that case, you cant just cancel out the -3.", "Solution_35": "okay. i get it now. thanks" } { "Tag": [], "Problem": "Not sure if this is the appropriate place to post this question. Does anyone know a good approximation to the trajectory of Halley's Comet?", "Solution_1": "See [url=http://en.wikipedia.org/wiki/Comet_Halley]here[/url]." } { "Tag": [], "Problem": "I want to know, in simple language, why any number raised to the zero is 1. I also want to know why any number raised to the one power is the base itself.\r\n\r\nSample:\r\n\r\n100^1 = 100...WHY?\r\n\r\nSample:\r\n\r\n100^0 = 1...WHY?\r\n\r\nPlease, do not explain using math textbook definitions.\r\nI can read a math book for myself. I am seeking a simple definition. Okay?", "Solution_1": "[quote=\"blueshark\"]I want to know, in simple language, why any number raised to the zero is 1. I also want to know why any number raised to the one power is the base itself.\n\nSample:\n\n100^1 = 100...WHY?\n\nSample:\n\n100^0 = 1...WHY?\n\nPlease, do not explain using math textbook definitions.\nI can read a math book for myself. I am seeking a simple definition. Okay?[/quote]First, why does $a^1=a$ for any real number $a$? This comes from the definition of $\\text{positive integer exponents}$. $a^n$ means the product of $n$ factors of $a$. So..\r\n\r\n$a^3=a\\cdot{a}\\cdot{a}\\\\\\\\a^2=a\\cdot{a}\\\\\\\\a^1=a$\r\n\r\nWhat else could $a^1$ be?\r\n\r\nNext, why does $a^0=1$ for any nonzero real number $a$ (it's true for negative $a$ right?)? It is defined this way for consistency:\r\n\r\nWe know that $a^ma^n=a^{m+n}$, since \\[a^ma^n=\\underset{m}{\\underbrace{a\\cdot{a}\\cdots{a}}}\\underset{n}{\\underbrace{a\\cdot{a}\\cdots{a}}}=\\underset{m+n}{\\underbrace{a\\cdot{a}\\cdots{a}}}=a^{m+n}.\\]This clearly works for positive integers $m$ and $n$, but we can extend it to $0$ (the product of no $a\\text{'s}$.) When, say, $m=0$ then we have $a^0a^n=a^{0+n}=a^n$, so when we divide both sides by $a^n$ we get $a^0=1$.", "Solution_2": "hey my method is....\r\n\r\n$a^0$=\r\n$a^{n-n}$\r\n$a^n/a^n\nlet $a^n$ be x\r\n\r\nhence x/x \r\n=1\r\n\r\nhence proved", "Solution_3": "I thank you both your explanation and examples." } { "Tag": [ "videos", "calculus", "integration" ], "Problem": "What factoring trick is this called? It looks like simmon's favorite factoring trick but its not. \r\n\r\n$AMC+AM+MC+CA=(A+1)(M+1)(C+1)-(A+M+C)-1$\r\n\r\nThanks :P", "Solution_1": "$(x+\\alpha)(x+\\beta)(x+\\gamma)=x^{3}+(\\alpha+\\beta+\\gamma)x^{2}+(\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha)x+\\alpha \\beta \\gamma .$\r\nPlugging $x=1,\\alpha =A,\\ \\beta =M,\\ \\gamma =C.$ to this equation gives your equation.", "Solution_2": "I wouldn't even call it a factoring trick. This is just a rewrite of\r\n\r\n$(A+1)(M+1)(C+1) = AMC+AM+MC+CA+A+M+C+1$\r\n\r\nWhich is just a standard expansion. (If you really wanted to, you could call this the \"three-variable\" case of Simon's.)", "Solution_3": "What's Simon's?", "Solution_4": "check out the video on aops wiki at the bottom of this page if u wish to know more :lol: \r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick", "Solution_5": "Thank you!\r\nI have just seen the video.\r\n\r\nThis trick is well known. :lol: \r\n\r\nAlternative approach\r\n\r\nFind all pairs of positive integers $(m,\\ n)$ such that $mn+4m-2n=99.$\r\n\r\n$mn+4m-2n=99\\Longleftrightarrow m(n+4)=2n+99$\r\n\r\n$\\Longleftrightarrow m=\\frac{2n+99}{n+4}$\r\n\r\n$\\Longleftrightarrow m=2+\\frac{91}{n+4}.$\r\n\r\nSince $n+4$ is positive integers such that $n+4\\geq 5$ and divisor of $91,$ yielding $n+4=7,\\ 13.\\therefore (m,\\ n)=(9,\\ 9),\\ (15,\\ 3).$\r\n\r\nBy the way, How about this problem?\r\n\r\n[color=blue]Find all pairs of positive integers[/color] $(m,\\ n)$ [color=blue]such that [/color]$6mn+21m+10n-305=0.$ :D", "Solution_6": "[quote=\"kunny\"]Thank you!\nI have just seen the video.\n\nThis trick is well known. :lol: \n\nAlternative approach\n\nFind all pairs of positive integers $(m,\\ n)$ such that $mn+4m-2n=99.$\n\n$mn+4m-2n=99\\Longleftrightarrow m(n+4)=2n+99$\n\n$\\Longleftrightarrow m=\\frac{2n+99}{n+4}$\n\n$\\Longleftrightarrow m=2+\\frac{91}{n+4}.$\n\nSince $n+4$ is positive integers such that $n+4\\geq 5$ and divisor of $91,$ yielding $n+4=7,\\ 13.\\therefore (m,\\ n)=(9,\\ 9),\\ (15,\\ 3).$\n\nBy the way, How about this problem?\n\n[color=blue]Find all pairs of positive integers[/color] $(m,\\ n)$ [color=blue]such that [/color]$6mn+21m+10n-305=0.$ :D[/quote]\r\nWell, your way can easily turn into Simon's Favorite Factoring Trick. Just subtract two from both sides of $m=2+\\frac{91}{n+4}$ and multiply everything by n+4, and that is the form of Simon's Favorite Factoring Trick. Now, for the problem you gave:\r\n[hide] $6mn+21m+10n-305=0\\Rightarrow 6mn+21m+10n=305$. We also have that $(3m+5)(2n+7)=6mn+10n+21m+35$. Thus, we know that $(3m+5)(2n+7)-35=305\\Rightarrow (3m+5)(2n+7)=340=2^{2}*5*17$. \n\nWe know that the factors of 340 are 1, 2, 4, 5, 10, 17, 20, 34, 68, 85, 170, and 340. We find that the factors that will make n integral for 2n+7=x (if x is a factor) are all of the odd factors, which are 1, 5, 17, and 85. However, 1 and 5 won't work because they will make n negative. \n\nWe can now say that 2n+7=17, so n=5. Now, this means that 3m+5=340/17=20. Thus, m=5. So, one pair is (5,5). Now, we can also say that 2n+7=85, thus n=34. Now, we know that 3m+5=340/85=4. This makes m negative, so that is not possible. \n\nFinally, we can conclude that $\\boxed{(5,5)}$ is the only solution. [/hide]" } { "Tag": [ "HCSSiM" ], "Problem": "Ok so this game goes like this.\r\nI start with a country, town, state, or province.\r\nYou have to name a country, town, state, or province that starts with the letter that my country ends in.\r\nExample:\r\nUSA\r\nAndorra\r\nAlbania\r\nAfghanistan\r\nNepal\r\nLebanon\r\n\r\n\r\nAnd no repeats\r\n\r\nCyprus\r\n\r\n(This is the 2012th topic in G&FF, the world is supposed to end in 2012, symbolic eh?)", "Solution_1": "saskatchewan", "Solution_2": "North Dakota", "Solution_3": "Arizona", "Solution_4": "Australia", "Solution_5": "Austria\r\n(Blah)", "Solution_6": "Antarctica", "Solution_7": "albania :D", "Solution_8": "Armenia\r\ntime for something other than A's.", "Solution_9": "afghanistan", "Solution_10": "Nantucket\r\n.", "Solution_11": "Texas\r\n(Dangit)", "Solution_12": "Sydney\r\n\r\nA side note, Antarctica is a continent not a country.", "Solution_13": "Yugoslavia", "Solution_14": "Alabama\r\n(oh no, not a's again!)", "Solution_15": "albania\r\nantarctica\r\nalberquerqe\r\nel alamein\r\nnubia\r\nalaska\r\nancorage", "Solution_16": "-_-;;\r\n\r\nonly one country, and you can't repeat..\r\n\r\nPLEASE read other posts before posting in an inconsiderate/improper manner....\r\n\r\nAlbania is STILL up for grabs...", "Solution_17": "Antarctica \r\nAustralia \r\nUnited States of America \r\nAlaska \r\nAnn Arbor \r\nRaleigh \r\nHaiti \r\nItaly \r\nYugoslavia \r\nAustria \r\nAntilles \r\nSaskatchewan \r\nNetherlands \r\nSri Lanka \r\nAbkhazia \r\nArgentina \r\nAtlanta \r\nAlabama \r\nAcapulco \r\nOrlando \r\nOman \r\nNairobi \r\nIceland \r\nDenmark \r\nKenya \r\nAlbania\r\nAfghanistan\r\n\r\n\r\nSorry, I just couldn't wait for someone else to post.", "Solution_18": "Antarctica \r\nAustralia \r\nUnited States of America \r\nAlaska \r\nAnn Arbor \r\nRaleigh \r\nHaiti \r\nItaly \r\nYugoslavia \r\nAustria \r\nAntilles \r\nSaskatchewan \r\nNetherlands \r\nSri Lanka \r\nAbkhazia \r\nArgentina \r\nAtlanta \r\nAlabama \r\nAcapulco \r\nOrlando \r\nOman \r\nNairobi \r\nIceland \r\nDenmark \r\nKenya \r\nAlbania \r\nAfghanistan \r\n[b]Norway[/b]", "Solution_19": "Antarctica\r\nAustralia\r\nUnited States of America\r\nAlaska\r\nAnn Arbor\r\nRaleigh\r\nHaiti\r\nItaly\r\nYugoslavia\r\nAustria\r\nAntilles\r\nSaskatchewan\r\nNetherlands\r\nSri Lanka\r\nAbkhazia\r\nArgentina\r\nAtlanta\r\nAlabama\r\nAcapulco\r\nOrlando\r\nOman\r\nNairobi\r\nIceland\r\nDenmark\r\nKenya\r\nAlbania\r\nAfghanistan\r\nNorway\r\n[b]Yemen[/b]", "Solution_20": "Antarctica \r\nAustralia \r\nUnited States of America \r\nAlaska \r\nAnn Arbor \r\nRaleigh \r\nHaiti \r\nItaly \r\nYugoslavia \r\nAustria \r\nAntilles \r\nSaskatchewan \r\nNetherlands \r\nSri Lanka \r\nAbkhazia \r\nArgentina \r\nAtlanta \r\nAlabama \r\nAcapulco \r\nOrlando \r\nOman \r\nNairobi \r\nIceland \r\nDenmark \r\nKenya \r\nAlbania \r\nAfghanistan \r\nNorway \r\nYemen\r\n[b]Nebraska[/b]", "Solution_21": "Antarctica\r\nAustralia\r\nUnited States of America\r\nAlaska\r\nAnn Arbor\r\nRaleigh\r\nHaiti\r\nItaly\r\nYugoslavia\r\nAustria\r\nAntilles\r\nSaskatchewan\r\nNetherlands\r\nSri Lanka\r\nAbkhazia\r\nArgentina\r\nAtlanta\r\nAlabama\r\nAcapulco\r\nOrlando\r\nOman\r\nNairobi\r\nIceland\r\nDenmark\r\nKenya\r\nAlbania\r\nAfghanistan\r\nNorway\r\nYemen\r\nNebraska\r\n[b]Azerbaijan[/b]", "Solution_22": "[hide=\"We need lives...\"]Antarctica\nAustralia\nUnited States of America\nAlaska\nAnn Arbor\nRaleigh\nHaiti\nItaly\nYugoslavia\nAustria\nAntilles\nSaskatchewan\nNetherlands\nSri Lanka\nAbkhazia\nArgentina\nAtlanta\nAlabama\nAcapulco\nOrlando\nOman\nNairobi\nIceland\nDenmark\nKenya\nAlbania\nAfghanistan\nNorway\nYemen\nNebraska\nAzerbaijan[/hide]\r\n\r\n[b]Nicaragua[/b]", "Solution_23": "Antarctica \r\nAustralia \r\nUnited States of America \r\nAlaska \r\nAnn Arbor \r\nRaleigh \r\nHaiti \r\nItaly \r\nYugoslavia \r\nAustria \r\nAntilles \r\nSaskatchewan \r\nNetherlands \r\nSri Lanka \r\nAbkhazia \r\nArgentina \r\nAtlanta \r\nAlabama \r\nAcapulco \r\nOrlando \r\nOman \r\nNairobi \r\nIceland \r\nDenmark \r\nKenya \r\nAlbania \r\nAfghanistan \r\nNorway \r\nYemen \r\nNebraska \r\nAzerbaijan\r\nNicaragua\r\n[b]Aruba[/b]", "Solution_24": "Antarctica \r\nAustralia \r\nUnited States of America \r\nAlaska \r\nAnn Arbor \r\nRaleigh \r\nHaiti \r\nItaly \r\nYugoslavia \r\nAustria \r\nAntilles \r\nSaskatchewan \r\nNetherlands \r\nSri Lanka \r\nAbkhazia \r\nArgentina \r\nAtlanta \r\nAlabama \r\nAcapulco \r\nOrlando \r\nOman \r\nNairobi \r\nIceland \r\nDenmark \r\nKenya \r\nAlbania \r\nAfghanistan \r\nNorway \r\nYemen \r\nNebraska \r\nAzerbaijan \r\nNicaragua \r\nAruba\r\n[b]Amsterdam[/b]", "Solution_25": "Antarctica \r\nAustralia \r\nUnited States of America \r\nAlaska \r\nAnn Arbor \r\nRaleigh \r\nHaiti \r\nItaly \r\nYugoslavia \r\nAustria \r\nAntilles \r\nSaskatchewan \r\nNetherlands \r\nSri Lanka \r\nAbkhazia \r\nArgentina \r\nAtlanta \r\nAlabama \r\nAcapulco \r\nOrlando \r\nOman \r\nNairobi \r\nIceland \r\nDenmark \r\nKenya \r\nAlbania \r\nAfghanistan \r\nNorway \r\nYemen \r\nNebraska \r\nAzerbaijan \r\nNicaragua \r\nAruba \r\nAmsterdam\r\n[b]Mongolia[/b]", "Solution_26": "Antarctica \r\nAustralia \r\nUnited States of America \r\nAlaska \r\nAnn Arbor \r\nRaleigh \r\nHaiti \r\nItaly \r\nYugoslavia \r\nAustria \r\nAntilles \r\nSaskatchewan \r\nNetherlands \r\nSri Lanka \r\nAbkhazia \r\nArgentina \r\nAtlanta \r\nAlabama \r\nAcapulco \r\nOrlando \r\nOman \r\nNairobi \r\nIceland \r\nDenmark \r\nKenya \r\nAlbania \r\nAfghanistan \r\nNorway \r\nYemen \r\nNebraska \r\nAzerbaijan \r\nNicaragua \r\nAruba \r\nAmsterdam \r\nMongolia\r\n[b]Albuquerque [/b]", "Solution_27": "[quote=\"ASPEED\"]Antarctica \nAustralia \nUnited States of America \nAlaska \nAnn Arbor \nRaleigh \nHaiti \nItaly \nYugoslavia \nAustria \nAntilles \nSaskatchewan \nNetherlands \nSri Lanka \nAbkhazia \nArgentina \nAtlanta \nAlabama \nAcapulco \nOrlando \nOman \nNairobi \nIceland \nDenmark \nKenya \nAlbania \nAfghanistan \nNorway \nYemen \nNebraska \nAzerbaijan \nNicaragua \nAruba \nAmsterdam \nMongolia\n[b]Albuquerque [/b][/quote]\r\nEraguay", "Solution_28": "[hide]Antarctica \nAustralia \nUnited States of America \nAlaska \nAnn Arbor \nRaleigh \nHaiti \nItaly \nYugoslavia \nAustria \nAntilles \nSaskatchewan \nNetherlands \nSri Lanka \nAbkhazia \nArgentina \nAtlanta \nAlabama \nAcapulco \nOrlando \nOman \nNairobi \nIceland \nDenmark \nKenya \nAlbania \nAfghanistan \nNorway \nYemen \nNebraska \nAzerbaijan \nNicaragua \nAruba \nAmsterdam \nMongolia\nAlbuquerque\nEraguay\n[/hide]\r\n\r\nI now invoke martial law; hide tags everybody. \r\n\r\nAnd you meant Uruguay :roll:", "Solution_29": "Since Eraguay isn't an actual place, I'm going back to Albuquerque. (Actually, I've never been to Albuquerque... :rotfl: )\r\n\r\n[hide]Antarctica \nAustralia \nUnited States of America \nAlaska \nAnn Arbor \nRaleigh \nHaiti \nItaly \nYugoslavia \nAustria \nAntilles \nSaskatchewan \nNetherlands \nSri Lanka \nAbkhazia \nArgentina \nAtlanta \nAlabama \nAcapulco \nOrlando \nOman \nNairobi \nIceland \nDenmark \nKenya \nAlbania \nAfghanistan \nNorway \nYemen \nNebraska \nAzerbaijan \nNicaragua \nAruba \nAmsterdam \nMongolia \nAlbuquerque\n[b]El Salvador[/b][/hide]" } { "Tag": [ "percent", "algebra", "system of equations" ], "Problem": "John now has two investments that produce a 150 dollars income each month. If\r\n1,000 dollars more is invested at 9 percent than at 10 percent per year, how much was invested\r\nat each percent? There were 12 months in that year.\r\n\r\nI just need to know how to change this from English to Algebra?", "Solution_1": "[quote=\"sharkman\"]John now has two investments that produce a 150 dollars income each month. If\n1,000 dollars more is invested at 9 percent than at 10 percent per year, how much was invested\nat each percent? There were 12 months in that year.\n\nI just need to know how to change this from English to Algebra?[/quote]\r\n\r\nEdit: Wait that is inconsistent, sorry. I was in a hurry\r\n\r\nBut I think the problem has extra information that gives no solution to this problem...", "Solution_2": "I am not so clear on this. Does this cover everything that the problem \r\nis asking? That is why I am stuck. Please, explain a little further as \r\nmy bulb is not to bright.\r\n\r\nThank you so much for your time.", "Solution_3": "The solution given makes sense. :maybe:", "Solution_4": "What is the answer?", "Solution_5": "[quote=\"sharkman\"]John now has two investments that produce a 150 dollars income each month.[/quote]\n\nIs this [i]on average[/i] over the course of one year?\n\n[quote=\"sharkman\"]There were 12 months in that year.[/quote]\r\n\r\nBecause this seems to suggest that that is the correct interpretation, but the problem is quite unclear.", "Solution_6": ":huh: I just looked at kohjhsd's answer and it looked correct, thought once I did the problem it does not seem to work, the system of equations did not yeild consistent results.\r\nAssuming that it is an average of 150 a month.", "Solution_7": "I'm sleepy but here's a shot:\r\n[hide]Let $x$ be the amount invested at 10 percent. We have:\n$\\frac{(x+1000)(1.09)+x(1.1)}{12}= 150 \\implies (x+1000)(1.09)+x(1.1) = 1800$\n$\\implies 2.19x = 710 \\implies x \\approx 324.2$. Then we have approximately $\\$324.20$ invested at $10\\%$ and $\\$1,324.20$ invested at $9\\%$.[/hide]", "Solution_8": "Oh, so 150 dollars was sum of the income of each investments :blush:" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Find the integral.\r\n$ \\int{\\frac{1}{1\\plus{}x^4}dx}$", "Solution_1": "Just so it's known, there was another topic on evaluating this indefinite integral [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200964]here[/url].", "Solution_2": "[quote=\"Patterns_34\"]Just so it's known, there was another topic on evaluating this indefinite integral [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200964]here[/url].[/quote]\r\nThank you.\r\nBut how I can use completing square to solve it?", "Solution_3": "Some application of elementary complex shows that if we put $ \\theta_k = \\tfrac{\\pi(2k + 1)}{n}$, then\r\n\r\n\\begin{eqnarray*} \\frac {1}{1 + x^n} & = & - \\frac {1}{n} \\sum_{k = 1}^{n} \\frac {x \\cos \\theta_k - 1}{x^2 - 2x \\cos \\theta_k + 1} \\\\\r\n& = & - \\frac {1}{n} \\sum_{k = 1}^{n} \\left[ \\frac {\\cos \\theta_k}{2} \\cdot \\frac {2 x - 2 \\cos \\theta_k}{x^2 - 2x \\cos \\theta_k + 1} - \\sin \\theta_k \\cdot \\frac {\\sin \\theta_k}{(x - \\cos \\theta_k)^2 + \\sin^2 \\theta_k} \\right], \\end{eqnarray*}\r\n\r\nNote that each summand is now integrable.\r\n\r\n\r\n[quote=\"Patterns_34\"]Just so it's known, there was another topic on evaluating this indefinite integral [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200964]here[/url].[/quote]\r\n\r\nAnyway, the method hash_include used is very impressive...!", "Solution_4": "[quote=\"sos440\"]Some application of elementary complex shows that if we put $ \\theta_k = \\tfrac{\\pi(2k + 1)}{n}$, then\n\n\\begin{eqnarray*} \\frac {1}{1 + x^n} & = & - \\frac {1}{n} \\sum_{k = 1}^{n} \\frac {x \\cos \\theta_k - 1}{x^2 - 2x \\cos \\theta_k + 1} \\\\\n& = & - \\frac {1}{n} \\sum_{k = 1}^{n} \\left[ \\frac {\\cos \\theta_k}{2} \\cdot \\frac {2 x - 2 \\cos \\theta_k}{x^2 - 2x \\cos \\theta_k + 1} - \\sin \\theta_k \\cdot \\frac {\\sin \\theta_k}{(x - \\cos \\theta_k)^2 + \\sin^2 \\theta_k} \\right], \\end{eqnarray*}\n\nNote that each summand is now integrable.\n\n\n[quote=\"Patterns_34\"]Just so it's known, there was another topic on evaluating this indefinite integral [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200964]here[/url].[/quote]\n\nAnyway, the method hash_include used is very impressive...![/quote]\r\nIs there a way you can do it which only need to use basic integration formulas to solve this specific question, instead to n's power?" } { "Tag": [ "USAMTS", "function" ], "Problem": "Hi,\r\nI am new to this site and to this competition. It is also very late, so I hope you will forgive me if I violate some sort of protocol in the way I posted this.\r\nI had a question about the USAMTS.\r\n\r\nOne of my solutions (completely ugly, but the best I could do) has a LOT of checking solutions. For that, can I simply say \"checked by TI-89\" and include as an appendix the precise formula I entered into the calculator, and get credit for the problem? That is, can I just say, \"we test the following values into the following equation looking for this, and, if we use a calculator in the following manner, we get these results\" and get credit for it? OTherwise, I'm probably going to have 4-6 pages to an otherwise 1.5 page problem. \r\nI was not sure if the section in the FAQ regarding computers was totally applicable.", "Solution_1": "That should be fine - as long as you are clear how you did the checking and that you covered all possibilities, it is acceptable under current rules.", "Solution_2": "Thank you. One more question: How important are Lemma's? I'm not sure how to use them, and really not sure if I should be starting to learn how to use them now.\r\n\r\nThanks\r\nsran.", "Solution_3": "[quote=\"sran\"]I had a question about the USAMTS.\n\nOne of my solutions (completely ugly, but the best I could do) has a LOT of checking solutions. For that, can I simply say \"checked by TI-89\" and include as an appendix the precise formula I entered into the calculator, and get credit for the problem? That is, can I just say, \"we test the following values into the following equation looking for this, and, if we use a calculator in the following manner, we get these results\" and get credit for it? Otherwise, I'm probably going to have 4-6 pages to an otherwise 1.5 page problem.\nI was not sure if the section in the FAQ regarding computers was totally applicable.[/quote]\n\nIf you wrote four pages of checking all cases, we graders would just scan the first few entries to check whether your test was an effective one. We would not read all four pages. While there is nothing bad about writing out all cases, except for paying extra postage to mail your submission, you may instead just show a few cases as examples of your test and then say that you continued in a like manner for the rest of the cases.\n\nAlas, if your answer is as ugly as you say, you won't get a commendation. Often, once a mathematician answers a problem by difficult and tedious methods, the answer itself suggests a more elegant solution. Perhaps if you ponder your answer, you could find an elegant solution and earn a commendation.\n\n[quote=\"sran\"]Thank you. One more question: How important are Lemma's? I'm not sure how to use them, and really not sure if I should be starting to learn how to use them now.[/quote]\r\n\r\nLemmas are useful if your argument has a big side proof that is not part of the main flow of the proof. It makes the explanation clearer to set it apart as a lemma. For example, in problem 1/1/15 of last year's USAMTS, I could say, \"Lemma: The function f is constant over all real numbers in the half-open interval [i/1000,(i+1)/1000) for all integers i.\", and then prove that lemma before returning to the main proof. You do not have to call it a lemma. I myself often write such a side proof in one paragraph, beginning with the sentence, \"I claim that truncating the value of x to three decimal places does not change the value of f(x).\", and writing the proof as the rest of the paragraph.\r\n\r\nRichard Rusczyk & Mathew Crawford provide a much more detailed description of solution-writing techniques with better examples at:\r\nhttp://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php\r\n\r\nErin Schram\r\nUSAMTS grader" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Find the exact length of the polar curve described by:\r\nr = 5 e^(-t)\r\non the interval (5/4)pi to 9pi\r\n\r\nI got:\r\n\u222b \u221a(r\u00b2 + (dr/d\u03b8)\u00b2) d\u03b8\r\n\u222b \u221a(25e^(2\u03b8) + 25e^(2\u03b8)) d\u03b8\r\n\u222b \u221a(2*25e^2\u03b8) d\u03b8\r\n\u222b \u221a2 * 5e^\u03b8 d\u03b8\r\n\u221a2 * 5e^\u03b8 {5\u03c0/4 to 9\u03c0}\r\n5\u221a2 * (1.903*10\u00b9\u00b2 - 50.75)\r\nabout equal to:\r\n 1.345 * 10\u00b9\u00b3\r\n\r\nwhich is wrong.\r\n\r\nAny help?? Thank you!!!!\r\n\r\nKent Merryfield:\r\nThat's what I had at first, but this gives a negative number (about -0.13932) and I figured negative number here didn't make sense.", "Solution_1": "How did you get that answer?", "Solution_2": "The curve spirals in, not out; the arclength should be short, not long.\r\n\r\nNote that if $ r\\equal{}5e^{\\minus{}\\theta},$ then $ \\frac{dr}{d\\theta}\\equal{}\\minus{}5e^{\\minus{}\\theta}$ and $ \\left(\\frac{dr}{d\\theta}\\right)\\equal{}25e^{\\minus{}2\\theta}.$\r\n\r\n$ r^2\\plus{}\\left(\\frac{dr}{d\\theta}\\right)\\equal{}50e^{\\minus{}2\\theta}.$\r\n\r\n$ \\sqrt{r^2\\plus{}\\left(\\frac{dr}{d\\theta}\\right)}\\equal{}5\\sqrt{2}\\,e^{\\minus{}\\theta}.$\r\n\r\nYou seem to have committed an exponent sign error in squaring.", "Solution_3": "The definite integral of a positive function must be positive. If you're getting something negative, then you're doing it wrong.\r\n\r\n$ \\int_{\\frac{5\\pi}4}^{9\\pi}5\\sqrt{2}\\,e^{\\minus{}\\theta}\\,d\\theta\\equal{} \\minus{}5\\sqrt{2}e^{\\minus{}\\theta}\\bigg{|}_{\\frac{5\\pi}4}^{9\\pi}\\equal{} 5\\sqrt{2}\\left(e^{\\minus{}5\\pi/4}\\minus{}e^{\\minus{}9\\pi}\\right)\\approx 0.139320351.$\r\n\r\nPerhaps you missed the minus sign that comes from the chain rule?\r\n\r\nNote also that the contribution of $ e^{\\minus{}9\\pi}$ is negligible. To 10-place accuracy this is the same as the arclength of the curve for $ \\theta\\in\\left[\\frac{5\\pi}4,\\infty\\right).$" } { "Tag": [ "geometry", "3D geometry", "sphere", "topology", "Functional Analysis", "real analysis", "advanced fields" ], "Problem": "Let $X$ be a real Banach space, $X^{*}$ its dual. By the Hahn-Banach theorem for any element $x$ of the unit sphere $S_{X}$ there exists $x^{*}\\in S_{X^{*}}$ such that $\\langle x^{*},x\\rangle=1$. In general we cannot choose $x^{*}$ so that it continuously depends on $x$ - consider $\\ell_{\\infty}$ or even its two-dimensional version $\\ell_{\\infty}^{2}$. Let's give ourselves a little more room in choosing $x^{*}$.\r\n\r\nProve: for any $\\delta\\in (0,1)$ there exists a norm-to-norm continuous map $\\phi\\colon S_{X}\\to S_{X^{*}}$ such that $\\langle \\phi(x),x\\rangle\\ge \\delta$ for all $x\\in S_{X}$.", "Solution_1": "[hide=\"Hint\"]The sets $U_{x^{*}}: =\\{x\\in S_{X}\\colon \\langle x^{*},x\\rangle>\\delta\\}$, where $x^{*}$ ranges over $S_{X^{*}}$, form an open covering of the metric space $S_{X}$.[/hide]", "Solution_2": "Alas, this is not too much of a hint for me :(. If there were a locally finite continuous partition of unity associated with this covering, it would finish the story in no time but I do not see how to get such a partition in general. What is clear is that it would suffice to take $X=\\ell^{\\infty}(S)$ where $S$ is some set. Then the covering mlok mentioned can be reduced to the covering by the sets $U^{\\pm}_{s}=\\{x\\,: \\,\\pm x_{s}>\\delta\\}$ but I do not even see how to make a desired partition of unity associated with this simpler covering even in the case when $S$ is countable. Seems like I'm missing something simple here :maybe: :blush: . Anyway, this was a very nice question to ask. Even if the solution for the \"continuous\" version turns out to be easy, one can then ask about the existence of a \"uniformly continuous\" or even \"Lipschitz\" mapping with these properties.", "Solution_3": "Isn't it true that for any open cover of a paracompact space there is a locally finite continuous partition of unity subordinated to it? :maybe: So says Proposition B.2 in the book I'm currently reading (\"Geometric nonlinear functional analysis\" by Benyamini and Lindenstrauss).\r\n\r\nI've been working on the uniformly continuous case for a while. It's easy to prove that \\[X\\text{ is superreflexive}\\implies \\exists \\phi\\in UC(S_{X})\\implies X\\text{ has type }>1\\] It's not easy to decide which converse fails (I might have a counterexample for the first one :maybe: ) Similarly, \r\n\\[X \\text{ is 2-smoothable}\\implies \\exists \\phi\\in Lip(S_{X})\\implies X\\text{ has type }=2\\] and this time I have no idea about either converse.", "Solution_4": "Yes, you're right. Stone's theorem... Seems I should refresh my memories of general topology :oops: :P." } { "Tag": [ "geometry", "function", "integration", "real analysis", "calculus", "calculus computations" ], "Problem": "I have just learned about calculating the area under the curve bounded by x-axis using Riemann Sum. And I'm stuck with the following problem:\r\nUsing Riemann Sum, calculating the area bounded by the curve and x-axis with x from 1 to 2: $ y\\equal{}1/x^2$.\r\n[b]Approach[/b]:\r\nI used the Riemann Sum and need to calculate the limit of $ 1/n. sigma(1/(1\\plus{}i/n)^2)$ with i from 1 to n. Probably it has something to do with the Riemann Zeta function, but I don't know how to calculate this limit.Wait! I used Squezee theorem and get the limit is 1? It should be 1/2 though! :(", "Solution_1": "$ U_n\\equal{}\\frac{1}{n}\\sum_{k\\equal{}0}^{n\\minus{}1} f(1\\plus{}\\frac{k}{n}) \\equal{}\\frac{1}{n}\\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{n^2}{(n\\plus{}k)^2} \\equal{}n\\sum_{k\\equal{}0}^{n\\minus{}1} \\frac{1}{(n\\plus{}k)^2}\\equal{} n\\sum_{k\\equal{}n}^{2n\\minus{}1}k^{\\minus{}2}$\r\n$ \\equal{}n[ \\zeta(2,n)\\minus{}\\zeta(2,2n)] \\equal{}\\frac{n}{\\Gamma (2)}\\left[ \\int_0^\\infty \\frac{t dt}{e^{nt}(1\\minus{}e^{\\minus{}t})}\\minus{}\\int_0^\\infty \\frac{t dt}{e^{2nt}(1\\minus{}e^{\\minus{}t})}\\right]$\r\n\r\n$ \\equal{}n\\int_0^\\infty \\frac{t}{1\\minus{}e^{\\minus{}t}}(e^{\\minus{}nt}\\minus{}e^{\\minus{}2nt})dt$ .. which looks fun\r\n\r\nbut then again.. if we were trying to find a better way to integrate $ \\frac{1}{x^2}$ we really shouldnt be looking at harder integrals.. should we?" } { "Tag": [], "Problem": "b\u1ea1n n\u00e0o c\u00f3 s\u00e1ch v\u1ec1 d\u1ed3n bi\u1ebfn, p\u00f3t l\u00ean cho minh v\u1edbi.\r\nt\u00f4i c\u1ea3m \u01a1n nhi\u1ec1u :roll:", "Solution_1": "[quote=\"nam_tran1237\"]b\u1ea1n n\u00e0o c\u00f3 s\u00e1ch v\u1ec1 d\u1ed3n bi\u1ebfn, p\u00f3t l\u00ean cho minh v\u1edbi.\nt\u00f4i c\u1ea3m \u01a1n nhi\u1ec1u :roll:[/quote]\r\nB\u1ea1n l\u00ean DDTH t\u00ecm b\u00e0i c\u1ee7a anh kimluan, c\u00f3 1 file v\u1ec1 d\u1ed3n bi\u1ebfn r\u1ea5t hay. D\u1ed3n bi\u1ebfn c\u00f3 nhi\u1ec1u s\u00e1ch l\u1eafm, b\u1ea1n c\u00f3 th\u1ec3 tham kh\u1ea3o STBDT c\u1ee7a anh PKH ho\u1eb7c SLVKP c\u1ee7a th\u1ea7y Thu\u1eadn. \u0110\u00e2y l\u00e0 pp d\u1ed3n bi\u1ebfn to\u00e0n mi\u1ec1n v\u00e0 pp d\u1ed3n bi\u1ebfn m\u1ea1nh :" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For $ abc\\equal{}1$, $ a, b, c>0$ prove the inequality:\r\n\r\n$ \\frac{a^{2}\\plus{}b^{2}}{c^{2}\\plus{}a\\plus{}b}\\plus{}\\frac{b^{2}\\plus{}c^{2}}{a^{2}\\plus{}b\\plus{}c}\\plus{}\\frac{c^{2}\\plus{}a^{2}}{b^{2}\\plus{}c\\plus{}a} \\geq 2$", "Solution_1": "Change $ a \\equal{} x^{3}$, $ b \\equal{} y^{3}$ and $ c \\equal{} z^{3}$. Obviously, $ x,y,z > 0$ and $ xyz \\equal{} 1$(I will use a,b,c instead of x,y,z and this a,b,c won't be the same as in the given condition). So we have to prove \r\n $ \\sum \\frac {a^{6} \\plus{} b^{6}}{c^{6} \\plus{} a^{4}bc \\plus{} ab^{4}c} \\geq 2$. If we use Cauchy-Shwarz inequality on fractions, we get that the LHS is more or equal to $ \\frac {4(a^{6} \\plus{} b^{6} \\plus{} c^{6})^{2}}{2(\\sum a^{6}b^{6}) \\plus{} 2(\\sum a^{10}bc) \\plus{} 2(a^{7}b^{4}c \\plus{} a^{4}b^{7}c \\plus{} ab^{4}c^{7} \\plus{} ab^{7}c^{4} \\plus{} a^{7}bc^{4} \\plus{} a^{4}bc^{7)}}$. So we need to prove that $ 2(\\sum a^{12}) \\plus{} 2(\\sum a^{6}b^{6})\\geq 2(\\sum a^{10}bc) \\plus{} (a^{7}b^{4}c \\plus{} a^{4}b^{7}c \\plus{} ab^{4}c^{7} \\plus{} ab^{7}c^{4} \\plus{} a^{7}bc^{4} \\plus{} a^{4}bc^{7})$.\r\nBy AM_GM inequality, we get: $ a^{12} \\plus{} 4a^{6}b^{6} \\plus{} a^{6}c^{6} \\geq 6a^{7}b^{4}c$. $ 4a^{12} \\plus{} a^{6}b^{6} \\plus{} a^{6}c^{6} \\geq 6a^{10}bc$. Writing the symmetric inequalities and adding, we get the desired result. Lasha Lakirbaia", "Solution_2": "...but how about this:\r\n\r\n$ \\frac{a^{2}\\plus{}b^{2}}{c^{2}\\plus{}a\\plus{}b}\\plus{}\\frac{b^{2}\\plus{}c^{2}}{a^{2}\\plus{}b\\plus{}c}\\plus{}\\frac{c^{2}\\plus{}a^{2}}{b^{2}\\plus{}c\\plus{}a} \\geq 2\\plus{} \\frac{2(a^{2}\\plus{}b^{2}\\plus{}c^{2}\\minus{}a\\minus{}b\\minus{}c)}{(a^{2}\\plus{}b^{2}\\plus{}c^{2})^{2}}$", "Solution_3": "How you got that?", "Solution_4": "I have used more powerful method (then Cauchy-Shwarz or AM_GM).", "Solution_5": "the solution could be finished in a slightly easier way, just as lasha did, to make it more convinient, we will consider $ a^3,b^3,c^3$ instead of $ a,b,c$.\r\n\r\nin each component of the sum, let's use the following inequality:\r\n\r\n$ \\frac{a^6\\plus{}b^6}{c(c^5\\plus{}a^4b\\plus{}ab^4)}\\geq\\frac{\\frac{a^6\\plus{}b^6}{c}}{a^5\\plus{}b^5\\plus{}c^5}$\r\n\r\nThe rest is a consequence of the Muirherd inequality.", "Solution_6": "[quote=\"Nak\"]...but how about this:\n\n$ \\frac {a^{2} \\plus{} b^{2}}{c^{2} \\plus{} a \\plus{} b} \\plus{} \\frac {b^{2} \\plus{} c^{2}}{a^{2} \\plus{} b \\plus{} c} \\plus{} \\frac {c^{2} \\plus{} a^{2}}{b^{2} \\plus{} c \\plus{} a} \\geq 2 \\plus{} \\frac {2(a^{2} \\plus{} b^{2} \\plus{} c^{2} \\minus{} a \\minus{} b \\minus{} c)}{(a^{2} \\plus{} b^{2} \\plus{} c^{2})^{2}}$[/quote]\r\nThe following inequality is also true\r\n$ \\frac {a^{2} \\plus{} b^{2}}{c^{2} \\plus{} a \\plus{} b} \\plus{} \\frac {b^{2} \\plus{} c^{2}}{a^{2} \\plus{} b \\plus{} c} \\plus{} \\frac {c^{2} \\plus{} a^{2}}{b^{2} \\plus{} c \\plus{} a} \\geq 2 \\plus{} \\frac {2(a^{2} \\plus{} b^{2} \\plus{} c^{2} \\minus{} a \\minus{} b \\minus{} c)}{3(a^{2} \\plus{} b^{2} \\plus{} c^{2})}$" } { "Tag": [ "probability", "counting", "distinguishability" ], "Problem": "8 dice are rolled. What is the probability that all of the numbers 1-6 appear at least once?", "Solution_1": "[quote=\"randomdragoon\"]8 dice are rolled. What is the probability that all of the numbers 1-6 appear at least once?[/quote]\r\n[hide]\n$1-\\frac{5}{6}^{8}$[/hide]?", "Solution_2": "[hide]\n$\\frac{\\binom{7}{5}}{\\binom{13}{5}}$[/hide]", "Solution_3": "both of the solutions are incorrect. bmps is just missing the parens i think. algosme, howd u get that???", "Solution_4": "(sorry, my 'g' key is broken, i have to copy and paste to make it work and i am too lazy for that)\r\n\r\nSo, i decided to think of each of the 6 numbers on the dice as slots and the 8 dice as balls. Whatever slot the ball drops in is the number the dice gets. This should wokr because each of the numbers appears with equal probability.\r\n\r\nSo it's the number of ways to have at least 1 ball in each slot over the number of ways to arrange the 8 balls randomly.", "Solution_5": "[hide]there are 28 ways to pick the individual dice for the rolls (123456xx, 12345x6x.. etc) and 6! ways to uniquely arrange the numbers 1 through 6 and a total number of 8^6 possabilities\n\ntherefore the answer is [b]6!(28)/8^6\n8^2*3*5*3*7/8^6\n315/8^4\n315/4096[/hide][/b]", "Solution_6": "[quote=\"jli\"]both of the solutions are incorrect. bmps is just missing the parens i think. algosme, howd u get that???[/quote]\r\nYa, It was a typo. Kinda obvious that it's a typo, my current answer is over one :D\r\nI meant\r\n[hide]$1-(\\frac{5}{6})^{8}$[/hide]\r\nWhy does everyone call me bmps?", "Solution_7": "It is bumpy.\r\n\r\nI don't think that is right the answer, because, you are saying 1- same number each time.\r\n\r\nThat doesn't inclued stuff like 11111112", "Solution_8": "[hide=\"my solution\"]\nThere are $\\binom{6}{2}=15$ different combinations of dice rolls that satisfy the conditions (1,2,3,4,5,6 and then any two other rolls). There are $\\binom{13}{5}=1287$ total combinations possible*. The probability is thus $\\frac{15}{1287}= \\frac{5}{429}$.\n\n\n*This fact is far from obvious... but the argument is similar to the balls-and-walls argument found at the end of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133029[/hide]", "Solution_9": "isn't it $\\binom{7}{2}$ since 4 separators enerate only 5 slots?", "Solution_10": "[quote=\"randomdragoon\"][hide=\"my solution\"]\nThere are $\\binom{6}{2}=15$ different combinations of dice rolls that satisfy the conditions (1,2,3,4,5,6 and then any two other rolls). There are $\\binom{13}{5}=1287$ total combinations possible*. The probability is thus $\\frac{15}{1287}= \\frac{5}{429}$.\n\n\n*This fact is far from obvious... but the argument is similar to the balls-and-walls argument found at the end of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133029[/hide][/quote]\r\n\r\nWould you explain how you got: $\\binom{6}{2}=15$?\r\nI think we need one more \"wall\"....", "Solution_11": "i keep on getting 7/11664...", "Solution_12": "[quote=\"usaha\"]i keep on getting 7/11664...[/quote]\r\n\r\ngreat......\r\n\r\num......\r\n\r\n\r\n\r\n\r\nwere not gonna know what your mistake was if you dont tell us how you did it", "Solution_13": "[quote=\"usaha\"]i keep on getting 7/11664...[/quote]\r\n\r\nYour denominator seems too large. Did you mistakenly assume that the dice are distinguishable? Otherwise I don't know how you can get a number so high.\r\n\r\nIn this problem, what's important is how many dice are getting what number, not which dice gets which number.", "Solution_14": "So what exactly is the answer?\r\n\r\nnow i get 7/429 because its just (7C5)/(13C5)", "Solution_15": "Oh right, my bad, it should be $\\binom{7}{2}$ in the numerator. :ewpu: \r\n\r\nusaha, correct :)" } { "Tag": [ "function", "logarithms", "calculus", "calculus computations" ], "Problem": "Can someone give me some hint on how to solve for t?\r\n\r\n$t = \\omega^{t}$\r\n\r\nI applied $ln$ to both sides but couldn't proceed.", "Solution_1": "It can't be solved completely in terms of the standard list of functions; the best we can do is something like $\\frac{\\ln t}{t}=\\ln \\omega$ or $t^{\\frac1t}=\\omega$. It isn't too hard to understand in this form." } { "Tag": [ "geometry", "rectangle", "perimeter", "trapezoid" ], "Problem": "1. The longest side of an isosceles right triangle has length $ \\sqrt [3]{2}$. Another side has length 2^{a}. Find the value $ a$ expressed as a common fraction.\r\n[hide]ANSWER: $ \\frac { \\minus{} 1}{6}$[/hide]\n\n2. Twelve square tiles, on inch on a side, are arranged in a rectangle, without overlapping. What is the number of inches in the sum of all possible distinct perimeters?\n[hide]ANSWER: 56[/hide]\n\n3. Which of these shapes will not tessellate the plane: an obtuse triangle, a trapezoid, a regular pentagon, or a regular hexagon?\n[hide]ANSWER: Regular Pentagon[/hide]", "Solution_1": "Problem 3 is ambiguous. If I gave you an arbitrary trapezoid or an arbitary obtuse triangle, it would not necessarily tesselate the plane.", "Solution_2": "[hide=\"I think I have Solution for 1\"]You can perceive that its an 45:45:90 triangle.\nTherefore formula can work.$ \\ (1: 1: \\sqrt {2})$\nSince $ \\sqrt {2}$ can be written as $ \\ 2^{\\frac 12}$ and so can $ \\sqrt [3]{2}$\nThus $ \\ 2^{\\frac12} * x \\equal{} 2^{\\frac 13}$\nThen you simplify to $ \\ x \\equal{} \\frac {2^{\\frac 13 \\minus{} \\frac 12}}2$ which give you \n$ \\boxed {x \\equal{} {\\frac { \\minus{} 1}6}}$[/hide]\n\n[hide=\"I think I have Solution for 2\"]You can decipher 12 tiles into 12 square units.\nSo, (1*12),(2*6),(4*3) are the pairs you can come up with.\nThus 2(13)+2(8)+2(7)= 26+16+14= [u][b]56 [/b][/u][/hide]\r\n\r\nOn question 3, what does tessellate mean??\r\n\r\nHope this helps.. :lol:" } { "Tag": [ "function", "induction", "trigonometry", "algebra unsolved", "algebra" ], "Problem": "Find all the continuous functions $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that:\r\n\r\n$ f(x\\plus{}y)\\plus{}f(x\\minus{}y)\\equal{}2f(x)f(y)$, for all $ x,y \\in \\mathbb{R}$", "Solution_1": "[quote=\"moldovan\"]Find all the continuous functions $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that:\n\n$ f(x \\plus{} y) \\plus{} f(x \\minus{} y) \\equal{} 2f(x)f(y)$, for all $ x,y \\in \\mathbb{R}$[/quote]\r\nLet $ P(x,y)$ the assertion $ f(x \\plus{} y) \\plus{} f(x \\minus{} y) \\equal{} 2f(x)f(y)$\r\n\r\n$ P(0,0)$ $ \\implies$ $ 2f(0) \\equal{} f(0)^2$ and so either $ f(0) \\equal{} 0$, either $ f(0) \\equal{} 1$\r\n\r\n$ f(0) \\equal{} 0$ and $ P(x,0)$ $ \\implies$ $ f(x) \\equal{} 0$ $ \\forall x$ and this is a first solution.\r\n\r\nSo we'll now consider $ f(0) \\equal{} 1$.\r\n$ P(0,x)$ $ \\implies$ $ f(x) \\plus{} f( \\minus{} x) \\equal{} 2f(x)$ and so $ f( \\minus{} x) \\equal{} f(x)$ and $ f(x)$ is even.\r\n$ P(x,x)$ $ \\implies$ $ f(2x) \\equal{} 2f(x)^2 \\minus{} 1$\r\n\r\nCase 1 : $ \\exists b$ such that $ f(b) < 1$.\r\n======\r\nSuppose now that $ \\exists b$ such that $ f(b) < 1$. Since $ f( \\minus{} x) \\equal{} f(x)$, wlog consider $ b > 0$.\r\nObviously, the sequence $ u_0 \\equal{} b < 1$, $ u_{n \\plus{} 1} \\equal{} 2u_n^2 \\minus{} 1$ contains negative numbers and so (since $ f(0) \\equal{} 1$ and $ f(x)$ is continuous), there exists $ c > 0$ such that $ f(c) \\equal{} 0$. So the set $ Z \\equal{} \\{x > 0$ such that $ f(x) \\equal{} 0\\}$ has an infimum \"$ a$\" and $ f(a) \\equal{} 0$ (since $ f(x)$ is continuous)\r\n\r\nSo we have $ f(0) \\equal{} 1$, $ f(a) \\equal{} 0$ and $ f(x) > 0$ $ \\forall x\\in[0,a)$\r\nThen $ P(x \\plus{} 2a,x)$ $ \\implies$ $ f(x \\plus{} 2a) \\equal{} \\minus{} f(x)$ and so $ f(x \\plus{} 4a) \\equal{} f(x)$ and so $ f(x)$ is a periodic function. \r\n\r\nIt's easy to show with induction over $ n$ (using $ P(\\frac {a}{2^n},\\frac {a}{2^n})$ and the fact that $ f(\\frac {a}{2^n})\\geq 0$ $ \\forall n\\geq 0$) that $ f(\\frac {a}{2^n}) \\equal{} \\cos(\\frac {\\pi}{2^{n \\plus{} 1}})$ $ \\forall n\\geq 0$\r\n\r\nIt's then easy to show with induction over $ p$ (using $ P(p\\frac {a}{2^n},\\frac {a}{2^n})$) that $ f(p\\frac {a}{2^n}) \\equal{} \\cos(p\\frac {\\pi}{2^{n \\plus{} 1}})$ $ \\forall n\\geq 0$ and $ \\forall p\\in[0,2^n]$\r\n\r\nAnd, since the set $ \\{p\\frac {a}{2^n}$, $ \\forall n\\geq 0$ and $ \\forall p\\in[0,2^n]\\}$ is dense in $ [0,a]$ and since $ f(x)$ is continuous :\r\n\r\n$ f(x) \\equal{} \\cos(\\frac {\\pi x}{2a})$ $ \\forall x\\in[0,a]$\r\nNow, since $ f(x \\plus{} 2a) \\equal{} \\minus{} f(x)$, since $ f(x \\plus{} 4a) \\equal{} f(x)$ and since $ f( \\minus{} x) \\equal{} f(x)$, we have :\r\n\r\n$ f(x) \\equal{} \\cos(\\frac {\\pi x}{2a})$ $ \\forall x\\in\\mathbb R$\r\n\r\nand it's easy to check that this function matches the original equation.\r\n\r\nCase 2 : $ f(x)\\geq 1$ $ \\forall x\\mathbb R$\r\n======\r\n$ f(1)\\geq 1$ and so $ \\exists u \\geq 0$ such that $ f(1) \\equal{} \\cosh(u)$\r\n\r\nIt's easy to show with induction over $ n$ (using $ P(\\frac {1}{2^n},\\frac {1}{2^n})$, and the fact that $ f(\\frac {1}{2^n})\\geq 1$ $ \\forall n\\geq 0$) that $ f(\\frac {1}{2^n}) \\equal{} \\cosh(\\frac {u}{2^n})$ $ \\forall n\\geq 0$\r\n\r\nIt's easy to show with induction over $ n$ (using $ P(2^n,2^n)$, and the fact that $ f(2^n)\\geq 1$ $ \\forall n\\geq 0$) that $ f(2^n) \\equal{} \\cosh(u2^n)$ $ \\forall n\\geq 0$\r\n\r\nIt's then easy to show with induction over $ p$ (using $ P(p2^k,2^k)$) that $ f(p2^k) \\equal{} \\cosh(pu2^k)$ $ \\forall k\\in\\mathbb Z$ $ \\forall p\\in\\mathbb N$\r\n\r\nAnd, since the set $ \\{p2^k$, $ \\forall k\\in\\mathbb Z$ $ \\forall p\\in\\mathbb N\\}$ is dense in $ \\mathbb R^ \\plus{}$ and since $ f(x)$ is continuous :\r\n\r\n$ f(x) \\equal{} \\cosh(ux)$ $ \\forall x\\in\\mathbb R \\plus{}$\r\n\r\nAnd, since $ f(0) \\equal{} 1$ and $ f( \\minus{} x) \\equal{} f(x)$, $ f(x) \\equal{} \\cosh(ux)$ $ \\forall x\\in\\mathbb R$\r\nand it's easy to check that this function matches the original equation.\r\n\r\nSynthesis :\r\n=========\r\nAll solutions are :\r\n$ f(x) \\equal{} 0$\r\n$ f(x) \\equal{} \\cos(\\alpha x)$\r\n$ f(x) \\equal{} \\cosh(\\alpha x)$\r\n\r\n(and the solution $ f(x) \\equal{} 1$ is obtained thru $ \\alpha \\equal{} 0$)" } { "Tag": [ "inequalities", "vector", "calculus", "calculus computations" ], "Problem": "Consider the space ${\\mathbf{R}}^{n}$(for $n\\ge1$) with the three metrics:\r\n\r\n1) $d_{1}((x_{1},.....,x_{n}),(y_{1},.....,y_{n}))=\\sum_{i=1}^{n}{|x_{i}-y_{i}|}$\r\n\r\n2)$d_{2}((x_{1},.....,x_{n}),(y_{1},.....,y_{n}))=\\sqrt{\\sum_{i=1}^{n}{(x_{i}-y_{i})^{2}}}$\r\n\r\n3)$d_{\\infty}((x_{1},.....,x_{n}),(y_{1},.....,y_{n}))= max_{i=1,2,...n}{\\{|x_{i}-y_{i}|\\}}$\r\n\r\na) Show that there is a constant $c>1$ such that for any two points $x,y$ $\\in \\mathbf{R}$ the following inequalities are satisfied: $\\frac{1}{c}d_{1}(x,y) $ 4\\sin^{2}\\left(3x \\plus{} \\frac {\\pi}{4}\\right) \\equal{} 1 \\plus{} 8\\sin 2x\\cos^2 2x$----> $ 4*\\frac{1\\minus{}cos(6x\\plus{}\\frac{\\pi}{2})}{2} \\equal{} 1 \\plus{} 8\\sin 2x\\cos^2 2x$-----> \r\n\r\n$ 2(1\\plus{}sin6x)\\equal{}1 \\plus{} 8\\sin 2x\\cos^2 2x$ ---> $ 2(1\\plus{}3sin2x\\minus{}4sin^{3}2x)\\equal{}1 \\plus{} 8\\sin 2x\\cos^2 2x$ ---> $ sin2x\\equal{}a$---> $ 2(1\\plus{}3a\\minus{}4a^{3})\\equal{}1\\plus{}8a(1\\minus{}a^{2})$ ----> $ 2\\plus{}6a\\minus{}8a^{3}\\equal{}1\\plus{}8a\\minus{}8a^{3}$ ---> $ a\\equal{}\\frac{1}{2}$ ---> \r\n\r\n\r\n$ sin2x\\equal{}\\frac{1}{2}$ ---> $ x\\equal{}(\\minus{}1)^{n}\\frac{\\pi}{12}\\plus{}\\frac{\\pi n}{2}$\r\n$ sin(3x\\plus{}\\frac{\\pi}{4})\\geq 0$ ----> $ \\minus{}\\frac{\\pi}{12}\\plus{}\\frac{2\\pi k}{3}\\leq x \\leq \\frac{\\pi}{4}\\plus{}\\frac{2\\pi k}{3}$ -----> \r\n\r\n$ x\\equal{}\\frac{\\pi}{12}\\plus{}\\pi n$" } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "$\\lim_{x \\to 0}\\frac{tg^{3}x-tg x^{3}}{x^{5}}$", "Solution_1": "Use the power series expansion for the tangent function.\r\n\r\n[hide=\"Answer\"]$\\tan x = x+\\frac{x^{3}}{3}+O(x^{5})$ and so \n$\\tan^{3}x = \\left(x+\\frac{x^{3}}{3}+O(x^{5})\\right)^{3}= x^{3}+x^{5}+O(x^{7})$\n$\\tan(x^{3}) = x^{3}+O(x^{9})$.\n\nFinally,\n\n$\\lim_{x\\to0}\\frac{\\tan^{3}x-\\tan(x^{3})}{x^{5}}= \\lim_{x\\to0}\\frac{x^{3}+x^{5}+O(x^{7})-x^{3}-O(x^{9})}{x^{5}}$\n$= \\lim_{x\\to0}\\frac{x^{5}+O(x^{7})}{x^{5}}= \\lim_{x\\to0}[1+O(x^{2})] = 1$.[/hide]", "Solution_2": "i want a solution without power series please. i'm not supposed to use this in my homework", "Solution_3": "And we are not supposed to do your homework :P. Anyway, the power series technique is the standard one to use here. Of course, you can do L'Hopital 5 times (it'll be a great mess but it will formally satisfy your demand for \"no power series\"). Otherwise I don't know: maybe there is some clever trick, but I do not see one.", "Solution_4": "yea, i thought there was some clever possible trick ... but i guess there isn't. power series are just fine then :P thanks ..", "Solution_5": "Maybe this idea would work:\r\n\r\n$\\lim_{x\\rightarrow 0}\\frac{(\\tan x-x)}{x^{3}}\\lim_{x\\rightarrow 0}\\frac{\\tan^{2}x+x\\tan x+x^{2}}{x^{2}}+\\lim \\frac{x^{3}-\\tan x^{3}}{x^{5}}=\\frac{1}{3}\\cdot 3+0=1$, \r\n\r\nyou calculate the first limit by using l'Hopital rule and the last limit is $0$ in view of the same rule.", "Solution_6": "[quote=\"didilica\"]Maybe this idea would work:\n\n$\\lim_{x\\rightarrow 0}\\frac{\\tan x-x}{x^{3}}\\lim_{x\\rightarrow 0}\\frac{\\tan^{2}x+x\\tan x+x^{2}}{x^{2}}+\\lim \\frac{x^{3}-\\tan x^{3}}{x^{5}}=\\frac{1}{3}\\cdot 3+0=1$, \n\nyou calculate the first limit by using l'Hopital rule and the last limit is $0$ in view of the same rule.[/quote]\r\n\r\nhey... :o nice trick !" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Hello! I've just been released after doing time... in the *insert unmentionable* army.\r\nSo here's my first contribution to this forum. (Finally!)\r\n\r\nI came up with this question for one of our training sessions a few months back. No one solved it. i can only think of two reasons why this was been so. 1. 'Valiowk' was too busy explaining her math project to someone. AND/OR 2. I screwed up, and the question is wrong. I looked over it but it still looks correct. But maybe I'm blind or something.\r\nSo here it is:\r\n\r\n\r\nA math contest was held in the planet Zorgland. Zorgland has 3 species \u2013 The Arv, Benz, and Cren. In this particular contest, 14 Arvs, 14 Benzs and 14 Crens took part. Each solved at most 4 questions, and every distinct triplet of Arv, Benz, and Cren solved at least 1 question in common. Show that there is a question which 3 Arvs, 3 Benzs, and 3 Crens solved.", "Solution_1": "Um... Any comments?\r\nEven a 'Hey! Your so-called question is a cheap rip-off of IMO 2001/3!' would be nice.", "Solution_2": "Hey! Your so-called question is a cheap rip-off of IMO 2001/3!", "Solution_3": "I just knew someone would say that.\r\nIMO 2001/3 v2.0, now in 3-D!" } { "Tag": [ "inequalities", "quadratics", "inequalities unsolved" ], "Problem": ":D Good day.!\r\n\r\nThis is the problem. I believe this has been used already as an Olympiad Problem. I just don't know whether this one was used as a Nat'l or an Int'l Problem... Here goes...\r\n\r\nHelp please.\r\n\r\nFor which m does the inequality mx^2 - 4mx + m^2 + 2m - 3 > 0 hold for all x?", "Solution_1": "Find the discriminant of the quadratic . What happens when it is negative ?\r\n\r\nAlso notice that its $mx^{2}$ and not $x^{2}$ .", "Solution_2": "$mx^{2}-4mx+m^{2}+2m-3=0 \\Rightarrow m(x-2)^{2}+(m^{2}-2m-3) > 0$\r\ni) $m \\geq 0, m^{2}-2m-3 > 0 \\Rightarrow m > 3$\r\nii) $m < 0$ , let $k =-m$, we have $k(x-2)^{2}< k^{2}+2k-3 \\Rightarrow 0 < (x-2)^{2}< \\frac{k^{2}+2k-3}{k}=-\\frac{m^{2}-2m-3}{m}$ for all possible value of $x$, then\r\n$\\frac{m^{2}-2m-3}{m}\\geq 0 \\Rightarrow m <-1$, but $(x-2)^{2}$ can take any positive real value, so that for any $m <-1$, we can take $x$ sufficiently big to reverse the ineq..\r\n\r\nHence, the only possible value of $m$ is $m > 3$\r\n\r\nI'm sorry if I wrote some mistakes :maybe:", "Solution_3": "Thanks for the answer. :D I really hope it's correct. Thanks guys." } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "X -3 -2 1 3\r\n\r\nY 1 2 5 7\r\n\r\n\r\n\r\nY = ????\r\n\r\nContinue the equation", "Solution_1": "[quote=\"Sinistra\"]X -3 -2 1 3\n\nY 1 2 5 7\n\n\n\nY = ????\n\nContinue the equation[/quote]\r\n[hide]\nYou add 4 to $x$ to get $y$ so $y=x+4$[/hide]", "Solution_2": "Well Done! :D", "Solution_3": "[hide]\ni think if you add 4 to x you get y[/hide]", "Solution_4": "I don't get it. would someone explain it to me? how could you tell?", "Solution_5": "[hide]kBabe-Just compare the x's to the y's and a pattern is evident.\n\n$x+4=y$[/hide]", "Solution_6": "whoa... I'm totally guessing, but are they x and y numbers suppose to line up? if they are then are you suppose to find the equation in slope-intercept form? I'm going to do the problem base on what i think it means. \r\n\r\n[hide]to find the slope it's $\\frac{difference of y}{difference of x}$ , if you random pick two numbers you can get $\\frac{1}{1} = 1$ so the slope is $ 1$. Then to find the y-intercept, you should notice that the difference of x and y is always $4$... so if x is $0$ then y is $4$. so the equation of $y = x +4$\n[/hide]\r\nand i know I didn't really need to type all that... but I was kinda bored", "Solution_7": "I think the x and y he posted are supposed to be in a table.\r\n\r\n[u] x | y[/u]\r\n-3 |1\r\n-2 |2\r\n 1 |5\r\n 3 |7\r\n\r\nIts a pretty bad table but i think this should help...", "Solution_8": "[color=green]X+4=Y[/color]\r\n\r\n :D :D", "Solution_9": "[hide]\n-3+4=-1, -2+4=2...\nSo the answer is $\\framebox{y=x+4}$\n[/hide]" } { "Tag": [ "national olympiad" ], "Problem": "Which countries' olympiads/TSTs are of about IMO difficulty (harder is preferred, although a tad easier isn't too bad)?", "Solution_1": "USA, China, Russia, Japan, Romania TST's are definitely IMO level. a rule of thumb would be, if the country does consistently well at the IMO, then its TST's are comparable in difficulty to the IMO.", "Solution_2": "Russian Olympiad.\r\n\r\nHere are a few problems solved by 0 or 1 Russian contestants. Some of the Russian contestants got 0 points in the following problems, but got perfect scores in IMO.\r\n\r\n2002 grade 11 pr 4 --- 0\r\n2002 grade 11 pr 7\r\n2002 grade 11 pr 8\r\n2005 grade 11 pr 8 \r\n2008 grade 10 pr 4 --- 0\r\n2008 grade 11 pr 4 --- 0\r\n2008 grade 11 pr 8\r\n\r\n* 2005 grade 11 pr 8 was solved by only 1 contestant, Vasiliy Astakhov, who was the author of IMO 2007 pr 3.", "Solution_3": "I believe that China Olympiad is also very difficult, who knows maybe it's more difficult than IMO" } { "Tag": [ "geometry", "circumcircle", "geometry solved" ], "Problem": "The excircles of a triangle $ABC$ touch the sides $BC,CA, AB$ at $A',B',C'$ respectively. The circumcircles of triangles $AB'C',A'BC',A'B'C$ meet the circumcircle of triangle $ABC$ at $A_1,B_1,C_1$ respectively for the second time. Prove that the triangle $A_1B_1C_1$ is similar to triangle whose vertices are the points of contact of the incircle $ABC$ with its sides.", "Solution_1": "We now that $BC'=CB'=P-a$\r\n $angleAC'A_1=angleAB'A_1$ =>$angleA_1C'B=angleA_1B'C$ and \r\n $angleABA_1=angleACA_1=angleB'CA_1$ => triangle $BC'A_1$\r\n is similiar to triangle $B'CA_1$ but $BC'=CB'$ \r\n => $1=BC'/CB'=A_1B/A_1C$ => $A_1B=A_1C$ similiar $B_1C=B_1A$ and $C_1A=C_1B$\r\n THE END :D \r\nI think do you finishing this solution :)", "Solution_2": "See also http://www.mathlinks.ro/Forum/viewtopic.php?t=35313 .\r\n\r\n darij" } { "Tag": [ "geometry", "3D geometry", "analytic geometry", "algebra", "system of equations" ], "Problem": "These are from Vestavia 01\r\n\r\n1. How many values of x satisfy the equation: log(x^2)=cos(x)\r\n\r\n2. Find the value of x in the following system of equations:\r\n3x+2y+z=3\r\n3y+2q+w=2\r\n3z+2w+x=1\r\n3w+2x+y=0\r\n\r\n3. Two circles have an external tangent with length 36 cm. The shortest distance between teh circles is 14 cm. Find the radius, in cm, of the larger circle if its radius is four times that of the smaller circle.\r\n\r\n4. If A is the fourth perfect number, B is the least integer number greater than one which is both a cube and a square, and C ist he twelfth triangular number, find the value of sqr[(A/B)-C]\r\n\r\n5. The graph of 2x^2-21xy+29x-62y+10y^2=0 is a(n)....\r\n\r\n6. Find the volume of a hexahedron with vertices (1,0,5) (4,0,2) (0,0,0) (-1,-3,-1) (2,3,3)\r\n\r\n7. Find the sum of the abscissas of all (x,y) coordinates which satisfy\r\n2x^2+xy-y^2+3y-2=0 and 3x^2+3xy+y^2-6x-5y+4=0\r\n\r\n8. If f(n) = n^8-2n^7+4n^6-8n^5+16n^4-32n^3+64n^2-128n+1 then what is f(1/128)?\r\n\r\n9. Find the distance between teh point (5,1,3) and the line defined by the parametric equations\r\nx=3, y=7, and z=1+t\r\n\r\n10. Two ships leave port at the same time. One travels 20 mph at a bearing of N32degreesW. The second ship travels 30mph at a bearing of S28W. Find the distance in miles between the ships after 2 hours.\r\n\r\nSorry I didn't use LATEX.\r\n\r\nThese are all individual questions I am having trouble with.\r\n\r\nPlease post solution as soon as you know how to do one.", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174875[/url]" } { "Tag": [ "function", "calculus", "derivative", "algebra", "polynomial", "Vieta", "system of equations" ], "Problem": "$ f{(x)} \\equal{} {x}^3 \\minus{} 3{x}^2 \\plus{} 2{x}$ \r\n\r\nThe equation $ f{(x)} \\equal{} k$ has exactly one positive solution and exactly one negative solution. Find $ k$.\r\n\r\n\r\n\r\nHelp needed :D", "Solution_1": "[hide=\"hint\"]\nthere must be a multiplicity since there is not a solution of 0[/hide]\r\n\r\ndont know if that wil help though...\r\nbut there is a nice calc solution", "Solution_2": "like \r\n\r\n$ ({x}^3 \\minus{} 3{x}^2 \\plus{} 2{x}) \\equal{} (ax \\minus{} k1)(bx \\plus{} k2)$ ?\r\n\r\n(I don't know how to do subscripts for the k1 k2 in latex).\r\n\r\nIt probably isn't as the coefficients don't work out.", "Solution_3": "What's the nice calc solution?????", "Solution_4": "it involves derivatives (might be too hard for intermediate forum)\r\n\r\nbut in regard to the multiplicity hint (multiplicity means a number is a double root of a polynomial-- it is a factor of the polynomial more than once) , i meant for you to think about $ (x \\plus{} k_1)^2(x \\plus{} k_2) \\equal{} x^3 \\minus{} 3x^2 \\plus{} 2x \\minus{} k$\r\n\r\nthen expand it and match coefficients and get equations and solve and find k \r\n:D", "Solution_5": "[quote=\"JasonSullivan\"]$ f{(x)} = {x}^3 - 3{x}^2 + 2{x}$ \n\nThe equation $ f{(x)} = k$ has exactly one positive solution and exactly one negative solution. Find $ k$.\n\n\n\nHelp needed :D[/quote]\r\n\r\nFirst, $ k$ is not $ 0$ (we can check that because the roots of $ x(x - 1)(x - 2) = 0$ are $ \\{0, 1, 2\\}$).\r\n\r\nGiven that, we have an equation:\r\n\\[ x^3 - 3x^2 + 2x - k = 0\r\n\\]\r\nWe factor it with the three roots:\r\n\\[ (x - r_1)(x - r_2)(x - r_3) = 0\r\n\\]\r\nNext, we know that since there are at least two real roots, all the $ r_i$ have to be real because if there were a complex root, the sum $ r_1 + r_2 + r_3$ would be complex, which violates Vieta's formulas because the coefficient of the $ x^2$ term is real.\r\n\r\nSo the last root either must be one of the original roots or 0. Plugging in $ x = 0$ gives $ x^3 - 3x^2 + 2x - k = 0 \\implies - k = 0 \\implies k = 0$.\r\n\r\nContradiction? Contradiction! So the last root cannot be 0. It follows that one of the roots must have a multiplicity of 2.\r\n\r\nSo we have two roots, and the equation turns into $ (x - r_1)^2(x - r_2) = 0$.\r\n\\[ (x - r_1)(x - r_1)(x - r_2) = x^3 + ( - r_2 - 2r_1)x^2 + (r_1^2 + 2r_1r_2)x - r_1^2r_2 = 0\r\n\\]\r\nGuess what? We can now solve for $ r_1$ and $ r_2$ using this system of equations:\r\n\\begin{align*} - r_2 - 2r_1 & = - 3 \\\\\r\nr_1^2 + 2r_1r_2 & = 2\\end{align*}\r\n\r\nAnd once we have those values (there will be two solutions), we choose the pair that has one negative and one positive and let $ k = r_1^2r_2$.\r\n\r\n[color=darkred]Edit: I said let $ k = -r_1^2r_2$... actually it was supposed to be the positive of that, since we already had a $ - k$ in the original equation.[/color]", "Solution_6": "Thanks to you both. I know we are not supposed to post calculus here but I can do it and would appreciate the second take on the problem. Thanks. :D", "Solution_7": "[hide=\"deadly calculus bomb\"]\nthe derivative of $ x^3\\minus{}3x^2\\plus{}2x\\minus{}k$ is $ 3x^2\\minus{}6x\\plus{}2$ for all k and we want to find where the derivative is zero, meaning where there MAY ba an extrema\nso $ 3x^2\\minus{}6x\\plus{}2\\equal{}0 \\implies x\\equal{}1\\pm\\frac{\\sqrt3}{3}$\nwhen $ x\\equal{}1\\minus{}\\frac{\\sqrt3}{3}$ the function is at its maximum, and if we substitute $ \\minus{}f(x)$ for k, we well end up with 2 solutions, but they will both be positive\nso we want the other extrema\nso $ x\\equal{}1\\plus{}\\frac{\\sqrt3}{3},f(x)\\equal{}2\\plus{}\\frac{10\\sqrt3}{9} \\implies k\\equal{}\\minus{}2\\minus{}\\frac{10\\sqrt3}{9}$\n[/hide]", "Solution_8": "why sub -f(x) for k when k=f(x)?\r\n\r\notherwise seems simpler than the other way :D", "Solution_9": "[quote=\"stevenmeow\"][hide=\"deadly calculus bomb\"]\nthe derivative of $ x^3 - 3x^2 + 2x - k$ is $ 3x^2 - 6x + 2$ for all k and we want to find where the derivative is zero, meaning where there MAY ba an extrema\nso $ 3x^2 - 6x + 2 = 0 \\implies x = 1\\pm\\frac {\\sqrt3}{3}$\nwhen $ x = 1 - \\frac {\\sqrt3}{3}$ the function is at its maximum, and if we substitute $ - f(x)$ for k, we well end up with 2 solutions, but they will both be positive\nso we want the other extrema\nso $ x = 1 + \\frac {\\sqrt3}{3},f(x) = 2 + \\frac {10\\sqrt3}{9} \\implies k = - 2 - \\frac {10\\sqrt3}{9}$\n[/hide][/quote]\r\n\r\nUmm... that doesn't work. Solving $ x^3 - 3x^2 + 2x = -2 - \\frac {10\\sqrt 3}9$ yields only one real solution, $ -.787576765619...$, and two complex solutions.\r\n\r\nThe answer can be retrieved from my post above. If we solve this system of equations: \\begin{align*}-r_2 - 2r_1 &= -3 \\\\\r\nr_1^2 + 2r_1r_2 &= 2\\end{align*} We can substitute $ r_2 = 3 - 2r_1$ to get this equation: \\[ r_1^2 + 2r_1(3 - r_1) = 2\\] And solving for $ r_1$: \\begin{align*}r_1^2 + 2r_1(3 - 2r_1) &= 2 \\\\\r\nr_1^2 + 6r_1 - 4r_1^2 &= 2 \\\\\r\n-3r_1^2 + 6r_1 - 2 &= 0 \\\\\r\n3r_1^2 - 6r_1 + 2 &= 0\\end{align*}\r\nSo $ r_1 \\&= \\frac {6 \\pm \\sqrt {36 - 4 \\cdot 3 \\cdot 2}}6 = \\frac {3 \\pm \\sqrt 3}3$. Backsubstituting, this means that $ r_2 = 3 - 2r_1 = 3 - 2\\left(\\frac {3 \\pm \\sqrt 3}3\\right) = 3 - 2 \\mp \\frac {2\\sqrt 3}3 = \\frac {3 \\mp 2\\sqrt 3}3$.\r\n\r\nSo our two possible pairs of solutions $ (r_1, r_2)$ are: \\[ (r_1, r_2) = \\left(\\frac {3 + \\sqrt 3}3, \\frac {3 - 2\\sqrt 3}3\\right) \\approx (1.58, -0.15) \\\\\r\n(r_1, r_2) = \\left(\\frac {3 - \\sqrt 3}3, \\frac {3 + 2\\sqrt 3}3\\right) \\approx (0.42, 2.15)\\] So it's the first combination we want, so letting $ k = r_1^2r_2$ yields $ k = -\\frac {2\\sqrt 3}9$.\r\n\r\nWe can check by plugging that value of $ k$ back into the original equation. I leave this up to you." } { "Tag": [ "HMMT", "Euler" ], "Problem": "Okay, about the K\u00f6nigsberg bridge problem, I know that it is impossible to cross all seven bridges without \"repeating\" going over a bridge. I want to know why, I've searched on Google, but I don't really get it. I found two ways that you can find out that's it's impossible. Net Theory, and graphing. Could you explain one or both of them? If you want to see the problem I drew it (And if you're too lazy to look though that again to find the problem, you have to cross all bridges without going on one twice, you must only use the bridges to cross the water).", "Solution_1": "Theorem(I don't know if this theorem exists or not, but I know for a fact that this is true): You can completely draw a graph without taking your pencil off the paper if and only if the graph has 0 or 2 vertices with odd degrees.\r\n\r\nIn the K\u00f6nigsberg Bridge, 4 vertices have degrees of 3, 3, 3, and 5.\r\n\r\nSince there are 4 vertices with odd degrees, it is impossible to cross all bridges without repeating any of them.\r\n\r\nApplication of this theorem was in somewhere between #34 and #36 on HMMT Guts Round...\r\nCould've gotten it, but had no time... :(", "Solution_2": "Yes, that's true. :) If you have no odd vertices, you can start anywhere, and if you have two odd vertices, you have to start at one of them and end at another (why?).", "Solution_3": "[quote=\"math154\"]Yes, that's true. :) If you have no odd vertices, you can start anywhere, and if you have two odd vertices, you have to start at one of them and end at another (why?).[/quote]\r\n\r\nBlah, forgot to mention that. :|", "Solution_4": "We were trying to see which figures were traceable using a rule... apparently I wrote it wrong... :blush: Then we're going to apply the rule to the K\u00f6nigsberg bridge tomorrow", "Solution_5": "Caveat: For no-odd vertex graphs, you can start anywhere, but you must end at the exact same point as where you started.", "Solution_6": "Out of curiousity, does this theorem have a name?\r\n\r\nIf so, what is it?", "Solution_7": "I'm not sure, but these are called [url=http://en.wikipedia.org/wiki/Eulerian_path]Eulerian paths[/url].", "Solution_8": "It's one of Euler's things... I think it's called Euler's tour, or Eulerian Circuit.", "Solution_9": "How do you know the DEGREES, did you use a protractor or something (3 is way too small for those though), and aren't there more than 4 vertices. ANd how did you come up with the theorems???", "Solution_10": "The degrees means the number of rays that touch that vertex right?\r\nAnd there are exactly four vertexes. Look at the picture. :)\r\nAnd about the theorems, I don't know... >.> Anyways, the two theorems that I mentioned should be the same thing...", "Solution_11": "this problem was the very first graph problem ever solved", "Solution_12": "Well Euler found out why you couldn't go over the bridges which lead to the graphing, which he was the first to find out so this was/is the first problem that was solved by graphing....", "Solution_13": "Do U know about Graph Theory? :mad:", "Solution_14": "Uh because I learned it? *is getting off topic*" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let a,b,c>0 Prove that\r\n$ \\sum\\frac{(4a\\plus{}b\\minus{}c)^2}{2a^2\\plus{}(b\\plus{}c)^2} \\geq 8$", "Solution_1": "Do you have a proof for that? then please post it at http://www.mathlinks.ro/viewtopic.php?t=66493 .\r\n\r\n darij" } { "Tag": [ "geometry", "perimeter" ], "Problem": "The perimeter of an equilateral triangle exceeds the perimeter of a square by $1989$ cm. The length of each side of the triangle exceeds the length of each side of the square by $d$ cm. The square has perimeter greater than 0. How many positive integers are NOT possible values for $d$?", "Solution_1": "[hide]\nThe side length of the square is $s$. The perimeter of the triangle is $3(s+d)$, and the perimeter of the square is $4s$. We set up equations to solve for $d$ and $s$.\n\n$3(s+d)-4s=1989$\n$3s+3d-4s=1989$\n$3d-s=1989$\n\nSince the perimeter of the square is greater than zero, $3d$ must be greater than $1989$.\n\n$3d-0>1989$\n$3d>1989$\n$d>663$\n\nOtherwise $s$ would need to be a negative number, which is impossible. So $d$ must be larger than $663$, which means that $663-1+1$ = $663$ values do not work.[/hide]", "Solution_2": "What's the source of this problem?", "Solution_3": "ubemaya, i remember seeing this question before, it's one of the mid 20's on the 1989 ahsme i believe." } { "Tag": [], "Problem": "A positive integer is equal to the sum of the squares of its four smallest divisors. What is the largest prime that divides this integer. A solution, not just answer please.\r\n\r\nThanks.", "Solution_1": "Anyone? The answer is\r\n\r\n[hide=\"answer\"]$13$[/hide]\r\n\r\nbut I don't have a solution.", "Solution_2": "A really cheap solution would be to assume the number is small and brute force :D", "Solution_3": "That's what I would have done during the competition (because this is a free response answer and an integer between 1 to 999). Believe it or not, this is from the Junior Division so brute force it probably the way. But I still want a solution.", "Solution_4": "Junior section of what? 1-999 sounds like the AIME.\r\n\r\nNevertheless, I think I've gotten the solution. In my solution, I would be determining this number which is the sum of the squares of its four smallest positive divisors, hence finding its largest prime divisor.\r\n\r\nFirst, we have 1 as the smallest divisor of the number (I'd call it x).\r\n\r\nx cannot be odd, or all of its 4 smallest divisors would be odd, and the sum of their squares even.\r\n\r\nx must thus be even, and its second smallest divisor must be 2.\r\n\r\nNow we check if x is a multiple of 4. If x is a multiple of 4, the 4 smallest divisors must be 1, 2, 4, p where p is any odd prime. It cannot be 1, 2, 4, 8 as the sum of the squares of these 4 numbers is odd. Now we have 21 + p^2 = 4pk, where k is the product of the rest of the factors of x. p must thus be 3 or 7 (since p has to be a factor of 21, and must be one of the 4 smallest divisors). By checking these two cases, we realise that x cannot be a multiple of 4 since 21 + p^2 would not be divisible by 4 in either case.\r\n\r\nSo now we have 1, 2 and some q as the 3 smallest divisors of x (where q is an odd prime). If the 4th smallest divisor is also odd, we would have an odd sum of the squares of the 4 smallest divisors. Hence, the 4th smallest divisor has to be 2q. Thus, we have 5 + 5q^2 = 2qy, where y is the product of the rest of the factors of x. Since 5 clearly divides x now, p has to be 3 or 5. Testing 3, we realise that it would not work. Therefore, q = 5 and x = 130.\r\n\r\nThe largest prime that divides 130 is 13, and therefore we are done." } { "Tag": [ "function", "algebra", "polynomial", "limit", "calculus", "real analysis", "real analysis solved" ], "Problem": "I think it's 1, but other might claim that it's 0. The reason I think it's 1:\r\nThe function [tex]f(z)=z^0[/tex] is continuous (in fact smooth) everywhere in [tex]\\mathbb{C}\\setminus\\{0\\}[/tex], and for any sequence [tex]c_n\\rightarrow0, f(c_n)\\rightarrow1[/tex].\r\n\r\nI am not very sure about the behavior of [tex]f(z)=0^z[/tex] near [tex]z=0[/tex]...seems to me that there might be some sort of branch cut on the negative real axis?\r\n\r\nAnybody?", "Solution_1": "$0^0$ is undefined.\r\n\r\nYou can also look at $F(z)=z^z$ and you will \"conclude\" that $0^0=1$ ;)", "Solution_2": "$0^0$ is undefined.\r\n\r\nIn the power series or polynomial expression $\\sum_na_nx^n$ we understand that $x^0=1$ for all $x,$ including $x=0.$ This is merely a notational convention and is not meant to imply anything about $0^0.$\r\n\r\nIn $\\lim_{x\\to a}f(x)^{g(x)},$ where $\\lim f(x)=0^+$ and $\\lim g(x)=0$, the overall limit could be any positive number or nonexistent; we could construct examples to do anything we want. The calculus shorthand phrase for this is \"$0^0$ is indeterminate.\"", "Solution_3": "I, and many authors, define $0^0$ to be 1. The key reason is the one Kent mentioned -- it makes dealing with polynomials and power series easier. I don't see why it's good to say $x^0$ is 1 when $x = 0$, but bad to say $0^0 = 1$ -- can't they both be viewed as notational conveniences? To me, the calculus argument merely shows $x^y$ is discontinuous at (0, 0) no matter what we do. But is continuity the final word? I figure it's better to be continuous in one variable than neither.\r\n\r\nAnyway, I don't want to get into an argument about it. I agree it's a matter of definitional convenience. We had a long discussion about it in a topic on a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11999]Limit Problem[/url]." } { "Tag": [], "Problem": "A magic triangle is a triangle with distinct numbers on each side, where the sum of the numbers on each side are the same. For example, [code] 5\n 4 2\n1 6 3[/code] is a magic triangle, since each side adds up to 10, and no numbers are repeated.\r\n\r\nProove that it is impossible to have a magic triangle with two numbers on each side.", "Solution_1": "Are you asking to prove that a triangle which looks like:\r\n[code]\n a\n b c\n[/code]\nCannot exist?\n[hide]\nI'm kinda new to the whole proof business, but I'll give it a shot:\n\nSay that there is such a triangle that exists:\n[code]\n a\n b c\n[/code]\na+c would have to equal a+b, b+a would have to equal b+c, and c+a would have to equal c+b. Every number is distinct, so it is not possible:\nIf a+c = a+b, by algebra:\na+c = a+b \na-a+c=b\nc=b\nTherefore all of the numbers cannot be distinct and it is impossible.\n[/hide]", "Solution_2": "yeah", "Solution_3": "Well,\n\n[hide]\n\na + b = a + c = b + c\n\n\n\na + b = a + c\n\nb = c\n\n\n\nExcept the numbers must be distinct, so...[/hide]" } { "Tag": [ "limit", "logarithms", "probability", "induction", "calculus", "calculus computations" ], "Problem": "find\r\n\r\n$(a)\\;\\;\\;\\;\\;\\lim_{n\\to 0}\\;\\sqrt[n]{\\frac{1+n}{1-n}}$\r\n\r\n$(b)\\;\\;\\;\\;\\;\\lim_{n\\to 0}\\;\\sqrt[n]{\\frac{(2n)!}{(n!)^{2}}}$\r\n\r\n\r\n\r\n :rotfl:", "Solution_1": "(b) Is $n$ approaches 0? I think $n$ should be infinity, if so, the answer is $\\frac{4}{e}$.", "Solution_2": "Yes, of course it should be as $n$ goes to $\\infty.$\r\n\r\nAnd this is just an application of this classical lemma :\r\nif $u$ is a sequence of positive reals, and if the sequence $\\left(\\frac{u_{n+1}}{u_{n}}\\right)$ converges towards some real $l,$ then the same goes for $(\\sqrt[n]{u_{n}}).$\r\n\r\nBy the way, [b]kunny[/b], for (b), this yields that the limit is actually $4.$\r\n\r\nBut since you wanted to see $e,$ let's add one more exercise :\r\ncompute the limit of $\\frac n{\\sqrt[n]{n!}}.$", "Solution_3": "First problem : obviusly. ( by Taylor)\r\nSeccond problem : Stirnling forever :wink:", "Solution_4": "$A.$ approaching $0$\r\n[hide]Let $y=\\lim_{n\\to 0}\\;\\sqrt[n]{\\frac{1+n}{1-n}}$\n\n$\\ln y=\\lim_{n \\to 0}\\frac{1}{n}\\cdot \\ln \\left( \\frac{1+n}{1-n}\\right)$\n$\\ln y=\\lim_{n \\to 0}\\frac{\\ln \\frac{1+n}{1-n}}{n}$; We have $\\frac{0}{0}$\n$\\ln y=\\lim_{n \\to 0}\\frac{\\frac{-2}{(n-1)(n+1)}}{1}$\n$\\ln y=\\frac{-2}{(0-1)(0+1)}$\n$\\ln y = 2$\n$y = \\boxed{e^{2}}$[/hide]\n\n$A.$ approaching $\\infty$\n[hide]If $\\lim_{n\\to \\infty}\\;\\sqrt[n]{\\frac{1+n}{1-n}}=L$ and $a_{n}=\\frac{1+n}{1-n}$ then $\\lim_{n \\to \\infty}\\left( \\frac{a_{n+1}}{a_{n}}\\right)=L$\n\n$\\lim_{n \\to \\infty}\\left( \\frac{\\frac{2+n}{1-(n+1)}}{\\frac{1+n}{1-n}}\\right)$\n$= \\lim_{n \\to \\infty}\\frac{n+2}{n}\\cdot \\frac{n-1}{n+1}$\n$= \\boxed{1}$[/hide]", "Solution_5": "Stirling's formula? Overkill.\r\n\r\nLook at the probability distribution of flipping $2n$ fair coins and counting the number of heads. The probability $p$ of exactly $n$ heads is $2^{-2n}\\binom{2n}{n}$. Clearly, $p\\le 1$. Also, $n$ heads are the most likely result, so $p\\ge\\frac1{2n+1}$.\r\nTaking $n$th roots, $(2n+1)^{-1/n}\\le p^{1/n}=\\frac14\\binom{2n}{n}^{1/n}\\le 1$ and $\\binom{2n}{n}^{1/n}\\to 4$.", "Solution_6": "[quote=\"jmerry\"]Stirling's formula? Overkill.\n\nLook at the probability distribution of flipping $2n$ fair coins and counting the number of heads. The probability $p$ of exactly $n$ heads is $2^{-2n}\\binom{2n}{n}$. Clearly, $p\\le 1$. Also, $n$ heads are the most likely result, so $p\\ge\\frac1{2n+1}$.\nTaking $n$th roots, $(2n+1)^{-1/n}\\le p^{1/n}=\\frac14\\binom{2n}{n}^{1/n}\\le 1$ and $\\binom{2n}{n}^{1/n}\\to 4$.[/quote]\r\nYes you are right. I agree with you. but\r\npeoples who more work with stirling formula often, for them this problem is obviusly:\r\nBy stirling we only must consider main expression:\r\n$(\\frac{(\\frac{2n}{e})^{2n}}{(\\frac{n}{e})^{2n}})^{\\frac{1}{n}}=4$", "Solution_7": "for part (b), you can also prove by induction that\r\n\\[\\frac{4^{n}}{2\\sqrt{n}}\\leq \\frac{(2n)!}{(n!)^{2}}\\leq \\frac{4^{n}}{\\sqrt{3n+1}}\\]\r\nand then use $\\lim_{n\\rightarrow \\infty}\\sqrt[n]{n}=1$. This can be done with logarithms+L'Hopital, or difference of powers+AM-GM in the form\r\n\\[\\sqrt[n]{n}-1=\\frac{n-1}{\\sum_{k=0}^{n-1}\\sqrt[n]{n^{k}}}\\leq \\frac{n-1}{1+(n-1)\\sqrt{n}}\\]", "Solution_8": "For part a): \r\nLet $h = \\frac{2n}{1-n}$ and note that:\r\n$\\lim_{n \\rightarrow 0}(\\frac{1+n}{1-n})^{\\frac{1}{n}}$\r\n$= \\lim_{n \\rightarrow 0}(1+\\frac{2n}{1-n})^{\\frac{1}{n}}$ \r\n$= \\lim_{h \\rightarrow 0}(1+h)^{\\frac{2+h}{h}}$ \r\n$= \\lim_{h \\rightarrow 0}(1+h)^{\\frac{2}{h}}\\times \\lim_{h \\rightarrow 0}(1+h)$\r\n$= [\\lim_{h \\rightarrow 0}(1+h)^{\\frac{1}{h}}]^{2}= e^{2}$", "Solution_9": "[quote=\"Mathmanman\"]But since you wanted to see $e$ let's add one more exercise: compute the limit of $\\frac{n}{\\sqrt[n]{n!}}$[/quote]\r\n\r\n$L = \\lim_{n\\to\\infty}\\frac{n}{\\sqrt[n]{n!}}= \\lim_{n\\to\\infty}\\sqrt[n]{\\frac{n^{n}}{n!}}= \\lim_{n\\to\\infty}\\frac{(n+1)^{n+1}n!}{(n+1)n!n^{n}}$\r\n$= \\lim_{n\\to\\infty}\\left(1+\\frac{1}{n}\\right)^{n}= e$." } { "Tag": [ "search", "inequalities proposed", "inequalities" ], "Problem": "Prove that $ (x^{2}\\plus{}2)(y^{2}\\plus{}2)(z^{2}\\plus{}2)\\geq 9(xy\\plus{}yz\\plus{}zx)$ for every $ x,y,z\\in R^{\\plus{}}$.", "Solution_1": "posted so many times...\r\nhttp://www.mathlinks.ro/viewtopic.php?search_id=1615461828&t=82832" } { "Tag": [ "integration", "calculus", "derivative", "analytic geometry", "function", "calculus computations" ], "Problem": "[b]1. The problem statement, all variables and given/known data[/b]\r\n\r\nProblem from Arnold's \"Mathematical Methods of Classical Mechanics\" on page 59.\r\n\r\nFind the differential equation for the family of all straight lines in the plane in polar coordinates.\r\n\r\n[b]2. Relevant equations[/b]\r\n\r\n$ \\Phi \\equal{} \\displaystyle\\int^{t_2}_{t_1} \\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}\\,dt$\r\n\r\n$ \\frac {d}{dt}\\frac {\\partial{L}}{\\partial{\\dot{q_i}}} \\minus{} \\frac {\\partial{L}}{\\partial{q_i}} \\equal{} 0$\r\n\r\n\r\n[b]3. The attempt at a solution[/b]\r\n\r\nL is the integrand.\r\n\r\nWe have two equations:\r\n\r\n$ \\frac {d}{dt}\\frac {\\dot{r}}{\\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}} \\equal{} \\frac {r\\dot{\\phi}^2}{\\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}}$\r\n\r\n$ \\frac {d}{dt}\\frac {r^2\\dot{\\phi}}{\\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}} \\equal{} 0$\r\n\r\nThe second gives: $ \\frac {r^2\\dot{\\phi}}{\\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}} \\equal{} c$\r\n\r\nIf c=0 we have the derivative of phi zero, that is phi would be constant and we are (essentially) done. If phi is constant, we have a bundle of lines passing through the origin. But what about r? Do we get problems in the origin? Polar coordinates are not defined there, are they? \r\nIf c is not zero, then the first equation can be rewritten as:\r\n$ \\frac {d}{dt}\\frac {\\dot{r}}{r^2\\dot{\\phi}} \\equal{} \\frac {\\dot{\\phi}}{r}$\r\n\r\nI cannot get an idea of how to solve it for r-dot AND phi-dot. Tried to differentiate the left side and see what happens, but somehow nothing attractive comes out. \r\nSo how to proceed? Is it the right way? Or is there anything I can't see at a glance that helps?", "Solution_1": "The lines not through the origin satisfy $ ax\\plus{}by\\equal{}1$ implying $ rf(\\theta)\\equal{}1$ for some function $ f$ such that $ f^{\\prime \\prime}(\\theta)\\equal{}\\minus{}f(\\theta)$. Now we have\r\n\\[ f(\\theta)\\equal{}\\frac{1}{r},f^\\prime (\\theta)\\equal{}\\minus{}\\frac{r^\\prime}{r^2},\\]\r\n\\[ f^{\\prime \\prime}(\\theta)\\equal{}\\frac{2(r^\\prime)^2}{r^3}\\minus{}\\frac{r^{\\prime \\prime}}{r^2}\\]\r\nSetting $ f^{\\prime \\prime}(\\theta)$ equal to $ \\minus{}f(\\theta)$ gives\r\n\\[ r^{\\prime \\prime}r\\equal{}2(r^\\prime)^2\\plus{}r^2\\]", "Solution_2": "[quote=\"stealth\"][b]1. The problem statement, all variables and given/known data[/b]\n\nProblem from Arnold's \"Mathematical Methods of Classical Mechanics\" on page 59.\n\nFind the differential equation for the family of all straight lines in the plane in polar coordinates.\n\n[b]2. Relevant equations[/b]\n\n$ \\Phi \\equal{} \\displaystyle\\int^{t_2}_{t_1} \\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}\\,dt$\n\n$ \\frac {d}{dt}\\frac {\\partial{L}}{\\partial{\\dot{q_i}}} \\minus{} \\frac {\\partial{L}}{\\partial{q_i}} \\equal{} 0$\n\n\n[b]3. The attempt at a solution[/b]\n\nL is the integrand.\n\nWe have two equations:\n\n$ \\frac {d}{dt}\\frac {\\dot{r}}{\\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}} \\equal{} \\frac {r\\dot{\\phi}^2}{\\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}}$\n\n$ \\frac {d}{dt}\\frac {r^2\\dot{\\phi}}{\\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}} \\equal{} 0$\n\nThe second gives: $ \\frac {r^2\\dot{\\phi}}{\\sqrt {{\\dot{r}}^2 \\plus{} r^2{}\\dot{\\phi}^2}} \\equal{} c$\n\nIf c=0 we have the derivative of phi zero, that is phi would be constant and we are (essentially) done. If phi is constant, we have a bundle of lines passing through the origin. But what about r? Do we get problems in the origin? Polar coordinates are not defined there, are they? \nIf c is not zero, then the first equation can be rewritten as:\n$ \\frac {d}{dt}\\frac {\\dot{r}}{r^2\\dot{\\phi}} \\equal{} \\frac {\\dot{\\phi}}{r}$\n\nI cannot get an idea of how to solve it for r-dot AND phi-dot. Tried to differentiate the left side and see what happens, but somehow nothing attractive comes out. \nSo how to proceed? Is it the right way? Or is there anything I can't see at a glance that helps?[/quote]\r\n\r\nYou've missed the point of the problem.\r\n\r\nYou do indeed have two differential equations, but the point is not to solve them necessarily but rather to expand them out and see what you get.\r\n\r\nYou have,\r\n(1) $ \\frac{d}{dt}\\frac{\\partial L}{\\partial \\dot{r}} \\minus{}\\frac{\\partial L}{\\partial r} \\equal{} 0$\r\n\r\n(2) $ \\frac{d}{dt}\\frac{\\partial L}{\\partial \\dot{\\phi}} \\minus{}\\frac{\\partial L}{\\partial \\phi} \\equal{} 0$\r\n\r\nActually do the indicated differentiations and leave the resulting expression equal to zero on their right hand sides. What you wind up with are homogeneous equations of motion." } { "Tag": [ "FTW", "blogs" ], "Problem": "Is there a special formula for problems like these?\r\n\r\n1. There are 24 diagonals in a polygon. How many sides does this polygon have?\r\n\r\n2. How many diagonals are in a polygon with 24 sides?\r\n\r\nI just want to know, because whenever I encounter problems like these, it takes me a while to figure them out, while on AOPS FTW, I noticed, people solve these in a very short time (not memmed). \r\n\r\nHELP PLEASE!! :D", "Solution_1": "To find the number of diagonals in an x sided figure the formula is:\r\n$ \\frac{x(x\\minus{}3)}{2}$\r\n\r\nThis is because in any figure for each point there can be 3 points to which a diagonal cannot be drawn. (the point itself and the adjacent points). And since you are counting each diagonal 2 times you must divide by 2", "Solution_2": "That's what I thought, too, but then for his question: How many sides are in a polygon with 24 diagonals, you don't get a whole number answer.", "Solution_3": "If a convex polygon has $ n$ sides, then it has $ \\frac{n(n\\minus{}3)}{2}$ diagonals. (Like what SuerNerd said). Another formula is also $ \\binom{n}{2}\\minus{}n$. This is because you can chose $ 2$ points to draw a line through, but $ n$ of them will be sides, not diagonals. For me, each take about the same time, so it's just your preference...\r\n[hide=\"1)\"]\nThere are no convex polygons with $ 24$ diagonals...\n[/hide]\n[hide=\"2)\"]\nWe just enter $ 24$ for $ n$ in our equations and solve...\n$ \\frac{24\\cdot21}{2}\\equal{}12\\cdot21\\equal{}\\boxed{252}$\nOR\n$ \\binom{24}{2}\\minus{}24\\equal{}\\frac{24\\cdot23}{2}\\minus{}24\\equal{}276\\minus{}24\\equal{}\\boxed{252}$\n[/hide]", "Solution_4": "Ok so you have 24 diagonals. There is a fourmula for this. It is x(x-3)/2\r\nso just plug the number 24 where the x goes. so now you have 24(21)/2\r\nIt has 252 sides. :)\r\nReventon", "Solution_5": "In that formula x stands for the number of sides not the number of diagonals... :) \r\nSo you would have to make an equation like this\r\n\r\n$ \\frac{x(x\\minus{}3)}{2} \\equal{} 24$\r\n\r\n$ x(x\\minus{}3) \\equal{} 48$\r\n\r\nWe can see that no number qualifies...", "Solution_6": "Yeah, there are no polygons with 24 diagonals... :ninja:", "Solution_7": "actually, they could not me whole number sides.", "Solution_8": "I've never seen a polygon with a noninteger number of sides. Perhaps you can draw one for me, say a polygon with 4.5 sides? :P", "Solution_9": "well ya good point izzy. :mad: :( :D", "Solution_10": "[asy]draw((0,0)--(1,0)--(1,1)--(0,1)--(1,0)--(0,-1)--(0,0));[/asy]\r\n\r\nDraw all diagonals and tell me how many there are. Then plug it back into your fancy formulas.", "Solution_11": "I believe it has been established that the formula applies for only convex polygons. I also believe that figure is neither convex nor a polygon.", "Solution_12": "I actually made a post about this on my wiki, and on my blog:\r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/User:Dojo#Diagonal_Forumla\r\n\r\nhttp://www.artofproblemsolving.com/Forum/weblog_entry.php?t=273655" } { "Tag": [ "function" ], "Problem": "What expression satisfy:\r\n\r\n$F(x)+F(\\frac{1}{1-x})=x$\r\n\r\n\r\na)$\\frac{x^3-x-1}{2x(x-1)}$\r\nb)$\\frac{x^3+x+1}{2x(x-1)}$\r\nc)$\\frac{x^3-x+1}{2x(x-1)}$\r\nd)$\\frac{x^3+x-1}{2x(x+1)}$\r\ne)$\\frac{x^3-x+1}{2x(x+1)}$", "Solution_1": "If you don't know how to do it, wouldn't that make it hard for [i]you[/i]? Thus why do you call it 'easy'? Also, the problem doesn't have to do with trig.\r\n\r\n[hide=\"Solution\"]\n\\begin{eqnarray}\nF(x)+F\\left(\\frac 1{1-x}\\right) &=& x\\\\\nF\\left(\\frac 1{1-x}\\right)+F\\left(\\frac{x-1}x\\right) &=& \\frac 1{1-x}\\\\\nF\\left(\\frac{x-1}x\\right)+F(x) &=& \\frac{x-1}x\n\\end{eqnarray}\n\n\\[(1)-(2)+(3)=2F(x)=x+\\frac 1{1-x}-\\frac{x-1}x\\Rightarrow F(x)=\\frac{x^3-x+1}{2x(x-1)}. \\ \\ \\boxed{C}\\] [/hide]", "Solution_2": "BTW this problem (without options of course) was probably one of the easiest german pre-TST problems", "Solution_3": "This is a Pre-Olympiad Problem???? :o \r\n\r\nThis should be in Getting Started!!!! :rotfl: :rotfl: :rotfl:" } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "homothety", "Gauss", "perpendicular bisector" ], "Problem": "For a triangle ABC, let I be the incenter. Let the reflection (line) of BC in AI be L.\r\nWhen midpoint of BC is M and tangent line of nine point circle at M is K,\r\nprove that L is parallel to K.\r\n\r\nI guess it is well known, isn't it ? afraid of its not being nice. Thank you for reading.", "Solution_1": "I'm not sure whether I saw this before, but anyway, here's how to do this:\r\n\r\nBoth lines are parallel to the tangent line to the circumcircle at the antipode of $ A$. Of course the tangent to the NPC is, because it is mapped to it by a homothecy with center $ H$ and factor $ 2$. The other line is too, because it is also perpendicular to $ AO$, since $ BC$ is perpendicular to $ AH$, the reflected line has to be perpendicular to $ AO$ because $ O$ and $ H$ are isogonal conjugates.\r\n\r\nNote that $ O$ and $ H$ are the circumcenter and orthocenter of $ ABC$.", "Solution_2": "Here is my solution\r\nCall P is midpoint of PO. By Gauss line theory, we have M, N, J is conlinear. Similar TST USA 2000, we get EF is perpendicular to MN, in the other hand, JE = JF. Hence, JMN is perpendicular bisector of MN\u2026\u2026.." } { "Tag": [], "Problem": "I can't for the life of me figure out how to do this:\r\n\r\n0.6x-10=1.4x-14 \r\n\r\nthe answer is 5 but I don't know how to get there. Please show your work. \r\n\r\nAlso -3(2x-1)<-4[2+3(x+2)]\r\n\r\nI just can't remember the order for brackets and para theses. \r\n\r\nPlease show work. or at least show me the second half. \r\n\r\n :idea:", "Solution_1": "For the first one:\r\n\r\n$ .6x\\minus{}10\\equal{}1.4x\\minus{}14\\Rightarrow 4\\equal{}.8x\\Rightarrow x\\equal{}5$\r\n\r\nAll I did was subtract $ .6x$ from each side and add 14 to each side.", "Solution_2": "hello, we want to solve the equation\r\n$ 0.6x\\minus{}10\\equal{}1.4x\\minus{}14 |\\minus{}0.6x$\r\n$ \\minus{}10\\equal{}0.8x\\minus{}14|\\plus{}14$\r\n$ 4\\equal{}0.8x|: 0.8$\r\n$ x\\equal{}5$\r\nSonnhard.", "Solution_3": "I still don't get it. \r\nHow can 0.8x divided by 4 equal 5?", "Solution_4": "hello, given the following inequation\r\n$ \\minus{}3(2x\\minus{}1)<4\\left(2\\plus{}3(x\\plus{}2)\\right)$\r\nWe have\r\n$ \\minus{}6x\\plus{}3<8\\plus{}12(x\\plus{}2)$\r\n$ \\minus{}6x\\plus{}3<8\\plus{}12x\\plus{}24 |\\plus{}6x$\r\n$ 3<8\\plus{}24\\plus{}18x|\\minus{}32$\r\n$ \\minus{}29<18x|: 18$\r\n$ \\minus{}\\frac{29}{18} 2$). We know that at least one of $ a$, $ b$, and $ c$ is greater than $ 3$. Thus, $ \\frac1a \\plus{} \\frac1b \\plus{} \\frac1c < \\frac13 \\plus{} \\frac13 \\plus{} \\frac13 \\equal{} 1$, a contradiction.\n\nWLOG, let $ a \\equal{} 2$. Then, we have a system of two equations with two unknowns:\n\n$ b \\plus{} c \\equal{} 8$ and $ \\frac1b \\plus{} \\frac1c \\equal{} \\frac12$\n\nSubstituting, we have:\n\n$ b \\equal{} 8 \\minus{} c\\Rightarrow \\frac1{8 \\minus{} c} \\plus{} \\frac1c \\equal{} \\frac12$\n\n$ 2(8 \\minus{} c \\plus{} c) \\equal{} c(8 \\minus{} c)$\n\n$ c^2 \\minus{} 8c \\plus{} 16 \\equal{} 0\\Rightarrow c \\equal{} 4$\n\nIt follows that the friends have 2, 4, and 4 coins, in some order. Therefore, the only possible values of the difference between the number of coins of $ A$ and $ B$ are 2 and 0.[/hide]\r\n\r\nThis is probably on the upper limit of difficulty of problems for HSB..." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let a,b,c>=0 and a^4+b^4+c^4=<2(a^2b^2+b^2c^2+c^2a^2). Prove that \r\n a^2+b^2+c^2=<2(ab+bc+ca).", "Solution_1": "[quote=\"Ulanbek_Kyzylorda KTL\"]Let a,b,c>=0 and a^4+b^4+c^4=<2(a^2b^2+b^2c^2+c^2a^2). Prove that \n a^2+b^2+c^2=<2(ab+bc+ca).[/quote]\r\nBecause $ \\sqrt a\\plus{}\\sqrt b\\equal{}\\sqrt{a\\plus{}b\\plus{}2\\sqrt{ab}}\\geq\\sqrt c.$ :wink:", "Solution_2": "$ a^4\\plus{}b^4\\plus{}c^4\\leq 2(a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)$\r\n$ \\Longleftrightarrow (a\\plus{}b\\plus{}c)(a\\plus{}b\\minus{}c)(a\\minus{}b\\plus{}c)(\\minus{}a\\plus{}b\\plus{}c)\\geq0$\r\n$ \\Longleftrightarrow a,b,c$ are left of side of triangle put that $ a\\equal{}x\\plus{}y$ , $ b\\equal{}y\\plus{}z$ and $ c\\equal{}z\\plus{}x$ with $ x,y,z\\geq 0$\r\nyour inequality is:\r\n$ (x\\plus{}y)^2\\plus{}(y\\plus{}z)^2\\plus{}(z\\plus{}x)^2\\leq 2[(x\\plus{}y)(y\\plus{}z)\\plus{}(y\\plus{}z)(z\\plus{}x)\\plus{}(z\\plus{}x)(x\\plus{}y)]$\r\n$ \\Leftrightarrow xy\\plus{}yz\\plus{}zx\\geq 0$\r\nobvious!" } { "Tag": [ "probability" ], "Problem": "Just curious as to peoples' scores", "Solution_1": "I got 30 total last year on State\r\n20 on Sprint\r\n10 on Target\r\nI got 11th\r\nThat was before i got specialized AOPS topic bokss\r\n\r\nNotes:\r\nI missed so many of the beginning problems\r\nI only missed two on the last ten\r\n4th place got 35\r\nI could've broke any tiebreakers\r\nI'm pretty sure i missed number 1 on sprint\r\nI missed 3 questions on the target because\r\n1. I couldn't understand the graph quick enough and I didn't feel like reading carefully\r\n2. I forgot to minus 1 from my answer\r\n3. I forgot to use complementary probability\r\nSprint tired me out literally\r\nMy hand was hurting at the end of the sprint round", "Solution_2": "27 last year. 17/10", "Solution_3": "I did it- but at home 2 days ago :P\r\n\r\n23 on Sprint.\r\n7 on Target\r\n\r\n23+7(2) = 37", "Solution_4": "I got a 42 (28 + 14) but that was probably because I did it at home for practice. If I was there at the actual competition my score probably would have been about 20 points less because of nerves.", "Solution_5": "I've only done the sprint since we go over the target round at school (our team made it to state for once) and I got 18. I need to read more carefully and not rush the arithmetic. I could have had a 23 if I'd made no careless errors. Nerves don't really affect me, with the wooden walls around me I forget I'm at a compeition.", "Solution_6": "32/18 :oops: /14 last yr, in KS.\r\n1st written was 35\r\n2nd was 34\r\n3rd-7th were 32. I got 7th :roll: , and cd up to 4th.", "Solution_7": "I would have taken 1st in Kansas then! Oh well I got first in WI anyway :P 37 (25,12)", "Solution_8": "Oh for last years test I actually did pretty well 26 sprint 16 target 26+16=42(that was practice at home) but on the real thing I do horrible. Plus when I did that test again like 5 months later I got a 28 on sprint.", "Solution_9": "second with 25/14. Apparently, 64-36 ISN'T 26...", "Solution_10": "Who got the perfect?", "Solution_11": "I assume someone who got an 8 who is trying to appear smarter than they actually are.", "Solution_12": "27/12=39. Lets just say I wasn't performing very well on target -_-. \r\n\r\nThat was barely good enough for 3rd in michigan, considering 7th was 38.", "Solution_13": "[quote=\"perfect628\"]Apparently, 64-36 ISN'T 26...[/quote]\r\n\r\n\r\nWhat are you talking about?? Anyway I do really bad on the real thing so practice scores dont even matter that much.", "Solution_14": "On the target problem about the sides of the right triangle, and lengths 6 and 8, I did $\\sqrt{8^{2}-6^{2}}$ to find the third side. I did 64-36 in my head, and somehow ended up with 26.", "Solution_15": "[quote=\"perfect628\"]On the target problem about the sides of the right triangle, and lengths 6 and 8, I did $\\sqrt{8^{2}-6^{2}}$ to find the third side. I did 64-36 in my head, and somehow ended up with 26.[/quote]\r\n\r\nOh okay.", "Solution_16": "I got 16 on sprint, 4 on target, making a total score of 24. i sucked so bad last year... i got a higher score in states when i was in [i]6th[/i] grade...", "Solution_17": "I got 25/16, good enough for first last year.", "Solution_18": "17/5=27 for me", "Solution_19": "i think you did better than that, bpms.\r\n\r\nlike a 39 maybe.\r\n\r\njorian", "Solution_20": "who put a 46?????????????\r\n\r\nyes this deserves a double post!\r\n\r\njorian", "Solution_21": "[quote=\"jhredsox\"]who put a 46?????????????\n\nyes this deserves a double post!\n\njorian[/quote]\r\nA liar.\r\nLast year the highest score was a 44 by both Nathan Benjamin and Karan Batra", "Solution_22": "[quote=\"bpms\"][quote=\"jhredsox\"]who put a 46?????????????\n\nyes this deserves a double post!\n\njorian[/quote]\nA liar.\nLast year the highest score was a 44 by both Nathan Benjamin and Karan Batra[/quote]\r\n\r\nlol, there were 2!!! :rotfl:", "Solution_23": "Eric (Pom) McCabe also got 44 I believe", "Solution_24": "who voted 45?\r\nActually I was mistaken, it is possible for someone to have gotten a 46, just extremely unlikely (the report covered most of the states and most of the others were mentioned on AoPS)", "Solution_25": "I FAILED states last year.", "Solution_26": "[quote=\"jhredsox\"]i think you did better than that, bpms.\n\nlike a 39 maybe.\n\njorian[/quote]\r\nI wish, that would've put me at 2nd.", "Solution_27": "last year, I was pretty bad...\r\n\r\n23 :blush: :( \r\n\r\nbut that was BEFORE I found the AoPS book and took the online lessons!" } { "Tag": [ "calculus", "integration", "logarithms", "function", "limit", "derivative" ], "Problem": "I tried finding a definite log integral on my ti-89, but it returned undef (probably because it was from 0 to something greater than 1). Does anyone know a way to find an approximate value? It doesnt have to be on an 89. Anything would be much appreciated.", "Solution_1": "An approximate value is infinity (actually -infinity since it's under the axis). If you go from 0 to anything greater than 0 it will give you infinity (undefined). It might help to take into consideration what the indefinite integral is, that $\\int \\ln(x) dx = x \\ln(x) - 1$. And of course when you plug in 0 to ln you're going to get undefined.\r\n\r\nAlso, calculus problems should be posted in the calculus sections, not in pre-olympiad.", "Solution_2": "sorry, by log integral, i meant $li(x)=\\displaystyle\\int_0^x\\displaystyle\\frac{dt}{\\ln t}$.\r\nThe only things i know about it are that its an upper bound for $\\pi(n)$, but that seems like a bad way to calculate it.\r\n\r\nAlso, I dont know how to move this, so if a moderator/admin or someone who does know how could move this it would be much appreciated.", "Solution_3": "[quote=\"gopherhole112\"]sorry, by log integral, i meant $li(x)=\\displaystyle\\int_0^x\\displaystyle\\frac{dt}{\\ln t}$.\nThe only things i know about it are that its an upper bound for $\\pi(n)$, but that seems like a bad way to calculate it.\n[/quote]\r\nActually, it's not an upper bound for the prime counting function (although it is for all practical purposes). Littlewood proved that the plots of $\\pi(x)$ and $\\text{li}(x)$ cross infinitely many times.\r\nMore information is at [url]http://mathworld.wolfram.com/LogarithmicIntegral.html[/url].", "Solution_4": "Ah, I was not familiar with the term log integral, I assumed it just meant the integral of a logarithm.", "Solution_5": "What!?! Where's the beautiful logic of mathematics? If you ask me, the log integral should the integral of a log!\r\n\r\nHokkage, I believe that you made a couple of mistakes:\r\n$\\int\\ln(x)dx=x\\ln(x)-x+C$\r\nAlso, just because the log function itself decreases without bound near 0, does [u]not[/u] mean that its integral is infinite (see [url]http://mathworld.wolfram.com/ImproperIntegral.html[/url]). As a matter of fact, $\\displaystyle\\lim_{x\\rightarrow\\ 0^+}(x\\ln(x)-x)=\\lim_{x\\rightarrow\\ 0^+}\\frac{\\ln(x)}{1/x}-0=\\lim_{x\\rightarrow\\ 0^+}\\frac{1/x}{-1/x^2}=\\lim_{x\\rightarrow\\ 0^+}-x=0$.\r\nSo, $\\displaystyle\\int_0^x\\ln(t)dt=x\\ln(x)-x-(0)$.", "Solution_6": "Danbert, $li(x)=\\displaystyle\\int_{\\mu}^x\\frac{dt}{\\ln(t)}$ for $x>1$.\r\n($\\mu=1.4513692...$)", "Solution_7": "Draperp, yes I did make a mistake in that integral (my calculus teacher would kill me). I forgot the formula and I didn't feel like doing parts so I guessed and checked, and I think I took the derivative wrong :oops: on that guess.\r\n\r\nAnd I forgot about impropers! Geeze. Summer is doing horrible things to my calculus. Thanks for pointing those out, maybe I won't go so fast next time.", "Solution_8": "[quote=\"draperp\"]Danbert, $li(x)=\\displaystyle\\int_{\\mu}^x\\frac{dt}{\\ln(t)}$ for $x>1$.\n($\\mu=1.4513692...$)[/quote]\r\n\r\nThat amounts to the same thing I wrote, except that my integral is improper.\r\n$\\int_0^\\mu\\frac{dt}{\\ln(t)}=0$\r\n\r\nAnd \"$x>1$\" should be \"$x>\\mu$\".", "Solution_9": "[quote=\"Danbert\"]And \"$x>1$\" should be \"$x>\\mu$\".[/quote]\r\nNo, $\\int_a^b$ is still okay even when $b1/9$, then by AM-GM $c+d>2/3$; also $d^2 \\geq cd>1/9$, so $d>1/3$, then $e>1/3$. Adding up gives $e+c+d>1$, absurd. On the other hand, if $be>1/9$ then $(b+c+d)e \\geq 3be>1/3>1/4$, so that, by AM-GM, $b+c+d+e>1$, absurd." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ P$ be the set of all points in $ R^n$ with rational coordinates.For points $ A,B\\in P$ ,one can move from $ A$ to $ B$ if $ AB \\equal{} 1$.Prove that: every point in $ P$ cab reached from any other point in $ P$ by a finite sequence of moves if only if $ n\\geq 5$", "Solution_1": "any ideas\r\nI have not found an idea which to slove problem?\r\ncan anyone help me?", "Solution_2": "http://www.mathlinks.ro/viewtopic.php?t=2635" } { "Tag": [ "LaTeX", "function", "algebra", "floor function", "number theory unsolved", "number theory" ], "Problem": "prove that n_c_7 -[n/7] is always divided by 7.\r\np.s-i dont have latex .you may have some problem in understanding the notaions.i hope you can understand n_c_7 and the greatest integer function is denoted of n/7 by[n/7].\r\nplease help.", "Solution_1": "It is a direct corollary of Lucas's Theorem.", "Solution_2": "please prove it.it'll be morehelpfull if it is written in LATEX." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "More specifically, anyone have the will and ability to sort those USAMO scores by score? \r\n\r\nthanks", "Solution_1": "you can sort it through excel...", "Solution_2": "well, that's clever :O\r\n\r\n\r\nthanks", "Solution_3": "I wrote a tight C++ program related to problem number 5 on the USAMO, and it appears that I must have gotten at least one point for it. :lol:", "Solution_4": "[quote=\"Xuan768\"]I wrote a tight C++ program related to problem number 5 on the USAMO, and it appears that I must have gotten at least one point for it. :lol:[/quote]\r\ncomputers are excellent at guessing", "Solution_5": "Sort of.\r\n\r\nYou really should have gotten into computer science, Mark." } { "Tag": [ "algebra", "polynomial", "algorithm", "induction", "combinatorics theorems", "combinatorics" ], "Problem": "Hello, I'm a italian student that I'm studing the AKS algorithm avaible on address http://www.math.princeton.edu/~annals/issues/2004/Sept2004/Agrawal.pdf.\r\n\r\nIn this paper is wrote :\r\n\r\n\"So there exist at least $l+1$ distinct polynomials of degree one in $G$. Therefore, there exists at least $\\binom{t+l}{t-1}$ distinct polynomials of degree < $t$ in $G$.\r\n\r\nThe question is : Why? help somebody me?\r\n\r\nThanks and excuse me for my english!", "Solution_1": "For $t = 1$, it counts the polynomial $f(x) = 1$. For $t = 2$, it counts that polynomial and the $l+1$ polynomials of degree 1. For $t \\geq 3$, it counts the polynomials which can be made by multiplying up to $t-1$ of our linear polynomials together.\r\n\r\nIn other words, there are $t+l \\choose t-1$ ways to choose up to $t-1$ objects from among $l+1$, if repitition is allowed but order isn't important.", "Solution_2": "Excuse me, but I have understood the theorem in this way:\r\n\r\nThe number of polynomial of degree less than $t$, is :\r\n\r\n$\\sum_{i=0}^{t-1}\\binom{l+1+i-1}{i}$ (from combination with ripetition).\r\n\r\nNow, I prove that $\\sum_{i=0}^{t-1}\\binom{l+i}{i}\\geq \\binom{t+l}{t-1}$ for induction on $t$\r\n\r\ncase base : $t=1$ then $\\binom{l}{0}\\geq \\binom{l+1}{0}$\r\ninductive reasoning : Let true the theorem for $t_{0}-1$ and prove for $t_{0}$\r\n\r\n$\\sum_{i=0}^{t_{0}}\\binom{l+i}{i}= \\sum_{i=0}^{t_{0}-1}\\binom{l+i}{i}+\\binom{l+t_{0}}{t_{0}}\\geq \\binom{t_{0}+l}{t_{0}-1}+\\binom{t_{0}+l}{t_{0}}= \\binom{t_{0}+1+l}{t_{0}}.$\r\n\r\nIs correct ?", "Solution_3": "Yes.", "Solution_4": "Thanks JBL!" } { "Tag": [ "induction", "function", "algorithm" ], "Problem": "a long time ago, I read the Art and Craft of Problem Solving by Zeitz, and I remember they had a problem:\r\n\r\nProve that any positive integer n can can be written as a difference of two subsets of distinct powers of 3.\r\n\r\nAnd if I remember correctly, he gave a proof using generating functions...does anybody know this proof?", "Solution_1": "Basically, prove by induction on n that\r\n$\\prod_{k=0}^n (x^{3^k}+1+x^{-3^k})=x^t+x^{t-1}+\\cdots+x^{-t}$,\r\nwhere $t=3^0+3^1+\\cdots+3^n$. In the limiting case, the LHS gives the generating function for expressing a number as a sum or difference of different powers of 3, and the RHS has each power of x exactly once. So, there is exactly one way to express each integer (not restricted to positive integers) in this way.\r\n\r\nIMO, a very nice proof!", "Solution_2": "[quote=\"zabelman\"]IMO, a very nice proof![/quote]Indeed! :cool:", "Solution_3": "that is cool, although i dont think you need induction because of the nice identity:\r\n\r\n$\\displaystyle (x^{3^n}+1+x^{3^{-n}}) = \\frac{x^{3^{n+1}}-1}{(x^{3^n}-1)(x^{3^n})}$\r\n\r\nthen you can cancel nicely in the numerator and denominator.", "Solution_4": "Could it also be proved by giving an algorithm to convert $n$ from base 3 to a sum or difference of powers of 3? Basically take\r\n\r\n$n=a_ka_{k-1}...a_1$ (base 3), where $a_i$ is 0, 1 or 2 for $i=1, 2, ..., n$, then take the smallest $i$ such that $a_i=2$ or $a_i=3$, substract 3 to give $b_i=-1$ or $b_i=0$ and add 1 to $a_{i+1}$. If $a_{i+1}=0$ and the new digit is $b_{i+1}=1$ you repeat for the next $i$, if $a_{i+1}=1$ or $2$, you do the same for digit $i+1$ and so on. Clearly all the digits to the right of $i+1$ are the ones we want and the algorithm must end (or else $k$ would be infinite), so it should work. ;)" } { "Tag": [ "function", "analytic geometry", "algebra unsolved", "algebra" ], "Problem": "Let $f$ be a strictly increasing function from $R$ to $R$ such that \r\n $ f(a+b-ab) = f(a) + f(b) - f(a)f(b)$ for every $a,b$\r\n Find $f$.", "Solution_1": "it isn't hard It was posted", "Solution_2": "Can you show me the solution ? :)", "Solution_3": "[quote=\"treegoner\"]Let $f$ be a strictly increasing function from $R$ to $R$ such that \n $ f(a+b-ab) = f(a) + f(b) - f(a)f(b)$ for every $a,b$\n Find $f$.[/quote]\r\n$f(x)=x$ ?? ;)", "Solution_4": "I don't think this is so easy.", "Solution_5": "[quote=\"treegoner\"]Let $f$ be a strictly increasing function from $R$ to $R$ such that \n $ f(a+b-ab) = f(a) + f(b) - f(a)f(b)$ for every $a,b$\n Find $f$.[/quote]\r\n\r\nIt is a disguised form of $h(AB) = h(A)h(B)$, with coordinates centered at 1 rather than 0.\r\n\r\nConsider the operation $a * b = a + b - ab$ (multiplication, written relative to 1) \r\nand let $g(x) = 1 - f(x)$.\r\n\r\n$ g(a * b) = g(a) g(b) $.\r\n\r\nNow let $A = 1-a$ and $B = 1-b$. Then $g (1 - AB) = g(1 - A) g(1-B)$.\r\nLet $h(x) = g(1-x) = 1 - f(1-x)$. \r\n$h(AB) = h(A)h(B)$\r\n\r\nWe have $h(x) = sgn(x) |x|^a$ for some $a > 0$ and this gives a family of\r\nsolutions for $f(x)$." } { "Tag": [ "parameterization", "integration", "limit", "probability and stats" ], "Problem": "Let X be an exponential random variable with parameter c. Find the expectation(if it exists) of 1/X.", "Solution_1": "$ E(1/X)\\equal{}\\int_0^{\\infty}\\frac1x\\cdot\\frac1ce^{\\minus{}x/c}\\,dx\\equal{}\\plus{}\\infty$ since the integrand behaves like a constant times $ \\frac1x$ as $ x\\to 0.$\r\n\r\nIn general: suppose $ X$ is a nonnegative random variable with density $ f(x)$ such that $ \\lim_{x\\to0\\plus{}}f(x)>0.$ Then $ E(1/X)\\equal{}\\plus{}\\infty.$", "Solution_2": "[quote=\"Kent Merryfield\"]In general: suppose $ X$ is a nonnegative random variable with density $ f(x)$ such that $ \\lim_{x\\to0 \\plus{} }f(x) > 0.$ Then $ E(1/X) \\equal{} \\plus{} \\infty.$[/quote]\r\n\r\nIt's really interesting and surprising to know. I m interested in knowing the proof. :roll:", "Solution_3": "replace $ \\frac{1}{c} \\exp(\\frac{\\minus{}x}{c})$ by $ f(x)$ in the proof of Kent Merryfield.", "Solution_4": "[quote=\"limsup\"]It's really interesting and surprising to know. I m interested in knowing the proof. :roll:[/quote]\r\n\r\nThat smiley indicates sarcasm, just so you know.", "Solution_5": "so take my answer as second order sarcasm ..." } { "Tag": [ "LaTeX", "Support" ], "Problem": "I usually create math documents using Microsoft Word. What do you think of this?", "Solution_1": "How? Does it let you use LaTeX? :huh:", "Solution_2": "I use texniccenter, but MathType is fine.", "Solution_3": "You can copy/paste LaTeX images into Word, but I normally use Equation Editor. My school has MathType, though.", "Solution_4": "I think that Math Type is better than LaTeX... it is like Windows and DOS :P", "Solution_5": "[quote=\"thatin\"]I think that Math Type is better than LaTeX... it is like Windows and DOS :P[/quote]\r\n\r\nProbably whatever thing you like most is best for you, but certainly latex and math type is not like windows and dos. Windows is a simpler, more intuitive environment which lets you do basicly the same things DOS did. Math Type is a simpler, more intuitive environment [i]for writing math equations[/i] which in the proccess of simplifyng loses a great amount of the possibilities latex offers you. Everything you can do wiht Math Type, if I remember correctly because I don't use it anymore, is well given with a few buttons. Understand you can do a lot of things with a few buttons and most of the time it is enough, but sometimes you want truly specific stuff. Well the thing with latex is that probably someone has wanted to do it before and there are gurus who write specific packages in order to be able to do exactly that. And in the extremely outlandish situation in which there is no package you can always make it yourself if you're in the mood to learn how to do it. Also latex is not simply for writing equations but a whole typesetting system for building entire documents, from letters to books, which I'm guessing math type is not.\r\nBut if math type serves you well that's what you should use. Nevertheless, mathematicians and physicists are switching to latex, I think. At work, it can be hard to overrun this tendencies.\r\n\r\n\r\nUrsula", "Solution_6": "ok! windows and unix then :) \r\nI must admit that I have never needed specific stuff.", "Solution_7": "[quote=\"lingomaniac88\"]I usually create math documents using Microsoft Word. What do you think of this?[/quote]\r\n\r\nI do too, occasionally. It depends on what the content is - I'd go with LaTeX on equation-heavy things. But the thing is, it's just so much easier than opening LaTeX. I mean most of the time I'd be done before I even had a heading done for the LaTeX article.\r\n\r\nBut I'd definitely recommend you learn LaTeX if you don't know it already. It's very handy for documents that you'd like to look good, where...you know... Sum[x = 0 => k][x^2] just wouldn't do.", "Solution_8": "You just have to make the head and all this stuff once, simple save some starting document somewhere.\r\nI'm even writing normal letters with $\\LaTeX$...", "Solution_9": "i created a header, it takes like fifteen secs to create a header then", "Solution_10": "You should try the new Office 2007. Granted, Excel 2007 is horrid in terms of usability, they've added equation support to Word.", "Solution_11": "Do you think it would be worthwhile for me to get the LaTeX typesetting stuff for my computer and use that as well?" } { "Tag": [ "geometry", "analytic geometry", "geometry proposed" ], "Problem": "Given a traingle $ ABC$ with area $ 1$. Let $ L,\\ M,\\ N$ be the points on the sides $ BC,\\ CA,\\ AB$ respectively and let $ P,\\ Q,\\ R$ be the points on the sides $ LM,\\ MN,\\ NL$ respectively. Find the maximum value of the following expression.\r\n\r\n\\[ \\sqrt [3]{\\triangle{CRL}} \\plus{} \\sqrt [3]{\\triangle{APM}} \\plus{} \\sqrt [3]{\\triangle{BQN}} \\plus{} \\sqrt [3]{\\triangle{BLP}} \\plus{} \\sqrt [3]{\\triangle{CMQ}} \\plus{} \\sqrt [3]{\\triangle{ANR}}\r\n\\]\r\n\r\nNote that $ \\triangle{XYZ}$ denotes the area of the triangle $ XYZ$.", "Solution_1": "I have get pretty expressions for the six areas with areal coordinates, but I wonder the role of the cubic roots in this problem.\r\nSome particular cases of the triangles LMN, PQR have appear in the old literature, but I have not found nothing so general like this.", "Solution_2": "[hide]Notice $ \\Delta CRL=\\frac{LC}{BC}\\Delta CRB=\\frac{RL}{NL}\\frac{LC}{BC}\\Delta CNB=\\frac{NB}{AB}\\frac{RL}{NL}\\frac{LC}{BC}$ since $ \\Delta ABC=1$. We similarly find for the other triangles \n\n$ \\Delta CMQ=\\frac{NA}{AB}\\frac{QM}{MN}\\frac{MC}{AC}$\n\n$ \\Delta APM=\\frac{LC}{BC}\\frac{PM}{LM}\\frac{AM}{AC}$ \n\n$ \\Delta ANR=\\frac{LB}{BC}\\frac{NR}{NL}\\frac{AN}{AB}$\n\n$ \\Delta BLP=\\frac{MC}{AC}\\frac{PL}{ML}\\frac{BL}{BC}$\n\n$ \\Delta BQN=\\frac{AM}{AC}\\frac{NQ}{MN}\\frac{NB}{AB}$\n\nNotice by AM-GM that \n\n$ \\sqrt[3]{\\Delta CRL}=\\sqrt[3]{\\frac{NB}{AB}\\frac{RL}{NL}\\frac{LC}{BC}}\\le\\frac{\\frac{NB}{AB}+\\frac{RL}{NL}+\\frac{LC}{BC}}{3}$\n\nWe find similar maximums for the other triangles. Add them up and we have\n\n$ \\sqrt[3]{\\Delta CRL}+\\sqrt[3]{\\Delta CMQ}+\\sqrt[3]{\\Delta APM}+\\sqrt[3]{\\Delta ANR}+\\sqrt[3]{\\Delta BLP}+\\sqrt[3]{\\Delta BQN}\\le \\frac{1}{3}(\\frac{NB}{AB}+\\frac{RL}{NL}+\\frac{LC}{BC}+\\frac{NA}{AB}+\\frac{QM}{MN}+\\frac{MC}{AC}+\\frac{LC}{BC}+\\frac{PM}{LM}+\\frac{AM}{AC}+\\frac{LB}{BC}+\\frac{NR}{NL}+\\frac{AN}{AB}+\\frac{MC}{AC}+\\frac{PL}{ML}+\\frac{BL}{BC}+\\frac{AM}{AC}+\\frac{NQ}{MN}+\\frac{NB}{AB})$\n\nA little simplifying gives that the right hand side is simply $ 3$. So the maximum value is $ 3$ which is attained when $ MNL$ is the midpoint triangle of $ ABC$ and $ PQR$ is the midpoint triangle of $ MNL$.\n[/hide]", "Solution_3": "Very nice solution, cosinator. Congratulations.", "Solution_4": "The expressions of the areas of six triangles using areal coordinates are the same as in the solution of cosinator, with a weird notation...that of cosinator is much better.", "Solution_5": "[quote=\"cosinator\"]Notice $ \\Delta CRL = \\frac {LC}{BC}\\Delta CRB = \\frac {RL}{NL}\\frac {LC}{BC}\\Delta CNB = \\frac {NB}{AB}\\frac {RL}{NL}\\frac {LC}{BC}$ since $ \\Delta ABC = 1$. We similarly find for the other triangles \n\n$ \\Delta CMQ = \\frac {NA}{AB}\\frac {QM}{MN}\\frac {MC}{AC}$\n\n$ \\Delta APM = \\frac {LC}{BC}\\frac {PM}{LM}\\frac {AM}{AC}$ \n\n$ \\Delta ANR = \\frac {LB}{BC}\\frac {NR}{NL}\\frac {AN}{AB}$\n\n$ \\Delta BLP = \\frac {MC}{AC}\\frac {PL}{ML}\\frac {BL}{BC}$\n\n$ \\Delta BQN = \\frac {AM}{AC}\\frac {NQ}{MN}\\frac {NB}{AB}$\n\nNotice by AM-GM that \n\n$ \\sqrt [3]{\\Delta CRL} = \\sqrt [3]{\\frac {NB}{AB}\\frac {RL}{NL}\\frac {LC}{BC}}\\le\\frac {\\frac {NB}{AB} + \\frac {RL}{NL} + \\frac {LC}{BC}}{3}$\n\nWe find similar maximums for the other triangles. Add them up and we have\n\n$ \\sqrt [3]{\\Delta CRL} + \\sqrt [3]{\\Delta CMQ} + \\sqrt [3]{\\Delta APM} + \\sqrt [3]{\\Delta ANR} + \\sqrt [3]{\\Delta BLP} + \\sqrt [3]{\\Delta BQN}\\le \\frac {1}{3}(\\frac {NB}{AB} + \\frac {RL}{NL} + \\frac {LC}{BC} + \\frac {NA}{AB} + \\frac {QM}{MN} + \\frac {MC}{AC} + \\frac {LC}{BC} + \\frac {PM}{LM} + \\frac {AM}{AC} + \\frac {LB}{BC} + \\frac {NR}{NL} + \\frac {AN}{AB} + \\frac {MC}{AC} + \\frac {PL}{ML} + \\frac {BL}{BC} + \\frac {AM}{AC} + \\frac {NQ}{MN} + \\frac {NB}{AB})$\n\nA little simplifying gives that the right hand side is simply $ 3$. So the maximum value is $ 3$ which is attained when $ MNL$ is the midpoint triangle of $ ABC$ and $ PQR$ is the midpoint triangle of $ MNL$.\n[/quote]\r\n\r\nSplendid! I have nothing to add more than this.\r\n\r\nkunny" } { "Tag": [ "number theory unsolved", "number theory", "109" ], "Problem": "Prove that if $ a^2\\plus{}ac\\minus{}c^2\\equal{}b^2\\plus{}bd\\minus{}d^2$, for the integers $ a>b>c>d>0$, then $ ab\\plus{}cd$ is a composite number.", "Solution_1": "Does anybody know the solution?", "Solution_2": "[quote=\"hollandman\"]If $a^2+ac-c^2=b^2+bd-d^2$ for the integers $a>b\\geq c\\geq d\\geq 0$, then $ab+cd$ is a composite number.[/quote]$(ab+cd)(3b+c-3a-d)^2$\n\n$=\\left[(3b+c-3a-d)^2+2(a-b)(4b+3c-4a-3d)\\right]\\left[(3b+c-3a-d)^2+5(2a-b-c)(2b-d-a)\\right]$\n\n$+\\left(b^2+bd-d^2-a^2-ac+c^2\\right)\\left(b^2+4bc-c^2-a^2-4ad+d^2\\right)$\n\n$=\\left[(3b+c-3a-d)^2+2(a-b)(4b+3c-4a-3d)\\right]\\left[(3b+c-3a-d)^2+5(2a-b-c)(2b-d-a)\\right],$\n\nwhere $d(4b+3c-4a-3d)$\n\n$=(a-b)(3a+3b+3c-4d)+3(b-c)(c-d)+3(b^2+bd-d^2-a^2-ac+c^2)$\n\n$=(a-b)(3a+3b+3c-4d)+3(b-c)(c-d)>0,$\n\n$d(2b-d-a)=(a-b)(a+b+c-d)+c(b-c)+b^2+bd-d^2-a^2-ac+c^2$\n\n$=(a-b)(a+b+c-d)+c(b-c)>0.$\n\nSee also : http://bbs.cnool.net/topic_show.jsp?thesisid=494&id=41856369" } { "Tag": [ "trigonometry", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $ (a_n)_{n\\geq 0}$ and $ (b_n)_{n\\geq 0}$ two sequences such that : \r\n\r\n$ a_0>0$, $ a_{n\\plus{}1}\\equal{}a_n\\cdot e^{\\minus{}a_n}$, $ \\forall n\\geq 0$\r\n\r\n$ b_0 \\in (0,1)$, $ b_{n\\plus{}1}\\equal{}b_n \\cdot \\cos \\sqrt{b_n}$, $ \\forall n\\geq 0$.\r\n\r\nCompute $ \\lim\\limits_{n\\to\\infty}\\frac{a_n}{b_n}$.", "Solution_1": "Note that both $ a_n$ and $ b_n$ approaches to zero. Then by Stolz's theorem,\r\n\r\n\\begin{eqnarray*}\r\n\\lim_{n\\to\\infty} \\frac{a_n}{b_n}\r\n& = & \\lim_{n\\to\\infty} \\frac{1 / b_n}{1 / a_n}\\\\\r\n& = & \\lim_{n\\to\\infty} \\frac{(1/b_{n+1}) - (1/b_n)}{(1/a_{n+1}) - (1/a_n)}\\\\\r\n& = & \\lim_{n\\to\\infty} \\frac{\\frac{\\sec(\\sqrt{b_n}) - 1}{b_n}}{\\frac{e^{a\r\n_n} - 1}{a_n}}\\\\\r\n& = & \\frac{1}{2}\\\\\r\n\\end{eqnarray*}", "Solution_2": "[quote=\"sos440\"]\\begin{eqnarray*} & = & \\lim_{n\\to\\infty} \\frac {\\frac {\\sec(\\sqrt {b_n}) - 1}{b_n}}{\\frac {e^{a _n} - 1}{a_n}} \\\\\n& = & \\frac {1}{2} \\\\\n\\end{eqnarray*}[/quote]How do you conclude that limit is $ \\frac {1}{2}$?", "Solution_3": "Denominator clearly converges to $ 1$. Also, note that $ (\\sec \\sqrt{x})' \\equal{} \\frac{\\sin\\sqrt{x}}{2\\sqrt{x}} \\sec^2 \\sqrt{x}$, which converges to $ \\frac{1}{2}$ as $ x \\to 0^\\plus{}$. Thus the limit is $ \\frac{1}{2}$.", "Solution_4": "[quote=\"JoeBlow\"]\nHow do you conclude that limit is $ \\frac {1}{2}$?[/quote]\r\n\r\nAt the numerator we have to compute $ \\lim\\limits_{n\\to\\infty} \\frac {1 \\minus{} \\cos \\sqrt {b_n}}{b_n}$.But using the relation $ 1 \\minus{} \\cos \\alpha \\equal{} 2\\sin ^2 \\frac {\\alpha}{2}$ and the well-known $ \\lim\\limits_{x\\to 0} \\frac {\\sin x}{x} \\equal{} 1$ we reach the same result." } { "Tag": [ "function", "jkjkjkjkjk" ], "Problem": "The sum of the last four digits of Marian's phone number is 30. How many such four-digit sequences are there?", "Solution_1": "{9, 9, 9, 3} : 4\r\n{9, 9, 8, 4} : 12\r\n{9, 9, 7, 5} : 12\r\n{9, 9, 6, 6} : 6\r\n{9, 8, 8, 5} : 12\r\n{9, 8, 7, 6} : 24\r\n{9, 7, 7, 7} : 4\r\n{8, 8, 8, 6} : 4\r\n{8, 8, 7, 7} : 6\r\n\r\n$ 4\\plus{}12\\plus{}12\\plus{}6\\plus{}12\\plus{}24\\plus{}4\\plus{}4\\plus{}6\\equal{}\\boxed{84}$", "Solution_2": "An alternative way to solve this is using... generating functions!\r\n\r\nIn general, if we are looking for the solutions to any $ a_1 \\plus{} a_2 \\plus{} a_3 \\plus{} a_4 \\plus{} \\cdots \\plus{} a_n \\equal{} k$ we can look for the coefficient of $ x^k$ in the product of the generating functions of $ a_1, a_2, a_3, a_4, \\cdots, a_n$.\r\n\r\nWe are looking for $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 30$. Each of a, b, c, d can be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Thus their generating function is $ (x^0 \\plus{} x^1 \\plus{} x^2 \\plus{} x^3 \\plus{} x^4 \\plus{} x^5 \\plus{} x^6 \\plus{} x^7 \\plus{} x^8 \\plus{} x^9)$. \r\n\r\nSo the problem is equivalent to finding the coefficient of $ x^{30}$ in $ (1 \\plus{} x \\plus{} x^2 \\plus{} x^3 \\plus{} x^4 \\plus{} x^5 \\plus{} x^6 \\plus{} x^7 \\plus{} x^8 \\plus{} x^9)^4$. This is equal to $ 84$.", "Solution_3": "[quote=\"Kouichi Nakagawa\"]{9, 9, 9, 3} : 4\n{9, 9, 8, 4} : 12\n{9, 9, 7, 5} : 12\n{9, 9, 6, 6} : 6\n{9, 8, 8, 5} : 12\n{9, 8, 7, 6} : 24\n{9, 7, 7, 7} : 4\n{8, 8, 8, 6} : 4\n{8, 8, 7, 7} : 6\n\n$ 4 \\plus{} 12 \\plus{} 12 \\plus{} 6 \\plus{} 12 \\plus{} 24 \\plus{} 4 \\plus{} 4 \\plus{} 6 \\equal{} \\boxed{84}$[/quote]\r\n\r\nThis is what I did too (though I missed the ones that start with 8 on my first attempt). But can anyone comment on the intuition behind why this works out to $ \\dbinom{9}{3}$?", "Solution_4": "No guys, let's be tricky.\r\n\r\n$ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 30\\implies(9 \\minus{} a) \\plus{} (9 \\minus{} b) \\plus{} (9 \\minus{} c) \\plus{} (9 \\minus{} d) \\equal{} 6$\r\n\r\n$ \\implies\\binom{6 \\plus{} 4 \\minus{} 1}{4 \\minus{} 1} \\equal{} \\binom93 \\equal{} \\boxed{84}$.", "Solution_5": "What's a 'distribution'? (the topic that this question is in) :?:", "Solution_6": "[quote=math154]No guys, let's be tricky.\n\n$ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 30\\implies(9 \\minus{} a) \\plus{} (9 \\minus{} b) \\plus{} (9 \\minus{} c) \\plus{} (9 \\minus{} d) \\equal{} 6$\n\n$ \\implies\\binom{6 \\plus{} 4 \\minus{} 1}{4 \\minus{} 1} \\equal{} \\binom93 \\equal{} \\boxed{84}$.[/quote]\n\nNice trick! \n\nCommenting so I can remember to jot this tricky thinking down later. :P", "Solution_7": "[quote=math154]No guys, let's be tricky.\n\n$ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 30\\implies(9 \\minus{} a) \\plus{} (9 \\minus{} b) \\plus{} (9 \\minus{} c) \\plus{} (9 \\minus{} d) \\equal{} 6$\n\n$ \\implies\\binom{6 \\plus{} 4 \\minus{} 1}{4 \\minus{} 1} \\equal{} \\binom93 \\equal{} \\boxed{84}$.[/quote]\n\nwhy does this work?", "Solution_8": "I'm not sure either.\nDoes it have anything to do with the thing about $n$ candies and $k$ kids (ways to distribute) is $\\binom{n + k - 1}{k-1}$??", "Solution_9": "Please hide your solutions.", "Solution_10": "The solutions are from 2009. I doubt the people that posted the solutions are going to see your post.", "Solution_11": "[quote=Samchooo]The solutions are from 2009. I doubt the people that posted the solutions are going to see your post.[/quote]\n\nit does not matter ", "Solution_12": "[quote=mathmaster010][quote=Samchooo]The solutions are from 2009. I doubt the people that posted the solutions are going to see your post.[/quote]\n\nit does not matter[/quote]\n\nactually, it does because there isn't a point of posting[quote=ojaswupadhyay]Please hide your solutions.[/quote]\n\nanymore.", "Solution_13": "[quote=Samchooo][quote=mathmaster010][quote=Samchooo]The solutions are from 2009. I doubt the people that posted the solutions are going to see your post.[/quote]\n\nit does not matter[/quote]\n\nactually, it does because there isn't a point of posting[quote=ojaswupadhyay]Please hide your solutions.[/quote]\n\nanymore.[/quote]\n\nokay", "Solution_14": "Other people can explain the solution, the original poster of the solution isn't the only one who understands it " } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $X(t)$ be a solution of the equation $X'=f(X)$,where $f: R\\to R$.Show that $X$ is monotonuous.", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67252[/url]" } { "Tag": [ "function", "vector", "algebra", "functional equation", "algebra proposed" ], "Problem": "A calculating ruler is a ruler for doing algebric calculations. This ruler has three arms, two of them are sationary and one can move freely right and left. Each of arms is gradient. Gradation of each arm depends on the algebric operation ruler does. For eaxample the ruler below is designed for multiplying two numbers. Gradations are logarithmic.\r\n[img]http://aycu05.webshots.com/image/5604/2000468517162383885_rs.jpg[/img]\r\nFor working with ruler, (e.g for calculating $x.y$) we must move the middle arm that the arrow at the beginning of its gradation locate above the $x$ in the lower arm. We find $y$ in the middle arm, and we will read the number on the upper arm. The number written on the ruler is the answer.\r\n1) Design a ruler for calculating $x^{y}$. Grade first arm ($x$) and ($y$) from 1 to 10.\r\n2) Find all rulers that do the multiplication in the interval $[1,10]$.\r\n3) Prove that there is not a ruler for calculating $x^{2}+xy+y^{2}$, that its first and second arm are grade from 0 to 10.", "Solution_1": "If we want to calculate the value of the monotonically increasing with respect to each variable function $t(x,y)$ when $x,y\\in [a,b]$ with such a rule we need to find three monotonic increasing funtions $f,g,h$, so that $f(x)+g(y)=h(t(x,y))$, where the value each one returns is the distance in the corresponding sub-rule between the origin and the tick signalling the If we want to calculate the value of the monotonically increasing with respect to each variable function $t(x,y)$ when $x,y\\in [a,b]$ with such a rule we need to find three monotonic increasing funtions $f,g,h$, so that $f(x)+g(y)=h(t(x,y))$, where the value each one returns is the distance in the corresponding sub-rule between the origin and the tick signalling the number we introduced as argument. Without loss of generality we can suppose $f(a)=g(a)=h(t(a,a))=0$.\r\n1) We can take $f(x)=h(c)=ln(lnx)$, $g(y)=lny$, so that $f(x)+g(y)=ln(ln(x))+lny=ln(yln(x))=ln(ln(x^{y}))=h(x^{y})$. I don't know if there is something more to be said.\r\n2) We have $f(x)+g(y)=h(xy)$, $x,y\\in [1,10]$. By the assumption made at the beginnign we can let $f(1)=g(1)=h(1)=0$. This leads by alternatively letting $x=1$ and $y=1$, to $f=g=h$, and hence Cauchy's functional equation $f(x)+f(y)=f(xy)$ which in this case, being $f(x)$ monotonic, has only solution $f(x)=c (lnx)$ for arbitrary constant $c$; here $c>0$. [ To see this result is just necessary to make the change $g(x)=f(e^{x})$ in orther to get the first Cauchy f.e.: $g(x)+g(y)=g(x+y)$ ].\r\n3) By alternatively letting $x=0$, $y=0$ we get that $f(x)=g(x)=h(x^{2})$, so we should have $h(x^{2})+h(y^{2})=h(x^{2}+xy+y^{2})$.\r\nOn one hand making the change $x=y=\\sqrt{z}$, leads to $2h(z)=h(3z)$. On the other $x=\\frac{y}{\\sqrt{3}}=\\sqrt z$ leads to $h(z)+h(3z)=h((4+\\sqrt{3})z) \\Rightarrow 3h(z)=h((4+\\sqrt{3})z)$. Calling $q(x)=ln (h(e^{x}))$ (also monotonically increasing), and letting $S$ the set of vectors such that $(z,w)\\in S\\leftrightarrow q(z)=w$ we get from the previous two equations: $(v\\in S \\leftrightarrow v+a\\in S) \\wedge (v\\in S \\leftrightarrow v+b\\in S)$ where $a=(ln3,ln2)\\;\\;b=(ln(4+\\sqrt{3}),ln3)$. But $a$ and $b$ are linearly indipendent, so $\\forall N\\in \\mathbb{R}^{+},\\;\\exists (a,b) \\in (S\\cap [0,ln 10]\\times \\mathbb{R})/b>N$ which is clearly impossible. (I know this last part should be written more in detail; sorry.)\r\n\r\nBeing so long, there must be many mistakes." } { "Tag": [ "invariant", "combinatorics proposed", "combinatorics" ], "Problem": "let $\\alpha\\geq 0$ be a given real. let $0\\leq p_i\\leq 1$, $i=1,...,2n$ (actually, we could replace $2n$ by an odd number, but that makes the problem statement a bit more complicated).\r\nprove that there exists $0\\leq x\\leq 1$ s.t.\r\n\\[\\sum_{i=1}^{2n} \\frac 1{|x-p_i|^{\\alpha}}\\leq 2(4n)^{\\alpha}\\sum_{i=1}^{n} \\frac 1{(2i-1)^{\\alpha}}\\].\r\n\r\nthe original problem treated the case of $\\alpha=1$ and gave a slightly worse bound.", "Solution_1": "nobody wants to solve it(or is it just too hard?)? you have to try it since the solution i found (after maybe 2 days) is just too beautiful.", "Solution_2": "Actually I don't know if anyone made the effort to read the formula :). Could you please post the original IMO training problem. I think it's reasonable to give others the chance to start with the beginning :). Thanks.\r\n\r\nEdit: Thanks :)", "Solution_3": "Okay, the original problem is IMO training 2005, pre-Rostock homework set (I) problem 6.2:\r\n\r\n[i]Given n real numbers $p_1$, $p_2$, ..., $p_n$ such that $0\\leq p_i\\leq 1$ for all i = 1, 2, ..., n.\nProve the existence of a real number x with $0\\leq x\\leq 1$ satisfying\n\n$\\sum_{i=1}^n\\frac{1}{\\left|x-p_i\\right|}\\leq 8\\cdot n\\cdot\\left(1+\\frac13+\\frac15+...+\\frac{1}{2n-1}\\right)$.[/i]\r\n\r\nActually, this is a weaker version of the particular case of Peter's problem above for $\\alpha=1$.\r\n\r\nBy the way, I am thinking about whether it would make sense to post all the homework problems so far (18 problems) on ML. Actually, many of them are boring or trivial (this one, of course, not).\r\n\r\n darij", "Solution_4": "actually, the bound of the original problem for $n$ numbers is the same as my bound for $2n$ numbers(i don't know if that's just accidentally), therefore the original problem isn't really my problem for $\\alpha=1$.", "Solution_5": "ok, seems like no one is interested in that problem, therefore i'll post the solution.\r\n\r\nassume all points are distinct - the rest is done via taking the limit (or appropriately changing some slight details of the proof).\r\n\r\nplace a balloon at each of the points $p_i$. then blow up all balloons at the same speed. it's allowed for the balloons to leave the interval $[0; 1]$ and they may change their midpoints - only at the beginning they're centered at the points $p_i$. if two balloons touch each other, the point where they touch is invariant but they are still getting bigger, if three balloons are in a row, the first touching the second touching the third, the midpoint of the second is invariant and the two outer ones are pushed away by the second one which gets larger and so on - so all in all they are acting the way one conjectures them to do.\r\n\r\nnow, after some time, all points of the interval $[0; 1]$ are covered. in that moment, stop to blow the balloons. let $x$ be (one of the) last point(s) which are uncovered up to the moment everything out of the interval $[0; 1]$ is covered. we have two observations:\r\n\r\n1) if $r$ is the radius of all the balloons at the end, then the midpoints of all the neighboring balloons have distance $2r$. furthermore, since $[0; 1]$ is covered, $r\\geq \\frac 1{4n}$. $x$ is some point where the balloons touch. all in all, something like that:\r\n0----x---1\r\nOOOOOOO\r\n\r\n2) if we replace the points $p_i$ by the midpoints of the balloons, the sum of the distances (to the corresponding powers) gets at most bigger - one easily sees that by looking at what happens if we do this instantly(use e.g. karamata).\r\n\r\ntherefore we only have to evaluate the sum we get in the end, but that's pretty easy. we may assume that $r=\\frac 1{4n}$ since for larger $r$ the sum gets smaller and we may assume that $x$ is exactly in the middle of all the balloons. this gives the desired bound.\r\n\r\nbtw, this shows that there is one $x$ for all $\\alpha$.", "Solution_6": "didnt get it", "Solution_7": "[quote=\"Pascual2005\"]didnt get it[/quote]\r\nMe too.Can you,peter,write it in a more slightly way", "Solution_8": "This problem and the \"baloon idea\" remembers me about a nice problem Lhvietbao told me: If $A_1,A_2,\\ldots,A_n$ are points in the plane then the set of points $P$ that satisfy $|PA_1PA_2 \\ldots PA_n|\\leq 1$ can be covered by a set of disks with sum of radii at most $C$, where $C$ is some constant (independent on $n$). I was told $C=e$ is good, but I myself did manage a bigger bound. Also I think the assertion holds for any dimension. You can try this one, Peter.\r\n\r\nI have an idea similar to Peter's, but worse for this problem.\r\n\r\nFor any point $A_i, 1\\leq i \\leq 2n$ consider segment $S_i$ with length $\\frac 1{4n}$ and midpoint $A_i$. Associate with $S_i$ the point $A_i$. then, perform the following step while possible:\r\n\r\n(*)If there is some collection of segments $S_1,S_2,\\ldots S_k$ with total length $\\frac k{4n}$(so $k$ associated points) and they can be covered by a single segment with legth $\\frac k{4n}$, then replace them by this new segment $S$, and associate all the ponits previously associated to $S_1,S_2,\\ldots, S_k$ to $S$.\r\n\r\nIt's easy to see that each segment will have length $\\frac{k}{4n}$ for some $k$ and will have exactly $k$ associated points. Clearly each segment contains all the points associated to it and and each $A_i$ is associated to exactly one segment (it may belong to more segments, but is associated just to one).\r\n\r\nNow for each segment $S$ consider the \"fat\" segment $S'$ sharing the midpoint with $S$ but twice bigger. The total length of all fat segments is $1$, so some point $X$ is not covered by them. \r\n\r\nNow let's estimate $\\prod XA_i$. Firstly, let's see how many $A_i$ can be at distance at most $r$ from $X$. Suppose $S_1,S_2,\\ldots,S_k$ with lengths $\\frac{r_1}{4n}, \\frac{r_2}{4n}, \\ldots \\frac{r_k}{4n}$. As the fat segment $S_i'$ doesn't contain $X$, it means that $X$ is at distance at least $\\frac{r_k}{8n}$ from the segment $S_i$. Therefore $\\frac{r_k}{8n}\\leq r$ and so the segment centered at $x$ and with length $6r$ contains $S_i$. Since we cannot perform (*) for $S_1,S_2,\\ldots,S_k$ we deduce that $\\frac{r_1}{4n}+\\ldots+\\frac{r_k}{4n}<6r$ so $r_1+r_2+\\ldots+r_n<24rn$ so there are less than $24rn$ points at distance at most $r$ from $x$. \r\n\r\nTherefore, at most one point can have distance no more than $\\frac 1{24n}$ from $X$, at most two $\\frac{2}{24n}$ and so on. Hence the sum seeked in problem is at most $\\sum_{i=1}^n (\\frac{24n}{i})^{\\alpha}=(24n)^{\\alpha}\\sum_{i=1}^n \\frac 1{i^\\alpha}$. We can actually replace $24$ by $\\frac{2}{(\\sqrt 2-1)^2}=11.65$(and this is the best constant for my method), but this is still way too big compared to Peter's bound(however better than the original one).\r\n\r\nI guess I understood Peter's idea (very nice!), but I doubt I can explain it better than him. \r\n\r\nCan you generalize the problem for larger dimensions?" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "$X$ consists of many pairwise disjoint Lebesgue-measurable sets. Every set in $X$ has a positive measure.\r\n\r\nShow that $X$ is countable.", "Solution_1": "Two things: when you say \"$X$ is countable\", you must mean that there are countably many sets in the union that makes up $X$, not that the whole set is countable. Second: we must assume that the context you implied by \"Lebesgue measurable\" was in $\\mathbb{R}$ (or $\\mathbb{R}^n$, or at the very least some $\\sigma$-finite measure space.)\r\n\r\nLet's say we're working on $\\mathbb{R}$ and $X=\\cup X_{\\alpha}.$ We start by assuming that each $X_{\\alpha}$ is contained in $[-n,n]$. (We make that assumption by intersecting each set with that interval, and discarding those whose intersections have measure zero.) Now, how many of the $X_{\\alpha}$ have measure that exceeds $\\frac1k$? Since their disjoint union must fit into a set of finite measure (the interval $[-n,n]$), only finitely many. Taking the union over $k$ leaves us with countably many sets. Now remove the restriction to the interval $[-n,n]$ by taking the union over $n$. That leaves us with a countable union of countable collections of sets, hence a countable collection." } { "Tag": [], "Problem": "Given a set of 1999 closed unit intervals on the real line, S is the set of points in exactly an odd number of such intervals.\r\n\r\nShow $|S| \\geq 1$, where we understand $|S|$ to be the length of the finite union of all it's disjoint intervals.\r\n\r\nI solved it, its good Q", "Solution_1": "Please note that this is problem from the May version of Olymon, which is not due until June 15:\r\n[url=http://www.math.toronto.edu/~barbeau/olymon57.pdf]www.math.toronto.edu/~barbeau/olymon57.pdf[/url]", "Solution_2": "It seems like May Olymon is too easy this month; in one day I already solved 6/7" } { "Tag": [ "inequalities", "three variable inequality" ], "Problem": "[size=200]Let a,b,c be real positive numbers and a+b+c=1, Prove that \n\n$ 5 \\geq \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1} \\geq \\frac {19}{6}$\n\nBQ[/size]", "Solution_1": "[quote=\"xzlbq\"][size=200]Let a,b,c be real positive numbers and a+b+c=1, Prove that \n\n$ 5 \\geq \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1} \\geq \\frac {19}{6}$\n\nBQ[/size][/quote]\r\n\r\n\r\ntry $ a \\equal{} \\frac {x}{x \\plus{} y \\plus{} z},b \\equal{} \\frac {y}{x \\plus{} y \\plus{} z},c \\equal{} \\frac {z}{x \\plus{} y \\plus{} z}$\r\n\r\n$ x,y,z > 0$\r\n\r\n\r\n$ \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1} \\geq \\frac {19}{6}$\r\n\r\n$ \\Longleftrightarrow169yzx \\plus{} 88z^3 \\plus{} 63y^3 \\plus{} 38zx^2 \\plus{} 120z^2x \\plus{} 244yz^2 \\plus{} 6x^2y \\plus{} 75y^2x \\plus{} 213y^2z\\geq 0$\r\n\r\n$ 5 \\geq \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1}$\r\n\r\n$ \\Longleftrightarrow 91yzx\\plus{}22x^3\\plus{}6y^3\\plus{}56zx^2\\plus{}35z^2x\\plus{}7yz^2\\plus{}65x^2y\\plus{}48y^2x\\plus{}14y^2z\\geq 0$", "Solution_2": "[size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers and $ a\\plus{}b\\plus{}c\\equal{}1$, research\n\n\n$ f(k_1,k_2,k_3) \\geq \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1} \\geq g(k_1,k_2,k_3)$\n\n [/size]\r\n BQ", "Solution_3": "[quote=\"xzlbq\"][size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers and $ a \\plus{} b \\plus{} c \\equal{} 1$, research\n\n\n$ f(k_1,k_2,k_3) \\geq \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1} \\geq g(k_1,k_2,k_3)$\n\n [/size]\n BQ[/quote]\r\n\r\nwhat ??", "Solution_4": "[size=200]if $ a\\plus{}b\\plus{}c\\equal{}1$,then:\n$ \\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{a} \\geq \\frac{1\\minus{}a}{1\\minus{}b}\\plus{}\\frac{1\\minus{}b}{1\\minus{}c}\\plus{}\\frac{1\\minus{}c}{1\\minus{}a}$[/size]\r\nBQ", "Solution_5": "Isn't this spamming?\r\nIsn't he/she should get banned or something his/her history is full of spamming.", "Solution_6": "[quote=\"hedeng123\"][quote=\"xzlbq\"][size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers and $ a \\plus{} b \\plus{} c \\equal{} 1$, research\n\n\n$ f(k_1,k_2,k_3) \\geq \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1} \\geq g(k_1,k_2,k_3)$\n\n [/size]\n BQ[/quote]\n\nwhat ??[/quote]\r\n\r\n[size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers and $ a \\plus{} b \\plus{} c \\equal{} 1$, research\n\n\n$ f(k_1,k_2,k_3) \\geq \\frac {a \\plus{} k_1}{b \\plus{} k_2} \\plus{} \\frac {b \\plus{} k_2}{c \\plus{} k_3} \\plus{} \\frac {c \\plus{} k_3}{a \\plus{} k_1} \\geq g(k_1,k_2,k_3)$\n\n [/size]\r\n BQ", "Solution_7": "[quote=\"xzlbq\"][size=200]Let a,b,c be real positive numbers and a+b+c=1, Prove that \n\n$ 5 \\geq \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1} \\geq \\frac {19}{6}$\n\nBQ[/size][/quote]due to the conditions, we have $ c \\plus{} 3\\ge b \\plus{} 2\\ge a \\plus{} 1\\ge1$. So the minimum is attained for $ c \\equal{} 0$ and the maximum for $ a \\equal{} 0$. For both cases, the problem is then reduced to a 2 variable problem. The same argument again shows: minimum for $ (1,0,0)$ and the maxmum for $ (0,0,1)$.\n\nEDIT: just saw\n[quote=\"xzlbq\"][quote=\"hedeng123\"][quote=\"xzlbq\"][size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers and $ a \\plus{} b \\plus{} c \\equal{} 1$, research\n\n\n$ f(k_1,k_2,k_3) \\geq \\frac {a \\plus{} 1}{b \\plus{} 2} \\plus{} \\frac {b \\plus{} 2}{c \\plus{} 3} \\plus{} \\frac {c \\plus{} 3}{a \\plus{} 1} \\geq g(k_1,k_2,k_3)$\n\n [/size]\n BQ[/quote]\n\nwhat ??[/quote]\n\n[size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers and $ a \\plus{} b \\plus{} c \\equal{} 1$, research\n\n\n$ f(k_1,k_2,k_3) \\geq \\frac {a \\plus{} k_1}{b \\plus{} k_2} \\plus{} \\frac {b \\plus{} k_2}{c \\plus{} k_3} \\plus{} \\frac {c \\plus{} k_3}{a \\plus{} k_1} \\geq g(k_1,k_2,k_3)$\n\n [/size]\n BQ[/quote]\r\n\r\nI suppose you mean: give the minimum and the maximum of $ \\frac {a \\plus{} k_1}{b \\plus{} k_2} \\plus{} \\frac {b \\plus{} k_2}{c \\plus{} k_3} \\plus{} \\frac {c \\plus{} k_3}{a \\plus{} k_1}$ in terms of $ k_1,k_2,k_3$. Why not say it like that :?:", "Solution_8": "[size=200]Let$ x_1,x_2,x_3,x_4$ be real positive numbers and$ x_1\\plus{}x_2\\plus{}x_3\\plus{}x_4\\equal{}1$, Prove that \n\n$ \\frac {x_1}{x_2} \\plus{} \\frac {x_2}{x_3} \\plus{} \\frac {x_3}{x_4} \\plus{} \\frac {x_4}{x_1} \\geq \\frac {1 \\minus{} x_1}{1 \\minus{} x_2} \\plus{} \\frac {1 \\minus{} x_2}{1 \\minus{} x_3} \\plus{} \\frac {1 \\minus{} x_3}{1 \\minus{} x_4} \\plus{} \\frac {1 \\minus{} x_4}{1 \\minus{} x_1}$\n\nBQ[/size]", "Solution_9": "[quote=\"xzlbq\"][size=200]if $ a \\plus{} b \\plus{} c \\equal{} 1$,then:\n$ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\geq \\frac {1 \\minus{} a}{1 \\minus{} b} \\plus{} \\frac {1 \\minus{} b}{1 \\minus{} c} \\plus{} \\frac {1 \\minus{} c}{1 \\minus{} a}$[/size]\nBQ[/quote]\r\n\r\nthis was an old problem I remember...", "Solution_10": "n variable?\r\nBQ", "Solution_11": "[quote=\"xzlbq\"][size=200]if $ a \\plus{} b \\plus{} c \\equal{} 1$,then:\n$ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\geq \\frac {1 \\minus{} a}{1 \\minus{} b} \\plus{} \\frac {1 \\minus{} b}{1 \\minus{} c} \\plus{} \\frac {1 \\minus{} c}{1 \\minus{} a}$[/size]\nBQ[/quote]\n\n[quote=\"xzlbq\"][size=200]Let$ x_1,x_2,x_3,x_4$ be real positive numbers and$ x_1 \\plus{} x_2 \\plus{} x_3 \\plus{} x_4 \\equal{} 1$, Prove that \n\n$ \\frac {x_1}{x_2} \\plus{} \\frac {x_2}{x_3} \\plus{} \\frac {x_3}{x_4} \\plus{} \\frac {x_4}{x_1} \\geq \\frac {1 \\minus{} x_1}{1 \\minus{} x_2} \\plus{} \\frac {1 \\minus{} x_2}{1 \\minus{} x_3} \\plus{} \\frac {1 \\minus{} x_3}{1 \\minus{} x_4} \\plus{} \\frac {1 \\minus{} x_4}{1 \\minus{} x_1}$\n\nBQ[/size][/quote]\r\n\r\nCouldn't you start a new thread for these? :mad:", "Solution_12": "[quote=\"xzlbq\"][size=200]Let$ x_1,x_2,x_3,x_4$ be real positive numbers and$ x_1 \\plus{} x_2 \\plus{} x_3 \\plus{} x_4 \\equal{} 1$, Prove that \n\n$ \\frac {x_1}{x_2} \\plus{} \\frac {x_2}{x_3} \\plus{} \\frac {x_3}{x_4} \\plus{} \\frac {x_4}{x_1} \\geq \\frac {1 \\minus{} x_1}{1 \\minus{} x_2} \\plus{} \\frac {1 \\minus{} x_2}{1 \\minus{} x_3} \\plus{} \\frac {1 \\minus{} x_3}{1 \\minus{} x_4} \\plus{} \\frac {1 \\minus{} x_4}{1 \\minus{} x_1}$\n\nBQ[/size][/quote]\r\n\r\n\r\ntsds 1 time. :lol:" } { "Tag": [ "function", "algebra", "domain", "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "1) Give an example to show that a sequence of functions may be uniformly continuous, pointwise equicontinuous, but not uniformly equicontinuous, when their domain is noncompact?\r\n\r\n2) Give an example of a sequence of smooth equicontinuous functions fn: [a,b] -> R (real line) whose derivatives are unbounded.\r\n\r\n3) If f: R->R is continuous and the sequence fn(x) = f(nx) is equicontinuous, what can be said about f???\r\n\r\nThanks in advance..", "Solution_1": "2) $f_n(x)=\\frac{x^n}{\\sqrt{n}}$ on $[0,1]$.", "Solution_2": "1) what is \" pointwise equicontinuous \"?\r\n3) what we need to say about $f$? :)", "Solution_3": "The term \"pointwise equicontinuous\" is not very common. If I understand it right, it means that we have a family of functions defined on some domain $D$ such that for every $\\varepsilon>0$ and every $x\\in D$, there exists $\\delta=\\delta(\\varepsilon,x)>0$ such that $|y-x|<\\delta$ implies that $|f(y)-f(x)|<\\varepsilon$ for all $f$ in our family.\r\n\r\nAs to the second question, \"What can be said about $f$?\" is just a politically correct form of \"Find all such $f$!\" :lol:", "Solution_4": "So there must be $f=const$ in 3) i think.\r\nSuppose $x < y$ and $f(x)\\neq f(y)$. Then $|f(x)-f(y)|=2\\varepsilon >0$. Then $\\exists \\delta >0$ such that\r\n$|x_1-x_2|<\\delta \\forall \\; n \\; \\Rightarrow |f_n(x_1)-f_n(x_2)|<\\varepsilon.$ Let $n>\\frac{|x-y|}{\\delta}$. Then $\\left|\\frac{x}{n} - \\frac{y}{n} \\right|< \\delta$ and we must have $|f_n(x/n)-f_n(y/n)|<\\varepsilon$. But \\[ 2\\varepsilon = |f(x)-f(y)| = |f_n(x/n) - f_n(y/n)| < \\varepsilon \\] Contradiction...\r\nOne more remark : we don't need condition that $f \\in C({\\mathbb R})$... :)" } { "Tag": [ "geometry", "perimeter", "calculus", "integration", "rotation", "geometric transformation", "reflection" ], "Problem": "In how many non-congruent triangles of perimeter 15 do all the sides have an integral length?\r\n\r\nNo Calculator", "Solution_1": "is it just me or is this one not that hard?\r\n[hide=\"I dont know why anyone would need a calculator\"]\n5-5-5, 6-5-4, 7-5-3, 6-6-3, 7-6-2, 7-7-1, 7-4-4\n\n7 triangles\n[/hide]", "Solution_2": "[quote=\"mathgeniuse^ln(x)\"]In how many non-congruent triangles of perimeter 15 do all the sides have an integral length?\n\nNo Calculator[/quote]\r\n\r\n[hide]5-5-5\n5-6-4\n4-7-4\n5-7-3\n3-6-6\n6-2-7\n1-7-7\n\n9 ways.[/hide]", "Solution_3": "Would you need to count scalene triangles twice? Because I don't think these are the same:\r\n\r\n[img]http://i9.photobucket.com/albums/a70/littlekasper/triangles.jpg[/img]", "Solution_4": "[quote=\"DanK\"]Would you need to count scalene triangles twice? Because I don't think these are the same:\n\n[img]http://i9.photobucket.com/albums/a70/littlekasper/triangles.jpg[/img][/quote]\r\n\r\nThose are congruent though.", "Solution_5": "[quote=\"boo\"][quote=\"mathgeniuse^ln(x)\"]In how many non-congruent triangles of perimeter 15 do all the sides have an integral length?\n\nNo Calculator[/quote]\n\n[hide]5-5-5\n5-6-4\n4-7-4\n5-7-3\n3-6-6\n6-2-7\n1-7-7\n\n9 ways.[/hide][/quote]\r\n\r\ncan you count?\r\nthat's 7", "Solution_6": "[quote=\"boo\"][quote=\"mathgeniuse^ln(x)\"]In how many non-congruent triangles of perimeter 15 do all the sides have an integral length?\n\nNo Calculator[/quote]\n\n[hide]5-5-5\n5-6-4\n4-7-4\n5-7-3\n3-6-6\n6-2-7\n1-7-7\n\n9 ways.[/hide][/quote]\r\n\r\nI think you mean 7 right.\r\n\r\nEitherway, that is the correct answer we were looking for.", "Solution_7": "Are they congruent? You can never rotate them so the sides are arranged the same way. Doesn't that constitute a different shape? And since congruent figures have identical size and shape, that would make them not congruent?", "Solution_8": "[quote=\"DanK\"]Are they congruent? You can never rotate them so the sides are arranged the same way. Doesn't that constitute a different shape? And since congruent figures have identical size and shape, that would make them not congruent?[/quote] You can reflect them though. As long as it doesn't require transformations.", "Solution_9": "Ok. Thanks.", "Solution_10": "As long as you rotate, reflect, or translate, that is okay." } { "Tag": [ "geometry", "AoPS Books", "AMC 10", "number theory", "prime factorization", "\\/closed" ], "Problem": "What classes would cover material that is not covered in AoPS 1 or 2. Which ones contain strategies that are not contained in 1 or 2?", "Solution_1": "what exactly does that mean...are you referring to strategies not included in those books but at the same level or a higher level like olympiad?", "Solution_2": "[quote=\"PenguinIntegral\"]What classes would cover material that is not covered in AoPS 1 or 2. Which ones contain strategies that are not contained in 1 or 2?[/quote]\r\n\r\nThe intro classes have considerable overlap with the AoPS books, but go into greater detail with an even higher focus on strategy. The non-intro classes contain a considerable amount of material that is not in the AoPS books.", "Solution_3": "[quote=\"rrusczyk\"][quote=\"PenguinIntegral\"]What classes would cover material that is not covered in AoPS 1 or 2. Which ones contain strategies that are not contained in 1 or 2?[/quote]\n\nThe intro classes have considerable overlap with the AoPS books, but go into greater detail with an even higher focus on strategy. The non-intro classes contain a considerable amount of material that is not in the AoPS books.[/quote]\r\n\r\nSo what would your reccomend for someone(me) who wants to get the most for their dollar? Are the intro courses worth it for someone with both Aops books(Just got the second one and the solutions for the first, still as brilliant as ever!)", "Solution_4": "you should try doing the \"are you ready\" pretests and \"do you need this\" post-tests to the classes to see where you stand.", "Solution_5": "[quote=\"ffdbzathf\"]you should try doing the \"are you ready\" pretests and \"do you need this\" post-tests to the classes to see where you stand.[/quote]\r\n\r\nThis is a good recommendation.\r\n\r\nAs for what best benefit for your dollar, it depends on what type of student you are. If you are the type who gets a ton out of 1-1 interaction and being able to bring your own questions to class, the Independent Study is your best bet. If you have mastered the Intro AoPS book, much of the Intro classes will be review for you (but certainly not all).\r\n\r\nWhat areas are you weak/strong in?", "Solution_6": "[quote=\"rrusczyk\"][quote=\"ffdbzathf\"]you should try doing the \"are you ready\" pretests and \"do you need this\" post-tests to the classes to see where you stand.[/quote]\n\nThis is a good recommendation.\n\nAs for what best benefit for your dollar, it depends on what type of student you are. If you are the type who gets a ton out of 1-1 interaction and being able to bring your own questions to class, the Independent Study is your best bet. If you have mastered the Intro AoPS book, much of the Intro classes will be review for you (but certainly not all).\n\nWhat areas are you weak/strong in?[/quote]\r\nI don't know, really. A few days ago, I went through the first fifteen questions of an amc 10, and I learned alot from the solutions of the three I missed.\r\n\r\nI remember learning two things. The first was that square has all the exponents of its prime factorization even(I promptly smacked myself on the head and exclaimed \"duh!\") and that recognizing factorizations is a very useful tool. I haven't found much use for the factorizations yet, but thinking of squares as prime factorizations led me to some unique solutions in the \"Using the integers\" sections of vol 1 of AoPS. \r\n(This was also before I got the solution manual, yesterday)\r\n\r\nI also spend generous amounts of time (>2 hours) thinking about hard problems. The technique used to solve it is usually simple when I look up the answer, and because I remeber some peticularitys of the problems, I sometimes use that method on other problems with the same structure.\r\n\r\nI like doing problems more than I like reading about how to do problems. If a new technique is introduced in a solution, I feel like I know it better because I have worked with the problem a lot and noted some things about it. That makes things stick better than reading a chapter of AoPS for me. I usually forget a lot.", "Solution_7": "If you are trying to choose a subject class, I'd fall back on ff's suggestion - use the pre & post tests as a guide.\r\n\r\nYour approach to learning (problems, problems, problems) is, of course, the most effective. Reading only primes the mind for discovery - you have to get in and think about problems to really understand the tools.", "Solution_8": "[quote=\"rrusczyk\"]If you are trying to choose a subject class, I'd fall back on ff's suggestion - use the pre & post tests as a guide.\n\nYour approach to learning (problems, problems, problems) is, of course, the most effective. Reading only primes the mind for discovery - you have to get in and think about problems to really understand the tools.[/quote]\r\nThanks for the complements on my learning method ;) \r\n\r\nI really have to talk with my parents on this one.", "Solution_9": "Hmm, I'm not sure I understand the merits on spending >2 hrs. on an AoPs/Contest problem. Usually I stop after 30-40 minutes if I'm not making progress. And when I look at the solution I usually find that I was thinking entirely in the wrong direction and probably wouldn't have thought of the solution anyway. This way I can gain some insight and move through problems moderately quickly, yet still only have to look up the solution to 1-2 of the earlier problems in each chapter of AoPs vol.2. But then again, I may be rushing for fear of the upcoming AIME...", "Solution_10": "Generally, I strongly agree with EFuzzy's approach to training - put a shot clock on yourself when working problems. Make it longer for harder problems. After that time, read the solution and learn from it. Then do the problem again in a couple weeks." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all real functions $f: \\mathbb R\\to \\mathbb R$ satisfying\n\\[f\\left(x^{2}+y\\right)=2f(x)+f(y^{2}),\\]\nfor all $x$ and $y$ in the domain.", "Solution_1": "did u mean this $f(x^{2}+y)=2f(x)+f(y^{2})$ :huh:", "Solution_2": "[quote=\"arkhammedos\"]did u mean this $f(x^{2}+y)=2f(x)+f(y^{2})$ :huh:[/quote]\r\n\r\nIf it is $P(x,y)$ : $f(x^{2}+y)=2f(x)+f(y^{2})$, then :\r\n$P(0,0)$ $\\implies$ $f(0)=3f(0)$ $\\implies$ $f(0)=0$\r\n$P(0,x)$ $\\implies$ $f(x)=f(x^{2})$\r\n$P(x,0)$ $\\implies$ $f(x^{2})=2f(x)$\r\nSo $2f(x)=f(x)$ and the unique solution is $f(x)=0$", "Solution_3": "sorry, is =\r\nanyway, thank for try helping." } { "Tag": [ "calculus", "integration", "function", "logarithms", "limit", "calculus computations" ], "Problem": "A little help?\r\n\r\nFor what value of a does the integral\r\n\r\n$\\int^{\\infty}_{1}(\\frac{ax}{x^{2}+1}-\\frac{1}{2x})dx$\r\n\r\nconverge? Evaluate the corresponding integral(s). (Give the answer as a function of a.)\r\n\r\nThanks.", "Solution_1": "The integrand is $\\frac{(2a-1)x^{2}-1}{2x(x^{2}+1)}$. If $2a-1\\ne 0$, this can be compared (by Limit Comparison test) to $1/x$, hence the integral diverges. If $2a-1=0$, the integral is convergent by comparison to $1/x^{3}$. \r\n\r\nThe antiderivative is $\\frac{a}{2}\\log(x^{2}+1)-\\frac{1}{2}\\log x$. When $a=1/2$, this simplifies to $\\frac14\\log\\frac{x^{2}+1}{x^{2}}$, which $\\to 0$ as $x\\to \\infty$. The integral is $-\\frac14\\log 2$. \r\n\r\n[quote]Give the answer as a function of a. [/quote]\r\n :what?:", "Solution_2": "I don't think we've learned those techniques in my class yet.\r\nThis is my answer so far, perhaps someone could check it for me. And tell me if I can even do it this way.\r\n\r\n$\\int^{\\infty}_{1}(\\frac{ax}{x^{2}+1}-\\frac{1}{2x})dx=[\\frac{a}{2}\\ln{x^{2}+1}-\\ln{2x}]^{\\infty}_{1}\\\\\\\\ =\\lim_{b\\rightarrow\\infty}[(\\frac{a}{2}\\ln{b^{2}+1}-\\ln{2b})-(\\frac{a}{2}\\ln{1^{2}+1}-\\ln{2(1)})]\\\\\\\\ =\\lim_{b\\rightarrow\\infty}[(\\frac{a}{2}\\ln{b^{2}+1}-\\ln{2b})-(\\frac{a-2}{2}\\ln{2})]$\r\n\r\nSo after I found this, I evaluated the function for several values of a, and found that when $1n$ we can change $ x_{i}\\equal{}0$ for $ i>n$.\r\nLet $ k\\equal{}[\\sqrt n]$, then obviosly $ x_{1}\\plus{}x_{2}\\plus{}...\\plus{}x_{n}\\le\\sum_{i\\equal{}1}^{k}(2i\\plus{}1)x_{i^{2}}\\le 3sum_{i\\equal{}1}^{k}ix_{i^{2}}\\le 3.$" } { "Tag": [ "FTW", "ratio", "blogs", "AMC", "AIME" ], "Problem": "[u]Private message[/u] me your answers to see how well you did! [b][color=red](Please [u]don't[/u] post your answers here.)[/color][/b]\r\n\r\n1. What is the maximum possible time allowance per question in a FTW game? (1 point)\r\n\r\n2. What is the maximum number of players in a normal game? (1 points)\r\n\r\n3. At what ratings do you receive the second and fourth rating badges, respectively? (4 points)\r\n\r\n4. An FTW normal game is Ranking-scored, has 10 questions, with 35 seconds per question. Find the difference between the maximum and minimum amount of time from when the \"Start game\" button is clicked to when the results screen is shown, assuming no lags or glitches. Show all your work. (12 points)\r\n\r\n5. Without glitches, how many seconds after time is up can you choose to \"Give up\"? Round your answer to the nearest tenth. Note that the answer may be zero or negative. (3 points)\r\n\r\n6. What is the most common answer to an FTW question? (2 points)\r\n\r\n7. When are you allowed to reveal someone else's real name and not have it be considered a ToS infringement? (2 points)\r\n\r\n8. Let $ \\frac {x}{y}$ be the ratio in lowest integer terms of the length of a LanguageFilter ban to the length of a FloodingFilter ban. Find $ x^3 \\minus{} y^3$. (3 points)\r\n\r\n9. What causes a Guest to have a five-digit or four-digit number? (3 points)\r\n\r\n10. How can a Guest play more games than the two that are allowed per day? (2 points)\r\n\r\n11. Who was the first moderator (not administrator) of FTW chat? (2 points)\r\n\r\n12. The 10-day period from March 21 to March 30 saw the joining to AoPS of four users that would eventually prove to be very influential to FTW, two of which joined on the same day. Name these four users. (8 points)\r\n\r\n13. The FTW Leaderboard's clock is set in what time zone? (2 points)\r\n\r\n14. In the year 2008, one user won over twice as many normal games and the user who placed second in normal games won. Name these two respective users (6 points), and for the nerdy bonus points (6 bonus points), how many games did each of them win? \r\n\r\n15. Name the first five users to pass 2000 rating, in order (Hint: One of the five is banned.). (10 points)\r\n\r\n[Freebee, hopefully]. Who programmed FTW? (5 points)\r\n\r\nDescribe the following FTW glitches. (3 points each)\r\n16. 8x\r\n17. Flashies\r\n18. Second Lobby\r\n19. Trapped Game\r\n20. Fall 2009 Rating Save Glitch\r\n\r\nWho do the following nicknames refer to? Provide the full formal username. (2 points each)\r\n21. Lonny\r\n22. Eli\r\n23. Levlev\r\n24. Ding\r\n25. Mabbitt\r\n26. Ylxy\r\n27. Rizzy\r\n\r\nAttribute these FTW chat quotes. (1 point each)\r\n28. \"I never said you never said [that] I never said I did.\"\r\n29. \"Not really, it went +2 +2 +2 -12 +2 +3.\"\r\n30. \"Sicily is a god dang multi like me?\"\r\n31. \"Special ways are izzy...i mean easy.\"\r\n32. \"I'm going to have fun in the shower with my friends.\"\r\n\r\nIn about a week, I will post a scoreboard up with the best scores! Yes, partial credit is given for some questions. \r\n[color=red]\nA few corrections (you can resubmit for questions with problems): \n\n#12: in 2008\n#20: the [i]second[/i] rating glitch[/color]\r\n\r\nAgain, just in case: \r\n[u]Private message[/u] me your answers to see how well you did! [b][color=red](Please [u]don't[/u] post your answers here.)[/color][/b]", "Solution_1": "[quote=\"phirekalk\"]\n15. Name the first five users to pass 2000 rating, in order (Hint: One of the five is banned.). (10 points) [/quote]actually, there are 2 banned.", "Solution_2": "I can't believe that there was no multis section in this, considering how 2/3 of these posts are from multis. And even more unbelievable is that r15s11z55y89w21 was the one who posted this quiz, and still no multi part.", "Solution_3": "A multis section would be a leak of classified intelligence and a breach of ethics. And dragon96, 2/3 (now 3/4) of the posts have users with legal multis, but all have some type of multi.", "Solution_4": "It is the end of Day 1 of the presentation of this test, and six people have submitted!\r\n\r\n[u]Leaderboard[/u]\r\n1. dragon96 - 63\r\n2. ilovepink - 37\r\n3. rizikin - 36\r\n4. k00lperson - 35\r\n5(tie). agentcx - 34\r\n5(tie). vahalla - 34", "Solution_5": "woah!! i actually got a 35!??????\r\nWOO HOO!!!!!!!!!!!!!!!!!", "Solution_6": "Yay, I got a 53! Also, for the first five people who reached 2000, I think there is only 1 multi, since it is only the first five people.", "Solution_7": "i'm amazed. \r\nThat\r\npinky beat me\r\nonly\r\nby\r\n2 points\r\nheh", "Solution_8": "Final Scoreboard: \r\n\r\n1. dragon96 - 63\r\n2. xxrxxhxx - 53\r\n3. ilovepink - 37\r\n4. rizikin - 36\r\n5. k00lperson - 35\r\n6(tie). agentcx - 34\r\n6(tie). vahalla - 34\r\n\r\nAnswers!\r\n\r\n1. What is the maximum possible time allowance per question in a FTW game? (1 point)\r\n[b]120 seconds[/b]\r\n\r\n2. What is the maximum number of players in a normal game? (1 points)\r\n[b]20[/b]\r\n\r\n3. At what ratings do you receive the second and fourth rating badges, respectively? (4 points) [b]1350 and 1750[/b]\r\n\r\n4. An FTW normal game is Ranking-scored, has 10 questions, with 35 seconds per question. Find the difference between the maximum and minimum amount of time from when the \"Start game\" button is clicked to when the results screen is shown, assuming no lags or glitches. Show all your work. (12 points)\r\n[b]346[/b]; [i]an explanation is on my [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=328532]blog[/url][/i]\r\n\r\n5. Without glitches, how many seconds after time is up can you choose to \"Give up\"? Round your answer to the nearest tenth. Note that the answer may be zero or negative. (3 points) \r\n[b]1 second[/b]; [i]you can give up any time before \"second 0\" is over[/i]\r\n\r\n6. What is the most common answer to an FTW question? (2 points)\r\n[b]4[/b]; [i]on harder problems, however, the answer is more likely to be 6, 8, 15, or 28, so guess those instead of 4[/i]\r\n\r\n7. When are you allowed to reveal someone else's real name and not have it be considered a ToS infringement? (2 points)\r\n[b]When it's part of their username[/b]\r\n\r\n8. Let $ \\frac {x}{y}$ be the ratio in lowest integer terms of the length of a LanguageFilter ban to the length of a FloodingFilter ban. Find $ x^3 \\minus{} y^3$. (3 points)\r\n[b]19[/b]; [i]a LanguageFilter ban is 15 minutes while a FloodingFilter ban is 10 minutes, these are in the ratio 3:2, and 3\u00b3-2\u00b3=27-8=19[/i]\r\n\r\n9. What causes a Guest to have a five-digit or four-digit number? (3 points)\r\n[b]Guest was created from a proxy[/b] OR [b]Guest was created while a \"second lobby\" formed[/b]\r\n\r\n10. How can a Guest play more games than the two that are allowed per day? (2 points)\r\n[b]Click on a game icon, wait for someone else to enter, join button automatically appears[/b]\r\n\r\n11. Who was the first moderator (not administrator) of FTW chat? (2 points)\r\n[b]iin77[/b]; [i]now inactive because he graduated from high school, note that levans is an administrator[/i]\r\n\r\n12. The 10-day period from March 21 to March 30 saw the joining to AoPS of four users that would eventually prove to be very influential to FTW, two of which joined on the same day. Name these four users. (8 points)\r\n[b]isabella2296, hurdler, AIME15, jjx1[/b]\r\n\r\n13. The FTW Leaderboard's clock is set in what time zone? (2 points)\r\n[b]Eastern Time[/b]\r\n\r\n14. In the year 2008, one user won over twice as many normal games and the user who placed second in normal games won. Name these two respective users (6 points), and for the nerdy bonus points (6 bonus points), how many games did each of them win? \r\n[b]isabella2296 - 1097 games won[/b]\r\n[b]iin77 - 540 games won[/b]\r\n\r\n15. Name the first five users to pass 2000 rating, in order (Hint: One of the five is banned.). (10 points)\r\n[b]AIME15, Mewto55555, Ttocs45, mousy, Jongy[/b]; [i]note that Jongy had username \"Jongao\" previously[/i]\r\n\r\n[Freebee, hopefully]. Who programmed FTW? (5 points)\r\n[b]levans[/b]\r\n\r\nDescribe the following FTW glitches. (3 points each)\r\n16. 8x\r\n[b]In FTW2, at the start, one of the bugs were that inputting one line will result in eight lines of your text appearing. [/b]\r\n17. Flashies\r\n[b]The game screen flashes, often with multiple questions, making it hard to read the question or submit an answer. [/b]; [i]neoflashies, which replaced flashies, occur less frequently but are also more detrimental to your game score, as you can't submit at all if your computer contracts neoflashies[/i]\r\n18. Second Lobby\r\n[b]Upon exiting a game, you find yourself in a different lobby that eventually everyone else also moves to upon refreshing. [/b]; [i]any game attempted to be made during the formation of a second lobby gets marked \"undefined\"[/i]\r\n19. Trapped Game\r\n[b]Game fails to close properly, and someone else joins causing the game to error and not be able to proceed to the final scoreboard. [/b]; [i]occurs much less frequently now, but an interesting rise in cases occurred last week[/i]\r\n20. Fall 2009 Rating Save Glitch\r\n[b]Leaving a losing game and quickly finishing another game with yourself saves your rating from the loss that would occur in that game.[/b]\r\n\r\nWho do the following nicknames refer to? Provide the full formal username. (2 points each)\r\n21. Lonny [b]vallon22[/b]\r\n22. Eli [b]AIME15[/b]\r\n23. Levlev [b]levans[/b]\r\n24. Ding [b]^_^[/b]\r\n25. Mabbitt [b]1=2[/b]\r\n26. Ylxy [b]PhireKaLk6781[/b]\r\n27. Rizzy [b]rizikin[/b]\r\n\r\nAttribute these FTW chat quotes. (1 point each)\r\n28. \"I never said you never said [that] I never said I did.\" [b]isabella2296[/b]\r\n29. \"Not really, it went +2 +2 +2 -12 +2 +3.\" [b]ilovepink[/b]\r\n30. \"Sicily is a god dang multi like me?\" [b]policecap[/b]\r\n31. \"Special ways are izzy...i mean easy.\" [b]Mewto55555[/b]\r\n32. \"I'm going to have fun in the shower with my friends.\" [b]AIME15[/b]", "Solution_9": "I don't remember saying #28 :o\r\n\r\nI also must say that I feel affronted I was not a part of the nickname section. :( \r\n\r\nCorrection on #31:\r\n\r\nThe sequence went:\r\n\r\nmewmew: special ways are izzy\r\nmewmew: i mean easy\r\nmewmew: lol i said izzy", "Solution_10": "Ylxy = PhireKalk9781\r\n\r\nHow?", "Solution_11": "It was a joke.\r\n[**:**] PhireKaLK6781: Why doesn't anyone call me ylxy?\r\n(later)\r\n[**:**] aleph0: Hi ylxy\r\n[**:**] PhireKaLK6781: LOL\r\n\r\n(or something like that)", "Solution_12": "I would like to point out that iin77 was not a moderator. He was deemed by levans as a muter.", "Solution_13": "[quote=\"isabella2296\"]I don't remember saying #28 :o[/quote]\n\n[16:22] isabella2296: I never said you never said thta I never said I did.\n[16:22] Mewto55555: ah well\n[16:22] isabella2296: pwnt!\n[16:22] Mewto55555: gj!\n[16:22] Mewto55555: you are very intelligent\n[16:23] Mewto55555: idk if anyone does\n\n[quote=\"isabella2296\"]I also must say that I feel affronted I was not a part of the nickname section. :( [/quote]\n\nI already have a freebee question. \n\n[quote=\"isabella2296\"]Correction on #31:\n\nThe sequence went:\n\nmewmew: special ways are izzy\nmewmew: i mean easy\nmewmew: lol i said izzy[/quote]\r\n\r\nThat's why I put the ellipses.", "Solution_14": "#20: I told you, this was discovered in Spring 2009...additionally, you described it incorrectly...:wink:\r\n#32: I don't recall saying this....", "Solution_15": "[quote=\"PhireKaLk6781\"]Final Scoreboard: \n10. How can a Guest play more games than the two that are allowed per day? (2 points)\n[b]Click on a game icon, wait for someone else to enter, join button automatically appears[/b]\n\n20. Fall 2009 Rating Save Glitch\n[b]Leaving a losing game and quickly finishing another game with yourself saves your rating from the loss that would occur in that game.[/b]\n[/quote]\r\n\r\nFor #10, you could also just refresh and get a different guest number. :) \r\n\r\nFor #20, at first I thought you were talking about username changes. Because you could just refresh your page, and your rating would go back to what it was before your username was changed.", "Solution_16": "@AIME15: I do remember you saying that once.", "Solution_17": "[quote=\"$ LaTeX$\"]For #10, you could also just refresh and get a different guest number. :) [/quote]\r\n\r\nThat would be a different guest.", "Solution_18": "@AIME\r\n[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=324312]This thingy[/url]", "Solution_19": "[quote=\"$ LaTeX$\"][quote=\"PhireKaLk6781\"]Final Scoreboard: \n10. How can a Guest play more games than the two that are allowed per day? (2 points)\n[b]Click on a game icon, wait for someone else to enter, join button automatically appears[/b]\n\n20. Fall 2009 Rating Save Glitch\n[b]Leaving a losing game and quickly finishing another game with yourself saves your rating from the loss that would occur in that game.[/b]\n[/quote]\n\nFor #10, you could also just refresh and get a different guest number. :) \n\nFor #20, at first I thought you were talking about username changes. Because you could just refresh your page, and your rating would go back to what it was before your username was changed.[/quote]\r\ni used that to get over 9000 rating.\r\nactually, i didn't, but i should've.", "Solution_20": "Lonny?\r\nok, i guess =)\r\n\r\nand BTW, when did you start planning to do this?", "Solution_21": "[quote=\"vallon22\"]and BTW, when did you start planning to do this?[/quote]\r\n\r\nI don't understand what you mean. Elaborate?", "Solution_22": "[quote=\"PhireKaLk6781\"][quote=\"vallon22\"]and BTW, when did you start planning to do this?[/quote]\n\nI don't understand what you mean. Elaborate?[/quote]\r\n\r\nI think he meant when you started to create the quiz.\r\n\r\nEdit: I meant \"I think he meant when you to think about creating a knowledge quiz on FTW.\"", "Solution_23": "[quote=\"$ LaTeX$\"][quote=\"PhireKaLk6781\"][quote=\"vallon22\"]and BTW, when did you start planning to do this?[/quote]\n\nI don't understand what you mean. Elaborate?[/quote]\n\nI think he meant when you started to create the quiz.[/quote]\r\n\r\nI'm quite sure he didn't mean that, given the time that something is posted is stamped on each post.", "Solution_24": "i mean when did you start planning to have a quiz like this\r\nyou have questions about 2008 leaderboard, so you've either been planning since 2008, or have an excellent memory for this like this", "Solution_25": "Oh, I came up with the idea of the quiz the day I made it. As for the 2008 Leaderboard, it's because I took a PrintScreen of that leaderboard for memories. :D", "Solution_26": "lol Smart idea to print screen the leaderboard, I miss the leaderboard in the \"good old days\" when Mewto55555 had a 1500 and he would make fun of me and say \"hey, mousy, why don't you play me? if you lose you only lose 2 points!\" lol\r\n\r\n*sniff* one of the five was banned off the leaderboard. Teehee sorry.\r\n\r\nI woulda gotten a 58 if I took the quiz. Nice Test.", "Solution_27": "iirc nt broke 2000?", "Solution_28": "No, he didn't." } { "Tag": [ "invariant" ], "Problem": "The three cups problem states that: Strting with three cups with one of them right side up and two of them upside down, the aim is to turn all the cups upside down in six moves, and you must turn exactly two cups over each turn.\r\n\r\nHow to prove that this is impossible?", "Solution_1": "[hide]No matter what you do, either one or all of the cups will be right side up, making it impossible to have them all upside down.[/hide]", "Solution_2": "[hide=\"To clarify\"] We can at any point overturn two right-side-up cups, a right-side-up cup and an upside-down cup, or two upside-down cups. These operations leave the parity of the number of right-side-up cups invariant, so this number can never be $ 0$ since it begins odd. [/hide]", "Solution_3": "Oh I see, how can we generalise this problem, so we can have: Strting with n cups with one of them right side up and n-1 of them upside down, the aim is to turn all the cups upside down in k moves, and you must turn exactly m cups over each turn. Such that the problem is still impossible. (what should k and m be?)" } { "Tag": [ "geometry", "geometric transformation", "reflection", "geometric sequence" ], "Problem": "How do you guys explain the equality $ (\\frac {1}{4}) \\plus{} (\\frac {1}{4})^{2} \\plus{} (\\frac {1}{4})^{3} \\plus{} ................. \\equal{} \\frac {1}{3}$ it seemed me very interesting when I saw it for the first time.", "Solution_1": "by Geometric progression we have\r\n$ (\\frac{1}{4})\\plus{}(\\frac{1}{4})^2\\plus{}...\\plus{}(\\frac{1}{4})^n\\equal{}\\frac{\\frac{1}{4}(1\\minus{}(\\frac{1}{4})^n)}{1\\minus{}\\frac{1}{4}}$\r\nit is simple when you consider \r\n$ lim((\\frac{1}{4})\\plus{}(\\frac{1}{4})^2\\plus{}...\\plus{}(\\frac{1}{4})^n)$", "Solution_2": "Elio, please give your posts more descriptive titles. Names like \"interesting\" and \"question\" are totally useless for determining the content of the post.", "Solution_3": "Or, let the sum in question equal k.\r\n\r\nMultiplying both sides by 4, we get 1+k=4k\r\n\r\nk=1/3", "Solution_4": "Not a completely correct argument.\r\n\r\n$ (1 \\plus{} 2 \\plus{} 4 \\plus{} \\ldots ) \\minus{} 2(1 \\plus{} 2 \\plus{} 4 \\plus{} \\ldots ) \\equal{} 1$, but $ 1 \\plus{} 2 \\plus{} 4 \\plus{} \\ldots \\neq \\minus{}1$. It's better to use the limit of partial sums, because otherwise the issue of convergence can mislead you. This example is silly, but your method might get values of sums where it is very difficult to determine convergence", "Solution_5": "[quote=\"Elio (n)\"]How do you guys explain the equality $ (\\frac {1}{4}) \\plus{} (\\frac {1}{4})^{2} \\plus{} (\\frac {1}{4})^{3} \\plus{} ................. \\equal{} \\frac {1}{3}$ it seemed me very interesting when I saw it for the first time.[/quote]\r\n\r\nthis logic explanation : :wink: \r\n\r\nwe know that $ lim(1/4)^{n \\plus{} 1} \\equal{} 0$ because $ \\minus{} 1 < \\frac {1}{4} < 1$ \r\n\r\nsupposing $ lim \\frac {1}{4}^{n \\plus{} 1} \\equal{} (\\frac {1}{4})^{n \\plus{} 1}$ and $ n$--> +00 \r\n\r\nso $ (\\frac {1}{4})^{n \\plus{} 1} \\equal{} 0$ ==> $ (1/4)^{n \\plus{} 1} \\minus{} 1 \\equal{} (\\frac {1}{4}) \\minus{} 1)(\\frac {4}{3})$\r\n\r\n==>$ (\\frac {1}{4}) \\minus{} 1) * (1 \\plus{} (1/4) \\plus{} (1/4)^2 \\plus{} .........(1/4)^{n} ) \\equal{} (\\frac {1}{4}) \\minus{} 1)(\\frac {4}{3})$\r\n\r\n$ 1 \\plus{} (1/4) \\plus{} (1/4)^2 \\plus{} .........(1/4)^{n} \\equal{} (4/3)$ that mean $ \\sum (1/4)^k \\equal{} (1/3)$", "Solution_6": "Not exactly Pre-Olympiad material. Nevertheless, I like to think of this identity in the following way. Consider an equilateral triangle with area $ 1$, and divide it up into four congruent equilateral triangles of area $ \\frac{1}{4}$. The middle triangle can further be divided into four congruent equilateral triangles of area $ \\frac{1}{4^2}$, and the middle triangle of that triangle can further be divided into four congruent equilateral triangles of area $ \\frac{1}{4^3}$, ad infinitum. The total area of all the triangles created in this process is\r\n\r\n$ \\frac{3}{4} \\plus{} \\frac{3}{4^2} \\plus{} \\frac{3}{4^3} \\plus{} ...$\r\n\r\nbut we know that the area of the original equilateral triangle is just $ 1$. :)", "Solution_7": "[quote=\"t0rajir0u\"]Not exactly Pre-Olympiad material. Nevertheless, I like to think of this identity in the following way. Consider an equilateral triangle with area $ 1$, and divide it up into four congruent equilateral triangles of area $ \\frac {1}{4}$. The middle triangle can further be divided into four congruent equilateral triangles of area $ \\frac {1}{4^2}$, and the middle triangle of that triangle can further be divided into four congruent equilateral triangles of area $ \\frac {1}{4^3}$, ad infinitum. The total area of all the triangles created in this process is\n\n$ \\frac {3}{4} \\plus{} \\frac {3}{4^2} \\plus{} \\frac {3}{4^3} \\plus{} ...$\n\nbut we know that the area of the original equilateral triangle is just $ 1$. :)[/quote]\r\ni like the idea it reflects the power of your immagination :)", "Solution_8": "[quote=\"Ibn Rochde\"][quote=\"t0rajir0u\"]Not exactly Pre-Olympiad material. Nevertheless, I like to think of this identity in the following way. Consider an equilateral triangle with area $ 1$, and divide it up into four congruent equilateral triangles of area $ \\frac {1}{4}$. The middle triangle can further be divided into four congruent equilateral triangles of area $ \\frac {1}{4^2}$, and the middle triangle of that triangle can further be divided into four congruent equilateral triangles of area $ \\frac {1}{4^3}$, ad infinitum. The total area of all the triangles created in this process is\n\n$ \\frac {3}{4} \\plus{} \\frac {3}{4^2} \\plus{} \\frac {3}{4^3} \\plus{} ...$\n\nbut we know that the area of the original equilateral triangle is just $ 1$. :)[/quote]\ni like the idea it reflects the power of your immagination :)[/quote]\r\nI think that idea might be present elsewhere... in fact, I see the animation right now. :P" } { "Tag": [ "function", "probability", "integration", "parameterization", "calculus", "probability and stats" ], "Problem": "Suppose we are given a probability distribution function $ P(\\xi \\leqslant t)\\equal{}\\int_{a}^{\\infty} \\sqrt{\\frac{2}{\\pi t}}e^{\\minus{}\\frac{x^2}{2t}}dx$. Help me to find the density function.", "Solution_1": "First of all may be you could chek that $ Lim_{t\\rightarrow \\plus{}\\infty} P(\\xi\\leq t)\\equal{}1$\r\nThis is not really obvious from your expression, I think that might have to mulitply with a copnstant (may depending on your parameter $ a$).\r\nYou should also check for the vailidity of your expression for example it is not well defined for negative $ t$.\r\n\r\nOnce you get there, apply derivation under integral sign theorem of your expression $ P(\\xi\\leq t )$ with respect to $ t$ to get your pdf." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that for all $ a,b,c > 0 , abc \\geq 1$\r\n$ \\frac{1}{1\\plus{}a\\plus{}b} \\plus{} \\frac{1}{1\\plus{}b\\plus{}c} \\plus{} \\frac{1}{1\\plus{}c\\plus{}a} \\leq 1$\r\n\r\n :?:", "Solution_1": "[quote=\"navidj\"]Prove that for all $ a,b,c > 0 , abc \\geq 1$\n$ \\frac {1}{1 \\plus{} a \\plus{} b} \\plus{} \\frac {1}{1 \\plus{} b \\plus{} c} \\plus{} \\frac {1}{1 \\plus{} c \\plus{} a} \\leq 1$\n\n :?:[/quote]\r\nAfter expanding,this inequality is equivalent to:\r\n$ 2 \\plus{} 2(a \\plus{} b \\plus{} c) \\le (a \\plus{} b)(b \\plus{} c)(c \\plus{} a)$\r\nBut $ (a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\equal{} (a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca) \\minus{} abc \\ge \\frac {8(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}{9} \\ge \\frac {8(a \\plus{} b \\plus{} c)}{3} \\ge 2(a \\plus{} b \\plus{} c) \\plus{} 2$\r\nThen,we have Q.E.D :)", "Solution_2": "nice and easy,\r\nBy Cauchy-Schwarz; $ (1\\plus{}a\\plus{}b)(c\\plus{}1\\plus{}1)\\geq (\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c})^2$ therfore, $ \\frac{1}{a\\plus{}b\\plus{}1}\\leq\\frac{c\\plus{}2}{(\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c})^2}$ thus, $ \\sum \\frac{1}{a\\plus{}b\\plus{}1}\\leq \\frac{a\\plus{}b\\plus{}c\\plus{}6}{(\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c})^2}$\r\nthen it's suffice to prove that, $ \\frac{a\\plus{}b\\plus{}c\\plus{}6}{(\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c})^2} \\leq 1$\r\n$ \\Leftrightarrow (\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c})^2\\geq a\\plus{}b\\plus{}c\\plus{}6$\r\n$ \\Leftrightarrow 2\\sum \\sqrt{ab} \\geq 6$ wich is AM-GM, ($ 2\\sum \\sqrt{ab} \\geq 6\\sqrt[3]{abc}\\geq 6$ because $ abc\\geq1$ )\r\nequality only if $ a\\equal{}b\\equal{}c\\equal{}1$" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "$2\\cdot sin\\left(\\frac{7x}{3} \\right)=1 \\leftrightarrow sin\\left(\\frac{7x}{3} \\right)=\\frac{1}{2}$\r\nso we got\r\n$x=\\frac{3}{7}\\cdot \\left(\\frac{\\pi}{6} + 2k\\pi\\right),k\\in %Error. \"varmathbb\" is a bad command.\n{Z}$\r\nis this right or wrong?\r\n\r\nat the answer section of my book i get somethin totally different:\r\n$(-1)^k \\cdot \\frac{\\pi}{14} + \\frac {3k\\pi}{7},k\\in %Error. \"varmathbb\" is a bad command.\n{Z}$", "Solution_1": "[hide]$2\\cdot sin\\left(\\frac{7x}{3} \\right)=1 \\Longleftrightarrow sin\\left(\\frac{7x}{3} \\right)=\\frac{1}{2}$\nAnd now $\\frac{7x}{3}=(-1)^{k}\\arcsin{\\frac{1}{2}}+k\\pi$ or $x=(-1)^{k}\\frac{\\pi}{14}+\\frac{3k\\pi}{7}$\n[/hide]", "Solution_2": "why ?", "Solution_3": "[u]Basic Trig equations[/u]\r\n\r\n[b]1.[/b] $\\sin x=a$, $a \\in [-1,1] \\Rightarrow x \\in{ (-1)^{k} \\arcsin a+k\\pi|k\\in Z}$\r\n\r\n[b]2.[/b] $\\cos x=a$, $a \\in [-1,1] \\Rightarrow x \\in{ +/- \\arccos a+2k\\pi|k\\in Z }$\r\n\r\n[b]3.[/b] $\\tan x=a$, $a \\in R \\Rightarrow x \\in{ \\arctan a+k\\pi|k\\in Z }$\r\n\r\n[b]4.[/b] $\\cot x=a$, $a \\in R \\Rightarrow x \\in{ arccot a+k\\pi|k\\in Z }$\r\n\r\ncot=ctg, tan=tg ;)" } { "Tag": [], "Problem": "I was wondering, is it possible to make a superconducting wire which has EXACTLY zero resistance?? I mean, it shouldn't be possible right?, since if a have a superconducting wire attached to a voltage source then having zero resisitance would give us an infinite current which we obviously cannot have in real life!\r\n\r\n-Swapnil", "Solution_1": "yes it is possible, in 1942 , karlmecling ornes discover it\r\nbut it can only occur in high temperature\r\nnow, the problem is how can we find out superconductor which can occur in room temperature using \" :rotfl: cooper\"", "Solution_2": "[quote=cuong]but it can only occur in high temperature[/quote]\r\n\r\nErr... by \"high\" you meant low? As far as I know, superconductivity has been achieved at very low temps first.", "Solution_3": "[quote=\"cuong\"]\nbut it can only occur in high temperature\n[/quote]\r\n\r\nErrr... bye \"high\" you mean low, right? Cause as far as I know, first superconductors were achieved at very low temperature.", "Solution_4": "Wait... What about there being INFINITE current if the wire has zero resistance? Isn't that impossible??", "Solution_5": "[quote=\"swapnillium\"]Wait... What about there being INFINITE current if the wire has zero resistance? Isn't that impossible??[/quote]why would the implication of 0 resistance be infinite current? electrons can't move at an infinite speed.", "Solution_6": "Because if we have a superconducting wire attached to a voltage source then having zero resisitance would give us an infinite current. That is, $v = i/R \\Rightarrow i = v/R$; so then (if $v$ is constant) as $R \\rightarrow 0, i \\rightarrow \\infty$", "Solution_7": "Ohm's law doesn't account for the effects of relativity, so it only applies for relatively low electron drift speeds. Hence, as that speed approaches a significant fraction of the speed of light, the predictions of Ohm's law are inaccurate. So, a wire of finite thickness cannot carry an infinite current.", "Solution_8": "to ANDERSON yes low temperature,sory for my mistake.\r\n\r\nohm law can not be apply here.OHM law is found by experiment with normal voltage and current, in this case R-----0,we can't apply OHM law to find out the voltage or the magnitude of the current.\r\n\r\nanother example is the mechanic mass of photon which velocity is c.We can't find its mass by EInstein equation:$m=\\frac{mo}{\\sqrt{1-\\frac{v^2}{c^2}}}$,the velocity of PHOTON is c so m tends to infinity which is wrong , so we can't apply it here.\r\nthe mechanic mass of photon is $\\frac{hf}{c^2}$ :D", "Solution_9": "So what exactly are the limitations of Ohm's law?\r\nBTW, thank you all for answering my questions. :)", "Solution_10": "anyone...??", "Solution_11": "I had asked my father about this question, he is researching it in HCM university of technology,but he didn't answer me :blush: .\r\n\r\nbtw,i also creat a technology to find the magnitude of current , i will check it.so wait for me ,the main idea is refered from an A level book :ninja:", "Solution_12": "anyone else...??" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "prove that:\r\n\r\n[tex] (a+b+c) - ( a^2b+ac^2+b^2c)+(ab+bc+ca)-3 \\leq \\ 0 [/tex]\r\n\r\nwhere abc=1", "Solution_1": "\\[(a+b+c) - ( a^2b+ac^2+b^2c)+(ab+bc+ca)-3 \\leq \\ 0 \\Leftrightarrow\\]\r\n\\[(\\frac{1}{c}-1)(a-1)+(\\frac{1}{a}-1)(b-1)+(\\frac{1}{b}-1)(c-1) \\ge 0.\\]\r\nCase 1: $a \\ge 1 \\ge b \\ge c$.\r\n$(\\frac{1}{a}-1)(b-1) \\ge 0$ and $(\\frac{1}{c}-1)(a-1) \\ge (\\frac{1}{c}-1)(1-c) \\ge (\\frac{1}{b}-1)(1-c)$.\r\n\r\nCase 2: $a \\ge b \\ge 1 \\ge c$.\r\n$(\\frac{1}{b}-1)(c-1) \\ge 0$ and $(\\frac{1}{c}-1)(a-1) \\ge (1-\\frac{1}{a})(a-1) \\ge (1-\\frac{1}{a})(b-1)$.\r\n\r\nCase 3: $a \\ge 1 \\ge c \\ge b$.\r\n$(\\frac{1}{a}-1)(b-1) \\ge 0$ and $(\\frac{1}{c}-1)(a-1) \\ge (1-c)(a-1) \\ge (\\frac{1}{b}-1)(1-c)$.\r\n\r\nCase 4: $a \\ge c \\ge 1 \\ge b$.\r\n$(\\frac{1}{a}-1)(b-1) \\ge 0$ and $(\\frac{1}{b}-1)(c-1) \\ge (a-1)(c-1) \\ge (1-\\frac{1}{c})(a-1)$.\r\n\r\nHere I assumed that a,b,c are positive. I think we can deal with the real a,b,c case by modifying the proof of positive a,b,c case.", "Solution_2": "[quote=\"sebescu\"]prove that:\n\n[tex] (a+b+c) - ( a^2b+ac^2+b^2c)+(ab+bc+ca)-3 \\leq \\ 0 [/tex]\n\nwhere abc=1[/quote]\r\n\r\nThis inequality is equivalent to $a^2b+b^2c+c^2a+3\\geq a+b+c+bc+ca+ab$ and was solved in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=42056 . Note that you need the condition that the numbers a, b, c are positive; else, the inequality is not generally true.\r\n\r\n darij" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "I searched for the following problem, but I couldn't find it on mathlinks. I would really like to see a proof of the following using inversive geometry.\r\n\r\nTwo circles $\\Gamma$ and $\\Gamma'$ intersect at $A$ and $D$. A line is tangent to $\\Gamma$ and $\\Gamma'$ at $E$ and $F$, respectively. A line passes through point $D$ and intersects circles $\\Gamma$ and $\\Gamma'$ at $B$ and $C$, respectively. Show that the circumcircles of $\\triangle BDE$ and $\\triangle CDF$ intersect again on the line $AD$.", "Solution_1": "Aren't the circumcircles of $\\triangle BDE$ and $\\triangle CDF$ just $\\Gamma$ and $\\Gamma'$ respectively?\r\nDid I misunderstand anything :maybe:" } { "Tag": [ "geometry", "geometric transformation", "reflection", "AMC", "AIME", "AMC 10", "AMC 10 B" ], "Problem": "Okay, I've taken the AMC 12A. I was wondering if it would be possible to take the 10B at a public school (with their permission of course) without paying the registration fee and whatnot. If I'm not mistaken, students can take tests at another school without registering. I was wondering if that would apply in my case.", "Solution_1": "According to our general principles, you should register as a \"Visitor\" in order to take the AMC 10 B or AMC 12B at a school which is not the school you regularly attend. In order to register as a visitor, you should call the AMC office at 402-472-2257 and request a visitor registration form. The \r\nform and accompanying instruction will have directions about which CEEB number to use. The fee for registering as a \"Visitor\" is $32.\nIf you do not register as a visitor, the following consequences may happen:\r\n1. The host school you visit will have a CEEB number on the school cover sheet, which precedes the scoring of the answer forms form that school. If your answer form in the midst of the host school's answer forms has the CEEB of your school we may believe that a serious mix-up has occurred \r\nwithin our office and it can cause us great confusion. This slows up the scoring of everyone's answer forms!\n2. If you take the AMC 12 B at a host school which is not your own school and use the host school's CEEB number, then we believe you are a student at that host school. If you score among the top three scores from that \r\nschool, then we automatically make you a member of the host school's team score. This is not an honest reflection of the school's score or standing. In addition, since Sliffe awards are partially based on the school's team score, the potential exists to select a Sliffe winner on the basis of a non-representative score. This is not fair to other teachers.\n3. If you take the AMC 12 B at a host school which is not your own school and use the host school's CEEB number, then we believe you are a student at that school. If you qualify for the AIME on the basis of what you \r\nscore on the AMC 12 B, we report you as an AIME qualifier to the host school, not your own school. We also send an AIME exam to that host school for you. That puts the burden on a teacher who does not know you to contact \r\nyou to make arrangments for the AIME. That is not a fair request to make of the hosting teacher, and it runs the risk of you rnot being notified \r\nthat you qualify for the AIME if the host teacher either does not have the time or the will to contact you.\nIf you register as a Visitor, then the $32 fee is used to extract your personal answer sheet from the host school's bundle and to process you separately, and contact you separately eliminating all three of the problems above.\r\n\r\nAMC Director\r\nSteve Dunbar", "Solution_2": "I won't bother with the 10B then. Thanks for the clarification on that.", "Solution_3": "I didn't catch all this visitor information on the AMC site or in the materials I received in the mail until AFTER I had already registered a group for the \"A\" tests in the name of my homeschool math club group, and indicated that my son would take a \"B\" test in the name of another homeschool group that has been organizing AMC 10/12 tests for several years across town. (My son took the 10B with the other group last year, so that is actually his \"previous school\" for the 10B, but my group is his \"previous school\" for the AMC 8, now three years running.) I am sorry that I missed that. I will have to see if I can unsnarl things." } { "Tag": [ "geometry", "perimeter" ], "Problem": "Determine all distinct triangles having one side of length 6,with the other two sides being integers,and the perimeter numerically equal to the area.", "Solution_1": "Yay for Heron :D\r\n\r\n[hide=\"a start\"]perimeter = a+b+c\ns = (a+b+c)/2\nArea = \u00ac/(s(s-a)(s-b)(s-c))\na = 6\n\nSo we want to find all integers (b,c) such that \n6+b+c = \u00ac/(s(s-6)(s-b)(s-c))\n(6+b+c)^2 = s(s-6)(s-b)(s-c)\n...\n:stink:\n\nJust by testing, I found one: (6,8,10) [/hide]" } { "Tag": [ "function", "algebra", "domain", "absolute value" ], "Problem": "could anyone give me hints on how to graph this one?\r\n\r\n$ y\\equal{}|x\\minus{}2| \\plus{} |x\\minus{}4| \\plus{} |2x\\minus{}6|$", "Solution_1": "See what happens at $ x\\equal{}2$, $ x\\equal{}3$, and $ x\\equal{}4$. In other words, set each of the expressions within the absolute value signs to zero.", "Solution_2": "yes, i know, i did that, these are the \"vertices\" of each of the 3 \"sub-graphs\". yet, i still dont understand what you are trying to tell me...:(", "Solution_3": "First, understand what a typical graph of the absolute value function looks like, which is like a \"V\" shape. Now set intervals at which each of the absolute values changes sign. Let me illustrate my idea here, this should be clear:\r\n\r\nI'll begin with a much simpler example and you apply the same technique to your problem.\r\n\r\nSuppose we want to graph $ y\\equal{}|x|\\plus{}|x\\minus{}1|$. The first thing you want to make sure is to know the proper definition of absolute value. You can't graph a function without knowing what it is!\r\n\r\nTo review that definition, [hide=\"click here\"]\\[ |x|\\equal{}\\begin{cases}x&\\text{if }x\\ge0\\\\\\minus{}x&\\text{if }x<0\\end{cases}\\][/hide]\r\n\r\nTake 123456789's hint, we look at the \"corners\" at $ x\\equal{}0$ and $ x\\equal{}1$.\r\n\r\nSo we have a domain to analyze the values of $ y$: (sorry I have to put this in code)\r\n\r\n[code]\n<------------------+-------------------+------------------>\nx 0 1[/code]\r\nSet intervals to make your \"piecewise\" graph:\r\n\r\nfor $ x<0$, $ y\\equal{}(\\minus{}x)\\plus{}(\\minus{}x\\plus{}1)\\equal{}1\\minus{}2x$\r\n\r\nfor $ 0\\le x<1$, $ y\\equal{}(x)\\plus{}(\\minus{}x\\plus{}1)\\equal{}1$\r\n\r\nfor $ 1\\le x$, $ y\\equal{}(x)\\plus{}(x\\minus{}1)\\equal{}2x\\minus{}1$.", "Solution_4": "ohhh..so it becomes a piecewise function? that's incredible...:)thanks", "Solution_5": "Every absolute value function is really a piecewise function. I always tell people that's by far the easiest way to deal with absolute values." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all pairs of prime numbers (p,q) such that :\r\n\r\n pq divides $(7^p - 2^p)(7^q - 2^q)$", "Solution_1": "Suppose that $p,q>7$ (if they are smaller or equal, you just go through all the cases).\r\nThen $7^p-2^p\\equiv 7-2 \\mod p$ so $p|7^q-2^q$ and $q|7^p-2^p$. Now we have that $p|7^{p-1}-2^{p-1}$ so $p|gcd(7^q-2^q,7^{p-1}-2^{p-1})$ and $gcd(7, 2)=1$ so \\[ p|gcd(7^q-2^q,7^{p-1}-2^{p-1})=7^{gcd(q, p-1)} - 2^{gcd(q, p-1)} \\] and like that \\[ q|7^{gcd(p, q-1)}-2^{gcd(p, q-1)}. \\] \r\nNow $qcd(p, q-1)$ and $gcd(q, p-1)$ can't be equal $1$, so it has to be $p|q-1$ and $q|p-1$... \r\nedit: so $p\\leq q-1 \\leq p-2$ which is imposible.\r\n\r\nbye", "Solution_2": "[quote] so it has to be $p|q-1$ and $q|p-1$ ... [/quote]\r\n... hence $p \\leq q-1$ and $q \\leq p-1$. Therefore $q \\leq p-1 \\leq q-2$, a contradiction. ;)\r\nI don't understand how does $p|q-1$ and $q|p-1$ imply that $p=q$ ? :?", "Solution_3": "It is obviosly one of them 5. If $p=5$, then $7^5-2^5=5^2*11*61$ give $q=5$ or $q=11$ or $q=61$.", "Solution_4": "[quote=\"{x}\"][quote] so it has to be $p|q-1$ and $q|p-1$ ... [/quote]\n... hence $p \\leq q-1$ and $q \\leq p-1$. Therefore $q \\leq p-1 \\leq q-2$, a contradiction. ;)\nI don't understand how does $p|q-1$ and $q|p-1$ imply that $p=q$ ? :?[/quote]\n\nI couldn't agree with you more, I made a mistake because I wrote this maybe little to fast whitout much of thinking...\n\nI will edit it now\n\n[quote] It is obviosly one of them 5. [/quote]\r\n\r\nI wonder why is this so obvious?", "Solution_5": "Let $p\\not =5,2,7 q\\not =5,2,7$, then $p|(7^q-2^q),q|(7^p-2^p)$. If $q\\not |(p-1)$ then exist k: kq=1(mod (p-1)). Therefore $p|(7^q-2^q)|(7^{kq}-2^{kq}=5(mod p)$ give contradition. If $q|(p-1),p|(q-1)$ we have contradition.\r\nIf p=2 or 7 give contradition $7|2^q$ or $2|7^q$.", "Solution_6": "Well, proof in 4 lines isn't so obvious :D\r\n\r\nbye" } { "Tag": [ "algebra", "polynomial", "trigonometry", "modular arithmetic", "algebra unsolved" ], "Problem": "Polynomial $ P(t)$ is such that for all real $ x$,\r\n\\[ P(\\sin x) \\plus{} P(\\cos x) \\equal{} 1.\r\n\\]\r\nWhat can be the degree of this polynomial?", "Solution_1": "(*) $P(\\sin x)+P(\\cos x)=1$\r\n\r\nFirst $P(-y)=P(\\sin (-x))=1-P(\\cos (-x))=1-P(\\cos x)=P(y)$ for all $y=\\sin x\\in [-1,1]$ using (*). \r\nThen $P(t)-P(-t)$ has infinitely many roots, and so $P(t)=P(-t)$ is even. \r\nTherefore $P(t)=Q(t^{2})$ for some polynomial $Q$.\r\nThen $1=P(\\sin x)+P(\\cos x)=Q(\\sin^{2}x)+Q(\\cos^{2}x)=Q({1\\over 2}+y)+Q({1\\over 2}-y)$ for all $y=\\sin^{2}x-{1\\over 2}\\in [-{1\\over 2},{1\\over 2}]$. \r\nSo the polynomial $Q({1\\over 2}+t)+Q({1\\over 2}-t)-1$ has infinitely many roots, and therefore the polynomial \r\n$Q({1\\over 2}+t)-{1\\over 2}=-(Q({1\\over 2}-t)-{1\\over 2})$ is odd. So $Q({1\\over 2}+t)-{1\\over 2}=tR(t^{2})$ for some polynomial $R$. \r\nBut now $Q(t)=(t-{1\\over 2})R((t-{1\\over 2})^{2})+{1\\over 2}$ and \r\n$P(t)=Q(t^{2})=(t^{2}-{1\\over 2})R((t^{2}-{1\\over 2})^{2})+{1\\over 2}$.\r\n\r\nConversely every so defined polynomial satisfies (*), \r\nwhere $\\deg P=2+4\\deg R$ runs through all natural numbers $\\equiv 2 \\pmod{4}$ for $R\\not \\equiv 0$,\r\nand $\\deg P=0$ for $R\\equiv 0$." } { "Tag": [ "integration", "trigonometry" ], "Problem": "A flat decorator is planning a hanger for clothes. He attaches two identical very strong but elastic metal fibres with a shape of a quarter circle, as in the picture below. He noticed that if he put the same amount of weight on the two hangers, one of them will bend lower than the other one. He couldnt tell why. Lets help him and tell him why one of them bends lower.", "Solution_1": "No one wants to help the flat decorator? :D :D \r\nThe inside wont be ready in time :D", "Solution_2": "No help from me, I'm afraid :( Could you post the solution?", "Solution_3": "ok here it is shortly:\r\nWe apply carefully a maximum of $G$ force to the fibres. then the endpoints of the fibre will be $\\delta h$ lower, than in the beginning. The work done by is the same as the elastic energy in the little bit bended fibre. Using the analogy with the elastic energy the energy in the fibre is proportional to the length of a little piece of fibre and the square of the torque at the end of it. \r\nSo:\r\n$\\frac{\\delta h_{a}}{\\delta h_{b}}=\\frac{\\int M(a)^{2}ds}{\\int M(b)^{2}ds}= \\frac{\\int_{0}^{\\frac{\\pi}{2}}\\sin^{2}\\phi d\\phi}{\\int_{0}^{\\frac{\\pi}{2}}(1-\\cos \\phi)^{2}d\\phi}=\\frac{\\pi}{3\\pi-8}=2,2$\r\nSo the bending in the a.) diagram is greater." } { "Tag": [ "AMC", "AIME", "articles", "percent", "USA(J)MO", "USAMO", "geometry", "\\/closed" ], "Problem": "We are now taking enrollments for our [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]2009-2010 Worldwide Online Olympiad Training[/url] (WOOT!)\r\n\r\nWOOT is a 7-month training program for national high school International Math Olympiad selection tests such as the AIME and USAMO. WOOT offers classes, articles, problem sets, practice tests, and an outstanding peer group. This year's schedule can be found [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOTSchedule.php]here[/url]. As in past years, generous sponsorship allows us to invite the 2009 MOSP students (those selected to train with the United States IMO team) to participate in WOOT for free. This year's sponsors are D. E. Shaw group, Jane Street Capital, and Two Sigma Investments, LLC. We also have received sponsorship from Science House to fund six top students from Central and South America. So, we expect to again have a truly outstanding group of students in WOOT!\r\n\r\nWe are offering an early-bird discount on enrollment. The registration fee is \\$695 if you enroll before August 15, 2009, and \\$845 thereafter. WOOT begins in mid-September and runs through mid-April, 2010. We will host two Math Jams to discuss WOOT in the fall, on August 4 and September 8, both at 7:30 PM ET.", "Solution_1": "Thanks a lot but I have a little question. Is the classes on Date 1 and Date 2? or Date OR Date 2?\r\nAnd what is the maximum participants number?\r\n\r\nThanks", "Solution_2": "No maximum participant number -- if we have way too many people enroll, we'll add another class date. The \"Date 1\" and \"Date 2\" classes are the same -- this is to give students multiple opportunities to fit WOOT into their schedules.", "Solution_3": "Math Jams are free, right?", "Solution_4": "[quote=\"myyellowducky82\"]Math Jams are free, right?[/quote]\r\n\r\nyes", "Solution_5": "What is the difference between the two math jams?", "Solution_6": "would someone who got a $ 5$ on AIME be ready for this?", "Solution_7": "[quote=\"PowerOfPi\"]What is the difference between the two math jams?[/quote]\n\nnothing, i think\nother than they are at different times\n\n[quote=\"nikeballa96\"]would someone who got a $ 5$ on AIME be ready for this?[/quote]\n\n[quote=\"WOOT page\"]Students who are not capable of consistently scoring 5 or higher on the American Invitational Mathematics Exam should strongly consider waiting until they have more experience before joining WOOT.\n[/quote]\r\n\r\ni'd say you should try", "Solution_8": "[quote=\"nikeballa96\"]would someone who got a $ 5$ on AIME be ready for this?[/quote]\r\n\r\nI know someone who took it as an 8th grader after getting a 5 on AIME in 7th grade...I plan on taking it, and I got a 6 (although this isn't really a good representative of my math skill).", "Solution_9": "I took it after getting a 4 on the AIME.\r\n\r\nI was ready for about half the material I think, more or less", "Solution_10": "[quote=\"PowerOfPi\"]What is the difference between the two math jams?[/quote]\r\n\r\nNothing -- there isn't any math in these two math jams; they will mainly be question&answer sessions about WOOT.\r\n\r\nTo those of you asking about \"Should I take this with a 4-6 on the AIME?\", the answer depends a lot on two things:\r\n\r\n1) What will you have done between the AIME and WOOT? If you take some AoPS classes, work through some books, or go to some summer programs, then you're probably ready. If you do nothing, then you may not be ready.\r\n\r\n2) Will you have plenty of time during the school year to investigate/ask questions about parts of WOOT you don't understand the first time you see them?", "Solution_11": "[quote=\"xpmath\"][quote=\"nikeballa96\"]would someone who got a $ 5$ on AIME be ready for this?[/quote]\n\nI know someone who took it as an 8th grader after getting a 5 on AIME in 7th grade...I plan on taking it, and I got a 6 (although this isn't really a good representative of my math skill).[/quote]\r\n\r\nWell this person did not do particularly well in woot :P.(trust me, i would know)", "Solution_12": "dang. so expensive. should've gone to MOP.", "Solution_13": "[quote=\"abacadaea\"]Well this person did not do particularly well in woot :P.(trust me, i would know)[/quote]\r\n\r\nHEY! I did decent! :P\r\n\r\nIf you make AIME and score at least a 3 you should consider trying WOOT. It has some hard problems that really make you think, and it has increased my mathematical maturity/confidence. This is my third year participating (scholarships FTW).", "Solution_14": "[quote=\"1=2\"][quote=\"abacadaea\"]Well this person did not do particularly well in woot :P.(trust me, i would know)[/quote]\n\nHEY! I did decent! :P\n\nIf you make AIME and score at least a 3 you should consider trying WOOT. It has some hard problems that really make you think, and it has increased my mathematical maturity/confidence. This is my third year participating (scholarships FTW).[/quote]\r\n\r\nI wasnt refering to you... :P", "Solution_15": "[quote=\"batteredbutnotdefeated\"]If you make USAMO you get to go to WOOT for free? How does that work? Do they send an invitation or something to USAMO qualifiers?[/quote]\r\n\r\nNo; only the MOP students are invited for free among the USAMO qualifiers.", "Solution_16": "If one does WOOT, would it still be useful to go through acops? Would it be best to go through it before, during, or after?", "Solution_17": "If you can go through it before WOOT, and are ready to do so, by all means, do go through it. If not, that's OK. It will be a fine complement to WOOT.", "Solution_18": "[quote=\"simo14\"]If one does WOOT, would it still be useful to go through acops? Would it be best to go through it before, during, or after?[/quote]\r\n\r\nWhat is \"acops\" ?", "Solution_19": "Art and Craft of Problem Solving. It's a book by Paul Zeitz that's about early olympiad level.", "Solution_20": "I went through part of ACoPS (the first few chapters about general problem solving skills) during part of WOOT (and went through several other math books as well) without a problem. It is a fine complement to WOOT, as rrusczyk said; I only have to add that, depending on how busy your life is, you may have to work very hard to add room to study this or any other book, especially in addition to WOOT. However, it is indeed quite possible, and I found it refreshing (since it was less intensive than working on IMO shortlist problems in the WOOT classroom, yet still had some very useful and applicable ideas).", "Solution_21": "If I get done AoPS Vol. 2 by the time WOOT starts, will WOOT be fit for me? By that time, I also plan to finish up Intermediate Algebra. Before that, I took an Intro to Counting and Probability course, as well as studied Intro to Number Theory and Geometry. so, i want to make sure this is right for me before asking my parents to sign me up for this.", "Solution_22": "[quote=\"qwertythecucumber\"]If I get done AoPS Vol. 2 by the time WOOT starts, will WOOT be fit for me? By that time, I also plan to finish up Intermediate Algebra. Before that, I took an Intro to Counting and Probability course, as well as studied Intro to Number Theory and Geometry. so, i want to make sure this is right for me before asking my parents to sign me up for this.[/quote]\r\n\r\nIf by \"getting done\" you mean \"worked on a lot of the problems and understand the material pretty well\", then yes, you should be ready to try WOOT in the fall. Some of it will still be quite hard for you, probably, but you should have the core knowledge necessary to fight through a lot of the hard parts and ask good questions when you're stuck.", "Solution_23": "Exactly how difficult does the course get? (in both classes and problems) also how does taking WOOT differ from studying out of a book? :huh:", "Solution_24": "I think the difficulty question was answered earlier in the thread.\r\n\r\nTo answer your second question: you get to work with some of the brightest minds in your age group and learn better ways to approach/solve problems from them and the instructors/handouts/lectures. You won't get this feature in any other program other than MOP (as far as I know). The instructors also have been USAMO winners and can give you various tips based on their experience. I highly doubt you would get this in any book. But of course to get all this, you must put time into it and allot time for yourself to work with others in the open classrooms.", "Solution_25": "[quote=\"sunehra\"]I think the difficulty question was answered earlier in the thread.[/quote]Ita.\n\n[quote]To answer your second question: you get to work with some of the brightest minds in your age group and learn better ways to approach/solve problems from them and the instructors/handouts/lectures. You won't get this feature in any other program other than MOP (as far as I know).[/quote] Other high level summer programs like the AMSP and AMY have this as well. \n\n[quote]The instructors also have been USAMO winners and can give you various tips based on their experience. I highly doubt you would get this in any book. But of course to get all this, you must put time into it and allot time for yourself to work with others in the open classrooms.[/quote]True, True, True, and QFT.", "Solution_26": "[quote=\"rrusczyk\"][quote=\"batteredbutnotdefeated\"]If you make USAMO you get to go to WOOT for free? How does that work? Do they send an invitation or something to USAMO qualifiers?[/quote]\n\nNo; only the MOP students are invited for free among the USAMO qualifiers.[/quote]\r\n\r\nDoes this work for every year's MOP, or only this year?", "Solution_27": "I believe it's every year. Correct me if I am wrong.", "Solution_28": "Are the woot lessons going to be the same as last year? Are the problem sets all the same?", "Solution_29": "The lessons and problem sets will be different this year than last. There will be some overlap in lessons from 3 years ago." } { "Tag": [ "inequalities", "trigonometry" ], "Problem": "Let be given a triangle $ABC$. Find the least value of the expression: \r\n\\[\\frac{\\sqrt{1+2\\cos^{2}A}}{\\sin B}+\\frac{\\sqrt{1+2\\cos^{2}B}}{\\sin C}+\\frac{\\sqrt{1+2\\cos^{2}C}}{\\sin A}\\]", "Solution_1": "[quote=\"April\"]Let be given a triangle $ABC$. Find the least value of the expression:\n\\[\\frac{\\sqrt{1+2\\cos^{2}A}}{\\sin B}+\\frac{\\sqrt{1+2\\cos^{2}B}}{\\sin C}+\\frac{\\sqrt{1+2\\cos^{2}C}}{\\sin A}\\]\n[/quote]\r\nAM-GM\r\n\\[LHS \\ge \\root 6 \\of{{{\\prod{1+2\\cos^{2}A}}\\over{\\prod{\\sin^{2}B}}}}\\]\r\nWe need prove\r\n\\[\\root 6 \\of{{{\\prod{1+2\\cos^{2}A}}\\over{\\prod{\\sin^{2}B}\\geq3\\sqrt 2\\]\r\n$Min=3\\sqrt 2$", "Solution_2": "[quote=\"kcbu\"][quote=\"April\"]Let be given a triangle $ABC$. Find the least value of the expression:\n\\[\\frac{\\sqrt{1+2\\cos^{2}A}}{\\sin B}+\\frac{\\sqrt{1+2\\cos^{2}B}}{\\sin C}+\\frac{\\sqrt{1+2\\cos^{2}C}}{\\sin A}\\]\n[/quote]\nAM-GM\n\\[LHS \\ge \\root 6 \\of{{{\\prod{1+2\\cos^{2}A}}\\over{\\prod{\\sin^{2}B}}}}\\]\nWe need prove\n\\[\\root 6 \\of{{{\\prod{1+2\\cos^{2}A}}\\over{\\prod{\\sin^{2}B}}}}\\geq3\\sqrt 2 \\]\n$\\min=3\\sqrt 2$[/quote]\r\nCould you show me the complete solution?\r\nThanks :lol:", "Solution_3": "We use some facts as following to infer the rezult:\r\n$\\sin A.\\sin B.\\sin C\\leq\\frac{3\\sqrt{3}}{8}$\r\n$(1+\\cos^{2}A)(1+\\cos^{2}B)(1+\\cos^{2}C)\\geq\\frac{125}{64}$", "Solution_4": "Please post the complete solution :)", "Solution_5": "we have a similar rezult relating the problem: Given a triangle $ABC$ then\r\n$\\frac{\\sqrt{1+8\\cos^{2}A}}{\\sin B}+\\frac{\\sqrt{1+8\\cos^{2}B}}{\\sin C}+\\frac{\\sqrt{1+8\\cos^{2}C}}{\\sin A}\\geq 6$", "Solution_6": "Okay, the general problem is:\r\n$\\frac{\\sqrt{1+k\\cos^{2}A}}{\\sin B}+\\frac{\\sqrt{1+k\\cos^{2}B}}{\\sin C}+\\frac{\\sqrt{1+k\\cos^{2}C}}{\\sin A}\\geq 3\\sqrt{\\frac{k+4}{3}}$ where $k$ is a positive integer number.", "Solution_7": "[quote=\"April\"]Please post the complete solution :)[/quote]\r\nSolution:\r\n$A=\\sqrt{\\frac{(2+1)(1+2\\cos^{2}A)}{\\sqrt{3}\\sin B}}+\\sqrt{\\frac{(2+1)(1+2cos^{2}B)}{\\sqrt{3}\\sin C}}+\\sqrt{\\frac{(2+1)(1+2\\cos^{2}C)}{\\sqrt{3}\\sin A}}$\r\n$\\geq\\frac{\\sqrt{2}+\\sqrt{2}\\cos A}{\\sqrt{3}\\sin B}+\\frac{\\sqrt{2}+\\sqrt{2}\\cos B}{\\sqrt{3}\\sin C}+\\frac{\\sqrt{2}+\\sqrt{2}\\cos C}{\\sqrt{3}\\sin A}$\r\n$=\\frac{\\sqrt{2}}{\\sqrt{3}}\\left(\\frac{1}{\\sin A}+\\frac{1}{\\sin B}+\\frac{1}{\\sin C}+\\frac{\\cos A}{\\sin B}+\\frac{\\cos B}{\\sin C}+\\frac{\\cos C}{\\sin A}\\right)$\r\n$=\\frac{\\sqrt{2}}{\\sqrt{3}}\\left(\\frac{1}{\\sin A}+\\frac{1}{\\sin B}+\\frac{1}{\\sin C}+\\frac{2\\cos^{2}\\frac{A}{2}-1}{\\sin B}+\\frac{2\\cos^{2}\\frac{B}{2}-1}{\\sin C}+\\frac{2\\cos^{2}\\frac{C}{2}}{\\sin A}\\right)$\r\n$=\\frac{\\sqrt{2}}{\\sqrt{3}}\\left(\\frac{\\cos^{2}\\frac{A}{2}}{\\sin\\frac{B}{2}\\cdot\\cos\\frac{B}{2}}+\\frac{\\cos^{2}\\frac{B}{2}}{\\sin\\frac{C}{2}\\cdot\\cos\\frac{C}{2}}+\\frac{\\cos^{2}\\frac{C}{2}}{\\sin\\frac{A}{2}\\cdot\\cos\\frac{A}{2}}\\right)$\r\n$\\geq\\frac{\\sqrt{2}}{\\sqrt{3}}.3\\sqrt[3]{\\cot\\frac{A}{2}\\cdot\\cot\\frac{B}{2}\\cdot\\cot\\frac{C}{2}}\\geq\\frac{\\sqrt{2}}{\\sqrt{3}}\\cdot 3\\cdot\\sqrt{3}=3\\sqrt{2}$\r\n\r\nDone :roll:", "Solution_8": "\\[A=\\sqrt{\\frac{(2+1)(1+2\\cos^{2}A)}{\\sqrt{3}\\sin B}}+\\sqrt{\\frac{(2+1)(1+2cos^{2}B)}{\\sqrt{3}\\sin C}}+\\sqrt{\\frac{(2+1)(1+2\\cos^{2}C)}{\\sqrt{3}\\sin A}}\\]\r\ndo you mean instead:\r\n\\[A=\\frac{\\sqrt{(2+1)(1+2\\cos^{2}A)}}{\\sqrt{3}\\sin B}+\\frac{\\sqrt{(2+1)(1+2cos^{2}B)}}{\\sqrt{3}\\sin C}+\\frac{\\sqrt{(2+1)(1+2\\cos^{2}C)}}{\\sqrt{3}\\sin A}\\]", "Solution_9": "Yes, it's my opinion. Sory for my typing mistake", "Solution_10": "Nice proof, do you have any ideas for the general case?\r\n\r\nNOTE: please keep your final step left unproven because that simple inequality is on the current application test for SIMUW", "Solution_11": "[quote=\"me@home\"]\nNOTE: please keep your final step left unproven because that simple inequality is on the current application test for SIMUW[/quote]\r\n\r\nWhat's is SIMUW?", "Solution_12": "Summer Math Institute at the University of Washington (in washington state)" } { "Tag": [ "analytic geometry", "calculus", "geometry", "integration", "calculus computations" ], "Problem": "Hey every1!!\r\n\r\nThese are some questions from multivariable calculus - polar coordinates. I've tried them out but I guess I need help.\r\n\r\n[b]1) Find the points of intersection of the curves with the the given polar equations: r = 1 + cos\u03b8, r = 1 - sin\u03b8 [/b]\r\n\r\n1 + cos\u03b8 = 1 - cos\u03b8\r\n(cos\u03b8 + sin\u03b8)^2 = 0\r\ncos^2\u03b8 + 2cos\u03b8sin\u03b8 + sin^2\u03b8 = 0\r\n1 + 2cos\u03b8sin\u03b8 = 0\r\nsin2\u03b8 = -1\r\n?????\r\n\r\n[b]2)Find the area bounded by the curve: a) r = 2 - 2sin\u03b8\nb) r = 3 + 2sin\u03b8[/b]\r\na) r = 2 - 2sin\u03b8\r\n 2 - 2sin\u03b8 = 0\r\n 1 - sin\u03b8 = 0\r\n sin\u03b8 = 1\r\n \u03b8 = \u03c0/2\r\n\r\nA = (integral from 0 to \u03c0/2) [1/2f(\u03b8)^2 d\u03b8]\r\n = (integral from 0 to \u03c0/2) [1/2 (2 - 2sin\u03b8)^2 d\u03b8]\r\n = (integral from 0 to \u03c0/2) [1/2 (4 + 4sin^2\u03b8 - 4sin\u03b8) d\u03b8]\r\n = (integral from 0 to \u03c0/2) [2\u03b8 + 2cos\u03b8 d\u03b8]\r\n????\r\n\r\nb) r = 3 + 2sin\u03b8\r\n 3 + 2sin\u03b8 = 0\r\n sin\u03b8 = -3/2\r\n????\r\n\r\nThanks alot for your time and help everyone! uhhh sorry but im not sure how to put in eqts nd stuff. :( hope u still understand tho\r\n\r\neagle", "Solution_1": "i think you should take a look in your results and analyse then before start to use some tricks in them, like the first problem. you got:\r\n$cos(x)+sin(x)=0$ it\u00b4s quite easy to realize that the values of x which satisfies this equality are 3pi/4 and 7pi/4.No extra algebra is necessary. About the secod one i am not so sure, polar cordinates it\u00b4s not my best... but i think using the general formula for finding areas is enough:\r\n\r\n$\\int_{0}^{2\\pi}(2-2sin(\\theta))^{2}.d\\theta$ so :\r\n$\\int_{0}^{2\\pi}4.d\\theta+\\int_{0}^{2\\pi}4sin^{2}(\\theta).d\\theta-\\int_{0}^{2\\pi}8sin(\\theta).d\\theta$ and so on...\r\nwe know that $\\int_{0}^{2\\pi}4sin^{2}(\\theta).d\\theta =\\int_{0}^{2\\pi}(4/2-4cos(\\theta)/2 *d\\theta =4\\pi$\r\nwe also know that $\\int_{0}^{2\\pi}8sin(\\theta).d\\theta =0$ and $\\int_{0}^{2\\pi}4.d\\theta =8\\pi$ so the area is $12\\pi$" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "$ \\int\\limits_{0}^{1} \\frac {4\\sqrt {1 \\minus{} x} \\minus{} \\sqrt {3x \\plus{} 1}}{(3x \\plus{} 1)^2\\left(4\\sqrt {1 \\minus{} x} \\plus{}\\sqrt {3x \\plus{} 1} \\right)} dx$\r\n\r\nSorry., i edited ! :blush:", "Solution_1": "Very easy. Maybe you made a typo.", "Solution_2": "[quote=\"Hong Quy\"]$ \\int\\limits_{0}^{1} \\frac {4\\sqrt {1 \\minus{} x} \\minus{} \\sqrt {3x \\plus{} 1}}{(3x \\plus{} 1)^2\\left(4\\sqrt {1 \\minus{} x} \\plus{} \\sqrt {3x \\plus{} 1} \\right)} dx$\n\nSorry., i edited ! :blush:[/quote]\r\nAns= $ \\frac{log5}{16}$", "Solution_3": "[quote=\"ith_power\"]\nAns= $ \\frac {log5}{16}$[/quote]\r\n\r\nhow did you get it? :huh:", "Solution_4": "Let $ \\sqrt{\\frac{3x\\plus{}1}{1\\minus{}x}} \\equal{}t$ \r\n my friend ith_power ! a how did you solve this problem?" } { "Tag": [], "Problem": "In the sequence 2001, 2002, 2003, ..., each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is 2001 + 2002 - 2003 = 2000. What is the 2004th term in this sequence?\r\n\r\nPlease Post Hints Only", "Solution_1": "[hide=\"Hint 1\"]Try to find an expression for $ n_i$ in terms of $ n_1,n_2$ and $ n_3$[/hide]\n\n[hide=\"Hint 2\"]My expression for $ n_i$ is different depending on whether $ i$ is odd or even[/hide]" } { "Tag": [ "AMC 10", "AMC" ], "Problem": "I have a question about the eligibility for AMC 10. Next year, I will be in grade 10, but I am going to take gr.11 course mathematics, so am I still eligible to write the AMC 10 (its not clear from the website)?\r\n\r\nThank you for your help.", "Solution_1": "AMC 10 and 12 are dependent on solely grade level.\r\nThus, regardless of what course you are taking, if you are 10th or below (which you are), you will be eligible to take the AMC 10.", "Solution_2": "[quote=\"klebian\"]AMC 10 and 12 are dependent on solely grade level.\nThus, regardless of what course you are taking, if you are 10th or below (which you are), you will be eligible to take the AMC 10.\n[/quote]\r\nWhile grade-level (and maximum age) are the only [url=http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/2008-AMC1012Bro.pdf]restrictions[/url] the AMC places on exam registrants, some schools may place additional restrictions on who can take the contests, which contest may be taken, costs paid by the student, and whether one can take the contest on the A date, B date, or both. \r\n\r\nA contest manager (typically a math teacher) will be placing orders for a specific number of 10A and 12A contests at most schools in early December (cheap A-date registration deadline: 16 Dec 2007.) Ideally, the contest manager will have a pretty good idea about who will be taking each contest so that enough exams are ordered. Make sure that your contest manager knows that you are interested in taking the AMC contest(s) and whether you are available on the A date (12 Feb 2008) or B date (27 Feb 2008).\r\n\r\nA few schools order AMC 10 & AMC 12 contests for both the A date and the B date but this is more expensive than ordering for just one date. So most schools order for just one date. It's important to make your contest manager aware of possible conflicts in early December. If your school orders contests for both dates, you could take the AMC 10 on one date and the AMC 12 on the other. Alternatively, you could also take the AMC 10 on both dates.\r\n\r\nSome schools restrict enrollment for the AMC 12 to juniors and seniors - all others take the AMC 10. If you want to take the AMC 12 prior to your junior year at such a school, make sure you convince the contest manager at your school that you are capable - well before the contest ordering deadline (i.e. early fall.) \r\n\r\nUnfortunately, most schools do not participate in the AMC contests. If your school does not participate, you can speak with a math teacher (two or three if necessary) about becoming the school contest manager. You'll need to find a way to pay for the contests. Assuming the school won't foot the bill, students who register will need to split the costs. For early [url=http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/08-AMC1012ARegisForms.pdf]A date registration[/url] you can get ten AMC 10 contest booklets for $\\$$14 plus $\\$$40 registration ($\\$$54 total which could be split by up to 10 students). You can reduce the per-student costs to a few dollars by getting more students to participate in either AMC 10 or AMC 12 on that date.\r\n\r\nOther possible obstacles to AMC contest participation: 1. a room and a proctor that will be available for ~90 minutes on the contest date(s). 2. excused class absences for participants, assuming the contest is given during class hours. 3. conflicts with other activities if the contest is administered after school hours.\r\n\r\nIf you can find enough interested students willing to pay for the contest, it's usually not very difficult to find a math teacher willing to administer it. Supporting a math team or extra-curricular math activities looks good on any teacher's resume, especially newer instructors." } { "Tag": [ "geometry", "trigonometry", "analytic geometry", "incenter", "inradius", "projective geometry", "power of a point" ], "Problem": "The incircle $ k$ of a non-isosceles $ \\triangle ABC$ touches the sides $ AB,BC,CA$ at $ C_{1},A_{1},B_{1}$ , respectively. If $ M$ and $ N$ are midpoints of $ B_{1}C_{1}$ and $ A_{1}C_{1}$ , respectively and $ MN \\cap BC\\equal{}P$ prove that $ C_{1}P\\parallel{}AA_{1}$.", "Solution_1": "Let $ MN\\cap BA\\equiv M_a$, $ A_1B_1\\cap AB\\equiv D$. Since $ M,N$ respectively are the midpoints of $ C_1B_1,C_1A_1$. But $ MN\\parallel{}A_1B_1$. Hence $ M_a$ will be midpoint of $ C_1D$. Now it is so obvious that $ PM_a\\parallel{}A_1D\\Longrightarrow \\frac {\\overline {PB}}{\\overline {PA_1}} \\equal{} \\frac {\\overline {M_aB}}{\\overline {M_aD}}$. Note that $ (DC_1BA) \\equal{} \\minus{} 1$, $ M_a$ is midpoint of $ DC_1\\Longrightarrow M_aD^2 \\equal{} \\overline {M_aB}\\cdot \\overline {M_aA}$. Therefore \r\n\r\n$ \\frac {\\overline {M_aB}}{\\overline {M_aD}} \\equal{} \\frac {\\overline {M_aD}}{\\overline {M_aA}} \\equal{} \\frac {\\overline {M_aB} \\minus{} \\overline{M_aD}}{\\overline {M_aD} \\minus{} \\overline {M_aA}} \\equal{} \\minus{} \\frac {\\overline {DB}}{\\overline {DA}} \\equal{} \\frac {\\overline {C_1B}}{\\overline {C_1A}}\\Longrightarrow \\frac {\\overline {PB}}{\\overline {PA_1}} \\equal{} \\frac {\\overline {C_1B}}{\\overline {C_1A}}$, which implies $ PC_1\\parallel{}AA_1$\r\n\r\nOur proof is completed then.", "Solution_2": "[color=darkblue]We have\n\n$ \\small{d(C,MN)=d(C,A_1B_1)+d(A_1B_1,MN)=}$ \n\n${CA_1 \\cos \\frac{C}{2}+\\dfrac{1}{2}d(C_1,A_1B_1)=(p-c)\\cdot \\cos \\frac{C}{2}+r \\cos \\frac{A}{2} \\cos \\frac{B}{2} }$\n\n$ \\small{A_1D=CD-A_1C=\\dfrac{d(C,MN)}{\\cos \\frac{C}{2}}-(p-c)=r \\dfrac{\\cos \\frac{A}{2} \\cos \\frac{B}{2}}{\\cos \\frac{C}{2}}=}$ \n\n${\\dfrac{\\sqrt{p(p-a)(p-b)(p-c)} \\cdot \\sqrt{\\frac{p(p-a)}{bc}} \\cdot \\sqrt{\\frac{p(p-b)}{ac}} }{p\\sqrt{\\frac{p(p-c)}{ab}} }}=\\dfrac{(p-a)(p-b)}{c}$\n\nWe need to prove\n\n$ \\dfrac{AC_1}{AB}=\\dfrac{A_1D}{A_1B} \\Longleftrightarrow \\dfrac{p-a}{c}=\\dfrac{\\dfrac{(p-a)(p-b)}{c}}{p-b}$,\n\nclearly true.[/color]", "Solution_3": "We'll use barycentric coordinates with respect to $\\triangle ABC.$\n\n$ I_c (a: b: \\minus{} c) \\ , \\ A_1 (0: p \\minus{} c: p \\minus{} b)$ \n\nLine $B_1C_1$ meets $ BC$ at $ X_a (0: p \\minus{} c: b \\minus{} p)$\n\n$ I_cA_1 \\equiv (c(p \\minus{} c) \\plus{} b(p \\minus{} b))x \\minus{} a(p \\minus{} b)y \\minus{} a(p \\minus{} c)z \\equal{} 0$\n\n$ Q \\equiv AB \\cap I_cA_1 \\equiv (a(p \\minus{} b): c(p \\minus{} c) \\plus{} b(p \\minus{} b): 0)$ \n\nThe incenter $ I \\ (a: b: c)$ clearly satisfies the equation of the line $I_cA_1$\n\n$ X_aQ \\equiv (p \\minus{} b)(c(p \\minus{} c) \\plus{} b(p \\minus{} b))x \\minus{} a(p \\minus{} b)^2y \\minus{} a(p \\minus{} c)(p \\minus{} b)z \\equal{} 0$ \n\n$ \\Longrightarrow Q,I, X_a$ are collinear.\n\n$ X_a, I_c$ and $ P$ are the poles of $ AA,MN$ and $ I_cA_1$ WRT $ (I)$ $\\Longrightarrow$ $ Q \\equiv {} AB \\cap I_cA_1$ is the pole of $ PC_1$ WRT $ (I).$ Since the line $ X_aQ$ goes through $ I$ then the polars of $ X_a, Q$ meet at infinite $ \\Longrightarrow$ $ PC_1 \\parallel AA_1.$", "Solution_4": "[quote=\"babylon\"]The incircle $ k$ of a non-isosceles $ \\triangle ABC$ touches the sides $ AB,BC,CA$ at $ C_{1},A_{1},B_{1}$ , respectively. If $ M$ and $ N$ are midpoints of $ B_{1}C_{1}$ and $ A_{1}C_{1}$ , respectively and $ MN \\cap BC = P$ prove that $ C_{1}P\\parallel{}AA_{1}$.[/quote]\r\nMy idea is to slog with trigonometry! :)\r\nHere it is enough to prove that, $ \\frac {AB}{AC_1}\\frac {BA_1}{PA_1} \\iff \\frac {c}{s - a} = \\frac {s - b}{PA_1}$\r\nNow we calculate $ PA_1$. Using sine law in $ \\triangle PNA_1$ we get, $ \\frac {PA_1}{\\cos \\frac A2} = \\frac {\\frac {C_1A_1}{2}}{\\cos \\frac C2}$\r\nAlso, using sine law in $ \\triangle BA_1C_1$ we get, $ \\frac {C_1A_1}{\\sin B} = \\frac {s - b}{\\cos \\frac B2}\\iff C_1A_1 = 2(s - b)\\sin \\frac B2$\r\nFinally, $ PA_1 = \\frac {(s - b) \\cos \\frac A2 \\sin \\frac B2}{\\cos \\frac C2}$\r\nNow setting the values for half angles we get,\r\n\\[{ \\frac {s - b}{PA_1} = \\frac {\\cos \\frac C2}{\\cos \\frac A2 \\sin \\frac B2} = \\frac {\\sqrt {\\frac {s(s - c)}{ab}}}{\\sqrt {\\frac {s(s - a)}{bc}} \\sqrt {\\frac {(s - a)(s - c)}{ac}}}} = \\frac {c}{s - a}\r\n\\]\r\nAs desired.\r\n\r\nEDIT: Fixed typo.", "Solution_5": "Dear Mathlinkers,\r\nlet L be the meet point of AA1 and MN.\r\nFor beginning synthetically this nice problem, we have to prove that C1L is parallel to BC and we are done.\r\nSincerely\r\nJean-Louis", "Solution_6": "Dear Mathlinkers,\r\nthink to draw the parallel to BC through A...\r\nSincerely\r\nJean-Louis", "Solution_7": "Let $ AA_1\\cap (I)\\equal{}\\{R\\}, Q$ be the midpoint of $ RA_1$.\r\nWe have $ \\angle NA_1P\\equal{}\\angle C_1RQ$\r\nBut $ PN//A_1B_1 \\Rightarrow \\angle NPA_1\\equal{}\\angle B_1A_1C\\equal{}\\angle A_1C_1B_1$\r\nBecause quadrilateral $ A_1C_1RB_1$ is harmonic, this result is well-known: $ \\angle A_1C_1B_1\\equal{}\\angle RC_1Q$\r\nTherefore, $ \\Delta PA_1N\\sim \\Delta C_1RQ\\Rightarrow \\Delta C_1PA_1$~$ \\Delta A_1C_1R$\r\n$ \\Rightarrow \\angle PC_1A_1\\equal{}\\angle C_1A_1R \\Rightarrow$ QED", "Solution_8": "The sketch of my proof (sorry because I don't have many time left)\r\nLet $ C_1P$ meet $ A_1B_1$ at $ R$.Using harmonic division,it is equivalent to prove that $ (A_1R,A_1P,A_1C,A_1A)$ is harmonic,which is true since $ AA_1,BB_1,CC_1$ are concurrent.", "Solution_9": "[quote=\"No Reason\"]The sketch of my proof (sorry because I don't have many time left)\nLet $ C_1P$ meet $ A_1B_1$ at $ R$.Using harmonic division,it is equivalent to prove that $ (A_1R,A_1P,A_1C,A_1A)$ is harmonic,which is true since $ AA_1,BB_1,CC_1$ are concurrent.[/quote]\r\nSorry for asking but $ A_1P\\equiv A_1C$, isn't it?", "Solution_10": "It's an typo,it must be $ A_1C_1$ not $ A_1C$ :P", "Solution_11": "[quote=\"jayme\"]Dear Mathlinkers,\nlet L be the meet point of AA1 and MN.\nFor beginning synthetically this nice problem, we have to prove that C1L is parallel to BC and we are done.\nSincerely\nJean-Louis[/quote]\r\n\r\nNice idea Jean-Louis! Let $ \\{T\\} \\equal{} AM\\cap A_1B_1$, then $ AT\\perp BT$. Therefore\r\n\\[ \\frac {AC_1}{BC_1} \\equal{} \\frac {AM}{MT} \\equal{} \\frac {AL}{LA_1},\r\n\\]\r\nwhich proves that $ C_1L\\parallel{}BC$ and the conclusion follows.", "Solution_12": "Dear Mathlinkers,\r\na purely synthetic proof:\r\n1. let U, V the meetpoints of the parallel to BC through A with A1B1, A1C1\r\nand L, X the meetpoints of the parallel to BC through C1 with AA1, A1B1\r\n2. According to Boutin's theorem,\tA is the midpoint of UV ; it follows that L is the midpoint of C1X \r\n(see for example, http://perso.orange.fr/jl.ayme vol. 1 A propos du th\u00e9or\u00e8me de Boutin)\r\n3. According to the Thales's theorem (the line joining two midpoints in a triangle), L, M, N are collinear.\r\nand we are done.\r\nSincerely\r\nJean-Louis", "Solution_13": "Let $ r$ be the inradius of $ \\triangle ABC$.\r\nNote that\r\n$ \\frac {BP}{PA_1} \\equal{} \\frac {[BNP]}{[PNA_1]} \\equal{} \\frac {BN \\cdot NP \\cdot sin BNP}{NP \\cdot NA_1 \\cdot sin PNA_1}$. But $ \\frac {BN}{NA_1} \\equal{} \\frac {BA_1}{r}$ and $ \\angle BNP \\equal{} \\angle MNI$, and $ \\angle PNA_1 \\equal{} 90 \\minus{} \\angle MNI$. So\r\n$ \\frac {BP}{PA_1} \\equal{} \\frac {BA_1}{r} \\cdot \\frac {sin INM}{cosINM} \\equal{} \\frac {BA_1}{r} \\cdot tan INM \\equal{} \\frac {BA_1}{r} \\cdot tan C_1AI* \\equal{} \\frac {BA_1}{r} \\cdot \\frac {r}{C_1A} \\equal{} \\frac {BA_1}{C_1A} \\equal{} \\frac {BC_1}{C_1A}$\r\nThus, $ \\frac {BP}{PA_1} \\equal{} \\frac {BC_1}{C_1A}$, so $ C_1P \\parallel AA_1$.\r\n\r\n*This is true because $ \\triangle IC_1M \\sim \\triangle IC_1A$, so $ IM \\cdot IA \\equal{} C_1I^2 \\equal{} IN \\cdot IB$, so $ AMNB$ is cyclic by power of a point, so $ \\angle BAM \\plus{} \\angle MNB \\equal{} 180$, so $ \\angle C_1AM \\equal{} \\angle MNI$.", "Solution_14": "[quote=\"livetolove212\"]Let $ AA_1\\cap (I) \\equal{} \\{R\\}, Q$ be the midpoint of $ RA_1$.\nWe have $ \\angle NA_1P \\equal{} \\angle C_1RQ$\nBut $ PN//A_1B_1 \\Rightarrow \\angle NPA_1 \\equal{} \\angle B_1A_1C \\equal{} \\angle A_1C_1B_1$\nBecause quadrilateral $ A_1C_1RB_1$ is harmonic, this result is well-known: $ \\angle A_1C_1B_1 \\equal{} \\angle RC_1Q$\nTherefore, $ \\Delta PA_1N\\sim \\Delta C_1RQ\\Rightarrow \\Delta C_1PA_1$~$ \\Delta A_1C_1R$\n$ \\Rightarrow \\angle PC_1A_1 \\equal{} \\angle C_1A_1R \\Rightarrow$ QED[/quote]\r\n\r\nQ: What does \"Harmonic Quadrilateral\" mean?" } { "Tag": [ "quadratics" ], "Problem": "For what values of $ k$ does $ (k\\plus{}1)\\cdot9^{x}\\minus{}4k\\cdot3^{x}\\plus{}k\\plus{}1\\equal{}0$ have 2 solutions for $ x$?", "Solution_1": "$ (k\\plus{}1)\\cdot9^{x}\\minus{}4k\\cdot3^{x}\\plus{}k\\plus{}1\\equal{}0$...\r\nUm...I changed $ 9^{x}$ to $ 3^{2x}$ to $ 9\\cdot3^{x}$. \r\nI distributed the 9, and ended up with $ (9k\\plus{}9)\\cdot3^{x}\\minus{}4k\\cdot3^{x}\\plus{}k\\plus{}1\\equal{}0$.\r\nSubtracting like terms gave me $ (5k\\plus{}9)\\cdot3^{x}\\equal{}\\minus{}k\\minus{}1$, \r\nleaving me with $ 3^{x}\\equal{}\\frac{\\minus{}k\\minus{}1}{5k\\plus{}9}$. What now? I think I went about this the wrong way.", "Solution_2": "[hide]Let $ 3^{x}\\equal{}y$. Thus, the equation becomes:\n\n$ (k\\plus{}1)y^{2}\\minus{}4ky\\plus{}k\\plus{}1\\equal{}0$\n\nFor there to be 2 roots of this equation, the discriminant must be positive, so we have:\n\n$ (4k)^{2}\\minus{}4(k\\plus{}1)(k\\plus{}1)>0$\n\n$ 16k^{2}\\minus{}4k^{2}\\minus{}8k\\minus{}4>0$\n\n$ 3k^{2}\\minus{}2k\\minus{}1>0$\n\n$ (3k\\plus{}1)(k\\minus{}1)>0$\n\n$ k<\\minus{}\\frac{1}3$ or $ k>1$[/hide]", "Solution_3": "Wow. It makes so much sense...:P", "Solution_4": "b-flat, that's a great start, but $ k$ cannot be $ \\minus{}1$ because the equation becomes linear and therefore only has 1 solution. Also, you need to show that $ y$ is positive because it equals $ 3^{x}$.", "Solution_5": "[quote=\"Icy\"]$ (k\\plus{}1)\\cdot9^{x}\\minus{}4k\\cdot3^{x}\\plus{}k\\plus{}1 \\equal{} 0$...\nUm...I changed $ 9^{x}$ to $ 3^{2x}$ to $ 9\\cdot3^{x}$. \nI distributed the 9, and ended up with $ (9k\\plus{}9)\\cdot3^{x}\\minus{}4k\\cdot3^{x}\\plus{}k\\plus{}1 \\equal{} 0$.\nSubtracting like terms gave me $ (5k\\plus{}9)\\cdot3^{x}\\equal{}\\minus{}k\\minus{}1$, \nleaving me with $ 3^{x}\\equal{}\\frac{\\minus{}k\\minus{}1}{5k\\plus{}9}$. What now? I think I went about this the wrong way.[/quote]\r\n\r\nI think your mistake is here:\r\n\r\n9^x = 3^(2x) = [b]9*3^x[/b]\r\n\r\nIt should be: 9^x = 3^(2x) = (3^x)^2, then substitute 3^x with y.", "Solution_6": "I don't think it was a mathematical error, I just used the wrong approach.", "Solution_7": "It is a mathematical error. $ 3^{2x}\\neq 9\\cdot 3^{x}\\equal{} 3^{x\\plus{}2}$.", "Solution_8": "Oh....haha. Sorry. :blush:" } { "Tag": [ "geometry", "circumcircle", "inequalities" ], "Problem": "There is a triangle ABC with lengths 4,7,9. For a point inside the triangle, X, Find the value of AX^2+BX^2+CX^2 such that it is minimum.\r\n\r\n(AX,BX,CX being the line segments)", "Solution_1": "Just as how a rectangle's minimum area is when it is a square, I think the minimum here is when x is the circumradius. That would make the answer we need $3d^2$ where $d$ is the circumradius, or distance from x to any vertex. There is a fun formula to find the circumradius of a circle, which is\r\n$\\frac{abc}{4\\sqrt{s(a+b-s)(b+c-s)(c+a-s)}}$ where $s$ is the semiperimeter. So we want\r\n$\\frac{3a^2b^2c^2}{16s(a+b-s)(b+c-s)(c+a-s)}$\r\nPlugging in values, our answer is $\\frac{1323}{20}$", "Solution_2": "sorry, but since i got this problem from TJHSST Inequality lesson, is there a way to use Cauchy-Schwarz, AM-GM chain, or any other inequality to solve this??? Maybe Mildorf can help, since he wrote this.", "Solution_3": "But the triangle is obtuse...which means that the circumradius lies outside the circle, and so it can't be X.\r\nYet I have no idead how to solve this...\r\nIf you put the point on the side with length 9 (let's say it's BC) and AX is an altitude, then the sum is 505/9..." } { "Tag": [ "geometry", "circumcircle", "geometry solved" ], "Problem": "Angle bisectors of angles A,B,C of triangle ABC meet its circumcircle in K,L,M respectively. Choose on AB a point R. For points P,Q holds the following: RP parallel to AK, BP perpendicular to BL, RQ parallel to BL and AQ perpendicular to AK. Prove that KP,LQ, MR are concurrent.\r\n\r\nI am sure it was posted here some time ago, but I can't find it...", "Solution_1": "Yes, it was...\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=118910\r\n\r\n Darij", "Solution_2": "That's exactly what I was looking for! Thanks :lol:" } { "Tag": [ "inequalities", "trigonometry" ], "Problem": "Prove that in any triangle $ ABC$ exists the strict inequality $ \\cos A + 2(\\cos B + \\cos C) < 3\\ .$\r\n\r\n[b]Remark.[/b] Obtain \"equality\" for $ \\{\\begin{array}{c}\r\nB=C\\searrow 0\\\\\\\r\nA\\nearrow\\pi\\end{array}\\ .$", "Solution_1": "Can we use casework?\r\nRephrase it as\r\n$ f\\equal{} 2.(cosA\\plus{}cosB\\plus{}cosC) \\minus{} cosA$\r\nFor A acute\r\nClearly by Jensen's$ (cosA\\plus{}cosB\\plus{}cosC)<\\equal{}3/2$\r\nClearly $ f<2.3/2 \\equal{}3$ (As cosA>0)\r\nFor A=90\r\nTST $ 2.(cosB\\plus{}cosC)<3$\r\n$ 2.sqrt(2).cos(B\\plus{}\\pi/4)<3$\r\nWhich is true.\r\n[b] I AM STUCK IN THE 3RD (AND MOST EASIEST CASE?) CAN SOMEONE HELP? [/b]", "Solution_2": "[quote=\"madness\"] [color=darkred]I AM STUCK IN THE 3RD ( AND [b]MOST EASIEST CASE[/b] ? ) CAN SOMEONE HELP ?[/color] [/quote]\r\n[color=darkblue]The third case is [b]most difficultly[/b] ! See the my remark from topic.[/color]", "Solution_3": "[hide]\nIf either $ B$ or $ C$ is obtuse, the inequality is obvious. Otherwise, use Jensen to get\n\\[ \\cos B\\plus{}\\cos C\\leq 2\\cos \\frac{B\\plus{}C}{2}\\equal{}2\\sin \\frac{A}{2}\\]\nLet $ x\\equal{}\\sin \\frac{A}{2}$. Then\n\\[ 1\\minus{}2x^2\\plus{}4x\\equal{}3\\minus{}2(x\\minus{}1)^2<3\\]\nsince $ 0<\\frac{A}{2}<\\frac{\\pi}{2}$ implies $ x\\neq 1$.\n[/hide]", "Solution_4": "\"If either or is obtuse, the inequality is obvious.\"\r\nHow is it so? Can you please provide a proof?\r\n :)", "Solution_5": "If $ B$ obtuse, then $ \\cos A,\\cos C\\leq 1$ and $ \\cos B<0$.", "Solution_6": "Why cant A be obtuse?", "Solution_7": "[hide=\"For Madness & Co.\"]\n[quote=\"Virgil Nicula\"] [color=darkred]$ \\triangle\\ ABC$ $ \\implies$ $ \\cos A + 2(\\cos B + \\cos C) < 3$ with \"equality\" iff $ \\{\\begin{array}{c} B = C\\searrow 0 \\\\\n\\ A\\nearrow\\pi\\end{array}$[/color] [/quote]\n[color=darkblue][b][u]Method I.[/u][/b] Prove easily that $ \\cos A = - 2\\cos B\\cos C + \\cos (B - C)\\ .$ Therefore,\n\n$ \\cos A + 2(\\cos B + \\cos C) = - 2\\cos B\\cos C + \\cos (B - C) + 2(\\cos B + \\cos C) =$\n\n$ 3 - \\underbrace {2(1 - \\cos B)(1 - \\cos C)}_{\\ge 0} - \\underbrace {[1 - \\cos (B - C)]}_{\\ge 0}\\le 3\\ .$\n\nWe'll have equality iff $ B - C = 0$ and $ B = 0$ or $ C = 0$, i.e. $ B = C\\searrow 0$ and $ A\\nearrow \\pi\\ .$\n\n[b][u]Method II.[/u][/b] $ \\cos A + 2(\\cos B + \\cos C) =$ $ 1 - 2\\sin^2\\frac A2 + 4\\sin\\frac A2\\cos\\frac {B - C}{2} =$\n\n$ 1 - 2(\\sin\\frac A2 - \\cos\\frac {B - C}{2})^2 + 2(1 - \\sin^2\\frac {B - C}{2}) =$ \n\n$ 3 - \\underbrace {2[(\\sin\\frac A2 - \\cos\\frac {B - C}{2})^2 + \\sin^2\\frac {B - C}{2}]}_{\\ge 0}\\le 3\\ .$\n\nWe'll have equality iff $ B = C$ and $ \\sin\\frac A2 = 1$ , i.e. $ A\\nearrow\\pi$ and $ B = C\\searrow 0\\ .$\n\n[b][u]Remark.[/u][/b] $ \\{\\begin{array}{c} (1 - \\cos x)(1 - \\cos y) = (\\cos\\frac {x + y}{2} - \\cos\\frac {x - y}{2})^2 \\\\\n \\\\\n\\cos^2\\frac {x + y}{2} + \\cos^2\\frac {x - y}{2} = 1 + \\cos x\\cos y\\end{array}$\n\n[b][u]Method III.[/u][/b] Distinguish two cases : $ A\\le 90^{\\circ}$ or $ A>90^{\\circ}\\ .$ Therefore :\n\n$ 1\\blacktriangleright$ $ A\\le 90^{\\circ}$ , i.e. $ \\cos A\\ge 0$ . I'll use the well-known inequality $ \\sum\\cos A\\le\\frac 32$ .\n\nThus, $ \\cos A+2(\\cos B+\\cos C)=2\\sum\\cos A-\\cos A\\le 3-\\cos A\\le 3\\ .$\n\n$ 2\\blacktriangleright$ $ A>90^{\\circ}$ , i.e. $ \\cos A<0$ . Thus, $ \\{B,C\\}\\subset (0,90^{\\circ})$ and $ 0<\\sin\\frac A2< 1$ $ \\implies$\n\n$ \\cos A+2(\\cos B+\\cos C)=$ $ 1-2\\sin^2\\frac A2+4\\sin\\frac A2\\cos\\frac {B-C}{2}\\le$\n\n$ 1-2\\sin^2\\frac A2+4\\sin\\frac A2=$ $ 3-2(1-\\sin\\frac A2)^2\\le 3\\ .$[/color][/hide]", "Solution_8": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2838176#p2838176]Mitrirovic,1967:[/url]\n\\[cosA+\\lambda (cosB+cosC)\\leq 1+\\frac{\\lambda^2}{2}.\\]" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "A cube (ABCDEFGH) has side length of 6. What is the distance between AC and DF?", "Solution_1": "what is the standard way of labeling a cube (two squares, ABCD and EFGH)?", "Solution_2": "Which is what I was wondering also...probably like A corresponds to E, B corresponds to F, C corresponds to G, D corresponds to H? Maybe? (corresponds is not the best word choice, but I think you can figure out what I am saying...)", "Solution_3": "[quote=\"Chinaboy\"]A cube (ABCDEFGH) has side length of 6. What is the distance between AC and DF?[/quote]\r\nIs that the distance between a face diagonal and a space diagonal that does not intersect it?", "Solution_4": "I'm pretty sure its the diagonal of the cube; otherwise, if its a diagonal of one of the faces, why would the problem include a cube and not just a square?", "Solution_5": "It could also be a face diagonal and a side. 2 skew lines? (If the top face ABCD and bottom EFGH are labeled in opposite directions, so A-E D-F C-G B-H correspond.)\n\n\n\nIf it's a face dia + space dia, I think [hide]we can draw the common perpendicular. That would be a vertical line through the center. Call it MN (M on AC, N on DF). \n\n\n\nThen draw AN. We know AN = DN and DN = 3 :rt3: . AM = 3 :rt2: . Using Pythagoras we get \n\n\n\n(AM) :^2: + (MN) :^2: = (AN) :^2: . \n\n\n\nSo MN = 3? Not sure if this is right.[/hide]", "Solution_6": "What I got last night, but I did it really quicky in my head....", "Solution_7": "Sorry for the confusion.\r\n\r\nE is under A\r\nF is under B\r\nG is under C\r\nH is under D\r\n\r\nthe two are indeed skew lines.\r\n\r\nOne is the space diagonal, one's a face diagonal." } { "Tag": [ "induction", "quadratics", "real analysis", "real analysis solved" ], "Problem": "Let two real numbers $p_1>0,q_1<0$. The equation $x^2+p_1x+q_1=0 $ have two real roots $p_2>0,q_2<0$.....The equation $ x ^2 +p_nx+q_n=0$ have two real roots $ p_{n+1}>0,q_{n+1}<0.....$.Prove that two sequences $\\{p_n\\},\\{q_n\\}$ have finite limit at $+\\infty$ and find theirs limit", "Solution_1": "Ok, I think I got it even the following is neither very elegant nor short.\r\nFirst I think that $q_1 < 0$ otherwise you wouldn't have $p_2 > 0$.\r\nMoreover, wed may note that by an easy induction, from $q_1 < 0$ that $q_n < \r\n0$ for all $n$ (that is 0 is never a solution of the quadratic).\r\n\r\nSolving the quadratic, it is easy to get :\r\n$p_{n+1}q_{n+1} = q_n$ (1) and $p_{n+1}+q_{n+1} = -p_n$ (2)\r\nthat is :\r\n$p_{n+1} = \\frac {-p_n + \\sqrt {p_n^2 - 4q_n} } 2$ (3) and $q_{n+1} = \\frac {-p_n - \\sqrt {p_n^2 - 4q_n} } 2$ (4)\r\n\r\nNow, we have \r\n$p_{n+1} - 1 = \\frac {-p_n - 2 + \\sqrt {p_n^2 - 4q_n} } 2 = \\frac {-2(p_n+q_n+1)} {p_n + 2 + \\sqrt {p_n^2 - 4q_n} } $.\r\nUsing (2), it leads to :\r\n$p_{n+1} - 1 = \\frac {2(p_{n-1} - 1)} {p_n + 2 + \\sqrt {p_n^2 - 4q_n} }$ (5)\r\n\r\nThus, $p_{n+1} - 1 $ and $p_{n-1} - 1 $ have the same sign.\r\nNow, if $p_1 \\geq 1$, then $p_{2n+1} \\geq 1$ for all $n$. It follows that :\r\n$|p_{2n+2} - 1 | = \\frac {2|p_{2n} - 1|} {p_{2n+1} + 2 + \\sqrt {p_{2n+1}^2 - 4q_{n2+1}} } \\leq \\frac {2|p_{2n} - 1|} {2p_{2n+1} + 2 } \\leq \\frac {|p_{2n} - 1|} {2 } $,\r\nfrom which we deduce easily that the sequence $\\{p_{2n} \\}$ is convergent to 1.\r\n\r\nThus, for $n$ sufficientely large, we have $p_{2n} > \\frac 1 2 $.\r\nUsing (5) again, we deduce in the same manner that \r\n$|p_{2n+1} - 1 | = \\frac {2|p_{2n-1} - 1|} 3$\r\nfrom which it follows that $\\{ p_{2n+1 \\}}$ converges to 1.\r\nThus $\\{ p_n \\}$ has limit 1, and from (2) we deduce that $\\{ q_n \\}$ has limit -2.\r\n\r\nIn fact, the reasoning above holds again if $p_2 \\geq 2$ (interchanging the parity).\r\n\r\nNow, assume that $00$ such that $e \\leq p_n$ for all $n$.\r\nUsing (5) and the reasoning above, we have $|p_{n+1} - 1| \\leq \\frac {|p_{n-1} - 1|} {1+e}$, and the conclusion is the same.\r\n\r\nThus, in any case, the sequence $\\{ p_n \\}$ has limit $1$, and the sequence $\\{ q_n \\} $ has limit $-2$.\r\n\r\nPierre." } { "Tag": [ "floor function", "number theory unsolved", "number theory" ], "Problem": "The questions on the programming contest asked for $N$th square free number, where $N=100,1000,10^6,10^9$.\r\nMy student solved this problem (by writing not too short program :) ) and obtained that $10^9$-th square free number is $1644934081$.\r\nAfter that I asked him to calculate $10^9\\cdot \\frac{\\pi^2}{6}\\approx 1644934066$ and said that it is an approximate answer. He was very surprized.\r\nJustify my observation!\r\n\r\n[b]Open question[/b] (at least for me)\r\nWhat is the estimation for difference $N$-th square free number - $N\\cdot \\frac{\\pi^2}{6}$?\r\n\r\nIs true that $N$-th square free number > $N\\cdot \\frac{\\pi^2}{6}$ ?", "Solution_1": "I think I've find a way to solve the problem but I'm not sure it's right:\r\n$M=N^{th}$ square free number\r\n\r\n$M = N+card(n, \\exists p>1\\ p^2|n, nn}\\mu(k)\\frac N{k^2}.$ Both sums can be estimated easily by taking absolute values; the first is less than $\\sqrt{N}$ and the second is less than $\\frac N{\\sqrt{N}}=\\sqrt{N}$. Thus $f(N)=\\frac{6N}{\\pi^2}+O(\\sqrt{N}).$\r\n\r\nSince $f$ and $f^{-1}$ differ by at most a constant multiple, the same estimate applies to your original problem.\r\nThese are usually gross overestimates; the normal values should be more like $\\sqrt[4]{N}.$ Your actual value at $10^9$ is even closer, but it is possible to get lucky. I wouldn't expect the direction of the error to be consistent, although sign changes would be very infrequent." } { "Tag": [], "Problem": "Carla must arrive at the airport that is 49 miles away in 70 minutes or less. What is the minimum average speed in miles per hours that Carla must maintain to get to the airport in 70 minutes or less? Express your answer as a whole number.", "Solution_1": "$ 70 minutes \\equal{} \\frac{7}{6}hours$\r\n$ v\\equal{}\\frac{d}{t}\\equal{}\\frac{49}{\\frac{7}{6}}\\equal{}\\boxed{42}$", "Solution_2": "Because of per hour, we convert 70 min to 60 min (an hour) :\n\n$70 minutes*\\frac{60}{70}=60$\n\n$49 minutes*\\frac{6}{7}=42$\n\n\nHence the answer is 42" } { "Tag": [], "Problem": "which window on an automatic grapher shows the graph of y =-7x +11 in three quagrant ?\r\n\r\na) -10 \u2264 x \u2264 10, -5 \u2264 y \u2264 10 b) -5 \u2264 x \u2264 5, -10 \u2264 y \u2264 5\r\n\r\nc) -10 \u2264 x \u2264 10 , -5 \u2264 y \u2264 15 d) -5 \u2264 x \u2264 10 , -5 \u2264 y \u2264 5", "Solution_1": "hello, i think it must be c),$ \\minus{}10\\le x\\le 10$,$ \\minus{}5\\le y \\le 15$\r\nSonnhard." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If a,b,c are sides of a triangle with a^n+b^n[/b] for all n>2 natural nos.", "Solution_1": "See http://www.mathlinks.ro/viewtopic.php?t=201445", "Solution_2": "the inequality is equavilent to :\r\n $ (a^2\\plus{}b^2)^n \\geq (a^n\\plus{}b^n)^2$ \r\n It's familar with", "Solution_3": "Hong Quy wrote;\r\n[quote]the inequality is equavilent to :\n(a^2+b^2)^n => (a^n+b^n)^2\nIt's familar with[/quote]\r\n\r\nNOT FAMILIAR TO ME!!! :(", "Solution_4": "it's the power mean inequality", "Solution_5": "[quote=\"not_trig\"]it's the power mean inequality[/quote]\r\n\r\nI have heard of this inequality, but what is the general statement of the inequality and the proof(s)? Thanks.", "Solution_6": "see \r\n\r\n[url]http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf[/url]" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Prove that :\r\n\\[ \\sum_{k\\equal{}0}^{n\\minus{}1} (\\minus{}1)^{k} (n\\minus{}k)^n\\binom{n}{n\\minus{}k} \\equal{} n!\\]", "Solution_1": "[b]Question:[/b] How many functions $ f : [n] \\to [n]$ are bijections?\r\n\r\n[b]Answer 1:[/b] $ n!$. This is the RHS.\r\n\r\n[b]Answer 2:[/b] For every subset $ S$ of $ [n]$ of size $ n\\minus{}k$, there are $ (n\\minus{}k)^n$ functions with range contained in $ S$ and $ {n \\choose n\\minus{}k}$ such subsets. To compute the number of functions with range exactly $ [n]$ is just inclusion-exclusion: start with the functions with range contained in $ [n]$, subtract the functions with range contained in $ [n\\minus{}1]$, etc. This is the LHS." } { "Tag": [ "function", "limit", "linear algebra", "matrix", "algebra", "domain", "induction" ], "Problem": "What condition(s) are necessary for newton's method to converge? My calc textbook says something about $ \\minus{}1< f(x)f''(x)/(f'(x))^2 < 1$, but where does this come from/how is it used?", "Solution_1": "The Newton's Method is a Iterative method. For the Iterative method ,if the Iterative function |g'(x)|<=1 .the Iterative sequence converge to the exact solution of the equation.\r\n\r\n\r\ng(x)=x-f(x)/f'(x) is the Iterative function of Newton's Method. you can refer to any numerical analysis textbook for the Iterative Method", "Solution_2": "thanks... I think it makes some sense now. I do not have access to any textbooks though (other than my calc book), so could you please explain where the condition $ |g'(x)| \\le 1$ comes from?\r\n\r\nAnother thing that I am interested in is how to determine which start values will produce a convergent sequence. Though most of the problems I have faced seem to work with nearly any start value, a few seem to require relatively good starting guesses to converge - such as $ y\\equal{}tan x \\minus{}x$ with $ x \\in (\\pi/2, 3\\pi/2)$.", "Solution_3": "When we use Iterative Methods (include Newton's Method) to solve the equation $ f(x)\\equal{}0$, we convert $ f(x)\\equal{}0$ to it's equivalent equation $ x\\equal{}g(x)$.such as $ g(x)\\equal{}x\\minus{}f(x)$,then, if $ x_0$ is a root of $ x\\equal{}g(x)$, namely $ x_0\\equal{}g(x_0)$ , so that \r\n$ x_0$ is a root of $ f(x)\\equal{}0$,because $ g(x_0)\\equal{}x_0\\minus{}f(x_0)\\Rightarrow f(x_0)\\equal{}0$. vice versa!\r\nThe function $ g(x)$ was called Iterative function ,For any initial value $ x_0$(here $ x_0$ is not root of $ x\\equal{}g(x)$ )can produce an Iterative sequence {$ x_n$}:\r\n$ x_1\\equal{}g(x_0)$...$ x_2\\equal{}g(x_1)$......$ x_n\\equal{}g(x_{n\\minus{}1})$ if $ {x_n}$ converge to $ x^*$ then $ \\lim_{n\\rightarrow \\plus{}\\infty}x_n\\equal{}\\lim_{n\\rightarrow\\infty}g(x_{n\\minus{}1}) \\Rightarrow x^*\\equal{}g(x^*)$ so, $ x^*$ is a root of $ x\\equal{}g(x)\\Leftrightarrow f(x)\\equal{}0$\r\n\r\n\r\nI am sorry ,the condition $ |g'(x)\\leq 1|$ is wrong .the correct condition is that : exist $ L$ ,s.t. $ |g'(x)|\\leq L<1$\r\n\r\n\r\nThe condition $ |g'(x)|\\leq L<1$ can ensure the Iterative sequence {$ x_n$}'s convergence for any initial value $ x_0$ :\r\nBecause :\r\n\\[ |x_{k\\plus{}i}\\minus{}x_{k\\plus{}i\\minus{}1}|\\equal{}|g(x_{k\\plus{}i\\minus{}1})\\minus{}g(x_{k\\plus{}i\\minus{}2})|\\equal{}|g'(\\xi)\\parallel{}x_{k\\plus{}i\\minus{}1}\\minus{}x_{k\\plus{}i\\minus{}2}|\\leq L|x_{k\\plus{}i\\minus{}1}\\minus{}x_{k\\plus{}i\\minus{}2}|\\leq\\cdots \\leq L^{k\\plus{}i\\minus{}1}|x_1\\minus{}x_0|\\]\r\nso \\[ |x_{k\\plus{}p}\\minus{}x_k|\\leq\\sum_{i\\equal{}1}^{p}|x_{k\\plus{}i}\\minus{}x_{k\\plus{}i\\minus{}1}|\\leq \\sum_{i\\equal{}1}^{p}L^{k\\plus{}i\\minus{}1}|x_1\\minus{}x_0|\\equal{}\\frac{L^k(1\\minus{}L^p)}{1\\minus{}L}|x_1\\minus{}x_0|\\]\r\nThe above expression tells a fact that {$ x_n$} is a Cauchy sequence . \r\n\r\n\r\nI am sorry!Because my English is very poor, so some syntax errors in my reply is inevitable,however i hope taking pleasure in help anybody!", "Solution_4": "xiangtianxu, thanks for the explanation! I only have a few things to say:\r\n\r\n-we are not necessarily using $ f(x)\\equal{}0$ for our iteration, we are just using a function with the same zeros. For example, with newton's method we are using $ g(x) \\equal{} x\\minus{}\\frac{f(x)}{f'(x)}$ - but $ \\frac{f(x)}{f'(x)}$ has the same zeros as $ f(x)$ as long as none of the zeros are shared with $ f'(x)$.\r\n\r\n-The existence of $ |L|<1$ proves that the method converges, but if such an $ L$ does not exist, the method could still converge for some $ x_0$ values... right? So my next question - how could we find/predict these values? Anyway, I suppose that this question may not have a definite answer.\r\n\r\n-what's a cauchy sequence?\r\n\r\nOn a side note, xiangtianxu, I think your English is pretty good. Though your grammar is imperfect, you still manage to communicate your ideas well, which is the main goal of language anyway.\r\n\r\na small note:\r\nwhen you say \"The condition $ |g'(x)| \\le L < 1$ can ensure,\" you are saying that the condition might or it might not imply that the sequence converges. I believe what you mean is that \"the condition $ |g'(x)| \\le L < 1$ will ensure the iterative sequence's convergence\" or that \"the condition $ |g'(x)| \\le L < 1$ ensures that the iterative sequence will converge\". The difference here is that either of these two statements assert that the condition implies the sequence's convergence, while the wording you used does not really say what the condition proves.\r\n\r\nOther than that, I think that everything you said is pretty clear.", "Solution_5": "A condition for the iterative scheme $ x_{n\\plus{}1} \\equal{} g(x_n)$ with $ f(a) \\equal{} 0$ to converge given an initial guess within some neighbourhood of $ a$ is that g is differentiable and $ |g'(x)| < 1$ for x within that interval.\r\n\r\nIf we want to use Newton's Method on f with $ f(a) \\equal{} 0$, we can let $ g(x) \\equal{} x \\minus{} \\frac{f(x)}{f'(x)}$ and it can be shown that there exists an interval around a for which $ |g'(x)| < 1$ provided that $ f''$ exists and is bounded in this interval, with $ f'(a) \\ne 0$, and so Newton's Method will converge.", "Solution_6": "TO facis:\r\n \"-what's a cauchy sequence? \"\r\nCauchy sequence's Definition:\r\n {$ x_n$} is a sequence,if $ \\forall \\epsilon>0,\\quad \\exists N_0\\in N$,while $ n,m>N_0$,then $ |x_n\\minus{}x_m|<\\epsilon$(apparently ,we can write m=n+p for $ p\\in N$)\r\nwe called {$ x_n$} as a Cauchy sequence.\r\n\r\nIn the real number field:The Cauchy sequence is a equivalence of the convergent sequence. (it hold on the general complete matric space).\r\nThen, the condition $ |g'(x)|\\leq L<1$ imply which for any initial value $ x_0$ in the function's(f) domain of definition .Certainly, the existence of $ g'(x)$ is required!\r\n\r\n\r\n\"The existence of proves that the method converges, but if such an does not exist, the method could still converge for some values... right? So my next question - how could we find/predict these values? Anyway, I suppose that this question may not have a definite answer\"\r\n\r\nI think that any methods should to hold some conditions.so ,Iteration can't solve some questions which haven't hold the conditions of Iteration.Newton's method else.\r\nBut,if a equation f(x) holds on the conditions of Iteration(Newton's method) ,we can use Iteration(Newton's method) to solve it.\r\n\r\nFor example:\r\n If f(x) is a defined in an interval [a,b] ,and hold on following conditions\r\n$ f(a)f(b)<0,\\quad f'(x)f''(x)\\neq 0 \\quad for \\quad x\\in [a,b],$\r\nwe can prove that for the starting number $ x_0\\in [a,b]$ if $ f(x_0)f''(x_0)>0$ then the Iterative sequence which was produced by the Newton's method is a monotonic sequence with convergence\u3002\r\nProof:\r\n\\[ \\because f(a)f(b)<0,f'(x)f''(x)\\neq 0\\]\r\n$ \\therefore f'(x)\\neq 0$ imply f(x) is a monotone function , therefore f(x)=0 has a unique $ x^*$ solution in [a,b].\r\n\r\n For the condition $ f'(x)f''(x)>0$,there are four cases should be considered.\r\n1.$ f'>0,f''>0$\r\n2.$ f'>0,f''<0$\r\n3.$ f'<0,f''>0$\r\n4.$ f'<0,f''<0$\r\nHere we just consider case 1, there is same way to handle the other cases.\r\n$ \\because f(x_0)f''(x_0)>0$,therefore $ f(x_0)>0$\r\n$ \\therefore x_0>x^*$ ,therefore $ x_1\\equal{}x_0\\minus{}\\frac{f(x_0)}{f'(x_0)}0$\r\n $ x_0>x_1>x^*$\r\nuse the induction,we can get that:\r\n $ x_{n\\minus{}1}>x_n>x^*$ it tells our a fact that the sequence {$ x_n$} is a monotone decreasing sequence with inf $ x^*$ \r\n$ \\therefore \\lim_{n\\rightarrow\\infty}x_n\\equal{}x^*$ \r\nThis proof already tells our a truth that how to choose the starting number when the f hold on the conditions\r\n\r\n\r\nSorry again: :) \r\nUse English to express my opinion is so difficult to me,there should have many syntax error in my state which would bring\r\nsome trouble for you.i honestly say :\"i am so sorry\". in the meantime, i hope you can point my error and correct it in paraphrase.\r\nThank you very much!" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "I don't know if it was proposed before.\r\n\r\n Two triangles, which lie on the same circle, intersect at points A,B,C,D,E,F.\r\n Then prove that lines AD, BE, CF intersect at one point.", "Solution_1": "Sketch :\r\nFrom Pascal theorem we have to demonstrate that $KN, LP, MQ$ are cevians in triangle $UVW$,\r\nequaling to demonstrate $\\frac{sin \\phi_1 }{sin \\phi_2} \\cdot \\frac{sin \\phi_3 }{sin \\phi_4} \\cdot \\frac{sin \\phi_5 }{sin \\phi_6}=1$.\r\nBut denoting arc angles as $\\alpha, \\beta , \\gamma, d, e, f$ (multiplyied by 2) and using\r\nCeva theorem in triangles $ABU$ and others we have \r\n$\\frac{sin f}{sin e} \\frac{sin \\phi_1 }{sin \\phi_2} \\frac{sin j}{sin b}= 1$,\r\nmultiplying we get the desired result.", "Solution_2": "Nice solution.\r\n In Coxeter's book I found that some of Pascal's lines are concurrent. Does it mean that this problem has been already proposed?" } { "Tag": [], "Problem": "In the $xy$-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}=25$?", "Solution_1": "[hide]The circle has center (6,8) with radius 5. (6,8) is the midpoint of the line from (0,0) to (12,16). The length from (0,0) to (6,8) is 10.\n\nIf we go around the circle, the straight parts of the path would be the leg of right triangle with other sides 5 and 10, which makes a 30-60-90 triangle, so the straight parts have length $5\\sqrt{3}$, and there are two of these. The arc part of the path has a angle of 60 degrees, so the length is $10\\pi/6$\n\n$\\boxed{10\\sqrt{3}+\\frac{5\\pi}{3}}$.[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that\r\n\\[ \\sum a\\sqrt{a^2\\plus{}bc} \\ge (2\\minus{}\\sqrt{2})(\\sum a^2\\plus{}\\sqrt{2}\\sum ab).\\]\r\n:)", "Solution_1": "Actually, for all $ p>1$, the inequality holds\r\n\r\n$ \\sum a\\sqrt {a^2\\plus{}(p^2\\minus{}1)bc} \\ge (2\\minus{}p)\\sum a^2\\plus{}2(p\\minus{}1)\\sum ab$.", "Solution_2": "[quote=\"Vasc\"]Actually, for all $ p > 1$, the inequality holds\n\n$ \\sum a\\sqrt {a^2 \\plus{} (p^2 \\minus{} 1)bc} \\ge (2 \\minus{} p)\\sum a^2 \\plus{} 2(p \\minus{} 1)\\sum ab$.[/quote]\r\nOf course, VasC. :) It follows from Schur's Inequality and my result:\r\n\\[ \\sum \\frac{1}{a\\plus{}\\sqrt{a^2\\plus{}kbc}} \\ge \\frac{9}{(1\\plus{}\\sqrt{k\\plus{}1})(a\\plus{}b\\plus{}c)}.\\]\r\n:)", "Solution_3": "But it is not shalex's one,it is from me .:(" } { "Tag": [ "inequalities" ], "Problem": "If $a,b,c$ are positive three real numbers such that $| a-b | \\geq c , | b-c | \\geq a, | c-a | \\geq b$ . Prove that one of $a,b,c$ is equal to the sum of the other two.", "Solution_1": "$a=-3,b=-4,c=-5$ also works . But none of them equal to the sum up to ther other two.", "Solution_2": "Try positive reals :-)", "Solution_3": "For non-negative reals it's easy:\r\n\r\nSquaring and rearranging the first inequality we get\r\n\r\n$(a-b-c)(a-b+c)\\geqslant 0$ or\r\n\r\n$(-a+b+c)(a-b+c)\\leqslant 0\\qquad(1)$\r\n\r\nSimilarly, from other two we get\r\n\r\n$(a-b+c)(a+b-c)\\leqslant 0\\qquad (2)$\r\n$(a+b-c)(-a+b+c)\\leqslant 0\\qquad(3)$\r\n\r\nIf neither of $-a+b+c,a-b+c,a+b-c$ is zero, the we have three numbers that are pairwise of opposite sign, which is impossible. Hence, at least one of them is zero.", "Solution_4": "Actually, it seems much easier than that: suppose without loss of generality that $a \\geq b \\geq c$. Then we have immediately that $b \\geq a + c$. From this it immediately follows that $c \\leq 0$. If we're looking only at the non-negative case, it's immediately clear that the only solution is $(a, a, 0)$. Also, if we don't restrict ourselves to the non-negative case, any solution of the form $(a, b, c)$ with $a \\geq b$ and $c\\leq b - a$ will work.", "Solution_5": "but in the question i think it was for all real\nbut it does not work for negative real", "Solution_6": "[quote=Rushil]If $a,b,c$ are positive three real numbers such that $| a-b | \\geq c , | b-c | \\geq a, | c-a | \\geq b$ . Prove that one of $a,b,c$ is equal to the sum of the other two.[/quote]\n[url=http://www.artofproblemsolving.com/community/c6h579719p3423804]Romania District Olympiad 2014[/url]", "Solution_7": "WLOG, We may assume that a>=b>=c.\nNow,a-b>=c, \n b-c>=a\n a-c>=b\nSumming up the first 2 equations we get 0>=c.\nBut since c is positive, c=0.\nAgain from the first 2 equations we get a>=b and b>=a. This implies that a=b.\nTherefore, we get a=b+c." } { "Tag": [ "induction", "number theory proposed", "number theory" ], "Problem": "Show that, for $n \\in N$ we have:\r\n\r\n$3^{n+1}||2^{3^n}+1$\r\n$5^{n+1}||4^{5^n}+1$\r\n\r\nmore generally:\r\n\r\n$p^{n+1}||(p-1)^{p^n}+1$\r\n\r\nwith p=2k+1 (k>0)", "Solution_1": "I have a question: does $\\|$ mean that $p^{n+1}$ is the power of $p$ with the maximal exponent which divides that number? (I've seen this notation before).", "Solution_2": "Induction after $n$ works perfectly fine:\r\n\r\n$p^n\\|(p-1)^{p^{n-1}}+1=a+1$. We now use the formula $a^p+1=(a+1)(a^{p-1}-a^{p-2}+\\ldots-a+1)$. It's pretty clear (also making use of the fact that $p$ is odd) that $a^{p-1}=M_{p^2}+1,\\ -a^{p-2}=M_{p^2}+1,\\ldots$, so we get $a^{p-1}-a^{p-2}+\\ldots-a+1=M_{p^2}+p$, so $p\\|\\frac{a^p+1}{a+1}$, and since $p^n\\|a+1$, we get $p^{n+1}\\|a^p+1$." } { "Tag": [ "Putnam", "greatest common divisor", "number theory unsolved", "number theory" ], "Problem": "1)\r\nprove that the number of 5-tuples of positive integers (a,b,c,d,e) satisfying\r\n1/a + 1/b + 1/c + 1/d + 1/e = 1/5\r\n is an odd integer\r\n\r\n2)let S denote the set of 6-tuples (a,b,c,d,e,f) satisfying \r\na 2 +b 2 +c 2 +d 2 +e 2 = f 2 \r\nconsider the set T = {abcdef } a,b,c,d,e,f beLonging to S.\r\nFind the greatest common divisor of the elements of T\r\n\r\n3)Let p be a prime > 3 .find the pairs of integers (a,b) satisfying\r\na 2 + 3ab + 2p(a+b) + p 2 = 0 \r\n\r\ncheers!!", "Solution_1": "I have a question regarding (1): by (a,b,c,d,e) we usually denote ordered 5-tuples. Is this the case here? I mean, if a=/=b, is (a,b,c,d,e) the same as (b,a,c,d,e) or not?", "Solution_2": "I think it should be ordered 5-tuples :? though the paper doesnt mention it explicitly it got me confused me in the real exam as well.it was the last question and the only one in which i dint make real progress even though i had a full hour at my disposal :( .that effectively kills my chances for selection this year,given the fact that they select only 3-4 students from 12th grade.(for me it's been a double blow,i was in the last 250 in national physics olympiad as well,but cudnt make it to the last 25 :( :( :(", "Solution_3": "Well, I think the 5-tuples could be found. We take min{a,b,c,d,e}. We can se that it must be <=25 and so on. I don't know if it works though, because there are WAY TOO MANY cases :? \r\n\r\nI think there's another idea here.", "Solution_4": "HEY\r\nlook up putnam 1997.it has got a similar problem", "Solution_5": "I think i get no.3\r\n\r\na^2+3ab+2p(a+b)+p^2=0\r\n4a^2+12ab+8p(a+b)+4p^2=0\r\n4a^2+12ab+9b^2+8p(a+b)+4p^2=9b^2\r\n(2a+3b)^2+4p(2a+3b)+4p^2=9b^2+4pb\r\n(2a+3b+2p)^2=9b^2+4pb\r\n(6a+9b+6p)^2=81b^2+36pb\r\n(6a+9b+6p)^2+4p^2=(9b+2p)^2\r\n(6a+4p)(6a+18b+8p)=-4p^2\r\n(3a+2p)(3a+9b+4p)=p^2\r\n3a+2p=(+/-)1, (+/-)p, (+/-)p^2...(1)\r\n3a+9b+4p=(-/+)p^2, (-/+)p, (-/+)1...(2)\r\n(2)-(1)\r\n2p+9b=(+/-)(p^2+1), (+/-)2p\r\n9b=(p-1)^2, -(p+1)^2, 0, -4p\r\n\r\nCase 1: 9b=(p-1)^2\r\nLet p=3t+1, then b=t^2\r\n3a+2p=-1 or -p^2\r\na=(-1-2p)/3 or (-p^2-2p)/3\r\na=-2t-1 or -3t^2-4t-1\r\n\r\nCase 2: 9b=-(p+1)^2, let p=3t-1, b=-t^2\r\n3a+2p=1 or p^2\r\na=(1-2p)/3 or (p^2-2p)/3\r\na=-2t+1 or 3t^2-4t+1\r\n\r\nCase 3: 9b=0\r\nb=0\r\n=>3a+4p=p, p=-a\r\n\r\nCase 4: 9b=-4p, no solution, as 9|p\r\n\r\nTherefore all solutions are:\r\n(a,b,p)=(-2t-1, t^2, 3t+1), (-3t^2-4t-1, t^2, 3t+1), where t E N such that 3t+1 is a prime\r\n (-2t+1, -t^2, 3t-1), (3t^2-4t+1, -t^2, 3t-1), where t E N such that 3t-1 is a prime.\r\n(-p, 0, p)\r\n\r\nI hope it's ok.", "Solution_6": "I think I've made some mistake. For ordered 5-tuples I get an EVEN number. I didn't count them, but I considered the cases: all of the 5 numbers are distinct, 2 of them are equal and the other 3 are different from the first 2 and distinct etc. If we have ordered 5-tuples we multiply each solution by some number, which is even for many of the cases I mentioned above, so we can eliminate them. \r\n\r\nCan someone else try to do this as well? I think I'm missing something.", "Solution_7": "Let me tell you what I mean:\r\n\r\nWe have 1 solution where all the numbers are the same:{25,25,25,25,25}.\r\n\r\nEach solution of the form {a,a,a,a,b} has to be counted 5 times, because we have 5C5 possible positions for b. I'll find the solutions of this sort later.\r\n\r\nEach solution of the form {a,a,a,b,b} has to be counted 5C2=10 times, which is even, so we can disregard it.\r\n\r\nEach solution of the form {a,a,a,b,c} has to be counted 20 times, which is also even.\r\n\r\nEach solution of the form {a,a,b,b,c} has to be counted 30 times (also even).\r\n\r\nEach solution of the form {a,a,b,c,d} has to be counted 60 times (even).\r\n\r\nFinally, each solution of the form {a,b,c,d,e} has to be counted 120 times (even).\r\n\r\nThis means that for the parity of the total number of solutions we only need to cionsider the solutions of the form (a,a,a,a,a) (1 such solution) and those of form (a,a,a,a,b). The parity of the total number of ordered 5-tuples is the parity of the number of solutions of the form {a,a,a,a,b}+1, which means that in order to have an odd number of solutions (in total) the number of sets of the form {a,b} which satisfy 4/a+1/b=1/5 has to be EVEN.\r\n\r\nNow let's solve the eqn 4/a+1/b=1/5. 1/5-1/b=4/a, so (b-5)/5b=4/a. It's easy to show that the only pair (a,b) is (45,9), so we have an ODD number of pairs (a,b) s.t. 4/a+1/b=1/5. This statement contradicts the last word of the last paragraph, so where's the mistake? Maybe they didn't want ORDERED 5-tuples?", "Solution_8": "ya MAYB they did not want ordered 5-tuples!!!(i.e after seeing your calculations)\r\ni am really sorry i can not make the question any more clearer.and i am really sorry about the fact that you (grobber,i mean) have wasted quite some time on it. (but i cant help it)\r\nif u \"wish\",try it for unordered 5-tuples\r\ncheers mate!!", "Solution_9": "Don't worry about it, it didn't take that much time. :D\r\n\r\nI doubt I'll try it for unordered 5-tuples, since I have no idea how to start :)" } { "Tag": [], "Problem": "According to CEE's website, the USABO Semifinalists (top 10%) will be announced on March 3 (Monday). From what I know, the list itself will be posted on CEE's website, and your teacher will be able to see your score (even if you did not make semifinals).\r\n\r\nLast year, the cutoff for Semifinals was a score of 30 out of 50 on the Open Exam.\r\n\r\nGood luck everyone!", "Solution_1": "haha...i didn't know that my teacher would be able to see my score as well. \r\nwow i'm really nervous. good luck to everyone who took the exam.", "Solution_2": "Are the semifinalists actually posted on the website? I can't find it anywhere and my school was off today... I want to know if I made it o.o", "Solution_3": "[quote=\"kops723\"]I can't find it anywhere and my school was off today... I want to know if I made it o.o[/quote]\r\n\r\nI think if you find the teacher that distributed the test, he/she might have your results.", "Solution_4": "omg! i got in XD\r\nhow did everyone else do?", "Solution_5": "I made it too! :)\r\n\r\nThe passing score this year is 29 out of 50 on the Open Exam. I believe the highest score in the nation is 43 (correct me if I'm wrong), which is the same as last year's highest. \r\n\r\nCorrection from previous post: Only your teacher/test distributor has your score right now. However, if your teacher/test distributor gives you his/her username and password for CEE, then you can check various other statistics (ie: how many students took the Open Exam, how many qualified for Semifinal Round, how many students qualified per school, etc). \r\n\r\n\r\nThe [b]Semifinal Exam[/b] will be administered between March 10th and March 21st (ask your test distributor when he/she will administer the test). According to CEE's website:\r\n[quote] The Semi-final is a two hour exam consisting of three sections. Part A of the Semi-Final Exam covers basic knowledge in a multiple choice format (approximately 40 minutes). Parts B and C consist of higher level questions with multiple answers, interpretation of graphs, analysis of data, case studies, fill in the blank questions and some open response questions (approximately 80 minutes). A student must pace him/herself over the three sections. [/quote]\r\n\r\nLast year, Part A was multiple choice, Part B was matching & fill-in-the-blank, and Part C was free response (mostly on genetics if I remember correctly). My teacher told me they change the free response emphasis each year (ie: genetics, plant biology, etc), but I am [i]not[/i] sure how accurate this is.", "Solution_6": "does anyone know where we can find practice questions? this is the first time i've qualified for semifinals so i'm new to this\r\n\r\ngood luck to everyone and congrats", "Solution_7": "Darn, I was expecting a 40+ and I ended up with a 33...\r\n\r\nBTW the first question about iron-deficiency was BS, it had two answers....", "Solution_8": "wow, i guessed on half the questions and i still got a 32 and got in..\r\ni was like blah i dont know anything about bio ill just see what happens\r\nand i got in. hm\r\n\r\napproximately how many students each year take USABO?\r\nand how many get into semifinals?", "Solution_9": "I think the number last year was about 8000... And 500 usually make it into semifinals. \r\n\r\nYeah, the first question threw me in a loop, too.\r\n\r\nBut I got in: 39 , woohoo!", "Solution_10": "I got in too! Quite excited--my AP bio class had barely finished genetics, like chapter 15 in Campbell, and I hadn't had general bio since 9th grade. \r\n\r\nIs there anyone here who's done the semifinal round before? I imagine that many people take AMC/AIME every year, but not many retake USABO since it is so year-specific.", "Solution_11": "[quote=\"Harmonikdiskorde\"]I got in too! Quite excited--my AP bio class had barely finished genetics, like chapter 15 in Campbell, and I hadn't had general bio since 9th grade. [/quote]\r\n\r\nCampbell is a good book but my 9th grade AP bio class uses the Solomam. I wanted to take the test by my teacher didn't distribute it :( .", "Solution_12": "I got in :)\r\n39 hehhehee\r\n\r\nSemifinals look impossible tho", "Solution_13": "[quote=\"7h3.D3m0n.117\"][quote=\"Harmonikdiskorde\"]I got in too! Quite excited--my AP bio class had barely finished genetics, like chapter 15 in Campbell, and I hadn't had general bio since 9th grade. [/quote]\n\nCampbell is a good book but my 9th grade AP bio class uses the Solomam. I wanted to take the test by my teacher didn't distribute it :( .[/quote]\r\n\r\n9th grade AP Bio? Alas, our schools must have different educational philosophies. At mine, AP sciences must be preceded by a first year, level honors science course. \"Biology I\" is the standard 9th grade science, and AP Bio is generally off-limits until 12th grade, per Madame Department Head. \r\n\r\nI remember a zebra-striped bio book in 9th grade. It actually tried to teach the details of the Krebs cycle, succinyl coA and all, a section that the teachers most eagerly skip.", "Solution_14": "[quote=\"Harmonikdiskorde\"]9th grade AP Bio? Alas, our schools must have different educational philosophies. At mine, AP sciences must be preceded by a first year, level honors science course. \"Biology I\" is the standard 9th grade science, and AP Bio is generally off-limits until 12th grade, per Madame Department Head. \n\nI remember a zebra-striped bio book in 9th grade. It actually tried to teach the details of the Krebs cycle, succinyl coA and all, a section that the teachers most eagerly skip.[/quote]\r\n\r\nIt's a new program designed to give the \"smarter\" kids in the class something challenging (which I would say has quite worked). We are taking the AP exam on May 12th, but I was nudging my teacher to register the school for the USABO, which apparently he never go to. \r\n\r\nWe had just finished evolution and genetics so I don't think we would have gotten in anyway, since we just started organismal during the test times." } { "Tag": [ "geometry", "inradius", "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "here's a nice problem. I found an idea but I couldn't solve the hole problem\r\nLet $ E$ be a Banach Space, and let $ \\Omega \\subset E$ be an open bounded set.\r\nsuppose that for all $ x,y \\in \\Omega$ there is a ball containd in $ \\Omega$ which contains $ x$ and $ y$\r\nshow that $ \\Omega$ is a ball", "Solution_1": "Let $ d$ be the diameter of $ \\Omega$, and $ r$ be its inradius (i.e. the supremum of the radii of all open balls contained in $ \\Omega$). Our assumption implies $ d \\equal{} 2r$. It follows that for any $ n\\ge 1$ there is a point $ x_n$ such that $ B(x_n,r\\minus{}1/n)\\subset \\Omega\\subset B(x_n,r\\plus{}1/n)$. The sequence $ x_n$ has a limit, or so I hope." } { "Tag": [ "geometry", "circumcircle", "congruent triangles", "Pythagorean Theorem" ], "Problem": "A ten foot ladder leans against a vertical wall. Its midpoint is twice the distance from the floor as from the wall. How high does the top of the ladder go on the wall?", "Solution_1": "[hide] 4*sqrt (5)[/hide]", "Solution_2": "solutions! post solutions, not answers. Or at least outline your method or phrase it as a hint or something.", "Solution_3": "how do i post geometric proof?", "Solution_4": "Explain the steps you took.\r\n\"Let the top of the ladder be point A, the bottom be point B, and the middle be point M. Let C be the corner of the wall. Draw perpendiculars from M to BC and CA, and let them intersect respectively at P and Q. We are given MP = 2MQ. We know \r\nMB = AM = 5. MP and AC are two parallel lines cut by the transversal AMB, so angle MAQ = angle BMP. Also, angle AQM = angle MPB because both are right. Thus, triangle AQM is similar to triangle MPB, and since they have the same hypotenuse, they are congruent.\"\r\n\r\nI mean, this was more technical really then you have to do -- you just have to explain everything you did. It's usually helpful to make an explicit labeling to begin with, and then after that just say everything you found and why you found it was true. I didn't finish the rest of the proof, but it can be written up the same way -- congruent triangles, so equal segments, then the pythagorean theorem, and then we're done.", "Solution_5": "Arg, I'm not very good with English so I thought the distance between the midpoint and the wall was twice the distance from the floor, but the answer is just 2*the answer I got ([hide]2:rt5:[/hide] so meh.\n\n\n\nI have to say it does get kind of annoying when people don't post solutions, especially on a constant basis \n\nYes, I'm talking about you, not you ml2, but you *looks at someone's name in AoPS*", "Solution_6": "Here is simplified version of JBL's description. (that one was kinda confusing)\r\n\r\n[img]http://bwsolo.com/prob1.jpg[/img]\r\n\r\nHeh hopefully this is right, tell me if im wrong lol", "Solution_7": "Wow, nice. Yes, you could post pictures, too :).", "Solution_8": "how can i post a picture?", "Solution_9": "Click on the \"Img\" button before and after you type in the address (URL) for the Image.\r\nYou can also highlight the URL and click on \"Img\", either way.\r\n\r\n\r\nWhat I'm curoous about is how you found the perfect pucture for this problem, do you host this website yourself?", "Solution_10": "Umm I made the image in photoshop and then put it on my webspace and linked it in tags", "Solution_11": "As JBL mentioned, you should label all the points you plan to use. A is the top of the ladder, B is the bottom, M is the midpoint, C is the right angle and such( refer to JBL's solution). MC is congruent to AM and BM because the circumradius of a right triangle is equal to half the hypotenuse. Using this you can determine that PC is congruent to BP and that ABC is a 30-60-90 degree triangle, and solve for AC.\r\n\r\n[/quote] If home is where the heart is and my heart is in this universe, what are you doing in my house?" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "[x] is defined to be the least integer which exceeds or is equal to x. Compute [((3^(1/2)) +1)^2m] for m=1,2,3,4,& 5. Show that 2^(m+1) divides these numbers. Prove \r\n2^(m+1) always divide [((3^(1/2)) +1)^2m].", "Solution_1": "The sequence in question is $ a_m \\equal{} (1 \\plus{} \\sqrt {3})^{2m} \\plus{} (1 \\minus{} \\sqrt {3})^{2m}$, which we can write as $ (4 \\plus{} 2 \\sqrt{3})^m \\plus{} (4 \\minus{} 2 \\sqrt{3})^m \\equal{} 2^m \\left( (2 \\plus{} \\sqrt{3})^m \\plus{} (2 \\minus{} \\sqrt{3})^m \\right)$. The sequence $ \\frac{a_m}{2^m} \\equal{} b_m$ satisfies the recurrence relation\r\n\r\n$ b_{m\\plus{}2} \\equal{} 4b_{m\\plus{}1} \\minus{} b_m$\r\n\r\nwith $ b_0 \\equal{} 2, b_1 \\equal{} 4$. Hence $ b_m$ is always even." } { "Tag": [ "Gauss", "probability", "function", "algorithm", "ARML", "calculus", "derivative" ], "Problem": "I know how to solve Two-Person Zero-Sum games, but how do you solve N-Person Zero-Sum games? For example, how would you find the optimal solution for each player in the following game:\r\n\r\nThree players simultaneously choose a number (with replacement) from the set $ \\{1,2,3\\}$. The person with the smallest [i]unique[/i] number recieves a payoff of 2, while the other two players each recieve a payoff of -1. When all players choose the same number, the payoff is 0 to each of them.", "Solution_1": "Winning this game depends on the strategies that your opponents use, so it cannot have an optimal solution.\r\n\r\nWith these kinds of games, the best you can generally do is make up a few strategies, see how they fare against each other, and tabulate the results. Of course, even doing that requires specifying some tournament rules.", "Solution_2": "This is not true. By the Minimax Theorem of von Neumann there always exists an optimal strategy in a Constant/Zero-Sum game.", "Solution_3": "No, that's not what the theorem says.", "Solution_4": "[quote=\"Mathworld and Wikipedia\"]Minimax Theorem: The fundamental theorem of game theory which states that every finite, zero-sum, two-person game has optimal mixed strategies. It was proved by John von Neumann in 1928. \n\nIn 1944 John von Neumann and Oskar Morgenstern proved that any zero-sum game involving n players is in fact a generalized form of a zero-sum game for two players.[/quote]\r\n\r\nThese two results imply that any constant/zero-sum game should have an optimal (mixed) strategy. In fact, I even know what the optimal mixed strategy is for this game, I just don't know how it was found.", "Solution_5": "You did not use the word \"mixed\" in your first post. It is a different question. Perhaps I misunderstood, but \"optimal for each player\" has a specific meaning.", "Solution_6": "\"Strategy\" is widely used to mean \"mixed strategy\" in the context of strategic games, for the reason that analyzing the pure strategies isn't interesting on its own (it's just some finite computation and it doesn't exhibit the same nice properties).\r\n\r\nGauss: I don't know how to do these in general except as differential equations. For the particular game you mentioned, the Nash Equilibrium is that each player picks 1 with probability $ \\frac{1}{2}$ and the other two with probability $ \\frac{1}{4}$, assuming I've correctly ruled out any equilibria on the boundary.", "Solution_7": "That's the solution that I have also, but I did it in a very ad hoc \"logic-it-out\" kind of way. I was hoping to find a procedure for how to solve N-Player games in general. What is your method with differential equations?", "Solution_8": "Well, it's a procedure, but to actually implement it with reasonable amounts of effort seems to require using the \"solve\" function on your favorite CAS. For this particular game, we have six variables (the probability that player $ i$ chooses $ j$ for $ i \\equal{} 1,2,3$, $ j \\equal{} 1,2$) with certain constraints (nonnegative and the sums of the appropriate pairs are at most 1); for each collection of variables, we can write down the utility to each player of choosing that strategy. At a Nash equilibrium, the partials of the utility to player $ i$ with respect to the variables he controls must be zero; this gives you six equations to solve, and Mathematica tells me that the solution I mentioned is unique. (I haven't tried to solve by hand, actually.)\r\n\r\nAs it turns out, there are some other Nash equilibria for this game on the boundary: for example, player 1 always picks 1 and player 2 and 3 never pick 1 is a Nash equilibrium, but it's clearly not optimal for players 2 and 3.\r\n\r\nI gather that there are algorithms for computing Nash equilibria, but I don't know anything about any of their details :(\r\n\r\nIs your \"logic-it-out\" approach simple enough to write down here?\r\n\r\nCompletely unrelatedly, I gather SC will be going to the new ARML site -- I hope it goes well, though I'm sorry I won't see you at Penn State.", "Solution_9": "My solution went like this:\r\n\r\nMy solution assumed that the equilibrium was symmetric with respect to all the players and involved experimenting with different mixes until I got close enough to prove the rest.\r\n\r\nAfter playing around with numbers I conjectured that the optimal strategy involved picking 1 50% of the time. To prove this I reasoned as follows: If each player plays a mixed strategy $ (p,q,r)$, where $ p < .5$, I can improve against them by playing the strategy $ (1,0,0)$. This is because my expected value when we all play $ (p,q,r)$ is zero, but when I play (1,0,0) against their mixed strategy my expected value is $ E \\equal{} 2(1\\minus{}p)^2 \\minus{} 2p(1\\minus{}p) \\equal{} 2(1\\minus{}p)(1\\minus{}2p) > 0$.\r\n\r\nIf we all play a mixed strategy $ (p,q,r)$ with $ p > .5$, I can improve depends on how they mix q and r. If $ p \\geq .5$ and $ q < r$, I can improve by playing $ (0,1,0)$. This is because my expected value against such a strategy is $ E \\equal{} 2(p^2 \\plus{} r^2) \\minus{} (2pq \\plus{} 2qr) \\equal{} 2p(p\\minus{}q) \\plus{} 2r(r\\minus{}q) > 0$. Likewise, if $ p \\geq .5$ and $ q > r$, I can improve by playing $ (0,0,1)$.\r\n\r\nThis shows that if there is a symmetric Nash equilibrium it must have $ p \\equal{} .5$ and $ q \\equal{} r$. If you continue to look at the expected values as above you can show that both of these are in fact sufficient for a Nash Equilibrium. So an equilibrium strategy is $ (.5, .25, .25)$. This method does not rule out the possibility of other Nash equilibria.\r\n\r\nA few days later I realized that this could be changed into a calculus and/or algebraic solution by setting the partial derivatives of the expected value equal to zero and proving that the solution was in deed a local minimum for E (but I was still working under the assumption that the equilibrium was symmetric for all players).\r\n\r\nMore recently, in a book called [u]Dynamic Noncooperative Game Theory[/u] by Tamer Basar and Geert Jan Olsder, I found an algebraic procedure for how to find all [i]inner[/i] mixed strategies for N-Person finite games in Normal Form. It leads to a system of nonlinear (quadratic) equations. The nasty looking, but simple, formula can be found on p.93 of their book.\r\n\r\nThey say, however, \"there is no simple method to determine the [boundary] Nash equilibrium solutions of N-Person finite games in normal form. One basically has to check exhaustively all possible combinations of N-tuples of strategies, to see which ones provide a Nash equilibrium. This enumeration, though straightforward, could at times be rather strenuous, especially when N, or $ m_i$ (the number of possible strategies for each player) are large. However, given an N-tuple of strategies asserted to be in Nash equilibrium, it is relatively simple to verify their equilibrium property...\" (p.89)", "Solution_10": "Thanks for the reference and the details. My knowledge of game theory is rather thin, so I was curious if you can answer the following (rather vague) question: why are zero-sum games special? I gather that solving them is relatively simpler than solving arbitrary strategic games -- is there a \"natural\" explanation for this phenomenon?", "Solution_11": "Here's a quote from some [url=http://www.math.ucla.edu/~tom/Game_Theory/Contents.html]lecture notes[/url] by Thomas S. Ferguson. They make a decent introduction to the subject of game theory, but focus mainly toward economics students.\r\n\r\n\"When the sum of the payoffs is no longer zero (or constant), maximizing one\u2019s own payoff is no longer equivalent to minimizing the opponent\u2019s payoff. The minimax theorem does not apply to bimatrix games. One can no longer expect to play \u201coptimally\u201d by simply looking at one\u2019s own payoff matrix and guarding against the worst case. Clearly, one must take into account the opponent\u2019s matrix and the reasonable strategy options of the opponent. In doing so, we must remember that the opponent is doing the same. The general-sum case requires other more subtle concepts of solution.\"\r\n\r\nAdditionally, non-constant sum games can be complicated by the possibility (if allowed) of cooperation/punishment." } { "Tag": [], "Problem": "Can we find solution to all the conflicts(war, etc) and problems prevalent in this world? Or is it too impractical?\r\nPlease express your views on this.", "Solution_1": "We cannot find solutions to all problems in the universe.\r\n\r\nAnd if we supposedly did find all solutions to all problems, life would have no purpose because there are no problems to solve, a vegetable life. Nothing can be made or created because new ideas would be created, potentially damaging ones. Feelings, including jealously, would also be aroused.\r\n\r\nSuch a utopia would only exist if no one had feelings, a nonexistential life.\r\n\r\nAnd while people might say that they have the solution to all the problems in the world and implement it, creativity and freedom would then be severely controlled, such as depicted in novels [u]Fahrenheit 451[/u] and [u]1984[/u]. Corruption would still occur.", "Solution_2": "We can't find solutions to all problems.\r\nSometimes when you fix one problem, you cause another.\r\nSo, we can usually make things better, but can't fix everything.", "Solution_3": "We couldn't even find Bin Laden -- how do you think we could end ignorance, abuse of power, crime, environmental destruction, poverty/malnutrition/hunger, disease, crime, hate, terrorism, and war (to name a few)?\r\n\r\nHere are a few more related questions:\r\n\r\n1. Do you think a problem-free state is stable?\r\n2. In particular, do you think that the individually-held notion of self-interest will ever be compatible with a problem-free world within any social context?\r\n3. If so, under what kind of social contexts/social conditioning would this be possible?\r\n\r\nThe authors of the books mentioned by [b]lightning[/b] attempted to answer that final question.", "Solution_4": "[quote=\"nobelium\"]Can we find solution to all the conflicts(war, etc) and problems prevalent in this world? Or is it too impractical?\nPlease express your views on this.[/quote]\n\nProbably not. As long as there is hegemony, and Weapons Of Mass Destruction (WMD)-- peace can't be attained. Also, peace in this world is contingent upon some countries or superpower(s) foreign policies. \n\nSo, As of now, your idea or notion of utopia is pretty far-fetched.\n\n[quote=\"TZF\"]We couldn't even find Bin Laden -- how do you think we could end ignorance, abuse of power, crime, environmental destruction, poverty/malnutrition/hunger, disease, crime, hate, terrorism, and war (to name a few)? [/quote]\r\n\r\nI really find it pretty unbelievable that the U.S. (or rather the CIA) couldn't find Bin Laden even after 8 years of relentless pursuit. It is well known that he resides in Pakistan or rather operates from Pakistan. \r\n\r\nhttp://www.dawn.com/wps/wcm/connect/dawn-content-library/dawn/news/pakistan/03-osama-bin-laden-operating-from-pakistan-ss-07\r\n\r\nSo, I think there are some vested interests which are protecting Bin Laden.", "Solution_5": "[quote=\"Gen8\"]As long as there is hegemony, and Weapons Of Mass Destruction (WMD)-- peace can't be attained.[/quote] This seems like overly strong conditions to reach the conclusion; I would have said, \"As long as there are preceived inequities, ambitious, charismatic people, and tools more deadly than a dinner knife, someone somewhere will be doing something unpeaceful.\"", "Solution_6": "[quote=\"JBL\"] This seems like overly strong conditions to reach the conclusion; I would have said, \"As long as there are preceived inequities, ambitious, charismatic people, and tools more deadly than a dinner knife, someone somewhere will be doing something unpeaceful.\" [/quote]\r\n\r\nDear [b] JBL [/b] :roll: , these aren't overly strong conditions. Elimination of WMD-- which implicitly means \"global nuclear disarmament\"-- is one of the most important preconditions for peace in this world. I reiterate again, as long as there are WMD--peace can't be attained. \r\n\r\nAnother precondition for world peace is a radical change in some countries' foreign policies. Some country or countries advocate nuclear disarmament whilst at the same time they possess thousands of nuclear weapons. Another important precondition is a radical change in people's mindsets-- which as an offshoot could result in the end of \"terrorism\". Again, a radical change in people's mindsets is contingent upon a radical change in some countries' (or country's) foreign policies. \r\n\r\nI also assert that hegemony should be curbed because it results in escalation of conflict thereby destabilizing world peace. But, I will acquiesce with some of the points (witty?) [b] JBL [/b] had mentioned.", "Solution_7": "You misunderstood his point -- he's saying that even if WMD's and such weighty issues never existed, there would still not be a \"problem free\" world.\r\n\r\nThis is because the problems of humanity are much more deeply rooted and fundamental -- namely, basic human nature and general patterns in equilibrium social structure (which result from basic human nature).", "Solution_8": "[quote=\"nobelium\"]Can we find solution to all the conflicts(war, etc) and problems prevalent in this world? Or is it too impractical?\nPlease express your views on this.[/quote]\r\n\r\nNot only is it impractical, it's meaningless. But you're right, we could achieve greater peace if these things were under better control.", "Solution_9": "I really hope this isn't spam but I'm a little scared because this is the first time I am posting in this forum.\r\n\r\nHonestly there is no possible way whatsoever to get rid of all potential problems in the world. There is always someone who is unsatisfied. (I recommend reading the book [u]The Giver[/u] by Lois Lowry for an example of a utopia that still has faults.)\r\n\r\nAlso, why should we get rid of all the problems in this world? In my opinion the only people that live on Earth that don't have problems are the people who are dead, buried in a cemetery, and in heaven. If you really don't have any problems, I really think you should get down on you knees, pray, and beg God for some problems because you're going to leaving soon.=]", "Solution_10": "i agree, i see no reason to be in a perfect world. though it seems nice to us right now, imagine being same as everyone else. This concept and its flaws are explained by writer Kurt Vonnegut in \"Harrison Bergeron\", a great story. basically, everyone has been equalized, except Diana Moon Glampers, who does the equalizing, and the methods of equalization are reall cruel. those with great physical strength must carry bags of birdshot, pretty people must wear ugly masks, singers must purposely sound awful, reporters must be monotonous, and the intelligent people wear headgear that sends sirens into their ears every 10 minutes to disrupt their thoughts. being in a perfect world would be boring, awful, and impossible. there will always be problems in this world, because imperfect people cannot create a perfect, problem-free society.", "Solution_11": "The world can never be problem free. If there were no problems, that in itself is a problem, because to have total peace would mean that feelings of people would have to be removed. As lightning said, Fahrenheit 451 and 1984 show the problematic \"problem-free\" societies of the future, where feelings and thoughts are suppressed. A real-world example would be the Soviet Union. The founders of the USSR set out to make a problem-free state, where everyone was equal, and it turned out to be a disaster. People revolted because they weren't allowed their full freedom. Suppression of freedom is necessary for a problem-free world, but then is has been proved no one can bear the thought of that.\r\n\r\nHJ :lol:", "Solution_12": "[quote=\"inoit\"]Honestly there is no possible way whatsoever to get rid of all potential problems in the world. There is always someone who is unsatisfied. (I recommend reading the book [u]The Giver[/u] by Lois Lowry for an example of a utopia that still has faults.)[/quote]\r\n\r\nLove that book! Gives great insight on trying to \"fix\" problems.", "Solution_13": "Obama won the peace award...And he is still sending troops to Afghanistan.", "Solution_14": "I think the answer should be NO. Different people have different points of view. Thus contradiction between always exists.\r\n\r\nJust like geometry. Given different systems of axioms, we can obtain different kind of geometry. They are not consistent with each other, but no one can say either of them is wrong.", "Solution_15": "[quote=\"infinite101\"]Obama won the peace award...And he is still sending troops to Afghanistan.[/quote]\r\n\r\nHe won the award before he did [i]anything[/i]." } { "Tag": [ "function", "Princeton", "college", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Find all functions $ f:R \\longrightarrow R$ satisfying the condition:\r\n$ [f(x)+f(y)].[f(u)+f(v)] = f(xu-yv)+f(xv+yu)$ for all $ x,y,u,v \\in R$", "Solution_1": "i think it is a hard problem too\r\nBut i think you shoulds write \"my notebook\"==> \"my teacher\"\r\nSolution will come soon", "Solution_2": "I think this problem is not too hard . I have a solution :\r\n $ f(x) = x^2 or f(x) = 0 or f(x) = 1/2 $", "Solution_3": "Wow, my friends (master and NTTu), I 'm sure that we both meet this one before (from our teacher), but I post it here so that we can receive many solutions,OK?", "Solution_4": "Isn't it IMO 2002.", "Solution_5": "Isn't it IMO 2002.", "Solution_6": "I'll try to be short, cause I'm pretty tired and the longer I write, the more I am bound to make mistakes. Set all 0, and we get $2f^2(0)=f(0)$. If $f(0)=\\dsp \\frac 12$ then by setting $u=v=0$ we get $f(x)+f(y)=1$ for all $x,y$. This ibviously leads to $f=\\dsp \\frac 12$\r\n\r\nIf $f(0)=0$ the either $f=0$ or $f \\neq 0$ (extraordinary logical argument, don't you think) In the latter case, make $y,v=0$ and we will get $f(xu)=f(x)f(u)$ for all $x,u$. Thus, $f$ is multiplicative, and as we assumed it non-zero, we will have $f(1)=1$. By switching the two pairs of variables $(x,y)$ and $(u,v)$ we will get $f(xu-yv)=f(yv-xu)$ for all $x,y,u,v$ and this implies that $f$ is even. Now, let $x$ and $y$ be non-zero and let $u=\\dsp \\frac 1x, v=\\dsp \\frac 1y$. Keeping multiplicativity in mind and denoting $a=\\dsp \\frac xy$, we get\r\n\r\n\\[f(a)+f(\\dsp \\frac 1a)+2=f(a+\\dsp \\frac 1a)\\]\r\n\r\nMultipliy this by $f(b)$ and we get\r\n\r\n\\[f(ab)+f(\\dsp \\frac ba)+2f(b)=f(ab+\\frac ba)\\]\r\n\r\nFor any positive $x,y$ we will find $a$ and $b$ s.t. $ab=x^2$, $\\dsp \\frac ba=y^2$, so the above relation becomes\r\n\r\n\\[f(x^2)+f(y^2)+2f(xy)=f(x^2+y^2) \\Rightarrow f^2(x)+f^2(y)+2f(x)f(y)=f(x^2+y^2) \\Rightarrow (f(x)+f(y))^2=f(x^2+y^2) \\eqno (1)\\]\r\n\r\nThis last equality and the even-ness of $f$ implies that $f$ is always non-negative. Thus, let $f(x)=g^2(x)$ where $g(x) \\geq 0$ only for positive $x$. $g$ is obviously multiplicative and relation (1) implies that $g^2(x)+g^2(y)=g(x^2+y^2) \\Rightarrow g(a)+g(b)=g(a+b)$ if we set $a=x^2$ and $b=y^2$. $g$ is thus aditive too, and since it is multiplicative, it will be non-zero and therefore positive for $x \\neq 0$. Thus $g$ is a positive aditive function, and thus it is increasing, and we know that this implies that it is linear. The fact that it is multiplicative implies that it is $x$. So $f(x)=x^2$.", "Solution_7": "Yes it's IMO 2002", "Solution_8": "As ussualy, Andrei forgot to check that the function does respect the functional equation, so, if he remembers IMO2002, he lost 1 point and gold medal? :)\r\nAnyway, this is ironic. Andrei, are you at Princeton now? How is there?" } { "Tag": [ "algebra", "polynomial", "abstract algebra", "calculus", "integration", "number theory", "Ring Theory" ], "Problem": "[i]Let $ K$ be a number field. Assume that $ \\mathcal{O}_K$ has only finitely many irreducibles $ p_1,\\cdots,p_n$. Then $ 1 \\plus{} p_1\\cdots p_n$ must be divisible by some irreducible $ q$, and this can't be any of $ p_1,\\cdots,p_n$. Contradiction. Of course the argument breaks down unless we can find at least one irreducible element in $ \\mathcal{O}_K$. But since not every element of $ \\mathcal{O}_K$ is a unit this is easy: let $ x$ be any non-unit and let $ p$ be some irreducible factor of $ x$. [/i]\r\n\r\n(a) Is the result about an infinity of irreducibles correct? Why or why not?\r\n(b) Is the proof given above correct? Why or why not?", "Solution_1": "[quote=\"Arne\"](a) Is the result about an infinity of irreducibles correct? Why or why not?[/quote]\nYes with a tiny amendment - [color=red]if I haven't overlooked the following (see below)[/color]. :whistling: Cover the case that $ u \\equal{} 1 \\plus{} p_1\\cdots p_n$ is a unit by taking some sufficiently high power of $ u$ (use that the group of units is finite by assumption). The other case that the element is a non-unit can be treated as proposed.\n\n[color=red]Correction: The given proof seems to be flawed if $ u$ is a unit. Then we have $ u^k\\equal{}1$ for some $ k>0$. This implies $ x \\cdot f(x) \\equal{}0$ with $ x\\equal{}p_1\\cdots p_n$ for a non-trivial polynomial $ f(X)\\in Z[X]$. Now it is not obvious that $ f(x)\\neq 0$ to have $ x\\equal{}0$ in order to get a contradiction. It seems to me that this cannot be deduced in the given setting. Unfortunately I do not have a counterexample at hand. [/color]\n\n[quote](b) Is the proof given above correct? Why or why not?[/quote]\r\nTrue since $ \\mathcal O_K$ is a noetherian integral domain.\r\n\r\n[color=red]Added sketch of proof: Let R be a noetherian integral domain. Assume that there is a non-unit $ a \\in R$ which is not divided by an irreducible element. Then construct inductively a sequence of elements $ (a_n)$ in $ R$ with $ a_0\\equal{}a$ and $ a_{n\\plus{}1}$ is a divisor of $ a_n$ where in the representation $ a_n\\equal{}a_{n\\plus{}1}b_{n\\plus{}1}$ both $ a_{n\\plus{}1}$ and $ b_{n\\plus{}1}$ are non-units. Then the family of ideals is increasing, hence stationary. This yields to a contradiction.[/color]\r\n\r\nDo you have any other doubts? :roll:", "Solution_2": "Well I'm confused. One of the books on algebraic number theory I'm using gives this as an exercise.\r\n\r\nThe question is: find the mistake in the proof.", "Solution_3": "Here a proof which is not flawed as the Euclidean type proof given in the exercise - just for completeness:\r\nIf there were only finitely many irreducibles in $ \\mathcal O_K$, there would be finitely many prime ideals in $ \\mathcal O_K$ such that their union would contain all irreducibles. Since every non-unit can be decomposed into irreducibles in a Noetherian ring, every prime integer (a real integer here) would be contained in the union. But this would imply that there would be only finitely many prime integers. AND this is absurd due to the Euclidean proof." } { "Tag": [ "inequalities", "linear algebra", "linear algebra unsolved" ], "Problem": "$A,B \\in M_n(R)$ such that $0m, hence,\r\n\r\npm-m[u]>[/u]m+1\r\n\r\nIt will delete all terms up to some term pm-m+1\r\n\r\n=(pm)x(pm-1)x...x(m+1)/(pm-m)!\r\n=(pm)x(pm-1)x...x(pm-m+1)\r\n\r\npm is our first term, and since all terms are being multiplied, p must be a factor.", "Solution_2": "Could you explain it better?\r\n\r\nI can't comprehend what you are trying to say.", "Solution_3": "[quote=\"mathgeniuse^ln(x)\"]Could you explain it better?\n\nI can't comprehend what you are trying to say.[/quote]\r\n\r\nI wish I knew how to use latex.\r\nAnyways,\r\n\r\n(pm)!/(m!(pm-m)!) \r\n=(pm)x(pm-1)x...x(m+1)/(pm-m)! \r\n\r\nIn this step, I know that pm>m\r\n\r\nSo, all terms in pm! get cancelled up to m+1\r\n\r\nThat leaves us with:\r\n\r\n=(pm)x(pm-1)x...x(m+1)/(pm-m)!\r\n\r\nWe know that since m>0 and pm>m, p>1.\r\n\r\nSo, we know that pm-m = m(p-1)[u]>[/u]m+1\r\n\r\nSo, All terms cancel up to pm-m\r\n\r\n=(pm)x(pm-1)x...x(m+1)/(pm-m)!\r\n=(pm)x(pm-1)x...x(pm-m+1)\r\n\r\nHence, since all of the terms are being multiplied, if any one of the terms contains a factor of p, p is a factor of (pm m)", "Solution_4": "[quote=\"neelnal\"](pm)!/(m!(pm-m)!)\n=(pm)x(pm-1)x...x(m+1)/(pm-m)!\n\nWe know pm>m, hence,\n\npm-m[u]>[/u]m+1\n\nIt will delete all terms up to some term pm-m+1\n\n=(pm)x(pm-1)x...x(m+1)/(pm-m)!\n=(pm)x(pm-1)x...x(pm-m+1)\n\npm is our first term, and since all terms are being multiplied, p must be a factor.[/quote]\r\nIf I understand you clearly, I'm not thinks you are right. I think what you have tried to say is that some of the term will be the same in the numerator and the deminstrator (after the first reduction) but you still have terms that aren't the same.\r\nanyway, here is my work \r\n[hide]$\\binom{pm}{m}=\\frac{(pm)!}{m!(pm-m)!}=\\frac{(m+1)\\cdots(pm)}{1\\cdot2\\cdots(pm-m)}$\nwe can say that $p>1$ and we want to show that there are more $p$ in the numerator than in the denominator-in order it will be a multiple of $p$. so, we want to show that $\\left\\lfloor{\\frac{pm-m}{p}}\\right\\rfloor<\\left\\lfloor{\\frac{pm}{p}}\\right\\rfloor-\\left\\lfloor{\\frac{m}{p}}\\right\\rfloor$ which is obviously true if $p\\nmid m$ if $p|m$ so we check $p^{2}$ and so on until $p^{n}\\nmid m$[/hide]" } { "Tag": [ "articles" ], "Problem": "Theres a new movie coming out on October 7th this year called Wallce and Gromit: The Curse of the Were-Rabbit. You might be familiar with this popular animated short. They take photographs for 1/24 of a second of the film, and repeat this process numerous times. If they did this process about 100,000 times, about how long is the movie?", "Solution_1": "[hide=\"ClIcK t0 rEaVeAl sTufF\"]\nif you divide 100,000 by 24 you will get the answer in seconds. So the film is $4166.\\bar{6}$ seconds\nSo if you divide that by 60 you will get the film in minutes. SO the film is $69.\\bar{4}$ minutes.\nThat can be converted to [b]1 hour and 9 minutes and about 27 seconds[/b]\nthat looks wrong though.[/hide]", "Solution_2": "thats actually correct. The article in the magazine said rougly 100,000 times. So its probably more, but thats the answer i was looking for ^.^", "Solution_3": "its about 1 hour 9 minutes and 27 seconds. It looks weird though and that is not an exact answer. Were you actually trying to figure out how long a movie you were going to watch was? :rotfl:" } { "Tag": [ "probability", "limit", "probability and stats" ], "Problem": "let $ [{{E_{n},n\\geq 1}}]$ be a set if increasing events.\r\n\r\n then prove that $ \\lim_{n\\rightarrow infinity}P(E_{n})$=$ P(\\lim_{n\\rightarrow infinity}E_{n})$\r\n\r\nwhere $ P(E_{n}$) denotes probability of the event $ E_{n}$.", "Solution_1": "The command \\to saves some writing and the command \\infty saves some ugliness ;)", "Solution_2": "[quote=\"JBL\"]The command \\to saves some writing and the command \\infty saves some ugliness ;)[/quote]\r\n\r\noh! thank u sir...", "Solution_3": "Let $ E: \\equal{}\\lim_{n\\to \\infty}E_n\\equal{}\\bigcup_{n}E_n$\r\n\r\nThen $ E\\equal{}E_1\\cup(E_2\\minus{}E_1)\\cup(E_3\\minus{}E_2)\\cup.\\, .\\, .$\r\n\r\n and $ P(E)\\equal{}P(E_1)\\plus{}P(E_2)\\minus{}P(E_1)\\plus{}P(E_3)\\minus{}P(E_2)\\plus{}.\\, .\\, .\\equal{}\\lim_{n\\to \\infty}P(E_n)$" } { "Tag": [ "function", "topology", "real analysis", "real analysis unsolved" ], "Problem": "The following is a technical issue I would like to use in another proof. Consider the Borel sets $B(\\mathbb{R},d)$ on the real line, where $d$ is the standard metric. Let $d'$ be the restriction of $d$ to $(a,b)\\times (a,b)$. Is it true that $B((a,b), d')$ is equal to the sets of $B(\\mathbb{R},d)$ contained in $(a,b)$. This seems likely, but I don't see how to show it.", "Solution_1": "[quote=\"Kalle\"]Let $d'$ be the restriction of $d$ to $(a,b)\\times (a,b)$.[/quote]\r\nIs this a typo, or did you really want to consider a product space? \r\n\r\nThe answer is yes, because the map $A\\mapsto A\\cap (a,b)$ [from the power set of $\\mathbb R$ to the power set of $(a,b)$] preserves all of the relevant structures: it sends open sets to open, unions to unions, complements to complements, etc.", "Solution_2": "[quote=\"mlok\"][quote=\"Kalle\"]Let $d'$ be the restriction of $d$ to $(a,b)\\times (a,b)$.[/quote]\nIs this a typo, or did you really want to consider a product space? \n\nThe answer is yes, because the map $A\\mapsto A\\cap (a,b)$ [from the power set of $\\mathbb R$ to the power set of $(a,b)$] preserves all of the relevant structures: it sends open sets to open, unions to unions, complements to complements, etc.[/quote]\r\nIt is not a typo. I was thinking of $d$ as a function from\u00a0$\\mathbb{R}\\times \\mathbb{R}$, because the \"input\" in the metric is a pair of real numbers. I am only trying to express that I want to consider $(a,b)$ as a subspace in the usual way. \r\n\r\nAnyway, the justification you give is more or less why I believed this result to be true. However, my definition of the borel sets in a metric space $(X,d)$ is the smallest $\\sigma$-algebra containing all open sets of $X$. I suppose that it would be possible to express this as \"finite compositions of countable unions and set differences\" or something like that. However, this seems to be very hard to carry out without doing a considerable amount of hand-waving. Isn't there a more succinct argument?", "Solution_3": "The intersection map $i$ sends $\\sigma$-algebras to $\\sigma$-algebras. So, $i(B(\\mathbb R))$ is a $\\sigma$-algebra that contains all open subsets of $(a,b)$. Conversely, the union of $B((a,b))$ with $B(\\mathbb R\\setminus (a,b))$ generates a $\\sigma$-algebra $B'$ that contains all open subsets of $\\mathbb R$ and satisfies $i(B')=B((a,b))$. \r\n\r\nThis works in more generality: for open subsets in a topological space." } { "Tag": [ "AMC", "AIME", "geometry", "geometric transformation", "reflection", "power of a point" ], "Problem": "Um, pardon me, but I thought I found really a much easier way to solve it.\r\n\r\nAren't QR and lets say ZR just secants through circle A? \r\nThe power of a point thereom for secants states:\r\nThe product of the external length times the total length of the secant is equal to the product of the external length of another secant times its total length.\r\n\r\nSo:\r\nPR = QR = x\r\nx(2x) = 2x :^2: \r\n\r\n2x :^2: = (8 + 6 + 2 + 4 + 6) (4 + 6)\r\n2x :^2: = 260\r\nx :^2: = 130", "Solution_1": "Were you responding to something? What are you talking about?", "Solution_2": "I'm sorry, I accidently posted this in response to the problem a bit further, accidental misclick.\r\n\r\nI'll repost it here:\r\nThe distance AB is 12. The circle center A radius 8 and the circle center B radius 6 meet at P (and another point). A line through P meets the circles again at Q and R (with Q on the larger circle), so that QP = PR. Find QP :^2: .", "Solution_3": "I'm sorry, I accidently posted this in response to the problem a bit further, accidental misclick.\r\n\r\nI'll repost it here:\r\nThe distance AB is 12. The circle center A radius 8 and the circle center B radius 6 meet at P (and another point). A line through P meets the circles again at Q and R (with Q on the larger circle), so that QP = PR. Find QP :^2: .", "Solution_4": "How did you get that huge sum? the 8+6+4+2+6 thing....", "Solution_5": "I think that Xbiz is assuming that R is also collinear with A and B,\r\nand if that is the case\r\n\r\n- that sum would be the total length of the secant.\r\n- the solution is coincidental\r\n\r\nmight be wrong though", "Solution_6": "Well I am sure you can just choose R wherever you want it to be... So I suppose you could define R to be collinear. Nice solution!", "Solution_7": "R is colinear with A and B ... but I'm not entirely sure how you would prove it. I made some comment about that in the thread that this was supposed to be part of. It may be a result of Zabelman's reflections, or it may not.\r\n\r\nRagingg -- you can't choose R whereever you want because it has to be such that RP = PQ, which doesn't happen for nearly every choice of R.\r\n\r\n\r\nXbizkitl -- how did you decide that BAR was a line?" } { "Tag": [ "function", "number theory proposed", "number theory" ], "Problem": "Find all pairs of positive integers $ (m,n)$ with $ m\\neq n$ such that $ m^{\\phi( n)} \\equal{} n^{\\phi( m)}$.", "Solution_1": "[quote=\"Max D.R.\"]Find all pairs of positive integers $ (m,n)$ with $ m\\neq n$ such that $ m^{\\phi( n)} \\equal{} n^{\\phi( m)}$.[/quote]\r\n\r\nClearly, $ m$ and $ n$ have the same set of prime divisors.\r\n\r\nThen, since $ \\varphi(m)\\equal{}m\\prod_{p|m}(1\\minus{}\\frac 1p)$ and $ \\varphi(n)\\equal{}n\\prod_{p|n}(1\\minus{}\\frac 1p)$, the equation becomes $ m^n\\equal{}n^m$\r\n\r\n\r\nThis equation is very classical and has general solution in $ \\mathbb R$ : $ (u,u)$ and $ (u^{\\frac{1}{u\\minus{}1}},u^{\\frac{u}{u\\minus{}1}})$\r\n\r\nThe set of non trivial solutions is reduced in $ \\mathbb N$ to $ u\\equal{}2$ and the couple $ (2,4)$" } { "Tag": [ "algorithm" ], "Problem": "I have only found one solution for this problem so far, could you please help me find out more solutions?\r\n\r\nWhen a certain positive integer N is divided by a positive integer d, the remaider is 7. If 2N+3 is divided by d, the remainder is 1. Find all possible values of d.\r\n\r\nThe only solution I got was D=16 and N=23, is there more solutions?", "Solution_1": "(1) N:d -> remainder=7\r\n(2) (2N+3):d -> remainder=1\r\n\r\n\r\nFrom (1) we get N=ad+7 where a is a positive integer and d>7 (because 7 is the remainder)\r\n\r\n\r\nLet's now find 2N+3\r\n\r\n2N+3=\r\n\r\n2(ad+7) +3 =\r\n\r\n2ad +14 +3 =\r\n\r\n2ad +17\r\n\r\nIf we try to divide this number by the number d, the only remainder will come from 17, because the mumber d divides the first term 2ad\r\n\r\nHence, the remainder of the division (2ad+17):d must be equal to the remainder of the division 17:d\r\n\r\nSo, 17 = rd+1 => 16=rd\r\n\r\nSo d divides the number 16.\r\n\r\nThe dividers of 16 are: 1,2,4,8,16\r\n\r\nBut d>7, so the only possible values for d are 8 and 16\r\n\r\ne.g. d=8 and N=15\r\n\r\nNotice that, once you choose the number d, there are infinite values for N, because if (1) holds then:\r\n\r\n2N+3 = 2ad+17 = 2ad+16+1 = 2ad+rd+1 = (2a+r)d+1 \r\n\r\n\r\n(e.g. for d=8, N may be 15,23,31,39,47,...)", "Solution_2": "Exactly the same question was solved earlier today on this board :lol: . [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42051]Here.[/url] Since both solutions are the same, we can be pretty sure the answer is correct :)" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry", "trigonometry", "algorithm", "AIME" ], "Problem": "So....people.\r\n\r\nWe finished the 37th annual USAMO!!! :P \r\n\r\nI didn't take the test last year, but was able to get 4 questions in a mock test...\r\n\r\nGot 3 for this year.\r\n\r\nSome people say that 2008 was harder than 2007....\r\n\r\nWhat do you guys think about the cut-off line?\r\n\r\n(I know it's not that meningful a thing to do, but predicting is fun :lol: ) \r\n\r\nMine:\r\n\r\nBlack - 24~26 (But, alas, there are SO many people who claim to have gotten 4 questions...that's scary.)\r\n\r\nBlue - 20~21 (I somewhat hoped for Blue, but...yeah, it's gonna be hard when you solved only 3 problems.)\r\n\r\nRed - Somewhere around 9 (Geez, I really hope I get in to Red...)", "Solution_1": "There are a LOT of 28s being thrown around, but I have a feeling a lot of them will be lower than that. Still, I think (these estimates may be a touch high)\r\n\r\nBlack cutoff: Probably around 28\r\nBlue cutoff: Probably around 22 or so\r\nRed cutoff: I have no idea.", "Solution_2": "[quote=\"physisnu\"]So....people.\n\nWe finished the 37th annual USAMO!!! :P \n\nI didn't take the test last year, but was able to get 4 questions in a mock test...\n\nGot 3 for this year.\n\nSome people say that 2008 was harder than 2007....\n\nWhat do you guys think about the cut-off line?\n\n(I know it's not that meningful a thing to do, but predicting is fun :lol: ) \n\nMine:\n\nBlack - 24~26 (But, alas, there are SO many people who claim to have gotten 4 questions...that's scary.)\n\nBlue - 20~21 (I somewhat hoped for Blue, but...yeah, it's gonna be hard when you solved only 3 problems.)\n\nRed - Somewhere around 9 (Geez, I really hope I get in to Red...)[/quote]\r\n\r\nWhy are you allowed to go to MOP? (Your location says Seoul).", "Solution_3": "In my opinion, it is much easier than last few years. I think black will >=29, blue will>=22. (Only my opinion) And my prediction of my score is around 25-27. so... :D", "Solution_4": "Yeah it kinda worries me that I can name 30 people who think they got > 28.", "Solution_5": "there's an 8th grader i know who thinks he got a 30\r\nand a 7th grader who claims a 23", "Solution_6": "I think blue will be about 3.5 problems and black 4.5 problems. There is no way a 21 can make blue, I think... 1, 4, 5 were all easy... But then, a lot of people who I trust disagree, so we will see.\r\n\r\nIn response to the last post, people taking the USAMO for the first time don't estimate their scores too well ;). I don't know that many people who legitimately claimed 4 problems, so I think it will be okay. But I definitely think there are more than 12 who got 4 problems.", "Solution_7": "I think I got 5 (12345)... would that be good enough for black MOP (I'm a senior)? It's definitely much better than I've done the last three years (3, 2, 2, respectively), so that led me to think scores would be higher this year than previous years. I definitely don't think this year was any harder than last year. It's hard to top a test on which Brian Lawrence got a 32.", "Solution_8": "wahhh what\r\nthis thread is making me really insecure now\r\ni guess im gonna be either barely in blue or miss by 1 or 2 points", "Solution_9": "[quote=\"jb05\"]I think blue will be about 3.5 problems and black 4.5 problems. There is no way a 21 can make blue, I think... 1, 4, 5 were all easy... But then, a lot of people who I trust disagree, so we will see.\n\nIn response to the last post, people taking the USAMO for the first time don't estimate their scores too well ;). I don't know that many people who legitimately claimed 4 problems, so I think it will be okay. But I definitely think there are more than 12 who got 4 problems.[/quote]\r\n\r\nThis post WILDLY exaggerates the number of high scorers.", "Solution_10": "Bleh whatever the case, I have talked to many many people who claim to have gotten 1,4,5, though perhaps I mostly talk to people who would make mop anyway... I dunno. I definitely think extra points above 21 will be necessary, if only one or two.", "Solution_11": "[quote=\"physisnu\"]Red - Somewhere around 9 (Geez, I really hope I get in to Red...)[/quote]\r\n\r\nWait, wuh? I got numbers 3, 4, and 5 kinda, with like 3 or 4 points on each. Would that qualify me for red? That would be awesome. I probably got like at most one point for my proofs for 1, 2, and 6, but I think my score could be around a nine. Not sure though...", "Solution_12": "Yeah I think too many people solved 4 problems so cutoff will be like 29 or 30. :(", "Solution_13": "While everyone exaggerates the scores, I don't think Red will be a 9 this year. Maybe 13 or 12...", "Solution_14": "[quote=\"Math Geek\"][quote=\"physisnu\"]Red - Somewhere around 9 (Geez, I really hope I get in to Red...)[/quote]\n\nWait, wuh? I got numbers 3, 4, and 5 kinda, with like 3 or 4 points on each. Would that qualify me for red? That would be awesome. I probably got like at most one point for my proofs for 1, 2, and 6, but I think my score could be around a nine. Not sure though...[/quote]\r\nYou're too young for red; it's only for freshmen. :wink:", "Solution_15": "Does anyone know how Evan O'Dorney did? He never replied my email.", "Solution_16": "[quote=\"nealth\"]Does anyone know how Evan O'Dorney did? He never replied my email.[/quote]\r\n\r\nAre you one of his friends?", "Solution_17": "[quote=\"Ignite168\"][quote=\"nealth\"]Does anyone know how Evan O'Dorney did? He never replied my email.[/quote]\n\nAre you one of his friends?[/quote]\r\n\r\nnealth = Delong\r\n\r\nMOPpers tend to want to know how each other did.", "Solution_18": "How was meeting Evan O'Dorney?\r\nWas it as great as imagined?", "Solution_19": "[quote=\"Ignite168\"]How was meeting Evan O'Dorney?\nWas it as great as imagined?[/quote]\r\n\r\nThat depends how great you'd imagine it to be. I personally didn't think it was as great as going to DC and meeting, say, Alex Zhai.", "Solution_20": "Gradually, I realized that this year is really going to surpass year 2002(black will reach 35 and blue will reach 28). There are already a bunch of people who solved more than 5 problems, and these people are always tend to estimate their score correctly.\r\n\r\nBy the way, how did you do. nealth? if you doesn't mind telling :D", "Solution_21": "I'm doubtful of that...\r\nI would be very surprised if black passed 30 or if blue passed 21 (or HM pass 23).\r\n\r\nI have not heard of many people who have solved \"more than 5\" problems. Heck, I have only heard of 12-14 who have solved $ \\geq 4$ problems.\r\n\r\nIn addition, if we go by the 5 returning red mop statistic, it seems like the cutoff would have to be around 21. I think I have the highest (of the red moppers with a possible exception for Vlad) score estimate, and that's 25.\r\n\r\nSo apparently we're thinking Vlad got about 21, which means that for there to be 5 returning red moppers, the cutoff'll have to be around 21.", "Solution_22": "Why don't we just wait like the 3 days until the results come out.", "Solution_23": "[quote=\"Phelpedo\"]Why don't we just wait like the 3 days until the results come out.[/quote]\r\nwe only have to wait 3 days?.. I thought it was 5.13 :D", "Solution_24": "If I remember correctly, MOP phone calls (it was an email last year I believe) went out and winners/HM were released 11 days after the USAMO (on a Sunday), and we got our scores 3 days later, on a Wednesday. I'm quite sure about the second one because I remember finding out the night I arrived at MathCounts.\r\n\r\nSo by that pattern, winners/HMs and MOPpers will be announced on 5/11, this Sunday, and scores will be available 5/14, this Wednesday. Obviously, the pattern may not follow exactly the same way as last year. I hear in past years, scores were available before MOP phone calls were made.\r\n\r\n(edit) all right, apparently in the USAMO manual it says both will happen on 5/13...that makes the above completely pointless.", "Solution_25": "[quote=\"CatalystOfNostalgia\"]If I remember correctly, MOP phone calls (it was an email last year I believe) went out and winners/HM were released 11 days after the USAMO (on a Sunday), and we got our scores 3 days later, on a Wednesday. I'm quite sure about the second one because I remember finding out the night I arrived at MathCounts.\n\nSo by that pattern, winners/HMs and MOPpers will be announced on 5/11, this Sunday, and scores will be available 5/14, this Wednesday. Obviously, the pattern may not follow exactly the same way as last year. I hear in past years, scores were available before MOP phone calls were made.\n\n(edit) all right, apparently in the USAMO manual it says both will happen on 5/13...that makes the above completely pointless.[/quote]\r\nNow u're making me very nervous~~ :D", "Solution_26": "[quote=\"Sean Gu\"][quote=\"Phelpedo\"]Why don't we just wait like the 3 days until the results come out.[/quote]\nwe only have to wait 3 days?.. I thought it was 5.13 :D[/quote]\r\n\r\nHence the \"like\" in his sentence.", "Solution_27": "[quote=\"davidyko\"][color=green]Please stop pyramid quoting. -mod[/color]\n\nHence the \"like\" in his sentence.[/quote]\r\noh...no I'm just figuring since I'll be out whatsoever,,,,it's better to know earlier~~ :lol:", "Solution_28": "This topic is getting very spammy very quickly, guys...", "Solution_29": "[quote=\"matt276eagles\"]Also, you'd be surprised how immature some very, very smart people are.[/quote]\r\n\r\nlol\r\n\r\n:(\r\n\r\n\r\n\r\noh, and i'm totally not making blue mop. i guess i'm gonna join the giant mob of people who got #1 and 4 right and pity points on the other four questions." } { "Tag": [ "percent", "Support" ], "Problem": "A rubber ball is dropped from a height of 27 feet onto a concrete floor. Each time it hits the floor, it bounces back up to a height 2/3 of the height from which it fell.\r\n\r\n 1. Calculate the height of each of the first four bounces.\r\n\r\n 2. Find a formula that gives the height of the nth bounce.\r\n\r\n 3. Use your formula to find the height of the 12th bounce to the nearest inch.\r\n\r\nExtra: Suppose a different ball is dropped from 40 feet, and on its tenth bounce it reaches a height of 1 foot. What percent of the height from which it fell does that ball bounce back up each time it hits the floor?\r\n\r\nExtension: Tell if you agree or disagree with the following statement and explain why. Support your opinion with any relevant mathematical ideas.", "Solution_1": "1. $18, 12, 8, \\frac{16}{3}$\r\n2. $f(n)=(\\frac{2}{3})^n\\cdot 27$\r\n3. $\\approx 2.49$\r\nExtra. $40\\cdot x^10=1\\Longrightarrow x^10=1/40\\Longrightarrow x=\\sqrt[10]{1/40}$\r\n\r\n[color=darkblue][size=75][/size][/color]", "Solution_2": "thanks for answering \r\n\r\ni cant believe only one person aswered", "Solution_3": "how did you get those answers? and are they right?", "Solution_4": "Yes, I think they're correct. This was from [b]May[/b] of [b]2005[/b]:D .", "Solution_5": "[quote=\"robinhe\"]This was from [b]May[/b] of [b]2005[/b]:D .[/quote]\r\n\r\nIn other words, don't bump topics this old. ;)" } { "Tag": [ "topology" ], "Problem": "I was wondering if any body could suggest a good book for elementry topology?", "Solution_1": "Everybody these days seems to recommend starting with the Munkres textbook. If you can't get that from a library, there are some inexpensive Dover reprints that are older and thus not as modern in approach, but still good.", "Solution_2": "I second the vote for Munkres. Also, Basic Topology by Armstrong is good.", "Solution_3": "I looked at one of the dover books and didnt really like it but ill try the Munkres book. Thank you.", "Solution_4": "There are some other books you might look at. [i]One is A Combinatorial Introduction to Topology[/i] by Michael Henle. I wouldn't use that alone, but in conjunction with other books. \r\n\r\n[i]Topology Now![/i] by Phil Straffin and Robert Messer just came out. I haven't seen it yet, but the table of contents combined with the excellent pedagogical reputation of the authors suggests it'll be great. The book's website is at \r\nhttp://www.albion.edu/math/ram/TopologyNow!/\r\n\r\nThere's another new topology text which either just came out or is about to come out, but I can't remember the title/author. (If someone else posts about it, I can confirm/deny, though...)\r\n\r\nFor less basic topology, the standard used to be Brown Munrkes but is now Hatcher's [i]Algebraic Topology[/i]. Red Munkres (now published with a green cover) is still the best point-set topology text, as tokenadult says." } { "Tag": [ "geometry", "summer program", "PROMYS", "Mathcamp" ], "Problem": "I'm from the 715 area code of the upper half of Wisconsin.\r\n\r\nSo what I really want to know is if anyone else is too, or at least from where people are planning to fly.", "Solution_1": "I'm coming from Long Island NY, but I'll probably have to fly out of NYC.", "Solution_2": "I hail from Westchester, NY (an overall large county, but north of NYC).\r\n\r\nIdeally, i'll be going to PROMYS, but I can't lose out going to Mathcamp either. Now I wait for the PROMYS decision...", "Solution_3": "I have a question on the flight plans. In fact it is about the AA company.\r\n\r\nIf I call the company to order my ticket, they only request to charge me the luggage fee and booking fee that time, right? They won't charge that 326 from me, but from Mathcamp (after I mail the Mathcamp my check). So I still need to mail the 326 to North Carolina? \r\n\r\nI want to make this sure so that I won't spend twice 326 on the ticket. :P", "Solution_4": "Hi,\r\n\r\n[quote=\"Go Beyond\"]I have a question on the flight plans. In fact it is about the AA company.[/quote]\r\n\r\nPlease send your travel questions to travel@mathcamp.org.", "Solution_5": "I will have my plane connection in Chicago flight #AA1611 at 8.35am\r\nAnybody can go with me?", "Solution_6": "My connection to seattle will be from Atlanta, on Delta flight #823, on the off chance that anyone uses the same flight.", "Solution_7": "For me, yet to be confirmed...\r\n\r\nMy friend has reserved his ticket already. I think he's taking Asiana OZ752/OZ272 from Singapore, transit in Seoul.", "Solution_8": "Is anyone coming from San Jose, CA?", "Solution_9": "From Colorado. 3 days till start!" } { "Tag": [ "ratio" ], "Problem": "In this game, someone posts up a word like \"good\", \"jump\", or \"book\". Then, the next user posts up as many synonyms as they can, earning the number of synonyms they post up as points.\r\n\r\nI'll start first:\r\n\r\n[b]wise[/b]", "Solution_1": "clever, smart, intelligent, knowing, knowledgeable, cunning,", "Solution_2": "New word, anyone? OK, I will..........lets do [b]capable[/b]", "Solution_3": "able, competent. good, qualified, proficient, skillfull\r\n\r\nNW\r\n\r\n[b]message[/b]", "Solution_4": "[b]reply, news, result, outcome on paper\n\nHow about \"ignorant\"?[/b]", "Solution_5": "From now on, everyone please post up scores.\r\n\r\n[b]careless, unmindful, inexperienced, oblivious[/b]\r\n\r\n[hide=\"Scores\"]\n\ninoit 6\ntennis123 6\nprofessordad 4\nmaybach 4\n\n[/hide]", "Solution_6": "Since professordad didn't post a new word, I will:\r\n[b]liberal(one of my favorite words and conservative is one of my least favorite:))[/b]", "Solution_7": "I bet you're conservative. :P\r\n\r\n/me looks up liberal just to make sure me knows what it means.\r\n\r\n/me does not look in a thesaurus.\r\n\r\n[b]open-minded, undedicated, hypocritical[/b]\r\n\r\nI don't know about the last 2, so you don't have to put 3 points if you don't want to.\r\n\r\n[b]justice[/b]\r\n\r\n:D\r\n[hide=\"Scores\"]\n\ninoit 6\ntennis123 6\nprofessordad 4\nmaybach 4\n1=2 ?\n\n[/hide]", "Solution_8": "code law rule truth\r\n[hide=\"scores\"]\ninoit 6 \ntennis123 6 \nprofessordad 4 \nmaybach 4 \nSuperNerd123 4\n1=2 3\n\n[/hide]\r\nNw = Echolocation", "Solution_9": "[b]locating, finding[/b] sigh, only two.\r\n\r\nNW:allocation", "Solution_10": "appropriation, quota, ration, share \r\nNw = Antidysenteric\r\n[hide=\"scores\"]\ninoit 6 \ntennis123 6 \nprofessordad 4 \nmaybach 4 \nSuperNerd123 8 \n1=2 3 \nAIME15USAMO 2[/hide]" } { "Tag": [ "vector", "trigonometry" ], "Problem": "Two positive point charges +q are on the y axis at y = +a and y = -a. A bead of mass m carrying a negtive charge -q slides without friction along a thread that runs along the x axis. (a) Show that for small displacements of x << a, the bead experiences a restoring force that is proprtional to x and therefore undergoes simple harmonic motion. (b) Find the period of the motion.", "Solution_1": "[quote=\"ffdbzathf\"]Two positive point charges +q are on the y axis at y = +a and y = -a. A bead of mass m carrying a negtive charge -q slides without friction along a thread that runs along the x axis. (a) Show that for small displacements of x << a, the bead experiences a restoring force that is proprtional to x and therefore undergoes simple harmonic motion. (b) Find the period of the motion.[/quote]\r\n\r\nThe collective E-field of the two point charges on the y-axis is given by:\r\n\r\n$\\vec{E}(x)= k \\{ q \\frac{\\vec{x}-a\\hat{y}}{|\\vec{x}-a\\hat{y}|^3}+q \\frac{\\vec{x}+a\\hat{y}}{|\\vec{x}+a\\hat{y}|^3}\\}$\r\n\r\nWhen $ a >> x$ this reduces to $\\vec{E}(x)= 2 k q \\frac{\\vec{x}}{a^3}$\r\n\r\nMultiplying this expression by $-q$ to obtain the coulomb force on the negative charge one obtains: $\\vec{F}(x)= -2 k q^2 \\frac{\\vec{x}}{a^3}$\r\n\r\nThis force is along the x-direction, due to the vector nature of $\\vec{x}$ itself, restorative because of the minus sign, and $\\propto \\vec{x}$; therefore, this coulomb force is an example of Hooke's law.\r\n\r\nThe angular frequency for this sort of arrangement is given by: $\\omega^2 =\\frac{K}{m}$\r\n\r\n$K= \\frac{2 k q^2}{a^3}$\r\n\r\n$\\therefore \\omega = \\frac{2\\pi}{T} = \\sqrt{\\frac{K}{m}}=\\sqrt{\\frac{2 k q^2}{m a^3}}$\r\n\r\n$T = 2 \\pi \\sqrt{\\frac{m a^3}{2 k q^2}}$", "Solution_2": "Wow, nice solution!", "Solution_3": "Let $A(+q),B(-q)$ and $P(-q)$ at distance of $OP=x$, $\\angle {APO}=\\angle {BPO}=\\theta$.For $a\\gg x$, we can consider that $PA=PB\\approx a$,\r\n\r\nyielding $\\cos \\theta \\approx \\frac{x}{a}$. Since we have $\\overrightarrow {E}=\\overrightarrow {E_{A}}+\\overrightarrow {E_{B}}$, we obtain $|\\overrightarrow {F_{P}}|\\approx 2|\\overrightarrow {E_{A}}|\\cos \\theta\\approx 2\\cdot k\\cdot\\frac{q^2}{a^2}\\cdot \\frac{x}{a}=\\frac{2kq^2}{a^3}x$, where $k$ is Coulomb's constant.Therefore the equation of motion for the bead at $P$ is $m\\frac{d^2 x}{dt^2}=-Kx$, where $K=\\frac{2kq^2}{a^3}$,yielding the desired period is that one which is derived by Dr.No." } { "Tag": [ "search" ], "Problem": "My wii isn't connected to the internet, but I wish it was.", "Solution_1": "Isn't online Wii free?", "Solution_2": "THE NON-WII-ERS MUST BE SUPREME!", "Solution_3": "[quote=\"7h3.D3m0n.117\"]Isn't online Wii free?[/quote]\r\n\r\nI don't know. \r\n\r\nThe wii isn't really that bad. It is really fun. I just got Wii Fit.", "Solution_4": "You can connect your wii to your internet. That can be done in setting. You can search out the wireless signal from your router, or you can attach the LAN to your wii with an adapter. The internet channel, which is an opera browser, used to be free, but you now have to pay 500 wii pts for it ( $ \\$$ 5 ).", "Solution_5": "What are wii points. I have never heard of these before.", "Solution_6": "I'm assuming they're similar to the Microsoft Points made by...Microsoft. Basically, you can buy them either online using a credit card, or at your local store on prepaid cards. You can use them to buy things such as downloadable games, or in this case, membership.", "Solution_7": "Hmm, that is interesting. I will have to take a look at that.", "Solution_8": "My wii is connected to the internet, but my dad is the only one who uses it... :mad:", "Solution_9": "What kinds of things does he do?", "Solution_10": "I'm getting a wii game for christmas but i'm not sure which game to get. I'll usually play by myself so should i choose mario party 8, super smash brothers brawl, or Ncaa football09??", "Solution_11": "SSBB. At least you can train for MLG in the long run...", "Solution_12": "I only have SSMB and Mario Party 8 out of the three jjx1 listed, but I haven't gotten enough time to play Mario Party 8 yet, but I have seen my brother play it. I think you should get SSMB.", "Solution_13": "too late. I got mario party 8 :P", "Solution_14": "I have Mario Kart wii. Whenever my connection fails during an online race, I automatically lose. Kinda like a forefit. It messes up my rating.", "Solution_15": "So get a better connection? Lol.", "Solution_16": "Online lags a lot.", "Solution_17": "I just got mine connected to the internet. What kinds of things can you do without any wii points?", "Solution_18": "[quote=\"Wickedestjr\"]I just got mine connected to the internet. What kinds of things can you do without any wii points?[/quote]\r\n\r\n\r\n\r\nWi-fi games (Call of duty, Super smash bros. brawl, Mario Kart, ect.ect) and you can download some of the other free channels such as voting channel, and the mii channel also has online capabilities. Other than that their really isn't alot to do with Wii online.", "Solution_19": "i have a wii, and it's really good. but i'm a casual gamer, and the whole wii system and almost all its games are casual. so the system isn't for people who like hardcore games, like Halo and GTA and stuff like that. but if you don't have enough games, it gets really boring after a while. :juggle:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Can we find $a, b, c, d \\in N$ s.t $a 0. Prove that: $\\sum \\frac{1}{7a^{2}+2bc}\\geq \\frac{3}{(a+b+c)^{2}}$\r\n\r\nI hope that every one can solve these problem with pleasure.", "Solution_1": "Here is the solution for problem $1$:\r\nWe have $a^{3}b^{3}+a^{3}c^{2}b=(a^{3}b)(b^{2}+c^{2})\\ge a^{5}b$\r\nTherefore $\\sum_{sym}a^{3}b^{3}+\\sum_{sym}a^{3}c^{2}b\\ge \\sum_{sym}a^{5}b$\r\nAlso, by Muirhead, $\\sum_{sym}a^{5}b\\ge \\sum_{sym}a^{4}b^{2}$\r\nTherefore:\r\n$(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3})=$\r\n$=a^{6}+b^{6}+c^{6}+\\sum_{sym}a^{4}b^{2}+\\sum_{sym}a^{5}b+\\sum_{sym}a^{3}b^{3}+\\sum_{sym}a^{3}c^{2}b$\r\n$\\ge a^{6}+b^{6}+c^{6}+2\\sum_{sym}a^{5}b+\\sum_{sym}a^{4}b^{2}$\r\n$\\ge a^{6}+b^{6}+c^{6}+3\\sum_{sym}a^{4}b^{2}$\r\n$=a^{6}+b^{6}+c^{6}+3\\sum_{cyc}a^{4}(b^{2}+c^{2})$\r\n$\\ge 4(a^{6}+b^{6}+c^{6})$ QED", "Solution_2": "[quote=\"rem\"]Here is the solution for problem $1$:\nWe have $a^{3}b^{3}+a^{3}c^{2}b=(a^{3}b)(b^{2}+c^{2})\\ge a^{5}b$\nTherefore $\\sum_{sym}a^{3}b^{3}+\\sum_{sym}a^{3}c^{2}b\\ge \\sum_{sym}a^{5}b$\nAlso, by Muirhead, $\\sum_{sym}a^{5}b\\ge \\sum_{sym}a^{4}b^{2}$\nTherefore:\n$(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3})=$\n$=a^{6}+b^{6}+c^{6}+\\sum_{sym}a^{4}b^{2}+\\sum_{sym}a^{5}b+\\sum_{sym}a^{3}b^{3}+\\sum_{sym}a^{3}c^{2}b$\n$\\ge a^{6}+b^{6}+c^{6}+2\\sum_{sym}a^{5}b+\\sum_{sym}a^{4}b^{2}$\n$\\ge a^{6}+b^{6}+c^{6}+3\\sum_{sym}a^{4}b^{2}$\n$=a^{6}+b^{6}+c^{6}+3\\sum_{cyc}a^{4}(b^{2}+c^{2})$\n$\\ge 4(a^{6}+b^{6}+c^{6})$ QED[/quote]\r\n\r\nSo nice a solution. I am not know much about Muirhead inequality, so my solution was not as beutiful as you. Thank 's alot.\r\nAnd how about the second problem, I think it's nice one. Or it so easy for you to prove? I am waiting for a nice solution.", "Solution_3": "Thank you.\r\nUnfortunately the second problem has been already posted on Aops and here is the link:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=frac+geq+2bc+7a+sum&t=33538[/url]\r\nThe solution by Barra is amazing:\r\n[quote=\"barra\"]The above inequality is equivalent to\n$\\sum \\frac{a^{2}+b^{2}+28ab}{(7a^{2}+2bc)(7b^{2}+2ac)}(b-a)^{2}\\\\+\\sum \\frac{42(a^{2}+ac+c^{2})}{(7a^{2}+2bc)(7b^{2}+2ca)(7c^{2}+2ab)}(a-c)^{2}(b-a)^{2}\\geq 0$[/quote]\r\nThis is just beautiful.", "Solution_4": "[quote=\"rem\"]Thank you.\nUnfortunately the second problem has been already posted on Aops and here is the link:\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=frac+geq+2bc+7a+sum&t=33538[/url]\nThe solution by Barra is amazing:\n[quote=\"barra\"]The above inequality is equivalent to\n$\\sum \\frac{a^{2}+b^{2}+28ab}{(7a^{2}+2bc)(7b^{2}+2ac)}(b-a)^{2}\\\\+\\sum \\frac{42(a^{2}+ac+c^{2})}{(7a^{2}+2bc)(7b^{2}+2ca)(7c^{2}+2ab)}(a-c)^{2}(b-a)^{2}\\geq 0$[/quote]\nThis is just beautiful.[/quote]\r\n\r\nUnfortunately, his solution is not correct (the very last post at the end of your link)", "Solution_5": "* $\\sum \\frac{1}{7a^{2}+2bc}\\geq \\frac{3}{(a+b+c)^{2}}$\r\n\r\nAssume that: $a \\geq b \\geq c$\r\nWe have the inequalaty with new form:\r\n$\\sum (a-b)^{2}\\frac{7a^{2}+7b^{2}+28ab-6ac-6bc}{ (a^{2}+2bc)(b^{2}+2ac)}\\geq 0$\r\n Have: $a \\geq b \\geq c$ so that :\r\n$7a^{2}+7b^{2}+28ab-6ac-6bc \\geq 0$\r\n$7a^{2}+7c^{2}+28ac-6bc-6ab \\geq 0$\r\nSo we need to prove:$(a-c)^{2}\\frac{7a^{2}+7c^{2}+28ab-6ac-6bc}{ (a^{2}+2bc)(c^{2}+2ab) }+$$(b-c)^{2}\\frac{7b^{2}+7c^{2}+28bc-6ab-6ac}{ (b^{2}+2ac)(c^{2}+2ba) }$\r\nBut we also have$(a-c)/(b-c) \\geq a/b$ with $a \\geq b \\geq c$\r\nSo we just only need to prove : $49a^{4}b^{2}+14a^{5}c+98a^{2}b^{2}c^{2}+14a^{3}c^{3}+154a^{3}b^{2}c+56a^{4}c^{2}-84a^{3}b^{3}-12a^{4}bc+154a^{2}b^{3}c-12a^{3}bc^{2}+49a^{2}b^{4}+14b^{5}c+14b^{3}c^{3}+56b^{4}c^{2}-12b^{4}ca-12ab^{3}c^{2}\\geq 0$\r\nThe last inequality is of course right.\r\nSo we can prove the problem.\r\n\r\n@: When I try to prove the last inequality, I try Mple software to factor it, but it 's not suceed. But we can easily see that it's so easy to prove, just only so \"ugly\". I post this here to find out another solution. And thanks you rem.", "Solution_6": "[quote=\"timlaiminh\"]In some of my country 's math forums, there are several problems that no body solves. Here are some of them:\nProblem1: Have: \n$a^{2}+b^{2}\\geq c^{2}, b^{2}+c^{2}\\geqq^{2}, c^{2}+a^{2}\\geq b^{2}$. \nProve that:\n$(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \\geq 4(a^{6}+b^{6}+c^{6})$ \n\n[/quote]\r\nApplying Mink's inequality, \\[(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \\ge (a^{2}+b^{2}+c^{2})^{3}.\\] It remains to show that: \\[(x+y+z)^{3}\\ge 4(x^{3}+y^{3}+z^{3}),\\] where $x,y,z$ are sides of a triangle. Done.", "Solution_7": "[quote=\"Lovasz\"][quote=\"timlaiminh\"]In some of my country 's math forums, there are several problems that no body solves. Here are some of them:\nProblem1: Have: \n$a^{2}+b^{2}\\geq c^{2}, b^{2}+c^{2}\\geqq^{2}, c^{2}+a^{2}\\geq b^{2}$. \nProve that:\n$(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \\geq 4(a^{6}+b^{6}+c^{6})$ \n\n[/quote]\nApplying Mink's inequality, \\[(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \\ge (a^{2}+b^{2}+c^{2})^{3}.\\] It remains to show that: \\[(x+y+z)^{3}\\ge 4(x^{3}+y^{3}+z^{3}),\\] where $x,y,z$ are sides of a triangle. Done.[/quote]\r\n\r\nCan you explain more about the last inequalty?\r\nIf we are expand it, what will happen? I think that it might be wrong.\r\nCan you explain it more clearly?", "Solution_8": "[quote=\"timlaiminh\"][quote=\"Lovasz\"][quote=\"timlaiminh\"]In some of my country 's math forums, there are several problems that no body solves. Here are some of them:\nProblem1: Have: \n$a^{2}+b^{2}\\geq c^{2}, b^{2}+c^{2}\\geqq^{2}, c^{2}+a^{2}\\geq b^{2}$. \nProve that:\n$(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \\geq 4(a^{6}+b^{6}+c^{6})$ \n\n[/quote]\nApplying Mink's inequality, \\[(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \\ge (a^{2}+b^{2}+c^{2})^{3}.\\] It remains to show that: \\[(x+y+z)^{3}\\ge 4(x^{3}+y^{3}+z^{3}),\\] where $x,y,z$ are sides of a triangle. Done.[/quote]\n\nCan you explain more about the last inequalty?\nIf we are expand it, what will happen? I think that it might be wrong.\nCan you explain it more clearly?[/quote]\r\nput $x=r+s, y=s+t, z=t+r$. expand it. we get \\[2\\sum r^{3}+6\\sum_{\\textrm{sym}}rs^{2}+12rst \\geq 2\\sum r^{3}+3\\sum_{\\textrm{sym}}rs^{2}\\] which is equivalent to \\[3(\\sum_{\\textrm{sym}}rs^{2}+4rst) \\geq 0\\] Also, you can call the mink's inequality to \"Holder's inequality\".\r\nah.... This inequality is not sharp. I think we can put some restrictions for it.", "Solution_9": "But I don't like Holder's inequality as its display." } { "Tag": [ "LaTeX" ], "Problem": "In LaTeX, is there a symbol for the long division sign? Also, can you underline equations? Thanks.", "Solution_1": "Please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227205]here[/url]." } { "Tag": [ "trigonometry", "geometry", "geometric transformation", "reflection", "symmetry", "algebra", "polynomial" ], "Problem": "For what values of $ x$ does the sequence $ f_{n}\\left(x\\right)\\equal{}\\sin 7^{n}\\pi x$ converge and what is its limit?", "Solution_1": "for which real numbers $ r$ does $ \\{7^nr\\}$ converge? ($ \\{\\cdot\\}$ denotes fractional part.) where's our resident expert on fractional parts of powers? :)", "Solution_2": "It converges if and only if it's eventually constant; all equilibria are unstable.", "Solution_3": "jmerry, is that a response to my question?", "Solution_4": "It's an answer to both versions, which are slightly different. The $ \\sin$ has reflection symmetry as well.\r\n\r\nThere is a polynomial $ p$ such that $ \\sin(7x)\\equal{}p(\\sin x)$. We're iterating $ p$, and all the usual tricks for dealing with iterations apply." } { "Tag": [ "Putnam", "combinatorics unsolved", "combinatorics" ], "Problem": "Can you draw circles on the plane so that every line intersects at least one of them but no more than 100 of them?", "Solution_1": "I believe so. Construct circles along the graph of $ y \\equal{} x^3$.", "Solution_2": "why do you need $ x\\equal{}\\minus{}y^3$?", "Solution_3": "Sorry; I was thinking of a Putnam problem where one solution required constructing circles near both axes, but clearly that's not necessary here.", "Solution_4": "but wait... something is weird here. it's too easy to be a M.S. problem... it might be a bad translation.\r\nAnother solution would be to draw unit circles along the x and y axis. (whoops! not that simple :lol: )\r\n\r\nHow about if we restrict that the circles must not intersect?", "Solution_5": "It's correct according to [url=http://www.math.u-szeged.hu/~mmaroti/schweitzer/]this site[/url], but I agree. Wouldn't the $ x$ and $ y$-axes themselves intersect the circles you describe?\r\n\r\nAnd the non-intersecting condition doesn't seem to add anything to the problem. One can just place a circle of some fixed radius $ r$ at the origin and then place circles of radius $ r$ tangent to your previous circles with centers on $ y \\equal{} x^3$.", "Solution_6": "I meant circles with no points in common..", "Solution_7": "The centers of the circles, instead of following a curve, should zigzag above and below the curve. For example, one can alternate placing circles on $ y \\equal{} x^3 \\minus{} 1$ and $ y \\equal{} x^3 \\plus{} 1$ of the appropriate radius.", "Solution_8": "I might be missing something but i don't see how to do that without making the circles tangent...", "Solution_9": "If I'm understanding you correctly, the only issue with adding the constraint that the circles can't be tangent is that there exist lines that pass between two consecutive circles. But in the zigzag construction the circles don't need to be tangent to guarantee that lines can't pass between them.\r\n\r\nIn the simplest case, two tangent unit circles can be replaced with three unit circles less than $ \\sqrt{3} \\minus{} 1$ apart pairwise; then the common external tangents of any two of them passes through the third. The zigzag construction should be done so that this always occurs.", "Solution_10": "I still feel like the argument is not full.\r\nWhen you add more and more circles you might not be able to keep the distance between them that small while on the two mentioned curves. (unless you increase their radius, in which case how do we know that there is no line intersecting infinitely many of them)", "Solution_11": "This problem recently appeared on MathOverflow as well (though my reaction was that Olympiad problems are 1) for AoPS, 2) for spending a month or more trying to solve them before posting). Since it would be nice to have a solution somewhere for those who already did 2) without any effect, I'm including it here. Note that you may learn quite a few things trying to solve the problem by yourself, so don't read the spoiler unless you decided to give up completely.\n[hide=\"Solution\"]\n[hide=\"Answer\"]\nIt is impossible. We will argue by contradiction. Assume that such a configuration exists. I'll be generous and count touching as an intersection in the \"should intersect at least one circle\" requirement but not in the \"should intersect not more than $100$ circles\" requirement. \n[/hide]\n\n[hide=\"Idea 1: almost circle free angle\"]\nLet $N$ be the maximal number of circles that can be intersected by one line and let $\\ell$ be the line intersecting exactly $N$ circles $C_1,\\dots,C_N$. Without loss of generality, $\\ell$ is the $y$-axis. Then every line passing through the origin and making sufficiently small angle with $\\ell$ intersects all these circles too and, thereby, no others. Hence, all other circles are entirely contained either in the angle $A=\\{(x,y):|y|\\le Bx\\}$ with some large $B>0$ or in the symmetric angle $-A$.\n[/hide]\n\n[hide=\"Idea 2: CAT scan\"]\nNote that if a circle is contained in the angle $A$ and intersects the line $x=r$, then it is also contained in the strip $Q^{-1}r0$. Now choose $R$ so large that no line at distance $R$ or more from the origin can intersect any of the circles $C_1,\\dots,C_N$ and consider the lines $x=r$ with $R0$. Since the sum of diameters of the circles contained in $I(R)$ is at least $R$, we conclude that the average number of circles contained in $I(R)$ a line $y=bx$ intersects is at least $c'=\\frac c{2B}$.\n[/hide]\n\n[hide=\"End of proof\"]\nNow just choose sufficiently large $R_1,\\dots,R_K$ so that $I(R_j)$ are disjoint. Then the average number of circles contained in $\\cup_j I(R_j)$ a line $y=bx$ intersects is at least $Kc'$. If $K$ is so large that $Kc'>100$, we are in trouble. \n[/hide]\n[/hide]", "Solution_12": "On the graph of x^3, eventually the graph starts getting straight up, then eventually there will be 100 circles on 1 line." } { "Tag": [ "induction", "number theory proposed", "number theory" ], "Problem": "Determine all positive integers a for which there exist infinitley many squarefree positive integers n such that n divides a^n-1.\r\n\r\nBomb", "Solution_1": "A very very beautiful problem, bomb! I think the answer is all numbers except 2 and 3. For 2 actually there are no numbers $n$ except 1 and for 3 it is not difficult to prove that if $n|3^n-1$ then $n$ is a multiple of 4. So, suppose $a>3$ and take $p_1$ an odd prime factor of $a-1$. Next, take $p_2$ an odd prime factor of $ a^{p_1}-1$ different from $p_1$, next $p_3$ an odd prime factor of $a^{p_1p_2}-1$ and so on. The idea is that we will always have the chance to choose these primes and this is a consequence of a more general property: \r\n If $p| x-1$ is an odd prime, then there is an odd prime dividing $ x^p-1$, different from $p$. \r\n The property follows by choosing any prime factor of $\\frac{x^p-1}{p(x-1)}$ (exercise!). \r\n There is stil some problem, namely when $a-1$ is a power of 2. But then $a^2-1$ is not a power of 2 and apply the same reasoning with this new value.", "Solution_2": "Harazi, nice proof!!! Your exercise is quite easy, expansion by binomial thorem does it,\r\n\r\nI proved the existence ogf such a prime by induction on coprime factors... while my other post on the primes of form nk+1 provides a stronger than needed identity..\r\n\r\nBomb" } { "Tag": [], "Problem": "In a convex polygon, exactly five of the interior angles are obtuse. The largest possible number of sides for this polygon is ?", "Solution_1": "Okay suppose that we have a $k+5$ sided polygon. There are $5$ angles less than $180$, there are $k$ angles less than $90$. Since it is convex, we have that all the angles are less than 180. \r\n\r\nSum of the angles = $180*(k+5-2)<180*5+k*90\\implies k<4$. Hence the max of $k$ is $3$, and the max number of sides is $3+5=\\boxed{8}$ sides.", "Solution_2": "[hide]Let n-gon $ Q$ have $ n$ sides, and $ n$ chillin' little angles of measure $ \\theta$. $ \\frac{2\\pi}{n}= \\theta$. Of the three angles constructing a theoretically isosceles triangle, $ \\theta$ is unique. $ \\theta$'s two sisters (who are identical) sum to be obtuse.\n\n\n$ \\theta = \\frac{2\\pi}{n}< \\frac{\\pi}{4}\\Rightarrow n = 8$[/hide]\r\nTa da? :huh:" } { "Tag": [ "pigeonhole principle", "geometry", "rectangle", "probability" ], "Problem": "Kevin is saying \"Susie sells seashells by the seashore\" in which that each word is rearranged in some way such that two consecutive words don't have the same starting letter. How many ways can this sentence be arranged?\r\n\r\n :D", "Solution_1": "0, by the Pigeonhole Principle (pair up the 6 words into 3 pairs).", "Solution_2": "hm doesnt\r\n\r\nussie ellss seashells by the seashore\r\n\r\nwork?\r\n\r\nstarting letters are U, E, S, B, T, S.", "Solution_3": "OH! I thought he said the words are rearranged... :(\r\n\r\nI foresee nasty casework... :|", "Solution_4": "i forsee constructive counting...", "Solution_5": "That will take a lot of casework to solve.", "Solution_6": "oh. could this one be solved with less casework? mark wanted to go to mcdonalds, and his house is on the bottom left corner of a 5x6 grid (each square in the grid has a side length that's a block). if mcdonalds is on the right upper corner and he wants to walk a path such that no more than two consecutive Norths are made and no more than three consecutive Easts are made (he cannot go backwards: down/left, and what i mean by norths and easts is like some path: NNNEEEEE is a path cuz it gets to mcdonalds.), what is the largest area of the polygon that his path forms with the left side of the 5x6 rectangle and the top side (this is kinda confusing but o well). oh yeah, he can only walk N or E, not like half N, half E, then one third N,....\r\n\r\nsecond problem: on the way back, mark walks some other way. what is the largest possible area of the polygon that can be formed by the two paths? (he can only walk no more than two consecutive S's and no more than three consecutive Ws, and no Ns or Es.)\r\n\r\n :D", "Solution_7": "if 8 kids are in a row and if two consecutive kids are wearing the same colored shirt, they would be in a \"funny\" arrangement. if each kid has 6 shirts: red, blue, green, yellow, orange, and black, then what is the probability that they will be arranged in a \"funny\" arrangement if they each randomly choose a shirt the night before?\r\n\r\n :D" } { "Tag": [ "projective geometry", "geometry proposed", "geometry" ], "Problem": "hello.\r\nin this picture :\r\nBD is diameter of ABCD and XM=YM . O is center of the circle.\r\nprove:LQ and PS and DM are crossover.", "Solution_1": "$ M$ can move on $ BD$ (don't need $ MX \\equal{} MY$)\r\nLet $ PQ\\cap AC \\equal{} \\{I\\}, LI\\cap (O) \\equal{} \\{S'\\}$\r\nWe will show that $ S'\\equiv S$:\r\nThis result is well-known:\" $ BD$ is the polar of $ I$\". Let $ PL\\cap QS' \\equal{} \\{N\\}$ then $ N$ lies on the polar of $ I$ so $ N,B,D$ are collinear.\r\nSince $ P \\equal{} AD\\cap NL, Q \\equal{} DC\\cap NS', I \\equal{} LS'\\cap AC$ are collinear then applying [b]Desargues[/b]'s theorem for triangle $ ADC$ and $ LNS'$ we have $ AL, ND, CS'$ are concurrent at $ M$.\r\nSo $ C,S',M$ are collinear or $ S'\\equiv S$.\r\nThus $ L,S,I$ are collinear\r\nLet $ J \\equal{} QL\\cap PS$ then $ J$ lies on the polar of $ I$ or $ J,B,D$ are collinear. \r\n$ \\rightarrow$ QED" } { "Tag": [ "algebra", "polynomial", "Rational Root Theorem" ], "Problem": "Solve the equat\u0131on\r\n$x^{3}+x^{2}-x-1=0$", "Solution_1": "[quote=\"inom\"]Solve the equat\u0131on\n$x^{3}+x^{2}-x-1=0$[/quote]\r\n[hide]$x=1$[/hide]", "Solution_2": "[quote=\"i_like_pie\"][quote=\"inom\"]Solve the equat\u0131on\n$x^{3}+x^{2}-x-1=0$[/quote]\n[hide]$x=1$[/hide][/quote]\n\nthat's incorrect :wink: \n\n[hide]moving the x to the other side of the equation, we can factor it as $x(x^{2}+x-1)=1$. this is equal true when $x=x^{2}+x-1=\\boxed{\\pm1}$[/hide]", "Solution_3": "I got the same thing, but how is $(-1)^{3}+(-1)^{2}-(-1)-1=0?$", "Solution_4": "the first term is of degree 3 :wink:", "Solution_5": "Oh, I forgot about odd powers. :oops: \r\nExcuse: I haven't eaten breakfast yet. :rotfl:", "Solution_6": "neither have I :wink:", "Solution_7": "[quote=\"inom\"]Solve the equat\u0131on\n$x^{3}+x^{2}-x-1=0$[/quote]\r\n\r\n[hide=\"classic\"]\nMove the one over to the other side and get $\\pm 1$ (It is a classic factorign sum)[/hide]", "Solution_8": "[hide]Factor this way:\n\n$x^{3}+x^{2}-x-1=x^{2}(x+1)-(x+1)=(x^{2}-1)(x+1)=0$ (do you see why?)\n\nThen it goes from there.[/hide]", "Solution_9": "[hide=\"solution\"]By inspection, x=1[/hide]", "Solution_10": "[quote=\"nutz_for2.718281828\"][hide=\"solution\"]By inspection, x=1[/hide][/quote]\n[hide=\"Incorrect\"]$\\pm1$[/hide]", "Solution_11": "Just by looking at it, I realize that 2 answers are\r\n\r\n[hide]1,-1[/hide]\r\n\r\nAfter doing a bit of work, I don't think that there are any more.", "Solution_12": "[hide=\"My Solution\"]\n\n$x^{3}+x^{2}-x-1=x^{2}(x+1)-(x+1)=(x+1)(x^{2}-1)=(x+1)^{2}(x-1)$ so the solution is $x=\\pm 1$[/hide]", "Solution_13": "The lesson to learn is that if you get an answer with \"inspection\" divide the equation by the solution...", "Solution_14": "what about $x=0$?", "Solution_15": "[quote=\"srulikbd\"]what about $x=0$?[/quote]\r\n$-1\\ne0$\r\n$x=0$ is a solution iff the constant term is $0$.", "Solution_16": "x^3 + x^2 - x- 1= 0\r\n\r\nSo, let us find all possible rational roots.\r\n\r\nax^3 + bx^2 + cx + d= 0\r\n\r\na: 1\r\nd: -1,1\r\n\r\nSo, rational roots are: -1/1, 1/1 or -1, 1\r\n\r\nPlug these into the equation and you get x= +/-1.", "Solution_17": "[quote=\"neelnal\"]x^3 + x^2 - x- 1= 0\n\nSo, let us find all possible rational roots.\n\nax^3 + bx^2 + cx + d= 0\n\na: 1\nd: -1,1\n\nSo, rational roots are: -1/1, 1/1 or -1, 1\n\nPlug these into the equation and you get x= +/-1.[/quote]\r\nThis is the Rational Root Theorem (aka RRT). Is it in high school math?", "Solution_18": "[quote=\"bpms\"]This is the Rational Root Theorem (aka RRT). Is it in high school math?[/quote]\r\nI learned it in Pre-Calc. :)" } { "Tag": [ "search", "combinatorics unsolved", "combinatorics" ], "Problem": "Let 4 pots: 18 litres pot with full of milk, 7,7,4 litres pot with empty of milk. Please find the minimum number steps of measure milk satisfy in each 7 litres pot we have 6 litres of milk.", "Solution_1": "[quote=\"nktp\"]Let 4 pots: 18 litres pot with full of milk, 7,7,4 litres pot with empty of milk. Please find the minimum number steps of measure milk satisfy in each 7 litres pot we have 6 litres of milk.[/quote]\r\n\r\nAt most 13 steps :\r\n\r\n00 : 18 0 0 0\r\n01 : 11 0 7 0\r\n02 : 11 0 3 4\r\n03 : 15 0 3 0\r\n04 : 15 0 0 3\r\n05 : 08 0 7 3\r\n06 : 08 0 6 4\r\n07 : 12 0 6 0\r\n08 : 05 7 6 0\r\n09 : 05 3 6 4\r\n10 : 09 3 6 0\r\n11 : 09 0 6 3\r\n12 : 02 7 6 3\r\n13 : 02 6 6 4", "Solution_2": "I also find 2 solution for this problem and this is 13 steps, too. But I can't show that is best number. Who can help me?", "Solution_3": "One way is to do a breadth-first search starting from the initial state to compute all possible outcomes reachable in at most 12 moves.\r\n\r\nFor example:\r\n[code]\n0: 18 0 0 0\n1: 11 0 7 0\n 14 0 0 4\n2: 04 7 7 0\n 07 0 7 4\n 11 0 3 4\n 14 0 4 0\n3: 00 7 7 4\n 04 3 7 4\n 07 4 7 0\n 10 0 4 4\n 11 4 3 0\n 15 0 3 0\n...\n[/code]\r\nIt will probably be quite a lot of work, but I doubt there is an easier way, at least for the general case, unless there is some special property intrinsic to this particular instance." } { "Tag": [ "number theory solved", "number theory" ], "Problem": "find all answer of$x^2+y^2+z^2=3xyz$.\r\n$(x,y,z\\in \\mathbb{N})$", "Solution_1": "This is a special case ($n=3$) of Markov' equation, which can be solved by descent.\r\nThis appeared several times on ML.\r\n\r\nPierre." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ AD$ be the altitude of an acute-angled triangle $ ABC$ and let $ H$ be any interior point on $ AD$. Lines $ BH$ and $ CH$, intersect $ AC$ and $ AB$ at $ E$ and $ F$, respectively. Prove that $ \\angle EDH\\equal{}\\angle FDH$.", "Solution_1": "Dear Mathlinkers,\r\nsee for example\r\nHonsberger R., Mathematical Gems, Two-Year College Mathematics Journal, vol. 14, 2, p. 154-155.\r\nSincerely\r\nJean-Louis" } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose $ \\alpha$ is a root of the irreducible polynomial $ f(x)=x^{3}-3x^{2}-4$. Find the minimal polynomial for $ 1/\\alpha$. Also, find the $ N(\\alpha^{3})$ and $ Tr(\\alpha^{3})$ (norm and trace of $ \\mathbb{Q}(\\alpha)$ over $ \\mathbb{Q}$, respectively).", "Solution_1": "[quote=\"baz\"]Suppose $ \\alpha$ is a root of the irreducible polynomial $ f(x)=x^{3}-3x^{2}-4$. Find the minimal polynomial for $ 1/\\alpha$.[/quote]\r\n\r\nThat means $ f\\left( \\frac{1}{x}\\right) =\\frac{1}{x^{3}}-\\frac{3}{x^{2}}-4$\r\n\r\nNow $ \\frac{1}{\\alpha}$ is a zero of $ f(1/x)$. Thus, it seems that $ 1-3x-4x^{2}$ is a polynomial having $ 1/\\alpha$ as a zero.\r\n\r\nEDIT: Fixed mistake mentioned by -oo-", "Solution_2": "@sylow: yes, $ 1-3x-4x^{2}$ is the minimal polynomial of $ 1/\\alpha$, but of course, $ 1/\\alpha^{3}-3/\\alpha^{2}-4=0$ is wrong.\r\n@baz: what have you tried so far, and what do you know about norm and trace? An approach would be using $ N(xy)=N(x) N(y)$ and calculating the minimal polynomial of $ \\alpha^{3}$.", "Solution_3": "Yes, I was able to see that the minimal polynomial for $ 1/\\alpha$ is $ x^{3}+(3/4)x-(1/4)$ (where my definition of minimal polynomial includes monic with minimal degree). Also, I know $ N(xy)=N(x) N(y)$ gives us $ N(\\alpha^{3})=N(\\alpha)^{3}=64$ since the field polynomial for $ 1/\\alpha$ equals $ f(x)$ in this case and the norm equals the negative of the constant term in the field polynomial. But, how does one find $ Tr(\\alpha^{3})$ without too much work? (I did not see if the other parts of the problem mentioned above provide any help toward the question.)", "Solution_4": "For those interested, I found :idea: that $ Tr(\\alpha^{3})\\equal{}Tr(3\\alpha^{2}\\plus{}4)\\equal{}3Tr(\\alpha^{2})\\plus{}4Tr(1)$. To find the $ Tr(\\alpha^{2})$, we find the minimal polynomial for $ \\alpha^{2}$. Notice that $ f(x)(x^{2}(x\\plus{}3)\\plus{}4)\\equal{}(x^{2}(x\\minus{}3)\\minus{}4)(x^{2}(x\\plus{}3)\\plus{}4)\\equal{}x^{4}(x^{2}\\minus{}9)\\plus{}4x^{2}(\\minus{}9)\\minus{}16$ kills $ \\alpha$ so that in turn\r\n$ x^{2}(x\\minus{}9)\\plus{}4x(\\minus{}9)\\minus{}16\\equal{}x^{3}\\minus{}9x^{2}\\minus{}36x\\minus{}16$ kills $ \\alpha^{2}$. Moreover, it is easy to see that this is its minimal polynomial and also its field polynomial (well, given some knowledge/theorems on the subject). So, the $ Tr(\\alpha^{2})\\equal{}9$ yields that $ Tr(\\alpha^{3})\\equal{}3Tr(\\alpha^{2})\\plus{}4Tr(1)\\equal{}3\\cdot 9\\plus{}4\\cdot 3\\equal{}39$. :D happy that this is done. \r\n\r\nThanks to the moderators and people here", "Solution_5": "[quote=\"baz\"]For those interested, I found :idea: that $ Tr(\\alpha^{3}) \\equal{} Tr(3\\alpha^{2}\\plus{}4) \\equal{} 3Tr(\\alpha^{2})\\plus{}4Tr(1)$. To find the $ Tr(\\alpha^{2})$, we find the minimal polynomial for $ \\alpha^{2}$. Notice that $ f(x)(x^{2}(x\\plus{}3)\\plus{}4) \\equal{} (x^{2}(x\\minus{}3)\\minus{}4)(x^{2}(x\\plus{}3)\\plus{}4) \\equal{} x^{4}(x^{2}\\minus{}9)\\plus{}4x^{2}(\\minus{}9)\\minus{}16$ kills $ \\alpha$ so that in turn\n$ x^{2}(x\\minus{}9)\\plus{}4x(\\minus{}9)\\minus{}16 \\equal{} x^{3}\\minus{}9x^{2}\\minus{}36x\\minus{}16$ kills $ \\alpha^{2}$. Moreover, it is easy to see that this is its minimal polynomial and also its field polynomial (well, given some knowledge/theorems on the subject). So, the $ Tr(\\alpha^{2}) \\equal{} 9$ yields that $ Tr(\\alpha^{3}) \\equal{} 3Tr(\\alpha^{2})\\plus{}4Tr(1) \\equal{} 3\\cdot 9\\plus{}4\\cdot 3 \\equal{} 39$. :D happy that this is done. \n[/quote]\r\n\r\n$ \\rm{SIMPLER}\\quad Tr(a^{2}) \\equal{} a^{2}\\plus{}b^{2}\\plus{}c^{2}\\equal{} (a\\plus{}b\\plus{}c)^{2}\\minus{}2(ab\\plus{}bc\\plus{}ca) \\equal{} 3^{2}\\minus{}2(0)$" } { "Tag": [ "\\/closed" ], "Problem": "[quote]Please enter the student's current Art of Problem Solving Username.\nThis is the same name that the student uses in the forums and/or has used in previous classes.\n\nEnter the student's username: \n\nIf the student does not have a username, click here to create a new account. [/quote]\r\n\r\nfamiliar?\r\nwhat do i do after i enter my username. when do we have to pay the money?", "Solution_1": "You enter your username there, then click Enroll on that page. Follow the directions on the pages that follow." } { "Tag": [ "calculus", "derivative", "function", "algebra", "binomial theorem", "calculus computations" ], "Problem": "Find a formula for $ f^{(n)}(x)$, where $ f(x)\\equal{}\\sqrt{x}$.", "Solution_1": "I have\r\n\r\n$ f^{(n)}(x)\\equal{}\\frac{(\\minus{}1)^{n\\plus{}1}(2n\\minus{}2)!}{2^{2n\\minus{}1}(n\\minus{}1)!x^{n\\minus{}\\frac{1}{2}}}$.\r\n\r\nIs there a better way to write this?", "Solution_2": "More find the ${ f^{(n)}(x)}$ where $ f(x)=x^a$", "Solution_3": "Does it look SOMETHING like the binomial theorem? I have a solution, but I'm not sure if it's right.\r\n\r\nMine looks VERY similar to the binomial theorem.", "Solution_4": "The binomial theorem, you say? Let's consider Newton's version of that.\r\n\r\nFor notation, we can make the following definition:\r\n\r\nIf $ \\alpha\\in\\mathbb{R}$ and $ n$ a nonnegative integer, then let\r\n\r\n$ \\binom{\\alpha}{n}\\equal{}\\frac{\\alpha(\\alpha\\minus{}1)(\\alpha\\minus{}2)\\cdots\r\n(\\alpha\\minus{}n\\plus{}1)}{n!}.$\r\n\r\nWe have this power series centered at zero:\r\n\r\n$ (1\\plus{}x)^{\\alpha}\\equal{}\\sum_{n\\equal{}0}^{\\infty}\\binom{\\alpha}{n}x^n$\r\n\r\nShift this and stretch it by letting $ x\\equal{}\\frac{y\\minus{}b}{b}$ so that $ 1\\plus{}x\\equal{}\\frac{y}{b}.$\r\n\r\n$ \\left(\\frac{y}{b}\\right)^{\\alpha}\\equal{}\r\n \\sum_{n\\equal{}0}^{\\infty}\\binom{\\alpha}{n}\\left(\\frac{y\\minus{}b}{b}\\right)^n$\r\n\r\n$ y^{\\alpha}\\equal{}\\sum_{n\\equal{}0}^{\\infty}\\binom{\\alpha}{n}b^{\\alpha\\minus{}n}(y\\minus{}b)^n$\r\n\r\nBut by Taylor's theorem, we should have $ f(y)\\equal{}\\sum_{n\\equal{}0}^{\\infty}\\frac{f^{[n]}(b)}{n!}(y\\minus{}b)^n$\r\n\r\nHence, if $ f(y)\\equal{}y^{\\alpha}$, then\r\n\r\n$ f^{[n]}(b)\\equal{}n!\\binom{\\alpha}{n}b^{\\alpha\\minus{}n}$\r\n\r\n$ \\equal{}\\alpha(\\alpha\\minus{}1)(\\alpha\\minus{}2)\\cdots(\\alpha\\minus{}n\\plus{}1)b^{\\alpha\\minus{}n}$\r\n\r\n(Which, of course, we could have gotten directly.)\r\n\r\nYou can also write $ \\binom{\\alpha}{n}$ in terms of Gamma functions - I leave that as an exercise.", "Solution_5": "Nifty!\r\n\r\nSo they are sort of related!\r\n\r\n:-)\r\n\r\nKM, your posts are always informative and helpful! Thank you for being such a great resource.", "Solution_6": "[quote=\"Kent Merryfield\"]\nHence, if $ f(y) \\equal{} y^{\\alpha}$, then\n\n$ f^{[n]}(b) \\equal{} n!\\binom{\\alpha}{n}b^{\\alpha \\minus{} n}$\n\n$ \\equal{} \\alpha(\\alpha \\minus{} 1)(\\alpha \\minus{} 2)\\cdots(\\alpha \\minus{} n \\plus{} 1)b^{\\alpha \\minus{} n}$\n\n(Which, of course, we could have gotten directly.)\n\nYou can also write $ \\binom{\\alpha}{n}$ in terms of Gamma functions - I leave that as an exercise.[/quote]\r\n\r\nThis is one approach to defining fractional derivatives! As above, we have that\r\n\r\n\\[ \\frac{d^\\mu}{dy^\\mu}\\left( y^\\lambda\\right) \\equal{} \\frac{\\Gamma (\\lambda \\plus{}1)}{\\Gamma (\\lambda \\minus{}\\mu \\plus{}1)}y^{\\lambda\\minus{}\\mu}\\]\r\n\r\nFor suitable $ \\lambda , \\mu \\in\\mathbb{C}$ (e.g. $ \\lambda \\neq \\minus{}1$), so that we have defined fractional derivatives (derivatives of non-integer order--even complex orders) of a power function. I suppose (though I haven't tried it) that one might begin a treatment of fractional calculus based on this via power series." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a,b,c >0 prove ab/(a+b+2c)+bc/(b+c+2a)+ca/(c+a+2b) <=(a+b+c)/4", "Solution_1": "Is really easy as SOS", "Solution_2": "please write your solution", "Solution_3": "Is there a nicer solution than SOS (sum of squares)?", "Solution_4": "[quote=\"Russian_Man\"]a,b,c >0 prove $ \\sum \\frac{ab}{a\\plus{}b\\plus{}2c} \\le \\frac{a\\plus{}b\\plus{}c}{4}$[/quote]\r\n $ a\\plus{}b\\plus{}2c\\equal{}x$\r\n $ a\\plus{}2b\\plus{}c\\equal{}y$\r\n $ 2a\\plus{}b\\plus{}c\\equal{}z$ \r\nthen we must prove $ \\sum \\frac{(\\frac{3(x\\plus{}y\\plus{}z)}{4}\\minus{}x\\minus{}z)(\\frac{3(x\\plus{}y\\plus{}z)}{4}\\minus{}x\\minus{}y)}{x}\\le \\frac{x\\plus{}y\\plus{}z}{16}$ or\r\n$ \\sum \\frac{(3y\\minus{}x\\minus{}z)(3z\\minus{}x\\minus{}y)}{x}\\le x\\plus{}y\\plus{}z$or\r\n$ \\sum \\frac{10yz\\minus{}3z^2\\minus{}3y^2}{x}\\le 4(x\\plus{}y\\plus{}z)$ or\r\n$ 5\\sum_{sym} x^2y^2\\le 2\\sum_{sym} x^2yz\\plus{}3\\sum_{sym}x^3y$ which is true", "Solution_5": "Let the left side be K, by Cauchy, we need to show\r\n$ K^2 \\le (ab \\plus{} bc \\plus{} ca)(\\sum \\frac {ab}{(a \\plus{} b \\plus{} 2c)^2}) \\le \\frac {(a \\plus{} b \\plus{} c)^2}{16}$\r\nBy AM-GM, $ (a \\plus{} b \\plus{} 2c)^2 \\ge 4(a \\plus{} c)(b \\plus{} c)$, so we have to show\r\n$ (ab \\plus{} bc \\plus{} ca)(\\sum \\frac {ab}{(a \\plus{} c)(b \\plus{} c)}) \\le \\frac {(a \\plus{} b \\plus{} c)^2}{4}$\r\n$ 4(ab \\plus{} bc \\plus{} ca)(a^2b \\plus{} ab^2 \\plus{} b^2c \\plus{} bc^2 \\plus{} c^2a \\plus{} ca^2) \\le (a \\plus{} b \\plus{} c)^2(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)$\r\nWhich I believe is true (Can someone check it?).\r\nEDIT: Whoops, it is false.\r\n\r\n[quote=\"dima ukraine\"]$ 5\\sum_{sym} x^2y^2\\le 2\\sum_{sym} x^2yz \\plus{} 3\\sum_{sym}x^3y$[/quote]\r\nWhat about x=y=1, z=0? I know this is impossible for the definitions of x,y,z, but how do you prove it?", "Solution_6": "[quote=\"Russian_Man\"]$ a,b,c >0$ prove $ ab/(a\\plus{}b\\plus{}2c)\\plus{}bc/(b\\plus{}c\\plus{}2a)\\plus{}ca/(c\\plus{}a\\plus{}2b) <\\equal{}(a\\plus{}b\\plus{}c)/4$[/quote]\r\nYes, we can kill it by SOS.\r\n$ \\sum_{cyc}\\frac{ab}{a\\plus{}b\\plus{}2c}\\leq\\frac{a\\plus{}b\\plus{}c}{4}\\Leftrightarrow\\sum_{cyc}\\left(\\frac{c}{4}\\minus{}\\frac{ab}{a\\plus{}b\\plus{}2c}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac{2c^2\\plus{}ac\\plus{}bc\\minus{}4ab}{a\\plus{}b\\plus{}2c}\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\left(\\frac{(c\\minus{}a)(2b\\plus{}c)}{a\\plus{}b\\plus{}2c}\\minus{}\\frac{(b\\minus{}c)(2a\\plus{}c)}{a\\plus{}b\\plus{}2c}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a\\minus{}b)\\left(\\frac{2c\\plus{}a}{b\\plus{}c\\plus{}2a}\\minus{}\\frac{2c\\plus{}b}{a\\plus{}c\\plus{}2b}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a\\minus{}b)^2(a\\plus{}b\\minus{}c)(a\\plus{}b\\plus{}2c)\\geq0.$\r\nLet $ a\\geq b\\geq c.$ Hence, $ S_b\\geq0,$ $ S_c\\geq0$ and $ b^2S_a\\plus{}a^2S_b\\geq0.$\r\nHence, $ \\sum_{cyc}(a\\minus{}b)^2S_c\\geq(a\\minus{}c)^2S_b\\plus{}(b\\minus{}c)^2S_a\\geq\\frac{a^2}{b^2}(b\\minus{}c)^2S_b\\plus{}(b\\minus{}c)^2S_a\\equal{}$\r\n$ \\equal{}\\frac{(b\\minus{}c)^2}{b^2}(a^2S_b\\plus{}b^2S_a)\\geq0.$ Done!", "Solution_7": "[b]Problem:[/b]\r\n\r\nGiven $ a,b,c \\geq 0$, Prove that:\r\n\\[ \\frac {ab}{a + b + 2c} + \\frac {bc}{b + c + 2a} + \\frac {ca}{c + a + 2b} \\leq \\frac {a + b + c}{4}\r\n\\]\r\nSolution (This is Can_Hang's solution, that i saw in other topic): \r\nBy Cauchy Schwarz:\r\n\\[ \\sum_{cyc}\\left(\\frac {ab}{a + b + 2c}\\right) \\le \\frac {1}{4}\\cdot{\\sum_{cyc}\\left(\\frac {ab}{a + c} + \\frac {ab}{b + c}\\right) = \\frac {1}{4}\\cdot{\\sum_{cyc}\\left(\\frac {ab}{a + c} + \\frac {bc}{a + c}\\right) = \\frac {1}{4}\\cdot\\sum_{cyc}\\left(\\frac {b(a + c)}{a + c}\\right) = \\frac {1}{4}\\cdot\\sum_{cyc}b = \\frac {a + b + c}{4}\r\n}}\\]", "Solution_8": "[quote=\"dgreenb801\"]\n$ 4(ab \\plus{} bc \\plus{} ca)(a^2b \\plus{} ab^2 \\plus{} b^2c \\plus{} bc^2 \\plus{} c^2a \\plus{} ca^2) \\le (a \\plus{} b \\plus{} c)^2(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)$\nWhich I believe is true (Can someone check it?).\n[/quote]\r\nYour statement is valid and we can prove it as follow:\r\nThis inequality is equivalent to\r\n\\[ 4(ab\\plus{}bc\\plus{}ca)[(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)\\minus{}3abc] \\le (a\\plus{}b\\plus{}c)^2[(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca) \\minus{}abc]\\]\r\n\\[ abc [12(ab\\plus{}bc\\plus{}ca) \\minus{}(a\\plus{}b\\plus{}c)^2] \\ge (a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)[4(ab\\plus{}bc\\plus{}ca)\\minus{}(a\\plus{}b\\plus{}c)^2]\\]\r\nIf $ 4(ab\\plus{}bc\\plus{}ca) \\le (a\\plus{}b\\plus{}c)^2$, then\r\n\\[ (a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)[4(ab\\plus{}bc\\plus{}ca) \\minus{}(a\\plus{}b\\plus{}c)^2] \\le 9abc[4(ab\\plus{}bc\\plus{}ca)\\minus{}(a\\plus{}b\\plus{}c)^2]\\]\r\nAnd we just need to prove that\r\n\\[ 9[4(ab\\plus{}bc\\plus{}ca)\\minus{}(a\\plus{}b\\plus{}c)^2] \\le 12(ab\\plus{}bc\\plus{}ca)\\minus{}(a\\plus{}b\\plus{}c)^2\\]\r\n\\[ (a\\plus{}b\\plus{}c)^2 \\ge 3(ab\\plus{}bc\\plus{}ca)\\]\r\nwhich is true.\r\nIf $ 4(ab\\plus{}bc\\plus{}ca) \\ge (a\\plus{}b\\plus{}c)^2$, applying Schur's Inequality, we have\r\n\\[ abc \\ge \\frac{[(a\\plus{}b\\plus{}c)^2\\minus{}(ab\\plus{}bc\\plus{}ca)][4(ab\\plus{}bc\\plus{}ca)\\minus{}(a\\plus{}b\\plus{}c)^2]}{6(a\\plus{}b\\plus{}c)}\\]\r\nWe can deduce our inequality to\r\n\\[ [(a\\plus{}b\\plus{}c)^2\\minus{}(ab\\plus{}bc\\plus{}ca)][12(ab\\plus{}bc\\plus{}ca)\\minus{}(a\\plus{}b\\plus{}c)^2] \\ge 6(a\\plus{}b\\plus{}c)^2(ab\\plus{}bc\\plus{}ca)\\]\r\nSetting $ p\\equal{}a\\plus{}b\\plus{}c,q\\equal{}ab\\plus{}bc\\plus{}ca$, then we may write the above inequality as\r\n\\[ (p^2\\minus{}q)(12q\\minus{}p^2) \\ge 6p^2q\\]\r\n\\[ \\minus{}p^4\\plus{}7p^2q \\minus{}12q^2 \\ge 0\\]\r\n\\[ (p^2\\minus{}3q)(4q\\minus{}p^2) \\ge 0\\]\r\nwhich is true since $ 4q \\ge p^2 \\ge 3q$.", "Solution_9": "shaaam, your solution is very nice and simple." } { "Tag": [ "calculus", "integration", "modular arithmetic" ], "Problem": "Find all integers solutions of $a^{3}+2b^{3}=4c^{3}$", "Solution_1": "[b]Solution.[/b] If $a_{1}^{3}+2b_{1}^{3}=4c_{1}^{3}$, then after checking parity, one obtains that $\\left(\\frac{a_{1}}{2},\\frac{b_{1}}{2},\\frac{c_{1}}{2}\\right)$ is also a solution. Similarly, one obtains that $\\left(\\frac{a_{1}}{2^{k}},\\frac{b_{1}}{2^{k}},\\frac{c_{1}}{2^{k}}\\right)$ is a solution for any integer $k$. This implies that the only integral solution to this equation is $(a,b,c)=(0,0,0)$. $\\Box$", "Solution_2": "A [b]brief interpretation[/b] by Infinite descent.\r\n\r\nLet $a=2a'$. $(\\because \\text{if}\\ a=2a'+1 \\ \\text{then}\\ a^{3}+2b^{3}\\not\\equiv 0 \\pmod{2})$\r\n$a^{3}+2b^{3}=8a'^{3}+2b^{3}=4c^{3}\\iff b^{3}= 2(2c^{3}-4a'^{3})$\r\n$\\therefore 2|b$\r\n\r\nLet $b=2b'$.\r\n$a^{3}+2b^{3}=8a'^{3}+16b'^{3}=4c^{3}\\iff c^{3}= 2(a'^{3}+2b'^{3})$\r\n$\\therefore 2|c$\r\n\r\nLet $c=2c'$.\r\n$a^{3}+2b^{3}=8a'^{3}+16b'^{3}=4c^{3}=32c'^{3}\\iff a'^{3}+2b'^{2}= 4c'^{3}$\r\n\r\nTherefore, $(a', b', c') \\ \\text{i.e.}\\ \\left ( \\frac{a}{2}, \\frac{b}{2}, \\frac{c}{2}\\right )$ is an answer if $(a, b, c)$ is an answer.\r\n\r\n\r\nIt is similar as follows.\r\n\r\nThus, $(a, b, c)=(0, 0, 0)$", "Solution_3": "The proof is nicer with contradiction:\r\n\r\nAssume for sake of contradiction that there is a solution set where $|a|+|b|+|c|>0$. By WOP there is a set where that value is minimal. \r\n\r\nNow go through the argument.\r\n\r\nContradiction; you a solution where that value is smaller. \r\n\r\nHence $|a|+|b|+|c|=0$ is the only other option; i.e. $a=b=c=0$.\r\n\r\nThis manner of arguing is much cleaner than the whole: \"a/2^k\" stuff." } { "Tag": [ "factorial", "calculus", "calculus computations" ], "Problem": "Can anyone find a closed form for $ \\sum_{k\\equal{}1}^n k!$?", "Solution_1": "Hum, we can find $ \\sum_{k\\equal{}1}^n k\\cdot k!$.", "Solution_2": "hello, it is\r\n$ \\sum_{k=1}^n k\\cdot k!=(n+1)!-1$.\r\nSonnhard.", "Solution_3": "That's right.", "Solution_4": "[quote=\"igiul\"]Can anyone find a closed form for $ \\sum_{k \\equal{} 1}^n k!$?[/quote]\r\n\r\nI think it's hard, and there are lot of other similar sums for example\r\n\r\n$ \\sum \\sqrt{k}$\r\n\r\nBut it might be possible to compute them.", "Solution_5": "See [url=http://mathworld.wolfram.com/FactorialSums.html]factorial sums[/url]." } { "Tag": [ "function", "logarithms" ], "Problem": ":blush: solve for x if 8x=4^x", "Solution_1": "It's a transcendental equation: but we can get something out of it. It's trivial to show that it has two roots. The bigger one is obviously $ 2$, and the other one can be found only numerically (correct me if I'm wrong, but you're dealing with Lambert W function here, so I don't expect anything elegant/without using Newton's method or sth like that.)\r\n\r\nMathematica's result was:\r\n\r\n$ 0.15495346619034528$\r\n\r\nand the one I got using second degree McLaurin approximation was\r\n\r\n$ \\frac {4 \\minus{} \\ln 2 \\minus{} \\sqrt {16 \\minus{} 8\\ln 2 \\minus{} (\\ln 2)^2}}{2(\\ln{2})^2}\\approx 0.15467723825798213$\r\n\r\nThe error is not too big ($ 0.17\\%$);)", "Solution_2": "For integers:\r\n\r\n[hide=\"solution\"]Taking the expression log 4, we get $ x\\equal{}\\log_{4}8\\plus{}\\log_{4}x$, or $ x\\minus{}\\log_{4}x\\equal{}3/2$\n\nSince this is increasing when $ x>1$, there should be either 1 or 0 values above 1 such that this expression is true. Let's try some small values of $ x$:\n$ 1\\minus{}\\log_{4}1\\equal{}1$\n$ 2\\minus{}\\log_{4}2\\equal{}3/2$\nBINGO!\n\nNow if $ x<1$, then we can let $ \\frac{1}{x}\\equal{}y>1$. Thus we have $ x\\minus{}\\log_{4}x\\equal{}\\frac{1}{y}\\plus{}\\log_{4}y$ for some positive $ y>1$. Since there is no rational $ y$ such that the expression equals two, I'm just going to state a bound: $ y\\in (8,16)$.[/hide]" } { "Tag": [ "calculus", "derivative", "function", "conics", "parabola", "calculus computations" ], "Problem": "Find the values of any relative extrema.\r\nf(x) = x^2 + 2x - 3", "Solution_1": "(This can be done with no calculus at all, but ...)\r\n\r\nShow us what you've done so far, please.", "Solution_2": "[quote=\"Kent Merryfield\"](This can be done with no calculus at all, but ...)\n\nShow us what you've done so far, please.[/quote]\r\n\r\nf'(x) = 2x + 2\r\n2x + 2 = 0\r\nx = - 1\r\nif this is correct what is ( or what are) answer", "Solution_3": "OK, why did you take the derivative and why did you set it equal to zero? What is the purpose of those actions?", "Solution_4": "Your extreme point is simply $ (\\minus{}1, f(\\minus{}1))$ but you must do either the first or second derivative test to determine whether you have a max, min or straddle point.", "Solution_5": "[quote=\"Kent Merryfield\"]OK, why did you take the derivative and why did you set it equal to zero? What is the purpose of those actions?[/quo\r\n\r\nTryed to find derivative....that isn't correct :(", "Solution_6": "[quote=\"Prime factorization\"]Your extreme point is simply $ ( \\minus{} 1, f( \\minus{} 1))$ but you must do either the first or second derivative test to determine whether you have a max, min or straddle point.[/quote]\r\nHow can i do that?", "Solution_7": "There were no computational errors in your post in which you found $ f'(x)$ and solved for when it was zero. \r\n\r\nAs I told a class of mine the other day after I handed a test back: being able to calculate does matter and will always matter. But being able to calculate will never be enough; it will never be anywhere close to enough. You have to know [i]what[/i] to compute, and you have to properly interpret the result of any such computation. \r\n\r\nThat's what's lacking here. You did do the correct computation, but showed no sign of knowing why you chose to do that. And then, you did not know how to interpret the result. Any computation whose result you can't interpret might as well not have been done.\r\n\r\nSome basic questions:\r\n\r\nWhat is the geometric meaning of the derivative of a function?\r\n\r\nWhat, exactly, is the relationship between zeros of the derivative and extreme values? There a one-way implication there: if (one thing) then (another thing), and the converse of that isn't true. You have to know that if-then statement.\r\n\r\nAnd that function is a pretty simple formula. Do you know what its graph looks like, generally speaking?", "Solution_8": "Well, your second derivative is 2, which is >0, thus you have a min point. Just to check, in the derivative you have the points (-2,-2) and (0,2) from plugging in. Since the first derivative goes from -ve to +ve relative to the critical point, the function itself goes from decreasing to increasing. Draw that on a graph, and you'll get a parabola opening upwards, which obviously has a min point as stated earlier.\r\n\r\nEdit: Just to add to what Jmerry said, also keep in mind that you have a critical point if the derivative leads to an indeterminate form. As an example |x| (which we know has a min) has derivative x for x > 0 and -x for x<0, but the derivative at x=0 doesn't exist." } { "Tag": [], "Problem": "$x^\\frac{3}{2}-x^{-\\frac{3}{2}}= \\frac{3}{2}$\r\n\r\n[size=59]Latexed my mod[/size]", "Solution_1": "[quote=\"ProblemSets\"]$x^\\frac{3}{2}-x^{-\\frac{3}{2}}= \\frac{3}{2}$\n\n[size=59]Latexed my mod[/size][/quote]\r\n\r\n[hide=\"solution\"]Let $x^\\frac{3}{2}=a$.\n\n$a-\\frac{1}{a}=\\frac{3}{2}$\n\n$\\Rightarrow 2a^{2}-3a-2=0$\n\n$\\Rightarrow 0=(2a+1)(a-2)$\n\n$\\Rightarrow a=2,-\\frac{1}{2}=x^\\frac{3}{2}$\n\nTherefore $x=2^{\\frac{2}{3}}$ and $\\frac{1}{2^\\frac{2}{3}}$[/hide]", "Solution_2": "i did an estimate on my calculator and was getting that x should be about 1.5ish...someplace around there. i typed your given answers and it did not satisfy the equation. am i crazy?", "Solution_3": "[quote=\"ProblemSets\"]i did an estimate on my calculator and was getting that x should be about 1.5ish...someplace around there. i typed your given answers and it did not satisfy the equation. am i crazy?[/quote]\r\nIt's about 1.58, so you might have made a mistake there. Anyways, if $x=2^\\frac{2}{3}$, then $x^\\frac{3}{2}=2^{(\\frac{3}{2})(\\frac{2}{3})}=2^{1}=2$. Thus, $\\frac{1}{x}=\\frac{1}{2}$, so $x-\\frac{1}{x}=2-\\frac{1}{2}=\\frac{3}{2}$, which is what we are given, so the two answers work.", "Solution_4": "[quote=\"ProblemSets\"]i did an estimate on my calculator and was getting that x should be about 1.5ish...someplace around there. i typed your given answers and it did not satisfy the equation. am i crazy?[/quote]\r\n\r\nyes. \r\n\r\njk...but when i checked on my computer calculator (dont have my ti-89 with me right now) i got 1.84-.54 which isnt [i]that[/i] close to 1.5...", "Solution_5": "i figured what i did wrong. i read 2^(3/2) as 2 and 2/3\r\n\r\n\r\nmy bad. sorry about that. my eyes are getting bad :)" } { "Tag": [ "ratio", "geometry" ], "Problem": "Can anyone help me out on how to solve this? It's driving me crazy. :rotfl: \r\n\r\nTwo circles are internally tangent. The smaller circle is also tangent to two \r\nperpendicular radii of the larger circle. What is the ratio of the area of the small circle to \r\nthe area of the large circle? Express your answer as a decimal rounded to the nearest \r\nhundredth.", "Solution_1": "[geogebra]63e178a92816325337e39522fe0cf04f5bee8a4f[/geogebra] \r\nSorry, I'm not too good with Geogebra.\r\nEDIT: Thanks for pointing that out. It's past 10:30 here, so at least I have an excuse (kind of)\r\nOk, I think this is right though.\r\nTo find the radius of the smaller circle, take the radius of the larger circle and divide by two. Now you have the length of the diagonal of the square. Dividing by the square root of two, we have the radius of the smaller circle. Thus the answer is $ (\\frac{1}{2\\sqrt{2}})^2=\\frac{1/8}$, or $ 0.13$ to the nearest hundredth.", "Solution_2": "Unfortunately your answer could not be correct. It's more complicated than that. Clearly the smaller circle is less than one quarter of the larger one.\r\n\r\nBut how to calculate???", "Solution_3": "Let the radius of the larger circle be $ x$ and the radius of the smaller circle $ r$. From 45-45-90 triangle relationships, and looking at the radii, we can conclude that $ r(1\\plus{}\\sqrt2)\\equal{}x$. We just square this ratio, then take the reciprocal to find the desired area ration, we find the answer is $ \\boxed{0.17}$", "Solution_4": "That's it! Thanks much! :P" } { "Tag": [ "geometry", "rectangle", "combinatorics open", "combinatorics" ], "Problem": "let us consider a square containing of square identical of format 4*4 how much rectangles can one find in the large square one", "Solution_1": "Do you mean how many rectangles can be found in a 4x4 grid?\r\nThe answer is $\\frac{25 * 16}{4}= 100$" } { "Tag": [ "modular arithmetic" ], "Problem": "Prove or disprove this statement:\r\nThere exist integer solution (x,y) except (4,6) such that\r\n$ x^2\\plus{}x^3\\equal{}2^y\\plus{}16$.", "Solution_1": "Here is my start. \r\n\r\n$ x^2(x\\plus{}1)\\equal{}2^y\\plus{}16$\r\n\r\nWe easily find that $ y\\ge6$\r\n\r\n1) x divisible by 4\r\n2) x+1 divisible by 16\r\n\r\nFirst case has one solution which is included into statement.\r\n\r\nBut how to solve second case?? :maybe: \r\n\r\nThanks in advance.", "Solution_2": "Idea for second case: We know that x+1 isn't divisible by 32, since the RHS isn't divisible by 32. Dividing both sides by 16 gives us\r\n\r\n$ x^2(oddinteger) \\equal{} 2^{y \\minus{} 4} \\plus{} 1$\r\n\r\nMaybe we could prove that no squares other than 1 divide the RHS...\r\nEDIT: That method fails because sometimes $ 2^{y\\minus{}4}\\equiv 8\\bmod{9}$ for some $ y$.", "Solution_3": "[hide]\nChecking $ y\\equal{}0$ gives us no solution, thus $ x$ must be even, so substitute $ x \\equal{} 2k$ to get:\n\n$ 4k^2 \\plus{} 8k^3 \\equal{} 2^y \\plus{} 16$\n\n$ k^2 \\plus{} 2k^3 \\equal{} 2^{y \\minus{} 2} \\plus{} 4$\n\n$ k^2(2k \\plus{} 1) \\equal{} 2^{y \\minus{} 2} \\plus{} 4 \\implies k \\equal{} 2m$ (checking $ y \\equal{} 2$ gives us no solution, thus $ y > 2$)\n\n$ 4m^2(4m \\plus{} 1) \\equal{} 2^{y \\minus{} 2} \\plus{} 4$\n\n$ m^2(4m \\plus{} 1) \\equal{} 2^{y \\minus{} 4} \\plus{} 1$\n\nquickly checking $ y \\equal{} 4$ gives no integer solution, thus RHS is odd $ \\implies m \\equal{} 2n \\plus{} 1$\n\n$ (4n^2 \\plus{} 4n \\plus{} 1)(8n \\plus{} 5) \\equal{} 2^{y \\minus{} 4} \\plus{} 1$\n\nchecking LHS we see that it's remainder is $ 1 \\pmod 4 \\implies y \\ge 6$\n\n$ y \\equal{} 6$ gives us solution so let's check $ y > 6$\n\n$ 2^{y \\minus{} 4} \\plus{} 1 \\equiv 1 \\pmod 8$\n\nKnowing that square of an odd number always gives remainder $ 1 \\pmod 8 \\implies (4n^2 \\plus{} 4n \\plus{} 1)(8n \\plus{} 5) \\equiv 5 \\pmod 8$ which is contradiction. Thus the only solution is $ (4, 6)$.[/hide]", "Solution_4": "Flame, you solved only first case of my problem (look at post #2, x is not neccessary even) :wink:", "Solution_5": "We would just do same as in my previous post, shouldn't be too hard to finish it. :)", "Solution_6": "I had to use divisibility of 32, then 64, then 128, then 256... but it still wasn't enough, so I don't think it's as easy as in the first case :)", "Solution_7": "Is it really to hard for pre-olympiad level?" } { "Tag": [ "function", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let M be a metric space with the discrete metric.\r\n1) show that every subset of M is clopen\r\n2) show that every function defined on M is continuous.\r\n\r\nI don't know where to start with this problem. I know the definition of closed and open sets, continuous function, but I just don't see how to approach this. Hope someone can give me some help.", "Solution_1": "1) If $ \\varepsilon<1$ and $ x\\epsilon M$ then the $ \\varepsilon$-Ball $ B_\\varepsilon(x)$ under the discrete metric is just the set $ \\{x\\}$. The fact that all subsets of M are open follows trivially from the definition of open sets under the metric topology.\r\n\r\n2) If $ f: M\\rightarrow Y$, then for any subset $ A\\subset Y$, and thus any open subset, $ f^{\\minus{}1}(A)$ is a subset of M, and therefore by 1), is open. So f satisfies the definition of continuity.", "Solution_2": "Thanks for your help.\r\n1)The singleton set {x} is closed, but I don't see how all subsets of M are open is trivial from the definition of open sets. I'm slow at this. If epsilon is greater than 1, then M_epsilon(x) is just M. We have a theorem saying that M_epsilon(x) is open for any epsilon greater than 0 and any x in M.\r\n\r\n2) I see that the pre-image of A is a subset of M, then by part 1) it is open. But why does this satisfy the definition of continuity. We have a theorem stating that if the pre-image of any open set V in N is open, then f is continuous. Did you apply that theorem here?", "Solution_3": "1) let $ X \\subset M$ and let $ x \\epsilon X$, then $ B_{\\varepsilon}(x)\\subset X$, so long as $ {0 < \\varepsilon < 1}$.\r\n\r\n2) In topology, that property is taken as the definition of continuity, not as a theorem. I was working under that definition." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Solve the equation in natural numbers\r\n$ 2^{x\\minus{}1}\\plus{}1\\equal{}y^{z\\plus{}1}$", "Solution_1": "We have $ y^{z \\plus{} 1} \\minus{} 2^{x \\minus{} 1} \\equal{} 1$ so by Mihailescu's theorem, the solutions where $ z\\geq 1$ and $ x\\geq 3$ are $ x \\equal{} 4,y \\equal{} 3,z \\equal{} 1$.\r\n\r\nAssume now that $ z \\equal{} 1$. \r\n\r\nThen $ y^2 \\minus{} 2^{x \\minus{} 1} \\equal{} 1$. But Ko Chao's result says that if $ x\\geq 4$ and $ y^2 \\minus{} 2^{x \\minus{} 1} \\equal{} 1$ has a solution then $ 2\\geq 2^{x \\minus{} 2} \\minus{} 2$. But this means that $ x \\equal{} 4$.\r\n\r\nSo let us check the cases $ x\\in \\{1,2,3\\}$. \r\n\r\nIf $ x \\equal{} 1$ then $ y^2 \\equal{} 2$ having no solutions in integers.\r\nIf $ x \\equal{} 2$ then $ y^2 \\equal{} 3$ having no solutions in integers.\r\nIf $ x \\equal{} 3$ then $ y^2 \\equal{} 5$ having no solutions in integers.\r\n\r\nSo $ (x,y,z) \\equal{} (4,3,1)$ is the only positive solution.", "Solution_2": "It is equivalent to $ 2^{x-1}=(y-1)(1+y+\\dots+y^z)$. Hence $ y=2^a+1$ for some $ a\\in\\mathbb N$. Then $ 2^{x-1-a}=1+y+\\dots+y^z\\equiv z+1\\pmod 2$. Thus $ z$ is odd. Therefore \r\n\\begin{align*}2^{x-1-a}&=1+y+\\dots+y^z=(1+y)(1+y^2+\\dots+y^{z-1})\\\\\r\n&=2(2^{a-1}+1)(1+y^2+\\dots+y^{z-1}).\\end{align*}\r\nHence $ 2^{a-1}+1$ must be a power of $ 2$, which is possible only when $ a=1$. Then $ y=3$ and we get\r\n\\[ 2^{x-4}=1+3^2+\\dots+3^{z-1}.\\]\r\nIf $ x=4$ we get $ z=1$. Otherwise assume that $ x>4$ and then we get $ 0\\equiv \\frac{z+1}{2}\\pmod 2$. Thus $ z\\equiv -1\\pmod 4$. Then\r\n\\[ 2^{x-4}=(1+3^2)(1+3^4+\\dots+3^{z-3}),\\]\r\nwhich is impossible as $ 5\\not |\\,2^{x-4}$ but $ 5$ divides $ 1+3^2=10$. Hence the only solution is $ (x,y,z)=(4,3,1)$." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Prove that there doen't exist function which satisfying all following conditions:\r\n$ f(f(1))\\equal{}5$, $ f(f(2))\\equal{}6$, $ f(f(3))\\equal{}4$, $ f(f(4))\\equal{}3$, $ f(f(n))\\equal{}n\\plus{}2$ for every $ n\\in N, n\\ge 5$", "Solution_1": "Let $ a\\equal{}f(3),b\\equal{}f(4)$. It is easy to chek, that $ a\\not \\equal{}3,4, \\ b\\not \\equal{}3,4, \\ a\\not \\equal{}b$ and $ f(f(a))\\equal{}b,f(f(b))\\equal{}a$, therefore $ a<5,b<5$.\r\nIt give $ a\\equal{}1,b\\equal{}2$ or $ a\\equal{}2,b\\equal{}1$ - contradition." } { "Tag": [ "inequalities", "geometry", "inequalities proposed" ], "Problem": "H is the orthocenter of a triangle ABC. Prove that\r\nAH*BH*CH <=(8*ha*hb*hc)/27\r\n\r\nHAPPY NEW YEAR!", "Solution_1": "It's easy to see (area computation) that HA/ha+HB/hb+HC/hc=2. We just apply AM-GM to this: 2/3>=[HA*HB*HC/(ha*hb*hc)] 1/3 iff 8/27 >= HA*HB*HC/(ha*hb*hc) iff HA*HB*HC<=(8*ha*hb*hc)/27.\r\n\r\nIt's obvious, for the same reasons, that if AD, BE, CF are cevians intersecting at X (D on BC, etc.), then XA*XB*XC<=(8*AD*BE*CF)/27, which is more general." } { "Tag": [ "trigonometry", "algebra", "polynomial", "quadratics", "calculus", "Kettering MO", "geometry" ], "Problem": "Today was the 5th Kettering Olympiad - and here are the problems, which are very good intermediate problems.\r\n\r\n1. Find all real $x$ so that $(1+x^2)(1+x^4)=4x^3$\r\n\r\n2. Mark and John play a game. They have $100$ pebbles on a table. They take turns taking at least one at at most eight pebbles away. The person to claim the last pebble wins. Mark goes first. Can you find a way for Mark to always win? What about John?\r\n\r\n3. Prove that\r\n\r\n$\\sin x + \\sin 3x + \\sin 5x + ... + \\sin 11 x = (1-\\cos 12 x)/(2 \\sin x)$\r\n\r\n4. Mark has $7$ pieces of paper. He takes some of them and splits each into $7$ pieces of paper. He repeats this process some number of times. He then tells John he has $2000$ pieces of paper. John tells him he is wrong. Why is John right?\r\n\r\n5. In a triangle $ABC$, the altitude, angle bisector, and median split angle $A$ into four equal angles. Find the angles of $ABC.$\r\n\r\n6. There are $100$ cities. There exist airlines connecting pairs of cities. \r\n\r\na) Find the minimal number of airlines such that with at most $k$ plane changes, one can go from any city to any other city.\r\n\r\nb) Given that there are $4852$ airlines, show that, given any schematic, one can go from any city to any other city.", "Solution_1": "1.\r\n[hide]$x=1$[/hide]", "Solution_2": "stop being so n00b fred, we all know it was Nick and John \r\n\r\nnah, j/k, it looks right to me besides that\r\n\r\nhopefully i didnt screw up too much", "Solution_3": "amirhtlusa, solutions are required.", "Solution_4": "[quote=\"amirhtlusa\"]1.\n[hide]$x=1$[/hide][/quote]How can you prove that is the only solution?", "Solution_5": "On 6.b., can there be several airlines connecting a pair of cities?\r\n[hide=\"6b\"]Since there are $100$ cities, and airlines connect $2$ of them, if there are $\\binom{100}{2}=\\frac{100\\times99}{2}=4950$ airlines, then all cities will be connected. So since there are $4952$ airlines, all cities must be connected, and one can travel from any city to another.[/hide]", "Solution_6": "On 6b) at most one airline connects each pair of cities. But check your work, eryaman.", "Solution_7": "[quote=\"tetrahedr0n\"]amirhtlusa, solutions are required.[/quote]\r\nisn't that obvious??? ;) \r\nhint\r\n[hide] rewrite the whole thing like this:\n$(1+x^2)((1+x^2)^2-2x^2)=4x^3$\n$(1+x^2)^3-(2x^2(1+x^2))=4x^3$\n$(1+x^2)^3-2x^2(1+x)^2=0$\nyou do the rest... :lol: [/hide]", "Solution_8": "Well, obvious or not, soutions are always required. \r\n\r\nBut can you explain the last step of your hint? Something just seems to misterously dissappear.", "Solution_9": "[quote=\"tetrahedr0n\"]Well, obvious or not, soutions are always required. \n\nBut can you explain the last step of your hint? Something just seems to misterously dissappear.[/quote]\r\nbring the $4x^3$ to the other side and then factor out a $2x^2$, you will end up with a $....2x^2(1+x^2+2x)$.\r\n :)", "Solution_10": "Ok, (your 3rd line confused me, it just had the $4x^3$ gone completely.) But write your complete solution amirhtlusa. Thats why I posted the problem - as a challenge for you guys to solve.", "Solution_11": "[quote=\"tetrahedr0n\"]On 6b) at most one airline connects each pair of cities. But check your work, eryaman.[/quote]Woopsies. I misread the question, thought it said 4952 instead of 4852.", "Solution_12": "[quote=\"tetrahedr0n\"]Ok, (your 3rd line confused me, it just had the $4x^3$ gone completely.) But write your complete solution amirhtlusa. Thats why I posted the problem - as a challenge for you guys to solve.[/quote]\r\n ;) \r\nhere is the third line:\r\n[hide]$(1+x^2)^3-(2x^2(1+x^2))=4x^3$\n$(1+x^2)^3-(2x^2(1+x^2))-4x^3=0$\n$(1+x^2)^3-2x^2(1+x^2+2x)=0$\n$(1+x^2)^3-2x^2(1+x)^2=0$[/hide]", "Solution_13": "I have a solution for problem 2.\r\n[hide=\"Problem 2\"]For Mark to win, at the end of the game, he must leave John with the last pebble. For there to be 1 pebble left for John, there must be 10 left when John draws pebbles. If John draws one, then Mark will draw 8, so there will be one left for John. If John draws 2, then mark will draw 7, so there will be one left for John, and so on until, if John draws 8, then Mark will draw 1, leaving the last pebble for John. So for there to be 10 left when John draws pebbles, John also must draw the 19, 28, 37, etc. pebbles. But this goes all the way up to 100, the total number, so there is no strategy for Mark to win. \n\nFor John to win, he must make Mark draw when there are 10 pebbles left. Going up to the pattern adding 9 each time again, Mark must draw when there are 91 pebbles. So if Mark draws 1 pebble, John draws 8, if Mark draws 2 pebbles, John draws 7, and so on until if Mark draws 8 pebbles, John draws 1. John must keep the numbers that Mark draws at 82, 73, 64, 55, 46, 37, 28, 19, 10, then 1 using the above method. [/hide]", "Solution_14": "[hide=\"Problem 4\"]\nHe starts with 7 sheets of paper. Each time he rips on into 7 pieces, the total number of pieces increases by 6, so he can never change the number of pieces he has in mod 6. Initially, he has 7 sheets, congruent to 1 mod 6, so he'll always have a number congruent to 1 mod 6. Because 2000 is congruent to 2 mod 6, he can never have 2000 sheets of paper.\n[/hide]", "Solution_15": "eryaman, for Mark to win, he has to take the last pebble, not vice-versa. \r\n\r\namirhtlusa, I'm with you so far. What's next?", "Solution_16": "[quote=\"tetrahedr0n\"]eryaman, for Mark to win, he has to take the last pebble, not vice-versa. [/quote]*needs new glasses* wow, second problem in a row I've misread. Just take my solution, and replace the word \"Mark\" with \"John, and \"John\" with \"Mark\".", "Solution_17": "[hide=\"Problem 1\"]\n$(1+x^2)(1+x^4)=4x^3$\nSince the left hand side must be positive, $x^3>0$.\n$1+x^2+x^4+x^6=4x^3$\n$\\frac{1}{x^3}+\\frac{1}{x}+x+x^3=4$\n$\\left(\\frac{1}{x^3}+x^3\\right)+\\left(\\frac{1}{x}+x\\right)=4$\nLet $y=\\frac{1}{x}+x$.\nThen $y^3=\\frac{1}{x^3}+\\frac{3}{x}+3x+x^3$.\n$(y^3-3y)+(y)=4$\n$y^3-2y-4=0$\n\nI don't know the method for depressed cubics, so must resort to guess and check. (Good thing $y=2$ gives a root)\n$(y-2)(y^2+2y+2)=0$\n$y-2$ then $y=2$\nOr $(y^2+2y+2)=0$ then $y=\\frac{-2\\pm\\sqrt{2^2-4(1)(2)}}{2(1)}$\nSince the latter produces imaginary roots, the only possibility is $y=2$.\n\n$2=\\frac{1}{x}+x$.\n$x^2-2x+1=(x-1)^2=0$\n\nSo $x=1$ is the only solution.\n[/hide]", "Solution_18": "I had essentially the same solution, except that I first I divided the given polynomial by $(x-1)^2$ then used the symmetric trick getting a quadratic in $x+1/x.$", "Solution_19": "Anyone use derivatives in 1? Find (1,4) is an inersection point; then show that 4x^3 increases/decreases more/less rapidly than (1+x^2)(1+x^4) on certain ntervals, proving that they never intersect again? I did that, and it took a long, long tme. \r\nI got 1, 2, 4, and 6a.\r\n\r\nOn 6a, I made a slight mistake in thinking that 1 change of flight meant only 1 flight, not 2. So I wrote for k=1, a= 4950 and for 1 2\u03c0, \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03bc\u03b5 \u03c4\u03b1 $ (7\\minus{}2\\pi) \\equal{} 0.7168...$ \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03b1 \u03ba.\u03c4.\u03bb. \u0391\u03bd \u0391 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03ba\u03b1\u03b9 \u0392 \u03c4\u03bf \u03c3\u03c5\u03bc\u03c0\u03bb\u03b7\u03c1\u03c9\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03c4\u03bf\u03c5, $ A \\cup B$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2. \u0391\u03bd $ A'$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03ae \u03c4\u03bf\u03c5 \u0391 \u03ba\u03b1\u03c4\u03ac \u03ad\u03bd\u03b1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf \u03c4\u03cc\u03c4\u03b5 $ A' \\cup B$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03bc\u03b5 1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03bd\u03b1 \u03bb\u03b5\u03af\u03c0\u03b5\u03b9. (\u0391\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03b3\u03c9\u03bd\u03af\u03b1 1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf.)", "Solution_9": "\u0394\u03b5\u03bd \u03c4\u03bf \u03c0\u03b9\u03ac\u03bd\u03c9 :blush: :maybe: \r\n\u03a0\u03c1\u03ce\u03c4\u03b7 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7:\r\n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03b1\u03bd \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ce\u03c2 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03b1\u03bd\u03ac \u03ad\u03bd\u03b1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03bb\u03bf \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03bb\u03cc\u03b3\u03c9 \u03c4\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03bf \u03c0 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c1\u03b7\u03c4\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03b4\u03b5 \u03b8\u03b1 \u03c3\u03c5\u03bc\u03c0\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c4\u03ad \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c6\u03bf\u03c1\u03ac;\r\n\u0394\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7: \r\n\u0391\u03bd \u03b7 \u03ad\u03bd\u03c9\u03c3\u03b7 \u03c4\u03c9\u03bd \u0391 \u03ba\u03b1\u03b9 \u0392 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2, \u03c0\u03bf\u03b9\u03bf\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03b9 2 \u03bd\u03ad\u03bf\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 \u03c0\u03bf\u03c5 \u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03af\u03b4\u03b9\u03bf \u03bc\u03ad\u03c4\u03c1\u03bf \u03bc\u03b5 \u03c4\u03bf\u03bd \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc;", "Solution_10": "Dem, \u03c0\u03bf\u03bb\u03cd \u03cc\u03bc\u03bf\u03c1\u03c6\u03b1. \u0393\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03c9 \u03c4\u03bf visualisation, \u03ba\u03ac\u03bd\u03c9 upload \u03bc\u03af\u03b1 \u03b5\u03b9\u03ba\u03cc\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03ce\u03c2 \"\u03bc\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9\" \u03b7 \u03bf\u03bc\u03ac\u03b4\u03b1 $ F_2$ \u03c0\u03bf\u03c5 \u03af\u03c3\u03c9\u03c2 \u03c6\u03b1\u03bd\u03b5\u03af \u03c7\u03c1\u03ae\u03c3\u03b9\u03bc\u03b7 \u03c3\u03b5 \u03cc\u03c3\u03bf\u03c5\u03c2 \u03b4\u03b5\u03bd \u03c4\u03b7\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03be\u03b1\u03bd\u03b1\u03c3\u03c5\u03bd\u03b1\u03bd\u03c4\u03ae\u03c3\u03b5\u03b9 (\u03c4\u03b1 $ x,y$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b1 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 $ a,b$ \u03c4\u03bf\u03c5 \u03c3\u03c7\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2).\r\n\r\nGeorge, \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ \\{e^{in}: n\\in\\mathbb{N}\\}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c0\u03c9\u03c2 \u03c4\u03bf \u03c0\u03b5\u03c1\u03b9\u03ad\u03b3\u03c1\u03b1\u03c8\u03b5 \u03bf \u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7\u03c2 \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03c7\u03c1\u03ae\u03c3\u03b9\u03bc\u03bf [i]\u03bc\u03cc\u03bd\u03bf[/i] \u03c3\u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03cc\u03c0\u03bf\u03c5 \u03b4\u03b5 \u03b4\u03b9\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bf \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03b1\u03bb\u03bb\u03ac \"\u03c7\u03ac\u03bd\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c7\u03c9\u03c1\u03af\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03c7\u03ac\u03bd\u03b5\u03b9\" (\u03b4\u03b7\u03bb. \u03c4\u03bf\u03c5 \u03b2\u03b3\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03bc\u03cc\u03bd\u03bf \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03ad\u03c2 \u03c4\u03c9\u03bd \u03ba\u03bf\u03bc\u03bc\u03b1\u03c4\u03b9\u03ce\u03bd \u03c3\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03c4\u03bf \u03c3\u03c0\u03ac\u03c3\u03b1\u03bc\u03b5). \u03a9\u03c2 \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03bf \u03bd\u03b1 \u03b3\u03af\u03bd\u03b5\u03b9 visualised \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03c5\u03ba\u03bd\u03cc \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf, \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03cc\u03c0\u03c9\u03c2 \u03bf\u03b9 \u03c1\u03b7\u03c4\u03bf\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c5\u03ba\u03bd\u03bf\u03af \u03c3\u03c4\u03bf $ \\mathbb{R}$.\r\n\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03c6\u03c4\u03b9\u03ac\u03be\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd \u03ad\u03bd\u03b1\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03b4\u03cd\u03bf \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5\u03c2, \u03b7 \u03b4\u03b9\u03b1\u03b4\u03b9\u03ba\u03b1\u03c3\u03af\u03b1 \u03c3\u03b5 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ac \u03bb\u03cc\u03b3\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c9\u03c2 \u03b5\u03be\u03ae\u03c2:\r\n\r\n0. \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ x$ \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf, \u03c6\u03c4\u03b9\u03ac\u03c7\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ M_x$ \u03c0\u03bf\u03c5 \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03b5\u03af\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c0\u03bf\u03c5 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03bf\u03c5\u03bd \u03c1\u03b7\u03c4\u03ae \u03b3\u03c9\u03bd\u03af\u03b1 \u03bc\u03b5 \u03c4\u03bf $ x$ (\u03b1\u03c2 \u03c0\u03bf\u03cd\u03bc\u03b5 \u03b3\u03c9\u03bd\u03af\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c1\u03b7\u03c4\u03cc \u03c0\u03bf\u03bb/\u03c3\u03b9\u03bf \u03c4\u03bf\u03c5 $ 2\\pi$, \u03b1\u03bd \u03ba\u03b1\u03b9 \u03b4\u03b5 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03b1 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03c3\u03cd\u03bc\u03b2\u03b1\u03c3\u03b7 \u03c0\u03c1\u03b9\u03bd). \u038c\u03c0\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c6\u03b1\u03bd\u03c4\u03b1\u03c3\u03c4\u03b5\u03af\u03c2, \u03ba\u03ac\u03b8\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c5\u03ba\u03bd\u03cc \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd, \u03b1\u03bd \u03b4\u03cd\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ x,y$ [i]\u03b4\u03b5\u03bd[/i] \u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03c5\u03bd \u03ba\u03b1\u03c4\u03ac \u03c1\u03b7\u03c4\u03ae \u03b3\u03c9\u03bd\u03af\u03b1, \u03c4\u03cc\u03c4\u03b5 \u03c4\u03b1 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 $ M_x, M_y$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03be\u03ad\u03bd\u03b1 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03bf\u03c5\u03c2.\r\n\r\n\u039c\u03b5 \u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c6\u03c4\u03b9\u03ac\u03be\u03b5\u03b9 \u03bc\u03af\u03b1 \u03c3\u03c5\u03bb\u03bb\u03bf\u03b3\u03ae \u03b1\u03c0\u03cc \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 (\u03c4\u03b1 $ M_x$), \u03cc\u03bb\u03b1 \u03be\u03ad\u03bd\u03b1 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03b7 \u03ad\u03bd\u03c9\u03c3\u03ae \u03c4\u03bf\u03c5\u03c2 \u03c6\u03c4\u03b9\u03ac\u03c7\u03bd\u03b5\u03b9 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf. \u038c\u03bc\u03bf\u03c1\u03c6\u03b1.\r\n\r\n1. \u03a4\u03ce\u03c1\u03b1, \u03b1\u03c0\u03cc \u03ba\u03ac\u03b8\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ M_x$ \u03b4\u03b9\u03ac\u03bb\u03b5\u03be\u03b5 \u03ad\u03bd\u03b1 (\u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03ad\u03bd\u03b1!) \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03ba\u03b1\u03b9 \u03b1\u03c2 \u03ba\u03b1\u03bb\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 $ A$ \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03ba\u03b1\u03c4'\u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf. \u0388\u03c4\u03c3\u03b9 \u03cc\u03c0\u03c9\u03c2 \u03c4\u03bf \u03c6\u03c4\u03b9\u03ac\u03be\u03b1\u03bc\u03b5, \u03c4\u03bf $ A$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03b5\u03b9 \u03c4\u03b7 \"\u03bc\u03b1\u03b3\u03b9\u03ba\u03ae\" \u03b9\u03b4\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03b6\u03b5\u03cd\u03b3\u03bf\u03c2 \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af\u03c9\u03bd \u03c4\u03bf\u03c5 \u03b4\u03b5 \u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9 \u03ba\u03b1\u03c4\u03ac \u03c1\u03b7\u03c4\u03ae \u03b3\u03c9\u03bd\u03af\u03b1 (!!!).\r\n\r\n2. \u03a3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2, \u03b1\u03bd \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03ad\u03c8\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf $ A$ \u03ba\u03b1\u03c4\u03ac \u03c1\u03b7\u03c4\u03ae \u03b3\u03c9\u03bd\u03af\u03b1 $ 2\\pi q, q\\in(0,1)\\cap\\mathbb{Q}$, \u03b8\u03b1 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c8\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b9\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ q\\cdot A$ \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03be\u03ad\u03bd\u03bf \u03bc\u03b5 \u03c4\u03bf $ A$.\r\n\r\n3. \u0391\u03c2 \u03b1\u03c0\u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03ce\u03c1\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c1\u03b7\u03c4\u03bf\u03cd\u03c2 \u03c3\u03c4\u03bf $ (0,1)$ \u03b5\u03ba\u03bc\u03b5\u03c4\u03b1\u03bb\u03bb\u03b5\u03c5\u03cc\u03bc\u03b5\u03bd\u03bf\u03b9 \u03c4\u03bf \u03b3\u03b5\u03b3\u03bf\u03bd\u03cc\u03c2 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf\u03b9 \u03c4\u03c9 \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2, \u03ba\u03b1\u03b9 \u03b1\u03c2 \u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 $ A_r = q_r\\cdot A$, \u03b4\u03b7\u03bb. \u03c4\u03bf $ A$ \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf \u03ba\u03b1\u03c4\u03ac \u03c1\u03b7\u03c4\u03ae \u03b3\u03c9\u03bd\u03af\u03b1 $ q_r$. \u0391\u03c5\u03c4\u03ac \u03c4\u03b1 \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1 \u03bc\u03b1\u03b6\u03af \u03bc\u03b5 \u03c4\u03bf $ A_0\\equiv A$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b1 (\u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1) \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03b9\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf. \u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2, \u03b7 \u03ad\u03bd\u03c9\u03c3\u03b7 $ A_0\\cup A_1 \\cup A_2\\cup\\ldots$ \u03c0\u03b1\u03c1\u03ac\u03b3\u03b5\u03b9 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf.\r\n\r\n4. \u03a0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b5 \u03c4\u03ce\u03c1\u03b1 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03ad\u03c8\u03bf\u03c5\u03bc\u03b5 \u03c0.\u03c7. \u03c4\u03bf $ A_r$ \u03ba\u03b1\u03c4\u03ac $ q_k - q_r$ \u03b8\u03b1 \u03c0\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf $ A_k$ (\u03b4\u03b7\u03bb. \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03ad\u03c8\u03b1\u03bc\u03b5 \u03c4\u03bf $ A$ \u03ba\u03b1\u03c4\u03ac $ q_r$ \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03c4\u03bf \u03b3\u03c5\u03c1\u03af\u03c3\u03b1\u03bc\u03b5 \u03c0\u03af\u03c3\u03c9 \u03ba\u03b1\u03c4\u03ac $ q_r - q_k$, \u03ad\u03c7\u03bf\u03bd\u03c4\u03b1\u03c2 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03ae \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03ae $ q_k$).\r\n\r\n5. \u03a0\u03ac\u03c1\u03b5 \u03c4\u03ce\u03c1\u03b1 \u03c4\u03b1 \u03bc\u03bf\u03bd\u03ac \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b7 \u03bc\u03af\u03b1 \u03bc\u03b5\u03c1\u03b9\u03ac \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03b6\u03c5\u03b3\u03ac \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b7.\r\n5\u03b1. \u03a0\u03b5\u03c1\u03af\u03c3\u03c4\u03c1\u03b5\u03c8\u03b5 \u03c4\u03bf $ A_2$ \u03ba\u03b1\u03c4\u03ac $ q_1 - q_2$, \u03c4\u03bf $ A_4$ \u03ba\u03b1\u03c4\u03ac $ q_2 - q_4$, \u03c4\u03bf $ A_6$ \u03ba\u03b1\u03c4\u03ac $ q_3 - q_6$ \u03ba.\u03bf.\u03ba. \u039c\u03b5 \u03b2\u03ac\u03c3\u03b7 \u03c4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03b8\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03be\u03b1\u03bd\u03b1\u03c0\u03ac\u03c1\u03b5\u03b9 [b]\u03cc\u03bb\u03b1 \u03c4\u03b1 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ac \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1[/b] $ A_1, A_2, A_3$. \u0388\u03c4\u03c3\u03b9 \u03c6\u03c4\u03b9\u03ac\u03c7\u03bd\u03b5\u03b9\u03c2 \u03c4\u03bf\u03bd \u03ad\u03bd\u03b1\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 [b]\u03bc\u03cc\u03bd\u03bf[/b] \u03c4\u03b1 \u03b6\u03c5\u03b3\u03ac \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1.\r\n5\u03b2. \u0395\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03b1\u03bd\u03ac\u03bb\u03bf\u03b3\u03b1, \u03ba\u03ac\u03bd\u03b5 \u03c4\u03b9\u03c2 \u03ba\u03b1\u03c4\u03ac\u03bb\u03bb\u03b7\u03bb\u03b5\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03ad\u03c2 \u03c3\u03c4\u03b1 \u03bc\u03bf\u03bd\u03ac \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03c6\u03c4\u03b9\u03ac\u03be\u03b5\u03b9 \u03ad\u03bd\u03b1\u03bd \u03ac\u03bb\u03bb\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf (!!!).\r\n\r\n(5*. \u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2, \u03b1\u03bd \u03c7\u03c9\u03c1\u03af\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf\u03cd\u03c2 \u03bc\u03b5 \u03b2\u03ac\u03c3\u03b7 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03bd \u03ac\u03bb\u03bb\u03bf\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc $ 3,4\\ldots$, \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 $ 3,4\\ldots$ \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5\u03c2)\r\n\r\nDoes this make better sense?\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_11": "[quote=\"gpapargi\"]\u0394\u03b5\u03bd \u03c4\u03bf \u03c0\u03b9\u03ac\u03bd\u03c9 :blush: :maybe: \n\u03a0\u03c1\u03ce\u03c4\u03b7 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7:\n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03b1\u03bd \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ce\u03c2 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03b1\u03bd\u03ac \u03ad\u03bd\u03b1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03bb\u03bf \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03bb\u03cc\u03b3\u03c9 \u03c4\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03bf \u03c0 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c1\u03b7\u03c4\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03b4\u03b5 \u03b8\u03b1 \u03c3\u03c5\u03bc\u03c0\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c4\u03ad \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c6\u03bf\u03c1\u03ac;\n\u0394\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7: \n\u0391\u03bd \u03b7 \u03ad\u03bd\u03c9\u03c3\u03b7 \u03c4\u03c9\u03bd \u0391 \u03ba\u03b1\u03b9 \u0392 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2, \u03c0\u03bf\u03b9\u03bf\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03b9 2 \u03bd\u03ad\u03bf\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 \u03c0\u03bf\u03c5 \u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03af\u03b4\u03b9\u03bf \u03bc\u03ad\u03c4\u03c1\u03bf \u03bc\u03b5 \u03c4\u03bf\u03bd \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc;[/quote]\r\n\r\n\u0391\u03bd \u03ba\u03b1\u03b9 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b5 \u03bf \u03a0\u03b1\u03bd\u03b1\u03b3\u03b9\u03ce\u03c4\u03b7\u03c2 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03bc\u03b9\u03b1 \u03b5\u03c0\u03b9\u03c3\u03ae\u03bc\u03b1\u03bd\u03c3\u03b7. \u0391\u03bd \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ce\u03c2 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b1\u03bd\u03ac \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b8\u03b1 \u03c3\u03c5\u03bc\u03c0\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c4\u03ad \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf 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(\u03a5\u03c0\u03bf\u03b8\u03ad\u03c4\u03c9 \u03c0\u03c9\u03c2 \u03b1\u03c1\u03c7\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03c1\u03ac\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03bf \u03ad\u03bd\u03b1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf.)", "Solution_12": "Thanks. \u0395\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd. \r\n\r\n\u03a9\u03c3\u03c4\u03cc\u03c3\u03bf \u03b7 \u03b4\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03bd\u03bf\u03bf\u03c4\u03c1\u03bf\u03c0\u03af\u03b1 \u03b8\u03b1 \u03bc\u03b5 \u03ad\u03c3\u03c4\u03b5\u03bb\u03bd\u03b5 \u03c0\u03c1\u03bf\u03c2 \u03c3\u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03b5\u03cd\u03b8\u03c5\u03bd\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bd\u03b1 \u03b4\u03c9 \u03c0\u03b9\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u00ab\u03ba\u03b1\u03ba\u03cc \u03b2\u03ae\u03bc\u03b1\u00bb. \u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03cc\u03c4\u03b1\u03bd 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\u03b2\u03ae\u03bc\u03b1\u03c4\u03b1. \u039d\u03b1 \u03ad\u03bd\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bb\u03ad\u03b3\u03b1\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03bf \u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7\r\n\r\n\r\n\r\n[quote=\"Demetres\"][quote=\"gpapargi\"]\u0394\u03b5\u03bd \u03c4\u03bf \u03c0\u03b9\u03ac\u03bd\u03c9 :blush: :maybe: \n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03b1\u03bd \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ce\u03c2 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03b1\u03bd\u03ac \u03ad\u03bd\u03b1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03bb\u03bf \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03bb\u03cc\u03b3\u03c9 \u03c4\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03bf 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\u03c0\u03bf\u03c4\u03ad \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c6\u03bf\u03c1\u03ac \u03b4\u03b9\u03cc\u03c4\u03b9 \u03bf \u03c0 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c1\u03b7\u03c4\u03bf\u03c2. \u0394\u03b5\u03bd \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03cc\u03bb\u03bf\u03bd \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf. \u03a0.\u03c7. \u03b4\u03b5\u03bd \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c0\u03bf\u03c5 \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af \u03c3\u03b5 0 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03b1. (\u03a5\u03c0\u03bf\u03b8\u03ad\u03c4\u03c9 \u03c0\u03c9\u03c2 \u03b1\u03c1\u03c7\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03c1\u03ac\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03bf \u03ad\u03bd\u03b1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf.)[/quote]\r\n\r\n\u0391\u03c2 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03be\u03b5\u03ba\u03b9\u03bd\u03ac\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03b4\u03b9\u03b1\u03b4\u03b9\u03ba\u03b1\u03c3\u03af\u03b1 \u03c0\u03bf\u03c5 \u03ba\u03ac\u03b8\u03b5 \u03b2\u03ae\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b1\u03c5\u03be\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 \u03ba\u03b1\u03c4\u03ac 1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf. \u039c\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c0\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae \u03b8\u03b1 \u03c0\u03b5\u03c1\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5;\r\n\u0395\u03b3\u03ce \u03bb\u03ad\u03c9 \u03cc\u03c7\u03b9 (\u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b5\u03bd\u03bd\u03bf\u03ce \u03c4\u03bf 0). \u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03b1\u03c0\u03cc \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c0\u03b5\u03c1\u03bd\u03ac\u03bc\u03b5 2 \u03c6\u03bf\u03c1\u03ad\u03c2, \u03b1\u03bb\u03bb\u03ac \u03ba\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03c4\u03b1 \u03b2\u03ae\u03bc\u03b1\u03c4\u03ac \u03bc\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b1. \u03a3\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03b1\u03bb\u03bb\u03ac \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1. \u038c\u03bc\u03c9\u03c2 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1. \u039f\u03c0\u03cc\u03c4\u03b5 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03bf\u03cd\u03c4\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc\u03c2. \u0391\u03bd \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03ae\u03c4\u03b1\u03bd \u03ac\u03c0\u03b5\u03b9\u03c1\u03ac \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03b8\u03b1 \u03ad\u03bb\u03b5\u03b3\u03b1 \u03cc\u03c4\u03b9 \u03c3\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf \u03b8\u03b1 \u03c0\u03b5\u03c1\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1. \u0391\u03bb\u03bb\u03ac \u03b4\u03b5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03bd\u03cc\u03b7\u03bc\u03b1 \u03bd\u03b1 \u03bc\u03b9\u03bb\u03ac\u03bc\u03b5 \u03b3\u03b9\u03b1 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf. \u03a0\u03c1\u03ce\u03c4\u03b1 \u03c6\u03c4\u03ac\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c4\u03bf \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03c3\u03c4\u03bf \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf. \u0391\u03c6\u03bf\u03cd \u03c3\u03c4\u03bf \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03b4\u03b5 \u03c6\u03c4\u03ac\u03bd\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c4\u03ad \u2026 \r\n\r\n\u039a\u03ac\u03c4\u03b9 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03bf \u03b4\u03b9\u03b1\u03b9\u03c3\u03b8\u03ac\u03bd\u03bf\u03bc\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u03b2\u03ae\u03bc\u03b1 1 \u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b1\u03c0\u03cc \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03b1. \u0391\u03bd \u03be\u03b5\u03ba\u03af\u03bd\u03b1\u03b3\u03b5\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2, \u03b4\u03b5 \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc\u03c2 \u03bf\u03cd\u03c4\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2. (\u0386\u03c0\u03b5\u03b9\u03c1\u03cc\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2 \u03b8\u03b1 \u03ad\u03c6\u03c4\u03b1\u03bd\u03b5 \u00ab\u03af\u03c3\u03b1 \u03af\u03c3\u03b1\u00bb \u03b3\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1.). \u039f\u03c0\u03cc\u03c4\u03b5 \u03b4\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03cc \u03bd\u03b1 \u03bc\u03b9\u03bb\u03ac\u03bc\u03b5 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c4\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf. \r\n\r\n\u0388\u03ba\u03b1\u03bd\u03b1 \u03bb\u03af\u03b3\u03bf brainstorming. \u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03b2\u03b3\u03ac\u03b6\u03bf\u03c5\u03bd \u03bd\u03cc\u03b7\u03bc\u03b1 \u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03bb\u03ad\u03c9.", "Solution_13": "[quote=\"gpapargi\"]Thanks. \u0395\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd. \n\n\u03a9\u03c3\u03c4\u03cc\u03c3\u03bf \u03b7 \u03b4\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03bd\u03bf\u03bf\u03c4\u03c1\u03bf\u03c0\u03af\u03b1 \u03b8\u03b1 \u03bc\u03b5 \u03ad\u03c3\u03c4\u03b5\u03bb\u03bd\u03b5 \u03c0\u03c1\u03bf\u03c2 \u03c3\u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03b5\u03cd\u03b8\u03c5\u03bd\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bd\u03b1 \u03b4\u03c9 \u03c0\u03b9\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u00ab\u03ba\u03b1\u03ba\u03cc \u03b2\u03ae\u03bc\u03b1\u00bb. \u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03cc\u03c4\u03b1\u03bd \u03b2\u03bb\u03ad\u03c0\u03b5\u03b9\u03c2 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03b1 \u03c0\u03bf\u03c5 \u03c0\u03b1\u03c1\u03b1\u03b2\u03b9\u03ac\u03b6\u03bf\u03c5\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u00ab\u03b1\u03c1\u03c7\u03ae \u03b4\u03b9\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7\u03c2\u00bb \u03b1\u03c5\u03c4\u03cc \u03b3\u03b9\u03b1 \u03bc\u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03c3\u03c4\u03b7 \u03b8\u03b5\u03bc\u03b5\u03bb\u03af\u03c9\u03c3\u03b7 \u03bc\u03b1\u03c2 \u03b4\u03b5\u03bd \u03c0\u03ac\u03b5\u03b9 \u03ba\u03b1\u03bb\u03ac.[/quote]\n\n\u039a\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ac \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 :)\n\n[quote]\u03a9\u03c3\u03c4\u03cc\u03c3\u03bf \u03ad\u03c7\u03c9 \u03b4\u03b9\u03ba\u03b1\u03af\u03c9\u03bc\u03b1 \u03bd\u03b1 \u03c1\u03c9\u03c4\u03ae\u03c3\u03c9 \u03b1\u03bd \u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03b6\u03c5\u03b3\u03af\u03b6\u03b5\u03b9 \u0392, \u03c0\u03cc\u03c3\u03bf \u03b6\u03c5\u03b3\u03af\u03b6\u03b5\u03b9 \u03c4\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03cd\u03b1\u03c3\u03bc\u03b1 \u03c3\u03c4\u03bf \u03c4\u03ad\u03bb\u03bf\u03c2. \u03a6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03c0\u03bf\u03c5 \u03ad\u03c1\u03c7\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03c0\u03bb\u03ae\u03c1\u03b7 \u03b1\u03bd\u03c4\u03af\u03b8\u03b5\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7 \u03c6\u03cd\u03c3\u03b7.[/quote]\n\n2B. \u03a0\u03bb\u03ae\u03c1\u03c9\u03c2 \u03b1\u03bd\u03c4\u03af\u03b8\u03b5\u03c4\u03bf \u03bc\u03b5 \u03c4\u03b7 \u03c6\u03cd\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03c4\u03b7 \u03b4\u03b9\u03b1\u03af\u03c3\u03b8\u03b7\u03c3\u03ae \u03bc\u03b1\u03c2 :)\n\n[quote]\u0395\u03af\u03bd\u03b1\u03b9 \u03bb\u03af\u03b3\u03bf \u03bf\u03bc\u03b9\u03c7\u03bb\u03ce\u03b4\u03b5\u03c2 (\u03b3\u03b9\u03b1 \u03bc\u03ad\u03bd\u03b1) \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c3\u03c4\u03bf \u03b2\u03ae\u03bc\u03b1 1 \u03c0\u03bf\u03c5 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\u03c4\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b1\u03c5\u03c4\u03cc\u03c2;[/quote]\r\n\r\nWell done. \u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03c1\u03b1, \u03c0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03c0\u03b5\u03c1\u03af\u03b5\u03c1\u03b3\u03bf \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03c3\u03c0\u03ac\u03b5\u03b9 \u03b7 \u03b4\u03b9\u03b1\u03af\u03c3\u03b8\u03b7\u03c3\u03ae \u03bc\u03b1\u03c2.\r\n\r\n\u0391\u03c0\u03cc \u03c6\u03c5\u03c3\u03b9\u03ba\u03ae \u03ac\u03c0\u03bf\u03c8\u03b7, \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5 \u03b8\u03b1 \u03b6\u03cd\u03b3\u03b9\u03b6\u03b5 $ 0$, \u03b4\u03b5 \u03b8\u03b1 \u03b6\u03cd\u03b3\u03b9\u03b6\u03b5 [b]\u03c4\u03af\u03c0\u03bf\u03c4\u03b1[/b]: \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03cc\u03c3\u03bf \u03c0\u03b1\u03b8\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03cc \u03c0\u03bf\u03c5 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b5\u03bd \u03bb\u03cc\u03b3\u03c9 \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf [b]\u03b1\u03c0\u03bb\u03ce\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bd\u03cc\u03b7\u03bc\u03b1 \u03bd\u03b1 \u03bc\u03b9\u03bb\u03ac\u03bc\u03b5 \u03b3\u03b9\u03b1 \u03c4\u03bf \"\u03b2\u03ac\u03c1\u03bf\u03c2\" \u03c4\u03bf\u03c5[/b] (!!!). \u039c\u03ac\u03bb\u03b9\u03c3\u03c4\u03b1, \u03b1\u03c5\u03c4\u03cc \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c6\u03b1\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03c0\u03bf\u03c5 \u03ba\u03b1\u03c4\u03b1\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03b9 \u03c4\u03bf \u03b5\u03bd \u03bb\u03cc\u03b3\u03c9 \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf: \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03bc\u03ad\u03c4\u03c1\u03bf (\u03b4\u03b7\u03bb. \u03ad\u03bd\u03b1\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03c1\u03ac\u03bc\u03b5 \u03c4\u03b7 \u03bc\u03ac\u03b6\u03b1 \u03ae \u03c4\u03bf\u03bd \u03cc\u03b3\u03ba\u03bf) \u03c3\u03c4\u03bf\u03bd $ \\mathbb{R}^n$ \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bc\u03ad\u03bd\u03b5\u03b9 \u03b1\u03bd\u03b1\u03bb\u03bb\u03bf\u03af\u03c9\u03c4\u03bf \u03b1\u03c0\u03cc \u03c3\u03c4\u03c1\u03bf\u03c6\u03ad\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03b1\u03c6\u03bf\u03c1\u03ad\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 [i]\u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5[/i] \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c4\u03bf\u03c5 $ \\mathbb{R}^n$. \u038c,\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03bd\u03b1 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\u03b1\u03af\u03c1\u03b5\u03c4\u03b1\u03b9: \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03b1\u03c3\u03c4\u03bf\u03cd\u03bd \u03c3\u03c4\u03b7 \u03c6\u03cd\u03c3\u03b7 (\u03cc\u03c7\u03b9 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03cc\u03c4\u03b9 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03b1\u03c3\u03c4\u03bf\u03cd\u03bd \u03bf\u03b9 \u03c1\u03b7\u03c4\u03bf\u03af \u03ae \u03ac\u03bb\u03bb\u03b1 \u03c0\u03c5\u03ba\u03bd\u03ac \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1, \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03ac \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03cc\u03c3\u03bf \u03c0\u03b1\u03b8\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03ad\u03c2 \u03b9\u03b4\u03b9\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2).\r\n\r\n\u03a9\u03c3\u03c4\u03cc\u03c3\u03bf, \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae\u03c2 \u03ac\u03c0\u03bf\u03c8\u03b7, \u03b7 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae \u03b1\u03c5\u03c4\u03bf\u03cd \u03c4\u03bf\u03c5 \u03c3\u03c5\u03bd\u03cc\u03bb\u03bf\u03c5 \u03b1\u03b3\u03b3\u03af\u03b6\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03c0\u03bf\u03bb\u03cd \u03bb\u03b5\u03c0\u03c4\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03b7\u03c2 \u03b8\u03b5\u03c9\u03c1\u03af\u03b1\u03c2 \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd. \u03a3\u03ba\u03ad\u03c8\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03b5\u03bc\u03b5\u03af\u03c2 \u03bb\u03ad\u03bc\u03b5 \"\u03b4\u03b9\u03ac\u03bb\u03b5\u03be\u03b5 \u03ad\u03bd\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b1\u03c0\u03cc \u03ba\u03ac\u03b8\u03b5 \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf $ M_x$\" \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b1\u03c6\u03ae\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b5\u03ba\u03b5\u03af (\u03bc\u03ac\u03bb\u03b9\u03c3\u03c4\u03b1, \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03b3\u03b9'\u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf \u03bb\u03cc\u03b3\u03bf \u03b4\u03b9\u03ac\u03bb\u03b5\u03be\u03b1 \u03c4\u03b7 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b7 \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7 ;) ). \u0388\u03bd\u03b1\u03c2 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03ae\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03be\u03b7\u03b3\u03b5\u03af\u03c2 \u03c4\u03b1 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c1\u03c9\u03c4\u03ae\u03c3\u03b5\u03b9: \"\u03bc\u03b5 \u03c0\u03bf\u03b9\u03cc\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03bd\u03b1 \u03b4\u03b9\u03b1\u03bb\u03ad\u03be\u03c9 \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1\"? \u0391\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7\u03bd \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7 \u03b4\u03b5\u03bd \u03c4\u03b7\u03bd \u03b1\u03c0\u03b1\u03bd\u03c4\u03ac\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03be\u03b5\u03c0\u03b5\u03c1\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c4\u03b1\u03c6\u03b5\u03cd\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b8\u03b5\u03c9\u03c1\u03af\u03b1\u03c2 \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd \u03c0\u03bf\u03c5 \u03bb\u03ad\u03b3\u03b5\u03c4\u03b1\u03b9 [url=http://en.wikipedia.org/wiki/Axiom_of_choice][b]\u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2[/b][/url] (\u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b8\u03b5\u03c9\u03c1\u03af\u03b1 \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd \u03ba\u03ac\u03c4\u03b9 \u03c3\u03b1\u03bd \u03c4\u03bf \u03b1\u03af\u03c4\u03b7\u03bc\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03ae\u03bb\u03c9\u03bd \u03b5\u03c5\u03b8\u03b5\u03b9\u03ce\u03bd \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03af\u03b1... \u03b1\u03c7, \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \"\u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ae\" \u03b1\u03be\u03b9\u03ce\u03bc\u03b1\u03c4\u03b1 :) ).\r\n\r\nLong story short, \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c0\u03c1\u03b1\u03ba\u03c4\u03b9\u03ba\u03ac \u03bb\u03ad\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03bc\u03af\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9, \u03b1\u03ba\u03cc\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9\u03c2 \u03bc\u03af\u03b1 \"\u03c3\u03c5\u03bd\u03c4\u03b1\u03b3\u03ae\" \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03ae\u03bd... \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b5 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03bf \u03bc\u03b5 \u03c4\u03b7\u03bd \"\u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ae\" \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03bc\u03af\u03b1 \u03c3\u03c5\u03bb\u03bb\u03bf\u03b3\u03ae \u03b1\u03c0\u03cc \u03bc\u03b7 \u03ba\u03b5\u03bd\u03ac \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1, \u03c4\u03cc\u03c4\u03b5 \u03c4\u03bf \u03ba\u03b1\u03c1\u03c4\u03b5\u03c3\u03b9\u03b1\u03bd\u03cc \u03c4\u03bf\u03c5\u03c2 \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03bc\u03b7 \u03ba\u03b5\u03bd\u03cc (\u03ba\u03b1\u03b9 \u03b1\u03c2 \u03bc\u03bf\u03c5 \u03c0\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03bf \u03c7\u03ad\u03c1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03ba\u03b1\u03c1\u03b4\u03b9\u03ac \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03c4\u03bf\u03c5 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 ;) ). \u0388\u03c4\u03c3\u03b9, \u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1, \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03cc\u03bd\u03c4\u03c9\u03c2 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03b7 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae \u03c4\u03c9\u03bd Banach-Tarski.\r\n\r\n(Triviapedia: \u03bc\u03b5\u03c1\u03b9\u03ba\u03ad\u03c2 \u03b4\u03b5\u03ba\u03b1\u03b5\u03c4\u03af\u03b5\u03c2 \u03bc\u03b5\u03c4\u03ac \u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf, \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03c7\u03c4\u03b7\u03ba\u03b5 \u03cc\u03c4\u03b9 \u03c7\u03c9\u03c1\u03af\u03c2 \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 [b]\u03b4\u03b5\u03bd[/b] \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 paradoxical decompositions. \u03a4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf \u03b4\u03b5\u03bd \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03b4\u03b5\u03c7\u03c4\u03bf\u03cd\u03bc\u03b5 \u03c4\u03bf \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03bf \u03b9\u03c3\u03c7\u03c5\u03c1\u03cc \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 dependent choice, \u03b1\u03bb\u03bb\u03ac \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03bb\u03af\u03b3\u03bf \u03c0\u03b9\u03bf \u03b9\u03c3\u03c7\u03c5\u03c1\u03cc ultrafilter lemma.)\r\n\r\n\u038c\u03c4\u03b1\u03bd \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03c3\u03c4\u03b7\u03ba\u03b5 \u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf, \u03bf\u03b9 Banach Tarski \u03b5\u03af\u03c7\u03b1\u03bd \u03b1\u03bd\u03b1\u03b3\u03bd\u03c9\u03c1\u03af\u03c3\u03b5\u03b9 \u03c4\u03b7 \u03c3\u03b7\u03bc\u03b1\u03c3\u03af\u03b1 \u03c4\u03bf\u03c5 \u03b1\u03be\u03b9\u03ce\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 \u03ba\u03b1\u03b9 \u03bb\u03ad\u03b3\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03b4\u03b7\u03bc\u03bf\u03c3\u03af\u03b5\u03c5\u03c3\u03b1\u03bd \u03c9\u03c2 \u03b5\u03c0\u03b9\u03c7\u03b5\u03af\u03c1\u03b7\u03bc\u03b1 \"\u03c0\u03bf\u03c5 \u03b1\u03be\u03af\u03b6\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03ad\u03be\u03bf\u03c5\u03bc\u03b5\" \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc. \u0391\u03bd\u03c4\u03af\u03b8\u03b5\u03c4\u03b1 \u03cc\u03bc\u03c9\u03c2, \u03bf\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf\u03b9 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03c4\u03bf \u03c5\u03c0\u03bf\u03b4\u03ad\u03c7\u03c4\u03b7\u03ba\u03b1\u03bd \u03c9\u03c2: \"wow, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03c4\u03bf axiom of choice? Cooool :coolspeak:\".\r\n\r\n\u0393\u03b5\u03bd\u03b9\u03ba\u03ac, \u03cc\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b5\u03b9\u03c2 \u03ac\u03c4\u03bf\u03c0\u03bf\u03bd \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2, \u03b5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03b9 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03b4\u03b5\u03bd \u03c4\u03bf \u03b1\u03c0\u03bf\u03b4\u03ad\u03c7\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7 \u03b4\u03bf\u03c5\u03bb\u03b5\u03b9\u03ac \u03c4\u03bf\u03c5\u03c2 \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc (\u03b1\u03bd \u03ba\u03b1\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03b3\u03af\u03bd\u03b5\u03b9 \u03c4\u03c1\u03bf\u03bc\u03b5\u03c1\u03ae \u03b4\u03bf\u03c5\u03bb\u03b5\u03b9\u03ac \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03bf\u03c5\u03bd \u03c4\u03b9 \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03ad\u03c3\u03bc\u03b1\u03c4\u03b1 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03bf\u03c5\u03bd \u03bc\u03b5 \u03ba\u03b1\u03b9 \u03c7\u03c9\u03c1\u03af\u03c2 \u03b1\u03c5\u03c4\u03cc). \u0391\u03c0\u03bb\u03ce\u03c2, \u03cc\u03c4\u03b1\u03bd \u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b1\u03bd\u03c4\u03b9\u03ba\u03cc \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 (\u03cc\u03c0\u03c9\u03c2 \u03c4\u03bf Hahn-Banach) \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03b1\u03c3\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03ae \u03c4\u03bf\u03bd \u03ac\u03bb\u03bb\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03c3\u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 (\u03c3\u03c5\u03bd\u03ae\u03b8\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b1 \u03c0\u03bf\u03bb\u03cd \"\u03b2\u03b1\u03b8\u03b9\u03ac\" \u03b8\u03b5\u03c9\u03c1\u03ae\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b2\u03b1\u03c3\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b5\u03ba\u03b5\u03af), \u03c8\u03ac\u03c7\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03bf\u03cd\u03bc\u03b5 \u03bc\u03ae\u03c0\u03c9\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b2\u03b1\u03b8\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc (\u03c0.\u03c7. \u03bc\u03ae\u03c0\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1).\r\n\r\n\u03a9\u03c3\u03c4\u03cc\u03c3\u03bf, \u03b1\u03be\u03af\u03b6\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03bf\u03bd\u03b9\u03c3\u03c4\u03b5\u03af \u03c4\u03bf \u03b5\u03be\u03ae\u03c2: \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 more or less \u03ad\u03bd\u03b1 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc rumination \u03c4\u03c9\u03bd \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ce\u03bd. \u03a3\u03c4\u03b7 \u03c6\u03cd\u03c3\u03b7, \u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03c3\u03c0\u03ac\u03bd\u03b5 \u03c0\u03bf\u03bb\u03cd \u03c0\u03c1\u03b9\u03bd \u03c6\u03c4\u03ac\u03c3\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b1\u03bd\u03b1\u03c1\u03c9\u03c4\u03b9\u03ad\u03c4\u03b1\u03b9 \u03b1\u03bd \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b5\u03af \u03c6\u03b9\u03bb\u03bf\u03c3\u03bf\u03c6\u03b9\u03ba\u03ac \u03bc\u03b5 \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03ae \u03cc\u03c7\u03b9. \u03a0.\u03c7. \u03c3\u03c4\u03bf \u03c3\u03c0\u03ac\u03c3\u03b9\u03bc\u03bf \u03c4\u03bf\u03c5 fractal \u03c0\u03bf\u03c5 \u03c0\u03c1\u03cc\u03c4\u03b5\u03b9\u03bd\u03b5 \u03bf Dem (\u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03c6\u03c5\u03c3\u03b9\u03ba\u03ac\" \u03b1\u03b4\u03cd\u03bd\u03b1\u03c4\u03bf), \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2, \u03b1\u03bb\u03bb\u03ac \u03c4\u03bf \u03cc\u03c4\u03b9 \u03c4\u03bf \"\u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\" \u03ba\u03b1\u03b9 \u03c4\u03bf \"\u03b1\u03c0\u03b5\u03b9\u03c1\u03bf\u03c3\u03c4\u03cc\" \u03b1\u03c0\u03bb\u03ce\u03c2 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c3\u03c4\u03b7 \u03c6\u03cd\u03c3\u03b7.\r\n\r\nHope this helps :) Cheerio,\r\n\r\nDurandal 1707", "Solution_14": "\u0391\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b5 \u03ae\u03b4\u03b7 \u03ba\u03b1\u03b9 \u03bf \u03a0\u03b1\u03bd\u03b1\u03b3\u03b9\u03ce\u03c4\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03cc\u03c4\u03b9 \u03b2\u03bb\u03ad\u03c0\u03ce \u03bb\u03ad\u03bc\u03b5 \u03c0\u03b1\u03c1\u03cc\u03bc\u03bf\u03b9\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03b1\u03bb\u03bb\u03ac \u03c4\u03bf \u03b1\u03c6\u03ae\u03bd\u03c9.\r\n\r\n[quote=\"gpapargi\"]\n[quote=\"Demetres\"][quote=\"gpapargi\"]\u0394\u03b5\u03bd \u03c4\u03bf \u03c0\u03b9\u03ac\u03bd\u03c9 :blush: :maybe: \n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03b1\u03bd \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ce\u03c2 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03b1\u03bd\u03ac \u03ad\u03bd\u03b1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03bb\u03bf \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03bb\u03cc\u03b3\u03c9 \u03c4\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03bf \u03c0 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c1\u03b7\u03c4\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03b4\u03b5 \u03b8\u03b1 \u03c3\u03c5\u03bc\u03c0\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c4\u03ad \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c6\u03bf\u03c1\u03ac;\n[/quote]\n\n\u0391\u03bd \u03ba\u03b1\u03b9 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b5 \u03bf \u03a0\u03b1\u03bd\u03b1\u03b3\u03b9\u03ce\u03c4\u03b7\u03c2 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03bc\u03b9\u03b1 \u03b5\u03c0\u03b9\u03c3\u03ae\u03bc\u03b1\u03bd\u03c3\u03b7. \u0391\u03bd \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ce\u03c2 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b1\u03bd\u03ac \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b8\u03b1 \u03c3\u03c5\u03bc\u03c0\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c4\u03ad \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c6\u03bf\u03c1\u03ac \u03b4\u03b9\u03cc\u03c4\u03b9 \u03bf \u03c0 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c1\u03b7\u03c4\u03bf\u03c2. \u0394\u03b5\u03bd \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03cc\u03bb\u03bf\u03bd \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf. \u03a0.\u03c7. \u03b4\u03b5\u03bd \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c0\u03bf\u03c5 \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af \u03c3\u03b5 0 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03b1. (\u03a5\u03c0\u03bf\u03b8\u03ad\u03c4\u03c9 \u03c0\u03c9\u03c2 \u03b1\u03c1\u03c7\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03c1\u03ac\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03bf \u03ad\u03bd\u03b1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf.)[/quote]\n\n\u0391\u03c2 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03be\u03b5\u03ba\u03b9\u03bd\u03ac\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03b4\u03b9\u03b1\u03b4\u03b9\u03ba\u03b1\u03c3\u03af\u03b1 \u03c0\u03bf\u03c5 \u03ba\u03ac\u03b8\u03b5 \u03b2\u03ae\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b1\u03c5\u03be\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 \u03ba\u03b1\u03c4\u03ac 1 \u03b1\u03ba\u03c4\u03af\u03bd\u03b9\u03bf. \u039c\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c0\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae \u03b8\u03b1 \u03c0\u03b5\u03c1\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5;\n\u0395\u03b3\u03ce \u03bb\u03ad\u03c9 \u03cc\u03c7\u03b9 (\u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b5\u03bd\u03bd\u03bf\u03ce \u03c4\u03bf 0). \u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03b1\u03c0\u03cc \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c0\u03b5\u03c1\u03bd\u03ac\u03bc\u03b5 2 \u03c6\u03bf\u03c1\u03ad\u03c2, \u03b1\u03bb\u03bb\u03ac \u03ba\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03c4\u03b1 \u03b2\u03ae\u03bc\u03b1\u03c4\u03ac \u03bc\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b1. \u03a3\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03b1\u03bb\u03bb\u03ac \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1. \u038c\u03bc\u03c9\u03c2 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1. \n[/quote]\n\n\u0393\u03b9\u03ce\u03c1\u03b3\u03bf, \u03b5\u03b4\u03ce \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1 \u03bb\u03ad\u03bc\u03b5. \u0394\u03b5\u03bd \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03bb\u03bf\u03bd \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf. 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\u03c0\u03c1\u03bf\u03c4\u03b9\u03bc\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03bf\u03c5\u03bd \u03bc\u03cc\u03bd\u03bf \u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc. \u03a5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03ac\u03bb\u03bb\u03bf\u03b9 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c4\u03bf \u03b4\u03ad\u03c7\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03cd\u03c4\u03b5 \u03b1\u03c5\u03c4\u03cc. \u03a3\u03b5 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03c0\u03b9\u03bf \u03b1\u03ba\u03c1\u03b1\u03af\u03b1 \u03ad\u03ba\u03b4\u03bf\u03c3\u03b7 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03ac\u03bb\u03bb\u03bf\u03b9 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b4\u03ad\u03c7\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c4\u03b7\u03bd \u03cd\u03c0\u03b1\u03c1\u03be\u03b7 \u03ac\u03c0\u03b5\u03b9\u03c1\u03c9\u03bd \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd.", "Solution_15": "[quote=\"Durandal\"]\u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c0\u03c1\u03b1\u03ba\u03c4\u03b9\u03ba\u03ac \u03bb\u03ad\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03bc\u03af\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9, \u03b1\u03ba\u03cc\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9\u03c2 \u03bc\u03af\u03b1 \"\u03c3\u03c5\u03bd\u03c4\u03b1\u03b3\u03ae\" \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03ae\u03bd... \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b5 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03bf \u03bc\u03b5 \u03c4\u03b7\u03bd \"\u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ae\" \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03bc\u03af\u03b1 \u03c3\u03c5\u03bb\u03bb\u03bf\u03b3\u03ae \u03b1\u03c0\u03cc \u03bc\u03b7 \u03ba\u03b5\u03bd\u03ac \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1, \u03c4\u03cc\u03c4\u03b5 \u03c4\u03bf \u03ba\u03b1\u03c1\u03c4\u03b5\u03c3\u03b9\u03b1\u03bd\u03cc \u03c4\u03bf\u03c5\u03c2 \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03bc\u03b7 \u03ba\u03b5\u03bd\u03cc (\u03ba\u03b1\u03b9 \u03b1\u03c2 \u03bc\u03bf\u03c5 \u03c0\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03bf \u03c7\u03ad\u03c1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03ba\u03b1\u03c1\u03b4\u03b9\u03ac \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03c4\u03bf\u03c5 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 ;) ).[/quote]\n\n\u0395\u03b4\u03ce \u03b3\u03b9\u03b1 \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd \u03b8\u03b1 \u03ad\u03bb\u03b5\u03b3\u03b1 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c0\u03b9\u03ba\u03af\u03bd\u03b4\u03c5\u03bd\u03bf \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03ba\u03ac\u03c4\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03bb\u03cc\u03b3\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03b5\u03af\u03c0\u03b1 \u03c0\u03c1\u03b9\u03bd (\u03b1\u03bd \u03ba\u03b1\u03b9... \u03b1\u03bd \u03b4\u03b5\u03bd \u03ae\u03c4\u03b1\u03bd \u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf \u03c3\u03c4\u03b7 \u03bc\u03ad\u03c3\u03b7 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ae\u03c1\u03c7\u03b5 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03bd\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03ce \u03c0\u03bf\u03bd\u03b7\u03c1\u03ac. \u0397 \u03c3\u03ba\u03ad\u03c8\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03ba \u03c4\u03c9\u03bd \u03c5\u03c3\u03c4\u03ad\u03c1\u03c9\u03bd.). \u0393\u03b9\u03b1 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b5\u03c2 (\u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b5\u03c2) \u03b5\u03bd\u03ad\u03c1\u03b3\u03b5\u03b9\u03b5\u03c2 \u03bc\u03cc\u03bb\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03c3\u03bf\u03c5 \u03c6\u03c4\u03ac\u03bd\u03b5\u03b9 \u03bf \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2. \u0393\u03b9\u03b1 \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b5\u03c2 \u03b5\u03bd\u03ad\u03c1\u03b3\u03b5\u03b9\u03b5\u03c2 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc\u03c2. \u0386\u03c1\u03b1 \u03b4\u03b5 \u03b3\u03af\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03cd\u03c4\u03b5 \u03bf\u03c1\u03b9\u03b1\u03ba\u03ac.\n\n[quote=\"Durandal\"]\n(Triviapedia: \u03bc\u03b5\u03c1\u03b9\u03ba\u03ad\u03c2 \u03b4\u03b5\u03ba\u03b1\u03b5\u03c4\u03af\u03b5\u03c2 \u03bc\u03b5\u03c4\u03ac \u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf, \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03c7\u03c4\u03b7\u03ba\u03b5 \u03cc\u03c4\u03b9 \u03c7\u03c9\u03c1\u03af\u03c2 \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 [b]\u03b4\u03b5\u03bd[/b] \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 paradoxical decompositions. \u03a4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf \u03b4\u03b5\u03bd \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03b4\u03b5\u03c7\u03c4\u03bf\u03cd\u03bc\u03b5 \u03c4\u03bf \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03bf \u03b9\u03c3\u03c7\u03c5\u03c1\u03cc \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 dependent choice, \u03b1\u03bb\u03bb\u03ac \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03bb\u03af\u03b3\u03bf \u03c0\u03b9\u03bf \u03b9\u03c3\u03c7\u03c5\u03c1\u03cc ultrafilter lemma.)\n\n\u038c\u03c4\u03b1\u03bd \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03c3\u03c4\u03b7\u03ba\u03b5 \u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf, \u03bf\u03b9 Banach Tarski \u03b5\u03af\u03c7\u03b1\u03bd \u03b1\u03bd\u03b1\u03b3\u03bd\u03c9\u03c1\u03af\u03c3\u03b5\u03b9 \u03c4\u03b7 \u03c3\u03b7\u03bc\u03b1\u03c3\u03af\u03b1 \u03c4\u03bf\u03c5 \u03b1\u03be\u03b9\u03ce\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 \u03ba\u03b1\u03b9 \u03bb\u03ad\u03b3\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03b4\u03b7\u03bc\u03bf\u03c3\u03af\u03b5\u03c5\u03c3\u03b1\u03bd \u03c9\u03c2 \u03b5\u03c0\u03b9\u03c7\u03b5\u03af\u03c1\u03b7\u03bc\u03b1 \"\u03c0\u03bf\u03c5 \u03b1\u03be\u03af\u03b6\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03ad\u03be\u03bf\u03c5\u03bc\u03b5\" \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc. \u0391\u03bd\u03c4\u03af\u03b8\u03b5\u03c4\u03b1 \u03cc\u03bc\u03c9\u03c2, \u03bf\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf\u03b9 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03c4\u03bf \u03c5\u03c0\u03bf\u03b4\u03ad\u03c7\u03c4\u03b7\u03ba\u03b1\u03bd \u03c9\u03c2: \"wow, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03c4\u03bf axiom of choice? Cooool :coolspeak:\".\n\n[/quote]\r\n\r\n\u039d\u03b1 \u03b5\u03b4\u03ce \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1, \u03b4\u03b5 \u03b8\u03b1 \u03ad\u03c0\u03c1\u03b5\u03c0\u03b5 \u03bd\u03b1 \u03c0\u03bf\u03c5\u03bd \u03bf\u03b9 \u03ac\u03bd\u03b8\u03c1\u03c9\u03c0\u03bf\u03b9 \u00ab\u03b8\u03b1 \u03ba\u03c1\u03b1\u03c4\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf dependence choice\u00bb \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03cc\u03bb\u03bf\u03b9 \u03c7\u03b1\u03c1\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf\u03b9; \u038c\u03bb\u03bf\u03c5\u03c2 \u03bc\u03b1\u03c2 \u03b3\u03bf\u03b7\u03c4\u03b5\u03cd\u03bf\u03c5\u03bd \u03c4\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03b1, \u03b1\u03bb\u03bb\u03ac \u03b7 \u03b3\u03bd\u03ce\u03bc\u03b7 \u03bc\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03be\u03b9\u03ce\u03bc\u03b1\u03c4\u03b1 \u03b8\u03ad\u03c4\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b7 \u03c6\u03cd\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03b5\u03bc\u03b5\u03af\u03c2 \u03b1\u03c0\u03bb\u03ac \u03c4\u03b1 \u03b1\u03bd\u03b1\u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03c5\u03bc\u03b5. \u0395\u03af\u03bd\u03b1\u03b9 \u03c9\u03c1\u03b1\u03af\u03b1 \u03c4\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03b1 \u03c0\u03bf\u03c5 \u03cc\u03bd\u03c4\u03c9\u03c2 \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03b5\u03ba\u03b5\u03af \u03ad\u03be\u03c9 \u03b1\u03bb\u03bb\u03ac \u03b7 \u03ba\u03b1\u03b8\u03b7\u03bc\u03b5\u03c1\u03b9\u03bd\u03ae \u03bc\u03b1\u03c2 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae/\u03b4\u03b9\u03b1\u03af\u03c3\u03b8\u03b7\u03c3\u03b7 \u03b4\u03b5 \u03bc\u03b1\u03c2 \u03b1\u03c6\u03ae\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03bf\u03c5\u03bc\u03b5 (\u03c0\u03c7 \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03b1 \u03c4\u03b7\u03c2 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03ae \u03c4\u03b7\u03c2 \u03ba\u03b2\u03b1\u03bd\u03c4\u03bf\u03bc\u03b7\u03c7\u03b1\u03bd\u03b9\u03ba\u03ae\u03c2). \u0394\u03b5 \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03ce \u03bc\u03b5 \u03c4\u03b7 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 \u03bd\u03b1 \u03ba\u03c1\u03b1\u03c4\u03ac\u03bc\u03b5 \u03b1\u03be\u03b9\u03ce\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03c4\u03bf \u03b5\u03c0\u03b9\u03c7\u03b5\u03af\u03c1\u03b7\u03bc\u03b1 \u03cc\u03c4\u03b9 \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03b5\u03bd\u03c4\u03c5\u03c0\u03c9\u03c3\u03b9\u03b1\u03ba\u03ac \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03b1\u03bd \u03c4\u03b1 \u03b4\u03b5\u03c7\u03c4\u03b5\u03af\u03c2.", "Solution_16": "[quote=\"gpapargi\"]\u039d\u03b1 \u03b5\u03b4\u03ce \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1, \u03b4\u03b5 \u03b8\u03b1 \u03ad\u03c0\u03c1\u03b5\u03c0\u03b5 \u03bd\u03b1 \u03c0\u03bf\u03c5\u03bd \u03bf\u03b9 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\u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf \u03c4\u03bf \"\u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\", \u03c0\u03b1\u03c1\u03ac \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1.\r\n\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf\u03bd\u03af\u03c3\u03c9 \u03bb\u03af\u03b3\u03bf \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf, \u03bf \u03c1\u03cc\u03bb\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 \u03c3\u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03bf\u03be\u03bf Banach-Tarski \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 \u03c4\u03b1 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03bc\u03b7 \u03bc\u03b5\u03c4\u03c1\u03ae\u03c3\u03b9\u03bc\u03b1\". \u0391\u03c5\u03c4\u03cc \u03b1\u03c0\u03cc \u03bc\u03cc\u03bd\u03bf \u03c4\u03bf\u03c5 \u03b4\u03b5 \u03bc\u03bf\u03c5 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c4\u03cc\u03c3\u03bf \u03b1\u03bd\u03c4\u03b9\u03b4\u03b9\u03b1\u03b9\u03c3\u03b8\u03b7\u03c4\u03b9\u03ba\u03cc: \u03bf\u03b9 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03cc\u03c3\u03bf (\u03bc\u03b1 \u03c4\u03cc\u03c3\u03bf) \u03c0\u03bf\u03bb\u03bb\u03bf\u03af, \u03c0\u03bf\u03c5 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03c6\u03c5\u03c3\u03b9\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03cc \u03bc\u03bf\u03c5 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03bb\u03ad\u03be\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1 \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03cc \u03c4\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03bc\u03b5\u03c4\u03c1\u03ae\u03c3\u03b5\u03b9\u03c2/\u03b6\u03c5\u03b3\u03af\u03c3\u03b5\u03b9\u03c2.\r\n\r\n\u0391\u03bd\u03c4\u03af\u03b8\u03b5\u03c4\u03b1, \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03be\u03b5\u03bd\u03bf\u03b4\u03bf\u03c7\u03b5\u03af\u03bf \u03c4\u03bf\u03c5 Hilbert \u03c0\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c4\u03c1\u03ad\u03c0\u03b5\u03b9 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03c4\u03bf\u03bd \"\u03b4\u03b9\u03c0\u03bb\u03b1\u03c3\u03b9\u03b1\u03c3\u03bc\u03cc\" \u03c4\u03c9\u03bd \u03cc\u03b3\u03ba\u03c9\u03bd \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03bf\u03c6\u03b5\u03af\u03bb\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03b9\u03b4\u03b9\u03b1\u03b9\u03c4\u03b5\u03c1\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf\u03bd \u03b1\u03bd\u03c4\u03b9\u03b4\u03b9\u03b1\u03b9\u03c3\u03b8\u03b7\u03c4\u03b9\u03ba\u03cc \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03ae\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03b1\u03c0\u03b5\u03af\u03c1\u03bf\u03c5, \u03cc\u03c7\u03b9 \u03c3\u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2.\r\n\r\nCheerio,\r\n\r\nDurandal 1707\r\n\r\nPS: BTW, Dem \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03b8\u03b1 \u03c7\u03c1\u03b5\u03b9\u03b1\u03c3\u03c4\u03b5\u03af \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 bump \u03c4\u03b7\u03bd $ F_2$ \u03b3\u03b9\u03b1\u03c4\u03af \u03c4\u03b7 \u03b2\u03bb\u03ad\u03c0\u03c9 \u03bd\u03b1 \u03c7\u03ac\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03cc\u03c3\u03b1 posts \u03ba\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03c1\u03af\u03bc\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bc\u03bf\u03c1\u03c6\u03bf \u03c6\u03b1\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf :)", "Solution_17": "\u038c\u03bd\u03c4\u03c9\u03c2 \u03c4\u03bf \u03be\u03b5\u03bd\u03bf\u03b4\u03bf\u03c7\u03b5\u03af\u03bf \u03c4\u03bf\u03c5 Hilbert \u03ad\u03ba\u03b1\u03bd\u03b5 \u03c4\u03bf \u03b4\u03b9\u03c0\u03bb\u03b1\u03c3\u03b9\u03b1\u03c3\u03bc\u03cc \u03c9\u03c3\u03c4\u03cc\u03c3\u03bf \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5 \u03bc\u03b5 \u03c7\u03ac\u03bb\u03b1\u03c3\u03b5 (\u03bf\u03cd\u03c4\u03b5 \u03c4\u03ce\u03c1\u03b1 \u03bf\u03cd\u03c4\u03b5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03c6\u03bf\u03c1\u03ac \u03c0\u03bf\u03c5 \u03c4\u03bf \u03b5\u03af\u03c7\u03b1 \u03b4\u03b5\u03b9 \u03c3\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03bc\u03b5 \u03c3\u03c0\u03b1\u03b6\u03bf\u03ba\u03b5\u03c6\u03b1\u03bb\u03b9\u03ad\u03c2). \u0391\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03b5\u03af\u03c0\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03c3\u03b1\u03bd 2 * \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf = \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf. \u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03b1\u03bd \u03c5\u03c0\u03ae\u03c1\u03c7\u03b5 \u03bc\u03ac\u03b6\u03b1 \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03c0\u03ac\u03bb\u03b9 \u03b4\u03b5 \u03b8\u03b1 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03bf\u03cd\u03c3\u03b5 \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1. \u0397 \u03bc\u03ac\u03b6\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03c4\u03b1\u03bd\u03b5\u03bc\u03b7\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c4\u03b1 \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03c4\u03b1 \u03b1\u03c0\u03b5\u03af\u03c1\u03c9\u03c2 \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03bd\u03cc\u03c2 \u03c3\u03c5\u03bd\u03cc\u03bb\u03bf\u03c5 \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03bc\u03ac\u03b6\u03b1. \r\n\u0395\u03bc\u03ad\u03bd\u03b1 \u03bc\u03b5 \u03c7\u03ac\u03bb\u03b1\u03c3\u03b5 \u03bf \u03c7\u03b5\u03b9\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c4\u03c9\u03bd \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03c9\u03bd, \u03b3\u03b9\u03b1 \u03ad\u03bd\u03b1 \u03c0\u03bf\u03bb\u03cd \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf \u03bb\u03cc\u03b3\u03bf: \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c7\u03b5\u03b9\u03c1\u03b9\u03c3\u03bc\u03cc \u03b4\u03b5\u03bd \u03b1\u03c1\u03ba\u03bf\u03cd\u03bd \u03bf\u03cd\u03c4\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b5\u03c2 \u03b5\u03bd\u03ad\u03c1\u03b3\u03b5\u03b9\u03b5\u03c2. \u038c\u03c4\u03b1\u03bd \u03bf \u03ac\u03bd\u03b8\u03c1\u03c9\u03c0\u03bf\u03c2 \u03ae \u03bf \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03ae\u03c2 \u03bb\u03ad\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b8\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03b5\u03bd\u03ad\u03c1\u03b3\u03b5\u03b9\u03b5\u03c2 \u03b5\u03c0\u2019 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03bd, \u03b5\u03bd\u03bd\u03bf\u03bf\u03cd\u03bc\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b5\u03c2. \u0395\u03b4\u03ce \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b5\u03b9\u03c3\u03ac\u03b3\u03c9 \u03c3\u03b9\u03c9\u03c0\u03b7\u03bb\u03ac \u03ad\u03bd\u03b1 \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03ad\u03c1\u03c7\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c6\u03c5\u03c3\u03b9\u03ba\u03ac \u03b1\u03af\u03c4\u03b9\u03b1. \u0394\u03b5\u03bd \u03c4\u03bf \u03ba\u03ac\u03bd\u03c9 \u03b1\u03c0\u03bb\u03ce\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03c1\u03c1\u03b9\u03c8\u03b7 \u03c4\u03bf\u03c5 \u03c0\u03b1\u03c1\u03b1\u03b4\u03cc\u03be\u03bf\u03c5. \u03a0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce \u03bd\u03b1 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03ac\u03c8\u03c9 \u03c4\u03b9 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03bc\u03bf\u03c5 \u03c6\u03c4\u03b1\u03af\u03b5\u03b9 \u03c3\u03c4\u03bf \u03b1\u03be\u03af\u03c9\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2. \r\n\r\n\u039a\u03b1\u03b9 \u03c4\u03bf \u03b2\u03ac\u03c1\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c6\u03c4\u03b9\u03b1\u03c7\u03c4\u03b5\u03af \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ae \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03bc\u03b5 \u03bc\u03ae\u03ba\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ae \u03c0\u03c5\u03ba\u03bd\u03cc\u03c4\u03b7\u03c4\u03b1, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c4\u03bf \u03bc\u03b7 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c3\u03c5\u03bd\u03b5\u03c7\u03ad\u03c2. \u0393\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b5\u03c3\u03c4\u03af\u03b1\u03c3\u03b1 \u03ba\u03b1\u03c4\u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03bd \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9 \u03c3\u03c4\u03bf \u03be\u03b5\u03bd\u03bf\u03b4\u03bf\u03c7\u03b5\u03af\u03bf. \r\n\r\nThanks \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac. \u03a3\u03ae\u03bc\u03b5\u03c1\u03b1 \u03ad\u03bc\u03b1\u03b8\u03b1 \u03ba\u03ac\u03c4\u03b9 \u03c3\u03b7\u03bc\u03b1\u03bd\u03c4\u03b9\u03ba\u03cc :omighty:" } { "Tag": [ "probability", "FTW" ], "Problem": "If there are three teams competing for three days, and every day, 2 out of the three teams are chosen to play, what is the chance that team A will get to play 2 out of the three days?\r\nThe answer sheet told me only the answer, and i want to figure out how to do it\r\n\r\nThis is what the answer sheet told me:\r\n[hide]20/27[/hide]\r\n\r\n :huh:", "Solution_1": "um lol i cant get it, i get lots of different answers trying it different ways but never the answer that you provided.", "Solution_2": "[hide]\nFirst, find out how many ways there are for Team A not to play at least 2 games.\nCase 1: Team A plays 1 day\n$ \\frac{2}{3}\\times\\frac{1}{3}\\times\\frac{1}{3}\\equal{}\\frac{2}{27}$\nThere are 3 different ways Team A can play 1 day, so the probability of them playing exactly 1 day is $ \\frac{6}{27}$.\nCase 2: Team A plays none of the 3 days.\nThe chances of this are $ \\frac{1}{3}\\times\\frac{1}{3}\\times\\frac{1}{3}\\equal{}\\frac{1}{27}$\n$ 1\\minus{}\\frac{6}{27}\\minus{}\\frac{1}{27}\\equal{}\\boxed{\\frac{20}{27}}$\n[/hide]", "Solution_3": "But the problem said \"what is the chance that team A will get to play 2 out of the 3 days?\"... not \"[i]at least[/i] 2 out of the three days\"...", "Solution_4": "The problem's faulty. It's from FTW, and it's supposed to be at least.", "Solution_5": "Im REALLY sorry... i forgot to read the whole question.\r\n\r\nit's [i]at least[/i] two out of three days\r\n\r\nagain, im REALLY sorry" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that all finite groups are isomorphic to some subgroup of $ A_N$. here N means all naturals.", "Solution_1": "Take the subgroup of $ A_{2n}$ (a subgroup of $ A_N$) which separately permutes $ (1,2,\\ldots,n)$ and $ (n\\plus{}1,n\\plus{}2, \\ldots 2n)$, with each permutation doing the same thing to each of those (thus even). This subgroup is isomorphic to $ S_n$. Apply Cayley's theorem.", "Solution_2": "Another one, Cayley theorem tell us that every group is isomorphic to a subgroup of $ S_{n}$ for some $ n$ but is easy to see that $ S_{n}$ is isomorphic to a subgroup of $ A_{n \\plus{} 2}$, (for the last statement just consider the morphism $ \\sigma \\rightarrow \\sigma (n \\plus{} 1$ $ n \\plus{} 2)$ and we are done).\r\n\r\nI edit: I'm sorry MellowMelon, at the moment I posted I didn't read your full solution and I wrote essentially the same." } { "Tag": [ "logarithms", "floor function" ], "Problem": "Given that $ \\log 2 \\equal{} 0.3010$, how many digits are in $ 5^{44}$?", "Solution_1": "[hide=\" huh --> ooooh\"]\n$ \\log 2 \\plus{} \\log 5 \\equal{} 1$[/hide]", "Solution_2": "[hide=\"oooh ---> duh\"]\nnumber of digits in $ n \\equal{} \\lfloor \\log n\\rfloor \\plus{} 1$\n[/hide]" } { "Tag": [], "Problem": "have a look at [url=http://www.mathlinks.ro/viewtopic.php?p=1061605#1061605]this problem[/url]", "Solution_1": "Do you consider the refraction coefficient of the water given???", "Solution_2": "yes take that to be 1.5", "Solution_3": "tell me if the picture i made is correct. angle a is $ \\frac{\\pi}{4}$ and we want to find segment h", "Solution_4": "yes that's rite i suppose" } { "Tag": [ "geometry", "3D geometry", "sphere", "inequalities", "vector", "geometry proposed" ], "Problem": "If $P_{i}$ (i=1,..,n) are points on a unit sphere then \r\n$\\sum_{i\\leq j}\\left|P_{i}P_{j}\\right|^{2}\\leq n^{2}$", "Solution_1": "i think you should have $n^{2}$ instead of $n$ in the RHS because as it is the problem is wrong.\r\nIn the case of $n^{2}$\r\nwe easily see by vectors that if $O$ is the center of the sphere\r\n$\\sum_{i\\le j}|P_{i}P_{j}|^{2}\\le n^{2}\\leftrightarrow \\sum (OP_{i}-OP_{j})^{2}\\le n^{2}$ \r\nand the later is equivalent to:\r\n$\\left(\\sum OP_{i}\\right)^{2}\\geq 0$ :wink:", "Solution_2": "Nice solution and you are right it is $n^{2}$ .Sorry for the typo." } { "Tag": [], "Problem": "How many 4 digit numbers are there such that all digits are unique, and such that the digits are in descending order? AKA: 9876.", "Solution_1": "what do we know. Look at the first digit. it can't be a 0, 1, 2 but i can be 3-9. which means there are 7 possible.\r\ndigit 2 can't be 0, 1, 9 but i can be 2-8. digit 3: 1-7, digit 4: 0-6.\r\nlet the 4 digits be a, b, c, d respectively.\r\na>b>c>d.\r\nthe main limiting factor is the less than since we have already found the range.\r\ncases:\r\na=9;b=8, c=7 then d can be 0-6 so 7\r\na=9;b=8, c=6 then d can be 0-5 so 6\r\na=9;b=8, c=5 then d can be 0-4 so 5\r\na=9;b=8, c=4 then d can be 0-3 so 4\r\na=9;b=8, c=3 then d can be 0-2 so 3\r\na=9;b=8, c=2 then d can be 0-1 so 2\r\na=9;b=8, c=1 then d can be 0 so 1\r\nthen repeat the addition leaving out the 7 then the 6 and so on\r\nso for a=9 the total is 84\r\na=8;56\r\na=7;35\r\na=6;20\r\na=5;10\r\na=4;4\r\na=3;1\r\ntotal = 210\r\n\r\nnote i am sure there is an easier way but this was relatively easy using ranges and looking at what happens when you decrease one of the beginning digits. You can also make a calc program to count all posibilities.", "Solution_2": "Hint:\r\n\r\n[hide]It's a one step problem. Just realize a certain correspondence and you have it![/hide]", "Solution_3": "is the number correct? i believe it is but i think i did way too much work.", "Solution_4": "The number is correct.\r\n\r\n:)\r\n\r\nHere's a hint - if you take the digits 3, 5, 2, 9, how many ways can you organize them so that the number satisfies the conditions?", "Solution_5": "Is the answer[hide]$10C5$[/hide]\nOops, I typed the wrong answer. I meant to say:\n[hide]$10C4$[/hide]", "Solution_6": "[hide=\"something\"]\n10C4 = 10 x 9 x 8 x 7 / 4 x 3 x 2 = 10 x 3 x 7 = 210, and there are 10 numbers, four unique digits, the digits must be positioned a certain way, so yeah :) .\n\nNote: Even if the problem said that the second-lowest number had to be first, highest number had to be second, second-highest number had to be third, and lowest number had to be last, the answer would still be 210. You see why, right?\n[/hide]", "Solution_7": "It is 210 right? cause that's what i got.", "Solution_8": "[hide]Yeah. It should be $\\binom{10}{4}$ because no matter which four numbers you choose, you can arrange them in exactly one way so that the digits are in decreasing order.[/hide]" } { "Tag": [ "LaTeX", "videos", "Alcumus", "probability", "MATHCOUNTS" ], "Problem": "Can someone teach me how to use choose. like 5 choose 2. I think you would use taht in prolems like there are 7 people and they want to make a 3 person comittee. How many different combinations are there?", "Solution_1": "nCp = n!/(p!(n-p)!))\r\n\r\nsorry idk how to use latex\r\n\r\nmaybe this will help:\r\n\r\nhttp://en.wikipedia.org/wiki/Combination", "Solution_2": "There are a lot of videos in Alcumus that discuss this. You could also buy the Intro to Counting and Probability textbook.", "Solution_3": "in $ \\text{\\LaTeX}$, we have $ \\binom{n}{p}=\\frac {n!}{(n-p)!p!}$\r\n\r\n$ \\binom52=\\frac{5!}{(5-2)!2!}$ or $ \\frac{5\\cdot4}{2}$ or $ \\boxed{10}$\r\n\r\n$ \\binom73=\\frac{7!}{(7-3)!3!}$ or $ \\frac{7\\cdot6\\cdot5}{3\\cdot2\\cdot1}$ or $ \\boxed{35}$.\r\n\r\nhope that helps, ask if you have questions.", "Solution_4": "This is used mostly in choosing groups of k from a set of n. For example, say there are 4 people:\r\n\r\nBOGTRO, jjx1, fantasylover, AIME15.\r\n\r\nWe want to choose 2 of them.\r\n\r\n4 choices for the first, 3 for the second.\r\n\r\nBut BOGTRO,jjx1=jjx1,BOGTRO, so we divide by 2!=2.\r\n\r\nFor more, see [url]http://en.wikipedia.org/wiki/Combination[/url].\r\n\r\nBe sure not to confuse these with [url]http://en.wikipedia.org/wiki/Permutations[/url].\r\n\r\nHope this helps.", "Solution_5": "Examples:\r\n\r\n(Permutation: Counting where order matters) How many different ways are there to select two people from a group of 5, where one person is prez and the other is vp (prez cannot be vp)?\r\n\r\nThere are 5 people for the first spot, and 4 for the second, so that gives a total of 20 ways.\r\n\r\nThe general notation for this is $ \\frac{n!}{r!}$, where $ n$ is the total number (in this case, 5) and $ r$ is the number or things/people you want to order (in this case, 2).\r\n\r\n(Combination: Counting where the order does not matter) levans wants to choose 3 mods out of the group of 5 people isabella2296, vallon22, jjx1, AIME15, and rrusczyk. How many differnt ways are there for him to do this?\r\n\r\nThere are 5 for the first, 4 for the second, and 3 for the third. However, the order makes no difference, so the answer would be $ \\frac{5\\times4\\times3}{3!}\\equal{}10$ (there are $ 3!$ ways to order 3 people). \r\n\r\nNote that choosing 3 mods out of a group of 5 is the same as choosing 2 people [b]not[/b] to be mods. Therefore (in combinations only), $ \\binom{n}{r}\\equal{}\\binom{n}{n\\minus{}r}$. The general formula is $ \\frac{n!}{r!\\cdot (n\\minus{}r)!}$", "Solution_6": "Speaking of permutations, $ P(5,2)\\equal{}20$, right? this problem showed up in a recent match (not mathcounts), and the judge said the answer was 60. :|", "Solution_7": "Um, it's $ 5 \\times 4\\equal{}20$, or $ \\frac{5!}{3!}\\equal{}20$.\r\n\r\nDragon, permutations are $ \\frac{n!}{(n\\minus{}r)!}$. Think about it :P" } { "Tag": [ "function", "limit", "inequalities", "ratio", "real analysis", "real analysis unsolved" ], "Problem": "Let $ \\alpha>1$. Can you construct a real function $ f$ such that for every $ t>0$ :\r\n$ \\lim_{\\delta \\to 0} \\sup_{\\| \\mathcal P _t \\| < \\delta} \\sum |f(x_{i+1})-f(x_i)|^{\\alpha}=t$\r\nwhere $ \\mathcal P_t$ is the set off all partitions: $ 0=x_0 0$ and define $ f_k$ as the sawtooth function with teeth of width $ 2\\ell_k$ and height $ \\ell_k^{1/\\alpha}$. Put $ f \\equal{} \\sum_k f_k$.\r\n[asy]size(200);\nD((0,2)--(0,0)--(10,0),black);\nfor(int k=0;k<10;++k) D((k,0)--(k+0.5,1)--(k+1,0));\nD((0,1)--(10,1),dashed);\nMP(\"\\ell_k^{1/\\alpha}\",(0,1),8,W);\nD((0,-0.3)--(0,0));\nD((0.5,-0.3)--(0.5,1));\nMP(\"\\ell_k\",(0.25,-0.1),8);[/asy]\r\nNote that $ |f_k(x) \\minus{} f_k(y)|\\le |x \\minus{} y|^{1/\\alpha}$ and, moreover, the equality is attained if one of the points $ x$ and $ y$ is at the base and the other one is at the top of the same tooth. In all other cases the inequality is strict and if the ratio $ \\frac {|x \\minus{} y|}{\\ell_k}$ is far from $ 1$, then it holds with a very small constant on the right (I leave it to you to make an exact statement out of this observation after you read the rest of the proof). Now we can argue that if $ |x \\minus{} y|$ is small, then $ |f(x) \\minus{} f(y)|\\le (1 \\plus{} \\varepsilon)|x \\minus{} y|^{1/\\alpha}$. Indeed, since $ \\ell_k$ decays very fast, there may exist only one $ \\ell_k$ comparable to $ |x \\minus{} y|$. For this $ k$ we use the estimate $ |f_k(x) \\minus{} f_k(y)|\\le|x \\minus{} y|^{1/\\alpha}$. For all other $ k$ we use the similar estimate with very small constants (we will have to ensure that certain series have small sums but it is not a big problem). That is enough to show that the variation is not greater than $ t$. To show that it is not less than $ t$, just use the bases and tips of teeth in $ f_k$ and observe that neither the previous, nor the subsequent functions can change the corresponding differences by any noticeable amounts if $ k$ is large enough." } { "Tag": [ "search", "AoPS Books" ], "Problem": "Due to the fact that my school's math department is extremely...well...basic, I am experiencing great difficulty in finding books and other resources that help me advance in math. Do you guys have any suggestions about certain resources (other than the AoPS books which I have and love) that are helpful?\r\n\r\nOh, and if this is the wrong spot to place something like this, I apologize. I'm new and still getting used to the forums.", "Solution_1": "[quote=\"horselvr101\"]Due to the fact that my school's math department is extremely...well...basic, I am experiencing great difficulty in finding books and other resources that help me advance in math. Do you guys have any suggestions about [b]certain resources[/b] (other than the AoPS books which I have and love) that are helpful?\n\nOh, and if this is the wrong spot to place something like this, I apologize. I'm new and still getting used to the forums.[/quote]Please specify what \"certain resources\" you are looking for. Also, if you got a fairly amount of online time, you can definitely search for any resource over the Internet that you want. Simply typing \"math competitions\" in Amazon.com already gives you a good start for what books consumers look for. More powerful search, try googling, view more AoPS, check out AoPS Resources section, ... I mean information nowadays comes virtually limitless and easy with the power of Internet. If you work hard, you can always find anything you want.", "Solution_2": "And there are loads of books available online (some are illegal but many are legal) :)", "Solution_3": "Books from the xyz press and the 100 problem books are good. (Mostly co-authored by Titu Andreescu) Some problems are AIME level but the difficult ones can get into Olympiad topics.", "Solution_4": "[url=http://artofproblemsolving.com/wiki/index.php/Math_books]Try these[/url]\n[url=http://artofproblemsolving.com/wiki/index.php/Math_textbooks]Or these[/url]\n[url=http://artofproblemsolving.com/wiki/index.php/Resources_for_mathematics_competitions]Don't forget these[/url]\n[url=http://artofproblemsolving.com/wiki/index.php/Olympiad_Books]And some of these olympiad level books[/url]\n\nA lot of books are present on multiple lists because their difficulty covers multiple levels and because redundancies haven't been consolidated.\n\nEDITI didn't notice this was a revived thread, but...I hope these links are still helpful :) :)" } { "Tag": [ "geometry", "logarithms", "calculus", "calculus computations" ], "Problem": "Given the curves $ f(x)\\equal{}kx$ and $ g(x)\\equal{}\\frac x{x^2\\plus{}1},$ then for $ k\\in(0,1]$ these curves enclose a region. Is this correct?\r\n\r\nHow would you find the area bounded by these curves?", "Solution_1": "That should be $ k \\in (0, 1)$, probably. Have you drawn the picture? What have you tried?", "Solution_2": "Area is $ k-1-\\ln{k}$ with $ k \\in (0, 1)$...\r\n\r\n(Note that the curves intersect at $ x=\\pm \\sqrt{\\frac{1-k}{k}}$. Then do routine integration.." } { "Tag": [ "calculus", "integration", "function", "Support", "logarithms", "Functional Analysis", "topology" ], "Problem": "You all know that $C([0,1])$ is a Banach space with the sup-norm and it is not with the p-integral norm: $\\|f\\|_p=\\left(\\int^1_0|f(x)|^pdx\\right)^\\frac1p$ for any $p\\ge 1$. But can you help me to construct a Cauchy sequence for any $p$ such that it converges to the function $g\\equiv0$ for any $q\\le p$ and does not converge if $q>p$?", "Solution_1": "Define a continuous function $\\varphi(x)$ on $[0,\\infty)$ such that $\\varphi(0)=0,$ the support of $\\varphi$ is $[0,1],$ and $\\varphi(x)>0$ for $x\\in(0,1).$ The exact details don't matter very much. Let $C_p=\\|\\varphi\\|_p.$\r\n\r\nNow fix a certain $a>0$ and let $g_n(x)=n^a\\varphi(nx).$ Note that $g_n$ tends to zero pointwise everywhere.\r\n\r\nThen $\\|g_n\\|_p=C_pn^{a-\\frac1p}$\r\n\r\nWe ought to be able to pick $a$ to do what you want.\r\n\r\nNo, wait - that doesn't seem to be what you want. But you can't have what you want.\r\n\r\nThe problem is that the zero function is in $L^p$ for every $p.$ A sequence that tends to zero in the $p$-norm must be Cauchy in the $p$-norm, because all convergent sequences are Cauchy.\r\n\r\nThe completeness of a metric space is about whether there exist enough elements to be the limits of the sequences that look like they ought to have limits. Since we're talking about the continuous functions not being complete in certain metric, try for something with a discontinuous limit.\r\n\r\nFor instance: for $n>2,$ let $g_n(x)$ be the piecewise linear continuous function whose graph connects the points $(0,0),\\left(\\frac12,0\\right),\\left(\\frac12+\\frac1n,1\\right),$ and $(1,1).$ This sequence is Cauchy for every $p<\\infty.$ Since there is no continuous function that can be its limit, it is not a convergent sequence in $C([0,1])$ considered as a metric space with the $p$-norm.", "Solution_2": "What I want is to show that the $p$-norm and $q$-norm are not equivalent if $p\\neq q$ (is this statement true?). So I want to find a sequence that is Cauchy and convegrent to $g\\equiv 0$ for any $q\\le p$ and [b]not Cauchy[/b] for any $q > p$. Something like $f_n(t)=\\begin{cases}\\sqrt[p]{n}(1-nt)&\\text{ if } t\\in[0,\\frac1n]\\\\0&\\text{ if }t\\in(\\frac1n,1]\\end{cases}$ but it converges for $qq.$", "Solution_4": "Ok, that's fine (and simple too), thanks." } { "Tag": [], "Problem": "An optically active compound A(assume that it is dextrorotatory) has the molecular formula C7H11Br. A reacts with hydrogen bromide, in the absence of peroxides to yield to isomeric products, B and C. B is optically active, c is not. B + K-O(CH3)3-> (+)A. C + K-O(CH3)3-> (+-) A. A + K-O(CH3)3 -> D\r\nD on ozonolysis gives 1,3 cyclopentadiene and formaldehyde. Provide stereochemical formulae for A,B,C,D and outline the reactions.", "Solution_1": "[quote=\"VARUNARASU\"]D on ozonolysis gives 1,3 cyclopentadiene and formaldehyde. [/quote]\r\n\r\nVARUN are u sure about this step. I think 1,3-cyclopentadiene can't be right.", "Solution_2": "sorry that was a typo. :oops: its dione, actually", "Solution_3": "[hide=\"Answer\"]A is 3-bromo-3-methyl-1-methylenecyclopentane. By reaction with HBr we get a meso compound C (cis-1,3-dibromo-1,3-dimethylcyclopentane) and the chiral compound B (trans-1,3-dibromo-1,3-dimethylcyclopentane). Compound D is 1,3-dimethylenecyclopentane.[/hide]", "Solution_4": "correct :D , this question was actually for hellever who wanted us to post organic problems on hydrocarbons, and the like." } { "Tag": [ "algebra", "polynomial", "function", "algebra proposed" ], "Problem": "Let$ a\\in R \\minus{} 0,b,c\\in R$Prove that Exist a polynomial $ (Px)$\r\nsuch that\r\n\\[ P(ax^2 \\plus{} bx \\plus{} c) \\equal{} a(P(x))^2 \\plus{} bP(x) \\plus{} c\r\n\\]\r\nHow many are polynomials that satisfies our condition?\r\n[hide=\"Result\"] only $ 1$[/hide]", "Solution_1": "Define $ f(x) \\equal{} ax^2 \\plus{} bx \\plus{} c$. Possible solutions to $ P(f(x)) \\equal{} f(P(x))$ are $ f^{(n)}(x)$, $ n \\minus{}$th composition of $ f$. $ f^{(0)}(x) \\equal{} x$.\r\nWe can show there are situations where there exist other solutions.\r\n\r\nNote this is not the case in http://www.mathlinks.ro/viewtopic.php?t=256092, where the equation is \r\n$ P(x^2 \\plus{} 1) \\equal{} P(x)^2 \\minus{} 1$, not\r\n$ P(x^2 \\plus{} 1) \\equal{} P(x)^2 \\plus{} 1$. \r\nThat is why it only has constant solutions.\r\n\r\nNow we discuss the case $ f(x) \\equal{} ax^2 \\plus{} bx \\plus{} c$. We can rewrite $ f(x) \\equal{} a(x \\plus{} d)^2 \\plus{} e$ with $ a\\ne0$.\r\n\r\n$ P(a(x \\plus{} d)^2 \\plus{} e) \\equal{} a(P(x) \\plus{} d)^2 \\plus{} e$\r\n$ P(ax^2 \\plus{} e) \\equal{} a(P(x \\minus{} d) \\plus{} d)^2 \\plus{} e$\r\n\r\nLet $ Q(x) \\equal{} P(x \\minus{} d) \\plus{} d$. We have, \r\n$ Q(ax^2 \\plus{} e \\plus{} d) \\equal{} aQ(x)^2 \\plus{} e \\plus{} d$.\r\n$ Q(\\frac {x^2}a \\plus{} e \\plus{} d) \\equal{} aQ(\\frac xa)^2 \\plus{} e \\plus{} d$.\r\n\r\nLet $ h(x) \\equal{} aQ(\\frac xa)$, we have\r\n$ h(x^2 \\plus{} s) \\equal{} h(x)^2 \\plus{} s$, where $ s \\equal{} a(e \\plus{} d)$.\r\n\r\nCase 1) $ s \\equal{} 0$\r\n$ P(x^2) \\equal{} P(x)^2$.\r\nSuppose $ \\deg P \\equal{} n$. Obviously $ P(x) \\equal{} x^n \\plus{} g(x)$ with $ m \\equal{} \\deg g < n$.\r\nNow $ P(x^2) \\minus{} x^{2n}$ has degree $ 2m$, while $ P(x)^2 \\minus{} x^{2n}$ has degree $ n \\plus{} m$.\r\nSo the only solution is $ x^n$. We see this is the case $ f(x) \\equal{} x^2$ but $ P(x)$ are not just $ f^{(n)}(x)$.\r\n\r\nCase 2) $ s > \\frac14$\r\nWe know $ h$ is either even or odd, the only odd function is $ h(x) \\equal{} x$ because $ h(x) \\equal{} x$ has infinitely many roots \r\n$ a_0 \\equal{} 0,a_n \\equal{} a_{n \\minus{} 1}^2 \\plus{} s$.\r\n\r\nIf $ h$ is even, we can show $ h(x) \\equal{} g(x)^2 \\plus{} s$ with $ g(x^2 \\plus{} s) \\equal{} g(x)^2 \\plus{} s$.\r\n\r\nHence, in this case, $ f^{(n)}(x)$ are the only possible solutions." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "If $ B \\equal{} PAP^{\\minus{}1}$, how to get the conclusion : $ tr(B) \\equal{} tr(A)$ ?", "Solution_1": "Use that $ tr(MN)\\equal{}tr(NM)$ (when defined of course)." } { "Tag": [ "geometry" ], "Problem": "This is probably obvious, but I can't get the answer for the life of me.\r\nIn triangle ABC, points D, E, F are marked on sides BC, CA, AB, respectivel, so that BD/DC = CE/EA = AF/FB = 2. Show that the triangle formed by the lines AD, BE, CF has area 1/7 that of triangle ABC.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?highlight=divide&t=40828", "Solution_2": "Don't use mass points.\r\n\r\nI like your post shobber, it emphasizes the importance of working through something menial to get something nice." } { "Tag": [ "inequalities", "geometry", "incenter", "geometry proposed" ], "Problem": "Given are $\\triangle ABC$ with incenter $I$ prove that:\r\n$IA+IB+IC\\ge 3r+\\frac{IA^{2}+IB^{2}+IC^{2}}{2R}$", "Solution_1": "let's put $p-a=x,p-b=y,p-c=z$\r\nrecall that :$|IA|=\\sqrt{\\frac{x(x+y)(x+z)}{x+y+z}},|IB|=\\sqrt{\\frac{y(y+x)(y+z)}{x+y+z}},|IC|=\\sqrt{\\frac{z(z+y)(x+z)}{x+y+z}}$\r\n$,r=\\sqrt{\\frac{xyz}{x+y+z}},R=\\frac{(x+y)(y+z)(z+x)}{4\\sqrt{xyz(x+y+z)}}$\r\nSo we have to prove:\r\n$\\sum_{cyc}\\sqrt{\\frac{x(x+y)(x+z)}{x+y+z}}\\geq 3\\sqrt{\\frac{xyz}{x+y+z}}+2\\sqrt{\\frac{xyz}{x+y+z}}\\left(\\sum_{cyc}\\frac{x}{y+z}\\right)$\r\nwhich is equivalent to:\r\n$\\sum_{cyc}\\sqrt{\\frac{(x+y)(x+z)}{yz}}\\geq 3+2\\sum_{cyc}\\frac{x}{y+z}$\r\nbut $\\sqrt{\\frac{(x+y)(x+z)}{yz}}\\geq \\frac{x+\\sqrt{yz}}{\\sqrt{yz}}\\geq 1+\\frac{2x}{y+z}$ and summing up we are done! :wink:", "Solution_2": "oh yes that is a good proof, here is my proof:\r\n$R(\\sum IA)=\\sum OA.IA\\ge\\sum \\vec{OA}\\vec{IA}=\\sum (\\vec{OG}+\\vec{GA})(\\vec{IG}+\\vec{GA})=3\\vec{OG}\\vec{IG}+\\sum GA^{2}(1)$\r\nnow we have $3\\vec{OG}\\vec{IG}=3\\frac{OG^{2}+IG^{2}-OI^{2}}{2}$\r\nwe have $IG^{2}\\ge 0$ and $OG^{2}-OI^{2}=R^{2}-\\frac{a^{2}+b^{2}+c^{2}}{9}-(R^{2}-2Rr)=2Rr-\\frac{a^{2}+b^{2}+c^{2}}{9}$ then\r\n$3\\vec{OG}\\vec{IG}\\ge 3Rr-\\frac{a^{2}+b^{2}+c^{2}}{6}(2)$ and $\\sum GA^{2}=\\frac{a^{2}+b^{2}+c^{2}}{3}(3)$ from $(1),(2),(3)$ we will have\r\n$\\sum IA)\\ge 3r+\\frac{a^{2}+b^{2}+c^{2}}{6R}\\ge 3r+\\frac{IA^{2}+IB^{2}+IC^{2}}{2R}$\r\nthus we will have strong inequalities and basic ineq can be wrote:\r\n$IA+IB+IC\\ge 3r+\\frac{3IG^{2}}{2R}+\\frac{a^{2}+b^{2}+c^{2}}{6R}$ :lol: :lol:" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Review the inequality \\[ (a+b+c)^2 \\ge 3(ab+bc+ca) \\]\r\nThere's nothing special, but now see another form and prove the following strange inequality\r\n\\[ 2^{ab}+2^{bc}+2^{ca} \\le 2+2^{9/4} \\]\r\nFor all $a,b,c \\ge 0$ and $a+b+c=3$.\r\n\r\nMoreover, find the maximum value of \r\n\\[ 4^{ab}+4^{bc}+4^{ca} \\]\r\nWith $a,b,c$ satisfy this condition.\r\nFrom that, find the best constant $k \\ge 0$ st the following inequality 's true\r\n\\[ k^{ab}+k^{bc}+k^{ca} \\le 3 \\]\r\nFor all $a,b,c \\ge 0$ and $a+b+c=3$.", "Solution_1": "[quote=\"hungkhtn\"] Prove the following inequality\n\\[ 2^{ab}+2^{bc}+2^{ca} \\le 6 \\]\nfor all $a,b,c \\ge 0$ and $a+b+c=3$[/quote]For $a=0$ and $b=c$, it is not true. :?:", "Solution_2": "Ok, sorry. I forget adding 2 to the left right hand. I write case k=2 just only for exams, because I've solved the general form. I have edited it, try to prove. Good luck.", "Solution_3": "It's very hard problem, anyone has any idea? :(", "Solution_4": "A very very hard and nice problem is as follow\r\n\r\nFind the best constant $k\\ge 0$ sthat for all $a,b,c \\ge 0,a+b+c=3$ then\r\n\\[ k^{a^2}+k^{b^2}+k^{c^2} \\ge 3k \\]\r\n\r\nI hope someone solved it completely in one week.", "Solution_5": "[quote=\"hungkhtn\"]A very very hard and nice problem is as follow\n\nFind the best constant $k\\ge 0$ sthat for all $a,b,c \\ge 0,a+b+c=3$ then\n\\[ k^{a^2}+k^{b^2}+k^{c^2} \\ge 3k \\]\n\nI hope someone solved it completely in one week.[/quote]\r\n\r\nThat is a strange and nice inequality. It is very creative of you, Hung. Let me give another:\r\n\r\nLet $a,b,c$ be real positive numbers such that $a^2+b^2+c^2=3$, find the smallest best positive constant $k$ such that\r\n\\[ k^{ab}+k^{bc}+k^{ca}\\geq 3k. \\]", "Solution_6": "I think better that before you show any problem, please one time think stragiely about it?", "Solution_7": "A long time and unsolved. So, I think I should give the hints : Using mixing Variable. ;)" } { "Tag": [ "Cauchy functional equation", "function", "functional equation", "algebra" ], "Problem": "Since I am stydying that at the moment, can somebody post problems or links to problems at IMO level that can be reduced to Cauchy function(function of the type f(x+y)=f(x)+f(y))and the way it is done(I mean functional equations at the first place)", "Solution_1": "Let the function $f(x)$ continuous in $[0;1]$ satisfying \r\n$f(0)=0, f(1)=1$ and\r\n $f(\\frac{x+y}{2})=(1-a)f(x)+f(y) \\forall x \\in R, a \\in (0;1)$\r\nand $x \\ge y$.\r\nFind $f(\\frac{1}{7})$", "Solution_2": "[size=75][b]Source de.rec.denksport , proposed by Gerhard J. Woeginger [/b][/size]\r\n[color=blue][b] \nTo find all functions $f: {\\mathbb Z}\\to {\\mathbb Z}$ which satisfy $f\\left(x+y+f(y)\\right) = f(x) + 2y\\; \\; \\; ,\\; \\; \\forall x,y \\in {\\mathbb Z} .$ [/b][/color]" } { "Tag": [], "Problem": "The same poll may apply on the native americans.", "Solution_1": "Absolutely", "Solution_2": "Same to PUERTO RICO!!!", "Solution_3": "[img]http://www.funnytimes.com/store/images/homelandsecurity.jpg[/img]\r\n\r\nInteresting way to view it.", "Solution_4": "keep in touch." } { "Tag": [], "Problem": "Inspirat de o problema de pe FORUMUL in lb. engleza, pe care tocmai am rezolvat-o, \r\nva propun spre rezolvare urmatoarea:\r\n\r\nFie A1B1C1 simetricul triunghiului ABC fa\u0163\u0103 de o dreapt\u0103 d din planul triunghiului. Paralelele duse prin v\u00e2rfurile A, B \u015fi C la dreapta d intersecteaz\u0103 dreptele B1C1, C1A1 \u015fi respective A1B1 \u00een punctele A2, B2 \u015fi respective C2. S\u0103 se arate c\u0103 punctele A2, B2 \u015fi C2 sunt coliniare. (Mihai Miculita ???, poate)\r\n\r\nv. ATTACHMENTUL, pentru un enunt scris in word in care gasiti si o figura facuta cu euklides.exe", "Solution_1": "In attachment gasiti o solutie sintetica a acestei probleme.\r\nInainte de a vizualiza aceasta solutie, incercati mai intai singuri!\r\n(Placerea de a gasi solutia singur, fiind incomparabil mai mare decat a citi solutia altcuiva)." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let$a,b,c,d$ be positive real \r\n and $abcd=1$\r\n prove that :$\\frac{1}{a(b+1)} \\frac{1}{b(c+1)} \\frac{1}{c(d+1)} \\frac{1}{d(a+1)} \\leq 2$or$\\geq 2$", "Solution_1": "tranthanhnam, did you have any mistakes ?", "Solution_2": "I am confused. o.o\r\n\r\n$(a+1)(b+1)(c+1)(d+1) \\le 8(abcd + 1) = 16$ by Chebyshev, so we have $\\frac{1}{abcd(a+1)(b+1)(c+1)(d+1)} \\ge \\frac{1}{16}$... did you make a typo?", "Solution_3": "Not hard. Take $a=y/x$ and so on... The inequality become the Nesbitt inequality with 4 variables.\r\n\r\n\\[ \\frac{x}{y+z}+\\frac{y}{z+t}+\\frac{z}{t+x}+\\frac{t}{x+y}\\ge 2 \\]", "Solution_4": "[quote=\"hungkhtn\"]Not hard. Take $a=y/x$ and so on... The inequality become the Nesbitt inequality with 4 variables.\n\n\\[ \\frac{x}{y+z}+\\frac{y}{z+t}+\\frac{z}{t+x}+\\frac{t}{x+y}\\ge 2 \\][/quote]\r\n\r\nWhat is Nesbitt inequality with $n$ variables?", "Solution_5": "I'm guessing it's probably $\\sum_{cyc} \\frac{a_1}{a_2 + a_3} \\ge \\frac{n}{2}$ from the cases $n = 3, 4$... normalizing $\\sum a_1 = 1$ and using Jensen's and Rearrangement should give a proof. I think.", "Solution_6": "[quote=\"t0rajir0u\"]I'm guessing it's probably $\\sum_{cyc} \\frac{a_1}{a_2 + a_3} \\ge \\frac{n}{2}$ from the cases $n = 3, 4$... normalizing $\\sum a_1 = 1$ and using Jensen's and Rearrangement should give a proof. I think.[/quote]\r\n\r\n But I don't think we have the general case of Nesbit as you say. As far as I remember, one guy, probably from Russia, have a very nice prove for the constant k!", "Solution_7": "[quote=\"t0rajir0u\"]I'm guessing it's probably $\\sum_{cyc} \\frac{a_1}{a_2 + a_3} \\ge \\frac{n}{2}$ from the cases $n = 3, 4$... normalizing $\\sum a_1 = 1$ and using Jensen's and Rearrangement should give a proof. I think.[/quote]\r\n\r\nThe Shapiro's inequality does not hold in general.", "Solution_8": "So, apparently, the cases are $n \\equiv 0 \\bmod 2, n \\le 12$ or $n \\equiv 1 \\bmod 2, n \\le 23$. Interesting :huh:", "Solution_9": "[quote=\"tranthanhnam\"]let$a,b,c,d$ be positive real and $abcd=1$\n\nprove that :$\\frac{1}{a(b+1)} \\frac{1}{b(c+1)} \\frac{1}{c(d+1)} \\frac{1}{d(a+1)} \\leq 2$or$\\geq 2$[/quote]\r\n\r\nI think it is $\\frac{1}{a(b+1)}+\\frac{1}{b(c+1)}+\\frac{1}{c(d+1)}+\\frac{1}{d(a+1)} \\geq 2$ with $abcd=1$\r\n\r\nCheck here: http://www.mathlinks.ro/Forum/viewtopic.php?t=17206\r\n\r\nBy the way, what is [b]Shapiro's inequality[/b]?", "Solution_10": "I think t0rajir0u is right.\r\n[url=http://planetmath.org/encyclopedia/ShapirosInequality.html]here[/url] you can see the Shapiro's inequality.\r\nbut no proof. :(" } { "Tag": [ "function", "factorial" ], "Problem": "given $ a_0\\equal{}a_1\\equal{}1$, find the generating function for \\[ a_n\\equal{}na_{n\\minus{}1}\\plus{}n(n\\minus{}1)a_{n\\minus{}2}\\]", "Solution_1": "It's sequence [url=http://www.research.att.com/~njas/sequences/A005442]A005442[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] -- presumably, you are looking for the [i]exponential[/i] generating function (since it's easy to see that the growth is at least as fast as factorial), which does have a nice form (contained in the link there).", "Solution_2": "denote $ {g_n}$ as $ g_n \\equal{} \\frac {\\alpha}{n!}a_n$ where $ \\alpha$is not equal to 0\r\nso $ g_0 \\equal{} \\alpha ; g_1 \\equal{} \\alpha$ \r\nwe have \r\n$ \\frac {n!}{\\alpha}g_n \\equal{} \\frac {n(n \\minus{} 1)!}{\\alpha}g_{n \\minus{} 1} \\plus{} \\frac {n(n \\minus{} 1)(n \\minus{} 2)!}{\\alpha}g_{n \\minus{} 2}$\r\n$ g_n \\equal{} g_{n \\minus{} 1} \\plus{} g_{n \\minus{} 2}$ where $ g_0 \\equal{} g_1 \\equal{} \\alpha$\r\nwhere we obtain the solution$ g_n \\equal{} x_1(\\frac {1 \\plus{} \\sqrt {5}}{2})^n \\plus{} x_2(\\frac {1 \\minus{} \\sqrt {5}}{2})^n$\r\nsince $ g_0 \\equal{} g_1 \\equal{} \\alpha , x_1 \\plus{} x_2 \\equal{} \\alpha ; and x_1 \\minus{} x_2 \\equal{} \\frac {\\alpha}{\\sqrt {5}}$\r\nso $ x_1 \\equal{} \\frac {\\alpha}{2}(1 \\plus{} \\frac {1}{\\sqrt {5}}) , x_2 \\equal{} \\frac {\\alpha}{2}(1 \\minus{} \\frac {1}{\\sqrt {5}})$\r\nand $ g_n \\equal{} \\frac {\\alpha}{2}(1 \\plus{} \\frac {1}{\\sqrt {5}}) (\\frac {1 \\plus{} \\sqrt {5}}{2})^n \\plus{} \\frac {\\alpha}{2}(1 \\minus{} \\frac {1}{\\sqrt {5}})(\\frac {1 \\minus{} \\sqrt {5}}{2})^n$\r\nand $ a_n \\equal{} \\frac {n!}{\\alpha} ( \\frac {\\alpha}{2}(1 \\plus{} \\frac {1}{\\sqrt {5}}) (\\frac {1 \\plus{} \\sqrt {5}}{2})^n \\plus{} \\frac {\\alpha}{2}(1 \\minus{} \\frac {1}{\\sqrt {5}})(\\frac {1 \\minus{} \\sqrt {5}}{2})^n )$ . \r\n$ \\equal{} \\frac {n!}{2}(1 \\plus{} \\frac {1}{\\sqrt {5}}) (\\frac {1 \\plus{} \\sqrt {5}}{2})^n \\plus{} \\frac {n!}{2}(1 \\minus{} \\frac {1}{\\sqrt {5}})(\\frac {1 \\minus{} \\sqrt {5}}{2})^n$", "Solution_3": "Why introduce $ \\alpha$ at all? If you replace it with 1 then it gives $ g_n \\equal{} g_{n \\minus{} 1} \\plus{} g_{n \\minus{} 2}$ with $ g_0 \\equal{} g_1 \\equal{} 1$, i.e., $ g_n$ is just the $ n$th Fibonacci number (or $ (n\\plus{}1)$th, depending how you index your Fibonacci numbers)." } { "Tag": [ "linear algebra", "matrix", "integration", "calculus", "topology", "complex numbers", "linear algebra unsolved" ], "Problem": "Let A,B,C be real matrices such that all eigenvalues of A and B have the real part strictly negative. Find all solutions X real matrix for which $AX+XB=C$. Express them in terms of A,B,C!", "Solution_1": "[quote=\"harazi\"]Let A,B,C be real matrices such that all eigenvalues of A and B have the real part strictly negative. Find all solutions X real matrix for which $AX+XB=C$. Express them in terms of A,B,C![/quote]\r\n\r\nThe operator (on matrices) $X \\to AX + XB$ has eigenvalues $a_i + b_j$ (sums of eigenvalues of A and B), and any matrix $X$ that sends a B-eigenvector to an A-eigenvector will be an eigenvector of this operator. This gives an implicit solution, but I think that there is a more direct expression as well. There is an old book by Wedderburn about matrix equations where he discusses this and similar problems, but I do not have access to it at the moment.", "Solution_2": "Could this be the book you are referring to?\r\n\r\nhttp://www.ajorza.org/papers/wedderburn/Wedderburn%20--%20Lectures%20on%20Matrices/\r\n\r\nAndrei", "Solution_3": "I think so --- but I could not find the solution of $AX + XB = C$ in that pdf file. It is possible that it's not in Wedderburn but some other source.", "Solution_4": "Consider the morphism of $M_n(C)$ $f(M)=AM+MB$ \r\nProve that $ker(f)=0$ so it is bijective\r\n\r\nCheck \r\n$X=-\\int_{0}^{+\\infty}e^{tA}Ce^{tB}dt$ is a solution of $AX+XB=C$", "Solution_5": "An amazing formula! But one must assume something about $A$ and $B$, no? \r\nker($f$) is 0 only when all $a_i + b_j$ (sums of eigenvalues of $A$ and $B$) are invertible. Also, the integral can only converge in some limited cases, such as when eigenvalues of $A,B$ have negative real part. Does this formula come from Lie algebras (something like Baker-Campbell-Hausdorff formula)?" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "[i] [u]Problem :[/u][/i]\r\n\r\n[i]Let $ a, b$ be positive integers such that :\n\n\n $ gcd(a \\ , \\ b) \\ \\equal{} \\ 1$ and $ max \\{ a \\ , \\ b \\} \\ \\geq \\ 2$\n\n\n [u]Prove that[/u] : $ \\forall n \\ \\in \\ N^{*}$ , we have :\n\n\n $ 2n | \\varphi(a^{n} \\ \\plus{} \\ b^{n} )$[/i]", "Solution_1": "$ gcd(a^k\\plus{}b^k,a^n\\plus{}b^n)\\equal{}a^{gcd(k,n)}\\plus{}b^{gcd(k,n)}$. Therefore exist prime $ p|a^n\\plus{}b^n$, suth that $ p\\not |a^d\\plus{}b^d$ for any divisor $ d C > 0$ and $a,b,c \\ge 0$\r\n\r\nThen\r\n\r\na) $f(a,b,c)$ is minimal when $a,b,c$ is propotional to three roots of polynomial: $x^3 - x^2 + C*x = 0$\r\n\r\nb) $f(a,b,c)$ is maximal when $a,b,c$ is propotional to three roots of polynomial: $x^3 - x^2 + C*x + \\alpha(C)= 0$\r\n\r\nI will give the value of $\\alpha(C)$ latter :D\r\n\r\nThis problem may be hard for some of you, I am not sure.", "Solution_1": "I wish I was a teacher who can give you this type of problem in an important contest for high-school students :lol: :P" } { "Tag": [ "function", "algebra open", "algebra" ], "Problem": "Find the function from naturals to naturals on $ x$, where the function is the $ n$ such that the expression $ a_1^x \\plus{} a_2^x \\plus{} \\cdots \\plus{} a_x^x \\minus{} na_1a_2\\cdots a_x$ is factorable among the integers.", "Solution_1": "For $ x\\equal{}1$, this is not even a well-defined function..", "Solution_2": "then for x greater than one." } { "Tag": [ "geometry", "ARML", "arithmetic sequence", "AMC" ], "Problem": "[b]1.[/b]\r\nThe 1st, 2nd and 3rd perfect squares are underlined in the following arithmetic sequence, [u]1[/u], 8, 15, 22, 29, [u]36[/u], 43, 50, 57, [u]64[/u], ...[b].[/b] Find the 100th perfect square in the sequence.\r\n\r\n[b]2.[/b]\r\nHow many subsets of S = {1, 2, 3, ..., 10} contain a 1, 2, or 3?\r\n\r\n[b]3.[/b]\r\nThree parallel lines [i]l[/i][size=59]1[/size], [i]l[/i][size=59]2[/size], and [i]l[/i][size=59]3[/size] are drawn through the vertices A, B, and C of a square ABCD. If the distance between [i]l[/i][size=59]1[/size], and [i]l[/i][size=59]2[/size] is 7 and between [i]l[/i][size=59]2[/size], and [i]l[/i][size=59]3[/size] is 12, find the area of ABCD.\r\n\r\n[b]4.[/b]\r\nFind the coefficient of x:^3: in the expansion of (1 + x + x:^2:)^12.\r\n\r\n[b]5.[/b]\r\nFind k given that\r\nf(0) = k; f(n) = f(n + 1) - 3n - 2; f(-50) = 4000.\r\n\r\n[b]6.[/b]\r\nOne of the sides of a triangle is divided into segments of 6 and 8 units by the point of tangency of the inscribed circle. If the radius of the incircle is 4, then what is the length of the shortest side of the triangle.", "Solution_1": "1. [hide]the sequence is given by 7n+1. If this is to equal m:^2:, then (m-1)(m+1)=7n, or basically (m-1)(m+1) is a multiple of 7. This will occur iff m:equiv:1,6 mod 7. So, the squares in the sequence are given by m:^2:, where m:equiv: 1,6 mod 7. We can see that the first few, 1:^2:, 6:^2:, and 8:^2:, indeed match the ones given in the example. In any case, the 100th such m will be -1+7*50=349, so it will be 349:^2:, which I do not care to do in my head, but I will anyway...um...121801?[/hide]\n\n2. [hide]there are 2^10 total subsets, and 2^7 do not contain 1, 2, or 3. we subtract to find 2^7(2^3-1)=128*7=896?[/hide]\n\n3. whatever", "Solution_2": "Yes I jacked (stole) this from the NC State Math Contest. By the way, 349:^2: is nowhere near 1219. Then again, even if you are to be correct on any of the problems, I wouldn't confirm it yet because I want to see others solutions to these problems also, and also, I don't have my work nor the solutions to these with me at the moment. For [b]3.[/b], I think I'll IM you my method for solving that problem later on, when you finish doing it. It didn't involve any extra lines being drawn, except for perpendicular lines between the pairs of parallel lines to show the distances. Your answers and solutions do seem very nice so far though Bob.", "Solution_3": "4. [hide]this can be either (x:^2:)^1*(x)^1 or (x)^3. The first coefficient is C(12,1)*C(11,1) and the second coefficient is C(12,3). Adding gives 1452?[/hide]\n\n5. [hide]f(n)=k+2n+3n(n-1)/2 (induction). thus 4000=f(-50)=k-100+3825 => k=275.[/hide]", "Solution_4": "is 3 from that NC state contest ? I had that same exact question at an ARML practice last week...", "Solution_5": "Yes problem number 3 is from the NC State Contest. It seems hard but it's really easy.", "Solution_6": "there are actually 2 answers as the problem is stated, i believe. At the practice they made sure to say the lines are drawn through a, b, c respectively so there could only be one answer.", "Solution_7": "I still would like to see more people try these problems...", "Solution_8": "is the answer 193?", "Solution_9": "I don't know if this is the quickest way, but oh well...\n\n[hide]\n\n6. we have sides a, b, c\n\n\n\na = 6 + 8 = 14\n\nb = 8 + x\n\nc = 6 + x\n\nr = 4\n\n\n\nr = (1/2) :sqrt: ((a+b-c)(a+c-b)(b+c-a)/(a+b+c))\n\n\n\nplugging them in and solving for x, we get x=7\n\n\n\nso the shortest is 13?[/hide]", "Solution_10": "Yes, 193 is the correct answer, but now show how you got it.", "Solution_11": "after drawing the two perpendicular lines between the three parallel lines to sho distances, can prove two right-angle triangles are congruent, each of the shorter sides' length is 7 and 12.", "Solution_12": "You are correct. That's what I did also :-)." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$x^4+y^4<4$\r\n$x^3+y^3>3$\r\nProve\r\n$x^2+y^2>2$", "Solution_1": "[quote=\"Beat\"]$x^4+y^4<4$\n$x^3+y^3>3$\nProve\n$x^2+y^2>2$[/quote]\r\n\r\n I love it because it very easy:\r\n$x^4+x^2 \\geq 2x^3$\r\nSimilarly for y\r\nSo $\\sum (x^4+x^2) \\geq \\sum 2x^3 \\geq 6$\r\n$x^2+y^2 \\geq 2(x^3+y^3)-x^4-y^4 \\geq 6-4=2$", "Solution_2": "by cauchy-schwarz we have that\r\n\r\n$4(x^2+y^2)>(x^4+y^4)(x^2+y^2)\\geq(x^3+y^3)^2> 9 \\\\\\Rightarrow x^2+y^2> \\frac{9}{4}>2$" } { "Tag": [ "complex numbers", "inequalities unsolved", "inequalities" ], "Problem": "[i][u] Problem [/u]: Let $ a \\ , \\ b \\ , \\ c \\ , \\ d$ be real numbers such that :\n\n $ ( 1 \\ \\plus{} \\ a^{2} )( 1 \\ \\plus{} \\ b^{2} )( 1 \\ \\plus{} \\ c^{2} )( 1 \\ \\plus{} \\ d^{2} ) \\ \\equal{} \\ 16$\n\n Prove that :\n\n\n $ \\minus{}3 \\ \\leq \\ ab \\ \\plus{} \\ ca \\ \\plus{} \\ ad \\ \\plus{} \\ cb \\ \\plus{} \\ bd \\ \\plus{} \\ cd \\ \\minus{} abcd \\ \\leq \\ 5$\n\n \n For my love [/i]\r\n\r\n[i] I hope she will be happy[/i] :oops:", "Solution_1": "OK.\r\nLet S = ab + bc + cd + da + ac + bd - abcd\r\nSo (S - 1)^2 = (1 - ab)(cd - 1) + (a + b)(c + d)\r\n\r\nBy the Cauchy - Schwarz, we have:\r\n (S - 1)^2 <= [(1 - ab)^2 + (a + b)^2] + [(1 - cd)^2 + (c + d)^2] = (1 + a^2)(1 + b^2)(1 + c^2)(1 + d^2) = 16\r\n\r\nAnd -4 <= S - 1 <= 4 or -3 <= S <= 5. This complete our proof.", "Solution_2": "[u] Solution :[/u]\r\n\r\n Using complex numbers , we can rewrite the condition :\r\n\r\n $ \\prod( i \\ \\plus{} \\ a) \\prod( a \\ \\minus{} \\ i) \\ \\equal{} \\ 16$\r\n\r\n Deduce $ \\prod( i \\ \\plus{} \\ a) \\prod( i \\ \\minus{} \\ a) \\ \\equal{} \\ 16$\r\n\r\n By Viette ' s theorem \r\n\r\n We have :\r\n \r\n $ ( i^{4} \\ \\minus{} \\ i^{3}\\sum a \\ \\plus{} \\ i^{2}\\sum ab \\ \\minus{} \\ i\\sum abc \\ \\plus{} \\ abcd )( i^{4} \\ \\plus{} \\ i^{3}\\sum a \\ \\plus{} \\ i^{2}\\sum ab \\ \\plus{} \\ i\\sum abc \\ \\plus{} \\ abcd) \\ \\equal{} \\ 16$", "Solution_3": "[i][u] Solution :[/u]\n\n Denote $ \\sum a \\ \\equal{} \\ a \\ \\plus{} \\ b \\ \\plus{} \\ c \\ \\plus{} \\ d$\n \n $ \\sum ab \\ \\equal{} \\ ab \\ \\plus{} \\ bc \\ \\plus{} \\ cd \\ \\plus{} \\ da \\ \\plus{} \\ ac \\ \\plus{} \\ bd$\n\n $ \\sum abc \\ \\equal{} \\ abc \\ \\plus{} \\ abd \\ \\plus{} \\ acd \\ \\plus{} \\ bcd$ \n Using complex numbers , we can rewrite the condition :\n\n $ \\prod( i \\ \\plus{} \\ a) \\prod( a \\ \\minus{} \\ i) \\ \\equal{} \\ 16$\n\n Deduce $ \\prod( i \\ \\plus{} \\ a) \\prod( i \\ \\minus{} \\ a) \\ \\equal{} \\ 16$\n\n By Viette ' s theorem \n\n We have :\n \n $ \\left( i^{4} \\ \\minus{} \\ i^{3}\\sum a \\ \\plus{} \\ i^{2}\\sum ab \\ \\minus{} \\ i\\sum abc \\ \\plus{} \\ abcd \\right) \\left( i^{4} \\ \\plus{} \\ i^{3}\\sum a \\ \\plus{} \\ i^{2}\\sum ab \\ \\plus{} \\ i\\sum abc \\ \\plus{} \\ abcd \\right) \\ \\equal{} \\ 16$\n \n So , we have \n\n $ \\left( 1 \\ \\plus{} \\ i\\sum a \\ \\minus{} \\sum ab \\ \\minus{} \\ i\\sum abc \\ \\plus{} abcd \\right) \\left( 1 \\ \\minus{} \\ i\\sum a \\ \\minus{} \\sum ab \\ \\plus{} \\ i\\sum abc \\ \\plus{} abcd \\right) \\equal{} \\ 16$ \n\n ( because $ i^{2} \\ \\equal{} \\ \\minus{} 1$ )\n\n $ \\left( 1 \\ \\minus{} \\ \\sum ab \\ \\plus{} \\ abcd \\right)^{2} \\minus{} \\left( i ( \\sum a \\ \\minus{} \\ \\sum abc ) \\right)^{2} \\ \\equal{} \\ 16$\n\n $ \\rightarrow \\left( 1 \\ \\minus{} \\ \\sum ab \\ \\plus{} \\ abcd \\right)^{2} \\plus{} \\left( \\sum a \\ \\minus{} \\ \\sum abc \\right)^{2} \\ \\equal{} \\ 16$\n\n $ \\rightarrow \\left( 1 \\ \\minus{} \\ \\sum ab \\ \\plus{} \\ abcd \\right)^{2} \\ \\leq 16$\n\n $ 4 \\ \\geq \\ | \\sum ab \\ \\minus{} \\ abcd \\ \\minus{} \\ 1 |$\n\n $ \\rightarrow \\minus{} 4 \\ \\leq \\ ab \\ \\plus{} \\ bc \\ \\plus{} \\ cd \\ \\plus{} \\ ac \\ \\plus{} ad \\ \\plus{} \\ bd \\ \\minus{} \\ abcd \\ \\minus{} 1 \\leq \\ 4$\n\n $ \\rightarrow \\minus{} 3 \\ \\leq \\ ab \\ \\plus{} \\ bc \\ \\plus{} \\ cd \\ \\plus{} \\ ac \\ \\plus{} ad \\ \\plus{} \\ bd \\ \\minus{} \\ abcd \\ \\leq \\ 5$\n\n\n Nice problem[/i]" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME" ], "Problem": "http://www.facebook.com/group.php?gid=11314939894", "Solution_1": "Oh man I totally want to facebook friend random AoPS people in this group haha.", "Solution_2": "So join! If you hurry up we'll have a majority of Illinoisians in the group :D", "Solution_3": "Nice group; I joined. Right now there are mostly people from Illinois. :maybe:", "Solution_4": "This is perfect. Now everyone else will be distracted with facebook while I swoop in and own the USAMO :)", "Solution_5": "I don't really understand the point of this group. -_-;;", "Solution_6": "lol i dont even have a facebook, waste of time...", "Solution_7": "Time you enjoyed wasting, was not wasted.\r\nWHO SAID IT? :D", "Solution_8": "John Lennon or Bertrand Russell, with the intention of wasting other ppl's time. :) \r\n\r\n@Hamster: I swear, ever since I chose my username, I notice the word a lot more. I don't have a facebook either. :lol:", "Solution_9": "I essentially only use Facebook to look up contact information and talk with my former math teacher.", "Solution_10": "thats kinda sad worthawholebean", "Solution_11": "nice group..", "Solution_12": "I joined facebook just to join this\r\n\r\nI will probably never go on again", "Solution_13": "Right now 44 members, 145 unreplied...\r\nWe should make an AIME group too. MORE PEOPLE", "Solution_14": "[quote=\"moogra\"]Right now 44 members, 145 unreplied...\nWe should make an AIME group too. MORE PEOPLE[/quote]\r\nIt's at 49 i think...", "Solution_15": "[quote=\"moogra\"]Right now 44 members, 145 unreplied...\nWe should make an AIME group too. MORE PEOPLE[/quote]\r\nlol yay now the 202-pointers won't be left out.. is it made yet?", "Solution_16": "Actually, I believe that I am the 66th member.\r\nYay" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "this is a variant of the question that was posed in this thread by carcul...\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=202038\r\n\r\n\r\nbtw/ the answer to that is $ \\boxed{\\dfrac{3\\,\\pi^3\\,\\sqrt {\\; 2\\; }}{128}}$ (provided my calculations are right)\r\n\r\n\r\nin this thread, i want to post a similar kind of question... :D \r\n\r\n\r\ncan you sum this series\r\n\r\n\r\n$ 1 \\plus{} \\dfrac{1}{3^3} \\minus{} \\dfrac{1}{5^3} \\minus{} \\dfrac{1}{7^3} \\plus{} \\dfrac{1}{9^3} \\plus{} \\dfrac{1}{{11}^3} \\plus{} \\dfrac{1}{{13}^3} \\minus{} \\dfrac{1}{{15}^3} \\minus{} \\dfrac{1}{{17}^3} \\minus{} \\dfrac{1}{19^3} \\plus{} \\dfrac{1}{21^3} \\plus{} \\dfrac{1}{{23}^3} \\minus{} \\dfrac{1}{{25}^3} \\minus{} \\dfrac{1}{{27}^3}\\;\\;\\; \\plus{} \\plus{} \\plus{} \\;\\;\\; \\minus{} \\minus{} \\minus{} \\;\\;\\;\\plus{} \\plus{} \\;\\;\\; \\minus{} \\minus{} \\;\\;\\; \\plus{} \\plus{} \\plus{} \\;\\;\\; \\minus{} \\minus{} \\minus{} \\;\\;\\;\\plus{} \\plus{} \\;\\;\\; \\minus{} \\minus{} \\;\\;\\;\\ldots \\;\\;\\;\\ldots \\;\\;\\;\\ldots$", "Solution_1": "[quote=\"misan\"]$ \\boxed{\\dfrac{3\\,\\pi^3\\,\\sqrt {\\; 2\\; }}{128}}$ (provided my calculations are right)[/quote]\n\nThe answer is right.\n\n[quote=\"misan\"]can you sum\n\n$ 1 \\plus{} \\dfrac{1}{3^3} \\plus{} \\dfrac{1}{5^3} \\minus{} \\dfrac{1}{7^3} \\minus{} \\dfrac{1}{9^3} \\minus{} \\dfrac{1}{{11}^3} \\plus{} \\dfrac{1}{{13}^3} \\plus{} \\dfrac{1}{{15}^3} \\plus{} \\dfrac{1}{{17}^3} \\minus{} \\dfrac{1}{19^3} \\minus{} \\dfrac{1}{21^3} \\minus{} \\dfrac{1}{{23}^3} \\;\\;\\; \\plus{} \\plus{} \\plus{} \\;\\;\\; \\minus{} \\minus{} \\minus{} \\;\\;\\;\\ldots$[/quote]\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=201988", "Solution_2": "oh ! sorry... :( Kouichi showed that i had already posted that question... and that too, just a few days ago... \r\n\r\n i have edited the 1st post and have now posed a different but related question... :huh:", "Solution_3": "Misan, have you have ever heard of \"hide the answer\"? Why are you always so eager to post answers and problems in an uncontroled way?", "Solution_4": "[quote=\"Carcul\"]Misan, have you have ever heard of \"hide the answer\"? Why are you always so eager to post answers and problems in an uncontroled way?[/quote]\r\n\r\nyou should notice i didn't really post the answer openly in [u]your[/u] thread. i will tend to hide an answer if i think it is necessary. personally, i don't find it the least appealing when people tend to hide every single answer that they post, especially after a thread has been in existence for a few days." } { "Tag": [], "Problem": "Can $ \\text{Cu}$ and $ \\text{HAc}$ interact with each other under high temprature? If they can, what is the end product?\r\n\r\nAlso, can $ \\text{FeS}_2$ interact with $ \\text{HAc}$ when the temp is high?\r\n\r\nThanks!", "Solution_1": "Predicting the outcome of high temperature reactions is not generally easy, unless some previous knowledge is used. Also, what is \"high temperature\" to you?" } { "Tag": [ "trigonometry" ], "Problem": "Solve the equation: $\\4\\sin^{2}\\left(x+\\frac{\\pi}{6}\\right)+\\sin 2x=1$", "Solution_1": "It is very easy.\r\nWe have:\r\n$sin2x=1-sin^{2}\\left(x+\\frac{\\pi}{6}\\right)=cos^{2}\\left(x+\\frac{\\pi}{6}\\right)=1+cos2\\left(x+\\frac{\\pi}{6}\\right)=1+cos\\left(2x+\\frac{\\pi}{3}\\right)=\\frac{1}{2}.cos2x-\\frac{\\sqrt{3}}{2}sin2x$\r\n$\\Leftrightarrow tan2x=\\frac{1}{2+\\sqrt{3}}=tg\\alpha$\r\nHence: $x=\\frac{\\alpha}{2}+k\\frac{\\pi}{2}(k\\in\\mathbb{Z})$" } { "Tag": [ "inequalities", "quadratics", "algebra", "inequalities unsolved" ], "Problem": "Find the greatest real number $ C$ such that, for all real numbers $ x$ and $ y\\not\\equal{} x$ with $ xy\\equal{}2$ it holds that:\r\n\r\n$ \\frac{((x\\plus{}y)^2\\minus{}6)((x\\minus{}y)^2\\plus{}8)}{(x\\minus{}y)^2} \\ge C.$\r\n\r\nWhen does equality occur?", "Solution_1": "Let $ a \\equal{} x \\plus{} y$ and $ b \\equal{} x \\minus{} y$ so that $ y \\ne x$ implies $ b \\ne 0$, and $ xy \\equal{} \\left(\\dfrac{a \\plus{} b}{2}\\right)\\left(\\dfrac{a \\minus{} b}{2}\\right) \\equal{} 2$ implies $ a^2 \\equal{} b^2 \\plus{} 8$.\r\nThe inequality $ \\frac {((x \\plus{} y)^{2} \\minus{} 6)((x \\minus{} y)^{2} \\plus{} 8)}{(x \\minus{} y)^{2}}\\ge C$ can be rewritten as $ \\dfrac{(a^2 \\minus{} 6)(b^2 \\plus{} 8)}{b^2} \\ge C$, which is equivalent to\r\n\\[ \\dfrac{(b^2 \\plus{} 2)(b^2 \\plus{} 8)}{b^2} \\ge C.\\]\r\nLet $ u \\equal{} f(b) \\equal{} \\dfrac{(b^2 \\plus{} 2)(b^2 \\plus{} 8)}{b^2}$.\r\nThis implies $ ub^2 \\equal{} b^4 \\plus{} 10b^2 \\plus{} 16$ and therefore\r\n\\[ b^4 \\plus{} (10 \\minus{} u)b^2 \\plus{} 16 \\equal{} 0.\\]\r\nSince $ b^2$ is a real number, and we have a quadratic equation in $ b^2$, the discriminant is nonnegative, so\r\n\\[ (10 \\minus{} u)^2 \\minus{} 64 \\ge 0\\]\r\n\r\n\\[ (u \\minus{} 10 \\minus{} 8)(u \\minus{} 10 \\plus{} 8) \\ge 0\\]\r\n$ u \\ge 18$ or $ u \\le 2$.\r\n$ u \\le 2$ cannot hold. If it does then the equation $ b^4 \\plus{} (10 \\minus{} u)b^2 \\plus{} 16 \\equal{} 0$ has no positive (or zero) roots in $ b^2$ by Descartes' Law of Signs, leading to a contradiction because $ b$ is a real number.\r\nThus $ u \\ge 18$. Equality holds when $ a \\equal{} 2 \\sqrt {3}$ and $ b \\equal{} 2$, or $ x\\equal{}\\sqrt{3}\\plus{}1$ and $ y\\equal{}\\sqrt{3}\\minus{}1$, so $ C \\equal{} 18$." } { "Tag": [ "integration", "real analysis", "real analysis solved" ], "Problem": "$E=\\{f [0,1]\\rightarrow R,C^{2} \\; f(0)=0\\; f(1)=1\\}$ \r\n\r\nFind the largest real number $u$ such that $u\\leq \\int_{0}^{1}|f''(x)-f(x)|dx $ for all $f\\in E$", "Solution_1": "I know it is your favorite kind of problems :) I was sure that I will see \"Created by Moubinool\" when I read the problem.", "Solution_2": "The infinum - in fact, the minimum - of that functional is zero.\r\n\r\nLet $f(x)=\\frac{\\sinh x}{\\sinh 1}.$" } { "Tag": [ "probability", "probability and stats" ], "Problem": "Hi all, I have a query in probability, this actually occurred to me while I was trying to solve another problem.\r\n\r\n[b] Question [/b]\r\n\r\nGiven an examination paper with 10 questions, with each question having 4 options, what is the probability of answering correctly:\r\n\r\n(a) A single individual question (It's 0.25, right? I mean, since there are 4 options and answering correctly means picking the correct option, which is 1/4 )\r\n\r\n(b) Assuming I randomly answer all the questions, then how many questions will I be able to answer correctly? (That is- If close my eyes, and answer all the questions, then, how many questions will I answer correctly?)\r\n\r\n(c) What's the probability of answering all the questions correctly?\r\n\r\nP.S.: I am rather naive in probability theory, so these questions may appear pretty silly.", "Solution_1": "a: right\r\nb: this is called a Binomial distribution: http://en.wikipedia.org/wiki/Binomial_distribution\r\nc: this is equal to $ \\frac {1}{4^n}$ where $ n$ is the number of question.Have you already learned the concept of \"independent random variables\" ?\r\n\r\nHow far have you been in your probability course ?", "Solution_2": "No, I don't pursue a degree in probability, my baccalaureate degree is in computer science, but, I had taken a course in probability at the undergraduate level. \r\n\r\nI might have learned about 'The independent random variable' in probability, but I need to refresh my knowledge of probability. \r\n\r\nI just dabble in probability for fun :lol:" } { "Tag": [ "inequalities", "algebra solved", "algebra" ], "Problem": "a,b,c>0 so that a+b+c=3\r\nprove that: \r\na^(1/2)+b^(1/2)+c^(1/2) >= ab+bc+ca", "Solution_1": "Id say that LHS<= a^2+ b^2+ c^2\r\nNow if we say x^2+ y^2+ z^2=3, then\r\nx+y+z >= x^4+ y^4+ z^4, or\r\nSum x*(x^3 - 1)>=0", "Solution_2": "Sorry for interruption, but here goes\r\nWLOG\r\nx>=y>=z\r\nx^3 - 1>=y^3 - 1>=z^3 - 1\r\nApllying the arranged sequences ineq (Also know as chebyshev) we get\r\n3*RHS>=(x+y+z)(x*3 + y^3 + z^3 -3 )>=0.\r\nWe just will prove that x^3+y^3+z^3>=x^2+y^2+z^2 = 3.\r\nI hope its right.\r\nI have to go.\r\nContinue Later", "Solution_3": "are you sure it's correct? I've not all understood.\r\n\r\nyou want to prove that\r\nx+y+z>=x^4+y^4+z^4\r\nor\r\nSum x(x^3-1) <= 0 wich is false, isn't it ?\r\n\r\nif x>=y>=z\r\nSum x(x^3-1) >= (1/3)(x+y+z)(x^3+y^3+z^3-3)\r\nbut\r\nx^3+y^3+z^3>=x^2+y^+z^2\r\nbecause Sum x(x^2-1)>=(1/3)(x+y+z)(x^2+y^2+z^2-3)=0\r\nthen Sum x(x^3-1)>=0\r\n\r\nwhere am I wrong ?", "Solution_4": "hey, here is my soln\r\n\r\nwe're going to prove that if x^2+y^+z^2=3 \r\nthen x+y+z>=(xy)^2+(yz)^2+(zx)^2\r\n\r\nfirst, we notice that \r\n9=(x^2+y^2+z^2)^2=x^4+y^4+z^4+2*( (xy)^2+(yz)^2+(zx)^2 )\r\nso\r\n(xy)^2+(yz)^2+(zx)^2 = ( 9 - (x^4+y^4+z^4) )/2\r\n\r\nHence we only have to prove that\r\n2(x+y+z)+x^4+y^4+z^4 >= 9\r\n\r\nbut x^4+2x >= 3*x^2 because x^4+2x-3*x^2=x(x+2)(x-1)^2\r\n\r\nthen\r\n2(x+y+z)+x^4+y^4+z^4 >= 3*(x^2+y^2+z^2)=9\r\n\r\nthat's all.\r\n\r\nbye." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let A be an nxn matrix in the field of complex numbers. Let A* denote the conjugate transpose of A. If AA*=i(A*A), is it true that det(A)=0 ?", "Solution_1": "Note that $\\text{tr}(A^{*}A)=\\text{tr}(AA^{*})=\\sum_{i,j}|a_{ij}|^{2}$; that's a nonnegative real number which is zero iff $A=0$.", "Solution_2": "So in general it is not true, am i right?", "Solution_3": "As [b]jmerry[/b] pointed out, $AA^{*}=i(A^{*}A)$ implies $A=0$, hence $\\det(A)=0$." } { "Tag": [ "geometry" ], "Problem": "$ P$ is a point inside $ \\triangle ABC$ such that $ \\angle PAC \\equal{} \\angle PBC$. The perpendicular from $ P$ to $ BC$ and $ CA$ meet the sides at $ L$ and $ M$ respectively. $ D$ is the midpoint of $ AB$. Prove that $ DL \\equal{} DM$.", "Solution_1": "[b]lemma:let $ M$ be the midpoint of side $ BC$ of $ \\triangle ABC$,now construct triangles $ ADB,AEC$ outside $ \\triangle ABC$ such that $ \\angle CAE \\equal{} \\angle BAD \\equal{} \\alpha$ and $ \\angle CEA \\equal{} \\angle BDA \\equal{} 90$.prove that $ MD \\equal{} ME$[/b]\r\n[hide=\"proof of the lemma\"]\nlet $ K,L$ be the midpoints of $ AC,AB$ respectively,so $ MK\\parallel AB$ and $ ML\\parallel AC$ so:\n\n$ \\angle MKC \\equal{} \\angle MLB \\equal{} \\angle BAC$ (*)\n\nalso we have:\n\n$ MK \\equal{} \\frac {c}2,ML \\equal{} \\frac {b}2$ (I)\n\nalso in right-angled triangle $ \\triangle ABD$,$ DL$ is the median thus $ DL \\equal{} LB \\equal{} LA \\equal{} \\frac {c}2$ and also $ \\angle DLB \\equal{} 2\\alpha$ in a similar way we get that $ KE \\equal{} \\frac {b}2$ and $ \\angle CKE \\equal{} 2\\alpha$ hence from (*),(I) we get that:\n$ MK \\equal{} DL,EK \\equal{} ML,\\angle MKE \\equal{} \\angle MLD$ therefore $ \\triangle MLD \\equal{} \\triangle MKE$ so $ ME \\equal{} MD$.\n--------------------------------------------------------------------------------[/hide]\r\n\r\nnow if we use the above lemma for triangle $ \\triangle PAB$ and points $ M,L$ we get that $ DM \\equal{} DL$ as wanted..." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "i know it's obvious but how do you proof that a if a monic polynom has no integer roots, it has no rational roots either ?", "Solution_1": "[quote=\"pureblack2003\"]i know it's obvious but how do you proof that a if a monic polynom has no integer roots, it has no rational roots either ?\n[/quote]\r\nI don't quite understand what you are asking, but isn't $P(X)=X-\\frac{1}{3}$ a counterexample?", "Solution_2": "Monic integer polynomial.\r\n\r\nWell, just assume a rational root to be $x=\\frac{p}{q}$ and set this into the equation $f(x)=0$, multiply by some power of $q$ and conclude that $q|1$, thus $q=\\pm 1$.", "Solution_3": "yea zetax that's i was talking about, thanks . i forgot to mention that it was an integer polynom :blush:\r\n\r\nso ... why does $q|1$ ? :oops:", "Solution_4": "If $P(X)=X^{n}+a_{n-1}X^{n-1}+...+a_{0}$ is an integer polynomial with the root $\\frac{p}{q}$ then $p^{n}+a_{n-1}qp^{n-1}+a_{n-2}q^{2}p^{n-2}+...+a_{0}q^{n}=0$ and if we take it modulo $q$ it gives that $(p,q)!=1$", "Solution_5": "This is part of the rational root theorem- one of those important theorems you learn in high school algebra.", "Solution_6": "yea that's how it's supposed to be .. but nobody taught me this. i knew about it , but what i wanted was a demonstration of it so i could understand it better ;)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "The sequence $ {c_n}$ is defined by $ c_1 \\equal{} c_2 \\equal{} 1$ and if n>2 $ c_{n} \\equal{} (n\\minus{}2) (c_{n\\minus{}1} \\plus{} c_{n\\minus{}2})$ Prove that $ c_{n}$ is congruent to -1 modulo n iff n is odd or n=2.", "Solution_1": "When $ n\\equal{}1$, $ c_n\\equal{}1$ which is not necessarily congruent to -1 modulo $ n$. Did I miss something?", "Solution_2": "okay excuse me if n is odd then c_n (modn)=-1 and if n is even and n>2 then c_n(modn) # -1", "Solution_3": "[quote=\"shoki\"]The sequence $ {c_n}$ is defined by $ c_1 \\equal{} c_2 \\equal{} 1$ and if n>2 $ c_{n} \\equal{} (n \\minus{} 2) (c_{n \\minus{} 1} \\plus{} c_{n \\minus{} 2})$ Prove that $ c_{n}$ is congruent to -1 modulo n iff n is odd or n=2.[/quote]\r\n$ c_n\\equal{}(n\\minus{}1)!$ so this congruence holds iff $ n$ is a prime .", "Solution_4": "in fact i wanted the answer to solve this problem\r\nthe sequence $ a_{n}$ is defined by $ a_1 \\equal{} 2$ and $ a_2 \\equal{} 1$ and \r\n$ a_{n \\plus{} 2} \\equal{} \\frac {n(n \\plus{} 1)a_{n \\plus{} 1} \\plus{} n^2a_{n} \\plus{} 5}{n \\plus{} 2} \\minus{} 2$\r\nfind all integer $ n$ for which $ a_{n}$ is an integer.\r\nthx TTsp... it was easy :(", "Solution_5": "[quote=\"shoki\"]in fact i wanted the answer to solve this problem\nthe sequence $ a_{n}$ is defined by $ a_1 \\equal{} 2$ and $ a_2 \\equal{} 1$ and \n$ a_{n \\plus{} 2} \\equal{} \\frac {n(n \\plus{} 1)a_{n \\plus{} 1} \\plus{} n^2a_{n} \\plus{} 5}{n \\plus{} 2} \\minus{} 2$\nfind all integer $ n$ for which $ a_{n}$ is an integer.\nthx TTsp... it was easy :([/quote]\r\nIt is not very hard to show that :$ a_n\\equal{}\\frac{(n\\minus{}1)!\\plus{}1}{n}$ ,so it is a integer iff $ n$ is a prime or equal 1 .", "Solution_6": "it is iran 2005 (i think)" } { "Tag": [ "inequalities", "function", "algebra", "polynomial" ], "Problem": "I have recently started to learn some of the basic inequalities (AM-GM, Cauchy, Jensen, etc.) but I am stuck on the concept of Homogenuity. Can someone please help me understand this or show me a good book/website/anything.\r\nSpecifically, in this question...\r\n\r\nFor positive reals $ a,b,c$ show that\r\n$ \\dfrac{a}{\\sqrt{a^2 \\plus{} 8bc}} \\plus{}\\dfrac{b}{\\sqrt{b^2 \\plus{} 8ac}} \\plus{}\\dfrac{c}{\\sqrt{c^2 \\plus{} 8ab}} \\ge 1$\r\n\r\nIf this equation is homogenous, why can you assume that $ a\\plus{}b\\plus{}c\\equal{}1$?", "Solution_1": "Let's say a,b,c worked in the given. If you plug in ak,bk, and ck, and simplified, you would get the same result (the ks would cancel), so it must also be true for ak,bk, and ck. Thus, we can scale a,b, and c to whatever we want, so we can scale them so their sum is one. For exmple, if we wanted to show it for a=3 b=5 c=6 we could instead use a=3/14 b=5/14 c=6/14", "Solution_2": "I see now\r\ncheers", "Solution_3": "I don't know what's homogenous means, can anyone explain for me?", "Solution_4": "A function $ f(a, b, c)$ is homogeneous of degree $ n$ if $ f(ta, tb, tc) \\equal{} t^n f(a, b, c)$. Homogeneity is a useful condition to look for because an equality $ f(a, b, c) \\equal{} 0$ holds if and only if it holds for $ ta, tb, tc$, and an inequality $ f(a, b, c) \\ge 0$ holds if and only if it holds for $ ta, tb, tc$ where $ t > 0$.", "Solution_5": "Though not as relevant in the case of inequalities, a further reason to like homogeneity is that we can always eliminate one variable; if $ f(a_1, a_2, \\ldots a_n)$ is homogeneous in degree $ d$, then dividing through by $ a_n^d$ gives a polynomial in $ a_1/a_n, a_2/a_n, \\ldots a_{n\\minus{}1}/a_n$ with degree at most $ d$.", "Solution_6": "how can you tell if an inequality is homogenuous?", "Solution_7": "Substitute each variable $ a_1, a_2 \\ldots a_n$ (in the problem above $ a,b,c$) with $ ta_1, ta_2 \\ldots ta_n$ for an arbitrary real $ t$, and try to show what you get is equivalent to the original inequality.", "Solution_8": "m new to these concepts of homogenization and normalization..... Can anyone provide me link that explains it from the scratch... It would be really appreciated........", "Solution_9": "How would the homogeneity solve this problem?", "Solution_10": "Edit:NVM........." } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AMC 10", "\\/closed" ], "Problem": "These questions may be difficult to answer, but I am curious about them, nevertheless:\r\nAbout how many people who take the AMC 12 Problem Series Class get a perfect score on the AMC 12 each year? A very nearly perfect score? Do you think that someone who can make the AIME would be able to get a near-perfect score by putting enough time and effort into this class?\r\n(I am also curious about the AIME and WOOT classes, and how many of the people taking those classes do extremely well on the AMC 12, although I understand that their performance may have more to do with their prior experience than the higher-level class that they are taking. But correct me if I'm wrong.)\r\n\r\nThanks,\r\nSomeone who is looking at taking the AMC 12 (instead of the AMC 10) to make the USAMO this year", "Solution_1": "As said in other threads, they don't keep statistics like that and even if they do, there is no guarantee that the data given to them was valid. (Anyone could lie about their score.) Your success on the AMC 12 depends on how much time and effort you put into the class. You should attempt to do all the weekly problems that are given and participate actively in class. As mentioned before, you won't do amazingly well on the AMC 12 just from attending the class, you must work hard. And in general you should devote time to practice problems (like old AMC 12 problems that can be found [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44]here[/url]) daily so that you can improve your speed on the problems. If you are aiming for a perfect score, then speed is key. You want to finish the first 20 problems as fast as possible, check them, and then be able to do the last five with plenty of time to spare. The last five usually require some extra thought.", "Solution_2": "[quote=\"sunehra\"]As said in other threads, they don't keep statistics like that and even if they do, there is no guarantee that the data given to them was valid. (Anyone could lie about their score.)[/quote]\r\n\r\nWell, the AoPS Community Awards page on the AoPS Wiki has a list of AoPSers who earned a perfect score on the AMC 12--that's why I asked specifically about perfect scores, to find out who on that list gained the most from which classes. I would like to extend my question to the AMC 10, as well, and even the AIME: In general, the people who score very highly (a perfect score is obviously very high) on those competitions take which classes? Or do they do most of their studying independently? Is having a thorough knowledge of AMC math and taking an AMC problem series enough to give one a good shot at a perfect score on test day, or would one need to take something higher level, like WOOT?\r\n\r\nThanks to all who respond. :)", "Solution_3": "People edit the wiki page and add their names on to it. Of course, you can't lie on the wiki page because the names of the perfect scorers are present on the official AMC website.\r\n\r\nAnd yes, there are people who have scored perfectly on the AIME that took the classes. Success in any math competition is all about the effort you put in. The classes simply give you some general and specific strategies and show how they are applied. It is also possible just to work through AoPS textbooks and get perfects. Again, it is all dependent on you. Also, you shouldn't aim for a perfect score, but a high score. A lot of the time, people get very close but one small stupid mistake messes them up. And a perfect score on the AIME means more than a perfect score on the AMC 10/12. That is due to the fact that the AMC 10/12 is a multiple choice test and AIME is short answer. Therefore, it is even possible to get a perfect through guessing on the AMC 10/12 but close to impossible to guess on AIME. WOOT prepares you for AIME and introduces Olympiad math.", "Solution_4": "If you can make the AIME, I suggest taking an AIME class, because the harder AMC 12 problems are about AIME 8-10 level." } { "Tag": [ "function" ], "Problem": "This program is similar to mathematica and math lab in the fact that it is very powerful and has 100s of functions at your disposal.\r\n\r\nThis tool is very powerful for calculations, quick graphs, etc etc.\r\n\r\nIf you register / login you can publish / save your programs and view them on a different computer when logged in.\r\n\r\nThe url is:\r\n\r\nhttp://nixelinc.org/\r\n\r\nIf you have any questions about how to use the calculator the administrators on the site would be glad to help you.", "Solution_1": "or you can download graphcalc\r\n\r\nworks well for me" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "geometry", "3D geometry", "sphere", "Asymptote" ], "Problem": "I don't get why people classified any number dividing by zero is undefined. My opinion to that is that dividing by zero yields infinity, and ironically, infinity is an undefined value, despite the fact that it is the largest number in the universe. However, I can prove that infinity is a neutral number, thus creating a circle of numbers... and due to the neutrality of that number, dividing by zero from both left hand side and right hand side will yield neutral infinity.\r\n\r\nSo everyone agrees that two parallel vertical lines have a same slope. Obviously, teachers define that \"undefined,\" but truly, I define that infinity. Anyone can see that by slightly tilting a vertical line to the left will yield a very steep negative slope, and vice versa to the right. Therefore this have to prove that negative infinity equals positive infinity :P\r\n\r\nAs the denominator of any fraction approaches zero from the positive side, it will yield positive infinity, and vice versa from the negative side. Since zero = zero and yield two \"apparent\" different solutions, the only thing is that positive infinity must equal negative infinity. Therefore, considering infinity neutral, thus, a circle of numbers are formed starting from zero ending at infinity.\r\n\r\nApplying the facts stated above, our coordinate grid should become a coordinate sphere, making any lines with asymptotes, actual lines crossing y = infinity.\r\n\r\nAny questions, contradictions?", "Solution_1": "You don't understand the way mathematicians define infinity. See [url=http://mathworld.wolfram.com/Infinity.html]Mathworld[/url] and [url=http://en.wikipedia.org/wiki/Infinity]Wikipedia[/url].\r\n\r\nThe real numbers are a [url=http://mathworld.wolfram.com/Field.html]field[/url], and in a field, every element [i]except[/i] $ 0$ has a multiplicative inverse. No element of the real numbers is a multiplicative inverse of $ 0$.\r\n\r\nSome of the concepts you are touching on: the [url=http://en.wikipedia.org/wiki/Point_at_infinity]point at infinity[/url] on the projective line and [url=http://mathworld.wolfram.com/ComplexInfinity.html]complex infinity[/url]. Your use of the word \"neutral\" is meaningless because you haven't defined it. The only thing that can really be gleaned from your argument is that the two-sided limit $ \\lim_{x \\to 0} \\frac{1}{x}$ doesn't exist.", "Solution_2": "I think a closer approximation to what he is doing is the one point compactification of the real numbers, rather than the plane, which forms a circle with infinity on top.", "Solution_3": "Oh, for crying out loud -- haven't we just gone through a \"making things up doesn't count as a proof that you can divide by zero\" thread a few days ago? The real numbers (and the complex numbers, and every other ring) do not contain a multiplicative inverse of 0. If you want the nice algebraic properties that come along with rings or fields (e.g. additive inverses, multiplicative inverses for things other than 0, multiplicative associativity, the distributive property, additive and multiplicative identities), you can't divide by 0. This is not some huge mystery: if we assume all those things I just listed (plus or minus a couple), we can prove (and here I mean prove in the rigorous sense, not the \"hey, if you tilt a vertical line just a little bit, you get a non-vertical line! Far out!\" sense) that multiplying 0 by anything gives 0 -- and this is equivalent to the fact that division by 0 is impossible. This is not deep, or troubling, or even particularly unusual.\r\n\r\nThe fact that the existence of a \"number\" named $ \\infty$ is consistent with certain things you might do (like, say, rotate a line in the plane and calculate its slope while you go) means nothing, because it's inconsistent with many much more important things. I will note one example of this: lines look \"about the same\" with very large positive or very large negative slopes and this was your entire motivation for claiming there is a \"neutral infinity.\" But not everything behaves this way: for example, $ 2^x$ behaves quite differently for very large positive and very large negative values of $ x$. Why, by your reasoning, don't I get to declare $ 0 \\equal{} \\infty$, too?", "Solution_4": "I don't want to start an argument, but this is the reason why certain graphs don't draw the asymptotes in the TI-83. We had this conversation in my math class, and it's a simple reasoning... I don't want to force this theory in anyone's mind, but I think it might be worth considering", "Solution_5": "You're attempting to argue based on a faulty piece of technology known as the TI-83? Please.\r\nHow many times do we, on this forum, have to say the infinity is [b]not a number?[/b] It is merely a convenient notation used.\r\nThe fact that the limit as $ x$ approaches $ 0$ of $ \\frac {1}{x}$ seems to yield different answers does not imply that the two answers are one and the same, it implies that the limit at that point simply does not exist. \r\nAnd what does \"neutral\" mean, anyway? If you mean that it is neither positive nor negative, then hate to break it to you, but there's only one number that satisfies that. You guessed it. $ 0$.\r\nAlso, by your definition, $ \\lim_{x\\to \\plus{} \\infty}{x} \\equal{} \\lim_{x\\to \\minus{} \\infty}{x}$, which is just preposterous.\r\n\r\nIt is a simple exercise in logic to see that dividing by 0 simply does not work.", "Solution_6": "I don't know what to do about it, because all I'm just doing is throwing a theory. I don't get why people have to respond in a rude manner. =/\r\n\r\nI do agree with all of you based that infinity is not a number. It's just a simple symbol for mathematicians to use that recognize that it is denoted the greatest number. It came to me when my math teacher said, \"Why do we have to do all this useless stuff? Because we have to know how technology was able to calculate many things including heart rates, etc.\" We were working on limits in class, and it just annoys me when he states that \"anything divided by zero yields undefined\". But that's the truth (in everyone else's eyes)\r\n\r\nMy proof is not solely based off of a calculator... its just that the calculator is able to prove my point.... if I have some good drawing diagram, I think people will see my point.\r\n\r\nOf course I agree with all the other concepts stated in this forum. I'm a liberal thinker and I consider what would happen if there is a different solution behind all this.\r\n\r\nBut the reason why I posted this is for people who are interested in understanding the unthinkable... I know reality contradicts my \"theory\", and I could care less what reality is... If I offend you even more, I take all my words back... and sorry for this nonsense.", "Solution_7": "Don't worry about the people being rude. They're just annoyed with how many threads on the same topic there are. You haven't really done anything wrong, except wonder the same thing as a bunch of people before you.\r\n\r\nSo, here's the thing. You can make such an infinity work. The question is, is it useful? There was a computer science professor a little while ago who decided to make a new number called nullity (symbol $ \\Phi$), and define $ \\frac {0}{0} \\equal{} \\Phi$, $ \\frac {1}{0} \\equal{} \\infty$, $ \\frac { \\minus{} 1}{0} \\equal{} \\minus{} \\infty$. Now, I don't think that anyone ever actually found an inconsistent statement in his system, but it was totally worthless. The largest criticism was that it was just the same as IEEE floating point arithmetic except that he wanted NaN to equal NaN, where IEEE clearly states that they are not equal. Also, a ton of his axioms had exceptions for $ 0, \\infty, \\minus{} \\infty,$ and $ \\Phi$.\r\n\r\nSo, give an example as to why your definition of infinity is more useful than others. There are good reasons for $ \\frac {c}{0}$ to be undefined - namely you would need a lot of exceptions in your axioms/theorems to handle such a value. Your particular infinity (as said above) arises in one-point compactifications. However, you should understand that there are other definitions of infinity which are equally useful, and you cannot lock yourself into just your own.\r\n\r\nAlso, $ \\lim_{x\\to\\infty}x \\equal{} \\lim_{x\\to\\minus{}\\infty}x$ is not preposterous. In fact, it's true in the one point compactification of the real line.", "Solution_8": "[quote=\"iironiic\"]But the reason why I posted this is for people who are interested in understanding the unthinkable...[/quote] It's not unthinkable -- it's not even extremely interesting. (Certainly not interesting enough to deserve all the words that have been wasted on it.) Unfortunately, I think there's a lot of \"confusing the question\" that's going on here, by you and Hamster. As far as graphs go, the reason we don't draw asymptotes for the exponential graph has literally nothing to do with the question at hand. I could equally well have chosen a hyperbola with one horizontal and one slant asymptote, for example. \r\n\r\nNow, about what Hamster has written: it is certainly true that there are systems in which something like \"division by 0\" is possible, or where the two ends of the real numbers wrap around and touch each other at a point at infinity. But [i]these systems aren't the real numbers[/i]. The real numbers have a lot of very, very nice properties: they are a totally ordered, archimedean field. In exchange for allowing the operation of division by 0, you have to throw out several field axioms, among them the existence of multiplicative and additive inverses and several other nice properties of the reals. Now, sometimes an extension like this is interesting: for example, when we extend the reals to the complex numbers we lose the ordering property, but in exchange we get lots and lots of nice things (e.g., the ability to solve all polynomial equations). That's a useful exchange, and the resulting system has interesting properties. The system of \"the reals plus some quantity for division by 0 minus all the field axioms I've mentioned\" just isn't interesting: you've thrown away far too many nice things, and all you've gotten in return is the ability to divide by 0. This is also true of the nullity nonsense. There is no great mystery about why division by zero isn't allowed, and if you had read some past threads you'd know all this and I wouldn't have to write it out again.\r\n\r\nAbout something else Hamster is going on about: the one-point compactification of the reals is a topological space, it has a point that conventionally is called $ \\infty$, but it doesn't have a natural algebraic structure. Its existence has no bearing whatsoever on questions of division by zero. It does capture the spirit of the geometric ideas the OP is thinking about, though, and I think it would have been nice if Hamster were clearer about the fact that the two ideas really aren't related to each other. Another associated geometric construction is projective space.\r\n\r\nAs an aside: there are extension fields of the real numbers that include infinite quantities. For example, the surreal numbers are such a field. (The piece of the reals you have to give up to get these fields is the archimedean axiom.) But, [i]you still can't divide by 0[/i] in these fields -- it's an operation that is just flat-out inconsistent with the field axioms.", "Solution_9": "[quote]Of course I agree with all the other concepts stated in this forum. I'm a liberal thinker and I consider what would happen if there is a different solution behind all this. [/quote]\r\n\r\nThanks God there is someone else like me here as well, liberal thinker.\r\n@iironiic sorry if people are abit rude,i say go for it , you might be right. dont worry if the facts others say doesnt fit your theory. change the fact!if the fact doesnt fit the theory change the fact!.\r\n\r\nhave a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184288", "Solution_10": "[quote=\"binomial_4eva\"]if the fact doesnt fit the theory change the fact!.[/quote]\r\n\r\nThis is the worst possible way to do mathematics. If you don't understand that by now, there's nothing I can say to change your mind.", "Solution_11": "[quote=\"t0rajir0u\"][quote=\"binomial_4eva\"]if the fact doesnt fit the theory change the fact!.[/quote]\n\nThis is the worst possible way to do mathematics. If you don't understand that by now, there's nothing I can say to change your mind.[/quote]\r\n\r\ni say Its the best way to do mathematics,the facts to you can sometimes can be fake facts which you think is fact. but it might not be a real fact.", "Solution_12": "I think we have established that such a theory would be useless in terms of real numbers, however there is no reason that you should give up on such thinking. It may end up creating intersesting mathematics. Also, as for ignoring already established facts, this should be avoided when making statements about well-established areas of math. This type of thinking is more fitting for separate unexplored areas where there is not yet a definitive set of axioms. Like many people here have said, a theory must serve a purpose without interfering with fact and reason.", "Solution_13": "[quote=\"JBL\"]\nAbout something else Hamster is going on about: the one-point compactification of the reals is a topological space, it has a point that conventionally is called $ \\infty$, but it doesn't have a natural algebraic structure. Its existence has no bearing whatsoever on questions of division by zero. It does capture the spirit of the geometric ideas the OP is thinking about, though, and I think it would have been nice if Hamster were clearer about the fact that the two ideas really aren't related to each other. Another associated geometric construction is projective space.[/quote]\r\n\r\nWell, in some sense the OP was essentially constructing the one-point compactification of the real line, since his vertical line argument essentially a homeomorphism from $ \\mathbb{R}/2\\pi\\mathbb{Z}$ (the circle of angles) to $ \\mathbb{R}\\cup\\{\\infty\\}$. And this vertical line can be viewed as dividing by zero as the OP suggested. However, you are correct in that this is not as related as I may have made it out to be.\r\n\r\nFor a serious attempt at trying to construct a system where you can divide by zero, look up James Anderson and his concept of \"nullity\" along with all the criticisms of it. Specifically. look at how many of his axioms have exceptions for his special numbers. That's why, although a consistent system with division by zero is possible, it's not particularly useful because all the limitations that you have to add make it essentially the same as where you can't divide by zero.", "Solution_14": "Yeah... I dunno how this could be applied to real life, but maybe one day, someone will make use to it... i don't know how, but I hope someone considers my theory and give feedbacks...\r\n\r\nTo be honest, I thought mathematicians are always open to new ideas, and not become dependent on the facts that they already know. I always thought that mathematicians are ready to analyze the unthinkable, and consider whether it's valid or not.\r\n\r\nMaybe how I worded my idea, may throw people off... I agree that infinity is a literally defined value (representing the largest value), but its value remains undefined (despite that no one knows the largest value) (oddest oxymoron). Every contradiction stated in this forum thus far is relied on what I believe to be \"lies.\"\r\n\r\nIf I offend you even more, I'll take it all back... Honestly, I should have not stated this, because apparently no one is considering my theory, which is backed up with true information as well... I should have find a better place to post this -.-;", "Solution_15": "[quote=\"iironiic\"]Yeah... I dunno how this could be applied to real life, but maybe one day, someone will make use to it... i don't know how, but I hope someone considers my theory and give feedbacks...\n\nTo be honest, I thought mathematicians are always open to new ideas, and not become dependent on the facts that they already know. I always thought that mathematicians are ready to analyze the unthinkable, and consider whether it's valid or not.\n\nMaybe how I worded my idea, may throw people off... I agree that infinity is a literally defined value (representing the largest value), but its value remains undefined (despite that no one knows the largest value) (oddest oxymoron). Every contradiction stated in this forum thus far is relied on what I believe to be \"lies.\"\n\nIf I offend you even more, I'll take it all back... Honestly, I should have not stated this, because apparently no one is considering my theory, which is backed up with true information as well... I should have find a better place to post this -.-;[/quote]\r\n\r\nThe main problem with this theory is that there is nowhere to go with it. As Hamster stated, there have been attempts by people to give meaning to the expression $ \\frac {x}{0}$, but their efforts have been largely worthless. Sure, you can make equations like $ x \\plus{} 6 \\equal{} x \\plus{} 8$ \"solvable\" in terms of \"nullity\", but there is no use for this. The axioms contain too many special cases and you are severely limited in what you can accomplish in such a system.\r\n\r\nIt's usually best not to make posts about division by zero on a forum because a.) on less intellectual forums, the arguments will go on forever, and b.) on more intellectual forums, like this one, people will become annoyed with you for bringing up a topic that has been beaten to death so many times. It's defined to be undefined; there's no use doing anything further in this area. \r\n\r\nIt may seem to be an interesting idea at first, but if you look a bit further into it, there's nothing there.", "Solution_16": "[quote=\"Hamster1800\"]And this vertical line can be viewed as dividing by zero as the OP suggested.[/quote] I'm with you until this moment: the one-point compactification certainly has topological meaning, but I completely fail to see the algebraic link you're drawing as anything more than the level of \"well, it would kind of make sense if you thought of it like this, right?\" This kind of thinking doesn't really have anything to do with mathematics.\n\n[quote]but I hope someone considers my theory and give feedbacks...\n...no one is considering my theory, which is backed up with true information as well...\n...the unthinkable, and consider whether it's valid or not...\nEvery contradiction stated in this forum thus far is relied on what I believe to be \"lies.\"[/quote]\nListen, if you hadn't earned rudeness before, you've certainly earned it now. I have devoted several posts in this thread and in others explaining in great detail what is wrong with what you're saying. You have given exactly zero indication of having read or comprehended anything I (or Hamster (who was writing somewhat in your defense) or others) have written on the actual mathematical content of what you're suggesting. Mathematics doesn't work by vague intuition, it works by formal proof and demonstration. If all you can do is vaguely mumble about how my (and others') detailed comments don't amount to consideration of your theory and are \"lies,\" all I can say is that I hope you cut this out and come back when you've grown up (mathematically or otherwise).\n\n[quote]I agree that infinity is a literally defined value (representing the largest value), but its value remains undefined (despite that no one knows the largest value)[/quote] Hamster, I hope you'll join me in agreeing that sentences like this are evidence that the OP really has no clue about the subject at hand. OP, please go read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1107529#1107529]this post[/url].", "Solution_17": "[quote]Listen, if you hadn't earned rudeness before, you've certainly earned it now. I have devoted several posts in this thread and in others explaining in great detail what is wrong with what you're saying. You have given exactly zero indication of having read or comprehended anything I (or Hamster (who was writing somewhat in your defense) or others) have written on the actual mathematical content of what you're suggesting. Mathematics doesn't work by vague intuition, it works by formal proof and demonstration. If all you can do is vaguely mumble about how my (and others') detailed comments don't amount to consideration of your theory and are \"lies,\" all I can say is that I hope you cut this out and come back when you've grown up (mathematically or otherwise). [/quote]\r\n\r\nOk, I'll take it all back... sorry about this. =/ I'm sorry for disappointing all of you. Next time, I should never mention my curiosity perhaps that it will disappoint everyone else... Peter, please delete this forum too (sorry about asking so much)", "Solution_18": "iironiic: It's not that you shouldn't share your curiosities, it's that you should think them through and research them before sharing them. You questioned why dividing by 0 is undefined, which is perfectly fine. What you did not do, however, is search for the reason. Instead you decided to offer a \"'fix\". Well, when you do such a thing, what you should do is to think through your fix and check if it actually has any usefulness. As has been stated many times in both this thread and others, dividing by 0 is just one of those things which is useless to do.\r\n\r\nAs an aside, IEEE floating point arithmetic actually does allow division by 0, with the result of $ \\infty$ or $ \\minus{}\\infty$ based on the sign of the dividend. It leaves $ \\frac{0}{0}$ to give a result of NaN, though. This was the basis for much of the criticism of Anderson's nullity. After considering his system more closely, people realized that it had exactly the same functionality as IEEE arithmetic, but with a small change in the definition of $ \\equal{}$ ($ \\Phi\\equal{}\\Phi$ to Anderson, but $ NaN\\neq NaN$ to IEEE).\r\n\r\nJBL: his map (which is actually from $ \\mathbb{R}/\\pi\\mathbb{Z}$, not $ \\mathbb{R}/2\\pi\\mathbb{Z}$) is taking the slope of a line, which is $ \\frac{\\sin{\\theta}}{\\cos{\\theta}}$, and if you naively plug in $ \\frac{\\pi}{2}$ you get $ \\frac{1}{0}$. Of course, I agree with you in that equating the two is not meaningful, but it can be done (especially since you aren't worried about the algebraic properties when you're doing this anyway). Also, while the sentence you pointed out does indeed indicate that he has not looked into what he's suggesting thoroughly, I don't believe that he deserves or \"has earned\" rudeness.", "Solution_19": "Here the reason such a number system would not be purposeful:\r\n\r\nInfinity does not have value, if it did, then it would be superficial because infinity plus one would be larger.\r\n\r\nTherefore, how could you define the following examples:\r\n\r\n$ \\infty\\minus{}\\infty$\r\n$ 0*\\infty$\r\n$ \\frac{\\infty}{\\infty}$\r\n$ \\frac{\\infty}{0}\\equal{}\\infty$\r\n\r\nIt is possible, but the system would not be closed under addition, subtraction, multiplication or division, not very usefull. These expressions above are called indeterminate forms, meaning they do not posses a numerical value. Once you get to higher mathematics, you will find that there are objects called zero-divisors that have no multiplicative inverse, just like zero in the real number systems.\r\n\r\nOne more thing, division is defined in terms of multiplication. Therefore, $ \\frac{a}{b}\\equal{}c$ can be seen as $ bc\\equal{}a$. There is no $ c$ such that $ c*0\\equal{}1$.", "Solution_20": "Why don't all you search about Cantor's infinity theory?\r\n\r\nI read at a cool book (Baldor's Algebra, very famous for spanish-speakers) something about the \"true value\" of an equation involving n/0 or 0/0 for a determinated value. If you find something like it, you should try to simplify and quit all the variables.\r\n\r\nAnd, if you say that you can divide for zero, then 1=2. I can prove it :mad:", "Solution_21": "[quote=\"JBL\"]I have devoted several posts in this thread and in others explaining in great detail what is wrong with what you're saying. You have given exactly zero indication of having read or comprehended anything I (or Hamster (who was writing somewhat in your defense) or others) have written on the actual mathematical content of what you're suggesting. Mathematics doesn't work by vague intuition, it works by formal proof and demonstration.[/quote]\r\n\r\nIt looks like he's not the only one who needs to read your posts more carefully. This thread should probably just be locked.\r\n\r\npacman: Cantor's notion of infinity is about cardinalities, not real numbers.", "Solution_22": "[quote=\"pacman2812\"]Why don't all you search about Cantor's infinity theory?\n\nI read at a cool book (Baldor's Algebra, very famous for spanish-speakers) something about the \"true value\" of an equation involving n/0 or 0/0 for a determinated value. If you find something like it, you should try to simplify and quit all the variables.\n\nAnd, if you say that you can divide for zero, then 1=2. I can prove it :mad:[/quote]\r\n\r\nshow it", "Solution_23": "[quote=\"Ihatepie\"][quote=\"pacman2812\"]Why don't all you search about Cantor's infinity theory?\n\nI read at a cool book (Baldor's Algebra, very famous for spanish-speakers) something about the \"true value\" of an equation involving n/0 or 0/0 for a determinated value. If you find something like it, you should try to simplify and quit all the variables.\n\nAnd, if you say that you can divide for zero, then 1=2. I can prove it :mad:[/quote]\n\nshow it[/quote]\r\n\r\nI think he means the traditional baloney proof in which $ a \\equal{} b$ and you divide by $ a \\minus{} b$, and thus $ 1 \\equal{}1$ by the division property of equality, apparently.", "Solution_24": "The flaw there is that the function $ f(x) \\equal{} 0 \\cdot x$ doesn't have an inverse function (so it doesn't constitute a reversible step valid in a proof), which is related (but not identical) to the notion that $ 0$ doesn't have a multiplicative inverse.", "Solution_25": "[quote=\"Temperal\"][quote=\"Ihatepie\"][quote=\"pacman2812\"]Why don't all you search about Cantor's infinity theory?\n\nI read at a cool book (Baldor's Algebra, very famous for spanish-speakers) something about the \"true value\" of an equation involving n/0 or 0/0 for a determinated value. If you find something like it, you should try to simplify and quit all the variables.\n\nAnd, if you say that you can divide for zero, then 1=2. I can prove it :mad:[/quote]\n\nshow it[/quote]\n\nI think he means the traditional baloney proof in which $ a \\equal{} b$ and you divide by $ a \\minus{} b$, and thus $ 1 \\equal{} 1$ by the division property of equality, apparently.[/quote]\r\n\r\nwell there are lots of other fake proofs. I would like to see his and show why it is wrong.", "Solution_26": "I'm not quite sure I follow the piling-on of pacman: pacman is absolutely correct that if you make as an axiom that $ a \\cdot 0 \\equal{} b\\cdot 0 \\implies a \\equal{} b$ (\"if you can divide by zero\") then everything falls apart. \r\n\r\n\r\nHamster: okay, fine, there's a sort-of reason to call the extra point $ \\infty$ and there's a sort-of reason to call it $ \\frac{1}{0}$, but (again, and I know you agree with this) these names have no algebraic meaning at all, only topological/geometric. (Actually, the sort-of reason that we have both these sort-of reasons is that the projective line is the same as the one-point compactification.)", "Solution_27": "JBL, how was anyone (except possibly Ihatepie, and that's only one person) \"piling on\"? I, in fact, agreed with his statement, and t0r's comment wasn't related to that.\r\n\r\nIhatepie, he mentioned himself that that was the fallacy in the 1=2 proof; I'm not sure why you would want to \"show why it is wrong\".", "Solution_28": "Oh, good grief! Why is it that these ridiculous \"divide by zero\" threads take so long to die?\r\n\r\nThe reason for the so called rudeness of more experienced posters, my young friend, is that you're not being clever, or innovative here; you're ignoring a logical inconsistency.\r\n\r\nSuppose that you have two numbers whose product is $ 1$ as in $ ab \\equal{} 1$.\r\n\r\nNow, let $ a \\equal{} 0$ so that $ ab \\equal{} 0\\cdot b \\equal{} 0$.\r\n\r\nIn such a case, $ ab \\equal{} 1 \\equal{} 0$ that's a contradiction. Therefore there is no such pair of numbers. It is self evident that $ 0$ exists; it follows therefore that $ b \\equal{}\\frac{1}{a}$ doesn't exist. Without $ b$, division by $ a$ has no meaning; it is undefined.", "Solution_29": "Ok, I take it all back... please let this board die... i'm sorry for mentioning it in the first place... my mistake...", "Solution_30": "[quote=\"Dr. No\"]Why is it that these ridiculous \"divide by zero\" threads take so long to die?[/quote]\r\nBecause people such as yourself revive them, even more than two weeks after the previous post?", "Solution_31": "[quote=\"JBL\"][quote=\"Dr. No\"]Why is it that these ridiculous \"divide by zero\" threads take so long to die?[/quote]\nBecause people such as yourself revive them, even more than two weeks after the previous post?[/quote]\r\nToo true, though I was referring to the fact that it took 28 posts before this one dissolved. I hadn't noticed the date on the last one -- my bad :oops:", "Solution_32": "[quote=\"iironiic\"]Ok, I take it all back... please let this board die... i'm sorry for mentioning it in the first place... my mistake...[/quote]\r\n\r\nI think maybe people were just a bit offended that you first worded the topic as a question, and then went on to discuss your idea as though it were a revolutionary theory that could reasonably contend with hundreds of years of mathematicians accepting otherwise... it's not. You seem to think there is some absolute mathematical truth out there and that it is important to find the truth and this seems to be your motivation for thinking about this. In reality, it isn't about absolute truth, but what logically follows from your axioms and if it yields anything useful.\r\n\r\nMath in high school tends to lead students to believe that real math develops in a similar way to the way it is taught. In school there is always a correct answer to each problem that someone else already knows because the textbooks are designed to have correct answers that can be found only requiring what has been taught... real mathematicians aren't sitting around thinking \"well what's actually right?\" ... it is more about the implications of your chosen \"self-evident truths\"" } { "Tag": [ "geometry", "parallelogram" ], "Problem": "$ 1.)$ Fie triunghiul $ ABC$ - isoscel cu $ m(\\hat{A}) \\equal{} 20^{\\circ}$ . Se duce din varful $ B$ , $ BB_1$ astfel incat $ m(\\widehat{B_1BC}) \\equal{} 60^{\\circ}$ si $ B_1 \\in [AC]$ . Din varful $ C$ se duce ceviana $ CC_1$ , astfel incat $ m(\\widehat{BCC_1}) \\equal{} 50^{\\circ}$ si $ C_1 \\in [AB]$ . Calculati $ m(\\widehat{AB_{1}C_{1}}) \\equal{} ?$\r\n\r\n$ 2.)$ Sa se arate ca triunghiul in care 2 bisectoare sunt congruente este isoscel .\r\n\r\n\r\n[b]OBS[/b] : Caut o solutie care sa foloseasca cunostintele unui elev de clasa a VI-a , a VII-a . Care nu stie lungimea bisectoarei sau alte teoreme :) .\r\n\r\nEdit : Da promath , era o gresala . Mi-am editat postul , scuze :blush: .", "Solution_1": "Problema 2 este cunoscta sub numelede teorema lui Steiner-Lehmus.\r\nO demonstratie simpla a acestei teoreme a fost data de catre Gilbert si Donnell intr-un\r\narticol cu titlul: \"The Steiner-Lehmus Theorem\" din aparut in \"American Mathematical Monthly\" nr.3/1965, p.79-80, \r\nAceasta demonstratie este reluata aproape identic in cartea lui Coxeter si Greitzer: GEOMETRY REVISITED(1967),\r\npe care o reproduc aici in mare: \r\nFie $ ABC$ un triunghi in care $ BM$ si $ CN$ sunt bisectoare, in plus: $ |BM| = |CN|.$ Trebuie sa aratam ca: $ m(\\^B) = m(\\^C).$\r\nSa presupunem ca: $ m(\\^B) < m(\\^C)$ si fie $ L\\in(BM)$ a.i. unghi$ LCN = \\frac {1}{2}.$unghi$ (B) =$unghi$ (LBN) \\Rightarrow$ NBCL-inscriptibil.\r\nInsa: $ |BL| < |CN|\\Rightarrow$arc.$ BL$ m(\\widehat{ABE}$ . Dar in triunghiurile $ BAD$ si $ ABE$ avem ($ AB = AB$ si $ BE=AD$) $ \\Rightarrow BD > AE$ $ (* * *)$.\n\n Pe de alta parte , avem :\n $ m(\\hat{A})>m(\\hat{B}) \\Rightarrow m(\\widehat{DAC})>m(\\widehat{EBC})$ si cum $ m(\\widehat{DAC})=m(\\widehat{EFD})$ , avem ca :\n\n $ \\{\\begin{array}{c} m(\\widehat{EFD}) > m(\\widehat{EBC}) \\\\ \\\\ m(\\widehat{EFB})=m(\\widehat{EBF}) \\end{array} \\Rightarrow m(\\widehat{BFD}) m(\\widehat{ABE}$ . Dar in triunghiurile $ BAD$ si $ ABE$ avem ($ AB = AB$ si $ BE = AD$) $ \\Rightarrow BD > AE$ $ (* * *)$.\n\n Pe de alta parte , avem :\n $ m(\\hat{A}) > m(\\hat{B}) \\Rightarrow m(\\widehat{DAC}) > m(\\widehat{EBC})$ si cum $ m(\\widehat{DAC}) = m(\\widehat{EFD})$ , avem ca :\n\n $ \\{\\begin{array}{c} m(\\widehat{EFD}) > m(\\widehat{EBC}) \\\\\n \\\\\nm(\\widehat{EFB}) = m(\\widehat{EBF}) \\end{array} \\Rightarrow m(\\widehat{BFD}) < m(\\widehat{FBD})$ . Atunci in triunghiul $ BDF$ avem $ BD < DF \\Leftrightarrow BD < AE$ . Contradictie cu $ (* * *)$ . Deci presupunerea facuta este falsa . Analog se demonstreaza si pentru $ m(\\hat{A}) < m(\\hat{B})$ . \n\n Deci $ m(\\hat{A}) = m(\\hat{B})$ .\n\n \n\n[/hide][/quote] pai da figura mie nu prea imi iese cum spui tu sau nu prea ma prind cum ar trebui sa semene", "Solution_8": "Aplica teorema medianei:\r\n\r\n$ m_a\\equal{}m_b \\Rightarrow m_a^2\\equal{}m_b^2 \\Rightarrow$\r\n\r\n$ \\frac{2(b^2\\plus{}c^2)\\minus{}a^2}{4}\\equal{}\\frac{2(a^2\\plus{}c^2)\\minus{}b^2}{4} \\Rightarrow 3a^2\\equal{}3b^2 \\Rightarrow a\\equal{}b$\r\n\r\nAsadar triunghiul dat este isoscel.", "Solution_9": "[b]Alta demonstratie elementara pentru:[/b] $m_b=m_c\\Rightarrow b=c$.\nIn $\\triangle{ABC}$, notez cu $B_1$ si cu $C_1$ mijloacele laturilor $[AC]$ si $[AB]$ si cu $\\{G\\}=[CC_1]\\cap [BB_1]$.\n[b]1).[/b] $|BB_1|=|CC_1|\\Rightarrow |GB|=\\frac{2}{3}.|BB_1|=\\frac{2}{3}.|CC_1|=|GC|\\Rightarrow\\angle{GBC}\\equiv\\angle{GCB}$.\n[b]2).[/b] $\\triangle{BCB_1}\\equiv\\triangle{BCC_1}(LUL)\\Rightarrow|B_1C|=|C_1B|$.\n[b]3).[/b] $|AC|=2.|B_1C|=2.|C_1B|=|AB|.$\n[b]sau si mai simplu:[/b]\n[b]1).[/b] $[B_1C_1]$ -linie mijlocie in $\\triangle{ABC}\\Rightarrow B_1C_1||BC\\Rightarrow BCB_1C_1$ trapez;\n[b]2).[/b] Cum $|BB_1|=|CC_1|\\Rightarrow BCB_1C_1$-trapez isoscel$\\Rightarrow |BC_1|=|CB_1|\\Rightarrow |AB|=|AC|$." } { "Tag": [ "geometry", "email", "blogs", "Alcumus", "videos", "trigonometry" ], "Problem": "hey guys \r\n\r\nthis is sen, i go to james ruse ag high school in sydney NSW\r\n\r\nim in year 10. =)", "Solution_1": "Hey Sen,\r\n\r\nThis is Charles, I go to Sydney Grammar, Yr 12\r\n\r\nIsn't this an busy community? :P", "Solution_2": "charles man how you going? \r\n\r\nlol relaxing or working hard?\r\n\r\nman i gotta study so hard.. to become senior.!!!! half yrlies comng up as well :mad:", "Solution_3": "Im Bonnie, Ruse, Year 10. =)\r\n\r\nWe need more people here. Where's everyone from camp? Good job with the forum. ^^", "Solution_4": "[quote=\"vulgarfraction\"]Good job with the forum. ^^[/quote]Thank sen, he brought the idea up ;)", "Solution_5": "sup im jarrah also from ruse.\r\n\r\ni'm really bad at maths by your standards - i do informatics because coding is what i do.", "Solution_6": "[quote=\"jaz\"]sup im jarrah also from ruse.\n\ni'm really bad at maths by your standards - i do informatics because coding is what i do.[/quote]\r\n\r\njarrah lacko?\r\n\r\nCongrats on making the Informatics team", "Solution_7": "hey jarrah lol strange seeing you on a maths forum =P", "Solution_8": "its because i need to get better at maths to do better in informatics.", "Solution_9": "ne topic in particular you want to learn in maths?", "Solution_10": "Gday guys... Im Julian Hua (AKA Captain Cool) Yr11 at Scotch College in Melbourne\r\n\r\nyeah, decided to be part of this Australian Forum... It's pretty hardcore sen. awesome stuff.\r\n\r\nI'll be posting sooner or later when i get roung to looking at the problems we got from math camp.\r\n\r\nAnyways, I'll catch you guys later...\r\n\r\n:D", "Solution_11": "aite thats kool. hope to see you posting stuff soon", "Solution_12": "*wave* Sean here. The Sydney Grammar dude. Year 10. Drowning in maths questions. And exam preparation. And internet forum/community catch-up.\r\n\r\n... yay diversions!", "Solution_13": "hey sean sup? hows ur canberra holidays? lol", "Solution_14": "*posts just so Sen isn't the only one saying hi to people*\r\n\r\nHey, Sean, if you keep swearing, the forum just dashes out what you say. ;) Caratheodory was a joke..... wasn't it... xP\r\n\r\nAnd, yeah, definitely keep in touch. :)", "Solution_15": "sdort;\n\nTo find the diagnostics go to the AoPS home page and hover your cursor over \"School\" located at the top on the green menu bar.\n\nSelect \"Schedule\" from the drop down menu and the Class List will appear.\n\nSelect a course your interested in and click on it - this will take you to your selected course.\n\nYou will see one or two lines describing the selected course you have chosen then under that it will say [b]\"Diagnostic Tests: Are You Ready? - Do You Need This?\"[/b] - Click on either of these, but attempt both because they both determine whether you do or don't need this level.\n\nAnd there you have it.\n\nI'm sorry to hear about your parents and their indifference towards your studies. Its a shame your so talented and not encouraged to capitalise on your strengths. Maybe you could get a part-time job or get paid tutoring kids younger than you after you strike a deal with their parents.\n\nThere is still Alcumus, the Videos and the Forums to get you busy learning and contributing on the AoPS website.\n\nMaybe you could twist your parents arm and buy just the AoPS textbooks and work through them. I did that in the beginning myself.\n\nGood luck sdort ! I wish you all the best.", "Solution_16": "Haha, quality idea, but I don't think that I am \"talented\" enough to actually use the knowledge and concepts that would be taught at the courses.\n\nConsidering the different curriculum this site is mainly based on, nor do I think that it would help me at schoo anyway.\n\nI'll have a look at the Diagnostics later when I have time to test myself, much is appreciated!\n\nFor now, Alcumus is quite fun to use as it specifies on what you want to learn which I believe is a great system, so I'll play around on it.", "Solution_17": "Hey Guys, any of you going to School of Excellence?", "Solution_18": "Hi i'm dom, and am hoping to do well in Grade 10 Australian Math Comp, in 2013. I'm from mackay, north queensland, where, saddly the concept of mathematics is unknown to the general population :)", "Solution_19": "Hello everyone, people call me Spiral. Glad to see some people from Australia here. :)", "Solution_20": "Anyone do the APMO?", "Solution_21": "I did the APMO, found it pretty hard but almost solved problem 2 and a bit of problem 1. Congratulations to Alex Gunning on getting a perfect score. :)", "Solution_22": "yeah, that is like impossible lol I managed to solve 3 questions...\n\nProblem $1$ mainly concerned trig or the medial triangle and problem $2$ is pretty cool in my opinion.\n\nProblem $5$ took too long for me... need to draw better diagrams I reckon.", "Solution_23": "http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=526918\n\nDude, just post it here...", "Solution_24": "Hi, my name is Amy. I thought I'd say hi and introduce myself because I haven't met anyone mathy yet. I live in the Gold Coast and will start attending QAHS next year. Hope everyone's having a great holiday!", "Solution_25": "Hi, I do not know if anyone is still on this forum but anyway. I am snooze (nick name) and live in Melbourne. I love maths", "Solution_26": "Hey, I'm paulkra (Paul), also in Melbourne. Just going into year 7, where there'll be AMC every year at my next school. It doesn't seem like many people see this, but I thought I'd pop in.", "Solution_27": "Hey Paul,\nI am snooze (my cat's name). What did you get in the amc this year? I failed! :( I got a distinction because I missed question 1 and filled in all the wrong numbers for the first half of the exam. (I was killing myself afterwards because I realized literally as I handed it in. I am going into year 9 next year. What school do you go to? I am home schooled. I am going to school in year 10. Hope to talk to you soon.\n~snooze", "Solution_28": "I was home schooled for a while, but I moved back to school just this year. My current school doesn't do the AMC, so I had to go across Melbourne to try it. I only got a distinction, but I'll probably be doing some more competitions. How long have you been doing AMC?\n", "Solution_29": "[quote=paulkra]I was home schooled for a while, but I moved back to school just this year. My current school doesn't do the AMC, so I had to go across Melbourne to try it. I only got a distinction, but I'll probably be doing some more competitions. How long have you been doing AMC?[/quote]\n\nA distinction is great for your first time! I have been doing the amc since primary school. Like year 4. There are quite a lot of Australian's on aops. Could everyone please start asking your friends (obviously that are interested in maths) to join. I have not done this yet (so I am probably sounding like a hypocrite) but I plan to when I come home in January from the holidays. Great to talk to you and I hope you are enjoying the ridiculously hot summer." } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "Let the sequence $(a_{n})$:$a_{0}=1, \\ \\ a_{n+1}=\\frac{1}{2}(a_{n}+\\frac{1}{3a_{n}})$ for $n=0,1,...$\r\nAnd let $A_{n}= \\frac{3}{3a_{n}^{2}-1}$.\r\nProve that $A_{n}$ is a perfect square and has at least $n$ distinct prime divisors.", "Solution_1": "This was included on a WOOT practice olympiad.\r\n\r\nOne of the easiest most straightforward olympiad problems I have seen, I must say.\r\n\r\n\r\nThe only strategy: just find the $A_{n}$ recursion. From there it is unimaginably easy and it just falls apart.\r\n\r\nSorry, but no more advice I can give you.", "Solution_2": "$a_{1}=\\frac{2}{3}$. If $a_{n}=\\frac{p_{n}}{3q_{n}}$, then $a_{n+1}=\\frac{p_{n+1}}{3q_{n+1}}, \\ p_{n+1}=p_{n}^{2}+3q_{n}^{2},q_{n+1}=2p_{n}q_{n}$.\r\nWe have $p_{1}^{2}-3q_{1}^{2}=1$. By induction $p_{n+1}^{2}-3q_{n+1}^{2}=(p_{n}^{2}+3q_{n}^{2})^{2}-12p_{n}^{2}q_{n}^{2}=(p_{n}^{2}-3q_{n})^{2}=1$. Therefore $(p_{n},q_{n})=1$ and $A_{n}=(3q_{n})^{2}$. Because $p_{m}|q_{n}$, when n>m and $(q_{n},p_{n})=1$ we get $(p_{m},p_{n})=1$. Therefore $q_{n}=2p_{n-1}q_{n-1}=2^{2}p_{n-1}p_{n-2}q_{n-2}=...=2^{n}p_{n-1}p_{n-2}...p_{2}$ and $A_{n}=(3q_{n})^{2}=2^{2n}3^{2}p_{2}^{2}...p_{n-1}^{2}$ had at least n distinct prime divisors.", "Solution_3": "thanks Rust anh me@home. It's only an easy problem. And Rust's solution is similarly with my solution. It's really nice." } { "Tag": [ "probability" ], "Problem": "Zane's parents have offered to take him and a friend to the concert this weekend. Zane knows that six of his friends would like to go to the concert. Each of his friends has equal probability of going or not going to the concert. What is the probability that at least one of Zane's friends can go with him? \r\n\r\nWhat do you think of this?", "Solution_1": "1\r\n\r\nDone.\r\n\r\nMaybe it's a trick question.", "Solution_2": "[quote=\"AznChristmas\"]1\n\nDone.\n\nMaybe it's a trick question.[/quote]\r\n\r\nThats what i thought at first too....but they said it wasnt. I'll wait untill more respones come and i'll reveal the answer", "Solution_3": "100% because their mom said that 1 could go", "Solution_4": "is the question asking for the probobility of one being chosen or just one going?", "Solution_5": "Ok I think the problem itself is messed up-\r\n\r\n1. First of all, it says Zane's parents have offered to take him and a friend to the concert, but the question asks- What is the probability that AT LEAST one of Zane's friends can go with him? \r\n\r\n2. Next it says- Zane knows that six of his friends would like to go to the concert. Right after that it says- Each of his friends has equal probability of going or not going to the concert. what?????", "Solution_6": "I agree that the wording is awful. Maybe the problem meant that each friend (independently) has a 50% chance of going, and the question is what is the probability that at least one of the friends goes.\r\n\r\nWhere did the problem come from?", "Solution_7": "[quote=\"Ravi B\"]I agree that the wording is awful. Maybe the problem meant that each friend (independently) has a 50% chance of going, and the question is what is the probability that at least one of the friends goes.\n\nWhere did the problem come from?[/quote]\r\n\r\nSo what do you say the answer is?", "Solution_8": "[hide=\"revelation\"]what if none go? if there is a 1/2 chance of a friend going, then there still will be a chance of none going.\n1/2^6=1/64\n64-1=63\n>>63/64<<[/hide]", "Solution_9": "Yes, that's the answer I intended for my rephrased question.", "Solution_10": "[quote=\"dwx314\"][hide=\"revelation\"]what if none go? if there is a 1/2 chance of a friend going, then there still will be a chance of none going.\n1/2^6=1/64\n64-1=63\n>>63/64<<[/hide][/quote]\r\n\r\nBut it says that 6 of his friends WOULD LIKE TO GO......", "Solution_11": "which one am i sopposed to believe when it says 6 of his friends would like to go...and each friend has equal probability of going and not going", "Solution_12": "[hide]Even though I usually disagree with dwx(my student :) ) I think I have to go with him on this one. While they all WANT to go, the problem says they only have a 50-50 chance of ACTUALLY being able to go. So dwx's work looks good to me. Does the answer you have say something different??[/hide]", "Solution_13": "[quote=\"Ravi B\"]\nWhere did the problem come from?[/quote]\r\n\r\nThis was from a Math Bee from North South (competition only for indian ppl :P )", "Solution_14": "Maybe all the friends would like to go, but whether a friend can actually go is up to the friend's parents, who like to flip coins to decide...\r\n\r\nHey, I'm an Indian too, but no longer a student.", "Solution_15": "[quote=\"frost13\"][hide]Even though I usually disagree with dwx(my student :) ) I think I have to go with him on this one. While they all WANT to go, the problem says they only have a 50-50 chance of ACTUALLY being able to go. So dwx's work looks good to me. Does the answer you have say something different??[/hide][/quote]\r\n\r\nThe answer is correct...but the solution they gave seems incorrect to me, They told me-\r\n\r\nThe probability of \r\n1 person going- 1/2\r\n2 peple going- 1/4\r\n3 people going- 1/8\r\n....\r\n6 people going- 1/64\r\n\r\nAdd all of those and get 63/64", "Solution_16": "Hey, one of my other team members took that contest(Last Saturday!)\r\n\r\nI think you can look at it like this. If mom says \"I'll flip a coin, and if it's heads you can go, if tails you can't\"\r\n\r\nIf the kids doesn't want to go, then the flip doesn't matter, the kid won't go. If the kid wants to go, then the flip of the coin determines the probability of the kid going.\r\n\r\nThey only tell you the kid wants to go, so that you know if their 50-50 chance comes up for being able to go, you know they won't decline the offer. If you didn't know they wanted to go, then technically you couldn't answer the question as those who could go might choose not to(but you wouldn't know :) )\r\n\r\nNot sure if that clears it up or makes it worse :) I hope better, but if worse, I apologize.", "Solution_17": "When they say the probability of person 2 going, they mean person 2 going but not person 1. Similarly, when they say person 3 going, they mean person 3 going but not persons 1 or 2.", "Solution_18": "[quote=\"Ravi B\"]When they say the probability of person 2 going, they mean person 2 going but not person 1. Similarly, when they say person 3 going, they mean person 3 going but not persons 1 or 2.[/quote]\r\n\r\nYeah...but they said only one friend can come", "Solution_19": "THen this problem is wrong in its own.\r\nIf only one can go, then how can the prob be 1/2 per person?\r\nContradiction...", "Solution_20": "[quote=\"frost13\"]Hey, one of my other team members took that contest(Last Saturday!)\n[/quote]\r\n\r\nThis was last years...", "Solution_21": "[quote=\"dwx314\"]THen this problem is wrong in its own.\nIf only one can go, then how can the prob be 1/2 per person?\nContradiction...[/quote]\r\n\r\nAnd the sad part is..this was on a competition", "Solution_22": "iT DEPENDS ON WHICH IS RIGHT.\r\niF THE ONLY ONE CAN GO PART IS TRUE, THAN ITS (1/6)^6 (like that)", "Solution_23": "I think the problem meant that many friends can go to the concert, but only one friend can go with Zane. Like I said, I think the wording is confusing, but I am trying to make a sensible problem out of it.", "Solution_24": "Wouldn't it be 6/7 because at least one is 6 because of the 6 firiends but there is a cance of aving 0 friends go. :)", "Solution_25": "[quote=\"GoBraves\"]Zane's parents have offered to take him and a friend to the concert this weekend. Zane knows that six of his friends would like to go to the concert. Each of his friends has equal probability of going or not going to the concert. What is the probability that at least one of Zane's friends can go with him?[/quote]\r\nYes, this problem is messed up.\r\n\r\nIt doesn't explicitly say that a maximum of one friend can go, but it rather implies that only one friend can go.\r\n\r\nThe phrase \"equal probability\" could mean $p(\\hbox{going})=p(\\hbox{not going})=\\frac{1}{2}.$ Or the phrase could mean $p(\\hbox{friend 1 going})=$ $p(\\hbox{friend 2 going})=$ $\\ldots=$ $p(\\hbox{friend 6 going}).$ There is clearly not enough information to get an answer, assuming the latter interpretation. Therefore, we must assume the former interpretation is what the problem intended.\r\n\r\nGiven the former interpretation, Zane's parents must have meant they will take Zane and *some* friends.\r\n\r\nThis would all be consistent with the *answer*, but not with the *solution* they gave you. The probabilities of zero through six friends going are: $\\frac{1}{64},$ $\\frac{6}{64},$ $\\frac{15}{64},$ $\\frac{20}{64},$ $\\frac{15}{64},$ $\\frac{6}{64},$ and $\\frac{1}{64}$. So the probability of one or more friends going is $\\frac{63}{64}$.\r\n\r\nWas this question translated to English from another language?", "Solution_26": "[quote=\"rcv\"][quote=\"GoBraves\"]Zane's parents have offered to take him and a friend to the concert this weekend. Zane knows that six of his friends would like to go to the concert. Each of his friends has equal probability of going or not going to the concert. What is the probability that at least one of Zane's friends can go with him?[/quote]\nWas this question translated to English from another language?[/quote]\r\n\r\nNo...this was written in english", "Solution_27": "What's the answer, then?", "Solution_28": "The given answer was 63/64" } { "Tag": [ "function", "calculus", "integration", "calculus computations" ], "Problem": "The number of billboards per mile along a 100 mile strech of an interstate highway approaching a certain city is modeled by the function B(x)=21-e^(0.03x), where x is the distance from the city in miles. About how many billboards are along that stretch of highway?\r\n\r\nI thought it would just be the integral of that function from 0 to 100, but I don't get the right answer. The right answer is suppose to be about 1464. Thanks.", "Solution_1": "$ \\int_0^{100}21\\minus{}e^{.03x}\\,dx\\equal{}2100\\minus{}\\frac1{.03}(e^3\\minus{}1),$ which does indeed round to 1464.\r\n\r\nI'd look at how you did the integral; there must be an error in there somewhere.", "Solution_2": "oh nvm I made a stupid mistake putting 0.003 instead of 0.03. Thanks :blush:" } { "Tag": [ "analytic geometry", "geometry", "graphing lines", "slope" ], "Problem": "[i]Find the vertices of a square which is centered at (2,3), has side length 4, and whose diagonals are parallel to the coordinate axes. [/i]\r\n\r\n[i]Find the equation of the line that passes through (2,4), and bisects the area of the square with vertices (0,0), (2,0), (2,2), and (0,2).[/i]\r\n\r\nI have no clue how to do either of those 2 problems, but I think the answer MIGHT be x=2.\r\nI'm probably way off.", "Solution_1": "[quote=\"batteredbutnotdefeated\"][i]Find the vertices of a square which is centered at (2,3), has side length 4, and whose diagonals are parallel to the coordinate axes. [/i][/quote]\n\nIf the diagonals are parallel to the coordinate axes, the square is like a diamond, and the center is the midpoint of the diagonal. \n\nIt is well known that the diagonal of any square is the side length of the square times $ \\sqrt{2}$. [hide=\"(Why)\"]Since all angles of the square are right angles, the diagonal makes a right triangle with two sides of the square, and the Pythagorean Formula gives $ s^2\\plus{}s^2\\equal{}d^2$, where $ s$ is a side, which is the leg of the right triangle, and $ d$ is a diagonal of the square, in this case the hypotenuse. Thus, solving for $ d$, $ d^2\\equal{}2s^2$, so $ d\\equal{}s\\sqrt{2}$.[/hide] Thus, if the square has length $ 4$, the diagonal has length $ 4\\sqrt{2}$. Because the center is the midpoint and thus by definition splits the diagonal into two congruent segments, each of these segments measures half of $ 4\\sqrt{2}$, or $ 2\\sqrt{2}$. Now, all you need to do is to calculate points along the axes $ 2\\sqrt{2}$ away from the center. These are $ (2\\minus{}2\\sqrt{2},3)$, $ (2\\plus{}2\\sqrt{2},3)$, $ (2,3\\minus{}2\\sqrt{2})$, and $ (2,3\\plus{}2\\sqrt{2})$. \n\n[quote=\"batteredbutnotdefeated\"][i]Find the equation of the line that passes through (2,4), and bisects the area of the square with vertices (0,0), (2,0), (2,2), and (0,2).[/i][/quote]\r\n\r\nLogically, in order to bisect the area of the square, a line will have to pass through the center. The center of this square is $ (1,1)$. Now, we have two points and we can form a line, since two points determine a line. Finding the slope, we use the slope formula $ m\\equal{}\\frac{y_2\\minus{}y_1}{x_2\\minus{}x_1}$. In this case $ m\\equal{}\\frac{4\\minus{}1}{2\\minus{}1}\\equal{}3$. Now, we can plug the slope and either of the two points we originally had into the point-slope formula and simplify that into slope-intercept form. \r\n\r\n$ y\\minus{}1\\equal{}3(x\\minus{}1)$\r\n$ y\\minus{}1\\equal{}3x\\minus{}3$\r\n$ y\\equal{}3x\\minus{}2$\r\n\r\nOR\r\n\r\n$ y\\minus{}4\\equal{}3(x\\minus{}2)$\r\n$ y\\minus{}4\\equal{}3x\\minus{}6$\r\n$ y\\equal{}3x\\minus{}2$", "Solution_2": "lol, now the answers seem obvious!\r\n\r\nThanks for enlightening me!" } { "Tag": [ "probability" ], "Problem": "If i have two decks of cards with no hearts at all, what is the probablity that i will draw an ace?", "Solution_1": "[hide]\nThere are a total of 2(52-13)=2*39=78 cards.\nthere are 2(4-1)=6 aces.\n$\\frac{6}{78}=\\frac{3}{39}=\\frac{1}{13}$\nDone.\n\n\n[/hide]", "Solution_2": "[hide]$52\\cdot \\frac{3}{4}=39$\nThere are now $39$ cards in the deck of cards, $3$ of them Aces. Therefore, the probability of drawing an Ace is $\\frac{3}{39}=\\frac{1}{13}$[/hide]" } { "Tag": [ "blogs", "\\/closed" ], "Problem": "First, where is the album??? Are the pictures still in it?\r\n\r\nAnd what happened to being able to collapse sections? Indeed, the same effect may be done by hiding forums, but then it's harder to see them again.\r\n\r\nFinally, where is the Blog Control Panel?", "Solution_1": "Albums will be back, no pictures will be lost. Have a bit of patience, it will be better than before.\r\n\r\nSame with collapsing categories.\r\n\r\nFinally the blog control panel lies within the Profile Button (in the header). Or, if you are browsing your blog, in the top right side of the blog." } { "Tag": [ "algebra", "functional equation", "real analysis", "real analysis unsolved" ], "Problem": "Consider the functional equation $2f(x)=f(x/2)+f(\\frac{1-x}{2})$. It is very easy to see that if $f$ is continuous at $0$ and bounded on $[-1,1]$, then $f$ is constant on $[-1,1]$. Ronx claims that if we drop the condition of boundedness, there are nonconstant solution continuous at $0$. Any construction? :maybe:", "Solution_1": "Take $f(x)$ arbitrary on $[0;\\frac14)$. Using functional equation it cat be extended on $[0;\\frac12]$. Then take $f(x)=f(1-x)$ on $[\\frac12;1]$ and $f(x)=f(-x)$ on $[-1;0]$" } { "Tag": [ "geometry", "trigonometry", "inequalities", "perpendicular bisector", "geometry unsolved" ], "Problem": "let $ Q$ is a quadrilateral whose side lengths are $ a,b,c,d$ in that order.show that the area of $ Q$ does not exceed $ (ac\\plus{}bd)/2$", "Solution_1": "Assume the vertices of the quadri are A,B,C,D (AB=a, BC=b etc). Consider quadrilateral ABCD' where D' is symmetric to D wrt perpendicular bisector of AC. We get quadri with side lenghts AB=a, BC=b, CD'=d and AD'=c. Now it suffices to use sine law ;)", "Solution_2": "Let e=AC and f=BD and let x the angle betwen e and f. By ptolomy inequality $ \\frac {ac \\plus{} bd}{2} \\ge \\frac {ef}{2} \\ge \\frac {ef \\sin x}{2} \\equal{} [ABCD]$ with equality when ABCD is cyclic with $ AC \\perp BD$." } { "Tag": [ "induction" ], "Problem": "Hello everyone. Kind of new here so I'm not sure if this is the right category for my question but here it is,\r\n\r\nHow do you prove that x^2 = 3n+2 has no integer solutions?\r\n\r\nThe progression I've made is first, x^2 = 4m or 4m+1 for some m, because either x is even, in which case x^2 is a multiple of 4, or it's odd, in which case it is 1 more than a multiple of 4 since\r\nx^2 = 4m+1\r\nx^2-1 = 4m\r\n(x-1)(x+1) = 4m\r\nWhich must be true since if x is odd then x-1 and x+1 are even.\r\n\r\nTo make 3n+2 a multiple of four or one more than a multiple of four,\r\nn=4b+2 --> 3(4b+2)+2 = 12b+8 = 4(b+2) which is a multiple of 4, x is even\r\nn=4b+1 --> 3(4b+1)+2 = 12b+5 = 12b+4+1 = 4(3b+1)+1 which is 1 more than a multiple of 4, x is odd\r\n\r\nThis narrows n down to the form 4b+2 or 4b+1, but I can't really progress further than that.\r\nAny ideas?", "Solution_1": "Try splitting it into cases when x is a multiple of 3, one more than a multiple of 3, or two more than a multiple of 3... i.e. set like x = 3k, x = 3k+1, x = 3k+2, plug it in and see what happens!", "Solution_2": "[quote=\"ThAzN1\"]Try splitting it into cases when x is a multiple of 3, one more than a multiple of 3, or two more than a multiple of 3... i.e. set like x = 3k, x = 3k+1, x = 3k+2, plug it in and see what happens![/quote]Great idea! I tried it out and I think it's solved, here's what I got:\r\nCase 1: x=3k\r\nx^2 = 9k^2 which is a multiple of 3, so it can't be in the form 3n+2\r\n\r\nCase 2: x=3k+1\r\nx^2 = 9k^2 + 6k + 1, and since 9k^2 and 6k are multiples of 3, then the whole thing is 1 more than a multiple of 3, so it's 3n+1, not 3n+2\r\n\r\nCase 3: x=3k+2\r\nx^2 = 9k^2 + 12k + 4 which again is one more than a multiple of 3 since 9k^2 + 12k is a multiple of 3 and adding 4 gives 1 more than a mutliple of 3.\r\n\r\nDone! Thanks for the help.", "Solution_3": "Awesome! :D", "Solution_4": "Its easy to show that the sum of odd numbers (squares) are not congruent to 2 mod 3. (You can use induction too, but its slightly harder) :D" } { "Tag": [ "quadratics", "modular arithmetic", "logarithms" ], "Problem": "So just to ask, is it possible to use square roots in modular arithmetic? Any higher roots? Would it be easier to use mods?", "Solution_1": "Yes you can use square roots, but it requires that the number in question that you're square rooting is a quadratic residue mod your modulus. For example, quadratic residues mod 5 are 1 and 4. This means $\\sqrt{-1} \\equiv \\sqrt4 \\equiv \\pm 2 \\pmod5$. But you can't perform $\\sqrt2 \\pmod5$.", "Solution_2": "You can even use logarithms if you want. They're called \"discrete logarithms\" in this context, and the fact that they're difficult to calculate is the basis of several popular encryption algorithms. :)" } { "Tag": [ "trigonometry", "calculus", "integration", "abstract algebra", "real analysis", "real analysis unsolved" ], "Problem": "No solution has been proposed so far for this problem ([b]U23[/b]) in the online publication. I am not sure if this problem belongs here.\r\n\r\nCompute the following sum:\r\n\r\n$\\sum_{k=0}^{n-1}\\frac{1}{1+8\\sin^{2}(\\frac{k\\pi}{n})}$", "Solution_1": "By comparing to an integral, it's close to $\\frac{n}{3}$.\r\n\r\nFor $n=1$, it's 1. For $n=2$, it's $\\frac{10}9$. For $n=3$, it's $1+\\frac27$. For $n=4$, it's $\\frac43+\\frac{8}{45}$. For $n=5$, it's a complicated irrational mess (about 1.77419). For $n=6$, it's $2+\\frac{4}{63}$.\r\n\r\nThere is almost certainly no closed form.", "Solution_2": "There exists a closed form for this. The idea is to use the Poisson kernel. Author's solution is quite tricky, but using the Poisson kernel it becomes pretty straightforward.", "Solution_3": "Harazi, can show us more details? :maybe:", "Solution_4": "Ok, I'll post tonight my solution, now I have a course.", "Solution_5": "Ok, here is the plan, you can easily complete the computations: recall that for $00$ satisfy $\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{24}$?", "Solution_7": "[quote=\"SnowStorm\"]Solution to Problem 4\n[hide]\nThere are 400(10) = 4000 five digit numbers that have 9 as the first digit and 6 as the third digit. One third of these will be divisible by three so two thirds will not be. So (2/3)4000 = 2666, I think.\n[/hide]\n[/quote]\r\n\r\nI dont think is 4000. It should be 1000, because there are three empty digit spots where we could have any digits from 0-9. thus, $10 \\cdot 10 \\cdot 10 = 1000$", "Solution_8": "[hide]There are 8 factors in 24. Every factor ads two different ways to make $\\frac 1{24}$. However, $\\frac 1{48} + \\frac 1{48}$ can only be counted once. The answer is therefore $2 \\cdot 8 - 1 = 15$.[/hide]\r\n\r\n[b][size=150]Problem 7[/size][/b]\r\n\r\nA circle is inscribed in an isoscoles right triangle with legs of length 5. Find the radius of the circle.", "Solution_9": "[hide=\"Back to problem 4\"]\nSo we have 9 _ 6 _ _ is not divisible by mod 3.\nThen the sum of the blanks must not be equivalent to 0 mod 3, and we can have all of the following in mod 3:\n0 0 1 (4*4*3)\n0 0 2 (4*4*3)\n0 1 0 (4*3*4)\n0 1 1 (4*3*3)\n0 2 0 (4*3*4)\n0 2 2 (4*3*3)\n1 0 0 (3*4*4)\n1 0 1 (3*4*3)\n1 1 0 (3*3*4)\n1 1 2 (3*3*3)\n1 2 1 (3*3*3)\n1 2 2 (3*3*3)\n2 0 0 (3*4*4)\n2 0 2 (3*4*3)\n2 1 1 (3*3*3)\n2 1 2 (3*3*3)\n2 2 0 (3*3*4)\n2 2 1 (3*3*3)\n\nThe products next to each configuration represents the number of different ways it can be satisfied. Note that the digits 0, 3, 6, 9 are equivalent to 0 mod 3, while 1, 4, 7 are equivalent to 1 mod 3, and 2, 5, 8 are equivalent to 2 mod 3.\n\nSo adding those together produces 666.\n[/hide]", "Solution_10": "Argh your right it should be (2/3)1000 = 666, for some reason I thought there were 400 for every 10 thousand but 90600 + 100 = 90700, heh", "Solution_11": "Problem 7\r\n[hide]\n[code]\n B\n | \\\n | \\\n | \\ sqrt(50)\n5 | O \\\n |_____\\\nA 5 C\nO is the center of the circle\n[/code]\narea(ABC) = 5*5/2 = 12.5\narea(ABC) = area(ABO) + area(COA) + area(BCO)\n= (1/2)*r*5 + (1/2)*r*5 + (1/2)*r*sqrt(50)\n12.5 = (5/2)r + (5/2)r + (sqrt(50)/2)r\n$\\frac{12.5}{5+5\\sqrt(2)/2}$ or 1.4644\n[/hide]\r\n[b][size=150]Problem 8[/size][/b]\r\nBrenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?", "Solution_12": "[hide=\"Problem 8\"]500. Easy: $2(100+150)$[/hide]\r\n\r\nHold on. Let me think of a good problem.", "Solution_13": "[size=150]Problem 9[/size]\r\n\r\nWhat is the area of the region bounded by the graph of $|x + y| + |x - y| = 4$?\r\n\r\nSource: Math League", "Solution_14": "[hide=\"problem 9\"]We can find four possible equations: $(x+y)+(x-y)=4$, $(x+y)-(x-y)=4$, $-(x+y)+(x-y)=4$, $-(x+y)-(x-y)=4$. For the first equation we get $x=2$, the second $y=2$, the third $y=-2$, and the fourth $x=-2$. The region bounded by these lines is a $4\\times4$ square, so the area is $16$. [/hide]\r\n\r\n[size=150]Problem 10[/size]\r\n\r\nWhat is the ordered triple $(x,y,z)$ which satisfies:\r\n$x+y+z=9$\r\n$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1$\r\n$xy+xy+yz=27$?\r\n\r\nSource: math league 1989. this is turning into a math league fest :lol:", "Solution_15": "Why is it not infinite?\n\nTake any such sum. There always exists a largest and a smallest number. Increase the larger number by one, decrease the smaller number by one, and clearly we have a new way to do it.", "Solution_16": "n? is the product of prime numbers less than n. Find n if n?=2n+16. Thanks u", "Solution_17": "@luonglengocidep you can just try some values and find that 7 works (Try $n?=2\\cdot3\\cdot5$).", "Solution_18": "@bestwillcui1 Can you explain the solution?", "Solution_19": "Well, $n?$ is the product of the prime numbers less than $n$, so the possible values are $2, 2\\times3, 2\\times3\\times5, 2\\times3\\times5\\times7,\\ldots$ You can try one of these values for $n?$ and solve for $n$ based on that, and if $n?$ is actually valid for the given $n$, then it is correct. If you try the first few, you get that when $n?=2\\cdot3\\cdot5=30$, $n=7$, which is valid. When $n?$ is large, then the value of $n$ will be too large.", "Solution_20": "@vincenthuang75025: Pretty sure it is infinite. No one ever said it wasn't ;-)\n\nAll problems so far are solved, so [hide=\"383\"][2014 TAMU] There are three mirrors in a line parallel to each other, and they are positioned such that their centers form a straight line. I shine a light at the mirror closest to me. Each mirror reflects off 3/4 of the light provided to it, and lets the remaining 1/4 go through. Please calculate the amount of light that goes through the third mirror.\n\nFor example, a possible sequence of the light that goes through mirror 3 is: go through 1, reflect off 2, reflect off 1, go through 2, reflect off 3, go through 2, reflect off 1, go through 2, reflect off 3, reflect off 2, go through 3.[/hide]", "Solution_21": "Short ans: [hide]40/91[/hide]\n\nSolution to come when I'm not so busy.", "Solution_22": "[hide=\"Solution383\"]\nConsider two mirrors. The light is reflected either 0,2,4,6... times and passes through the each mirror exactly once. So the fraction of light that passes through two mirrors is\n$\\frac{1}{16}\\left(1+\\frac{9}{16}+\\frac{9}{16}^2+...\\right)=\\frac{1}{7}$\nThus 1/7 of the original light passes through two mirrors and 6/7 of it is reflected. Now think of the bottom two mirrors in our original problem as a single mirror that reflects 6/7 of the original light and allows 1/7 to pass through. Now we don't have to worry about what goes on between the bottom two mirrors. The light is reflected either 0,2,4,6... times and passes through each mirror exactly once (remember that the bottom two mirrors are now a single mirror). The fraction of light that passes through these two mirrors is\n$\\frac{1}{4}*\\frac{1}{7}\\left[1+\\left(\\frac{3}{4}*\\frac{6}{7}\\right)+\\left(\\frac{3}{4}*\\frac{6}{7}\\right)^2+...\\right]=\\frac{1}{10}$\nSo our answer is $\\frac{1}{10}$\n[/hide]\n[hide=\"Problem384\"]\nFind the last 3 digits of $\\dbinom{99}{19}$\n[/hide]", "Solution_23": "Why would there be such a problem?\n\nEDIT: he deleted his post", "Solution_24": "Problem 27.\nAnswer: [b]1402[/b]\nProblem 28. Min: $\\frac{x^4+1}{x^3-x}$; with x>1", "Solution_25": "Phantranhuongth, I think you looked at the wrong page of the thread; we are on problem 384 currently. As for your problem, just take the derivative of the function, and confirm it is an absolute minimum.", "Solution_26": "[quote=\"WorstIreliaNA\"][hide=\"Solution383\"]\nConsider two mirrors. The light is reflected either 0,2,4,6... times and passes through the each mirror exactly once. So the fraction of light that passes through two mirrors is\n$\\frac{1}{16}\\left(1+\\frac{9}{16}+\\frac{9}{16}^2+...\\right)=\\frac{1}{7}$\nThus 1/7 of the original light passes through two mirrors and 6/7 of it is reflected. Now think of the bottom two mirrors in our original problem as a single mirror that reflects 6/7 of the original light and allows 1/7 to pass through. Now we don't have to worry about what goes on between the bottom two mirrors. The light is reflected either 0,2,4,6... times and passes through each mirror exactly once (remember that the bottom two mirrors are now a single mirror). The fraction of light that passes through these two mirrors is\n$\\frac{1}{4}*\\frac{1}{7}\\left[1+\\left(\\frac{3}{4}*\\frac{6}{7}\\right)+\\left(\\frac{3}{4}*\\frac{6}{7}\\right)^2+...\\right]=\\frac{1}{10}$\nSo our answer is $\\frac{1}{10}$\n[/hide]\n[hide=\"Problem384\"]\nFind the last 3 digits of $\\dbinom{99}{19}$\n[/hide][/quote]\n[hide=\"progress384\"]\nFirst, the problem is screaming at us to find the last three digits of $\\dbinom{100}{20}$. It suffices to find it mod $8$ and $125$. These cubes of primes remind us of Wolstenholme's theorem. By Wolstenholme's we find\n\\[\\dbinom{100}{20}\\equiv \\dbinom{25}{5}=5\\cdot 23\\cdot 22\\cdot 21 \\equiv 2\\pmod{8}\\]\nand \n\\[\\dbinom{100}{20}\\equiv \\dbinom{20}{4}=5\\cdot 19\\cdot 3\\cdot 17\\equiv 95\\pmod{125}\\]\nand by CRT we have \n\\[\\dbinom{100}{20}\\equiv 970\\pmod{1000}.\\]\nThus we just want \n\\[\\dfrac{20\\cdot 970}{100}\\pmod{1000}.\\]\nBut that only gives that the last two digits are $94$, but I can't finish\n\n[/hide]", "Solution_27": "[quote=\"mathtastic\"][quote=\"WorstIreliaNA\"][hide=\"Solution383\"]\nConsider two mirrors. The light is reflected either 0,2,4,6... times and passes through the each mirror exactly once. So the fraction of light that passes through two mirrors is\n$\\frac{1}{16}\\left(1+\\frac{9}{16}+\\frac{9}{16}^2+...\\right)=\\frac{1}{7}$\nThus 1/7 of the original light passes through two mirrors and 6/7 of it is reflected. Now think of the bottom two mirrors in our original problem as a single mirror that reflects 6/7 of the original light and allows 1/7 to pass through. Now we don't have to worry about what goes on between the bottom two mirrors. The light is reflected either 0,2,4,6... times and passes through each mirror exactly once (remember that the bottom two mirrors are now a single mirror). The fraction of light that passes through these two mirrors is\n$\\frac{1}{4}*\\frac{1}{7}\\left[1+\\left(\\frac{3}{4}*\\frac{6}{7}\\right)+\\left(\\frac{3}{4}*\\frac{6}{7}\\right)^2+...\\right]=\\frac{1}{10}$\nSo our answer is $\\frac{1}{10}$\n[/hide]\n[hide=\"Problem384\"]\nFind the last 3 digits of $\\dbinom{99}{19}$\n[/hide][/quote]\n[hide=\"progress384\"]\nFirst, the problem is screaming at us to find the last three digits of $\\dbinom{100}{20}$. It suffices to find it mod $8$ and $125$. These cubes of primes remind us of Wolstenholme's theorem. By Wolstenholme's we find\n\\[\\dbinom{100}{20}\\equiv \\dbinom{25}{5}=5\\cdot 23\\cdot 22\\cdot 21 \\equiv 2\\pmod{8}\\]\nand \n\\[\\dbinom{100}{20}\\equiv \\dbinom{20}{4}=5\\cdot 19\\cdot 3\\cdot 17\\equiv 95\\pmod{125}\\]\nand by CRT we have \n\\[\\dbinom{100}{20}\\equiv 970\\pmod{1000}.\\]\nThus we just want \n\\[\\dfrac{20\\cdot 970}{100}\\pmod{1000}.\\]\nBut that only gives that the last two digits are $94$, but I can't finish\n\n[/hide][/quote]\n\nNoting that $\\binom{100}{20} \\frac{1}{5} = \\binom{99}{19}$, maybe find $\\binom{100}{20}$ mods $8$ and $625$. This way when we divide by $5$ we get the answer mod $125$ and CRT will give the desired.", "Solution_28": "[quote=\"vincenthuang75025\"][quote=\"mathtastic\"][quote=\"WorstIreliaNA\"][hide=\"Solution383\"]\nConsider two mirrors. The light is reflected either 0,2,4,6... times and passes through the each mirror exactly once. So the fraction of light that passes through two mirrors is\n$\\frac{1}{16}\\left(1+\\frac{9}{16}+\\frac{9}{16}^2+...\\right)=\\frac{1}{7}$\nThus 1/7 of the original light passes through two mirrors and 6/7 of it is reflected. Now think of the bottom two mirrors in our original problem as a single mirror that reflects 6/7 of the original light and allows 1/7 to pass through. Now we don't have to worry about what goes on between the bottom two mirrors. The light is reflected either 0,2,4,6... times and passes through each mirror exactly once (remember that the bottom two mirrors are now a single mirror). The fraction of light that passes through these two mirrors is\n$\\frac{1}{4}*\\frac{1}{7}\\left[1+\\left(\\frac{3}{4}*\\frac{6}{7}\\right)+\\left(\\frac{3}{4}*\\frac{6}{7}\\right)^2+...\\right]=\\frac{1}{10}$\nSo our answer is $\\frac{1}{10}$\n[/hide]\n[hide=\"Problem384\"]\nFind the last 3 digits of $\\dbinom{99}{19}$\n[/hide][/quote]\n[hide=\"progress384\"]\nFirst, the problem is screaming at us to find the last three digits of $\\dbinom{100}{20}$. It suffices to find it mod $8$ and $125$. These cubes of primes remind us of Wolstenholme's theorem. By Wolstenholme's we find\n\\[\\dbinom{100}{20}\\equiv \\dbinom{25}{5}=5\\cdot 23\\cdot 22\\cdot 21 \\equiv 2\\pmod{8}\\]\nand \n\\[\\dbinom{100}{20}\\equiv \\dbinom{20}{4}=5\\cdot 19\\cdot 3\\cdot 17\\equiv 95\\pmod{125}\\]\nand by CRT we have \n\\[\\dbinom{100}{20}\\equiv 970\\pmod{1000}.\\]\nThus we just want \n\\[\\dfrac{20\\cdot 970}{100}\\pmod{1000}.\\]\nBut that only gives that the last two digits are $94$, but I can't finish\n\n[/hide][/quote]\n\nNoting that $\\binom{100}{20} \\frac{1}{5} = \\binom{99}{19}$, maybe find $\\binom{100}{20}$ mods $8$ and $625$. This way when we divide by $5$ we get the answer mod $125$ and CRT will give the desired.[/quote]\n\nProblem 385:\n[hide]Find all solutions to the equation: ${1}/{a^2} + {1}/{b^2} = {1}/{c^2}$ for all integers ${c} \\le {100}$[/hide]", "Solution_29": "[hide=S385]\n$c^2(a^2+b^2)=a^2b^2$, so $a^2+b^2=d^2$ for some $d$. Then, $cd=ab$, so $d$ divides $ab$. Suppose $(a,b,d)$ is generated from the primitive pythagorean triple $(x,y,z)$. As $gcd(z,y)=gcd(z,x)=1$, and we need $d|ab$, then $(a,b,d)=(kxz, kyz, kz^2)$. $c\\le 100$, so $kxy\\le 100$. We can have $(x,y)=(3,4);(5,12)$, which yields the solutions $(a,b,c)=(15,20,12);(30,40,24);(45,60,36);(60,80,48);(65,156,60);(75,100,60);(90,120,72);(105,140,84);(120,160,96)$\n[/hide]\n[hide=P386]\nFind the sum of all natural numbers $k\\le 100$ for which there exists a solution to the equation $\\sin(\\alpha k)=\\sin(\\alpha)$ for $0\\le\\alpha\\le\\frac{\\pi}{2}$" } { "Tag": [], "Problem": "A rise of $ 600$ feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from $ 3\\%$ to $ 2\\%$ is approximately:\r\n\r\n$ \\textbf{(A)}\\ 10000 \\text{ ft.} \\qquad\\textbf{(B)}\\ 20000 \\text{ ft.} \\qquad\\textbf{(C)}\\ 30000 \\text{ ft.} \\qquad\\textbf{(D)}\\ 12000 \\text{ ft.} \\qquad\\textbf{(E)}\\ \\text{none of these}$", "Solution_1": "[hide]\nWe can make a right triangle where the rise and run of the track are legs and the length of track is the hypotenuse.\n\nWhen the grade is $ 3\\%$, $ \\frac{rise}{run}\\equal{}\\frac{600}{run}\\equal{}.03 \\Leftrightarrow run\\equal{}20000$ and the length of track is $ \\sqrt{600^2\\plus{}20000^2}\\approx 20009$.\n\nWhen the grade is $ 2\\%$, $ \\frac{rise}{run}\\equal{}\\frac{600}{run}\\equal{}.02 \\Leftrightarrow run\\equal{}30000$ and the length of track is $ \\sqrt{600^2\\plus{}30000^2}\\approx 30006$.\n\n$ 30006\\minus{}20009\\approx 10000$, so the answer is $ \\boxed{A}$.\n\n[/hide]" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find the functions $f:\\mathbb{Z}\\times \\mathbb{Z}\\to\\mathbb{R}$ such that\r\n\r\na) $f(x,y)\\cdot f(y,z) \\cdot f(z,x) = 1$ for all integers $x,y,z$;\r\n\r\nb) $f(x+1,x)=2$ for all integers $x$.", "Solution_1": "Cute but easy :\r\n\r\nFor all $x$, we have $f(x,x) \\cdot f(x,x) \\cdot f(x,x) = 1$ so that $f(x,x)=1.$\r\nIt follows that for all $x,y$, we have $f(x,y) \\cdot f(y,x) = 1.$\r\nThus $f(y,x) = f(y,z) \\cdot f(z,x)$ and $f(y,x) = \\frac 1 {f(x,y)}.$\r\n\r\nNow, for all $x,y$. Let $y= x + p$ with $p >0.$\r\nThus $f(y,x) = f(x+p,x) = f(x+p),x+p-1) \\cdot f(x+p-1,x+p-2) \\cdots f(x+1,x) = 2^{y-x}.$\r\n\r\nConversely $f(x,y) = 2^{x-y}$ is a solution of the problem.\r\n\r\nPierre." } { "Tag": [], "Problem": "What is the value of $ (8\\minus{}4)!\\div(8\\minus{}3)!$? Express your answer as a common fraction.", "Solution_1": "\\begin{align*}(8 - 4)!\\div(8 - 3)! & = 4!\\div 5! \\\\\r\n& = \\frac {4!}{5!} \\\\\r\n& = \\frac {4!}{5(4!)} \\\\\r\n& = \\boxed{\\frac {1}{5}}\\end{align*}" } { "Tag": [ "FTW" ], "Problem": "What reduced common fraction is equivalent to $ 20 \\frac{5}{6} \\%$?", "Solution_1": "It says the correct answer is 1/48 in FTW but I got 5/24, so should I report error?", "Solution_2": "Yes you should.\r\n\r\nAlthough I doubt you can, now." } { "Tag": [ "floor function", "number theory unsolved", "number theory" ], "Problem": "Find the maximum value of $ c$ such that $ \\{n\\sqrt3\\} > \\frac {c}{n\\sqrt3}$ holds for every positive integer $ n$.", "Solution_1": "If $ (x,y)$ is a positive integer solution to the pell-type equation $ x^2 \\minus{} 3y^2 \\equal{} \\minus{} 2$ then $ (2x \\plus{} 3y,x \\plus{} 2y)$ is also a solution (Note that $ (1,1)$ is a solution giving the base case).\r\nWe can find arbitrarily large solutions.\r\nLet $ (x,y)$ be such a solution. \r\nThen $ \\sqrt {3} \\equal{} \\frac {\\sqrt {x^2 \\plus{} 2}}{y}$\r\nNow if we take $ n \\equal{} y$ then $ n\\sqrt {3} \\equal{} \\sqrt {x^2 \\plus{} 2}$\r\n$ \\{n\\sqrt {3}\\} \\equal{} \\{\\sqrt {x^2 \\plus{} 2}\\} \\equal{} \\sqrt {x^2 \\plus{} 2} \\minus{} x \\equal{} \\frac {2}{x \\plus{} \\sqrt {x^2 \\plus{} 2}}$.\r\nIf we take $ x$ an arbitrarily large solution then it becomes a arbitrarily small positive real.\r\nHence [b]the required maximum value is [/b]$ 0$", "Solution_2": "In fact, $ \\{n\\alpha\\}$ is dense in $ ]0,1[$ if $ \\alpha$ is irrational (Kronecker's Theorem). That means we can find $ n$ such that $ \\{n\\alpha\\}$ becomes each time closer to any number in $ [0,1]$, and it includes $ 0$. Of course $ \\{n\\alpha\\}$ is always bigger than $ 0$, because $ \\alpha$ is irrational ($ n\\alpha$ is never integer), so the maximum $ c$ is $ 0$ in the problem.", "Solution_3": "I'm sorry, the RHS should be $ \\frac{c}{n\\sqrt3}$. It has been edited now.", "Solution_4": "This solution is quite similar to the previous. But for the sake of rigorousness and better understanding I'll rewrite everything.\r\n\r\nSince $ 3x^2 - y^2 = 3(2x + y)^2 - (3x + 2y)^2$, and $ 3\\cdot 1^2 - 1^2 = 2$ there are infinitely may integer solutions to $ 3x^2 - y^2 = 2$. \r\nNote that for those $ x$ we have $ y = \\lfloor x\\sqrt {3}\\rfloor$. \r\n\r\nThus there is an infinite strictly increasing sequence of positive integers $ \\{a_k\\}_{n\\in \\mathbb N}$ which satisfies \r\n$ 3a_k^2 - \\lfloor a_k\\sqrt {3} \\rfloor ^2 = 2$ for all $ k\\in \\mathbb N \\cdots \\boxed{1}$\r\n\r\nSince squares are $ \\not \\equiv 2\\mod 3$, there are no positive integer solutions to $ 3x^2 - y^2 = 1$\r\nHence there are no positive integer $ n$ for which $ 3n^2 - \\lfloor n\\sqrt {3} \\rfloor ^2 = 1\\cdots \\boxed{2}$\r\n\r\nLet $ f(n) = n\\sqrt {3}\\{n\\sqrt {3}\\}$\r\nWe need to find the minimum of $ f(n)$ where $ n$ runs over the positive integers.\r\nFor some $ n$, define $ n'$ to be the smallest positive integer $ \\ge n$ which satisfies $ 3n'^2 - \\lfloor n'\\sqrt {3}\\rfloor^2 = 2$ ($ \\boxed{1}$ shows $ n'$ must exist)\r\n\r\n[u][b]Lemma 1[/b][/u] $ f(n) \\ge f(n')$\r\n[b]Proof:[/b]\r\n$ f(n) = n\\sqrt {3}\\{n\\sqrt {3}\\} = \\frac {1}{2}\\left(1 + \\frac {3n^2 - \\lfloor n\\sqrt {3}\\rfloor^2}{(n\\sqrt {3} + \\lfloor n\\sqrt {3}\\rfloor )^2}\\right)(3n^2 - \\lfloor n\\sqrt {3}\\rfloor^2)$\r\nFrom $ \\boxed{2}$ we have $ 3n^2 - \\lfloor n\\sqrt {3}\\rfloor^2\\ge 2 = 3n'^2 - \\lfloor n'\\sqrt {3}\\rfloor^2$, and from $ n'\\ge n$ we have $ n\\sqrt {3} + \\lfloor n\\sqrt {3}\\rfloor \\le n'\\sqrt {3} + \\lfloor n'\\sqrt {3}\\rfloor$. \r\n$ f(n) = \\frac {1}{2}\\left(1 + \\frac {3n^2 - \\lfloor n\\sqrt {3}\\rfloor^2}{(n\\sqrt {3} + \\lfloor n\\sqrt {3}\\rfloor )^2}\\right)(3n^2 - \\lfloor n\\sqrt {3}\\rfloor^2)\\ge$\r\n$ \\frac {1}{2}\\left(1 + \\frac {3n'^2 - \\lfloor n'\\sqrt {3}\\rfloor^2}{(n'\\sqrt {3} + \\lfloor n\\sqrt {3}\\rfloor )^2}\\right)(3n'^2 - \\lfloor n'\\sqrt {3}\\rfloor^2) = f(n')$\r\n\r\nFrom [b]Lemma 1[/b] we need to find the minimum of $ f(n)$ where $ n$ runs over those positive integers for which $ 3n^2 - \\lfloor n\\sqrt {3}\\rfloor^2 = 2$.\r\nFor those $ n: \\lfloor n\\sqrt {3}\\rfloor = \\sqrt {3n^2 - 2}$\r\nSo we need to find the minimum of \r\n$ f(n) = n\\sqrt {3}\\{n\\sqrt {3}\\} = n\\sqrt {3}(n\\sqrt {3} - \\lfloor n\\sqrt {3}\\rfloor ) = n\\sqrt {3}(n\\sqrt {3} - \\sqrt {3n^2 - 2}) =$\r\n\r\n$ = 3n^2 - \\sqrt {(3n^2 - 1)^2 - 1} = 1 + (3n^2 - 1) - \\sqrt {(3n^2 - 1)^2 - 1} =$\r\n$ = 1 + \\frac {1}{(3n^2 - 1) + \\sqrt {(3n^2 - 1)^2 - 1}}$\r\nWhich is strictly decreasing.\r\nBut from $ \\boxed{1}$ we can take $ n\\to \\infty$ for which we find the minimum of $ f(n)$ to be $ 1$\r\nHence [b]the required maximum value of $ c$ is $ 1$[/b]" } { "Tag": [ "symmetry", "inequalities unsolved", "inequalities" ], "Problem": "", "Solution_1": "hi !\r\nlet : a>=b>=c and x>=y>=z\r\nwe have by chebycev : (x*a/y+z) +(y*b/x+z)+(z*c/x+y)>=1/3(a+b+c)*(x/y+z + y/x+z + z/x+y)>=(a+b+c)/2", "Solution_2": "noths, you can't assume $ a \\ge b \\ge c$ and $ x \\ge y \\ge z$ simultaneously.\r\n\r\nCase $ abc \\equal{} 0$, there is no minimum (if one of the variables tends to infinity, $ E$ tends to $ 0$).\r\n\r\nCase $ \\sqrt {a}$, $ \\sqrt {b}$ and $ \\sqrt {c}$ are sides of a triangle, with Cauchy we get\r\n\r\n$ E \\plus{} a \\plus{} b \\plus{} c \\equal{} \\sum \\frac {a(x \\plus{} y \\plus{} z)}{y \\plus{} z} \\equal{} (x \\plus{} y \\plus{} z)\\sum \\frac {a}{y \\plus{} z} \\equal{} \\frac {1}{2}\\sum{(y \\plus{} z)}\\sum \\frac {a}{y \\plus{} z} \\ge \\frac {1}{2}(\\sum \\sqrt {a})^2 \\Rightarrow E \\ge \\sqrt {bc} \\plus{} \\sqrt {ca} \\plus{} \\sqrt {ab} \\minus{} \\frac {a \\plus{} b \\plus{} c}{2}$\r\n\r\nI still don't know the other cases...", "Solution_3": "[quote=\"feliz\"]noths,[b] you can't assume $ a \\ge b \\ge c$ and $ x \\ge y \\ge z$ simultaneously[/b].\n\nCase $ abc \\equal{} 0$, there is no minimum (if one of the variables tends to infinity, $ E$ tends to $ 0$).\n\nCase $ \\sqrt {a}$, $ \\sqrt {b}$ and $ \\sqrt {c}$ are sides of a triangle, with Cauchy we get\n\n$ E \\plus{} a \\plus{} b \\plus{} c \\equal{} \\sum \\frac {a(x \\plus{} y \\plus{} z)}{y \\plus{} z} \\equal{} (x \\plus{} y \\plus{} z)\\sum \\frac {a}{y \\plus{} z} \\equal{} \\frac {1}{2}\\sum{(y \\plus{} z)}\\sum \\frac {a}{y \\plus{} z} \\ge \\frac {1}{2}(\\sum \\sqrt {a})^2 \\Rightarrow E \\ge \\sqrt {bc} \\plus{} \\sqrt {ca} \\plus{} \\sqrt {ab} \\minus{} \\frac {a \\plus{} b \\plus{} c}{2}$\n\nI still don't know the other cases...[/quote]\r\n\r\nplz tell me why ?", "Solution_4": "Yes. Actually, you couldn't even have assumed $ x \\ge y \\ge z$. When $ a \\not \\equal{} b$, there is no symmetry between $ x$ and $ y$. For example, if we want to maximize $ 0x \\plus{} 2y$ knowing that $ x^2 \\plus{} y^2 \\equal{} 2$, obviously we cannot suppose $ x \\ge y$; the maximum occurs for $ x \\equal{} 0$ (I could have used $ x \\plus{} 2y$, but $ 0$ makes the argument clearer). Anyway, in general, to maximize $ ax \\plus{} by \\plus{} cz$, we can't assume both things because $ a \\ge b$ breaks the symmetry between $ x$ and $ y$.\r\n\r\nCase $ a, b, c > 0$ and $ \\sqrt {a} \\ge \\sqrt {b} \\plus{} \\sqrt {c}$, define $ A \\equal{} b \\plus{} 2\\sqrt {bc} \\plus{} c \\le a$, $ B \\equal{} b$ and $ C \\equal{} c$. Then, with the argument above, we get\r\n\r\n$ E \\ge \\sum \\frac {Ax}{y \\plus{} z} > \\sqrt {BC} \\plus{} \\sqrt {CA} \\plus{} \\sqrt {AB} \\minus{} \\frac {A \\plus{} B \\plus{} C}{2}$\r\n\r\nMaking $ x \\rightarrow 0$, $ y \\rightarrow \\sqrt {c}$ and $ z \\rightarrow \\sqrt {b}$, the RHS approaches indefinitely to the LHS. So there is no minimum, but we get values very close to $ 2\\sqrt {bc}$." } { "Tag": [], "Problem": "Read the topic question\r\n\r\nElectric <- drums included\r\n\r\nAcoustic", "Solution_1": "Electric...\r\nEspecially if there are drums included!\r\n :D", "Solution_2": "SPANISH FLAMENCO :)", "Solution_3": "I'm not sure what the poll is asking. I voted acoustic because currently i own only an acoustic guitar, but I'm buying an electric very soon.", "Solution_4": "Ooops. I thought you meant electric or acoustic pianos... though I really am not sure if you call a regular piano \"acoustic.\"", "Solution_5": "I think we were talking about guitars so I voted acoustic...", "Solution_6": "both", "Solution_7": "I own both but my style of playing is much more electric. BTW question can anyone bend even half on an acoustic? because I sure can't.", "Solution_8": "[quote=\"juicybooty911\"]both[/quote]\r\nI agree. With guitars I llike electric. With drums acoustic. They are both good. :D", "Solution_9": "Electric guitar. Played on the electric piano.", "Solution_10": "[quote=\"Tunio\"]Ooops. I thought you meant electric or acoustic pianos... though I really am not sure if you call a regular piano \"acoustic.\"[/quote]\r\n\r\ndian dian dian tj.. um i just voted acoustic cuz i only have an acoustic guitar... and it feels a lot more comfortable than electric (obviously cuz i dont play an electric)", "Solution_11": "I voted acoustic.\r\nWas I supposed to have a reason to pick one over the other?", "Solution_12": "electroacustic. :) \r\n\r\nacustic with some electric elements and effects...", "Solution_13": "i couldn't find the poll question, but i think electric is better. electric has a stronger tone, as in more poingnant. acoustic is more dulled, but i do like both. when i play appregios i think acoustic is better b/c of the resonanting effect and the notes blend better, but i solo more.", "Solution_14": "If we're talking about guitars, the the electric one! :D \r\n\r\nObviously with a great amp! ;)", "Solution_15": "only flamenco guitar...\r\n\r\nand only paco delucia... :D" } { "Tag": [ "topology", "geometry", "3D geometry", "real analysis", "real analysis unsolved" ], "Problem": "Prove that if $ s>d$, then the Hausdorff outer measure $ H^{s*}(A)\\equal{}0$ for every set $ A\\subset \\mathbb{R}^d$", "Solution_1": "It's enough to prove this for the unit cube, because the outer measure is subadditive. Divide the cube into $ N^d$ equal subcubes and see what happens." } { "Tag": [ "LaTeX", "geometry", "angle bisector" ], "Problem": "Note the diagram is not up to scale (sorry for the bad pictures :oops: ). Please help. :(", "Solution_1": "[quote=\"piperson\"]Let quadrilateral $CEDF$ be cyclic. $CE$ is perpendicular to CF. $ED$ is perpendicular to $FD$. let $M$ be the midpoint of $DE$. prove $\\measuredangle EAJ=\\measuredangle FAJ$.\n\nNote the diagram is not up to scale (sorry for the bad pictures :oops: ). Please help. :([/quote]\r\n\r\nI've corrected the LaTeX code at the end.\r\n\r\nAs far as I can see, if this is a correct statement, this can't hold since $DM=ME$ but $DA\\neq AE$.", "Solution_2": "Edited Now :oops:", "Solution_3": "Still missing a condition - we have no clue on the position of $J$.\r\n\r\nIf $T$ is still the midpoint of $DE$, then $AT$ can't be the bisector of $\\angle EAD$, because by Angle Bisector Law, we must have $DT: TE=DA: AE$, but $DT: TE=1$ whilst $DA: AE<1$ (a side and the hypotenuse of a right triangle)", "Solution_4": "Edited again :( :roll:", "Solution_5": "You keep editing the problem, but to no avail.\r\n\r\n(1) You say $J$ is the midpoint of $FE$, but the midpoint of $FE$ is the center of the circle.\r\n\r\n(2) If you meant \"$J$ is the midpoint of $FC$\", then it still can't hold, for the same reason as in my previous post ($FJ=JC$ and $FA>AC$)\r\n\r\nWhy don't you try to find the exact statement of the problem and stop this mindless \"editing\"? Or if you want to create the problem yourself, take time to really think it over.\r\n\r\n[u]TO MODS[/u]: If this keeps going on like this, perhaps this topic should be locked.", "Solution_6": "Edited for the last time.", "Solution_7": "Anyone :maybe:", "Solution_8": "I don't see a diagram :maybe: :( :| :huh: :wink: \r\n\r\n :maybe: :(", "Solution_9": "Now there is :wink:", "Solution_10": "Anyone willing to help :maybe:", "Solution_11": "Please help :( :help:" } { "Tag": [ "geometry", "rectangle", "analytic geometry" ], "Problem": "I am having trouble with this cs problem from sec. 2.1 of the USACO training gateway: \r\nN opaque rectangles (1 <= N <= 1000) of various colors are placed on a white sheet of paper whose size is A wide by B long. The rectangles are put with their sides parallel to the sheet's borders. All rectangles fall within the borders of the sheet so that different figures of different colors will be seen. \r\n\r\nThe coordinate system has its origin (0,0) at the sheet's lower left corner with axes parallel to the sheet's borders. \r\n\r\nPROGRAM NAME: rect1\r\nINPUT FORMAT\r\nThe order of the input lines dictates the order of laying down the rectangles. The first input line is a rectangle \"on the bottom\". Line 1: A, B, and N, space separated (1 <= A,B <= 10,000) \r\nLines 2-N+1: Five integers: llx, lly, urx, ury, color: the lower left coordinates and upper right coordinates of the rectangle whose color is `color' (1 <= color <= 2500) to be placed on the white sheet. The color 1 is the same color of white as the sheet upon which the rectangles are placed. \r\n\r\n\r\nSAMPLE INPUT (file rect1.in) \r\n20 20 3\r\n2 2 18 18 2\r\n0 8 19 19 3\r\n8 0 10 19 4\r\n\r\nOUTPUT FORMAT\r\nThe output file should contain a list of all the colors that can be seen along with the total area of each color that can be seen (even if the regions of color are disjoint), ordered by increasing color. Do not display colors with no area. \r\nSAMPLE OUTPUT (file rect1.out)\r\n1 91\r\n2 84\r\n3 187\r\n4 38\r\n\r\nFirst of all, it comes right after the text about graphs so I initially (and it seems mistakenly) thought this problem would be related to the preceding text, though this does not seem to be the case to me. First I thought that maybe I could set up an array with an element for every square on the grid and then go through after each rectangle is placed and put it's color in the corresponding element. But this method doesn't work, because the grid can be 10,000 by 10,000. Next I thought that I would compare each rectangle, but this method becomes very complicated after a while (besides it seems pretty slow from my basic analysis). Could someone give me a hint on how to solve or at least point me in a helpful direction?\r\n\r\nAlso, on the USACO training gate a little message appeared off to the side that stated that the dates of future contests will be posted in early October. Seems a bit past early October. Has anyone seen any dates yet?\r\n\r\nThanks", "Solution_1": "Hi\r\n\r\nI also had lots of problems on this one. One possible approach is to maintain all the rectangles as non-overlapping rectangles. You need to be able to add rectangles to this. In the end, you can count all the colors easily.\r\n\r\nAnother approach(my friend's) is to divide the whole thing into say 100x100 pieces and solve those in the way you suggested initially. This way you might not be able to clear all test cases, but if you vary the numbers(200x200 or 300x300...), it will pass all cases for some number(I dont remember what my friend used).\r\n\r\nRegarding the USACO contests, looking at past years' schedule, i guess the first contest is supposed to be in mid November.", "Solution_2": "[quote=\"themonster\"]\nAlso, on the USACO training gate a little message appeared off to the side that stated that the dates of future contests will be posted in early October. Seems a bit past early October. Has anyone seen any dates yet?\n[/quote]\r\n\r\nThe first contest is tentatively Nov. 5-8. Final schedule is in the works. It should be firmed up and on the website soon.", "Solution_3": "When the rectangles overlap, make new rectangles.\r\n\r\nA lot of times, the text right before the problem doesn't include the best method." } { "Tag": [ "calculus" ], "Problem": "hi\r\nhere's a question...the general term of a series is t(n)=1/(a*n^4 + 1).....find the sum of this infinite series...\r\nShyam[/img]", "Solution_1": "Moderator, please move this to a calculus forum." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $a,b,c\\in\\mathbb {R^*}$ satisfing the condition:$a.b.c=1$.Prove that:\\[\\frac{1}{b+c+1}+\\frac{1}{a+c+1}+\\frac{1}{b+a+1}\\le \\frac{2}{b+c+2}+\\frac{2}{a+c+2}+\\frac{2}{b+c+2}\\]", "Solution_1": "We have already discussed it several times. Of course, new solutions (non brutee-force?) are welcomed.", "Solution_2": "We prove stronger inequality:\r\n$\\sum\\frac{1}{a+b+1}\\leq\\sum\\frac{1}{a+2}$" } { "Tag": [], "Problem": "(1) When acetone is treated with conc. H2SO4 and distilled, what compound do we get.\r\n\r\n(2) Which reagent(s) can be used to convert acetaldehyde into 2,2-dihydroxymethyl-1,3-propandiol?\r\n\r\n(3) When tert-butyl trifluoromethyl ketone is subjected to Baeyer Villiger oxidation (using per acid, say PBA), where is the O inserted? The tert-butyl is the bulkier of the 2, but the trifuoromethyl carbon is very highly electropositive, so which will have higher migratory aptitude?\r\n\r\n(4) What is the reaction, if any of chloroform or chloral with water, individually?\r\n\r\n(5) Explain what could be the intermediates in the following 3 reactions?\r\nIn the reaction involving SeO2, can the O be inserted in between the 2 carbonyls? Referring to the mechanism, it would involve anegative charge on the caronyl. Also what is the final step of the sequnce?\r\nIn the second reaction, what coulbe the role of quinone?", "Solution_1": "[quote=\"hell_ever\"]\n(2) Which reagent(s) can be used to convert acetaldehyde into 2,2-dihydroxymethyl-1,3-propandiol?\n\n[/quote]\r\n\r\ntreat acetaldehyde with excess of HCHO! (first 3 moles--->aldol condensation....next canizzaro :D )\r\n\r\nEDIT: Stupid 10 minute rule! :mad: ...i am editing this post and posting ans for first question!\r\n\r\n1 ans- 1,3,5-trimethyl benzene(mesitylene )", "Solution_2": "For 5(b)\r\n\r\nQuinone acts as a very good hydrogen acceptor (guess why! :wink: )\r\n\r\nThe resulting oxidised product is", "Solution_3": "3) hey! the migratory aptitude depends on the electron releasing ability (the higher the ER ability, greater the migratory apti.) (i.e, the ease of migration of substituents depends on their ability to stabilize a positive charge in the transition state) \r\n\r\nIs $ CF_3$ grp an electron releasing grp? :| :?:", "Solution_4": "5) the first intermediate is a 1,2 -cyclopentanediketone(well, my IUPAC naming sux..........i think it is 2-oxocyclopentanone right?)\r\n\r\nThe second intermediate is the anhydride because when 1,2-diketones are subjected to baeyer-villager, anhydrides are formed........next hydrolysis ofc to form the corresponding dicarboxylic acid......next no idea! :(\r\n\r\nEDITING again! :mad: for 4th quest., isnt it chloral hydrate?(extensive intramolecular H-bonding....hence very stable) :maybe:", "Solution_5": "for 5 C, i think its just mixed aldol condensation(after all sulphur and oxygen have similiar properties)(only one possibility of carbanion.......so easy :D ) and maybe next step a $ SH_2$ grp is removed just like dehydration!", "Solution_6": "[quote=\"Da Vinci\"]for 5 C, i think its just mixed aldol condensation(after all sulphur and oxygen have similiar properties)(only one possibility of carbanion.......so easy :D ) and maybe next step a $ SH_2$ grp is removed just like dehydration![/quote]\r\n\r\n\r\nDo you mean intermolecular aldol condensation? Why should it be mixed? :?:", "Solution_7": "[quote=\"Da Vinci\"]3) hey! the migratory aptitude depends on the electron releasing ability (the higher the ER ability, greater the migratory apti.) (i.e, the ease of migration of substituents depends on their ability to stabilize a positive charge in the transition state) \n\nIs $ CF_3$ grp an electron releasing grp? :| :?:[/quote]\n\nIt is a very powerful e withdrawing group. But when a group migrates, doent it migrate with its pair of electrons? So shouldnt in fact the group be a powerful e withdrawing group rather than e releasing?\n\n\nEDIT: It is indeed a stupid rule. :P There is possible formation of ammonium salt and then amide after step 4 in 5a.... :maybe: \n\nEDIT: Thanks for all the help. It was really very helpful. :)\n\n\nEDIT 3: (phew) \n[quote=\"Da Vinci\"]For 5(b)\n\nQuinone acts as a very good hydrogen acceptor (guess why! :wink: )\n\nThe resulting oxidised product is[/quote]\r\n\r\n\r\nI am not thinking very well lately, so can you explain the shift of the double bond?", "Solution_8": "[quote=\"hell_ever\"]\n\nI am not thinking very well lately, so can you explain the shift of the double bond?[/quote]\n\ni am afraid i cant! :oops: \nIts an experimentally observed fact that $ \\beta \\minus{} \\gamma$ unsaturated alcohols form $ \\alpha \\minus{} \\beta$ unsaturated ketones when subjected to oppeneur oxidation! :P \n\n\n[quote=\"hell_ever\"]\nSo shouldnt in fact the group be a powerful e withdrawing group rather than e releasing? \n\n[/quote]\n\nIf its $ e^\\minus{}$ withdrawing how can it stabilise an electron deficient oxygen? :| \n\nAlso, if you believe that electron withdrawing ability increases the mig. apti.,(which i think isnt the case) why did you have a doubt between t-butyl and $ CF_3$ grps?\nclearly, t-butyl is an ERG! :maybe: \n\n\n[quote=\"hell_ever\"]\nDo you mean intermolecular aldol condensation? Why should it be mixed?\n\n[/quote]\r\n\r\nsorry da! :D \r\nwhatever i said doesnt make sense to me also now! :o i must have typed that in sleep! :P \r\nlemme try it again!", "Solution_9": "which round in chem olympiad are u both aiming for ???", "Solution_10": "i think beyond IChO :) :rotfl: :D" } { "Tag": [ "Asymptote", "articles", "LaTeX" ], "Problem": "I noticed that you created an article about the Asymptote package cse5.\r\n\r\nShouldn't that be included here: [[Asymptote: Macros and Packages]]?", "Solution_1": "No. CSE5 is far more vast and it needs its own place. In plus it was developed by a member of this community so it's more important than other packages :P\r\nI'm just trying to make that resource actually useful. In particular unless you are doing 10 asy pictures a day, even if you have it installed, etc. know the basic principles it still takes one a lot of time to create a picture when he needs one. That's why I made those to entries, and I will continue to add to them as soon as I have the time.", "Solution_2": "Also, should not\r\n\r\nAsymptote: * and LaTeX: *\r\n\r\nbe moved to\r\n\r\nAsymptote:* and LaTeX:* (eliminating the space)\r\n\r\nfor consistence with other namespaces?" } { "Tag": [ "geometry", "rectangle", "probability", "analytic geometry", "absolute value", "probability and stats" ], "Problem": "X and Y are two independent Uniformly distributed random variables. X is Uniformly distributed between 0 and 1. Y is Uniformly distributed between 0 and 0.5.\r\n\r\nFind P(|X-Y| < 0.5).\r\n\r\nThe absolute value sign confuses me. Any ideas?\r\n\r\nThanks!\r\n-PD", "Solution_1": "Any ideas? DRAW THE PICTURE!!!\r\n\r\nThis is a rectangle, and the distribution is such that probability is proportional to area. Figure out what the set you want looks like. Ask whether it might be more convenient to compute the probability of the complement. (It is.) Use what you know about the areas of triangles. Don't compute any integrals.", "Solution_2": "I guess the brute force method was inevitable. Thanks, Dr. Merryfield!\r\n\r\n-PD", "Solution_3": "What I'm wondering actually is whether the areas where one variable or the other, is non-existent, counts as well into the probability. \r\n\r\nWe know X is a square from 0 to 1, both x and y-coordinate-wise, and Y is a rectangle with 090^{o}$", "Solution_1": "Let be O the circumcenter of ABC, and E the midpoint of OH. Let R, r the circumradius and inradius, respectively. It is well know that OH = 3OG, OI^2 = R^2 - 2rR, IE = (R/2) -r.\r\nFrom this we have\r\nIH = 2IE - IO\r\nIG = (2IE + IO)/3\r\nThen the scalar product of the vectors IH . IG equals (4.IE^2 - IO^2)/3 = - 2r (R-2r)/3 <0 and therefore cos (GIH) <0.\r\n\r\nThis problem was proposed but not used in the IMO 1990, Beijeing, China, proposed by France (FR1)" } { "Tag": [ "blogs", "\\/closed" ], "Problem": "Visit my blog and everything is way huge, however my font size is down to like 10/12 that range. It was fine yesterday...", "Solution_1": "For some reason your blog font size was missing. I've made it 8 now (you can change it to whatever you like though)." } { "Tag": [ "trigonometry", "algebra", "polynomial" ], "Problem": "if\r\n$x+\\frac{1}{x}=-1$\r\nfind\r\n$x^{999}+\\frac{1}{x^{999}}$", "Solution_1": "[hide]Let $x=\\cos\\theta+i\\sin\\theta$\n\nThen we know that\n\n$x+\\frac{1}{x}=2\\cos\\theta$.\n\nSo $\\cos{\\theta}=-0.5$\n\nNow, we also know that\n\n$x^{999}=\\cos{999\\theta}+i\\sin{999\\theta}$\nand\n$x^{-999}=\\cos{999\\theta}-i\\sin{999\\theta}$\n\nSo, those two add up to $2\\cos{999\\theta}$, which equals to 2, in all cases, since $\\cos{999\\theta}=1$, for all values of $\\theta$ satisfying the original equation for $\\cos{\\theta}=-0.5$.\n\n$2$[/hide]", "Solution_2": "[quote=\"ashrafmod\"]If $x+\\frac{1}{x}=-1$ find $x^{999}+\\frac{1}{x^{999}}$[/quote]\r\nAssume $x^{n}+\\frac{1}{x^{n}}=\\alpha.$ Denote by $T_{n}(x)$ the Chebychev polynomial of degree $n .$ We note that\r\n\\[(*)\\; \\; \\; \\left\\{ \\begin{array}{c}T_{n}(y)=\\frac{\\left(y+\\sqrt{y^{2}-1}\\right)^{n}+\\left(y-\\sqrt{y^{2}-1}\\right)^{n}}{2}=\\\\ \\\\ =\\cos{\\left(n\\cdot \\arccos{y}\\right) }\\; ,\\; \\; \\mbox{ when}\\; \\; |y|\\le 1\\; . \\end{array}\\right. \\]\r\nFrom $(*)$ we give \\[T_{n}\\left(\\frac{A+B}{2\\sqrt{AB}}\\right)=\\frac{A^{n}+B^{n}}{2\\left(AB\\right)^\\frac{n}{2}}\\] which imply ( $A\\leadsto x\\; ,\\; \\; B: \\leadsto \\frac{1}{x}$ )\r\n\\[\\begin{array}{c}A^{n}+B^{n}=x^{n}+\\frac{1}{x^{n}}=2\\cdot T_{n}\\left(\\frac{\\alpha}{2}\\right)\\\\ \\end{array}\\; . \\]\r\nFor instance, when $\\alpha=-1 \\; ,\\;$ because $T_{n}(-z)=(-1)^{n}T_{n}(z)$ , one finds\r\n\\[\\begin{array}{|c|}\\hline \\\\ x^{n}+\\frac{1}{x^{n}}=2\\cdot T_{n}\\left(-\\frac{1}{2}\\right)=2(-1)^{n}T_{n}\\left(\\frac{1}{2}\\right)=2(-1)^{n}\\cos\\left(\\frac{n\\pi}{3}\\right) \\\\ \\\\ \\hline \\end{array}\\; , \\]\r\nor (e.g. using first equality from $(*)$ )\r\n\\[\\begin{array}{|c|}\\hline \\hline \\\\ x^{n}+\\frac{1}{x^{n}}=2(-1)^{n}\\cdot T_{n}\\left(\\frac{1}{2}\\right)= =(-1)^{n}\\frac{\\left(1+i\\sqrt{3}\\right)^{n}+\\left(1-i\\sqrt{3}\\right)^{n}}{2^{n}}\\\\ \\\\ \\hline\\hline \\end{array}\\; . \\]\r\nFinally ($n\\leadsto 999$)\r\n\\[\\begin{array}{|c|}\\hline \\hline \\\\ x^{999}+\\frac{1}{x^{999}}=-2 \\frac{\\left(1+i\\sqrt{3}\\right)^{999}+\\left(1-i\\sqrt{3}\\right)^{999}}{2^{999}}=-2\\cdot \\cos{\\left(\\frac{999\\cdot \\pi}{3}\\right)}=\\\\ \\\\ =-2\\cdot\\cos{(333\\cdot \\pi)}=1 \\\\ \\\\ \\\\ \\hline\\hline \\end{array}\\; . \\]\r\n[b][color=red]Happy New Year ! [/color][/b]", "Solution_3": "Another way to look at it is to use the recursion:\r\n\r\n$(a^{n+1}+b^{n+1})=(a+b)(a^{n}+b^{n})-(ab)(a^{n-1}+b^{n-1})$\r\n\r\ntake $a=x$ and $b=\\frac{1}{x}$ for this particular problem.\r\n\r\nAlso, if $r$ is a solution, then\r\n\r\n$r+\\frac{1}{r}=-1$\r\n$r^{2}+r+1=0$\r\n$r^{3}-1=0$\r\n\r\nThen the equation obviously reduces... $(r^{3})^{333}+(r^{3})^{-333}=(1)^{333}+(1)^{-333}=1+1=2$", "Solution_4": "[quote=\"Altheman\"]Another way to look at it is to use the recursion:...\nThen the equation obviously reduces... $(r^{3})^{333}+(r^{3})^{-333}=(1)^{333}+(1)^{-333}=1+1=2$[/quote]\r\nInded, it's a more ellementary method ! Best wishes,proposer .", "Solution_5": "Obviously , $x\\neq0$ so $x^{2}+x+1=0$\r\nWe get , $x=\\frac{-1}{2}-i\\frac{\\sqrt{3}}{2}$ ,or $x=e^{-\\frac{2\\pi}{3}i}$ and,$x=\\frac{-1}{2}+i\\frac{\\sqrt{3}}{2}$,or $x=e^{\\frac{2\\pi}{3}i}$;In the two cases and via $Moiver$ formula we have ,$x^{9}99+\\frac{1}{x^{9}99=2}$." } { "Tag": [ "geometry", "circumcircle", "modular arithmetic", "combinatorics unsolved", "combinatorics" ], "Problem": "There is a frog in every vertex of a regular 2n-gon with circumcircle($n \\geq 2$). At certain time, all frogs jump to the neighborhood vertices simultaneously (There can be more than one frog in one vertex). We call it as $\\textsl{a way of jump}$. It turns out that there is $\\textsl{a way of jump}$ with respect to 2n-gon, such that the line connecting any two distinct vertice having frogs on it after the jump, does not pass through the circumcentre of the 2n-gon. Find all possible values of $n$.", "Solution_1": "[b]Claim:[/b] $2n = 4 \\pmod 8$.\r\n\r\nI label the vertices with classes $\\pmod{2n}$. Suppose the frogs follow the way of jump described (i.e.: after the jump no opposite vertices are both occupied).\r\n\r\n[i][u]Sufficiency[/i][/u]\r\n\r\nConsider the following w.o.j.: the frog in vertex $k$ jumps in $k+1$ if $k = 0,1 \\pmod 4$, otherwise it jumps in $k-1$. This is easily seen as an example of a good way of jump. []\r\n\r\n[i][u]Necessity[/i][/u]\r\n\r\n[b]Subclaim[/b] [i]If, after the jump, a frog lands in $k$, then another frog must land also in $k+n-2$.[/i]\r\n\r\nIn fact, the opposite of $k$ is $k+n$ and must be empty. The frog starting in $k+n-1$ must now find itself in $k+n-2$. []\r\n\r\nWLOG we can suppose that after the jump, vertex 0 is occupied. Now apply repeatedly the subclaim: every multiple of $n-2$ must be occupied and, this is equivalent to ask that every multiple of $\\gcd(2n, n-2)$ must be occupied.\r\n\r\nTwo cases: (even) $2n = 0 \\pmod 8$ and (odd) $2n = 2 \\pmod 4$.\r\n\r\n[b]Even case:[/b] $\\gcd(2n, n-2) = 2$ [because $4 = 2n-2(n-2)$, but $4 \\nmid n-2$]. So every even node must be occupied. In particular, $n$ is occupied. But then $0$ and $n$ are opposite vertices both occupied, and this is a contradiction. []\r\n\r\n[b]Odd case:[/b] $\\gcd(2n, n-2) = 1$ [$n-2$ is odd]. So every single node must be occupied and clearly this contradicts the hypotesys. []", "Solution_2": "We will show that $v_2(n)=1$.\nKey observations:\n[b]1.[/b] Any frog has only $2$ vertices to hop onto (since it has $2$ adjacent vertices). So only $2$ vertices are possible for a frog, and also a vertex can have only $2$ possible frogs.\n[b]2.[/b] There will be $n$ vertices and $2n$ frogs so by [b]1[/b] each vertex must have exactly $2$ frogs.\nHence, we assign to each vertex a pair: $(x_i,x_{i+1})$ to the [hide=odd vertices]\nLabel the vertices $v_1, v_2, \\cdots v_{2n}$. The the odd vertices are $v_1, v_3, v_5 \\cdots v_{2n-1}$ and the rest are even.\nThis is made more clear by the diagrams below.\n[/hide] and $(y_i, y_{i+1})$ to the even vertices, where $1 \\le i \\le n$ (and $x_{n+1}=x_1, y_{n+1}=y_1$).\n[b]3.[/b] By [b]2[/b], no vertex can be occupied by only $1$ frog. Hence if a vertex $(a,b)$ is chosen, then the 2 vertices of the form $(a,x), (b,y)$ cannot be chosen.\n\n[b]Contradiction when $n$ is odd[/b]\nStart with $(x_1,x_2)$ and start moving clockwise. By [b]3[/b], $(x_2,x_3)$ cannot be chosen. Hence $(x_3, x_4)$ must be chosen. (since we can't leave out $x_3$). Hence $(x_4, x_5)$ cannot be chosen. Hence $(x_5, x_6)$ mst be chosen and so on. So we get the chain (of the vertices we choose):\n$$(x_1, x_2) \\implies (x_3, x_4) \\implies (x_5, x_6) \\implies \\cdots \\implies (x_n, x_1)$$\nwhich contradicts our point [b]3[/b] as we cannot chose both $(x_1, x_2)$ and $(x_n, x_1)$. $\\blacksquare$\n\n[b]Note[/b]: Since $n$ is odd, the antipode of $(x_1, x_2)$ is an even vertex, and so won't bother our chain. Also, since $n$ is odd, hence $(x_n, x_1)$ is a member of our chain. \n[hide=Diagram]\n[asy]\n /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */\nimport graph; size(10cm); \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 = -180.5556147171591, xmax = 178.61017945740556, ymin = -38.903026657714214, ymax = 230.72568568296495; /* image dimensions */\npen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); \n\ndraw((-39.241507915278135,25.78735718949745)--(-2.677024375985958,36.541617053995175)--(20.583068850171806,66.73406022838508)--(21.654206732294128,104.83219962344998)--(0.12725100604783535,136.28384089840648)--(-35.77523291545517,149.07552608819023)--(-72.33971645474733,138.3212662236925)--(-95.59980968090511,108.12882304930261)--(-96.67094756302744,70.03068365423769)--(-75.14399183678115,38.579042379281205)--cycle, linewidth(2) + rvwvcq); \n /* draw figures */\ndraw((-39.241507915278135,25.78735718949745)--(-2.677024375985958,36.541617053995175), linewidth(2) + rvwvcq); \ndraw((-2.677024375985958,36.541617053995175)--(20.583068850171806,66.73406022838508), linewidth(2) + rvwvcq); \ndraw((20.583068850171806,66.73406022838508)--(21.654206732294128,104.83219962344998), linewidth(2) + rvwvcq); \ndraw((21.654206732294128,104.83219962344998)--(0.12725100604783535,136.28384089840648), linewidth(2) + rvwvcq); \ndraw((0.12725100604783535,136.28384089840648)--(-35.77523291545517,149.07552608819023), linewidth(2) + rvwvcq); \ndraw((-35.77523291545517,149.07552608819023)--(-72.33971645474733,138.3212662236925), linewidth(2) + rvwvcq); \ndraw((-72.33971645474733,138.3212662236925)--(-95.59980968090511,108.12882304930261), linewidth(2) + rvwvcq); \ndraw((-95.59980968090511,108.12882304930261)--(-96.67094756302744,70.03068365423769), linewidth(2) + rvwvcq); \ndraw((-96.67094756302744,70.03068365423769)--(-75.14399183678115,38.579042379281205), linewidth(2) + rvwvcq); \ndraw((-75.14399183678115,38.579042379281205)--(-39.241507915278135,25.78735718949745), linewidth(2) + rvwvcq); \n /* dots and labels */\ndot((-39.241507915278135,25.78735718949745),dotstyle); \nlabel(\"$(y_3 , y_4)$\", (-44.89336951430181,14.34440458440733), NE * labelscalefactor); \ndot((-2.677024375985958,36.541617053995175),dotstyle); \nlabel(\"$(x_3 , x_4)$\", (-3.1772291144232,22.484139296578775), NE * labelscalefactor); \ndot((20.583068850171806,66.73406022838508),dotstyle); \nlabel(\"$(y_2 , y_3)$\", (27.685931669226832,62.84365724442886), NE * labelscalefactor); \ndot((21.654206732294128,104.83219962344998),dotstyle); \nlabel(\"$(x_2 , x_3)$\", (22.937753087126826,108.29050938738611), NE * labelscalefactor); \ndot((0.12725100604783535,136.28384089840648),dotstyle); \nlabel(\"$(y_1 , y_2)$\", (1.5709494676768048,139.83198139705047), NE * labelscalefactor); \ndot((-35.77523291545517,149.07552608819023),dotstyle); \nlabel(\"$(x_1 , x_2)$\", (-45.23252512730895,158.1463844994362), NE * labelscalefactor); \ndot((-72.33971645474733,138.3212662236925),dotstyle); \nlabel(\"$(y_5 , y_1)$\", (-91.0185328832733,146.61509365719334), NE * labelscalefactor); \ndot((-95.59980968090511,108.12882304930261),dotstyle); \nlabel(\"$(x_6 , x_1)$\", (-118.82929314985903,112.69953235647897), NE * labelscalefactor); \ndot((-96.67094756302744,70.03068365423769),dotstyle); \nlabel(\"$(y_4 , y_5)$\", (-118.49013753685189,62.84365724442886), NE * labelscalefactor); \ndot((-75.14399183678115,38.579042379281205),dotstyle); \nlabel(\"$(x_4 , x_5)$\", (-88.30528797921615,20.449205618535913), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]\n[/hide]\nThe figure shows the case when $2n=10$.\n\n[b]Contradiction when $v_2(n) \\geq 2$[/b]\nAs above, we get\n$$(x_1, x_2) \\implies (x_3, x_4) \\implies (x_5, x_6) \\implies \\cdots \\implies (x_{n-1}, x_{n})$$\nThe issue here is that if the antipode of $(x_1, x_2)$ is of the form $(x_{n/2+1}, x_{n/2+2})$, then $n/2+1$ is odd since $n/2$ is even. Hence, the antipode is a part of this chain, a contradiction. $\\blacksquare$\n\n[hide=Diagram]\n[asy] \n/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */\nimport graph; size(10cm); \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 = -289.08081682802305, xmax = 443.7246821153303, ymin = 41.65393379090981, ymax = 591.7770420628335; /* image dimensions */\npen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); \n\ndraw((-94.74924462635187,159.16385133845765)--(-10.977808213587494,178.83487689287196)--(34.34802698119563,251.97974321249555)--(14.677001426781317,335.7511796252599)--(-58.46786489284226,381.077014820043)--(-142.23930130560663,361.40598926562876)--(-187.56513650038977,288.2611229460052)--(-167.89411094597546,204.48968653324079)--cycle, linewidth(2) + rvwvcq); \n /* draw figures */\ndraw(shift((-10.977808213587494,178.83487689287196))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\ndraw((-94.74924462635187,159.16385133845765)--(-10.977808213587494,178.83487689287196), linewidth(2) + rvwvcq); \ndraw((-10.977808213587494,178.83487689287196)--(34.34802698119563,251.97974321249555), linewidth(2) + rvwvcq); \ndraw((34.34802698119563,251.97974321249555)--(14.677001426781317,335.7511796252599), linewidth(2) + rvwvcq); \ndraw((14.677001426781317,335.7511796252599)--(-58.46786489284226,381.077014820043), linewidth(2) + rvwvcq); \ndraw((-58.46786489284226,381.077014820043)--(-142.23930130560663,361.40598926562876), linewidth(2) + rvwvcq); \ndraw((-142.23930130560663,361.40598926562876)--(-187.56513650038977,288.2611229460052), linewidth(2) + rvwvcq); \ndraw((-187.56513650038977,288.2611229460052)--(-167.89411094597546,204.48968653324079), linewidth(2) + rvwvcq); \ndraw((-167.89411094597546,204.48968653324079)--(-94.74924462635187,159.16385133845765), linewidth(2) + rvwvcq); \ndraw(shift((34.34802698119563,251.97974321249555))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\ndraw(shift((-142.23930130560663,361.40598926562876))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\ndraw(shift((-187.56513650038977,288.2611229460052))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\n /* dots and labels */\ndot((-94.74924462635187,159.16385133845765),dotstyle); \nlabel(\"$(x_3 , x_4)$\", (-113.31821368202519,126.76732035373574), NE * labelscalefactor); \nlabel(\"$(y_2 , y_3)$\", (-15.749209573420057,153.75449170292444), NE * labelscalefactor); \nlabel(\"$(x_2 , x_3)$\", (36.84117561987065,258.93526208950607), NE * labelscalefactor); \ndot((14.677001426781317,335.7511796252599),dotstyle); \nlabel(\"$(y_1 , y_2)$\", (17.465770548658284,342.6646911472454), NE * labelscalefactor); \ndot((-58.46786489284226,381.077014820043),dotstyle); \nlabel(\"$(x_1 , x_2)$\", (-66.95563726162416,397.3310125981661), NE * labelscalefactor); \nlabel(\"$(y_4 , y_1)$\", (-182.51608893635506,375.1876925167805), NE * labelscalefactor); \nlabel(\"$(x_4 , x_1)$\", (-237.18241038727567,276.9267096556319), NE * labelscalefactor); \ndot((-167.89411094597546,204.48968653324079),dotstyle); \nlabel(\"$(y_3 , y_4)$\", (-210.195239038087,181.43364180465645), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]\n[/hide]\nThe figure shows the case when $2n=8$.\n\n[b]Construction when $v_2(n)=1$[/b]\nIt is evident by the above arguments that we must have $2|n$ and $2 \\nmid n/2$, i.e. $v_2(n)=1$. \nWe now present our construction for $2n=12$, and in general, the idea is just to choose alternate pairs of vertices. The chosen vertices are marked.\n[hide=Construction]\n[asy]\n /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */\nimport graph; size(10cm); \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 = -4608.4679468105105, xmax = 7766.992511744151, ymin = -920.2182568903293, ymax = 8370.141577435446; /* image dimensions */\npen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); \n\ndraw((-981.0559417675616,1503.644810336426)--(-125.45697915408921,1931.444291643163)--(401.61371726742163,2729.7289914873395)--(458.9279800367176,3684.5991692155876)--(31.12849872998072,4540.19813182906)--(-767.1562011141956,5067.2688282505715)--(-1722.026378842444,5124.583091019867)--(-2577.6253414559164,4696.78360971313)--(-3104.6960378774274,3898.498909868954)--(-3162.0103006467234,2943.6287321407053)--(-2734.210819339987,2088.029769527234)--(-1935.9261194958106,1560.9590731057222)--cycle, linewidth(2) + rvwvcq); \n /* draw figures */\ndraw(shift((-981.0559417675616,1503.644810336426))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\ndraw((-981.0559417675616,1503.644810336426)--(-125.45697915408921,1931.444291643163), linewidth(2) + rvwvcq); \ndraw((-125.45697915408921,1931.444291643163)--(401.61371726742163,2729.7289914873395), linewidth(2) + rvwvcq); \ndraw((401.61371726742163,2729.7289914873395)--(458.9279800367176,3684.5991692155876), linewidth(2) + rvwvcq); \ndraw((458.9279800367176,3684.5991692155876)--(31.12849872998072,4540.19813182906), linewidth(2) + rvwvcq); \ndraw((31.12849872998072,4540.19813182906)--(-767.1562011141956,5067.2688282505715), linewidth(2) + rvwvcq); \ndraw((-767.1562011141956,5067.2688282505715)--(-1722.026378842444,5124.583091019867), linewidth(2) + rvwvcq); \ndraw((-1722.026378842444,5124.583091019867)--(-2577.6253414559164,4696.78360971313), linewidth(2) + rvwvcq); \ndraw((-2577.6253414559164,4696.78360971313)--(-3104.6960378774274,3898.498909868954), linewidth(2) + rvwvcq); \ndraw((-3104.6960378774274,3898.498909868954)--(-3162.0103006467234,2943.6287321407053), linewidth(2) + rvwvcq); \ndraw((-3162.0103006467234,2943.6287321407053)--(-2734.210819339987,2088.029769527234), linewidth(2) + rvwvcq); \ndraw((-2734.210819339987,2088.029769527234)--(-1935.9261194958106,1560.9590731057222), linewidth(2) + rvwvcq); \ndraw((-1935.9261194958106,1560.9590731057222)--(-981.0559417675616,1503.644810336426), linewidth(2) + rvwvcq); \ndraw(shift((458.9279800367176,3684.5991692155876))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\ndraw(shift((31.12849872998072,4540.19813182906))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\ndraw(shift((-2577.6253414559164,4696.78360971313))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\ndraw(shift((-3104.6960378774274,3898.498909868954))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\ndraw(shift((-1935.9261194958106,1560.9590731057222))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */\n /* dots and labels */\nlabel(\"$(x_4 , x_5)$\", (-1079.299808488029,1066.399569317697), NE * labelscalefactor); \ndot((-125.45697915408921,1931.444291643163),dotstyle); \nlabel(\"$(y_3 , y_4)$\", (147.72884887574773,1744.186827671024), NE * labelscalefactor); \ndot((401.61371726742163,2729.7289914873395),dotstyle); \nlabel(\"$(x_3 , x_4)$\", (451.56451641344484,2842.6696256919336), NE * labelscalefactor); \nlabel(\"$(y_2 , y_3)$\", (509.99445247838656,3800.920577156982), NE * labelscalefactor); \nlabel(\"$(x_2 , x_3)$\", (77.61292559781762,4653.997643705135), NE * labelscalefactor); \ndot((-767.1562011141956,5067.2688282505715),dotstyle); \nlabel(\"$(y_1 , y_2)$\", (-728.7201920983784,5273.35496599352), NE * labelscalefactor); \ndot((-1722.026378842444,5124.583091019867),dotstyle); \nlabel(\"$(x_1 , x_2)$\", (-1944.0628622491668,5401.900825336393), NE * labelscalefactor); \nlabel(\"$(y_6 , y_1)$\", (-3439.8692255116757,5004.577260094787), NE * labelscalefactor); \nlabel(\"$(x_6 , x_1)$\", (-3942.3666756701746,3871.0365004349123), NE * labelscalefactor); \ndot((-3162.0103006467234,2943.6287321407053),dotstyle); \nlabel(\"$(y_5 , y_6)$\", (-3977.4246373091396,2725.8097535620495), NE * labelscalefactor); \ndot((-2734.210819339987,2088.029769527234),dotstyle); \nlabel(\"$(x_5 , x_6)$\", (-3381.4392894467337,1568.897019476198), NE * labelscalefactor); \nlabel(\"$(y_4 , y_5)$\", (-2189.4685937219224,1113.1435181696506), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]\n[/hide]", "Solution_3": "Assume that jumping is over and the condition is satisfied..Label the vertices $A_1 \\dots A_{2n}$ with $L$ or $R$ depending on the direction the frog on them has jumped(left or right).Note that at least half of the vertices are empty and every vertex has either 0,1 or 2 frogs,so by PHP,exactly $n$ vertices are empty and so each vertex contains either 0 or 2 frogs.This means that\n $$ \\forall i \\leq 2n :A_i \\neq A_{i+2} \\implies 4|2n \\implies \\text{n is even}$$\nthat's because if the neighbors of $A_i$ have same labels,then $A_i$ contains exactly one frog,So the labels are $RRLLRR \\dots RRLL$ in this order.Note that if every block $RR$ is antipode of a $LL$,the condition would be satisfied,so $n=4k+2$ works and also,if every $RR$ is antipode of a $RR$ then the condition will not satisfied;meaning that $n$ is not divisible by 4,so answer is $n=4k+2$.", "Solution_4": "Claim: $n\\equiv 2 \\pmod 4$.\n\nThere are $2n$ frogs and $2n$ vertices or $n$ pairs of diametrically opposite points. If the frogs end up in more than $n$ vertices, then by PHP, we can find a pair of diametrically opposite vertices each with at least one frog.\n\nNumber the vertices: $1,2,\\dots ,2n$ in anticlockwise direction.\n\nWe have $2n$ frogs, at most $n$ vertices and at each vertex, at most $2$ frogs (from the neighbours). Thus $n$ vertices have exactly $2$ frogs.\n\nWithout loss of generality, vertex $1$ has $2$ frogs and frog from vertex $1$ jumped to vertex $2$. Notice that each vertex contains either two frogs or zero. Now vertex $2$ also has $2$ frogs or frog from vertex $3$ jumped to vertex $2$. As frog from vertex $2$ jumped to vertex $1$, frog from vertex $4$ jumped to vertex $5$ and so on. Also notice that frog from vertex $1$ did not jump to vertex $2n$ thus vertex $2n$ has $0$ frogs.\n\nConcluding we get that vertex $k$ has two frogs if and only if $k\\equiv 1,2\\pmod 4$. We have that vertex $2n$ has $0$ frogs or $2n\\equiv -1,0\\pmod 4$. Thus $n$ is even, say $n=2m$. Mark 'X' on a vertex if it has 2 frogs and 'O' if it has 0.\n\nIf $n+1$ labelled vertex is marked $X$ then we get two diametrically opposite points with at least one frog on them. Thus we need $n+1\\equiv 3,0\\pmod 4$ or $2m\\equiv 2,3\\pmod 4$ or $m$ is odd. Thus $n = 4m+2$.\n\nConstruction for $n \\equiv 2\\pmod 4$: \nThe frogs on vertex $4k+1$ and $4k+2$ swap after jump. Frog on $4k+3$ jump on the left vertex or $4k+2$ and frog on vertex $4k+4$ jump on the right vertex or $4k+5$. ", "Solution_5": "$n=4k+2$\nFirst note that after the move each vertex has atmost 2 frogs since there are two neighbouring vertices.Call a vertex rich if it has frogs after the move and poor otherwise.Since no two opposite vertices are rich there are atmost $n$ vertices but since each vertex can accommodate atmost $2$,we conclude all rich vertices have $2$ frogs each\n\nClaim $1$:$n$ is even.\nSuppose not.\nConsider a bipartite graph of rich and poor vertices.Draw an edge between $2$ vertices if they are adjacent.So every vertex has degree $2$.Notice that every poor vertex has to be connected to a rich vertex,since every frog is jumping from its place.Also a poor vertex cannot be connected to $2$ rich vertices which will imply that it is jumping at $2$ distinct places.Therefore every poor vertex is connected to exactly one poor vertex and every rich vertex is connected to exactly one rich vertex.But this means there are $\\frac {n}{2}$ edges among poor vertices,contradiction to our assumption.\n\nClaim $2$:$n\\neq 4k$\nNow a poor vertex $a_1$ is connected to a rich vertex $b_1$which is connected to rich vertex $b_2$ which is connected to poor vertex $a_2$.So they must lie in this order on the polygon(say in clockwise direction).Now to the right of $a_2$ there is a poor vertex $a_3$.So this process of formation of quadruple of $(a_{2k+1},a_{2k+2},b_{2k+1},b_{2k+2})$ will continue.But since the vertex numbers of opposite vertices are congruent mod $4$,we get that both are rich or both are poor.This is a contradiction.\nFor $n=4k+2$ the construction is simple.$4m+1$ and $4m+3$ jump to $4m+2$ and $4m+2$ and $4m+4$ jump on $4m+3$ where $m$ takes values from $0$ to $2k$", "Solution_6": "Note : I\u2019ll be sometimes calling \u201cinhabited\u201d vertices R or red, similarly with \u201cuninhabited\u201d (blue)\nWe claim that $n=4k+2$ is the only possible value. Call a vertex \u201cinhabited\u201d if there\u2019s at least one frog on it (after the algorithm is applied), and call a vertex \u201cuninhabited\u201d if not. \n\nClaim 1: every vertex has 2 or 0 frogs on it \nProof : initially, there are $2n$ frogs. Now, $a_{1}+a_{2}+ \\dots + a_{i} = 2n \\leq 2i \\leq 2n$. This occurs when $i=n, $ and there are two frogs on all inhabited vertices $\\Box$\n\nClaim 2: at least 2 inhabited vertices are adjacent. \nProof : The frog on any inhabited vertex must jump to an adjacent vertex. \n\nclaim 3: the vertices must occur as $RRBBRRBB \\dots$\nProof : Let the vertices be labelled in the format $A_i$, such that the $i$s increase by one as we move clockwise. (Diagram below). Let $A_1, A_2$ be adjacent R\u2019s. $A_3$ should be $B$ as $A_2$s frog must have gone to $A_1$ and $A_3$s frog to $A_2$. $A_4$s frog going to $A_3$ isnt possible coz then it has only 1 frog. Similarly, $A_4$ is uninhabited too. Now $A_4$ s frog must have gone to $A_5$ so it is inhabited. Similarly $A_5$s frog went to $A_6$ so it is inhabited too. $A_{4n}$ and $A_{4n-1}$ have to be uninhabited coz $A_1$s frog went to $A_2$. So basically, the string $RRBB$ is repeating. this implies the thing has $4k$ vertices (as if there are $4k+2$, it will end with $AAAA$, which as we said is not possible). Now, let $\\text{number of vertices}=4t$. Note that $A_{2t+1}, A_{2t+2}$ are B. We see that $A_i$ is R if $i$ is $1,2 \\mod 4$, B otherwise. So $2t+1 \\equiv{3} \\pmod{4} \\implies t \\equiv{1} \\pmod{2}$. So number of vertices $=2n=4t=4(2k+1) \\implies n \\equiv{2} \\pmod{4}$\n\n\n\nConstruction: label vertices as $a_i$s mod 4, that is $1,2,3,4,1,2,3,4, \\dots$. Consider consecutive vertices $4,1,2,3$. Transfer frogs from $4,2$ to $1$, and from $1,3$ to $2$ for all $\\blacksquare$", "Solution_7": "Solved with [size=150][color=#f00]None[/color][/size] since I am lonely and just got kicked out of my last Math group.\nWe claim is that the answer is just $n \\equiv 2\\pmod 4$.\nActually,one of our insights is that the simultaneous condition is not of any concern.We can just move the frogs one by one.\n\nSo,we just place a $1$ at each of the vertices and then $1'$ for a frog that has moved.\nNow, we just start moving with the first frog and see that the diametrically opposite vertex just gets fixed and we circle it.\nNow, the $1s$ adjacent to it have to move in a specific direction and thus we gain some insight.\nWith some casework by observing the patterns in the occurrence of $1's$ ,circles and empty points ,we are done.", "Solution_8": "The answer is $n \\equiv 2 \\pmod 4$.\nAll the congruences below are taken modulo $2n$.\n\nAfter the frogs move, we color each vertex which has at least one frog.\n[b]CLAIM:[/b] there are exactly $n$ colored vertices, each of them has exactly two frogs.\n\n[i]Proof:[/i] First notice that the total number of diameters which connect the vertices of the 2n-gon is $n$, and each diameter has at most one colored vertex, hence $n$ is the maximum number of colored vertices. Also it is easy to see that each colored vertex has at most two frogs, hence the claim is proved.\n\nLet $A_1A_2...A_{2n}$ be the 2n-gon, and take the indices modulo $2n$. Assume that $A_i$ is colored, then $A_i$ has two frogs from $A_{i-1}$ and $A_{i+1}$. Now it is easy to see that $A_{i-1}$ or $A_{i+1}$ is colored. Assume WLOG that $A_{i+1}$ is colored, then we can see a pattern of colored vertices, that is, each one of $A_{i+4k}$ and $A_{i+1+4k}$ is colored.\n\n$\\boxed A$ Since $A_{i-1}$ isn't colored, then:\n$$i-1 \\not\\equiv{i+1+4k} \\Rightarrow 2\\not\\equiv{4k} \\Rightarrow 2\\mid n$$\n$\\boxed B$ $A_{i+n}$ is the antipode of $A_i$, then:\n$$i+n \\not\\equiv i+4k \\Rightarrow n\\not\\equiv 4k \\Rightarrow 4 \\nmid n$$\nHence $v_2(n)=1$, and this actually works due to $A$ and $B$. $\\blacksquare$" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let x,y,z be positive real numbers such that x^4+y^4+z^4=1.\r\nDetermine with proof the minimum value of x^3/1-x^8 + y^3/1-y^8 + z^3/1-z^8", "Solution_1": "let\u00b4s work with each term individually.\r\nwe have that $ \\frac{x^4}{x(1\\minus{}x^8)}$ attains it\u00b4s minimal value when it\u00b4s denominator is in its maximum.so let\u00b4s see when $ x(1\\minus{}x^8)$ is maximum.let $ k$ be this maximum.we have that :\r\n$ 8x^8(1\\minus{}x^8)^8\\leq{8k^8}$.by AM-GM,we\u00b4ll get that $ 8x^8(1\\minus{}x^8)\\leq{(\\frac{8}{9})^9}$,so we\u00b4ll\r\nhave that $ 8k^8\\equal{}{(\\frac{8}{9})^9}$.so $ k\\equal{}\\frac{8}{9\\sqrt[4]{3}}$.now all you have to do is compute:\r\nthe min is obtained when the denominetor is $ k$.direct substitution yields that the minimun is\r\n$ \\dfrac{9\\sqrt[4]{3}}{8}$. Done. :D", "Solution_2": "[quote=\"v235711\"]let\u00b4s work with each term individually.\nwe have that $ \\frac {x^4}{x(1 \\minus{} x^8)}$ attains it\u00b4s minimal value when it\u00b4s denominator is in its maximum.so let\u00b4s see when $ x(1 \\minus{} x^8)$ is maximum.let $ k$ be this maximum.we have that :\n$ 8x^8(1 \\minus{} x^8)^8\\leq{8k^8}$.by AM-GM,we\u00b4ll get that $ 8x^8(1 \\minus{} x^8)\\leq{(\\frac {8}{9})^9}$,so we\u00b4ll\nhave that $ 8k^8 \\equal{} {(\\frac {8}{9})^9}$.so $ k \\equal{} \\frac {8}{9\\sqrt [4]{3}}$.now all you have to do is compute:\nthe min is obtained when the denominetor is $ k$.direct substitution yields that the minimun is\n$ \\dfrac{9\\sqrt [4]{3}}{8}$. Done. :D[/quote]\r\n\r\n'we have that $ \\frac {x^4}{x(1 \\minus{} x^8)}$ attains it\u00b4s minimal value when it\u00b4s denominator is in its maximum.so let\u00b4s see when $ x(1 \\minus{} x^8)$ is maximum.'\r\n\r\nBut why is it?", "Solution_3": "[quote=\"Stephen\"][quote=\"v235711\"]let\u00b4s work with each term individually.\nwe have that $ \\frac {x^4}{x(1 \\minus{} x^8)}$ attains it\u00b4s minimal value when it\u00b4s denominator is in its maximum.so let\u00b4s see when $ x(1 \\minus{} x^8)$ is maximum.let $ k$ be this maximum.we have that :\n$ 8x^8(1 \\minus{} x^8)^8\\leq{8k^8}$.by AM-GM,we\u00b4ll get that $ 8x^8(1 \\minus{} x^8)\\leq{(\\frac {8}{9})^9}$,so we\u00b4ll\nhave that $ 8k^8 \\equal{} {(\\frac {8}{9})^9}$.so $ k \\equal{} \\frac {8}{9\\sqrt [4]{3}}$.now all you have to do is compute:\nthe min is obtained when the denominetor is $ k$.direct substitution yields that the minimun is\n$ \\dfrac{9\\sqrt [4]{3}}{8}$. Done. :D[/quote]\n\n'we have that $ \\frac {x^4}{x(1 \\minus{} x^8)}$ attains it\u00b4s minimal value when it\u00b4s denominator is in its maximum.so let\u00b4s see when $ x(1 \\minus{} x^8)$ is maximum.'\n\nBut why is it?[/quote]\r\n\r\ni calculated this maximun denominator without interfering the value of the numerator." } { "Tag": [], "Problem": "OK I'm having trouble with this problem. Bare with me because I don't know how to type square roots\r\n\r\nrationalize the denominator:\r\n\r\n\r\nThe square root of 9 / the square root of 20x^2y, this reduces to; 3 / the square root of 20x^2y. this is where I'm stuck, I don't know what to multiply it by.\r\n\r\nEdit: Ok I looked at the answer in the back of my book and the answer is 3 times the square root of 5y / 10xy. Now you can get this answer by multiplying the equation by the square root of 5y / the square root of 5y, correct? if so why do you multiply it by that?", "Solution_1": "So, you were able to get to $ \\frac{3}{\\sqrt{20x^2y}}$. Now how do you solve this?\r\n\r\nI think the easiest way to do this is to start by simplifying the denominator before you rationalize it. In other words:\r\n$ \\sqrt{20x^2y}\\equal{}\\sqrt{(4x^2)(5y)}\\equal{}2x\\sqrt{5y}$\r\n\r\nNow, we have $ \\frac{3}{2x\\sqrt{5y}}$. To rationalize the denominator, it should be clearer now that you should multiply by $ \\frac{\\sqrt{5y}}{\\sqrt{5y}}$ to get:\r\n\r\n$ \\frac{3}{2x\\sqrt{5y}}\\cdot \\frac{\\sqrt{5y}}{\\sqrt{5y}} \\equal{} \\frac{3\\sqrt{5y}}{2x\\cdot 5y} \\equal{} \\frac{3\\sqrt{5y}}{10xy}$", "Solution_2": "Ok thanks that makes sense now. I couldn't find an example similar to this in the book." } { "Tag": [ "inequalities", "factorial", "induction", "number theory unsolved", "number theory" ], "Problem": "If $ n \\leq m$ s.t $ m,n \\in \\mathbb{Z}^\\plus{}$ .\r\n\r\nProve or disprove :\r\n\r\n$ 2^n n! \\leq \\frac{(m\\plus{}n)!}{(m\\minus{}n)!} \\leq (m^2\\plus{}m)^n$.\r\n\r\n :D", "Solution_1": "$ \\frac {(m \\plus{} n)!}{(m \\minus{} n)!} \\equal{} (m \\minus{} n \\plus{} 1)\\times....\\times(m \\minus{} n \\plus{} 2n)\\geq1\\times2\\times...\\times2n\\geq2\\times4\\times...\\times2n \\equal{} 2^n n!$\r\n and for $ 1\\leq{k}\\leq{n}$ we have:$ (m \\minus{} n \\plus{} k)(m \\minus{}n\\plus{} 2n \\minus{} k \\plus{} 1)\\leq{m(m \\plus{} 1)}$if multiple this inquality for $ k \\equal{} 1,...,n$ we get\r\n$ \\frac {(m \\plus{} n)!}{(m \\minus{} n)!} \\equal{} (m \\minus{} n \\plus{} 1)\\times....\\times(m \\minus{} n \\plus{} 2n)\\leq{(m^2 \\plus{} m)^n}$", "Solution_2": "[quote=\"ali666\"]$ \\frac {(m \\plus{} n)!}{(m \\minus{} n)!} \\equal{} (m \\minus{} n \\plus{} 1)\\times....\\times(m \\minus{} n \\plus{} 2n)\\geq1\\times2\\times...\\times2n\\geq2\\times4\\times...\\times2n \\equal{} 2^n n!$\n and for $ 1\\leq{k}\\leq{n}$ we have:$ (m \\minus{} n \\plus{} k)(m \\minus{} n \\plus{} 2n \\minus{} k \\plus{} 1)\\leq{m(m \\plus{} 1)}$if multiple this inquality for $ k \\equal{} 1,...,n$ we get\n$ \\frac {(m \\plus{} n)!}{(m \\minus{} n)!} \\equal{} (m \\minus{} n \\plus{} 1)\\times....\\times(m \\minus{} n \\plus{} 2n)\\leq{(m^2 \\plus{} m)^n}$[/quote]\r\n\r\nFaster than I thought .....\r\n\r\n\r\nNice work dude :wink:", "Solution_3": "Induction kills ti easily...especially left side... :D", "Solution_4": ":D :wink:" } { "Tag": [], "Problem": "For how many positive integers $ n$ will $ \\frac{60}{n}$ also be an integer?", "Solution_1": "factors of $ 60$. $ 60\\equal{}2^2\\cdot3\\cdot5$, which has $ (2\\plus{}1)(1\\plus{}1)(1\\plus{}1)\\equal{}3\\cdot2\\cdot2\\equal{}\\boxed{12}$ factors. thus, the answer is $ 12$." } { "Tag": [ "geometry", "\\/closed" ], "Problem": "Is there a way to upload images to the forum without having them attached at the end of a post? Sometimes when I'm solving a geometry problem, I make a diagram and want to include it next to my solution rather than at the end. This would make it possible to hide a picture rather than having it display at the end.\r\n\r\n(I believe this works: post the image as an attachment, click submit, edit, delete the attachment, and use the img tags to include it within the text body. But it's a painful workaround...)", "Solution_1": "At this moment there is no way other than the one you suggested (which itself is a bug! and will be fixed - so don't do it anymore :) ) but maybe we'll figure out something in the future. \r\n\r\nYou can make an attachement in a future post and then edit your previous one to display the link.", "Solution_2": "I have same question with mathfanatic.\r\n\r\nI see the images within the posts, not as the attached image. How do you do it? I saw those in this site so this IS possible. ;)", "Solution_3": "Or you could sign up with a free direct-linkable host such as http://www.photobucket.com and upload your diagrams there...", "Solution_4": "[quote=\"LynnelleYe\"]Or you could sign up with a free direct-linkable host such as http://www.photobucket.com and upload your diagrams there...[/quote]\r\n\r\nYes... hot-linking works very well, except the pictures get deleted if you don't use your account often enough, i think....", "Solution_5": "Hmm I advise you not to do that because of many reasons. Mostly because there might be different issues with browsers. After all, up to one month ago there was no way to hide anything but text in spoilers, and everybody was happy :) \r\n\r\nThere will be a proper solution found soon, but I repeat, try [u][b]NOT[/b][/u] to link images to external free-advertising site." } { "Tag": [], "Problem": "Chess problem marathon!!!!!!!!!\r\n\r\nPost a problem after you solve it.\r\n\r\nCompositions only, please.\r\n\r\nDont know how to post the picture, sorry.\r\n\r\nW:Kg8 Qh7 Rc1 Re7 Na4 Nb7 Ba2 Bb2 Pb3 Pg7\r\nB:Kd5 Qh1 Ra5 Rh5 Nd4 Bd1 Bg1 Pa7 Pa6 Pe3 Ph2\r\n\r\nWhite mates in 2\r\n\r\nBy Godfrey Heathcote\r\n[i]Hampstead and Highgate Express,[/i] 1905-06\r\n(1st Prize)[/img]", "Solution_1": "Is this the position you're referring to?\r\n[hide=\"If so, here is a possible solution\"][list]1. Ng3+, Kc6\n2. Qg6 1-0 (unless you count the knight block from black to e6 as an extra move)[/list][/hide]\nA way to create a chess diagram and display it: [list]Go to [url=http://www.apronus.com/chess/wbeditor.php]www.apronus.com[/url], \nCreate a chess diagram (click \"make diagram\" to get a jpeg of the diagram after placing the pieces), \nSave it to your desktop, \nAdd it as an attachment to your post.[/list]\r\n\r\nAlso, there used to be a chess marathon [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193263&start=60]here[/url], although it hasn't been active since December 2008.", "Solution_2": "Sorry, that does count as 3 moves.\r\n\r\nThis is a chess composition, so the point is not to win; it is to do what the instructions say.\r\n\r\nHint: the KEY first move is not a check", "Solution_3": "Nice problem! I have not idea why I did not consider a [hide] capture[/hide] to be the first move until like after spending 30 minutes on this.\n\n[hide=\"solution\"]\n1.Nxa5 \n\nif \n\n1.... Nxb3\n2.Rxd1#\n\nif \n\n1... Ne6\n2.Rd7#\n\nif black makes another move, then\n2.Rc5#[/hide]\r\n\r\nnew problem (much easier):\r\n\r\nWhite mates in 2:\r\n\r\nW: Kb8, Qb3,Re1\r\nB:Kd8, Re8, pd7 pe7 (yes this is pretty ridiculous but...)\r\n\r\nhopefully I remembered this correctly..", "Solution_4": "[hide=\"At hurdler s Solution\"]If 1. Nxa5, what about 1. ... Kd6 from black? That white rook at e7 would be hanging, and gives black an extra move (and even with 2. Ba3+, black can just block with the h-rook).[/hide]\n[hide=\"As for hurdler s Problem\"]We can just finish this in one move with 1. Qb6++ (++ is another way of saying mate).\n\nBut if we want to extend this to 2 moves,\n1. Qe6.\n[list]If black moves the rook, white finishes with 2. Qxe7++.\nIf 2. dxe6, Rd1++\nIf 2. d6 or d5, Qc8++\n[/list][/hide]\r\n[img]http://www.chessvideos.tv/bimg/zow0ceba8u8.png[/img]\r\n\r\n\r\nA New Problem:\r\nBlack to move, and mate white (are we allowed to not specify how many moves to mate?).\r\n[img]http://www.chessvideos.tv/bimg/4a32618rld0k0.png[/img]", "Solution_5": "sorry hurdler, but what about Nxa5 Bc2?\r\n\r\n(this is still for the first porblem)\r\n\r\nHere is a BIG hint. [hide]1.Rcc7[/hide]", "Solution_6": "That big hint is more of a giveaway.\r\n[hide=\"To put the first problem to rest\"]After that move (1. Rcc7), white can just finish the next move with 2. Nc3++, with a few exceptions:\n\nIf 1. ... Ne2, 2. Qxh5++\nIf 1. ... Rc5, 2. Rxc5++\nIf 1. ... Rxa4, 2. Rc5++[/hide]", "Solution_7": "Geez i hope i remember Pattern's problem correctly...i saw ir once at home...but now some reason the image is a \"x\".\r\n\r\n[hide]1...Qh8\r\n2.Kh8 Rh2\r\n3.Kh2 (or Nf3) Nf3\r\n4.Kg1 Rh8[/hide]\r\n\r\nIs this the right position i have in mind?", "Solution_8": "[quote=\"Patterns_34\"][hide=\"At hurdler s Solution\"]If 1. Nxa5, what about 1. ... Kd6 from black? That white rook at e7 would be hanging, and gives black an extra move (and even with 2. Ba3+, black can just block with the h-rook).[/hide]\n[hide=\"As for hurdler s Problem\"]We can just finish this in one move with 1. Qb6++ (++ is another way of saying mate).\n\nBut if we want to extend this to 2 moves,\n1. Qe6.\n[list]If black moves the rook, white finishes with 2. Qxe7++.\nIf 2. dxe6, Rd1++\nIf 2. d6 or d5, Qc8++\n[/list][/hide]\n[img]http://www.chessvideos.tv/bimg/zow0ceba8u8.png[/img]\n\n\nA New Problem:\nBlack to move, and mate white (are we allowed to not specify how many moves to mate?).\n[img]http://www.chessvideos.tv/bimg/4a32618rld0k0.png[/img][/quote]\r\n\r\n\r\n1...rh2. Since black threatens mate on g2, white''s only choices are qf3 or kh2 (notice there are no checks for white). If qf3, obviously nf3 mate. If 2.kh2 rh8. AFter 3.nh5, just r takes night and mate on h1. Otherwise, kG1. Then, qg3+ and after fg3, nf3 +++ (lol it's checkmate (++) and double check (+) to give (+++)) ." } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "Let $ \\alpha\\equal{}\\{v_1,v_2\\}$ be a basis for $ V$ and $ \\beta^*\\equal{}\\{(v_1\\plus{}v_2)^*,(v_1\\minus{}v_2)^*\\}$ where $ v_1^*,v_2^*$ is the canonical basis of the dual. Prove that $ \\beta^*$ is basis of $ V^*$ and $ (v_1\\plus{}v_2)^*\\ne v_1^*\\plus{}v_2^*.$", "Solution_1": "There's a flaw in the notation: $ *$ is not a meaningful operator when applied to individual vectors. It is only meaningful applied to bases, where we say that $ v_i^*$ is the linear functional which is $ 1$ at $ v_i$ and zero at $ v_j$ for $ j\\neq i$.\r\nBy this definition, the $ *$ of a basis is automatically a basis for the dual space. Now, if we take $ w\\equal{}v_1^*\\plus{}v_2^*$ coming from the standard basis, we get that $ w(v_1)\\equal{}1\\plus{}0$ and $ w(v_2)\\equal{}0\\plus{}1$, so $ w(v_1\\plus{}v_2)\\equal{}2$ and $ w(v_1\\minus{}v_2)\\equal{}0$. That takes the wrong value at $ v_1\\plus{}v_2$ to be $ (v_1\\plus{}v_2)^*$ with respect to any basis." } { "Tag": [ "AMC", "AIME", "function", "AMC 12", "AMC 12 B", "AMC 10", "AMC 10 B" ], "Problem": "Does anyone know the cutoff for the AMC 12 B... it was hard", "Solution_1": "It won't be determined until they score the tests.", "Solution_2": "Also: the initial inquiries were fine, but from this point forward, please don't ask us about the qualifying scores for the AIME. \r\n\r\nWe don't know. We don't have any inside information. When the AMC office knows, they will let us know. For the A test, it took about a week. It might take less time for the B, it might take more. Please don't complain about it. When we know, we'll know. Absolutely do not contact the AMC to ask them about it.", "Solution_3": "We need to tabulate and verify a large number of answer forms in order to determine the 1% (on the 2007 AMC 10 B) and the 5% (on the 2007 AMC 12 B) levels. Before we can do that, we have to receive, log-in, sort and open envelopes, and then run the answer forms through the scanners. It will take us about two weeks before I can confidently report an AIME qualification cut-off score on the 2007 AMC 10 B and 2007 AMC 12 B. As I write this on Thursday, 1 day after the contest date, we have received 22 packages out of an expected 1,500. Please be patient.\r\n\r\nI can tell you however that the cut-off score on the 2007 AMC 10 B will be no higher than 120, and on the 2007 AMC 12 B will be no higher than 100, just as we say on the contest cover page.\r\n\r\nSteve Dunbar\r\nDirector, American Mathematics Competitions", "Solution_4": "[quote=\"AMCDirector\"]We need to tabulate and verify a large number of answer forms in order to determine the 1% (on the 2007 AMC 10 B) and the 5% (on the 2007 AMC 12 B) levels. Before we can do that, we have to receive, log-in, sort and open envelopes, and then run the answer forms through the scanners. It will take us about two weeks before I can confidently report an AIME qualification cut-off score on the 2007 AMC 10 B and 2007 AMC 12 B. As I write this on Thursday, 1 day after the contest date, we have received 22 packages out of an expected 1,500. Please be patient.\n\nI can tell you however that the cut-off score on the 2007 AMC 10 B will be no higher than 120, and on the 2007 AMC 12 B will be no higher than 100, just as we say on the contest cover page.\n\nSteve Dunbar\nDirector, American Mathematics Competitions[/quote]\r\n\r\nSo if we get over $100$ on the 2007 AMC 12 B, we are sure to do the AIME?", "Solution_5": "100 or more, yeah, and 100 can't really be done, so sure.", "Solution_6": "Can someone pm me the questions please because our test director took our booklets", "Solution_7": "[quote=\"usaha\"]Can someone pm me the questions please because our test director took our booklets[/quote]\r\n\r\nI'd have to request the same. We actually had school cancelled on Wednesday, and only by chance did we manage to convince our principal to administer the test to us on Wednesday during the day. Our test booklets are history though.\r\n\r\n(or, when will AoPS post the problems in the contest resources section?)", "Solution_8": "what school do you go to? My school was cancelled but we got to convince our prinicipal to let us take it (i live in indiana", "Solution_9": "Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135055]this post[/url].", "Solution_10": "No i want the test, not the answers", "Solution_11": "[quote=\"usaha\"]No i want the test, not the answers[/quote]\r\n\r\nYeah.. we're requesting questions, not answers.. since I didn't get to take any paper out of the testing room.\r\n\r\nI also live in Indiana, where we all experienced such a wonderful fog day...", "Solution_12": "carmel? We got to convince the principal and like 40 out of trhe 100 people schelduled took it(im estimating)", "Solution_13": "If you want specific questions, I can post them for you, but I am not about to type out the entire test.", "Solution_14": "uh... 6-11 I don't remember what the questions were.", "Solution_15": "6. Triangle $ABC$ has side lengths $AB=5$, $BC=6$, and $AC=7$. Two bugs start simultaneously from $A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $D$. What is $BD$?\r\n\r\n(A) 1 (B) 2 (C) 3 (D) 4 (E) 5\r\n\r\n7. All sides of the convex pentagon $ABCDE$ are of equal length, and $\\angle A=\\angle B=90^{\\circ}$. What is the degree measure of $\\angle E$?\r\n\r\n(A) 90 (B) 108 (C) 120 (D) 144 (E) 150\r\n\r\n8. Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$?\r\n\r\n(A) 2 (B) 3 (C) 4 (D) 5 (E) 6\r\n\r\n9. A function $f$ has the property that $f(3x-1)=x^{2}+x+1$ for all real numbers $x$. What is $f(5)$?\r\n\r\n(A) 7 (B) 13 (C) 31 (D) 111 (E) 211\r\n\r\n10. Some boys and girls are having a car wash to raise money for a class trip to China. Initially 40% of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then 30% of the group are girls. How many girls were initially in the group?\r\n\r\n(A) 4 (B) 6 (C) 8 (D) 10 (E) 12\r\n\r\n11. The angles of quadrilateral $ABCD$satisfy $\\angle A=2\\angle B=3\\angle C=4\\angle D$. What is the degree measure of $\\angle A$, rounded to the nearest whole number?\r\n\r\n(A) 125 (B) 144 (C) 153 (D) 173 (E) 180", "Solution_16": "Thanks a lot! ...I guess it's good that I at least got the first 12 right..", "Solution_17": "I thought the amc12b was easier than the amc12a. Maybe it would be 100 this time.", "Solution_18": "[quote=\"moogra\"]I thought the amc12b was easier than the amc12a. Maybe it would be 100 this time.[/quote]\r\n\r\nI didn't find much difference. Then again, I didn't try to do all of the problems. The first 20 problems on both tests were similar difficulty, but this test seemed to have more algebra/counting problems that the A test." } { "Tag": [ "Duke", "college" ], "Problem": "What's the next big tournament?", "Solution_1": "Here's the link to the majority of Georgia competitions:\r\nhttp://www.gctm.org/math_competitions.htm\r\n\r\nI believe the next one would be Cobb County, which I'm pretty sure we're going to. Anyone else?", "Solution_2": "There's one in Alabama that GACS is going to soon", "Solution_3": "I might be coming to Cobb. No guarantees whatsoever, though (auditions, auditions). Then there's Duke (which is admittedly a North Carolina competition, but still), but again, I'm not sure if I'll be able to make it. I'll be at UGA, though, which is the week after Duke. (Be there, everybody!)", "Solution_4": "yeah i'm not sure about Duke; it's so far away!", "Solution_5": "I am going to Duke and UGA.", "Solution_6": "when is duke? I'll be in NC in I think Charlotte Nov. 11-12, 2006 so JIC...", "Solution_7": "it is that weekend, here is the link for more info\r\n\r\nhttp://www.math.duke.edu/dumu/contests/2006/index.html", "Solution_8": "Thanks! Too bad I don't think I'll make it then...", "Solution_9": "are you all going to vestavia hills math tournament on December 9th, 2006?", "Solution_10": "[quote=\"now a ranger\"]are you all going to vestavia hills math tournament on December 9th, 2006?[/quote]\r\nI'm no one, and I can't speak for anyone else, but I know Rockdale Magnet (recognize the name at all? I can't remember what place we got at Vestavia last year) is only sending 4 people this year. We took 8 or 12, maybe more, last year on a charter bus and everything... but this year they're kinda sneaking off in secret by car. Literally, in secret--apparently, if the whole team knew, too many people would ask if they could go, and they'd all be turned down. \r\n\r\nYeah... Vestavia... bad memories... Hope it all goes well for you, though.", "Solution_11": "Well, it was supposed to be a secret. But everyone knows about it now--funny how that works, isn't it? Yes, Rockdale is taking four people, but they may not all be from Rockdale and we're going for fun because I begged our coach to let us go.", "Solution_12": "yeah, I hope I can get into the top 10, score of 60+ at least.", "Solution_13": "[quote=\"cleansmyheart\"]Well, it was supposed to be a secret. But everyone knows about it now--funny how that works, isn't it? Yes, Rockdale is taking four people, but they may not all be from Rockdale and we're going for fun because I begged our coach to let us go.[/quote]\r\n\r\nSorry, Amy, it probably wasn't my right to say the little I knew about it, but now a ranger just seems to be posting everywhere--thought I'd direct a message at him. Thanksgiving Break is boring. X.x" } { "Tag": [ "induction", "geometry", "3D geometry", "algebra open", "algebra" ], "Problem": "Prove by mathematical induction the formula for sum of the cubes of the first n positive integers:\r\n1^3 + 2^3 + 3^3 + ... n^3 = (n(n+1)/2)^2. Write down explicitly the first five equations.", "Solution_1": "very simple :D \r\nyou can prove it yourself\r\n\r\n(i don't think this is an open problem. please post it at Algebra Solved Problems)", "Solution_2": "For $n=1$, it is clearly true.\r\n\r\nAll you have to do is show that \\[\\left[\\frac{n\\left(n+1\\right)}{2}\\right]^{2}+\\left(n+1\\right)^{3}=\\left[\\frac{\\left(n+1\\right)\\left(n+2\\right)}{2}\\right]^{2},\\] which will complete the proof by induction." } { "Tag": [], "Problem": "Hi,\r\n\r\nI have no idea about these problems.Can some one give me a hint.\r\n\r\nhttp://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=2612\r\nhttp://www.spoj.pl/problems/LIAR/\r\n\r\nThanks.", "Solution_1": "The ACM one looks quite easy, just visualize the sentences as vertices of a graph and draw edges between them. There will be two types of edges - type 1 means that the vertices must have the same boolean value and type 2 means that they have the opposite boolean value. Rest is up to you.", "Solution_2": "The other one is actually even simpler, brute force should work all right. Suppose that person 1 says the truth - then you know exactly who is a liar and who's not, you only need to check that this setup works ($O(N^{2})$). Do this for every person. And then try the possibility that they're all liars. And that's it.", "Solution_3": "Your replies were really helpful.\r\nEspecially for Acm-Icpc problem After viewing it in terms of graph i could see components and cycles.I got it Accepted.\r\n\r\nNow in this problem\r\nhttp://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=3794\r\nI was able to find maximal independent set size.I am not getting how to check \r\nwhether a unique solution exists or not.\r\n\r\nThanks a lot for the replies.", "Solution_4": "Well.... I do have an idea but I have this feeling that there exists a much simpler solution :).\r\n\r\nSo anyway, you could try to calculate for every person the max. size of the independent set if he is invited and the max. size if he isn't invited. If these two numbers are different for every person, the solution is unique, otherwise it isn't.", "Solution_5": "Hi ,\r\nI used the approach that u told.It's giving TLE.I feel it should pass because it takes O(n^2) to find MIS (when invited/no invited) on all nodes.\r\n\r\nAny pointers regarding this.", "Solution_6": "Hm, finding the MIS for one person should take O(n), so the whole program should run in O(n^2). Is that what you meant?", "Solution_7": "Yes i meant that.", "Solution_8": "Then it should definitely work. Perhaps you get TLE because your program gets stuck somewhere? I would try some random inputs and debugging, can't think of anythink else...", "Solution_9": "I figured the problem with my implementation.\r\nThanks 4 ur replies.. :lol:" } { "Tag": [ "induction", "function", "algorithm", "topology", "calculus", "calculus computations" ], "Problem": "Show that a real solution exist for $ e^{\\minus{}x}\\equal{}x$ analytically.", "Solution_1": "What do you mean analytically? $ f(x) \\equal{} e^{\\minus{}x} \\minus{} x$ is continuous, $ f(0) \\equal{} 1 > 0$, $ f(1) \\equal{} e^{\\minus{}1} \\minus{} 1 < 0$, so there exists a solution in $ (0,1)$. Since $ e^{\\minus{}x}$ is decreasing, while $ x$ is increasing, moreover this solution is unique. How else?", "Solution_2": "Your answer is analytic opposed to the intuitive approach of the intersection of $ e^\\minus{}x$ with $ x$. However, can you explain some more. Thanks", "Solution_3": "mavropnevma is relying on the Intermediate Value Theorem (for continuous functions). Have you seen it before? If not, you may wish to google it.", "Solution_4": "The Intermediate Value Theorem would have been my first choice, as well. But let's take a look at my second choice.\r\n\r\nDefine a sequence as follows: $ x_0\\equal{}0,$ and after that, $ x_{n\\plus{}1}\\equal{}e^{\\minus{}x_n}.$\r\n\r\nNote that $ x_2\\equal{}e^{\\minus{}1}>0$ and I'm willing to argue (by induction) that $ x_2\\le x_n\\le x_1\\equal{}1$ for all $ n\\ge 2.$ \r\n\r\nThat means we are iterating the function $ f(x)\\equal{}e^{\\minus{}x}$ on an interval $ [a,b]$ for some $ 01,$ then $ e^{\\minus{}x}1.$ And if $ xe^{\\minus{}e^{\\minus{}1}}>e^{\\minus{}1}$ (the latter since $ e^{\\minus{}1}<1$), so you can't have $ x\\equal{}e^{\\minus{}x}$ for any of those, either.", "Solution_7": "Thank you very much." } { "Tag": [ "floor function" ], "Problem": "Find real numbers $ x,$ such that $ x^2\\minus{}\\lfloor x\\rfloor^2\\equal{}1$", "Solution_1": "Hello,\r\n\r\nI assume I did a mistake here, but I cannot find it :blush: . :o \r\n\r\nWe need to prove $ x^{2} \\minus{} \\lfloor x\\rfloor^{2} \\equal{} 1$\r\n\r\nAnd after that : $ \\lfloor x\\rfloor^{2} \\equal{} x^{2} \\minus{} 1$\r\n\r\nSince $ \\lfloor x\\rfloor^{2}$ is an integer, so must $ x^{2} \\minus{} 1$ be an integer.\r\n\r\nAnd from there $ x^{2}$ must be an integer, so we can write:\r\n\r\n$ x^{2} \\equal{} x^{2} \\minus{} 1$. So no solutions. :blush: :ninja:", "Solution_2": "$ \\lfloor x \\rfloor ^2 \\neq \\lfloor x^2 \\rfloor$.\r\n\r\nFor example, $ \\sqrt{2}$ is certainly a solution of the original equation.", "Solution_3": "[hide=\"Your mistake\"]\n$ x^2\\in \\mathbb Z$ doesn't imply $ x\\in\\mathbb Z$.[/hide]\n\n[hide]\nThe solution is $ x \\equal{} \\sqrt {k^2 \\plus{} 1}$ for any positive integer $ k$.\n\nLet $ \\lfloor x\\rfloor \\equal{} k$. We can write the equation as: $ x^2 \\equal{} k^2 \\plus{} 1$. Taking the square root of both sides yields $ x \\equal{} \\sqrt {k^2 \\plus{} 1}$. Now we verify that the solution works. We have $ k < \\sqrt {k^2 \\plus{} 1} < \\sqrt {k^2 \\plus{} 2k \\plus{} 1} \\equal{} k \\plus{} 1$, so $ \\lfloor\\sqrt {k^2 \\plus{} 1}\\rfloor \\equal{} k$. Therefore:\n\\[ x^2 \\minus{} \\lfloor x\\rfloor^2 \\equal{} (k^2 \\plus{} 1) \\minus{} k^2 \\equal{} 1.\\][/hide]" } { "Tag": [ "function", "IMO Shortlist", "graph theory", "combinatorics", "Extremal combinatorics" ], "Problem": "Let $n\\geq 3$ be a fixed integer. Each side and each diagonal of a regular $n$-gon is labelled with a number from the set $\\left\\{1;\\;2;\\;...;\\;r\\right\\}$ in a way such that the following two conditions are fulfilled:\n\n[b]1.[/b] Each number from the set $\\left\\{1;\\;2;\\;...;\\;r\\right\\}$ occurs at least once as a label.\n\n[b]2.[/b] In each triangle formed by three vertices of the $n$-gon, two of the sides are labelled with the same number, and this number is greater than the label of the third side.\n\n[b](a)[/b] Find the maximal $r$ for which such a labelling is possible.\n\n[b](b)[/b] [i]Harder version (IMO Shortlist 2005):[/i] For this maximal value of $r$, how many such labellings are there?\n\n[hide=\"Easier version (5th German TST 2006) - contains answer to the harder version\"]\n[i]Easier version (5th German TST 2006):[/i] Show that, for this maximal value of $r$, there are exactly $\\frac{n!\\left(n-1\\right)!}{2^{n-1}}$ possible labellings.[/hide]\n\n[i]Proposed by Federico Ardila, Colombia[/i]", "Solution_1": "This problem was created by federico Ardila, Colombian IMO team leader an exmember of the IMO AB.", "Solution_2": "This problem follows easy by induction. In a) is easy to guess that the answer is n-1, but guess the answer in b) is practicaly imposible. \r\nIn b) we can get: If f(n) is the number of labellings \r\n \r\n 2f(n)=\u2211 [C(n-1;a)C(n-1;b)f(a)f(b)] for a+b=n when C(n-1;i) is the binomial coeficient\r\n\r\n There is some method to get the value of f(a) ? \r\n There is other solution to this problem or the only solution is guess the answer and prove by induction?", "Solution_3": "I haven't checked what you've done, but I suspect it can be done by considering the generating function $ F(x) \\equal{} \\sum_{n \\geq 1} \\frac {f(n)x^n}{n!(n \\minus{} 1)!}$. (Both the form of the recursion and answer indicate that this, or something very similar, will work.) Then the expression you've written is very close to a convolution (i.e. the product $ F(x)\\cdot F(x)$). Also, [[LaTeX]] is enabled on this forum, and it would be much easier to read what you write if you use it. (See e.g. the link in the previous sentence or [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690[/url].)", "Solution_4": "[hide]\nFor (a), it follows easily by induction that the maximal $ r$ is $ n\\minus{}1$.\n\nFor (b), note that it is impossible by (a) for a $ n\\minus{}1$-labeling of an $ n$-gon to have more than one $ 1$. Suppose it did. Let $ AB$ be one of these edges. Clearly no other edge connected to $ A$ is labeled $ 1$, and for every vertex $ C$, $ AC\\equal{}BC$. Then, remove vertex $ A$. Clearly, each label $ {1, 2, ... , n\\minus{}1}$ still remains on some edge, and the labeling remains good. This contradicts (a).\n\nThus there is a unique edge labeled $ 1$ for each $ n\\minus{}1$-labeling of $ n$. Now, choose this edge $ AB$ in $ \\binom{n}{2}$ ways and collapse $ AB$ into a single vertex $ A'$. Label $ A'C$ the label that $ AC$ and $ BC$ were both labeled. Next decrease every label of the graph by $ 1$. Clearly we have an $ \\binom{n}{2}$-to-1 mapping from the $ n\\minus{}1$-labelings of $ n$ to the $ n\\minus{}2$-labelings of $ n\\minus{}1$. By induction the answer is $ \\frac{n!(n\\minus{}1)!}{2^{n\\minus{}1}}$.[/hide]", "Solution_5": "[quote=\"m.candales\"]In b) we can get: If f(n) is the number of labellings \n \n 2f(n)=\u2211 [C(n-1;a)C(n-1;b)f(a)f(b)] for a+b=n when C(n-1;i) is the binomial coeficient[/quote]\nThis is slightly off... consider a vertex $V$ with $k\\ge2$ of its edges labeled $n-1$, which clearly must exist since some triangle must have two sides labeled $r=n-1$. Let the $k$ vertices connected to $V$ be group $A$ and the other $n-k$ vertices (including $V$) be group $B$. It's easy to check that all edges between $A$ and $B$ must be labeled $n-1$, and all other edges are labeled at most $n-2$. By part (a), group $A$ has at most $k-1$ different labels and group $B$ has at most $n-k-1$ labels. Hence to get the labels from $1$ to $n-1$ in $A\\cup B$, we must choose $k-1$ labels from $\\{1,2,\\ldots,n-2\\}$ to be the labels in group $A$, so the remaining $n-k-1$ will be those in group $B$. Thus we have the recursion\n\\begin{align*}\n2f(n) &= \\sum_{k=1}^{n-1}\\binom{n}{k}\\binom{n-2}{k-1}f(k)f(n-k)\\\\\n&= n!(n-2)!\\sum_{k=1}^{n-1}\\left(\\frac{f(k)}{k!(k-1)!}\\right)\\left(\\frac{f(n-k)}{(n-k)!(n-k-1)!}\\right).\n\\end{align*}After this if we let $f(x)=x!(x-1)!g(x)$ then we have\n\\[2(n-1)g(n)=\\sum_{k=1}^{n-1}g(k)g(n-k).\\]From here it's pretty easy to guess the answer after a few small cases (this is basically what [b]JBL[/b] had...).", "Solution_6": "The polygon doesn't matter, it is a graph theory problem. To distinguish, the vertices have numbers ($n$ vertices numbered $1$ through $n$) and the edges have \"colors\" ($r$ colors, from $1$ to $r$). For part (a), the easy part: Suppose the answer is $R(n)$. Take any edge with color $r$. Say this edge is $AB$. It is easy to see that for any other vertex $C$, exactly one of $CA$ and $CB$ has color $r$. Call $S_A$ the set of vertices $C$ such that $CA$ is color $r$, and $S_B$ the analogous set. It is easy to see $\\{1,...,n\\} = S_A \\cup S_B$ is a partition. Suppose $|S_A|=a$, $|S_B|=b$. Clearly $a+b=n$. Also, $R(n)=R(a)+R(b)+1$. From induction it is easy to see $R(n)=n-1$. \n\nNow for the difficult part, part (b). Here is the idea. In the original graph, suppose we draw a circle/oval around the whole graph. Then, we draw a circle/oval around $S_A$ and another one around $S_B$. Clearly $S_A$ is also partitioned into two \"smaller\" graphs. Draw a circle/oval around each of these. Do this again and again until you are finished, and each vertex has a circle drawn around it, and every circle is \"paired\" with one other circle, both inside a bigger circle.\n\nWe want to count the number of possible final drawings.\n\nBut what if we draw them in reverse? First draw a circle around each vertex. Then, draw a circle that contains exactly two circles (in this case, these smaller circles will have only a vertex inside). Then draw a circle that contains two circles. Then draw a circle that contains two circles. Etcetera... At the end, we clearly have drawn $2n-1$ circles in total. Now, take the moment when we draw the circles around each vertex. After the moment, we will label the first circle we draw with a $1$. We will label the second circle we draw with a $2$. Etcetera... Then we will label the last circle we draw (the one around the entire graph) with $n-1$. \n\nThe trick is that, if a circle is labeled $x$, then the graph inside the circle will contain only edges with color $\\le x$, and the only edges with color $x$ will be edges between vertices from one of the circle's \"sub-circles\" to the other \"sub-circle\" (remember each circle has two sub-circles). For example, the biggest circle is labeled $n-1$, and the edges labeled $n-1$ go from $S_A$ (a subcircle) to $S_B$ (another subcircle).\n\nTherefore, the process of drawing circles (the order is important!!!) completely determines the final configuration of colors. \n\nNow, say the answer for part (b) is $B(n)$. Then, draw the first $n$ circles (around the vertices), and then draw the $n+1$-th circle. There are ${n}\\choose{2}$ ways of doing this. Then, we can consider this $n+1$-th circle as normal circle (a circle that only has one vertex inside), and then proceed as if we only had $n-1$ vertices, pretending like this $n+1$-th circle didn't have two vertices inside it. Then,\n\n$B(n) =$ ${n}\\choose{2}$$B(n-1)$. \n\nFrom this, it is very easy to conclude $B(n)=\\frac{n!(n-1)!}{2^{n-1}}$.\n\nThis is a really beautiful and simple problem in my opinion.", "Solution_7": "Excellent problem.\n\nNote that we may simply let the regular polygon be a complete graph on $n$ vertices; let the vertices be $A_1,A_2,\\cdots, A_n.$ Let $f(A_iA_j)$ return the label of the edge connecting $A_i$ and $A_j.$\n\n[b]a)[/b] We claim that $r=n-1\\ \\forall\\ n.$ First, we show that we cannot have $r\\geq n$ for any $n.$ Assume this is true for $n=3,4,\\cdots, k-1.$ Assume for the sake of contradiction that it is also possible for $n=k.$ Then by the first restriction, there must exist an edge with label $k$, WLOG let it be $A_{k-1}A_k.$\n\nObviously the claim holds for $n=3.$\n\nThen for $1\\leq i\\leq k-2,$ exactly one of $j\\in\\{k-1,k\\}$ must satisfy $f(A_iA_j)=k$ by the second condition. Let $X_1,X_2$ be the set of indices such that $j=k,k-1$ respectively. Let $x_1=|X_1|,x_2=|X_2|$ and WLOG $x_1\\geq x_2.$ \n\nCase 1: $x_2=0.$ Then $f(A_{k-1}A_i)\\neq k\\ \\forall\\ i,$ and no two $i,j\\in X_1$ can have $f(A_iA_j)=k$ without violating the second condition (we get a triangle with equal labels on all sides). Hence we may ignore $A_k,$ since all other edges (which aren't connected to $A_k$ ) must have label less than $k.$ Hence we must distribute labels from $1,2,\\cdots, k-1$ to the remaining edges formed by $A_1,A_2,\\cdots, A_{k-1},$ but this is impossible by the induction hypothesis.\n\nCase 2: $x_2\\neq 0.$\n\nClaim: If this case is doable, it is always doable by letting $f(A_iA_j)=k$ for all $i\\in X_1,j\\in X_2.$\n\nProof: This is a very natural claim. Note that since no edge with both vertices in one of $X_1,X_2$ can have label $k.$ Hence by replacing any edge between the two sets with label not $k,$ we can replace the label with $k$ without causing a contradiction (no triangles are formed, and the other conditions are easy to check).\n\nNow it suffices to label $X_1\\cup \\{n-1\\}$ and $X_2\\cup \\{n\\},$ but each of these can give us less than $x_1,$ $x_2$ colors, meaning their union gives us less than $n$ colors. Hence we may conclude $r0$, but they can equal $1$). By inductive hypothesis, we can use at most $x-1$ labels for $A$ and $y-1$ labels for $B$, so adding $r$, it follows that there are at most $n-1$ labels for $G$, proving part (a).\\\\\nWe use the same setup for part (b). Note that equality only holds for part (a) if $A$ and $B$ don't share colors. Note that once we've chosen $A$ and $B$, the actual identities of vertices $v$ and $w$ doesn't matter. Thus, if we suppose that $|A|=k$, we have $\\binom{n}{k}$ ways of choosing $A$ (determining $B$). Once we do this, there are $(n-1)-1$ (subtracting $r$) labels we can possibly use for $A$, of which we have to pick $k-1$, so there are $\\binom{n-2}{k-1}$ ways to do this. Since $A$ and $B$ are indistinguishable though, we must divide by two. Thus, if $f(n)$ denotes the answer, we have the recursion\n$$f(n)=\\frac{1}{2}\\sum_{k=1}^{n-1} \\binom{n}{k}\\binom{n-2}{k-1}f(k)f(n-k).$$\nBy using strong inductive hypothesis (since $k,n-k 0,$ find $ m$ so that $ \\sum_{n \\equal{} m\\plus{}1}^{\\infty}|\\widehat{g}(n)|^{2}<\\epsilon^{2}.$\r\n\r\nFor $ N$ such that $ N^{2}> m,$\r\n\r\n$ \\left|\\frac{1}{N}\\sum_{n \\equal{} 1}^{N^{2}}\\widehat{g}(n)\\right|\\le\\frac{1}{N}\\sum_{n \\equal{} 1}^{m}|\\widehat{g}(n)|\\plus{}\\frac{1}{N}\\sum_{n \\equal{} m\\plus{}1}^{N^{2}}|\\widehat{g}(n)|$\r\n\r\nThe first term on the right is $ \\frac{1}{N}$ times a fixed finite sum and goes to zero as $ N\\to\\infty.$ We use the Cauchy-Schwartz inequality on the second term:\r\n\r\n$ \\frac{1}{N}\\sum_{n \\equal{} m\\plus{}1}^{N^{2}}|\\widehat{g}(n)|\\le\\frac{1}{N}\\sqrt{N^{2}\\minus{}m}\\left[\\sum_{n \\equal{} m\\plus{}1}^{N^{2}}|\\widehat{g}(n)|^{2}\\right]^{\\frac{1}{2}}$\r\n\r\n$ \\le\\frac{N}{N}\\left[\\sum_{n \\equal{} m\\plus{}1}^{\\infty}|\\widehat{g}(n)|^{2}\\right]^{\\frac{1}{2}}<\\epsilon.$\r\n\r\nHence $ \\limsup_{N\\to\\infty}\\left|\\frac{1}{N}\\sum_{n \\equal{} 1}^{N^{2}}\\widehat{g}(n)\\right| <\\epsilon,$ \r\n\r\nwhich means that the limit is zero.\r\n\r\nNote that we might as well have been working in $ \\ell^{2}$ the whole time - everything was given from the beginning in terms of a particular orthonormal basis." } { "Tag": [], "Problem": "In $\\triangle ABC, \\angle A = 100^\\circ, \\angle B = 50^\\circ, \\angle C = 30^\\circ, \\overline{AH}$ is an altitude, and $\\overline{BM}$ is a median. Then $\\angle MHC =$\n\n[asy]\ndraw((0,0)--(16,0)--(6,6)--cycle);\ndraw((6,6)--(6,0)--(11,3)--(0,0));\ndot((6,6));\ndot((0,0));\ndot((11,3));\ndot((6,0));\ndot((16,0));\nlabel(\"A\", (6,6), N);\nlabel(\"B\", (0,0), W);\nlabel(\"C\", (16,0), E);\nlabel(\"H\", (6,0), S);\nlabel(\"M\", (11,3), NE);[/asy]\n\n$\\text{(A)} \\ 15^\\circ \\qquad \\text{(B)} \\ 22.5^\\circ \\qquad \\text{(C)} \\ 30^\\circ \\qquad \\text{(D)} \\ 40^\\circ \\qquad \\text{(E)} \\ 45^\\circ$", "Solution_1": "[hide=\"Answer\"]$\\triangle AHC$ is a 30-60-90 triangle, so we have $AC=2x$, $AM=CM=AH=x$, $HC=x\\sqrt{3}$. Because $AH=AM$, $\\triangle AHM$ is isosceles, and $\\angle HAM=60^\\circ$, so $\\triangle AHM$ is equilateral. Therefore, $HM=x$. Finally, we draw the altitude from $M$ to $HC$, which must split $HC$ in half because $\\triangle MHC$ is isosceles. Therefore, we see that both triangles must be 30-60-90 triangles, so $\\angle MHC=30^\\circ\\Rightarrow \\boxed{C}$.[/hide]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Suppose $n$ lines in plane are such that no two are parallel and no three are concurrent. For each two lines their angle is a real number in $[0,\\frac{\\pi}2]$. Find the largest value of the sum of the $\\binom n2$ angles between line.\r\n[i]By Aliakbar Daemi[/i]", "Solution_1": "Actually,If we denote by $\\angle(l_{1},l_{2})$ the angle between $l_{1},l_{2}$,and $0\\leq \\angle(l_{1},l_{2})<\\frac{\\pi }{2}$ \r\nand $\\measuredangle(l_{1},l_{2})$ the measured angle between $l_{1},l_{2}$($0\\le\\measuredangle(l_{1},l_{2})<\\pi$)\r\nthen$\\measuredangle(l_{1},l_{2})\\leq \\angle(l_{1},l_{2})$\r\nIF the n lines are $l_{1},...l_{n}$\r\nTHen\r\nFor$n=2m+1$\r\n\\[\\sum_{1\\leq ia):\npi(b/2)^2/2-pi(a/2)^2/2 = (pi)b^2/8-(pi)a^2/8 = (pi/8)(b^2-a^2)\nWe must now add/subtract the difference to the area of each half of the large circle to find the shaded/clear area. Since we assumed that the area of semicircle b is greater than that of semicircle a, we must subtract the difference from the area of the large half-circle to determine the shaded area, which contains semicircle a:\nshaded area = (pi/8)(a^2+2ab+b^2)-(pi/8)(b^2-a^2) = (pi/8)(2a^2+2ab)\nConversely, we must add the difference to the area of the large half-circle in order to find the clear area, which contains semicircle b:\nclear area = (pi/8)(a^2+2ab+b^2)+(pi/8)(b^2-a^2) = (pi/8)(2b^2+2ab)\nTherefore, the ratio of the shaded area to the clear area is the following:\n(2a^2+2ab)/(2b^2+2ab) = (a^2+ab)/(b^2+ab) = [a(a+b)]/[b(a+b)] = [b]a/b[/b][/hide]", "Solution_2": "[hide=\"Answer\"]$A_{shaded}=\\frac{\\pi (a+b)^2}{2}+\\frac{\\pi a^2}{2}-\\frac{\\pi b^2}{2}=\\pi (a^2+ab)$\n$A_{unshaded}=\\frac{\\pi (a+b)^2}{2}+\\frac{\\pi b^2}{2}-\\frac{\\pi a^2}{2}=\\pi (b^2+ab)$\n$\\frac{A_{shaded}}{A_{unshaded}}=\\frac{a^2+ab}{b^2+ab}=\\frac{a}{b}\\Rightarrow \\boxed{B}$[/hide]", "Solution_3": "It doesn't seem like ya'll are accounting for the area that lies outside of the circle with diameter a+b.\r\nIf you are, please explain how you controlled for that.", "Solution_4": "There is no area that lies outside of the circle. Both of the smaller circles are internally tangent to the larger circle. They can't possibly have any area outside it.", "Solution_5": "[hide]The entire circle has radius $\\frac{a+b}{2}$. The area of the shaded region is the area of the big semicircle minus the area of semicircle b, plus semicircle a's area. The area of the white is the big semicircle minus semicircle b plus semicircle a. \n\nShaded: $\\frac{(\\frac{a+b}{2})^2}{2}\\pi-\\frac{(\\frac{b}{2})^2}{2}\\pi+\\frac{(\\frac{a}{2})^2}{2}\\pi=\\frac{a(a+b)}{4}\\pi$. \n\nWhite: $\\frac{(\\frac{a+b}{2})^2}{2}\\pi-\\frac{(\\frac{a}{2})^2}{2}\\pi+\\frac{(\\frac{b}{2})^2}{2}\\pi=\\frac{b(a+b)}{4}\\pi$.\n\nRatio: $\\frac{\\frac{a(a+b)}{4}\\pi}{\\frac{b(a+b)}{4}\\pi}=\\frac{a(a+b)}{b(a+b)}=\\frac{a}{b}$. \n[/hide]", "Solution_6": "Oops, I didn't even realize I was using the diameter instead of the radius. :roll: Oh well, it still works. :P", "Solution_7": "[quote=\"JesusFreak197\"]Oops, I didn't even realize I was using the diameter instead of the radius. :roll: Oh well, it still works. :P[/quote]\r\n\r\nI was thinking that your solution was so easy... until I realized that there was a mistake (but since you always calculated the area from the diameter, the change wasn't visible).", "Solution_8": "the reason that the smaller circles cannot lie outside the larger circles like that is because they have a smaller radius of curvature... if you think about it too, when you move along the larger circle, the smaller circle pulls in faster than the larger circle...i also believe you can show this with power of a point...not sure though" } { "Tag": [ "ratio", "projective geometry", "geometry unsolved", "geometry" ], "Problem": "two triangles $ABC$ and $DEF$ are such that $AD$ , $BE$, $CF$ are concurrent at $O$.let $DE, AB$ intersect at $Z$ ,$AC,DF$ intersect at $Y$ and $BC,FE$ intersect at $X$ prove that $A\\{XYZO\\}=D\\{XYZO\\}$ without using desargues theorem.", "Solution_1": "please help", "Solution_2": "I think that, by the notation $A\\{XYZO\\},$ you mean the double ratio (=cross ratio) of the pencil $A.XYZO,$ which let us to change in to $(A.XYZO).$\r\n\r\nSo, if this is correct, we can solve your problem as follows:\r\n\r\nWe denote the points $S\\equiv AD\\cap EF$ and $T\\equiv AD\\cap BC.$\r\n\r\nBecause of the pencil $A.XYZO$ is intersected from the segment line $BC,$ we have that $(A.XYZO) = (X,C,B,T)$ $,(1)$\r\n\r\nThe segment line $EF,$ intersects the pencil $D.XYZO$ and so, we have $(D.XYZO) = (X,F,E,S)$ $,(2)$ \r\n\r\nThe segment lines $BC,$ $EF,$ intersect the pencil $O.XCBA$ and so, we have that $(X,C,B,T) = (X,F,E,S)$ $,(3)$\r\n\r\nFrom $(1),$ $(2),$ $(3)$ $\\Longrightarrow$ $(A.XYZO) = (D.XYZO)$ $,(4)$ and the proof is completed.\r\n\r\nKostas Vittas.\r\n\r\nPS. From $(4),$ we conclude that the points $X,$ $Y,$ $Z,$ are collinear (because of the pencils $A.XYZO,$ $D.XYZO,$ have the segment line $AD,$ as their common ray) and so, we have a proof ( well known I think) of the Desarques's theorem, based on the double ratio theory.", "Solution_3": "thank you very much ! vittasko" } { "Tag": [ "limit", "calculus" ], "Problem": "Compute the following:\r\n\r\n$ \\lim_{r \\to 0}\\frac {1 \\minus{} \\frac {1}{(1 \\plus{} \\frac {r}{k})^{ nk}}}{r}$\r\n\r\nAs a side note, this formula has some financial implications (it is not in the exact same form, but it is similar), such as computing prices of bonds given certain yields to maturity. $ n$ represents the number of years until maturity, $ k$ represents the frequency of payment per year (i.e. if $ k \\equal{} 2$, one receives payments every 6 months) and $ r$ represents the yield to maturity percentage. In this case, the limit is evaluating what the price of the bond would be if $ r \\equal{} 0$.\r\n\r\n\r\nHint:\r\n[hide]Binomial Expansion[/hide]", "Solution_1": "First I simplified the expression\r\n\r\nLim [(1 + r/k)^-nk - 1]/[ r(1 + r/k)^-nk]\r\n\r\nthen using l\u00b4hospitale\u00b4s \r\n\r\nit simplified to lim 1/r which doesn't exist...\r\n \r\nAre you sure that the limit was supposed to approach 0?", "Solution_2": "I apologize - I made a mistake with the signs. The new expression should now be correct (hopefully :D).\r\n\r\nl'Hopital's rule does make this problem fairly straightforward I suppose, but I would think it makes it a bit more messy and not as elegant. If you solve it using that rule, also consider the hint I gave for another approach.", "Solution_3": "[quote=\"pkothari13\"]Compute the following:\n\n$ \\lim_{r \\to 0}\\frac {1 \\minus{} \\frac {1}{(1 \\plus{} \\frac {r}{k})^{ nk}}}{r}$\n\nAs a side note, this formula has some financial implications (it is not in the exact same form, but it is similar), such as computing prices of bonds given certain yields to maturity. $ n$ represents the number of years until maturity, $ k$ represents the frequency of payment per year (i.e. if $ k \\equal{} 2$, one receives payments every 6 months) and $ r$ represents the yield to maturity percentage. In this case, the limit is evaluating what the price of the bond would be if $ r \\equal{} 0$.\n\n\nHint:\n[hide]Taylor series[/hide][/quote]\r\n\r\nTaylor series goes into the calculus forum not around here....", "Solution_4": "Sorry about that - I guess while technically it is a Taylor Series, this should be under the realm of Intermediate Topics because it is a [b]binomial [/b]expansion. That gives most of the problem away lol, but oh well." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "I came up with this new problem when I try to select icons on my desktop. I myself isn't a nice combinatorician, so can anybody help me to solve this?\r\n\r\nFor a $ n\\times n$ table, find the maximum number $ a$ of cells that can be coloured such that the following is satisfied:\r\nFor all $ 0\\leq x\\leq a$, there always exist a rectangular region in the $ n\\times n$ table, that contains $ x$ coloured cells.", "Solution_1": "I should add one more point here, a rectangular region means a region of reatangular shape, with sides parallel to either one edge of the $ n\\times n$ table.\r\n\r\nAlso, a region of zero size is also considered to be a reactangular region." } { "Tag": [ "arithmetic sequence" ], "Problem": "In how many ways can we pick three different numbers out of the group\r\n\r\n1,2,3,..., 100\r\n\r\nsuch that the largest number is larger than the product of the two smaller ones? (The order in which we pick the numbers does not matter.)\r\n\r\n\r\nI'm confused, please help...thanks! :lol:", "Solution_1": "I present two approaches, one can work them out in detail.\r\n\r\nApproach 1: Let $ xy$ and equality occurs when $a_1=a_2=\\ldots=a_n$.\r\n\r\n$\\text {RMS-AM-GM-HM}$\r\n\r\n$\\sqrt{\\frac{a_1^2+a_2^2+\\ldots+a_n^2} {n}} \\geq {\\frac{a_1+a_2+\\ldots+a_n} {n}} \\geq \\sqrt[n]{a_1 \\cdot a_2 \\cdot \\ldots \\cdot a_n} \\geq {\\frac{n} {\\frac{1} {a_1}+\\frac{1} {a_2}+\\ldots+\\frac{1} {a_n}}}$\r\n\r\nNote that RMS-AM is just power mean with $x=2$ and $y=1$.\r\n\r\nWhile I'm at it, does anyone know what RMS (or power mean) is used for in real life?", "Solution_4": "Wow, they sure raised the standard for getting started... you now have to know the power-mean inequality?... and RMS-AM-GM-HM...", "Solution_5": "[url=http://mathcircle.berkeley.edu/BMC6/ps0506/inequalities.pdf]This[/url] says it's the Power Mean Inequality. :D", "Solution_6": "Does anyone have proofs for these that don't use Lagrange Multipliers?" } { "Tag": [ "geometry", "3D geometry", "icosahedron", "USAMTS", "geometry unsolved" ], "Problem": "Faces ABC and XY Z of a regular icosahedron are parallel, with the vertices labeled\r\nsuch that AX, BY , and CZ are concurrent. Let S be the solid with faces ABC, AY Z,\r\nBXZ, CXY , XBC, Y AC, ZAB, and XY Z. If AB = 6, what is the volume of S?\r\n\r\n\r\n[b]( GHAAQ)[/b]", "Solution_1": "This is a current [url=http://www.usamts.org/]USAMTS[/url] problem." } { "Tag": [], "Problem": "A teacher asks his pupil to find the greates angle of a triangle with sides 21, 41 and 50. The pupil \"solves\" the problem as follows : \r\n\r\n[i]Let C be the vertex that corresponds with the greatest angle. \nThen sin( \\frac 13$.now if we observe $1\\leq x\\frac 13$. This means that when you choose $x$ randomly, the expected number of $d_i$ which are in $C$ is $>\\frac n3$, and this is what we need. \r\n\r\nI read an article which said that in order for this to work, the constraints on the sets should enforce a somewhat uniform distribution, as is the case here.\r\n\r\nAnyway, I don't know that much on the subject. [url=http://encyclopedia.thefreedictionary.com/Probabilistic%20method]Here[/url]'s a link to (sort of) a definition.", "Solution_5": "Sorry, perhaps I'm just a little too dumb for this forum, but I don't understand sam-n's solution, for I don't know, what $b_i$ means. As far as I can see, it is not defined anywhere. And is A supposed to be a set of consecutive positive integers or can the $a_i$ be chosen arbitrarily from the set of natural numbers (provided that there are no repetitions, i.e. there are no two numbers $a_i, a_j \\in A, \\, i \\ne j$ with $a_i=a_j$)?", "Solution_6": "My guess is (actually, I'm pretty sure) that $b_i=a_i$, and $a_i$ need not be consecutive, just distinct (well, that's included in the definition of a set).", "Solution_7": "Yes, I didn't read sam-n's post while posting my previous mesage, sorry.\r\nAs fas as I understand, sam_n says for $p=3k+1$ the probability of\r\n $a_i d$ $mod$ $p$ \r\nto lie \r\nin $k+1 \\cdots 2k$ is $\\frac{1}{3}$ for any fixed $d$.\r\nIf we let d rund through any of $1..p-1$ we select the set $B_d$ to be all $a_i$ with \r\n$a_i d \\in \\{k+1 \\cdots 2k\\}$.\r\nNow since $d$ runsthrough $ALL$ residues (the uniformity harazi was speaking)\r\none of $B_d$ will have at least the expected cardinality which is $\\frac{n}{3}$.\r\nHence we choose $B_d$ to be our set.\r\nThanks, Sam_n!" } { "Tag": [ "geometry" ], "Problem": "does anyone have the result for this competition?", "Solution_1": "Ya, Don Mills owned! :gleam:", "Solution_2": "[quote=\"11th dimension\"]Ya, Don Mills owned! :gleam:[/quote]\r\n\r\n\r\ndo you have the result that you can post it here though?", "Solution_3": "Those are machine marked, and only teachers have it. The hand marked results are not out yet. last year's results came out much faster. I wonder what the markers are doing...?\r\n:play_ball:", "Solution_4": "the [url=http://www.ecf.utoronto.ca/apsc/davinci/2005_exam/results.html]result[/url] is out", "Solution_5": "You know what I just noticed? They put Don Mills Collegiate in NORTH YORK, Ontario. I wonder if that's a mistake because I've always consider DMCI as in the area of Don Mills. They also put Lin Fei in grade 11 and I heard a while back when he was arguing about how his age/grade always seems to be posted wrong. LOL, it must suck being in such a position (in my opinion, ofcourse).", "Solution_6": "[quote=\"d.lam.86\"]You know what I just noticed? They put Don Mills Collegiate in NORTH YORK, Ontario. I wonder if that's a mistake because I've always consider DMCI as in the area of Don Mills. They also put Lin Fei in grade 11 and I heard a while back when he was arguing about how his age/grade always seems to be posted wrong. LOL, it must suck being in such a position (in my opinion, ofcourse).[/quote]\r\n\r\nLol, after you get used to it, it feels fine.\r\nAlso, Peng who got first in Da Vinci (congrats by the way :first: ) he is in grade 11 and they put him as gr 12 winner. And for me, who should have won the gr 10, they put me as gr 11 winner (again.....).\r\nIn conclusion, the results are so messed up.", "Solution_7": "[quote=\"11th dimension\"][quote=\"d.lam.86\"]You know what I just noticed? They put Don Mills Collegiate in NORTH YORK, Ontario. I wonder if that's a mistake because I've always consider DMCI as in the area of Don Mills. They also put Lin Fei in grade 11 and I heard a while back when he was arguing about how his age/grade always seems to be posted wrong. LOL, it must suck being in such a position (in my opinion, ofcourse).[/quote]\n\nLol, after you get used to it, it feels fine.\nAlso, Peng who got first in Da Vinci (congrats by the way :first: ) he is in grade 11 and they put him as gr 12 winner. And for me, who should have won the gr 10, they put me as gr 11 winner (again.....).\nIn conclusion, the results are so messed up.[/quote]\r\n\r\nThey should get editors. I'd love to be an editor; I practically editted my school yearbook in record time. I've got keen eyes at spelling mistakes, punctuation errors, etc. Maybe I should put that on my r\u00e9sum\u00e9: \"keen eyes\"! :rotfl:", "Solution_8": "crap... that looks like a really cool contest. I'd have done well, since I do electronics, physics, and chemistry...\r\n\r\nWhat's the answer to the one with the 2 cylinders rolling down the ramps? Is the hollow O-shaped one slower than the filled-in circle?", "Solution_9": "its rigged for people that live in toronto - at the science center, everyone visits that display where they actually have a metal ring and a wooden cylinderblock." } { "Tag": [ "geometry", "3D geometry", "perimeter" ], "Problem": "In the cube pictured, $ M$ and $ N$ are trisection points of $ \\overline{AB}$, $ P$ and $ Q$ are trisection points of $ \\overline{EF}$, and each edge is 3 units in length. What is the number of units, rounded to the nearest tenth, in the perimeter of $ \\bigtriangleup MNQ$?\n[asy]\nimport olympiad; size(200); import geometry; import graph; import three; import graph3; import solids;\n\ndraw((0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)--cycle, linewidth(2pt));\n\ndraw((0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)--cycle, linewidth(2pt));\n\ndraw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)--cycle,linewidth(2pt));\n\ndraw((1,0,1)--(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle,linewidth(2pt));\n\ndraw((1,0,1)--(1,1,0)--(0,1,1)--(1,0,1)--cycle,linewidth(2pt));\n\ntriple P = (1,1/3,0); triple Q = (1,2/3,0); triple M = (2/3,0,1); triple N = (1/3,0,1);\n\ndot(Label(\"$P$\",align=SW),P); dot(Label(\"$Q$\",align=SW),Q); dot(Label(\"$M$\",align=NW),M); dot(Label(\"$N$\",align=NW),N);\n\nlabel(\"$A$\", (0,0,1), N); label(\"$B$\",(1,0,1),N); label(\"$E$\",(1,0,0),W); label(\"$F$\",(1,1,0),E); draw(M--Q--N); label(\"$3$\",((1,1,1)--(1,1,0)),E);\n[/asy]", "Solution_1": "MN = 1\r\nMQ = sqrt(1^2+3^2+2^2)=sqrt(14)\r\nNQ = sqrt(2^2+3^2+2^2)=sqrt(17)\r\n\r\nthen just add them up.\r\n\r\n8.8647630123916019354051585882906...\r\n\r\nso answer is [b]8.9[/b]\r\n\r\nseeing that this is a target round I believe this should have a no-calc answer, but this can be done by just estimating", "Solution_2": "Actually, on target round the use of calculators is permitted.", "Solution_3": "Wait I don't get how to find MQ and NQ", "Solution_4": "For MQ, you find MB and BQ, then use the Pythagorean theorem. For NQ, you find NB and BQ, then use the Pythagorean theorem. If you want me to post a solution, just tell me." } { "Tag": [ "linear algebra", "matrix", "vector", "complex numbers", "linear algebra unsolved" ], "Problem": "Let $ A$ a $ n \\times n$ matrix and $ C$ a $ m \\times n$ matrix.\r\n\r\nWe say vector $ x \\in \\mathbb{R}^n$ is $ x \\geq 0$ iff $ x_i \\geq 0$ $ \\forall i \\in \\{1,2,...,n\\}$.\r\n\r\nDefine $ S(C)\\equal{}\\{x \\in \\mathbb{R}^n | Cx \\geq 0 \\}$.\r\n\r\nWe say that a $ n \\times n$ matrix $ P$ is positive definite over $ S(C)$ iff $ x^TPx > 0$ $ \\forall x \\in S(C)\\minus{}\\{0\\}$.\r\n\r\nProblem:\r\ngive a necessary and sufficient condition on $ A$ so that $ \\exists$ a simmetric and positive definite matrix $ P$ s.t. $ A^TP \\plus{} PA$ is negative definite over $ S(C)$.\r\n\r\nHints:\r\n\r\n1 - An only necessary condition is that $ A$ hasn't eigenvectors in $ S(C)$ that correspond to positive eigenvalues (if the eigenvalue is complex, the positiveness is referred to its real part).\r\n\r\n2 - If $ S(C) \\equiv \\mathbb{R}^n$, then we have a solution of $ A^TP \\plus{} PA$ iff all eigenvalues of $ A$ are negative (again negativeness is referred to real part if the eigenvalue is complex).", "Solution_1": "No ideas :?: \r\n :ninja:", "Solution_2": "come on guys :!: \r\n\r\n :ninja:", "Solution_3": "Can't believe no answers..." } { "Tag": [ "counting", "distinguishability", "function", "combinatorics theorems", "combinatorics" ], "Problem": "Why are there $\\binom{n+k-1}{k-1}$ ways to select $n$ objects from $k$ types?", "Solution_1": "aright wouldn't it be there are n + 2 places to place k - 1 partitions? that would separate it into k types? Maybe your right tho", "Solution_2": "That's not quite the right statement for that formula $ {n \\plus{} k \\minus{} 1 \\choose k \\minus{} 1}$ describes the number of ways to place $ n$ indistinguishable objects [i]into[/i] $ k$ distinguishable categories. The result can be understood as follows: we can separate indistinguishable objects into distinguishable categories by placing dividers between the objects. We need $ k\\minus{}1$ dividers to do this, and once we have put them in we have $ n \\plus{} k \\minus{} 1$ symbols, some of which are dividers and some of which are objects. The number of ways to arrange them all is clearly $ {n \\plus{} k \\minus{} 1 \\choose k \\minus{} 1}$.\r\n\r\nSee also a solution by generating functions [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=184510]here[/url].", "Solution_3": "Ah i understand- thanks t0ra.\r\nSo if there are 8 types of dogs and we want to choose 19 it would be 26C7?" } { "Tag": [ "inequalities", "linear algebra", "matrix", "function", "quadratics", "linear algebra unsolved" ], "Problem": "Let $A,B\\in S_n(R)$ be 2 symmetric positive definite matrix and $\\alpha \\in (0,1)$.Show that:\r\n$det(\\alpha A+(1-\\alpha)B)\\geq(det(A))^{\\alpha}(det(B))^{1-\\alpha}$.", "Solution_1": "Since $A,B$ are positive definite symmetric, there exists matrix $P$ such that $P^*AP=diag(\\lambda_1,...,\\lambda_n)=D$ and $P^*BP=I$. Thus \r\n$det(\\alpha A +(1-\\alpha)B)=det(B)det(P^*(\\alpha A +(1-\\alpha)B)P))=det(B) \\prod_{i=1}^n \\left (\\alpha\\lambda_i+1-\\alpha \\right)$ (*). \r\nNow the convexity comes in the inequality $\\alpha \\lambda+1-\\alpha \\geq \\lambda^{\\alpha}$, because the function $f(t)=\\lambda^t$ is convex and $f(\\alpha x+(1-\\alpha)y) \\leq \\alpha f(x)+(1-\\alpha)f(y)$ and taking $x=1,y=0$ we get the required inequality $\\alpha \\lambda+1-\\alpha \\geq \\lambda^{\\alpha}$ for $0<\\alpha<1$ and $\\lambda>0$. Using this inequality and (*) we obtain that $det(\\alpha A +(1-\\alpha)B) \\geq det(B)det(D)^{\\alpha}=det(B)det(A)^{\\alpha}det(B)^{-\\alpha}=det(A)^{\\alpha}det(B)^{1-\\alpha}$.", "Solution_2": "i think the matrix $P$ in this case isn't orthogonal my method is to consider the positive symetric matrix $C$ such that $A=C^2$ and let $Z=C^{-1}$ thus we have $det(\\alpha A+(1-\\alpha) B)=det(A) det(\\alpha I+(1-\\alpha)ZBZ)$\r\nand we have $ZBZ$ is symetric positive the conclusion follows thanx to the convexity of $t\\to t^{\\alpha}$ as you have proved :).", "Solution_3": "[quote=\"eugene\"]Since $A,B$ are positive definite symmetric, there exists matrix $P$ such that $P^*AP=diag(\\lambda_1,...,\\lambda_n)=D$ and $P^*BP=I$. [/quote] Not in general. Only if $AB=BA$. However, your argument shows that we may assume that $A$ is diagonal.", "Solution_4": "[quote=\"hpe\"][quote=\"eugene\"]Since $A,B$ are positive definite symmetric, there exists matrix $P$ such that $P^*AP=diag(\\lambda_1,...,\\lambda_n)=D$ and $P^*BP=I$. [/quote] Not in general. Only if $AB=BA$. However, your argument shows that we may assume that $A$ is diagonal.[/quote]\r\nNot. What i said that if $A,B$ are hermitian(symmetric) and $A$ is positive definite then there exists a matrix $P$(not necessary unitary) such that $P^*AP=I$ and $P^*BT$-diagonal\r\nProof: It is known that positive definite quadratic form is congruent with identity matrix, so there is $Y$ such that $Y^*AY=I$.Now the matrix $C=Y^*BY$ is hermitian and there exists unitary matrix $U$ such that $U^*CU$ is diagonal.Now we can just take $P=YU$.\r\n\r\nOr maybe there is a bug because of the basic field $\\mathbb{R}$?", "Solution_5": "you're right eugene :blush: actually it seems that i was so sleepy that i didn't noticed that $det(B)det(P^*P)=1$", "Solution_6": "[quote=\"eugene\"]What i said that if $A,B$ are hermitian(symmetric) and $A$ is positive definite then there exists a matrix $P$(not necessary unitary) ...[/quote]You're right, Eugene :oops:" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "An acute-angled triangle $ ABC$ is inscribed in a circle with center $ O$.\r\nLet $ D$ be the intersection of the bisector $ A$ with $ BC$ and suppose that the perpendicular to $ AO$ though $ D$ meets the line $ AC$ in a point $ P$ interior to the segment $ AC$.show that triangle $ ABP$ is isosceles.", "Solution_1": "It is rather easy.\r\nDenote H be the ortho projection of A on BC. $ \\angle OAD\\equal{}\\angle HAD$, so $ \\angle ADP\\equal{} \\angle ADB$, hence result.", "Solution_2": "[quote=\"mr.danh\"]It is rather easy.\nDenote H be the ortho projection of A on BC. $ \\angle OAD \\equal{} \\angle HAD$, so $ \\angle ADP \\equal{} \\angle ADB$, hence result.[/quote]\r\nWhy $ \\angle OAD \\equal{} \\angle HAD$ :blush: \r\nthanks", "Solution_3": "[quote=\"anonymous111\"][quote=\"mr.danh\"]It is rather easy.\nDenote H be the ortho projection of A on BC. $ \\angle OAD \\equal{} \\angle HAD$, so $ \\angle ADP \\equal{} \\angle ADB$, hence result.[/quote]\nWhy $ \\angle OAD \\equal{} \\angle HAD$ :blush: \nthanks[/quote]\r\n$ \\angle BAH \\equal{} 90^0 \\minus{} \\angle ABC \\equal{} \\frac {180^0 \\minus{} 2\\angle ABC }{2} \\equal{} \\angle CAO$\r\n\r\n$ \\Leftrightarrow \\angle HAD \\equal{} \\angle BAD \\minus{} \\angle BAH \\equal{} \\angle DAC \\minus{} \\angle CAO \\equal{} \\angle OAD.$", "Solution_4": "[quote=\"apollo\"]An acute-angled triangle $ ABC$ is inscribed in a circle with center $ O$.\nLet $ D$ be the intersection of the bisector $ A$ with $ BC$ and suppose that the perpendicular to $ AO$ though $ D$ meets the line $ AC$ in a point $ P$ interior to the segment $ AC$.show that triangle $ ABP$ is isosceles.[/quote]\r\n\r\n I think I have seen this problem in an Italian olympiad with a solution in som issue in Crux. Tryin to solve the problem I had found the same solution as mr. Danh .The solution in Crux was a little different .The problem is really easy , but basic.It uses the fact that $ AO, AD$ are isogonal wrt $ \\angle A$, where $ AD$ is an altitude.\r\n\r\n[b] Babis[/b]", "Solution_5": "Dear Mathlinkers,\r\nI present a proof based on well-knowed theorems.\r\nLet 1 the incircle, A', B' the points of contact of 1 wrt BC, CA.\r\n1. It is knowed that DP is tangent to 1.\r\n2. According to Newton's theorem applied to the tangential quadrilateral ABDP, AD, BP and A'B' are concurrent.\r\n3. Let U this point of concurrence.\r\n4. According to \"An unlikely concurrence\" (Honsberger R., Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, New Mathematical Library, 1995 p. 31), AU is perpendicular to BU.\r\n5. In the triangle ABP, the A-bissector being also the A-altitude, ABP is A-isoceles.\r\nSincerely\r\nJean-Louis" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "sole e besiyar sade, fekr konam male IMO taiwan bashe.\r\n\r\nhameye zoj hayer moratab e $(a,b)$ az aadad e sahih va mosbat ra biyabid, ke $a^2b+a+b$ bar $ab^2+b+7$ bakshpazir bashad.", "Solution_1": "$ab^{2}+b+7\\mid a^2b+a+b\\Rightarrow ab^{2}+b+7\\mid a^2b^2+ab+b^2 , ab^{2}+b+7\\mid a^2b^2+ab+7a\\Rightarrow ab^{2}+b+7\\mid (a^2b^2+ab+b^2)-(a^2b^2+ab+7a)\\Rightarrow ab^{2}+b+7\\mid b^2-7a$\n\u062d\u0627\u0644\u0627 \u062d\u0627\u0644\u062a \u0628\u0646\u062f\u06cc \u0645\u06cc \u06a9\u0646\u06cc\u0645:\n$1)b^2-7a> 0\\Rightarrow b^2< ab^2+b+7< b^2-7a0$ such that $|A(x)|\\geq B\\sqrt{x}$ - it is an exercise in Narkiewicz's \"Teoria Liczb" } { "Tag": [ "analytic geometry" ], "Problem": "How do you find the center of intersection of the two lines each of the form ax+by+c=0.\r\n\r\nEg: the center of intersection of the lines $ 3x \\plus{} \\frac{7}{2}y \\plus{} \\frac{5}{2} \\equal{} 0$ and $ \\frac{7}{2}x \\plus{} 2y \\plus{} \\frac{5}{2} \\equal{} 0$ is $ (\\minus{}\\frac{3}{5},\\minus{}\\frac{1}{5})$.\r\n\r\nBut how to obtain that coordinate ,,, ?\r\n \r\nI think I have learnt it before ,,, but I can't seem to recall ,,,\r\n\r\nYour guide will be appreciated ,,, !", "Solution_1": "hello, multiplying the second equation by $ - \\frac {7}{4}$ we get\r\n\\begin{eqnarray*}\r\n3x + \\frac {7}{2}y + \\frac {5}{2} &=& 0\\\\\r\n- \\frac {49}{8}x - \\frac {7}{2}y - \\frac {35}{8}& =& 0\r\n\\end{eqnarray*}\r\nadding both we have $ 3x - \\frac {49}{8}x + \\frac {5}{2} - \\frac {35}{8} = 0$, from here we get $ x = - \\frac {3}{5}$\r\nplugging this in the first equation and solving this for $ y$ we get $ y = - \\frac {1}{5}$.\r\nSonnhard.", "Solution_2": "oh yea ,,, ! now i remember ,,, it's solving by elimination ,,, !\r\ngosh ,,, i sometimes get these small problems confused from somewhere deep in some complicated examples ,,, !\r\n\r\nThank anyway ,,, !\r\n\r\n\r\n[size=75]-Topic closed-[/size]", "Solution_3": "Just a formula I made up:\r\n\r\nIf the lines $ ax\\plus{}by\\plus{}c\\equal{}0$ and $ dx\\plus{}ey\\plus{}f\\equal{}0$ meet at point $ (x,y)$, then:\r\n\r\n$ x\\equal{}\\frac{fb\\minus{}ce}{ae\\minus{}db}$\r\n\r\n$ y\\equal{}\\frac{\\frac{\\minus{}afb\\plus{}ace}{ae\\minus{}db}\\minus{}c}{b}$\r\n\r\nThat is rather long though.", "Solution_4": "[quote=\"PowerOfPi\"]Just a formula I made up:\n\nIf the lines $ ax \\plus{} by \\plus{} c \\equal{} 0$ and $ dx \\plus{} ey \\plus{} f \\equal{} 0$ meet at point $ (x,y)$, then:\n\n$ x \\equal{} \\frac {fb \\minus{} ce}{ae \\minus{} db}$\n\n$ y \\equal{} \\frac {\\frac { \\minus{} afb \\plus{} ace}{ae \\minus{} db} \\minus{} c}{b}$\n\nThat is rather long though.[/quote]\r\n\r\nYou can get $ y\\equal{}\\frac{cd \\minus{} fa}{ae \\minus{} bd}$ by replacing the corresponding coefficients in your first equation with the y- coeffiecient. Another way is to use the matrices, and calculate the determinants, and divide to get x and y. It's in Art Of Problem Solving Volume 1: The basics, and I think Introduction to algebra, and probably a lot more other books(they're all awesome).", "Solution_5": "lolx ,,, no need for that formula ,,, !\r\n\r\nas you know our mind have only the limited cells to remember all maths formulas ,,, ! :P", "Solution_6": "such a formula, though welly constructed, I believe, has no use.\r\n\r\nElimination is just as fast, and I would prefer not to plug in so many terms...\r\n\r\nNo offense, of course :lol:" } { "Tag": [], "Problem": "Joe multiplies a number by 4, adds 1, and then divides by 3, getting a result of 7. Sue divides the same original number by 3, adds 1, and multiplies by 4. What result does she get? Express your answer as a common fraction.", "Solution_1": "Let the number be x.\r\nWe know that 4x+1=21, so x=5.\r\nPlug it in, and (5/3)+1=8/3.\r\n8/3*4=32/3.[/b]" } { "Tag": [ "summer program", "Mathcamp" ], "Problem": "Does the amount of money we said we can pay on the scholarship application have any role in whether or not one gets admitted? Like, if someone said that they can only pay X dollars, and Mathcamp can't provide that much, is it possible that the applicant would be denied on those grounds?\r\n\r\nJust curious.", "Solution_1": "No, admission is totally need-blind.", "Solution_2": "OK, thanks." } { "Tag": [ "algorithm" ], "Problem": "A hacker is attacking a computer network. Each computer on the network is connected to various other computers. The hacker releases a worm on a source computer. When a worm infects a computer, that computer can propagate a copy of the worm to a connected computer at a rate of once per second. Suppose the hacker knows the layout of the network; i.e., he knows which computers are connected to which. How should the worm proceed to infect the whole network as quickly as possible? \r\n\r\nAlso, if possible, describe an algorithm for computing the minimum time till total network infection. (Admittedly a computer science background will help here, but it is not absolutely necessary.)", "Solution_1": "Common sense tells me to attack the computer that is connected to the most other computers (i.e. the server or root of the network), because the server can then infect lots of computers.\r\n\r\nWe really can't do this without more information, I think.", "Solution_2": "suppose the number of computers is $n$\r\n\r\nthe number of people is $p$\r\n\r\nthe gravity is $g$\r\n\r\nthe velocity is $v$\r\n\r\netc\r\n\r\n\r\nif you dont have more information, creat this assuming logical things\r\nin physics we do it a lot\r\n\r\n\"can we approximate the variable velocity to a constant velocity?\"\r\n\r\nif the result does not change a lot the result of the result, so, we can do this\r\n\r\n\r\nand, this problem can be found on the \"Difficult Problems and DIfficult People\" amazing magazine, edition 24" } { "Tag": [], "Problem": "prove that all of integer numbers such that $y^{2}=x^{3}+1$\r\nare (-1,0), (0,1), (2,3)", "Solution_1": "the equation is quitefamous, and the solution uses elliptic curves of course. THis is a special case of Catalan's Conjecture.\r\n\r\n(haha, the identical problem was discussed in SIMUW 2005 Elliptic curves class by Prof. Robert Pollak)" } { "Tag": [ "function", "induction", "limit", "algebra proposed", "algebra" ], "Problem": "Let $\\mathcal{F}$ be the set of all functions $f : (0,\\infty)\\to (0,\\infty)$ such that $f(3x) \\geq f( f(2x) )+x$ for all $x$. Find the largest $A$ such that $f(x) \\geq A x$ for all $f\\in\\mathcal{F}$ and all $x$.", "Solution_1": "Let $\\mathcal{G}$ be the set of all $\\alpha\\in (0,\\infty)$ with $f(x)\\geq \\alpha x$ for all $f\\in\\mathcal{F}$ and all $x\\in (0,\\infty)$.\r\nThen $f(x) \\geq f( f({2x\\over 3}) )+{x\\over 3}>{x\\over 3}$ for all $f\\in\\mathcal{F}$ and all $x\\in (0,\\infty)$, so ${1\\over 3}\\in\\mathcal{G}$ and $\\mathcal{G}$ is not empty.\r\nAlso $(0,\\infty)\\to (0,\\infty),x\\mapsto{x\\over 2}$ is obviously in $\\mathcal{F}$. So ${1\\over 2}$ is an upper bound for $\\mathcal{G}$.\r\nTherefore $A: =\\sup \\mathcal{G}\\leq{1\\over 2}$ exists and again $A\\in\\mathcal{G}$.\r\nThen for all $f\\in\\mathcal{F}$ and all $x\\in (0,\\infty)$ we have \r\n$f(x) \\geq f( f({2x\\over 3}) )+{x\\over 3}\\geq A f({2x\\over 3})+{x\\over 3}\\geq A^{2}{2x\\over 3}+{x\\over 3}={2A^{2}+1\\over 3}x$.\r\nSo ${2A^{2}+1\\over 3}\\in\\mathcal{G}$ and therefore ${2A^{2}+1\\over 3}\\leq A$ and $0\\geq 2A^{2}-3A+1=(A-1)(2A-1)$.\r\nThis and $A\\leq{1\\over 2}$ give $A={1\\over 2}$.", "Solution_2": "My solution :D\r\nBy induction on $n$ we have $f(x)\\geq u_{n}x\\forall n\\in\\mathbb{N}$, where $u_{1}=\\frac{1}{3},u_{n+1}=\\frac{2u_{n}^{2}+1}{3}(n=1,2,...)$. Now, I see $\\lim_{n\\to\\infty}u_{n}=\\frac{1}{2}$.", "Solution_3": "My solution is [url=https://artofproblemsolving.com/community/q1h532939p3077799]here[/url]" } { "Tag": [], "Problem": "Find the sum of all k digit palindromic numbers if k is odd.", "Solution_1": "[hide=\"hint:\"]what is the average of all $k$ digit palindromes?[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$a,b,c\\ge 0$ and a+b+c=3. Prove that $4(a+b+c)^{3}\\ge 27(ab^{2}+bc^{2}+ca^{2}+abc)$", "Solution_1": "[quote=\"Beat\"]$a,b,c\\ge 0$ and a+b+c=3. Prove that $4(a+b+c)^{3}\\ge 27(ab^{2}+bc^{2}+ca^{2}+abc)$[/quote]\r\nWhy $a+b+c=3$ $?$ This superfluous.", "Solution_2": "Let $a=\\min\\{a,b,c\\}.$ Then $4(a+b+c)^{3}\\ge 27(ab^{2}+bc^{2}+ca^{2}+abc)\\Leftrightarrow$\r\n$\\Leftrightarrow9a(a^{2}+b^{2}+c^{2}-ab-ac-bc)+(b+4c-5a)(a+c-2b)^{2}\\geq0.$ :)\r\nSee also the Sung-yoon Kim's idea:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960", "Solution_3": "Prove me that if a+b+c=3 and a,b,c>0 then $ab^{2}+bc^{2}+ca^{2}+abc\\le4$ in a different from arqady`s method", "Solution_4": "$x+y+z+\\sqrt[3]{xyz}\\leq 4.$ :D", "Solution_5": "how to prove this kunny?", "Solution_6": "The problem is: If the sum of x,y,z non-negatives is 1 then prove x^2y+y^2z^+z^2x+xyz<=4/27\r\nWe should know that (see in the solved problems) x^2y+y^2z+z^2x<=4/27 and that was a quite hard question :([/hide]", "Solution_7": "assuming that $a>b>c$\r\n$a^{2}b+b^{2}c+c^{2}a-ab^{2}-bc^{2}-ca^{2}=(b-c)(a-c)(a-b)>0$\r\nThen $2(ab^{2}+bc^{2}+ca^{2}+abc)\\leq (a+b)(a+c)(b+c)\\leq (\\frac{2a+2b+2c}{3})^{3}=8$" } { "Tag": [ "\\/closed" ], "Problem": "[quote=\"Internet Explorer\"] phpBB : Critical Error \n\nCould not connect to the database \n\n[/quote]\r\n\r\nAnybody know what happened?...i got in now...but yea...was the server down for a bit?", "Solution_1": "Same here.\r\n[quote=\"forum\"]phpBB : Critical Error \n\nCould not connect to the database [/quote]\r\n\r\nNow I encounter even stronger error messages than the \"Debug Mode\" errors :)", "Solution_2": "eh...\"strong\" error messages?...lol", "Solution_3": "Well the first time I saw the \"Debug\" error I thought \"there's blue screen of death in AoPS too!? They're everywhere! Oh the horror, THE HORROR!\" :) \r\n\r\nYeah, I know, I'm such a drama king.\r\n\r\nAnyways, so after I saw the critical error I was like \"oh (enter the person/object you worship here :) )! Not again!!!\" :D \r\n\r\nBut, yeah, i got over it :P", "Solution_4": "lol...i just thought there was something wrong with the server for once (it doesnt happen often on math websites)" } { "Tag": [ "number theory" ], "Problem": "A man who lives within half a kilometer from my house has proved the Beal's Conjecture and Fermat's Last Theorem using number theory only!\r\nHis lecture in the Hawaii International Conference on Statistics is available on the web at http://www.hicstatistics.org\r\nNow where should I post this sensational news?", "Solution_1": "[quote=\"bubka\"]A man who lives within half a kilometer from my house has proved the Beal's Conjecture and Fermat's Last Theorem using number theory only!\nHis lecture in the Hawaii International Conference on Statistics is available on the web at http://www.hicstatistics.org\nNow where should I post this sensational news?[/quote]\r\n\r\n[url=http://www.norvig.com/beal.html]Beal's Conjecture[/url]\r\n\r\nWhat you say sounds like not true..." } { "Tag": [], "Problem": "So I thought that I would make it but as of now I still have not gotten a letter from the AAPT. Is anyone in a similiar position to mine? Also, if you already got the letter, hom much of the semifinal exam did you solve?", "Solution_1": "So would anyone like to respond?", "Solution_2": "There are so many topics regarding the USAPhO that perhaps no one noticed yours. I would suggest you choose one topic for your discussion instead of generating new topics for every stage of the USAPhO.", "Solution_3": "yes that wud b the best idea :oops:", "Solution_4": "But not too easy. As you can see, there are so many USAPhO-related topics, they are going to have a tough time deciding which one is the best for their discussion. :rotfl:", "Solution_5": "they shud better discuss it in their Natinal forums :D :rotfl:", "Solution_6": "Kamil might be angry Rituraj! He expected some information about the letter and we (not even remotely related to USAPhO) are raining spam in his topic.\r\n :rotfl:", "Solution_7": "No Kamil, i hav not got the letter (i m in a similar position...)\r\n\r\n[hide]this shud help :P\n[/hide]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let $ n$ and $ k$ be given natural numbers, and let $ A$ be a set such that \\[ |A| \\leq \\frac{n(n+1)}{k+1}.\\] For $ i=1,2,...,n+1$, let $ A_i$ be sets of size $ n$ such that \\[ |A_i \\cap A_j| \\leq k \\;(i \\not=j)\\ ,\\] \\[ A= \\bigcup_{i=1}^{n+1} A_i.\\] Determine the cardinality of $ A$. \r\n\r\n[i]K. Corradi[/i]", "Solution_1": "I 'm a little bit confused about this problem. Is there any difference between the two $ A$?", "Solution_2": "Let's assume $ A\\equal{}\\{1,2,\\dots,m\\}$.\r\nLet $ b_i$ ($ i\\equal{}1,2,...,m$) denote the number of occurrences of $ i$ among all the sets $ A_j$ ($ j\\equal{}1,2,...,n\\plus{}1$)\r\nwe have \r\n$ k\\binom{n\\plus{}1}{2}\\geq \\sum_{i>j}\\left|A_i\\cap A_j\\right|\\equal{}\\sum_{i\\equal{}1}^m \\binom{b_i}{2}$\r\n\r\nSo, $ kn(n\\plus{}1)\\geq \\sum_{i\\equal{}1}^m (b_i\\minus{}\\frac12)^2\\minus{}\\frac{m}{4}\\geq \\frac{1}{m}(\\sum_{i\\equal{}1}^m(b_i\\minus{}\\frac12))^2\\minus{}\\frac{m}4$\r\nAs, $ \\sum_{i\\equal{}1}^m b_i\\equal{}n(n\\plus{}1)$\r\nwe get $ kn(n\\plus{}1)\\geq \\frac{1}{4m}\\left((2n^2\\plus{}2n\\minus{}m)^2\\minus{}m^2\\right)\\equal{}\\frac{n(n\\plus{}1)(n^2\\plus{}n\\minus{}m)}{m}$\r\n$ \\Rightarrow mk\\geq n(n\\plus{}1)\\minus{}m$\r\nor $ m\\geq \\frac{n(n\\plus{}1)}{k\\plus{}1}$\r\nso we conclude that $ |A|\\equal{}\\frac{n(n\\plus{}1)}{k\\plus{}1}$", "Solution_3": "Main part is proof\n\\[ |A| \\geq \\frac{n(n+1)}{k+1}.\\]", "Solution_4": "$ |A|=\\frac{n(n+1)}{k+1}$ This is equivalent to: every element of $A= \\bigcup_{i=1}^{n+1} A_i$ belongs to the same number of subsets $A_1,A_2,...,A_{n+1}$. Can anyone give a construction for this?", "Solution_5": "@above I doubt what you meant to say/ask. The deduction in #3 already conclude that the answer is $\\frac{n(n+1)}{k+1}$ and no construction is needed since the problem already given the existence of $A$." } { "Tag": [ "videos" ], "Problem": "THE BEST THING EVA: [youtube]iRcI-mFr6aQ[/youtube]", "Solution_1": "Haha yeah I remember that this was played on one of the \"Funniest Home Videos\" (or something close to it).", "Solution_2": ":rotfl: :rotfl: :rotfl:", "Solution_3": ":10: :thumbup: :rotfl:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $g$ be a Fibonacci primitive root ($\\mod p$). i.e. $g$ is a primitive root\r\n($\\mod p$) satisfying $g^2 \\equiv g + 1 \\mod p$. Prove that:\r\n\r\n(a) Prove that $g - 1$ is also a primitive root ($\\mod p$).\r\n\r\n(b) If $p = 4k+3$, then $(g-1)^{2k+3} \\equiv g-2 \\mod p$ and deduce\r\nthat $g - 2$ is also a primitive root ($\\mod p$).\r\n\r\nP.S. a) is quite easy and i don't have a solution for b)", "Solution_1": "For (b) we have $X = (g-1)^{2k+3} = (g-1)^{(p+3)/2}$. So $X^2 = (g-1)^{p+3} = (g-1)^4$ and $X = \\pm (g-1)^2$. Now $X$ is an odd power of a primitive root, and so not a square. Since $+1$ is a square and $-1$ isn't, we have $X = -(g-1)^2 = -g^2 + 2g - 1 = g-2$. Now $p-1$ and $(p+3)/2$ can't have any factor in common, since such a factor would divide $2.(p+3)/2 - (p-1) = 4$ and $(p+3)/2 = 4k+3$ is odd. So $X$ is a primitive root, as it's a primitive root raised to a power coprime to $p-1$." } { "Tag": [ "probability" ], "Problem": "Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.", "Solution_1": "[hide]Multiply 10/10(1) * 9/10 * 8/10 * 7/10 * 6/10 * 5/10 * 4/10 * 3/10* 2/10 * 1/10. Simplify fractions to get 567/1562500[/hide]", "Solution_2": "[quote=\"vishalarul\"]Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.[/quote]\r\n[hide]Isn't it just $\\frac{10}{10}\\cdot\\frac{9}{10}\\cdot\\frac{8}{10}\\cdot\\frac{7}{10}\\cdot\\frac{6}{10}\\cdot\\frac{5}{10}\\cdot\\frac{4}{10}\\cdot\\frac{3}{10}\\cdot\\frac{2}{10}\\cdot\\frac{1}{10}=\\frac{567}{1562500}$.[/hide]\r\nI know I did this wrong. I don't think it should be a big denominator.", "Solution_3": "wait... it should be\r\nthe first marble he picks doesn't matter\r\nthe second can only be 9/10 (not the one he already picked)\r\nthe third can only be 8/10 (not the two he already picked)\r\n....\r\nthe tenth can only be 1/10 (not the nine he already picked)\r\n\r\nso it is that big fraction thingy....\r\nunless im being dumb", "Solution_4": "[quote=\"vishalarul\"]Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.[/quote]\r\n[hide]The first marble can be any of the $10$. The second can be any of the other $9$. The third can be any of the other $8$, and so forth.\n\n$\\frac{10!}{10^{10}}=\\frac{9\\cdot7\\cdot3\\cdot3}{5^{6}\\cdot10^{2}}=\\boxed{\\frac{567}{1562500}}$[/hide]\r\n\r\n@ davidlizeng - With replacement.", "Solution_5": "Just wondering... Why do people like to answer the same solved problem again and again and again?", "Solution_6": "Because people don't look at others' answers and/or solutions and post their own.", "Solution_7": "[quote=\"vishalarul\"]Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.[/quote]\r\n[hide]\n$\\frac{10!}{10^{10}}$=$\\frac{567}{1562500}$\n\n[/hide]", "Solution_8": "[quote=\"i_like_pie\"]Because people don't look at others' answers and/or solutions and post their own.[/quote]\r\n\r\nAnd maybe because they want a higher post count.", "Solution_9": "[quote=\"i_like_pie\"]Because people don't look at others' answers and/or solutions and post their own.[/quote]\r\n\r\nno, it just helps to write out answers and people want to solve them on their own\r\n\r\nu don't HAVE to look, that's why there's a hide button...\r\n\r\njorian", "Solution_10": "i think u of all people want a higher post count.", "Solution_11": "[quote=\"jhredsox\"][quote=\"i_like_pie\"]Because people don't look at others' answers and/or solutions and post their own.[/quote]\n\nno, it just helps to write out answers and people want to solve them on their own\n\nu don't HAVE to look, that's why there's a hide button...\n\njorian[/quote]\r\n\r\nThe hide button is there so you can solve the problem yourself, not so you can just type up what they wrote, even if you don't know what they said. The purpose of the thread is discussion of the problem. Let's say that I were to have a conversation with some people. Person X starts with a comment that brings up a topic that we can discuss. Person Y makes a comment about the topic. But if I say exactly what person Y said, the conversation goes no where. If I disagree with person Y, however, then it contributes something to the conversation.\r\n\r\nSee my point?", "Solution_12": "What's with this higher post count theory? Is there a prize or something.\r\n\r\nSeriously, is there any legit reasoning to that claim or is it just a rumor that could possibly be true.\r\n\r\n+1 [jk]", "Solution_13": "[quote=\"vishalarul\"]Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.[/quote]\r\n\r\n[hide]He will pick the first marble and there is a 100% chance of him not repeating. for the second it is 9/10, then 8/10, then 7/10... 1/10\n\nmultiplying those would mean\n\n$\\frac{9!}{10^{9}}$[/hide]" } { "Tag": [ "inequalities", "search" ], "Problem": "Is this too difficult for this forum? I know that it is definitely too hard for HSB...\r\n\r\n$\\textrm{Given }x>\\frac{1}{2}\\textrm{ and }y>\\frac{1}{3}\\textrm{. Find the smallest value of}$\r\n\\[\\frac{4x^{2}}{3y-1}+\\frac{9y^{2}}{2x-1}\\]\r\n[hide=\"My start\"]I tried two things, and they failed (at least for me)\n1. AM-GM\n2. Combine into 1 fraction, try to factor[/hide]", "Solution_1": "I'm new at inequalities, but here are my thoughts:\r\n\r\n[hide]The denominators are positive because of the constraints given. The numerators are trivially positive. So the expression will always be greater than zero. At least we have a lower bound.\n\nPerhaps if you could get $\\frac{4x^{2}}{3y-1}>\\frac{4x^{2}}{3y}$ and $\\frac{9y^{2}}{2x-1}>\\frac{9y^{2}}{2x}$\nso that your have another bound $\\frac{4x^{2}}{3y}+\\frac{9y^{2}}{2x}=\\frac{8x^{3}+27y^{3}}{6xy}$\n$=\\frac{(2x)^{3}+(3y)^{2}}{(2x)(3y)}$\n\nThis has been mathnerd314's random speculations.\n\nGraagh, I'm really bad at inequality problems :wallbash:[/hide]", "Solution_2": "let $a=2x-1$ and $b=3y-1$\r\n\r\nthen we have\r\n\r\n$\\frac{(a+1)^{2}}{b}+\\frac{ (b+1)^{2}}{a}$\r\n\r\neventhough we are shunned for calculus...it is quite easy to solve with calc...i think the min is when a=b=1...\r\n\r\ni think you can contrive some algebraic justification using that info...", "Solution_3": "if you have something like:\r\n\r\n$\\frac{(a^{2}+1)^{2}}{b^{2}}+\\frac{(b^{2}+1)^{2}}{a^{2}}$, you can do direct algebra probably...because then a,b are reals", "Solution_4": "[quote=\"Altheman\"]let $a=2x-1$ and $b=3y-1$\n\nthen we have\n\n$\\frac{(a+1)^{2}}{b}+\\frac{ (b+1)^{2}}{a}$\n\neventhough we are shunned for calculus...it is quite easy to solve with calc...i think the min is when a=b=1...\n\ni think you can contrive some algebraic justification using that info...[/quote]\r\nyes , minimum occurs when a=b=1 :) \r\nnow with the same denote , its equivalent to finding min of A=$\\frac{(a+1)^{2}}{b}+\\frac{ (b+1)^{2}}{a}$ \r\non the other hand , $A \\geq \\frac{(x+2)^{2}}{x}$ where $x=a+b$\r\nnow $(x+2)^{2}\\geq 8x$ then $A \\geq 8$", "Solution_5": "[quote=\"vishalarul\"]Is this too difficult for this forum? I know that it is definitely too hard for HSB...\n\n$\\textrm{Given }x>\\frac{1}{2}\\textrm{ and }y>\\frac{1}{3}\\textrm{. Find the smallest value of}$\n\\[\\frac{4x^{2}}{3y-1}+\\frac{9y^{2}}{2x-1}\\]\n[hide=\"My start\"]I tried two things, and they failed (at least for me)\n1. AM-GM\n2. Combine into 1 fraction, try to factor[/hide][/quote]\r\nWhat an easy problem for so many posts!\r\nTake $a,b$ as before. \\[\\frac{(a+1)^{2}}{b}+\\frac{(b+1)^{2}}{a}=\\frac{(a-1)^{2}+4a}{b}+\\frac{(b-1)^{2}+4b}{a}\\geq 4\\left(\\frac{a}{b}+\\frac{b}{a}\\right)\\geq 8\\]\r\nEquality for the last is at $a=b=1$. Checking, this is the equality all along. Thus $x=1,y=1.5$ or something like that.", "Solution_6": "We have\r\n$\\frac{4x^{2}}{3y-1}+\\frac{9y^{2}}{2x-1}\\ge \\frac{(2x+3y)^{2}}{2x+3y-2}$\r\nBut we al so have\r\n$\\frac{(2x+3y)^{2}}{8(2x+3y)-16}-1= \\frac{(2x+3y-4)^{2}}{8(2x+3y)-16}\\ge 0$\r\nSo that\r\n$\\frac{4x^{2}}{3y-1}+\\frac{9y^{2}}{2x-1}\\ge 8$\r\n[hide=\"inequality occurs\"]Inequality occurs if and only if $x=1$ and $y=\\frac{2}{3}$ not when $x=1$ and $y=1.5$[/hide]", "Solution_7": "[quote=\"tunganh\"]We have\n$\\frac{4x^{2}}{3y-1}+\\frac{9y^{2}}{2x-1}\\ge \\frac{(2x+3y)^{2}}{2x+3y-2}$[/quote]\r\n\r\nHow did you get that?", "Solution_8": "$\\frac{(a+1)^{2}}{b}+\\frac{(b+1)^{2}}{a}-\\frac{(a+b+2)^{2}}{a+b}=\\frac{(a-b)^{2}(a+b+1)^{2}}{ab(a+b)}$\r\n\r\n$\\implies \\frac{(a+1)^{2}}{b}+\\frac{(b+1)^{2}}{a}-\\frac{(a+b+2)^{2}}{a+b}\\geq 0$\r\n\r\n$\\implies \\frac{(a+1)^{2}}{b}+\\frac{(b+1)^{2}}{a}\\geq \\frac{(a+b+2)^{2}}{a+b}$\r\n\r\nFrom the first line, it's clear that equality occurs when $a-b=0$", "Solution_9": "[quote=\"vishalarul\"][quote=\"tunganh\"]We have\n$\\frac{4x^{2}}{3y-1}+\\frac{9y^{2}}{2x-1}\\ge \\frac{(2x+3y)^{2}}{2x+3y-2}$[/quote]\n\nHow did you get that?[/quote]\nCauchy Engel Form. Search mathlinks (or derive it yourself - its equiv. to cauchy (hint) )\n[quote=\"tunganh\"]\n[hide=\"inequality occurs\"]Inequality occurs if and only if $x=1$ and $y=\\frac{2}{3}$ not when $x=y=1$ or $x=1$ and $y=1.5$[/hide][/quote]\r\nSorry I meant 2/3 but since I was doing it all in my head late at night I wrote 3/2. Both solutions are just as correct :D" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "1) Find all integers (m,n) satisfying : \r\n $ (m^2+n)(n^2+m) = (m+n)^3 $\r\n \r\n\r\n 2) Find all integers (x,y) satisfying:\r\n $ (x+y^2)(y+x^2)= (x-y)^3 $", "Solution_1": "For the first problem.\r\n[tex] (m^2+n)(n^2+m)=(m+n)^3 [/tex]\r\n[tex] \\Leftrightarrow mn(mn+1-3m-3n)=0 [/tex]\r\nThis is easy, [tex] 3|m|+3|n| \\geq |mn|-1 [/tex].\r\nFor the second one:\r\nThe equation is equal to:\r\n[tex] 2y^2+(x^2-3x)y+3x^2+x=0 [/tex] or [tex] y=0 [/tex]\r\nAnd we must find $ x $ satisfying [tex] x^4-6x^3-15x^2-8x=z^2 [/tex]( if you don't want you can do by gen method). This is easy, right? :)" } { "Tag": [ "conics", "parabola", "analytic geometry", "real analysis", "real analysis unsolved" ], "Problem": "Write the equation of the parabola given the focus at (-3,-2) and the directrix is the line y=-6. Draw a sketch of what is given to help in choosing the correct formula. Be sure you have the correct formula to fill in. leave the equation in standard form.\r\n\r\n\r\nPlease list all steps and coordinates, so I can go back and teach myself.\r\n\r\n\r\n\r\nThank you.", "Solution_1": "Let $F(-3,-2), P(x,\\ y)$, by the defnition of a parabola, $\\overline{PF}$=the distance between the point $P$ and the directrix $y=-6\\Longleftrightarrow |y+6|=\\sqrt{(x+3)^{2}+(y+2)^{2}}\\Longleftrightarrow (y+6)^{2}-(y+2)^{2}=(x+3)^{2}$, yielding the desired equation of the parabola, $(x+3)^{2}=4\\cdot 2(y+4)$." } { "Tag": [ "AMC", "AIME", "geometry", "USA(J)MO", "USAMO", "AMC 10" ], "Problem": "I'm in middle school so I contacted the nearby high school that offers the AMC and the lady there who organizes it said that every school has a seperate number and that they only administer the AMC to their students. I was almost sure that the people here said earlier at some point that the AMC should be taken at a high school for middle schoolers. Can someone clarify?", "Solution_1": "Last year, our high school had a girl from the middle school take the AMC 12 at our school, and then end up taking the AIME at our school...", "Solution_2": "I took the AMC 10 last year as an eighth grader and made AIME so I'm sure middle school people can take it at the nearby high school.", "Solution_3": "Is there any link on the AMC website that says something about how if you are in middle school you need to find a high school?\r\nAlso, when are we supposed to get our AMC 10 invitation for being a good scorer on the 8.", "Solution_4": "Just ask around, I'm sure you'll find something", "Solution_5": "well, my middle school has a AMC 10/12 club organized by volunteer parents that let you take AMC 10A and your choice of AMC 10B or AMC 12B. \r\n :lol:", "Solution_6": "i think i'm taking 10a maybe", "Solution_7": "[quote=\"Go Around the Tree\"]i think i'm taking 10a maybe[/quote]\r\nThat, my friend, would be spam.\r\n\r\nAnyway, did you try any local universities? There are probably plenty of high schools in your area taking the AMCs, there's bound to be at least one letting you in. Also, you can talk to your middle school coach to see if he/she will accomidate it, if it's not too late.\r\n\r\nAlso, I never got an AMC 10 invitation, and I got a 25 on the 8.", "Solution_8": "[quote=\"13375P34K43V312\"][quote=\"Go Around the Tree\"]i think i'm taking 10a maybe[/quote]\nThat, my friend, would be spam.\n\nAnyway, did you try any local universities? There are probably plenty of high schools in your area taking the AMCs, there's bound to be at least one letting you in. Also, you can talk to your middle school coach to see if he/she will accomidate it, if it's not too late.\n\nAlso, I never got an AMC 10 invitation, and I got a 25 on the 8.[/quote]\r\n\r\n\r\nPlease refrain from speaking,13375P34K43V312. You are being very annoying and disruptive. Try being a polite girl. \r\n\r\nAnyways, bpms I'm pretty sure most schools offer it. You should just ask someone to join in. I did that since my high school doesn't offer it ( Mountview High School).", "Solution_9": "I was able to take the AMC at my school when I was in eighth grade, so I don't think there's any rule against that.", "Solution_10": "There's certainly no rule against middle schoolers taking it at a high school, as many others have said. I'm another who took it at the local high school when I was in 8th grade. It is possible, though, that your local high school is going to be a pain in the you-know-where and not let you take it; there's no rule that says they *have* to let you. In that case, look for another high school, as there's also no rule that says you have to take it at the local school. If you take it at another school, though, make sure you talk to them in advance about how to get your results. And also keep in mind that you'll probably have to take AIME and maybe even USAMO at that school if you make it.", "Solution_11": "Sly Si is correct on all accounts.\r\n\r\nNo high school is obligated to take you. Local rules differ by high school and local circumstances. It also depends on the willingness of the Contest Adminstrator to do the extra work required (see below about the AIME, etc.)\r\n\r\nYou should ask around and find a willing high school, or local college or University (see the AMC website for a list of colleges and universities adminstering the AMC B contests) or a local home school group.\r\n\r\nBe prepared in advance to get your score from the administering site and contest manager. Also be prepared in advance to take the AIME and potentially the USAMO again at the administering site. The AMC has no way of individually contacting and reporting back to your everyday school if your everyday school does not adminster the contests.\r\n\r\n\r\nSteve Dunbar\r\nAMC Director", "Solution_12": "[quote=\"purple plume\"][quote=\"13375P34K43V312\"][quote=\"Go Around the Tree\"]i think i'm taking 10a maybe[/quote]\nThat, my friend, would be spam.\n\nAnyway, did you try any local universities? There are probably plenty of high schools in your area taking the AMCs, there's bound to be at least one letting you in. Also, you can talk to your middle school coach to see if he/she will accomidate it, if it's not too late.\n\nAlso, I never got an AMC 10 invitation, and I got a 25 on the 8.[/quote]\n\n\nPlease refrain from speaking,13375P34K43V312. You are being very annoying and disruptive. Try being a polite [b]girl[/b]. \n\nAnyways, bpms I'm pretty sure most schools offer it. You should just ask someone to join in. I did that since my high school doesn't offer it ( Mountview High School).[/quote]\r\n\r\n :rotfl: :rotfl: :rotfl:", "Solution_13": "If you are within reasonable driving distance of the Twin Cities metropolitan area of Minnesota, I am always very happy to accommodate you in my homeschoolers testing group. We have a full set of test kits for this year, and not all have been claimed yet.", "Solution_14": "[quote=\"tokenadult\"]If you are within reasonable driving distance of the Twin Cities metropolitan area of Minnesota, I am always very happy to accommodate you in my homeschoolers testing group. We have a full set of test kits for this year, and not all have been claimed yet.[/quote]\r\nI'm pretty sure he lives in ...", "Solution_15": "[quote=\"lotrgreengrapes7926\"][quote=\"tokenadult\"]If you are within reasonable driving distance of the Twin Cities metropolitan area of Minnesota, I am always very happy to accommodate you in my homeschoolers testing group. We have a full set of test kits for this year, and not all have been claimed yet.[/quote]\nI'm pretty sure he lives in[/quote]\r\nYou are correct. Thanks for the offer tokenadult. Can you edit your post, I do not want everyone to know where I live." } { "Tag": [], "Problem": "[url=http://www.savetoby.com/]http://www.savetoby.com/[/url]\r\n\r\n[i]Moderator Note: This is NOT appropriate for viewers \"of all ages\". If you are under 13 years of age or if you are a parent with a child under 13 years old beside you, I [b]strongly[/b] discourage you from visiting the site.[/i]", "Solution_1": "That guy has a great way to make money, but if he doesn't get it, will he really eat the bunny?", "Solution_2": "wow, I cannot beleive this is true.", "Solution_3": "Lol, awesome.\r\n\r\nI wonder if he really has gotten that much money or if he just says so to get others to be more inclined to help Toby. :D", "Solution_4": "I wonder if he only raises 49,999$ will he still eat him? :)", "Solution_5": "OK.... That is a little twisted. Smart! But twisted...", "Solution_6": "probing furthur onto the site, you find out that paypal has stopped accepting donations because of numerous PETA complaints", "Solution_7": "thats pretty crazy...", "Solution_8": "weird....", "Solution_9": "it almost makes you want to buy a shirt just for the utter... \"WHAT?\"-ness of it. :D even though i love my bunny to death and think that's awful. (his name is finley. we found him in our driveway not long after easter about two years ago. CURSE those **** people who think pets are disposable).", "Solution_10": "who shall donate?", "Solution_11": "Well, unfortunately, I just found out that it isn't true. \r\n\r\n[url=http://www.snopes.com/critters/crusader/savetoby.asp]Oh Snopes, always ruining my fun[/url]", "Solution_12": "yeah, but you could still possibly buy shirts....althought that could be fake too\r\n\r\nI dont want to try.", "Solution_13": "who would buy a \"save the rabbit\" shirt if the rabbit didn't need to be saved?", "Solution_14": "Maybe I should try this...........\r\n[img]http://www.goldfiles.com/images/screenshots/s1_frogblender.png[/img][img]http://www.goldfiles.com/images/screenshots/s1_hamstermicro.png[/img]\r\nI bet I could earn money twice as fast with twice the number of animal lives at stake. (note to self-get better pictures)", "Solution_15": "Yay JoeCartoon! That site is not nearly as well-known as it should be!", "Solution_16": "I definitely could earn some money off this!\r\n[img]http://blog.trg.ru/img/shaved_cat.jpg[/img]" } { "Tag": [ "factorial" ], "Problem": "In the figure to the right, the largest circle has a radius of six meters. Five congruent smaller circles are placed as shown and are lined up in east-to-west and north-to-south orientations. What is the radius in meters of one of the five smaller circles?", "Solution_1": "2.\r\n\r\nThe message is too small. Please make the message longer before submitting.", "Solution_2": "There's supposed to be a picture. :wink:", "Solution_3": "u dont need it though :P", "Solution_4": "It would be helpful to explain HOW it's 2. :| \r\n\r\n\r\n :D Please explain how.", "Solution_5": "Fine izzy... GOD!\r\n\r\nThe large circle has diameter 12. Since there are 3 circles lined up, divide by 3. Each circle has diameter 4, thus, they have radius 2.\r\n\r\nThere... you wanted it.", "Solution_6": "Umm...please explain how people are supposed to know how to solve it when all they have is an answer? They can see the answer themselves by reviewing the questions...what they want is an explanation. :)", "Solution_7": "If they want an explaination, they should ask for it. I believe mz94 already brought this up to you, izzy. :D", "Solution_8": "Ok.\r\n\r\n\r\nPlease explain why a person would go to the Review Questions, go to a problem, and go to the TROUBLE of Asking On the Forum, when about 2 inches above the \"Ask on Forum\" button is the answer. People want to know HOW it's whatever it is. *sigh*", "Solution_9": "yea they wanna know how the problem is done \r\n\r\nthats right izzy\r\n\r\nthey just dont wanna know certain random facts like factorials unless they specifically ask it in the problem or ask it on teh problem in teh forum\r\n\r\n\r\nthe picture is basically 2 concentric circles, the larger w/ radius 6, the smaller having 4 circles tangent to it that are congruent to it and such that a certain diameter of teh circle will pass thru 3 of the circles' centers", "Solution_10": "Wh00ps, my bad. :blush: :blush: :blush:" } { "Tag": [ "number theory", "prime numbers" ], "Problem": "i don't remember the problem too well, but i'll try.\r\n\r\nthere are three distinct prime numbers, e, f, and g. If $ e\\equal{}f^2 \\plus{} g^2$, what is the greatest possible value $ e$ such that e is less than 100?\r\n\r\n[hide=\" Answer\"]\n53[/hide]\r\n\r\nAre there any quicker ways of doing this besides guess and check? \r\n\r\nThanks,\r\nLHM", "Solution_1": "I would go through the possibilities of $ f$ and $ g$ and use those. It's not too bad given 5-10 seconds :)", "Solution_2": "[hide]All primes except 2 are odd, so you know either $ f^2$ or $ g^2$ must be even. The only even prime is 2, so one of them has to be $ 2^2 \\equal{} 4$.\n\nSo now you've got $ 100 \\ge 4 \\plus{} g^2$\n\nso $ 96 \\ge g^2$\n\nThat means $ g$ can be: 2, 3, 5, or 7. To maximize $ e$ you need to choose the greatest of these, which is 7.\n\n$ e \\equal{} 2^2 \\plus{} 7^2 \\equal{} 4 \\plus{} 49 \\equal{} \\boxed{53}$[/hide]\r\n\r\nyeah I guess to think _all_ of this out during a cd would take too much time.. I think it gets faster once you grasp it well." } { "Tag": [ "inequalities", "ratio", "invariant", "function", "Cauchy Inequality", "inequalities proposed" ], "Problem": "Let $a,b,c \\ge 0$ then prove \r\n\r\n\r\n $\\sum_{cyc}^{a,b,c} \\frac{a^2+b}{b+c} \\ge\\frac{\\displaystyle \\sum_{cyc}^{a,b,c} a +3}{2}$\r\n\r\n[hide=\"My solution\"]\nSubstitued $\\begin{cases} a+b=x\\\\[.1in] b+c=y\\\\[.1in]c+a=z \\end{cases} \\longrightarrow \\begin{cases}a=\\frac{x+z-y}{2} \\\\[.1in]b= \\frac{y+x-z}{2} \\\\[.1in]c= \\frac{z+y-x}{2} \\end{cases}$\n\n$\\longrightarrow$\n\n\n\n$\\sum_{cyc}^{x,y,z}\\frac{(\\displaystyle\\frac{x+z-y}{2})^2+\\frac{y+x-z}{2}}{x}\\ge \\frac{\\displaystyle\\frac{x+y+z}{2}+3}{2}=\\frac{x+y+z+6}{4}$ \n\n$\\longrightarrow\n\\sum_{cyc}^{x,y,z}\\displaystyle\\frac{(x+z-y)^2}{x} +\\sum_{cyc}^{x,y,z} 2(1+\\frac{y}{x}-\\frac{z}{x}) \\ge x+y+z+6$\n\n\n$\\longrightarrow\n \\sum_{cyc}\\displaystyle\\frac{(x+z-y)^2}{x}+\\sum_{cyc}^{x,y,z}\\frac{y}{x} \\ge (x+y+z)+\\sum_{cyc}^{x,y,z}\\frac{z}{x}$\n\nWhich we devide to :\n$1.\\sum_{cyc}^{x,y,z}\\displaystyle\\frac{(x+z-y)^2}{x} \\ge x+y+z$[hide=\"Solution\"]\nFrom cauchy inequality \n$(\\sum_{cyc}^{x,y,z} \\displaystyle \\frac{(x+z-y)^2}{x})( \\sum_{cyc}^{x,y,z} x) \\ge (\\sum_{cyc}^{x,y,z} (x+z-y)=\\sum_{cyc}^{x,y,z} x)^2$ \n\n$\\longrightarrow \\sum_{cyc}^{x,y,z}\\displaystyle\\frac{(x+z-y)^2}{x} \\ge x+y+z$[/hide]\n$2.\\sum_{cyc}^{x,y,z}\\frac{y}{x} \\ge \\sum_{cyc}^{x,y,z}\\frac{z}{x}$[hide=\"rearrangement\"]\nAssume $x\\ge y \\ge z \\longrightarrow \\frac1z \\ge \\frac1y \\ge \\frac1x $ \nBy rearrangement :\n$\\sum_{cyc}^{x,y,z}\\frac{y}{x} \\ge \\sum_{cyc}^{x,y,z}\\frac{z}{x}$[/hide]\n[/hide]", "Solution_1": "[quote=\"lomos_lupin\"]Let $a,b,c \\ge 0$ then prove \n\n\n $\\sum_{cyc}^{a,b,c} \\frac{a^2+b}{b+c} \\ge\\frac{\\displaystyle \\sum_{cyc}^{a,b,c} a +3}{2}$[/quote]\r\n\r\nThe inequality is wrong (take a = 0.0001, b = 0.0002, c = 0.0003). So is the solution (you can't assume $x\\geq y\\geq z$ in the proof of the inequality 2 since it is not symmetric). But some additional conditions can make the inequality correct. Here is a proof under the condition $a+b+c\\geq\\frac15$:\r\n\r\n[color=blue][b]Theorem 1.[/b] Let a, b, c be three nonnegative reals such that $a+b+c\\geq\\frac15$. Then,\n\n$\\frac{a^2+b}{b+c}+\\frac{b^2+c}{c+a}+\\frac{c^2+a}{a+b}\\geq\\frac{a+b+c+3}{2}$.[/color]\r\n\r\n[i]Proof of Theorem 1.[/i] First we show a helpful inequality:\r\n\r\n[color=blue][b]Theorem 2.[/b] For any three nonnegative reals a, b, c, we have\n\n$(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)\\geq 3(c^2b+a^2c+b^2a)$.[/color]\r\n\r\n[i]Proof of Theorem 2.[/i] Here is the ugliest possible proof of a 3-variable inequality.\r\n\r\nSince the inequality in question is cyclic, we can WLOG assume that a is the smallest among the numbers a, b, c. Then, $a\\leq b$ and $a\\leq c$, so that $b-a\\geq 0$ and $c-a\\geq 0$. Denote b - a = X and c - a = Y; then, $X\\geq 0$ and $Y\\geq 0$. Now, some boring computation shows that\r\n\r\n$(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)-3(c^2b+a^2c+b^2a)$\r\n$=a((b-c)^2+(c-a)^2+(a-b)^2)+(X^3+2X^2Y-3XY^2+Y^3)$.\r\n\r\nNow, we have to show that $(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)\\geq 3(c^2b+a^2c+b^2a)$, i. e. that $(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)-3(c^2b+a^2c+b^2a)\\geq 0$. Since $a((b-c)^2+(c-a)^2+(a-b)^2)$ is trivially $\\geq 0$, this will be proven once we show that $X^3+2X^2Y-3XY^2+Y^3\\geq 0$. But this is true, since AM-GM yields $X^3+X^2Y+X^2Y+\\frac{Y^3}{4}+\\frac{Y^3}{4}+\\frac{Y^3}{4}+\\frac{Y^3}{4}\\geq 7\\sqrt[7]{X^3\\cdot X^2Y\\cdot X^2Y\\cdot\\frac{Y^3}{4}\\cdot\\frac{Y^3}{4}\\cdot\\frac{Y^3}{4}\\cdot\\frac{Y^3}{4}}$, what simplifies to $X^3+2X^2Y+Y^3\\geq 7\\sqrt[7]{\\frac{1}{4^4}}XY^2$. But $7\\sqrt[7]{\\frac{1}{4^4}}>3$, and thus $X^3+2X^2Y+Y^3\\geq 3XY^2$, so $X^3+2X^2Y-3XY^2+Y^3\\geq 0$, and Theorem 2 is proven.\r\n\r\nNow, we equivalently transform the inequality in question:\r\n\r\n$\\frac{a^2+b}{b+c}+\\frac{b^2+c}{c+a}+\\frac{c^2+a}{a+b}\\geq\\frac{a+b+c+3}{2}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ (\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b})+(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b})\\geq\\frac{a+b+c}{2}+\\frac32$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ (\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b})-\\frac{a+b+c}{2}\\geq\\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b})$.\r\n\r\nFirst note that, if the right hand side of this inequality, $\\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b})$, is < 0, then the inequality is trivial, since the left hand side, $\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b})-\\frac{a+b+c}{2}$, is always $\\geq 0$ (in fact,\r\n\r\n$\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b})-\\frac{a+b+c}{2}=\\frac{(a+b+c)((b+c)(b-c)^2+(c+a)(c-a)^2+(a+b)(a-b)^2)}{2(b+c)(c+a)(a+b))}$\r\n$\\geq 0$).\r\n\r\nHence, in the following, we will assume that the right hand side, $\\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b})$, is $\\geq 0$. Now,\r\n\r\n$(\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b})-\\frac{a+b+c}{2}-5(a+b+c)(\\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b}))$\r\n$=\\frac{(a+b+c)((a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)-3(c^2b+a^2c+b^2a))}{(b+c)(c+a)(a+b)}\\geq 0$,\r\n\r\nsince $(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)\\geq 3(c^2b+a^2c+b^2a)$ by Theorem 2. Thus,\r\n\r\n$(\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b})-\\frac{a+b+c}{2}\\geq 5(a+b+c)(\\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b}))$.\r\n\r\nOn the other hand, $a+b+c\\geq\\frac15$ implies $5(a+b+c)\\geq 1$, what, together with $\\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b})\\geq 0$, yields\r\n\r\n$5(a+b+c)(\\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b}))\\geq\\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b})$.\r\n\r\nThus, $(\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b})-\\frac{a+b+c}{2}\\geq \\frac32-(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b})$, so that Theorem 1 is proven.\r\n\r\n Darij", "Solution_2": "Perfect ,I couldnt expect a better answer(you are good) ,I am going to show darij`s solution to my teacher and say [b]At least check your problems before giving them to us[/b] :mad: \r\n\r\nI managed to reach untill here : \r\n\r\n\r\n$\\sum_{cyc}^{,x,y,z} \\frac{a^2+b}{b+c} - \\sum_{cyc}^{x,y,z} \\ge 3 - \\sum_{cyc}^{x,y,z} \\frac{b}{b+c}$\r\nAnd the case that [b]LHS[/b] be negative ,But i have a question :\r\nHow can you do these computations ,I couldnt even think about doing these computations ,cause i will get lost in them . :( :(,look at your solution man ,You perfectly just played with these numbers,Welldone.", "Solution_3": "Well, let me explain some things about the proof.\r\n\r\nAt first, look at your inequality:\r\n\r\n$\\frac{a^2+b}{b+c}+\\frac{b^2+c}{c+a}+\\frac{c^2+a}{a+b}\\geq\\frac{a+b+c+3}{2}$.\r\n\r\nIt contains some terms of degree 1 and some terms of degree 0. So it is natural to bring all the degree 1 terms on the one side and the degree 0 terms on the other side:\r\n\r\n$\\left(\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b}\\right)-\\frac{a+b+c}{2}\\geq\\frac32-\\left(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b}\\right)$.\r\n\r\nNow, if you decrease the numbers a, b, c, keeping their ratios b / c, c / a, a / b fixed, then the degree 0 terms remain invariant, but the degree 1 terms decrease and get arbitrarily close to 0. Thus, if such an inequality is generally true, then we must have\r\n\r\n$0\\geq\\frac32-\\left(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b}\\right)$.\r\n\r\nBut this is nonsense, since a simple calculation shows that\r\n\r\n$\\frac32-\\left(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b}\\right)=\\frac{\\left(b-c\\right)\\left(c-a\\right)\\left(a-b\\right)}{2\\left(b+c\\right)\\left(c+a\\right)\\left(a+b\\right)}$,\r\n\r\nand if this is $\\leq 0$ for some number triple (a, b, c), then it is $\\geq 0$ for the permutation (a, c, b) of this number triple.\r\n\r\nThis shows that your inequality, in the form given, cannot be true. So it needs an additional condition. I have looked at conditions of the form $a+b+c\\geq k$ for a constant k, and found that $k=\\frac15$ is more or less the optimal constant (there exists a slightly smaller constant, but if I am not mistaken, it is a root of an ugly irreducible polynomial). Now, of course, having the condition $a+b+c\\geq\\frac15$, we can homogenize the inequality into the form\r\n\r\n$\\left(\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b}\\right)-\\frac{a+b+c}{2}\\geq 5\\left(a+b+c\\right)\\left(\\frac32-\\left(\\frac{b}{b+c}+\\frac{c}{c+a}+\\frac{a}{a+b}\\right)\\right)$,\r\n\r\nin which it contains only terms of degree 1 and thus is homogeneous. Now, after a simple computation (I have done it using computer algebra, but it is well manageable at hand: only b + c, c + a and a + b ocur as denominators, and lots of things are divisible by a + b + c, so the terms will not get very long) we arrive at Theorem 2.\r\n\r\nRemains to prove Theorem 2. Well, define the function\r\n\r\n$f\\left(a;\\;b;\\;c\\right)=\\left(a^3+b^3+c^3\\right)+2\\left(b^2c+c^2a+a^2b\\right)-3\\left(c^2b+a^2c+b^2a\\right)$.\r\n\r\nThen, we have to show that $f\\left(a;\\;b;\\;c\\right)\\geq 0$ for any three nonnegative a, b, c. Now, if we decrease the numbers a, b, c by a given positive number d, then the value of the function f decreases:\r\n\r\n$f\\left(a;\\;b;\\;c\\right)-f\\left(a-d;\\;b-d;\\;c-d\\right)=d\\left(\\left(b-c\\right)^2+\\left(c-a\\right)^2+\\left(a-b\\right)^2\\right)\\geq 0$.\r\n\r\nHence, if we succeed to prove that $f\\left(a-d;\\;b-d;\\;c-d\\right)\\geq 0$ for some d, then we automatically get $f\\left(a;\\;b;\\;c\\right)\\geq 0$. But in fact, if we WLOG assume that a is the smallest of the numbers a, b, c, and take d = a, setting b - a = X and c - a = Y, this comes down to showing that $f\\left(0;\\;X;\\;Y\\right)\\geq 0$ for nonnegative reals X and Y; since $f\\left(0;\\;X;\\;Y\\right)=X^3+2X^2Y-3XY^2+Y^3$, it remains to show that $X^3+2X^2Y-3XY^2+Y^3\\geq 0$. Now this is straightforward.\r\n\r\n Darij" } { "Tag": [ "combinatorics solved", "combinatorics", "Sequences" ], "Problem": "To every natural number k \\geq 2 corresponds a sequence a_n(k) according to the following rules:\r\n\r\na_0 = k, a_1 = \\gamma (a_0), a_2 = \\gamma (a_1), ..., a_n = \\gamma (a_(n-1)) in which \\gamma (a) is the number of different divisors of a. \r\n\r\nFind all k for which the sequence a_n(k) does not contain the square of integer numbers !", "Solution_1": "For k= 2: an=2 is not perfect square.\r\n\r\nIt is clear if n>2 then \\gamma (n) < n ==> for some m am=2. Let m be the minimal such number ==> am-1=p>2 where p is prime number ==> p is odd number ==> if m>2 then am-2 is a perfect square.\r\n\r\nThus k must be prime number." } { "Tag": [ "ARML" ], "Problem": "I also posted this on the mathcounts.org forum, but then I decided I'd like to be able to see replies in less than say, a week (any of you who frequent that forum undoubtedly know about the screening).\r\n\r\nIs there anyone who wouldn't mind kind of giving me a general run-down of ARML? The website is very unhelpful. I believe Washington State does send an ARML team sometimes, but I have not yet had a chance to get involved. I'm just curious because just about everyone at MOP had gone to ARML, and it seems like something I should get involved in as well.\r\n\r\nIn particular, what are the different rounds and how do they work? I know what the round names are (from the problems of arml.com), but I don't know much about them. Also, how does the individual competition work? Obviously since the indiv. test is out of 8, there are lots of ties, and I know that there is some sort of \"ARML Countdown\" do determine the champion, but do they rank everyone top to bottom somehow as well?", "Solution_1": "[color=cyan]Okay, this is fun. A team is 15 people. The competition is four rounds:\nTeam, Power, Individual, Relay.\nTeam: 20 minutes. Calculators are okay. The entire team may work together. 10 short-answer questions ranging from very easy to quite difficult. 4pts / question\nPower: the highlight of ARML. 1 hour. The entire team may work together. Calculators are okay, but rarely useful. A series of proofs which build upon each other. Very much fun. Earlier problems are often just computation. Later problems can be very difficult proofs. 40 points total.\nIndividual: 8 questions given in pairs, with 10 minutes per pair. No calculators. No team work. Each question is worth 1 point, making the entire deal worth 120, or half of the overall total for the competition.\nRelay: The relay portion of the contest is really bizarre. Every team is broken up into 5 groups of 3 people. Within each group, each person gets a separate question. The answer to the first question is necesary for the second question, the asnwer for the second question is necessary for the third, and only the third answer counts for points. For each group that gets it right within 3 minutes, 4 points are awarded. There is also an opportunity to hand in an answer after 6 minutes, for which 2 points are awarded. The whole procedure is carried out twice. Thus the entire round is worth at most 5*4*2 = 40 points. I enjoy the relays. NYC A is not any good at them.\n\nAt the very end of the competition, the number of 8's and 7's determines how many people will compete in a tiebreaker round, which is carried out simultaneously at the three ARML sites. The fastest correct answer among the 8s earns first place; if there are fewer than 5 or 10 8's, the sevens also take part in the tiebreaker, and the highest scorer among the 7's earns a place after the lowest-ranking 8. There is no ranking system for individual scorers not competing in the tiebreaker.\n\nOkay, I think that's it. I have no idea if Washington sends a team, but there is a list of teams which competed for the past few years on the AMRL website.[/color]", "Solution_2": "What individual won this years ARML?", "Solution_3": "Anders Kaseorg." } { "Tag": [], "Problem": "Given that $ a \\otimes b \\equal{} (a^2 \\plus{}b) \\div 2$. What is the value of $ 5 \\otimes 3$?", "Solution_1": "plugging in $ 5$ and $ 3$ for $ a$ and $ b$, we get\r\n\r\n$ (5^2 \\plus{} 3)/2 \\equal{} 28/2 \\equal{} \\boxed{14}$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x,y,z>0$ prove that\r\n\\[ \\frac{x}{4x^2\\plus{}y^2\\plus{}z^2}\\plus{}\\frac{y}{x^2\\plus{}4y^2\\plus{}z^2}\\plus{}\\frac{z}{x^2\\plus{}y^2\\plus{}4z^2}\\leq\\frac{3(x\\plus{}y\\plus{}z)}{10(xy\\plus{}yz\\plus{}zx)}\\]", "Solution_1": "[quote=\"encyclopedia\"]Let $ x,y,z > 0$ prove that\n\\[ \\frac {x}{4x^2 \\plus{} y^2 \\plus{} z^2} \\plus{} \\frac {y}{x^2 \\plus{} 4y^2 \\plus{} z^2} \\plus{} \\frac {z}{x^2 \\plus{} y^2 \\plus{} 4z^2}\\leq\\frac {3(x \\plus{} y \\plus{} z)}{10(xy \\plus{} yz \\plus{} zx)}\n\\]\n[/quote]\r\nSorry but it is not true, try $ a \\equal{} b \\equal{} 0,4$ and $ c \\equal{} 1$. \r\n\r\nLet $ x,y,z \\ge 0$ then\r\n\r\n$ \\frac {x}{{4x^2 \\plus{} y^2 \\plus{} z^2 }} \\plus{} \\frac {y}{{x^2 \\plus{} 4y^2 \\plus{} z^2 }} \\plus{} \\frac {z}{{x^2 \\plus{} y^2 \\plus{} 4z^2 }} \\le \\frac {{x \\plus{} y \\plus{} z}}{{2\\left( {xy \\plus{} yz \\plus{} zx} \\right)}}$\r\n\r\nBy AM-GM we have \r\n\r\n$ \\frac {x}{{4x^2 \\plus{} y^2 \\plus{} z^2 }} \\le \\frac {x}{{2x^2 \\plus{} 2xy \\plus{} 2zx}}$\r\n\r\n$ \\Rightarrow \\sum {\\frac {x}{{4x^2 \\plus{} y^2 \\plus{} z^2 }}} \\le \\sum {\\frac {x}{{2x^2 \\plus{} 2xy \\plus{} 2zx}}} \\equal{} \\frac {3}{{2\\left( {x \\plus{} y \\plus{} z} \\right)}}$\r\n\r\nIt suffices us to show that \r\n\r\n$ \\frac {3}{{2\\left( {x \\plus{} y \\plus{} z} \\right)}} \\le \\frac {{x \\plus{} y \\plus{} z}}{{2\\left( {xy \\plus{} yz \\plus{} zx} \\right)}}$\r\n\r\nwhich is obviously :).", "Solution_2": "no i think that true for $ a\\equal{}b\\equal{}0.4$ and $ c\\equal{}1$\r\nby caushy shwartz :\r\n$ (4x^2\\plus{}y^2\\plus{}z^2)(\\frac{1}{4}\\plus{}1\\plus{}1)\\leq(x\\plus{}y\\plus{}z)^2$\r\n$ \\iff$ $ \\frac{1}{4x^2\\plus{}y^2\\plus{}z^2}\\leq\\frac{4}{9(xy\\plus{}yz\\plus{}zx)}$\r\nwe have to prove now :\r\n$ \\frac{4(x\\plus{}y\\plus{}z)}{9(x\\plus{}y\\plus{}z)^2}\\leq\\frac{3(x\\plus{}y\\plus{}z)}{10(xy\\plus{}yz\\plus{}zx)}$\r\n\r\n$ \\iff$ $ 27(x\\plus{}y\\plus{}z)^2\\geq40(xy\\plus{}yz\\plus{}zx)$\r\n$ \\iff$ $ 27(x^2\\plus{}y^2\\plus{}z^2)\\plus{}14(xy\\plus{}yz\\plus{}zx)\\geq0$\r\n\r\nwe are done :wink:", "Solution_3": "[quote][b]mehdi cherif wrote:[/b]\n$ (4x^2 \\plus{} y^2 \\plus{} z^2)(\\frac {1}{4} \\plus{} 1 \\plus{} 1)\\leq(x \\plus{} y \\plus{} z)^2$\n$ \\iff$ [/quote]it is >= and not =< :wink: .", "Solution_4": "My inequality isn't flase but not nice, I will post solution after three days!", "Solution_5": "[quote=\"Honey_S\"]\n$ \\sum {\\frac {x}{{2x^2 \\plus{} 2xy \\plus{} 2zx}}} \\equal{} \\frac {3}{{2\\left( {x \\plus{} y \\plus{} z} \\right)}}$[/quote] \r\nYou are wrong.", "Solution_6": "[quote=\"Sudoku\"][quote=\"Honey_S\"]\n$ \\sum {\\frac {x}{{2x^2 \\plus{} 2xy \\plus{} 2zx}}} \\equal{} \\frac {3}{{2\\left( {x \\plus{} y \\plus{} z} \\right)}}$[/quote] \nYou are wrong.[/quote]\r\nI think it's true. :wink:" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Find the maximum value of $ b \\minus{} \\frac{1}{a^2 \\plus{} b^2}$ when $ a,b > 0$ verify$ a^3 \\plus{} b^5 \\leq a^2 \\plus{} b^2.$", "Solution_1": "I must be misreading this but\r\n\r\n$ b \\minus{} \\frac {1}{a^2\\plus{}b^2}$ will not have a maximum, because as b gets large, $ \\frac{1}{a^2\\plus{}b^2}$ will get smaller.\r\n\r\nThe inequality is false too. The only way that that would make sense is if $ a,b < 0$, then the question would make sense", "Solution_2": "[quote=\"SimonM\"]I must be misreading this but\n\n$ b \\minus{} \\frac {1}{a^2 \\plus{} b^2}$ will not have a maximum, because as b gets large, $ \\frac {1}{a^2 \\plus{} b^2}$ will get smaller.\n\nThe inequality is false too. The only way that that would make sense is if $ a,b < 0$, then the question would make sense[/quote]\r\nMaybe i'm wrong too,but i think that means:\r\n$ \\frac{b\\minus{}1}{a^2\\plus{}b^2}$,not $ b\\minus{}\\frac{1}{a^2\\plus{}b^2}$", "Solution_3": "Still doesn't make any difference to the problem in terms of an answer" } { "Tag": [], "Problem": "Two angles of a triangle measure $ 55^{\\circ}$ and $ 47^{\\circ}$. What is the number of degrees in the measure of the supplement of the remaining angle of the triangle?", "Solution_1": "The supplement of the remaining angle is equal to the sum of the 2 other angles, so it's $ 55\\plus{}47 \\equal{} \\boxed{102}$.", "Solution_2": "We first note that the supplement of the remaining angle is equal to the sum of the other two angles (the supplement is 180-$ n$, where n is the degree measure of the other angle, and the sum of the angles of any triangle is $ 180^{\\circ}$). So, $ 55^{\\circ}\\plus{}47^{\\circ}\\equal{}\\boxed{102^{\\circ}}$." } { "Tag": [ "calculus", "integration", "inequalities", "inequalities proposed" ], "Problem": "$ \\bigstar$ Let $ a,b,c\\ge 0$. Prove that:\r\n\\[ \\frac {4}{3a} \\plus{} \\frac {4}{3b} \\plus{} \\frac {4}{3c} \\plus{} \\frac {3}{2a \\plus{} b} \\plus{} \\frac {3}{2b \\plus{} c} \\plus{} \\frac {3}{2c \\plus{} a}\\ge \\frac {3}{a \\plus{} b \\plus{} c} \\plus{} \\frac {6}{c\\plus{}2a}\\plus{}\\frac {6}{a \\plus{} 2b} \\plus{} \\frac {6}{b \\plus{} 2c} \r\n\\]\r\nFor Argady :wink:", "Solution_1": "[quote=\"zaizai-hoang\"]$ \\bigstar$ Let $ a,b,c\\ge 0$. Prove that:\n\\[ \\frac {4}{3a} \\plus{} \\frac {4}{3b} \\plus{} \\frac {4}{3c} \\plus{} \\frac {3}{2a \\plus{} b} \\plus{} \\frac {3}{2b \\plus{} c} \\plus{} \\frac {3}{2c \\plus{} a}\\ge \\frac {3}{a \\plus{} b \\plus{} c} \\plus{} \\frac {6}{c \\plus{} 2a} \\plus{} \\frac {6}{a \\plus{} 2b} \\plus{} \\frac {6}{b \\plus{} 2c}\n\\]\nFor Argady :wink:[/quote]\r\nBecause $ \\sum_{cyc}(4x^3\\plus{}3x^2y\\minus{}6x^2z\\minus{}xyz)\\geq0$ is true.", "Solution_2": "[quote=\"arqady\"][quote=\"zaizai-hoang\"]$ \\bigstar$ Let $ a,b,c\\ge 0$. Prove that:\n\\[ \\frac {4}{3a} \\plus{} \\frac {4}{3b} \\plus{} \\frac {4}{3c} \\plus{} \\frac {3}{2a \\plus{} b} \\plus{} \\frac {3}{2b \\plus{} c} \\plus{} \\frac {3}{2c \\plus{} a}\\ge \\frac {3}{a \\plus{} b \\plus{} c} \\plus{} \\frac {6}{c \\plus{} 2a} \\plus{} \\frac {6}{a \\plus{} 2b} \\plus{} \\frac {6}{b \\plus{} 2c}\n\\]\nFor Argady :wink:[/quote]\nBecause $ \\sum_{cyc}(4x^3 \\plus{} 3x^2y \\minus{} 6x^2z \\minus{} xyz)\\geq0$ is true.[/quote]\r\nCan you explain me? mr.arqady?", "Solution_3": "[quote=\"Sunjee\"]\nCan you explain me? mr.arqady?[/quote]\r\nOf cause! \r\nThe harasi's integral method:\r\nLet $ t > 0$ and $ x \\equal{} t^a,$ $ y \\equal{} t^b$ and $ z \\equal{} x^c.$\r\nHence, $ \\sum_{cyc}(4t^{3a} \\plus{} 3t^{2a \\plus{} b} \\minus{} 6t^{2a \\plus{} c} \\minus{} t^{a \\plus{} b \\plus{} c})\\geq0,$ which gives the following inequality:\r\n$ \\int_0^1\\left(\\sum_{cyc}(4t^{3a \\minus{} 1} \\plus{} 3t^{2a \\plus{} b \\minus{} 1} \\minus{} 6t^{2a \\plus{} c \\minus{} 1} \\minus{} t^{a \\plus{} b \\plus{} c \\minus{} 1})\\right)dt\\geq0,$\r\nwhich gives the original inequality." } { "Tag": [ "linear algebra", "matrix", "complex numbers", "linear algebra unsolved" ], "Problem": "$A,B$ -diagonalizable matrices from $M_n(C)$, such that $e^A=e^B$ prove that $A=B$", "Solution_1": "wich contest ?", "Solution_2": "Is there a result like if $A,B$ diagonalisable in $M_n(C)$ and $f$ injective \r\nanalytic $f(A)=f(B)$ then $A=B$ ?", "Solution_3": "The result is false. Let\r\n\r\n\\[A_n=\\left[\\begin{array}{cc}0&-2n\\pi \\\\ 2n\\pi&0\\end{array}\\right], n\\in\\mathbb{Z}\\]\r\nThen each $A_n$ is diagonalizable over the complex numbers, but $e^{A_n}=I$ for each $n$.", "Solution_4": "[quote=\"Moubinool\"]Is there a result like if $A,B$ diagonalisable in $M_n(C)$ and $f$ injective \nanalytic $f(A)=f(B)$ then $A=B$ ?[/quote]\r\n\r\nDear Kent do you heard some result like this ?\r\nthx", "Solution_5": "2Moubinool: 1). It is some kinda training/select contest. 2). I don't know anything about you generalization-maybe you know something about truth or non-trus of this fact-in the contest was proposed problem,posted by me\r\n2grobber: Maybe my question stupid:as far as i understand you proved that each eigenvalue of A is equal to some eigenvalue of B-how it will do if for example every\r\n$a_i=b_1$ -in that case it's not neccesary to be equal A and B --- it seems to mke that it's stupid, i think that i don't understand you solution", "Solution_6": "I had based my.. \"solution\" on the following false fact: $diag(e^{a_i})=S\\cdot diag(e^{b_i})\\cdot S^{-1}\\Rightarrow a_i=b_i,\\ \\forall i$, but we can simply take $a_i$ as we please ($a_1\\ne a_2$), and take $S$ to be the matrix having $1$ on $(1,2),(2,1), (i,i),\\ \\forall i\\ge 3$ and $0$ everywhere else.\r\n\r\nTo eugene: that was no solution: I was wrong, and Kent has given a counterexample.", "Solution_7": "No, Moubinool, I don't know any result like that. The best I can say is that I haven't yet spotted any obvious reason for it to be false (now that you've included the word \"injective.\")", "Solution_8": "Does what I wrote up there count for a counterexample to Moubinool's question?", "Solution_9": "ok, sorry everyone for posting wrong problem\r\nthanks to Kent for counterexample", "Solution_10": "If $A,B \\in M_n(R)$ diagonalisable such that $e^A=e^B$\r\nthen $A=B$ \r\n\r\nThis one was an Oral Examination Ecole Polytechnique \r\n\r\nRMS 4 (Revue de Mathmatiques Spciales)\r\nAnne 2002-2003 \r\npage 307", "Solution_11": "[quote]If $ A,B \\in M_n(R) $ diagonalisable such that $ e^A=e^B $\nthen $ A=B $ [/quote]\r\n\r\nFor this to have any chance of being true, we must assume that by \"diagonalisable\" you mean that $\\exists P$ such that $P^{-1}AP=D$ where $P$ and $D$ have only real entries and $D$ is diagonal. That rules out my previous counterexample which has imaginary eigenvalues.\r\n\r\nI'd like to claim that this is true. I am missing one step: I need that $e^A=e^B$ implies that $e^Ae^{-B}=e^{A-B}.$ (Which would follow from $AB=BA.$) If I had that, I'd be done.", "Solution_12": "I think the bad solution I had given for the first one works for this one. Assume WLOG that $A=diag (a_i)$ and $B=S\\cdot diag(b_i)\\cdot S^{-1}$ where $S=(s_{ij})$ is invertible. The condition $e^A=e^B$ translates to $diag(e^{a_i})=S\\cdot diag(e^{b_i})\\cdot S^{-1}\\iff diag(e^{a_i})\\cdot S=S\\cdot diag(e^{b_i})$, and this is equivalent to $(e^{a_i}-e^{b_j})s_{ij}=0,\\ \\forall i,j$, and from here and the injectivity of $e^x$ on $\\mathbb R$ we easily get $(a_i-b_j)s_{ij}=0,\\ \\forall i,j$, which is equivalent to $A=diag(a_i)=S\\cdot diag(b_i)\\cdot S^{-1}=B$." } { "Tag": [ "Putnam", "geometry", "probability", "conics", "parabola", "symmetry", "integration" ], "Problem": "A dart, thrown at random, hits a square target. Assuming that any two parts of the target of equal area are equall likely to be hit, find the probability that hte point hit is nearer to the center than any edge.", "Solution_1": "I've never liked this question very much -- it's just kind of ugly.\r\n\r\nThe basic approach is to temporarily ignore the fact that you have a square and just ask about the point being closer to the center than to one side. Then you realize that you have the classical description of a parabola, and coordinatize and integrate and add or subtract a few triangles as necessary.", "Solution_2": "Ohhh... the center is a focus, and the sides are directrixes (directrici?). Makes sense.", "Solution_3": "[quote=\"justdudxit\"]Ohhh... the center is a focus, and the sides are directrixes (directrici?). Makes sense.[/quote]\r\n\r\nRight, exactly. Up until that point, the idea is very nice (and who'd have thought you'd ever again use the fact that a parabola is the locus of points equidistant from a fixed line and point?). But the actual calculation after that point is not so pretty.", "Solution_4": "In fact, using your brains you can really avoid the calculation. ;) Take a square with sidelength $2$ for easiest calculations. Draw the diagonals and consider the lower part of the four. Obviously the parabola here is $y=\\frac12(x^{2}-1)$ (filling in 2 well-chosen points gives you the unique one -- you don't even need a pencil for that). \r\n\r\nTo find the requested probability, it's sufficient (by symmetry) to calculate the area above the parabola in the right half of the lower part: \\[\\text{probability}= 2 \\int^{\\sqrt{2}-1}_{0}-x-\\frac12(x^{2}-1) dx =\\frac{4\\sqrt{2}-5}{3}.\\]" } { "Tag": [ "induction", "inequalities", "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "Find the greatest real number $ \\alpha$ for which there exists a sequence of infinitive integers $ (a_n)$, ($ n \\equal{} 1, 2, 3, \\ldots$) satisfying the following conditions:\r\n1) $ a_n > 1997n$ for every $ n \\in\\mathbb{N}^{*}$;\r\n2) For every $ n\\ge 2$, $ U_n\\ge a^{\\alpha}_n$, where $ U_n \\equal{} \\gcd\\{a_i \\plus{} a_k | i \\plus{} k \\equal{} n\\}$.", "Solution_1": "Is $ \\alpha\\equal{}\\frac{1}{2}$ correct answer?I've proved that $ \\alpha\\leq\\frac{1}{2}$ but I don't know whether this value can be achieved... :blush:\r\nI tried to construct an example,but I failed...", "Solution_2": "I think $ \\alpha\\equal{}1$ is a solution. Take $ a_n\\equal{}1998n$. Then if $ i\\plus{}j\\equal{}n$ then $ a_i\\plus{}a_j\\equal{}1998(i\\plus{}j)\\equal{}1998n$. So $ U_n\\equal{}1998n\\equal{}a_n$.", "Solution_3": "[quote=\"cosinator\"]I think $ \\alpha \\equal{} 1$ is a solution. Take $ a_n \\equal{} 1998n$. Then if $ i \\plus{} j \\equal{} n$ then $ a_i \\plus{} a_j \\equal{} 1998(i \\plus{} j) \\equal{} 1998n$. So $ U_n \\equal{} 1998n \\equal{} a_n$.[/quote]\r\nThere is a serious mistake in the statement of the problem above,there should be $ a_n> 1997^{n}$ instead of $ a_n>1997\\cdot n$.", "Solution_4": "Oh, then it's my mistake. Sorry!", "Solution_5": "Note that $\\alpha=0$ works for any sequence with $a_n>1997^n$ for all $n$, so we can assume $\\alpha>0$.\n\nIf $\\alpha\\ge1$ then $a_n\\le a_1+a_{n-1}$ shows that $a_n=O(n)$, contradicting the fact that $a_n>1997^n$ for all $n$, so $\\alpha<1$. Now $a_{2n}\\le a_n+a_n$ shows that $a_{2^k}\\le 2^{\\alpha^{-1}}a_{2^{k-1}}^{\\alpha^{-1}}$, so by induction\n\\[a_{2^k}\\le2^{\\alpha^{-1}+\\cdots+\\alpha^{-k}}a_1^{\\alpha^{-k}}\\implies 1997^{(2\\alpha)^k}\\le 2^{1+\\alpha+\\cdots+\\alpha^{k-1}}a_1<2^{1/(1-\\alpha)}a_1.\\]Thus $2\\alpha\\le1$.\n\nTo find a construction for $\\alpha=1/2$, we use \"sum-to-product\" identities (motivated by Chebyshev polynomials, Fibonacci numbers, etc.): set $\\alpha=N+\\sqrt{N^2-1}$ (so $\\alpha^{-1}=N-\\sqrt{N^2-1}$) for some large integer $N$ and define $b_n=\\alpha^n+\\alpha^{-n}$ for all $n\\ge0$ (clearly these are all positive integers); letting $a_n=\\alpha^{2n}+\\alpha^{-2n}$ for each $n\\ge1$, we have\n\\[a_i+a_j=\\alpha^{2i}+\\alpha^{2j}+\\alpha^{-2i}+\\alpha^{-2j}=(\\alpha^{i+j}+\\alpha^{-i-j})(\\alpha^{i-j}+\\alpha^{j-i})=b_{i+j}b_{|i-j|}.\\]For $N$ sufficiently large, clearly $a_n>\\alpha^{2n}>1997^n$ for each $n$, and\n\\begin{align*}\n\\gcd\\{a_i+a_j\\mid i+j=n\\} \\ge b_n &= \\alpha^n+\\alpha^{-n} \\\\\n&= \\sqrt{\\alpha^{2n}+\\alpha^{-2n}+2} > \\sqrt{\\alpha^{2n}+\\alpha^{-2n}} = a_n^{1/2},\n\\end{align*}as desired.\n\n[b]Edit:[/b] Cleared up some things in the first part.", "Solution_6": "There is you only took the case n>=1 and find a contradiction. It does not mean that n should be smaller than 1/2. Can you explain it?" } { "Tag": [ "AMC" ], "Problem": "Just wondering whast everyone got\r\n\r\nWe got 129+130.5+144=403.5", "Solution_1": "how are team scores calculated? i am ignorant yes.. is it like top 3 of each school", "Solution_2": "yeah top 3 of your school on the same test (i.e. can't mix A and B)", "Solution_3": "[quote=\"nat mc\"]Just wondering whast everyone got\n\nWe got 129+130.5+144=403.5[/quote]\r\n\r\nwho got the 144???", "Solution_4": "probalby david benjamin?", "Solution_5": "yeah david did", "Solution_6": "o.O Phil, Carmel almost beat you guys. If I hadn't screwed up #1, we would have.\r\n\r\n135+135+132=402. (I think since I haven't really checked, if we didn't get that, then someone probably made a careless mistake).", "Solution_7": "$150+150+126 = 426$. :D" } { "Tag": [ "geometry", "power of a point", "radical axis", "cyclic quadrilateral", "geometry open" ], "Problem": "I found very interesting Miquel theorems. Most interesting for me was this one: http://www.gogeometry.com/miquel_pentagram1.htm\r\nInspired of that theorem using geogebra I found the configuration on the picture.\r\n\r\nTo be easier for you - take a look at the file attached.\r\nIt is given a pentagram (five rays star). The vertices of the pentagram lies on a circle. There are constructed five smaller circles.\r\nProve that the five lines, each formed by inersecting point of two small circles and opposite star vertex, intersects at a common point. \r\n\r\nMy questions are:\r\n1. Is it a famous theorem or a problem from math competition?\r\n2. How to prove that statement?\r\n3. What is its level of difficulty?\r\n4. Is it a beatiful problem?\r\n\r\nI'm sorry for the quality of the picture if the problem is not clear I can restate it in a better way.", "Solution_1": "It is a nice problem. I don't know if this will work, but\r\n\r\nlet $ A,B,....E$ be the points of the large circle, and $ A_1,B_1..E_1$ be the opposite points on the smaller circle. By radical axis, it would suffice to show that $ ABA_1B_1$ is cyclic. (If the concurrency is true, then this is clearly true by power of a point).", "Solution_2": "[b]Solution: [/b]\r\nFirst we will show that $ EB_1A_2B$ is cyclic. $ (1)$\r\n$ \\Leftrightarrow \\angle EBA_2 \\plus{} \\angle EB_1A_2 \\equal{} 180^o$\r\n$ \\Leftrightarrow \\angle EBA_2 \\plus{} \\angle B_2C_2B \\plus{} \\angle B_2DA_2 \\equal{} 180^o$ (right)\r\nTherefore $ (1)$ is true. Similarly we get $ EB_1A_2E_2B, AD_1E_2A_1D$,... are cyclic.\r\nWe will show that $ \\angle A_1B_1B \\equal{} \\angle A_1AB$\r\n$ \\Leftrightarrow \\angle DB_1B \\minus{} \\angle DB_1A_1 \\equal{} \\angle DAB \\minus{} \\angle DAA_1$\r\n$ \\Leftrightarrow \\angle DC_2B \\minus{} \\angle DA_2A_1 \\equal{} \\angle DAB \\minus{} \\angle DE_2A_1$\r\n$ \\Leftrightarrow \\angle DAB \\plus{} \\angle EBA \\minus{} \\angle DE_2A_1 \\minus{} \\angle A_2A_1E_2 \\equal{} \\angle DAB \\minus{} \\angle DE_2A_1$\r\n$ \\Leftrightarrow \\angle EBA \\equal{} \\angle A_2A_1E_2 \\equal{} \\angle A_2CE_2.$\r\nThe last is right because $ ABCE$ is a cyclic quadrilateral.\r\nSo $ ABA_1B_1$ is a cyclic quadrilateral.\r\nSimilarly we can show that $ AEA_1E_1$ is a cyclic quadrilateral. On the other side, $ EB_1E_1B$ is also a cyclic quadrilateral then by the theorem about radical center, we obtain $ AA_1, BB_1, CC_1$ are concurrent. \r\nSimilarly our problem is solved. \r\n\r\n[b]Remark:[/b] \r\nDenote $ X,Y,Z$ be the second intersections of $ (B_2A_2D), (EB_2C_2), (CE_2A_2)$ and $ (O)$. Show that $ DX, YZ, AB$ are concurrent.\r\n\r\nHAPPY LUNAR NEW YEAR! :)", "Solution_3": "Very very good idea and solution. Just to remark there is small technical issue. From the quadrilaterals cyclic mentioned it follows that AA1, BB1 and EE1 meet at common point - but this does not change the result, we can take suitable other cyclic quadrilaterals and show by analogy that AA1, BB1, CC1; AA1, BB1,DD1 meet also at common point and we are done.", "Solution_4": "I would like to thank to livetolove212 for the excellent solution. I think he is a great mathematician! There must me more people like him.\r\nIf the problem is new you can call\r\n\"Pentagram open problem 1\" and\r\n\"Pentagram open problem 2\" with the names:\r\nMirchev-Kazakov and Mirchev-Lynh theorems by your choice.\r\nI would like to mention - both problem can be solved using complex numbers.\r\ngeorgi111 is very busy person but some day he can post solutions using complex numbers.\r\nIf you want I can publish more open problems. Some of them are related to the pentagrams and pentacles.\r\nFor all mathlinkers - you can feel free to give an opinion for my last unsolved open problem: \r\nhttp://www.mathlinks.ro/viewtopic.php?t=313384\r\n\r\nHAPPY LUNAR NEW YEAR! :wink:" } { "Tag": [ "probability", "videos", "AoPSwiki", "articles", "Alcumus" ], "Problem": "I have seen PIE in many solutions and i cant understand excactly what it is. Could someone help me. Thx", "Solution_1": "PIE counting is when you overcount something then you subtract those .\r\n\r\nExample:\r\n\r\nHow many numbers less than $ 50$ are multiples of $ 2$ and $ 3$ but not $ 4$?\r\n\r\nSolution: We count $ 6, 12, 18,..., 48$ or $ \\frac{50}{6} \\est 8$. But we also counted some multiples of $ 4$. So we subtract the multiples of $ 12$, which are $ 4$. So the answer is $ 8 - 4 = \\boxed{4}$.", "Solution_2": "I know it as the Inclusion-Exclusion Principle so I guess PIE stands for the Principle of Inclusion and Exclusion.", "Solution_3": "[quote=\"KingJamesMvP\"]I have seen PIE in many solutions and i cant understand excactly what it is. Could someone help me. Thx[/quote]\r\n\r\nPIE is Principle of Inclusion/Exclusion. It is in Intermediate Counting/Probability but don't just memorize the formula because a lot of upper level contest problems require:\r\n\r\n1) A [b]thorough[/b] understanding of PIE, and\r\n2) A good [b]insight[/b] on how to use it!", "Solution_4": "Look [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]here[/url], [url=http://www.artofproblemsolving.com/Alcumus/Videos/Video.php?video_id=37&m=af6aa14999718745da45a47a8075b91b]here[/url] and [url=http://www.artofproblemsolving.com/Alcumus/Videos/Video.php?video_id=71&m=86ef46c49a5de3839873fbebe49af597]here[/url]. The first one is an AoPSWiki article, and the other two are Alcumus Videos." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "With $ a,b,c > 0$ Prove that $ \\sum_{cyc}\\frac {\\sqrt {a^2 \\plus{} bc}}{(a \\plus{} b)\\sqrt {a \\plus{} b}} \\geq \\frac {3\\sqrt {3}}{2\\sqrt {a \\plus{} b \\plus{} c}}$", "Solution_1": "[quote=\"2424\"]With $ a,b,c > 0$ Prove that $ \\sum_{cyc}\\frac {\\sqrt {a^2 \\plus{} bc}}{(a \\plus{} b)\\sqrt {a \\plus{} b}} \\geq \\frac {3\\sqrt {3}}{2\\sqrt {a \\plus{} b \\plus{} c}}$[/quote]\r\nIT's very interesting! :P", "Solution_2": "Do you meant:\r\n$ (a\\plus{}b\\plus{}c)^3(a^2\\plus{}bc)(b^2\\plus{}ca)(c^2\\plus{}ab) \\geq\\ \\frac{27}{64}(a\\plus{}b)^3(b\\plus{}c)^3(c\\plus{}a)^3$\r\n :maybe:", "Solution_3": "[quote=\"nguoivn\"]Do you meant:\n$ (a \\plus{} b \\plus{} c)^3(a^2 \\plus{} bc)(b^2 \\plus{} ca)(c^2 \\plus{} ab) \\geq\\ \\frac {27}{64}(a \\plus{} b)^3(b \\plus{} c)^3(c \\plus{} a)^3$\n :maybe:[/quote]\r\nyes,It's true and interesting! :P \r\nI have nice sulotion for it!\r\n :P", "Solution_4": "@nguoivn:[hide]E nghe noi anh va anh Can co newbook sap ra.E dang to mo ve nhung ki thuat moi su dung AM-GM cua anh.\nKhoang bao jo thi 2 anh cho xuat ban quyen do ak?[/hide]", "Solution_5": "[quote=\"2424\"][quote=\"nguoivn\"]Do you meant:\n$ (a \\plus{} b \\plus{} c)^3(a^2 \\plus{} bc)(b^2 \\plus{} ca)(c^2 \\plus{} ab) \\geq\\ \\frac {27}{64}(a \\plus{} b)^3(b \\plus{} c)^3(c \\plus{} a)^3$\n :maybe:[/quote]\nyes,It's true and interesting! :P \nI have nice sulotion for it!\n :P[/quote]\r\nCan you share it for every body? Thank you :) \r\nIndeed, I don't see any proof without pqr :oops: \r\n@2424: [hide]Khoang? dau` thang' 9 cuon \"Bat dang thuc va n~ loi giai dep\" cua tui anh se~ ra mat' ban. doc. em a. Tui. anh vua` moi ki' hop dong` chinh' thuc' hom qua ^^![/hide]", "Solution_6": "[quote=\"nguoivn\"][quote=\"2424\"][quote=\"nguoivn\"]Do you meant:\n$ (a \\plus{} b \\plus{} c)^3(a^2 \\plus{} bc)(b^2 \\plus{} ca)(c^2 \\plus{} ab) \\geq\\ \\frac {27}{64}(a \\plus{} b)^3(b \\plus{} c)^3(c \\plus{} a)^3$\n :maybe:[/quote]\nyes,It's true and interesting! :P \nI have nice sulotion for it!\n :P[/quote]\nCan you share it for every body? Thank you :) \nIndeed, I don't see any proof without pqr :oops: \n@2424: [hide]Khoang? dau` thang' 9 cuon \"Bat dang thuc va n~ loi giai dep\" cua tui anh se~ ra mat' ban. doc. em a. Tui. anh vua` moi ki' hop dong` chinh' thuc' hom qua ^^![/hide][/quote]\n@2424:[hide]Bai nay e dung mot bai cua e da post len dien dan nhung chua ai jai,bdt nay chi la truong hop dac biet cua bdt do voi k=2 thui a ak.E muon xem a co cach jai bat ngo nao bang AM-GM cho no k thui a ak.:).E se post loi jai cua minh sau vai hum nua :).E ngoi wan nen ngai post loi giai lem.anh ah,khi nao quyen do ra thi anh bao vs e 1 tieng dc k ak?E rat to mo ve AM-GM cua anh tn do :roll: Thanks anh Quoc Anh nhiu nha! :) [/hide]" } { "Tag": [ "limit", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "evaluate the limit ${\\lim_{n\\to \\infty }\\, \\left(\\sum_{i=1}^{n}\\left(\\frac{1}{n i}+\\frac{1}{n}+\\frac{1}{i}\\right)^{i+n}\\right)}$\r\n\r\n[hide=\"Hint\"]\nRemember Euler' Limit\n[/hide]", "Solution_1": "Let $a_{i}= \\left ( \\frac{a}{ni}+\\frac{1}{n}+\\frac{1}{i}\\right )^{i+n}$\r\n\r\n$\\boxed{\\text{Case}\\ i=1}$\r\n$\\lim_{n \\to \\infty}a_{1}= \\lim_{n \\to \\infty}\\left ( \\frac{a}{n}+\\frac{1}{n}+1 \\right )^{n+1}= \\lim_{n \\to \\infty}\\left ( \\frac{a+1}{n}+1 \\right )^{n+1}= e^{a+1}$\r\n\r\n$\\boxed{\\text{Case}\\ i\\ge2}$\r\n$\\lim_{n \\to \\infty}a_{i}= \\lim_{n \\to \\infty}\\left ( \\frac{a}{ni}+\\frac{1}{n}+\\frac{1}{i}\\right )^{n+i}= \\lim_{n \\to \\infty}\\left ( \\frac{a+i+n}{ni}\\right )^{n+i}= \\lim_{n \\to \\infty}\\left ( \\frac{\\frac{a}{n}+\\frac{i}{n}+\\frac{n}{n}}{\\frac{ni}{n}}\\right ) = \\lim_{n \\to \\infty}\\left ( \\frac{\\frac{a}{n}+\\frac{i}{n}+1}{i}\\right )^{n+i}= 0$\r\n\r\n$\\therefore \\ \\lim_{n \\to \\infty}\\left ( \\sum_{i=1}^{n}\\left ( \\frac{a}{ni}+\\frac{1}{n}+\\frac{1}{i}\\right )^{i+n}\\right ) = \\lim_{n \\to \\infty}\\left ( a_{1}+\\sum_{i=2}^{n}a_{i}\\right ) =e^{a+1}$\r\n\r\n\r\nThis Problem is case of $a=1$.\r\nThus, $\\boxed{\\boxed{\\lim_{n \\to \\infty}\\left ( \\sum_{i=1}^{n}\\left ( \\frac{1}{ni}+\\frac{1}{n}+\\frac{1}{i}\\right )^{i+n}\\right ) = e^{2}}}$", "Solution_2": "correctly :)\r\n\r\njust a small remark. you forgot $n+1$ in this line.\r\n\r\n$\\lim_{n \\to \\infty}\\left ( \\frac{\\frac{a}{n}+\\frac{i}{n}+\\frac{n}{n}}{\\frac{ni}{n}}\\right ) = \\lim_{n \\to \\infty}\\left ( \\frac{\\frac{a}{n}+\\frac{i}{n}+1}{i}\\right )^{n+i}= 0$\r\n\r\nbut the solution is correct, so this does not matter", "Solution_3": "You're right, I made a typing error:(\r\nThanks for the input.", "Solution_4": "I don't think that you can pass to the limit inside the sum\r\nyou have that $ a_{i}\\to 0,i\\ge 2$ but you have infinitely many $ a_{i}$s when you take the limit as $ n\\to\\infty$ and you get smth like $ \\infty\\cdot 0$.", "Solution_5": "Kouichi Nakagawa is rigth, we can do this yes.\r\n\r\nsee\r\n\r\n$ {i\\geq 2}$\r\n$ {\\lim_{n\\to \\infty }\\, n \\left(\\frac{\\frac{i}{n}+\\frac{1}{n}+1}{i}\\right)^{i+n}=0}$\r\n\r\nyour indertemination does not exist, because the limit is zero.", "Solution_6": "You can not take the limit into the sum and after that sum up infinitely (because the sum is from 2 to $ n\\to\\infty$ ) many zeros and say that it is zero.\r\nWhat if, for example $ a_{i,n}=\\frac1{i\\log n}$ for $ i=1,2,...\\ldots,n$\r\nthen $ \\lim_{n\\to\\infty}a_{i,n}=0,\\forall i\\ge 1$\r\nbut $ \\lim_{n\\to\\infty}\\sum_{i=1}^{n}a_{i,n}=1$", "Solution_7": "see my previos post.\r\n\r\nthat limit is greater than the sum.\r\n\r\ni didnt want to post it. because my last limit explain everything..\r\n\r\n\r\n$ {i\\leq n\\land i\\geq 2}$\r\n$ {\\left(\\frac{1}{2}+\\frac{3}{2 n}\\right)^{n+2}\\geq \\left(\\frac{1}{n i}+\\frac{1}{n}+\\frac{1}{i}\\right)^{i+n}\\geq \\left(\\frac{1}{n^{2}}+\\frac{2}{n}\\right)^{2 n}}$\r\n\r\n$ {n\\left(\\frac{1}{2}+\\frac{3}{2 n}\\right)^{2+n}\\geq \\sum_{i=2}^{n}\\left(\\frac{1}{n}+\\frac{1}{n i}+\\frac{1}{i}\\right)^{n+i}\\geq n\\left(\\frac{1}{n^{2}}+\\frac{2}{n}\\right)^{2 n}}$\r\n\r\n$ {\\lim_{n\\to \\infty }\\, n \\left(\\frac{1}{2}+\\frac{3}{2 n}\\right)^{n+2}=0}$\r\n$ {\\lim_{n\\to \\infty }\\, n \\left(\\frac{1}{n^{2}}+\\frac{2}{n}\\right)^{2 n}=0}$\r\n\r\nSo, just the first term matters.\r\n\r\nQ.E.D" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "In the sequence $ (a_n)$ with general term $ a_n \\equal{} n^3 \\minus{} (2n \\plus{} 1)^2$, does there exist a term that is divisible by 2006?", "Solution_1": "$ n\\equal{}361$ Is a possible solution.\r\nIn fact there are many more, because of the chinese reminder theorem.\r\nIf you want to get all the solutions work $ mod 2, 17, 59$ and you will get all solutions. :) \r\n\r\nDaniel" } { "Tag": [ "MATHCOUNTS", "algorithm" ], "Problem": "This is a problem from the All-Time Greatest MathCounts Problems:\r\nIn a 4x4 chessboard, what is the maximum number of squares that a knight can visit, starting from the top left corner without going to any square more than once?\r\n\r\nThis problem, in my opinion leaves a lot to luck. I still can't understand the book's solution, so could anyone explain clearly how to get the answer? Is there some [b]MathCounts-level[/b] algorithm to do this? I know that this is a really common question in mathematics, so I really need help. Thanx! :D", "Solution_1": "Hmmmm...I remember our mathCounts teacher told us to figure out a pattern that would get the most squares.", "Solution_2": "I think this is luck. I mean, after a few tries i would just put down the largest number i got.", "Solution_3": "[hide]\nhm, is it 13?\n\ni dont think its pure luck, it has a bit of logic in it.[/hide]", "Solution_4": "the answer's 15. i remember this problem very well.", "Solution_5": "Yup I just got 15.\r\n\r\nYeah i dont think theres a straightforward method to this problem... does anyone have a logical approach?", "Solution_6": "i think my reasoning is very off, but isn't it kinda obvious the answer is 15. i mean, if the knight starts at square 1 in the top left corner, he can just keep going all the way across the first row, then down and all the way across the second row, till he gets to the last square at the bottom left corner, for a total of 15 squares. i thought it would be (total number of squares)-1. \r\nim probably wrong.", "Solution_7": "[quote=\"mihail911\"]i think my reasoning is very off, but isn't it kinda obvious the answer is 15. i mean, if the knight starts at square 1 in the top left corner, he can just keep going all the way across the first row, then down and all the way across the second row, till he gets to the last square at the bottom left corner, for a total of 15 squares. i thought it would be (total number of squares)-1. \nim probably wrong.[/quote]\r\nFirst a knight doesnt move that way...(it doesnt move horizontally or vertically by 1 step)_\r\n\r\nSecond, if it did move the way you are tlaking about, then it would be 16 squares, because the first square he is on he visits too\r\n\r\nSo yes you are totally wrong..the only reason why you think is 15 because indianamath already posted that answer, and you are trying to find a solution thjat complements the answer..", "Solution_8": "Ha, the first two topics are \"Knights tour\" and \"Knights not getting much of a tour\" :lol: \r\n\r\nanyways, if the grid is\r\n1 2 3 4\r\n5 6 7 8\r\n9 10 11 12\r\n13 14 15 16\r\nit has to start 1,10,16,7, then it's random luck time.\r\nit turns out to be 1,10,16,7,9,15,8,2,11,13,6,12,3,5,14.\r\nI remember putting 14 for this :oops:", "Solution_9": "[quote=\"emerson woerner\"][quote=\"mihail911\"]i think my reasoning is very off, but isn't it kinda obvious the answer is 15. i mean, if the knight starts at square 1 in the top left corner, he can just keep going all the way across the first row, then down and all the way across the second row, till he gets to the last square at the bottom left corner, for a total of 15 squares. i thought it would be (total number of squares)-1. \nim probably wrong.[/quote]\nFirst a knight doesnt move that way...(it doesnt move horizontally or vertically by 1 step)_\n\nSecond, if it did move the way you are tlaking about, then it would be 16 squares, because the first square he is on he visits too\n\nSo yes you are totally wrong..the only reason why you think is 15 because indianamath already posted that answer, and you are trying to find a solution thjat complements the answer..[/quote]\r\n\r\nI am such a moron!\r\nwhen the question asked for knight, i didn't know it meant the actual chess figure..then yes, ur right!. srry, for the stupid mistake! :wallbash_red:", "Solution_10": "I got the right answer on the the first try.\r\n\r\nPlaying chess has its rewards..." } { "Tag": [ "modular arithmetic", "real analysis", "real analysis unsolved" ], "Problem": "Prove that the sequence $ |sin n|+|sin n^3|$ has no limit.", "Solution_1": "Upps, sorry I do have a solution for this one, so it would be better to move it to proposed section.", "Solution_2": "It follows rather easily from the fact that $\\{n\\},\\{n^3\\}$ are uniformly distributed in $(0,\\pi)$, where $\\{\\}$ is the \"fractional part\" $\\pmod \\pi$. Did you use weaker results to prove it?" } { "Tag": [ "inequalities", "inequalities theorems" ], "Problem": "I was working through some inequality sheets and came across the term $ inf [tx^{2}\\plus{}(y^{2})/t]$. What does $ inf$ mean please? $ max$ which i know stands for maximum was used in a similar situation if that helps clarify things further.", "Solution_1": "$ inf\\equal{}infimum$, it means that if $ a$ is $ inf$ of $ S$ and $ a'>a$ then there is an element of $ S$, may be $ x_{0}$, such that $ x_{0} 99 -> ... -> 91 -> 90, as 90 can only go to 9, not 91. \n\n\nThis is now a simpler counting problem. \n\nConsider only the remainder of the word after the B (18 letters). Then there are 2^18 possibilities. But some do not work. They fall into two nonoverlapping categories. \n\n\n1. Starts with 9 A's but has no 10 consecutive A's. \n2. Has 10 consecutive A's. \n\n1) The word is of form AAAAAAAAABXXXXXXXX, where each X can be anything without violating any conditions. This makes 2^8 words of this form. \n\n\n2) Either the word starts with 10 A's (2^8 words), or has a BAAAAAAAAAA (B, 10A's) in the word. There are 8 ways to choose where this goes in the word (pick the number of letters before it from 0 to 7), and 2^7 ways to pick the other letters (7 spots remain), making 8 * 2^7 = 2 ^ 10 words. \n\n\nTherefore, there are 2^18 - 2 * 2^8 - 2^10 = 2^18 - 3 * 2^9 = 509 * 2^9 words. The sum of its prime divisors is 509 + 2 = 511 as desired.\n[/quote]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let be given positive numbers a;b;c satisfying the condition $abc>1$.Prove that:\r\n$\\frac{a}{\\sqrt{b+\\sqrt{ac}}}+\\frac{b}{\\sqrt{c+\\sqrt{ab}}}+\\frac{c}{\\sqrt{a+\\sqrt{bc}}}\\geq \\frac{3}{\\sqrt{2}}$", "Solution_1": "[quote=chien than]Let be given positive numbers a;b;c satisfying the condition $abc>1$.Prove that:\n$\\frac{a}{\\sqrt{b+\\sqrt{ac}}}+\\frac{b}{\\sqrt{c+\\sqrt{ab}}}+\\frac{c}{\\sqrt{a+\\sqrt{bc}}}\\geq \\frac{3}{\\sqrt{2}}$[/quote]\n[url]https://artofproblemsolving.com/community/c4h2565013p21994721[/url]\n" } { "Tag": [ "inequalities" ], "Problem": "If $x , y , z >0$ and $x+y+z = 1$ , find the maximum value of the expression\r\n $A = xy+yz+zx-3xyz$\r\n\r\n [u]Babis[/u]", "Solution_1": "[hide=\"NOT 2/9\"]\n... but $\\boxed{\\frac{1}{4}}$, attained when $a=b=\\frac{1}{2}$ and $c=0$. To prove the inequality, WLOG let $a\\geq b\\geq c$ and rewrite as\n\\[ab(1-3c)+c(a+b)\\leq \\frac{1}{4}\\]\nAs $a\\geq b\\geq c$, $c\\leq \\frac{1}{3}$ so $1-3c$ is nonnegative. Also, $a+b=1-c$ and $ab\\leq \\frac{(a+b)^{2}}{4}=\\frac{(1-c)^{2}}{4}$. This gives us\n\\[ab(1-3c)+c(a+b)\\leq\\frac{(1-c)^{2}(1-3c)}{4}+c(1-c)\\]\nso it remains to prove\n\\[(1-c)^{2}(1-3c)+4c(1-c)\\leq 1\\]\nBut the left hand side is $(1-c)((1-c)(1-3c)+4c)=(1-c)(1+3c^{2})=1-c+3c^{2}-3c^{3}$. The above inequality then follows from\n\\[1-(1-c)^{2}(1-3c)-4c(1-c)=c(3c^{2}-3c+1)=c\\left( \\frac{3}{4}(2c-1)^{2}+\\frac{1}{4}\\right) \\geq 0\\]\n[/hide]", "Solution_2": "[hide=\"I dont know if this can go anywhere, but with some manipulation I get that...\"]\n$(x+y+z)^{3}=1 \\implies 3A-1= \\sum x^{3}-4 \\sum x^{2}-3 \\sum xy$\nand\n$(x+y+z+1)^{3}=8 \\implies 6A-7= 2 \\sum x^{3}-6 \\sum x^{2}-3 \\sum x-12xyz$ \n(All sums here are symmetric.)[/hide]", "Solution_3": "[quote=\"scorpius119\"][hide=\"NOT 2/9\"]\n... but $\\boxed{\\frac{1}{4}}$, attained when $a=b=\\frac{1}{2}$ and $c=0$. To prove the inequality, WLOG let $a\\geq b\\geq c$ and rewrite as\n\\[ab(1-3c)+c(a+b)\\leq \\frac{1}{4}\\]\nAs $a\\geq b\\geq c$, $c\\leq \\frac{1}{3}$ so $1-3c$ is nonnegative. Also, $a+b=1-c$ and $ab\\leq \\frac{(a+b)^{2}}{4}=\\frac{(1-c)^{2}}{4}$. This gives us\n\\[ab(1-3c)+c(a+b)\\leq\\frac{(1-c)^{2}(1-3c)}{4}+c(1-c) \\]\nso it remains to prove\n\\[(1-c)^{2}(1-3c)+4c(1-c)\\leq 1 \\]\nBut the left hand side is $(1-c)((1-c)(1-3c)+4c)=(1-c)(1+3c^{2})=1-c+3c^{2}-3c^{3}$. The above inequality then follows from\n\\[1-(1-c)^{2}(1-3c)-4c(1-c)=c(3c^{2}-3c+1)=c\\left( \\frac{3}{4}(2c-1)^{2}+\\frac{1}{4}\\right) \\geq 0 \\]\n[/hide][/quote]\r\nUmm... didn't the question say $c>0$?", "Solution_4": "[quote=\"bigboy82892\"][quote=\"scorpius119\"][hide=\"NOT 2/9\"]\n... but $\\boxed{\\frac{1}{4}}$, attained when $a=b=\\frac{1}{2}$ and $c=0$. To prove the inequality, WLOG let $a\\geq b\\geq c$ and rewrite as\n\\[ab(1-3c)+c(a+b)\\leq \\frac{1}{4}\\]\nAs $a\\geq b\\geq c$, $c\\leq \\frac{1}{3}$ so $1-3c$ is nonnegative. Also, $a+b=1-c$ and $ab\\leq \\frac{(a+b)^{2}}{4}=\\frac{(1-c)^{2}}{4}$. This gives us\n\\[ab(1-3c)+c(a+b)\\leq\\frac{(1-c)^{2}(1-3c)}{4}+c(1-c) \\]\nso it remains to prove\n\\[(1-c)^{2}(1-3c)+4c(1-c)\\leq 1 \\]\nBut the left hand side is $(1-c)((1-c)(1-3c)+4c)=(1-c)(1+3c^{2})=1-c+3c^{2}-3c^{3}$. The above inequality then follows from\n\\[1-(1-c)^{2}(1-3c)-4c(1-c)=c(3c^{2}-3c+1)=c\\left( \\frac{3}{4}(2c-1)^{2}+\\frac{1}{4}\\right) \\geq 0 \\]\n[/hide][/quote]\nUmm... didn't the question say $c>0$?[/quote]\n\n[hide=\"Well...\"]in that case we can consider $x=y=\\frac{1-\\epsilon }{2}$ and $z=\\epsilon$ (for some reason i used $a,b,c$ in my solution...) as $\\epsilon$ approaches 0, the expression approaches 1/4. So there would technically be [b]no[/b] maximum, but 1/4 would still be the best bound.\n[/hide]\r\n\r\n(of course, that wouldn't be necessary if it just said that the variables were nonnegative rather than positive.)", "Solution_5": "[quote=\"scorpius119\"][quote=\"bigboy82892\"][quote=\"scorpius119\"][hide=\"NOT 2/9\"]\n... but $\\boxed{\\frac{1}{4}}$, attained when $a=b=\\frac{1}{2}$ and $c=0$. To prove the inequality, WLOG let $a\\geq b\\geq c$ and rewrite as\n\\[ab(1-3c)+c(a+b)\\leq \\frac{1}{4}\\]\nAs $a\\geq b\\geq c$, $c\\leq \\frac{1}{3}$ so $1-3c$ is nonnegative. Also, $a+b=1-c$ and $ab\\leq \\frac{(a+b)^{2}}{4}=\\frac{(1-c)^{2}}{4}$. This gives us\n\\[ab(1-3c)+c(a+b)\\leq\\frac{(1-c)^{2}(1-3c)}{4}+c(1-c) \\]\nso it remains to prove\n\\[(1-c)^{2}(1-3c)+4c(1-c)\\leq 1 \\]\nBut the left hand side is $(1-c)((1-c)(1-3c)+4c)=(1-c)(1+3c^{2})=1-c+3c^{2}-3c^{3}$. The above inequality then follows from\n\\[1-(1-c)^{2}(1-3c)-4c(1-c)=c(3c^{2}-3c+1)=c\\left( \\frac{3}{4}(2c-1)^{2}+\\frac{1}{4}\\right) \\geq 0 \\]\n[/hide][/quote]\nUmm... didn't the question say $c>0$?[/quote]\n\n[hide=\"Well...\"]in that case we can consider $x=y=\\frac{1-\\epsilon }{2}$ and $z=\\epsilon$ (for some reason i used $a,b,c$ in my solution...) as $\\epsilon$ approaches 0, the expression approaches 1/4. So there would technically be [b]no[/b] maximum, but 1/4 would still be the best bound.\n[/hide]\n\n(of course, that wouldn't be necessary if it just said that the variables were nonnegative rather than positive.)[/quote]\r\n\r\n [b]YES ! [/b]\r\nI must change the conditiion to $x , y , z \\geq 0$. Really sorry , but I saw the problem under these conditions and I had not solved the problem , before posting it.Thank you !" } { "Tag": [ "LaTeX" ], "Problem": "Hi. Just wondering how to type:\r\n\r\n1) The modular congruence sign.\r\n2) The not-equal sign\r\n3) More emotions in my signature.\r\n\r\nThanks you.", "Solution_1": "\\equiv $ \\equiv$\r\n\r\n\\neq $ \\neq$\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/posting.php?f=123&mode=smilies[/url]\r\n(These are the available emoticons for use in messages, signatures, etc. Click the \"more emoticons\" link on the left, underneath the visible emoticons, to see this list. Then, click on the emoticon you with to insert. )\r\n\r\nEDITS: sorry, grammar fail :oops:", "Solution_2": "[quote=\"topofmath\"]Hi. Just wondering how to type:\n\n1) The modular congruence sign.\n2) The not-equal sign\n3) More emotions in my signature.\n\nThanks you.[/quote]\r\n\r\nNote: For other symbols in $ \\text{\\LaTeX}$ that you don't know you can find them [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]here[/url]." } { "Tag": [], "Problem": "Let $ P$ be a 30-sided polygon inscribed in a circle. Find the number of triangles whose vertices are the vertices of $ P$ such that any two vertices of each triangle are separated by at least three other vertices of $ P$.\r\n\r\nOfficial Solution:\r\nLet $ A$ be a vertex of $ P$. First we shall count the number of such triangles having $ A$ as a vertex. After taking away $ A$ and 3 consecutive vertices of $ P$ on each side of $ A$, we are left with 23 vertices from which we can choose two vertices in such a way that, together with $ A$, a desired triangle can be formed. There are $ \\binom{2\\plus{}(23\\minus{}3\\minus{}2)}{2}\\equal{}\\binom{20}{2}$ ways to do so. Hence there are $ 30\\binom{20}{2}\\times\\frac{1}{3}$ (because each triangle is counted thrice) = 1900 such triangles.\r\n\r\nWhat I would like to know is how they get $ \\binom{2\\plus{}(23\\minus{}3\\minus{}2)}{2}\\equal{}\\binom{20}{2}$ ways to choose the remaining two vertices after one is fixed..", "Solution_1": "I don't get it either. Also, what's wrong with this method: Choose 3 out of 21 vertices (12C3) and \"add\" in 3 more between each pair.", "Solution_2": "Ubemaya, I think your method fails because the points are distinct so you can't arbitrarily take 9 out if you don't know where they are supposed to be. But I'm not sure, since I probably would have believed you if we didn't know your answer was wrong.\r\n\r\nAs for the author's question, I'm not sure exactly how they got that, but here is a different calculation that gets the same result (assuming you haven't already found one):\r\n\r\nLet the $ 23$ points be OOOO...OOOO. Pick the \"left\" point first. If it is the first point, there are $ 19$ choices for the \"right\" point (just count). If it is the second, there are $ 18$. Third has $ 17$, and so on. So there are $ 19\\plus{}18\\plus{}17\\plus{}\\cdots\\plus{}1 \\equal{}\\frac{19\\cdot 20}{2}\\equal{}\\binom{20}{2}$.", "Solution_3": "I used the same method as yours initially, but then i double counted because i didn't assign \"left\" and \"right\" points. Thanks =)" } { "Tag": [ "AMC 10", "AMC" ], "Problem": "anybody know if there are other taiwan amc exams besides the 2002 one? If not, why did they only do that for one year?", "Solution_1": "Answer: There are no other contests prepared specfically for Taiwan other than the one given in January 2002. After that contest was used by students in Taiwan, we released it on the Web for practice purposes and we designated it as the 2002 AMC 10 P and 2002 AMC 12 P, P for Practice.\r\n\r\nThe reason is that when the Taiwan Secondary Education Association originally contracted to participate in the AMC contests, they did not believe that it was possible to fit the February administrations of the regular A and B dates into their academic calendar. This was partly because of their academic calendar and partly because of where the Chinese New Year fell that year. \r\n\r\nSince then, both sides have agreed that it is easier to administer the AMC 10 and 12 contests at the same time. \r\n\r\nSteve Dunbar\r\nAMC Director" } { "Tag": [ "LaTeX", "Olimpiada de matematicas" ], "Problem": "Estos son los problemas de la final de la olimpiada de Uruguay en mi nivel\r\n(no est\u00e1n ordenados por dificultad)\r\n\r\n1. Demostrar que en cualquier conjunto de 11 enteros positivos siempre es posible seleccionar 6 de sus elementos tales que su suma sea m\u00faltiplo de 6.\r\n\r\n2. Sea $\\triangle ABC$ tri\u00e1ngulo is\u00f3sceles en $A$. Llamamods $D$ al punto medio de $AC$, $E$ a la proyecci\u00f3n ortogonal de $D$ sobre $BC$ y $F$ al punto medio de $DE$. \r\nDemostrar que $BF$ y $AE$ son perpendiculares si y solo s\u00ed $\\triangle ABC$ es equil\u00e1tero.\r\n\r\n3. Encontrar todos los enteros positivos $a, b, c$ tales que $MCM(a,b,c)=a+b+c$\r\n\r\n4. Sean los polinomios $P(x)=6x^2-24x-4a$ y $Q(x)=x^3+ax^2+bx-8$\r\nEncontrar todos los valores reales de $a$ y $b$ para que $P(x)$ y $Q(x)$ tengan toas sus ra\u00edces reales no \r\nnegativas.\r\n\r\n[i]Editado por el Moderador: Le puse Latex![/i]", "Solution_1": "ok\r\n\r\n1. entre 5 enteros positivos siempre hay 3 que suman multiplo de 3. Entre 3 enteros positivos siempre hay 2 que suman multiplo de 2.\r\n\r\nCon los 11 numeros puedo armar tres grupos de tres numeros cada uno la suma de cada uno multiplo de 3 (usando la parte 1)\r\n\r\nEntre estos tres grupos puedo escoger 2 con suma multiplo de 2 (usando parte 2)\r\n\r\nEstos dos grupos tienen en total 6 numeros y su suma es multiplo de 6.\r\n\r\n3. Sea $a$ el mayor de los numeros, los 3 numeros iguales no es solucion. entonces $a$ divide a $b+c$ de donde $a=b+c$. Asi pues se obtiene que $a+b+c=2a$ sea $b$ el segundo mayor. entonces $b$ divide a $2a$ o $2(b+c)$ entonces $b$ divide a $2c$ se sigue que $b=2$c o $b=c$. Asi pues las potenciales triplas a servir son: $c,c,2c$ o $c,2c,3c$ se ve que solo la segunda sirve.", "Solution_2": "Generalizaci\u00f3n del problema 1:\r\nPara cada $n$ natural encuentra el menor n\u00famero $k$ tal que es posible garantizar que entre cualesquiera $k$ enteros, es posible escoger $n$ cuya suma es m\u00faltiplo de $n$.\r\nEse problema apareci\u00f3 en los selectivos de M\u00e9xico para la IMO o la Ibero (no recuerdo cual). Nadie lo pudo terminar. Seguro ya lo postearon por all\u00ed, pero ser\u00eda interesante intentarlo un rato ac\u00e1.", "Solution_3": "Y la respuesta es $2n-1$ este es un celebre teorema de Erdos.\r\n\r\ncalramente $2n-2$ no son suficientes, $n-1$ $1's$ y $n-1$ $0's$.\r\n\r\nla prueba es un corolario de lo siguiente: Sea $p$ un numero primo, entonces entre $2p-1$ enteros puedo seleccionar $p$ tales que su suma es multiplo de $p$.\r\n\r\nLa prueba es: sea $A=(a_1,a_2...,a_{2p+1})$ el conjunto de numeros, para un conjunto $T$ sea $S(T)$ la suma de los elementos de $T$ y sea $B$ el conjunto de subconjuntos de tama\u00f1o $p$ de $A$.\r\n\r\nConsideremos la siguiente suma, modulo $p$ \r\n\r\n$\\sum_{T\\in B} (S(T)^{p-1}-1)$\r\n\r\nsi $S(T)$ es diferente de $0$ para todo $T$ en $B$, entonces la suma es $0$ por Fermat.\r\n\r\nPero por otro lado, podemos ver esta suma tambien de la siguiente manera: cada producto asi: $a_1^{t_1}a_2^{t_2}...a_k^{t_k}$ Con $t_1+t_2+...t_k=p-1$ y $k 0$, then\r\n\r\n$ a^2(b \\plus{} rd) \\plus{} b^2(c \\plus{} ra) \\plus{} c^2(d \\plus{} rb) \\plus{} d^2(a \\plus{} rc)$\r\n\r\n$ \\leq\\frac {r^2}{2(r^2 \\minus{} r \\plus{} 1)^{\\frac {3}{2}} \\minus{} (r \\plus{} 1)(r \\minus{} 2)(2r \\minus{} 1)}(a \\plus{} b \\plus{} c \\plus{} d)^3$\r\n\r\n$ \\minus{} \\frac {37r^4 \\minus{} 32r^3 \\plus{} 118r^2 \\minus{} 32r \\plus{} 37}{32(r^2 \\minus{} r \\plus{} 1)^{\\frac {3}{2}} \\minus{} (r \\plus{} 1)(r \\minus{} 5)(5r \\minus{} 1)}(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$,\r\n\r\nwith equality if $ a \\equal{} b \\equal{} c \\equal{} d$.:yinyang: \r\n\r\n\r\nFor $ r\\to 0$, we obtain \r\n\r\n$ 27(a^2b \\plus{} b^2c \\plus{} c^2d \\plus{} d^2a)\\leq4(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} 37(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$,\r\n\r\nwith equality if $ a \\equal{} b \\equal{} 0, c \\equal{} 2d$.\r\n\r\n[hide=\"Other special cases.\"]$ r \\equal{} \\frac {8}{3}$:\n\n$ 25[a^2(3b \\plus{} 8d) \\plus{} b^2(3c \\plus{} 8a) \\plus{} c^2(3d \\plus{} 8b) \\plus{} d^2(3a \\plus{} 8c)]$\n\n$ \\leq36(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} 301(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$,\n\nwith equality if $ a \\equal{} b \\equal{} 0, 3c \\equal{} 2d$;\n\n\n$ r \\equal{} \\frac {8}{5}$:\n\n$ 243[a^2(5b \\plus{} 8d) \\plus{} b^2(5c \\plus{} 8a) \\plus{} c^2(5d \\plus{} 8b) \\plus{} d^2(5a \\plus{} 8c)]$\n\n$ \\leq 400(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} 3241(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$,\n\nwith equality if $ a \\equal{} b \\equal{} 0, 5c \\equal{} 4d$;\n\n\n$ r \\equal{} \\frac {15}{7}$:\n\n$ 432[a^2(7b \\plus{} 15d) \\plus{} b^2(7c \\plus{} 15a) \\plus{} c^2(7d \\plus{} 15b) \\plus{} d^2(7a \\plus{} 15c)]$\n\n$ \\leq 1225(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} 10096(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$,\n\nwith equality if $ a \\equal{} b \\equal{} 0, 7c \\equal{} 5d$;\n\n\n$ r \\equal{} \\frac {15}{8}$:\n\n$ 49[a^2(8b \\plus{} 15d) \\plus{} b^2(8c \\plus{} 15a) \\plus{} c^2(8d \\plus{} 15b) \\plus{} d^2(8a \\plus{} 15c)]$\n\n$ \\leq 144(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} 1177(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$,\n\nwith equality if $ a \\equal{} b \\equal{} 0, 4c \\equal{} 3d$;\n\n\n$ r \\equal{} \\frac {21}{5}$:\n\n$ 64[a^2(5b \\plus{} 21d) \\plus{} b^2(5c \\plus{} 21a) \\plus{} c^2(5d \\plus{} 21b) \\plus{} d^2(5a \\plus{} 21c)]$\n\n$ \\leq 225(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} 1936(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$,\n\nwith equality if $ a \\equal{} b \\equal{} 0, 5c \\equal{} 3d$.[/hide]", "Solution_20": "[quote=\"Ji Chen\"]$ \\frac {(a \\plus{} b \\plus{} c \\plus{} d)^3}{4} \\minus{} \\sum_{cyc}{a^{2}(b \\plus{} c)} \\minus{} 2\\sum_{cyc}{abc} \\equal{} \\sum_{cyc}{\\frac {a}{4}(a \\minus{} b \\minus{} c \\plus{} d)^2}\\geq 0$.[/quote]\r\n$ \\frac {m(a \\plus{} b \\plus{} c \\plus{} d)^3}{4} \\plus{} (1 \\minus{} 3m)\\sum_{sym}{bcd} \\minus{} \\sum_{cyc}{a^{2}(b \\plus{} mc)}$\r\n\r\n$ \\equal{} \\frac {a \\plus{} c}{4m}[(ma \\minus{} b \\minus{} mc \\plus{} d)^2 \\plus{} (m \\minus{} 1)(3m \\plus{} 1)(b \\minus{} d)^2]$\r\n\r\n$ \\plus{} \\frac {b \\plus{} d}{4m}[(a \\plus{} mb \\minus{} c \\plus{} md)^2 \\plus{} (m \\minus{} 1)(3m \\plus{} 1)(a \\minus{} c)^2]\\geq 0$\r\n\r\nfor $ m\\geq 1$.", "Solution_21": "[quote=\"Ji Chen\"][quote=\"Ji Chen\"]$ \\frac {(a \\plus{} b \\plus{} c \\plus{} d)^3}{4} \\minus{} \\sum_{cyc}{a^{2}(b \\plus{} c)} \\minus{} 2\\sum_{cyc}{abc} \\equal{} \\sum_{cyc}{\\frac {a}{4}(a \\minus{} b \\minus{} c \\plus{} d)^2}\\geq 0$.[/quote]\n$ \\frac {m(a \\plus{} b \\plus{} c \\plus{} d)^3}{4} \\plus{} (1 \\minus{} 3m)\\sum_{sym}{bcd} \\minus{} \\sum_{cyc}{a^{2}(b \\plus{} mc)}$\n\n$ \\equal{} \\frac {a \\plus{} c}{4m}[(ma \\minus{} b \\minus{} mc \\plus{} d)^2 \\plus{} (m \\minus{} 1)(3m \\plus{} 1)(b \\minus{} d)^2]$\n\n$ \\plus{} \\frac {b \\plus{} d}{4m}[(a \\plus{} mb \\minus{} c \\plus{} md)^2 \\plus{} (m \\minus{} 1)(3m \\plus{} 1)(a \\minus{} c)^2]\\geq 0$\n\nfor $ m\\geq 1$.[/quote]\r\nIt is really very very nice :o How could you find it? :o", "Solution_22": "[quote=\"hungkhtn\"]If $ a,b,c,d$ are non-negative with sum $ 4$, then\n\n$ a^{2}(b \\plus{} c) \\plus{} b^{2}(c \\plus{} d) \\plus{} c^{2}(d \\plus{} a) \\plus{} d^{2}(a \\plus{} b) \\plus{} 8abcd \\le 16$.[/quote]\n[quote=\"Vasc\"]Actually, the following stronger inequality holds\n\n$ \\sum a^{2}(b \\plus{} c) \\plus{} 2\\sum abc \\le 16$.[/quote]\r\nFor $ a,b,c,d\\geq 0$ and $ a \\plus{} b \\plus{} c \\plus{} d > 0$, the following inequality \r\n\r\n$ a^2(b \\plus{} mc) \\plus{} b^2(c \\plus{} md) \\plus{} c^2(d \\plus{} ma) \\plus{} d^2(a \\plus{} mb)$\r\n\r\n$ \\leq\\frac {m}{4}(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} n(bcd \\plus{} cda \\plus{} dab \\plus{} abc) \\minus{} \\frac {16(3m \\minus{} 1 \\minus{} n)abcd}{a \\plus{} b \\plus{} c \\plus{} d}$\r\n\r\nholds if and only if $ \\left(\\frac {16}{27}\\leq m\\leq1 \\wedge f(m,n)\\leq0\\right)\\vee(m\\geq1 \\wedge n\\leq 3m \\minus{} 1)$, \r\n\r\nwhere the irreducible polynomial \r\n\r\n$ f(m,n) \\equal{} 4mn^7 \\minus{} m(27m \\plus{} 16)n^6 \\plus{} 2(39m^3 \\plus{} 55m^2 \\plus{} 24m \\minus{} 8)n^5$\r\n\r\n$ \\minus{} (125m^4 \\plus{} 306m^3 \\plus{} 277m^2 \\minus{} 56m \\minus{} 16)n^4$\r\n\r\n$ \\plus{} 4m(30m^4 \\plus{} 111m^3 \\plus{} 155m^2 \\plus{} 9m \\minus{} 33)n^3$\r\n\r\n$ \\minus{} m(69m^5 \\plus{} 356m^4 \\plus{} 870m^3 \\plus{} 28m^2 \\minus{} 403m \\plus{} 104)n^2$\r\n\r\n$ \\plus{} 2m(11m^6 \\plus{} 75m^5 \\plus{} 318m^4 \\plus{} 278m^3 \\minus{} 777m^2 \\plus{} 415m \\minus{} 64)n$\r\n\r\n$ \\minus{} m(3m^7 \\plus{} 26m^6 \\plus{} 173m^5 \\plus{} 396m^4 \\plus{} 669m^3 \\minus{} 2086m^2 \\plus{} 1331m \\minus{} 256)$.\r\n\r\n[b]Remark[/b] $ f\\left(\\frac{16}{27},\\frac{17}{54}\\right)\\equal{}f(1,2)\\equal{}0$.\r\n\r\n[hide=\"Special case.\"]$ m \\equal{} \\frac {16}{27},n \\equal{} \\frac {17}{54}$:\n\n$ a^2(27b \\plus{} 16c) \\plus{} b^2(27c \\plus{} 16d) \\plus{} c^2(27d \\plus{} 16a) \\plus{} d^2(27a \\plus{} 16b)$\n\n$ \\leq 4(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} \\frac {17}{2}(bcd \\plus{} cda \\plus{} dab \\plus{} abc) \\minus{} \\frac {200abcd}{a \\plus{} b \\plus{} c \\plus{} d}$,\n\nwith equality if $ a \\equal{} b \\equal{} 0, c \\equal{} 2d$.[/hide]", "Solution_23": "[quote=\"Ji Chen\"][quote=\"hungkhtn\"]If $ a,b,c,d$ are non-negative and $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 3$, then\n\n$ ab(a \\plus{} 2b \\plus{} 3c) \\plus{} bc(b \\plus{} 2c \\plus{} 3d) \\plus{} cd(c \\plus{} 2d \\plus{} 3a) \\plus{} da(d \\plus{} 2a \\plus{} 3b) \\le 6\\sqrt {3}$.[/quote]Actually, the following stronger inequality holds \n\n$ a^2(b \\plus{} 2d) \\plus{} b^2(c \\plus{} 2a) \\plus{} c^2(d \\plus{} 2b) \\plus{} d^2(a \\plus{} 2c)$\n\n$ \\leq \\frac {2}{3\\sqrt {3}}(a \\plus{} b \\plus{} c \\plus{} d)^3 \\plus{} \\left(3 \\minus{} \\frac {32}{3\\sqrt {3}}\\right)(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$,\n\nwith equality if $ a \\equal{} b \\equal{} c \\equal{} d$. [/quote]\r\n$ 2(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} \\left(32 \\minus{} 9\\sqrt {3}\\right)(bcd \\plus{} cda \\plus{} dab \\plus{} abc)$\r\n\r\n$ \\minus{} 3\\sqrt {3}\\left[a^2(b \\plus{} 2d) \\plus{} b^2(c \\plus{} 2a) \\plus{} c^2(d \\plus{} 2b) \\plus{} d^2(a \\plus{} 2c)\\right]$\r\n\r\n$ \\equal{} 8a(a \\minus{} b \\plus{} c \\minus{} d)^2 \\plus{} (16 \\minus{} 9\\sqrt {3})a\\left[(a \\minus{} c)^2 \\plus{} (b \\minus{} d)^2\\right]$\r\n\r\n$ \\plus{} \\left(113 \\minus{} 65\\sqrt {3}\\right)(b \\minus{} a)(a \\minus{} d)^2$\r\n\r\n$ \\plus{} \\frac {13 \\plus{} 4\\sqrt {3}}{44}(d \\minus{} a)\\left[b \\plus{} \\left(17 \\minus{} 11\\sqrt {3}\\right)c \\plus{} 2\\left(25 \\minus{} 14\\sqrt {3}\\right)d \\plus{} \\left(39\\sqrt {3} \\minus{} 68\\right)a\\right]^2$\r\n\r\n$ \\plus{} \\left[2b \\plus{} \\left(2 \\plus{} \\sqrt {3}\\right)c \\plus{} \\left(\\frac {87}{11}\\sqrt {3} \\minus{} \\frac {453}{44}\\right)d \\minus{} \\left(\\frac {98}{11}\\sqrt {3} \\minus{} \\frac {277}{44}\\right)a\\right]\\left[b \\plus{} \\left(1 \\minus{} \\sqrt {3}\\right)c \\plus{} 2\\left(2 \\minus{} \\sqrt {3}\\right)d \\plus{} 3\\left(\\sqrt {3} \\minus{} 2\\right)a\\right]^2$\r\n\r\n$ \\geq0$.\r\n\r\n[hide=\"Remark.\"]$ 3 \\minus{} \\frac {32}{3\\sqrt {3}} \\equal{} \\minus{} 3.15\\cdots$;\n\n$ 16 \\minus{} 9\\sqrt {3} \\equal{} 0.411\\cdots$;\n\n$ 113 \\minus{} 65\\sqrt {3} \\equal{} 0.416\\cdots$;\n\n$ \\frac {87}{11}\\sqrt {3} \\minus{} \\frac {453}{44} \\equal{} 3.40\\cdots$.[/hide]", "Solution_24": "[quote=\"hungkhtn\"]Ok. This one is left for you, VASC. It is not as easy as before.\n\n\n[b]Problem. (hungkhtn dedicates to VASC)[/b]\nIf $ a,b,c,d$ are non-negative and $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 3$ then\n\\[ ab(a \\plus{} 2b \\plus{} 3c) \\plus{} bc(b \\plus{} 2c \\plus{} 3d) \\plus{} cd(c \\plus{} 2d \\plus{} 3a) \\plus{} da(d \\plus{} 2a \\plus{} 3b) \\le 6\\sqrt {3}.\n\\]\nIt must be very hard, I think.[/quote]\r\nIt is not very hard, I think. Denote $ f(a,b,c,d)\\equal{}LHS$, we have\r\n\\[ f(a,b,c,d)\\minus{}f(a\\plus{}c,b,0,d)\\equal{}c(b\\minus{}d)(a\\plus{}c\\minus{}b\\minus{}d)\\]\r\nand\r\n\\[ f(a,b,c,d)\\minus{}f(0,b,a\\plus{}c,d)\\equal{}\\minus{}a(b\\minus{}d)(a\\plus{}c\\minus{}b\\minus{}d)\\]\r\nFrom now, we have\r\n\\[ f(a,b,c,d) \\le \\max\\{f(a\\plus{}c,b,0,d),f(0,b,a\\plus{}c,d)\\}\\]\r\nSimilarly,\r\n\\[ f(a,b,c,d) \\le \\max\\{f(a\\plus{}c,b\\plus{}d,0,0),f(a\\plus{}c,0,0,b\\plus{}d),f(0,b\\plus{}d,a\\plus{}c,0),f(0,0,a\\plus{}c,b\\plus{}d)\\}\\]\r\nMoreover, it is easy to check that\r\n\\[ \\max\\{f(a\\plus{}c,b\\plus{}d,0,0),f(a\\plus{}c,0,0,b\\plus{}d),f(0,b\\plus{}d,a\\plus{}c,0),f(0,0,a\\plus{}c,b\\plus{}d)\\} \\le 6\\sqrt{3}\\]\r\nEquality holds for $ a\\equal{}d\\equal{}0,b\\equal{}3\\minus{}\\sqrt{3},c\\equal{}\\sqrt{3}.$ :)", "Solution_25": "[quote=Ji Chen]\nAnother proof. $ (a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} 4\\sum_{cyc}{a^{2}(b \\plus{} c)} \\minus{} 8\\sum_{cyc}{abc}$\n$ \\equal{} 4a[(a \\minus{} c)^2 \\plus{} (b \\minus{} d)^2] \\plus{} (c \\minus{} a)(a \\minus{} b \\minus{} c \\plus{} d)^2 \\plus{} (b \\plus{} d \\minus{} 2a)(a \\plus{} b \\minus{} c \\minus{} d)^2\\geq 0$,\nwhich is clearly true for $ a \\equal{} \\min\\{a,b,c,d\\}$. :blush: [/quote]\nShorter\n\\[(a \\plus{} b \\plus{} c \\plus{} d)^3 \\minus{} 4\\sum_{cyc}{a^{2}(b \\plus{} c)} \\minus{} 8\\sum_{cyc}{abc} = (c+a)(d+a-b-c)^2+(b+d)(a+b-c-d)^2 \\geqslant 0.\\]\n" } { "Tag": [], "Problem": "In a diagram, DE is parallel to AC. \r\nBD = 4, DA = 6\r\nand EC = 8. Find BC to the nearest tenth.", "Solution_1": "Where's $B$ supposed to be?", "Solution_2": "Assuming that B is the vertex of triangle BAC, and that D is on side Ba and E is on side BC, I get the followong answer.\r\n\r\n[hide]Using similiar triangles we have the following equation $\\frac{2}{5}=\\frac{x}{x+8}$. Solving for x we get $\\frac{8}{3}$. Then add that to 8 to get $\\frac{32}{3}$. This to the nearest tenth is $10.67$.[/hide]", "Solution_3": "If i assume that ABC is a triangle nd D is on AB nd E is on BC.Then ABC nd DBE are similar triangles.\r\nSo BD:DA=BE:EC then calculating BE=16/3 So,BC=BE+EC=40/3\r\n>tarik adnan", "Solution_4": "[quote=\"ckck\"]Assuming that B is the vertex of triangle BAC, and that D is on side Ba and E is on side BC, I get the followong answer.\n\n[hide]Using similiar triangles we have the following equation $\\frac{2}{5}=\\frac{x}{x+8}$. Solving for x we get $\\frac{8}{3}$. Then add that to 8 to get $\\frac{32}{3}$. This to the nearest tenth is $10.67$.[/hide][/quote]\r\n\r\nThat's hundredth. He said tenth.", "Solution_5": "To i_like_pie,\r\n\r\nI am not seeking angle B. The question is asking for side BC, which the other two guys answered.\r\n\r\nThanks!", "Solution_6": "If you ever post a problem that came with a diagram, be sure to attach it.", "Solution_7": "Good idea BUT HOW do I attach a diagram here?" } { "Tag": [ "trigonometry", "geometry", "AMC", "AIME", "analytic geometry", "number theory", "\\/closed" ], "Problem": "How fast would one expect to finish these two books (thoroughly and able to comprehend/grasp the most out of the book), assuming that the person is in Precalculus, and has experience with trigonometry and geometry? When I say this, I mean spending one hour solid a day.", "Solution_1": "It depends a lot on that person's ability.\r\n\r\nI think one big aspect of \"finishing\" the books is doing most if not every problem provided. Only by doing the problems will you reinforce the chapter's contents.\r\n\r\nSo, the time it takes depends on your own speed.", "Solution_2": "Definitely make sure you go all the way through it at least once, though. I stopped at Chapter 12 and haven't started back since, and it kills me on some topics..", "Solution_3": "Yeah, I really got stuck on chapter 12 specifically. I got to it.. hit a road block, so I tried it for a few days, and then moved on when I realized I still didn't get it. Analytic Geometry is rough lol. \r\n\r\nIts not about working an hour a day and trying to finish the books. You can read through the entire book in a few days, and not really gain anything. You learn more by reading through the book very slowly. If you get stuck on a concept, leave it for a few days.. move on or just let it simmer for a while in your head. Come back to the chapter a little bit later and see if it looks any better then. Worst case, you can just ask someone on the forum.\r\n\r\nAnd another thing... SOLVE THE PROBLEMS. Especially if you're not in a crunch for time, you should solve the problems. \r\n\r\nI personally would spend at least a few months finishing the books. Volume 1 should probably only take like a month and a half if you work constantly but Volume 2 should take at least a solid three months because it involves math most people would find... impossible.", "Solution_4": "Would you probably get Art and Craft of Problem Solving while you do AoPS or after?", "Solution_5": "Art and Craft is only useful if you're trying to learning how to solve proofs and AIME questions 11-15. I would only get it if you're perfectly comfortable with all but the last few chapters of Art of Problem Solving Volume 2.", "Solution_6": "[quote=\"Pakman2012\"]Art and Craft is only useful if you're trying to learning how to solve proofs and AIME questions 11-15. I would only get it if you're perfectly comfortable with all but the last few chapters of Art of Problem Solving Volume 2.[/quote]\r\n\r\nWhen you say perfectly comfortable, do you mean that you can solve all the problems quite rapidly and efficiently? :huh:", "Solution_7": "As in, you can use most of the concepts fairly efficiently.. on most of the problems. You're not gonna be able to get all of the problems, without some help from the solutions manual or what not.", "Solution_8": "Thanks :D Any other suggestions? :P", "Solution_9": "Practice tests, practice tests, practice tests.\r\n\r\nPractice problems, practice problems, practice problems.\r\n\r\nLike thats basically all of what got me to do well on AIME, along with learning quite a few important concepts in AoPS classes... specifically Intermediate Counting. Tough course... especially if you slack like me, but it paid dividends when I figured out how to solve questions I normally wouldn't be able to on AMC and AIME, largely because I'd put in the time with the transcripts. \r\n\r\nNumber theory classes on AoPS helped me quite a bit... at least it helped my confidence when approaching number theoretical problems. Even if I didn't have to use any high powered tools.. it helped knowing that I had them haha. \r\n\r\nProblem series classes might be the best way to learn for a minimalist haha. From what I heard, the AIME Problem Series was really good. AMC 12 problem series DEFINITELY raised my score by like.. 15 points purely on the merit that I'd seen more of the difficult questions. I'm not saying that would happen to everyone.. but it was well worth it.", "Solution_10": "Where should someone with an AIME score of 2 start?", "Solution_11": "[quote=\"cooljoe\"]Where should someone with an AIME score of 2 start?[/quote]\r\n\r\nSomebody scoring a 2 on the AIME probably still has a lot to learn from AoPS Volume 1 and the Introductory subject books." } { "Tag": [ "linear algebra", "matrix", "analytic geometry", "vector", "function", "algebra", "polynomial" ], "Problem": "Let $ M$ be a square matrix with exactly one $ 1$ in each column and all other entries $ 0$. Show that the only possible eigenvalues of $ M$ are $ 0$ and roots of unity.", "Solution_1": "Let's work with $ A \\equal{} M^T$, so each row has exactly one $ 1$ (they have the same eigenvalues). Then supposing $ \\alpha\\neq 0$ an equation of the form $ Ax \\equal{} \\alpha x$ , $ x\\neq 0$ can be translated to: $ \\forall i$ ,there is a $ j$ such that $ x_i \\equal{} x_j/\\alpha$, where $ i,j\\in\\{1,\\dots n\\}$. Take $ x_i\\neq 0$, denote it with $ y_0$. So we can build an infinite sequence $ y_0, y_1\\dots$ from the coordinates of $ x$ (a finite set), satisfying $ y_i \\equal{} y_{i \\plus{} 1}/\\alpha$, and they all are nonzero, for $ y_0$ is nonzero. This means there will be $ i < j$, such as $ y_i \\equal{} y_j$, which means $ \\alpha^{j \\minus{} i} \\equal{} 1$.", "Solution_2": "Let $ MX\\equal{}\\lambda X$ for some nonzero $ \\lambda$ and nonzero $ X$; Let $ X^T\\equal{}(x_1 x_2....x_n)$.\r\nLet there be $ r$ nonzero rows in $ M$. It is clear that, if the $ k^{th}$ row is zero, then, $ x_k\\equal{}0$.\r\nRemove all the zero rows from $ M$ and the corresponding zero entries in $ \\lambda X$. Now, the $ n\\minus{}r$ columns, multiplied with the $ n\\minus{}r$ zero elements of $ X$ in the product $ MX$ can also be removed. This leaves us with an $ r*r$ matrix, say, $ B$, and, the column vector $ Y$ comtaining the left over elements, say $ (x_1....x_r)$ in $ X$., satisfying $ BY\\equal{}\\lambda Y$\r\n\r\nNow, this $ B$ has the property that, every columnn has exhactly one 1, and rest all zeroes, and, every row has exhactly one 1 and rest all zeroes (that's because, there are no zero rows, and there are exhatly $ r$ 1's in the matrix). Hence, $ BY$ is a permutation of the elements, $ (x_1...x_r)$. Now, since, not all of $ x_1,...x_r$ are zero, there exists atleast one nonzero expression among: the sum of $ x_1....x_r$; the sum of products taken two at a time;...........; the product of $ x_1......x_r$. That will give us, some integer power of $ \\lambda$ equal to 1.", "Solution_3": "Let $ f$ be the linear function associated to $ M$. If $ e_i$ are the vectors of the basis then $ f(e_i)=e_j$. Considering the orbit of the $ e_i$ under the action $ e_i\\rightarrow{e_j}$ or $ i\\rightarrow{j}$, we change the order of these vectors. $ e_1=1\\rightarrow{f(e_1)=e_j=2}$ if $ j>1$, else $ =1$ and STOP. If $ j>1$ then $ 1\\rightarrow{2}\\rightarrow{f(e_j)=e_k=3}$ if $ e_k$ has not been used, else $ =$ the new value of $ e_k$ and STOP; and so on. \r\nWhen the STOP occurs we have $ 1\\rightarrow{2}\\cdots\\rightarrow{k}\\rightarrow{j}$ with $ j\\leq{k}$. \r\nWe consider one $ e_i$ that has not been used. Then $ e_i=k+1\\rightarrow\\cdots$ and when the STOP occurs we obtain:$ k+1\\rightarrow{k+2}\\cdots\\rightarrow{l}\\rightarrow{j}$ with $ j\\leq{l}$.\r\nWhen the process finishes then we obtain a matrix similar to $ M$ that is a blockmatrix in a triangular form (Darij will not be happy). The elements of the diagonal are companion matrices s.t. their last column is zero or has exactly one term equal to $ 1$ (and the other are $ 0$).\r\nThe charac polynomial of $ M$ is the product of the charac polynomials of these matrices; these polynomials are in the form $ x^k$ or $ x^k-x^q$ with $ q 1$.\r\n\r\nProof: Suppose $|\\epsilon| \\leq 1$ for $\\displaystyle q_{1}\\epsilon^{n-1}+q_{2}\\epsilon^{n-2}+...+q_{n-1}\\epsilon=-q_n$\r\n\r\n$\\displaystyle q_{1}\\epsilon^{n}+q_{2}\\epsilon^{n-1}+...+q_{n-1}\\epsilon^{2}-q_{1}\\epsilon^{n-1}-q_{2}\\epsilon^{n-2}+...-q_{n-1}\\epsilon=q_n-q_n\\epsilon$\r\n\r\n$\\displaystyle q_{1}\\epsilon^{n}+(q_{2}-q_{1})\\epsilon^{n-1}+...+(q_{n}-q_{n-1})\\epsilon=q_n$\r\n\r\n$\\displaystyle |q_n|\\leq |q_{1}\\epsilon^{n}|+|(q_{2}-q_{1})\\epsilon^{n-1}|+...+|(q_{n}-q_{n-1})\\epsilon|\\leq$\r\n\r\n$\\displaystyle |q_{1}|+|(q_{2}-q_{1})|+...+|(q_{n}-q_{n-1}|=q_n$, which holds iff $|\\epsilon|=1$ and $q_{i}=q_{j}$ - contradiction.\r\n\r\nThen $g(x)=u_{1}(x)v_{1}(x)$ and one from them WLOG $u_{1}(x)$ has roots with modulo greater than 1.\r\n\r\nBut $g(0)=u_{1}(0)v_{1}(0)=-1$ and $|u_{1}(0)|=|\\epsilon_{i_{1}}\\cdot \\epsilon_{i_{2}}\\cdot ...\\epsilon_{i_{l}}|> 1$ - contradiction." } { "Tag": [ "Divisors" ], "Problem": "How many positive integers have exactly three proper divisors, each of which is less than 50?", "Solution_1": "[hide=\"Solution\"]To have 3 proper divisors, we need 1 and two other divisors. The ways in which this can occur is if we have a number which is the product of two primes, or the square of a prime number.\n\nCase 1: There are 15 primes less than 50, and we must choose two at a time. The number of these is thus ${15 \\choose 2}=105$\n\nCase 2: There are 4 prime squares less than 50, which are the squares of 2, 3, 5, and 7. These will be our other proper divisors.\n\nTherefore, our final answer is 105+4=$109$[/hide]", "Solution_2": "Ah, this surprisingly was a problem many people got wrong last year because they forgot to consider a few cases... :roll:", "Solution_3": "Did I get it wrong? :?\r\n\r\nEdit: Okay. :)", "Solution_4": "Oh no! I was just saying for everyone doing this before looking at the spoilers! Good solution.", "Solution_5": "[hide=\"Answer\"]Since $n$ is not a proper divisor of $n$, we must have $(a+1)(b+1)...-1=3\\Rightarrow (a+1)(b+1)...=4$ for $n=p_1^ap_2^b...$. Therefore, we can have either two primes multiplied together or one perfect cube where the base is a prime. The largest perfect cube with the other divisors less than $50$ is $343$, which has $1$, $7$, $49$ as divisors, so our options are $2$, $3$, $5$, $7$. With primes, we have $1$, $a$, $b$, as divisors of $ab$, and the first three must be less than $50$. There are $15$ primes less than $50$, so we have $\\binom{15}{2}=105$, and our answer is $105+4=\\boxed{109}$.[/hide]", "Solution_6": "Actually, it's prime cubes rather than prime squares. But the prime still must be less than or equal to 7, so the answer doesn't change.", "Solution_7": "wat r proper divisors?\r\n\r\ni know tat for the 15 choose 2 thing.. for a, b being primes, the factors r 1 * ab and a * b...and also for a*a, the factors are 1 * a^2 and a * a\r\n\r\nhow is that 3 for each?\r\none has...\r\n1 ab\r\na b\r\nthe other...\r\n1 a^ 2\r\na a\r\n\r\ndo u not count the 1's? and count the a's two times?", "Solution_8": "A proper divisor is any integer that evenly divides a number integer $n$, excluding $n$ itself. For example, the proper divisors of $18$ are $1,2,3,6,9$.\r\n\r\nSo, in this case, you have a multiple $ab$ of two primes $a$ and $b$. $ab$ is not a proper divisor of $ab$, but $1,a,b$ are, so all numbers of that form will have three proper divisors. Also, for any perfect cube $m^3$, you have $1,m,m^2$ as proper divisors, so those also have three; however, you can only use prime values of $m$. If you used composite, say $m=4$, you would have $1,2,4,8,16,32$ as proper divisors of $64$. ;)", "Solution_9": "How are 4, 9, 25, and 49 counted?\r\n\r\n4 has only 1 and 2 as proper divisors \r\n\r\n9 has only 1 and 3\r\n\r\n25 has only 1 and 5\r\n\r\n49 has only 1 and 7", "Solution_10": "[quote=\"LawOfSigns\"]How are 4, 9, 25, and 49 counted?\n\n4 has only 1 and 2 as proper divisors \n\n9 has only 1 and 3\n\n25 has only 1 and 5\n\n49 has only 1 and 7[/quote]\r\nThey're not; as I pointed above, it's the cubes of primes that work (that is, 8,27,125,343). Strangely enough, the primes being cubed are still 2,3,5,7, so the answer is still the same.", "Solution_11": "EDIT: disregard", "Solution_12": "Can we change the official solution? I got really confused after reading it.", "Solution_13": "Call the positive integer $n$. \n\n$n$ must have $4$ positive integer divisors. \n\nCase 1: $n=p^3$ for some prime $p$. \nThen $p$ and $p^2$ are less than $50$. This gives $4$ ways. \n\nCase 2: $n=pq$ for primes $p$ and $q$. \nThen both primes $p$ and $q$ are less than $50$. This gives $\\binom{15}{2}=105$ ways as there are $15$ primes less than $50$. \n\n\nThus, the answer is $4+105=\\boxed{109}$. " } { "Tag": [], "Problem": "Download c\u00e1c \u0111\u1ec1 thi TOEFL c\u00e1c n\u0103m\r\n\r\n[url=http://luyenthianhvan.blogspot.com/]http://luyenthianhvan.blogspot.com/[/url]", "Solution_1": "ummm no\r\n\r\nuntill you explain what you want us to download, i stronly urge reveryone to not click on the link.", "Solution_2": "Looking at the title, I'm pretty sure it's online TOEFL Test. (I'm only guessing so do not think that I know Vietnamese)", "Solution_3": "What does a vietnamese keyboard look like? hmm....", "Solution_4": "[quote=\"Zmastr\"]What does a vietnamese keyboard look like? hmm....[/quote]\r\nIt totally likes any other standard keyboards,if somesone want to type in Vietnamese s(he) has to set up a software.for example if I type\"aa\"(after turning off this software) ,it will change to the symbol\" \u00e2\"\r\nI wondered that the title of topic was written in Vietnamese :D .It means that\" download some problems of TOEFL test in years \".I am sure that this links he post is very safe(no virus,spyware,..etc),If you care about TOEFL test you can try to visit this links.\r\nI am a beginer to learn ENglish so :\r\n[b]If anyone find any grammatical mistakes in upper sentences ,plz correct if possible :) thanks so much :D [/b]", "Solution_5": "It should have been more like this:\r\n----------------------------\r\n\r\nIt [i]looks like[/i] any other standard [i]keyboard[/i][i];[/i] if [i]someone[/i] [i]wanted[/i] to type in Vietnamese [i]s/he[/i] [i]would have[/i] to set up software. [i]For[/i] example, if I type \"aa\" (after turning off [shouldn't this be on?] this software), it will change [i]into[/i] the symbol \"\u00e2\".\r\nI wondered [i]why[/i] the title of [i]the[/i] topic was written in Vietnamese :D . [i]It means[/i] \"download some problems of TOEFL test in years\" [Not sure what you meant here, but I think you mean \"download some TOEFL test problems from past years\"]. I am sure that [i]the link[/i] he [i]posted[/i] is very safe (no [i]viruses[/i], spyware,... etc). If you care about TOEFL tests you can try to visit [i]the link[/i].\r\nI am a beginer to learn English so [There are a variety of ways to say this in a more fluent way; \"I have just begun to learn English\" is probably the best.]:\r\n[b]If anyone [i]finds[/i] any grammatical mistakes in [i]the sentences above[/i] , plz correct if possible :) thanks so much :D [/b]\r\n\r\n-----------------------------------\r\n\r\nMost of the problems are with fluency, and not with grammar. A test grader probably wouldn't have made so many corrections. There is no space before a comma, but there is one after a comma. The same goes for periods.\r\n\r\nYour English is very good, in my opinion. You are not a beginner. :) \r\n\r\nI probably made a little dialectical mistake myself somewhere in there :oops: .\r\n\r\n-----------------------------\r\n\r\nAbout the link - I clicked it and there is not really a \"download\", it's just a little quiz thingy.", "Solution_6": "600 b\u00e0i test m\u1edbi nh\u1ea5t c\u1ee7a http://luyenthianhvan.org\r\n\r\nhttp://luyenthianhvan.org/2008/10/toeic-test-vocabulary.html\r\n\r\nTr\u00ean trang n\u00e0y c\u00f3 \u0111\u1ea7y \u0111\u1ee7 c\u00e1c t\u00e0i li\u1ec7u toeic, toefl, anh v\u0103n tr\u00ecnh \u0111\u1ed9 A,B,C, \u00f4n t\u1eadp ng\u1eef ph\u00e1p, ki\u1ec3m tra tr\u1ef1c tuy\u1ebfn, v\u00e0 c\u00f2n nhi\u1ec1u, nhi\u1ec1u \u0111i\u1ec1u h\u1ea5p d\u1eabn tr\u00ean \u0111\u1ea5y n\u1eefa. C\u00e1c b\u1ea1n t\u1eeb t\u1eeb kh\u00e1m ph\u00e1 nha. Ch\u00fac c\u00e1c b\u1ea1n th\u00e0nh c\u00f4ng." } { "Tag": [ "function", "topology", "real analysis", "real analysis theorems" ], "Problem": "Definition of metric space says:\r\n\r\nA metric space is a set $ X$ together with a real valued function $ d: X \\times X \\rightarrow R$ with property ...\r\n\r\nMy question is, since $ X$ can be any set (no restriction specified), assume $ X \\equal{} (1,3)$\r\n\r\nFYI, $ (1,3)$ means all the value between 1 and 3 excluding 1 and 3.\r\n\r\nBut I found that X must be open and closed...(since empty set is open and closed)\r\n\r\nDoes that mean (1,3) is open and closed too in this metric space? I'm confused, since in usual sence, (1,3) is open and not closed.\r\n\r\nI got confused when I was working on definition of dense set. (A is dense in X if cl(A)=X. But we only need to prove X is in cl(A), since the fact that cl(A) is in X is obvious. But why is it obvious? Because X is closed(?).)", "Solution_1": "Oh, the definition of openness depends on metric d...is this where I made mistake I think?\r\n\r\nDoes this mean that metric space itself always should be open and closed?", "Solution_2": "Yes and yes. Openness and closedness of a set are not intrinsic properties -- they depend on whether you are considering the set as part of some larger metric space, and if so, what this larger metric space looks like. Every metric space is both open and closed in itself.\r\n\r\nTo the last question of the first proof, it's obvious because you're taking the closure of A [i]in X[/i], i.e., your definition of the closure is something like, \"a point x in X is in the closure of A if ....\"" } { "Tag": [ "function", "number theory", "Diophantine equation" ], "Problem": "I'm looking for a good math software that will help me do those hard problems. They say that the best are Mathematica and Maple. There are also Matlab, Mathcad, etc. What would you recommend?", "Solution_1": "\"Those hard problems?\"", "Solution_2": "Maple's Trademark is \"Command the Brillance of a thousand mathematicians. Goto for a review: [url] so I guess having a thousand mathematicians would help you do those hard problems.\r\n[url]http://www.deskeng.com/articles/03/july/cover/index.htm[/url].\r\nIf you are setting for low standards. Download the FREE Virtual TI by Rusty Wagner from [url]ticalc.org[/url]. It does not have any TI-OS-ROMS. Download a TI-89 one at [url]http://ti89games.free.fr/Html/emu.htm#2[/url]. There are some more on [url]http://electronika2.tripod.com/indice_tir.htm[/url]. Cool Fact! THe Virtual TI simulates everything on your caluclator FLASH APPS, games, everything.", "Solution_3": "In theory you have to own a calculator to use the ROMS since TI has a copyright on them. I didn't download them because I own a TI-89 and TI-83+ and I used the VTI to transfer to my computer.", "Solution_4": "I have no idea if your last two posts had anything to do with my question. They confused me more, not less.\r\n\r\nWhat type of problems are you talking about? All of those softwares are very limited in their ability to, say, prove a geometric theorem. However, they are all probably quite good about, say, calculating some sum of many many terms to a great degree of accuracy. What exactly are you expecting your program to do for you?", "Solution_5": "THese software have the capabilities to do those function problems, and those find the values of solutions to equation, and those geometric problems.\r\nI am mainly refering to \"hard\" problems as in college level classes such as Number THeory, Linear Algebra.\r\nAND the Virtual TI is just something useful for people who want to use their calculator on the computer just like \"mathematical software as in Maple\"", "Solution_6": "I'm pretty sure all those programs can do linear algebra and number theory problems without too much trouble.", "Solution_7": "Eh, what types of number theory problems are we talking about here? I mean, perhaps something like a diophantine equation (brute-forcing it), but no proofs I don't think.", "Solution_8": "Sometimes computers can prove things. Their proofs are a bit strange though. http://www.math.upenn.edu/~wilf/AeqB.pdf is a nice example of things computers can prove." } { "Tag": [ "LaTeX" ], "Problem": "If $ \\frac {a \\plus{} b}{2} \\equal{} 2\\sqrt {ab}$, where $ a > b > 0$, find a possible value of $ \\frac {a}{b}$.\r\n\r\nI managed to get $ a^2 \\minus{} 2ab \\plus{} b^2 \\equal{} 0$, but that doesn't seem very useful.\r\n\r\nWhat if we wanted to find all possible values of $ \\frac {a}{b}$?", "Solution_1": "Well now you can solve for $ a$ in terms of $ b$ or $ b$ in terms of $ a$ from that quadratic. Then, once you plug it in, you get a general expression for $ a/b$.", "Solution_2": "You needn't get $ a^2\\minus{}2ab\\plus{}b^2\\equal{}0$.\r\nWe have $ \\frac{a\\plus{}b}{2}\\equal{}2\\sqrt{ab}\\Rightarrow 4a\\minus{}4\\sqrt{ab}\\plus{}b\\equal{}3a$\r\n$ \\Rightarrow 2\\sqrt{a}\\minus{}\\sqrt{b}\\equal{}\\sqrt{3a}$ (because $ a>b>0$)\r\n$ \\Rightarrow \\sqrt{a}(2\\minus{}\\sqrt{3})\\equal{}\\sqrt{b}$\r\n$ \\Rightarrow \\frac{\\sqrt{a}}{\\sqrt{b}}\\equal{}\\frac{1}{2\\minus{}\\sqrt{3}}$\r\n$ \\Rightarrow \\frac{a}{b}\\equal{}9\\equal{}(2\\plus{}\\sqrt{3})^2\\equal{}7\\plus{}4\\sqrt{3}$\r\n :wink:[/u]", "Solution_3": "[quote=\"ThetaPi\"]If $ \\frac {a \\plus{} b}{2} \\equal{} 2\\sqrt {ab}$, where $ a > b > 0$, find a possible value of $ \\frac {a}{b}$.\n\nI managed to get $ a^2 \\minus{} 2ab \\plus{} b^2 \\equal{} 0$, but that doesn't seem very useful.\n\nWhat if we wanted to find all possible values of $ \\frac {a}{b}$?[/quote]\r\n\r\nIf you have $ a^2 \\minus{} 2ab \\plus{} b^2 \\equal{} 0$, then $ (a\\minus{}b)^2\\equal{}0 \\implies a\\minus{}b\\equal{}0 \\implies a\\equal{}b \\implies \\frac{a}{b}\\equal{}1$\r\n\r\nI don't see how you got that though, since $ \\frac {a \\plus{} b}{2} \\equal{} 2\\sqrt {ab}\\implies a\\plus{}b\\equal{}4\\sqrt{ab} \\implies a^2\\plus{}2ab\\plus{}b^2\\equal{}16ab \\implies a^2\\minus{}14ab\\plus{}b^2\\equal{}0$", "Solution_4": "Kyoshiro's answer is correct, except for the nine. Where did that nine come from? O.o", "Solution_5": "[i]now a^2-2ab+b^2=0\ndivide through by b^2\nthen let a/b=x\nyou'll arive at x^2-2x+1=0\nx=1,so a/b=1 :D [/i]", "Solution_6": "[quote=\"metafor\"]Kyoshiro's answer is correct, except for the nine. Where did that nine come from? O.o[/quote]\r\nOh! Thanks.\r\nWell, It's just an incidental idea! :lol: \r\nWhen I saw $ \\minus{}4\\sqrt{ab}$, I thought of equality: $ 4a\\minus{}4\\sqrt{ab}\\plus{}b\\equal{}(2\\sqrt{a}\\minus{}\\sqrt{b})^2$", "Solution_7": "[quote=\"21 7 15\"][i]now a^2-2ab+b^2=0\ndivide through by b^2\nthen let a/b=x\nyou'll arive at x^2-2x+1=0\nx=1,so a/b=1 :D [/i][/quote]ThetaPi incorrectly got $ a^2\\minus{}2ab\\plus{}b^2\\equal{}0$. The rest is just bogus. Kyoshiro's answer is correct.", "Solution_8": "But then thetapi's equation implies a = b because if it were then a + b /2 = sqrt{ab} by AM-GM not 2sqrt{ab}\r\nSorry im feeling 2 lazy 4 LaTeX\r\nEDIT: Didnt read 1000th users comment", "Solution_9": "if 7+4sqr rt(3) is one solution\r\n\r\ndoes 7-4sqr rt(3) work as well?\r\nbut a and b have to be imaginary #s", "Solution_10": "[quote=\"lgd0612\"]if 7+4sqr rt(3) is one solution\n\ndoes 7-4sqr rt(3) work as well?\nbut a and b have to be imaginary #s[/quote]Hmm $ 7\\minus{}4\\sqrt3$ is not a possible value because it is less than 1 because we have the condition $ a>b>0\\Rightarrow\\frac{a}{b}>1$.", "Solution_11": "ah i was careless\r\n\r\ni thought 7-4sqr rt(3) was going to be a negitive #\r\n :oops:", "Solution_12": "Yepp, I did some algebra error then. What's wrong with me lol.", "Solution_13": "[hide=\"Another solution\"]Let $ \\frac{a}{b}\\equal{}k\\ (k>1)$\n$ \\Rightarrow a\\equal{}kb \\Rightarrow b\\plus{}kb\\equal{}4b\\sqrt{k}$\n$ \\Rightarrow k\\minus{}4\\sqrt{k}\\plus{}1\\equal{}0 \\Rightarrow k\\equal{}7\\plus{}4\\sqrt{3}$[/hide]" } { "Tag": [ "geometry", "ratio", "angle bisector" ], "Problem": "If AD and BF are bisectors,AB=4 ,BC=3 and CA=2 then find $ S_{AFD}: S_{ABC}$.", "Solution_1": "hello, i have got\r\n$ \\frac{S_{AFD}}{S_{ABC}}\\equal{}\\frac{4}{21}$\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, i have got\n$ \\frac {S_{AFD}}{S_{ABC}} \\equal{} \\frac {4}{21}$\nSonnhard.[/quote]\r\nYes,that's the answer.How did you got it?", "Solution_3": "Can you post your solution? :blush:", "Solution_4": "If I have provided an outline to the solution because the solution itself is somewhat long.\r\n\r\n[hide=\"Outline\"]\nNote: $ [ABC]$ denotes the area of triangle $ \\triangle ABC$.\n\nLet the intersection of the cevians be $ E$. Let $ a \\equal{} [CFD]$, $ b \\equal{} [EFD]$, $ c \\equal{} [AFE]$, $ d \\equal{} [BED]$, $ e \\equal{} [AEB]$. \n\n1. Find the ratios $ AF: FC$ and $ BD: DC$ from the Angle Bisector Theorem. (If the given cevians aren't angle bisectors, then I assume they are medians, in which case the two ratios are just $ 1: 1$.)\n\n2. Use the fact that the ratio of the areas of two triangles that share the same height is equal to the ratio of the corresponding bases of the triangles. (Hint: Take a look at $ \\triangle ADB$ and $ \\triangle ADC$ considering the bases to lie along $ \\overline{BC}$. There is only more pair of such triangles, see if you can find them.)\n\n3. Compute $ [ABC]$ using Heron's Formula.\n\n4. Use the ratios from Step 2 to come up with 4 equations. The fifth equation will be that the sum of the areas of all the triangles mentioned before is whatever you got in Step 3.\n\n5. Finally, find $ b \\plus{} c$ and divide that by the number you got in Step 3 to get the answer.\n[/hide]\r\n\r\nIf you still need help, let me know and I'll add some more detail." } { "Tag": [ "LaTeX", "function" ], "Problem": "How do you make the $ \\$$ signs for the wiki? I have a wiki that I would like to use the $ \\$$ signs instead of the tags. Thanks.", "Solution_1": "I don't think there is a way. When the administrators enabled dollar signs in addition to tags, the $ \\$$ thing got messed up.", "Solution_2": "dollarsign", "Solution_3": "[quote=\"#H34N1\"]How do you make the $ \\$$ signs for the wiki? I have a wiki that I would like to use the $ \\$$ signs instead of the tags. Thanks.[/quote]\r\n\r\n@Archimedes1, solafidefarms, I think he's asking about how to let $ \\$$ turn into tags on a different wiki, not how to type $ \\$$ signs on this wiki. That would probably require some programming. I imagine it's possible to just do it by adding some JavaScript to the personal [i]monobook.js[/i] file. For example, add an onSubmit() or onClick() handler to the \"Save page\" button on the wiki, or add a separate tab that would convert all $ \\$$ into tags (see Wikipedia > User:Omegatron/monobook.js/addlink.js). Then if there are an even number of $ \\$$ (using regular expressions or something), convert alternatively each to or ; otherwise alert an error. \r\n\r\nOr if you have \"special\" abilities on the wiki (like administrator), I guess just ask our admins how they managed to do it (I'm assuming Valentin did it?)", "Solution_4": "The client-side version would not be a good solution, as any subsequent edits would have the tags instead of the dollar signs.\r\n\r\nA simple thing to do would be to interpret all the dollar signs as math tags with a pre-save-in-db function in the MediaWiki file that recieves the editing form." } { "Tag": [ "algebra", "polynomial", "Support", "function", "real analysis", "real analysis unsolved" ], "Problem": "Given $A,B \\subset \\mathbb{R}^{n}$, define the Minkowski sum $A+B= \\{a+b \\mid a \\in A, b \\in B \\}$. Suppose that $A$ is compact and $B$ is a closed ball of radius $r$. \r\n(1) Prove that if $A$ is convex, then the volume of $A+B$ is a polynomial in $r$. \r\n(2) Is the converse to (1) true?", "Solution_1": "it has already been posted but received no answer : http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial&t=90537\r\nAny new ideas ?", "Solution_2": "Your idea of approximation with polytopes works (at least for (1)), but I hope that there is an approximation-free approach, maybe using the support function $h_{A}(x)=\\sup\\{\\langle x,a\\rangle \\colon a\\in A\\}$. This function determines a convex body, and it converts Minkowski sums into ordinary sums. Is there a formula for volume of $A$ in terms of $h_{A}$?", "Solution_3": "i posted a generalized version here:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial&t=13874\r\nand you can find a complete solution :\r\nRolf Schneider,convex bodies ,The Brunn Minkowsky theory,cambridge university press (1993)." } { "Tag": [ "calculus", "derivative", "geometry", "trapezoid", "3D geometry", "sphere", "function" ], "Problem": "Hi guys, I have been practicing on some maximization-minimization problems and I have been stuck on solving those problems in which only variables are introduced (i.e.: How can you use the local extreme theorems when there are only variables? Here are a few questions that have boggled me: \r\n\r\n1. Find the dimensions of the trapezoid of greatest area that can be inscribed in a semicircle of radius r. (I can't figure out the general position of the trapezoid so that I can assign variables, how do I work with unknown r?). \r\n\r\n2. Find the volume of the largest right circular cylinder that can be inscribed in a sphere of radius r. \r\n\r\n3. A solid is formed by adjoining a hemisphere to each end of a right circular cylinder. The total volume of the figure is 12 cubic inches. Find the radius of the cylinder that produces the minimum surface area.\r\n\r\nCan someone please explain all the steps needed in these problems, especially the first two? Any help is GREATLY APPRECIATED. Thanks.", "Solution_1": "[quote=\"keta\"]Hi guys, I have been practicing on some maximization-minimization problems and I have been stuck on solving those problems in which only variables are introduced (i.e.: How can you use the local extreme theorems when there are only variables? Here are a few questions that have boggled me: \n\n1. Find the dimensions of the trapezoid of greatest area that can be inscribed in a semicircle of radius r. (I can't figure out the general position of the trapezoid so that I can assign variables, how do I work with unknown r?). \n\n2. Find the volume of the largest right circular cylinder that can be inscribed in a sphere of radius r. \n\n3. A solid is formed by adjoining a hemisphere to each end of a right circular cylinder. The total volume of the figure is 12 cubic inches. Find the radius of the cylinder that produces the minimum surface area.\n\nCan someone please explain all the steps needed in these problems, especially the first two? Any help is GREATLY APPRECIATED. Thanks.[/quote]\r\n\r\nThe hard part of most differentiation problems lies in finding the quantity to be minimized or maximized.\r\n\r\n1. A point (rcost,rsint) 'generates' a trapezoid with larger base 2r and smaller base 2rcost and height rsint.\r\n\r\nThus A(t) = 1/2(2r+2rcost)*(rsint)\r\n\r\nThe rest is trivial.\r\n\r\n2. A sphere is identical with respect to all dimensions. Thus let (rcost,rsint) be an arbitrary point on the sphere and see what the volume of such a CIRCULAR cylinder will be. Now minimize that.\r\n\r\n3. You have a quantity in two variables and a constraint. This constraint gives you either height at a function of radius or vice versa. Now you can reduce to one variable." } { "Tag": [ "calculus", "integration", "special factorizations" ], "Problem": "Find the largest possible integer $ n$ such that $ n^2 \\plus{} 2008n$ is a perfect square.\r\n\r\nUgh Im so used to there being no solution that I never actually learned how to solve it if there is one.\r\n\r\nSo by completing the square you get:\r\n$ (n \\plus{} 1004)^2 \\equal{} q^2 \\plus{} 1004^2$\r\n$ n\\equal{}\\sqrt{q^2\\plus{}1004^2} \\minus{}1004$\r\nand Im pretty sure you can ignore the 1004, so basically what your trying to find is the largest value of $ q$ for which $ \\sqrt{q^2\\plus{}1004^2}$ is an integer. How would you do that?", "Solution_1": "$ q^2 \\plus{} 1004^2 \\equal{} k^2$\r\n$ (k\\plus{}q)(k\\minus{}q)\\equal{}1004^2$\r\n$ k\\minus{}q\\equal{}2, k\\plus{}q\\equal{}502\\cdot 1004 \\Rightarrow q\\equal{}502^2\\minus{}1$", "Solution_2": "[hide=\"What I got\"]I'm getting $ 251$:\n\nLet $ n^2+2008n=q^2$. We have the following:\n\\begin{eqnarray*} n^2+2008n & = & q^2 \\\\\n2008n & = & q^2-n^2 \\\\\n2008n & = & (q+n)(q-n) \\\\\n2^3 \\cdot 251 \\cdot n & = & (q+n)(q-n) \\end{eqnarray*}\nSo we have two (integral) factors of $ 2008n$ that differ by $ 2n$. The possibilities are the following (and what $ n$ would be if the two factors differed by $ 2n$, and there are two possibilities for each because either one could be bigger):\n\\[ 1 \\times 2008n \\longrightarrow n=\\frac{1}{2006} \\text{ or } \\frac{1}{2010} \\\\ \\\\\n2 \\times 1004n \\longrightarrow n=\\frac{1}{501} \\text{ or } \\frac{1}{503} \\\\ \\\\\n4 \\times 502n \\longrightarrow n=\\frac{1}{125} \\text{ or } \\frac{1}{126} \\\\ \\\\\n8 \\times 251n \\longrightarrow n=\\frac{8}{249} \\text{ or } \\frac{8}{253} \\\\ \\\\\n251 \\times 8n \\longrightarrow n=\\frac{251}{10} \\text{ or } \\frac{251}{6} \\\\ \\\\\n502 \\times 4n \\longrightarrow n=251 \\text{ or } \\frac{251}{3} \\\\ \\\\\n1004 \\times 2n \\longrightarrow n=251 \\\\ \\\\\n2008 \\times n \\longrightarrow n=-2008 \\text{ or } \\frac{2008}{3}\\]\nWe see that the largest integral value of $ n$ (out of $ 251$ and $ -2008$) is $ \\boxed{251}$. If we plug that into the original expression, we get $ 251^2+2008 \\cdot 251 = 567009=753^2$.\n\nI probably did something wrong, though[/hide]", "Solution_3": "[quote=\"PowerOfPi\"]\nSo we have two (integral) factors of $ 2008n$ that differ by $ 2n$. The possibilities are the following (and what $ n$ would be if the two factors differed by $ 2n$, and there are two possibilities for each because either one could be bigger):\n[/quote]\r\n\r\nI dont understand this part and how you got the list of values of n. :oops:", "Solution_4": "[quote=\"athunder\"][hide=\"Solution #1\"] Therefore, $ z^2 \\plus{} 1004^2$=$ (n \\plus{} 1004)^2$\n\nSo basically we have a number $ z$ when added to $ 1004^2$ becomes a perfect square.[/quote]\nI think you lost a \"squared\" between these two steps...\n\n[quote=\"Thunder365\"]I dont understand this part and how you got the list of values of n. :oops:[/quote]\n[hide=\"Another explanation\"]\nThink of it like this problem, where $ a$ and $ b$ are positive integers:\n$ a^2\\equal{}24\\plus{}b^2$\nWhere we get $ (a\\minus{}b)(a\\plus{}b)\\equal{}24$. We find factors of $ 24$ that have a product of $ 24$, and set the two expressions equal to the factors. Of course, this takes some case work, but we can find all the possible pairs $ (a,b)$. \n\nNow back to the problem, which is essentially the same as above, except that we have a variable on the RHS.\n\nWe had two expressions, $ k\\plus{}n$ and $ k\\minus{}n$, that had a product of $ 2008n$. We know $ k$ is an integer because its square is a perfect square. We know $ n$ is an integer because we are given that. Thus, the two expressions have integer values and a product of $ 2008n$. The difference between the expressions is $ 2n$.\n\nThe above paragraph just concludes stuff about the two expressions.\n\nNow, we find the factors of $ 2008n$ (and treat $ n$ as a number, not a variable), so we can set them equal to the two expressions. These pairs of factors are what I listed before the arrows, in the list.\n\n(Note that these things I am about to say were not said in the post, but this is how I got the list.) Now, since the expressions differ by $ 2n$, we can solve for $ n$. For example, for the first case, I let $ 1 \\pm 2n\\equal{}2008n$ and solved, getting two values of $ n$, which I listed beside the factors. Note that for the second last case, one value was omitted because you get $ 1004\\equal{}0$, so there was no solution for that case.\n\nI hope that explains things.\n[/hide][/hide]" } { "Tag": [ "function", "inequalities" ], "Problem": "solve in IR:\r\n\r\n$\\displaystyle \\sqrt{\\frac{x-1977}{23}}+\\sqrt{\\frac{x-1978}{22}}+\\sqrt{\\frac{x-1979}{21}}=\\sqrt{\\frac{x-23}{1977}}+\\sqrt{\\frac{x-22}{1978}}+\\sqrt{\\frac{x-21}{1979}}$", "Solution_1": "I don't have a clean solution started yet but I think it's obvious that one such x is [hide=\"answer\"]2000[/hide]", "Solution_2": "And if my preliminary work is correct, there is a way to show that it is the only one. Show that this is the only intersection of the two functions (LHS and RHS).", "Solution_3": "[color=blue]Substract 3 from each side and put them in factor, I get the solution of this equation, but i am not sure whether it has other solutions or not. However I can get one: 2000. :blush: :D [/color]", "Solution_4": "Set $y=2000+x$ and $g(y,t)=\\sqrt{1+x/t}$. It's clear that we need to show that $x=0$ is the only solution to\r\n\r\n$g(y,21)+g(y,22)+g(y,23)=g(y,1977)+g(y,1978)+g(y,1979)$\r\n\r\nIt is not difficult to show that $g(y,21)>g(y,1978)$ for $x>0$ and the inequality is reversed for $x<0$. The answer follows easily from this." } { "Tag": [], "Problem": "The class has 32 pupils , 7 of them play football , 16 of them play volleyball , 16 of them play basketball . 4 of them play football and volleyball , 6 of them play football and basketball , 3 of play volleyball and basketball , 1 pupil plays the three sports . How many pupils in that class don't play any sport?", "Solution_1": "Let $ n$ be the pupils that don't play any sport. By the Principle of Inclusion-Exclusion, we have\r\n\r\n$ 7 \\plus{} 16 \\plus{} 16 \\minus{} 4 \\minus{} 6 \\minus{} 3 \\plus{} 1 \\plus{} n \\equal{} 32 \\implies 27 \\plus{} n \\equal{} 32 \\implies n \\equal{} \\boxed{5}$.", "Solution_2": "[quote=\"Elio (n)\"]The class has 32 pupils , 7 of them play football , 16 of them play volleyball , 16 of them play basketball . 4 of them play football and volleyball , 6 of them play football and basketball , 3 of play volleyball and basketball , 1 pupil plays the three sports . How many pupils in that class don't play any sport?[/quote]\r\n\r\nI think there is something wrong with the problem,because we can find the number of those who only play football is -2", "Solution_3": "[quote=\"sunehra\"]Let $ n$ be the pupils that don't play any sport. By the Principle of Inclusion-Exclusion, we have\n\n$ 7 \\plus{} 16 \\plus{} 16 \\minus{} 4 \\minus{} 6 \\minus{} 3 \\plus{} 1 \\plus{} n \\equal{} 32 \\implies 27 \\plus{} n \\equal{} 32 \\implies n \\equal{} \\boxed{5}$.[/quote]\r\nI think we have:\r\n$ 7\\plus{}16\\plus{}16\\minus{}4\\minus{}6\\minus{}3\\minus{}1\\plus{}n \\equal{} 32\\implies 27\\plus{}n \\equal{} 32\\implies n \\equal{}\\boxed{7}$", "Solution_4": "cgcgbcbc is right. Problem's impossible." } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "Can somebody show me how to solve a continued radical.\r\n\r\nFor example, [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/870d2aa1f3a203f1eac62458c5e2af27.gif[/img]", "Solution_1": "Hi Derek,\r\n\r\n[quote=\"Derek\"]Can somebody show me how to solve a continued radical.\n\nFor example, [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/870d2aa1f3a203f1eac62458c5e2af27.gif[/img][/quote]\r\n[hide]Let $S=\\sqrt{3+\\sqrt{3+\\sqrt{3+\\dotsb}}}$\n\nThen $S^2-3=S,$\nand using the quadratic formula, we can find the exact value of S.[/hide]", "Solution_2": "[quote=\"Derek\"]Can somebody show me how to solve a continued radical.\n\nFor example, [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/870d2aa1f3a203f1eac62458c5e2af27.gif[/img][/quote]\r\n\r\nRight.\r\n\r\nSet that equation to S, and you have $S = \\sqrt{3 + S}$, which is how Mathew Got his equation.", "Solution_3": "Oh ok, I see now.\r\n\r\nThank you for your help.", "Solution_4": "[hide]so the asnwer would be \\displaystyle $S=\\frac{1+\\sqrt{13}}{2}=2.302775638...$[/hide]", "Solution_5": "[quote=\"jli\"][hide]so the asnwer would be 2.302775638[/hide][/quote]The answer is given by the positive solution to $S^2-3=S,$ as given above. This number is $S=\\frac{1+\\sqrt{13}}{2a}$, which is not rational because 13 is not a perfect square. The number you gave is rational. Therefore your answer isn't right." } { "Tag": [], "Problem": "In maple i have defined a string:\r\nA:=\"gfhgfnsdklqsemhpjsdmoqezhjfjjfdjgfdjkhqhqsdy\"\r\n\r\nI want to see wich words of 3 consecutive letters apears more than one time in the string.\r\nFor example the first word is \"gfh\", the second \"fhg\" ...\r\n\r\nWhat i tried to do is defining a row with those 3-characters string :\r\nT:=n->[substring(A,n..n+2)]:\r\n\r\nBut if i do T[1]; I onle get T_1 so i think that is not correct. \r\n\r\nplease ask questions if I am not clear.\r\n\r\nThanks in advance", "Solution_1": "Hello Amazigh!\r\n\r\nTry T(1) instead of T[1]...\r\n\r\nJust in case, the answer to your original question is \"fdj\". Try the following Maple input:\r\n\r\nA:=\"gfhgfnsdklqsemhpjsdmoqezhjfjjfdjgfdjkhqhqsdy\";\r\nListTools[FindRepetitions]([substring(A,i..i+2)$i=1..length(A)-2]);\r\n\r\nHave a nice day!\r\nSergey Markelov\r\nmarkelov@mccme.ru" } { "Tag": [ "linear algebra", "matrix", "trigonometry" ], "Problem": "A and B are two M2(R) matrices. we know $A^{2}=B^{2}=I_{2}$ and $AB \\neq BA$. is this enough to show that $(BA)^{n}\\neq I_{2}$ , no matter what $n\\in\\mathbb{N}$", "Solution_1": "The answer is \"No\". Just take $A=\\left[\\begin{matrix}a&1-a^{2}\\\\1&-a\\end{matrix}\\right]$ and $B=\\left[\\begin{matrix}b&1-b^{2}\\\\1&-b\\end{matrix}\\right]$. Then the element in the left upper corner of $AB$ is $ab+(1-a^{2})$, which is different from the corresponding element in $BA$ if $a^{2}\\ne b^{2}$. Now, the trace of $BA$ is $2-(a-b)^{2}$, which can be made any number less than $2$ you wish. In particular, you can make it $2\\cos \\frac{2\\pi}n$ if $n\\ge 3$, which will result in the matrix $BA$ being similar to $\\left[\\begin{matrix}e^{2\\pi i/n}&0\\\\0&e^{-2\\pi i/n}\\end{matrix}\\right]$, so $(BA)^{n}=I_{2}$. Of course, $(BA)^{1}\\ne I_{2}$ because otherwise $B$ and $A$ would commute. I leave the remaining question of whether one can achieve $(BA)^{2}=I_{2}$ to somebody else ;)." } { "Tag": [ "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "Is there another way to solve this limit without using L'hopital's Rule????\r\n\r\nI've tried everything I can think of to manipulate this into function in order to be able to plug in the limit without getting something indeterminate. \r\n\r\nLim x->pi/4 (1-Tanx)/(Sinx-Cosx)\r\n\r\nPlease help....... :(", "Solution_1": "How about some algebra?\r\n\r\n$\\frac{1-\\tan x}{\\sin x-\\cos x}=\\frac1{\\cos x}\\cdot\\frac{\\cos x-\\sin x}{\\sin x-\\cos x}=-\\frac1{\\cos x}$.", "Solution_2": "Thanks!!!\r\n\r\nFunny, I got all the way to (1/cos)(cos-sin)/(sin-cos) but just didn't see that (cos-sin)/(sin-cos) was just -1" } { "Tag": [ "geometry", "projective geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "S={1;...15}.Assume that there are exist n subset A_1...A_n satisfying:\r\n$|A_i|=7$\r\n$|A_i\\cap A_j|\\leq\\ 3$\r\nFor every subset M of S;|M|=3.there are exist A_k such that $M\\in A_k$\r\nFind the least value of n", "Solution_1": "This is the number of covering designs C(15,7,3)=15 which can be obtained as a projective geometry of dimension 3 over GF(2).\r\nSee Section 3 from the paper [url=http://www.ccrwest.org/cover/cover.pdf]\"New Constructions for Covering Designs\"[/url] at [url=http://www.ccrwest.org/cover.html]La Jolla Covering Repository[/url].", "Solution_2": "Do you have solution for this particular case?My english is so poor,i can't understand this paper", "Solution_3": "[quote=\"Bluesea\"]Do you have solution for this particular case?My english is so poor,i can't understand this paper[/quote]\r\n\r\n[url=http://www.ccrwest.org/cover/t_pages/t3/k7/C_15_7_3.html.gz]This is it[/url]." } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Compute the limits: a) $ \\int_{-1}^{1} \\frac{1}{(e^{x}+1)(x^{2}+1)} dx.$\r\n\r\nb) $\\int_{-a}^{a} \\frac{1}{(x^{4}+1)(e^{x}+1)} dx,$ with $a>0$ and $a \\neq 1.$", "Solution_1": "Change variable $y=-x$ gives $\\int_{-1}^{1}\\frac{e^x}{(e^x+1)(x^2+1)}dx$\r\n\r\nTwice of the first integral gives $2\\int_{0}^{1}\\frac{dx}{x^2+1}$\r\n\r\nFor the second integral sme technic", "Solution_2": "The same trick works for the following problem, which was a competition problem of USSR in 1977.\r\n\r\nProve that the value of $\\int_0^\\infty \\frac{dx}{(x^2+1)(x^\\alpha+1)}$ is independent of $\\alpha$", "Solution_3": "[quote=\"liyi\"]The same trick works for the following problem, which was a competition problem of USSR in 1977.\n\nProve that the value of $\\int_0^\\infty \\frac{dx}{(x^2+1)(x^\\alpha+1)}$ is independent of $\\alpha$[/quote]\r\n\r\nFor this one the change variable is $x=1/y$", "Solution_4": "perhaps the handbook of Alan Jeffery helps", "Solution_5": "Don't need this book the change variable x=1/y \r\n\r\n$\\int_{0}^{+\\infty}\\frac{y^\\alpha}{(y^2+1)(y^\\alpha +1)}dy$\r\n\r\nTwice of the given integral is $\\int_{0}^{+\\infty}\\frac{1}{(y^2+1)}dy$", "Solution_6": "[quote=\"Moubinool\"][quote=\"liyi\"]The same trick works for the following problem, which was a competition problem of USSR in 1977.\n\nProve that the value of $\\int_0^\\infty \\frac{dx}{(x^2+1)(x^\\alpha+1)}$ is independent of $\\alpha$[/quote]\n\nFor this one the change variable is $x=1/y$[/quote]\r\nYes, you are right. It was my mistake... :blush: I didn't read your solution to the previous two integrals carefully..." } { "Tag": [ "geometry", "3D geometry", "sphere", "percent", "ratio", "search" ], "Problem": "An infinite number of congruent spheres are packed in space hexagonally*. To the nearest tenth, what percent of the space is in the interior of one of these spheres?\r\n\r\n\r\n*say you have 4 balls, you can stack them into a pyramid. Pack the rest of the space similiarly.", "Solution_1": "The ratio you are talking about is $\\frac{\\pi}{\\sqrt{18}} = 0.74048\\ldots$. Calculating the ratio is easy. Getting bounds on this ratio is hard. In fact, it is interesting to note that while it seems that this is the best possible packing of spheres for $3$ dimensions, its proof still remains as an open problem.", "Solution_2": "Are you sure it's still open?\r\n\r\nThis problem is Kepler's conjecture, and as far as I know, Hales has proved it. \r\nBut, yes, I think that no \"formal\" proof has been given yet. (Hales used computers.)", "Solution_3": "Hmmm.. A search on [url=http://en.wikipedia.org/wiki/Kepler%27s_conjecture]wikipedia[/url] reveals that indeed the problem was proven. :blush: \r\n\r\nUnfortunately, with 250 pages and 3 gigabytes of programs, data and results, it's kinda hard to accept it as a proof. In my opinion, Hales' proof is more of a verification, an example of an enormous amount of brutal whacking on a seemingly 'innocent' question.\r\n\r\nThis problem can actually be generalised to $n$ dimensions, where we want to find the maximum 'density' of spheres that can be packed in an $n$ dimensional space. The last I read, the exact maximum bound is an open problem for all dimensions $n \\geq 3$ (definitely not up-to-date). If we don't even have a 'nice' proof for $3$ dimensions, then the proof for higher dimensions seem hopeless without depending on linear programming methods. :( :(", "Solution_4": "I have read (somewhere...) that 10,11 and 13 dimensions seem to be vaguely special cases in that the densest (known)\r\npacking for most dimensions are very similar (cross-sections of each other) except these dimensions- though how much of this\r\nproof and how much computer I dont know", "Solution_5": "The so-called special cases are dimensions $8$ and $24$. They are known as the $E_8$ lattice and the Leech lattice respectively, and they are highly symmetric in their respective dimensions." } { "Tag": [ "probability", "function", "algorithm" ], "Problem": "In my opinion i think we can find a formula for predicting chances and randomness it is impossible, what do you think ?\r\n\r\nimpossible? why ?\r\npossible? how ?", "Solution_1": "The questions of whether randomness exists in the real world and whether it exists in an abstract mathematical sense are not the same. The first question is debatable (and in fact it is still hotly debated, especially by philosophers and physicists). However, I don't see how you could say that randomness and probability theory don't exist in an abstract mathematical sense. The theory is there.", "Solution_2": "Also, that theory of randomness has a great many practical uses. Random models describe the motion of large numbers of particles very well- that's the core of thermodynamics. The motion of any particular particle is determined by its initial state and its collisions, but there's no way to get enough of that information to build a useful nonrandom model of a container of gas.\r\n\r\nRandom models extract useful information out of extreme complexity. It doesn't really matter whether the process being modeled is random or not.", "Solution_3": "It's an open question in computational theory whether pseudorandom generators (deterministic functions whose output is \"random\" in the sense that there is no efficient algorithm that can predict the outcomes more than epsilonically better than random guessing) exist. The notion of randomness that they are trying to mimic, of course, is of a function whose outputs are independent and uniformly distributed. Calling something random in this sense is [i]equivalent[/i] to saying that you can't predict it.", "Solution_4": "i disagree, u lot are putting restrictions, we have to be far-sighted. you think you know everything.in my opinion humans brains are too small+ feeble for finding that complexity in randomness if we think negative, i say its possible by time.\r\nthere is a definite formula for any random thing. no formula is complete. if you think a formula is complete then we can add something else to it like +1 or ^2 or do many things which would be true for another thing.we cant stop doing this can we ? that formula wouldnt be wrong.\r\n\"useful information out of extreme complexity\" how can you judge that easy and be so sure that its impossible, without being able to know that complexity completly.\r\n\r\nsorry ppl dont get it bad , i am just saying my opinion.", "Solution_5": "Statistical mechanics and thermodynamics work. Trying to model a volume of air as about $10^{24}$ particles interacting through various forces... doesn't, because we can't measure that much data, store that much data, or run computations of that size, much less solve the equations. I'm not talking about human limits here- I'm talking about computer limits.\r\n\r\nAlso, embracing random models is not \"thinking negative\". Random models do have a great deal of order and predictability to them, often much more than deterministic but chaotic models." } { "Tag": [ "geometry", "trapezoid", "geometry proposed" ], "Problem": "[b][size=100][color=DarkBlue]A triangle $ \\bigtriangleup ABC$ is given with circucircle $ (O)$ and let $ (K)$ be, an arbitrary circle taken as cord its side-segment $ BC.$ Let be the point $ D\\equiv BC\\cap AK,$ where $ K$ is the center of $ (K)$ and we denote as $ E,\\ F,$ the points of intersection of $ (K),$ from the circles $ (O_{1}),\\ (O_{2})$ with diameters $ BD,\\ CD,$ respectively. Prove that $ PQ\\parallel BC,$ where $ P\\equiv AB\\cap DE$ and $ Q\\equiv AC\\cap DF.$[/color][/size][/b]\r\n\r\nKostas Vittas.", "Solution_1": "[quote=\"vittasko\"][b][size=100][color=DarkBlue]A triangle $ \\bigtriangleup ABC$ is given with circucircle $ (O)$ and let $ (K)$ be, an arbitrary circle taken as cord its side-segment $ BC.$ Let be the point $ D\\equiv BC\\cap AK,$ where $ K$ is the center of $ (K)$ and we denote as $ E,\\ F,$ the points of intersection of $ (K),$ from the circles $ (O_{1}),\\ (O_{2})$ with diameters $ BD,\\ CD,$ respectively. Prove that $ PQ\\parallel BC,$ where $ P\\equiv AB\\cap DE$ and $ Q\\equiv AC\\cap DF.$[/color][/size][/b]\n\nKostas Vittas.[/quote]\r\nLet $ J\\equal{}DP\\cap AO_1, R\\equal{}DQ\\cap AO_2$\r\nBecause $ KO_1\\perp BE$ and $ \\angle BED\\equal{}90^o$ then $ KO_1//DE$, similarly $ KO_2//DF$\r\nUsing Thales's theorem we have $ \\frac{AJ}{AO_1}\\equal{}\\frac{AD}{AK}\\equal{}\\frac{AR}{AO_2}$\r\nThen $ JR//O_1O_2$\r\nApplying Menelaus's theorem for triangles $ JO_1D$ and $ RDO_2$ we get $ \\frac{AJ}{AO_1}.\\frac{BO_1}{BD}.\\frac{PD}{PJ}\\equal{}\\frac{AR}{AO_2}.\\frac{CO_2}{CD}.\\frac{QR}{QD}$\r\n$ \\Rightarrow \\frac{PD}{PJ}\\equal{}\\frac{QR}{QD}\\Rightarrow PQ//JR\\rightarrow$ QED", "Solution_2": "Since $ BE$ and $ CF$ are radical axes of $ (K),(O_1)$ and $ (K),(O_2),$ then $ KO_1$ and $ KO_2$ are respectively parallel to $ DP$ and $ DQ$ and go through the midpoints $ M,N$ of $ PB$ and $ QC.$ Hence, by Menelaus' theorem we obtain: \n\n$ \\frac {KD}{KA} \\equal{} \\frac {BM}{MA} \\equal{} \\frac {CN}{NA} \\ \\Longrightarrow \\ MN \\parallel BC$ \n\nTherefore, $ MN$ becomes the midline of the trapezoid $ PQCB$ $\\Longrightarrow$ $ PQ \\parallel BC$" } { "Tag": [ "MATHCOUNTS", "AMC", "AIME", "percent", "AMC 8" ], "Problem": "I started this thread because i tend to make a LOT of careless mistakes, especially on sprints and other tests where speed is an issue (well it is for me so don't brag). I just wanted to know if there are any tips people can give me to improve my accuracy. I've tried almost everything and i'm getting desperate :(", "Solution_1": "I used to be just like you. I averaged like 20 on any sprint i took, no matter the difficulty. I practiced doing problems quickly, and I can now do them pretty well.", "Solution_2": "If you don't know how to do a problem or want to do it faster, post the question on AoPS and someone's going to help you.", "Solution_3": "It's not that i don't know how to do them. I wouldn't be as disappointed if i didn't know how to do it but got it wrong than if i knew how to do it but didn't read the question or something like that...this totally isn't making sense. \r\n\r\nI think time is a problem for me because i worry too much about getting things wrong so i slow down on others. Is there a way to prepare or focus yourself or something so that you won't have to worry?? Maybe i'm psyching myself out... :maybe:", "Solution_4": "the way i check to see if i made a careless mistake is to just see if my answer makes sense and if you are given an algebra problem try plugging your answer into the equation/expression.\r\ni should have used these tips on the school round \r\n\r\ni got a 36/46, when i could easily could have gotten a 40 or 41.\r\n :mad:", "Solution_5": "I should've had a 46 but because of carless mistakes my score dropped to 22 :mad: :mad:", "Solution_6": "bpms, don't become ubemaya. please.\r\n\r\nespeacially when you criticize others about it. please.\r\n\r\n:lol:\r\n\r\njorian", "Solution_7": "he wasnt criticizing anyboday :maybe: \r\n\r\nwow...that mus be A LOT of careless mistakes . i used to make about 5 careless ones on each test in the biggining of the year but i got better just beforre the real thing. i dunn, try underlining the question and all the important stuff. \r\n\r\nwhat kind of careles mistakes do you make?", "Solution_8": "[quote=\"jhredsox\"]bpms, don't become ubemaya. please.\n\nespeacially when you criticize others about it. please.\n\n:lol:\n\njorian[/quote]\r\n\r\nare you trying to make a point that mistakes aren't careless or something? you're usually not the one to spam.\r\n\r\njorian", "Solution_9": "you should look at my \r\nmathcounts preparation guide for state thread\r\nfor help in reducing stupid mistakes", "Solution_10": "[quote=\"jhredsox\"][quote=\"jhredsox\"]bpms, don't become ubemaya. please.\n\nespeacially when you criticize others about it. please.\n\n:lol:\n\njorian[/quote]\n\nare you trying to make a point that mistakes aren't careless or something? you're usually not the one to spam.\n\njorian[/quote]\r\n\r\n\r\nyou're spamming jhredsox \r\n\r\na careless mistake still fits under the description of a MISTAKE . Everyone makes mistakes. \r\n\r\nto make a mistake = to understand, interpret, or evaluate wrongly; misunderstand; misinterpret.\r\n\r\nI think careless mistakes are just ways to make an excuse to increase your self esteem, especially ones with low self-esteem.", "Solution_11": "Well there's one way to cure it. when you do it, find out what you did wrong. don't just say, ok, i just did this wrong. actually do it out. don't be afraid to go into algebra and stuff, and check your work. when you check your work, there's not much sense in just looking over your work. actually do it out again, and when possible, try to find an alternate way to do the problem.", "Solution_12": "I missed AIME 2 times this year because of careless mistakes (at least for the first one). The second one, I didn't mark a correct answer--lame i know.\r\n\r\nA way to aviod mistakes is to check your answers, and sometimes to do them twice", "Solution_13": "Sry to get off topic, but... wait missed that #1 on AIME last year? That was like a mathcounts problem..\r\n\r\nChecking your work is a good idea on a previous problem if you think you cant do anymore on teh sprint. Instead of wasting like 3 mins trying to figure out a problem, use those three minutes to check your work on another problem that you werent 100 percent sure about. But in the end, its how many problems you can do, nto the accuracy(especially true on sprint). Ill take a score of getting 25 right, 5 wrong over a score of gettin 22 right, 0 wrong anyday...\r\n\r\n\r\nBut I wouldnt totally recommend doing problems 2x because on mathcounts, there isnt enough time..but check over the ones that you think you might have made a computational or even reading mistake on when you cant figuer out the remaining ones on the test, as its pointless to waste time looking at a rpoblem you cant do", "Solution_14": "I should have had a 36 at Chapters (reflecting my skill level), but due to careless mistakes I got a 44 :mad: :mad: :mad: .\r\n\r\nBTW, accuracy counts just as much as how many you can do IMSAMO.", "Solution_15": "do every problem in one minute, and then you have 10minutes, left. and then start checking from the front, not from the back...this is how I got a perfect score on the school, (but not on the chapter) :blush:", "Solution_16": "Wow.. it also sucks a lot when you type the number wrong in the calculator.. like I did on #1 on an AMC8 one year.. i was enraged..", "Solution_17": "[quote=\"Ubemaya\"]I should have had a 36 at Chapters (reflecting my skill level), but due to careless mistakes I got a 44 :mad: :mad: :mad: .\n\nBTW, accuracy counts just as much as how many you can do IMSAMO.[/quote]\r\n\r\nOh boy! Mr. Sarcasm.\r\n\r\n :roll:", "Solution_18": "@undefined117: ya, my math teacher said she cried when she looked at ur pamphlet!\r\n\r\nand this thread is getting [i]really[/i] off topic... It'll probably get locked, but at least it served it's purpose.\r\n\r\nThanks!", "Solution_19": "[quote=\"undefined117\"]Wow.. it also sucks a lot when you type the number wrong in the calculator.. like I did on #1 on an AMC8 one year.. i was enraged..[/quote]\r\n\r\nlucky for you amc is banning calculators", "Solution_20": "back on topic: \r\n\r\nI took a practice 2006 State Sprint and my score dropped 8 points because for some of them\r\n\r\n1) arithmetic error (simple)\r\n2) counted incorrectly the number of tiems so and so\r\n3) being dumb and forgot how to do a simple problem\r\n\r\n\r\ni believe the best way to avoid careless mistakes is to double check...usually I only double check ones that I might have doubts on...if it is a simple answer try not to check it until the end, when you have completed the test", "Solution_21": "if nobody made careless mistakes then there would be like 20 46's.", "Solution_22": "[quote=\"cognos599\"]if nobody made careless mistakes then there would be like 20 46's.[/quote]\r\n\r\nHaha if nobody made careless mistakes the US wouldn't be in war with Iraq now", "Solution_23": "[quote=\"7h3.D3m0n.117\"]back on topic: \n\nI took a practice 2006 State Sprint and my score dropped 8 points because for some of them\n\n1) arithmetic error (simple)\n2) counted incorrectly the number of tiems so and so\n3) being dumb and forgot how to do a simple problem\n\n[/quote]\r\n\r\nthat sounds exactly like something i would do! except not on the practices, but the real thing :( .", "Solution_24": "hehe I usually don't get my MATHCOUNTS paper back so i never know which ones i got wrong due to error", "Solution_25": "[quote=\"Ubemaya\"]I should have had a 36 at Chapters (reflecting my skill level), but due to careless mistakes I got a 44 :mad: :mad: :mad: .\n\nBTW, accuracy counts just as much as how many you can do IMSAMO.[/quote]\r\n\r\nwho are you ubemaya", "Solution_26": "[quote=\"Ubemaya\"]I should have had a 36 at Chapters (reflecting my skill level), but due to careless mistakes I got a 44 :mad: :mad: :mad: .\n\nBTW, accuracy counts just as much as how many you can do IMSAMO.[/quote]\r\n\r\nAre you so mad that you reversed the 44 and the 36?\r\n\r\n\r\nI had a lot of time left after I did the chapter, so I did the whole sprint again. I got 25 out of 30.\r\n\r\nThe targets were really fun problems. REALLY fun problems!!!", "Solution_27": "personally i thought the targets were really easy. i mean most of us in the hall were done by thirty seconds or something...", "Solution_28": "[quote=\"Arvind_sn\"]personally i thought the targets were really easy. i mean most of us in the hall were done by thirty seconds or something...[/quote]\r\n\r\n\r\nI agree. Except most people in the hall also didn't use that extra time to [u]check.[/u]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Is the number $2^{10}+5^{12}$ prime or composite?", "Solution_1": "composite, since we have \\[ 2^{10}+5^{12}=(2^5+5^6)^2-2(2^5)(5^6)=(2^5+5^6-(2 \\cdot 5)^3)(2^5+5^6+(2\\cdot 5)^3) \\]", "Solution_2": "\\begin{eqnarray*}2^{10}+5^{12} &=& 2^{10}+2^65^6+5^{12}-2^65^6\\\\ &=& (2^5)^2+2\\cdot 2^5\\cdot 5^6+(5^6)^2-(2^35^3)^2\\\\ &=& (2^5+5^6)^2-(2^35^3)^2\\\\ &=& (2^5+5^6+2^35^3)(2^5+5^6-2^35^3)\\end{eqnarray*}\r\n\r\nTherefore it's composite." } { "Tag": [], "Problem": "If 27,783= 3^x*7^y find x*y.", "Solution_1": "You should read the Introduction to Counting and Probability. It explains everything you want to know.", "Solution_2": "Will do. :]", "Solution_3": "3^4*7^3 is the ans :P", "Solution_4": "alright well lets start with the 3's instead of dividing by threes lets see if we can speed it up by dividing by nines.\r\nSeeing that the sum of the digits of the original number is a a multiple of nine, we divide by nine to get 3087, seeing how the sum of its digits is 18 we divide by nine again to yield 343 note that 343 is just $ 7^3$ So basically we divided by nine twice or instead we can say we divided by 3 4 times and 7 three times so 3*4 = 12\r\ndang beaten to it.", "Solution_5": "yes well you actually explained it :P\r\n\r\ngj though, and sorry" } { "Tag": [ "probability", "combinatorics unsolved", "combinatorics" ], "Problem": "16 players take part in a tennis tournament. The order of the matches is chosen at random. There's always a player better than another one. In a match, the better wins.\r\nCalculate:\r\na) The probability that all the 4 best players reach the semi-final\r\n\r\nb) The probability that the sixth player reach the semifinal.\r\n\r\nNote: The order of the matches doesn't mind players'skill. The winner of the first match plays with the winner of the second, etc.\r\n\r\nEnjoy!", "Solution_1": "In other words, we have a single-elimination bracket-style tournament. \r\n\r\n[hide=\"(a)\"] can only happen if each of the 4 best players lies in a different quadrant of the original setup, the probability of which is $ 1\\cdot \\frac{12}{15}\\cdot \\frac{8}{14}\\cdot \\frac{4}{13}$.[/hide]\n\n[hide=\"(b)\"] can only happen if the three other players in the same quarter of the bracket with player 6 are ranked lower, the probability of which is $ \\frac{10}{15}\\cdot \\frac{9}{14}\\cdot \\frac{8}{13}$. This is nearly twice as likely as in part (a)[/hide]", "Solution_2": "The b) is ok...can you explain better the part a)?\r\nAnother question..shouldn't I count the probability 4! times?", "Solution_3": "The answer I gave is larger than $ \\frac{1}{24}$, so you certainly shouldn't multiply it by 4!. In fact, the factor of 4! is included in my answer: write $ 1 = \\frac{16}{16}$ and the numerator of the fraction becomes $ 4! \\cdot 4^{4}$.", "Solution_4": "I found the condition a) \"they lie in different quadrants\" but i can't understand how to express it.\r\nIn particular where you multiply 12/15 by 8/14 and 4/13...\r\nThen, call the 4 best players A B C D, how can i express that all these ways are right:\r\n1A 2B 3C 4D\r\n1B 2A 3C 4D\r\n1D 2A 3B 4C\r\netc...\r\nForgive me, i'm a beginner :P", "Solution_5": "The first player has 16 spaces to go, so $ \\frac{16}{16}$. The second player has 12 spaces remaining in other quadrants out of 15 total, so $ \\frac{12}{15}$. The third player has 8 spaces remaining in other quadrants out of 14 total, so $ \\frac{8}{14}$. The fourth player has 4 spaces remaining in other quadrants out of 13 total, so $ \\frac{4}{13}$.\r\n\r\nDoing it your way, there are $ 4!$ ways to assign each player to a different quadrant. In addition, there are $ 4^{4}$ ways to assign each player to a position within the quadrant. This gives us $ 4! \\cdot 4^{4}$ arrangements, each of which occurs with probability $ \\frac{1}{16\\cdot15\\cdot14\\cdot13}$.", "Solution_6": "ok..solved! thank you very much..." } { "Tag": [ "inequalities", "function", "inequalities proposed", "n-variable inequality" ], "Problem": "Prove that if $x_1, x_2 \\cdots ,x_n \\ge 0$ add up to $1$, then \n\\[\\sqrt{1-x_1}+\\sqrt{1-x_2}+\\cdot+\\sqrt{1-x_n} \\ge n-1\\]", "Solution_1": "It's not very clear !!!", "Solution_2": "I assumed you meant x_i >= 0. That expression is a concave function in each variable, so it reaches its minimum when the variables take the ends of the interval [0,1] as values, so its minimum is reached when n-1 variables are 0 and one of them is 1. It's not very rigorous, but I think this argument works.", "Solution_3": "It is really the easiest.\r\n\r\nOfcourse we have to assume that x_i>=0 (If not so, we can take x1=x2=...=x_n-1=1, xn=2-n and the inequality not holds).\r\nThen we have 1-xi <=1 => sqrt(1-xi) >= 1-xi. Add up we have the result!\r\n\r\nSometimes, the result can be received in such way!\r\n\r\nNamdung", "Solution_4": "Well, harazi did name it 'the EASIEST inequality'... :D" } { "Tag": [ "trigonometry", "limit", "geometry proposed", "geometry" ], "Problem": "Determine the extreme values of \r\n\r\n$ \\sin^2 A \\plus{} \\sin^2 B \\minus{} \\sin^2 C$", "Solution_1": "[quote=\"Ligouras\"][color=darkred]In any triangle $ ABC$ exists the full-relation $ \\sin^2 A \\plus{} \\sin^2 B \\minus{} \\sin^2 C\\in\\left[ \\minus{} \\frac 14\\ ,\\ 2\\right)$ .[/color] [/quote]\r\n\r\n$ \\mathrm{\\underline{Proof}.}\\ f(A,B,C)\\equiv\\sin^2 A \\plus{}$ $ \\sin^2 B \\minus{} \\sin^2 C\\equal{}\\sin ^2(B\\plus{}C) \\plus{} \\sin (B \\plus{} C)\\sin (B \\minus{} C)\\equal{}$\r\n\r\n$ \\equal{} \\sin A\\left[\\sin (B \\plus{} C) \\plus{} \\sin (B \\minus{} C)\\right] \\equal{}$ $ 2\\sin A\\sin B\\cos C\\implies \\boxed {f(A,B,C) \\equal{} 2\\sin A\\sin B\\cos C}$ .\r\n\r\n$ \\blacktriangleright\\ f(A,B,C) < 2\\ \\ ;\\ \\ \\left(\\forall \\right)x\\in\\left(0,\\frac {\\pi}{2}\\right)\\ ,\\ g(x)\\equiv f\\left(\\frac {\\pi}{2},\\frac {\\pi}{2} \\minus{} x,x\\right) \\equal{} 1 \\plus{} \\cos 2x$ and $ \\lim_{x\\searrow 0}\\ g(x) \\equal{} 2$ . \r\n\r\n$ \\blacktriangleright\\ f(A,B,C) \\equal{} \\cos C[\\cos (A \\minus{} B) \\plus{} \\cos A] \\equal{}\\left[\\cos C \\plus{} \\frac 12\\cdot \\cos (A \\minus{} B)\\right]^2 \\plus{}$ $ \\frac 14\\cdot\\sin^2(A \\minus{} B) \\minus{} \\frac 14\\ge \\minus{} \\frac 14$ \r\n\r\n$ \\mathrm{\\ with\\ equality\\ iff\\ }A \\equal{} B \\equal{} 30^{\\circ}$ . $ \\mathrm{In\\ conclusion\\ ,}\\ \\min_{\\triangle ABC}\\ f(A,B,C) \\equal{} \\minus{} \\frac 14\\ \\ \\ \\mathrm{and}\\ \\ \\sup_{\\triangle ABC}\\ f(A,B,C) \\equal{} 2$ .", "Solution_2": "$sin^2 A + \\sin^2 B - \\sin^2 C\\le |sin^2 A + \\sin^2 B - \\sin^2 C|$\n${{= 2|\\sin A\\sin B\\cos C}|\\le 2|\\sin A||\\sin B||\\cos C}|\\le 2$\n$\\implies sin^2 A + \\sin^2 B - \\sin^2 C<2.$\n$sin^2 A + \\sin^2 B - \\sin^2 C\\ge -\\frac 14 :$\n1995, I prove\n$a^2 - b^2 - c^2 \\leq R^2 \\iff sin^2 B+sin^2 C-\\sin^2 A\\ge -\\frac 14$\n$\\iff cosAsinBsinC\\ge-\\frac 18.$\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=480082]\u300aHunan mathematics communication\u300b(China Changsha)No.3(1995)[/url]\n1996, I prove\n$cos^2 A+\\lambda(sin^2 B+\\sin^2 C)\\ge \\lambda-\\frac {\\lambda^2}{4}\\implies sin^2 B+sin^2 C-\\sin^2 A\\ge -\\frac 14.$\n\u300aMiddle school Mathematical Monthly\u300b(china Suzhou)No.9(1996)" } { "Tag": [ "probability", "geometry", "rectangle", "analytic geometry", "similar triangles", "AMC" ], "Problem": "1. The number of 5-digit numbers in which every two neighbouring digits differ by 3 is\r\n(A) 40\r\n(B) 41\r\n(C) 43\r\n(D) 45\r\n(E) 50\r\n\r\n2. A $3 \\times 3$ square is divided into nine $1 \\times 1$ unit squares. Different integers from 1 to 9 are written into these nine unit squares. Consider the pairs of numbers in the squares sharing a common edge. What is the largest number of pairs where one number is a factor of the other number?\r\n(A) 7\r\n(B) 8\r\n(C) 9\r\n(D) 10\r\n(E) 12\r\n\r\n3. Each point on the four sides of a 1m $\\times$ 1m square is coloured one of $n$ colours so that no two points that are exactly 1m apart are coloured the same. What is the smallest $n$ for which such a colouring can be made?\r\n\r\n4. A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer?", "Solution_1": "For number one, i made a little tree and got 47, must've made a silly mistake. :( \r\n\r\nFor number two, I get something like 9, with this arrangement\r\n\r\n7 5 4\r\n9 1 8\r\n3 6 2\r\n\r\nIt obviously can't be 12 because that's how many edges there are.", "Solution_2": "On 3: I would say that the answer is 5. Am I correct?\r\n\r\n\r\noops,duh, I was thinking of a cube...", "Solution_3": "[quote=\"solafidefarms\"]On 3: I would say that the answer is 5. Am I correct?[/quote]\r\n\r\nHmmm.... Do the points have to be in the center of a side? Otherwise the answer is 1.", "Solution_4": "For #1, I got (D) 45...\r\nI drew a tree diagram, just like 236factorial did. Is that correct?", "Solution_5": "Okay, I thought 8 - 2 was 3 for a second, so I also got 45 as the answer (D). For number one, that is.", "Solution_6": "for number 1 i get D\r\nfor number 2 i get B\r\n\r\nI am not sure what i did though.", "Solution_7": "[quote=\"math92\"]for number 1 i get D\nfor number 2 i get B\n\nI am not sure what i did though.[/quote]\r\n\r\nplease don't post answers without solutions, especially if someone's already given the same answer ;)", "Solution_8": "[quote=\"Iversonfan2005\"][quote=\"math92\"]for number 1 i get D\nfor number 2 i get B\n\nI am not sure what i did though.[/quote]\n\nplease don't post answers without solutions, especially if someone's already given the same answer ;)[/quote]\r\n\r\nIt would be nice if you gave some answers to begin with :roll:", "Solution_9": "Well number one is definitely (D) and I got that one wrong... I miscounted and got 47 too, but I ran out of time and just put 50. Dang! I don't know any other solution, other than brute counting and the tree diagram though, with which you get 45. \r\n\r\nAnyway, 236Factorial, you're my lifesaver! I answered 9 for question 2 and I thought I got it wrong all along, but now it turns out there [i]is[/i] a way to get 9 (I hope there's no way to get 10). \r\n\r\nNumbers 3 and 4 are among the last 5 questions, in which competitors nominate an integer answer between 0 and 999 (and they're really difficult), so I don't really know the answers. If I have the time, I'll scan the question booklet and post it in AoPS. :) There's even an interesting question on how to cut the cake in the best possible way to avoid your brother or sister getting the larger piece.", "Solution_10": "[quote=\"aidan\"]I hope there's no way to get 10. [/quote]\r\n\r\nI can't see anything that will involve 10. 5 and 7 are prime, and the only number from 1 and 9 they can share a factor with is 1. The 5 and seven will be sharing sides with at least 4 different numbers (put 5 on the corner and 7 right next to it. We counted the one that 5 and 7 share). Say one of these numbers is 1. Only one of the numbers can be touching it adjacently, so there must be at least three out of the twelve possible meetings where one is not a factor of the other. So the maximum is 9.", "Solution_11": "[quote]Only one of the numbers can be touching it adjacently, so there must be at least three out of the twelve possible meetings where one is not a factor of the other. So the maximum is 9.[/quote]\r\n\r\nExcellent observation... didn't think of that. Here's one more question I need someone's opinion on:\r\n\r\n14. Two dice are thrown at random. The probability that the two numbers obtained are the two digits of a perfect square is\r\n\r\n(A) $\\frac {1}{9}$\r\n\r\n(B) $\\frac {2}{9}$\r\n\r\n(C) $\\frac {7}{36}$\r\n\r\n(D) $\\frac {1}{4}$\r\n\r\n(E) $\\frac {1}{3}$", "Solution_12": "So the total number of options are \r\n$6\\cdot6 = 36$\r\nWe know our perfect square must be (inclusive) 11-16,21-26,31-36,41-46,51-56,61-66\r\n\r\nWe see 16, 25, 36, and 64 work.\r\n\r\nWe might jump the gun and go! \"Ooh! 1/9\" And get it wrong\r\n\r\nBut we have to realize if we roll 61, 52, 63, or 46 we also get the correct answer\r\n\r\nSo we have $4\\cdot2 = 8 $ways to get 2 digits of perfect square so\r\n\r\n$\\frac{8}{36} = \\boxed{\\frac{2}{9}}$", "Solution_13": "[hide]There are 36 total possibilities. The possible ones are (1,6), (6,1), (2,5), (5,2), (3,6), (6,3), (6,4), (4,6). That's 8/36=2/9, \n\nRemember that the dice only goes up to 6. [/hide]", "Solution_14": "[quote]We might jump the gun and go! \"Ooh! 1/9\" And get it wrong [/quote]\r\n\r\n :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: \r\n\r\nI think I should use that on my friends.... I said 2/9 and they kept saying 1/9.", "Solution_15": "My name is Louis and my father has cooked me an L-shaped cake for my birthday. He says that I must cut it into three pieces with a single cut, so that my brother and sister can have a piece too. So, I have to cut it\r\n\r\n \r\n\r\n\r\n\r\n\r\nHe says that I have to be polite and let them have the first choice of the pieces, but I just know that they\u2019ll be greedy and leave the smallest possible piece for me. So I want to cut the cake so that my little piece will be as big as possible. If I do this, how big, in square centimeters, will my piece be?", "Solution_16": "Hey everybody, I'm new to this and it is my first post!\r\n\r\nI also did the intermediate test on Thursday over here in New Zealand and found some questions hard.\r\n\r\nFor #30, does anyone have an answer for it:\r\n\r\n[i]A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer.[/i]\r\n\r\nAnd for #23, the area of the shaded rectangle - I wrote (D) between 7/16 and 1/2, but I'm not sure about it. A friend of mine wrote (E) more than 1/2 and we all thought he was crazy, but he might end up with the last laugh!\r\n\r\nAnd, finally, could someone fill me in on #17. The 12cm^2 square was folded and you needed to find the length of the fold line...\r\n\r\nThanks,\r\nJordan.", "Solution_17": "Welcome... it's a nice community over here. :) I have no idea at all about #30. \r\n\r\nThe answer for #23 is definitely (D) as the area is found to be 15/32 square units. There are 4 similar triangles in the diagram, if you have your question paper. \r\n\r\nFor #17, the answer is 4. It is said that the total visible area of the paper is half shaded and half white. So you can form the equation: $12 - x^2 = \\frac {x^2}{2}$, where $x = US$. So $x^2 = 8$ and $x = 2 \\sqrt {2}$. The the fold line is $\\sqrt {2} \\times 2 \\sqrt {2} = 4$.", "Solution_18": "[quote=\"aidan\"]My name is Louis and my father has cooked me an L-shaped cake for my birthday. He says that I must cut it into three pieces with a single cut, so that my brother and sister can have a piece too. So, I have to cut it\n\n \n\n\n\n\nHe says that I have to be polite and let them have the first choice of the pieces, but I just know that they\u2019ll be greedy and leave the smallest possible piece for me. So I want to cut the cake so that my little piece will be as big as possible. If I do this, how big, in square centimeters, will my piece be?[/quote]\r\nI got the answer to be 80, is that right? I used some little coordinate type things.", "Solution_19": "[quote=\"nat mc\"][quote=\"aidan\"]My name is Louis and my father has cooked me an L-shaped cake for my birthday. He says that I must cut it into three pieces with a single cut, so that my brother and sister can have a piece too. So, I have to cut it\n\n \n\n\n\n\nHe says that I have to be polite and let them have the first choice of the pieces, but I just know that they\u2019ll be greedy and leave the smallest possible piece for me. So I want to cut the cake so that my little piece will be as big as possible. If I do this, how big, in square centimeters, will my piece be?[/quote]\nI got the answer to be 80, is that right? I used some little coordinate type things.[/quote]\r\n\r\nwell, this is my solution:\r\n[hide]\nfrom the diagram(in the middle), we know that in order to cut three pieces in one slice, the line must touch the turning pt. therefore, it will make a right triangle with sides $30 - x$ and $20 - y$, and two trapezoids. therefore the total area is:\n\n$\\displaystyle \\frac{(30-x)(20-y)}{2} + (y + 10)*5 + (x + 20) * 5 = 10*30 + 10*10 = 400$\n\nsimplifying get:\n\n$-20y -10x +xy = -100$\n\nadding 200 and factor yields:\n\n$(20-x)(10-y) = 100$\n\nsince we want $y$ to be as large as possible, $x = 0$, and $y = 5$. plug in and we get $(y+10)*5 = 75$\n[/hide]", "Solution_20": "I got 80, by making the top most portion being a triangle with base 8 and height 20, area 80. the cut goes through the corner, and the right most portion is a 10 by 10 square minus a triangle base 4 height 10, to get area 100-20=80", "Solution_21": "[quote=\"Maths_Crazy\"]\n[i]A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer.[/i]\n[/quote]\r\n\r\nI think it's 13. $1^2+2^2+5^2+10^2=130$. That's guess and check...", "Solution_22": "^ Oh man, I guessed 11!", "Solution_23": "[quote=\"nat mc\"]I got 80, by making the top most portion being a triangle with base 8 and height 20, area 80. the cut goes through the corner, and the right most portion is a 10 by 10 square minus a triangle base 4 height 10, to get area 100-20=80[/quote]\r\nHere is the solution for that. I tried it your way, then I tried it this way. You want the two smallest areas to be equal in this case, so that his isn't really the smallest. so we have\r\n$\\frac{20*2a}2=100-\\frac{10a}2$\r\n$20a=100-5a$\r\n$25a=100$\r\n$a=4$\r\n$100-5*4=\\boxed{80}$", "Solution_24": "Ok... good. I answered 80 after solving some equation but some friends claimed it was 75. Made me worried for a moment. \r\n\r\nI got question 30 wrong too... I guessed 5! Even worse...", "Solution_25": "Couldn't you use a calculator in the AMC? Wait, are you in australia?", "Solution_26": "I think 130 is the only integer that equals to the sum of the squares of its four smallest divisors. Here is my solution for #4, but i m not sure if it good. I'd be good to see someone else give a solution.\r\n[hide=\"edited solution\"]Let the positive integer that equals to the sum of the squares of its four smallest positive divisors be \u201cn\u201d. n must be divisible by 1. Also it must be divisible by 2, since the sum of four odd numbers squared is still even. The sum of the squares of the two smallest divisors is 5, therefore the next two are an odd and an even divisor. This means that if the odd divisor is \u2018x\u2019, and the even divisor is \u20182x\u2019, unless the even divisor is a power of 2. \nTherefore n=1^2+2^2+x^2+(2x)^2=5+5x^2=5(x^2+1). Hence 5 is a divisor of n thus 3 can\u2019t equal \u2018x\u2019 since 5 would be the next smallest divisor. Therefore x=5 and 2x=10, and n= 1^2+2^2+5^2+10^2=130. \nThus the largest prime that divides n is 13. \nExamining the case in which the even divisor is a power of 2, then n must be divisible by 4 and 8. Therefore the odd divisor of n must be 3,5,7. However none of these satisfy the conditions. [/hide]", "Solution_27": "[quote=\"236factorial\"]Couldn't you use a calculator in the AMC? Wait, are you in australia?[/quote]\r\n\r\nNo you can't.\r\n\r\nYes, I was doing the Australian MC, however, I live in New Zealand, slightly south-east of Australia (if you didn't know already)!", "Solution_28": "[quote=\"p.c\"]and n= 12+22+52+102=230. \nThus the largest prime that divides n is 13.[/quote]\r\nI assume you mean $1^2+2^2+5^2+10^2=130$", "Solution_29": "The code probably got messed up. Anyway:\r\n\r\n[quote]Also it must be divisible by 2, since the sum of four odd numbers squared is still even.[/quote]\r\n\r\nWhy is this so? Someone explain... don't really get this.", "Solution_30": "[quote=\"aidan\"]The code probably got messed up. Anyway:\n\n[quote]Also it must be divisible by 2, since the sum of four odd numbers squared is still even.[/quote]\n\nWhy is this so? Someone explain... don't really get this.[/quote]\r\n\r\nAn odd number squared is odd. four odds added together is even.", "Solution_31": "Anyone got an answer for #12...the question with the triangle inside a 6x6 square.", "Solution_32": "[quote]An odd number squared is odd. four odds added together is even.[/quote]\r\n\r\nGet this now... thanks. Anyway the answer for #12 is $10.5cm^2$. You minus off the area of the right-angles triangles from the rectangle around triangle we need.", "Solution_33": "OK thanks, what would #15 be?\r\n\r\nP.S. Sorry for asking for the answers - I'm trying to figure out my mark (currently 69 out of 120).", "Solution_34": "No problem... #15 is $(E)$, $\\sqrt {2} : 1$ We set up equations as follows: $\\frac {2x}{y} = \\frac {y}{x}$, as the two rectangles are similar. Cross multiply to get $y^2 = 2x^2$. Then ${\\frac {y}{x}}^2 = 2$, so $\\frac {y}{x} = \\sqrt {2}$. We can substitute back to get $x = 1$.", "Solution_35": "OK, thanks.\r\n\r\nFinal question, #25.\r\n\r\nWhat is the maximum n of reflex angles in a polygon?", "Solution_36": "The answer is $n - 3$. In a polygon with $n$ sides, the total number of degrees in the polygon is $180(n - 2)$. Now a reflex angle is an angle $\\geq 180$ and $\\leq 360$. So there can at most be $n - 3$ angles greater than 180 degrees in any polygon (if there were $n - 2$ reflex angles, then every angle would have to be 180, and the shape wouldn't be a polygon).", "Solution_37": "[quote=\"aidan\"]The answer is $n - 3$.[/quote]\r\n\r\nYyyyeeeessss!!\r\n\r\n74 out of 120 then for me...", "Solution_38": "[quote]74 out of 120 then for me...[/quote]\r\n\r\nBut you have to remember that how well you do depends on how well everyone does in New Zealand...", "Solution_39": "Ok guys, it's been a while (because it takes ages for the test to get marked and returned), but guess what I got?\r\n\r\n(Highlight below)\r\n\r\n[color=white][size=117][b]HIGH DISTINCTION[/b][/size]\n\nI actually ended up with 75/120, enough to be in the top 1% of candidates in New Zealand.[/color]\r\n\r\nI would like to thank all of you guys for answering the many questions I had after the test.\r\n\r\nCheers,\r\nJordan." } { "Tag": [ "arithmetic sequence" ], "Problem": "Let $ n \\ge 3$ be a prime number and $ a_{1} < a_{2} < \\cdots < a_{n}$ be integers. Prove that $ a_{1}, \\cdots,a_{n}$ is an arithmetic progression if and only if there exists a partition of $ \\{0, 1, 2, \\cdots \\}$ into sets $ A_{1},A_{2},\\cdots,A_{n}$ such that\r\n\\[ a_{1} \\plus{} A_{1} \\equal{} a_{2} \\plus{} A_{2} \\equal{} \\cdots \\equal{} a_{n} \\plus{} A_{n},\\]\r\nwhere $ x \\plus{} A$ denotes the set $ \\{x \\plus{} a \\vert a \\in A \\}$.", "Solution_1": "The problem is from Romanian TST 1998 and was proposed by Vasile Pop.\n\nYou can find it [url=http://projectpen.wordpress.com/2008/04/17/08s-an-arithmetic-partition-of-the-positive-integers/]here[/url]. I remember I spend so much time solving and writing the second solution in [url=http://projectpen.files.wordpress.com/2008/04/pen08s-draft.pdf]the pdf[/url]. Good old days when I had time to write for PEN :). Actually, not even now am I sure if that solution is fully corect. I really wonder if anyone has solved it during the TST. Valentin, any information? :)" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $ a,b,c$ are positive numbers, then :\r\n\r\n$ \\frac {a^{2} \\minus{} bc}{4(a^{2} \\plus{} b^{2}) \\plus{} c^{2}} \\plus{} \\frac {b^{2} \\minus{} ca}{4(b^{2} \\plus{} c^{2}) \\plus{} a^{2}} \\plus{} \\frac {c^{2} \\minus{} ab}{4(c^{2} \\plus{} a^{2}) \\plus{} b^{2}}\\geq 0$\r\n\r\n500th post :D", "Solution_1": "[quote=\"alex2008\"]If $ a,b,c$ are positive numbers, then :\n\n$ \\frac {a^{2} \\minus{} bc}{4(a^{2} \\plus{} b^{2}) \\plus{} c^{2}} \\plus{} \\frac {b^{2} \\minus{} ca}{4(b^{2} \\plus{} c^{2}) \\plus{} a^{2}} \\plus{} \\frac {c^{2} \\minus{} ab}{4(c^{2} \\plus{} a^{2}) \\plus{} b^{2}}\\geq 0$\n\n[/quote]\r\nIt's true for all real $ a,$ $ b$ and $ c$ such that $ a^2\\plus{}b^2\\plus{}c^2\\neq0.$" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Hi!\r\n\r\nWhat is a monotone class?\r\n\r\nMike", "Solution_1": "A collection $C$ if subsets of a given set is called a monotone class if it is closed under increasing countable unions and decreasing countable intersections." } { "Tag": [ "vector", "geometry", "algebra", "polynomial", "geometry proposed" ], "Problem": "we have a regular $ p \\minus{} \\text{gon}$ $ A_1A_2\\dots A_p$ and $ O$ be its center.we have vectors $ \\overrightarrow \\alpha_1,\\dots,\\overrightarrow \\alpha_n$ that vector $ \\overrightarrow \\alpha_i$ is parallel to $ OA_i$. if $ \\overrightarrow \\alpha_1 \\plus{} \\dots \\plus{} \\overrightarrow \\alpha_p \\equal{} 0$\r\nthen prove :$ |\\overrightarrow \\alpha_1| \\equal{} \\dots \\equal{} |\\overrightarrow \\alpha_p|$ where $ p$ is a prime number.", "Solution_1": "[quote=\"Amir.S\"]we have a regular $ p \\minus{} \\text{gon}$ $ A_1A_2\\dots A_p$ and $ O$ be its center.we have vectors $ \\overrightarrow \\alpha_1,\\dots,\\overrightarrow \\alpha_n$ that vector $ \\overrightarrow \\alpha_i$ is parallel to $ OA_i$. if $ \\overrightarrow \\alpha_1 \\plus{} \\dots \\plus{} \\overrightarrow \\alpha_p \\equal{} 0$\nthen prove :$ |\\overrightarrow \\alpha_1| \\equal{} \\dots \\equal{} |\\overrightarrow \\alpha_p|$ where $ p$ is a prime number.[/quote]\r\n\r\nI think you want to add the condition that the lengths of $ \\alpha_i$ all are rational... And then again, it is not a geometry problem, but a consequence of the fact that cyclotomic polynomials are irreducible in $ \\mathbb{Q}\\left[X\\right]$.\r\n\r\n darij" } { "Tag": [], "Problem": "determine the mathematical word whose letters have been rearranged:\r\n1. bdeiiiilnsv\r\n2. dehlnoopry\r\n3. aacglnoot\r\n4. ceegimort\r\n5. ceimoopst\r\n6. acdiiimnnrst\r\n7. accdiinnoortt\r\n8. accenost\r\n9. ddeeennpt\r\n10. aeeilnnoptx\r\n11. aaeeltuv\r\n12. ddehinor\r\n13.yhrynageunir. :P HAVE FUN!", "Solution_1": "errr...didn't you also post this in fun and games?", "Solution_2": "[quote=\"eryaman\"]errr...didn't you also post this in fun and games?[/quote]\r\nYES, i just want to get more replies................ :lol:", "Solution_3": "i know none of them...\r\nperhaps i won't recognize them even if you put them in right orders...", "Solution_4": "We are told that this is from the CPSC. Thread locked." } { "Tag": [ "Alcumus", "Gamebot" ], "Problem": "Five couples were at a party. If each person shook hands exactly once with everyone else except his/her spouse, how many handshakes were exchanged? (Note: One obviously doesn't shake hands with oneself.)", "Solution_1": "There were 10 people in total, who shook hands with 8 other people, to make a total of $ 10 \\times 8 \\equal{} 80$. However, we're overcounting - that is, we counted Person A shaking hands with Person B and Person B shaking hands with Person A twice - so we have to divide by two to get a total answer of $ \\boxed{40}$.", "Solution_2": "Another way to do it:\nThe total number of handshakes was 10 choose 2 which is 45.\n\nBut the couples did not shake hands.\n\nSince there are 5 couples, the answer is $ \\boxed{40} $" } { "Tag": [ "geometry" ], "Problem": "A quadrilateral has the following vertices: (2,0), (4,5), (5,2), (1,4). \nHow many square units are in its area?", "Solution_1": "[geogebra]939681c0825b248237a7d3f479b6ec5d2ade7b58[/geogebra] \r\n\r\nAnswer: 20-3-2-1.5-1.5=12\r\n\r\nMany ways: shoelace, area chasing, Pick's theorem, etc.", "Solution_2": "Unless my Shoelace Theorem fails, isn't this:\r\n\r\n$ \\frac{1}{2} | (2 \\times 5 + 4 \\times 2 + 5 \\times 4 + 1 \\times 0 ) - (0 \\times 4 + 5 \\times 2 + 5 \\times 4 + 1 \\times 2 ) | = \\frac{1}{2} | 38-32 |=\\boxed{3}$...\r\n\r\nBut that doesn't seem right, because just graphing it looks bigger...", "Solution_3": "You have to put the first point twice...\r\n\r\nSo:\r\n\r\n$ (2,0)$\r\n$ (1,4)$\r\n$ (4,5)$\r\n$ (5,2)$\r\n$ (2,0)$\r\n\r\nWhich yields an answer of $ 12$." } { "Tag": [ "articles" ], "Problem": "I noticed the \"Featured Article\" box is gone from the front page. Did WotW replace it?", "Solution_1": "[quote=\"Temperal\"]I noticed the \"Featured Article\" box is gone from the front page. Did WotW replace it?[/quote]\r\n\r\nI guess so: [url]http://www.artofproblemsolving.com/Wiki/index.php?title=Main_Page&diff=prev&oldid=15807[/url]", "Solution_2": "I think it should be returned. WotW is an entirely different concept from featured articles, almost the opposite, really." } { "Tag": [], "Problem": "A grocer bought 15 dozen oranges at $1.00 a dozen. She threw away 20 rotten oranges,\r\nand then sold the rest at 8 oranges for 85 cents. How much profit did the grocer make,\r\nin dollars and cents?", "Solution_1": "[hide]$1 a dozen, 15 dozens, 15$.\n\n15 * 12 =180 oranges\n\nThrow away 20,\n\n180-20=160 oranges.\n\nSell them at 8 for 85 cent.\n\n160/8= 20, 20 * 85=$17\n17-15= $2 profit[/hide]", "Solution_2": "let's do it without triggering LATEX...\r\n[hide]she brought everything for 15 dollars. after throwing 20 away, she had 160 oranges. so she had 20 8-orange packs, selling for 17 dollars. \nShe make a 2 dollar profit. [/hide]", "Solution_3": "15*12=180=15d\r\n15\\12=125 cents\r\nso an orange is 125 cents\r\n180-20=160\r\nshe had left oranges worth 2 dollars\r\n20*85c= 1700c=17d\r\nshe gained 17 dollars by selling them but lost 15 dollars by buying them\r\nso she gained a total of 2 dollars and zero cents" } { "Tag": [], "Problem": "[color=darkred]$1.\\blacktriangleright$ Se dau doua drepte concurente neperpendiculare $d_{1}$ , $d_{2}$ si constantele reale pozitive $k_{1}$ , $k_{2}$ , $k\\ .$ Sa se determine locul geometric al punctelor $L$ din planul $\\pi=(d_{1},d_{2})$ pentru care $k_{1}\\cdot \\delta_{d_{1}}(L)+k_{2}\\cdot \\delta_{d_{2}}(L)=k\\ .$ Sa se stabileasca relatia intre $k_{1}$ , $k_{2}$ , $k$ si masura unghiului ascutit dintre cele doua drepte ca locul geometric sa fie nevid ([u]relatia de compatibilitate[/u]).\n[b]Aplicatie.[/b] Sa se arate ca intr-un patrulater circumscriptibil mijloacele diagonalelor si centrul cercului inscris sunt coliniare ([u]dreapta lui Newton[/u]).[hide=\"Raspuns.\"]Un paralelogram.[/hide]\n$2.\\blacktriangleright$ Se dau doua puncte fixe $A$ , $B$ si constantele reale nenule $k_{1}$ , $k_{2}$ , $k$. Sa se determine locul geometric al punctelor $L$ pentru care $k_{1}\\cdot LA^{2}+k_{2}\\cdot LB^{2}=k$. Sa se stabileasca relatia de compatibilitate. Discutie.\n[b]Aplicatie.[/b] Se considera patrulaterul convex $ABCD$. Sa se determine locul geometric al punctelor $L$ pentru care $LA^{2}+2\\cdot LB^{2}+3\\cdot LC^{2}=6\\cdot LD^{2}+k\\cdot [ABCD]$. [hide=\"Raspuns.\"] $k_{1}+k_{2}=0\\Longrightarrow$ O dreapta perpendiculara pe $AB$;\nin caz contrar $\\Longrightarrow$ un cerc cu centrul pe dreapta $AB$.[/hide]\n$3.\\blacktriangleright$ Se da un triunghi $ABC$, o dreapta (un cerc) $d\\ (w)$, un punct fix $F$ si un punct mobil $M\\in d\\ (w)$. Sa se determine locul geometric al punctului $L$ pentru care $\\triangle FML\\sim\\triangle ABC$.\n[b]Aplicatie.[/b] Fiind date doua drepte $d_{1}$ , $d_{2}$ si un cerc $w$, sa se construiasca cel putin un triunghi $XYZ$ asemenea cu un triunghi dat (in particular $XYZ$- triunghi echilateral) avand $X\\in d_{1}$ , $Y\\in d_{2}$ si $Z\\in w$.[hide=\"Raspuns.\"]Reuniunea a doua drepte (a doua cercuri).[/hide]\n[b]Notatii[/b].\n$\\delta_{d}(X)$ - distanta punctului $X$ la dreapta $d$;\n$[XYZT]$- aria patrulaterului convex $XYZT$.[/color]", "Solution_1": "1.Notam distanta de la $L$ la dreapta $d_{1}$ cu $LL_{1}$ si distanta de la $L$ la dreapta $d_{2}$ cu $LL_{2}$.\r\nDaca $LL_{2}=0$ rezulta $L\\equiv M\\in d_{2}\\Rightarrow LL_{1}\\equiv MM_{1}=\\frac{K}{K_{1}}$\r\nAnalog\r\nDaca $LL_{1}=0$ rezulta $L\\equiv N\\in d_{1}\\Rightarrow LL_{2}\\equiv NN_{2}=\\frac{K}{K_{2}}$\r\nVom arata ca segmentul $MN$ apartine locului geometric.\r\nFie $L\\in MN$. Rezulta usor $\\frac{LL_{1}}{MM_{1}}=\\frac{NL}{MN}\\Rightarrow LL_{1}=\\frac{NL}{NM}\\cdot \\frac{K}{K_{1}}\\ (1)$\r\nAnalog\r\n$\\frac{LL_{2}}{NN_{2}}=\\frac{ML}{MN}\\Rightarrow LL_{2}=\\frac{ML}{NM}\\cdot \\frac{K}{K_{2}}\\ (2)$\r\nDin $(1)$ si $(2)$ avem $K_{1}\\cdot LL_{1}+K_{2}\\cdot LL_{2}=\\frac{K}{NM}(ML+NL)=\\frac{K}{NM}\\cdot NM=K$\r\nNotam $M'$ si $N'$ simetricele lui $M$ si $N$ fata de intersectia $O$ a dreptelor $d_{1}$ si $d_{2}$. Evident ca $M'N'$ apartine locului geometric.\r\nLuam $L'\\in MN'$. Ducem analog perpendicularele pe $d_{1}$ si $d_{2}$ si exact ca mai inainte aratam ca si punctele segmentului $MN'$ apartin locului geometric si evident ca si punctele segmentului $M'N$ apartin locului geometric. Asadar locul geometric este constituit din laturile paralelogramului $MN'M'N$.\r\nSolutia este valabila pentru orice $K_{1},\\ K_{2},\\ K$ pozitive si nu depinde nici de unghiul ascutit dintre cele doua drepte. Daca $K_{1}=K_{2}$ paralelogramul devine dreptunghi.", "Solution_2": "2.[u]Cazul I[/u] $k_{1}k_{2}>0$ si pentru compatibilitate $k$ are semnul lui $k_{1}$ si $k_{2}$.\r\nSe reduce la $k_{1},\\ k_{2},\\ k$ pozitive.\r\nImpartim segmentul fix $AB$ in $k_{1}+k_{2}$ parti de lungime\r\n$p$. Fie $D\\in AB$ astfel incat $AD=k_{2}p$ si $DB=k_{1}p$ Conform\r\nteoremei lui Stewart, pentru orice $L$ avem:\r\n$pk_{1}LA^{2}+pk_{2}LB^{2}=pLD^{2}(k_{1}+k_{2})+k_{1}k_{2}(k_{1}+k_{2})p^{3}$\r\nDeci\r\n$k_{1}LA^{2}+k_{2}LB^{2}=LD^{2}(k_{1}+k_{2})+k_{1}k_{2}(k_{1}+k_{2})p^{2}$\r\nRezulta\r\n$LD=\\sqrt{\\frac{k-k_{1}k_{2}(k_{1}+k_{2})p^{2}}{k_{1}+k_{2}}}$ $D$ este fix, iar $LD$ este constant, rezulta locul geometric este\r\ncercul de centru $D$ si raza\r\n$\\sqrt{\\frac{k-k_{1}k_{2}(k_{1}+k_{2})p^{2}}{k_{1}+k_{2}}}$\r\n\r\n\r\n[u]Cazul II[/u] $k_{1}k_{2}<0$\r\nPresupunem $k_{1}>0$ si notam $k_{2}=-|k_{2}|$\r\nDe asemenea presupunem $|k_{2}|>k_{1}$. Generalitatea nu se restrange pentru ca daca conditia nu este indeplinita atunci relatia din enunt se inmulteste cu $(-1)$.\r\nSe imparte segmentul AB in $|k_{2}|-k_{1}$ parti, o parte notandu-se cu $p$. Se ia $D\\in AB$ astfel incat $B\\in [AD]$ si $BD=k_{1}p$\r\nSe scrie relatia lui Stewart in triunghiul $ADL$ si se obtine $D$ fix si $DL$ constant, deci locul geometric este cercul de centru $D$ si raza $DL$.\r\n\r\nPentru $k_{1}=|k_{2}|$ relatia din enunt devine $LA^{2}-LB^{2}=\\frac{k}{k_{1}}$, caz pe care il rezolvam cu teorema lui Pitagora, obtinand ca loc geometric o perpendiculara pe $AB$ la stanga sau la dreapta mijlocului, dupa cum $\\frac{k}{k_{1}}$ este negativ sau pozitiv. Pentru $k=0$ se afla mediatoarea segmentului.", "Solution_3": "[u]Aplicatia de la 1.[/u]\r\nFie $ABCD$ patrulaterul circumscriptibil, $M$ si $N$ mijloacele diagonalelor $BD$ si $AC$, iar $O$ centrul cercului inscris.\r\nVom demonstra ca $M,\\ N,\\ O$ apartin locului geometric de la 1., dreptele $d_{1}$ si $d_{2}$ fiind $AD$, respectiv $BC$.\r\n$k_{1}=AD$, $k_{2}=BC$ si $k=S_{ABCD}$\r\nPentru $M$ avem:\r\n$AD\\cdot d(M,AD)+BC\\cdot d(M,BC)=S_{ABD}+S_{BCD}=S_{ABCD}$\r\nPentru $N$ avem:\r\n$AD\\cdot d(N,AD)+BC\\cdot d(N,BC)=S_{ADC}+S_{ABC}=S_{ABCD}$\r\nPentru $O$ avem:\r\n$AD\\cdot d(O,AD)+BC\\cdot d(O,BC)=r(AD+BC)=r(AD+BC+AB+CD)\\cdot \\frac{1}{2}=S_{ABCD}$", "Solution_4": "[u]Aplicatia de la 2.[/u]\r\nFie $D_{1}\\in [AB]$ astfel incat $\\frac{AD_{1}}{AB}=\\frac{2}{3}$\r\nDin Stewart in triunghiul $LAB$, dupa simplificarea cu $\\frac{AB}{3}$ avem:\r\n$LA^{2}+2LB^{2}=3LD_{1}^{2}+\\frac{2}{3}AB^{2}$\r\nInlocuim in relatie si avem:\r\n$3LD_{1}^{2}+3LC^{2}-6LD^{2}=k\\cdot [ABCD]-\\frac{2}{3}AB^{2}$\r\nLuam $D_{2}$ mijlocul lui $CD_{1}$ si cu teorema medianei avem:\r\n$3(\\frac{4LD_{2}^{2}+CD_{1}^{2}}{2})-6LD^{2}=k\\cdot [ABCD]-\\frac{2}{3}AB^{2}$\r\nDeci\r\n$LD_{2}^{2}-LD^{2}=\\frac{k\\cdot [ABCD]-\\frac{3}{2}CD_{1}^{2}-\\frac{2}{3}AB^{2}}{6}$\r\nLocul geometric este deci o dreapta perpendiculara pe $DD_{2}$.", "Solution_5": "[quote=\"Virgil Nicula\"][color=darkred]$2.\\blacktriangleright$ Se dau doua puncte fixe $A$ , $B$ si constantele reale nenule $k_{1}$ , $k_{2}$ , $k$. Sa se determine locul geometric al punctelor $L$ pentru care $k_{1}\\cdot LA^{2}+k_{2}\\cdot LB^{2}=k\\ \\ (*)$. Sa se stabileasca relatia de compatibilitate. Discutie.\n[b]Aplicatie.[/b] Se considera patrulaterul convex $ABCD$. Sa se determine locul geometric al punctelor $L$ pentru care $LA^{2}+2\\cdot LB^{2}+3\\cdot LC^{2}=6\\cdot LD^{2}+k\\cdot [ABCD]$. [hide=\"Raspuns.\"] $k_{1}+k_{2}=0\\Longrightarrow$ O dreapta perpendiculara pe $AB$;\nin caz contrar $\\Longrightarrow$ un cerc cu centrul pe dreapta $AB$.[/hide][/color][/quote]\r\n[color=darkblue][b]Demonstratie.[/b] Pentru $k_{1}+k_{2}=0$ relatia $(*)$ devine \"diferenta $LA^{2}-LB^{2}$ este constanta\", ceea ce inseamna ca locul geometric cautat este o dreapta perpendiculara pe dreapta $AB$. Presupunem $k_{1}+k_{2}\\ne 0$ si notam punctul $C\\in AB$ pentru care $k_{1}\\cdot \\overline{CA}+k_{2}\\cdot\\overline{CB}=0\\ \\ (1)$. Se observa usor ca relatia $(1)$ inseamna $(2)\\ \\{\\begin{array}{c}(k_{1}+k_{2})\\cdot\\overline{CA}=-k_{2}\\cdot\\overline{AB}\\\\\\ (k_{1}+k_{2})\\cdot\\overline{CB}=-k_{1}\\cdot \\overline{AB}\\end{array}$.\nIn triunghiul $ALB$ aplicam relatia Stewart cevienei $[LC$ :\n$LA^{2}\\cdot\\overline{BC}+LB^{2}\\cdot\\overline{CA}+LC^{2}\\cdot\\overline{AB}+\\overline{BC}\\cdot\\overline{CA}\\cdot\\overline{AB}=0$.\nFolosind relatiile $(2)$ obtinem :\n$(k_{1}+k_{2})\\cdot (k_{1}\\cdot LA^{2}+k_{2}\\cdot LB^{2})=(k_{1}+k_{2})\\cdot LC^{2}+k_{1}k_{2}\\cdot AB^{2}$, adica lungimea segmentului $LC$ este constanta si punctul $C$ este fix, ceea ce inseamna ca locul geometric cautat in acest caz este un cerc cu centrul pe dreapta $AB$. Acest cerc exista daca si numai daca se verifica urmatoarea relatie de compatibilitate : $\\boxed{\\ k(k_{1}+k_{2})\\ge k_{1}k_{2}\\cdot AB^{2}\\ }$.[/color]", "Solution_6": "3.Fie $\\frac{FL}{FM}=\\frac{AC}{AB}=K$, constant.\r\nAvand pe FM luam un L' pe FM astfel incat $FL'=FM\\cdot K$ si apoi rotim pe FL' cu unghiul A. Astfel L' se suprapune peste L.\r\nObservam ca L' descrie prin omotetie o dreapta d', paralela cu d, sau un cerc C', omotetic cu C.\r\nRoting figura respectiva cu unghiul A, obtinem locul geometric cautat.\r\nIn cazul dreptei locul geometric va fi o dreapta care face unghiul A cu dreapta data.\r\nIn cazul cercului locul geometric va fi un cerc cu raza de k ori mai mare, iar unghiul format de dreptele care unesc centrele cercurilor cu punctul F va fi de asemenea A.", "Solution_7": "[u]Aplicatia 3[/u]\r\nLuam varful $Z'$ pe cerc, punct fix.\r\nDaca $Y'$ descrie dreapta $d_{1}$ atunci locul geometric al lui $X'$ astfel incat $X'Y'Z'\\sim XYZ$ este, conform problemei 3, o dreapta ce face unghiul $XZY$ cu $d_{1}$ si care se gaseste conform solutiei respective. Atunci $X'$ se va gasi la intersectia dintre dreapta respectiva si $d_{2}$\r\nAvand $X'$ si $Z'$ gasim pe $Y'\\in d_{1}$" } { "Tag": [], "Problem": "[ab/c(c+a)] + [bc/a(a+b)] + [ca/b(b+c)] >= (a/a+c) + (b/b+c) + (c/c+b)", "Solution_1": "[quote=\"Seo-Yeon, Sohn\"]$ \\frac{ab}{c(c\\plus{}a)} \\plus{} \\frac{bc}{a(a\\plus{}b)} \\plus{} \\frac{ca}{b(b\\plus{}c)}\\ge \\frac{a}{a\\plus{}c} \\plus{} \\frac{b}{b\\plus{}c} \\plus{} \\frac{c}{c\\plus{}b}$\n[/quote]", "Solution_2": "you are right, ElChapin...\r\n\r\nhow about you to try to prove it?? \r\n\r\ni tried it ,, but i can't", "Solution_3": "Is there a typo...like...$ \\frac{b}{b\\plus{}c} \\rightarrow \\frac{b}{a\\plus{}b}$?", "Solution_4": "[quote=\"Zellex\"]Is there a typo...like...$ \\frac {b}{b \\plus{} c} \\rightarrow \\frac {b}{a \\plus{} b}$?[/quote]\r\n\r\n sry, it is typo..." } { "Tag": [ "AMC" ], "Problem": "Hi, are there any awards or scholarships that are currently awarded to the top scorers on the two tests?", "Solution_1": "[quote=\"keta\"]Hi, are there any awards or scholarships that are currently awarded to the top scorers on the two tests?[/quote]\r\n\r\nThere were scholarships in 2001 and 2002, but they lost the funding. None before or since, although I suppose it could happen again if they got some funding for it.", "Solution_2": "There are substantial prizes for high performance on the USAMO. The Clay Math Award carries a prize of $\\$5000 (given for most elegant proof), while the third, second, and first prizes are somewhere around $\\$10,000, $\\$15,000, and $\\$20,000(maybe 25,000 - not sure)." } { "Tag": [ "geometry", "trapezoid" ], "Problem": "A tiling is assigned a product number based on the number of triangles and quadrilaterals that are visible in the tiling. A tiling's product number is equal to three times the number of triangles plus four times the number of quadrilaterals. What is the product number assigned to the pictured tiling?\n[asy]import olympiad; size(100); import geometry_dev; import graph; defaultpen(linewidth(0.8));\ndraw(origin--(1.5,0)--(1.5,1)--(0,1)--cycle); draw(origin--(0.75,1)--(1.5,0));[/asy]", "Solution_1": "There are three triangles and three quadrilaterals (a trapezoid on either side plus the rectangle). 3(3) +3(4)=9+12= 21." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "$ f$ is a real-valued function on the reals such that $ | f(x) |\\geq 1$ and $ f(x \\plus{} \\frac {13}{42}) \\plus{} f(x) \\equal{} f(x \\plus{} \\frac{1}{6}) \\plus{} f(x \\plus{} \\frac {1}{7})$ for all $ x$. Show that there is a real number $ c > 0$ such that $ f(x \\plus{} c) \\equal{} f(x)$ for all $ x$.", "Solution_1": "[quote=\"tdl\"]$ f$ is a real-valued function on the reals such that $ | f(x) |\\geq 1$ and $ f(x \\plus{} \\frac {13}{42}) \\plus{} f(x) \\equal{} f(x \\plus{} \\frac {1}{6}) \\plus{} f(x \\plus{} \\frac {1}{7})$ for all $ x$. Show that there is a real number $ c > 0$ such that $ f(x \\plus{} c) \\equal{} f(x)$ for all $ x$.[/quote]\r\nMust be $ |f(x)|\\leq 1$" } { "Tag": [ "limit", "floor function", "inequalities", "calculus", "calculus computations" ], "Problem": "Denote with $ f(k)$ greatest odd divisor (divider) of $ k\\in\\mathbb{N}$. Calculate\r\n(a) $ L: \\equal{}\\lim_{n}\\frac{1}{n}\\sum_{k\\equal{}1}^{n}\\frac{f(k)}{k}$\r\n(b)$ \\lim_{N\\rightarrow\\infty}\\sum_{n\\equal{}1}^{N}(\\sum_{k\\equal{}1}^{n}\\frac{f(k)}{k}\\minus{}nL)$", "Solution_1": "For $ 2^{m}\\leq n <2^{m+1}$ let $ \\lfloor x\\rfloor$ denote the largest integer smaller or equal to $ x$ and set $ T_{n}^{\\pm}=\\frac{2}{3}n\\left(1-4^{-(1+m)}\\right)\\pm\\left(1-2^{-(1+m)}\\right)$ and calculate:\r\n\r\n$ S_{n}: =\\sum_{k=1}^{n}\\frac{f(k)}{k}=\\sum_{i=0}^{m}\\sum_{j=0}^{\\lfloor\\frac{n-2^{i}}{2^{i+1}}\\rfloor}\\frac{f(2^{i}(2j+1))}{2^{i}(2j+1)}=\\sum_{i=0}^{m}\\sum_{j=0}^{\\lfloor\\frac{n-2^{i}}{2^{i+1}}\\rfloor}2^{-i}=\\sum_{i=0}^{m}2^{-i}\\lfloor\\frac{n+2^{i}}{2^{i+1}}\\rfloor$\r\n\r\ni.e. from the inequalities $ \\frac{n-2^{i}}{2^{i+1}}<\\lfloor\\frac{n+2^{i}}{2^{i+1}}\\rfloor\\leq\\frac{n+2^{i}}{2^{i+1}}$ we can deduce $ T_{n}^{-}nL$ and $ R_{N}=\\sum_{n=1}^{N}(S_{n}-nL)$ is an increasing sequence.\r\n\r\nFor the special case of $ N=2^{M}$ we compute \r\n\r\n$ R_{N}=\\frac{1}{3}\\sum_{n=1}^{2^{M}}\\sum_{k=0}^{m_{n}}\\epsilon_{k,n}2^{-k}=\\sum_{k=0}^{M}2^{-k}\\left|\\left\\{n=1,\\ldots,2^{M}|k\\leq m_{n}\\right\\}\\right|=\\\\ \\;\\;\\;\\;=\\sum_{k=0}^{M}2^{-k}(2^{M}-2^{k}+1)\\end{array}$\r\n\r\nwhich converges to infinity\r\n\r\nP.S. (hopefully no typos)", "Solution_2": "I spotted a typo, although it does not change the conclusions; the last lines should correctly read\r\n\r\n$ R_{N}\\equal{}\\frac{1}{3}\\sum_{n\\equal{}1}^{2^{M}}\\sum_{k\\equal{}0}^{m_{n}}\\epsilon_{k,n}2^{\\minus{}k}\\equal{}\\\\ \\;\\;\\;\\equal{}\\frac{1}{3}\\sum_{k\\equal{}0}^{M}2^{\\minus{}k}\\left|\\left\\{n\\equal{}1,\\ldots, 2^{M}|k\\leq m_{n},\\epsilon_{k,n}\\equal{}1\\right\\}\\right|\\equal{}\\\\ \\;\\;\\;\\equal{}\\frac{1}{3}\\sum_{k\\equal{}0}^{M}2^{\\minus{}k}\\left|\\left\\{n\\equal{}1,\\ldots,2^{M}|\\epsilon_{k,n}\\equal{}1\\right\\}\\right|\\equal{}\\\\ \\;\\;\\;\\equal{}\\frac{1}{3}\\left[2^{\\minus{}M}\\plus{}\\sum_{k\\equal{}0}^{M\\minus{}1}2^{\\minus{}k}2^{M\\minus{}1}\\right]\\equal{}\\frac{1}{3}[2^{M}\\plus{}2^{\\minus{}M}\\minus{}1]$\r\n\r\nwhich converges to infinity, too" } { "Tag": [ "geometry", "circumcircle", "inradius", "inequalities", "AMC" ], "Problem": "Consider acute $\\triangle ABC$. Let $R$, $r$, and $h_c$ denote the circumradius, inradius, and height to side $AB$ from point $C$ respectively. \r\n\r\n\r\nProve $R+r \\leq h_c$", "Solution_1": "Is this really true? Let $C = \\frac{\\pi}{2} - 2\\epsilon$, $A = B = \\frac{\\pi}{4} + \\epsilon$ and $\\epsilon \\rightarrow 0$. Then we have $h_c = \\frac{c}{2} = R$, but $r = \\frac{a+b-c}{2} = \\frac{c\\sqrt{2} - c}{2}$ so that $R + r = \\frac{\\sqrt{2}}{2}c > \\frac{c}{2} = h_c$.", "Solution_2": ":? Reverse inequality is wrong too. :?", "Solution_3": "woops" } { "Tag": [ "algebra unsolved", "algebra", "2-variable inequality", "Diophantine equation" ], "Problem": "Determine all pairs of positive numbers $a,b$ such that\n\n\\[9(a+b) +\\frac{1}{a}+\\frac{1}{b} \\geq 10 +\\frac{a}{b} + \\frac{b}{a}\\]\n\n[i]unknown source[/i]", "Solution_1": "Dear Manlio,\r\n\r\nWhy did you send this one to Number theory? Maybe you forgot something here? \r\n\r\nNamdung", "Solution_2": "well I'll move it to algebra, it seems like an algebra problem, if anything was forgotten by manlio I'll move it back! :D :D", "Solution_3": "Sorry for my banal error.", "Solution_4": "I dont see the meaning of this problem. But I have do solve it.\r\n\r\n1) a = b=1/3; \r\n2) b>=1 or b<=1/9 and a >= positive root of\r\n (9b-1)x^2 + (9b^2-10b+1)x + b-b^2 = 0\r\n\r\nNamdung" } { "Tag": [], "Problem": "Find, as a proper base ten fraction, the sum of the following infinite series of repeating decimals, all written in base eight:\r\n\r\nS = .07* + .007* + .0007* + ... (where * indicates a repeating digit)", "Solution_1": ".07*=.07+.007+.0007+...=.1=1/8. Similarly, .007*=.007+.0007+.00007+...=1/64, etc.\r\n\r\nThus the sum is\r\n\r\n1/8+1/64+1/512+...=(1/8)/(1-1/8)=1/7.", "Solution_2": "I did it slightly different: \r\n8S=1+S (.7* in base 8 is 1, just like .9* in base 10 is 1)\r\n\r\nSolvins for S easily gives you 1/7." } { "Tag": [ "inequalities" ], "Problem": "Find the minimum value of $ \\sum_{k \\equal{} 1}^{2n}|x \\minus{} k|$, where $ x\\in\\mathbb{R}$.", "Solution_1": "[hide=\"yay\"]\nLet $ E(n) \\equal{} \\min\\sum_{k\\equal{}1}^{n} |x\\minus{}k|$.\n\nFor odd $ n \\equal{} 2k\\plus{}1$ we have $ E(n) \\equal{} \\frac{n^2\\minus{}1}{4}$.\n\nFor even $ n \\equal{} 2k$ we have $ E(n) \\equal{} \\frac{n^2}{4}$.\n\nProof: For odd $ n$, we pair up the integers in the interval $ [1,n]$ as $ (1,n), (2,n\\minus{}1), (3,n\\minus{}2), \\cdots, (k,k\\plus{}2)$ with $ k\\plus{}1$ left over. Consider the $ k\\minus{}1$ intervals $ [1,n]$, $ [2,n\\minus{}1]$, $ \\cdots$. Obviously if $ x$ is in one of these intervals $ [a,b]$, then the sum $ |x\\minus{}a|\\plus{}|x\\minus{}b|$ is constant and less than the sum when $ x \\notin [a,b]$. Hence the sum attains a minimum when $ x$ is in all of the intervals - that is, when $ x \\in [k,k\\plus{}2]$. But we have the additional $ |x\\minus{}(k\\plus{}1)|$ term. Hence the minimum occurs when $ x \\equal{} k\\plus{}1$. The sum is relatively easily computed as $ E(n) \\equal{} 2\\sum_{i\\equal{}1}^k i \\equal{} k^2\\plus{}k \\equal{} \\frac{n^2\\minus{}1}{4}$.\n\nFor even $ n$, we have the intervals $ [1,n], [2,n\\minus{}1], \\cdots, [k,k\\plus{}1]$. By similar reasoning the minimum occurs when $ x \\in [k,k\\plus{}1]$. Obviously $ E(n) \\equal{} \\sum_{i\\equal{}1}^k 2i\\minus{}1 \\equal{} k^2 \\equal{} \\frac{n^2}{4}$.\n[/hide]" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Let $ A,B,C,D,E$ be points on a circle such that $ \\angle ABC\\equal{}\\angle BCD\\equal{}\\angle CDE\\equal{}45^{\\circ}$. Prove that the region above the zig-zag line $ ABCDE$ and the region below it have equal areas.", "Solution_1": "Let $ X \\equiv CB \\cap AE , Y \\equal{} CD \\cap AE.$ Then $ \\triangle CAE$ is right with $ CA \\equal{} CE \\equal{} L.$ Now, we shall prove that the area of $ \\triangle CXY$ equals the sum of the areas of $ \\triangle ABX$ and $ \\triangle EDY.$ Since these triangles are similar, because of $ AB \\parallel CY$ and $ ED \\parallel CX,$ we shall show $ XY^2 \\equal{} AX^2 \\plus{} EY^2.$ $ \\triangle CXY \\sim \\triangle AYC \\sim \\triangle EXC$ yields\n\n$ \\frac {AY}{L} \\equal{} \\frac {L}{EX} \\Longrightarrow AY \\cdot EX \\equal{} L^2 \\equal{} \\frac {_1}{^2}AE^2$ \n\n$2(AX \\plus{} XY)(EY \\plus{} XY) \\equal{} (XY \\plus{} AX \\plus{} EY)^2 \\Longrightarrow XY^2 \\equal{} AX^2 \\plus{} EY^2.$" } { "Tag": [ "geometry", "Science Olympiad" ], "Problem": "Whats yours?\r\n\r\nI started playing FICS chess today and i got my rating to 1372.", "Solution_1": "why don't you play at yahoo?", "Solution_2": "I have been playing on yahoo chess, but recently it has not working....in the middle of games, i get disconnected and my rating drops. :( With FICS theres a lot less glitches.", "Solution_3": "Is this blitz or standard? (blitz is deflated and standard is inflated on FICS..)", "Solution_4": "I play on USCL. My blitz is 2100.", "Solution_5": "you need to specify what rating? \r\n\r\nyahoo-1400\r\nFICS- umm 1400?\r\nUSCF- 1100 lol", "Solution_6": "USCF- 1300", "Solution_7": "USCF-1223 :D \r\nFICS-1300-ish\r\n\r\nWhat's sad is that my little brother's rating is higher than mine :rotfl:", "Solution_8": "lol I'm 1493 USCF\r\n\r\n1832 Yahoo\r\n\r\n1980 Blitz Fics", "Solution_9": "yahoo-1400ish, yeah I know I suck cuz I gave up in chess :(", "Solution_10": "I once had a USCF but haven't kept it up\r\nI do play actively on Gameknot.com (good well reputed chess website) with rating around 1700, although I haven't played very many games (it will be going up... I should be around 1900)", "Solution_11": "Cool, what's your handle on gameknot? We can play sometimes. My is chess55555, and my rating is 1880 right now. :)", "Solution_12": "my national is 1765 before the update... \r\n\r\nmy online is 2200\r\n\r\nmy gameknot.com is 1781 \r\n\r\ni drew a 2150 otb today :D", "Solution_13": "FICS: 1350-1400, fluctuates\r\nUSCF: 1100", "Solution_14": "For someone new at playing chess competitively, (I've played several times with friends/family) what is the suggested website?", "Solution_15": "By ......playing rating tmt of course! :D\nAre you new in chess?", "Solution_16": "I am new to this rating thing.. you mind explaining. ??", "Solution_17": "You havent played in competitions then??\nAnyway i will try to explain.Rating is just used to describe the stength of a player.For example,a player of rating 2000 is (considered)stronger than the one having 1200.To get the rating you have to score at least some points( i guess min2.5 of 9,I dont know about recent rules) against rated players in a rating tmt(most tmt are rating tmt though).Then you earn rating based on your performance.From then you can increase your rating by beating other rated players.The more your opponents rating the more you gain.If you lose against a rated then you lose some rating depending upon his rating.To become an international master(IM) you need more than 2400,and more than 2600 to become Grand master(GM).At the point of writing, the top rated player in the world is Magnus carlsen with rating above 2800.\nSo that is all about rating. ", "Solution_18": "USCF is 1517, online I suck (~1200)", "Solution_19": "I was reading this thread, and saw this:\n[quote=BOGTRO]USCF - 1815\n\nICC Standard - about 1825\n\nICC Blitz - about 1750 (Yeah I know, crashed)\n\nDraw a GM OTB on Monday =)\n\n\nbtw Alex Lenderman is on ICC as 'manest'\n\nMy account is 'Cryptochess'[/quote]\nI burst out laughing....and then looked at the timestamp...\n", "Solution_20": "Yep, what a difference 7 years makes :) ", "Solution_21": "Yow.. I'm maybe 800. -.-'", "Solution_22": "I have 384 on chess.com.\n\nI hate timed chess.", "Solution_23": "[quote=Bob_Smith]I have 384 on chess.com.\n\nI hate timed chess.[/quote]\n\nI was 600 when I stopped. I think I can maybe get to 800, now.\nI also hate timed chess, but I hate it even more when you have to give up because your opponent is taking 3 days between each move, and then your rating goes down. :(", "Solution_24": "I'm a 124.", "Solution_25": "[quote=lkarhat]I'm a 124.[/quote]\n\n:D", "Solution_26": "Umm... 500? Uhh...", "Solution_27": "My grandpa was around 1800", "Solution_28": "Online 1900\n\n", "Solution_29": "[s]2000 uscf[/s]" } { "Tag": [], "Problem": "Let $ a$, $ b$, $ c$, $ d$ be four positive reals such that $ d\\equal{}a\\plus{}b\\plus{}c\\plus{}2\\sqrt{ab\\plus{}bc\\plus{}ca}$. Prove that $ a\\equal{}b\\plus{}c\\plus{}d\\minus{}2\\sqrt{bc\\plus{}cd\\plus{}db}$.", "Solution_1": "\\begin{eqnarray*} d & = & a + b + c + 2\\sqrt {ab + bc + ca} \\\\\r\nd - a - b - c & = & 2\\sqrt {ab + bc + ca} \\\\\r\na^2 + b^2 + c^2 + d^2 + 2ab + 2bc + 2ca - 2ad - 2bd - 2cd & = & 4ab + 4bc + 4ca \\\\\r\na^2 + b^2 + c^2 + d^2 - 2ab + 2bc - 2ca - 2ad + 2bd + 2cd & = & 4bc + 4bd + 4cd \\\\\r\n\\left(a - b - c - d\\right)^2 & = & \\left(2\\sqrt {bc + bd + cd}\\right)^2 \\\\\r\nd > a + b + c > a - b - c & \\Rightarrow & a - b - c - d < 0 \\\\\r\na - b - c - d & = & - 2\\sqrt {bc + bd + cd} \\\\\r\na & = & b + c + d - 2\\sqrt {bc + bd + cd} \\end{eqnarray*}\r\nqed." } { "Tag": [ "algorithm", "number theory", "relatively prime" ], "Problem": "Prove that $(f_n, f_{n-1})=1$, where $f_n$ is the nth Fibonacci number.\r\n\r\nI'm stuck :(.", "Solution_1": "we have $f_n=f_{n-1}+f{n-2}$ and $f_0=f_1=1$ and we have one more thing $f_n= \\frac{(x_1)^n-(x_2)^n}{\\sqrt{5}}\\leq 1$ and $x_1=\\frac{1+\\sqrt{5}}{2}$ and ${x_2=\\frac{1- \\sqrt{5}}{x}}$.\r\nand i think with these we can prove it easily. :D", "Solution_2": "Hint. .....If two numbers have a common factor, their difference also is divisible by that factor ... [hide]Think if $f_{100}$ and $f_{99}$ had common factor (say $d$ ) won't it also divide $f_{98}$ .. How about $f_{97}$ ? ... and remember $f_1$ has no divisors except 1 :)[/hide]", "Solution_3": "[quote=\"jensen\"]we have $f_n=f_{n-1}+f{n-2}$ and $f_0=f_1=1$ and we have one more thing $f_n= \\frac{(x_1)^n-(x_2)^n}{\\sqrt{5}}\\leq 1$ and $x_1=\\frac{1+\\sqrt{5}}{2}$ and ${x_2=\\frac{1- \\sqrt{5}}{x}}$.\nand i think with these we can prove it easily. :D[/quote]\r\n\r\nThis won't help. Look at Gyan's hint.", "Solution_4": "Other similar, well known , and \"easy\" to prove problems are of the type:\r\n \r\n 1. $f_n$ will divide $f_m$ if $n$ divides $m$. \r\n 2. If $f_n$ is prime, then $n$ is prime. ([b]Well , except for the case n=4)[/b] (Reverse of course need not be true.) \r\n 3. For n>4, if $n$ is prime, $f_{n+1}$ is never a prime. \r\n\r\nAnd one not so easy to prove ('cause if you show this you'd be famous, .. I don't think it has been proven yet) :) \r\n\r\n4. There are infinite fibonacci primes.", "Solution_5": "Heres a pretty cool theorem you could use, (if you can prove it!)\r\n\r\n$\\gcd{(F_n,F_m)}=F_{\\gcd{(n,m)}}$", "Solution_6": "if you use the reverse of the euclidean algorithm, and add one each time, you generate the fibbonacci sequence", "Solution_7": "[hide=\"Solution\"] We can use induction. We know that 1 and 1, the first two Fibonacci numbers, are relatively prime. Now we show that if $F_{n},F_{n+1}$ are relatively prime, then $F_{n+1},F_{n+2}$ are also relatively prime.\n\nLet $m$ be any divisor of $F_{n+1}$. We know that $F_{n+1}\\equiv 0\\mod{m}$, and $F_{n}\\equiv a_{m}\\mod{m}$, for some $a\\not\\equiv 0\\mod{m}$. Adding the two to generate $F_{n+2}$ in mod $m$, we see that $F_{n+2}\\equiv a_{m}\\mod{m}$, so $m$ is not a divisor of $F_{n+2}$. Since this works for any divisor of $F_{n+1}$, we know that $F_{n+1},F_{n+2}$ have no common divisors and are therfore relatively prime.\n\nQ.E.D.[/hide]" } { "Tag": [], "Problem": "Simplify the sum\r\n$i+2i^2+3i^2+\\ldots2007i^{2007}$", "Solution_1": "$S: =i+2i^2+3i^3+...+2007i^{2007}$\r\n\r\n$Si=i^2+2i^3+3i^4+...+2006i^{2007}+2007i^{2008}$\r\n\r\n$S-Si=(i-2007i^{2008})+i^2+i^3+...+i^{2007}$\r\n\r\n$S= \\frac{2007i^{2008}+1}{i-1}$\r\n\r\n$=-1004(i+1)$", "Solution_2": "that's a nice solution. if you do it the long way you run into some combinatorics that could lead to an error", "Solution_3": "[hide]We will first find $i+2i^2+...+2008i^{2008}$ and subtract the last term, which is 2008. \n\nIf group them into 4's: \n\n$(i+2i^2+3i^3+4i^4)+(5i^5+6i^6+7i^7+8i^8)+....+2007i^{2007}+2008i^{2008}=(-2i+2)+(-2i+2)+...+(-2i+2)$.\n\nThere are 502 goups, so we have the sum $1004(1-i)$. \n\nAs said in the beginning, we must subtract 2008, so the answer is \n\n$1004-1004i-2008=-1004i-1004=-1004(1+i)$. \n\n[/hide]", "Solution_4": "[quote=\"drunner2007\"]Simplify the sum\n$i+2i^2+3i^2+\\ldots2007i^{2007}$[/quote]\r\n\r\nThe solution by [b]Altheman[/b] is great! I would like to suggest yet another way of computing the given complex sum that uses the idea of differentiation. It might look a bit cumbersome at first glance but knowing this method will help you tackle a bunch of similar problems in a more general way. Anyway, here is the method and the solution.\r\n\r\nWe know \r\n$1 + x + x^2 + x^3 + ... + x^n = \\frac{x^{n+1} - 1}{x-1}$\r\n\r\nNow differentiating both sides of the equation w.r.t. $x$, we obtain\r\n$1 + 2x + 3x^2 + ... + nx^{n-1} = \\frac{(x-1)(n+1)x^n - (x^{n+1}-1)}{(x-1)^2}$\r\n\r\nMultiplying both sides of the equation with $x$, we obtain\r\n$x + 2x^2 + 3x^3 + ... + nx^n = \\frac{x}{(x-1)^2}.[(n+1)x^{n+1}-(n+1)x^n-x^{n+1}+1]$\r\n\r\n$\\therefore x + 2x^2 + 3x^3 + ... + nx^n = \\frac{x}{(x-1)^2}.[nx^{n+1}-(n+1)x^n+1]$\r\n\r\nNow put $x=i$ and $n=2007$ in the above equation to get\r\n$i + 2i^2 + 3i^3 + ... + 2007i^{2007} = \\frac{i}{(i-1)^2}.[2007i^{2008}-2008i^{2007}+1]$\r\n$= -\\frac12.[2007+2008i+1] = -1004(1+i)$" } { "Tag": [ "modular arithmetic", "number theory" ], "Problem": "find the remainder when 2000! is divided 2003?", "Solution_1": "Hint: By Willson's theorem we have $2002!\\equiv-1(\\mod 2003)$.", "Solution_2": "I hope this is right...\r\n\r\n[hide=\"Solution\"]By Wilson's theorem, we have $2002!\\equiv-1 (\\text{mod 2003})$. This equals $2001!\\equiv 1(\\text{mod 2003})$.\nThis can be transformed into $3\\cdot 23\\cdot 29\\cdot 2000!\\equiv 2004(\\text{mod 2003})=23\\cdot 29\\cdot 2000!\\equiv 668(\\text{mod 2003})$\n$\\Longrightarrow 23\\cdot 29\\cdot 2000!\\equiv-668334(\\text{mod 2003})\\Longrightarrow 2000!\\equiv 1001(\\text{mod 2003})$. Our answer is then $\\boxed{1001}$. [/hide]", "Solution_3": "[quote=\"Xaero\"]I hope this is right...\n\n[hide=\"Solution\"]By Wilson's theorem, we have $2002!\\equiv-1 (\\text{mod 2003})$. This equals $2001!\\equiv 1(\\text{mod 2003})$.\nThis can be transformed into $3\\cdot 23\\cdot 29\\cdot 2000!\\equiv 2004(\\text{mod 2003})=23\\cdot 29\\cdot 2000!\\equiv 668(\\text{mod 2003})$\n$\\Longrightarrow 23\\cdot 29\\cdot 2000!\\equiv-668334(\\text{mod 2003})\\Longrightarrow 2000!\\equiv 1001(\\text{mod 2003})$. Our answer is then $\\boxed{1001}$. [/hide][/quote]\n\n[hide] After you get $2001! \\equiv 1 \\pmod{2003}$, we can simply divide by $-2$ to get $2000! \\equiv \\frac{2004}{-2}\\equiv-1002 \\equiv 1001 \\pmod{2003}$.[/hide]", "Solution_4": "Yes that is correct.\r\n\r\nHow would you do it if it said the remainder when 1000! is divided by 2003?", "Solution_5": "[quote=\"mathgeniuse^ln(x)\"]remainder when 1000! is divided by 2003?[/quote]\r\n\r\n[hide]We have $1=-2002,2=-2001,...,1000=-1003,1001=-1002$ in $\\mathbb{Z}_{2003}$[/hide]", "Solution_6": "Well, for $1000! \\pmod{2003}$, we'll still consider wilson's theorem such that $2002!\\equiv-1 \\pmod{2003}$. Since we're are in $\\mod 2003$, $2002!\\equiv 1001! \\cdot-(1001!) \\equiv-(1001!)^{2}$. This will give two answers $1000!\\pmod{2003}=2001$ or $1000!\\pmod{2003}=2$ because it is a quadratic." } { "Tag": [ "geometry", "parallelogram", "geometric transformation", "vector", "rotation", "angle bisector", "geometry solved" ], "Problem": "let $ABCD$ be a cyclic quadrilateral and $P$ be a point such that \r\n$\\angle APD+\\angle BPC=180$ find \r\n$\\angle PAD+\\angle PCB$", "Solution_1": "Are you sure this is the statement? I don't think it's a fixed value.. :?", "Solution_2": "I saw this problem in a magazine, I think it should be right !", "Solution_3": "But think about it: we can fix $A,D$, and move $B,C$ on the circle so that $\\angle BPC+\\angle APD=\\pi$. The triangle $PBC$ won't have all its angles fixed, in general.", "Solution_4": "[quote=\"payman_pm\"]let $ABCD$ be a cyclic quadrilateral and $P$ be a point such that \n$\\angle APD+\\angle BPC=180$ find \n$\\angle PAD+\\angle PCB$[/quote]\r\n\r\nIndeed, < PAD + < PCB is not a fixed value. But the following modified version of the problem is correct:\r\n\r\n[color=blue]Let ABCD be a parallelogram and P a point inside it such that < APD + < BPC = 180\u00b0. Prove that < PAD + < PCB = < BAD.[/color]\r\n\r\nThe [i]solution[/i] is canonical: Since the quadrilateral ABCD is a parallelogram, we have the vectorial equality $\\overrightarrow{AB}=\\overrightarrow{DC}$. Now, let Q be a point such that $\\overrightarrow{PQ}=\\overrightarrow{AB}=\\overrightarrow{DC}$ (this point Q can be constructed as the image of the point P under the translation along the vector $\\overrightarrow{AB}=\\overrightarrow{DC}$). Since $\\overrightarrow{PQ}=\\overrightarrow{AB}$, the quadrilateral APQB is a parallelogram, and thus QB || PA and < PQB = < PAB.\r\n\r\nSince $\\overrightarrow{PQ}=\\overrightarrow{AB}=\\overrightarrow{DC}$, the points Q, B and C are the images of the points P, A and D under a given translation. Since translations preserve angles, we thus have < BQC = < APD. Thus, the equation < APD + < BPC = 180\u00b0 becomes < BQC + < BPC = 180\u00b0. Hence, the quadrilateral BPCQ is cyclic, and it follows that < PQB = < PCB. Comparing this with < PQB = < PAB, we get < PCB = < PAB, and thus < PAD + < PCB = < PAD + < PAB = < BAD. Problem solved.\r\n\r\n Darij", "Solution_5": "No,you are actually wrong...the problem statement is very corect...and here is the sollution.\r\nSo, if we name M the intersection between [BC] and [AP] and Q1 the intersection of [BD] and [AM] and Q2= the int between [CE] and [AM] and also <(BAM)=x,<(MAC)=y,<(ABP)= :alpha: <(PBM)= :beta: <(PCM)= :gamma: ,<(ACP)= :lambda and <(PMB)= :theta: , it is obvious that <(ABQ)=:alpha/2 and <(APQ2)= :lambda:/2.Thus,we shall prove that Q1==Q2.\r\nAQ1/AP=sin(:alpha:/2)/sin:alpha:\r\nBQ2/BP=sin(:lambda:/2)/sin: lambda:\r\nAQ1/Q1M=AQ1/Q1M and also AQ2/Q2M=AQ2/Q2M\r\nso AQ1/sin( : alpha :/2)=AB/sin(:alpha:/2+APB)\r\nQ1M/sin(:alpha:/2+:beta:)=BM/sin(180-APB)=BQ/sin :theta: \r\nbut it is kmowm that sin(180-APB)=sin APB, so\r\nAQ1/Q1m=AB/BM*sin(:alpha:/2)/sin(:alpha:/2+APB)*sin(:alpha:/2+:beta:)/sin(APB) and so on,because at this point the sloveent is obvious...still thinikng of a 100% sollution of my own still.confused", "Solution_6": "Hmm... sounds like nonsense... :?\r\n\r\n Darij", "Solution_7": ":lambda", "Solution_8": "[quote=\"pestich\"]This problem is from Polish olympiad of 1971 problem #4, \n and your saw the official solution with lambdas and what not. \n Only 2% of participants solved it.\n \n\n For APD + BPC =180 P is on AC so that AC is angle bisector of BPD.\n Diagonal BD makes angle X with diagonal AC and this angle is\n exactly PAD + PCB as half sum of arcs AB and CD.[/quote]\r\n\r\nPlease explain why you think that P must lie on AC (actually, my sketch proves this wrong)...\r\n\r\n Darij", "Solution_9": "When P is on AC then APB+BPC = 180\r\n If AC is angle bisector of BPD then APB = APD and APD+BPC=180 \r\n as required.\r\n\r\n\r\n\r\n Maj. Pestich\r\n\r\n\r\n P.S.\r\n Are you pulling my leg (and everything that goes with it) , Darij ? \r\n To repay me for all the contraversial nonsense I put up here?", "Solution_10": "[quote=\"pestich\"]For APD + BPC =180 P is on AC so that AC is angle bisector of BPD.\n Diagonal BD makes angle X with diagonal AC and this angle is\n exactly PAD + PCB as half sum of arcs AB and CD.\n\n\n\n Maj. Pestich[/quote]\r\n\r\nI do not understand why $P$ must lie on the angle bisector $\\overline{AC}$ of $\\angle{BPD}$. Obviously we can rotate rays $PB$ and $PC$ about the point $P$, which would move $P$ off of the diagonal.", "Solution_11": "[quote=\"pestich\"]When P is on AC then APB+BPC = 180\n If AC is angle bisector of BPD then APB = APD and APD+BPC=180 \n as required.[/quote]\r\n\r\nBut, please, Pestich, you showed that if P is a point on the diagonal AC such that AC bisects < BPD, then < APD + < BPC = 180\u00b0, and now you claim the converse, namely that if < APD + < BPC = 180\u00b0, then P lies on AC and AC bisects < BPD. That's just wrong! You could easily see that there are infinitely many points P inside the quadrilateral ABCD satisfying < APD + < BPC = 180\u00b0, but only one point P on the diagonal AC such that AC bisects < BPD (at least, in the non-degenerate case). Hence your argument cannot be true.\r\n\r\n Darij", "Solution_12": ">=", "Solution_13": "Calm down guys. Pestich a small request from me would be, if possible, to give a little bit more detailed proofs. I admit that I have tried to read some of your geo proofs, but it is really hard. You leap over huge steps and sometimes we can't make the connections. ;)" } { "Tag": [], "Problem": "How many even four-digit counting numbers can be formed by \nchoosing digits from the set $ \\{1, 2, 3, 4, 5\\}$ if digits can be repeated?", "Solution_1": "5x5x5x2=250", "Solution_2": "The first three-digits can be any one of the $ 5$ digits, as they will not break restrictions. The last digit, however, must be either $ 2$ or $ 4$ because the number must be even." } { "Tag": [], "Problem": "In how many ways can we give 6 computers to 3 schools such that each school receives at least a computer??\r\nConsider the cases:1)the computers are distinct( example of arrangement: computer 1 to school 3,computer 2 to school 2,...,computer 6 to school 1)\r\n2)the computers are not distinct(example of arrangement:2 computers at school 1,2 computers at school 2,2 computers at school 3)\r\n\r\nGeneralization.", "Solution_1": "[hide=\"hint\"]Say each school gets 1, then the problem reduces to how many ways can you give out 3 computers to 3 school.[/hide]", "Solution_2": "[quote=\"xpmath\"][hide=\"hint\"]Say each school gets 1, then the problem reduces to how many ways can you give out 3 computers to 3 school.[/hide][/quote]\n\nThen would it be\n\n[hide]\nIf the computers are distinct;\nLets call the 3 computers 1, 2 and 3 respectively.\nThen all the possible ways to arrange them between the schools would be: \n\n1, 1, 1\n0, 1, 2\n0, 2, 1 \n1, 2, 0, \n1, 0, 2\n2, 1, 0,\n2, 0, 1\n3, 0, 0\n0, 3, 0\n0, 0, 3\n\nAdding all them up gives us 10 possibilities. \n\n\n[/hide]", "Solution_3": "Faster way-\r\n[hide]Let the computers be $ a_1,a_2,$ and $ a_3$, so $ a_1\\plus{}a_2\\plus{}a_3\\equal{}3$, and the number of solutions by balls and urns is 10.[/hide]", "Solution_4": "Or, if you think about, we have the five spots in between the 6 computers, and we need to choose 2 of them to insert dividers, so $ \\binom{5}{2}\\equal{}10$.\r\n\r\nExample: Let x=computer, |=divider\r\n\r\nxx|x|xxx" } { "Tag": [ "function", "algorithm" ], "Problem": "I'm new to computer programming, and I was wondering about the very first USACO problem in the training pages: Your Ride Is Here\r\n\r\n[size=100][code] It is a well-known fact that behind every good comet is a UFO. These UFOs often come to collect loyal supporters from here on Earth. Unfortunately, they only have room to pick up one group of followers on each trip. They do, however, let the groups know ahead of time which will be picked up for each comet by a clever scheme: they pick a name for the comet which, along with the name of the group, can be used to determine if it is a particular group's turn to go (who do you think names the comets?). The details of the matching scheme are given below; your job is to write a program which takes the names of a group and a comet and then determines whether the group should go with the UFO behind that comet. \n\n Both the name of the group and the name of the comet are converted into a number in the following manner: the final number is just the product of all the letters in the name, where \"A\" is 1 and \"Z\" is 26. For instance, the group \"USACO\" would be 21 * 19 * 1 * 3 * 15 = 17955. If the group's number mod 47 is the same as the comet's number mod 47, then you need to tell the group to get ready! (Remember that \"a mod b\" is the remainder left over after dividing a by b; 34 mod 10 is 4.) \n\n Write a program which reads in the name of the comet and the name of the group and figures out whether according to the above scheme the names are a match, printing \"GO\" if they match and \"STAY\" if not. The names of the groups and the comets will be a string of capital letters with no spaces or punctuation, up to 6 characters long.[/code][/size]\n\nHow do you read each letter one by one? And for converting each letter to the corresponding integer, I was going to cast each one as an int and do the necessary subraction/addition stuff, but was wondering if there was a shorter way (a method that does that?). I've looked online at the Sun webpage at the InputStreamReader class, but I couldn't find anything :maybe: [/code]", "Solution_1": "well... i dont know what do u use c++ or java.\r\n\r\nbut actually i advise u to take the word as char input. thn loop through the letters, convert them to integer by using their ascii value. then multiply and mod them... thats all.", "Solution_2": "Hello.\r\n\r\nWhat you can do is this:\r\n[code]\nint main(){\nchar word[5];\nofstream fout (\"filename.out\");\nifstream fin (\"filename.in\");\n\nfin >> word[0] >> word[1] >> word[2] >> word[3] >> word[4];\n\n}\n\n[/code]\nThis should work. You could also do:\n[code]\n#include \n#include \n#include \n\nusing namespace std;\n\nint main(){\nString word;\nofstream fout (\"filename.out\");\nifstream fin (\"filename.in\");\n\nfin >> word;\n\n}\n[/code]\r\nand from here use the string length function that will give you the characters from n to k (specified in the argument). If you do n to n+1, that will return that character. I don't remember the exact syntax, but I hope this will help.\r\nEDIT: This is c++.", "Solution_3": "With C++, you'd use cin.get(char& c).\r\n\r\nSo, as an example:\r\n[hide=\"C++ Solution\"]\n[code]\n#include \nusing namespace std;\n\nint word_total()\n{\n char in;\n int k, total=1;\n cin.get(in);\n while(!cin.fail() && in!='\\n')\n {\n k = (int) in - (int) 'A' + 1;\n total *= k%47;\n total = total%47;\n cin.get(in);\n }\n return total;\n}\n\nint main()\n{\n int word1, word2;\n word1 = word_total();\n word2 = word_total();\n if(word1 == word2)\n cout << \"GO\";\n else\n cout << \"STAY\";\n return 0;\n}\n[/code][/hide]\r\n\r\nBut you mentioned Sun, so I assume you're using Java. I haven't used Java in a while, but I think what you're looking for is InputStreamReader::read (http://java.sun.com/j2se/1.5.0/docs/api/java/io/InputStreamReader.html#read()). The algorithm will be the exact same as in the C++ example (read a character, add it to the total, total=total%47, and repeat until the character read is a newline).", "Solution_4": "Thanks a lot of the replies :)", "Solution_5": "[quote]But you mentioned Sun, so I assume you're using Java. I haven't used Java in a while, but I think what you're looking for is InputStreamReader::read (http://java.sun.com/j2se/1.5.0/docs/api/java/io/InputStreamReader.html#read()). The algorithm will be the exact same as in the C++ example (read a character, add it to the total, total=total%47, and repeat until the character read is a newline).\n[/quote]\r\n\r\nso it is impossible\r\n :P", "Solution_6": "metehan says [quote=\"metehan\"][quote]But you mentioned Sun, so I assume you're using Java. I haven't used Java in a while, but I think what you're looking for is InputStreamReader::read (http://java.sun.com/j2se/1.5.0/docs/api/java/io/InputStreamReader.html#read()). The algorithm will be the exact same as in the C++ example (read a character, add it to the total, total=total%47, and repeat until the character read is a newline).\n[/quote]\n\nso it is impossible\n :P[/quote]\r\n\r\nbut i don't understand why it is impossible.. :!:", "Solution_7": "Hashus, if you can't understand this, please don't write anything to this forum. :mad:", "Solution_8": "metehan, I don't see why it's impossible either. Please edify us with why \"it\" is impossible...", "Solution_9": "I would recommend reading the \"Submitting Solutions\" text in the training pages if you haven't already. However, to answer your immediate question for JAVA:\r\n\r\n[code]\n/*\nID: your_id_here\nLANG: JAVA\nTASK: ride\n*/\nimport java.io.*;\nimport java.util.*;\n\npublic class ride {\n public static void main (String [] args) throws IOException {\n// Use BufferedReader rather than RandomAccessFile; it's much faster\n BufferedReader f = new BufferedReader(new FileReader(\"ride.in\")); // input file name goes here\n PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter(\"ride.out\")));\n\n String s = f.readLine(); // gets entire line\n char[] c = s.toCharArray();\n int[] arr = new int[c.length];\n for(int i=0;ii = 0 ni follows with n \\geq 3:\r\n\r\na.) n - 1 = 2 => n = 3 or\r\nb.) n - 1 = p => due to n odd is p even: contradiction or\r\nc.) n - 1 = 2p => sum = 1; hence k = 1: contradiction => hence n = 3.\r\n\r\n2.) suppose k is not prime, e.g. k = rs with r,s \\in N, r,s > 1; then \r\n2p = 3k - 1 = 3rs - 1 = (3r ) s - 1 = (3r -1)* \\sum s-1 i = 0 3ri with 3r - 1 \\geq 8 even, which means p has non-trivial divisors: contradiction\r\n\r\nTherefore the original equation can have solutions at most in the case if k is prime. (e.g. for 2,5,11,17,19 1/2*(3^k-1) is no prime but for k = 3,7,13).\r\n\r\nSo the partial result: 2p + 1 = nk with p prime; k \\geq 2 is solvable at most if n = 3 and p,k odd primes; but it is not solvable for odd primes k. Does anybody know whether there are finitely or infinitely many solutions ?\r\n\r\nPS: Why don't the guest do not register or if registered yet then log in please. :)", "Solution_5": "Guest say:\r\n\r\n--------------------------------------------------------------------------------\r\n \r\n \r\nI think this problem is not yet solved. \r\nThe case that needs considering is 2p = 3^k - 1, where k is odd prime, which divides p-1. \r\nfor example one solution is p = 13; k = 3. \r\n \r\n \r\n--------------------------------------------------------------------------------\r\nYou see:\r\nif p=2:n=5,k=1. \r\n*if p=3^e+...+3+1: \r\n=>+ n=3,k=e+1. \r\n+ n=2.p+1,k=1; \r\n*For another case:n=2.p+1,k=1.\r\n--->>p=13->e=2->k=3." } { "Tag": [], "Problem": "Does anyone play connect 5 (gomoku)?\r\n\r\nI mean ya the rules are pretty simple so you guys may not think it takes much strategy, but if you're really good at it, it is a pretty fun strategy game, and it doesn't take hours like chess, or a specialized board and game pieces.", "Solution_1": "Yeah, I did use to play connect 5.\r\nUnfortunately, it seemed to me that the first player has a substantial advantage.\r\n\r\nSo now I mostly play Connect6, sometimes connect 5 agains classmates." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $f: E\\to E$ be an isometry with $E$ complete and $f(E)$ totally bounded. Prove that $f$ is a bijection and that $E$ is compact.\r\n[hide]Injection is trivial. I can think of two approaches:1.(the way I tried!) First we show that $f$ is surjective. Then according to a theorem $E$ being complete and totally bounded, will be compact.\n2.We prove $E$ is compact and according to a problem on the forum $f$ will be surjective.[/hide]", "Solution_1": "(2) makes more sense. Since $f(E)$ is totally bounded, so is $E$, because $f$ is an isometry. Hence $E$ is compact." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Is it possible to list the non-negative integers as $n_1$, $n_2$, $n_3$, ..., with each non-negative integer appearing precisely once in the list, in such a way that for each $i = 1$, 2, 3, ..., we have $n_{i + 1} = n_i + 2$, $n_{i + 1} = n_i - 2$, $n_{i + 1} = 2n_i$, or $n_{i + 1} = n_i/2$?", "Solution_1": "to start with note that from $0$ one has only 1 option; $0\\rightarrow 2$;likewise from 2 there are only 2 options; $2\\rightarrow 1,2\\rightarrow 0$,so the following is forced on us: $a_0=0,a_1=2,a_2=1,a_3=3$.\r\nnow we make the following observation: from $2n+1$ the following sequence of 'moves' is possible:\r\n \\[2n+1\\rightarrow 4n+2\\rightarrow 4n\\rightarrow 4n-2\\rightarrow\\cdots\\rightarrow 2n+2\\rightarrow 4n+4\\rightarrow 4n+6\\rightarrow 2n+3\\rightarrow 2n+5\\rightarrow\\cdots\\rightarrow 4n+5\\rightarrow 4n+7\\] which implies that if we have a sequence containing all the integers $\\leq 2n+1$,with $2n+1$ as the lastin sequence, we can extend it to contain all the integers till $4n+7$, which is again odd and so this can be extended further again.\r\nsince we start out having $0,1,2,3$ as the first 4 terms with $a_3=3$, it follows that we can get such a sequence." } { "Tag": [ "algebra", "polynomial" ], "Problem": "$(1+x^2)(1-x^3)$ equals\n\n$ \\text{(A)}\\ 1 - x^5\\qquad\\text{(B)}\\ 1 - x^6\\qquad\\text{(C)}\\ 1+ x^2 -x^3\\qquad \\\\ \\text{(D)}\\ 1+x^2-x^3-x^5\\qquad \\text{(E)}\\ 1+x^2-x^3-x^6 $", "Solution_1": "[hide](1+x^2)(1-x^3)\n1-x^3+x^2-x^5\n-x^5-x^3+x^2+1\nD.[/hide]\r\nYou forgot to put the answer in descending order. :P", "Solution_2": "[quote=\"Fish_Factorial\"]You forgot to put the answer in descending order. :P[/quote]\r\n\r\nBut he did put it in ascending order :D", "Solution_3": "[hide]\n$(1+x^2)(1-x^3)=1-x^3+x^2-x^5=-x^5-x^3+x^2+1=1+x^2-x^3-x^5$ So, $\\boxed{D}$.[/hide]", "Solution_4": "[hide]\n$( 1 + x^2)(1 - x^3)=1 + x^2 - x^3 - x^5$\n\nthe answer is d\n[/hide]", "Solution_5": "[hide]\\[ (1+x^2)(1-x^3)=1-x^3+x^2-x^5 \\]\n\\[ Answer=D \\][/hide]", "Solution_6": "[quote=\"Silverfalcon\"]$(1+x^2)(1-x^3)$ equals\n\nA. $1 - x^5$\nB. $1 - x^6$\nC. $1+ x^2 -x^3$\nD. $1+x^2-x^3-x^5$\nE. $1+x^2-x^3-x^6$[/quote]\r\n\r\n[hide]D where did you get this? [/hide]", "Solution_7": "silverfalcon said:\r\n\r\n[quote]Source: USA AMC 12 1987, Problem 1 [/quote]\r\n\r\n\r\n[hide]i got d[/hide]", "Solution_8": "[hide]Answer D by foil[/hide]\r\nI always keep thinking there's some kind of trick on these polynomial questions, but the subject topic sort of gave it away. :)", "Solution_9": "[hide]$\\boxed {D}$[/hide]", "Solution_10": "[hide]D[/hide]\r\n\r\nThis is totoally random but doesn't $\\text\\ FOIL$ stand for something?\r\nI'm pretty sure it does but I can't remember what it stands for.\r\nIf not then why do they call it $\\text\\ FOIL$?", "Solution_11": "[b]F[/b]irst\r\n[b]O[/b]uter\r\n[b]I[/b]nner\r\n[b]L[/b]ast", "Solution_12": "this is kind of random, but why do 10 ppl have to put up the same answer to a problem?! if it has been solved by a few ppl u dont have to solve it to!! :|", "Solution_13": "[quote=\"hoshattack\"]this is kind of random, but why do 10 ppl have to put up the same answer to a problem?! if it has been solved by a few ppl u dont have to solve it to!! :|[/quote]I guess people have a knack for 'showing' to others that they can solve the problem too, and by 'showing' I mean it literally :roll:", "Solution_14": "but still, once it has been solved anybody could copy the same thing. And this forum is not about showing off, but LEARNING", "Solution_15": "[quote=\"Silverfalcon\"]$(1+x^2)(1-x^3)$ equals\n\nA. $1 - x^5$\nB. $1 - x^6$\nC. $1+ x^2 -x^3$\nD. $1+x^2-x^3-x^5$\nE. $1+x^2-x^3-x^6$[/quote]\r\n\r\n[hide]D[/hide]", "Solution_16": "oh im sorry. i didnt read the posts right above me. :blush:", "Solution_17": "Hey buzzer and Jose, you guys should post a solution at least even if there are many others already.", "Solution_18": "First:\r\n$1\\times1$\r\nOuter:\r\n$1\\times-x^3$\r\nInner:\r\n$x^2\\times1$\r\nLast:\r\n$x^2\\times-x^3$\r\nthen add the four together.\r\n$1+x^2-x^3-x^5$\r\nThe answer is $\\boxed{D}$\r\n\r\nJust to clarify for everyone what foil is and how to do this problem :)", "Solution_19": "The answer is D due to FOIL.\r\n \r\nFirst \r\nInner\r\nOuter \r\nLast" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO" ], "Problem": "I'm asking this for my brother because he never posts on this forum:\r\n\r\nWould the Intro books be good for preparing for the AIME if you know the general ideas but not the details?\r\n\r\nThanks in advance.", "Solution_1": "I don't know if the Intro books are great for AIME;if you just want to make AIME, the challenge problems are good.", "Solution_2": "The Intro books are probably good for the first 3 to 5 problems on the AIME.", "Solution_3": "[quote=\"xpmath\"]I don't know if the Intro books are great for AIME;if you just want to make AIME, the challenge problems are good.[/quote]\r\nHis goal is to make the USAMO\r\n\r\nYeah, I seem to notice maybe 1 or 2 AIME problems at the end of the Intro books, but they seem as if they might be # 2 or 3 or something.", "Solution_4": "To make USAMO, the Intermediate C+P and Vol. 2 would be the best. And old problems.", "Solution_5": "I'm not exactly sure about those ,but I do know that the 21st Century Contests CD can be a great help. It contains AIME competitions all the way back from 2001.", "Solution_6": "[quote=\"gauss1181\"]I'm not exactly sure about those ,but I do know that the 21st Century Contests CD can be a great help. It contains AIME competitions all the way back from 2001.[/quote]\r\n\r\nwhich is what XPmath said about old problems. :wink:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $\\{a_k\\}_{k\\geq 0}$ be a sequence given by $a_0 = 0$, $a_{k+1}=3\\cdot a_k+1$ for $k\\in \\mathbb{N}$.\r\n\r\nProve that $11 \\mid a_{155}$", "Solution_1": "What is $a_0$? 0? 11? 121?\r\n\r\nUsing $b_n =a_n + \\frac 1 2$, we deduce that $b_{n+1} = 3b_n$ so that $b_n = 3^nb_0$.\r\nTherefore $a_n = 3^n(a_0 + \\frac 1 2 ) - \\frac 1 2$\r\n\r\nIt follows that $a_{155} = 3^{155}a_0 + \\frac {3^{155} - 1} 2$\r\n\r\nBut, it is easy to see that $3^5 = 1$ mod[11]. Since $3^{155} - 1$ is even, we deduce that $a_{155} = a_0$ mod[11].\r\n\r\nThus, we are back to my initial question : what is $a_0$?\r\n\r\nPierre.", "Solution_2": "0. weird, apparently it vanished.\r\n\r\nanyway, I think I proved it for a higher power of 11... any try? :P", "Solution_3": "[quote=Peter]anyway, I think I proved it for a higher power of 11... any try? :P[/quote]\nPut $x=11$\n$$3^5=2x^2+1\\implies 3^{155}-1=(2x^2+1)^{31}-1= \\left(\\sum_{k=0}^{31} \\binom{31}{k} (2x^2)^{31-k} \\right) -1 \\implies v_{11}(3^{155}-1)=2$$\n:) so $11^2$ also works\n\n", "Solution_4": "That was computable. Let's try making it a bit generalized in various cases, one of these cases being below.\nProve that for any integer $k$, $a_{110k+54}$ is divisible by $11$", "Solution_5": "Divide by $ 3^{k+1}$ and put $ b_n= \\frac {a_n} {3^n} $ to get $ b_{n+1} = b_n + \\frac 1 {3^{n+1}} $\nThis directly gives $a_n= \\frac {3^n - 1}{2}$\nNow we can generalize anyway we want.\nAs $Ord_{11} 3 = 5 $ we have $11|a_n$ for all $n$ divisible by 5\n@above I think you meant $110k+55$", "Solution_6": "\\begin{align*}\na_{155} &= \\frac{3^{155}-1}{2}\\\\\n&= 44953600931767927210351145067881142268656481697351341318544905244162412253\\\\\n&= 11 \\times 4086690993797084291850104097080103842605134699759212847140445931287492023\n\\end{align*}" } { "Tag": [ "search", "\\/closed" ], "Problem": "Everytime I log on, I check the \"remember me\" option, but when I come back, it still says that I haven't logged in.\r\n\r\nEven though it says \"View posts since last visit (0)\", when I click on it, it tells me \"Search found 22 matches\". Before the update, only the new/edited posts are displayed, but now, it seems rather random.", "Solution_1": "About the remember me option, you must delete your cookies (if you can locate the mathlinks cookie, just that) and do the operation again. It will work this time. \r\n\r\nAbout the unread flags, yes, I know, it's a bug that I coudn't fix up to know, but I am working on it. \r\n\r\nKeep the reports coming in." } { "Tag": [ "function", "limit", "trigonometry", "algebra", "domain", "real analysis", "topology" ], "Problem": "If $\\forall \\epsilon>0 \\exists \\delta(\\epsilon): \\text{ } \\forall x,y \\in (0,+\\infty) : |x-y|<\\delta \\Rightarrow |f(x)-f(y)|<\\epsilon$ ( i dont know how it calls, equally continuous???) then what can we state about f(+0), f(+oo)?", "Solution_1": "[quote]i dont know how it calls[/quote]\r\ni'm not sure if this is what you're asking for, but here it is\r\n\r\nJust say i have a function $f$, and i think the limit of the function at point $a$ is $L$.\r\n\r\nEpsilon delta basically makes a box. the two vertical lines are at $a + \\delta$ and $a - \\delta$. the horizontal lines are at $L + \\epsilon$ and $L - \\epsilon$. If the limit is really the limit, then no matter how small epsilon and delta are, the limit will be in the center of the box.\r\n\r\nFor every value of epsilon greater than zero, a box can be created that the limit is inside of\r\n\r\nfor all $\\epsilon$ $|L - f(x)| < \\epsilon$ there exists $0 < |x - a| < \\delta$\r\n\r\nthis [url=http://www2.scc-fl.edu/lvosbury/CalculusI_Folder/EpsilonDelta.htm]link[/url] animates that.\r\n\r\n\r\nf has been defined such that a limit exists for every value between 0 and positive infinity, which means continuity", "Solution_2": "That's uniform continuity; the function is uniformly continuous.\r\n\r\nAt zero, the limit exists and is finite. At $\\infty$, anything can happen.", "Solution_3": "i find that problem in the book Sadovnichiy Podkolzin \"USSR students olympiad\"\r\nthere was a solution like that:\r\nsince function is uniformly continuous we applying the Cauchy criteria get that the limit at +0 exist\r\ncan u tell me why it is so?\r\nfor the +oo its said that sinx is uniformly continuous and doesn't have a limit at +oo", "Solution_4": "Choose $\\delta_n$ decreasing so that $|f(x)-f(y)|<\\frac1n$ when $|x-y|<\\delta_n$. Then $|f(c)-f(\\delta_n)|<\\frac1n$ whenever $0f(x)=f(x-q)+q\\geq q$ (since $x-q>0$), which is an obvious contradition. Similiarly, if there is $x$ such that $f(x)>x$. Again choose rational $q$ such that $f(x)>q>x$. Now $r 0\\ \\exists \\delta > 0$ such that $|x-c|<\\delta \\Rightarrow |f(x) - L| < \\epsilon$. \r\n\r\nIf you look at the limit of $\\ln(f(x))$, you will have $|\\ln(f(x)) - \\ln(L)|$, and if $f(x)$ gets close to $L$, then $\\frac{f(x)}{L}$ will get close to 1, so $|\\ln(f(x)) - \\ln(L)|$ also gets small. So $\\ln(L)$ is the limit we're looking for, and we see that the log of the limit equals the limit of the log. :) I hope this is clear.", "Solution_12": "If $\\lim_{x \\to a}g(x) = b$ and $\\lim_{x \\to b} f(x)= \\ell=f(b)$ then $\\lim_{x \\to b} f(g(x))= \\ell$ \r\n\r\nwhich is a bit more general than assuming continuity.\r\n\r\n :)\r\n\r\nEdit : I messed it up a bit ... now it is correct i think.... :D", "Solution_13": "If you don't assume continuity, you have to add an additional condition: $g(x) \\neq b$ for some neighbourhood $V \\setminus \\{ a \\}$." } { "Tag": [ "function", "invariant", "complex numbers", "combinatorics proposed", "combinatorics" ], "Problem": "I really enjoyed this one :).\r\n\r\nWe are working with words made up of the letters $a$ and $b$. We can change the words either by inserting a sequence of the type $www$, where $w$ is a word, anywhere inside a given word, or by deleting such a sequence from a word. \r\n\r\nCan we reach $ba$ if we start with $ab$?", "Solution_1": "wow, i have reached a nice idea :heli: :trampoline: .\r\nI dont know if it is true but here it is:\r\n\r\nI think the answer is $NO$.\r\nLet us consider $a=f(x)$ and $b=g(x)$,such that \r\n\r\n$(f\\circ f \\circ f)(x)=x,(g \\circ g \\circ g)(x)=x$\r\n\r\n\r\nNow we have $ab=(f\\circ g)(x)$ and if we add $aaa,bbb$ somewhere then we wont have any change cause $aaa=(f\\circ f\\circ f)(x)=x=(g\\circ g\\circ g)(x)=bbb$.\r\n\r\nSo\r\n\r\nNow all that we should do is find two functions like $f,g$ such that \r\n\r\n$f(g(x)) \\neq(f(g(x))$.\r\n\r\nAm i right :) :) ?", "Solution_2": "[color=red]Here are two functions satisfying my condition.[/color]\r\n\r\nLet $\\alpha_1,\\alpha_2$ be two other 3th roots of unit in complex numbers beside of $1$\r\n\r\nLet $f(x)=x^{\\alpha_1}$ and $g(x)=\\alpha_2 x$\r\n\r\nSo we have $(f \\circ f \\circ f)(x)=x^{\\alpha_1^3}=x$\r\n\r\nand we have $(g \\circ g \\circ g )(x)={\\alpha_2}^ 3x$\r\n\r\nAnd we know that\r\n\r\n$f(g(x))={\\alpha_2x}^{\\alpha_1} \\neq \\alpha_1x^{\\alpha_2}=g(f(x))$", "Solution_3": "I think there might be something we could do.\r\nFor any sequence $S=a_1 a_2\\ldots a_n$ $f : S\\rightarrow \\mathbb{Z}_3$ so that $f(S)=[\\sum_{i=1}^n a_i\\cdot i] $ mod $3$, where you assign to $a$ and $b$ different values mod $3$. Now $f(S_1)\\neq f(S_2)$ and I hope $f$ is invariant under that transformation :?", "Solution_4": "Yes, I think that works. I think I had essentially the same invariant, but I worded it differently :). \r\n\r\nlomos_lupin, now that I understand what you wanted to do, I don't think that's enough. The triplicates $aaa,bbb$ are not the only ones. You also have $ababab$, for example, and infinitely many others. So $(f\\circ f\\circ f)(x)=(g\\circ g\\circ g)(x)=x,\\ \\forall x$ does not suffice. You also want, for instance, $(f\\circ g\\circ f\\circ g\\circ f\\circ g)(x)=x,\\ \\forall x$, and so on. \r\n\r\nBasically, what you're trying to do is construct the Burnside group $B(2,3)$, i.e. a group with two generators in which every element (except for the identity) has order $3$. This group has $27$ elements, so it's not an easy task to construct its multiplication table.", "Solution_5": "[quote=\"grobber\"]\n\nBasically, what you're trying to do is construct the Burnside group $B(2,3)$, i.e. a group with two generators in which every element (except for the identity) has order $3$. This group has $27$ elements, so it's not an easy task to construct its multiplication table.[/quote]\r\n\r\nwell, this is precisely what i had in mind as soon as i saw the question!", "Solution_6": "[quote]\nlomos_lupin, now that I understand what you wanted to do, I don't think that's enough. The triplicates $aaa,bbb$ are not the only ones. You also have $ababab$, for example, and infinitely many others. So $(f\\circ f\\circ f)(x)=(g\\circ g\\circ g)(x)=x,\\ \\forall x$ does not suffice. You also want, for instance, $(f\\circ g\\circ f\\circ g\\circ f\\circ g)(x)=x,\\ \\forall x$, and so on. \n\nBasically, what you're trying to do is construct the Burnside group $B(2,3)$, i.e. a group with two generators in which every element (except for the identity) has order $3$. This group has $27$ elements, so it's not an easy task to construct its multiplication table.[/quote]\r\n\r\nMaybe i am missing something really easy but\r\nWould you please ,explain a little ,grobber,?\r\nIn the problem we can insert $aaa,bbb$ so i made the condition $fff(x)=x=ggg(x)$\r\nto be free to add $f \\circ f \\circ f$ or $g \\circ g \\circ g$ where ever we want.\r\nBut why ababa :?: :?: :? :? \r\n\r\n\r\nHmm.what is Burnsid group :fool: ? :blush:", "Solution_7": "The problem says that you can insert or delete [b]any[/b] triplicate. $ababab$ is also a triplicate: it's $ab$ repeated three times. So is $abaabaaba$, and so is $aaabaaabaaab$, and so on. You need to acount for these as well with your functions. Taking $f^3(x)=g^3(x)=x$ does not suffice; you need extra conditions.\r\n\r\nAs for the other question, I suggest you google it. I bet you'll find much more accurate answers than what I could provide :).", "Solution_8": "Let $f(w)=(s_{ab}, s_{ba}, s_{aa}, s_{bb})$ where $s_{ij}$ is the number of adjacent letters $i,j$ in the word $w$ modulo 3.\r\n\r\nSo $f(ab)=(1,0,0,0)$ and $f(ba)=(0,1,0,0)$.\r\n\r\nIt can be easily verified that if word $w'$ is obtained from $w$ by inserting or deleting some triplicated word, then\r\neither $f(w')=f(w)$ or $f(w')=f(w)\\pm (2,2,1,1)$ (all operations are performed modulo 3).\r\n\r\nSince $f(ba)-f(ab)=(0,1,0,0)-(1,0,0,0)=(2,1,0,0)$ is not a multiple of $(2,2,1,1)$, the word $ba$ [b]cannot[/b] be obtained from $ab$ with insertions/deletions of triplicated words.", "Solution_9": "Nice solution ,Opposite to my stupid nonsenses :( .\r\nWelldone." } { "Tag": [], "Problem": "xe^(3x)-1=3\r\nxe^(2x)+5=7", "Solution_1": "$xe^{3x}=4$\r\n$xe^{2x}=2$\r\n\r\n$x=4/e^{3x}$\r\n$x=2/e^{2x}$\r\n\r\n$2e^{2x}=e^{3x} \\rightarrow e^{2x}(e^{x}-2)=0 \\rightarrow x=ln2$" } { "Tag": [ "vector", "advanced fields", "advanced fields unsolved" ], "Problem": "N.B. I can't type the symbol for vector, so instead i will type vect.\r\n\r\nSo how do you solve :\r\n\r\n | R - (vect.R' - vect.a)/|vect.R - vect.a| |\r\n\r\nthe term (vect.R' - vect.a)/|vect.R - vect.a| is less than one due to physics.", "Solution_1": "I need help expanding the expression i posted.", "Solution_2": "How would you expand 1-x? It seem already a good expression. :)" } { "Tag": [ "geometry", "geometric transformation", "reflection", "power of a point" ], "Problem": "In the diagram, $ P$ is any point on a diameter $ AB$ of a circle; $ QPR$ is a chord such that $ /angle APQ \\equal{} 45$, prove that $ AB^2 \\equal{} 2PQ^2 \\plus{} 2PR^2$", "Solution_1": "Very nice!! :D \r\n[hide]Let O be the center of the circle. Reflect R over AB to get S, on the circle. Then QP^2+PS^2=QS^2. We have to show QOS=90, or QOPS is cyclic. But $ \\angle OQP \\equal{} \\angle ORP \\equal{} \\angle OSP$![/hide]", "Solution_2": "How does angle QOS being 90 degrees have to do with QOPS being cyclic?", "Solution_3": "He used the fact that $ QOPS$ is cyclic (proven by the last statement) to deduce that $ m\\angle QOS\\equal{}90^\\circ$. This is because $ m\\angle QOS\\equal{}m\\angle QPS\\equal{}90^\\circ$ as shown by $ QP^2\\plus{}PS^2\\equal{}QS^2$.", "Solution_4": "Oh, I knew that QOS and QPS are both right angles, but I didn't realize that QS is actually the diameter for the circle that circumscribes the quadrilateral. :wink:", "Solution_5": "Another way you can do it is to draw $ QB$ and $ QA$ so that $ 2072$ hence $ x,y,z \\le 12$. Also note that $ 7 \\mid 2072$. We cannot have $ 7 \\mid x,y,z$ cause then would $ x \\equal{} y \\equal{} z \\equal{} 7$... That means that one number is divisble by $ 7$, but the others are not. Say that $ x$ is that number.\r\nThen we have $ 7^3 \\plus{} y^3 \\plus{} z^3 \\equal{} 2072$ or $ y^3 \\plus{} z^3 \\equal{} 2072\\minus{}7^3 \\equal{} 2072\\minus{}343 \\equal{} 1729$. Because of symmetry we can assume that $ y^3 \\equiv \\minus{}z^3 \\equiv 1 \\bmod 7$.\r\nThus we have $ y \\in \\{1, 2, 4, 8, 9, 11\\}$\r\n\r\nThen we can just check:\r\n$ y \\equal{} 1 : z \\equal{} (1729\\minus{}1)^{\\frac{1}{3}} \\equal{} 12$\r\n$ y \\equal{} 2 : z \\equal{} (1729\\minus{}8)^{\\frac{1}{3}} \\equal{} Non\\quad Integer$\r\n$ y \\equal{} 4 : z \\equal{} (1729\\minus{}64)^{\\frac{1}{3}} \\equal{} Non\\quad Integer$\r\n$ y \\equal{} 8 : z \\equal{} (1729\\minus{}512)^{\\frac{1}{3}} \\equal{} Non\\quad Integer$\r\n$ y \\equal{} 9 : z \\equal{} (1729\\minus{}729)^{\\frac{1}{3}} \\equal{} 10$\r\n$ y \\equal{} 11 : z \\equal{} (1729\\minus{}1331)^{\\frac{1}{3}} \\equal{} Non\\quad Integer$\r\n\r\nSo the only solutions to $ (x,y,z)$ are permutations of $ (7,1,12)$ and $ (7,9,10)$", "Solution_7": "Yep as mornik says, because $ a^{3}\\equiv 1,\\minus{}1,0 mod7$ and $ 7|2072$. It's obviously that a,b,c<13 and there are 2 options :\r\n1)$ x\\equiv y \\equiv z \\equiv 0 (mod7)$ hence x=y=z=7, but $ 7^{3}\\plus{}7^{3}\\plus{}7^{3}\\neq 2072$ so it doesn't work\r\n2) $ x^{3}\\equiv 1 mod7$. $ y^{3}\\equiv \\minus{}1 mod7$, $ z^{3}\\equiv 0 mod7$\r\nWe know a,b,c<13 and $ 7|z^{3}$,$ 7|z , z\\equal{}7$\r\n$ x^{3}\\plus{}y^{3}\\equal{}1729$\r\nNow just check what pass ! greetings", "Solution_8": "isnt 1729 the great ramanujan number!", "Solution_9": "Yup, http://en.wikipedia.org/wiki/1729_(number) . It is, among other things, the smallest number expressible as the sum of two cubes in two different ways (Mathias_DK calculations have shown that $ 9^3\\plus{}10^3\\equal{}1729$ and $ 1^3\\plus{}12^3\\equal{}1729$)." } { "Tag": [ "geometry", "inradius", "IMO Shortlist", "Triangle", "incenter" ], "Problem": "For a given triangle $ ABC$, let $ X$ be a variable point on the line $ BC$ such that $ C$ lies between $ B$ and $ X$ and the incircles of the triangles $ ABX$ and $ ACX$ intersect at two distinct points $ P$ and $ Q.$ Prove that the line $ PQ$ passes through a point independent of $ X$.", "Solution_1": "The problem follows nicely by a \"lemma\" proposed by Virgil Nicula here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=132338[/url]\r\n\r\nNow, returning to the problem:\r\nLet the incircles of $\\triangle{ABX}$ and $\\triangle{ACX}$ touch $BX$ at $D$ and $F$, respectively, and let them touch $AX$ at $E$ and $G$, respectively. \r\n\r\nClearly, $DE \\| FG$. \r\nIf the line $PQ$ intersects $BX$ and $AX$ at $M$ and $N$, respectively, then $MD^{2}= MP\\cdot MQ = MF^{2}$, i.e., $MD = MF$ and analogously $NE = NG$. \r\n\r\nIt follows that $PQ \\| DE$ and $FG$ and equidistant from them. The midpoints of $AB, AC$, and $AX$ lie on the same line $m$, parallel to $BC$.\r\n\r\nApplying the \"lemma\" to $\\triangle{ABX}$, we conclude that $DE$ passes through the common point $U$ of $m$ and the bisector of $\\angle{ABX}$. \r\nAnalogously, $FG$ passes through the common point $V$ of $m$ and the bisector of $\\angle{ACX}$. Therefore $PQ$ passes through the midpoint $W$ of the line segment $UV$ . Since $U, V$ do not depend on $X$, neither does $W$.", "Solution_2": "Let the incircle of $\\triangle{ABC}$ meet $BC,CA,AB$ at $D_1,E_1,F_1$ and have inradius $r$. Then setting $u=AE_1=AF_1$ and so on, we have $r^2=uvw/(u+v+w)$. Similarly, let $\\triangle{ABX}$'s incircle meet $BX,XA,AB$ at $D_2,E_2,F_2$ and let $\\triangle{ACX}$'s incircle meet $CX,XA,AC$ at $D_3,E_3,F_3$. By simple length chasing, we find that $D_2D_3=E_2E_3=CD_1=CE_1=w$. Now define $M=BX\\cap PQ$ and $N=AX\\cap PQ$. Then $MD_2^2=MQ\\cdot MP=MD_3^2$, and so $MD_2=MD_3=NE_2=NE_3=w/2$. Clearly we also have $D_3E_3 \\| MN \\| D_2E_2$, so\n\\[\\angle{NMX}=\\frac\\pi2-\\frac{\\angle{AXB}}{2}=\\frac{\\angle{D_1BF_1}}{2}+\\frac{\\angle{E_2AF_2}}{2},\\]and the tangent addition formula combined with some simple ratios and lengths gives us\n\\begin{align*}\n\\tan\\angle{NMX}=\\frac{\\frac{r}{v}+\\frac{s}{u+v-sv/r}}{1-\\frac{r}{v}\\frac{s}{u+v-sv/r}}=\\frac{r(u+v)-sv+sv}{v(u+v-sv/r)-rs}\n=\\frac{r^2(u+v)}{v(r(u+v)-sv)-r^2s}\n&=\\frac{uvw(u+v)}{v(u+v+w)(r(u+v)-sv)-uvws}\\\\\n&=\\frac{uw(u+v)}{r(u+v)(u+v+w)-s(v+u)(v+w)}\\\\\n&=\\frac{uw}{r(u+v+w)-s(v+w)}.\n\\end{align*}Also,\n\\[BM=BD_2+MD_2=\\frac{sv}{r}+\\frac{w}{2}.\\]Now consider the coordinate plane with $x$-axis $BM$ and origin $B=0$. The line $MPQN$ is\n\\[y=\\frac{uw}{r(u+v+w)-s(v+w)}\\left(x-\\frac{sv}{r}-\\frac{w}{2}\\right).\\]But it's easy to check that the point\n\\[\\left(\\frac{w}{2}+\\frac{v(u+v+w)}{v+w},\\frac{uvw}{r(v+w)}\\right)\\]always lies on this line, so we're done.", "Solution_3": "[hide=\"Diagram\"]\n[asy]\nimport olympiad;\nunitsize(2cm);\npair A=(3,4);\npair B=(0,0);\npair C=(4,0);\npair X=(10,0);\npath C1=incircle(A,B,X);\npath C2=incircle(A,C,X);\npair O1=incenter(A,B,X);\npair O2=incenter(A,C,X);\npair B1=foot(O1,B,X);\npair B2=foot(O1,A,X);\npair C_1=foot(O2,C,X);\npair C_2=foot(O2,A,X);\npair M1=(B1+C_1)/2;\npair M2=(B2+C_2)/2;\npair D=(A+B)/2;\npair E=(A+C)/2;\npair F=(A+X)/2;\npair k[]=intersectionpoints(D--F,M1--M2);\npair K=k[0];\npair p[]=intersectionpoints(C1,C2);\npair P=p[0];\npair Q=p[1];\ndraw(A--B--C--X--A--C);\ndraw(D--F);\ndraw(M1--M2);\ndraw(C1); draw(C2);\ndot(A); dot(B); dot(C); dot(X); dot(P); dot(Q); dot(B1); dot(B2); dot(C_1); dot(C_2); dot(M1); dot(M2); dot(D); dot(E); dot(F); dot(K);\nlabel(\"$A$\",A,dir(90));\nlabel(\"$B$\",B,dir(-90));\nlabel(\"$C$\",C,dir(-90));\nlabel(\"$X$\",X,dir(-90));\nlabel(\"$P$\",P,dir(-135));\nlabel(\"$Q$\",Q,dir(110));\nlabel(\"$B_1$\",B1,dir(-90));\nlabel(\"$B_2$\",B2,dir(90));\nlabel(\"$C_1$\",C_1,dir(-90));\nlabel(\"$C_2$\",C_2,dir(90));\nlabel(\"$M_1$\",M1,dir(-45));\nlabel(\"$M_2$\",M2,dir(90));\nlabel(\"$D$\",D,dir(135));\nlabel(\"$E$\",E,dir(45));\nlabel(\"$F$\",F,dir(90));\nlabel(\"$K$\",K,dir(140));\n[/asy]\n[/hide]\n\nLet the incircle of $ABX$ be tangent to $BX$ at $B_1$ and $AX$ and $B_2$. Also, let the incircle of $ACX$ touch $CX$ at $C_1$ and $AX$ at $C_2$. Let $M_1$ be the midpoint of $B_1$ and $C_1$, and $M_2$ be the midpoint of $B_2$ and $C_2$. Since $PQ$ is a radical axis, it passes through $M_1$ and $M_2$. Thus, it is sufficient to prove that, as $X$ varies, there is a constant point which $M_1M_2$ always passes through.\n\nLet $D$, $E$, and $F$ be the midpoints of $AB$, $AC$, and $AX$ respectively. Since $B$, $C$, and $X$ are collinear, the points $D$, $E$, and $F$ are collinear also. I will prove that $K=DE\\cap M_1M_2$ does not change as $X$ changes, this will show that all $M_1M_2$ pass through a common point. It is sufficient to prove that $DK$ is a constant.\n\nSince $B_1,B_2, C_1,$ and $C_2$ are points of tangency of incircles, $B_1X=B_2X=\\frac{BX+AX-AB}{2}$ and $C_1X=C_2X=\\frac{CX+AX-AC}{2}$. Thus, \\[M_1X=M_2X=\\frac{BX+CX+2AX-AC-AB}{4}\\]Since $DF\\|BX$, $M_2FK\\sim M_2XM_1$. Since $M_1X=M_2X$, $M_2F=KF$. We can now compute the length of $DK$ (notice the signed lengths):\n\\begin{align*}\nDK&=DF-KF\\\\\n &=\\frac{BX}{2}-M_2F\\\\\n &=\\frac{BX}{2}-M_2X+FX\\\\\n &=\\frac{BX}{2}-\\frac{BX+CX+2AX-AC-AB}{4}+\\frac{AX}{2}\\\\\n &=\\frac{BX-CX+AC+AB}{4}\\\\\n &=\\frac{BC+AC+AB}{4}\n\\end{align*}\nThis is independent of $X$, so all $M_1M_2$ (and so $PQ$) intersect at a common point, as desired.", "Solution_4": "another solution :\nLet , U is exenter of triangle ABC opposite to A , I is incenter of ACX , incircle of CXA is tangent to XB at point T , incircle of ABX is tangent to BX at point K , incircle of BCA is tangent to BC at point L\nEasy to see that KT = LC\nA - Excircle of ABC is tangent to XB at point H\nLet line thru L' = midpont of LC and || to CI intersect line thru H' midpoint of BH and perpendicular to BU at point D , let line thru point D and perpendicular to XI intersect XB at point N\nEasy to see that angle NDL' = IAC , angle L'DH' = CAU , angle DH'N = AUI , so H'L'ND ~ UCIA , UC/CI = H'L'/L'N = HC/CT = H'L'/CT , L'N = CT , so N is midpoint of KT , line ND = line PQ , D is fixed . done", "Solution_5": "Let $\\omega_1$ and $\\omega_2$ be the incircles of $\\triangle ABX$ and $\\triangle ACX$, respectively. Denote by $(P, \\omega)$ the power of $P$ with respect to circle $\\omega$. Define a function $f: \\mathbb{R}^2 \\to \\mathbb{R}$ by\\[f(P) = (P, \\omega_1) - (P, \\omega_2).\\]This function is linear. We now use barycentrics with respect to $\\triangle ABC$. Let $R = (x:y:z)$ be the constant point that lies on $PQ$. Since $R$ lies on the radical axis of $\\omega_1$ and $\\omega_2$, we have $f(R) = 0$. Let $BC = a$, $CA = b$, $AB = c$, $AX = p$, and $CX = q$. We claim that the point $R = (2a: a-b-c: a+b+c)$ works, which is independent of the position of $X$. We can compute\n\\begin{align*}\nf(A) &= \\left(\\frac{a+c+p+q}{2}-a-q\\right)^2 - \\left(\\frac{b+p+q}{2}-q\\right)^2 \\\\&= \\frac{1}{4}(-a+b+c+2p-2q)(-a-b+c), \\\\ f(B) &= \\left(\\frac{a+c+p+q}{2}-p\\right)^2 - \\left(\\frac{b+p+q}{2}-p+a\\right)^2 \\\\&= \\frac{1}{4}(3a+b+c-2p+2q)(-a-b+c), \\\\ f(C) &= \\left(\\frac{a+c+p+q}{2}-c-q\\right)^2 - \\left(\\frac{b+p+q}{2}-p\\right)^2 \\\\&= \\frac{1}{4}(a+b-c)(a-b-c+2p-2q).\n\\end{align*}\nBy linearity, we have $f(R) = xf(A)+yf(B)+zf(C)$. But now it's straightforward to check that indeed $2af(A)+(a-b-c)f(B)+(a+b+c)f(C) = 0$, implying that $R$ always lies on the radical axis of $\\omega_1$ and $\\omega_2$, as desired.\n\n[hide=\"Remark\"]\n$R$ lies on the line passing through the midpoints of segments $AB$ and $AC$, which has been pointed out above; this is because $R$ lies on the line $x = y+z$.\n[/hide]", "Solution_6": "Short Solution:\n\nTake $B'$ and $C'$ fixed points that lie on $MN$ (line connecting midpoints of $AB$ and $AC$) such that $AB'B = AC'C = 90$. Let $K$ be the midpoint of $B'C'$. I claim $K$ is the desired point. First I prove a lemma\n\nLEMMA In triangle $XYZ$, let $X'$ and $Z'$ be the points of tangency of the incircle (centered at $I$) with $YX$ and $YZ$ respectively and let $W$ be the intersection of the bisector of angle $Z$ and $X'Z'$. Then, if $M$ and $N$ are the midpoints of $XY$ and $XZ$, we have $MNW$ are collinear and $XWZ=90$.\nProof: Angle chasing\n\nTherefore, $B'$ lies on the line of the points if tangency of the incircle of $ABC$ with $AX$ and $BC$, and analogously $C'$. From this we see clearly $K$ lies on the radical axis, and we're done.", "Solution_7": "Let $\\omega$ be the circle centered at the midpoint of the two incircles with radius the average of the other two. Let $Y$ be the point on $BC$ such that $\\omega$ is the incircle of $AYX$. Note that\n\\[\nAY+AX-YX=\\frac{AX+AC-CX}{2}+\\frac{AX+AB-BX}{2}\n\\]\nso\n\\[\n2AY+BY-CY=AB+AC\n\\]\nThere are finitely many points $Y$ on $BC$ satisfying this, and $Y$ as a function of $X$ is continuous, so $Y$ must be the same for all $X$. Note that $PQ$ meets $XY,XA$ at their tangency points with $\\omega$, so it suffices to show that for all $X$ varying on ray $BY$ past $Y$, there exists a constant point on all $X$-touch chords of $AYX$. But by a well known lemma, the intersection of the bisector of $AYX$ and the circle with diameter $AY$ suffices.", "Solution_8": "Let the $A$-midline of $\\triangle ABC$ intersect the internal bisector of $\\angle B$ at $V$ and the external bisector of $\\angle C$ at $W$. Let $L$ be the midpoint of $\\overline{VW}.$ We will show that $L \\in \\overline{PQ}$ to prove the result.\n\nNote that $\\overline{PQ}$ is the mid-parallel of the $X$-touch chord of the incircles of $\\triangle ABX$ and $\\triangle ACX$. Point $V, W$ lie on the $X$ touch chord of $\\triangle ABX$ and $\\triangle ACX$, respectively (from the \"right angles on the intouch chord\" lemma). Evidently, $L$ being the midpoint of $\\overline{VW}$ lies on $\\overline{PQ}$ as desired.", "Solution_9": "[color=#00f]Lemma:In $\\triangle ABC$ the $B$-bisector $A$-midline and the $C$-touch chord of the incircle are concurrent[/color]\n[i]Proof:[/i]\nTake the polar dual and it suffices to show that the line connecting $C$ and orthocenter of $\\triangle CIB$ is perpendicular to $AI$ which is immediate.\n\nLet the incircles of $\\triangle ABX,\\triangle $ touch AX,BC in $E,F,E_1,F_1$.Now let $EF,E_1F_1\\cap \\text{A-midline}=\\{R,S\\}$.By lemma $R$ lies on the B-angle bisector and\n$S$ on C- outer angle bisector and hence they're fixed.$PQ||EF||E_1F_1$ and $PQ$ bisects $EE_1$ and hence it bisects $RS$ $\\implies$ it passes thru the midpoint of $RS$ which is fixed.$\\blacksquare$", "Solution_10": "Really nice problem!\n\nLet the incircle of $\\Delta ABX, \\Delta ACX$ centered at $I_1, I_2$ and touches $BC$ at $E, F$ resp. Let $I_B$ be the center of $B$-excircle of $\\Delta ABC$, which touches $BC$ at $T$. Let $K, M$ be the midpoint of $AT, EF$ and finally let $R$ be the point on $BC$ such that $AR\\perp I_1I_2$.\n\nNote that $M$ lies on radical axis of those incircles, which is $PQ$. Now we claim that $K$ also lies on $PQ$, which is the fixed point. It suffices to show that $MK\\perp I_1I_2$, which is parallel to $AR$. Or equivalently, $M$ is the midpoint of $RT$.\n\nTo prove this, note by symmetry that $XR=XA$. So by side-chasing,\n\\begin{align*} ET &= CT - CF \\\\\n&= \\frac{AB + AC - BC}{2} - \\frac{AC + CX - AX}{2} \\\\\n&= \\frac{AB + AX - BC - CX}{2} \\\\\n&= \\frac{AB + AX - BX}{2} \\\\\n\\end{align*}\nand\n\\begin{align*}FR &= XR - EX \\\\\n&= XA - \\frac{XA + XB - AB}{2} \\\\\n&= \\frac{XA + AB - XB}{2} \\\\\n\\end{align*}\nso $ET=FR$ which implies $M$ is midpoint of $RT$ and we are done.\n\n[b]Note :[/b] The most effective way to claim the fixed point is to plug $X=C$ and $X={\\infty}_{BC}$ and intersect the radical axii.", "Solution_11": "Nice algebra problem! Solved with [b]eisirrational[/b].\n\nWe use barycentric coordinates on $\\triangle ABC$. Let $CX = d$ and $AX = x$. Moreover, let the incircles of $ABX$ and $ACX$ be tangent to line $AX$ at $M_1$, $N_1$, and to line $BX$ at $M_2$, $N_2$, respectively.\n\nI claim that the point $R = \\left( \\frac{1}{2}, \\frac{a-b-c}{4a}, \\frac{a+b+c}{4a} \\right)$ is the desired point. It suffices to show that $R$ lies on the radical axis of the two incircles, i.e. $R$, the midpoint of $\\overline{M_1N_1}$, and the midpoint of $\\overline{M_2N_2}$ are collinear. Call these midpoints $S_1$ and $S_2$.\n\nWe first compute some lengths. By equal tangents, we have that $N_1X = N_2X = \\frac{x+d-b}{2}$ and $M_1X = M_2X = \\frac{a+d+x-c}{2}$, so $S_1X = S_2X = \\frac{2x+2d+a-b-c}{4}$. The coordinates of $S_2$ is readily calculated to be (with directed ratios)\n$$(0:CS_2:BS_2) = (0:CX-XS_2:BX-XS_2) = \\left(0,\\frac{a-b-c+2x-2d}{4a},\\frac{3a+b+c-2x+2d}{4a} \\right).$$\nThe coordinates of $S_1$ are a bit harder, but still not too bad. Note the coordinates of $X$ are $(0:CX:BX) = \\left( 0, -\\frac{d}{a}, \\frac{a+d}{a} \\right)$, so\n$$S_1 = XS_1 \\cdot (1,0,0) + AS_1 \\cdot \\left( 0, -\\frac{d}{a}, \\frac{a+d}{a} \\right) = \\left( \\frac{2x+2d+a-b-c}{4x}, \\frac{-2xd+2d^2+ad-bd-cd}{4ax}, \\frac{2ax+2dx-3ad-d^2-a^2+ab+bd+ac+cd}{4ax}\\right).$$\nTo show that $X,S_1,S_2$ are collinear, we simply show that the displacement vectors $\\overrightarrow{XS_1}$ and $\\overrightarrow{XS_2}$ are proportional. In fact, because all of these coordinates are homogenized, we only to verify the proportion for the first two coordinates. Looking at the first coordinates, we receive a ratio of\n$$\\frac{\\frac{1}{2} - \\frac{2x+2d+a-b-c}{4x}}{\\frac{1}{2}} = \\frac{-2d-a+b+c}{2x}.$$\nBut for the second coordinates we also have\n$$\\frac{\\frac{a-b-c}{4a} - \\frac{2xd+2d^2+ad-bd-cd}{4ax}}{\\frac{a-b-c}{4a} - \\frac{a-b-c+2x-2d}{4a}} = \\frac{(d-x)(2d+a-b-c)/4ax}{(2x-2d)/4a} = \\frac{-2d-a+b+c}{2x}.$$\nWe thus have the desired.", "Solution_12": "[img]https://lh3.googleusercontent.com/-45dm7aNud6c/Xg7HbI2IrhI/AAAAAAAAFgY/AGXpRORVABEOpvje3wJgMpcR99dnnewOgCK8BGAsYHg/s0/2020-01-02.jpg[/img]\nLet $I_1$ be the incenter of $ABX$ and let $I_2$ be the incenter of $ACX$. Let $D$ and $F$ be the contact points of the incircle of $ABX$ with $BX$ and $AX$ respectively, and let $E$ and $G$ be the contact of the incircle of $ACX$ with $CX$ and $AX$ respectively.\n\nLet $P$ and $Q$ be the intersections of $DF$ and $EG$ with the $A$-midline respectively. The key claim is that $P$ and $Q$ are fixed as $X$ varies. This follows from the so called Iran lemma, as $Q$ is the intersection of the internal $C$-angle bisector with the midline and $Q$ is the intersection of the external $B$-angle bisector with the midline.\n\nNote that the midpoints of $DE$ and $FG$ are on the radical axis of the two circles, so the intersection of the radical axis with $PQ$ is simply the midpoint of $PQ$, which is a fixed point, as desired.", "Solution_13": "Let the incircle of $\\triangle ABX$ touch $XA$ and $XB$ at $S$ and $T$, and let the incircle of $\\triangle ACX$ touch $XA$ and $XC$ at $U$ and $V$. Let the incenters of $\\triangle ABX$ and $\\triangle ACX$ be $I$ and $J$ respectively, and let $Y$ and $Z$ be the foots of perpendicular from $A$ to $\\overline{BI}$ and $\\overline{CJ}$. Let $M$ be the midpoint of segment $YZ$. We claim that $M$ is the desired fixed point. By the Iran lemma, it follows that $Y$ and $Z$ lie on $\\overline{UV}$ and $\\overline{ST}$ respectively. It is also easy to see that $UV\\parallel ST$ since both of them are perpendicular to $\\overline{IX}$. Moreover, the radical axis passes through the midpoints of segments $US$ and $VT$, so it is actually the midline of $\\overline{UV}$ and $\\overline{ST}$, and this passes through $M$ as desired.", "Solution_14": "First ISL G7.\nHere's solution using famous [color=#f00][b]Iran Lemma[/b][/color]:\nLet $I_{1}$ and $I_{2}$ be the incenters of $\\triangle{ABX}$ and $\\triangle{ACX}$, respectively. Denote by $T_{1}$, $T_{2}$ touchpoints of incircle of $\\triangle{ABX}$ with $\\overline{BX}$ and $\\overline{AX}$, respectively. Similarly, let incircle of $\\triangle{ACX}$ touches $\\overline{CX}$ and $\\overline{AX}$ at $T_{3}$ and $T_{4}$, respectively. Let $M$ and $N$ be the midpoints of $\\overline{AB}$ and $\\overline{AX}$, respectively.\nBy Iran Lemma we have $T = T_{1}T_{2} \\cup BI_{1} \\cup MN$ and $S = T_{3}T_{4} \\cup CI_{2} \\cup MN$ are fixed, since $MN$, $BI_{1}$, and $CI_{2}$ are fixed.\nSimple [hide=angle chasing]$\\measuredangle{T_{2}T_{1}T_{3}} = \\frac{\\measuredangle{T_2XT_1}}{2}\\ = \\measuredangle{XT_4T_3} = \\measuredangle{T_2T_4T_3}$ [/hide] gives us $T_2T_1T_3T_4$ is cyclic. From angle chasing we also get, that $T_1T_2 \\parallel T_3T_4$. Hence, $TT_1T_3S$ is parallelogram. It's well-known, that $PQ$ bisects $\\overline{T_1T_3}$; since $TT_1T_3S$ is parallelogram , it's also pass through the midpoint of $\\overline{TS}$, which is fixed. $\\blacksquare$", "Solution_15": "[hide=Solution(with geogebra)][quote=ISL 2004 G7]For a given triangle $ ABC$, let $ X$ be a variable point on the line $ BC$ such that $ C$ lies between $ B$ and $ X$ and the incircles of the triangles $ ABX$ and $ ACX$ intersect at two distinct points $ P$ and $ Q.$ Prove that the line $ PQ$ passes through a point independent of $ X$.[/quote]\nDenote the incircle of $\\triangle ABX$ as $\\omega_1$ and that of $\\triangle ACX$ as $\\omega_2$.Say $\\ell$ be the $A-\\text{midline}$ of all the 3 triangles.Say $\\omega_1$ meets $XA$ and $XB$ at $U_1$ and $V_1$ respectively and say $\\omega_2$ touches $XA$ and $XB$ at $U_2$ and $V_2$ respectively.\n[b]Claim :[/b]$U_1V_1 \\cap \\ell,u_2v_2 \\cap \\ell$ are fixed.\n[b]Proof :[/b]As the point where the internal angle bisector of $\\angle ABC$ and the external angle bisector of $\\angle ACB$(or the internal angle bisector of $\\angle ACX$) meet $\\ell$ are fixed (as the midline can be determined by the midpoints of $AB$ and $AC$ irrespective of $X$).By the Well-known Iran-Incentre Lemma we are done $\\blacksquare$\nNow Say $U_1V_1 \\cap \\ell= L_1,U_2V_2 \\cap \\ell=L_2$ and say $L_3$ is the midpoint of $L_1L_2$,clearly $L_3$ is fixed.We claim that $P-Q-L_3$ or $L_3$ is the required point !\nFor the proof note that $PQ ||U_1V_1||U_2V_2$,say $PQ \\cap XA=U_3,PQ \\cap XB=V_3$.\nSince $PQ$ is the radical axis of $\\omega_1$ and $\\omega_2$,$U_3$ and $V_3$ become the midpoints of $U_1U_2$ and $V_1V_2$.The conclusion follows trivially $\\blacksquare$[/hide]\n\n", "Solution_16": "Let incircles of $ABX,ACX$ meet $AX$ again at $D_1,D_2$ and $BX$ at $E_1,E_2$ respectively. Let also line homothetic to $BC$ wrt $A$ with coefficient $\\frac{1}{2}$ intersect internal bisector of and $ABC$ and external bisector of $ACB$ at $X,Y$ respectively, so by Iran lemma $X\\in D_1E_1,Y\\in D_2E_2.$ But obviously $D_1XD_2Y$ is an isosceles trapezoid, therefore $PQ$ passes through midpoint of $XY,$ which is fixed.", "Solution_17": "Let $M,N$ be midpoints of $AB,AC$. Incircle of triangle $ABX$, which has center $I$ touches $XA,XB$ at $K,L$. Let incircle of triangle $ACX$ touches $AX,CX$ at $J,G$. Let $D=PQ\\cap MN, E=PQ\\cap BC, F=BI\\cap MN$ and $R=PQ\\cap AX$. From Iran Lemma $F\\in KL$. Since $RK=RJ$ and $El=EG$, we get $KL||RE \\implies FDEL$ is parallelogram \n$\\implies FD=EL=\\frac{LG}{2}=\\frac{XL-XG}{2}=\\frac{(XA+XB-AB)-(XC+XA-AC)}{4}=\\frac{BC+AC-AB}{4}$.\n Also $MF=MB=\\frac{AB}{2} \\implies MD=MF+FD=\\frac{(BC+AC-AB)+2AB}{4}=\\frac{AB+BC+CA}{4}$, which is constant when $ABC$ is fixed. So $D$ lies on fixed line $MN$ and length of $MD$ is constant, which implies $D$ is fixed point as $X$ varies. So we are done.", "Solution_18": "Let the incircle of $ABX$ intersect $BX$ at $D$ and $AX$ at $E$, and let $M$, $N$ be the midpoints of $AB$ and $AX$ respectively. Let $I$ be the incenter, then we claim that $BI$, $MN$, $DE$ are concurrent.\n[center][img width=45]https://media.discordapp.net/attachments/1091890356796280852/1095004634659172352/Screenshot_2023-04-10_at_11.09.13.png?width=1692&height=1112[/img][/center]\nLet $S$ be the intersection point of $DE$ and $MN$. Let $F$ be on $DE$ such that $AF\\parallel BX$. Let $G$ be $AS$ intersection with $BX$. Let $R$ be the contact point on $AB$. Clearly, $AFGD$ is a parallelogram with center $S$. We have \\[\\angle AEF = \\angle DEX = \\angle EDX = \\angle AFE\\] so $AE=AF$. Thus, $DG=AF=AE=AR$. Also, $BD=BR$ so $AB=AG$. Since $AS=SG$, $\\triangle ABS\\cong \\triangle GBS$ so $S$ is on $AI$ as desired.\n[center][img width=45]https://media.discordapp.net/attachments/1091890356796280852/1095004634894057572/Screenshot_2023-04-10_at_11.12.41.png?width=1898&height=1048[/img][/center]\nNow, similarly, if $J$, $K$ are the points of contact of the incircle of $ACX$ on $CX$ and $AX$, respectively then $JK$ passes through $T$, a fixed point on the midline and the $C$-angle bisector. Let $U$ be the midpoint of $ST$, also a fixed point with respect to $ABC$. \n\nSince $XI\\perp DE$ and $XI\\perp JK$, $DE\\parallel JK$. Now, since $PQ$ is the radical axes of the two circles, it bisects $EK$ and $DJ$. Thus, $PQ$ is the line right in the middle of $DE$ and $JK$, so it passes through $U$, as desired.", "Solution_19": "nice problem! glad i solved it in ~45 minutes? this is the last problem im doing on configgeo im done\n\nDefine the points as in the diagram, with all the lines given by Iran Lemma; note that $YJ\\perp CD\\perp VW\\implies YJ\\parallel VW\\parallel PQ$. On the other hand, since PQ is radax which bisects the common tangent RK, PQ is the midline of JYWV, which passes through the midpoint of YV. We conclude. :surf:\n\nalso note that my diagram is extremely overkill i didnt need ALL of those info but its helpful to list out when you dont know what to do", "Solution_20": "Consider an arbitrary point $X$ on ray $BC$. Let $I$ be the incenter of $\\triangle ABC$ with $\\triangle DEF$ being the intouch triangle of $\\triangle ABC$. Further, let $M_B$ and $M_C$ be the midpoints of $AC$ and $AB$ respectively. Let $M_{CD}$ and $M_{CE}$ be the midpoints of $CD$ and $CE$. We claim that all lines $PQ$ pass through $Z=\\overline{M_{CD}M_{CE}} \\cap \\overline{M_CB_C}$.\\\\\n\nLet $\\omega$ be the incircle of $\\triangle ABC$ and $\\omega_B$ and $\\omega_C$ the incircles of $\\triangle ABX$ and $\\triangle ACX$ respectively. Let $O_B$ and $O_C$ be the centers of $\\omega_B$ and $\\omega_C$ respectively. Now, let $G$ and $H$ be the tangency points of the incircle of $\\triangle ABX$ with sides $AX$ and $BX$. Let $M$ be the midpoint of $AX$. \n\nClearly, $B-I-O_B$ and $C-O_C-O_B$ due to the common tangents. Further, $M_C-M_B-M$. By Iran Lemma on $\\triangle ABC$, we see that $\\overline{BI},\\overline{M_BM_C}$ and $\\overline{DE}$ concur (say at $N$). Now, clearly $N$ also lies on $BI$ and $M_CM_B$ due to the above collinearities.Further notice that by Iran Lemma on $\\triangle ABX$, we must have that lines $\\overline{GH},\\overline{M_CM}$ and $\\overline{BO_B}$ concur. But clearly the latter two of these lines intersect at $N$. Thus, $GH$ also must pass through $N$.\\\\\n\nNow, simply notice that $\\overline{ED}\\parallel\\overline{M_{CD}M_{CE}}$ by Midpoint Theorem and $\\overline{GH} \\parallel \\overline{PQ}$ since both these lines are perpendicular to $XO_C$. Thus, the intersection of $EF$ and $GH$ and the intersection of $M_{CD}M_{CE}$ and $PQ$ must form two triangles which are similar. Now, we prove the following. Let $D',E'$ be the tangency points of $\\omega_C$ with $AX$ and $BX$. Then, \n\n[b][color=#f00]Claim : $GD'=CD$.[/color][/b]\n[i]\nProof : [/i] Note that $XD'$ and $XG$ are tangents to $\\omega_B$ and $\\omega_C$ respectively. Then, let $s_1$ and $s_2$ denote that semiperimeters of $\\triangle ABX$ and $\\triangle ACX$ and $s$ denote the semiperimeter of $\\triangle ABC$. We have,\n \\begin{align*}\n GD' &= XG-XD'\\\\\n &= s_1-AB - (s_2-AC)\\\\\n &= s_1-s_2 + AC - AB\\\\\n &= \\frac{AX+XB + AB - AX - CX - AC}{2} + AC - AB\\\\\n &= \\frac{BC + AB - AC}{2} + AC - AB\\\\\n &= \\frac{AB+AC+BC}{2}-AB\\\\\n &= s-AB\\\\\n &= CD\n \\end{align*}\n Thus, indeed $GD'=CD$ as claimed.\n\nNow, let $R = \\overline{PQ} \\cap \\overline{BX}$. It is well known that the radical axis bisects the common tangent. Thus, $RD'=\\frac{GD'}{2}=\\frac{CD}{2}=DM_{CD}$\nThus, the previously described similar triangles must infact also be congurent. This means, that the intersections of $EF$ and $GH$ and of $M_{CD}M_{CE}$ and $PQ$ must lie on a line parallel to $BC$. But clearly the former intersection point is $N$ which lies on $\\overline{M_CM_B}$ and thus $Z'= M_{CD}M_{CE} \\cap PQ $ must also lie on $\\overline{M_CM_B}$. This means, $Z'=Z$.\\\\\n\nThus, all lines $PQ$ must pass through a common point, this point being $Z=\\overline{M_{CD}M_{CE}} \\cap \\overline{M_CB_C}$.", "Solution_21": "Define line $\\ell$ as the $A$-midline and the touch points of the incircles of $\\triangle ABX$ and $\\triangle ACX$ to $AX$, $BX$ to be $J_1$, $J_2$ and $K_1$, $K_2$. If we suppose $J = J_1J_2 \\cap \\ell$ and $K = K_1K_2 \\cap \\ell$, we know\n[list]\n[*] $J$ and $K$ are fixed as they lie on the $B$-angle bisector and $C$-external angle bisector, respectively, through Iran Lemma.\n[*] $PQ$ bisects both $J_1K_1$ and $J_2K_2$ by power of a point, and $J_1J_2 \\parallel PQ \\parallel K_1K_2$.\n[/list]\n\nThus $PQ$ passes through the midpoint of $JK$, which is fixed. $\\blacksquare$", "Solution_22": "I'm surprised no one found this clean linpop solution!\n\nLet $\\omega_1$ and $\\omega_2$ denote the incircles of $\\triangle ABX$ and $\\triangle ACX$ respectively. Define $f(\\bullet) = \\text{Pow}(\\bullet, \\omega_1) - \\text{Pow}(\\bullet, \\omega_2)$. We claim that the intersection of the $A$-midline and the radical axis of $\\omega_1$ and $\\omega_2$ is fixed. In order to prove this, it suffices to show that the powers of $f((A+B)/2)$ and $f((A+C)/2)$ are independent of $X$, because it follows that the point on the $A$-midline that $f$ vanishes on is independent of $X$. This can be done by bashing. \n\nHere's an example of what the bash would look like; consider the example of proving $f((A+B)/2)=(f(A)+f(B))/2$ constant (proving $f((A+C)/2)$ is analogous). Let all lengths be signed. Then using the intouch points, we calculate \\[ f(A) = \\frac{1}{4} \\left[ (AB+AX-BX)^2 - (AC+AX-CX)^2\\right] \\] and \\[ f(B) = \\frac{1}{4} \\left[ (AB+BX-AX)^2 - (2BC+AC+CX-AX)^2\\right]. \\] Now we can bash out $f(A)+f(B)$, and utilizing the fact that $BC+CX=BX$, we compute this quantity to be independent of $X$, as desired. ", "Solution_23": "This problem is so cute!\n\nI claim that the fixed point $K$ lies on the $A$-midline $\\overline{MN}$. Let $E$ and $F$ are the tangency points of the incircle of $ABX$ to $\\overline{BX}$ and $\\overline{AX}$, respectively, and define $G$ and $H$ similarly. Let $I_1$ and $I_2$ be the incenters of triangles $ABX$ and $ACX$, and recall that $R = \\overline{BI_1} \\cap \\overline{MN}$ lies on the tangent chord $\\overline{EF}$. Similarly, $S = \\overline{CI_1} \\cap \\overline{MN}$ lies on the tangent chord $\\overline{GH}$. Note that $R$ and $S$ are fixed points.\n\nNow, by radical axis, $\\overline{PQ}$ is the midline of trapezoid $EGHF$, i.e. $K = \\overline{PQ} \\cap \\overline{MN}$ is the midpoint of $\\overline{RS}$. It follows that $K$ is the desired fixed point.\n\n" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "(a,b)=1. a,b E Z , and a,b is not 0 \r\na^2+6ab+b^2 is square. find a,b", "Solution_1": "there is no sollution in non-zero numbers to:\r\n$a^{2}+6ab+b^{2}=n^{2}$ because if we supposed so then\r\n$(a^{2}+6ab+b^{2})(a-b)^{2}=(a+b)^{4}-(4ab)^{4}=m^{2}$ fot some $m\\in \\mathbb{Z}$ which is impossible unless $a=b$ or $ab=0$ \r\nin the case $a=b$ we get $n^{2}=8a^{2}$ and $\\sqrt{2}\\in \\mathbb{Q}$ :D", "Solution_2": "[quote=\"Albanian Eagle\"]$(a^{2}+6ab+b^{2})(a-b)^{2}=(a+b)^{4}-(4ab)^{4}=m^{2}$ [/quote]\r\nI think it should be\r\n$(a^{2}+6ab+b^{2})(a-b)^{2}=(a+b)^{4}-(4ab)^{2}=m^{2}$ :P" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "if $ a,b,c > 0$ and $ a\\plus{}b\\plus{}c\\equal{}1$ prove that $ \\frac{1}{a\\plus{}b^2\\plus{}c^3}\\plus{}\\frac{1}{b\\plus{}c^2\\plus{}a^3}\\plus{}\\frac{1}{c\\plus{}a^2\\plus{}b^3} \\leq 4\\plus{}\\sqrt{\\frac{1}{a^3}\\plus{}\\frac{1}{b^3}\\plus{}\\frac{1}{c^3}}$", "Solution_1": "[quote=\"apollo\"]if $ a,b,c > 0$ and $ a\\plus{}b\\plus{}c\\equal{}1$ prove that $ \\frac{1}{a\\plus{}b^2\\plus{}c^3}\\plus{}\\frac{1}{b\\plus{}c^2\\plus{}a^3}\\plus{}\\frac{1}{c\\plus{}a^2\\plus{}b^3} \\leq 4\\plus{}\\sqrt{\\frac{1}{a^3}\\plus{}\\frac{1}{b^3}\\plus{}\\frac{1}{c^3}}$[/quote]\nBecause $(a+b^2+c^3)(a^3+b^2+c)\\geq(a^2+b^2+c^2)^2$ and $\\sqrt{\\frac{1}{a^3}\\plus{}\\frac{1}{b^3}\\plus{}\\frac{1}{c^3}}\\geq9$." } { "Tag": [ "calculus", "integration", "probability", "probability and stats" ], "Problem": "The continuous random variables X and Y have joint pdf f(x,y) where\r\nf(x,y)=k(x^2 +2xy) for 0 < x < 1 and 0 < y < 1,\r\n\r\ni)Find k.\r\n\r\nDo i integrate first with respect to x and then w.r.t y? Then do i put the answer equal to 1 and find k? When i do this i get k=6/5.\r\n\r\nii)Find the marginal pdf of X and of Y.\r\n\r\nis the marginal pdf of X the integral from 0 to 1 of f(x,y) wrt y?\r\nthen the marginal pdf of Y is calculated similarly to this - integral from 0 to 1 of f(x,y) wrt x?\r\n\r\niii)find prob[Y<0.5|X<0.5].\r\niv)find prob[Y<0.5|X=0.5].\r\nv)find prob[X 0$, then\r\n$ d_{x_0}(y) = \\frac {f(x_0,y)}{\\int_{y \\in [0;1]} f(x_0,y) dy}$\r\ncan be seen as the probability distribution of $ Y$ conditionally to the fact that $ X = x_0$.\r\nHence: $ E[Y | X = x_0] = \\int_{y \\in [0;1]} y d_{x_0}(y) dy$\r\n\r\nAnother way to do that, if you know the notion of conditional expectation, is to write\r\n$ E[Y | X] = f(X)$\r\nand then $ E[Y | X = x_0] = f(x_0)$" } { "Tag": [ "probability", "search", "combinatorics unsolved", "combinatorics" ], "Problem": "Let A and B be two vertices of a tree with 2006 edges. We move along the edges starting from A and would like to get to N without turning back. At any vertex we choose the next edge among all possible edges(i.e. excluding all the one wich we arrived) with equal probability.Over all possible trees and choices of vertices A and B. Find the minimum probability of getting from A to B.", "Solution_1": "What is the source of this problem?\r\n\r\n[hide=\"All the setup\"]Well, the nice property of trees is that there is a unique such path between any $A$ and $B$. So, for a given tree $T$, suppose this path is $A = a_{0}, a_{1}, \\ldots, a_{n}= B$ and that the degree $d(a_{i}) = d_{i}$. Then our probability of winning is $(d_{0}\\cdot (d_{1}-1) \\cdot (d_{2}-1) \\cdots (d_{n-1}-1))^{-1}$. We want to choose a tree to minimize this probability. Note that if there exists an edge joining two vertices, neither of which are an $a_{i}$, we can decrease the success probability by breaking off this portion of the tree and reattaching it to one of the $a_{i}$. Repeating this process as much as possible shows that the minimum occurs for some tree in which every vertex either is one of the $a_{i}$ or is connected only to one of the $a_{i}$. Thus, what we are really looking for is an integer $n$ and integers $d_{0}, d_{1}, \\ldots, d_{n}$ such that $1 \\leq d_{0}, d_{n}$, $2 \\leq d_{2}, \\ldots, d_{n-1}$ such that $(d_{0}-1)+(d_{1}-2)+(d_{2}-2)+\\ldots+(d_{n-1}-2)+(d_{n}-1) = 2006-n$ and $(d_{0}\\cdot (d_{1}-1) \\cdot (d_{2}-1) \\cdots (d_{n-1}-1))$ is maximized.\n\nIt's clear that we want to set $d_{n}= 1$. Then, taking $c_{n}= d_{0}$ and $c_{i}= d_{i}-1$, we can re-phrase this as a search for an integer $n$ and $n$ integers $c_{1}, \\ldots, c_{n}$, such that $c_{1}+\\ldots+c_{n}= 2006$ and $c_{1}\\cdot c_{2}\\cdots c_{n}$ is maximized.[/hide]" } { "Tag": [ "probability", "geometry", "3D geometry", "geometric transformation", "rotation" ], "Problem": "20. Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?\r\n\r\n(A) $ \\frac{1}{4}$ (B) $ \\frac{5}{16}$ (C) $ \\frac{3}{8}$ (D) $ \\frac{7}{16}$ (E) $ \\frac{1}{2}$\r\n\r\n\r\n[b]HINTS ONLY[/b]\r\n\r\nThanks.", "Solution_1": "Well since you want a hint, how many ways can you paint the cube? How many satisfy the given conditions?", "Solution_2": "$ 2^6$ ways to paint the cube.\r\n\r\nI'm not sure how many ways satisfy the given conditions.", "Solution_3": "Well for it to satisfy it, how many ways can you choose the 4 faces?", "Solution_4": "[quote=\"mewto55555\"]Well for it to satisfy it, how many ways can you choose the 4 faces?[/quote]\r\n\r\nhow do i find that? (i need to brush up on my counting skills)", "Solution_5": "I'll go through my entire solution, and have hide tags between steps.\r\n\r\n[hide=\"Step 1\"]\nWe want to find all ways to get 4 sides. It is clear that the other 2 sides are opposite each other\n[hide=\"Next\"]\nWe have 3 sets of opposite sides. Therefore the number of ways to choose the 2 sets that we make the same color is\n[hide=\"Next\"]\n3c2 or 3. We also have 2 ways to choose the color for this. Thus we have \n\n6 ways to choose and color the sides. For the other 2 sides\n\n2x2=4 ways so there are a 24 ways\n$ \\frac {24}{64} = \\frac {3}{8}$[/hide][/hide][/hide]\r\n\r\nIs this correct?", "Solution_6": "You force me to look at the solution now.\r\n\r\nThe answer is: [hide]B[/hide].\r\n\r\nYour step 3 is right and the other steps following it are wrong.", "Solution_7": "[hide=\"Hints\"]\nThere are $ 2^6 \\equal{} 64$ choices total--these include rotations of all cubes, which is important.\n[hide=\"What cubes have this property?\"]\nThe only ways to have this possible is to have at least four of one color, so\ni. RRRRRR or BBBBBB\nii. RRRRRB or BBBBBR\niii. RRRRBB or BBBBRR\nNow calculate the number of possibilities (including rotations!) of these figures.\n[hide=\"Case i\"]\nObviously, there are exactly $ 2$ cubes for this (only $ 1$ rotation for each RRRRRR and BBBBBB).\n[/hide]\n[hide=\"Case ii\"]\nIn RRRRRB, there are $ 6$ places for the blue cube, and then the reds are fixed.\n\nThe same holds for BBBBBR, so there are $ 12$ possibilities in all for this case.\n[/hide]\n[hide=\"Case iii\"]\nIn RRRRBB, the blue faces must be opposite each other, and the reds are forced.\n\nThere are $ 3$ ways for this (why aren't there $ 6$ ways?).\n\nThis is the same for BBBBRR, so there are $ 6$ total possibilities for this case.\n[/hide]\n[hide=\"Answer\"]\n$ \\frac {2 \\plus{} 12 \\plus{} 6}{64} \\equal{} \\frac5{16}\\implies\\boxed B$.\n[/hide]\n[/hide]\n[/hide]", "Solution_8": "How did I overcount?", "Solution_9": "You're counting each of RRRRRR and BBBBBBB three times, i.e. for each of these cubes, there are three choices of the pair of sides, then the rest is fixed.", "Solution_10": "Oh ok so since I counted those 3 times each instead of 1 time each you subtract 4 and get 20 possibilities." } { "Tag": [], "Problem": "Ten distinct points are identified on the circumference of a\ncircle. How many different convex quadrilaterals can be\nformed if each vertex must be one of these 10 points?", "Solution_1": "For each set of 4 points, there is one possible convex quadrilateral that can be formed. Therefore, there are $ \\binom{10}{4} \\equal{} \\boxed{210}$." } { "Tag": [ "LaTeX" ], "Problem": "Can someone help me out? I downloaded the MikTex stuff, but then I go through the setup, and it wants me to choose a folder to run the setup from, but it doesn't seem to work no matter what folder i choose. It always gives me something like \"this is not a valid directory\"", "Solution_1": "[quote=\"captcha000\"]Can someone help me out? I downloaded the MikTex stuff, but then I go through the setup, and it wants me to choose a folder to run the setup from, but it doesn't seem to work no matter what folder i choose. It always gives me something like \"this is not a valid directory\"[/quote]You should choose the folder where your download has been stored. Or maybe you've not downloaded the whole thing, just the webinstalled. In that case you won't find any local repositories. You can install it online, using the internet package repository (if you have a cable connection of course).", "Solution_2": "The Mik$\\text{\\TeX}$ download is a bit strange. You have to download the installer first and then the program. So it's a two-pass installation.", "Solution_3": "i got it to work, thanks!" } { "Tag": [ "articles" ], "Problem": "I need some information about air resistance...specially in projectile motion...\r\nAny ideas?", "Solution_1": "i read an article on Projectile motion with air resistance last night, which involves lots of Calc and stuff (it is from an old College math journal)\r\ni can post some stuff from that here, ifyou want, \r\nbut specifically what are youlooking for??", "Solution_2": "I would appreciate that\r\n :lol:" } { "Tag": [ "geometry", "3D geometry", "sphere", "tetrahedron", "perimeter" ], "Problem": "This problem is similar to an old MOEMS problem, but I added some more to it:\r\n\r\nGiven a circle, square, equilateral triangle, and regular pentagon that all lie in the same plane, what is the maximum number of intersection points that can be formed? \r\n\r\nAnd here's an even better one:\r\n\r\nEDIT: Given a sphere, a cube, and a tetrahedron, what is the maximum number of intersection points that can be formed?\r\n\r\nEDIT: An intersection point is defined as a point which is common to the perimeters of at least two geometrical objects", "Solution_1": "The circle can intersect each polygon twice on each side: 6 points for the triangle, 8 for the square, 10 for the pentagon. The triangle can have 6 points with the square and 6 for the pentagon. The square can have 8 points with the pentagon. These numbers give a total of $ 6 \\plus{} 8 \\plus{} 10 \\plus{} 6 \\plus{} 6 \\plus{} 8 \\equal{} \\boxed{44}\\text{intersection points}$.\r\n\r\nAs to your challenge problem with the 3-D objects, how can one have an intersection between a sphere and a cube? If they have intersection points, then either the sphere is tangent to the cube at one point, or the sphere intersects the cube at an infinite number of points on its surface.\r\n\r\nThus, their is no limit to how many intersection points they share.\r\n\r\nAlso, your definition for intersection points does not apply to a circle, whose perimeter contains no line segment. I think the definition should be \"Any point that is common to the perimeters of two or more figures.\"", "Solution_2": "Oh....when I drew a sphere and a cube I got that they could intersect at a maximum of $ 24$ points ($ 3$ at each vertex of the cube)\r\n\r\nIs this correct?", "Solution_3": "Oh, I see what you're saying.....If you count intersection points as a point where the edge of a polyhedron meets the face of another 3-D figure, then it sorta works to be 24. Thanks." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "If $ 09$ than $ 2^{n}\\plus{}2^{16}\\plus{}2^{9}\\equal{}2^{9}(2^{n\\minus{}9}\\plus{}2^{16}\\plus{}1)$ which obviously can't be square\r\n\r\nif\u02d8$ n\\leq9$ than $ 2^{16}\\plus{}2^{9}\\plus{}1\\equal{}(2^{8}\\plus{}1)^{2}<2^{16}\\plus{}2^{9}\\plus{}2^{n}\\leq2^{16}\\plus{}2^{10}<2^{16}\\plus{}2^{10}\\plus{}4\\equal{}(2^{8}\\plus{}2)^{2}$\r\n\r\nobivously no solutions for $ n$", "Solution_3": "[quote=\"Jure the frEEEk\"]if $ n > 9$ than $ 2^{n} \\plus{} 2^{16} \\plus{} 2^{9} \\equal{} 2^{9}(2^{n \\minus{} 9} \\plus{} 2^{16} \\plus{} 1)$ which obviously can't be square\n\nif\u02d8$ n\\leq9$ than $ 2^{16} \\plus{} 2^{9} \\plus{} 1 \\equal{} (2^{8} \\plus{} 1)^{2} < 2^{16} \\plus{} 2^{9} \\plus{} 2^{n}\\leq2^{16} \\plus{} 2^{10} < 2^{16} \\plus{} 2^{10} \\plus{} 4 \\equal{} (2^{8} \\plus{} 2)^{2}$\n\nobivously no solutions for $ n$[/quote]\r\nOh yes use this we can solve general problem ." } { "Tag": [], "Problem": "1. How many stereoisomers are possible for Farnesol (structure and information [url=http://en.wikipedia.org/wiki/Farnesol]here[/url])?\r\n\r\n2. What product(s) might be formed when farnesol is treated with sulphuric acid?", "Solution_1": "[hide=\"answer\"]\n1.4 stereoisomers (2 geometrical centres).\n2. allylic rearrangement to give conjugated diene:\n3,7,11-trimethyl 1,3,6,10-dodecatetraene.\n[/hide]", "Solution_2": "Your answer for 2 is wrong.", "Solution_3": "[hide=\"possible answer\"]\n1,2,2-trimethyl-1-(4'-methyl-3'-cyclohexenyl)-cyclopentan-3-ol\n1-methyl-2-(1'-hydroxy-1'-methyl ethyl)-1-(4'-methyl 3'-cyclohexenyl)-cyclobutane[/hide]", "Solution_4": "[quote=\"mythili\"][hide=\"possible answer\"]\n1,2,2-trimethyl-1-(3'-cyclohexenyl)-cyclopentane\n1-methyl-2-isopropyl-1-(3'-cyclohexenyl)-cyclobutane[/hide][/quote]\r\nwhat is the mechanism and are my eyes playing tricks on me or is there really one carbon atom less in the products.", "Solution_5": "sorry wrong naming :blush: there will be a 4' methyl at 3'cyclohexenyl.\r\nthe mechanism is first the protonation of the $ \\ce{OH}$ followed by ring closing steps...", "Solution_6": "i very much doubt if cyclization can take place as the hydrogens are not very acidic. but will have to wait for carcul to post his answer. :)", "Solution_7": "well this is an inspired procedure from the the closing of nerol to give $ \\alpha$ terepinol under similar conditions...well i have edted some parts of the answer see the edited answer ....i posted that in a hurry :blush:", "Solution_8": "Mythili's answer is also wrong. Just one ring will form predominantly.", "Solution_9": "is atleast a part of the answer right or is it a one big ring :(", "Solution_10": "What rings form more easily in cyclizations?", "Solution_11": "6 or 5 membered rings :maybe:", "Solution_12": "Yes, that's right.", "Solution_13": "well i had both of them in one of the answers...well aren't my answers even partially right :rotfl:" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Suppose X={x_1,x_2,...,x_n} and Y={y_1,y_2,...,y_n} are two sequences of natural number and satisfying: \r\n Multiply_{k=1}^{n}{x_k}=Multiply_{k=1}^{n}{y_k}=N,\r\nwhere N is a constant.\r\n Suppose Var(X),Var(Y) are variances and Mean(X), Mean(Y) are means of two sequences.\r\n I guess that: if Var(X) \\endtemplate", "Solution_1": "The default maximum number of columns in bmatrix (and pmatrix etc) is 10 and you have 11 columns. All you need to do is, before the bmatrix environment, add the line\r\n\\setcounter{MaxMatrixCols}{11}" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Quite easy\r\n$ a,b,c\\in Z$\r\n$ \\frac{\\sqrt{3}a\\plus{}b}{\\sqrt{3}b\\plus{}c}$ is a rational number\r\nProve that\r\n$ a\\plus{}b\\plus{}c|a^{2}\\plus{}b^{2}\\plus{}c^{2}$\r\n2)Find all prime $ p,q$\r\n$ p^{2}\\minus{}p\\plus{}1 \\equal{} q^{3}$\r\n3)$ n > 1$\r\n $ a > 0$\r\np is a prime(p>2)\r\n$ a^{p}\\equal{} 1(\\text{mod}p^{n})$\r\nProve that \r\n$ a \\equal{} 1(mod p^{n\\minus{}1})$\r\nSolve equation\r\n$ x^{n}\\equal{} p^{z}\\plus{}1$(n>1)", "Solution_1": "[hide=\"1\"]\n$ \\frac{\\sqrt{3}a\\plus{}b}{\\sqrt{3}b\\plus{}c}\\equal{}\\frac{(\\sqrt{3}a\\plus{}b)(\\sqrt{3}b\\minus{}c)}{3b^{2}\\minus{}c^{2}}$ \nequating the radical part to $ 0$ we get \n$ b^{2}\\equal{} ac\\rightarrow a^{2}\\plus{}b^{2}\\plus{}c^{2}\\equal{} (a\\plus{}b\\plus{}c)(a\\minus{}b\\plus{}c)$\n[/hide]" } { "Tag": [ "function", "limit", "calculus", "calculus computations" ], "Problem": "Let $f(x)$ be differentiable function such that $f'(0)=k.$\r\nDefine the sequence $\\{a_{n}\\}$ as follows.\r\n\\[a_{1}=1,\\ \\ a_{n}=\\lim_{x\\to\\infty}x^{2}\\left(f\\left(\\frac{a_{n-1}}{x}\\right)-f(0)\\right)^{2}\\ (n=2,\\ 3,\\ \\cdots). \\]\r\nFind the range of the value of $k$ for which the sequence $\\{a_{n}\\}$ converges.", "Solution_1": "Suppose $a_{n-1}\\neq0$. Then\r\n\r\n$a_{n}$\r\n\r\n$=\\lim_{x\\to\\infty}x^{2}\\left(f\\left(\\frac{a_{n-1}}{x}\\right)-f(0)\\right)^{2}$\r\n\r\n$=\\lim_{h\\to 0}(a_{n-1})^{2}\\left( \\frac{f(h)-f(0)}{h}\\right)^{2}$ where $h=\\frac{a_{n-1}}{x}$\r\n\r\n$= \\left( ka_{n-1}\\right)^{2}$\r\n\r\nIf $a_{n-1}=0$, Then $a_{n}= 0$.\r\n\r\n\r\n\r\nYou can check that $a_{n}$ converges if and only if $|k|\\leq1$" } { "Tag": [ "algorithm" ], "Problem": "hey this might seem easy but i am lost:)\r\ngiven a circle with N numbers written on it.\r\nwrite a O(N) algorithm which finds the consecutive block with largest sum\r\n\r\nwell i can't figure out how to use dp for this\r\n\r\nbest regards\r\nvasu", "Solution_1": "Hint: split it into 2 cases: one where the block of numbers does not 'wrap around' from N to 1, and one where it does.\r\n\r\n[edit]Actually, theres another way to do it.. involving going around the circle twice, storing the maximum sum so far as well as the length of the sequence ending there, stopping when that length gets bigger than N.", "Solution_2": "well as u said earlier \r\ni did this \r\nleave the first element\r\nfind the max consecutive block using dp.call it \"maxbest\". and also find the total sum including the first number\r\nalso find the min consectuive block leaving the first number,call it minbest.\r\nnow ur answer is max(maxbest,sum-minbest)\r\nright:)" } { "Tag": [ "trigonometry" ], "Problem": "Solve in reals:\r\n\r\n1.$ \\frac {x^2}{(x\\plus{}1)^2}\\plus{}x^2\\equal{}3$\r\n\r\n2.$ 2^{\\frac {2\\sin x}{\\sin x\\plus{} \\cos x}} \\equal{} 1 \\plus{} 2^{\\frac {\\sin x\\minus{} \\cos x}{\\sin x\\plus{} \\cos x}}$", "Solution_1": "[quote=\"hell_ever\"]2.$ 2^{\\frac {2\\sin x}{\\sin x \\plus{} \\cos x}} \\equal{} 1 \\plus{} 2^{\\frac {\\sin x \\minus{} \\cos x}{\\sin x \\plus{} \\cos x}}$[/quote]\r\n$ \\Leftrightarrow 2.2^{\\frac {\\sin x \\minus{} \\cos x}{\\sin x \\plus{} \\cos x}} \\equal{} 1 \\plus{} 2^{\\frac {\\sin x \\minus{} \\cos x}{\\sin x \\plus{} \\cos x}}\\Leftrightarrow 2^{\\frac {\\sin x \\minus{} \\cos x}{\\sin x \\plus{} \\cos x}} \\equal{} 1$\r\n$ \\Leftrightarrow \\sin x \\minus{} \\cos x\\equal{}0\\Leftrightarrow \\tan x\\equal{}1\\Leftrightarrow x\\equal{}\\frac{\\pi}{4}\\plus{}k\\pi$", "Solution_2": "nice work tdl :)", "Solution_3": "[quote=\"hell_ever\"]Solve in reals:\n\n1.$ \\frac {x^2}{(x \\plus{} 1)^2} \\plus{} x^2 \\equal{} 3$\n\n[/quote]\r\n\r\n\r\nLet A = x/x+1\r\nB = x\r\n\r\n$ AB \\equal{} A \\minus{} B \\equal{} X^2/X \\plus{} 1$\r\n\r\n$ A^2 \\plus{} B^2 \\equal{} 3$\r\n\r\nSquaring equation 1\r\n\r\n$ 3 \\minus{} 2AB \\equal{} A^2B^2$\r\n\r\nSolving AB = 1, -3. \r\n\r\nDue to obvious reasons -3 is not possible.\r\n\r\n$ (A \\plus{} B)^2 \\minus{} 2 \\equal{} 3$\r\n\r\nA + B = Root(5) \r\n\r\n$ x^2 \\minus{} Root(5)x \\plus{} 1 \\equal{} 0$\r\n\r\nSolving x = $ \\frac {Root(5) \\plus{} 1}{2}$\r\n$ \\frac{Root(5) \\minus{} 1}{2}$\r\n\r\nI might have made some stupid calculation error. I least bother. Hopefully you got the method.\r\n\r\nNice one tdl!", "Solution_4": "[quote=\"hell_ever\"]Solve in reals:\n\n1.$ \\frac {x^2}{(x \\plus{} 1)^2} \\plus{} x^2 \\equal{} 3$\n\n2.$ 2^{\\frac {2\\sin x}{\\sin x \\plus{} \\cos x}} \\equal{} 1 \\plus{} 2^{\\frac {\\sin x \\minus{} \\cos x}{\\sin x \\plus{} \\cos x}}$[/quote]\r\n\r\nAnother method for question 1:\r\n[hide]\nCompleting the squares:\nAdd and subtract $ (2 \\cdot x \\cdot \\frac {x}{x\\plus{}1}) \\implies (\\frac {x}{x\\plus{}1} \\minus{} x)^2 \\plus{} 2 \\cdot x \\cdot \\frac{x}{x\\plus{}1} \\equal{} 3$\n$ \\implies (\\frac {x^2}{x\\plus{}1})^2 \\plus{} \\plus{} 2 \\cdot x \\cdot \\frac{x}{x\\plus{}1} \\equal{} 3$\nSolve from now on :) \n[/hide]" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ S \\equal{} \\{1,2,4,..,2^n,..\\}$ and consider the fraction ring $ Z_S$. Show that $ Z_S/Z$ is an Artinian non-Noetherian Z-module.", "Solution_1": "Every proper submodule is either $ 0$ or is generated by $ 2^{ \\minus{} n}$." } { "Tag": [ "topology", "vector", "geometry", "3D geometry", "sphere" ], "Problem": "call $ X$ the space of unit length vectors in the tangent space to the sphere $ S^2$ (with respect to the standard metric, but that's not important), and $ Y$ the union of $ X$ and the set of tangent vectors of length $ \\le 1$ to a fixed point of $ S^2$ (say it's the north pole).\r\n\r\n1. show that $ X$ is homeomorphic to $ \\mathbb{RP}^3$.\r\n2. compute $ \\pi_1(Y)$.", "Solution_1": "Should $ X$ be unit length vectors in the tangent bundle, or are we fixing some point there as well?", "Solution_2": "no special point there, just the unit tangent bundle.", "Solution_3": "bonus question: what is the homotopy type of $ {\\rm Fr}(S^n)$, the (total space of) the frame bundle of $ S^n$?\r\nor, if you prefer, what is the total space of the space of orthormal oriented frames over $ S^n$?" } { "Tag": [ "algebra", "polynomial", "function", "IMO Shortlist" ], "Problem": "Let $ n$ be an even positive integer. Prove that there exists a positive inter $ k$ such that\r\n\r\n\\[ k \\equal{} f(x) \\cdot (x\\plus{}1)^n \\plus{} g(x) \\cdot (x^n \\plus{} 1)\\]\r\n\r\nfor some polynomials $ f(x), g(x)$ having integer coefficients. If $ k_0$ denotes the least such $ k,$ determine $ k_0$ as a function of $ n,$ i.e. show that $ k_0 \\equal{} 2^q$ where $ q$ is the odd integer determined by $ n \\equal{} q \\cdot 2^r, r \\in \\mathbb{N}.$ \r\n\r\nNote: This is variant A6' of the three variants given for this problem.", "Solution_1": "[quote=\"\"ISL questions and solutions from 1995-2001\" by A.Ajorloo\"]\nnote that $ n$ is an even number so $ (x + 1)$ divides $ 1 - x^n$,so there exists a polynomial $ a(x)$ with integer coefficients such that:\n\n$ (1 + x)\\cdot a(x) = 1 - x^n = 2 - (1 + x^n)$\n\n$ \\Rightarrow 2 = (1 + x)\\cdot a(x) + (1 + x^n)$\n\n$ \\Rightarrow 2^n = (1 + x)^n\\cdot\\left(a(x)\\right)^n + (1 + x^n)\\cdot b(x)$\n\nwhere $ b(x)$ is a polynomial with integer coefficients,as wanted.\nnow let $ f(x),g(x)$ be two polynomials with integer coefficients such that:\n\n$ k_0 = f(x)\\cdot (x + 1)^n + g(x)\\cdot(x^n + 1)\\textrm{ (1)}$\n\nnow let $ n = q2^r$ where $ q$ is an odd integer,we'll show that $ k_0 = 2^q$.put $ t = 2^r$ and note that $ x^n + 1 = (x^t + 1)\\cdot Q(x)$ in which:\n\n$ Q(x) = x^{t(q - 1)} - x^{t(q - 2)} + \\ldots - x^t + 1$\n\nso $ Q( - 1) = 1$,now the roots of the equation $ x^t + 1 = 0$ are:\n\n$ w_m = \\cos\\left(\\frac {(2m - 1)\\pi}{t}\\right) + \\sin\\left(\\frac {(2m - 1)\\pi}{t}\\right)\\textrm{ , } m = 1,2,\\ldots ,t$\n\nnow according to $ (1)$ we get that:\n\n$ f(w_m)\\cdot (w_m + 1)^n = k_0\\textrm{ }1\\leq m\\leq t\\textrm{ (2)}$\n\nnow note that $ (w_1 + 1)(w_2 + 1)\\cdots (w_t + 1) = 2$,now we can see that:\n\n$ f(w_1)\\cdots f(w_m) = F$\n\nwhere $ F$ is an integer,now according to $ (2)$ we get that:\n\n$ 2^nF = k_0^t$,so $ k_0$ is divisible by $ 2^q$,now from the fact that $ Q( - 1) = 1$ we get that:\n\n$ Q(x) = (x + 1)\\cdot c(x) + 1$\n\nwhere $ c(x)$ has integer coefficients,thus:\n\n$ (x + 1)^n\\cdot\\left(c(x)\\right)^n = \\left( - 1 + Q(x)\\right)^n = 1 + Q(x)\\cdot d(x)\\textrm{ (3)}$\n\nin which $ d(x)$ is a polynomial with integer coefficients,now note that for a fixed number $ m$,the powers $ w_m^{2j - 1}$ for $ 1\\leq j\\leq t$ are different and also form all the roots of $ x^t + 1 = 0$,so:\n\n$ (1 + w_m)(1 + w_m^3)\\cdots (1 + w_m^{2t - 1}) = 2$\n\nand\n\n$ (1 + w_m^{2j - 1}) = (1 + w_m)\\left(1 - w_m + w_m^2 - \\ldots + w_m^{2j - 2}\\right)$\n\nso there exists a polynomial $ h(x)$ with integer coefficients,independent from $ m$,such that:\n\n$ (1 + w_m)^t\\cdot h(w_m) = 2$\n\nso $ (x + 1)^t\\cdot h(x) - 2$ is divisible by $ x^t + 1$ and we can write:\n\n$ (x + 1)^t\\cdot h(x) = 2 + (x^t + 1)\\cdot g(x)$\n\nin which $ g(x)$ is a polynomial with integer coefficients,so:\n\n$ (x + 1)^n\\cdot\\left(h(x)\\right)^q = 2^q + (x^t + 1)\\cdot w(x)\\textrm{ (4)}$\n\nin which $ w(x)$ is a polynomial with integer coefficients,now using $ (3),(4)$ together,we obtain that:\n\n\\begin{eqnarray*}(x + 1)^n\\cdot\\left(c(x)\\right)^n\\cdot (x^t + 1)\\cdot w(x) & = & (x^t + 1)\\cdot w(x) + (x^t + 1)\\cdot w(x)\\cdot Q(x)\\cdot d(x) \\\\\n& = & (x + 1)^n\\cdot\\left(h(x)\\right)^q - 2^q + (x^n + 1)\\cdot w(x)\\cdot d(x) \\\\ & \\Rightarrow & 2^q = p(x)\\cdot (x + 1)^n + q(x)\\cdot (x^n + 1)\\end{eqnarray*}\n\nin which $ p(x),q(x)$ have integer coefficients,so $ k_0\\leq 2^q$ but we had $ 2^q\\mid k_0$,hence $ k_0 = 2^q$.\n\nQED\n\n\n[/quote]", "Solution_2": "I like this problem it is really nice :) \r\nsolution is good I like it :lol:", "Solution_3": "I understood the whole solution, except one small step: why is $ f(w_{1})\\cdots f(w_{m}) = F $ an integer?\n\nCan someone explain me this step with more detail?\n\nThanks in advance\nThinkertoys", "Solution_4": "$f(w_1) f(w_2) \\cdots f(w_n)$ is an integer by the fundamental theorem of symmetric polynomials. We can express $f(w_1) f(w_2) \\cdots f(w_n)$ as a symmetric polynomial with integer coefficients evaluated at the symmetric sums of $w_1, w_2, \\ldots, w_n$, which, by Vieta's formulas, are themselves integers. \n\nAnyway, here is another way to show that a minimum of $2^q$ can be attained. \n[hide]\nFirst, we show that the minimum, $k$, is a power of 2. Because $x+1$ divides $x^n - 1$, we have $x^n - 1 = a(x) (x+1)$, or $2 = (x^n+1) - a(x)(x+1)$. Raising both sides to the $n$th power, we have \n\\[ 2^n = a(x)^n (x+1)^n + b(x)(x^n + 1). \\, \\, \\,\\, \\, \\,\\, \\, \\, (1)\\]\n for some polynomial $b(x)$ (which we obtain from binomially expanding the sum and factoring out $x^n+1$ from all but the $a(x)^n (x+1)^n$ term.) It follows that some minimal $k$ with\n\\[ k = f(x) (x+1)^n + g(x)(x^n + 1) \\, \\, \\,\\, \\, \\,\\, \\, \\, (2) \\]\nexists. Now, if we subtract (2) $\\lfloor 2^n/k \\rfloor$ times from (1), the left-hand side must be 0 (or else the minimality of $k$ is contradicted), so $k$ must divide $2^n$, i.e,. $k$ must be a power of 2. \n---------------------------------------------------------------------------------------------------------------------------\nWe now show that when $q=1$, we must have $k=2$. First, by substituting $x=1$, it is easy to see that $k$ must be even. Now suppose for the sake of contradiction that $k=2^c$ for some $c \\geq 2$. Let $h(x) = -g(x)$. Our new equation is\n\\[ 2^c = f(x) (x+1)^{2^r} - h(x) (x^{2^r}+1). \\, \\, \\,\\, \\, \\,\\, \\, \\, (3) \\]\nIt is clear that $\\deg f = \\deg h$. Let $f(x) = (x^{2^r} + 1)u(x) + v(x)$, with $\\deg r(x) < 2^r$. Because we can replace $f(x)$ and $h(x)$ with $f(x)-u(x)(x^{2^r}+1) = v(x)$ and $h(x) + u(x)(x+1)^{2^r}$, respectively, we can suppose without loss of generality that $\\deg f = \\deg h < 2^r$. \n\nConsider this equation in the finite field $F_2$, i.e., the finite field of order 2. $f$ and $h$ cannot be the zero polynomial in $F_2$, or else we can divide (3) by 2 to obtain a smaller $k$. Equation (3) taken in $F_2$ yields\n\\[ f(x)(x+1)^{2^r} = h(x)(x^{2^r}+1). \\, \\, \\,\\, \\, \\,\\, \\, \\, (4) \\]\nSince $(x+1)^{2^r} = x^{2^r} + 1$ in $F_2$, we have \n\\[ (x^{2^r} + 1)(f(x) - h(x)) = 0 \\]\nin $F_2$. Since $x^{2^r} + 1$ is nonzero in $F_2$, we must have $f(x)-h(x) = 0$ in $F_2$, i.e., we must have $f(x) = h(x) + 2p(x)$ for some integer polynomial $p$. Substituting this into (3), we have\n\\[ 2^c = h(x)((x+1)^{2^r} - (x^{2^r} + 1)) + 2p(x)(x+1)^{2^r}. \\, \\, \\,\\, \\, \\,\\, \\, \\, (5) \\]\nLet $e(x) =((x+1)^{2^r} - (x^{2^r} + 1))/2$. Note that $e$ is an integer polynomial, and that $e(1) = 2^{2^r - 1} - 1$ is odd. Dividing (5) by 2 gives \n\\[ 2^{c-1} = h(x)e(x) + p(x)(x+1)^{2^r}. \\, \\, \\,\\, \\, \\,\\, \\, \\, (6) \\]\nBecause $c \\geq 2$, $2^{c-1}$ is even. Thus, if we consider (6) in $F_2$, we get \n\\[ p(x)(x+1)^{2^r} = (-h(x))(e(x)). \\]\nSince $e(1)$ is odd, the polynomials $(x+1)^{2^r}$ and $e(x)$ are coprime in $F_2$. Thus, $(x+1)^{2^r}$ divides $-h(x)$ in $F_2$, so there exist polynomials $c(x)$ and $d(x)$ such that $h(x) = c(x)(x+1)^{2^r} + 2d(x)$. But this implies that either $\\deg h \\geq 2^r$ or $h$ is the zero polynomial in $F_2$, both of which are contradictions. Thus, our assumption that $c \\geq 2$ is false, so we must have $k=2$. \n---------------------------------------------------------------------------------------------------------------------------\nWe now prove the result for all $q$. Suppose we have polynomials $f(x)$ and $g(x)$ such that \n\\[ 2 = f(x) (x+1)^{2^r} + g(x)(x^{2^r} + 1). \\]\nRaise both sides to the $q$th power to get\n\\[ 2^q = f_0(x)(x+1)^{q 2^r} + g_0(x)(x^{2^r}+1), \\]\nwhere $f_0(x) = f(x)^q$ and $g_0(x)$ is obtained from binomially expanding and factoring $x^{2^r} + 1$ out of all the terms but $f_0(x) (x+1)^{q 2^r}.$ Now replace $x$ with $x^q$ to get \n\\[ 2^q = f_0(x^q) (x^q + 1)^{q 2^r} + g_0(x^q) (x^{q 2^r} + 1) = \\left( f_0(x^q) \\frac{x^q+1}{x+1}\\right)^{q 2^r} (x+1)^{q 2^r} + g_0(x^q) (x^{q 2^r} + 1), \\]\nso our construction is complete. \n[/hide]", "Solution_5": "As above, I'll prove that $k = 2^q$, where $n = q \\cdot 2^r$, and $q$ is odd, and $r$ is an integer greater than or equal to 0. There are many examples of the construction for $k = 2^q$ above, so I'll just give my solution for proving the lower bound: $k \\ge 2^q$\n\nObviously, we want that for any integer $x$, $\\gcd((x+1)^n, x^n+1) \\ge k$. This looks bad initially, because for any integers $x$, $\\gcd((x+1)^n, x^n+1) \\le 2$. So instead of using only integers for $x$, why not use [i]algebraic integers?[/i] Remember that the algebraic integers forma ring, i.e. the sum or product of any two algebraic integers is an algebraic integer. Our algebraic integers here will just use roots of unity. Let $\\omega = e^{\\frac{\\pi i}{2^r}}$. Note that $\\omega^n + 1 = 0$. Let's turn to evaluating $(\\omega + 1)^n$.\n\n$\\omega + 1 = (\\cos \\left( \\frac{\\pi }{2^r} \\right) + 1) + i \\sin \\left( \\frac{\\pi }{2^r} \\right) = 2 \\cos \\left( \\frac{\\pi }{2^{r+1}} \\right) e^{\\frac{\\pi i}{2^{r+1}} }.$ Therefore, $(\\omega+1)^n = 2^n \\cos ^n \\left( \\frac{\\pi }{2^{r+1}} \\right) i^q$, by simple computations.\n\nLet's ignore the $i^q$ at the end, and just focus on $2^n \\cos ^n \\left( \\frac{\\pi }{2^{r+1}} \\right)$. The previous expression equals $\\left( 2^{2^r} \\cos ^{2^r} \\left( \\frac{\\pi }{2^{r+1}} \\right) \\right)^q$. Let $\\alpha = \\frac{1}{2} \\left( 2^{2^r} \\cos ^{2^r} \\left( \\frac{\\pi }{2^{r+1}} \\right) \\right) = 2^{2^r-1} \\cos ^{2^r} \\left( \\frac{\\pi }{2^{r+1}} \\right)$. I claim that $\\alpha$ is an algebraic integer. We prove this by induction on $r$. The base case $r = 0$ is obviously true.\n\nNote that \\[2^{2^r-1} \\cos ^{2^r} \\left( \\frac{\\pi }{2^{r+1}} \\right) = 2^{2^r-1} \\left(\\frac{1+\\cos \\left(\\frac{\\pi}{2^r} \\right)}{2} \\right)^{2^{r-1}} = 2^{2^{r-1}-1} ( 1+\\cos \\left(\\frac{\\pi}{2^r} \\right) )^{2^{r-1}}, \\] using the cosine double angle identity. The fact that the previous expression is an algebraic integer then comes from implicitly expanding the binomial theorem and using the inductive hypothesis. Therefore, $\\left( 2^{2^r} \\cos ^{2^r} \\left( \\frac{\\pi }{2^{r+1}} \\right) \\right)^q = (2 \\alpha)^q = 2^q \\alpha^q.$ Therefore, $(\\omega+1)^n = 2^q \\cdot \\alpha^q \\cdot i^q$. Since $f(\\omega) (\\omega+1)^n + g(\\omega) (\\omega^n+1) = f(\\omega)(\\omega+1)^n = k$, and $f(\\omega)$ is an algebraic integer, and $(\\omega+1)^n = 2^q \\cdot \\alpha^q \\cdot i^q$, $k$ must be a multiple of $2^q$. I'll leave the construction to the other posts. Mine was the same.", "Solution_6": "First, we prove that if the expression is an integer, it's divisible by $2^q$. Let $\\omega=e^{\\pi i / n}$, so the roots of $x^n+1$ are $\\omega,\\omega^3,\\cdots \\omega^{2n-1}$. Now if $k=f(\\omega)(\\omega+1)^n$, then for any conjugate $\\omega'=\\omega^{2j-1}$ of $\\omega$, $k=f(\\omega')(\\omega'+1)^n$ as well. So\n\\begin{align*}\nk^{2^r}&=f(\\omega^{q})(\\omega^{q}+1)^n f(\\omega^{3q})(\\omega^{3q}+1)^n \\cdots f(\\omega^{(2^{r+1}-1)q})(\\omega^{(2^{r+1}-1)q}+1)^n\\\\\n&=h(-1)^n f(\\omega^q)f(\\omega^{3q})\\cdots f(\\omega^{(2^{r+1}-1)q})\\\\\n&=2^n f(\\omega^q)f(\\omega^{3q})\\cdots f(\\omega^{(2^{r+1}-1)q})\n\\end{align*}\nwhere $h(x)=x^{2^r}+1$. Now $ f(\\omega^q)f(\\omega^{3q})\\cdots f(\\omega^{(2^{r+1}-1)q})$ is both an algebraic integer and rational, so it is an integer. Thus $2^{2^r q}=2^n\\mid k^{2^r}$, implying $2^q\\mid k$.\n\nNow we show that $k=2^q$ can be constructed in the given form. It suffices to find $f\\in \\mathbb{Z}[x]$ such that $2^q=f(\\omega) (\\omega+1)^n$. Take $f$ such that\n\\[\nf(x)\\cdot (x+1)^n=\\left((x^q+1)(x^{3q}+1)\\cdots (x^{(2^{r+1}-1)q}+1)\\right)^{q}\n\\]\n(Observe that the right hand side has $2^r\\cdot q = n$ factors of $(x+1)$.) As before, then, \n\\[\nf(\\omega)\\cdot (\\omega+1)^n = h(-1)^q=2^q\n\\]" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Given an example of fields $ F\\subseteq K\\subseteq L$ where $ L/K$ and $ K/F$ are normal but $ L/F$ is not normal.", "Solution_1": "An extension is normal if and only if it is the splitting field of a given polynomial. So let $ F\\equal{}\\mathbb{Q}$, $ K\\equal{}F(\\sqrt 2)$ and $ L\\equal{}K(\\sqrt[4] 2)$. $ L\\mid F$ is not normal since the splitting field of $ X^4\\minus{}2$ is a not a subfield of $ \\mathbb{R}$." } { "Tag": [], "Problem": "Hi , all .\r\nI like here .", "Solution_1": "Hi madison22, I am a new member too! :P", "Solution_2": "Welcome to the awsome(if not mostly in-active) canadian forum!", "Solution_3": "Hey, it's not inactive, I'm here! :D\r\n\r\nWelcome. I live in Saskatchewan. It's still cold here :(", "Solution_4": "ya its snowing here in waterloo :(" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "What biggest area can have a triangle, which radius of the circumscribed circle is $ R$, and the length of one of medians is $ m$?", "Solution_1": "I have an idea... The the biggest triangle will be equilateral.\r\nWho can solve it? Help please" } { "Tag": [], "Problem": "In case you haven't heard the story, a railroad worker named Phineas Gage experienced major brain injury when a metal rod blasted through his skull on September 13, 1848. Other than the loss of functionality of his left eye, he made a full physical recovery. However, it was reported that his personality had changed, speculated to have been caused by damage to his frontal lobe.\r\n\r\nThus, I wonder exactly what free will/thought is. If brain damage can completely change a person's personality, then can personality be correlated with brain condition? What I wonder is if there really is unique qualities to each organism that couldn't be explained by brain conditions. It appears as though we are somewhat like machines. We perceive and respond in such a way that is written in our brains.\r\n\r\nSimilarly, what exactly is love? If certain chemical changes did not occur in our bodies, would there be any reason for people to fall in love? I can't imagine any other reason why love occurs.\r\n\r\nIn conclusion, what is thought? How can we define it? Does free will really exist? I suppose that we shall never be able to know, but please feel free to discuss. However, please avoid bringing religion into the equation because that's not what I'm talking about.", "Solution_1": "[quote=\"12markkram34\"] Similarly, what exactly is love? [/quote]\r\n\r\nLove is often characterized by an intense feeling of mutual attraction, tender feeling of caring for another person, etc. Probably, IMO, love is extreme affection for a person or an individual. \r\n\r\nI guess there are many aspects of love. One such aspect is-- teenage love--which most often arises (in most cases) due to strong mutual physical attraction. 'Love' in this context is intense interpersonal attraction.\r\n\r\nAnother aspect of love is parent-child relationship -- maternal or paternal love. Similarly love (or affection) tends to exist between siblings. 'Love' in this context is intense affection. \r\n\r\nFor example, there could also be religious love. Some people might construe extreme religious love as 'fanaticism'. Similarly, some people might construe 'lust' as love. The answer to a question like \"What is Love\" depends upon the perception of an individual or person. Many people have varied or different opinion(s) on love.\r\n\r\nI personally feel love is an abstract concept just like truth. I think it's pretty difficult to articulate the varied philosophical definitions of love." } { "Tag": [ "MATHCOUNTS" ], "Problem": "i know this has probably been asked before, but where can i get past mathcounts tests (before 2008 tests)online?", "Solution_1": "Hmmm...you can try asking around. Or, you can buy them online from MATHCOUNTS.", "Solution_2": "There is a thread in the stickies about this." } { "Tag": [ "calculus", "derivative", "vector", "real analysis", "real analysis unsolved" ], "Problem": "$ D_x^{\\alpha}\\left(e^{ixy}\\right)\\equal{}?,x,y\\in\\mathbb{R}^n$", "Solution_1": "Are you asking for the $ \\alpha$th derivative with respect to $ x$? Not difficult:\r\n$ \\frac{\\partial f}{\\partial x}\\equal{}iye^{ixy}$\r\n$ \\frac{\\partial^{2}f}{\\partial x^{2}}\\equal{}i^{2}y^{2}e^{ixy}\\equal{}\\minus{}y^{2}e^{ixy}$\r\n\r\nSo we drag down $ \\alpha$ powers of $ i$ and $ y$, so $ D_{x}^{\\alpha}f\\equal{}(iy)^{\\alpha}e^{ixy}$.", "Solution_2": "What does $ e^{i x y}$ mean when $ n > 1$?", "Solution_3": "The way in which this notation makes the most sense: the context is Fourier transforms on $ \\mathbb{R}^n.$\r\n\r\nThat should really be $ e^{ix\\cdot y}$ - that is, that's a dot product in the exponent.\r\n\r\nIn the symbol $ D_x^{\\alpha},$ $ \\alpha$ is a multi-index. That is, $ \\alpha \\equal{} (\\alpha_1,\\alpha_2,\\dots,\\alpha_n)$ and $ D_x^{\\alpha}\\equal{}\\frac{\\partial^{|\\alpha|}}{\\partial x_1^{\\alpha_1}\\partial x_2^{\\alpha_2}\\cdots\\partial x_n^{\\alpha_n}}.$\r\n\r\n(Here, $ |\\alpha|\\equal{}\\alpha_1\\plus{}\\alpha_2\\plus{}\\cdots\\plus{}\\alpha_n$ is the degree of the multi-index.)\r\n\r\nWith that understanding, we do have $ D_x^{\\alpha}e^{ix\\cdot y}\\equal{}(iy)^{\\alpha}e^{ix\\cdot y}\\equal{}i^{|\\alpha|}y^{\\alpha}e^{ix\\cdot y}.$\r\n\r\nOf course, by $ y^{\\alpha}$ we mean $ y_1^{\\alpha_1}y_2^{\\alpha_2}\\cdots y_n^{\\alpha_n}.$", "Solution_4": "How should it make sense when $ \\alpha$ is not integer vector, I mean $ \\alpha_i$ may be real number for some $ i$?", "Solution_5": "I suppose using $ \\alpha$ for this purpose presupposes it being integral.", "Solution_6": "Thank you very much for your effective discussions. :lol:" } { "Tag": [ "limit", "analytic geometry", "trigonometry", "calculus", "calculus computations" ], "Problem": "How to compute the limit $ \\lim_{(x,y)\\to(0,0)}\\frac {x^3 \\plus{} y^3}{x^2 \\plus{} y^2}$ ?", "Solution_1": "move to polar coordinates: $ x\\equal{}r\\cos\\theta, y\\equal{}r\\sin\\theta$, and:\r\n$ \\lim_{(x,y)\\to(0,0)}\\frac {x^3 \\plus{} y^3}{x^2 \\plus{} y^2}\\equal{}\\lim_{r \\to 0} r ((\\cos\\theta)^3\\plus{}(\\sin\\theta)^3)\\equal{}0$ (because $ r \\to 0$ and $ (\\cos\\theta)^3\\plus{}(\\sin\\theta)^3$ is bounded)", "Solution_2": "Thank you very much, Bambaman! :D \r\n\r\nAfter seeing your solution, I thought out a more general method to solve this kind of problems.\r\n\r\nWith the original limit, think about a similar but a little bit more complicated one:\r\n\r\n$ \\lim_{\\mathbf x\\to\\mathbf0} \\frac{x_1^3\\plus{}x_2^3\\plus{}\\cdots\\plus{}x_n^3}{x_1^2\\plus{}x_2^2\\plus{}\\cdots\\plus{}x_n^2}$ with $ \\mathbf x\\equal{}(x_1, x_2,\\cdots, x_n)\\in\\mathbb R^n$.\r\n\r\nIn this case, we can't use the usual \"polar coordinates\". (Well, maybe we can introduce one.)\r\n\r\nNevertheless, we can let $ x_i\\equal{}\\|\\mathbf x\\|\\xi_i$ for $ i\\equal{}1,2,\\cdots, n$. Then\r\n\r\n$ |\\xi_i|\\equal{}\\frac{|x_i|}{\\|\\mathbf x\\|}\\equal{}\\frac{|x_i|}{\\sqrt{x_1^2\\plus{}x_2^2\\plus{}\\cdots\\plus{}x_n^2}}\\leq\\frac{|x_i|}{\\sqrt{x_i^2}}\\equal{}1.$\r\n\r\nBy bambaman's way, we find the limit is $ 0$.", "Solution_3": "[quote=\"ifai\"]\n$ \\lim_{\\mathbf x\\to\\mathbf0} \\frac {x_1^3 \\plus{} x_2^3 \\plus{} \\cdots \\plus{} x_n^3}{x_1^2 \\plus{} x_2^2 \\plus{} \\cdots \\plus{} x_n^2}$ with $ \\mathbf x \\equal{} (x_1, x_2,\\cdots, x_n)\\in\\mathbb R^n$.[/quote]\r\nlet $ f(x_1,....,x_n)\\equal{}\\frac {x_1^3 \\plus{} x_2^3 \\plus{} \\cdots \\plus{} x_n^3}{x_1^2 \\plus{} x_2^2 \\plus{} \\cdots \\plus{} x_n^2}$.\r\n\r\nwe have $ |x_i|^3 \\le (x_1^2 \\plus{} x_2^2 \\plus{} \\cdots \\plus{} x_n^2)^{\\frac{3}{2}}$ .\r\n\r\nso $ |f(x_1,....,x_n)| \\le \\frac {|x_1|^3 \\plus{} |x_2|^3 \\plus{} \\cdots \\plus{} |x_n|^3}{x_1^2 \\plus{} x_2^2 \\plus{} \\cdots \\plus{} x_n^2}$\r\n\r\n$ \\le n\\frac {(x_1^2 \\plus{} x_2^2 \\plus{} \\cdots \\plus{} x_n^2)^{\\frac{3}{2}} }{x_1^2 \\plus{} x_2^2 \\plus{} \\cdots \\plus{} x_n^2}$\r\n\r\n$ \\equal{}n\\sqrt{x_1^2 \\plus{} x_2^2 \\plus{} \\cdots \\plus{} x_n^2}$\r\n\r\nwhich has limit $ 0$." } { "Tag": [ "Columbia", "Putnam", "real analysis" ], "Problem": "I know there's been discussions on Waterloo vs Toronto, but how does University Of British Columbia stack up against them in terms of math and physics?", "Solution_1": "To my knowledge UBC has no real record of mathematical or physics related achievments. SFU on the other hand, has outdone UBC in previous years. SFU has even produced a putnam fellow! ( A extremely rare child-prodigy :D )", "Solution_2": "I go to SFU, but that being said, here is my opinion of UBC.\r\n\r\nUBC is a nice university. It's undergrad courses are nothing special (I won large entrance scholarships to both universities, and satisfied myself that there was little to no difference between the two for undergrad). Their selection of undergrad courses is solid - they have a large enough faculty that they're not really deficient in anything. I don't like their choice of textbooks for most courses - they seem to take books almost exclusively published by Prentice-Hall. While you can get away with this up to the point, it is nice to take courses where they use the 'classics' - Rudin, Ahlfors, Lang, Dummit and Foote, etc.\r\n\r\nHowever, their selection of graduate courses is fantastic for a department of their size. I will probably be spending most of my third and fourth years of university at UBC taking graduate courses. They have a ton of 'core' graduate courses, plus a bunch of selected topics courses in just about every sub-discipline of mathematics. If you're interested in it, they also have some pretty interesting course on financial mathematics (and I don't mean watered-down 'economics' math).\r\n\r\nAs for the Putnam, they have turned in a nice showing in recent years (several team top 10s) mainly due to Daniel Brox, who was an IMO gold medalist. As far as I know, he's graduated now, so that will probably not be the case in the immediate future. But they have a good group of people who are willing to help prepare for the Putnam, and a 'problem solving' course (IIRC, it was taught out of Wilf's Generatingfunctionology).\r\n\r\nI don't know anything about their physics department, other than that it's probably pretty good. They are also pretty good for CS, but SFU is better there (one of the few departments that I could say that about).\r\n\r\nIMHO, UBC is the third best mathematics university in Canada. But it's a distant third.", "Solution_3": "Hey, thanks for the responses! I feel a little foolish for not asking this question first, but what constitutes a good math university? I'm interested in becoming a professor, so obviously it needs the prestige to get me into Grad School. Is it the quality of teaching, the diversity of courses, their difficulty, the research oppurtinities? \r\n\r\nI feel passionately about research, but thats one of those things that depends on you more than anything else. A good putnam training program is a big plus, and the environment has to suit my style. But what distinguishes say UOT or SFU from UBC?\r\n\r\nThanks a lot!\r\n\r\n\r\nP.S. blahblahblah, you've got me all excited, how are you able to graduate courses in your third year? I'm more of interested in Pure math and theoritical physics. I like that UBC has a particle accelator :P" } { "Tag": [ "trigonometry", "function" ], "Problem": "Hey I need some help.\r\nWe are supposed to simplify these expressions.\r\n\r\n[u]Tanx [/u]\r\nsecx + 1\r\n\r\ncos^2(x/2) - sin^2(x/2)\r\n\r\nAny help is appreciated.", "Solution_1": "[hide=\"1\"]\nThe solution to this problem illustrates a common precept of trig problems: [i]when in doubt, write everything in terms of $\\sin$ and $\\cos$.[/i]\n\n$\\frac{\\tan x}{\\sec x + 1} = \\frac{ \\frac{\\sin x}{\\cos x} }{\\frac{1}{\\cos x} + 1} = \\frac{\\sin x}{1 + \\cos x} = \\frac{\\sin x (1 - \\cos x)}{1 - \\cos^2 x} = \\frac{1 - \\cos x}{\\sin x}$, which we can rewrite as $\\csc x - \\cot x$, if this is what you mean by 'simplification.'\n\nThere is, however, another way to rewrite this function, but it is a bit more convoluted and is a less common identity (though this is probably the expected identity). We will use the common identities $\\sin 2x = 2 \\sin x \\cos x$ and $\\cos 2x = \\cos^2 x - \\sin^2 x$.\n\n$\\frac{1 - \\cos x}{\\sin x} = \\frac{1 - \\cos^2 \\frac{x}{2} + \\sin^2 \\frac{x}{2} }{2 \\sin \\frac{x}{2} \\cos \\frac{x}{2} } = \\frac{\\sin^2 \\frac{x}{2} }{\\sin \\frac{x}{2} \\cos \\frac{x}{2} } = \\tan \\frac{x}{2}$.\n[/hide]\n\n[hide=\"2\"]\nAs before, we will use the common identity $\\cos 2x = \\cos^2 x - \\sin^2 x$. This gives us $\\cos^2 \\frac{x}{2} - \\sin^2 \\frac{x}{2} = \\cos x$.\n[/hide]", "Solution_2": "i would have done the write in terms of cos and sin but i like the other method better, it gets a more \"exact\" answer.", "Solution_3": "Hey thanks a lot guys, I have one more problem I need help on.\r\n\r\nWe need to solve for x in this problem.\r\nI know that 0 is an answer, I just need the other answers.\r\n\r\n[b]Tan2x + tanx = 0[/b]", "Solution_4": "[hide]Since $\\tan{2x}=\\frac{2\\tan{x}}{1-\\tan^2{x}}$: \n\n$\\frac{2\\tan{x}}{1-\\tan^2{x}}+\\tan{x}=0$. \n\nMultiply by $1-\\tan^2{x}$: \n \n$2\\tan{x}+\\tan{x}-\\tan^3x=0$. \n\nCombine like terms, multiply by -1, and factor: \n\n$\\tan{x}(\\tan^2x-3)=0$. \n\nSo now we know that one solution is when x=0. The other solutions are the solutions to the equation \n\n$\\tan^2x-3=0$. \n$\\tan{x}=\\pm\\sqrt3$. \n$x=\\frac{5\\pi}{3}, \\frac{\\pi}{3}$. [/hide]", "Solution_5": "Congratulations. You received help with math homework for the first and last time on AoPS.\r\n\r\nTopic locked.", "Solution_6": "[quote=\"chess64\"]Congratulations. You received help with math homework for the first and last time on AoPS.\n\nTopic locked.[/quote]Ouch. :stretcher:\r\n\r\nI think it's ok to help people on a few problems. Just because somebody posts 2 homework problems doesn't mean you should lock the topic, as long as they don't post periodically.\r\n\r\nTherefore, topic temporarily unlocked. :)", "Solution_7": ":D Well okay, I personally don't like it... but I guess it's just the first time.\r\n\r\nSorry about that! :)", "Solution_8": "Sorry if I broke any rules, I didn't see anywhere where we couldn't do what I did.\r\nIf I broke a rule, please link me to the rule, or a forum where I can post those problems.\r\n\r\nAnyway, I need help on these 2 problems. (last ones I promise!)\r\n\r\nsin2x=cos2x\r\n\r\ncos4x=1-3cos2x\r\n\r\nThanks", "Solution_9": "1. Um...intersect x^2+y^2=1 (unit circle) w/ x=y take arcsin then divide by two?", "Solution_10": "the second one\r\n\r\n$cos(4x)=1-3cos(2x)$\r\n$2cos^2(2x)-1=1-3cos(2x)$\r\n\r\nfrom here just solve like a normal quadradic." } { "Tag": [ "blogs", "\\/closed" ], "Problem": "Signatures don't work on blogs. Is this just for me?", "Solution_1": "I thought if you selected \"Attach Signature,\" it would work, but it doesn't. :huh:", "Solution_2": "It's going to stay that way for now (due to different reasons). Later on I'll add an option to enable/disable signatures in your own blog.", "Solution_3": "Will the signature thing be applied to contirbutors as well?", "Solution_4": "What? No, you will have an option to enable/disable general use of signatures in your blog so that anybody or nobody can post with signatures." } { "Tag": [], "Problem": "What is the ten-thousandths digit of the product $ 3.57 \\times 2.43$?", "Solution_1": "In this case, you can count 2 digits after decimal+2 digits after decimal=4 digits after decimal\r\nso you know ten-thousandth digit is last one. just the long multiplication thingy and you know last digit is $ 3*7$'s last digit, which happens to be $ 1$" } { "Tag": [ "limit", "inequalities", "parameterization", "integration", "absolute value", "real analysis", "real analysis unsolved" ], "Problem": "If f(x) is continuos and bounded on $(0,\\infty)$ then for any $a>0$ \r\n$\\lim_{n\\to+\\infty}e^{-an}\\sum_{k=1}^{\\infty}f(x+\\frac{k}{n})\\frac{(an)^{k}}{k!}= f(x+a)$", "Solution_1": "I think that the problem should read:\r\n\r\n$\\lim_{n\\rightarrow \\infty}e^{-an}\\sum_{k=0}^{\\infty}f(x+\\frac{k}{n})\\frac{(an)^{k}}{k!}=f(x)$.\r\n\r\nAnd here is the proof:\r\nLet $b_{n}=f(x+\\frac{k}{n})-f(x)$. then observe that this sequence is bounded and converges to 0, so we prove more generally that the following holds:\r\nIf $b_{n}\\rightarrow 0$ then \r\n$\\lim e^{-an}\\sum_{k=0}^{\\infty}b_{k}\\frac{(an)^{k}}{k!}=0$.\r\n\r\nSince $b_{n}\\rightarrow 0$ then there is $N$ such that $b_{n}\\leq \\epsilon$ for all $n\\geq N$.\r\nalso $b_{n}$ is bounded therefore we get that $|b_{n}|\\leq M$.\r\n\r\nNow, \r\n$|e^{-an}\\sum_{k=0}^{\\infty}b_{k}\\frac{(an)^{k}}{k!}|\\leq e^{-an}M\\sum_{k=0}^{N-1}\\frac{(an)^{k}}{k!}+e^{-an}\\epsilon \\sum_{k=N}^{\\infty}\\frac{(an)^{k}}{k!}$\r\n$\\leq e^{-an}\\sum_{k=0}^{N}\\frac{(an)^{k}}{k!}+\\epsilon$.\r\nLetting $n$ converge to $\\infty$ we get that the limit of the absolute value of the left hand side inequality is less than epsilon and since epsilon is arbitrary taken the result follows.\r\n\r\nA question:\r\nDoes the reverse conclusion holds, namely if the sum converges to 0 does it follow that $b_{n}$ converges to 0? The answer is no.", "Solution_2": "[b]Didilica[/b], I think, the statement in 1st post is true. Your proof is not correct since $b_{n}$ depends on $n$ and $k$\r\nWe just need to prove that there's $b_{n}=\\overline{\\overline{o}}(n)$ such that\r\n\\[\\sum_{|k-an|>b_{n}}\\frac{(an)^{k}}{k!}=\\overline{\\overline{o}}(e^{an})\\]\r\nI think smth like $b_{n}=n^{\\tfrac23}$ (but I'm not sure)", "Solution_3": "If we take $e^{imx}$ it seems true. If we approximate f(x) by Fourier series and then formally interchange sums and limit statement holds. But of course this is not strict.. :|", "Solution_4": "Here's simple proof of my statement.\r\nLet $\\xi_{1},\\xi_{2},\\ldots,\\xi_{n}$ be identically distributed independent variates with Poisson distribution with parameter $a$, i.e. $P\\{\\xi_{i}=k\\}=\\frac{a^{k}}{k!}e^{-a},\\ k=0,1,2,\\ldots$ (I'm extremely sorry for my poor English, but I hope you understand what I mean). Take $S_{n}=\\xi_{1}+\\xi_{2}+\\ldots+\\xi_{n}$. Then $P\\{S_{n}=k\\}=\\frac{(an)^{k}}{k!}e^{-an}$ and the wellknown theorem says that\r\n\\[P\\{a_{n}<\\frac{S_{n}-na}{\\sqrt{n\\sigma^{2}}} 5$. Hence $p$ has two factors each greater than $1$, meaning it is not prime. Contradiction. So $p - 4 \\neq x^4$ for $p > 5$.[/hide]", "Solution_2": "how did u know how to factor it? experience?\r\nand I tried, and I have a mistake somewhere but I couldn't find it:\r\n$p-4=x^4,p=x^4+4$ of course $x$ cannot be even, so it's odd $x=2k+1, p=4z+1+4,p=4z+5,p=4z+1$ and there is a theorem that states that there are infinetely many primes in that form :?", "Solution_3": "the substitution that you made is nonlinear, i.e. \r\n\r\n$x=4(4k^4+8k^3+6k^2+2k)+1+4$ (1)\r\n\r\n$x=4(z+1)+1$\r\n\r\nyou cannot use dirchlet's theorem for substutions like that...i have asked exactly what those are, but i have not gotten a conclusive answer, but i do know that kind of substution does not necessarily work", "Solution_4": "[quote=\"srulikbd\"]how did u know how to factor it? experience?\nand I tried, and I have a mistake somewhere but I couldn't find it:\n$p-4=x^4,p=x^4+4$ of course $x$ cannot be even, so it's odd $x=2k+1, p=4z+1+4,p=4z+5,p=4z+1$ and there is a theorem that states that there are infinetely many primes in that form :?[/quote]\r\nLook at the first problem [url=http://www.artofproblemsolving.com/Forum/topic-75738.html]Here.[/url]\r\n\r\nIt is a special factorzation you see a bunch. I think there is a general form.", "Solution_5": "So what is the case where you can factor the expression like Paladin8 did. I know that 4 and 16 work, but what are the other values that work?", "Solution_6": "You cleverly add zero to get $a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$.\r\n\r\nHow can you go to nats and not know that? :?", "Solution_7": "I saw that that factorization is called the identity of Sophie Germain in the PSS..." } { "Tag": [], "Problem": "The product of $ n$ number is $ n$, the sum of this numbers is 0. Prove that $ n$ is divisible by 4.", "Solution_1": "No one solve this?", "Solution_2": "I guess because the problem isn't currently understandable", "Solution_3": "[u][b]Interpretation:[/b][/u] Consider $ n$ numbers such that $ \\prod_{j \\equal{} 1}^n a_j \\equal{} n$ and $ \\sum_{j \\equal{} 1}^n a_j \\equal{} 0$. Prove that this is possible iff. $ 4 | n$.\r\n\r\n[u][b]Potential counterexample:[/b][/u] $ \\left \\{ \\minus{} \\frac {\\sqrt [3]{12}}{2}, \\; \\minus{} \\frac {\\sqrt [3]{12}}{2}, \\sqrt [3]{12} \\right \\}$\r\n\r\n[b]Edit:[/b] [i]thanks![/i]", "Solution_4": "[quote=\"The Zuton Force\"][u][b]Interpretation:[/b][/u] Consider $ n$ numbers such that $ \\prod_{j \\equal{} 1}^n a_n \\equal{} n$ and $ \\sum_{j \\equal{} 1}^n a_n \\equal{} 0$. Prove that this is possible iff. $ 4 | n$.\n\n[u][b]Counterexample:[/b][/u] $ \\left \\{ \\minus{} \\frac {\\sqrt [3]{12}}{2}, \\; \\minus{} \\frac {\\sqrt [3]{12}}{2}, \\sqrt [3]{12} \\right \\}$[/quote]It would not be a counterexample if $ a_j\\in\\mathbb{Z}$, and your index is wrong ;)", "Solution_5": "[hide=\"Hint\"]Parity[/hide]\n[hide=\"Solution\"]Suppose $ n$ is odd. Then for the product of the numbers to be $ n$, all numbers must be odd. But the sum of an odd number of odd numbers is odd, so it can't be $ 0$. Contradiction.\n\nSuppose $ n$ is a multiple of $ 2$ but not a multiple of $ 4$. Then for the product of the numbers to be $ n$, exactly one of the numbers is even. But the sum of an even number and an odd number of odd numbers is odd, so it can't be $ 0$. Contradiction.[/hide]" } { "Tag": [ "search", "inequalities unsolved", "inequalities" ], "Problem": "$ x_1, x_2, ..., x_n$ are positive reals that satisfies the condition $ x_1 \\plus{} x_2 \\plus{} ... \\plus{} x_n \\equal{} \\frac {1}{x_1} \\plus{} \\frac {1}{x_2} \\plus{} ... \\plus{} \\frac {1}{x_n}$.\r\n\r\nProve that \r\n\r\n$ \\frac{1}{n \\minus{} 1 \\plus{} x_1} \\plus{} \\frac {1}{n \\minus{} 1 \\plus{} x_2} \\plus{} ... \\plus{} \\frac {1}{n \\minus{} 1 \\plus{} x_n}\\le1$", "Solution_1": "Can someone please post a solution to this problem?\r\nThank you! :blush:", "Solution_2": "It looks similar to this: http://www.mathlinks.ro/viewtopic.php?search_id=843194689&t=238609" } { "Tag": [], "Problem": "Fie triunghiul ABC inscris in cercul C(O,R).Daca $\\ R_1$, $\\ R_2$ si $\\ R_3$ sunt razele cercurilor circumscrise triunghiurilor OBC, OCA si OAB, demonstrati ca\r\n$\\frac{3}{R}$>$\\frac{1}{R_1}+\\frac{1}{R_2}+\\frac{1}{R_3}$", "Solution_1": "[quote=\"stancioiu sorin\"]Fie triunghiul ABC inscris in cercul C(O,R).Daca $\\ R_1$, $\\ R_2$ si $\\ R_3$ sunt razele cercurilor circumscrise triunghiurilor OBC, OCA si OAB, demonstrati ca[/quote]\r\nce trebuie sa demonstram?", "Solution_2": "[quote=\"csabafarkas\"][quote=\"stancioiu sorin\"]Fie triunghiul ABC inscris in cercul C(O,R).Daca $\\ R_1$, $\\ R_2$ si $\\ R_3$ sunt razele cercurilor circumscrise triunghiurilor OBC, OCA si OAB, demonstrati ca[/quote]\nce trebuie sa demonstram?[/quote]\r\nscuze n-am vazut :blush:" } { "Tag": [ "induction", "modular arithmetic" ], "Problem": "prove that $\\frac{n^{3}+2n}{3}$ is an interger for all intergers $n=1,2,3...k,k+1$ :)", "Solution_1": "Why $n=1,2,3...k,k+1$? Surely it works for all positive integers.", "Solution_2": "[hide]\nrewrite as $\\frac{n(n^2+2)}{3}$. If $n\\equiv 0 \\mod 3$ the expression is clearly an integer, if \n$n\\equiv 1 \\vee 2 \\mod 3 \\implies n^2\\equiv 1 \\mod 3 \\implies n^2+2 \\equiv 0 \\mod 3$[/hide]", "Solution_3": "[hide]with n=1 we have\n (1^3+2)/3=3:3=1(in [b]Z[/b])(right)\nSuppose: n=k(right).That is\n(k^3+2k)/3=x(in [b]Z[/b]) \nwith n= k+1 \n((k+1)^3+2.(k+1))/3= (k^3+3k(k+1)+2k+3)/3=y(in [b]Z[/b])[/hide]", "Solution_4": "[hide]$\\ n^3 +2n= (n^2+2)n$\nIf $\\ n\\equiv1V2\\mod 3$ then $\\ n^2\\equiv1\\ mod 3$ so $\\ n^2+2\\equiv0\\mod 3$=> $\\ (n^2+2)n\\equiv0\\ mod 3$.\nIf $\\ n\\equiv0\\mod 3$=> $\\ (n^2+2)n\\equiv0\\ mod 3$.\nSo that $\\frac{n^3+2n}{3}$ is an integer.[/hide]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n :rotfl: Math is my life :!:", "Solution_5": "[quote=\"salah\"]prove that $\\frac{n^{3}+2n}{3}$ is an interger for all intergers $n=1,2,3...k,k+1$ :)[/quote]\r\n\r\n$=\\frac{(n-1)(n)(n+1)}{3}+n$ which is clearly an integer since a multiple of 3 comes along every 3 consecutive integers", "Solution_6": "[hide]Our base case is n=1: $\\frac{1^2+2\\cdot1}{3}=1$, which is an integer. Now for the inductive step. Assume $\\frac{n^3+2n}{3}$ is an integer, is true. Thus,\n$\\frac{(n+1)^3+2(n+1)}{3}=\\frac{n^3+3n^2+3n+1+2n+2}{3}=\\frac{n^3+2n}{3}+(n^2+n+1)$ is an integer, is also true. So, by mathematical induction, $\\frac{n^3+2n}{3}$ is an integer for all natural number n.\n\n--------\n\n$n\\equiv 0,1,2 \\pmod{3}$.\nSo checking all three cases,\n$0^1+2\\cdot0\\equiv 0 \\pmod{3}$\n$1^2+2\\cdot1\\equiv 0 \\pmod{3}$\n$2^2+2\\cdot2\\equiv 0 \\pmod{3}$, we see that $n^2+2n$ is always divisible by 3, so $\\frac{n^2+2n}{3}$ is always an integer.\n[/hide]", "Solution_7": "[quote=\"Altheman\"]\n\n$=\\frac{(n-1)(n)(n+1)}{3}+n$ which is clearly an integer since a multiple of 3 comes along every 3 consecutive integers[/quote] :thumbup:" } { "Tag": [ "inequalities", "geometry", "perimeter", "trigonometry", "circumcircle" ], "Problem": "[i]Let $ a, b, c$ be the lengths of the sides of a triangle. Let $ s \\equal{} a \\plus{} b \\plus{} c$ be the perimeter of the triangle. Let $ D$ be the area of the triangle. Then prove that\n\\[ \\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}<\\frac{s}{2D}\\]\nNotice that you must prove strict inequality.[/i]\r\n[hide]\nWe establish the tighter inequality:\\[ \\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\leq\\frac{\\sqrt{3}}{4}\\frac{s}{D}<\\frac{s}{2D}\\]Then since $ s>0$, we will have proven strict inequality.\n\nMultiply through by the area: \\[ \\sum_{cyc}\\frac{1}{a}\\leq\\frac{\\sqrt{3}}{4}\\frac{s}{D}\\iff\\sum_{cyc}\\frac{D}{a}\\equal{}\\sum_{cyc}\\frac12bc\\frac{\\sin a}{a}\\leq\\frac{s\\sqrt{3}}{4}\\]Now direct application of the Sine Law shows $ \\frac{\\sin a}{a}\\equal{}\\frac{1}{2R}$ and same cyclically, where $ R$ is the circumradius of the triangle.\\[ \\iff3\\sum_{cyc}ab\\leq(a\\plus{}b\\plus{}c)(3sqrt{3})R\\]Furthermore, \\[ 3\\sum_{cyc}ab\\leq\\left(\\sum_{cyc}a\\right)^2\\]so it only remains to prove \\[ \\sum_{cyc}\\frac{a}{2R}\\equal{}\\sum_{cyc}\\sin a\\leq\\frac{3\\sqrt{3}}{2}\\]But since $ \\frac{d^2}{dx^2}\\sin x\\leq 0$ on $ \\displaystyle\\left[0,\\frac{\\pi}{2}\\right]$, this follows from Jensen as \\[ 3\\sin\\left(\\frac{a\\plus{}b\\plus{}c}{3}\\right)\\equal{}3\\sin\\left(\\frac{\\pi}{2}\\right)\\equal{}\\frac{3\\sqrt{3}}{2}\\]$ \\square$\n\nNote: The original weak inequality is proven the same way, except in the last step one doesn't need Jensen - only the fact that $ a\\leq2R$.\n[/hide]", "Solution_1": "i think i have a 2 line soltuion for the origial question. we have to prove\r\n$ \\frac 1a\\plus{}\\frac 1b\\plus{}\\frac 1c<\\frac 1r$\r\n\r\nBut $ \\frac 1r\\equal{}\\frac {1}{h_a}\\plus{}\\frac {1}{h_b}\\plus{}\\frac {1}{h_c}$**\r\n\r\nAnd the inequality is now obvious using $ a>h_b, b>h_c, c>h_a$\r\n\r\n**Proof\r\ns=semi-perimeter\r\n$ 2D\\equal{}\\frac {2s}{1/r}\\equal{}\\frac {a}{1/h_a}\\equal{}\\frac {b}{1/h_b}\\equal{}\\frac {c}{1/h_c}\r\n\\equal{}\\frac {a\\plus{}b\\plus{}c}{1/h_a\\plus{}1/h_b\\plus{}1/h_c}$", "Solution_2": "Very nice solution, hell_ever!\r\nIt fits the specific problem very well. I attacked it from the angle of proving it in the equality case, then weakening.\r\nI never even knew that identity with $ \\frac{1}{r}$... anyway I think your solution has to change significantly to generalize to the stronger equality case:\\[ \\sum\\frac{1}{a}\\leq\\frac{\\sqrt{3}}{2}\\sum\\frac{1}{h_a}\\]" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Inspired by http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265075.\r\n\r\nLet $ G$ be a non abelian finite group s.t. every proper subgroup is abelian.\r\nShow that $ G$ contains a normal subgroup of prime index.", "Solution_1": "[b]Claim[/b]: If all subgroups of the finite group $ G$ are abelian, then $ G$ contains a normal subgroup of prime index.\r\n\r\n[i]Proof[/i]: If $ G$ is a minimal counterexample with regard to $ |G|\\equal{}n$, then $ G$ must be simple. For $ A,B$ maximal subgroups of $ G$ we therefore have $ A\\cap B\\equal{}1$ and $ A,B$ are self-normalizing. Let $ A_1,\\ldots,A_m$ be representatives of all conjugacy classes of maximal subgroups with orders $ a_1,\\ldots,a_m$. Then we have the identity $ n\\minus{}1\\equal{}\\sum_{i\\equal{}1}^m\\frac{n}{a_i}(a_i\\minus{}1)\\Longrightarrow (m\\minus{}1)n\\plus{}1\\equal{}n\\sum_{i\\equal{}1}^m\\frac{1}{a_i}\\leq \\frac{nm}{2}$ which implies $ m\\equal{}1$. But this yields the contradiction $ a_1(n\\minus{}1)\\equal{}n(a_1\\minus{}1)$." } { "Tag": [ "complex analysis" ], "Problem": "Hi, how will u show that under the transformation $w=\\frac{1}{z}$, the image of the lines $y=x-1$ and $y=0$ are the circle $u^2 + v^2 -u -v=0$ and the line $v=0$, respectively.\r\nThx", "Solution_1": "Using $w=u+iv$ and $z=x+iy$ then you could slove it... \r\n\r\n$w=\\frac{1}{z}$ then $z=\\frac{1}{w}=\\frac{1}{x+iy}=...$" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "[i]Mediovagio[/i] is a computer game that consists in a $3 \\times 3$ table in which each of the nine cells has a integer number from $1$ to $n$. When one clicks a cell, the numbers in the clicked cell and in the cells that share an edge with it are increased by $1$ and the sum is evaluated${}\\bmod n$. Determine the values of $n$ for which it's possible, with a finite number of clicks, obtain any combination of numbers from an given initial combination.\r\n\r\nEDIT: I corrected the statement.", "Solution_1": "I think the orginal statement of this problem was: \r\n\r\n\"...cell and in the cells that share an EDGE with it are increased....\"\r\n\r\nBut, anyway, there is another problem!" } { "Tag": [ "search" ], "Problem": "Prove that: $\\sum\\limits_{k=1}^{\\infty}\\frac{1}{k^2}=\\frac{\\pi^2}{6}$", "Solution_1": "this is definely not indermediate. at least i dont know an elementary proof. i saw a proof with the taylor series of sin x i think. maybe it helps u.", "Solution_2": "This has been posted before. Do a search on it...and this is indeed not appropriate for intermediate.", "Solution_3": "oh!!! i made mistake. Because i don't know that.", "Solution_4": "Here's a link to a previous discussion about this: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=sum+reciprocals&t=35545\r\n\r\nIt's ok mathcat :)" } { "Tag": [ "function", "LaTeX", "integration", "algebra proposed", "algebra" ], "Problem": "find all f:R\\ritharrow R continuos functions satisfaies :[img][url=http://imageshack.us][img]http://img372.imageshack.us/img372/5703/equationfonctionelavecintegral1.jpg[/img][/url][/img]", "Solution_1": "I don't have a CLUE what the formula means...\r\n\r\nand you should REALLY try to learn LaTeX, and stop posting these attachments, since I had to edit way to many of your posts already.", "Solution_2": "dear Arne\r\ni'm sorry because i'm not better in latex ; but you can click on the image to view it at its original size", "Solution_3": "I still don't understand and I don't think many others do.\r\n\r\nPLEASE learn LaTeX, it's not hard. There are plenty of instructions on this site.", "Solution_4": "TeX version:\r\nFind all $f:\\mathbb R \\rightarrow \\mathbb R $, for which $\\forall (x,y) \\in \\mathbb R^2: f(x^2)-f(y^2)=\\int_{r=x+y^2}^{y+x^2}f(r)dr$", "Solution_5": "thanks you Kondr :lol:" } { "Tag": [ "linear algebra", "matrix", "vector", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": "If A is an nxn matrix over F with n distinct eigenvalues, show that every nxn matrix over F commuting with A is diagonalizable and is of the form f(A) for some f in F\uf05bx. As a hint, it's recommended to get use of Lagrange's Interpolation Formula.", "Solution_1": "Hi. Check post 9 in http://www.mathlinks.ro/Forum/viewtopic.php?t=73310", "Solution_2": "You can show that for any matrix A n*n with n distincts eigenvalues, the dimension of the vector space of all the matrix which commute with A, C(A), is n.\r\nThen note that the dimension of the polynoms in A is n.\r\nThen conclude.", "Solution_3": "Thanks, i've got it! So, it's almost unnecessary to deal with Lagrange's interpolation in this problem. And here's another problem which has made me a little confused (but this time it's easier than the previous one):\r\n\r\nLet A be an nxn matrix over afield F and f be a polynomial in F\uf05bx. How can we show that f(A) is diagonalizable if A is diagonalizable?(I tried to show it by using the fact that minimal polynomial of A has distinct roots.) Or is it more effective to try to prove the contrapositive? And also, how can we show that if a family of simultaneously diagonalizable nxn matrices over a field F contains a matrix A with distinct eigenvalues then every eigenvector of A is an eigenvector of all members of this family?", "Solution_4": "For the first point, try to look at the eigenvalues of P(A) and vector spaces associated.\r\n\r\nFor the other point, if u already have seen the demo of A,B diagonalisable, AB=BA, then A,B simultaneously diagonalisable then u should see how to prove it.", "Solution_5": "Thanks, I think i've achieved to prove the first question. And you're right, second question is directly related to the former." } { "Tag": [], "Problem": "Eli and Bishoy like to have ninja fights. Eli uses ultimate llama penguin power while Bishoy throws textbooks. Since Bishoy doesn't use good powers, he only has a $.00002$ chance of winning each ninja fight. Eli and Bishoy fight 4100 times. The number of fights Bishoy would be expected to win can be expressed as $a.b \\times 10^{c}$. Find $a+b+c$.", "Solution_1": "[quote=\"ragnarok23\"]Eli and Bishoy like to have ninja fights. Eli uses ultimate llama penguin power while Bishoy throws textbooks. Since Bishoy doesn't use good powers, he only has a $.00002$ chance of winning each ninja fight. Eli and Bishoy fight 4100 times. The number of fights Bishoy would be expected to win can be expressed as $a.b \\times 10^{c}$. Find $a+b+c$.[/quote]\r\n[hide]$.00002=\\frac{1}{50000}$\nSo Bishoy is expected to win $\\frac{41}{500}=.082$ times that is $8.2*10^{-2}$. So the answer is 8.[/hide]", "Solution_2": "[hide]$.00002=\\frac{1}{50000}$\n\nThen, you multiply 4100 by that to get: $\\frac{41}{500}$. Expressing that as the form of $a.b\\times10^{c}$, I get $8.2*10^{-2}$. $8+2+(-2)=8+2-2=\\boxed{8}$.[/hide]", "Solution_3": "There's a slightly easier way; try making everything in scientific notation so that it becomes easier at the end.", "Solution_4": "first of all, my head is not deranged. my psychiatrist said it was \"special\". =]\r\n\r\nand heres how i did it:\r\n\r\n[hide].00002=[b]2 X 10^-5[/b]\n4100=[b]4.1 X 10^3[/b]\n\nso multiply those and its 8.2 X 10^-2. so 8+2-2=[b]8[/b][/hide]\r\n[/b]", "Solution_5": "[hide] You can express the first number as $2*10^{-5}$. then, you multiply 2 by 4100 (the amount of times they fight) and you wind up with 8200. 8200 can be expressed as $8.2*10^{3}$. Multiply 10^{-5} into the numbers you just got (10^-5 is appended to 2, so you have to multiply it back in), to get:\n$8.2*10^{3}*10^{-5}$, which makes\n$8.2*10^{-2}$. then, you want the sum of $a.b*10^{c}$, so you simply add 8, which is A, 2, which is B, and -2, which is C. \n$2-2+8=8$, and thats the answer. \n$\\boxed{8}$\n[/hide]" } { "Tag": [], "Problem": "If I go shopping at Wal-Mart, buy an item, and pay with my new $\\$$10 bill, I recieve some change. The digits in my change are equal to the digits in the price of the item, but are in a different order. How many different possible prices could the item cost?", "Solution_1": "Maybe the fact that this is a reworded 2006 Nationals problem will interest you...", "Solution_2": "Ok, I'll try it.\r\n\r\nWhen adding up to a power of 10, all corresponding digits must add up to 9 except the last 2. So since you want it to be a permutation of the other, something other than a 5 could not work becuase it would have to add up to 10 not 9. So the last digit is 5. Now the other two digits must add up to 9, so there are 10 ways (: 9.05,8.15,7.25,6.35,5.45,4.55,3.65,2.75,1.85,0.95)", "Solution_3": "Hehe, you forgot one thing, the change and the amount he paid are both above $\\$$1, therefore, the answer is 8.", "Solution_4": "Well, in the Nationals question, the answer was 8, because of that restriction. However, in my reworded problem, the problem is changed a bit, so 10 is the correct answer!" } { "Tag": [ "geometry", "USAMTS", "email", "ARML", "inequalities", "articles", "number theory" ], "Problem": "what kind of prices did you guys want??? I chose Mathmatica, the geometry problems, and the book on Diophatine Equations (I'm bad at number theory). I'm so happy because I got Mathmatica for free!!! :lol: Hopefully I can get better with the diophatine book tho.", "Solution_1": "Eh, prices?", "Solution_2": "So when do silver medalists get to choose their prizes?", "Solution_3": "From the USAMTS email:\r\n\r\n[quote]The scoring for the final round of the USAMTS is now available at http://www.usamts.org. Starting next week, winners will be able to choose their prizes. The prize levels and the date upon which winners can choose prizes are below:\n\nGold 96-100 April 17\nSilver 76-95 April 20\nBronze 60-75 April 22\nHonorable Mention 40-59 April 25[/quote]", "Solution_4": "Looks like I already have everything except \"The Elegent Universe\" (this is my third year of USAMTS and I've also won a lot of books in other math competitions). I don't know what to do for my other two prizes. :?", "Solution_5": "Wow...choose \"Music of the Primes\" and the ARML practice problems and give it to me :P", "Solution_6": "Eh, I don't think there'll be any more Mathematicas left by the time it's my turn to choose...", "Solution_7": "I chose 360 Mathematical Challenges, the book of KoMaL problems, and the Code Book. I already owned a large fraction of the available prizes (including, unfortunately, Mathematica for Students - it would have been great to have gotten it for free!), so my selection was pretty limited, but I'm still pleased with the prizes available. If only they still had Geometer's Sketchpad available this year....", "Solution_8": "I got the olympiad problems book, the olympiad geo book, and the diophantine equations book.\r\n\r\nI already had mathematica ... two copies, actually -- one from ARML and one from HCSSiM.\r\n\r\nJB", "Solution_9": "Mathematica... jeeze! Here I was planning to ask for a student copy for my birthday (April 23rd), and I find that, assuming there are still copies left when silver winners get to choose, I'll be getting a copy for free from USAMTS.\r\n\r\nOut of curiosity, were the copies donated by Wolfram Research, or did the USAMTS get enough funding to afford Mathematica copies? Either way, this has easily been the best math competition I've ever been in. Great questions, Great registration fee, and now, Great prizes too. Hopefully, the mention of \"Oh yeah, and top scorers got free copies of Mathematica last year\" will entice some boarderline folks to join the contest.\r\n\r\nAs a side note, even if Mathematica is out of stock by April 20th, I'm still looking forward to the Diophantine Equations book. Interestingly enough, my interest in Diophantines was sparked by round 3 problem 4, which was my personal favorite among the ones I solved correctly (Best question overall is easily round 3 problem 5, on rotating tables, IMO).", "Solution_10": "Can somebody please post the name and author of the Diophantine Equations book?\r\n\r\nI didn't select it, but I'm actually thinking about buying it now.", "Solution_11": "It says\r\n\r\nAn Introduction to Diophantine Equations (Andreescu/Andrica) - 198 pages. Various approaches to solving difficult Diophantine equations. Difficulty 8-10/10.\r\n\r\nOH NO wait, I wonder what happened to the old/new inequalities book they had last year...", "Solution_12": "I have the Diophantine Equations book - it's really good, but also gets really complicated, so you have to be pretty experienced in number theory to understand some of the more advanced sections. For those who are interested, the ISBN is 9739238882.", "Solution_13": "Zantrang --\r\n\r\nIn our intro to number theory we covered basic diophantine equations (3x + 4y = 5, etc), modular congruences, quadredic residues/quadratic reciprocity, fermat's last thm/Euler's generalization and Wislon's thm, among other things.\r\n\r\nDo you think I've larned enough that the book won't fly over my head??\r\n\r\nOoh and PS, what's the name of the really good inequality book everyone here loves from last year?\r\n\r\nJB", "Solution_14": "Yeah, I SUCKS at inequalities...\r\n\r\nfeel like getting that book now...", "Solution_15": "[quote=\"jackb1115\"]Zantrang --\n\nIn our intro to number theory we covered basic diophantine equations (3x + 4y = 5, etc), modular congruences, quadredic residues/quadratic reciprocity, fermat's last thm/Euler's generalization and Wislon's thm, among other things.\n\nDo you think I've larned enough that the book won't fly over my head??\n\nOoh and PS, what's the name of the really good inequality book everyone here loves from last year?\n\nJB[/quote]\r\nYou've covered linear Diophantine equations and modular arithmetic - that should give you a good starting point. I don't think the book will be over your head. Also, the book gets progressively more difficult, so if you spend time on th easier parts and work your way up, then it should never be overwhelming.\r\n\r\nAs for the inequality book, it's Old and New Inequalities by Titu Andreescu et al. (there are three more authors I don't want to type out now). The ISBN is 9739417353, and it's published by Gil publishing house.", "Solution_16": "Seems Mathematica is still there...I decided to choose it because of this discussion, but what is it?", "Solution_17": "i got the diophantine book and the geometry book, they sound cool. anybody know what the t-shirts are like?", "Solution_18": "T-shirts are awesome...last years had the problem on the front and this crazy solution by Zach Abel on the back. I chose large even though I'm a small...I'll use it as a night gown or something...", "Solution_19": "[quote=\"filletwho\"]Seems Mathematica is still there...I decided to choose it because of this discussion, but what is it?[/quote]\r\n\r\nIt's pretty much a super expensive awesome math program (as far as I know anyway... I don't have it, though).", "Solution_20": "It's an officially super-expensive math program, but if you do high school contest math (or camps), you're likely to get at least a few free copies... their business model is to get young mathematicians hooked, not try to make them convince their parents to pay an absurd amount.", "Solution_21": "I got the Elegant Universe and Mathematica. I already have a copy of Mathematica but it's still more valuable than any of the other prizes on there...and it beats getting a fourth copy of Count Down :P", "Solution_22": "I got mathematica and the 360 problems, because that's the only one i couldn't find on Amazon. :D", "Solution_23": "I couldn't find Diophantine on Amazon either. so I got mathematica, 360 problems, and diophantine eqs. Very happy with the prizes :lol: I was hoping to maybe win mathematica at ARML, but then this was a nice surprise! Thanks much.", "Solution_24": "What's the name and author of the geometry book?", "Solution_25": "I was so excited Mathematica was still available that I forgot to read what it said after it about Macintosh and Windows users, and something about email...What exactly did it say?", "Solution_26": "Filletwho, when I saw the special instructions for non-PC Mathematica users, I just took a screenshot, to make sure I didn't do something stupid. Anyway, so long as you need the info on what it said, and so long as a few others are asking for authors and such, here's the full list of prizes/descriptions that was available to a Silver Winner early on thursday morning:\r\n\r\n1. T-shirt of sizes Small, Medium, Large, or Extra Large.\r\n\r\n2. Two of (2 for silver winners. I understand that gold got 3, and bronze/HM traditionally receive 1):\r\n\r\nThe Code Book by Simon Singh - A sweeping history of the subject of encryption, with technical and mathematical explanations and fascinating personalities\r\n\r\nThe Music of Primes by du Sautoy - The extraordinary history behind one of mathematics great unsolved mysteries, the Riemann Hypothesis\r\n\r\n360 Problems for Mathematica Contests by Andreescu/Andrica - Contains 360 mostly very challenging Olympiad problems grouped by subject. Contains solutions. 280 pages. Difficulty: 9/10\r\n\r\nChallenging Problems in Geometry by Salkind/Posamentier - Interesting Geometry problems of Varying difficulty and topic. ~240 pages. Difficulty (5-8)/10.\r\n\r\nC2K2: Century 2 of Komal - 190 pages. A variety of problems and articles in mathematics (and a little physics) from the highly regarded Hungarian journal Komal. Difficulty (7-9)/10.\r\n\r\nAn introduction to Diophantine Equations (Andreescu/Andrica) - 198 pages. Various approaches to solving difficult Diophantine equations. Difficulty: (8-10)/10.\r\n\r\nThe Elegant Universe by Brian Greene - A layman's guide to fascinating physics concepts such as string theory, relativity, and quantum physics.\r\n\r\nCount Down - National Book Award finalist Steve Olson chronicals the 2001 United States IMO team. Book signed by the author.\r\n\r\nMathematica for Students - Powerful mathematical programming tool. Students choosing this prize agree to allow the USAMTS to release their address and email to Mathematica. (MAKE SURE YOUR ADDRESS AND EMAIL ON THE PROFILE ARE CORRECT!!) If you use an operating system besides Windows XP and choose this prize, email your username and operating system to usamts@usamts.org.\r\n\r\nWhew", "Solution_27": "I got 360 Problems and Mathematica.\r\n\r\nThe 360 Problems looked good right away. Then I wasn't sure if I wanted Challenging Problems in Geometry or Mathematica. I'm not really sure how much I'm going to use Mathematica (if at all), but on Amazon, Mathematica was like $\\$$150 and Challenging Problems in Geometry was $\\$$10. It may be sad, but that's how I made my decision. :D", "Solution_28": "I have Mathematica 4.2, 5 and Maple 6, 8 on my computer, but I don't know how to use them.........", "Solution_29": "[quote=\"eyefragment\"]Filletwho, when I saw the special instructions for non-PC Mathematica users, I just took a screenshot, to make sure I didn't do something stupid. Anyway, so long as you need the info on what it said, and so long as a few others are asking for authors and such, here's the full list of prizes/descriptions that was available to a Silver Winner early on thursday morning:\n\n1. T-shirt of sizes Small, Medium, Large, or Extra Large.\n\n2. Two of (2 for silver winners. I understand that gold got 3, and bronze/HM traditionally receive 1):\n\nThe Code Book by Simon Singh - A sweeping history of the subject of encryption, with technical and mathematical explanations and fascinating personalities\n\nThe Music of Primes by du Sautoy - The extraordinary history behind one of mathematics great unsolved mysteries, the Riemann Hypothesis\n\n360 Problems for Mathematica Contests by Andreescu/Andrica - Contains 360 mostly very challenging Olympiad problems grouped by subject. Contains solutions. 280 pages. Difficulty: 9/10\n\nChallenging Problems in Geometry by Salkind/Posamentier - Interesting Geometry problems of Varying difficulty and topic. ~240 pages. Difficulty (5-8)/10.\n\nC2K2: Century 2 of Komal - 190 pages. A variety of problems and articles in mathematics (and a little physics) from the highly regarded Hungarian journal Komal. Difficulty (7-9)/10.\n\nAn introduction to Diophantine Equations (Andreescu/Andrica) - 198 pages. Various approaches to solving difficult Diophantine equations. Difficulty: (8-10)/10.\n\nThe Elegant Universe by Brian Greene - A layman's guide to fascinating physics concepts such as string theory, relativity, and quantum physics.\n\nCount Down - National Book Award finalist Steve Olson chronicals the 2001 United States IMO team. Book signed by the author.\n\nMathematica for Students - Powerful mathematical programming tool. Students choosing this prize agree to allow the USAMTS to release their address and email to Mathematica. (MAKE SURE YOUR ADDRESS AND EMAIL ON THE PROFILE ARE CORRECT!!) If you use an operating system besides Windows XP and choose this prize, email your username and operating system to usamts@usamts.org.\n\nWhew[/quote]\r\n\r\nAha, that's good, I use XP. Smart idea...", "Solution_30": "Eh forgot to choose on April 20...and Mathematica is still here! :) \r\n\r\nAnyway no more Diophantine Equations book, so I chose 360 problems. (I already have the Geo book and Count Down :P )", "Solution_31": "Did anyone besides me choose C2K2? I looked it up, and it has some pretty interesting stuff in it (I also got Mathematica. Diophantine Equations was gone when I chose...).", "Solution_32": "[quote=\"E^(pi*i)=-1\"]Did anyone besides me choose C2K2? I looked it up, and it has some pretty interesting stuff in it (I also got Mathematica. Diophantine Equations was gone when I chose...).[/quote]\r\nI did too - we'll see how it turns out being. I've seen a lot of interesting stuff from KoMaL.", "Solution_33": "I don't think anyone has mentioned the certificates. Those are the greatest certificates I've ever seen.", "Solution_34": "wait, they passes out the certificates already?", "Solution_35": "[quote=\"pkerichang\"]wait, they passes out the certificates already?[/quote]\r\nI haven't gotten one; maybe he's referring to the certificates from last year? Also, congrats on having your problem 5 solution picked for the official solution list. I'm still working on achieving that goal...", "Solution_36": "YEAH!!!!I'M HAPPY TOO :first:", "Solution_37": "Hurrah! I got 360 Challenging Problems and Mathematica (for Linux). I figure IF I ever win another copy of Mathematica, it'll be a Windows version, so might as well get the Mathematica for Linux now instead of having 2 windows copies. Besides which my platform of choice for mathematical thought is Linux :D.", "Solution_38": "Is mathematica 5.1 much different from 4.2? I figured I'd go for two books because I already have mathematica, but it's version 4.2...\r\nWhy does mathematica cost so much? Sure it's extremely powerful and very useful, but...dah!!" } { "Tag": [ "calculus", "derivative", "geometry", "3D geometry", "calculus computations" ], "Problem": "The suface area of a cube is changing at a rate of $10cm.^{2}\\: / sec$ when an edge of the cube is $5\\: cm$. At what rate is the volume of the cube changing when an edge of the cube is $5\\: cm$?", "Solution_1": "The surface area $S$ of the cube is $S = 6a^{2}$ and its volume $V$ is $V = a^{3}$. You have $\\frac{dS}{dt}= 10$ and you want $\\frac{dV}{dt}$. First you have to relate $V$ with $S$:\r\n\r\n$a = \\left(\\frac{S}{6}\\right)^{1/2}\\Rightarrow V = \\left(\\frac{S}{6}\\right)^{3/2}$.\r\n\r\nNow differentiate:\r\n\r\n$\\frac{dV}{dt}= \\frac{3}{2}\\cdot \\frac{1}{6}\\cdot \\frac{dS}{dt}\\cdot \\left(\\frac{S}{6}\\right)^{1/2}= \\frac{a}{4}\\cdot \\frac{dS}{dt}$.\r\n\r\nIn the present case, $a = 5 cm$ and so\r\n\r\n$\\frac{dV}{dt}= \\frac{5 cm}{4}\\cdot 10 cm^{2}/s = 12.5 cm^{3}/s.$", "Solution_2": "thanks a lot nice solution" } { "Tag": [ "probability", "combinatorics proposed", "combinatorics" ], "Problem": "Does there exist two unfair dices such that probability of their sum being $j$ be a number in $\\left(\\frac2{33},\\frac4{33}\\right)$ for each $2\\leq j\\leq 12$?", "Solution_1": "[quote=\"Omid Hatami\"]Does there exist two unfair dices such that probability of their sum being $j$ be a number in $\\left(\\frac2{33},\\frac4{33}\\right)$ for each $2\\leq j\\leq 12$?[/quote]\r\n\r\nVery nice problem !\r\n\r\nLet $a_{n}$ the probability of value $n\\in\\{1,2,3,4,5,6\\}$ for die 1 and $b_{n}$ the probability of value $n\\in\\{1,2,3,4,5,6\\}$ for die 2.\r\nLet $s_{n}$ the probability of value $n\\in\\{2,3,4,5,6,7,8,9,10,11,12\\}$ for sum of two dice.\r\n\r\nIf $a_{1}b_{6}+a_{6}b_{1}\\geq 2a_{1}b_{1}$ , then $s_{7}\\geq 2s_{2}$ and this is in contradiction with $s_{2}$ and $s_{7}$ in $\\left(\\frac2{33},\\frac4{33}\\right)$. \r\nHence : (I1) $0\\leq a_{1}b_{6}+a_{6}b_{1}< 2a_{1}b_{1}$\r\n\r\n\r\nIf $a_{1}b_{6}+a_{6}b_{1}\\geq 2a_{6}b_{6}$ , then $s_{7}\\geq 2s_{12}$ and this is in contradiction with $s_{12}$ and $s_{7}$ in $\\left(\\frac2{33},\\frac4{33}\\right)$.\r\nHence : (I2) $0\\leq a_{1}b_{6}+a_{6}b_{1}< 2a_{6}b_{6}$\r\n\r\n\r\nThen, the product (I1)*(I2) gives $(a_{1}b_{6}+a_{6}b_{1})^{2}< 4a_{1}b_{1}a_{6}b_{6}$ $\\Leftrightarrow$ $(a_{1}b_{6}-a_{6}b_{1})^{2}< 0$, which is impossible.\r\n\r\nSo, it does not exist two unfair dices with the requested property.\r\n\r\n-- \r\nPatrick", "Solution_2": "slightly simpler, but no better solution:\r\n\r\nUsing Patrick(pco)'s notation $a_{n}$ is probabilities for side n of die 1 and $b_{n}$ for side n of die 2 and $s_{m}$ for the sum m.\r\n\r\nNow $s_{7}= a_{1}b_{6}+...+a_{6}b_{1}> a_{1}b_{6}+a_{6}b_{1}$ \r\n\r\nApplying $AM \\ge GM$ we get\r\n\r\n$s_{7}> 2\\sqrt{a_{1}b_{1}a_{6}b_{6}}= 2\\sqrt{s_{2}s_{12}}\\ge 2 \\min \\{s_{2}, s_{12}\\}$", "Solution_3": "Are we certain that the dice have to be labelled only with the numbers from 1 to 6?", "Solution_4": "That would be the meanest [b]trap[/b] in a problem... ever..." } { "Tag": [ "combinatorics solved", "combinatorics", "Sequences", "duplicate" ], "Problem": "What is the largest number of subsequences of the form $n, n+1, n+2$ that a sequence of 2001 positive integers can have? For example, the sequence {1, 2, 2, 3, 3} of 5 terms has 4 such subsequences.\n\n[color=#cc0000][mod says: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=17453& ][/color]", "Solution_1": "for a solution just check out:\r\n\r\nhttp://www.kalva.demon.co.uk/short/soln/sh01c1.html :D" } { "Tag": [ "MATHCOUNTS", "AMC", "AIME", "percent", "calculus" ], "Problem": "Contest Format: \r\nIversonfan2005 and myself are writing/sponsoring a multiple choice math contest called the Masters Invitational Math Contest. It is on a high school level, but anyone can participate. There will be awards on for a high-scorer on every grade level, and an award for the top three scorers overall. \r\n\r\nDifficulty: \r\nThe contest will be 50 questions, multiple choice, ranging from National MathCounts to upper level AIME difficulty. We would like every AoPSer to participate, but do not be discouraged by harder problems towards the end. Although we encourage middle-schoolers to participate, middle schoolers are not expected to be able to anwer the higher numbered questions. \r\n\r\nScoring: \r\nEach correct answer will earn 5 points, a question left blank will earn 1 point, and an incorrect answer will earn 0 points. Scores will range from 0 - 250. \r\n\r\nTime Limit: \r\nDue to our inability to accurately track time, all participants will have 3 weeks from the time of contest posting to turn in their answers. EDIT: We have extended the deadline to May 16th.\r\n\r\nParticipation: \r\nIf you want to participate, just PM your answers to [b]MIMC Director[/b]. Include your grade and state. Answers are [b]due May 16th[/b]. Answers can be enclosed in a PM, or attached in a MS Word document.\r\n\r\n[size=134][color=red]CONTEST IS OVER:[/color][/size]\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88519&postorder=asc&start=20]MIMC Answers[/url]: View answers here, and challenge problems that you disagree with.\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88650]MIMC Solutions[/url]: Post solutions here, or view/request solutions to problems you missed.\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88651&sid=16b02e38d43c4bdf86641fc667836e39]MIMC Winners[/url]: Check soon for the Top 16, as well as winning States/Regions. Plus other awards!", "Solution_1": "Challenge Procedure:\r\n\r\nBefore we send out scores and compile results, I will post the answer key here. Any challenged should be PMed to MIMC Director, and our team will review the challenge, and tell you the result. After 3 days, the challenge window will be closed, and results will be posted, and scores sent out.", "Solution_2": "[size=150][b]Deadline Extension![/b][/size]\r\n\r\nThe deadline has been extended to May 16th!\r\nIf you already turned in your answers, you may send in a new answer key. Be sure to send ALL your answers again. Also, make sure you ask us to throw your old answer key away.", "Solution_3": "What if I have already PMed mysmartmouth my answers (I saw you want the answers to go to MIMC Director), should I send them right now to MIMC Director and then again if there are any changes?", "Solution_4": "MIMC Director & mysmartmouth are the same person. However, mysmartmouth's inbox is full of MIMC stuff, which can't be deleted yet. So I received your answers nat mc (and everyone else who sent them to me or iversonfan), but from now on, please send all answers to MIMC Director.\r\n\r\nAlso: Please note that this thread is strictly for important questions, concerns, and announcements. Any off-topic posts will be deleted.", "Solution_5": "Thanx for the extension -- I needed it :D", "Solution_6": "sorry, i just sent my answers to iversonfan and then I saw this", "Solution_7": "Answers sent to iversonfan2005 will be forwarded to MIMC Director. Answers sent to me will be accepted, but from now on contact MIMC Director for all things MIMC, including answers. (MIMC Director is my other account.)", "Solution_8": "I just thought of something: would the results include the percents of people answer each of the questions correctly like the AMC statistics? Some problems are very hard and I would like to see how difficult each question was for AoPS community. That would be great if we can see that kind of statistics.", "Solution_9": "[quote=\"10000th User\"]I just thought of something: would the results include the percents of people answer each of the questions correctly like the AMC statistics? Some problems are very hard and I would like to see how difficult each question was for AoPS community. That would be great if we can see that kind of statistics.[/quote]\r\n\r\nwe will", "Solution_10": "I feel so proud of the MIMC team. We managed to correct 3 mistakes in the answer key before almost any grading happened. I am now reasonably confident that there are no errors in the answer key.", "Solution_11": "I will be at National MathCounts until the 14th. So don't expect any response from me. During this time, continue to PM [b]answers [/b]to MIMC Director. However, PM all [b]questions [/b]to iversonfan2005. Thanks!", "Solution_12": "[quote=\"10000th User\"]I just thought of something: would the results include the percents of people answer each of the questions correctly like the AMC statistics? Some problems are very hard and I would like to see how difficult each question was for AoPS community. That would be great if we can see that kind of statistics.[/quote]\r\n\r\nIn order for us to do this, we need some more participation! Most people have not turned in answers; we'll wait and see. Perhaps we could do it with the data we have now, but it would not be very statistically accurate.", "Solution_13": "Who should I PM if I think the choices for an answer is wrong? I keep getting something 1 off of one of the choices for #22, should I just guess the choice it's one off from?", "Solution_14": "[quote=\"Ignite168\"]Who should I PM if I think the choices for an answer is wrong? I keep getting something 1 off of one of the choices for #22, should I just guess the choice it's one off from?[/quote]\r\n\r\nThere's nothing wrong with the test questions. Everything has been revised, and number 22 has the right answer choice. ;)", "Solution_15": "The test questions are now all fine, as well as the answer key. This was confirmed by several proofers, as well as a perfect score of an exceptionally smart student. No problems require calculus, though some may appear to. If you are having problems with a question, redownload the PDF. Maybe you have an old (flawed) version.", "Solution_16": "PERFECT SCORE!??!?\r\n\r\nOK I'm going to start doing this now.", "Solution_17": "Well, the deadline is up. However, you have until Saturday to get your answers in, because the statistician is on vacation, and we want as much participation as possible.", "Solution_18": "So this is a deadline extention to 5/20?", "Solution_19": "[quote=\"nat mc\"]So this is a deadline extention to 5/20?[/quote]\r\n\r\ni guess yeah", "Solution_20": "Will solutions be given?", "Solution_21": "[url]http://www.artofproblemsolving.com/Forum/files/mimc_rules___regulations_2.0_178.pdf[/url]\r\n\r\nRead the second page.", "Solution_22": "PERFECT SCORE?!?!?! :o \r\n\r\nomg that has to beyond godlike! I wouldn't be surprised if the 16th place score was <200... unless I just did relatively bad here. Of course, that is possible... :roll:", "Solution_23": "Yes, this is a deadline extension basically. Aren't we nice? Anyway, about no calculus: some people told me that the problem in question can be solved without calculus. Personally, I cannot solve this problem at all, but then again, I can't solve many of these...", "Solution_24": "can someone explain number 37 please?", "Solution_25": "It is pretty self explanitory.", "Solution_26": "WHOA! Please do not discuss he contest questions until after it is over. Based on feedback, reworks, and a few near-perfect scores, we know that there are no flaws in the questions. Question 37 is very tricky, but it is possible. We've had 2 people PM and say it was impossible or un-understandable, then later PM and say never mind. So all those posts discussing the problems have been deleted.", "Solution_27": "whoa... sorry...", "Solution_28": "So ... when are the results going to be available? Or, are we waiting for the questions appeal period to end?", "Solution_29": "[quote=\"Iversonfan2005\"][quote=\"nat mc\"]So this is a deadline extention to 5/20?[/quote]\n\ni guess yeah[/quote]\r\n\r\nThe deadline has been extended, 10000th.", "Solution_30": "I didn't know the extension until the 20th. I got a PM from Iversonfan2005 that there was an extension through the 16th. And I handed in my answers... can I hand in again? I might be able to solve a few more problems that I left blank.", "Solution_31": "yes, but after sending it again, you have to ask us to burn your old scores...literally", "Solution_32": "You should probably change your signature then...", "Solution_33": "Oh that's just great .And i ws wondering why haven't the results come out yet.I just worked huriedly on it for 45 mins on the 16th and submitted what,like 10 answes man.Really, iverson,should have changed your sign earlier :D", "Solution_34": "Well, you can submit new answers if you want! However, as soon as it hits 12:00 AM of May 21st, I will post the answers. Then we will have a 2-day challenge window. Then we will post winners on the 24th. Yay!", "Solution_35": "I finally found a sweet method for \"Find the number of divisors of 30!\". :D", "Solution_36": "Where are the answers?", "Solution_37": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88519&postorder=asc&start=20]MIMC Answers[/url]: View answers here, and challenge problems that you disagree with.\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88650]MIMC Solutions[/url]: Post solutions here, or view/request solutions to problems you missed.\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88651&sid=16b02e38d43c4bdf86641fc667836e39]MIMC Winners[/url]: Check soon for the Top 16, as well as winning States/Regions. Plus other awards!" } { "Tag": [ "function", "limit", "integration", "calculus", "inequalities", "logarithms", "calculus computations" ], "Problem": "f(x) is positive and continuous function and $ 0\\le f(x) \\le 1$\r\n\r\n$ \\lim_{ n \\to \\infty }{(\\int_{0}^{1}\\sqrt[n]{f(x)}dx})^{n}$", "Solution_1": "the answer is $ \\mathrm{max}f(x),x \\in [0,1]$\r\ninfact those $ [0,1]$ can be replaced by $ [a,b]$\r\nprovided the function remains continious and positive\r\nhave a look at my solution [url=http://www.mathlinks.ro/viewtopic.php?t=190763]here[/url]", "Solution_2": "pardesi, you are mistaking $ \\left(\\int f^{1/n}\\right)^n$ for $ \\left(\\int f^{n}\\right)^{1/n}$. They are quite different, to say the least ;)", "Solution_3": "sorry :rotfl: :rotfl: :blush: but isn't the same method applicable :maybe:", "Solution_4": "Well, we know that if $ 0 < p \\leq 1$ then:\r\n\r\n$ \\lim_{n\\rightarrow \\infty}\\sqrt[n]{p}\\equal{}1$\r\n\r\nNow, for large enough $ n$ we can say that:\r\n\r\n$ \\int^1_0 \\sqrt[n]{f(x)} \\,dx \\equal{} \\sqrt[n]L$\r\n\r\nalso, using the above equality, and using it on every point of the function we get:\r\n\r\n$ \\int^1_0 \\sqrt[n]{f(x)} \\,dx \\equal{} \\int^1_0 1 \\,dx \\equal{} 1$\r\n\r\nand so $ L\\equal{}1$. Can it be done this way?", "Solution_5": "No, the limit is not $ 1$. It's the zero-mean: the geometric mean.\r\nI leave it to others to come up with the appropriate definition in an integral context.", "Solution_6": "[quote=\"pardesi\"]sorry :rotfl: :rotfl: :blush: but isn't the same method applicable :maybe:[/quote]\r\nno. consider holder's inequality:\r\n$ \\int_{0}^{1}fg\\leq \\left(\\int_{0}^{1}f^{p} \\right)^{1/p}\\left(\\int_{0}^{1}g^{q}\\right)^{1/q}$\r\nand let $ g: \\equal{} 1$ $ p \\equal{} n$ (for big $ n$), $ f \\equal{} F^{1/p}$\r\nthen :\r\n$ \\left( \\int_{0}^{1}F^{1/n} \\right)^{n}\\leq \\int_{0}^{1}F \\leq \\max_{[0\\leq x \\leq 1]}F$", "Solution_7": "I think it should be $ \\exp \\left( \\int_0^1 \\ln f(x) \\, dx \\right)$. As for a proof, let's just say I'm not in the mood for one right now.", "Solution_8": "[quote=\"perfect_radio\"]I think it should be $ \\exp \\left( \\int_0^1 \\ln f(x) \\, dx \\right)$. As for a proof, let's just say I'm not in the mood for one right now.[/quote]\r\nit looks like true.\r\nby Jensen inequality we have :\r\n$ \\exp(\\frac {1}{b \\minus{} a}\\int_{a}^{b}\\ln f(x)dx)\\leq \\frac {1}{b \\minus{} a}\\int_{a}^{b}f(x)dx$\r\nput $ a \\equal{} 0$, $ b \\equal{} 1$ $ f(x) \\equal{} F^{1/n}$ then :\r\n$ \\exp( \\int_{0}^{1}\\ln F)\\leq \\left( \\int_{0}^{1}F^{1/n}dx\\right)^{n}\\leq \\int_{0}^{1}F$", "Solution_9": "[b]Problem :[/b] [color=blue] let $ f,g \\in L^{0}(E)$ and $ f,g \\geq 0$ , if $ \\int_{E}g(x)dx \\equal{} 1$ and $ \\int_{E}g(x)f(x)dx < \\infty$ then:\n$ \\left(\\int_{E}f^{p}(x)g(x)dx \\right)^{1/p} \\to \\exp \\left(\\int_{E}g(x) \\ln f(x) dx \\right)$ as $ p \\to \\plus{} 0$[/color]\r\n[b]proof :[/b]\r\n1. integral $ \\int_{E}g(x)\\ln f(x)dx$ exist. \r\n2. if $ \\int_{E}g(x)\\ln f(x)dx < \\infty$ then we use inequality : $ |\\frac {e^{pt} \\minus{} 1}{p}|\\leq |t| \\plus{} e^{|t|}$\r\n$ (t \\in R, 0 < p < 1)$ thus \r\n$ \\int_{E}\\frac {f^{p}(x) \\minus{} 1}{p}g(x)dx \\to \\int_{E}g(x) \\ln f(x)dx$ as $ p \\to \\plus{} 0$\r\n3. if $ \\int_{E}g(x)\\ln f(x)dx \\equal{} \\minus{} \\infty$, then consider function:\r\n$ f_{\\delta} \\equal{} \\max(\\delta, f)$where $ 0 < \\delta < 1$ and consider :\r\n$ \\left(\\int_{E}f^{p}(x)g(x)dx \\right)^{1/p} \\leq \\left(\\int_{E}f_{\\delta}^{p}(x)g(x)dx \\right)^{1/p}$\r\nfirst take upper limit with $ p \\to \\plus{} 0$ and after take limit with $ \\delta \\to \\plus{} 0$ Done.", "Solution_10": "I think I finally managed to salvage my original idea which lead me to the answer.\r\n\r\nSo let's take $ f : [0,1] \\to \\left[ a, \\infty \\right)$ (where $ a > 0$), $ f \\in \\mathcal R[0,1]$. By replacing $ f(x)$ with $ c \\cdot f(x)$, we can suppose that $ \\sup_{x \\in [0,1]} f(x) < 1.$\r\n\r\nThere is a partition $ P = [x_0,x_1,\\ldots,x_k]$ of $ [0,1]$ such that $ U(P,f) - L(P,f) < \\epsilon$. Let $ M_i = \\sup_{t \\in [x_{i - 1},x_i]} f(t)$, $ m_i = \\inf_{t \\in [x_{i - 1},x_i]} f(t)$ and $ \\Delta x_i = x_i - x_{i - 1}$. We have that $ M_i^\\frac1{n} - m_i^\\frac1{n} = \\left( M_i - m_i \\right) \\cdot \\frac1{n} \\cdot \\chi_{i,n}^{\\frac1{n} - 1}$, where $ \\chi_{i,n} \\in \\left[ m_i, M_i \\right]$. The $ \\limsup$ of the last term is $ \\frac1{m_i} < \\infty$, so $ \\lim_{n \\to \\infty} \\frac {M_i^\\frac1{n} - m_i^\\frac1{n}}{M_i - m_i} = 0$. Thus, for all huge $ n$ we have that $ U \\left( P, f^\\frac1{n} \\right) - L \\left( P, f^\\frac1{n} \\right) < \\epsilon$. It follows that\r\n\\[ \\left| \\int_0^1 f^\\frac1{n} (x) \\m dx - \\sum_{i = 1}^k f^\\frac1{n} \\left( x_i \\right) \\cdot \\Delta x_i \\right| < \\epsilon\r\n\\]\r\nfor all huge $ n$. Now, both the integral and the sum are $ \\leq \\sup_{x \\in [0,1]} f(x) < 1$, so we can conclude that\r\n\\[ \\left| \\left( \\int_0^1 f^\\frac1{n} (x) \\, dx \\right)^n - \\left( \\sum_{i = 1}^k f^\\frac1{n} \\left( x_i \\right) \\cdot \\Delta x_i \\right)^n \\right| < \\epsilon\r\n\\]\r\nfor all humongous $ n$ (via another MVT argument).\r\n\r\nNow, I'll also estimate $ \\left| \\left( \\sum_{i = 1}^k f^\\frac1{n} \\left( x_i \\right) \\cdot \\Delta x_i \\right)^n - \\exp \\left( \\int_0^1 \\ln(f(x)) \\, dx \\right) \\right|$. Both terms can be written as $ \\exp \\left( \\text{something smaller than 0} \\right)$, so that their difference is $ \\leq \\left| \\frac {\\ln \\left( \\sum_{i = 1}^k f^\\frac1{n} \\left( x_i \\right) \\cdot \\Delta x_i \\right)}{\\frac1{n}} - \\int_0^1 \\ln (f(x)) \\, dx \\right|$. As $ n \\to \\infty$, the left term goes to $ \\sum_{i = 1}^k \\Delta x_i \\cdot \\ln \\left( f \\left( x_i \\right) \\right)$ (it is basically the definition of a derivative). We note that $ \\ln M_i - \\ln m_i \\leq \\frac {M_i - m_i}{a}$, so $ U \\left( \\ln \\circ f, P \\right) - L \\left( \\ln \\circ f, P \\right) < \\frac {\\epsilon}{a}$. Thus, for epic $ n$, we have that\r\n\\[ \\left| \\left( \\int_0^1 f^\\frac1{n} (x) \\, dx \\right)^n - \\exp \\left( \\int_0^1 \\ln(f(x)) \\, dx \\right) \\right| < \\epsilon \\cdot \\left( 2 + \\frac1{a} \\right) .\r\n\\]\r\nThis took me a looot to write, so it better be good :oops: \r\n\r\nNote that we can obviously replace $ \\frac1{n}$ by $ p \\to 0$. Also, it should be possible (but I'm not sure it is) to use a density argument to prove the general case.", "Solution_11": "May I ask how the limit contains a ln f in the integrand when it is said that 0<=f<=1?", "Solution_12": "[quote=\"Extremal\"][b]Problem :[/b] [color=blue] let $ f,g \\in L^{0}(E)$ and $ f,g \\geq 0$ , if $ \\int_{E}g(x)dx \\equal{} 1$ and $ \\int_{E}g(x)f(x)dx < \\infty$ then:\n$ \\left(\\int_{E}f^{p}(x)g(x)dx \\right)^{1/p} \\to \\exp \\left(\\int_{E}g(x) \\ln f(x) dx \\right)$ as $ p \\to \\plus{} 0$[/color]\n[b]proof :[/b]\n1. integral $ \\int_{E}g(x)\\ln f(x)dx$ exist. \n2. if $ \\int_{E}g(x)\\ln f(x)dx < \\infty$ then we use inequality : $ |\\frac {e^{pt} \\minus{} 1}{p}|\\leq |t| \\plus{} e^{|t|}$\n$ (t \\in R, 0 < p < 1)$ thus \n$ \\int_{E}\\frac {f^{p}(x) \\minus{} 1}{p}g(x)dx \\to \\int_{E}g(x) \\ln f(x)dx$ as $ p \\to \\plus{} 0$\n3. if $ \\int_{E}g(x)\\ln f(x)dx \\equal{} \\minus{} \\infty$, then consider function:\n$ f_{\\delta} \\equal{} \\max(\\delta, f)$where $ 0 < \\delta < 1$ and consider :\n$ \\left(\\int_{E}f^{p}(x)g(x)dx \\right)^{1/p} \\leq \\left(\\int_{E}f_{\\delta}^{p}(x)g(x)dx \\right)^{1/p}$\nfirst take upper limit with $ p \\to \\plus{} 0$ and after take limit with $ \\delta \\to \\plus{} 0$ Done.[/quote]\r\n\r\nStep 2 doesn't make sense :|", "Solution_13": "[quote=\"\u00a7outh\u00a7tar\"]...\nStep 2 doesn't make sense ... :|[/quote]\r\nwhy do you think so ?\r\n\r\nwe know that :\r\n\\[ \\lim_{p \\to 0} \\frac {f^{p}(x) - 1}{p} - \\ln f(x) = 0\r\n\\]\r\nthus \r\nhence for all $ \\epsilon$ there exist $ \\delta$ :\r\n\\[ |\\frac {f^{p}(x) - 1}{p} - \\ln f(x)| < \\epsilon\r\n\\]\r\n(for all $ p < \\delta$ ) we have :\r\n\\[ |\\int_{E} g(\\frac {f^{p} - 1}{p} - \\ln f)dx| < |\\epsilon \\int_{E}g(x)dx| <\\epsilon\r\n\\]\r\nin my case $ \\int_{E}g = 1$ :lol: .\r\nDone.\r\nnow it's easy to note that when $ p \\approx 0$ then \r\n\r\n$ \\int_{E}f^{p}g \\approx 1$ since $ \\int_{E} g = 1$\r\nthus we have\r\n\\[ \\frac {1}{p}\\ln (\\int_{E} f^{p} g d - 1 + 1 ) = \\frac {1}{p} \\ln (\\int_{E} (f^{p} - 1)g d + 1) \\approx \\frac {1}{p}\\int_{E}f^{p}g\r\n\\]" } { "Tag": [ "linear algebra", "matrix", "LaTeX", "vector", "induction", "inequalities", "linear algebra unsolved" ], "Problem": "Prove that : \r\nThe eigenvalues of a real symmetric matrix are real.", "Solution_1": "Here's the simplest argument I know. Use $A^*$ for the conjugate transpose of a matrix, and note that if $A$ is real symmetric, then $A^*=A.$ (The proof depends only on $A^*=A$ and is thus also valid for Hermitian matrices.)\r\n\r\nSuppose $Ax=\\lambda x$ where $x\\in\\mathbb{C}^n$ and $x\\ne 0.$\r\n\r\nOn the one hand, $x^*Ax=x^*(Ax)=x^*(\\lambda x)=\\lambda x^*x.$\r\n\r\nOn the other hand, $x^*Ax=x^*A^*x=(Ax)^*x=(\\lambda x)^*x=\\overline{\\lambda}x^*x$\r\n\r\nEquating the two sides and dividing by the nonzero $x^*x,$ we get\r\n\r\n$\\lambda=\\overline{\\lambda}$ or $\\lambda$ is real.\r\n\r\nAs a corollary: suppose $x=u+iv$ where $u$ and $v$ are real vectors. Then\r\n\r\n$A(u+iv)=\\lambda(u+iv)$ and since we know $\\lambda$ is real, we can split that into its real and imaginary parts as $Au=\\lambda u$ and $Av=\\lambda v.$ From that we can assume WLOG that $x=u$ is real and thus $A$ has real eigenvectors.\r\n\r\nWe have to work harder than that to prove that $A$ is diagonalizable.\r\n\r\n[color=green][Edit. OK, \\overline.][/color]", "Solution_2": "As a little $\\LaTeX$ note, \\overline{} produces better results than \\bar{}\r\n\r\n\r\ne.g, $\\overline{\\lambda}$ and $\\bar{\\lambda}$", "Solution_3": "I'll go ahead and mention that the slickest proof of the diagonalizability follows from Schur's Lemma, which is truly a wonderful theorem.\r\n\r\n[b]Schur's Lemma:[/b] Suppose $A$ is a complex $n\\times n$ matrix. Then there exists a unitary matrix $U$ (that is, $U^*U=I$) such that $U^*AU=T,$ where $T$ is upper triangular and the elements of $T$ on the main diagonal are the eigenvalues of $A,$ with their algebraic multiplicities.\r\n\r\nThe proof of Schur's Lemma is straightforward. I won't give it in detail, but the outline is something like this:\r\n\r\n(1) One eigenvalue and eigenvector exists, by the Fundamental Theorem of Algebra.\r\n\r\n(2) One vector can be extended to an orthonormal basis, by Gram-Schmidt. Multiply front and back by the resulting matrix and its adjoint (conjugate transpose).\r\n\r\n(3) That takes care of one column. Continue the argument by induction, noting that the product of finitely many unitary matrices is a unitary matrix.\r\n\r\nNow, assume that Schur's Lemma has been proved.\r\n\r\n[b]Theorem: unitary diagonalizability of Hermitian matrices.[/b]\r\n\r\nSuppose $A^*=A.$ Then there exists a unitary $U$ such that $U^*AU=D,$ where $D$ is real and diagonal.\r\n\r\nProof: by Schur's Lemma, there exists unitary $U$ such that $U^*AU=T.$\r\n\r\nBut then $T^*=(U^*AU)^*=U^*A^*(U^*)^*=U^*AU=T.$\r\n\r\nThat makes $T$ both upper triangular and Hermitian; that can only be if $T$ is diagonal and real.\r\n\r\n[b]Corollary:[/b] A real symmetric matrix can be orthogonally diagonalized.\r\n\r\nWe have by the main result that the eigenvalues are real. That means we can choose the eigenvectors to be real, and use the real version of Gram-Schmidt in the proof of Schur's Lemma. That gives us a real $U$ such that $U^TU=I$ - that is, a real orthogonal matrix.", "Solution_4": "By the way, as an aside, are there any texts that you particularly like for linear algebra? I've gotten most of my linear algebra out of a mix of Hoffman and Kunze and Dummit and Foote, but I'm not a huge fan of either.", "Solution_5": "[quote=\"Kent Merryfield\"] Schur's Lemma, which is truly a wonderful theorem ...\n[/quote]\r\nIt has the additional advantage that the Schur factorization can be stably computed, while the Jordan form sometimes can't." } { "Tag": [ "LaTeX" ], "Problem": "Is there a command similar to \\boxed that circles things?", "Solution_1": "\\textcircled{[i]number[/i]}\r\n$ \\textcircled{1}\\textcircled{2}\\textcircled{3}\\textcircled{4}\\textcircled{5}\\textcircled{6}\\textcircled{7}\\textcircled{8}\\textcircled{9}\\textcircled{0}$\r\nthough I think \\textcircled{\\footnotesize{[i]number[/i]}} looks better\r\n$ \\textcircled{\\footnotesize{1}}\\textcircled{\\footnotesize{2}}\\textcircled{\\footnotesize{3}}\\textcircled{\\footnotesize{4}}\\textcircled{\\footnotesize{5}}\\textcircled{\\footnotesize{6}}\\textcircled{\\footnotesize{7}}\\textcircled{\\footnotesize{8}}\\textcircled{\\footnotesize{9}}\\textcircled{\\footnotesize{0}}$\r\n\r\nIf you'd prefer the larger number then Scott Pakin has produced the command \\centercircle{[i]number[/i]}. Just put this code at the beginning of your document:\r\n[code] \\newsavebox{\\circlebox} \n \\savebox{\\circlebox}{\\fontencoding{OMS}\\selectfont\\Large\\char13} \n \\newlength{\\circleboxwdht} \n\n\n \\newcommand{\\centercircle}[1]{% \n \\setlength{\\circleboxwdht}{\\wd\\circlebox}% \n \\addtolength{\\circleboxwdht}{\\dp\\circlebox}% \n \\raisebox{0.5\\dp\\circlebox}{% \n \\parbox[][\\circleboxwdht][c]{\\wd\\circlebox}{\\centering#1}}% \n \\llap{\\usebox{\\circlebox}}% \n } [/code]" } { "Tag": [ "MATHCOUNTS", "AMC", "AIME" ], "Problem": "Okay, now that everyone is done with the mock test, it is time to release the final results! :D Sadly, only 5 people did mine, so I would just like to thank all 5 of those people for participating.\r\n\r\n[hide=\"AND THE WINNER OF THE 2008 GAUSS1181 COMPETITION #1 IS...\"][size=200]shentang!!!!!!!!!!!!!![/size] :winner_first: [/hide]\n\nOnce again, thank you, thank you to everyone who did the mock test!!!!!!!!!\n\nAnswers to the mock test will be coming soon! But right now I've attached an Excel file of the statistics, such as who got certain problems right and what score everyone got.\n\nIf you want to see the mock test problems, go to the thread in this Middle School Tournaments forum called \"My mock test.\"\n\n[hide=\"Interesting facts about this competition!\"]No one got Target #8, Sprints #19, 29, or 30 right.\nFantasyLover was the only person to get Sprint #12 right.\nshentang was the only person to get Sprints #13, 17, 18, 24, and 25 right!\nsuma_milli didn't do the sprint round, so she got a 0 (I guess she wouldn't if she really took the sprint round!)[/hide]\r\n\r\nAlso of note: PLEASE DO NOT MESS UP THE POLL!!! IN ORDER TO AVOID ANY LIES OR FOOLISH SILLINESS, I ASK THAT YOU PLEASE EXPLAIN WHY YOU VOTED A CERTAIN CHOICE IF YOU DID VOTE!!!!!!!!!", "Solution_1": "Can you copy paste them on to a word document, my computer says that your file does not exist.\r\n\r\nYay I got first!! :winner_first: first time ever.", "Solution_2": "What is the Mock Test. I said that I didn't know what it was.", "Solution_3": "@Wicked: See my thread in the Middle School Tournaments forum, \"My mock test\"\r\n\r\n@shentang: Do you not have excel, or why doesn't the file show up?", "Solution_4": "[quote=\"gauss1181\"]@Wicked: See my thread in the Middle School Tournaments forum, \"My mock test\"\n\n@shentang: Do you not have excel, or why doesn't the file show up?[/quote]\r\n\r\nidk why the file doesn't show up, I am not that good with computers.", "Solution_5": "gauss1181, is the mock test just a few math questions you wrote?", "Solution_6": "Yeah, I made them myself. :D That's why they have a bitty humor in them :D", "Solution_7": "I think some citations (cough, cough) are required...\r\n\r\nEdit: if you didn't understand that, I meant the poll choices here are clearly a mimic of another poll (cough cough)...", "Solution_8": "I thought it was something really special like the AIME or something. lol :rotfl:", "Solution_9": "Did you, Yongyi, put in as easy as A-B-C??!?!?!?!??!?! :o :o :huh:", "Solution_10": "I put in \"I didn't see the mock questions, so I don't know.\"\r\n\r\nAnd you should add a note saying your poll options were \"borrowed\" (being generous) from my mock poll.", "Solution_11": "[quote=\"Yongyi781\"]I put in \"I didn't see the mock questions, so I don't know.\"\n\nAnd you should add a note saying your poll options were \"borrowed\" (being generous) from my mock poll.[/quote]\r\nOkay. This poll was being made in acknowledgments to Yongyi's mock test poll on difficulty. :D \r\n\r\nthank you\r\nthat is all\r\n :D :D \r\n\r\nbah out of smilies", "Solution_12": "I thought it was something between States and Nationals level, so I voted States.", "Solution_13": "Do you guys mind if I start a new mock test?", "Solution_14": "Why would we mind?", "Solution_15": "No, I'm just concerned if people would do it...not like this one where only 5 people did it...", "Solution_16": "Could you pm my score to me? I do not have excel on my computer.", "Solution_17": "either most of them suck or the test was international+ or something (just looking at the scores)", "Solution_18": "Whoa I thought this ended December 7th.", "Solution_19": "This Gauss's first mock test...check the dates.", "Solution_20": "@theprodigy - Please look at the last post before posting. You posted two months after the last post. \r\n\r\n :spam:" } { "Tag": [ "calculus", "integration", "logarithms", "floor function", "inequalities unsolved", "inequalities" ], "Problem": "Let $ d$ be a positive integer. Prove that if $ n>1.5\\left(\\frac{d}{\\ln 2}\\right)^d$ then $ n<2\\lfloor n^{1/d}\\rfloor^d$", "Solution_1": "Maybe number theory forum is more appropriate ?" } { "Tag": [ "Putnam", "college contests" ], "Problem": "Is $\\sqrt2$ the limit of a sequence of numbers of the form $^{3}\\sqrt{n}-^{3}\\sqrt{m}$ ($n, m = 0, 1, 2...$)?", "Solution_1": "That should be $n,m\\in \\mathbb{N}$; they can be large.", "Solution_2": "Edited. Sorry about the mixup", "Solution_3": "yes.\r\nthe solution that i remmember involeves using the fact that for large n the expression $\\sqrt[3]{n+1}-\\sqrt[3]{n}$ approaches zero. which shows that the set of numbers $\\sqrt[3]{n}-\\sqrt[3]{m}$ is closed under multipication...which leads to the fact that the set is dense in $[0,\\infty)$. then you use the same argument for a diffrent set (which i cant quite remmember), to show that its also dense for $(-\\infty, 0)$" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "if for all sequance $ x_{i} \\geq 0$ such as :\r\n$ \\sum_{i \\equal{} 1}^{\\infty} x_{i}^{p} < \\infty$ and $ \\sum_{i \\equal{} 1}^{\\infty}x_{i}y_{i} < \\infty$ where $ p > 1$ and $ y_{i}\\geq 0$ is fixed sequance,\r\n then \r\n$ \\sum^{\\infty}_{i \\equal{} 1}y_{i}^{p'} < \\infty$\r\nwhere $ 1/p \\plus{} 1/p' \\equal{} 1$", "Solution_1": "this is because the dual of $ l^p$ is $ l^{p'}$.", "Solution_2": "Thanks but.. :oops: , Ok.\r\nLet me try to change the question .\r\nsuppose :\r\n$ \\sum_{k \\equal{} 1}^{\\infty} y_{k}^{p'} \\equal{} \\infty$ where $ y_{i}\\geq 0$\r\n then construct nonegative sequance $ x_{i}$ such that \r\n$ \\sum_{k \\equal{} 1}^{\\infty}x_{i}^{p} < \\infty$ and $ \\sum_{i \\equal{} 1}^{\\infty}x_{i}y_{i} \\equal{} \\infty$ :)\r\n$ 1/p\\plus{}1/p'\\equal{}1$", "Solution_3": "why not: $ x_k\\equal{}y_k^{p'\\minus{}1}$ ?", "Solution_4": "[quote=\"alekk\"]why not: $ x_k \\equal{} y_k^{p' \\minus{} 1}$ ?[/quote]\r\nsince $ \\sum x_{i}^{p} \\equal{} \\sum y_{i}^{p(p'\\minus{}1)} \\equal{} \\sum y_{i}^{p'} \\equal{} \\infty$\r\nbut we need $ \\sum x_{i}^{p}<\\infty$", "Solution_5": "I solved it :trampoline: .\r\nit's enought to choose such sequance :\r\n$ x_{n} \\equal{} \\frac{y_{n}^{p'/p}}{y_{1}^{p'}\\plus{}y_{2}^{p'}\\plus{}...\\plus{}y_{n}^{p'}}$\r\nif some one can't understand why it works I can post full solution" } { "Tag": [], "Problem": "I thought this was interesting -- http://www.numberspiral.com/", "Solution_1": "So this only works when the perfect squares are lined up in that way?\r\n\r\nWhat if the spiral was made tighter or looser?" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "*/This one is just time consuming, that's all... :huh: \r\n\r\nThe diagonal of a square of a cube is 2^1/2 inches long. How many of those cubes would fit in a rectangular wardrobe with its height being 7 feet more than the width of the wardrobe and has a length that\u2019s 9 diameters longer than the width? The wardrobe must stay intact, and assume that the cubes are stiff and cannot be bent or shaped into any other form.\r\n\r\nP.S. I made these up myself, so feel free to offer any comments, questions, or concerns. :)\r\n\r\n-Answer will come out next week along with a new question- :lol:", "Solution_1": "[quote=\"GavMasta993628\"] and has a length that\u2019s 9 diameters longer than the width? [/quote]\r\n\r\nSo what is this \"diameter\"?", "Solution_2": "yeah, what circle are you talking about?", "Solution_3": "or sphere.\r\n\r\nhe might have meant some sort of unit of length but was thinking about something else at that moment" } { "Tag": [ "probability" ], "Problem": "Vicki has 6 cards. Four are Red on each side, one is Red on 1 side and Green on the other, and the sixth card is Red on 1 side and Blue on the other. Vicki selects one card at random and then looks at one side of that card. If the side of the card at which she is looking is Red, find the probability that the other side of this card is Green. Express your answer as a common fraction reduced to lowest terms.", "Solution_1": "[hide=\"Isnt it just...\"] $\\frac{1}{8+1+1}= \\frac{1}{10}$ ? [/hide]", "Solution_2": "you're good!!! Can you explain how you did it?\r\nHere's the \"real\" solution:\r\n\r\n[hide]Let $RR$, $RG$ and $RB$ be the obvious events of what card is selected. \nThen $P$(side seen is R)$=P(RR) \\cdot 1+P (RG) \\cdot \\frac{1}{2}+P(RB) \\cdot \\frac{1}{2}= \\frac{2}{3}+\\frac{1}{12}+\\frac{1}{12}= \\frac{5}{6}$. \nBy Baye's Theorem, $P(RG|R seen ) = \\frac{P (R seen |RG) \\cdot P(RG)}{P(R seen )}=\\frac{(1/2)(1/6)}{5/6}=\\frac{1}{10}$[/hide]", "Solution_3": "[hide]Can't you just say there are 10 red sides and in only one case the other side is green...[/hide]" } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "Solve for $z$:\r\n$z+\\frac{1}{z} = 2+4i$", "Solution_1": "What is $i$ ? :?", "Solution_2": "sqrt of -1\r\n\r\nits the basis of complex numbers :D", "Solution_3": "Is it[hide]$2+\\sqrt{2}+i(3-\\sqrt{2})$ and $i(1+\\sqrt{2})-\\sqrt{2}$. \nSolution, Kalva Style:\n :D \nI multiplied both sides by $z$ then used the quadratic formula. Then I got something involving $\\sqrt{1-i}$, which I simplified by: $(a+bi)^2=1-i$ then solving for $a,b$. I probably messed up somewhere though cause by answer is pretty complicated...[/hide]", "Solution_4": "[hide=\"solution\"]\n$z-2=4i-\\frac{1}{z}$\n$z-2=\\frac{4iz-1}{z}$\n$z^2-2z=4iz-1$\n$z^2-z(2+4i)+1=0$\nsolving this we get,\n$\\frac{2+4i+\\sqrt{4-16+16i-4}}{2}$\n$1+2i+2\\sqrt{i-1}$\nSo the solutions are,\n$1+2i+2\\sqrt{i-1}$ and $1+2i-2\\sqrt{i-1}$\n[/hide]\r\nCorrect?", "Solution_5": "Actually I was unsure about the use of quadratic formula with complex number, so I posted this problem.\r\n\r\nThanks for your replies. \r\n\r\nBy the way, eryaman, is it needed to simplify $\\sqrt{1-i}$?", "Solution_6": "[quote=\"warut_suk\"]Actually I was unsure about the use of quadratic formula with complex number, so I posted this problem.\n\nThanks for your replies. \n\nBy the way, eryaman, is it needed to simplify $\\sqrt{1-i}$?[/quote]Honestly I don't know. I just did it because I knew how...it does make the solution look a bit cleaner without square roots though." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For every $ n\\in N$ and $ 0 < 2\\alpha < \\frac {\\pi} {n}$ prove :\r\n\r\n$ \\sqrt [n]{\\prod^{n}_{i \\equal{} 1} cos\\left(\\frac {n \\plus{} 1 \\minus{} i} {i} \\alpha \\right)} \\leq cos \\left(\\frac {\\sum^n_{i \\equal{} 1} \\{C^i_n [(i \\minus{} 1)!]^2\\}} {n.n!} \\alpha \\right)$\r\n\r\nSolution :\r\n$ f(x)\\equal{}\\minus{}ln (cos x) \\equal{}> f'(x)\\equal{}tg x > 0$ for $ 0 f(x)$ is increasing \r\n$ f''(x)\\equal{}\\frac {1}{cos^2x} > 0$ for every x $ \\equal{}> f(x)$ is convex\r\n\r\n$ n\\alpha \\geq \\alpha$\r\n$ n\\alpha \\plus{}\\frac {n\\minus{}1}{2} \\alpha \\geq \\alpha \\plus{} \\frac {\\alpha } {2}$\r\n...\r\n$ \\sum^{k}_{i \\equal{} 1} \\left(\\frac {n \\plus{} 1 \\minus{} i} {i} \\alpha \\right)\\geq \\sum^{k}_{i \\equal{} 1} \\left(\\frac {\\alpha} {i} \\right)$\r\nfor $ 1\\leq k\\leq n$ . \r\n\r\nWhen $ f(x)$ is increasing and convex and using Karamata , we have (1) :\r\n$ \\sum ln\\left[ cos\\left(\\frac {n \\plus{} 1 \\minus{} i} {i} \\alpha \\right) \\right]\\leq \\sum ln\\left[ cos\\left(\\frac {\\alpha } {i} \\right) \\right]$\r\n\r\nUsing Jensen we have (2) :\r\n$ \\sum ln\\left[ cos\\left(\\frac {\\alpha } {i} \\right) \\right]\\leq n.ln\\left[ cos\\left(\\sum \\frac {\\alpha } {i} \\right)/n \\right]$\r\n\r\nFinally from (1) and (2) we have :\r\n$ \\sum ln\\left[ cos\\left(\\frac {n \\plus{} 1 \\minus{} i} {i} \\alpha \\right) \\right]\\leq n.ln\\left[ cos\\left(\\sum \\frac {\\alpha } {i} \\right)/n \\right]$\r\n<=> $ \\sqrt [n]{\\prod^{n}_{i \\equal{} 1} cos\\left(\\frac {n \\plus{} 1 \\minus{} i} {i} \\alpha \\right)} \\leq cos \\left(\\frac {\\sum^n_{i \\equal{} 1} \\{C^i_n [(i \\minus{} 1)!]^2\\}} {n.n!} \\alpha \\right)$\r\n\r\n :maybe: :roll:", "Solution_1": "it is wrong $ n\\equal{}2$ $ \\alpha\\equal{}60^{0}$", "Solution_2": "but $ 120^{\\circ}>\\frac{\\pi}{2}$ ;)", "Solution_3": "For positive a,b and c :\r\n$ \\frac{\\left( a\\plus{}b\\plus{}c\\right) \\left( a^5\\plus{}b^5\\plus{}c^5\\right)}{abc}\\plus{}3abc\\geq a(a\\plus{}b)(b\\plus{}c)\\plus{}b(b\\plus{}a)(b\\plus{}c)\\plus{}c(c\\plus{}a)(c\\plus{}b)$", "Solution_4": "I think it's better is...\r\n$ \\frac {\\left( a \\plus{} b \\plus{} c\\right) \\left( a^5 \\plus{} b^5 \\plus{} c^5\\right)}{abc} \\plus{} 3abc\\geq a(a \\plus{} b)(b \\plus{} c) \\plus{} b(b \\plus{} c)(c \\plus{} a) \\plus{} c(c \\plus{} a)(a \\plus{} b)$", "Solution_5": "I think the following one is stronger:\r\n\r\n$ \\frac {\\left( a \\plus{} b \\plus{} c\\right) \\left( a^5 \\plus{} b^5 \\plus{} c^5\\right)}{abc} \\plus{} 3abc\\geq a(a \\plus{} b)(a \\plus{} c) \\plus{} b(b \\plus{} c)(b \\plus{} a) \\plus{} c(c \\plus{} a)(c \\plus{} b)$\r\n\r\n[hide=\"Proof\"]It's equivalent to $ (a\\plus{}b\\plus{}c)(a^5\\plus{}b^5\\plus{}c^5)\\plus{}3a^2b^2c^2\\ge abc\\sum a(a\\plus{}b)(a\\plus{}c)$.\n\nVerify that $ \\sum a(a\\plus{}b)(a\\plus{}c)\\equal{}(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2)\\plus{}3abc$. Thus we only need to prove that $ (a\\plus{}b\\plus{}c)(a^5\\plus{}b^5\\plus{}c^5)\\ge abc(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2)$\n\n$ \\iff a^5\\plus{}b^5\\plus{}c^5\\ge abc(a^2\\plus{}b^2\\plus{}c^2)$, just rearrangement.\n[/hide]", "Solution_6": "yes i envisaged this . i just confused the letters , when i typed it. \r\n\u041c\u0443 proof is :\r\n$ \\frac {1}{2}[6,0,0] \\plus{} [5,1,0]\\ge \\frac {1}{2}[4,1,1] \\plus{} [3,2,1]$\r\nbut yours may be better than that." } { "Tag": [ "probability" ], "Problem": "Express as a common fraction the probability that a randomly chosen positive factor of 12 is a prime number.", "Solution_1": "12 has 6 factors:\r\n1, 2, 3, 4, 6, and 12\r\n\r\nOnly '2' is prime, so the probability is $ \\boxed{\\frac{1}{6}}$", "Solution_2": "You forgot 3 :wink: \r\n\r\nSo the correct probability is $ \\frac{2}{6}\\equal{}\\boxed{\\frac13}$", "Solution_3": "[hide]$2,3$ are the prime factors of $12. 2/6=1/3$[/hide]" } { "Tag": [ "geometry", "geometry solved" ], "Problem": "let ABC be a triangle an P be a arbitary point out of it.draw line (d) such that cut ABC to \r\ntwo parts which have same area!", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=26105 .\r\n\r\n darij", "Solution_2": "nothing" } { "Tag": [ "quadratics" ], "Problem": "can you write a quadratic equation that has the soulutions of -3 and 4/5?", "Solution_1": "Working backwards:\r\n(x +3)(x-4/5)\r\nso x 2 -x4/5 +3x -12/5\r\nso x 2 +3.2x - 12/5?", "Solution_2": "Yes, I agree with that. That's also an excellent way to derive formulas from some graphs. \r\n\r\nThere are formulas too, but it's much slower and are usually used for harder formulas. I won't display in this section though.", "Solution_3": "[quote=\"PenguinIntegral\"]Working backwards:\n(x +3)(x-4/5)\nso x 2 -x4/5 +3x -12/5\nso x 2 +3.2x - 12/5?[/quote]\r\n\r\nThe last step is wrong, unfortunately........... :(", "Solution_4": "Really? I wasn't even paying attention to the work, just the steps", "Solution_5": "Most ppl wouldn't pay attention to the work, cuz it was such an easy question thus EXTREMELY difficult to make mistakes. :D", "Solution_6": "[hide]\n(x+3)(x-4/5)\n\nx2 + 3x - 4/5x - 12/5\nx2 + 11/5x - 12/5?[/hide]", "Solution_7": "I know I didn't do some of the fractions right. I just put down what I thought it was. Its the method that matters..." } { "Tag": [ "probability" ], "Problem": "Here is the question:\r\n\r\nIf the famous baseball player from the San Francisco Midgets, Larry Ponds, has a $ \\frac25$ chance of earning a walk on each plate appearence what is the probability that he will earn a walk on exactly once on his next two appearences?\r\n\r\nClick[hide=\"here\"]$ \\frac{12}{25}$[/hide] to see the answer.[/list]", "Solution_1": "there are two ways he could get exactly one walk.\r\n\r\nHe could do:\r\n\r\nWalk,Non-walk\r\nNon-walk, Walk\r\n\r\nFor the first situation, the chance that he gets a walk is 2/5, and for turn two, the chance he doesn't get a walk is 3/5, so 2/53/5=6/25.\r\n\r\nFor the second situation, it is the opposite of the first. The chance that he gets a non-walk is 3/5, and for turn two, the chance that he gets a walk is 2/5\r\n\r\nso 3/5*2/5=6/25, and add the two together to get 12/25.", "Solution_2": "[quote=\"ajain\"]Here is the question:\n\nIf the famous baseball player from the San Francisco Midgets, Larry Ponds, has a $ \\frac25$ chance of earning a walk on each plate appearence what is the probability that he will earn a walk on exactly once on his next two appearences?\n\nClick[hide=\"here\"]$ \\frac {12}{25}$[/hide] to see the answer.[/list][/quote]\r\n\r\nWe have $ 2$ Ways to get exactly $ 1$ walk. Let $ n$ be a walk and $ m$ be a \r\n\r\nnon-walk. Then, the two cases are $ n,m$ and $ m,n$. Both cases have \r\n\r\nprobability $ \\frac{6}{25}$, so $ \\frac{12}{25}$.", "Solution_3": "It's a nice problem, i'll use it in a funny contest in Vn! Thank you, BOGTRO!" } { "Tag": [ "geometry", "IMO" ], "Problem": "Just some background, I'm no longer in high school, not the most rigorous prover. I was just looking at 1991 IMO A1 out of interest because I'm bored. And came up with the following solution and was wondering what anyone out there thinks of this solution and what score you guys think this would have gotten on the actual exam.\r\n\r\nProblem: \r\nA1. Given a triangle ABC, let I be the incenter. The internal bisectors of angles A, B, C meet the opposite sides in A', B', C' respectively. Prove that: \r\n 1/4 < AI\u00b7BI\u00b7CI/(AA'\u00b7BB'\u00b7CC') \u2264 8/27. \r\n\r\nSolution:\r\nLets call the sides of ABC, a, b, and c respectively. \r\nLets call the quantity we want to solve for S. \r\nAfter a little geometry I come up with S = (a+b)(b+c)(c+a)/(a+b+c). \r\nWithout loss of generality lets say a+b+c = 1. \r\n\r\nLets also say that \r\n a+b=x, \r\n b+c=y, \r\n a+c=z. \r\n\r\nSo then S = xyz and we are trying to find the min and max of S given the constraint that x+y+z=2 and 1/21$.\n$(\\text{a})$ Find $2004$ perfect powers in arithmetic progression.\n$(\\text{b})$ Prove that perfect powers cannot form an infinite arithmetic progression.", "Solution_1": "a) Let's go for an induction procedure :\r\nFirst note that 1 and 4 are the first two terms of the arithmetic progression $1+3n$ where $n=0,1,2,...$.\r\nNow suppose that for some $k$, you have an arithmetic progression $x+nr$ and that for $n=0,1,...,k-1$ we have $x+nr = a_n^{b_n}$ for some positive integers $a_n,b_n$ and $b_n >1$.\r\nLet $q= lcm \\{b_1,...,b_n \\}$.\r\nLet $t$ be any positive integer. Then, using that $t^q(x+nr) = t^qx + n(t^qr)$, we see that $ \\{t^q(x+nr) \\}$ is an arithmetic progression.\r\nMoreover for $n=0,1,...,k-1,$ we have let $q=b_nc_n$, we then have $t^q(x+nr) = t^{b_nc_n}a_n^{b_n} = (t^{c_n}a_n)^{b_n}$ is a perfect power.\r\nNow choose $t=x+kr$, clearly $t^q(x+kr)$ is a perfect power too, and so we have an arithmetic progression formed with perfect power of kength $k+1$.\r\n\r\nIt follows that we may find arithmetic progressions of perfect powers with arbitrary long, but finite, length.\r\n\r\nb) Suppose that $x+nr$ is an infinite arithmetic progresion form with perfect powers.\r\nLet $d = gcd(x,r)$, $x=dy$ and $r=ds$, where $y,s$ are coprimes.\r\nThen for each $n$, we have $x+nr = d(y+ns)$.\r\nNow, from the well-known theorem of Dirichlet about primes contained in the arithmetic progressions, we know that there are an infinite number of primes of the form $y+ns$. Thus, we may find $n$ such that $y+ns$ is a prime $p$ which does not divide $d$.\r\nIn that case $d(y+ns)$ is a term of the sequence, it is divisible by $p$ but not by $p^2$. Therefore it cannot be a perfect power. A contradiction.\r\n\r\nPierre.", "Solution_2": "It was in our exams last year (Iran 2003)", "Solution_3": "for a) you can find number $d$ such that all numbers $d, 2d... nd$ are all perfect powers. Solution is boring, based on CRT. First you have to choose all primes less or equal than $n$ and carefully choose their exponents in factorization of $d$. Choosing is similar like http://www.artofproblemsolving.com/Forum/viewtopic.php?p=117503&sid=c792bd67b26d3da8c862eb112a28e1fb#p117503 - here of course you need to use $n^k$ primes- there only $ n^2$ were enough.", "Solution_4": "[hide=\"a\"]We inductively construct a sequence $x_1,x_2,\\cdots ,x_n$ of natural numbers in arithmetic progression such that $x_1=y_1^{t_1},\\cdots,x_n=y_n^{t_n}$, with the $y_i,t_i$ natural numbers greater that one.\n\n[b]Base case[/b] - $n=2$:\nWe put $x_1=4=2^2$, $x_2=9=3^2$.\n\n[b]Induction step[/b]:\nLet $x_{n+1}=2x_n-x_{n-1}$ (so that $x_1,x_2,\\cdots,x_{n+1}$ are in arithmetic progression), $k=x_{n+1}^{\\prod_{j=1}^nt_j}$.\nThen we have that $x'_i=kx_i$ are $n+1$ naturals in arithmetic progression (we multiplied an arithmetic progression by a constant), and each is a perfect power (because if $i\\leq n$, $x'_i$ is the product of two $t_i$-th perfect powers, and if $i=n+1$ we have that $x'_{n+1}=(x_{n+1})^{1+{\\prod_{j=1}^nt_j}}$)\n\nThis concludes the proof that for any natural number $n$ there is such a sequence, as desired.[/hide]\n[hide=\"b\"]\nSuppose that there is such an arithmetic progression $x_n$. Then we have that $\\sum_{n=1}^{\\infty}\\frac{1}{x_n}=\\infty$.\n\nBut on the other hand $\\sum_{n=1}^{\\infty}\\frac{1}{x_n}\\leq 1+\\sum_{a=2}^{\\infty}\\sum_{b=2}^{\\infty}\\frac{1}{a^b}=1+\\sum_{a=2}^{\\infty}\\frac{1}{a(a-1)}=2$, contradiction, as desired.\n[/hide]" } { "Tag": [ "algebra", "functional equation", "algebra solved" ], "Problem": "Determine all pairs (a,b) of nonnegative real numbers such that the functional equation f(f(x)) + f(x) = ax + b has a unique continuous solution f : R -> R.", "Solution_1": "Well, it is easy to see that if we look for solutions of the form f(x) = rx + t, we have r = (-1 \\pm \\sqrt (1 + 4a) )/2 and t(r+2) = b.\r\nThus, if a =/= 2, then r =/= - 2 and there are at least two continuous solutions. Thus, we must have a = 2 and then r = -2).\r\nMoreover, in that case, if b = 0 there still are two continuous solutions(the same as above).\r\nThus, we must have b > 0 (since b is nonegative).\r\nIn that case, there is at least one continuous solution, which is f(x) = x + b/3.\r\n\r\nIt remains to prove that it is the only one....if true.\r\n\r\nPierre.", "Solution_2": "I think I have a solution to this one. Let g(x)=f(x)-x-b/3. One can see that f doesn't have fixed points , so g is bounded (superior or inferior, it depends on how f(x)>x for all x or f(x)\\frac{1}{2}$ and $\\frac{4}{5}>\\frac{3}{4}$, and this works all the way up to $\\frac{100}{101}>\\frac{99}{100}$, I guess this isn't really a formal proof...\r\nJudging from your first hint, am I supposed to multiply these two expressions together? The cancel out quite nicely.\r\n\r\nEdit: oops, I didn't see those two posts above while making this post.", "Solution_13": "Ah, good catch! That's the problem there.\r\n\r\nWhat would this look like using products?", "Solution_14": "probability, can you tell me your thought process in coming with the solution?", "Solution_15": "If you multiply the fraction by the the fraction 2*4*6*...*100/2*4*6*...100, cant you simplify it greatly.", "Solution_16": "$\\frac{1}{2}\\cdot \\frac{3}{4}\\cdots \\frac{99}{100} < \\frac{2}{3}\\cdot \\frac{4}{5}\\cdots \\frac{98}{99}$\r\n\r\n$\\Big( \\frac{1}{2}\\cdot \\frac{3}{4}\\cdots \\frac{99}{100} \\Big)^2$\r\n\r\n$< \\Big( \\frac{1}{2}\\cdot \\frac{3}{4}\\cdots \\frac{99}{100} \\Big) \\Big( \\frac{2}{3}\\cdot \\frac{4}{5}\\cdots \\frac{98}{99} \\Big)$\r\n\r\n$= \\frac{1}{100}$\r\n\r\nSee if you can do the rest.", "Solution_17": "Sorry, I didn't make my post very clear. I understand how to do the problem now, what I was asking was what you thought when you read the question and how you came up with the idea of using the second sequence.", "Solution_18": "First of all, I've seen this problem before. When I first solved it, it was with some guidance. But it should pop into your mind pretty early to try somehow multiplying by that second sequence to simplify things. Then if you're lucky/experienced you'll think of the rest.", "Solution_19": "telescoping comes to mind...so you think of those missing intermediate terms", "Solution_20": "very similar problem; prove that 1/15<1/2*3/4*...*99/100<1/12", "Solution_21": "Actually The answer can be easily shown between $1/12$ and $1/13$ \r\n\r\nIn fact: \r\n\r\n $\\frac 1 2 \\times \\frac 3 4 \\times \\dots \\times \\frac {2n-1}{2n} $ can easily be shown between\r\n\r\n$\\sqrt {\\frac {1}{n \\pi}} $ and $\\sqrt {\\frac{2}{(2n+1)\\pi}}$\r\n\r\nFor more detail see the post [url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=%281%2F2%29%2A%283%2F4%29&t=15144]here[/url]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that :\r\n$\\frac{a^{4}b}{2a+b}+\\frac{b^{4}c}{2b+c}+\\frac{c^{4}}{2c+a}\\geq 1$\r\nwhere a;b;c are positive numbers satisfying the condition $ab+bc+ca \\leq 3abc$.\r\nWhen does equality occur?", "Solution_1": "[quote=\"chien than\"]Prove that :\n$\\frac{a^{4}b}{2a+b}+\\frac{b^{4}c}{2b+c}+\\frac{c^{4}}{2c+a}\\geq 1$\n[/quote]\r\n\r\nI think it must be \r\n$\\frac{a^{4}b}{2a+b}+\\frac{b^{4}c}{2b+c}+\\frac{c^{4}{a}}{2c+a}\\geq 1$\r\nI will use well known inequality $\\frac{a^{3}}{x}+\\frac{b^{3}}{y}+\\frac{c^{3}}{z}\\geq \\frac{(a+b+c)^{3}}{3(x+y+z)}$\r\n$3 \\geq \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\geq \\frac{9}{a+b+c}\\Longrightarrow a+b+c \\geq 3$\r\n$\\frac{a^{4}b}{2a+b}+\\frac{b^{4}c}{2b+c}+\\frac{c^{4}a}{2c+a}=\\frac{a^{3}}{\\frac{2}{b}+\\frac{1}{a}}+\\frac{b^{3}}{\\frac{2}{c}+\\frac{1}{b}}+\\frac{c^{3}}{\\frac{2}{a}+\\frac{1}{c}}\\geq \\frac{(a+b+c)^{3}}{9(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})}\\geq \\frac{27}{9.3}=1$\r\nEquality iff $a=b=c=1$", "Solution_2": "[quote=\"cefer\"]\nI will use well known inequality $\\frac{a^{3}}{x}+\\frac{b^{3}}{y}+\\frac{c^{3}}{z}\\geq \\frac{(a+b+c)^{3}}{3(x+y+z)}$\n[/quote]\r\nIt's the Holder's inequality! :P\r\nNice proof, cefer. :)", "Solution_3": "[quote=\"arqady\"][quote=\"cefer\"]\nI will use well known inequality $\\frac{a^{3}}{x}+\\frac{b^{3}}{y}+\\frac{c^{3}}{z}\\geq \\frac{(a+b+c)^{3}}{3(x+y+z)}$\n[/quote]\nIt's the Holder's inequality! :P\nNice proof, cefer. :)[/quote]\r\n\r\nsorry, but is isn't that power mean inequality(by some transformation), i don't know the holder's inequality you were referring to :blush:", "Solution_4": "[quote=\"andy_tok\"] i don't know the holder's inequality you were referring to :blush:[/quote]\r\nsee http://en.wikipedia.org/wiki/Holder%27s_inequality :D" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "f:[0,oo]->[0,1) continuous and (x_n), x_0=1, x_(n+1)=x_nf(x_n)\r\na)Prove that x_n is convergent and find it's limit\r\nb)Compute: lim{n->oo} \\sum {k=1,n}{x_k(1-f(x_k))}", "Solution_1": "???? If x_0= 0 then x_n = 0 for each n.\r\n\r\nPierre.", "Solution_2": "a) All x_n \\geq 0 \r\nx_(n+1)-x_n = x_n(f(x_n)-1) \\leq 0\r\nx_n decreasing positive => lim x_n = L \\geq 0 exist\r\n\r\nSuppose that L>0, f is continous => L=Lf(L) => f(L)=1 contradiction since the value of f is in [0;1[ \r\nSo lim x_n = 0\r\n\r\nb) \\sum {k=1,n}x_k(1-f(x_k)) =\r\n\r\n\\sum {k=1,n}x_k - \\sum {k=1,n} x_k.f(x_k) =\r\n\r\n\\sum {k=1,n}x_k - \\sum {k=1,n}x_(k+1) =\r\n\r\nx_1 - x_(n+1) ---> x_1=f(1)" } { "Tag": [ "AMC", "AIME", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "Find $ x,y,u,vt\\in R$ such that they are satisfying simultan the conditions:\r\n$ u\\plus{}v\\equal{}2$\r\n$ ux\\plus{}vy\\equal{}1$\r\n$ ux^2\\plus{}vy^2\\equal{}\\minus{}1$\r\n$ ux^3\\plus{}vy^3\\equal{}\\minus{}5$ (Exam 2006,UPB)", "Solution_1": "1990 AIME number 15 is a very similar question, and the solution method that works there works here, too:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=80286\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=68753\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=6820\r\n\r\nThis also has nothing to do with calculus ;)", "Solution_2": "I was too lazy to do a single non-trivial computation (not even to solve a 3 by 3 linear system), so I came up with the following \"solution\".\r\n\r\nThe curve $ t\\mapsto(1,t,t^2,t^3)$ has the nice property that every 4 distinct points on it are linearly independent in $ \\mathbb R^4$ (the corresponding determinant is a Vandermonde determinant). The equations tell us that a certain non-zero point belongs to the plane generated by 2 points on that curve. Note that the point itself doesn't lie on the line passing through any point on the curve ($ 1\\cdot( \\minus{} 1)\\ne 2\\cdot( \\minus{} 5)$). Thus, the 2 points in question are unique (otherwise the corresponding planes would either have the trivial intersection or intersect by a line containing the common point on the curve). Thus, the solution is essentially unique (up to the order of points). Looking into the crystal ball attentively, we see that $ (2,1, \\minus{} 1, \\minus{} 5) \\equal{} 3\\cdot(1,1,1,1) \\plus{} ( \\minus{} 1)\\cdot(1,2,4,8)$ (I understand that this point in my argument is somewhat weak, but it is still formally correct :P). The end. \r\n\r\nOf course, the recursion trick is better in the sense that the crystal ball is not needed for solving the problem, but I decided to post this anyway." } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix" ], "Problem": "can anybody tell me what's the minimal polynomial for $ T: \\mathbb R_2[x]\\to\\mathbb R_2[x]$ where $ T(a\\plus{}bx\\plus{}cx^2)\\equal{}b\\plus{}2cx\\plus{}2(a\\plus{}bx\\plus{}cx^2)$ ?\r\n\r\ni already have the answer, i wanted to know if it's the correct one or not.", "Solution_1": "$ T$ has the following matrix: $ A\\equal{}\\begin{bmatrix}0&1&0\\\\0&0&2\\\\2&2&2\\end{bmatrix}.$\r\n\r\n(Some details depend on what order you choose to write the basis in.)\r\n\r\nNow, $ A^2\\equal{}\\begin{bmatrix}0&0&2\\\\ *&*&*\\\\ *&*&*\\end{bmatrix}.$ \r\n\r\nJust from looking at the top rows, we see that $ \\{I,A,A^2\\}$ is a linearly independent set. Hence, $ A$ satisfies no degree two (or lower) polynomial. That means that the minimal polynomial must be the characteristic polynomial, which we just compute directly:\r\n\r\n\\[ x^3\\minus{}2x^2\\minus{}4x\\minus{}4\\]\r\n\r\nNote that this has no rational roots.", "Solution_2": "for the basis $ \\{1,x,x^2\\}.$", "Solution_3": "Right. But in any other basis, the matrix would be similar and the minimal polynomial would be the same.", "Solution_4": "ohhh... can you please explain how you got that matrix?" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "We usually measure angles in degrees but we can use any other unit we choose. For example, if we choose $ 30^{\\circ}$ as a new unit, then the angles of a $ 30^{\\circ}, 60^{\\circ},90^{\\circ}$ triangle would be equal to $ 1, 2, 3$ new units respectively. \r\n\r\nA Triangle $ ABC$ has another triangle $ DEF$ inscribed in it with $ D$ on $ BC$, $ E$ on $ CA$ and $ F$ on $ AB$.\r\n\r\n\r\n$ \\angle BAC \\equal{} a, \\angle ABC \\equal{} b, \\angle ACB \\equal{} c, \\angle FDE \\equal{} d, \\angle DEF \\equal{} e, \\angle EFD \\equal{} f, \\angle AFE \\equal{} g, \\angle AEF \\equal{} h, \\angle BFD \\equal{} i, \\angle FDB \\equal{} j, \\angle EDA \\equal{} k, \\angle DEA \\equal{} l$. \r\n\r\nAll the angles are whole number multiples of some new (unknown) unit; their sizes $ a, b, c, d, e, f, g, h, i, j, k, l$ with respect to this new angle unit are all distinct.\r\n\r\nFind the smallest possible value of $ a \\plus{} b \\plus{} c$ for which such an angle unit can be chosen, and find the corresponding values of $ a$ to $ l$.", "Solution_1": "Has anyone tried this?", "Solution_2": "$ a\\plus{}b\\plus{}c\\equal{}a\\plus{}g\\plus{}h\\equal{}b\\plus{}i\\plus{}j\\equal{}c\\plus{}k\\plus{}l\\equal{}d\\plus{}e\\plus{}f\\equal{}g\\plus{}f\\plus{}i\\equal{}h\\plus{}e\\plus{}l\\equal{}j\\plus{}d\\plus{}k$\r\n\r\n$ a\\plus{}b\\plus{}c\\plus{}d\\plus{}e\\plus{}f\\plus{}g\\plus{}h\\plus{}i\\plus{}j\\plus{}k\\plus{}l \\equal{}1\\plus{}2\\plus{}3\\plus{}...\\plus{}11\\plus{}12$." } { "Tag": [], "Problem": "Out of my own curiosity, what kinds of levels of expertise do you guys have in computer science coming out of high school? (or still in it, if you're there)\r\n\r\nWhat kinds of capabilities do kids have who, say, get a 4 or 5 on Computer Science AP? Can you program a game or other small piece of software on your own? Or do you just know theory?\r\n\r\nI'm asking because I've had some discussions with some corporate folks about launching a new competition involving high schoolers and computer programming...the idea we're kicking around is giving teams of high schoolers - say, teams of 5, like the AHSIMC, or perhaps as many as 10 - a month to program a game.\r\n\r\nThe best game, as voted on by a panel of selected judges from within the industry, would win.\r\n\r\nEach member of the winning team would get a top-of-the-line PC from Alienware or another high-end computer firm.\r\n\r\nComments? Is this type of programming knowhow required for this type of competition within current school curriculum, or only something students would know how to do based on outside personal experience?\r\n\r\nAny insight you can provide would be great.", "Solution_1": "Three members of my AHSIMC team, including myself, got 5s on computer science AB. Another got a 4 on the A. I haven't done any programming lately, though. One guy made a sort of Monopoly game for his computer science final project. It was pretty cool. It even had a cheating feature. He's probably good enough to make the IOI (I think that's what it's called) team, but he's too lazy. He did all of our team's brute-force stuff.", "Solution_2": "I my school, the only computer classes are for Word and Powerpoint, not really any programming. I know some, from what I've learned, but I can't program a game, the graphice and the user interface part would be really hard for me.", "Solution_3": "Questions:\r\n\r\nDoes the game have to work on a certain OS?\r\nDoes it have to be a certain programming language?" } { "Tag": [], "Problem": "If $ x$, $ y$, and $ z$ are positive numbers satisfying \\[x \\plus{} 1/y \\equal{} 4,\\quad y \\plus{} 1/z \\equal{} 1,\\quad\\text{and}\\quad z \\plus{} 1/x \\equal{} 7/3,\\] then $ xyz \\equal{}$\n\n$ \\textbf{(A)}\\ 2/3 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 4/3 \\qquad \\textbf{(D)}\\ 2 \\qquad \\textbf{(E)}\\ 7/3$", "Solution_1": "Ok, I stink at tex, so it'll be a little messy, but here goes nothing...\r\n\r\n[hide=\"solution\"]\nFirst you can multiply the original equations together:\n4*1*7/3=(x+1/y)(y+1/z)(z+1/x)\n\nAfter multiplying this out, we get:\n28/3 = xyz+x+y+z+1/x+1/y+1/z+1/xyz\n\nThen, we can alter the original equations slightly:\n\nx+1/y=4\nx=4-1/y\n\nDoing the same to the other equations, we get:\n\nx=4-1/y\ny=1-1/z\nz=7/3-1/x\n\nThen, we can substitute these into the product of these equations:\n\n28/3 = xyz+4+1+7/3+1/x+1/y+1/z-1/x-1/y-1/x+1/xyz\n28/3 = xyz+1/xyz+22/3\n2 = xyz+1/xyz\nThen, we just substitute u in for xyz\n2 = u+1/u\n2u = u^2+1\nu^2-2u+1=0\n\nThis is easily factorable:\n(u-1)^2=0\nThen take the square root:\nu-1=0\nu=1\nxyz=1\nSo the answer is B.[/hide]", "Solution_2": "[hide=\"Quicker?\"]\nAdding the equations, we see that\n\n$ x\\plus{}y\\plus{}z\\plus{}\\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z}\\equal{}\\frac{22}{3}$.\n\nMultiplying the equations, we find\n\n$ x\\plus{}y\\plus{}z\\plus{}\\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z}\\plus{}xyz\\plus{}\\frac{1}{xyz}\\equal{}\\frac{28}{3}$.\n\nSubtracting the first equation from the second, \n\n$ xyz\\plus{}\\frac{1}{xyz}\\equal{}2$.\n\nAM-GM implies that $ xyz\\equal{}\\frac{1}{xyz}$, so $ xyz\\equal{}1$. Answer is $ B$.\n[/hide]", "Solution_3": "It would be nice to demonstrate that $ x,y,z$ actually exist. \r\n\r\nSince $ xyz\\equal{}1$, we can eliminate $ z$, and get\r\n\r\n\\[ x\\plus{}\\frac{1}{y}\\equal{}4\\qquad y\\plus{}xy\\equal{}1\\qquad \\frac{1}{xy}\\plus{}\\frac{1}{x}\\equal{}\\frac{7}{3}\\]\r\n\r\nThe third equation can be rewritten as $ 1\\plus{}y\\equal{}\\frac{7}{3}xy$. Using that with equation $ 2$ gives us a linear system in $ (y,xy)$. It follows that $ (y,xy)\\equal{}\\left(\\frac{2}{5},\\frac{3}{5}\\right)$. So $ x\\equal{}\\frac{xy}{y}\\equal{}\\frac{3}{2}$. Then $ z\\equal{}\\frac{1}{xy}\\equal{}\\frac{5}{3}$. That means that a satisfactory triple is \\[ (x,y,z)\\equal{}\\left(\\frac{3}{2},\\frac{2}{5},\\frac{5}{3}\\right)\\] we can see that these values satisfy the original equation.", "Solution_4": "Altheman's solution is also far more obvious on a 75-minute test.", "Solution_5": "[quote=\"E^(pi*i)=-1\"]\n[hide]\nAM-GM implies that $ xyz \\equal{} \\frac {1}{xyz}$, so $ xyz \\equal{} 1$. \n[/hide]\n[/quote]\r\n\r\nWhy is this true? Also, is it true for all xyz, or just in this situation?", "Solution_6": "Since $ x$, $ y$, and $ z$ are all positive, we know that their product will be positive. So we can apply AM-GM:\r\n\r\n$ \\frac {xyz \\plus{} \\frac {1}{xyz}}{2}\\geq \\sqrt {xyz\\left(\\frac {1}{xyz}\\right)} \\equal{} 1$.\r\n\r\nWe already know that $ xyz \\plus{} \\frac {1} {xyz} \\equal{} 2$, so equality holds. AM-GM only has equality when everything is equal, so $ xyz \\equal{} \\frac {1}{xyz}$.", "Solution_7": "AM and GM stand for arithmetic mean and geometric mean, btw.", "Solution_8": "[quote=\"E^(pi*i)=-1\"] AM-GM only has equality when everything is equal, so $ xyz \\equal{} \\frac {1}{xyz}$.[/quote]\r\nOh, I never knew that. Thanks.", "Solution_9": "The third equation gives z = (7/3) - (1/x). Then substituting this into the second equation, y = 1 - (1/z) = (4x-3)/(7x-3). Finally substituting in the first equation, x + (7x-3)/(4x-3) = 4 which gives 4x^2 -12x +9 = 0, that is, (2x-3)^2 = 0, so that x = 3/2. Then y=2/5 and z = 5/3.", "Solution_10": "__________________", "Solution_11": "AM-GM only works with positive numbers.", "Solution_12": "$x+\\frac{1}{y}=4\\ \\cdots (1),\\ y+\\frac{1}{z}=1\\ \\cdots (2),\\ z+\\frac{1}{x}=\\frac 73\\ \\cdots (3)$\n\nFrom $(1),\\ (2)$, $x=\\frac{4y-1}{y},\\ z=\\frac{1}{1-y}\\ \\cdots (4)$, thus from $(3)$, we have $\\frac{1}{1-y}+\\frac{y}{4y-1}=\\frac 73\\ \\therefore 25y^2-20y+4=0\\Longleftrightarrow (5y-2)^2=0$, yielding $y=\\frac 25.$\nThen from $(4)$, we get $x=\\frac 32,\\ z=\\frac 53\\Longrightarrow xyz=1.$", "Solution_13": "$\\frac{28}{3}=\\left(x+\\frac{1}{y}\\right)\\left(y+\\frac{1}{z}\\right)\\left(z+\\frac{1}{x}\\right)=$ $xyz+x+y+z+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}+\\frac{1}{xyz} = $ $xyz+\\frac{1}{xyz} + \\frac{22}{3}$. Hence, $xyz+\\frac{1}{xyz} = 2$ and $xyz=1$." } { "Tag": [ "calculus", "integration", "function", "real analysis", "real analysis unsolved" ], "Problem": "Discuss the convergence of the squence $ \\int_0^1 x^{\\minus{}1/n}|f(x)|dx$ for $ f \\in L^1[0,1]$. \r\n\r\nAnyone having any idea? Thanks.", "Solution_1": "It is a non-increasing sequence of non-negative functions ;)", "Solution_2": "That is right. But how can you use the Lebesgue dominated convergence theorem if you have no dominator? Or am I missing something?", "Solution_3": "There are two cases: either one of the integrals is finite (then it provides the majorant), or all of them are infinite (then there isn't much more to say)." } { "Tag": [], "Problem": "Add parenthesis to make this equation true.\r\n\r\n$\\frac{9 + 10 - 17 + 3}{4 - 5 + 3}= 5$", "Solution_1": "[hide]\r\n\r\n( 9+10-17+3) / (4-5+3) = 5\r\n\r\nis there really a way to do this? my thinking was that since its all addition, the paranthesis dont matter except for 2 spots: around the 17 and 3 at the top or around the 5 and 3 at the bottom. none of those work! i hope this isn't oneof those riddle things :)", "Solution_2": "Oops...there isn't supposed to be a 9 in the numerator.\r\nHere is the new problem:\r\n$\\displaystyle \\frac{10 - 17 + 3}{4 - 5 + 3}= 5$", "Solution_3": "(10-(17+3))/(4-5+3) = -5\r\n\r\nhmm are you sure there's an answer even now? :lol: \r\n\r\nthe best you can get is -5 I think", "Solution_4": "[hide]yes i agree with Barnacle, i got $\\frac{[10-(17+3)]}{(4-5+3)}=-5$[/hide]" } { "Tag": [ "algebra", "polynomial", "geometry", "limit", "trigonometry", "function" ], "Problem": "can someone tell me what pi equals as a fraction? Or am I just asking for too much? PLEASE REPLY!", "Solution_1": "uh, pi is irrational.", "Solution_2": "$\\pi$ is irrational as well as transcendental, which means that it cannot be expressed as a fraction or as a root of a polynomial with integer coefficients.\r\nHowever, there are two good fraction approximations of $\\pi$: $22/7$, and $355/113$. \r\nAnd I'm assuming you know the first few digits, at least.", "Solution_3": "There's the Gregory-Leibniz Series:\r\n$\\pi$ $=$ $\\frac{4}{1}-\\frac{4}{3}+\\frac{4}{5}-\\frac{4}{7}+\\frac{4}{9}-\\frac{4}{11}...$\r\nNot a fraction but it has fractions :oops: .\r\nSource: http://en.wikipedia.org/wiki/Pi", "Solution_4": "Pi is cirumference divided by diameter and the area of a circle divided by its radius squared. Also, pi is about 3.141592653589793238462643383279502884197169399375105820974944.\r\nAnd I think pi can be a continued fraction.", "Solution_5": "Well isn't there a limit definition? for example e can be expressed by \r\n\r\n[img]http://www.artofproblemsolving.com/Wiki/images/math/2/1/5/2158d71ea12e95f58290eb18897a0c3d.png[/img]\r\n\r\nbut that is beside the point\r\n\r\n$\\pi$ is irrational but you can (to a certain nubmer of digits) make a fraction", "Solution_6": "[quote=\"7h3.D3m0n.117\"]Well isn't there a limit definition? for example e can be expressed by \n\n[img]http://www.artofproblemsolving.com/Wiki/images/math/2/1/5/2158d71ea12e95f58290eb18897a0c3d.png[/img]\n\nbut that is beside the point\n\n$\\pi$ is irrational but you can (to a certain nubmer of digits) make a fraction[/quote]\r\nThere is a limit defition. \r\n$\\pi=\\lim_{n\\rightarrow\\infty}{\\left(\\frac12n\\sin\\frac{360^\\circ}n\\right)}$\r\n\r\nThere is also the Wallis product\r\n\r\n$\\pi=2\\cdot\\frac{2\\cdot2\\cdot4\\cdot4\\cdot6\\cdot6\\cdot8\\cdot8\\cdots}{1\\cdot3\\cdot3\\cdot5\\cdot5\\cdot7\\cdot7\\cdot9\\cdots}$\r\n\r\nOr $\\pi=2\\prod_{n=1}^\\infty\\left(\\frac{4n^{2}}{4n^{2}-1}\\right)$", "Solution_7": "Also, applying the Zeta function, which is $\\zeta(x)=\\sum_{i=1}^\\infty \\frac{1}{i^{x}}$, results in some patterns involving pi when $x$ is even.\r\n\r\n$\\zeta(2)=\\sum_{i=1}^\\infty \\frac{1}{i^{2}}=\\frac{\\pi^{2}}{6}$\r\n$\\zeta(4)=\\sum_{i=1}^\\infty \\frac{1}{i^{4}}=\\frac{\\pi^{4}}{90}$\r\n\r\nI have no idea why.", "Solution_8": "you guys are all heretics\r\n\r\n$\\pi=3$", "Solution_9": "there is no definite fraction, but most people use 22/7 or the easy to remember 113355 fraction (335/113).\r\nlike davidyko said :rotfl:", "Solution_10": "I use $\\frac{1087}{346}$ because I like to be original!", "Solution_11": "[quote=\"davidyko\"]However, there are two good fraction approximations of $\\pi$: $22/7$, and $355/113$. [/quote]\n\nOnly two?\n\nThere are in fact an infinite series of such [b]convergents[/b] $\\frac{p_{n}}{q_{n}}$, which have some fascinating properties. \n\nhttp://mathworld.wolfram.com/PiContinuedFraction.html\n\n[quote=\"8parks11\"]I use $\\frac{1178}{375}$ because I like to be original![/quote]\r\n\r\nUnfortunately, this is not a convergent, and is in fact much worse than $\\frac{355}{113}$.", "Solution_12": "I changed it!\r\nit can't be better than the \"official\" good ones\r\nbecause I just found two random digits", "Solution_13": "[quote=\"t0rajir0u\"][quote=\"davidyko\"]However, there are two good fraction approximations of $\\pi$: $22/7$, and $355/113$. [/quote]\n\nOnly two?\n\nThere are in fact an infinite series of such [b]convergents[/b] $\\frac{p_{n}}{q_{n}}$, which have some fascinating properties. \n\nhttp://mathworld.wolfram.com/PiContinuedFraction.html\n\n[quote=\"8parks11\"]I use $\\frac{1178}{375}$ because I like to be original![/quote]\n\nUnfortunately, this is not a convergent, and is in fact much worse than $\\frac{355}{113}$.[/quote]\r\n\r\nHmm, now that is interesting.\r\nI guess I meant \"there are only two reasonably accurate fraction approximations of $\\pi$ that I can think of right now\". :lol:", "Solution_14": "Or, you could check my signature :D \r\n\r\n(only true for the 83+ line of calculators)", "Solution_15": "see\r\n\r\nhttp://en.wikipedia.org/wiki/Continued_fraction\r\n\r\nin Continued fraction expansions of \u03c0", "Solution_16": "I usually approximate with $\\frac{104348}{33215}$", "Solution_17": "why not just use (3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609)/100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.\r\n\r\nits always been accurate enough for all my uses", "Solution_18": "OK, I think the question has been answered and this doesn't need to become a \"Who can name the most accurate representation of Pi Contest.\"\r\n\r\nLocked." } { "Tag": [ "geometry", "trigonometry", "3D geometry", "perimeter", "circumcircle", "inequalities", "ratio" ], "Problem": "1.[EASY] THe MIDpoints P, Q, R, S of the sides of quadrilateral ABCD are connected to form quadrilateral PQRS. Also, THe MIDpoints W, X, Y, Z of the sides of quadrilateral PQRS are connected to form quadrilateral WXYZ. If the area of quad ABCD is one-half the product of the diagonals. What kind of quad is WXYZ?\r\n\r\n2. [EASY] What is the area of the largest semincircle that can be drawn in an isoscles right triangle with hypotenuse 4?\r\n\r\n3. [EASY] Lines SP and UE are secants of a circle that intersect at R, an interior point of the circle. IF measure of angle SPE is 26 deg, and the measure of minor arc UP is 46 deg., find the measure of angle SRU.\r\n\r\n4. [EASY] the sine value of each internal angle of a heptagon is greater than 0.5. WHat is the maximum number of acute angles that the heptagon can have?\r\n\r\n5. [Moderate] The two tangent lines to a circle from a point make an angle of 45 deg., Of the dianmeter of the circle is 8, find the area of the region bounded by the two tangent lines and the circle.\r\n\r\n6. [Moderate] 2009 unit cubes are glued together to form a solid cube. What is the maximum surface area of the figure.\r\n\r\n7. [Moderate] B and X are points on angle CAX (w;c measure 45 deg), such the B is between A and C. If AB=8, BC=10, and D lies on the other ray, how far shud B be from A so that angle CDB is a maximum.\r\n\r\n8. [EASY] A circle of radius 1 is inscribed in a 3-60-90 triangle. Find the perimiter of the triangle.\r\n\r\n9. [EASY] Sqaure ADGJ is inscribed in a unit circle. Points B, C, E, F, H, I, K, and L are selcted on the circumference of the unit circle such that a regular dodecagon ABCDEFGHIJKL is formed. Find the area of the region outside the sqaure but inside the dodecagon.\r\n\r\n10. [MOderate] Find all values of x if 10, 11, and !x+12! (abs value i mean)\r\n are the lengths of the sides of the triangle.\r\n\r\n11. [HARD] IN triangle ABC, AB=6, BC=8, and AC=7. D is on side AC such that AD =3, and E is on side AB while F is on side BC. IF the areas of triangle ADE, CDF, AND quadrilateral BEDF are equal, find BE+DF.", "Solution_1": "#11 \r\nfirst use heron's foruma find the area of ABC\r\nsqr root [(21/2)(9/2)(7/2)(5/2)] = 21*sqr rt(15)/4\r\n\r\nthe triangle ABC is divided in to 3 parts ADE = CDF = BEDF\r\nso divide the area by 3 \r\n\r\nADE = CDF = 7*sqr rt(15)/4 (these are the areas of the 2 little triangles)\r\n\r\nNow use Law of Cosine to find angle A and angle C\r\n\r\nA = Arc cos (1/4), C = Arc cos (11/16)\r\n\r\nSin(A)= sqr rt(15)/4, Sin(B) = 3*sqr rt(15)/16\r\n\r\nUse the foruma: \r\n\r\nArea of a triangle = 1/2*a*b*sin(C)\r\n\r\nADE = 1/2 AE*AD*sin(A)\r\nAD = 3 (Given), Sin(A)= sqr rt(15)/4, area of ADE = 7*sqr rt(15)/4\r\n\r\nAE = 14/3\r\nAB = 6,\r\n\r\nBE = AB - AE =4/3\r\n(we just found the first part of the problem)\r\n\r\n\r\ndo the same thing with the other little triangle CDF\r\n\r\nCDF= 1/2 CF*CD*sin(C)\r\nCD = 7-3=4, Sin(C)= 3*sqr rt(15)/16, area of CDF = 7*sqr rt(15)/4\r\n\r\nCF = 14/3 \r\n\r\nThen use law of cosine\r\nDF^2 = CD^2 + CF^2 - 2(CD)(CF)*cos(C)\r\nDF = sqr rt(109)/3\r\n\r\nDF+BE = (4+sqr rt(109))/3\r\n\r\n(i am not sure if the numbers are right...)", "Solution_2": "Key lemma for #7 (well-known): The point D on AX which will maximize $ \\angle BDC$ is the point of tangency of the circle which passes through B and C and is tangent to AX.\r\nProof:For any other point K on AX, the circumcircle of KBC will be larger because the circle will cut through the line, not just touch it, and thus $ \\angle BKC$ will be smaller since it is a bigger circle subtending the same size chord.\r\nI actually forgot the lemma but rediscovered it once I wanted to find the locus of points such that BDC is constant, which is a circle.", "Solution_3": "[hide=\"#8\"]\n\nI take it you mean 30-60-90 right triangle. Let the shorter leg be $ s$, the longer leg be $ s\\sqrt3$, and the hypotenuse $ 2s$. We can express the area in two ways:\n\n$ \\frac {s^2\\sqrt3}{2}$ by $ bh/2$ and $ \\frac {s \\plus{} s\\sqrt3 \\plus{} 2s}{2} \\equal{} \\frac {s(3 \\plus{} \\sqrt3)}{2}$ by $ rs$\n\nIf we set these equal and solve for s, we have $ s \\equal{} 1 \\plus{} \\sqrt3$. So, the perimeter is $ 6 \\plus{} 4\\sqrt3$\n\n[/hide]\n\n[hide=\"#10\"]\n\nBy the triangle inequality, we have \\[ 10\\plus{}11>|x\\plus{}12|\\] \\[ 10\\plus{}|x\\plus{}12|>11\\] \\[ 11\\plus{}|x\\plus{}12|>10\\]\n\nSimplifying the first gives $ \\minus{}33\\minus{}11$ and $ x<\\minus{}13$. Since the third works for all real x, we ignore this one. Combining the inequalities gives $ \\minus{}33 correct answer", "Solution_5": "More to Go!!", "Solution_6": "#2\r\n[hide]You can position the semicircle such that it touches the legs of the triangle and its center lies on the midpoint of the hypotenuse. Drawing the line from this midpoint to the tangent point on one of the legs, you get similar triangles in ratio $ 1: 2$. The leg of the triangle has length $ 2\\sqrt {2}$, so the leg of our smaller, similar triangle is $ \\sqrt {2}$. The area of this semicircle is $ \\boxed{\\pi}$.[/hide]\n#5\n[hide]Let the tangent points to the circle be $ A$ and $ B$ and the center of the circle be $ O$. Also, let the tangents meet at $ E$. Draw $ AB$. We now have an isosceles triangle with $ < E \\equal{} 45^{\\circ}$. Drawing the altitude from $ A$ to $ BE$ to $ F$, and letting it have length $ \\beta$, we get that $ AE \\equal{} \\sqrt {2}\\beta$. We then find that $ BF \\equal{} \\beta\\sqrt {2} \\minus{} \\beta$. Using right triangle $ \\triangle {AFB}$, we get that $ AB \\equal{} \\beta\\sqrt {4 \\minus{} 2\\sqrt {2}}$. Now, let's look at triangle $ AOB$.We have that $ < AOB \\equal{} 135^{\\circ}$. Draw the altitude from $ A$ to the extension of $ OB$ to the point $ G$. $ < GOA \\equal{} 45^{\\circ}$! So, since $ AO \\equal{} 4, AG \\equal{} OG \\equal{} 2\\sqrt {2}$. By the Pythagorean Theorem, and then setting it equal to our other expression for $ AB$ we then find that $ /beta \\equal{} \\frac {4\\sqrt {2 \\plus{} \\sqrt {2}}}{\\sqrt{4 \\minus{} 2\\sqrt {2}}} \\equal{}$. Finally, we can multiply this by $ \\sqrt {2}$ to get our tangents. After some simplifying and some algebra, we get that this equals $ 1 \\plus{} \\sqrt {2}$.Now, we just have to find the areas of the two triangles, which is just $ 4 \\plus{} 4\\sqrt {2}$. The area of the circle is just $ 16\\pi$. The area of their overlap is $ \\frac {135}{360} \\bullet 16\\pi \\equal{} 6\\pi$. Adding and subtracting the necessary results, we get that the area of the total figure is $ \\boxed {10\\pi \\plus{} 4 \\plus{} 4\\sqrt {2}}$.[/hide]\r\nAnd all of that without any trig =]\r\n\r\nEDIT: fixed latex", "Solution_7": "Correct for number 2. Thanks!\r\n\r\n#5 should be 16(sqrt2)+16-6pi" } { "Tag": [ "LaTeX" ], "Problem": "If we were to simplify the following...\r\n\r\n\r\n(1+3+...+(2n-1)) / (2+4+...+2n)\r\n\r\n\r\nwould this simplify to\r\n\r\nn/(n+1) or (n-1)/n\r\n\r\nOne strategy leads me to the first answer and the other to the second answer.\r\n\r\nTHANKS!", "Solution_1": "[quote=\"ProblemSets\"]If we were to simplify the following...\n\n\n(1+3+...+(2n-1)) / (2+4+...+2n)\n\n\nwould this simplify to\n\nn/(n+1) or (n-1)/n\n\nOne strategy leads me to the first answer and the other to the second answer.\n\nTHANKS![/quote]\r\n\r\n[hide]$\\frac{1+3+...+(2n-1)}{2+4+...+2n}$\n\n$\\frac{n^{2}}{2(1+2+...+n)}$\n\n$\\frac{n^{2}}{n(n+1)}$\n\n$\\frac{n}{n+1}$[/hide]", "Solution_2": "[hide]We Know That (1+3+...+(2n-1))----> n*(2n)/2---->n\u00b2\n\nANd (2+4+...+2n)---->n*2(n+1)/2--->n(n+1) \n\nSO (1+3+...+(2n-1)) / (2+4+...+2n) ---->n/n+1[/hide]", "Solution_3": "yeah\r\nits clearly\r\n[hide]\n(n^2)/(n^2+n)=n/(n+1)\n\n[/hide]\r\nI SO WISHED I could have the time to learn Latex", "Solution_4": "[quote=\"ProblemSets\"]If we were to simplify the following...\n\n\n(1+3+...+(2n-1)) / (2+4+...+2n)\n\n\nwould this simplify to\n\nn/(n+1) or (n-1)/n\n\nOne strategy leads me to the first answer and the other to the second answer.[/quote]\r\n\r\nA very simple way to check your answer: Plug in $n = 1$. The expression is $\\frac{1}{2}$, which invalidates the second answer." } { "Tag": [], "Problem": "Find the sum of the first $ 30$ odd numbers", "Solution_1": "The formula for the sum of the first $ n$ odd numbers is $ n^2$.\r\nThus, we have $ 30^2$ or $ 900$." } { "Tag": [ "LaTeX" ], "Problem": "I am using texniccenter and miktex on windows. IS there any way to get the font that it uses to use on mS word. The TTF file is fine.", "Solution_1": "yes, but don't expect it to be easy. See [url=http://www.radamir.com/tex/ttf-tex.htm]Using TrueType fonts with TeX (LaTeX) and pdfTeX (pdfLaTeX)[/url]", "Solution_2": "No. I mean like the default font in latex. I want the TTF for that or the font name at least so I can use the latex font in MS Word.", "Solution_3": "The default font is probably Computer Modern. What do you mean by [quote]The TTF file is fine[/quote] - what TTF file? The cm fonts come in lots of different sizes eg ftp://ftp.dante.de/tex-archive/fonts/cm/ps-type1/bakoma/ttf/ and the UM fonts in um.zip at http://www.truetex.com/ Notice that there are also separate versions of the fonts for bold and italics.", "Solution_4": "Nono. i mean if I type a report in latex (no math though) and don't use math mode. Then I get a font not Times New Roman. I want that", "Solution_5": "Yes, the default text fonts are Computer Modern. It's not a single font but a family of fonts. You'll see the fonts I mentioned above in the LaTeX log for your document.\r\n\r\nFor example, cmr10 is for upright text 10 point and cmmib10 is bold face math font as you'll see on page 5 of [url=http://www.latex-project.org/guides/fntguide.pdf]LaTeX2e Font Selection[/url]", "Solution_6": "Thanks. Got it.", "Solution_7": "What do you want to do with Word that you can't with LaTeX? Why would you want to use Word anyway?" } { "Tag": [ "linear algebra", "linear algebra theorems" ], "Problem": "well , linear algebra will be the course taught for us fromm the upcoming semester starting from january , \r\n\r\ni want to know which is the most widely used book for this branch of mathematics", "Solution_1": "Look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=291885." } { "Tag": [ "blogs", "geometry", "rectangle", "combinatorics proposed", "combinatorics" ], "Problem": "2.A room in the shape of a rectangular has been covered by some rectangular-shaped carpets such that each point in the floor is covered by exactly one carpet. Prove that the sum of the width of the carpets is not smaller than the width of the room.\r\n\r\n(Note. The width of a rectangular is it's smallest side's lenght)\r\n\r\nhere's it's link in my blog:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/weblog_entry.php?p=749422#749422", "Solution_1": "Let p1,...,pn and q1,...,qn be lines which contain horizontal and vertical \r\nsides of rectangles (p1 and pn as well as q1 and qn contain sides of big rectangle). \r\nThen one of 2 statements must be true:\r\n(1)\tin sequence p1,...pn, distances between two consecutive lines \r\n are parts of smaller sides of some little rectangles\r\n(2)\tthe same thing about q1,...,qn.\r\nIn any case the conclusion follows immediately.\r\nI havent explained it in details, but do you think its correct?" } { "Tag": [ "trigonometry", "email" ], "Problem": "[b] Inegalitati trigonometrice:[/b]", "Solution_1": "Bun si care-i scopul? Ele pot fi gasite in cartea mea \"Inegalitatii alese in matematica\", Editura Niculescu.\r\nO mica observatie. Inegalitatea lui Ballieu ( data in 1937) are loc si intr-un context mai general si anume:\r\n\r\n Daca $t\\in (0,1]$ sau $t=2$, atunci in orice triunghi are loc inegalitatea \\[ 2^{t-1}\\cdot\\sin^{t}\\frac{A}{2}\\leq\\frac{a^{t}}{b^{t}+b^{t}}. \\]\r\n\r\nUite o aplicatie asa de inceput.... :) \r\n\r\nSa se demonstreze ca in orice triunghi $ABC$ are loc inegalitatea \\[ \\frac{R}{r}\\geq\\frac{(a+b)(b+c)(c+a)}{4abc}. \\]", "Solution_2": "Caesar, scopul este de a informa utilizatorii acestui site cu privire la anumite inegalitati trigonometrice, pe care eu le-am considerat a fi utile in concursurile de matematica din Romania si nu numai.Da, e adevarat ca acestea se gasesc si in cartea, impropriu zis (mai sunt si altii:prof.I.V.Maftei, George Popescu...), dar poate nu toti au cartea .Am aflat(din G.M.) ca tu esti de origine din Constanta(tatal tau cred ca e profesor).Ai putea sa-mi dai si mie niste numere de telefon(sau adrese de email) ale Revistei de matematica si informatica din Constanta?As dori sa trimit si eu probleme propuse de mine(evident corecte-sper).Poti sa ma ajuti?\r\nP.S.:vezi si sectiunea de pe forumul romanesc. :)" } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "Trying many different ways, there doesn't seem to be a way to involve a dollar sign in a hide tag without arousing $ \\text{\\LaTeX}$ coding.", "Solution_1": "Backslashes are banned in the code; \"hide=string\" requires that string to avoid several special characters, and for good reason. The dollar sign works just fine in hidden text." } { "Tag": [ "topology" ], "Problem": "Hi all, \r\n\r\nThis is exercise 6 in section 1.3 of Allen Hatcher's \"Algebraic Topology.\" \r\n\r\nLet $ X$ be a shrinking wedge of circles (that is, a union of circles $ C_n$ of radius $ 1/n$ with center $ (1/n,0)$ for $ n\\equal{}1,2, \\ldots$), and let $ \\tilde{X}$ be the covering space of $ X$ that is a linear chain of $ X$s with a line connecting them, meeting each copy of $ X$ at the wedge point. The problem is to construct a $ 2$-sheeted cover $ Y \\rightarrow \\tilde{X}$ such that the composite $ Y \\rightarrow \\tilde{X} \\rightarrow X$ is not a covering space. \r\n\r\nFirst, I am having trouble seeing that $ \\tilde{X}$ is a covering space of $ X$. Where do the line segments connecting each copy of $ X$ map to under the covering map? \r\n\r\nMy intuition is that the space $ Y$ that we're looking for looks like $ \\tilde{X}$ but with another copy of $ X$ for each copy of $ X$ in $ \\tilde{X}$, glued to the wedge point. This would seem to be $ 2$-sheeted, but I can't visualize the composition. \r\n\r\nIf anyone has trouble seeing the spaces $ X, \\tilde{X}$, and $ Y$, I will try to clarify further.\r\n\r\nAny help would be greatly appreciated. Thanks.", "Solution_1": "Okay, I don't think that $ Y$ works because the fiber above an $ x \\in \\tilde{X}$ on a line segment connecting two copies of $ X$ has cardinality $ 1$. Should I somehow introduce two line segments, like two copies of $ \\tilde{X}$ glued together in some way? Every time I try this though, there is still a point whose fiber has $ 1$ element." } { "Tag": [ "probability", "AMC", "AIME", "induction", "limit", "ratio" ], "Problem": "Suppose you flip a coin $n$ times such that two consecutive heads never appear. Let $p_n$ be the probability that, given such a sequence of $n$ flips, it ends in a tail. Find $p_n$ as $n$ approaches infinity.\r\n\r\n[hide]\nAIME 1990 Problem 9[/hide]", "Solution_1": "[hide]\nLet F_1 = 1, F_2 = 2, F_n = F_(n-2) + F_(n-1). We have\n\np_1 = F_1/F_2\np_2 = F_2/F_3\np_3 = F_3/F_4\np_4 = F_4/F_5\n\n(easily verified) So I would conjecture that p_n = F_n/F_(n+1) and lim(n to infinity) p_n = 1/phi = 2/(1 + sqrt(5)).\n\nNow I just have to prove that conjecture :D. Induction probably works...\n[/hide]", "Solution_2": "Ummm, its just the last two numbers that matter? Oh hold on thats wrong...", "Solution_3": "Well the maximum number of heads is n/2, so you could set up a summation. I think that it would be easier to find the probability that it ends in a head, and then subtract that from 1.", "Solution_4": "[quote=\"nr1337\"][hide]\nLet F_1 = 1, F_2 = 2, F_n = F_(n-2) + F_(n-1). We have\n\np_1 = F_1/F_2\np_2 = F_2/F_3\np_3 = F_3/F_4\np_4 = F_4/F_5\n\n(easily verified) So I would conjecture that p_n = F_n/F_(n+1) and lim(n to infinity) p_n = 1/phi = 2/(1 + sqrt(5)).\n\nNow I just have to prove that conjecture :D. Induction probably works...\n[/hide][/quote]\r\n\r\nYes. That is correct. But I don't think anybody can really tell what you are doing or how you did it :P", "Solution_5": "Heh, okay I'll try again :)\r\n\r\n[hide]\nSo first I try to calculate some small values of $p_n$. We can easily see that $p_1 =\\frac{1}{2}$, $p_2 = \\frac{2}{3}$, $p_3 = \\frac{3}{5}$, $p_4 = \\frac{5}{8}$ just by listing out possibilities. We define a sort of shifted Fibonacci Sequence by $F_1 = 1$, $F_2 = 2$, and $F_n = F_{n-1} + F_{n-2}$. We can see that, for all $1\\leq n \\leq 4$, $p_n = \\frac{F_n}{F_{n+1}}$. So we conjecture that this is true for all $n$. Ummmm..... \"we leave it to the reader to prove this.\" :D Now we have $\\lim_{n \\to \\infty} p_n = \\lim_{n \\to \\infty} \\frac{F_{n}}{F_{n+1}} = \\frac{1}{\\phi}$ where $\\phi$ is the golden ratio $\\frac{1+\\sqrt{5}}{2}$.[/hide]", "Solution_6": "[quote=\"nr1337\"]Heh, okay I'll try again :)\n\n[hide]\nSo first I try to calculate some small values of $p_n$. We can easily see that $p_1 =\\frac{1}{2}$, $p_2 = \\frac{2}{3}$, $p_3 = \\frac{3}{5}$, $p_4 = \\frac{5}{8}$ just by listing out possibilities. We define a sort of shifted Fibonacci Sequence by $F_1 = 1$, $F_2 = 2$, and $F_n = F_{n-1} + F_{n-2}$. We can see that, for all $1\\leq n \\leq 4$, $p_n = \\frac{F_n}{F_{n+1}}$. So we conjecture that this is true for all $n$. Ummmm..... \"we leave it to the reader to prove this.\" :D Now we have $\\lim_{n \\to \\infty} p_n = \\lim_{n \\to \\infty} \\frac{F_{n}}{F_{n+1}} = \\frac{1}{\\phi}$ where $\\phi$ is the golden ratio $\\frac{1+\\sqrt{5}}{2}$.[/hide][/quote]\n\nCan you think of a proof for it?\n\n[hide=\"hint\"]\nFor a sequence of length $n$, look at the last coin of a sequence of length $n-1$.[/hide]", "Solution_7": "I got a proof now; I just needed a piece of cake to get my brain working again :D :D\r\n\r\n[hide]\nSuppose p_n = F_n/F_(n+1) for some n. Let p_(n+1) = a/b. We will prove that a = F_(n+1) and b=F_(n+2). Let's start with b.\n\nLet R be a sequence of coin flips satisfying the condition of no consecutive heads. Suppose R has n flips and is of the form ...T (T tails, H heads). From that, we can construct 2 more sequences of coin flips of the forms ...TH and ...TT. If R is of the form ...H, we can only construct 1 more sequence, ...HT. There are F_n sequences R of the form ...T and F_(n+1) - F_n = F_(n-1) sequences of the form ...H, so there are a total of b= 2F_n + F_(n-1) = F_n + F_n + F_(n-1) = F_n + F_(n+1) = F_(n+2) sequences with no consecutive heads. Going back to our sequence R with n flips, we knew that there are F_n sequences R of the form ...T and F_(n-1) of the form ...H. So we should have F_n sequences of length n+1 of the form ...TT and F_(n-1) of the form ...HT, for a total of a = F_n + F_(n-1) = F_(n+1). p_(n+1) = a/b = F_(n+1)/F_(n+2). By induction, we are done.\n[/hide]\r\n\r\nThat was kind of hard to explain....", "Solution_8": "Yeah that's basically it. Simply put...\r\n\r\nLet $h_n$ be the number of working sequences of length $n$ that end in a head and $t_n$ be the number of working sequences of length $n$ that end in a tail.\r\n\r\nNotice that $t_{n+1} = h_n+t_n$ because you can add a tail to either previous sequence and you won't violate the condition. Also notice that $h_{n+1}=t_n$ because you can add a head to a previous sequence iff the previous sequence ended in a tail. The base case is easy to find ($h_1=1, t_1=1$), so we see that $t_n = F_{n+1}$ and $h_n = F_n$. So $p_n = \\frac{t_n}{t_n+h_n} = \\frac{F_{n+1}}{F_n+F_{n+1}} = \\frac{F_{n+1}}{F_{n+2}}$ and $\\lim_{p \\rightarrow \\infty}{p_n} = \\frac{1}{\\phi}$.", "Solution_9": "This is a classic Fibonacci problem. I don't see how they would have converted the answer into an integer though. :?", "Solution_10": "[quote=\"probability1.01\"]This is a classic Fibonacci problem. I don't see how they would have converted the answer into an integer though. :?[/quote]\r\n\r\nGo look at the problem - this is a variation/extension." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Show that any number greater than n^4/16 (n is positive integer) can be written in at most one way as the product of two of its divisors having difference not exceeding n.", "Solution_1": "I don't undestand. Let $p>\\frac{n^{4}}{16}$ is prime.\r\nIf $p=d_{1}*d_{2}(d_{1}\\ge d_{2})$ then $d_{1}=p,d_{2}=1,d_{1}-d_{2}=p-1>n$ if n>2.", "Solution_2": "But a question says:\"can be written in at most one way\"\r\nYou have found one way but you should prove that you can not find two or more ways to decompose that number in a such way.", "Solution_3": "I will show that if $m$ can be written in two ways as the product of two of its divisors having difference not exceeding $n$ then $m\\leq n^{4}/16$\r\n\r\nLet $m=ab=cd$ and $a>c\\geq d>b$\r\n\r\nAs $ab=cd$, we can let $a=ef, b=gh, c=eg, d=fh$ with $e>h$ and $f>g$\r\n\r\nDenote $u=a-b$ and $v=c-d$ thus $0\\leq v\\leq u\\leq n$\r\n\r\nthen $16efgh\\leq (e+h)^{2}(f+g)^{2}\\leq (e^{2}-h^{2})^{2}(f^{2}-g^{2})^{2}=(u^{2}-v^{2})^{2}\\leq n^{2}$" } { "Tag": [], "Problem": "A triangle and a square are drawn on a piece of paper. What are the least and greatest number of regions into which they divide the paper?", "Solution_1": "the least is 3 since each one has to make a new one. the most intersections between theirs lines is 3, The most times their line could intersect is 7, creating 8 regions total." } { "Tag": [], "Problem": "N\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03c5\u03bd \u03bf\u03b9 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03bf\u03b9 \u03c0\u03bf\u03c5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03c5\u03bd \u03c4\u03b7\u03bd \u03c3\u03c7\u03b5\u03c3\u03b7\r\n$x^{2}+y^{2}=(x-y)^{3}$\r\n\r\n :D :D", "Solution_1": "\u03a4\u03b7 \u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03ae \r\n$(x-y)^{2}(x-y-1)=2xy$ \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 x-y kai x-y-1 \u03c0\u03c1\u03ce\u03c4\u03bf\u03b9 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03bf\u03c5\u03c2 .\r\n\u039a\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03b3\u03b9\u03b1 x>4y \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $(x-y)^{2}>2xy$ \u03ac\u03c1\u03b1 \r\n4y>x>y \r\nH \u03b9\u03b4\u03ad\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ae..... :wink:", "Solution_2": "\u0395\u03b3\u03c1\u03b1\u03c8\u03b1 \u03bc\u03b9\u03b1 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b5\u03c3\u03c4\u03b5\u03b9\u03bb\u03b5 \u03ba\u03b1\u03bd\u03b5\u03b9\u03c2.\u0397 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03b1^2+\u03b2^2=2\u03b2^3(\u03b5\u03b8\u03b5\u03c3\u03b1 \u03c7+\u03c8=\u03b1, \u03c7-\u03c8=\u03b2) \u03b1\u03c1\u03b1 \u03bf \u03b2 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9 \u03c4\u03bf\u03bd \u03b1 \u03b1\u03c1\u03b1 \u03b5\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 \u03b2\u03c4=\u03b1 \u03c4\u03bf\u03c4\u03b5 \u03c4^2+1=2\u03b2 \u03b5\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 \u03c4=2\u03ba+1 \u03c4\u03bf\u03c4\u03b5 \u03b2=\u03ba^2+(\u03ba+1)^2 \u03ba\u03b1\u03b9 \u03b1=(2\u03ba+1)[\u03ba^2+(\u03ba+1)^2] \u03bf\u03c0\u03bf\u03c4\u03b5 \u03bf\u03b9 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b7\u03c2 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03b9 \u03c7=[\u03ba^2+(\u03ba+1)^2](\u03ba+1) \u03ba\u03b1\u03b9 \u03c8=[\u03ba^2+(\u03ba+1)^2]\u03ba.", "Solution_3": "[quote=\"\u0391\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7\u03c2 6\"] \u03bf\u03c0\u03bf\u03c4\u03b5 \u03bf\u03b9 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b7\u03c2 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03b9 \u03c7=[\u03ba^2+(\u03ba+1)^2](\u03ba+1) \u03ba\u03b1\u03b9 \u03c8=[\u03ba^2+(\u03ba+1)^2]\u03ba.[/quote]\r\n\r\n\u0391\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7 6 \u03b5\u03b9\u03c3\u03b1\u03b9 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2?\u03b5\u03b2\u03b1\u03bb\u03b1 \u03bf\u03c0\u03c5 \u03c7 \u03ba\u03b1\u03b9 y \u03b1\u03c5\u03c4\u03b5\u03c3 \u03c4\u03b9\u03c3 \u03c4\u03b9\u03bc\u03b5\u03c3 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b9\u03ba\u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 ..\u03b1\u03bb\u03bb\u03b1 \u03b4\u03b5\u03bd \u03bc\u03bf\u03c5 \u03b2\u03b3\u03b7\u03ba\u03b5 \u03c4\u03b1\u03c5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1.... \u0398\u03b1 \u03c4\u03bf \u03be\u03b1\u03bd\u03b1\u03b4\u03c9,\u03bc\u03b7\u03c0\u03c9\u03c2 \u03b5\u03c7\u03c9 \u03ba\u03b1\u03bd\u03b5\u03bd\u03b1 \u03bb\u03b1\u03b8\u03bf\u03c2 ,\u03b1\u03bb\u03bb\u03b1 \u03b4\u03b5\u03c2 \u03c4\u03bf \u03ba\u03b1\u03b9 \u03b5\u03c3\u03c5....\r\nSilouan,\u03b1\u03c5\u03c4\u03b7 \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03b1 \u03b4\u03b5\u03bd \u03c0\u03bf\u03bb\u03c5\u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1 \u03c4\u03b7\u03bd \u03ba\u03b5\u03bd\u03c4\u03c1\u03b9\u03ba\u03b7 \u03b9\u03b4\u03b5\u03b1..\u039c\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5\u03b9\u03c2 \u03bb\u03b9\u03b3\u03bf \u03c0\u03b9\u03bf \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03b1?\r\n\u039c\u03b5 \u03c4\u03b7\u03bd \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b4\u03b5\u03bd \u03b5\u03c7\u03c9 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03b5\u03b9 \u03c0\u03b5\u03c1\u03b5\u03c4\u03b5\u03c1\u03c9 \u03b1\u03c0\u03bb\u03b1 \u03bc\u03bf\u03bd\u03bf \u03bc\u03b9\u03b1 \u03c6\u03bf\u03c1\u03b1 \u03b3\u03b9 \u03b1\u03c5\u03c4\u03bf +\u03bb\u03b5\u03c9 \u03bf\u03c4\u03b9 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03c7\u03c9 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03c9 \u03bb\u03b1\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b9\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03bc\u03bf\u03c5...\r\n\u03a0\u03b1\u03bd\u03c4\u03c9\u03c2 \u03b5\u03b9\u03c7\u03b1 \u03b2\u03c1\u03b5\u03b9[hide]$(x,y)=(k^{2}+1)(k+1)/4,(k^{2}+1)(k-1)/4$kai $-(k^{2}+1)(k+1)/4,-(k^{2}+1)(k-1)/4$\n\u0398\u03b1 \u03c4\u03bf \u03be\u03b1\u03bd\u03b1\u03b4\u03c9...[/hide]\r\n :D :D\r\n\r\nEdit:sorry,\u03c0\u03b1\u03c1\u03b1\u03bb\u03b5\u03b9\u03c8\u03b7 \u03bc\u03bf\u03c5...\u03ba=\u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03bf\u03c2....", "Solution_4": "\u03a1\u03b1\u03bb\u03bb\u03b7 \u03c4\u03b9\u03c2 \u03b9\u03b4\u03b9\u03b5\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03b2\u03c1\u03b7\u03ba\u03b1\u03bc\u03b5 (x,y)=...... A\u03c0\u03bb\u03b1 \u03b5\u03b3\u03c9 \u03c0\u03b1\u03c1\u03b5\u03bb\u03b7\u03c8\u03b1 \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c8\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 (-x,-y) \u03c0\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03b5\u03c2.\u0395\u03be\u03b1\u03bb\u03bb\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03b7 \u03c4\u03b1\u03c5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1 \u03b5\u03c0\u03b1\u03bb\u03b7\u03b8\u03b5\u03c5\u03b5\u03c4\u03b1\u03b9. :wink:", "Solution_5": "\u039d\u03b1\u03b9,\u03b5\u03c7\u03b5\u03b9\u03c2 \u03b4\u03b9\u03ba\u03b9\u03bf \u03bf\u03b9 \u03b9\u03b4\u03b9\u03b5\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9,\u03b1\u03c0\u03bb\u03b1 \u03b5\u03b3\u03c9 \u03b5\u03ba\u03b1\u03bd\u03b1 \u03b5\u03bd\u03b1 \u03bb\u03b1\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b9\u03c2 \u03c0\u03c1\u03b1\u03be\u03b5\u03b9\u03c2 \u03bf\u03c4\u03b1\u03bd \u03c4\u03bf \u03b5\u03bb\u03b5\u03b3\u03c7\u03b1... :o \r\n\u0395\u03b3\u03c9 \u03b5\u03b9\u03c0\u03b1 \u03bf\u03c4\u03b9 \u03c4\u03bf \u03b1 \u03bc\u03b5\u03bb\u03bf\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf \u03ba\u03b1\u03b9 \u03b1\u03c1\u03b1 $x>y$,\u03b5\u03b8\u03b5\u03c3\u03b1 $x=y+a$ \u03bc\u03b5 $a>0$ k\u03b1\u03b9 \u03bc\u03b5\u03c4\u03b1 \u03c0\u03b7\u03c1\u03b1 \u03c4\u03c1\u03b9\u03c9\u03bd\u03c5\u03bc\u03bf \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c7.\u0397 \u03b4\u03b9\u03b1\u03ba\u03c1\u03b9\u03bd\u03bf\u03c5\u03c3\u03b1 \u03b4\u03b9\u03bd\u03b5\u03b9 $a=(k^{2}+1)/2$ \u03bc\u03b5 \u03ba \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03bf \u03ba\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b9\u03ba\u03b1 \u03c0\u03c1\u03bf\u03ba\u03c5\u03c0\u03c4\u03bf\u03c5\u03bd \u03bf\u03b9 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03b2\u03b3\u03b1\u03b9\u03bd\u03bf\u03c5\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9...\r\n\r\n :D :D", "Solution_6": "\u0398\u03b1 \u03bc\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c4\u03c1\u03b5\u03c8\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b5\u03b9\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03b3\u03c9 \u03bc\u03b9\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b1\u03b3\u03b7 \u03c4\u03b7\u03c2 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7\u03c2 \u03c4\u03bf\u03c5 \u03a1\u03b1\u03bb\u03bb\u03b7.(\u03b4\u03b5\u03bd \u03b5\u03c7\u03c9 \u03b2\u03c1\u03b5\u03b9 \u03c4\u03b9\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03b1\u03c0\u03bb\u03b1 \u03c0\u03c1\u03bf\u03b5\u03ba\u03c5\u03c8\u03b5). \u039d\u03b1 \u03bb\u03c5\u03b8\u03b5\u03b9 \u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03c7^2+y^2+z^2=(x-y+z)^3 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03bf\u03c5\u03c2.\u0391\u03bc\u03b2\u03c1\u03bf\u03c3\u03b9\u03b5,\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd,\u03a1\u03b1\u03bb\u03bb\u03b7......\u03ba\u03b1\u03b9 \u03bf\u03bb\u03bf\u03b9 \u03c3\u03b1\u03c2 \u03c0\u03b5\u03c1\u03b9\u03bc\u03b5\u03bd\u03c9 \u03c4\u03b9\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03b7 \u03c4\u03b9\u03c2 \u03b9\u03b4\u03b5\u03b5\u03c2 \u03c3\u03b1\u03c2. :wink:", "Solution_7": "\u03a6\u03b1\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bf\u03c4\u03b9 \u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c4\u03b5\u03b9\u03bd\u03b1 \u03b2\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03bc\u03b9\u03b1 \u03b1\u03ba\u03bf\u03bc\u03b7 \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03b7 (\u03c4\u03b7\u03bd \u03bc\u03b5\u03c4\u03b5\u03b2\u03bb\u03b7\u03c4\u03b7 z) \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03b1\u03b6\u03b5\u03b9 \u03c0\u03bf\u03bb\u03bb\u03b5\u03c2 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03b9\u03b5\u03c2 \u03c3\u03b5 \u03c3\u03c7\u03b5\u03c3\u03b7 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03b7\u03bd \u03c4\u03bf\u03c5 \u03a1\u03b1\u03bb\u03bb\u03b7.K\u03b1\u03bc\u03b9\u03b1 \u03b9\u03b4\u03b5\u03b1?", "Solution_8": "\u0391\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7 6,\u03ba\u03b1\u03bb\u03b7 \u03c3\u03b1\u03bd \u03c3\u03ba\u03b5\u03c8\u03b7,\u03b1\u03bb\u03bb\u03b1 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03bc\u03b5\u03c4\u03b5\u03b2\u03bb\u03b7\u03c4\u03b5\u03c2 \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03bc\u03bf\u03bd\u03bf \u03c3\u03c7\u03b5\u03c3\u03b7....!!!!! :maybe: \r\n\u039c\u03b7\u03c0\u03c9\u03c2 \u03b1\u03bd \u03c7\u03c1\u03b5\u03b9\u03c3\u03b7\u03bc\u03bf\u03c0\u03bf\u03b9\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c3\u03c7\u03b5\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b5\u03b9\u03c7\u03b5\u03c2 \u03c0\u03b5\u03b9 \u03c3\u03b5 \u03b5\u03bd\u03b1 \u03b1\u03bb\u03bb\u03bf \u03c4\u03bf\u03c0\u03b9\u03ba...$3(x^{3}+y^{3}+z^{3})\\geq (x+y+z)(x^{2}+y^{2}+z^{2})$?????\u03bf\u03c0\u03bf\u03c4\u03b5 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03b8\u03b9\u03c3\u03c4\u03c9\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf $x^{2}+y^{2}+z^{2}=(x-y+z)^{3}$ \u03c0\u03b1\u03b9\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 <\u03ba\u03b1\u03c4\u03b9> \u03c4\u03c1\u03b9\u03c4\u03bf\u03c5 \u03b2\u03b1\u03b8\u03bc\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03c5\u03c4\u03b5\u03c1\u03bf \u03b1\u03c0\u03bf <\u03ba\u03b1\u03c4\u03b9> \u03bc\u03b5\u03b3\u03b1\u03bb\u03c5\u03c4\u03b5\u03c1\u03bf\u03c5\u03c5 \u03b2\u03b1\u03b8\u03bc\u03bf\u03c5(4\u03bf\u03c5),\u03bf\u03c0\u03bf\u03c4\u03b5 \u03c4\u03b1 x,y,z \u03c6\u03c1\u03b1\u03c3\u03c3\u03bf\u03bd\u03c4\u03b1\u03b9.\u0391\u03c5\u03c4\u03bf \u03bc\u03bf\u03bd\u03bf \u03c3\u03b1\u03bd \u03b9\u03b4\u03b5\u03b1....\u0394\u03b5\u03bd \u03c4\u03bf \u03b5\u03c7\u03c9 \u03b4\u03bf\u03c5\u03bb\u03b5\u03c8\u03b5\u03b9 \u03b1\u03ba\u03bf\u03bc\u03b7...\r\n\r\n :D :D", "Solution_9": "\u039a\u03b1\u03bb\u03b7 \u03b7 \u03b9\u03b4\u03b5\u03b1 \u03c3\u03bf\u03c5 \u03a1\u03b1\u03bb\u03bb\u03b7.\u03a8\u03b1\u03be\u03c4\u03bf \u03bf\u03bc\u03c9\u03c2. :) \u03a0\u03b5\u03c1\u03b9\u03bc\u03ad\u03bd\u03c9 \u03c4\u03b7\u03bd \u03b5\u03c0\u03b9\u03bb\u03c5\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7\u03c2.\u0391\u03bd \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c4\u03b7\u03bd \u03b5\u03bb\u03c5\u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9 \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03c4\u03bf \u03c0\u03b5\u03b9. :wink:" } { "Tag": [ "geometry", "perimeter" ], "Problem": "Two congruent squares are placed so as to form an 8-point star with a regular octagon as their common area. If a side of one of the squares is of length 10 + (5 sqrt 2), what is the perimeter of the octagon in units?\r\n\r\nI don't understand the question, but the answer is\r\n[hide]\n40 sqrt 2\n[/hide]", "Solution_1": "Imagine an {8/4} star.\r\n[url=http://en.wikipedia.org/wiki/File:Regular_Star_Polygons.jpg]Look at the second figure in row 8[/url]\r\n\r\nThe isosceles leg portion of the square (jutting out of the octagon) is rt(2) smaller, but there are 2 of them, so the total length is rt(2) larger.\r\n\r\nThus we are looking at x(1+rt(2))=10+5rt(2), for which x=5rt(2).\r\n\r\nThere are eight sides so 8x=40rt(2), our answer.", "Solution_2": "So imagine\r\n[asy]draw((0,1)--(1,1)--(1,0)--(0,0)..cycle);\ndraw((.5,-.2)--(1.2,.5)--(.5,1.2)--(-.2,.5)..cycle);[/asy]\r\n\r\nwith the octagon in the middle being regular." } { "Tag": [ "inequalities", "function", "inequalities proposed" ], "Problem": "Let be $ a,b,c \\in \\mathbb{R}_+^*$ Prove that: \r\n\r\n1)$ a\\sqrt{b+c}+b\\sqrt{c+a}+c\\sqrt{a+b}\\le \\sqrt{2(a+b+c)(ab+bc+ca)}$ ;\r\n\r\n2)$ \\sum_{cyc}\\frac{a}{\\sqrt{b+c}}\\ge (a+b+c)\\sqrt{\\frac{a+b+c}{2(ab+bc+ca)}}$", "Solution_1": "consider the concave function $ f$ , such that $ f(x) \\equal{} \\sqrt {x}$\r\n\r\n$ af(b \\plus{} c) \\plus{} bf(c \\plus{} a) \\plus{} cf(a \\plus{} b) \\le (a \\plus{} b \\plus{} c)f \\bigg(\\frac {2(ab \\plus{} bc \\plus{} ca)}{a \\plus{} b \\plus{} c} \\bigg)$\r\n\r\n$ a \\sqrt {b \\plus{} c} \\plus{} b \\sqrt {c \\plus{} a} \\plus{} c \\sqrt {a \\plus{} b} \\le \\sqrt {2(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}$", "Solution_2": "[quote=\"alex2008\"]Let be $ a,b,c \\in \\mathbb{R}_ + ^*$ Prove that: \n\n1)$ a\\sqrt {b + c} + b\\sqrt {c + a} + c\\sqrt {a + b}\\le \\sqrt {2(a + b + c)(ab + bc + ca)}$ ;\n\n2)$ \\sum_{cyc}\\frac {a}{\\sqrt {b + c}}\\ge (a + b + c)\\sqrt {\\frac {a + b + c}{2(ab + bc + ca)}}$[/quote]\r\nBy Cauchy-Schwarz inequality ,we have:\r\n1)$ LHS \\le \\sqrt{(a+b+c)(a(b+c)+b(c+a)+c(a+b))} =RHS$\r\n2)$ LHS \\ge \\frac{(a+b+c)^2}{a\\sqrt{b+c}+b\\sqrt{c+a}+c\\sqrt{a+b}} \\ge RHS$\r\nThen,we have Q.E.D", "Solution_3": "we can use jensen inequality , the convex function $ f(x) = \\frac {1}{\\sqrt x}$\r\n\r\n$ \\sum_{cyc} af(b + c) \\geq (a + b + c)\\sqrt {\\frac {a + b + c}{2(ab + bc + ca)}}$" } { "Tag": [ "Alcumus", "videos", "algebra", "binomial theorem", "Support" ], "Problem": "Hey guys,\r\n I'm just wondering what do I do!\r\nI have been using Alcumus for quite sometime.\r\nI didn't \"speed\" through the first sixty questions or anything, but what happened was:\r\nI was going on with questions until it came back to one category that it said I had \"mastered\" just because I got all of those questions right during the first 60 questions, fine. Those were easy. But now I am getting towards the stuff that I do not have access to, and naturally, I am getting them wrong. So my question is:\r\nHow do I move forward and pass my current lessons if I don't get anymore questions from them.\r\nAND, won't I always be in a rut? If I am missing those new questions now, won't they eventually count towards my score (if I somehow get this fixed) and mess me up then?\r\n\r\nThanks in advanced.", "Solution_1": "you're confusing me...\r\ni think i know what you mean\r\nyou should maybe review the videos if you still have them\r\nor study the types of problems/learn them if you're getting them wrong???\r\ndon't know what you mean so i hope this helps???\r\ndon't really know what you're asking.... :arrow: :?:", "Solution_2": "Basically, I've gotten to the point where it is asking questions that have not been taught to me yet.\r\nFor example, a bunch of Binomial Theorem questions have been thrown my way, but none of those videos are available to me. And it's gotten to the point where I am no longer being asked the questions for things I've learned. For example, If it only asks me about Binomial Theorem, but I haven't learned it, how do I progress (answer them correctly) if I never learn how to!", "Solution_3": "hmmm....\r\nit seems to me (through what i have read on this forum) that alcumus only gives you videos once you have answered a few of those questions correctly\r\nthis doesn't seem to work out that well.....\r\n\r\ni think alcumus should provide video lessons as soon as the first problem of that category is given\r\nex: as soon as a binomial theorem question is asked, the video lesson should be availiable\r\n\r\nis it possible for someone to program alcumus so that this can be done?", "Solution_4": "You are given access to videos just as soon as you are ready to work on problems from the associated topics. You do not have to answer any particular number of those problems before being granted access to videos.", "Solution_5": "I tink taht we shood be given the video once we solve a problem of taht lesson. we can understand things better. also, we learn, then we do the problems, isnt that wat the book is doing, FIRST explaining, then giving u problems? I believe that we shood get the videos as soon as we get the problem corresponding to the video. \r\nAt least, i learn better this way......\r\n\r\nI like the videos too. :)", "Solution_6": "You can certainly choose not to watch the videos until you get a problem in the topic correct. That's the benefits of technology, you get to choose!", "Solution_7": "but, dont we haveta pass the topic BEFORE we get the videos??? \r\nI haveta first solve some prblms correct, THEN i can see the videos.....\r\nNow im confused...... :huh:" } { "Tag": [ "vector" ], "Problem": "Find an equation for the plane through the point (4,1,3) and perpendicular to the line x=2+5t, y=1-3t, z=2t.\r\n\r\nThe answer is 5x-3y+2z=23 but i'm not sure how to get this... I keep coming up with -11x-11y+6z=-29.", "Solution_1": "Rather than me write out a solution, which may or may not help you, I think it would be more beneficial if you put up your solution and then we try to find your mistake.", "Solution_2": "hello, make the ansatz $ (\\vec{x}\\minus{}\\vec{x_P})\\circ \\vec{n}\\equal{}0$ where $ \\vec{n}$ is the normal vector and $ \\circ$ stands for the dot product.\r\nSonnhard." } { "Tag": [ "LaTeX", "USAMTS", "geometry" ], "Problem": "How do you draw geometric figures using LaTeX? For example, if I'm doing the USAMTS, how would I send in a copy of my figure (or can I just describe what lines, triangles, etc. I drew on the original figure provided for the problem)? What program could you use to draw it if not LaTeX?", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217[/url]\r\n\r\nI recommend the geometry program Cinderella ([url]http://cinderella.de/tiki-index.php[/url]), which feels a lot like Geometer's Sketchpad, has a free educational version and can export directly to PNG or PDF.", "Solution_2": "Is the educational version the demo version? If it is, how will they make it inconvient for you unless you but the full version? If I can make do with that program I would not need to buy one.", "Solution_3": "This might be a little off topic, but what is a good Drawing program that is easy-to-use and will last me a long time?\r\n\r\nI am thinking about getting a program for drawing on usamts, but I want some extra reach with my dollar. Is Cinderella the best choice?", "Solution_4": "I use geogebra. It might take a little bit of getting used to, though (but that applies to all programs like this)." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove this inequality with $5n \\geq m \\geq 0$ \r\n$\\sum \\frac{a}{\\sqrt{mb^{2}+mc^{2}-na^{2}}}\\geq \\frac{3}{\\sqrt{2m-n}}$", "Solution_1": "no one?????????\r\nis there any problem in my inequality?" } { "Tag": [ "search", "real analysis", "real analysis unsolved" ], "Problem": "f(x) is continuous on [a,b], f(a)0}\\frac{f(x\\plus{}t)\\minus{}f(x\\minus{}t)}{t}\\equal{}g(x)\\]\r\nexist and is not infinity.\r\nProve that there exists c in [a,b] s.t. g(c)>=0.", "Solution_1": "$ lim_{t \\minus{} > 0}\\frac {f(x \\plus{} t) \\minus{} f(x \\minus{} t)}{t} \\equal{} lim_{t \\minus{} > 0}(\\frac {f(x \\plus{} t) \\minus{} f(x)}{t} \\minus{} (\\frac {f(x \\minus{} t) \\minus{} f(x)}{t}) \\equal{} lim_{t \\minus{} > 0}\\frac {f(x \\plus{} t) \\minus{} f(x)}{t} \\plus{} lim_{t \\minus{} > 0}\\frac {f(x \\minus{} t) \\minus{} f(x)}{ \\minus{} t} \\equal{} 2 f'(x)$\r\n\r\nso $ 2 f'(x) \\equal{} g(x)$\r\n\r\n1* if exist x,y in [a,b] s.t. f(x)=f(y) using Rolle's theorem we have that exist c in (x,y) such that f'(c)=0 \r\nso exist c in [a,b] such that g(c)=2f'(c)>=0\r\n\r\n2* if don't exist x,y in [a,b] s.t. f(x)=f(y) that mean ('cause f is continuous) that f is monotonious and because f(a)=0 so exist c in [a,b] such that g(c)=2f'(x)>=0\r\n\r\n\r\n1*,2* -> exist c in [a,b] s.t. g(c)>=0 q.e.d \r\n\r\n\r\nsorry for my bad english :roll:", "Solution_2": "The problem with the above solution is that we don't know if $ f'$ exists.", "Solution_3": "Compare to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1870450144&t=155036]this[/url]" } { "Tag": [ "function", "limit", "algebra unsolved", "algebra" ], "Problem": "Find all continuous functions $\\large f\\colon\\mathbb{R}\\to\\mathbb{R}$, such that\r\n $f (x+y+f(xy))=xy+f(x+y)$\r\nfor all $\\large x,y \\in R.$", "Solution_1": "[hide=\"Check it!\"]Given $y=0$ then $f(x+f(0))=f(x)$ \nSo $f(x)$ is a periodic funtion.\n$y=\\frac{x}{x-1}$\n$t=\\frac{x^{2}}{x-1}$\n$x+y=xy=t$\n$f(t+f(t))=t+f(t)$\nThen if $f(t)\\neq-t$ for $x$ is sufficiently great, $f(x)=x$ holds onwards. Then $f(x)=x$ for all $x \\in\\mathbb{R}$\nSo the equation has two roots $f(x)=x$ and $f(x)=-x$[/hide]", "Solution_2": "It give only one solution. $f(x)=-x$ is solution too.", "Solution_3": "Thanks Rust, I edited my previous post. Can you check it again?", "Solution_4": "Your solution is not correct and complete.", "Solution_5": "Can you give yours?", "Solution_6": "[b](*)[/b] $f(x+y+f(xy))=xy+f(x+y)$\r\n$a=x+y,b=xy$ gives\r\n[b](**)[/b] $f(a+f(b))=b+f(a)$ for all $a,b\\in\\mathbb{R}$ with $a^{2}\\geq 4b$.\r\n\r\n[b]1)[/b] $f$ is injective:\r\nIf $f(b)=f(c)$ for some $b,c\\in\\mathbb{R}$, \r\nthen take $a\\in\\mathbb{R}$ with $a^{2}\\geq 4b,4c$,\r\nand then we get $b+f(a)=f(a+f(b))=f(a+f(c))=c+f(a)$ and $b=c$ using (**).\r\n\r\n[b]2)[/b] $f(0)=0$ and $f(f(b))=b$ for all $b\\leq 0$:\r\n$a=b=0$ in (**) gives $f(f(0))=f(0)$ and $f(0)=0$ using 1).\r\nThen for $b\\leq 0$ taking $a=0$ in (**) gives $f(f(b))=b$.\r\n\r\n[b]3)[/b] $f(a+b)=f(a)+f(b)$ for all $a,b\\in\\mathbb{R}$ with $b\\leq 0$ and $a^{2}\\geq 4f(b)$:\r\nReplacing $b$ with $f(b)$ in (**) gives the result using 2).\r\n\r\n[b]4)[/b] $f(b+c)=f(b)+f(c)$ for all $b,c\\in\\mathbb{R},\\ b,c\\leq 0$:\r\nTake $a\\in\\mathbb{R}$ large enough with $a^{2}\\geq 4f(b+c)$, $(a+b)^{2}\\geq 4f(c)$ and $a^{2}\\geq 4f(b)$.\r\nThen 3) gives $f(a)+f(b+c)=f(a+b+c)=f(a+b)+f(c)=f(a)+f(b)+f(c)$\r\nand $f(b+c)=f(b)+f(c)$.\r\n\r\nNow $\\alpha: =-f(-1)\\not =0$ by 1) and 2). Then\r\n[b]5)[/b] $f(x)=\\alpha x$ for all $x\\leq 0$:\r\nFor all $m,n\\in\\mathbb{N}$ we have\r\n$f(-{m\\over n})={1\\over n}(\\underbrace{f(-{m\\over n})+\\cdots+f(-{m\\over n})}_{n\\text{ times}})={1\\over n}f(\\underbrace{-{m\\over n}-\\cdots-{m\\over n}}_{n\\text{ times}})={1\\over n}f(-m)\\\\\\hspace*{0.65in}={1\\over n}f(\\underbrace{-1-\\cdots-1}_{m\\text{ times}})={1\\over n}(\\underbrace{f(-1)+\\cdots+f(-1)}_{m\\text{ times}})={m\\over n}f(-1)=\\alpha (-{m\\over n})$\r\nwhich gives the result, as $f$ is continuous.\r\n\r\nAnd now\r\n[b]6)[/b] $f(x)={1\\over \\alpha}x$ for all $x\\in\\mathbb{R}$:\r\nFor $z\\in\\mathbb{R}$ we have $\\lim_{x\\to-\\infty}(x+{z\\over x})=-\\infty$.\r\nSo take $x\\in\\mathbb{R}_{-}$ small enough with $x+{z\\over x}\\leq 0$ and $x+{z\\over x}+{1\\over \\alpha}z\\leq 0$.\r\nThen taking $y={z\\over x}$ in (*) we get using 5)\r\n$f(x+{z\\over x}+f(z))=z+f(x+{z\\over x})=z+\\alpha(x+{z\\over x})=\\alpha(x+{z\\over x}+{1\\over \\alpha}z)=f(x+{z\\over x}+{1\\over \\alpha}z)$\r\nand 1) gives $f(z)={1\\over \\alpha}z$ for all $z\\in\\mathbb{R}$.\r\n\r\nThen 5) and 6) give $f(-1)=-\\alpha=-{1\\over \\alpha}$ and $\\alpha=\\pm 1$.\r\nSo the continuous solutions of (*) are $f(x)=x$ and $f(x)=-x$.\r\n(ugly proof, I know [img]http://www.emoticons.online.fr/smileys/Default/Various_Artists-doubt.gif[/img])", "Solution_7": "There are uncorrect moments.\r\n[quote=\"olorin\"][b](*)[/b] $f(x+y+f(xy))=xy+f(x+y)$\n$a=x+y,b=xy$ gives\n[b](**)[/b] $f(a+f(b))=b+f(a)$ for all $a,b\\in\\mathbb{R}$ with $a^{2}\\geq 4b$.[/quote]\nCorrect $a^{2}\\ge 4|b|$ or $-a^{2}\\le 4b\\le 4b$.\nTherefore uncorrect \n[quote]2)[/b] $f(0)=0$ and $f(f(b))=b$ for all $b\\leq 0$:\n$a=b=0$ in (**) gives $f(f(0))=f(0)$ and $f(0)=0$ using 1).\nThen for $b\\leq 0$ taking $a=0$ in (**) gives $f(f(b))=b$.[/quote]\r\nand etc.", "Solution_8": "[quote=\"Rust\"]Correct $a^{2}\\ge 4|b|$ [/quote]\r\nNo, if $a^{2}\\geq 4b$, then $x={a+\\sqrt{a^{2}-4b}\\over 2}$ and $y={a-\\sqrt{a^{2}-4b}\\over 2}$ exist in $\\mathbb{R}$ with $x+y=a$ and $xy=b$." } { "Tag": [ "LaTeX", "geometry", "geometric transformation", "reflection", "trigonometry", "trapezoid", "trig identities" ], "Problem": "Given ABCD is a regulated quadrilateral inscribed in a circle (O) (which has $AB\\cdot CD=BC\\cdot DA$).\r\nLet K, H be the midpoints of AC and BD respectedly. Prove that AK+CK=BH+DH.\r\n\r\n[color=red][Moderator edit: LaTeXed the equation. Please use the standard notation $x\\cdot y$ for the product of two numbers x and y, rather than the obsolete notation x.y (note that LaTeX has the command \\cdot for a dot like $\\cdot$).][/color]", "Solution_1": "That is equivalent to prove that $AC=BD$, which is not true (at least it doesn't seem to be )... or perhaps you mean that we have to prove $AH+CH=BK+DK$.", "Solution_2": "He's referring to a harmonic cyclic quadrilateral. It has lots of nice properties. I'm sure he meant $AH+CH=BK+DK$, as RaMlaF said.\r\n\r\nI think similar problems have been discussed before. For instance, I'll use the fact that $AC,BD$ bisect the angles $\\angle BKD,\\angle AHC$ without proof (take it as a nice problem :)). From this we can deduce that it suffices to show that the chords obtained by producing $AH,DK$ are equal, which, in turn, would be derived from $\\angle DAH=\\angle ADK\\ (*)$, so let's focus on proving $(*)$.\r\n\r\nNow I'll use another property of the harmonic quadrilaterals: the tangents through $A,C$ to the circle concur in $P\\in BD$, and the symmetric situation also holds. From here we find $K$ to be the image of $O$, the center of the circle, through the inversion of pole $P$ and power $PB^2=PD^2$, so $BDKO$ is cyclic, meaning that (look at the second equality) $\\angle DKA=\\frac{\\angle DKB}2=\\frac{\\angle DOB}2=\\angle DCB$, and since $\\angle DAK=\\angle DBC$, we find $DKA,DCB$ to be similar. In the same way we show that $AHD,ABC$ are similar, so $\\angle DAH=\\angle CAB=\\angle BDC=\\angle ADK$, which is precisely $(*)$.", "Solution_3": "Sorry for my stupid mistake. I really mean AH+CH=BK+DK :D", "Solution_4": "But Grober, can you explain to me what we should do after we have $\\angle DAH=\\angle ADK$? :?", "Solution_5": "I don't see so clearly what to do with $\\angle DAH = \\angle ADK$ ... but i have another solution (without magic properties ... :lol: just kiddin'): \r\nLet $D'$ be the reflection of $D$ with respect to $A$. Then $\\angle BAD = \\angle BCD$ and $\\frac {BA}{AD'} = \\frac{BA}{AD}=\\frac{BC}{CD}$ since $ABCD$ is an harmonic quadrilateral. Therefor $BAD'\\simeq BCD$, with $BD'=\\frac{BD.AB}{BC}$ And, since $DH=BH$ and $DA=DA'$ we have $BD'=2AH$ so $AH=\\frac{BD.AB}{2BC}$. Similarly, $CH=\\frac{BD.BC}{2AB}$. And doing the same with $BK$ and $DK$ the problem reduces to show that $BD(\\frac{AB}{BC}+\\frac{BC}{AB})=AC(\\frac{BC}{CD}+\\frac{CD}{BC}) \\iff BD(\\frac{AB}{BC}+\\frac{CD}{AD})=AC(\\frac{AB}{AD}+\\frac{CD}{BC})$ (because $ABCD$ is harmonic) $\\iff BD(AB.AD+BC.CD)=AC(AB.BC+CD.AD) \\iff \\frac{AC}{BD}=\\frac{AB.AD+BC.CD}{AB.BC+CD.AD}$. And we see that $(AB.AD+BC.CD).\\sin BCD = (AB.BC+CD.AD).\\sin CDA = 2.Area(ABCD)$. Using that, and the Extended Law of Sines with $\\sin BCD=\\frac{BD}{2R}$ and so on, we obtain the desired result.", "Solution_6": "Since $\\angle DAH=\\angle DAK$, let $P,Q$ be the intersections of $DK,AH$ with the circle. Then $DPQA$ is an isosceles trapezoid, meaning that $DK,AH$ produced form equal chords of the circle. This is what we wanted.", "Solution_7": "YEAH, I finally find another simply solutions :D \r\nUse the condition $AB\\cdot CD=AD\\cdot BC$ and the Ptoleme theorem we have the triangle DAH and BAC are similar. \r\n$DH=\\frac{2R\\cdot S(ABD)}{AC\\cdot BD}$\r\nAnd qith the same way we will get the result\r\nSorry if I did make mistakes because I must be very hurry now :D", "Solution_8": "this problem is not difficult\r\nto prove the equality you can use this\r\nthe tangent to the circle at B,D and the line AC are concurrent after you use ptoleme theorem; :D :D" } { "Tag": [], "Problem": "I found this in a lateral thinking puzzle(I currently can't find it so these aren't exact replicates, but here are the problems. It's kinda like a dummy test\r\n1. What is 14 times 2 divided by a half?\r\n2. How many seconds are in a year?\r\n3. What is the largest island in the world before Australia was discovered?\r\n4. If it takes 20 men 5 days to dig 1 hole, how many days does it take 10 men to dig 1/2 of a hole?\r\n5. Who starred with Jack Nicolson in the movie \"Who flew over the Robin's Nest\"?\r\nCan't remember any more............... :( :blush: \r\nHere are the answers:(post results here)\r\n[hide]\n1. 56\n2. 12.......there's January the second, Febuary the second and so on....\n3. Greenland...Australia is a continental landmass.\n4. You can't dig half of a hole.\n5. Such a movie was never created(It was \"Who flew over the cukoos nest)\n[/hide]\r\nHere's the scoring ranges\r\n5 right: Genius\r\n4 right: Smart Alec\r\n3 right: Average\r\n2 right: Wally\r\n1 right: Super Wally\r\n0 right: Ultra Wally\r\n\r\nNote: You DON't want a \"wally\" ranking.", "Solution_1": "yay. i'm a super wally", "Solution_2": "I got average yay.\r\n\r\nAnd btw #2 is just stupid :wink:", "Solution_3": "I have some other WALLY test questions, also from a lateral thinking puzzles book. :)\r\n\r\n1. How can you stand behind someone while he or she stands behind you?\r\n2. What looks like a horse, moves like a horse, and is as big as a horse but weighs nothing?\r\n3. Who is bigger: Mr. Bigger or Mr. Bigger's son?\r\n4. What has four fingers and a thumb but isn't a hand?\r\n5. What multiplies by division?\r\n6. What's white when it's dirty, and black when it's clean?\r\n7. What gets higher as it falls?\r\n8. Why did the overweight actor fall through the theater floor?\r\n9. How did an actor get his name up in lights in every theater in the country?\r\n10. Where would you find a square ring?\r\n11. How do you make a slow horse fast?\r\n12. Why did Sam wear a pair of pants with three large holes?\r\n13. Where do you find a no-legged dog?\r\n14. Approximately how many house bricks does it take to complete a brick house in England?\r\n15. What cheese is made backward?\r\n16. Take away my first letter; I remain the same.\r\nNow, take away my fourth letter; I remain the same.\r\nNow, take away my last letter; I remain the same.\r\nWhat am I?\r\n17. If a white man threw a black stone into the Red Sea, what would it become?\r\n18. If an atheist died in church, what would they put on his coffin?\r\n19. Who was it that went into the lion's den unarmed and came out alive?\r\n20. A man rode down the street on a horse, yet walked. How come?\r\n21. How can you eat an egg without breaking the shell?\r\n22. In China they hang many criminals, but they will not hang a man with a wooden leg. Why?\r\n23. A circular field is covered in thick snow. A black cow with white spots is in the middle. Two white cows with black spots are on the edge of the field. What time is it?\r\n24. What was the problem with the wooden car with wooden wheels and a wooden engine?\r\n\r\nOkay, that was a lot. I won't post the answers yet, but I know them if you want to ask me. :)", "Solution_4": "Is 13, \"at a concession stand\"?", "Solution_5": "[hide=\"answers?\"]1. You stand back to back\n2. The horse's shadow\n3. Mr. Bigger's son\n4. A glove\n5. A single-celled organism\n6. A blackboard\n7. Snow\n8. It was a stage he was going through\n9. He changed his name to Exit\n10. At a boxing arena\n11. Give it no food\n12. To get his feet in it\n13. Where you left it\n14. One\n15. Edam\n16. A mailman\n17. Wet\n18. A lid\n19. The lion\n20. The horse was named \"yet\"\n21. Get someone else to break the shell\n22. They hang people with a rope and a noose\n23. Wintertime\n24. It wooden go[/hide]\r\n\r\nOkay, okay, I have the book. :D", "Solution_6": "I don't have the book.\r\n\r\n[hide=\"my answers\"]1. Face him.\n2. Horse's shadow.\n3. Both of them.\n4. Blender that hasn't been cleaned out yet.\n5. Black adder.\n6. Black cat with a barrel of flour.\n7. Pothead fatally falling off a cliff.\n8. Trap door.\n9. He was overweight and he kept falling through the trap door.\n10. On a square wife. Or husband. It doesn't matter.\n11. Get off of it.\n12. All pants have three large holes.\n13. Chiropractor.\n14. The same number of bricks it takes to complete it in the USA.\n15. eseehc\n16. Pen pal that lost all his letters.\n17. Wet.\n18. Dirt.\n19. A lion.\n20. He walked to the stable, he walked into the house, etc.\n21. Have somebody else cook it, he breaks the shell not you.\n22. ???\n23. Time to bring the cows in.\n24. Answer 1: It needed a wooden road. Good luck finding that. Answer 2: It had a wooden windshield.[/hide]" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Let X,p be a metric space where for each uncountable subset A of X there is an accumulation point in X. Show that X,p is separable.", "Solution_1": "For every $k$, let $N_{k}\\subset X$ be a maximal set s.t. the distance between every two points in $N_{k}$ is at least $1/k$ ($N_{k}$ exists by Zorn's Lemma). The set $N_{k}$ has no accumulation points. Therefore, it is at most countable. Now consider the set \r\n\\[D = \\bigcup_{k}N_{k}.\\]\r\nThe set $D$ is a countable dense set in $X$. Thus $X$ is separable." } { "Tag": [ "ARML" ], "Problem": "$ (1\\minus{}\\frac{1}{3^{2}})(1\\minus{}\\frac{1}{4^{2}})(1\\minus{}\\frac{1}{5^{2}})...(1\\minus{}\\frac{1}{1991^{2}})\\equal{}\\frac{x}{1991}$.\r\nFind x.", "Solution_1": "Since $ 1\\minus{}\\frac{1}{a^2}\\equal{}1^2\\minus{}(\\frac1a)^2\\equal{}(1\\minus{}\\frac1a)(1\\plus{}\\frac1a)$, we have \\[ \\left(1\\minus{}\\frac{1}{3^2}\\right)\\left(1\\minus{}\\frac{1}{4^2}\\right)\\left(1\\minus{}\\frac{1}{5^2}\\right)\\cdots\\left(1\\minus{}\\frac{1}{1991^2}\\right)\\\\\\equal{}\\left(1\\minus{}\\frac{1}{3}\\right)\\left(1\\plus{}\\frac{1}{3}\\right)\\left(1\\minus{}\\frac{1}{4}\\right)\\left(1\\plus{}\\frac{1}{4}\\right)\\left(1\\minus{}\\frac{1}{5}\\right)\\left(1\\plus{}\\frac{1}{5}\\right)\\cdots\\left(1\\minus{}\\frac{1}{1991}\\right)\\left(1\\plus{}\\frac{1}{1991}\\right)\\\\\\equal{}\\left(\\frac23\\right)\\left(\\frac43\\right)\\left(\\frac34\\right)\\left(\\frac54\\right)\\left(\\frac45\\right)\\left(\\frac65\\right)\\cdots\\left(\\frac{1990}{1991}\\right)\\left(\\frac{1992}{1991}\\right)\\equal{}\\left(\\frac23\\right)\\left(\\frac{1992}{1991}\\right)\\\\\\equal{}\\frac{1328}{1991},\\] so $ x\\equal{}\\boxed{1328}$. :)", "Solution_2": "Slightly different approach...\r\n\r\n[hide]Well, we can write the equation as:\n\n$ \\frac{(3\\minus{}1)(3\\plus{}1)(4\\minus{}1)(4\\plus{}1)(5\\minus{}1)(5\\plus{}1)...(1991\\plus{}1)(1991\\minus{}1)}{3^2\\cdot4^2\\cdot5^2\\cdot...\\cdot1991^2}$\n\nThis simplifies to:\n\n$ \\frac{2\\cdot3\\cdot4^2\\cdot5^2\\cdot...\\cdot1990^2\\cdot1991\\cdot1992}{3^2\\cdot4^2\\cdot5^2\\cdot...\\cdot1991^2}$\n\n$ \\equal{}\\frac{2\\cdot1992}{3\\cdot1991}$\n\n$ \\equal{}\\frac{1328}{1991}$\n\nSo $ x\\equal{}1328$.[/hide]" } { "Tag": [], "Problem": "If $w^9=1$ and $w\\not=1$, then find the sum: $1+w+w^2+...+w^8$\r\n[hide=\"hint\"]You can do this in your head... also, for more information $w$ is a complex number[/hide]\n\n[hide=\"-\"]\n[hide=\"Warning... here is the solution... but still attempt the problem, I am looking to see what are the reactions by people who see this problem for the first time...DO NOT CLICK, attempt first\"]\nIf $S$ is the sum we are trying to find, $S*w=w+w^2+...w^9=w+w^2+...1=S$. Since $S=Sw, w\\not=1, S=\\boxed{0}.$[/hide][/hide]", "Solution_1": "[quote=\"#20002 pwns\"]If $w^9=1$ and $w\\not=1$, then find the sum: $1+w+w^2+...+w^8$\n[hide=\"hint\"]You can do this in your head... also, for more information $w$ is a complex number[/hide]\n\n[hide=\"-\"]\n[hide=\"Warning... here is the solution... but still attempt the problem, I am looking to see what are the reactions by people who see this problem for the first time...DO NOT CLICK, attempt first\"]\nIf $S$ is the sum we are trying to find, $S*w=w+w^2+...w^9=w+w^2+...1=S$. Since $S=Sw, w\\not=1, S=\\boxed{0}.$[/hide][/hide][/quote]\n[hide]$1+w+w^2+...w^8=\\frac{1-w^9}{1-w}=0$[/hide]" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "pyramid" ], "Problem": "On the Scholastic Aptitude Test (an American college entrance exam), students were once asked to determine the number of faces of the polyhedron obtained by gluing a regular tetrahedron to a square pyramid along one of the triangular faces. The answer expected by the test authors was 7, since the two polyhedra have 9 faces, 2 of which are moved by gluing, but a student taking the exam pointed out that this is incorrect.\r\nWhat is the correct answer?", "Solution_1": "[hide=\"This is a classic\"] When you glue the two figures together, 2 pairs of the triangular faces (\"pair\" meaning one face from the tetrahedron and one lateral face from the square pyramid) align in such a way as to create 2 uniform face, due to the angles involved , i.e., the internal angles of the two figures add to 180 degrees. I figured this out by actually using paper to build the figures, I think. [/hide]" } { "Tag": [ "probability" ], "Problem": "If 2 standard 6-sided dice are rolled, then what is the probability that at least 1 of them shows a prime number and that their sum is also prime?", "Solution_1": "[quote=\"banned UBER\"]If 2 standard 6-sided dice are rolled, then what is the probability that at least 1 of them shows a prime number and that their sum is also prime?[/quote]\r\nMake a chart. Cross out all the numbers that aren't prime.You're left with \r\n6,5,11,\r\n5,6,11,\r\n6,1,7,\r\n5,2,7,\r\n4,3,7,\r\n3,4,7,\r\n2,5,7,\r\n1,6,7,\r\n4,1,5,\r\n3,2,5,\r\n2,3,5,\r\n1,4,5,\r\n2,1,3,\r\n1,2,3,\r\n1,1,2,\r\nCrossing out (6,1,7), (1,6,7), (4,1,5), (1,4,5), (1,1,2), you're left with 10/36, 0r 5/18 chance of doing what banned UBER said" } { "Tag": [ "modular arithmetic", "geometry", "inequalities", "algebra", "polynomial", "power of a point", "AMC" ], "Problem": "Discuss the solutions here and request anything that you would like to be explained.\r\n\r\nI didn't do this one but as a moderator, it's my job to set these stuff up to prevent multiple threads.", "Solution_1": "Thanks.\r\n\r\n[hide=\"24\"]\n$\\left(a-\\frac{b}{2}\\right)^2+\\frac{3}{4}\\left(b-\\frac{2c}{3}\\right)^2+\\frac{2}{3}\\left(c-\\frac{3d}{4}\\right)^2+\\frac{5}{8}\\left(d-\\frac{4}{5}\\right)^2=0$ :) :)\n[/hide]", "Solution_2": "That's cool but I still can't figure out how to do 25.", "Solution_3": "I'm having a boredom attack during my MOP free time right now, so I'll just post bunches of solutions... I'll have to get back to a couple of these (for diagrams, and I can't remember how I did the uber problem! I can brute force it, but I remember having a nicer way...)\r\n\r\n[hide=\"1\"]2007+xy = 2007 + (4)(5) = 2027. (C)\n[/hide]\n[hide=\"2\"]-13-12-11-...-0+1+2+...+13+14. Everything cancels but 14. (B)\n[/hide]\n[hide=\"3\"] The minute hand is 1/3 of the way around, or 120 degrees. Between the 3 and the 4, there are 30 degrees, and the hour hand is 1/3 of the way between. This is 90+30/3 = 100 degrees past the hour: 120-100=20. (C)\n[/hide]\n[hide=\"4\"]$3+3^2+3^3+3^4+3^5 = 363$. (E)\n[/hide]\n[hide=\"5\"]2007 = 2 mod 5. By Euler's theorem, $2^4 \\equiv 1 \\pmod 5,$ so $2^{2007} \\equiv 2^3 \\equiv 3 \\pmod 5$. (D)\n[/hide]\n[hide=\"6\"]Simply count the ones that don't have that letter and subtract from the total: $6^3-5^3 = 91$. (B)\n[/hide]\n[hide=\"7\"]If you put everything over a common denominator of $abcdef$, you get the sum of six odd number in the numerator (which is even) over the product of them (which is odd) and thus cannot be 1. (A)\n[/hide]\n[hide=\"8\"]Just do synthetic division and you'll notice the pattern... 37*73 - 37*73 = 0. (B)\n[/hide]\n[hide=\"9\"]Just using the 3-4-5 triangle, it can be at most 12. $ab$ must be divisible by four since squares are $0$ or $1 \\pmod 4$. Also, $ab$ must be divisible by 3 since squares are $0$ or $1 \\pmod 3$. So the answer is 3*4=12. (B)\n[/hide]\n[hide=\"10\"]\nHaha, this was funny. Our number has to be more than one hundred digits (obviously). And it must be divisible by 3, so the number of digits must be divisible by 3 (since it's all 1s). 300 is the only answer that fits. (C).\n[/hide]\n[hide=\"11\"]\nDo this a different way: assume you have 7 students and you want to pick where to put five such that none are next to each other. There are ${8 \\choose 5} = 56$ ways to do this.\n[/hide]\n[hide=\"12\"]\nPlug in $x_1=0$, and use the ball-and-urn formula to get $165$ such solutions. Plug in $x_1=1$ and use ball-and-urn to get $9$ solutions; add to get 174. (D)\n[/hide]\n[hide=\"13\"]I just listed out the possibilities... I'll get back to this one.\n[/hide]\n[hide=\"14\"]Suppose the diagonal of length 6 divides the other into lengths $x$ and $y$. So the area is $\\frac{6x + 6y}{2} = 3(x+y) = 21$. (A)\n[/hide]\n[hide=\"15\"]$f(x) = ax+10$ from the first equation. From the second, we can solve for $a$ to get $a=2$. So we end up with a triangle with vertices $(-2,0)$, $(0,0)$, and $(0,10)$. 10*2/2 = 10. (C)\n[/hide]\n[hide=\"16\"]Note to self: don't make this harder than it needs to be! There are 18! ways to arrange them, and pretend there is a divider every 2. The sum of the sum of the digits is 18! mod 9, which is 0. Since the sum obviously can't be 0, it must be 9. (E)\n[/hide]\n[hide=\"17\"]\nPlug in x=1 and get 5f(1) = 4, so f(1) = 4/5. (A)\n[/hide]\n[hide=\"18\"]\nI can't explain this well without a diagram, but I'll just say you use power of a point to find the length of the tangents from A to the incircle and use the Pythagorean Theorem. (B)\n[/hide]\nOk, I'm going to stop trying to explain these geometry problems without diagrams. Someone else can do that, or I'll make diagrams when I get home...\n[hide=\"20\"]Crap, I forgot how I did this one... getting back to it...\n[/hide]\n[hide=\"21\"]AM-GM says that it will have it's minimum when $a^4$, $b^4$, and $c^4$ are equal. And lo-and-behold, (2,2,2) satisfies the conditions! 3*2^4 = 48. (B)\n[/hide]\nAnother geometry problem with no diagram... skip!\n23's already been posted...\nalong with 24....\n[hide=\"25\"]\n$\\frac{a}{b+c} = \\frac{(a+b+c)-(b+c)}{b+c} = \\frac{7}{b+c} - 1.$ Add them cyclicly to get $7 \\left(\\frac{1}{b+c} + \\frac{1}{a+c} + \\frac{1}{a+b}\\right) - 3 = \\frac{19}{10}.$ (A).\n[/hide]\r\ndone!", "Solution_4": "[hide=\"Solution to 21\"]\n\nThis is an excellent problem. I use Cauchy's Inequality here but I don't know if this was necessary because normally, AMC problems do not use Cauchy.\n\nBy Cauchy:\n\n$(1^2+1^2+1^2)(a^2 + b^2 + c^2) \\geq (a + b + c)^2$\n\nSo, the minimum value of $a^2 + b^2 + c^2 = 12$ after doing some algebra and using $a+b+c = 6$.\n\nNow, note that:\n\n$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)$\n\nSince we minimized $a^2+b^2+c^2,$ we can now see that $ab+ac+bc = 12$.\n\nNow, observe these two squares:\n\n$(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$\n\n$(ab+ac+bc)^2 = a^2 b^2 + a^2 c^2 + b^2 c^2 + 2abc(a+b+c)$\n\nUsing what we know, we can get $a^2 b^2 + a^2 c^2 + b^2 c^2 = 48$. So, $a^4 + b^4 + c^4 = 12^2 - 2(48) = \\fbox{48}$. [/hide]\r\n\r\nAs I said, nice job on making this problem. By the way, I'm pretty sure that #24 is not original. I think this is in Mathematical Circles (Russain Experience).", "Solution_5": "[quote=\"Silverfalcon\"]By the way, I'm pretty sure that #24 is not original. I think this is in Mathematical Circles (Russain Experience).[/quote]\r\n\r\n:D Yeah, I couldn't think of any more hard problems.", "Solution_6": "[quote=\"Silverfalcon\"][hide=\"Solution to 21\"]\n\nThis is an excellent problem. I use Cauchy's Inequality here but I don't know if this was necessary because normally, AMC problems do not use Cauchy.\n\nBy Cauchy:\n\n$(1^2+1^2+1^2)(a^2 + b^2 + c^2) \\geq (a + b + c)^2$\n\nSo, the minimum value of $a^2 + b^2 + c^2 = 12$ after doing some algebra and using $a+b+c = 6$.\n\nNow, note that:\n\n$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)$\n\nSince we minimized $a^2+b^2+c^2,$ we can now see that $ab+ac+bc = 12$.\n\nNow, observe these two squares:\n\n$(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$\n\n$(ab+ac+bc)^2 = a^2 b^2 + a^2 c^2 + b^2 c^2 + 2abc(a+b+c)$\n\nUsing what we know, we can get $a^2 b^2 + a^2 c^2 + b^2 c^2 = 48$. So, $a^4 + b^4 + c^4 = 12^2 - 2(48) = \\fbox{48}$. [/hide]\n\nAs I said, nice job on making this problem. By the way, I'm pretty sure that #24 is not original. I think this is in Mathematical Circles (Russain Experience).[/quote]\r\n\r\nI solved that faster than I solved #3 just by plugging in 2 for all the variables :rotfl:", "Solution_7": "[hide=\"23\"]\n$f(x)=\\frac{x+1}{x}$\n\n$\\iff x\\cdot f(x)-x-1=0$\n\n$g(x): =x\\cdot f(x)-x-1$\n\nThe roots of $g(x)$ are $1,2,\\ldots,2006$. We can write\n\n$C(x-1)(x-2)\\cdots(x-2006)=x\\cdot f(x)-x-1$\n\nSubstituting $x=0$, we get\n\n$C(2006!)=-1$\n\n$\\therefore C=\\frac{-1}{2006!}$\n\n$\\Rightarrow x\\cdot f(x)-x-1=-\\frac{1}{2006!}(x-1)(x-2)\\cdots(x-2006)$\n\nAt $x=2007$, $2007f(2007)-2008=-\\frac{1}{2006!}(2006)(2005)\\cdots(1)$\n\n$f(2007)=1$\n\nOh dang I screwed this one up too...\n\nThe right answer is not even on there :(\n[/hide]", "Solution_8": "[quote=\"SplashD\"][quote=\"Silverfalcon\"][hide=\"Solution to 21\"]\n\nThis is an excellent problem. I use Cauchy's Inequality here but I don't know if this was necessary because normally, AMC problems do not use Cauchy.\n\nBy Cauchy:\n\n$(1^2+1^2+1^2)(a^2 + b^2 + c^2) \\geq (a + b + c)^2$\n\nSo, the minimum value of $a^2 + b^2 + c^2 = 12$ after doing some algebra and using $a+b+c = 6$.\n\nNow, note that:\n\n$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)$\n\nSince we minimized $a^2+b^2+c^2,$ we can now see that $ab+ac+bc = 12$.\n\nNow, observe these two squares:\n\n$(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$\n\n$(ab+ac+bc)^2 = a^2 b^2 + a^2 c^2 + b^2 c^2 + 2abc(a+b+c)$\n\nUsing what we know, we can get $a^2 b^2 + a^2 c^2 + b^2 c^2 = 48$. So, $a^4 + b^4 + c^4 = 12^2 - 2(48) = \\fbox{48}$. [/hide]\n\nAs I said, nice job on making this problem. By the way, I'm pretty sure that #24 is not original. I think this is in Mathematical Circles (Russain Experience).[/quote]\n\nI solved that faster than I solved #3 just by plugging in 2 for all the variables :rotfl:[/quote]\r\n\r\nOops, I made that one too brute-forcable :(", "Solution_9": "[quote=\"chess64\"][hide=\"23\"]\n$f(x)=\\frac{x+1}{x}$\n\n$\\iff x\\cdot f(x)-x-1=0$\n\n$g(x): =x\\cdot f(x)-x-1$\n\nThe roots of $g(x)$ are $1,2,\\ldots,2006$. We can write\n\n$C(x-1)(x-2)\\cdots(x-2006)=x\\cdot f(x)-x-1$\n\nSubstituting $x=0$, we get\n\n$C(2006!)=-1$\n\n$\\therefore C=\\frac{-1}{2006!}$\n\n$\\Rightarrow x\\cdot f(x)-x-1=-\\frac{1}{2006!}(x-1)(x-2)\\cdots(x-2006)$\n\nAt $x=2007$, $2007f(2007)-2008=-\\frac{1}{2006!}(2006)(2005)\\cdots(1)$\n\n$f(2007)=1$\n\nOh dang I screwed this one up too...\n\nThe right answer is not even on there :(\n[/hide][/quote]\r\n\r\nHaha, I was wondering about that... :)", "Solution_10": "[quote=\"chess64\"]\n$C(x-1)(x-2)\\cdots(x-2006)=x\\cdot f(x)-x-1$\n\nSubstituting $x=0$, we get\n\n$C(2006!)=-1$\n[/quote]\r\nHow did you get this? $f(0)$ is undefined.", "Solution_11": "$f(k)=1+1/k$ only for $k=1,2,\\ldots,2006$. It's a polynomial that just happens to satisfy that equation for those specific values. But since it's a polynomial it's defined on all reals.\r\n\r\nEDIT: And if you're asking how I got that $0\\cdot f(0)-0-1=-1$, it's because of the 0 multiplied by $f(0)$.", "Solution_12": "Did you guys like my $\\textit{uber}$ problem? :D", "Solution_13": "That was the problem I was unable to solve as well as 23. I don't know what other question I got wrong.\r\n\r\nSilverFalcon, for #21, couldn't u just use AM-GM, since a^4+b^4+c^4 is always positive as well as abc?", "Solution_14": "Hmm, turns out I wasn't using the latest version of the test ... none of the choices for Question 3 on the version I downloaded was 20. No wonder I couldn't answer that one :P", "Solution_15": "[quote=\"nr1337\"]\n[hide=\"25\"]\n$\\frac{a}{b+c} = \\frac{(a+b+c)-(b+c)}{b+c} = \\frac{7}{b+c} - 1.$ Add them cyclicly to get $7 \\left(\\frac{1}{b+c} + \\frac{1}{a+c} + \\frac{1}{a+b}\\right) - 3 = \\frac{19}{10}.$ (A).\n[/hide]\n[/quote]\r\n\r\nWow, that pwns the way I got the answer. \r\n\r\nI decided to let a=4 b=2 and c=1\r\n\r\nThat worked surprisingly well.", "Solution_16": "[quote=\"chess64\"][quote=\"SplashD\"][quote=\"Silverfalcon\"][hide=\"Solution to 21\"]\n\nThis is an excellent problem. I use Cauchy's Inequality here but I don't know if this was necessary because normally, AMC problems do not use Cauchy.\n\nBy Cauchy:\n\n$(1^2+1^2+1^2)(a^2 + b^2 + c^2) \\geq (a + b + c)^2$\n\nSo, the minimum value of $a^2 + b^2 + c^2 = 12$ after doing some algebra and using $a+b+c = 6$.\n\nNow, note that:\n\n$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)$\n\nSince we minimized $a^2+b^2+c^2,$ we can now see that $ab+ac+bc = 12$.\n\nNow, observe these two squares:\n\n$(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$\n\n$(ab+ac+bc)^2 = a^2 b^2 + a^2 c^2 + b^2 c^2 + 2abc(a+b+c)$\n\nUsing what we know, we can get $a^2 b^2 + a^2 c^2 + b^2 c^2 = 48$. So, $a^4 + b^4 + c^4 = 12^2 - 2(48) = \\fbox{48}$. [/hide]\n\nAs I said, nice job on making this problem. By the way, I'm pretty sure that #24 is not original. I think this is in Mathematical Circles (Russain Experience).[/quote]\n\nI solved that faster than I solved #3 just by plugging in 2 for all the variables :rotfl:[/quote]\n\nOops, I made that one too brute-forcable :([/quote]\r\nYeah, I did the same thing. Problems like that (and I've done a couple of similar ones recently) almost always seem to work out that way.", "Solution_17": "I would like to see a solution to 22. It was very confusing, you didn't specify which point had the right angle! Although, I assumed it was C since that made it have nice integer side lengths. However, I solved it that way twice and didn't get integer answers. It was very simple, just using coordinates...", "Solution_18": "[quote=\"mathgeniuse^ln(x)\"]That was the problem I was unable to solve as well as 23. I don't know what other question I got wrong.\n\nSilverFalcon, for #21, couldn't u just use AM-GM, since a^4+b^4+c^4 is always positive as well as abc?[/quote]\r\n\r\nYou can't and in this case, it was lucky.\r\n\r\nNote that the problem stated that $a,b,c$ are REAL not necessarily positive. In order for AM-GM to work, all three variables must be positive.\r\n\r\nThat's why I disagree with your idea and nr1337's solution about using AM-GM.", "Solution_19": "About no 16,\r\n\r\nI informally took the test (I only discovered this mock exam an hour ago), and I am wondering about question 16.\r\n\r\nMy experience from reading Discrete and Combinatorial Mathematics says that the way to solve the problem would be to solve for:\r\n\r\n$\\left(\\begin{array}{c}18 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}16 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}14 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}12 \\\\ 2 \\end{array} \\right) ...$\r\n\r\nOr\r\n\r\n$\\frac{18!}{2^9}$\r\n\r\nWhy is this different? Do I have the wrong idea? Obviously the answer is different on the test, because that one is based off of just plain 18! (or is it the same? I have no idea... I'd just like a bit of help).", "Solution_20": "[quote=\"futuremontrealian\"]About no 16,\n\nI informally took the test (I only discovered this mock exam an hour ago), and I am wondering about question 16.\n\nMy experience from reading Discrete and Combinatorial Mathematics says that the way to solve the problem would be to solve for:\n\n$\\left(\\begin{array}{c}18 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}16 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}14 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}12 \\\\ 2 \\end{array} \\right) ...$\n\nOr\n\n$\\frac{18!}{2^9}$\n\nWhy is this different? Do I have the wrong idea? Obviously the answer is different on the test, because that one is based off of just plain 18! (or is it the same? I have no idea... I'd just like a bit of help).[/quote]\r\n\r\nI used the same method, and you are basically right, but you still overcounted by $9!$ for the different orders to choose the pairs. Still, the key to this problem is that the sum of digits is equivalent to the number $\\pmod{9}$, which is clearly $0$ because the number is divisible by 9. Then the sum of digits is 9, because it obviously can't be 0. This isn't such a great problem because you can use $18!$ and still get the right answer, but oh well... :)", "Solution_21": "[quote=\"Silverfalcon\"][quote=\"mathgeniuse^ln(x)\"]That was the problem I was unable to solve as well as 23. I don't know what other question I got wrong.\n\nSilverFalcon, for #21, couldn't u just use AM-GM, since a^4+b^4+c^4 is always positive as well as abc?[/quote]\n\nYou can't and in this case, it was lucky.\n\nNote that the problem stated that $a,b,c$ are REAL not necessarily positive. In order for AM-GM to work, all three variables must be positive.\n\nThat's why I disagree with your idea and nr1337's solution about using AM-GM.[/quote]\r\n\r\n$a^4, b^4, c^4$ are always positive\r\nBy am-gm\r\n$a^4 + b^4 + c^4 \\ge 3 (abc)^{ \\frac 4 3}$\r\n$a^4 + b^4 + c^4 \\ge 48$", "Solution_22": "[hide=\"another sol for 25\"]Let what we have to find =x\n\nAdd 3 to both sides(on l.h.s,add them as +1,+1,+1 to each of the fractions and u'll obtain):-\n\n$\\frac{7}{b+c}+\\frac{7}{c+a}+\\frac{7}{a+b}=x+3$\n\n$\\implies 49/10=x+3$[/hide]\r\n\r\nnice questions chess!!! It wasn't easy to complete it in the time limit,had to think for some :D", "Solution_23": "[quote=\"matt276eagles\"][quote=\"futuremontrealian\"]About no 16,\n\nI informally took the test (I only discovered this mock exam an hour ago), and I am wondering about question 16.\n\nMy experience from reading Discrete and Combinatorial Mathematics says that the way to solve the problem would be to solve for:\n\n$\\left(\\begin{array}{c}18 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}16 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}14 \\\\ 2 \\end{array} \\right) \\left(\\begin{array}{c}12 \\\\ 2 \\end{array} \\right) ...$\n\nOr\n\n$\\frac{18!}{2^9}$\n\nWhy is this different? Do I have the wrong idea? Obviously the answer is different on the test, because that one is based off of just plain 18! (or is it the same? I have no idea... I'd just like a bit of help).[/quote]\n\nI used the same method, and you are basically right, but you still overcounted by $9!$ for the different orders to choose the pairs. Still, the key to this problem is that the sum of digits is equivalent to the number $\\pmod{9}$, which is clearly $0$ because the number is divisible by 9. Then the sum of digits is 9, because it obviously can't be 0. This isn't such a great problem because you can use $18!$ and still get the right answer, but oh well... :)[/quote]\r\n\r\nOh! I forgot about that mod 9 thing. I will have to keep that in mind for future sum of digits problems. Good solution.", "Solution_24": "[quote=\"Xantos C. Guin\"][quote=\"nr1337\"]\n[hide=\"25\"]\n$\\frac{a}{b+c} = \\frac{(a+b+c)-(b+c)}{b+c} = \\frac{7}{b+c} - 1.$ Add them cyclicly to get $7 \\left(\\frac{1}{b+c} + \\frac{1}{a+c} + \\frac{1}{a+b}\\right) - 3 = \\frac{19}{10}.$ (A).\n[/hide]\n[/quote]\n\nWow, that pwns the way I got the answer. \n\nI decided to let a=4 b=2 and c=1\n\nThat worked surprisingly well.[/quote]\r\n\r\nI magically found a very simple answer...\r\n\r\n$(a+b+c)(\\frac{1}{b+c}+\\frac{1}{a+c}+\\frac{1}{a+b})=1+\\frac{a}{b+c}+1+\\frac{c}{a+b}+1+\\frac{b}{a+c}=7(\\frac{7}{10})$\r\nAnswer = $\\frac{49}{10}-3\\rightarrow \\frac{19}{10}$", "Solution_25": "Oh yeah. You're right. I forgot about raising to 4th power. Haha.. I did way too much work.", "Solution_26": "[quote=\"Adunakhor\"]Hmm, turns out I wasn't using the latest version of the test ... none of the choices for Question 3 on the version I downloaded was 20. No wonder I couldn't answer that one :P[/quote]\r\n\r\nSame happened to me, I asked on the forum what were the answer choices, and some guy just said \"its just changed 940 to 320.\" -,- he didnt give me the answer choices so I got a retarded one wrong\r\n\r\nI also wrote down the wrong answer thinking the right answer for another problem.\r\n\r\nAlso when i saw #21 i skipped it because it looked like a big expansion problem but then when i went back after i finished the test and turned in my answers i was like OMG. 2 2 2 because i didnt read the actual sum and product thing, i just didn't think it would be so nice.", "Solution_27": "[hide=\"13\"]$\\frac{1}{x}+\\frac{1}{y}=\\frac{x+y}{xy}=\\frac{1}{2007}$\n$\\implies xy-2007x-2007y=0$\n$\\implies (x-2007)(y-2007)=2007^2$ from Simon's Favorite Factoring Trick.\nThe prime factoriztation of $2007^2=3^4\\times223^2$. $2007^2$ has $5\\times3=15$ divisors.\nThere are 7 ordered pairs. C.[/hide]", "Solution_28": "[quote=\"pianoforte\"][hide=\"13\"]$\\frac{1}{x}+\\frac{1}{y}=\\frac{x+y}{xy}=\\frac{1}{2007}$\n$\\implies xy-2007x-2007y=0$\n$\\implies (x-2007)(y-2007)=2007^2$ from Simon's Favorite Factoring Trick.\nThe prime factoriztation of $2007^2=3^4\\times223^2$. $2007^2$ has $5\\times3=15$ divisors.\nThere are 7 ordered pairs. C.[/hide][/quote]\r\n\r\nHeh, I made that way harder than it needed to be :D", "Solution_29": "futuremontrealian wrote: \r\nAbout no 16, \r\n\r\nI informally took the test (I only discovered this mock exam an hour ago), and I am wondering about question 16. \r\n\r\nMy experience from reading Discrete and Combinatorial Mathematics says that the way to solve the problem would be to solve for: \r\n\r\n \r\n\r\nOr \r\n\r\n \r\n\r\nWhy is this different? Do I have the wrong idea? Obviously the answer is different on the test, because that one is based off of just plain 18! (or is it the same? I have no idea... I'd just like a bit of help). \r\n\r\n\r\nI used the same method, and you are basically right, but you still overcounted by 9! for the different orders to choose the pairs. Still, the key to this problem is that the sum of digits is equivalent to the number , which is clearly because the number is divisible by 9. Then the sum of digits is 9, because it obviously can't be 0. This isn't such a great problem because you can use 18! and still get the right answer, but oh well... (sorry, too lazy to do quote)\r\n\r\n--------------------------------------\r\n\r\nAre you sure I overcounted? I understand now why it doesn't matter, I'm just curious about the 9! thing. I will gladly admit a flaw in my logic, just tell me what's going wrong so I can fix it.\r\n\r\n18C2 is $\\frac{18!}{16!2!}$ Order doesn't matter here because person a paired with person b is the same as the opposite.\r\n\r\nThis simplifies to $\\frac{18 x 17}{2!}$ The same thing is going to happen with 16c2 and 14c2 and the rest, so the end result will be\r\n\r\n$18 x 17 x 16 x15...$ or $18!$ over $2!2!2!2!2!2!2!2!2!$ I hope that's nine twos.\r\nThat's what gave me $\\frac{18!}{2^{9}}$\r\n\r\nWhere'd I go wrong (if I did)?\r\n\r\nNote: those x's should be read \"times\" :) I'm too lazy to type \\ast six million times.", "Solution_30": "[quote=\"futuremontrealian\"]futuremontrealian wrote: \nAbout no 16, \n\nI informally took the test (I only discovered this mock exam an hour ago), and I am wondering about question 16. \n\nMy experience from reading Discrete and Combinatorial Mathematics says that the way to solve the problem would be to solve for: \n\n \n\nOr \n\n \n\nWhy is this different? Do I have the wrong idea? Obviously the answer is different on the test, because that one is based off of just plain 18! (or is it the same? I have no idea... I'd just like a bit of help). \n\n\nI used the same method, and you are basically right, but you still overcounted by 9! for the different orders to choose the pairs. Still, the key to this problem is that the sum of digits is equivalent to the number , which is clearly because the number is divisible by 9. Then the sum of digits is 9, because it obviously can't be 0. This isn't such a great problem because you can use 18! and still get the right answer, but oh well... (sorry, too lazy to do quote)\n\n--------------------------------------\n\nAre you sure I overcounted? I understand now why it doesn't matter, I'm just curious about the 9! thing. I will gladly admit a flaw in my logic, just tell me what's going wrong so I can fix it.\n\n18C2 is $\\frac{18!}{16!2!}$ Order doesn't matter here because person a paired with person b is the same as the opposite.\n\nThis simplifies to $\\frac{18 x 17}{2!}$ The same thing is going to happen with 16c2 and 14c2 and the rest, so the end result will be\n\n$18 x 17 x 16 x15...$ or $18!$ over $2!2!2!2!2!2!2!2!2!$ I hope that's nine twos.\nThat's what gave me $\\frac{18!}{2^{9}}$\n\nWhere'd I go wrong (if I did)?\n\nNote: those x's should be read \"times\" :) I'm too lazy to type \\ast six million times.[/quote]\r\n\r\nThere are 9 pairs of people. The order of the pairs does not matter. Call the 9 pairs of one possible system of pairing A, B, C, ..., J\r\n\r\nABCDEFGHIJ has the same people paired with each other as JABCDEFGHI. There are 9! ways to order the 9 pairs, so we must divide by 9! to correct the overcounting.", "Solution_31": "But why is that so? ABCDEFGHIJ has a with b and c with d, etc, whereas a is with j on the second list. I'm going to look this up again in Grimaldi's book.\r\n\r\n---edit:\r\nwait I get it now! the 9! is because it doesn't matter what order you choose the pairs themselves in, so long as the people in them are the same! hazzah!", "Solution_32": "Yeah, those letters represented pairs of people, not individuals. That's why there were only 9 of them :P", "Solution_33": "So, how did people like the uber problem? I didn't get that question." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "AMC 10", "AMC 10 A", "AMC 10 B" ], "Problem": "To all sophomores, freshmen, and middle schoolers:\r\n\r\nWhat was your USAMO index (AMC 10(A/B) + 10*AIME)?", "Solution_1": "$ 135 \\text{ AMC 10B} \\plus{} 10(9 \\text { AIME I}) \\equal{} 225$. :P", "Solution_2": "$ 135 (\\text{AMC 10B}) \\plus{} 10\\cdot8 (\\text{AIME}) \\equal{} 215$\r\nSo close...", "Solution_3": "144\r\n11\r\nso 254", "Solution_4": "130.5 / 8 = 210.5", "Solution_5": "117/8, so 197", "Solution_6": "$ 132 \\left(\\text{AMC 10A}\\right) \\plus{} 10\\cdot7 \\left(\\text {AIME I}\\right) \\equal{} \\boxed{202}$\r\nStupid mistakes...", "Solution_7": "[quote=\"davidyko\"]$ 135 \\text{ AMC 10B} \\plus{} 10(9 \\text { AIME I}) \\equal{} 225$. :P[/quote]\r\n$ 127.5 \\text{ AMC 10A} \\plus{} 10(10 \\text { AIME I}) \\equal{} 227.5$. :P\r\n\r\nHmm... AIME does make a big difference. (BTW, I could've got a 145.5 on the 10A if I hadn't made any stupid mistakes...)", "Solution_8": "144+10(9) = 234. 2 careless mistakes on AIME, 1 on AMC 10. Maybe I'll stiill make USAMO though.", "Solution_9": "[quote=\"pythag011\"]144+10(9) = 234. 2 careless mistakes on AIME, 1 on AMC 10. Maybe I'll stiill make USAMO though.[/quote]You definitely will make USAMO...", "Solution_10": "[quote=\"vishalarul\"][quote=\"davidyko\"]$ 135 \\text{ AMC 10B} \\plus{} 10(9 \\text { AIME I}) \\equal{} 225$. :P[/quote]\n$ 127.5 \\text{ AMC 10A} \\plus{} 10(10 \\text { AIME I}) \\equal{} 227.5$. :P\n\nHmm... AIME does make a big difference. (BTW, I could've got a 145.5 on the 10A if I hadn't made any stupid mistakes...)[/quote]\r\n\r\nLawl, go and die, Vishal. Just kidding. :wink: \r\nI feel like my 10B score was vaguely unsatisfying. Maybe because I got a higher score two years ago.", "Solution_11": "$ 145.5 \\text{ AMC 10A} \\plus{} 10(9 \\text { AIME I}) \\equal{} 235.5.$\r\n\r\nStupid mistakes on AIME, changed correct answers to incorrect answers, due to doubts... :( \r\nHopefully, I'll make USAMO...", "Solution_12": "ehh i completely and utterly failed the 10, (better on the 12, but still failed :D ) so\r\n$ 118.5\\plus{}60\\equal{}178.5$...i doubt i'll make USAMO", "Solution_13": "$ 144 (\\text{AMC 10A}) \\plus{} 10\\cdot8 (\\text{AIME}) \\equal{} 224$\r\n\r\nHope I didn't add wrong :D\r\n\r\n$ 144 \\plus{} 80 \\equal{} 224$ right?", "Solution_14": "I failed the 10.\r\n\r\n$ 120 \\plus{} 7 \\cdot 10 \\equal{} \\boxed{190}$.", "Solution_15": "126 (AMC 12)/11 AIME -> 236.. I failed miserably on the AMC..", "Solution_16": "[quote=\"FloodFilter:\"][quote=\"miller4math\"]ANYONE here in 7th grade and has decent scores?\noh well, i got a 138/9... so 228\nI [i]PROBABLY[/i] made USAMO\nim the only optimistic one around here these days...\n\nEDIT: heh spelled optimistic wrong[/quote]\n\nJohnny (his name hehe i spoiled it) is uber pro! :mad: He can totally destroy all the middle schoolers in california except Allissa Zhang (did i spell it right). As for pythag, I bet he's pro too. Anyways, Johnny knows a bunch of crazy trig formulas so you can't blame him (even though he got the trig problem on AIME wrong :huh: )[/quote]\r\n\r\nYou're in California???(So am I).", "Solution_17": "144/8 = 224 :(, but the AIME I i took was unofficial. gonna take the AIME II this wednesday.", "Solution_18": "you people are making me feel stupid...\r\n123/between 6 and =between 183 and 213\r\nbut im an 8th grader. :D", "Solution_19": "139.5+8*10=219.5\r\n\r\nI really am nervous about this with all the \"easy AIME\" rumors that have been going around... I dunno if i'll make it...\r\n\r\nEDIT: The AMC looks at both AMC scores right? I put my slightly worse 12A score because I couldn't confirm my 10B score.", "Solution_20": "Stop making us all feel bad! You're in 8th grade and probably qualified for the USAMO by like 20 points.", "Solution_21": "[quote=\"Zmastr\"]Stop making us all feel bad! You're in 8th grade and probably qualified for the USAMO by like 20 points.[/quote]\r\n\r\n :) Thanks. Now I have a little more hope :ninja:", "Solution_22": "i would any day trade my 126/60 for your 139.5/80. :roll:", "Solution_23": "Wow, I was looking at the poll and I was thinking it was a poll for the index cutoff for about 10 seconds.\r\nWhich kind of scared me considering the number of people who put super-high cutoffs.\r\n\r\nBut anyways, 138 (10A) + 9*10 = 228.", "Solution_24": "[quote=\"alkjash\"]Wow, I was looking at the poll and I was thinking it was a poll for the index cutoff for about 10 seconds.\nWhich kind of scared me considering the number of people who put super-high cutoffs.\n\nBut anyways, 138 (10A) + 9*10 = 228.[/quote]\r\n\r\nAHAHA same. I was like \"Holy S*** 250???\" before actually figuring out what the poll was for.\r\nAnyways,\r\nAMC10 A - 144(pretty good, I think) and\r\nAIME I - 7(phail XD)\r\n= 214 index", "Solution_25": "same here too.\r\n\r\nAMC 10a -- 120 \r\n :mad: \r\n\r\n :mad: \r\n\r\n :mad: \r\nAIME I -- 70\r\nfor a total of 190.", "Solution_26": "[quote=\"Moldytape\"]\n\nI probably won't make the =<10th grade index though, since the floor will probably be a 7, and with my bad AMC10 score, :([/quote]\r\n\r\nyou think you have it bad, think of the =<10th graders who are forced to take the amc12.", "Solution_27": "[url=http://www.imo-official.org/participant_r.aspx?id=21095]david yang[/url] versus [url=http://carp.di.unipi.it/]johnny ho[/url]???", "Solution_28": "David Yang fo sho", "Solution_29": "Woah, what's with the major revive?" } { "Tag": [ "inequalities" ], "Problem": "Let $ x_1,x_2,...,x_n$ be positive numbers. Prove that\r\n\r\n[list]$ \\frac{x_1^2}{x_2}\\plus{}\\frac{x_2^2}{x_3}\\plus{}...\\plus{}\\frac{x_{n\\minus{}1}^2}{x_n}\\plus{}\\frac{x_n^2}{x_1} \\geq x_1\\plus{}x_2\\plus{}...\\plus{}x_n$[/list]", "Solution_1": "[quote=\"4865550150\"]Let $ x_1,x_2,...,x_n$ be positive numbers. Prove that\n\n[list]$ \\frac {x_1^2}{x_2} \\plus{} \\frac {x_2^2}{x_3} \\plus{} ... \\plus{} \\frac {x_{n \\minus{} 1}^2}{x_n} \\plus{} \\frac {x_n^2}{x_1} \\geq x_1 \\plus{} x_2 \\plus{} ... \\plus{} x_n$[/list][/quote]\r\nIt is straight B-C-S(known as Andreescu inequality)....\r\nJust multiply both sides with:\r\n$ (x_2\\plus{}x_3\\plus{}...\\plus{}x_n\\plus{}x_1)$", "Solution_2": "you can also directly use CBS.Titu Andreescu is another way to express CBS...but a very useful one" } { "Tag": [ "trigonometry", "algebra", "polynomial", "algebra open" ], "Problem": "Let $P,Q$ be nonconstant real polynomials. Prove that every $x\\in\\Bbb R$ is an \r\naccumulation point of the sequence $a_n=P(n)\\sin Q(n)$. At least prove that the \r\nsequence $(a_n)$ always has a finite accumulation point, or prove the general statement for some special polynomials $P$ and $Q$ (the case $P(n)=Q(n)=n$ is already hard enough).", "Solution_1": "??? Why do you say that the case $P(n)=Q(n) = n$ is hard enough? As far as I know it is an open question to know if the sequence $U_n = n \\cdot sin(n)$ is dense in $R$...\r\nHave you some results on it ?\r\n\r\nPierre.", "Solution_2": "That problem is open? I think it was posted on ML, and not as an open problem..", "Solution_3": "I answered here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=108295#108295\r\n\r\nPierre.", "Solution_4": "Is it my misunderstanding or did you say that it's unknown whether 0 is an accumulation \r\npoint of $n\\sin n$?", "Solution_5": "Why? Can you prove it?\r\n\r\nPierre.", "Solution_6": "I can't prove it, but it was my exam problem, that's why I was confused.", "Solution_7": "I've heard of teachers which give some unsolved problems in exams, just in case a brilliant student has a good idea without being discouraged by the fact that they're unsolved :). Maybe you have such a teacher.", "Solution_8": "Oh, I think all 7 Clay's Millenium Problems will be in ML Open Questions soon :)" } { "Tag": [], "Problem": "If the 25th day of the year 2003 falls on a Saturday, on what day of the week did the 284th day of the year 2003 fall?", "Solution_1": "$ 25 \\equiv 284 \\mod 7$ so it falls on a [b]Saturday[/b]." } { "Tag": [ "geometry", "parallelogram" ], "Problem": "How would you proof the theorem: If a quadrilateral has both pairs of opposite angles congruent, then it's a parallelogram. I can't seem to make any progress", "Solution_1": "If a quarilateral ABCD with $\\angle A \\cong \\angle C $ and $\\angle B \\cong \\angle D $ then $ m\\angle A+m\\angle B+m\\angle C+m\\angle D=360^\\circ$, so $ 2m\\angle A + 2m\\angle C=360^\\circ $ which means that $m\\angle A + m\\angle B=180^\\circ$. By one of the basic axioms of geometry if a line interesects two other lines and two interior, same side angles that add up to 180 are created, then the two intersected lines are parallel. Thus AD||BC. The same argument works for AB and DC, so the figure is a parallelogram.", "Solution_2": "Thanks for your help." } { "Tag": [], "Problem": "What do you think about that story ?\r\n\r\nShould he have suffered more? Or was that what he deserved? \r\n\r\n\r\nPlease post your thoughts and opinions. :lol:", "Solution_1": "there's a topic about this in round table" } { "Tag": [ "search", "advanced fields", "advanced fields unsolved" ], "Problem": "Find explicit form for $x$ using $y$ such that $x^y=y^x$ nad $x,y \\in \\mathbb{C}$.", "Solution_1": "steel nothing", "Solution_2": "Pose y/x=k and modify the equation to keep only k.\r\n\r\nThe detailed process is on http://www.les-mathematiques.net , search y^x=x^y in archives.", "Solution_3": "The link doesn't work because you have a comma at the end.", "Solution_4": "can you give more specifik link, i can't get it", "Solution_5": "[url]http://les-mathematiques.u-strasbg.fr/phorum/read.php?f=2&i=97844&t=96315#reply_97844[/url]\r\n\r\nIf you have any problem with the french just ask me.", "Solution_6": "if I unedrstood it, it's only for $(x,y)\\in R^2$. I need for complex...." } { "Tag": [ "calculus", "derivative", "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $f:R->R$ be a $n$-times bounded derivable function such that $ f^{(n)}$ is Lipschitz. Then $f,f',...,f^{(n)}$ are bounded.", "Solution_1": "Since the boundedness of $f$ and $f^{(n)}$ implies the boundedness of all the other derivatives, as was discussed before, it suffices to prove the boundedness of $f^{(n)}$, given that it's Lipschitz and $f$ is bounded.\r\n\r\nThe boundedness of $f^{(n)}$ follows from this (which is pretty obvious and easy to prove):\r\n\r\nIf $g$ is a differentiable function s.t. for all $M>0$ and all $t$ (no matter how large) we can find an interval of length $t$ on which $|g'|\\ge M$, then $g$ has the same property.\r\n\r\nIf $f^{(n)}$ were unbounded then we could apply the sentence to $g=f^{(n-1)},g=f^{(n-2)},\\ldots,g=f$, until we reach the conclusion that $f$ must be unbounded, a contradiction.\r\n\r\nI hope I didn't say anything stupid." } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "trigonometry", "geometry proposed" ], "Problem": "Point $C$ is a midpoint of $AB$. Circle $o_1$ which passes through $A$ and $C$ intersect circle $o_2$ which passes through $B$ and $C$ in two different points $C$ and $D$. Point $P$ is a midpoint of arc $AD$ of circle $o_1$ which doesn't contain $C$. Point $Q$ is a midpoint of arc $BD$ of circle $o_2$ which doesn't contain $C$. Prove that $PQ \\perp CD$.", "Solution_1": "Suppose $CP\\cap AD=X,CQ\\cap BD=Y$, then the result follows from $\\triangle CXY\\sim\\triangle CQP$.", "Solution_2": "My solution was based on a well-known:\r\n\r\n$CD \\perp PQ$ iff $DP^2-DQ^2=CP^2-CQ^2$. Then two Ptolemy's, two cosine's theorem and everything comes out nicely. :)", "Solution_3": "Take $P',Q'$ by reflection of $P,Q$ w.r.t $O_1,O_2$, now:$O_1O_2//P'Q'//PQ$.As $O_1O_2$ orthogonal $CD$ then $O_1O_2$ orthogonal $CD$. QED", "Solution_4": "[quote=\"juancarlos\"]Take $P',Q'$ by reflection of $P,Q$ w.r.t $O_1,O_2$, now:$O_1O_2//P'Q'//PQ$.As $O_1O_2$ orthogonal $CD$ then $O_1O_2$ orthogonal $CD$. QED[/quote]\r\n\r\n\r\n And before you constructed P' and Q' segment PQ was not // to O_1O_2 ???\r\n\r\n\r\n \r\n\r\n T.Y.\r\n\r\n M.T.", "Solution_5": "Dear Armpist:\r\nSorry, I had a big confusion to try to solve it", "Solution_6": "I'm sorry, but I can't see why $PQ$ is parallel to $O_1O_2$...", "Solution_7": "A beaautifull solution using the ortological triangles th can also be done.\r\n\r\nConsider the triangles $ABD$ and $PQC$. Perpendiculars from $P,Q,C$ are perpendicular bisectors of sides of the triangle $ABD$ hence intersect each other in it's circumcenter. Thus, perpendiculars from $A,B,D$ on edges of $PQC$ intersect each other in one point. \r\n\r\nTaking $T\\in (CD$ such that $CT=AC$ we can easily show that perpendiculars on $PC$ and $QC$ through $A$ and $B$ pass through point $T$. Since the three perpendiculars are concurent, follows that $TD\\perp PQ$ which finishes the proof. :lol:", "Solution_8": "to xirti: your solution is very nice!\r\ni have another solution:\r\nlet $M = AD \\cap CP, N = BD \\cap CQ$.\r\nwe know $PD^2 = PM.PC$ and CM.CP = CD.CA; thus $CP^2 - DP^2 = CP^2 - PM.PC = CP.CM = CA.CD$.\r\nanalog we have $CQ^2 - DQ^2 = CB.CD$, thus $CP^2 - DP^2 = CQ^2 - DQ^2$, then $CD \\perp PQ$", "Solution_9": "dears ,let me do an inversion with center $C$ and radius $CA^2=CB^2$.\r\n\r\ncircles $o_1$ and $o_2$ goes through $C$,then the inversive form of them is two line.\r\n\r\n$CP$ is bisector of $\\angle DCA$. and $CQ$ is bisector of $\\angle DCB$. ($\\angle DCP+\\angle DCQ=180$.\r\n\r\nthen in our inversive shape, $P'$ and $Q'$ are the intersections of $CP$ and $CQ$ with lines $A'D'$ and $B'D'$\r\n\r\n respectively.\r\n\r\nnow if $CD$ is prependicular to $PQ, CD'$ should goes through the circumcircle of triangle $P'CQ'$.\r\n\r\nand because of $\\angle P'CQ'=90$ ,$CD'$ should bisect $P'Q'$. \r\n\r\nlet me check it...\r\n\r\n\r\nproblem: $D'A'B'$ is a triangle.and $C$ is mid point of $A'B'$ the bisectors of angles $D'CA'$ and $D'CB'$, intersects\r\n\r\n$D'A'$ and $\u0650DB'$ at$P',Q'$ respectively.\r\n\r\n\r\n\r\nprove that:\r\n\r\n $D'C$ biscts $P'Q'$\r\n\r\nok, now this problem is easier then before.\r\n\r\n\r\n $\\fac{A'C}: {CD'}$=$\\frac{A'P'}{P'D'}$\r\n\r\n\r\n $\\fac{B'C}: {CD'}$=$\\frac{B'Q'}{Q'D'}$\r\n\r\n\r\nand:$\\frac{A'P'}{P'D'}$=$\\frac{B'Q'}{Q'D'}$\r\n\r\nit means $P'Q'$ is paralell to $A'B'$.\r\n\r\n$D'C$ is the median of $A'B'$ .and because of the symilarity of triangles $D'P'Q'$ and $D'A'B'$,\r\n\r\n$D'C$ also bisects $P'Q'.$\r\n\r\n\r\n\r\nand the problem is solved.\r\n\r\n(NOTE:did u know that the circumcircle of triangle $PCQ$ is tanjent to $AB$?)", "Solution_10": "megus, what grade is this problem for...?", "Solution_11": "there are no grade divisions in Polish Mathematical Olympiad, everyone has the same set of problems.", "Solution_12": "the second round of iran mathematical olympiad will begin at 18 apr 2006\r\n\r\ni hope the geometry problems on the exam ,be nice and as enjoyable as poland olympiad. ;)", "Solution_13": "[hide=\"Here is a short metrical proof.\"][quote][color=red][b]Lemma.[/b] Let $ABC$ be a triangle. Denote the circumcircle $w$ of the given triangle and the middlepoint $X$ of the arc $\\stackrel{\\frown}{BC}\\subset w$ which doesn't contain the vertex $A$. Then there is the relation [/color]$XA^2-XB^2=bc\\ .$[/quote][b]The first proof.[/b] Denote the intersection $D\\in AX\\cap BC$. From the similitude $\\triangle ABX\\sim \\triangle ADC$ we obtain the relation $AD\\cdot AX=bc$ and from the similitude $\\triangle ABX\\sim\\triangle BDX$ we obtain the relation $XB^2=XA\\cdot XD$. Thus, $XA^2-XB^2=$ $XA^2-XA\\cdot XD=$ $XA(XA-XD)=$ $AX\\cdot AD=bc\\ .$ Therefore, $XA^2-XB^2=bc\\ .$\n\n[b]The second proof.[/b] In the circle $w=C(O,R)$ we have $XA=2R\\sin \\left(B+\\frac A2\\right)$ and $XB=2R\\sin \\frac A2$. Therefore, $XA^2-XB^2=$ $4R^2\\left[\\sin^2\\left(B+\\frac A2\\right)-\\sin^2\\frac A2\\right]=$ $4R^2\\sin B\\sin C=bc\\ .$\n[u][i]Remark.[/i][/u] If denote the middlepoint $Y$ of the arc $\\stackrel{\\frown}{BC}\\subset w$ which contains the vertex $A$, then there is and the relation $YB^2-YA^2=bc$ because $AX^2+AY^2=$ $BX^2+BY^2=4R^2\\ .$ In the proof of the lemma I used the well-known relation $\\sin^2x-\\sin^2y=$ $\\sin (x+y)\\sin (x-y)$ and the [u]theorem of the Sinus[/u].\n\n[b]The third proof.[/b] Denote the intersection $D\\in BC\\cap AX$and the incircle $C(I,r)$. Therefore, $XA^2-XB^2=XA^2-XI^2=$ $(XA-XI)(XA+XI)=$ $IA(AX+IX)=$ $AI\\cdot AX+\\rho (I)=$ $2Rr+AX\\cdot \\frac{b+c}{2p}AD=$ $2Rr+\\frac{bc(b+c)}{2p}=$ $2Rr+bc-\\frac{abc}{2p}=$ $2Rr+bc-\\frac{4RS}{2p}=$ $2Rr+bc-2Rr=bc\\ .$\n[u][i]Remark.[/i][/u]. I denoted the power $\\rho (I)$ of the point $I$ w.r.t. the circumcircle and I used the well-known relations: $XB=XC=XI$, $\\rho (I)=2Rr$, $\\frac{IA}{b+c}=\\frac{ID}{a}=\\frac{AD}{2p}$, $AX\\cdot AD=bc$ (from the similitude $\\triangle ABX\\sim\\triangle ADC$), $abc=4RS$ and $S=pr\\ .$\n\n[b]The fourth proof.[/b] $XA^2-XB^2=\\frac{b^2c^2}{AD^2}-\\frac{a^2}{4\\cos^2\\frac A2}=$ $b^2c^2\\cdot \\frac{(b+c)^2}{4bcp(p-a)}-\\frac{a^2bc}{4p(p-a)}=$ $\\frac{bc\\left[(b+c)^2-a^2\\right]}{4p(p-a)}=bc\\ .$\n[u][i]Remark.[/i][/u] I used the well-known relations $AX\\cdot AD=bc$,\n${AD=\\frac{2}{b+c}\\cdot \\sqrt{bcp(p-a)}}$ and $\\cos \\frac A2=\\sqrt{\\frac{p(p-a)}{bc}}\\ .$\n\n[color=darkred][b][u]The proof of the proposed problem.[/u][/b] Apply the above lemma for the triangles $DAC$ and $CBD$ $\\Longrightarrow$ $PC^2-PD^2=CD\\cdot CA$ and $QC^2-QD^2=CD\\cdot CB$. But $CA=CB$. Thus, [/color]$PC^2-PD^2=QC^2-QD^2\\Longleftrightarrow$ $PQ\\perp CD\\ .$[/hide]", "Solution_14": "Let $CP$ meet $AD$ at $S$ and $CQ$ meet $BD$ at $K$. $\\frac{DS}{SA} = \\frac{DC}{CA} = \\frac{DC}{CB} = \\frac{DK}{KB}$ so $SK || AB$. Note that $\\angle PCQ = 90$ so we need to prove $\\angle DCQ = \\angle BCQ = \\angle QPC$. Note that $\\angle BCQ = \\angle CKS$ so we need to prove $PQKS$ is cyclic. It's well known that $CS.CP = CD.CB = CD.CA = CK.CQ$ so $PQKS$ is cyclic.\nwe're Done.", "Solution_15": "A computational solution: it suffices to show $\\frac{OO_1}{OO_2} = \\frac{r_1}{r_2}$, where $O$ is the circumcenter of $ABD$ and $O_1, O_2$ are the centers of the two circles. Then the result will follow as $\\overline{PQ} \\perp \\overline{O_1O_2}$.\n\nBy a computation, $$OO_1 = R_{ABD}\\left(\\cos \\angle DBA - \\frac{\\cos \\angle DCA}{\\sin \\angle DCA} \\cdot \\sin \\angle DBA \\right).$$ Then the desired ratio equals $$\\frac{OO_1}{OO_2} = \\frac{\\cos \\angle DBA \\sin \\angle DCA - \\cos \\angle DCA \\sin \\angle DBA}{\\cos \\angle DAB \\sin \\angle DCB - \\cos \\angle DCB \\sin \\angle DAB} = \\frac{\\sin \\angle BDC}{\\sin \\angle ADC} = \\frac{AB}{BD} = \\frac{r_1}{r_2},$$ as needed." } { "Tag": [ "calculus", "integration", "geometric series" ], "Problem": "$\\text{How come }\\sqrt{6\\cdot\\sum_{j=1}^{\\infty}{\\left(\\frac{1}{j^{2}}\\right)}}= \\pi \\text{ ?}$", "Solution_1": "This should not be in this forum. I will post a calculus solution, but would a moderator move this to a suitable place?\r\n\r\nFirst, prove that\r\n\\[\\int_{0}^{1}\\int_{0}^{1}\\frac{1}{1-xy}dxdy=\\sum_{n=1}^{\\infty}\\frac{1}{n^{2}}=\\zeta(2)\\]\r\nusing a geometric series expansion.\r\nBy substituting $x=\\frac{\\sqrt{2}(\\alpha-\\beta)}{2}$ and $y=\\frac{\\sqrt{2}(\\alpha+\\beta)}{2}$, you compute the integral to obtain $\\zeta(2)=\\frac{\\pi^{2}}{6}$. Rearranging the identity\r\n\\[\\sum_{n=1}^{\\infty}\\frac{1}{n^{2}}=\\frac{\\pi^{2}}{6},\\]\r\nthe desired result follows." } { "Tag": [ "logarithms" ], "Problem": "Are these two terms differnet ...?if yes what are the conditions for a reaction to be feasible and that to be sponataneous :?:", "Solution_1": "Yes i think so bcos $ \\Delta$G can be < 0 but $ \\Delta$S can be < than 0 :maybe: thus making it feasible but non spontaneous :maybe:", "Solution_2": "the condition for the reaction to be spontaneous is $ \\Delta$G < 0", "Solution_3": "but for every spontaneous process isnt there increase in entropy? :maybe: that means, $ \\Delta$S > 0 doesnt it? :huh:\r\n\r\nEDIT: Maybe they are seen from 2 different perspectives but mean the same :maybe:", "Solution_4": "As I have always been taught it:\r\n\r\nFor a process occuring at an absolute temperature $ T$ with a change in enthalpy $ \\Delta H$ and a change in entropy $ \\Delta S$, the associated Gibbs Free Energy $ \\Delta G$ is given by:\r\n\r\n$ \\Delta G \\equal{} \\Delta H \\minus{} T\\Delta S$\r\n\r\nAs long as $ \\Delta G < 0$ (which means that $ \\Delta H > T\\Delta S$), the process is spontaneous.", "Solution_5": "My point is very clear:\r\nSpontaneity=>$ \\Delta G < 0$ or thermodynamic feasibility\r\nFeasibility=>$ E_a$ is very small or kinetic feasibility", "Solution_6": "Yes i just checked out, Kinetics VS Thermodynamics :)", "Solution_7": "Feasibility = possible process that can be reversible or irreversible.\r\nSpontaneity = the process is possible and irreversible.\r\n\r\nThose arguments above with the Gibbs free energy are valid as long as we are only considering processes at contant T and P. For different conditions, other thermodynamic potentials should be minimized.", "Solution_8": "is there any specific condition carcul say at constant temperature and pressure by which we can say a reaction is feasible", "Solution_9": "For a reaction to occur at constant T and P we must have $ \\Delta G < 0$.", "Solution_10": "taht i think refers for it to happen spontaneously :) \r\nwhat about the feasibility my thought was...\r\nwe have $ \\Delta G^{0} \\equal{} \\minus{}RT \\ln K$ so for a reaction to be feasible and not only spontaneous in real sense $ K > > 1$ so we must have $ \\Delta G^{0}$ at that temp. $ < < 0$", "Solution_11": "I think the term feasible refers to the yield. Its more like \"Is this reaction feasible for labaratory purpose\" and so.", "Solution_12": "yes that's what i wrote above and the necessary condition for that :)", "Solution_13": "[quote=\"Pardesi\"]so for a reaction to be feasible and not only spontaneous in real sense[/quote]\r\n\r\nThat's not also a good criterion. There are reactions for which K >> 1 and that simply do not occur at an appreciable rate. What you are looking for is this: for a reaction to be feasible it must have a reasonable rate constant at room temperature or nearby.", "Solution_14": "yes had thought of that but...i wanted the feasibilty criterion at a partciular temp...so in some sense the term should not be exavcly 'feasible' rather relatively feasible :)", "Solution_15": "So feasibility is associated with kinetics (i.e., a feasible process is kinetically favorable)?\r\n\r\nAnd spontaneity is associated with thermodynamics (i.e., a spontaneous process is thermodynamically favorable)?", "Solution_16": "Yes, that's right." } { "Tag": [ "FTW" ], "Problem": "Well, I have seen any threads here about the US Open Tennis Tournament, so I guess I should start one :)\r\n\r\nWhat have you thought of the matches? What have you enjoyed the most? Have you been there, and if so, what were your experiences? etc, etc\r\n\r\n\r\n\r\nMy experiences:\r\n\r\nI went there on Friday, September 1.\r\nI first saw the Vaidisova vs. Jankovich, which was a good close match (not to mention sitting in the front row, which you can do without paying any extra in the side courts).\r\n\r\nLater, saw the Peer vs. Schiavone match, which was definitely the best one of the day. Peer was down 5 1 in the first set and managed to tie it up, winning *5 match points* against her in the process. She then proceeded to actually win the match in a tie-breaker.\r\n\r\nAfter that, I saw the Navratilova/Bryan vs. Morariu/Bryan match, which was great.\r\n\r\nEncounters with players:\r\n- Jankovich almost crashed into me\r\n- Schiavone served a ball and it hit me in the cheek (hard)\r\n- Got Bob Bryan's autograph\r\n- Was within 2 feet of Martina Navratilova, but got no autograph\r\n\r\nI must say, it's must more fun to actually be there. :yup:", "Solution_1": "what channel? i didn't even watch junior nationals in swimming..but i have a guess ESPN?", "Solution_2": "Well, I was watching it on NBC, USA Network, and CBS.", "Solution_3": "GRAAAAARGH YOU ARE SO LUCKY\r\nyou actually went there????\r\nAHHHH\r\n\r\nI only saw one match...my mom wouldn't let me watch them because I had to do homework... but I watched Agassi's second round match!\r\nDefinitely the best match I ever saw(but then again, I didn't watch a lot of matches.just part of wimbledon2006 and some usopen series 2006).", "Solution_4": "I'm in awe of Agassi. and Baghdatis.\r\n\r\nWOW.", "Solution_5": ":( :( \r\nAgassi just lost :(", "Solution_6": "[quote=\"junggi\"]:( :( \nAgassi just lost :([/quote]\r\n\r\nYes, his last match. Agassi is oficially retired from tennis.", "Solution_7": "davenport vs. S. Williams\r\n\r\nshould be a good match.", "Solution_8": "What about Hewitt vs. Gasquet? Really interesting match tomorrow", "Solution_9": "Farewell, Agassi :(", "Solution_10": "[quote=\"Jos\u00e9\"]What about Hewitt vs. Gasquet? Really interesting match tomorrow[/quote]\r\nthat match is today.\r\nI wanna see the Murray v. Davydenko match and Haas v. Safin", "Solution_11": "I went to the US open for the first weekend (kid's day/practice day) and opening monday and tuesday. AMAZING experience. I saw the Agassi-Pavel match which was really good. Got lots of autographs from players such as Mauresmo and Sharapova, and i am looking forward to going back for the finals weekend.\r\n\r\nI was wondering who everyone is cheering for.\r\nMy favorites are Blake, Nadal, Mauresmo, and Sharapova.", "Solution_12": "OH MY GOD\r\nI am so jealous of everyone who got to go to US Open and get autographs and stuff... :( \r\n\r\nMy favorites are \r\nmen's:Federer,Murray,Haas and Ferrero.\r\nwomen's:Henin-Hardenne and Sharapova.", "Solution_13": "Henin-Hardenne! Go Belgium! Go Belgium! :)", "Solution_14": "OMG AGASSI LOST!", "Solution_15": "I was soooooooooooo sad when Agassi lost. :( He also gave a very touching speech.\r\nI am actually quite surprised how well Janovic is doing. I hope she continues to do well.", "Solution_16": "Henin-Hardenne will lose to Davenport.\r\nMy favourites:\r\nmen: Federrer, Safin\r\nwomen: Sharapova, Davenport", "Solution_17": "Do you really think henin will lose to Davenport? I hope so, but most people are expecting henin to win. I hope they will at least Play today. There are WAY to many rain delays. Who will win the Sharapova vs. Golovin? I think that will be a very good match. I like them both a lot, but think Sharapova will prevail in three sets.", "Solution_18": "I think henin will beat davenport and go on to win the open :ninja:", "Solution_19": "Federer and Mauresmo or Sharapova will win the tournament, that's my bet!", "Solution_20": "omfg Nadal lost today to Youzhny.\r\nWow Youzhnyc took out the 3 top spaniards in a row :o :o", "Solution_21": "I can't believe Nadal LOST! Have you seen Youzhny's ritual after winning? It is cool", "Solution_22": "nooo....I just saw the news, b/c I was at school.. :( \r\noh wow I don't think I ever saw Robredo or Ferrer play :( \r\nanyways, I was kinda hoping Nadal will lose. Now Federrer will win for sure :D\r\n\r\nhaha I knew Henin would beat davenport... :ninja:", "Solution_23": "Have you heard of the new stuffed toy called the feder-bear? It is sme fundraiser for Unicef or something.\r\nAnyway..... Davydenko beat Haas! I am pretty sad, because i was hoping Haas would play federer in the semi's (unless of course Blake beats him :P ) and beat him. He has a fairly good chance. I heard the Roddick- Hewitt was a good match- but, i couldn't watch it because of HOMEWORK! I hope Roddick beats Youzhny.", "Solution_24": "Sharapova v. Henin-Hardenne in the finals :) \r\nshould be a good match...but I was hoping for Mauresmo Henin rematch for some reason. Probably b/c I didn't get to see the wimbledon finals :(", "Solution_25": "I REALLY hope Maria (the pretty girl) will beat henin. I will be cheering her on, but i have the WORST seats. Section v in Promenade or something. I hope Navratilova and Bob win also.", "Solution_26": "Of course Henin will win... :D", "Solution_27": "[quote=\"Arne\"]Of course Henin will win... :D[/quote]\r\nof course... :D \r\n\r\nfederrer won :D", "Solution_28": "Now Rodick vs. Youzhny....I want Rodick to win. Big Serve ftw! :P \r\n\r\nUnfortunately, Rodick had a mere 70% first serve the first set...", "Solution_29": "[quote=\"calc rulz\"]Unfortunately, Rodick had a mere 70% first serve the first set...[/quote] Is that bad?? :maybe: \r\nPS. It's Roddick. :D", "Solution_30": "It's not THAT bad but it would be a lot better if it's higher(huge first serves :D )edit:isnt 70% good?\r\n\r\nWow tv broadcast says henin will become #1 with usopen win! :D \r\ngo henin!", "Solution_31": "whoahoho...\r\nsharapova wins US open! :clap: :clap2: \r\nNow how does that change the rankings....\r\n$\\#4\\rightarrow\\#2$??", "Solution_32": "Yes! Sharapova won! Now, she's number 3, number 2 is Henin, and 1 is Mauresmo.", "Solution_33": "I stayed up late last night and watched Federer vs. Roddick (the match started at 23:30 or so in my country). I was very sad that Federer won. Again. Roddick made two very good sets (2 & 3), but I guess he couldn't handle Federer all the way.\r\n\r\nNow only the Masters are left. Will he win there also? I hope not.", "Solution_34": "He'll win again too. He's, far far far away, the best player nowadays (maybe the greatest ever?)" } { "Tag": [ "vector", "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Show algebraically that if u and v are non-zero vectors in R^2 which are not\r\nparallel to each other, then any vector \u2329x, y) \u2208 R^2 can be written as a linear\r\ncombination of these two vectors.\r\n\r\nIt just seems obvious to me but i don't know how to prove", "Solution_1": "Rephrasing this: let $ u$ and $ v$ be two given linearly independent vectors in $ F^2$, and let $ w$ be any other vector in $ F^2$. Show that $ u,v,w$ are linearly dependent.\r\n\r\nThis is a case of the theorem that says dimension makes sense; we want to show that there can't be three linearly independent vectors in the two-dimensional space $ F^2$.\r\nThe standard tool is Gaussian elimination. You're working with a non-square matrix, so there isn't room for enough pivots and you must get a zero row/column. If you track the operations to get there, you can see how that zero is a linear combination of the original vectors.\r\n\r\n(Note: that $ F$ is an arbitrary field. You're only doing arithmetic on the entries, so the only relevant properties of $ \\mathbb{R}$ are that it's a field. I prefer stating these basic theorems in the general form, so you don't have to redo them when you deal with other fields later.)" } { "Tag": [ "algorithm", "logarithms", "advanced fields", "advanced fields unsolved" ], "Problem": "Take a look at the following algorithm that approximates the Graph coloring algorithm (in which we should color vertices with minimal number of colors possible, and where a legal coloring is such that no two adjacent vertices have the same color).\r\n\r\ninput - graph G\r\n1. find the I - biggest independent set in G and color it with some unique color.\r\n2. G <- G - I\r\n3. goto 1\r\n\r\n\r\nCan you please help me finding the relation between the optimal solution and the approximation? what is the guarantee factor?\r\n\r\nthanx", "Solution_1": "Given this 3-colorable graph\r\n\r\n[img]http://www.math.tamu.edu/~chillar/images/chaograph.gif[/img]\r\n\r\nyour algorithm may choose $ I\\equal{}\\{3,4,5,7\\}$ on the first step and end up using four colors. So the guarantee factor is at least $ 4/3$. \r\n\r\n(Image taken from [url=http://www.math.tamu.edu/~chillar/]Chris Hillar's webpage[/url])", "Solution_2": "If the graph is $ k$-colorable and has $ n \\ge 2$ vertices, then your algorithm will use at most $ k \\ln n$ colors. That's because, at each step, the independent set will remove at least a fraction $ \\frac{1}{k}$ of the vertices.", "Solution_3": "And the $ \\ln n$ factor is inavoidable unless you improve your algorithm somehow. Indeed, consider the regular $ n \\equal{} 2^m$-gon and number its vertices from $ 0$ to $ n$. Inscribe the regular $ 2^{m \\minus{} 1}$-gon into it joining all even vertices in a cycle. Then inscribe the regular $ 2^{m \\minus{} 2}$-gon into it joining all vertices with indices divisible by $ 4$ and so on. Here is the picture for $ n \\equal{} 16$.\r\n[asy]size(150); int k,m; for(k=1; k<16; k*=2) for(m=0; m<16; m+=k) D(dir(360/16*m)--dir(360/16*(m+k))); for(k=1; k<16; k+=2) D(dir(360/16*k));[/asy]\r\nSuppose that you chose the red vertices for the first step (clearly, it is [b]the[/b] maximal independent set). Then, after the first step, you are down to the same picture with $ m \\minus{} 1$ and you are welcome to do this unwise choice again. So, you'll get about $ m$ colors. On the other hand, 3 colors will suffice for this graph (just use induction)." } { "Tag": [ "ratio", "geometry" ], "Problem": "What is the ration of the area of the square to the area of the circle drawn in the diagram? Kindly explain the answer. Thanks.", "Solution_1": "[hide=\"I got\"]\n$\\frac{64}{25 \\pi}$\n\n(See graph)\n\nSet EC=x, FE=y.\nFB=FE(Both are radii), FG=EC, BG=2x-y. So\n\\[y^2=x^2+(2x-y)^2\\]\nSolve it and we have 4y=5x. The area of circle is $y^2 \\pi$ and the area of square is $4a^2$.\n[/hide]", "Solution_2": "I did it a similar way you did BornforMath but I got $64/100pi$. Here's how I did it: I let the radius equal to $2$ and 1/2 the side length of the square $x$. Then I said that the right triangle would have hypotenuse of $2$ and one side length of $x$ so the other would be the square root of $4-x^2$. So then $2x=sqrt4-x^2+2$ since it's a square and I got $x=4/5$ so the area of the square would be $64/25$. Maybe I made an error in finding $x$?", "Solution_3": "Let's start from here:\r\n\r\n$2x = \\sqrt{4 - x^2} + 2$\r\n\r\n$(2x - 2)^2 = 4 - x^2$\r\n\r\n$4x^2 - 8x + 4 = 4 - x^2$\r\n\r\n$5x^2 - 8x = 0$\r\n\r\n$x(5x - 8) = 0$\r\n\r\n$x = \\frac{8}{5}$\r\n$A_{square} = 4x^2 = \\frac{256}{25}$", "Solution_4": "[quote=\"dakyru\"]Let's start from here:\n\n$2x = \\sqrt{4 - x^2} + 2$\n\n$(2x - 2)^2 = 4 - x^2$\n\n$4x^2 - 8x + 4 = 4 - x^2$\n\n$5x^2 - 8x = 0$\n\n$x(5x - 8) = 0$\n\n$x = \\frac{8}{5}$\n$A_{square} = 4x^2 = \\frac{256}{25}$[/quote]\r\n\r\nYup, I screwed up on my calculations, $x$ should have been $8/5$ not $4/5$, thanks darkyru" } { "Tag": [ "MATHCOUNTS", "calculus", "trigonometry", "articles" ], "Problem": "If an arrow is shot at a target, it is moving right.\r\nBut if you pause time at any given moment how is the arrow \r\nmoving :?: \r\nDiscuss the question. \r\nYou can also ask other questions or post other problems\r\nBest of Luck to all you ppl going to states! :lol:", "Solution_1": "wasn't that from the art of problem solving book 1?\r\ni remember seeing that somewhere...i thought about it... :D", "Solution_2": "yeah\r\nI'm pretty sure that it isn't moving\r\nit all is based on you perception of \r\nmovement, and the different forms of physics.\r\nYou do Mathcounts?", "Solution_3": "yeah, im taking it tomorrow (i live in California)", "Solution_4": "States? \r\nMine are on March 23rd :thumbup: \r\nOur coach had us learn everything\r\nwe did everything we got Calculus, Trigonometry, Algebra 2, etc.", "Solution_5": "Awww this is such a basic paradoxical question. Shows up in any random book that talks about paradoxes or puzzling stuff. I think it has something to do with an infinite number of actions of zero length can be done in a finite time or something but you don't have to listen to that I'm not good at this philosophical stuff.\r\n\r\nAnyway hey are you Kevin Li from Texas? You know like in 7th grade and went to San Marcos last summer", "Solution_6": "Nope wrong guy dude\r\nNationals are in Texas I think though\r\nI've lived in PA my whole life.\r\n\r\nthe answer has to do with something you can't picture but doesn't exist like ripping a hole so fast that you puncture the atoms and it creates a vacuum, bad example. :blush:", "Solution_7": "kevin li from texas is actually budi713", "Solution_8": "[quote=\"Kevinli99\"]States? \nMine are on March 23rd :thumbup: \nOur coach had us learn everything\nwe did everything we got Calculus, Trigonometry, Algebra 2, etc.[/quote]\r\n\r\nis it just me, or does that seem a tad pointless? well the trig could come in handy...but calculus??", "Solution_9": "[quote=\"Walk Around The River\"][quote=\"Kevinli99\"]States? \nMine are on March 23rd :thumbup: \nOur coach had us learn everything\nwe did everything we got Calculus, Trigonometry, Algebra 2, etc.[/quote]\n\nis it just me, or does that seem a tad pointless? well the trig could come in handy...but calculus??[/quote]\r\n\r\nHey you never know when calculus can come in handy in Mathcounts. Just another way to brute force an easy problem to check if it's right", "Solution_10": "We can not possibly take a picture at every moment, and we are not responsible for explaining what cannot be observed. Took that argument from a Scientific American article on Zeno's paradoxes.", "Solution_11": "kevin li is from college station, texas he's my best friend and he's going to nationals.are you kevin chen from sugar land?[/hide]" } { "Tag": [ "geometry", "3D geometry", "probability" ], "Problem": "A large cube is dipped into red paint and then divided into 125 smaller congruent cubes. One of the smaller dubes is then randomly selected. What is the probability that the cube selected will have at least $ 25\\%$ of its surface area painted red? Express your answer as a common fraction.", "Solution_1": "This will mean it has at least 2 red faces.\r\n\r\nLets use complementary counting. WE have 125 total cubes and we want to eliminate ones with 0 or 1 face painted.\r\n\r\n125-27(inside 3x3x3)-6x9(6 faces, each with 9 central cubes with only one painted face)=98-54=44.", "Solution_2": "Assuming that's correct, the answer is then $ \\frac{44}{125}$ :)" } { "Tag": [ "Duke", "college", "FTW", "ARML" ], "Problem": "dude people wear your ARML shirt\r\n\r\n(no not the dirty ARML shirt or the inexplicably red ARML shirt, the real ARML shirt)", "Solution_1": "why wouldn't I?\r\n\r\nGood luck to everyone going!\r\n\r\nW00T!!! I ordered turkey? anyone else?", "Solution_2": "ham ftw\r\n\r\noh and spread the word about ARML shirtingness tis an awesome ARML shirt", "Solution_3": "oh im wearing a shirt even better than an ARML shirt. Ima wear a DUKE shirt. yea i know i'm so original :rotfl:", "Solution_4": "dude no, that's silly\r\n\r\nwear the ARML shirt\r\n\r\n(were you on the ARML teams)", "Solution_5": "nah I wasn't that great at math last year. Hopefully I can make it this year :)", "Solution_6": "oh ok this is an acceptable reason for not wearing the ARML shirt", "Solution_7": "Man I've got to do that *changes*", "Solution_8": "yes in fact I am wearing the ARML shirt RIGHT NOW", "Solution_9": "[quote=\"MysticTerminator\"]oh ok this is an acceptable reason for not wearing the ARML shirt[/quote]\r\n\r\nWhat are you talking about? That's a horrible reason.\r\n\r\nSorry guys! When you leave for Duke at 6am and go to sleep at 9 the night before, it's kind of hard to get messages. Trust me, I seriously considered wearing that shirt. And I [i]obviously[/i] will for Princeton.", "Solution_10": "dude I did not tell jeremy nor noah yet they both wore it\r\n\r\nit is just something you KNOW to do\r\n\r\nalso\r\n\r\nwearing the red arml shirt is simply NOT ACCEPTABLE", "Solution_11": "yeah I know\r\n\r\nWe've got to wear the same thing again this year" } { "Tag": [ "modular arithmetic", "calculus", "integration" ], "Problem": "In year N, the 300th day of the year is a Tuesday. In year N+1, the 200th day is also a Tuesday. On what day of the week did the 100th day of year N-1 occur? \r\n\r\n(A) Thursday\r\n(B) Friday\r\n(C) Saturday\r\n(D) Sunday\r\n(E) Monday\r\n\r\nPlease post a solution as well as an answer.\r\n\r\n[hide]I first got (B) Friday, then got (D) Sunday, and then (A) Thursday. I still don't know what I did.[/hide]", "Solution_1": "[quote=\"SplashD\"]In year N, the 300th day of the year is a Tuesday. In year N+1, the 200th day is also a Tuesday. On what day of the week did the 100th day of year N-1 occur? \n\n(A) Thursday\n(B) Friday\n(C) Saturday\n(D) Sunday\n(E) Monday\n\nPlease post a solution as well as an answer.\n\n[hide]I first got (B) Friday, then got (D) Sunday, and then (A) Thursday. I still don't know what I did.[/hide][/quote]\n\n[hide] The 300th day of year N and the 200th day of year N+1 are either 265 or 266 days apart. Since $266\\equiv0\\pmod7$, year N is a leap year. Therefore, year N-1 is not a leap year and 300th day of N and the 100th day of N-1 are 565 days apart. $565\\equiv5\\pmod7$. $\\boxed{Sunday}$ is the 5th day from Tuesday. :D [/hide]", "Solution_2": "[hide]\nThe first part of the problem shows that year $N$ is a leap year, so year $N-1$ is not a leap year.\n\nTherefore, $565 \\mod {7}\\equiv 5 \\mod {7}$\n\n$5$ days before Tuesday is $\\boxed {Thursday}$[/hide]", "Solution_3": "[quote=\"surge\"][hide]\nThe first part of the problem shows that year $N$ is a leap year, so year $N-1$ is not a leap year.\n\nTherefore, $565 \\mod {7}\\equiv 5 \\mod {7}$\n\n$5$ days before Tuesday is $\\boxed {Thursday}$[/hide][/quote]\r\n\r\nOh. Yeah. :blush:", "Solution_4": "[quote=\"surge\"][hide]\nThe first part of the problem shows that year $N$ is a leap year, so year $N-1$ is not a leap year.\n\nTherefore, $565 \\mod {7}\\equiv 5 \\mod {7}$\n\n$5$ days before Tuesday is $\\boxed {Thursday}$[/hide][/quote]\r\n\r\nactually year N+1 is a leap year\r\nbut it doesnt change the answer\r\ni think that's right", "Solution_5": "No... we wouldn't be able to tell if $N+1$ is a leap year...\r\n\r\nRemember it's going by days, not months (so February 29th cannot be distinguished).", "Solution_6": "actually you can tell N+1 is a leap year\r\n\r\n(365-100)=265 which is not a multiple of seven but (366-100)=266 is :D", "Solution_7": "Right, but going from the 300th day of year $N$ to the 200th day of year $N+1$ you cannot pass the 366th day of year $N+1$, but rather you pass the 366th day of $N$. Therefore, $N$ is a leap year and $N+1$ is not.", "Solution_8": "[quote=\"surge\"]No... we wouldn't be able to tell if $N+1$ is a leap year...[/quote]\r\n\r\nyou just contradicted yourself by saying year $N$ is a leap year and year $N+1$ isn't :P", "Solution_9": "That isn't contradiction (tsk.. tsk... an integral word in the realm of math). If I said year $N+1$ is a leap year and it is not a leap year then it would be contradiction.\r\n\r\nYear $N+1$ IS NOT a leap year. Year $N$ is a leap year, check lotr's solution and my original solution." } { "Tag": [], "Problem": "If $n$ is positive integer, prove that $(n+1)(n+2)...(2n)$ is divisible by $2^n$", "Solution_1": "Base case: $n=1$\r\nThen the product is just $(1+1)=2$, which is divisible by $2^1=2$.\r\n\r\nInductive step:\r\nAssume $2^{n-1}|(n-1+1)(n-1+2)...(2(n-1))$.\r\nThen $2^{n-1}|n(n+1)...(2n-2)$.\r\nWant to show: $2^n|(n+1)(n+2)...(2n)$.\r\nThe product can also be written $(2n)(n+1)(n+2)...(2n-2)(2n-1)$.\r\nSince $2^{n-1}|n(n+1)...(2n-2)$, then $2^{n-1}|n(n+1)...(2n-2)(2n-1)$.\r\nHowever, the product still has an extra factor of two at the very beginning, so $2^n|(2n)(n+1)...(2n-2)(2n-1)$ and we are done." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "$ \\left|\\begin{array}{cccc} 0&x&y&z\\\\x&0&z&y\\\\y&z&0&x\\\\z&y&x&0\\end{array}\\right|\\equal{}\\left|\\begin{array}{cccc}0&1&1&1\\\\1&0&z^2&y^2\\\\1&z^2&0&x^2\\\\1&y^2&x^2&0\\end{array}\\right|$", "Solution_1": "$ \\left|\\begin{array}{cccc}0 & x & y & z \\\\\r\nx & 0 & z & y \\\\\r\ny & z & 0 & x \\\\\r\nz & y & x & 0\\end{array}\\right| \\equal{} (xyz) \\left|\\begin{array}{cccc}0 & x & y & z \\\\\r\n1 & 0 & z/x & y/x \\\\\r\n1 & z/y & 0 & xy \\\\\r\n1 & y/z & x/z & 0\\end{array}\\right| \\equal{} \\left|\\begin{array}{cccc}0 & xz & xy & yz \\\\\r\n1 & 0 & z & y^2/x \\\\\r\n1 & z^2/y & 0 & x \\\\\r\n1 & y & x^2/z & 0\\end{array}\\right|$\r\n\r\n$ \\equal{} \\frac {1}{xyz} \\left|\\begin{array}{cccc}0 & xyz & xyz & xyz \\\\\r\n1 & 0 & z^2 & y^2 \\\\\r\n1 & z^2 & 0 & x^2 \\\\\r\n1 & y^2 & x^2 & 0\\end{array}\\right| \\equal{} \\left|\\begin{array}{cccc}0 & 1 & 1 & 1 \\\\\r\n1 & 0 & z^2 & y^2 \\\\\r\n1 & z^2 & 0 & x^2 \\\\\r\n1 & y^2 & x^2 & 0\\end{array}\\right|$, for $ x,y,z \\neq 0$." } { "Tag": [], "Problem": "Determine all pairs of positive integers such that\r\n$ \\frac{1}{x}\\plus{}\\frac{x}{y}\\plus{}\\frac{1}{xy}\\equal{}1$", "Solution_1": "[hide]$ y\\plus{}x^{2}\\plus{}1 \\equal{} xy$\n$ x^{2}\\plus{}1 \\equal{} y(x\\minus{}1)$, thus\n$ x\\minus{}1|x^{2}\\plus{}1$. Also\n$ x\\minus{}1|x^{2}\\minus{}1$, so \n$ x\\minus{}1|2$.\nIf x=3, y=5,\nif x=2, y=5.[/hide]", "Solution_2": "[quote=\"galois01\"]Determine all pairs of positive integers such that\n$ \\frac{1}{x}\\plus{}\\frac{x}{y}\\plus{}\\frac{1}{xy}\\equal{} 1$[/quote]\r\n\r\n[hide=\"solution\"]\nCombine all the fractions to get:\n\n$ \\frac{y\\plus{}x^{2}\\plus{}1}{xy}\\equal{}1$.\n\nSolving for $ y$ gives:\n\n$ y\\equal{}\\frac{x^{2}\\plus{}1}{x\\minus{}1}$.\n\nWe can also rewrite the fraction:\n\n$ y\\equal{}x\\plus{}1\\plus{}\\frac{2}{x\\minus{}1}$.\n\n$ x$ is an integer so $ x\\plus{}1$ is also an integer. Thus $ \\frac{2}{x\\minus{}1}$ must also be an integer. However the only values for $ x$ that will make it an integer is when $ x\\equal{}2$ or $ 3$.\n\nTherefore the only solutions are $ \\boxed{(2,5)\\text{ and }(3,5)}$.[/hide]", "Solution_3": "Kondr and jli your solutions are correct :D" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Let $ABDC$ be a tangential quadrilateral with an incircle $(I)$ and $X$ the intersection of the diagonals $AC, BD$. A normal to the diagonal $BD$ through the point $X$ intersects the angle bisectors $AI, CI$ at points $P, R$. Show that $XP = XR$.", "Solution_1": "I think it's been solved before on the forum.\r\n\r\nI'll use something which I won't prove. It's a pretty nice problem in itself: Let $M$ be the projection of $I$ on $BD$. Then $\\angle AMB=\\angle CMD$. Let's refer to this result as $(*)$.\r\n\r\nFrom Menelaus, we see that our conclusion is equivalent to $\\frac{PA}{AI}\\cdot\\frac{IC}{CR}=1\\ (\\#)$. Now let $U=MA\\cap PR,V=MC\\cap PR$. It's easy to see, from the fact that $IM\\|PR$, that $(\\#)$ is equivalent to $\\frac{UA}{AM}\\cdot\\frac{MC}{CV}=1$. This, however, is easy to derive from $(*)$.", "Solution_2": "LOL, grobber - looks like it's easier to you to write down a solution than to find the link... :D\r\n\r\nActually, the previous discussion of the problem was at http://www.mathlinks.ro/Forum/viewtopic.php?t=20538 .\r\n\r\n darij" } { "Tag": [], "Problem": "What kind of cell phone do you like/have/would like to have :)? I'm satisfied about my Sagem MyX-7. Lately I use it more than my PC :)", "Solution_1": "I have one, but I don't know anything about it. I rarely use it. \r\n\r\nYou have a ton of good topics today :D", "Solution_2": ":blush: really? Thanks :)", "Solution_3": "i like all of the nextel phones because they look cool. I wish i had one of them or the motorolla razor.", "Solution_4": "i have a funny looking t mobile with horrible connection becuz t mobile sucks", "Solution_5": "Bah. I still wanna RAZR.", "Solution_6": "Yeah! That's a nice phone. I would like a Motorola mpx300 :D", "Solution_7": "i have a Siemens CF62. its a very convenient phone. you can use it in any country. the motorola RAZR looks awesome. but, what happens when you drop a thin phone like that? ;)", "Solution_8": "Today I went out with some friends and I saw one. It's so thin ... I took it in my hand but that didn't last too much `cuz I was afraid I could broke it :P. Nokia N90 is awesome :o", "Solution_9": "i dont want a RAZR since the battery sucks :P", "Solution_10": "[quote=\"aznmonkey1992\"]i dont want a RAZR since the battery sucks :P[/quote]\r\n\r\nThe put it on the charger when you go to sleep like every other day. You have to do that with most phones anyways.", "Solution_11": "[quote=\"MithsApprentice\"][quote=\"aznmonkey1992\"]i dont want a RAZR since the battery sucks :P[/quote]\n\nThe put it on the charger when you go to sleep like every other day. You have to do that with most phones anyways.[/quote]\r\n\r\nMy phone is strange. When I use it, it sort of \"recharges\" itself. wHen I dont' use it, the batteries drain.", "Solution_12": "Eh, I only use my phone at school. And sometimes on vacations and stuff. Not too good, but it's light weight and small, which is just fine for me.", "Solution_13": "[quote=\"yif man12\"]Eh, I only use my phone at school. [/quote]\r\n\r\nI thought cell phones were illegal in school (if they ring or something). They are in mine.", "Solution_14": "Not really in school. Just calling my parents about pick-up times and things.", "Solution_15": "I want one of those awesome Nextel phones...so awesome ;)", "Solution_16": "I share a phone with my mom, it's a siemens C56 I think... A candy bar phone, I like it. I'm not all that picky though, my brother gave me his old cellphone that I love!!!! I'ts a nokia color screen and you can open it up and it's like a keyboard to send e-mails and txt messages and such. I love it!", "Solution_17": "[quote=\"236factorial\"]I thought cell phones were illegal in school (if they ring or something). They are in mine.[/quote]\r\n\r\nIn our distrcit, you can have them as long as they don't ring. i.e. put it on silent and send text messages.", "Solution_18": "whats a razr", "Solution_19": "I have the really basic nextel phone, the one they're offering free with a contract. It's not really that special, but it's battery lasts for a while if I don't do anything like go on the internet.\r\n\r\nCourse, we should probably be calling nextel something like Sprint Nextel, because that's the name of the new company.", "Solution_20": "[quote=\"noneoftheabove\"]whats a razr[/quote]\r\n\r\nThis is a Motorola v3 Razr :D \r\n\r\n[img]http://www.cacell.co.za/images/v3-large.gif[/img]", "Solution_21": "what are the pros and cons cuz from what i can understand all it does is look cool", "Solution_22": "[quote=\"noneoftheabove\"]what are the pros and cons cuz from what i can understand all it does is look cool[/quote]\r\npros: looks cool\r\ncons: price\r\n\r\ni have a motorola v180 flip phone.", "Solution_23": "lol i have a something phone that i barely know how to work...\r\ni mean...all you need it for is calling...", "Solution_24": "i wouldnt want a phone just because it looked cool i want a fun cuz it works well and the features are cool", "Solution_25": "my mom is getting the black razr.", "Solution_26": "[quote=\"furious\"]pros: looks cool\ncons: price\n\ni have a motorola v180 flip phone.[/quote]\r\n\r\nApparently it also gets dirty quite easily. But I guarantee you there's more pros than that. For one, it has some fairly nice features. And that amazing keypad isn't bad at all.", "Solution_27": "how much does it cost becuz u could get some cell phones for free as long as u buy a certain plan", "Solution_28": "Some Cingular agents offer it for free. T-Mobile also has the phone and you can also get it free with them. But you have to get one of the more expensive plans with T-Mobile.", "Solution_29": "[quote=\"MithsApprentice\"]\n\n\n\n And that amazing keypad isn't bad at all.[/quote]\r\n\r\n\r\nUm, sorry... Amazing? How does it work, it looks like a piece of paper. There's a nokia that I like, I have no idea what the model is...", "Solution_30": "[quote=\"Black_stripe*36\"]How does it work, it looks like a piece of paper. [/quote]\r\n\r\nTouchpad may have been a more appropriate description. But you know what I mean.", "Solution_31": "we bought a certain type of phone only so that we could have 3 phones for free but the connection is horrible" } { "Tag": [], "Problem": "In how many ways can eight horses finish a race provided that there can be ties?", "Solution_1": "[hide=\"Ties of more than 2? \"]\nNo tie-pairs up to 4 tie-pairs.\n$ _{8}P_{8}+_{7}P_{7}+_{6}P_{6}+_{5}P_{5}+_{4}P_{4}$\n\nTies including three and tie-pairs\n$ _{6}P_{6}+_{5}P_{5}+_{4}P_{4}+_{4}P_{4}+_{3}P_{3}$\n\nTies with four and others\n$ _{5}P_{5}+_{3}P_{3}+_{4}P_{4}+_{3}P_{3}+_{2}P_{2}$\n\nTies with five and others\n$ _{4}P_{4}+_{2}P_{2}+_{3}P_{3}$\n\nTies with six and two\n$ _{3}P_{3}+_{2}P_{2}$\n\nTies with seven\n$ _{2}P_{2}$\n\nAll tie\n$ 1$\n\n$ \\sum_{n=1}^{8}n!+\\sum_{x=4}^{6}\\sum_{n=2}^{x}n!+\\sum_{n=3}^{4}n!$\n\nOr if only tie of two, then\n$ \\sum_{n=4}^{8}n! \\ \\square$\n[/hide]", "Solution_2": "http://mathworld.wolfram.com/BellNumber.html", "Solution_3": "Can anyone explain the recursion to me?" } { "Tag": [ "rotation", "number theory", "modular arithmetic" ], "Problem": "The integers from 1 to 15 are written in numerical order in pencil going clockwise around a circle. A student begins moving clockwise around the circle erasing every third integer that has not yet been erased until only the integer 11 remains. Which integer did the student erase first?\r\n\r\n\r\n\r\nSigh, I'm dumb. :D", "Solution_1": "OK....well there is actually a solution for this. Look at problem 15. I think you can solve this with modular arithmetic but I don't know how. But here's a stopgap solution I guess(Problem 15):", "Solution_2": "This is a particular case of the famous [b]Josephus Flavius problem[/b], see [url]http://en.wikipedia.org/wiki/Josephus_problem[/url] and many other links (enter the bold words in Google).", "Solution_3": "Hint: Work Backwards\r\n\r\n[hide=\"Solution\"]\nI don't know why this works, but if you start at 11 and do whatever the problem says, you arrive at the answer, which is 9.\n[/hide]", "Solution_4": "alternatively, you can just pick any number to erase first, let's pick 3\r\nafter you are done, you will end up with a number that would be the answer if our first number erased was 3\r\nin this case, it is 5\r\n\r\nwe know the actual number left is 11, which is 6 more than our answer\r\n\r\nthus, the first number the student erased must be 3+6, or 9", "Solution_5": "[quote=\"vallon22\"]we know the actual number left is 11, which is 6 more than our answer\n\nthus, the first number the student erased must be 3+6, or 9[/quote]\r\n\r\nIs this always true? I'm a bit skeptical at how convenient that is. :huh:", "Solution_6": "it's actually the way i solved it the first time i saw this problem\r\n\r\ni'm pretty sure it always works\r\n\r\nlet's call the first number we erase \"x\"\r\nthen the second number is \"x+3\"\r\nthe third number is \"x+6\", etc (all mod 15, in this case)\r\n\r\nno matter what number we start with, we should get the same answer, since the intervals between the numbers are always the same", "Solution_7": "Think of it this way. You assign letters to the spots (ABC...O), so if you start at C, you end up at E. With the numbers, rotate the whole thing 6 numbers counterclockwise such that 9 is in the C spot and 11 is in the E spot. It will now work." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "sphere", "algebra unsolved", "algebra" ], "Problem": "This is from a Brazilian Olympiad.\r\nFind the maximum volume of a tetrahedron that can be subscribed in a ellipsoid of equation $ \\dfrac{x^2}{9} \\plus{} \\dfrac{y^2}{16} \\plus{} \\dfrac{z^2}{25} \\equal{} 1$\r\nI have the solution here. But I can't understand. The man calculates the volume of the tetrahedron in a sphere of unitary radius. Then he finds the volume of this tethahedron subscribed in this sphere. Soon after, he does a linear transformation to transform the sphere into the ellipsoid. Like this:\r\n\r\nT=\r\n\\[ \\left( \\begin{array}{ccc} 3 & 0 & 0 \\\\\r\n0 & 4 & 0 \\\\\r\n0 & 0 & 5 \\end{array} \\right)\r\n\\]\r\nThen he multiplies the volume of the tethahedron by the determinant of the transformation. Why? \r\nI didn't understand the meaning of this transformation. Is that $ T: \\Re^3 \\minus{} > \\Re^3$ ?", "Solution_1": "Maybe you should see this: http://en.wikipedia.org/wiki/Determinant\r\n\r\nlook at the applications section. It gives a geometric definition for the determinant of a linear transformation." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Given [tex]A=\\{1;2;...;m\\}[/tex]\r\nand [tex]A_1;A_2;..;A_n\\subset A ;A_i\\neq A_j\\forall i\\neq j[/tex]\r\nProve that :exists [tex] x_1;x_2;..;x_n [/tex] such that [tex]x_i\\in A_i \\forall i=i;2;..;n[/tex] and [tex]x_i\\neq x_j \\forall i\\neq j [/tex]\r\nif only if [tex]\\forall I\\subset \\{1;2;...;n\\}[/tex] then [tex]|\\cup\\limit_{i\\in I}A_i|\\ge |I|[/tex]", "Solution_1": "This is Hall's marriage theorem. You could Google it. I'm sure tons of links will turn up." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $G$ be a finite group and intersection of its p-Sylow subgroups is trivial($p\\,|\\,|G|$). Let $G$ has abelian p-Sylow subgroups. Prove that there are $R$ and $Q$ such that $P\\bigcap Q=1$ where $P$ and $Q$ are p-Sylow subgroups of $G$.", "Solution_1": "Don't you mean : \r\n\r\nLet $G$ be a finite group and intersection of its p-Sylow subgroups is trivial($p\\,|\\,|G|$). Let $G$ has abelian p-Sylow subgroups. Prove that there are $P$ and $Q$ such that $P\\bigcap Q=1$ where $P$ and $Q$ are p-Sylow subgroups of $G$?\r\n\r\n :maybe:", "Solution_2": "Yes it is. I'm sorry" } { "Tag": [ "calculus", "LaTeX", "videos", "geometry", "3D geometry", "blogs", "MATHCOUNTS" ], "Problem": "This game is simple.\r\n\r\nFirst, I'll say *inserts coin* and then you say what comes out of the vending machine\r\n\r\nhere i go (hope i dont die)\r\n\r\n*inserts coin*", "Solution_1": "lots of math homework pops out and you spend your time not on the good computer\r\n\r\n*inserts coin*", "Solution_2": "Receives an alternate reality portal.\r\n\r\nInserts a mouse.", "Solution_3": "Nothing comes out.\r\n\r\n*Insert coin*", "Solution_4": "a soda comes out and sprays you with root beer\r\n\r\n*inserts coin*", "Solution_5": "Receives a vending machine that dispenses vending machines that dispense...\r\n\r\nInserts 999 Hyrule Rupees.", "Solution_6": "I remember this game!\r\n\r\nrecieves 999 root beers that were shaken a little too much.\r\n\r\n\r\ninserts an elephant and a donkey.", "Solution_7": "Recieves elephent and donkey ah........ waste.\r\n\r\nPuts in teacher.", "Solution_8": "receives nothing of value\r\n\r\nputs in euro", "Solution_9": "Receives a death-metal listening, circuit debator, sri lankan, serial killer(for all of you people satisfying some of these, don't be insulted, I'm just trying to describe my friend).\r\n\r\n*inserts TWO coins*", "Solution_10": "Recuevers TWO root beers that spray all over you.\r\n\r\n\r\n*Inserts a computer.*", "Solution_11": "Receives the \"I love you\" virus and 3 keys from the keyboard\r\nALT CTRL DEL\r\n\r\n*inserts coin*", "Solution_12": "recieves a pressed penny. :]\r\n\r\ninserts history homework.", "Solution_13": "Receives C+\r\n\r\nInserts 99 mario coins, and a 1-up mushroom.", "Solution_14": "recieves nintendo ds that is broken\r\n\r\n*inserts coin*", "Solution_15": "Receives a black hole.\r\n\r\nInserts MOOFINS! (fake muffins!!)", "Solution_16": "Receive a moderator who locks this thread.", "Solution_17": "Inserts a request PM to demon that asks to lock this hate-filled thread.", "Solution_18": "Receives a \"NO!\" for 1=2\r\n\r\nInserts 7h3.D3m0n.117", "Solution_19": "Receives a pile of poo.\r\n\r\nInserts a coin that I pull out after I insert it.", "Solution_20": "Receives icy pi.\r\n\r\nInserts kimmystar94", "Solution_21": "Receive a request PM to demon that asks to lock this hate-filled thread.", "Solution_22": "Receives nothing.\r\nInserts a lock on the thread.", "Solution_23": "Receives a key to unlock it.\r\n\r\nInserts icy pi", "Solution_24": "Receives fiery pi.\r\nInserts a computer.", "Solution_25": "receives error message..\r\ninserts a PM for this to be locked.", "Solution_26": "Receives a shut up sign for kimmystar94.\r\n\r\nInserts kimmystar94's heart, lungs, and brain. :D", "Solution_27": "Receives a rabbit.\r\nInserts spam.", "Solution_28": "Receives kimmystar94.\r\n\r\nInserts a shameful guy :blush:", "Solution_29": "Yeah, this is getting kind of spammy, sorry I didn't see the posts earlier. \r\n\r\nSigh. It is with great regret that I lock G&FF's current longest running marathon (not to mention it was my sister's creation). \r\n\r\n :(" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a_1,a_2,..,a_n\\in R$. and $ \\max |a_i \\minus{} a_j| \\equal{} 1$ ($ 1\\leq i,j\\leq n$)\r\nFind the minimum of $ A \\equal{} \\sum^n_{i \\equal{} 1}|a_i|^3$", "Solution_1": "nobody done?? :o" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "find all functions from the nonegative integers into themselves, such that: $2f(m^2+n^2)=f^2(m)+f^2(n)$ and for $m\\geq n$ $f(m^2)\\geq f(n^2)$.", "Solution_1": "In fact, you can solve it without using the second condition (that's what I did for Crux some years ago :) )\r\n\r\nPierre." } { "Tag": [ "induction", "ratio", "number theory", "relatively prime", "number theory unsolved" ], "Problem": "$ F_{n}$ is the Fibonacci sequence $ F_{0}= F_{1}= 1$, $ F_{n+2}= F_{n+1}+F_{n}$. Find all pairs $ m > k \\geq 0$ such that the sequence $ x_{0}, x_{1}, x_{2}, ...$ defined by $ x_{0}= \\frac{F_{k}}{F_{m}}$ and $ x_{n+1}= \\frac{2x_{n}-1}{1-x_{n}}$ for $ x_{n}\\not = 1$, or $ 1$ if $ x_{n}= 1$, contains the number $ 1$[/quote] :D", "Solution_1": "assume $ x_{0}\\not=1$ otherwise we are done.\r\nthen $ x_{0}>1$ and $ x_{1}<0$. then a trivial induciton gives us $ x_{n}<0$ so the only solution is $ x_{0}=1$ which directly implies $ m=1$ and $ k=0$...", "Solution_2": "[quote=\"goc\"]Then $ x_{0}>1$ and $ x_{1}<0$.[/quote]\r\n\r\nNo, $ x_{0}<1$.\r\n\r\nm=2n+1, k=2n\r\n\r\nWe see that $ x_{n}=\\frac{x_{n+1}+1}{x_{n+1}+2}$. Then we can easily show by induction that if $ x_{n}=1$, $ x_{n-i}=\\frac{F_{2i}}{F_{2i+1}}$. In particular, $ x_{0}=\\frac{F_{2n}}{F_{2n+1}}$ for any n.\r\n\r\nNow suppose that some other ratio of Fibonacci numbers is of this form, i.e. that $ \\frac{F_{2n}}{F_{2n+1}}=\\frac{F_{k}}{F_{m}}$. Since $ F_{2n}$ and $ F_{2n+1}$ are relatively prime, $ F_{k}>F_{2n}$. Note that the ratio of consecutive terms $ \\frac{F_{n+1}}{F_{n}}$ is strictly increasing. Then $ F_{m}=F_{k}\\frac{F_{2n+1}}{F_{2n}} (x^2 \\plus{} 2mx)^2 \\plus{} 4x$. Contradiction. [/hide]", "Solution_3": "WLOG assume that $ x\\geq y$ then we have:\r\n\r\n$ x^2 2*x and\r\n\r\nx > 2*y\r\nwhich doesn't make sense. Also, I have no idea how you solved for x. I got something completely different. Moreover, the solutions that you \"happened upon\", when plugged back into the equation don't return any of the values of y and x that you posted.\r\n\r\nAnyway, I thought maybe you mistyped and that m,n could be any integer. but even that doesn't make sense. Every number y cannot be represented as \r\n\r\ny = 2*m*x + m^2 (with m,n any integer)\r\n\r\nTake y = 6\r\n\r\nWe get\r\n\r\nm^2 + 2*m*x -6 = 0\r\n\r\nSolving this as a quadratic in m, we get (after simplification)\r\n\r\nm = -x +- sqrt(x^2 + 6) \r\n\r\nbut the quantity x^2 + 6 never is equal to a perfect square because this quantity oscillates between 2 and 3 mod 4, while perfect squares must be 0, or 1 mod 4.\r\n\r\nI have no idea what happened.", "Solution_5": "[quote=\"daydrummer347\"]hsbhatt, I think something went massively awry in your solution. Your basis, the fact that y = 2*m*x + m^2 (for natural numbers m and n) and the other equation for x imply that,\n\ny > 2*x and\n\nx > 2*y\nwhich doesn't make sense.[/quote]\r\n\r\nHe got his two initial equations by letting $ x^{2}\\plus{}y\\equal{}(x\\plus{}m)^{2}$ and $ y^{2}\\plus{}x\\equal{}(y\\plus{}n)^{2}$ and then simplifying. The contradiction you point out is true, but it doesn't invalidate the solution - actually it provides a quicker way to finish the problem. He could have just noticed the contradiction and finished there without solving the equations. Then the solution would be basically equivalent to Babek's.\r\n\r\nI didn't check whether he solved the equation correctly, so you may be right about that.", "Solution_6": "hey sorry abt that.\r\n\r\nyeah. i had done a similar problem recently, the question being find x,y such that $ x^2\\plus{}3y$ and $ y^2\\plus{}3x$ are both perfect squares. I seem to have typed it out wrong. But otherwise its fine" } { "Tag": [ "geometry", "AMC 10" ], "Problem": "I noticed that not so many people like geometry... I think it's more fun that other types of math, because you don't have to go squaring both sides of an equation a million times to solve for x. SO I wonder why people hate Geometry so much... I think I know why. \r\n\r\nThe main reason why I don't like geometry is because there are way too many proofs. Other than that, I like geometry.", "Solution_1": "Same.\r\n\r\nI hate geometry.\r\n\r\nAlthough I got an AMC 10 question right by measuring out with my graph paper, b/c it was draw to scale :D", "Solution_2": "I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's.", "Solution_3": "I fail at geometry.", "Solution_4": "What do you mean? We don't hate geometry!", "Solution_5": "I chose \"Some of the above\" for everything except proofs. Because proofs are part of every Olympiad-level problem.", "Solution_6": "And now we can't bring out programmable graphing calculators with all our formulas :(", "Solution_7": "I don't know how to construct ANYTHING.", "Solution_8": "I am not a good artist. I suck at graphing. I suck at diagrams.", "Solution_9": "I love geometry!", "Solution_10": "Proofs = GOOD. PROOFS = EXTREMELY GOOD. Just not Geometry proofs that teachers make you do. Teachers are bad at making you do proofs in good ways. So everything in geometry is taught the wrong way. I don't hate geometry...I hate school geometry.", "Solution_11": "I hate not drawing diagrams properly.", "Solution_12": "Whoa. I can't believe that the major reason why people hate geometry (not icluding those who love it) is because they can't draw a perfect diagram. I thought proofs were the main reason.", "Solution_13": "The only possible reason I could have for disliking geometry is that I'm not particularly good at it contest proofs in geometry, but that's hardly a reason. I admire those who can do them well :)", "Solution_14": "Depends what you mean by Geometry. I like geometry when it's taught/self-taught the AoPS way. It also depends what you mean by proofs - two-column proofs or real proofs? I despise two-column proofs but I like real proofs. They're hard, I'm bad at them, but they're also fun.", "Solution_15": "[quote=\"pythag011\"]Proofs = GOOD. PROOFS = EXTREMELY GOOD. Just not Geometry proofs that teachers make you do. Teachers are bad at making you do proofs in good ways. So everything in geometry is taught the wrong way. I don't hate geometry...I hate school geometry.[/quote]\r\n\r\nI completely agree!", "Solution_16": "If you do a two-column proof of an IMO problem using basic stuff you would have to use quite a number of pages. The two-column proof is inefficient.", "Solution_17": "The two column proof, in my opinion, is an excellent way to initiate a person into proofs. I also think that, when studying basic geometry, it is important to see how the fundamental postulates and theorems are rigorously used to prove new results. It's worth entertaining even if you already know the stuff.", "Solution_18": "I like geometry, so there's obviously nothing I hate about it! :)" } { "Tag": [ "LaTeX" ], "Problem": "hi\r\nI have a large (too large) table built with tabular. I centered it using \\begin{center} and in ajusting the margins and it worked. Now I want to add a caption to my table so...\r\n\r\n\\newenvironment{changemargin}[2]{%\r\n \\begin{list}{}{%\r\n \\setlength{\\topsep}{0pt}%\r\n \\setlength{\\leftmargin}{#1}% \r\n \\setlength{\\rightmargin}{#2}% \r\n \\setlength{\\listparindent}{\\parindent}% \r\n \\setlength{\\itemindent}{\\parindent}% \r\n \\setlength{\\parsep}{\\parskip}%\r\n }%\r\n\\item[]}{\\end{list}}\r\n\r\n\\begin{changemargin}{-4cm}{-4cm}\r\n\\begin{table}[htb]\r\n\\caption{Matrice des reponsabilit\u00e9s}\r\n\\begin{center}\r\n\\begin{tabular}{c|l|c|c|c|c|}\r\n...\r\n\\end{...bla}\r\n\r\nbut now my table is left align with no consideration to my margin changes and it doesn't fit in the page on the right.\r\n\r\nthank you\r\n\r\nledurt", "Solution_1": "well, i have an answer that worked really well.\r\n\r\n\r\n\\begin{table}\r\n\\caption{...}\r\n\\end{table}\r\n\\begin{tabular}\r\n...\r\n\\end{tabular}\r\n\r\nso case closed!\r\n\r\nledurt" } { "Tag": [], "Problem": "[color=red][b]\u039f \u03c7\u03ce\u03c1\u03bf\u03c2 \u03b1\u03c5\u03c4\u03cc\u03c2 \u03b8\u03b1 \u03bc\u03b5\u03af\u03bd\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b3\u03b9\u03b1 \u039f\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2.[/b][/color] .\r\n\r\n\u0386\u03bb\u03bb\u03b5\u03c2 \u03c4\u03c5\u03c7\u03cc\u03bd \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ad\u03c2 \u03b1\u03c0\u03bf\u03c1\u03af\u03b5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03bc\u03c0\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03b5\u03bc\u03b2\u03cc\u03bb\u03b7\u03bc\u03b1 \u03c3\u03c4\u03b1 \u03bc\u03b7\u03bd\u03cd\u03bc\u03b1\u03c4\u03b1.\r\n\r\n \u0393\u03b9\u03b1 \u03bb\u03c5\u03ba\u03b5\u03b9\u03b1\u03ba\u03ad\u03c2 - \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ad\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 , \u03cc\u03c3\u03bf\u03b9 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf :\r\n\r\nhttp://clubs.pathfinder.gr/MATHEMATICA\r\n\r\n \u0394\u03ce\u03c3\u03b1\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b5\u03c5\u03ba\u03b1\u03b9\u03c1\u03af\u03b1 \u03b3\u03b9\u03b1 \u03b1\u03c1\u03ba\u03b5\u03c4\u03bf\u03cd\u03c2 \u03bc\u03ae\u03bd\u03b5\u03c2 \u03bc\u03b5 \u03c4\u03b1 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03ac \u03c3\u03c4\u03af\u03ba\u03b9\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c0\u03bb\u03b1\u03c4\u03cd\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03ba\u03bf\u03b9\u03bd\u03cc , \u03b1\u03bb\u03bb\u03ac \u03b7 \u03b1\u03bd\u03c4\u03b1\u03c0\u03cc\u03ba\u03c1\u03b9\u03c3\u03b7 \u03ae\u03c4\u03b1\u03bd \u03bc\u03b9\u03ba\u03c1\u03ae. \u03a6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b7 \u03b4\u03b9\u03ac\u03b8\u03b5\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03bc\u03b5\u03bb\u03ad\u03c4\u03b7 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ce\u03bd \u03bc\u03b5 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b9\u03ba\u03bf\u03b9\u03bd\u03c9\u03bd\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1 \u03bc\u03b9\u03ba\u03c1\u03ae \u03c3\u03c4\u03b7\u03bd \u03c0\u03b1\u03c4\u03c1\u03af\u03b4\u03b1 \u03bc\u03b1\u03c2. \u0391\u03c5\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03c6\u03c5\u03c3\u03b9\u03ba\u03cc, \u03b1\u03c6\u03bf\u03cd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03bf \u03b5\u03cd\u03ba\u03bf\u03bb\u03bf \u03c3\u03c7\u03bf\u03bb\u03b5\u03af\u03bf \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c6\u03c1\u03bf\u03bd\u03c4\u03b9\u03c3\u03c4\u03ae\u03c1\u03b9\u03bf. \r\n\r\n \u0395\u03b4\u03ce \u03cc\u03bc\u03c9\u03c2 \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03ba\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b1\u03bd\u03b7\u03c3\u03c5\u03c7\u03af\u03b5\u03c2 \u03c4\u03bf\u03c5 \u03ba\u03bf\u03b9\u03bd\u03bf\u03cd \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c0\u03b5\u03c1\u03b5\u03bc\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5. \u0397 \u03c0\u03c1\u03bf\u03c4\u03b5\u03c1\u03b1\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2. \u039d\u03b1 \u03c3\u03b1\u03c2 \u03b8\u03c5\u03bc\u03af\u03c3\u03c9 \u03c9\u03c3\u03c4\u03cc\u03c3\u03bf \u03cc\u03c4\u03b9 \u03c4\u03b1 \u03c3\u03c4\u03af\u03ba\u03b9\u03c2 \u03b3\u03af\u03bd\u03b1\u03bd \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7 \u03b3\u03b5\u03bc\u03af\u03b6\u03b5\u03b9 \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bc\u03b5 \u03b4\u03b9\u03ac\u03c3\u03c0\u03b1\u03c1\u03c4\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03bd\u03b1 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03b4\u03cd\u03bd\u03b1\u03c4\u03b7 \u03b7 \u03b5\u03cd\u03c1\u03b5\u03c3\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03bb\u03ae\u03c2 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7\u03c2 \u03c4\u03b7 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae \u03c0\u03bf\u03c5 \u03c4\u03b7 \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2. \u039a\u03b1\u03b9 \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 \u03cc\u03bb\u03bf\u03b9 \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03ae\u03c3\u03b5\u03b9 , \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c4\u03b9\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 , \u03ad\u03c3\u03c4\u03c9 \u03bb\u03af\u03b3\u03b5\u03c2 , \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03c3\u03c5\u03b3\u03ba\u03b5\u03bd\u03c4\u03c1\u03ce\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03be\u03b5\u03c7\u03c9\u03c1\u03b9\u03c3\u03c4\u03cc \u03c7\u03ce\u03c1\u03bf.\r\n\u03a4\u03bf \u03be\u03ad\u03c1\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bc\u03b5 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c6\u03c9\u03bd\u03ae \u03ba\u03b1\u03b9 \u03b7 \u03ac\u03c0\u03bf\u03c8\u03b7 \u03c3\u03b1\u03c2 . \u039f \u03b4\u03b9\u03ba\u03cc\u03c2 \u03bc\u03bf\u03c5 \u03c1\u03cc\u03bb\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03c9\u03c4\u03af\u03c3\u03c4\u03c9\u03c2 \u03bd\u03b1 \u03ba\u03c1\u03b1\u03c4\u03ae\u03c3\u03c9 \u03c4\u03bf \u03c7\u03ce\u03c1\u03bf \u03bc\u03b1\u03ba\u03c1\u03c5\u03ac \u03b1\u03c0\u03cc \u03ad\u03c1\u03b9\u03b4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03bd\u03b5\u03c0\u03af\u03c4\u03c1\u03b5\u03c0\u03b5\u03c4\u03b5\u03c2 \u03b5\u03bd\u03ad\u03c1\u03b3\u03b5\u03b9\u03b5\u03c2 \u03ba\u03b1\u03b8\u03ce\u03c2 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03ce \u03b4\u03b9\u03b1\u03ba\u03c1\u03b9\u03c4\u03b9\u03ba\u03ac \u03c4\u03b7\u03bd \u03b1\u03bb\u03bb\u03b7\u03bb\u03bf\u03b3\u03c1\u03b1\u03c6\u03af\u03b1.\r\n \r\n[b]\u0391\u03b3\u03b1\u03c0\u03b7\u03c4\u03bf\u03af \u03c6\u03af\u03bb\u03bf\u03b9 , [/b]\r\n \u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac , \u03bf\u03b9 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03af \u03ba\u03b1\u03b9 \u03c4\u03bf \u03bc\u03ad\u03bb\u03bb\u03bf\u03bd \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd \u03c3\u03b5 \u03c3\u03b1\u03c2. \u03a4\u03c1\u03b1\u03b2\u03ac\u03c4\u03b5 \u03bc\u03c0\u03c1\u03cc\u03c2 \u03ba\u03b1\u03b9 \u03c7\u03b1\u03c1\u03ac\u03be\u03c4\u03b5 \u03c4\u03bf \u03b4\u03b9\u03ba\u03cc \u03c3\u03b1\u03c2 \u03b4\u03c1\u03cc\u03bc\u03bf !\r\n \r\n [color=red][b] \u039a\u03b1\u03bb\u03ae \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 !!![/b][/color]\r\n\r\n [i]( \u0391! \u0391\u03bd \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b5\u03af\u03c4\u03b5 , \u03bd\u03b1 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03b9\u03ba\u03b9\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03ba\u03bb\u03ac\u03b4\u03bf\u03c5\u03c2 \u03c4\u03c9\u03bd 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(\u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03cc\u03c4\u03b1\u03bd \u03bc\u03c0\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03ba\u03b1\u03b8\u03b1\u03c1\u03ac \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bd\u03b1 \u03b2\u03bb\u03ad\u03c0\u03b5\u03b9\u03c2 \u03b1\u03bd \u03c0\u03ae\u03c1\u03b5 \u03b7 \u03c4\u03ac\u03b4\u03b5 \u03bf\u03bc\u03ac\u03b4\u03b1 \u03c4\u03bf\u03bd \u03c4\u03ac\u03b4\u03b5 \u03c0\u03b1\u03af\u03c7\u03c4\u03b7 \u03ae \u03b1\u03c3\u03c4\u03b5\u03b9\u03b1\u03ba\u03b9\u03b1 \u03ba\u03c4\u03bb) .\r\n\u0391\u03c5\u03c4\u03cc \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c6\u03c5\u03c3\u03b9\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03cc . \u0394\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bf\u03cd\u03c4\u03b5 \u03c4\u03bf \u03b1\u03b8\u03bb\u03b7\u03c4\u03b9\u03ba\u03cc \u03c3\u03ac\u03b9\u03c4 \u03bd\u03b1 \u03bc\u03b9\u03bb\u03ac\u03b5\u03b9 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\u0391\u03bb\u03bb\u03ac \u03b1\u03c6\u03bf\u03cd \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03b1\u03bb\u03ac\u03be\u03b5\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1 \u03c4\u03ad\u03bb\u03bf\u03c2 \u03c0\u03ac\u03bd\u03c4\u03c9\u03bd)", "Solution_3": "\u0392\u03b1\u03c3\u03b9\u03ba\u03ac, \u03b8\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03c1\u03c9\u03c4\u03ae\u03c3\u03c9 \u03ba\u03ac\u03c4\u03b9.\r\n\r\n\u03a4\u03b1 topics \u03c0\u03bf\u03c5 \u03b5\u03af\u03c7\u03b1\u03bd \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03b7\u03b8\u03b5\u03af \u03c9\u03c2 stickies \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c0\u03bb\u03ad\u03bf\u03bd \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc; (\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03bf\u03ba, \u03c4\u03b1 \u03b2\u03b3\u03ac\u03bb\u03b1\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \"\u03c0\u03b5\u03c1\u03af\u03bf\u03c0\u03c4\u03b7\" \u03b8\u03ad\u03c3\u03b7, \u03b1\u03bb\u03bb\u03ac, \u03ba\u03b1\u03c4\u03b1\u03c1\u03b3\u03ae\u03b8\u03b7\u03ba\u03b1\u03bd \u03c4\u03b5\u03bb\u03b5\u03af\u03c9\u03c2; )\r\n\r\n\r\n\r\n\u03a4\u03ad\u03bb\u03bf\u03c2, \u03bc\u03b5 \u03c3\u03c5\u03b3\u03c7\u03c9\u03c1\u03b5\u03af\u03c4\u03b5 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03bf\u03c5\u03c3\u03af\u03b1 \u03b1\u03c0\u03cc \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc, \u03b1\u03bb\u03bb\u03ac \u03ad\u03c7\u03c9 \u03b1\u03bd\u03b1\u03bb\u03ac\u03b2\u03b5\u03b9 \u03c4\u03bf \u03c3\u03c4\u03ae\u03c3\u03b9\u03bc\u03bf \u03c4\u03bf\u03c5 \u03b1\u03c3\u03cd\u03c1\u03bc\u03b1\u03c4\u03bf\u03c5 \u03bc\u03b7\u03c4\u03c1\u03bf\u03c0\u03bf\u03bb\u03b9\u03c4\u03b9\u03ba\u03bf\u03cd \u03b4\u03b9\u03ba\u03c4\u03cd\u03bf\u03c5 \u03c3\u03c4\u03bf \u039a\u03b9\u03bb\u03ba\u03af\u03c2 (Kilkis wireless metropolitan network - http://www.kilwmn.gr \u03b3\u03b9\u03b1 \u03c0\u03bb\u03b7\u03c1\u03bf\u03c6\u03bf\u03c1\u03af\u03b5\u03c2 ) \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 \u03c7\u03c1\u03cc\u03bd\u03bf :( \r\n\r\n\u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03b5\u03bc\u03c6\u03b1\u03bd\u03ce\u03c2 \u03c0\u03ac\u03b5\u03b9 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b1 ! \u039c\u03c0\u03c1\u03ac\u03b2\u03bf \u03c3\u03b5 \u03b1\u03c5\u03c4\u03bf\u03cd\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03bf \u03ba\u03b1\u03c4\u03ac\u03c6\u03b5\u03c1\u03b1\u03bd \u03ba\u03b1\u03b9 \u03b1\u03bd\u03b5\u03b2\u03ac\u03b6\u03bf\u03c5\u03bd \u03c4\u03bf\u03bd \u03c0\u03ae\u03c7\u03c5 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ce\u03c2 ! :) \r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2 , \u03b1\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 [b]\u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bc\u03ae\u03bd\u03b1[/b] \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03af \u03c3\u03c4\u03bf site \u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd [b]\u039f\u03ba\u03c4\u03ce\u03b2\u03c1\u03b9\u03bf[/b], \u03b1\u03c2 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03b5\u03af\u03bb\u03b5\u03b9 \u03ad\u03bd\u03b1 pm \u03ae e-mail \u03c3\u03c4\u03bf stfu.gr@gmail.com ! \r\n\r\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce,\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2", "Solution_4": "[quote=\"mostel\"]\u0392\u03b1\u03c3\u03b9\u03ba\u03ac, \u03b8\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03c1\u03c9\u03c4\u03ae\u03c3\u03c9 \u03ba\u03ac\u03c4\u03b9.\n\n\u03a4\u03b1 topics \u03c0\u03bf\u03c5 \u03b5\u03af\u03c7\u03b1\u03bd \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03b7\u03b8\u03b5\u03af \u03c9\u03c2 stickies \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c0\u03bb\u03ad\u03bf\u03bd \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc; (\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03bf\u03ba, \u03c4\u03b1 \u03b2\u03b3\u03ac\u03bb\u03b1\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \"\u03c0\u03b5\u03c1\u03af\u03bf\u03c0\u03c4\u03b7\" \u03b8\u03ad\u03c3\u03b7, \u03b1\u03bb\u03bb\u03ac, \u03ba\u03b1\u03c4\u03b1\u03c1\u03b3\u03ae\u03b8\u03b7\u03ba\u03b1\u03bd \u03c4\u03b5\u03bb\u03b5\u03af\u03c9\u03c2; )\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce,\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2[/quote]\r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf \u039d\u0391\u0399 ! \r\n \r\n\u0391\u03c5\u03c4\u03cc \u03b5\u03af\u03c0\u03b1\u03c4\u03b5 \u03c3\u03c4\u03b1 \u03bc\u03b7\u03bd\u03cd\u03bc\u03b1\u03c4\u03ac \u03c3\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03ae\u03b4\u03b7 \u03c4\u03bf \u03ad\u03c0\u03c1\u03b1\u03be\u03b1. \u0395\u03b4\u03ce \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c9\u03c3\u03c4\u03cc\u03c3\u03bf \u03bd\u03b1 \u03c4\u03bf \u03be\u03b1\u03bd\u03b1\u03c3\u03ba\u03b5\u03c6\u03c4\u03bf\u03cd\u03bc\u03b5 , \u03b1\u03bd \u03c7\u03c1\u03b5\u03b9\u03b1\u03c3\u03c4\u03b5\u03af.\r\n \u0398\u03b1 \u03c6\u03c4\u03b9\u03ac\u03be\u03c9 \u03cc\u03bc\u03c9\u03c2 \u03c4\u03b1 \u03c3\u03c4\u03af\u03ba\u03b9\u03c2 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03b1\u03ba\u03ac \u03ba\u03b5\u03c6\u03ac\u03bb\u03b1\u03b9\u03b1 , \u03ad\u03bd\u03b1 \u03b1\u03bd\u03ac \u03ba\u03b5\u03c6\u03ac\u03bb\u03b1\u03b9\u03bf , \u03b4\u03b9\u03cc\u03c4\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03c4\u03bf site \u03b8\u03b1 \u03b3\u03b5\u03bc\u03af\u03c3\u03b5\u03b9 \u03bc\u03b5 \u03c3\u03ba\u03cc\u03c1\u03c0\u03b9\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 , \u03c0\u03b1\u03c1\u03b5\u03bc\u03b2\u03bf\u03bb\u03ad\u03c2 , \u03b5\u03c1\u03c9\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 , \u03ac\u03c3\u03c7\u03b5\u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03ba\u03bb\u03c0 \u03ba\u03b1\u03b9 \u03cc\u03bb\u03bf \u03c4\u03bf \u03ba\u03b1\u03bb\u03cc \u03ba\u03b1\u03b9 \u03b1\u03be\u03b9\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03b9\u03bc\u03bf \u03c5\u03bb\u03b9\u03ba\u03cc \u03b8\u03b1 \u03b5\u03be\u03b1\u03c6\u03b1\u03bd\u03b9\u03c3\u03c4\u03b5\u03af . \u03a4\u03bf \u03c0\u03ac\u03b8\u03b1\u03bc\u03b5 \u03ae\u03b4\u03b7 \u03c0\u03ad\u03c1\u03b9\u03c3\u03c5 \u03b1\u03c5\u03c4\u03cc \u03ba\u03b1\u03b9 \u03b8\u03c5\u03bc\u03ac\u03c3\u03c4\u03b5 \u03c4\u03b9 \u03b1\u03b3\u03ce\u03bd\u03b1 \u03ba\u03ac\u03bd\u03b1\u03bc\u03b5 \u03bc\u03ad\u03c7\u03c1\u03b9 \u03bd\u03b1 \u03c6\u03c4\u03b9\u03ac\u03be\u03bf\u03c5\u03bc\u03b5 \u03b5\u03ba\u03ad\u03b9\u03bd\u03b1 \u03c4\u03b1 '' \u0393\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03af\u03b1 , \u0391\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 1 - 20 '' , \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c0\u03ac\u03b5\u03b9 \u03b7 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\u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03bf\u03c5\u03bd. \u0388\u03c4\u03c3\u03b9 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03bf\u03cd\u03bd \u03c3\u03c9\u03c3\u03c4\u03ac \u03c0\u03c1\u03cc\u03c4\u03c5\u03c0\u03b1 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b5\u03b3\u03ba\u03b1\u03c4\u03b1\u03bb\u03b5\u03af\u03c0\u03bf\u03c5\u03bd \u03c3\u03c4\u03b7 \u03bc\u03ad\u03c3\u03b7 \u03c4\u03bf\u03bd \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03bf \u03b4\u03c1\u03cc\u03bc\u03bf \u03c4\u03b7\u03c2 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae\u03c2 \u03b5\u03c0\u03b9\u03c3\u03c4\u03ae\u03bc\u03b7\u03c2. \u03a3\u03c5\u03bd\u03bf\u03c0\u03c4\u03b9\u03ba\u03ac , \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03b8\u03b1 \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03c3\u03b5\u03b9 \u03bd\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 , \r\n [b]\u03b1) [/b]\u03b1\u03bd \u03b4\u03b5\u03bd \u03c6\u03cd\u03b3\u03b5\u03c4\u03b5 \u03b5\u03c3\u03b5\u03af\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03bf \u03c6\u03c4\u03b9\u03ac\u03be\u03b1\u03c4\u03b5 \u03bc\u03b5 \u03c4\u03bf \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd \u03ba\u03b1\u03b9 \u03c4\u03bf \u03bc\u03b5\u03c1\u03ac\u03ba\u03b9 \u03c3\u03b1\u03c2\r\n [b]\u03b2) [/b]\u03b1\u03bd \u03bc\u03b5 \u03c4\u03b7 \u03c3\u03c5\u03bc\u03c0\u03b5\u03c1\u03b9\u03c6\u03bf\u03c1\u03ac \u03c3\u03b1\u03c2 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b9\u03bd\u03ae\u03c3\u03b5\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03c4\u03bf\u03c5\u03c2 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03bd\u03b1 \u03ad\u03c1\u03b8\u03bf\u03c5\u03bd \u03ba\u03bf\u03bd\u03c4\u03ac \u03c3\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03b4\u03b9\u03b4\u03b1\u03c7\u03b8\u03bf\u03cd\u03bd \u03b1\u03c0\u03cc \u03b5\u03c3\u03b1\u03c2\r\n [b]\u03b3) [/b]\u03b1\u03bd \u03bf\u03b9 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03bf\u03bd\u03af\u03ba\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03c5\u03c1\u03af\u03c9\u03c2 \u03b1\u03c5\u03c4\u03bf\u03af \u03c0\u03bf\u03c5 \u03c3\u03c0\u03bf\u03c5\u03b4\u03ac\u03b6\u03bf\u03c5\u03bd \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03ae \u03c3\u03b5 \u03c0\u03bf\u03bb\u03c5\u03c4\u03b5\u03c7\u03bd\u03af\u03b1 \u03bc\u03c0\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc , \u03b5\u03af\u03c4\u03b5 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03b5\u03af\u03c4\u03b5 \u03bb\u03cd\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c4\u03c1\u03ad\u03c0\u03bf\u03c5\u03bd \u03c4\u03bf\u03c5\u03c2 \u03bd\u03b5\u03cc\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03bf\u03cd\u03bd \u03cc\u03bb\u03bf \u03ba\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b2\u03b1\u03b8\u03b5\u03b9\u03ac \u03bc\u03b5 \u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2.\r\n [b] \u03b4)[/b] \u0391\u03bd \u03ba\u03b1\u03c4\u03b1\u03c6\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b2\u03b3\u03ac\u03bb\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03b2\u03ac\u03bb\u03bb\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03ad\u03c1\u03b3\u03bf \u03bc\u03b1\u03c2.\r\n\r\n\r\n \r\n \u0391\u03c5\u03c4\u03ac \u03b3\u03b9\u03b1 \u03c4\u03ce\u03c1\u03b1 . \u03a4\u03b1 \u03be\u03b1\u03bd\u03b1\u03bb\u03ad\u03bc\u03b5\r\n\r\n\r\n \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2", "Solution_5": "\u0393\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c0\u03b1\u03c1\u03b5\u03be\u03b7\u03b3\u03b7\u03b8\u03ce \u03b3\u03c1\u03ac\u03c6\u03c9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03bc\u03cd\u03bd\u03b7\u03bc\u03b1 . \r\n\u0395\u03b3\u03ce \u03b5\u03af\u03bc\u03b1\u03b9 \u03c5\u03c0\u03ad\u03c1 \u03c4\u03bf\u03c5 stickies \u03c0\u03bf\u03c5 \u03c6\u03c4\u03b9\u03ac\u03be\u03b1\u03c4\u03b5 \u03b3\u03b9\u03b1 \u039f\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b1 \u039d\u03ad\u03c9\u03bd \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03b5\u03af\u03bc\u03b1\u03b9 \u03b4\u03af\u03c0\u03bb\u03b1 \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03c0\u03bf\u03c1\u03af\u03b1 \u03bc\u03b9\u03ba\u03c1\u03bf\u03cd \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b5\u03af \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03b7 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2 (\u03ad\u03c3\u03c4\u03c9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c4\u03bf\u03c5 Cauchy)\r\n\u0391\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b5\u03b4\u03ce \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd (\u03b5\u03ba \u03c4\u03c9\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03ac\u03c4\u03c9\u03bd ) \u03bd\u03b1 \u03bb\u03cd\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ad\u03c2 \u03b1\u03c0\u03bf\u03c1\u03af\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03b3 \u03bb\u03c5\u03ba\u03b5\u03af\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03bb\u03cd\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b3\u03b9\u03b1 \u03c6\u03bf\u03b9\u03c4\u03b7\u03c4\u03ad\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bd \u03bd\u03b1 \u03c0\u03b5\u03c1\u03ac\u03c3\u03bf\u03c5\u03bd \u03c4\u03bf \u03bc\u03ac\u03b8\u03b7\u03bc\u03b1 . \r\n\u0395\u03bb\u03c0\u03af\u03b1\u03c9 \u03bd\u03b1 \u03be\u03b5\u03ba\u03b1\u03b8\u03ac\u03c1\u03b9\u03c3\u03b1 \u03c4\u03b7 \u03b8\u03ad\u03c3\u03b7 \u03bc\u03bf\u03c5", "Solution_6": "[quote=\"silouan\"]\u0393\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c0\u03b1\u03c1\u03b5\u03be\u03b7\u03b3\u03b7\u03b8\u03ce \u03b3\u03c1\u03ac\u03c6\u03c9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03bc\u03cd\u03bd\u03b7\u03bc\u03b1 . \n\u0395\u03b3\u03ce \u03b5\u03af\u03bc\u03b1\u03b9 \u03c5\u03c0\u03ad\u03c1 \u03c4\u03bf\u03c5 stickies \u03c0\u03bf\u03c5 \u03c6\u03c4\u03b9\u03ac\u03be\u03b1\u03c4\u03b5 \u03b3\u03b9\u03b1 \u039f\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b1 \u039d\u03ad\u03c9\u03bd \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03b5\u03af\u03bc\u03b1\u03b9 \u03b4\u03af\u03c0\u03bb\u03b1 \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03c0\u03bf\u03c1\u03af\u03b1 \u03bc\u03b9\u03ba\u03c1\u03bf\u03cd \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b5\u03af \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03b7 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2 (\u03ad\u03c3\u03c4\u03c9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c4\u03bf\u03c5 Cauchy)\n\u0391\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b5\u03b4\u03ce \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd (\u03b5\u03ba \u03c4\u03c9\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03ac\u03c4\u03c9\u03bd ) \u03bd\u03b1 \u03bb\u03cd\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ad\u03c2 \u03b1\u03c0\u03bf\u03c1\u03af\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03b3 \u03bb\u03c5\u03ba\u03b5\u03af\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03bb\u03cd\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b3\u03b9\u03b1 \u03c6\u03bf\u03b9\u03c4\u03b7\u03c4\u03ad\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bd \u03bd\u03b1 \u03c0\u03b5\u03c1\u03ac\u03c3\u03bf\u03c5\u03bd \u03c4\u03bf \u03bc\u03ac\u03b8\u03b7\u03bc\u03b1 . \n\u0395\u03bb\u03c0\u03af\u03b1\u03c9 \u03bd\u03b1 \u03be\u03b5\u03ba\u03b1\u03b8\u03ac\u03c1\u03b9\u03c3\u03b1 \u03c4\u03b7 \u03b8\u03ad\u03c3\u03b7 \u03bc\u03bf\u03c5[/quote]\r\n\r\n[b]\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad, [/b]\r\n\u03bc\u03b7\u03bd \u03b1\u03bd\u03b7\u03c3\u03c5\u03c7\u03b5\u03af\u03c2 \u03b3\u03b9\u03b1\u03c4\u03af \u03ae\u03c3\u03bf\u03c5\u03bd\u03b1 \u03c3\u03b1\u03c6\u03ae\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03ae. \u0394\u03b5\u03bd \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b5\u03be\u03b7\u03b3\u03b5\u03af \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03cc\u03bb\u03bf\u03b9 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\u03ac\u03bb\u03bb\u03bf \u03ba\u03b1\u03b9 \u03b3\u03c1\u03ac\u03c8\u03c4\u03b5 \u03ba\u03ac\u03c4\u03b9 . \u039c\u03af\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03c1\u03c7\u03ae \u03ad\u03b3\u03b9\u03bd\u03b5 .\u0391\u03c2 \u03bc\u03b7\u03bd \u03ba\u03bf\u03bb\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03b4\u03b9\u03b1\u03b4\u03b9\u03ba\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac \u03ba\u03b1\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1. \u03a4\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03b1 \u03be\u03b5\u03ba\u03b1\u03b8\u03b1\u03c1\u03af\u03c3\u03b5\u03b9 . \u0397 \u03b6\u03c9\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ae \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b1 . \u0397 \u03c0\u03b1\u03c4\u03c1\u03af\u03b4\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03b3\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03c4\u03ad\u03bb\u03bc\u03b1. \u03a7\u03c9\u03c1\u03af\u03c2 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae \u03c3\u03ba\u03ad\u03c8\u03b7 , \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5 \u03b8\u03b1 \u03b3\u03af\u03bd\u03b5\u03b9 \u03c0\u03bf\u03c4\u03ad ! \r\n \u03a0\u03ac\u03bc\u03b5 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd ! \r\n\r\n[b] \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2[/b]\r\n\r\n [b]\u0386\u03bd\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03ba\u03b1\u03bb\u03cc \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c3\u03b1\u03c2 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ae\u03c3\u03c9:[/b] \r\n\r\n [color=blue]\u03a3\u03b5 \u03bc\u03b5\u03c1\u03b9\u03ba\u03ad\u03c2 \u03bc\u03ad\u03c1\u03b5\u03c2 \u03b8\u03b1 \u03b2\u03ac\u03bb\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03bf \u03c7\u03ce\u03c1\u03bf \u03bc\u03b1\u03c2 \u03c4\u03b7\u03bd \u03b2\u03af\u03b2\u03bb\u03bf \u03c4\u03b7\u03c2 \u03b8\u03b5\u03c9\u03c1\u03af\u03b1\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd. \u0388\u03bd\u03b1 \u03c3\u03c0\u03ac\u03bd\u03b9\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf (\u03c3\u03b5 pdf) \u03ba\u03b1\u03b8\u03b1\u03c1\u03ac \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03b1\u03ba\u03bf\u03cd \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03ae\u03c1\u03b1. \u0398\u03b1 \u03bb\u03b1\u03c4\u03c1\u03ad\u03c8\u03b5\u03c4\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 ! \u0398\u03b1 \u03c4\u03bf \u03c3\u03c4\u03b5\u03af\u03bb\u03bb\u03c9 \u03c3\u03c4\u03bf \u03a3\u03c4\u03ad\u03bb\u03bb\u03b9\u03bf.[/color]" } { "Tag": [ "function", "algebra open", "algebra" ], "Problem": "find all the Functions, for that all x and y are real, \r\n\r\nf(x.f(y))=f(x.y)+x\r\n\r\nwhen f:R---> R", "Solution_1": "Taking $ x \\equal{} 1$ yields $ f(a) \\equal{} a \\plus{} 1$ for $ a\\in f\\left( \\mathbb{R}\\right)$. Taking $ y \\equal{} 0$ gives $ f(cx) \\equal{} x \\plus{} c$ for some constant $ c$ and all real $ x$. Since $ f\\left( \\mathbb{R} \\right)\\neq \\emptyset$, we must have $ c \\equal{} 1$ and hence $ f(x) \\equal{} x \\plus{} 1$ for all real $ x$.\r\n\r\nA problem of this level would more likely belong in the pre-olympiad forum." } { "Tag": [], "Problem": "What is the greatest number of points of intersections can be made with two circles and a square?\r\n[hide=\"CLUE\"][size=59] Use a pattern :cool: [/size][/hide]", "Solution_1": "Two circles cut at at most two places. (Can i assume they arn't the same circle?) The maximum crossing points of a square and a circle is 8. Therefore, the absolute maximum is $8+8+2=18$ A quick sketch shows this will work. (draw a square and a circle inside it that cuts it in eight places. Draw another circle that is nearly exactly the same place, of equal radius. QED)", "Solution_2": "Since nobody is answering I'll Just post the answer.\r\n\r\nWhat seamusoboyle said is the practicly the same thing I was going to say you just need to put the two circles where they cross over one another but they still make 8 oints of intersection with the sqaure", "Solution_3": "It looks like an illusion\r\n\r\nlooking from the outer side of the square, it's bent\r\n\r\nbut\r\n\r\nif you follow the lines, they are all straight\r\n\r\nmy eyes hurt now....\r\n\r\nby the way,\r\n\r\nhow do you use patterns for this problem?", "Solution_4": "I believe its 18.", "Solution_5": "Woah, that illusion is so cool. The circles look like ovals, the square looks like it's curved. Wow! :rotfl:", "Solution_6": "Its not supposed to be an Illusion and I was wrong about using a pattern" } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "Here go three famous problem:\r\n1. Characterize all functions f: R -> R such that f(x) + f(y) = f(x+y).\r\n\r\n2. Does there exist a function f: R -> R which is continuous at all irrationals but discontinuous at the rationals?\r\n\r\n3. Does there exist a function f: R -> R which is everywhere continuous but nowhere differentiable?\r\n\r\nAnd here is a problem posed by Gabriel Carroll. I would imagine it has been considered before, but I have not seen it:\r\nDoes there exist a continuous f: R -> R such that f(x) is rational iff x is irrational?", "Solution_1": "[quote=\"gauss202\"]\n\n3. Does there exist a function f: R -> R which is everywhere continuous but nowhere differentiable?\n[/quote]\r\n\r\nVan Der Waerden function\r\nFor t \\in R , =d(t;Z)=inf{|t-m|; m\\in Z}\r\nConsider f(t)=\\sum_{n \\geq 0}<(10^n).t>/10^n f is defined R->R\r\ncontinous, 1-periodic, nowhere differentiable.\r\n\r\nAnother example Weierstrass function\r\nf(t)=\\sum_{n \\geq 0}2^(-n).cos(((2004)^n)t)\r\n\r\nMr.Omarjee\r\nSTT teacher", "Solution_2": "For the Carroll one, consider f(x)*x.", "Solution_3": "[quote=\"gauss202\"]Here go three famous problem:\n3. Does there exist a function f: R -> R which is everywhere continuous but nowhere differentiable?\n[/quote]\r\nyou can find the answer to this question nearly in all textbooks on mathematical analysis.", "Solution_4": "I never knew the Weierstrass function was so simple in form. I remember reading that he shocked the entire mathematical world by discovering this.\r\n\r\nI think problem 2 was solved b4 on this forum.", "Solution_5": "For problem 2, yes, there is : \r\nx irrational ==> f(x)=0\r\nx=p/q rational , (p,q) coprimes ===> f(p/q) = 1/q", "Solution_6": "I think that's called Riemann's function. I was wrong, I thought it was another problem. I was just wondering: what is f(0)? Are you sure the soln works?", "Solution_7": "Yes, Diogene's function does in deed work! Though he needs to make an appropriate definition for f(0). Can you see why this function works? It is very neat! I don't know what the name of this function is though, if it has a name. I've never heard it called the Riemann function, but then again there are so many functions known as the Riemann function it's hard to keep track!\r\n\r\nAs for the Weierstrass function, it is in deed very interesting. I did my senior thesis as an undergraduate on this subject. Using wavelets to study the regularity (i.e. degree of \"non-differentiability\") of these types of functions.\r\n\r\nThere was actually an even more famous function before this one, called the Riemann function (yes another one!! lol!!) defined by f(t) = \\sum {n \\geq 1}[sin (pi*n^2*t)/n^2], which was shown to be continuous everywhere, but differentiable only at rational points, p/q, with both odd numerator and denominator!! This remarkable result was shown by Gerver in 1971 when he was still an undergraduate!!\r\n\r\nSo all of the questions have been appropriately solved except for the first one - which is the easiest, but also very interesting in my opinion.\r\n\r\nCongradulations to Mindspa on the Carroll one! That was a very elegant and clever solution.", "Solution_8": "Gerver also discovered the function, or it was known before, but he was the first to prove (or to even try?!) that it is continous everywhere and differentiable only on rationals? :?", "Solution_9": "No, it was known before Gerver. It was proposed by Riemann as a function not differentiable on a set dense in the reals. But this was first proven and published about by Weirstrass (hence it is also sometimes known as the Weierstrass function - how confusing!). However his proof did not say at which points, if any, it was differentiable.\r\n\r\nThe problem was since worked on by many people, including Hardy, who showed that it was not differentiable at any irrational points. However it was not until Gerver that the problem of it's differentiability was completely solved. His proof was long, but more or less elementary (meaning it did not require any hard results). Quite a remarkable feat for an undergraduate! I have no idea what he has gone on to do in his career.\r\n\r\nBut it just shows that some unsolved problems are not out of the reach of undergraduate students!", "Solution_10": "Yeah, I think I was hasty about the function Diogene posted. I thought it wouldn't work because it wasn't defined properly in 0, but all we need to do is make f(0)=a=/=0 and it's done.\r\n\r\nThe problem that was solved b4 and which I was speaking abt was this: does there exist a function f:[0,1]->R s.t the pts of discontinuity are the irrational points of [0,1] ? I think the answer for this one was no." } { "Tag": [ "calculus", "derivative", "function", "real analysis", "real analysis unsolved" ], "Problem": "suppose that a differentiable real function has a periodical nonconstant derivative f:R--->R .prove that f has at least positive period", "Solution_1": "[quote=\"drapt@\"]suppose that a differentiable real function has a periodical nonconstant derivative f:R--->R .prove that f has at least positive period[/quote]\r\nI dont understand the problem. Consider $ .5x^2$ its derivative is $ x$ which is a non-periodic function? :?:", "Solution_2": "[quote=\"sylow_theory\"]Consider $ .5x^2$ its derivative is $ x$ which is a non-periodic function? :?:[/quote]\r\n\r\nand so this is not a counterexample for this :wink:" } { "Tag": [ "analytic geometry" ], "Problem": "The rocket MaoMao is on the point (3, -5) and needs to get to (10,1) by only moving up and right. How many possible paths are there along gridlines?\r\n\r\nI've seen all kinds of these problems, with pictures too, but do not know how to approach these.", "Solution_1": "[hide]he goes to the right 7 times and up 6 times, and since any order works thats 13!/(6!7!) = 13C6[/hide]\r\n\r\nextension: what if he cannot pass through the point (6,3)?", "Solution_2": "[quote=\"nateharman1234\"]\n\nextension: what if he cannot pass through the point (6,3)?[/quote]\r\n\r\nDid you mean some other coordinate? If so, I guess you would just subtract the triangle formed from that point you can go to, and 10,1.", "Solution_3": "[quote=\"nateharman1234\"]he goes to the right 7 times and up 6 times, and since any order works thats 13!/(6!7!) = 13C6 [/quote]\n\nWe can think of this as having to arrange 7 R's and 6 U's to form a 13 letter word where R represents a right and U represents a up.\n\n[quote=\"nateharman1234\"]extension: what if he cannot pass through the point (6,3)?[/quote]\r\nThe ans will be the same as he will never go to (6,3) as the y coord > given y coord and he is taking the shortest path." } { "Tag": [ "Asymptote", "abstract algebra" ], "Problem": "I am having trouble with asymptote, and I was hoping someone could help me. I have installed TeXnic, GSview, GhostScript, and Asymptote. I have done what it says in the asymptote getting started in AoPSWiki. When I paste the commands into a new file (which I saved as picture.asy) and run asymptote, a black box pops up, stays blank, flashes an error message (which I think says: error: could not load module picture.asy) and immediately closes. If I open asymptote and tell it to draw a line, a GSview window opens and I see the line. Will someone please help me be able to use Asymptote in Texnic? Thank you very much.", "Solution_1": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148557]this thread[/url] for ways to view error messages.", "Solution_2": "Thank you very much!" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $\\triangle ABC$.\r\nThe inequality $\\sum \\frac{1}{a}\\cdot \\sum a^{2}\\ge 3\\cdot \\sum a$ is abviously.\r\nProve that a reinforcement of this inequality:\r\n$\\sum \\frac{1}{a}\\cdot \\sum a^{2}\\ge 3\\cdot \\sum a+\\sum \\frac{a}{bc}\\cdot (b-c)^{2}\\ge \\sum \\frac{1}{a}\\cdot \\sum ab\\ge 3\\cdot \\sum a.$", "Solution_1": "The original inequality is equivalent to $\\sum a(b+c-a)(b-c)^2 \\ge 0$ which I think may not be true.. :huh:", "Solution_2": "I am sorry, you didn't watch condition: $\\triangle ABC$ ! This inequality is a manufacture of the my inequality (1) from the post \"uncommon\". See \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=44897", "Solution_3": "[quote=\"levi\"]I am sorry, you didn't watch condition: $\\triangle ABC$ ! This inequality is a manufacture of the my inequality (1) from the post \"uncommon\". See \nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=44897[/quote]\r\noops, I really missed the condition a,b,c are sidelengths of a triangle.\r\nThat means I've solved it. :D", "Solution_4": "Actually,Levi about the first one,here is better one,but trivial:\r\n\r\n $\\sum\\frac{1}{a}\\cdot \\sum ab \\geq 3\\cdot \\sum a $ ;) so now you may reconsider a new improvment for the second one. ;)\r\n\r\n\r\n\r\n By the way, this is true for any $a,b,c>0$", "Solution_5": "May be true the following inequality [b](Try you !)[/b]:\r\n$\\sum \\frac{1}{a}\\cdot \\sum ab\\le 3\\cdot \\sum a+\\sum \\frac{a}{bc}\\cdot (b-c)^{2}.$", "Solution_6": "Well,if we assume that it is true, then it's for $a,b,c$ sides of a triangle.I don't know,I'll give it a try, but not now.\r\n Tomorrow I am going to do that. Now I'm very sleepy. :sleeping:", "Solution_7": "I improved the initial message. Caesar, I solve the your assumpsion.", "Solution_8": "[quote=\"Cezar Lupu\"][size=117][color=darkred]Well,if we assume that it is true, then it's for $a,b,c$ sides of a triangle.\nI don't know,I'll give it a try, but not now. Tomorrow I am going to do that. Now I'm very sleepy.[/color][/size] :sleeping:[/quote]\r\n\r\n[color=darkblue]See http://www.mathlinks.ro/Forum/viewtopic.php?t=47305\n\nThe my $\\triangle ABC$-inequality (proposed in Aug. 07 2005)\n \n$\\sum\\frac{1}{a}\\cdot\\sum bc\\le3\\sum a+\\sum\\frac{a}{bc}\\cdot(b-c)^{2}$\n\nyou didn't prove so far, Aug 17 2006 ! [b]Please, try again ... to-morrow ![/b][/color]", "Solution_9": "Actually, just expand it out. \\[3\\sum{a}+\\sum(\\frac{a}{bc}(b-c)^{2})=3\\sum{a}+\\sum(\\frac{ab}{c}+\\frac{ac}{b}-2a)\\] \\[=\\sum{a}+\\sum(\\frac{ab}{c}+\\frac{ac}{b})=\\sum{a}+2\\sum\\frac{ab}{c}\\] Whereas \\[\\sum{\\frac{1}{a}}\\sum ab=2\\sum{a}+\\sum\\frac{ab}{c}\\] So we then are talking about a comparison between $a+b+c$ and $\\frac{ab}{c}+\\frac{bc}{a}+\\frac{ca}{b}$. But then \\[(ab)^{2}+(bc)^{2}+(ca)^{2}\\ge (ab)(bc)+(bc)(ca)+(ca)(ab)=abc(a+b+c)\\] \\[\\implies \\frac{ab}{c}+\\frac{bc}{a}+\\frac{ca}{b}\\ge a+b+c\\] which tells us \\[3\\sum{a}+\\sum(\\frac{a}{bc}(b-c)^{2}) \\ge\\sum{\\frac{1}{a}}\\sum ab\\] ...\r\n\r\nAfter that, the last of the three inequalities should be easy, by the same principle, and the first should be easy also. Essenaitlly, it all reduces to \\[\\sum(a+\\frac{a^{2}}{b}+\\frac{a^{2}}{c}) \\ge \\sum(a+2\\frac{ab}{c}) \\ge \\sum(2a+\\frac{ab}{c}) \\ge \\sum(3a)\\] which all follows from the fact that $\\sum{a}\\le \\sum{\\frac{ab}{c}}$." } { "Tag": [], "Problem": "If $a,b,c$ are the three sides of a right triangle , and $a^{n}+b^{n}+c^{n}=2(a^{2n}+b^{2n}+c^{2n})$ . And $n \\geq 2$ , $n$ is a integer , can you tell me $n = ?$", "Solution_1": "I think you can use subst $a^{n}=x^{n}+y^{n}$,etc." } { "Tag": [ "induction", "geometric series" ], "Problem": "After watching the show THE HAPPENING, thers this maths teacher who gave a maths quiz, its about if u get 2 dollars on the first day of April, and the 2nd u day u get 2 times the amount of the first day and continue doing that for the next 30 days. \r\nI found out its actualy 2^1+2^2+2^3+2^4................+2^30. How do i sum up 2^1+2^2+2^3+2^4...................+2^30[/code]", "Solution_1": "I think you have to do it manually.\r\n\r\n$ 2^1 \\plus{} 2^2 \\plus{} 2^3 \\plus{} 2^4....\\plus{} 2^{29} \\plus{} 2^{30}$\r\n\r\nis\r\n\r\n$ 2 \\plus{} 4 \\plus{} 8 \\plus{} 16 ....\\plus{} 536870912 \\plus{} 1073741824$", "Solution_2": "No need for that. Notice that it's actually $ 2^{31} \\minus{} 1$.\r\n\r\nWhy? $ a^n \\minus{} b^n \\equal{} (a \\minus{} b)(a^{n \\minus{} 1} \\plus{} a^{n \\minus{} 2}b \\plus{} ... \\plus{} ab^{n \\minus{} 1} \\plus{} b^{n \\minus{} 1})$. Now plug in $ a \\equal{} 2$, $ b \\equal{} 1$, $ n \\equal{} 31$.\r\n\r\nEdit: Actually, you have an extra dollar in this calculation, (can you see why-it's the $ b^{n \\minus{} 1}$ member) so the result is actually $ 2^{31}\\minus{}2$.", "Solution_3": "Which, by the way, is $ \\$2147483646$.", "Solution_4": "i dun under stand the a^n-b^n formula.....so deep", "Solution_5": "There are unfortunately only two ways to simplify this calculation: first one is this ($ a^n\\minus{}b^n$), and the second one is by geometric series (probably too advanced for your level), so I used this approach, as it is how the [url=http://en.wikipedia.org/wiki/Second_half_of_the_chessboard]wheat and chessboard problem[/url] (more famous version of your problem) was explained to me in middle school. \r\n\r\nI don't see a nice approach to the explaining of the formula. I hope someone else can explain it...", "Solution_6": "Well, if you don't want to be so rigorous, you can notice that:\r\n\r\n\\begin{align*}\r\n2 &= 2 = 2^2 - 2 \\\\\r\n2 + 4 &= 6 = 2^3 - 2 \\\\\r\n2 + 4 + 8 &= 14 = 2^4 - 2 \\\\\r\n2 + 4 + 8 + 16 &= 30 = 2^5 - 2 \\\\\r\n2 + 4 + 8 + 16 + 32 &= 62 = 2^6 - 2 \\\\\r\n2 + 4 + 8 + 16 + 32 + 64 &= 126 = 2^7 - 2\r\n\\end{align*}\r\n\r\nYou might notice that pattern and deduce that $ 2 + 4 + 8 + \\dots + 2^n = 2^{n + 1} - 2$.\r\n\r\nIf you want to be more rigorous, you can use induction:\r\n\r\n[b]Base case[/b]\r\n$ n = 1$: $ 2 = 2^2 - 2$. Check.\r\n\r\n[b]Inductive hypothesis[/b]\r\n$ 2 + 4 + 8 + \\dots + 2^n = 2^{n + 1} - 2$\r\n\r\n[b]Inductive step[/b]\r\nAssume $ k = n$. Then for $ k + 1$:\r\n\r\n$ 2 + 4 + 8 + \\dots + 2^k + 2^{k + 1} = 2^{k + 1} - 2 + 2^{k + 1} = 2 \\cdot 2^{k + 1} - 2 = 2^{k + 2} - 2$\r\n\r\nTherefore, if we proved that IF it works for $ n$, then it works for $ n + 1$. Since we proved our base case for $ n = 1$, then it works for 2, 3, 4, etc...", "Solution_7": "Geometric series is nice to learn btw too... it comes up a lot even in the middle school forums (although some of the stuff in the MC forum arguably could go in the Intermediate HS section lol...) Here is one case:\r\n\r\n$ \\displaystyle 1\\plus{} r\\plus{}r^2\\plus{}r^3\\plus{}r^4\\plus{}\\ldots\\plus{}r^n \\equal{} \\frac{r^{n\\plus{}1} \\minus{} 1}{r\\minus{}1}$", "Solution_8": "This is simply a specific case of the finite geometric series formula, with $ a\\equal{}2$ and $ r\\equal{}2$. Then, just plug in:\r\n\\[ S\\equal{}\\frac{a\\left(1\\minus{}r^{n\\plus{}1}\\right)}{1\\minus{}r}\\]", "Solution_9": "For a proof of this specific case define the sum to be [b]S[/b].\r\nThen [b]Sr=r+r^2+r^3+...+r^n+r^(n+1)[/b]....1\r\nSubtracting [b]S[/b] from 1 we get,\r\n[b]Sr-r=S(r-1)=r+r^2+r^3+...+r^n+r^(n+1)-(1+r+r^2+r^3+...r^n)\n =r^(n-1)-1.[/b].\r\nTherefore we get [b]S=r^(n-1)-1/(r-1)[/b]", "Solution_10": "Here's another way to solve it:\r\n\r\nLet $ x \\equal{} 2^0 \\plus{}2^1 \\plus{} 2^2 \\plus{} 2^3 \\plus{} ... \\plus{} 2^n$\r\n\r\nWe multiply by $ 2$ on both sides.\r\n\r\n$ 2x \\equal{} 2(2^0 \\plus{} 2^1 \\plus{} 2^2 \\plus{} 2^3 \\plus{} ... \\plus{} 2^n$\r\n\r\nThis is the same as:\r\n\r\n$ 2x \\equal{} 2(2^0) \\plus{} 2(2^1) \\plus{} ... \\plus{} 2(2^n)$\r\n\r\nWhen you multiply numbers with the same bases, you add the exponents. $ a^x \\times a^y \\equal{} a^{x \\plus{} y}$.\r\n\r\n$ 2x \\equal{} 2^{0\\plus{}1} \\plus{} 2^{1 \\plus{} 1} \\plus{} 2^{2 \\plus{} 1} \\plus{} ... \\plus{} 2^{n \\plus{} 1}$\r\n\r\nwhich simplifies to:\r\n\r\n$ 2x \\equal{} 2^1 \\plus{} 2^2 \\plus{} 2^3 \\plus{} 2^4 \\plus{} ... \\plus{} 2^{n \\plus{} 1}$\r\n\r\nWhat's the difference between this and our first equation? We notice that $ 2^0$ is missing, and we have an extra term - $ 2^{n \\plus{} 2}$. So we can make this a new equation. \r\n\r\n$ x \\minus{} 2^0 \\plus{} 2^{n \\plus{} 1} \\equal{} 2x$\r\n\r\nAny number to the $ 0$th power is equal to $ 1$(except 0.)\r\n\r\n$ x \\minus{} 1 \\plus{} 2^{n \\plus{} 1} \\equal{} 2x$\r\n\r\nNow we can solve.\r\n\r\n$ 2^{n \\plus{} 1} \\minus{} 1 \\equal{} x$\r\n\r\nGreat! So now we know the answer! The original problem was up to $ 2^{30}$, so we can substitute this.\r\n\r\n$ 2^{30 \\plus{} 1} \\minus{} 1 \\equal{} x$\r\n\r\n$ 2^{31} \\minus{} 1 \\equal{} x$\r\n\r\nAnd that is our answer.", "Solution_11": "Unfortunately, it doesn't equal $ 2147483646$ which is the right answer.\r\nIt equals $ 2147483647$ (this also happens to be the max int)", "Solution_12": "My favorite way of looking at it (very easy to understand):\r\n1+1+2+4+8+16+32...\r\nThe first 1+1 makes 2, plus the next 2 makes 4, plus the next 4 makes 8, etc. This pattern keeps continuing and as you can see, it always makes double the last number. So for your problem, you'd have to subtract the first two ones from the final answer.\r\nUsing this method, find $ 3 \\plus{} 2\\cdot3 \\plus{} 2\\cdot9 \\plus{} 2\\cdot27$...\r\n\r\nStarting out with 2^n and successively subtracting all the powers of 2 below it is another way to see it. Let's say you start out with 16. Subtracting 8, you have the other 8 left. Subtracting 4, you have the other 4 left. Subtracting 2, you have the other two left. Subtracting the last one gives 1 left over. So 8+4+2+1=16-1.\r\n\r\nAnother way to look at it is using base. \r\n1+2+4+8+16...=1111111... (base 2)=1000000...-1(base 2)=$ 2^n \\minus{} 1$\r\n\r\nI believe there is a combinatorial argument too, since $ 2^n$ is the number of subsets of n distinct objects, since each can either be included or not. Can anyone come up with a combinatorial proof?\r\n\r\nThis is the basis of the double-or-nothing strategy at the casino. First you bet 1. If you lose, you bet 2, net 1 in return if you win. If you lose again, you bet 4, again netting 1 if you win. Then you bet 8, 16, etc until you win. Once you win, no matter how long it takes, you will always end up netting 1 in the end (unless you keep losing and run out of money and go broke!).", "Solution_13": "[quote=\"moogra\"]Unfortunately, it doesn't equal $ 2147483646$ which is the right answer.\nIt equals $ 2147483647$ (this also happens to be the max int)[/quote]\r\n\r\nOh, because I thought the problem was $ 2^0 \\plus{} \\text{etc.}$. It really doesn't have the $ 2^0$, I misread it. That's why it's 1 off." } { "Tag": [ "function", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello everybody!\r\n\r\nI have this exercise with a hint, in elementary group theory. I proved the hint quite easily but can't, for the life of me, see how it helps prove the statement in the exercise. So I'll be really grateful for any further hints or general direction. \r\n\r\nLet G be a final group with the property $ (ab)^3 \\equal{} a^3b^3$ for all $ a,b \\in G$. Prove that if $ gcd(o(G),3)\\equal{}1$ (where $ o(G)$ is the order of the group G), than G is abelian.\r\nHint: Prove that $ a^2b^3 \\equal{} b^3a^2$ for all $ a,b \\in G$ and that the function $ f: G \\to G$ defined by $ f(a) \\equal{} a^3$ is isomorphism.\r\n\r\nI can't see how the hint helps here, I'm sure I'm missing something simple :( \r\n\r\nMany thanks,\r\nPerrin", "Solution_1": "Let $ |G|\\equal{}m$.Since $ (m,3)\\equal{}1$ it follows that $ f$ is izomorphism (1).Since $ f(x)\\equal{}x^3$ is surjective morfism we have that $ x^2 \\in Z(G)$ $ \\forall x\\in G$ (2). ( where $ Z(G)\\equal{}\\{g \\in G | gx\\equal{}xg ,\\forall x \\in G\\}$)\r\n\r\n(You said you managed to prove (1) and (2), so I think you are interested in the next part)\r\n\r\nSince $ (ba)^3\\equal{}b(ab)^2a\\equal{}ba(ab)^2$ (since $ (ab)^2$ commutes with any element of $ G$, in particular with $ a$ ), from where we get that $ (ba)^2\\equal{}(ab)^2$.From this we have that $ (ba)^3\\equal{}(ab)(ab)ba\\equal{}abab^2a\\equal{}aba^2b^2\\equal{}a^3b^3\\equal{}(ab)^3$.So we got that $ (ba)^3\\equal{}(ab)^3$.But since $ f$ is injective we can conclude that $ ab\\equal{}ba$, for any arbitrary $ a,b \\in G$.", "Solution_2": "Thanks! All that was needed is playing a bit with those equalities ($ (ab)^3$ etc.). I always get stuck on the last, most simple step. Thank you very much for your help! :)" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "For each $ n\\in N$ find the largest number $ C(n)$ such that the inequality\r\n\r\n$ (n \\plus{} 1)\\sum_{j \\equal{} 1}^{n}a_j^2 \\minus{} (\\sum_{j \\equal{} 1}^{n}a_j)^2\\geq C(n)$\r\n\r\nholds for all $ n$-tuples $ (a_1,a_2,...,a_n)$ of pairwise different integers.", "Solution_1": ":maybe: \r\n\r\nI think it is: $ \\frac{n(n\\plus{}1)^2(n\\plus{}2)}{12}$\r\n\r\n\r\n :)", "Solution_2": "Can you prove it? :)", "Solution_3": "ElChapin, I think you would be right if the problem was about nonzero integers.\r\n\r\nHint:\r\n\r\nAssume WLOG that $ a_1\\frac{(q-1)^2+2}{2}$ for $q\\geq 4$ we conclude there exists node $B$ which connected to at least two nodes from $A_1$, ..., $A_{q-1}$. It follows that by excluding nodes $A$ and $B$ and all their neighbourhoods we will remove at most $2q-2$ nodes. So remaining graph will consist of at least $\\frac{(q-2)^2+3}{2}$ nodes. Thus we will be able to find a triangle or $q-2$ disconnected nodes. Both variants leads to contradiction.", "Solution_3": "Ok, now for $\\frac {p \\cdot 2^{\\frac p 2}} {e \\cdot 2^{\\frac 1 2} } 2 x )\nb)the product of the members of any subset of A isn't square .\nc)if n be in A and p be prime number s.t n is divisible by p then p is in X\n( could u post splendid solution beacaus eI saw boring one)", "Solution_1": "The answer is (2n-1)(2n-2) ... (2n-2m-1)/m! for m \\leq n, and 0 otherwise.\r\n\r\nSuppose m\\leq n. Because all the elements of A are square free, we may write a vector in Z2n for each of them, where i-th position is 1 iff the i-th prime in X divides the number.\r\n\r\nSince no product of the numbers in a subset of A amounts to a perfect square, this translates to no sum of vectors in a subset is 0. Therefore they are all linearly independent. Also, any m linearly independent vectors satisfy this. If m>n it's clear that we have 0 possibilities (since the m vectors form a basis of a subspace of Z2n, which has dimension n).\r\n\r\nThe first vector can be chosen in 2n-1 ways. Assume that we have chosen the first k. They span a subspace of dimension k of Z2n, with 2k elements. There are 2n-2k vectors left and we can choose any of them. In the end we divide by m! because we count the same m-uple of linearly independent vectors m! times, once for each permutation." } { "Tag": [ "search" ], "Problem": "1. Find all positive integers m, n, where n is odd, that satisfy\r\n1/m+4/n=1/12.\r\n\r\nSome help please, I started by letting n=2k+1, this gives an equation that can be continually cancelled down working mod 2, but now i can't see where to go.", "Solution_1": "Just solve it as if there wasn't any restriction, and then only include the allowed.", "Solution_2": "$\\frac{1}{m}+\\frac{4}{n}=\\frac{1}{12}$\r\n$n+4m=\\frac{mn}{12}$\r\n$mn-48m+12n=0$\r\nWe search for u and v, for which uv=mn-48m+12n+k:\r\n$mn-48m+12n-576=-576$\r\n$(m+12)(n-48)=-576$\r\nn is odd, so $n-48=\\pm 1, \\pm 3, \\pm 9$ ect.", "Solution_3": "[quote=\"krassi_holmz\"]$\\frac{1}{m}+\\frac{4}{n}=\\frac{1}{12}$\n$n+4m=\\frac{mn}{12}$\n$mn-48m+12n=0$\nWe search for u and v, for which uv=mn-48m+12n+k:\n$mn-48m+12n-576=-576$\n$(m+12)(n-48)=-576$\nn is odd, so $n-48=\\pm 1, \\pm 3, \\pm 9$ ect.[/quote]\r\n\r\nhmm you screwed up...\r\n\r\n$\\frac{1}{m}+\\frac{4}{n}=\\frac{1}{12}$\r\n$n+4m=\\frac{mn}{12}$\r\n$mn-48m-12n=0$\r\n$mn-48m-12n+576=576$\r\n$(m-12)(n-48)=576$", "Solution_4": "Yes. :D \r\nI was hurrying.\r\nBut the method is the same..." } { "Tag": [ "geometry", "rectangle", "perpendicular bisector", "geometry proposed" ], "Problem": "Let $ABCD$ be a rectangle.Let $E$ be a point on $BD$ such that $\\hat{DAE}=15^{o}$.$EF \\perp AB (F \\in AB)$$EF =\\frac{1}{2}AB;AD= \\alpha$.Caculate $\\hat{EAC}=?;EC=?$", "Solution_1": "[quote=\"chien than\"][color=darkred]Let $ABCD$ be a rectangle. Let $E$ be a point on $BD$ such that $\\widehat{DAE}=15^{\\circ}.$ Let $F$ be the projection of $E$ on $AB$ and assume that $EF =\\frac{1}{2}AB.$ Caculate $\\widehat{EAC}$[/color] [/quote]\r\n[color=indigo][b]Solution. [/b] Denote $B'$ be the intersection of perpendicular bisector of $AE$ with $AB$\n\nWe have: $\\angle EAB'=\\angle AEB'=75^\\circ\\Longrightarrow\\angle AB'E=30^\\circ$\nThus $AB'=EB'=2EF=AB,$ so $B\\equiv B'$\n\nAnd hence: $\\angle EAC=\\angle DAC-\\angle DAE=60^\\circ-15^\\circ =45^\\circ$[/color]" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "If positive integers a,b,c are such that b divides a^3, c divides b^3, and a divides c^3, prove that abc divides (a+b+c)^13.", "Solution_1": "Quite easy. \r\nThe key is that a divides c 3 which divides b 9 \r\nThus, a divides b 9 . Simplifying the (a+b+c) 13 under modulo a, we get that it is congruent to 0. Similarly for the others. It remains to consider the cases of common factors...", "Solution_2": "It is easy to see that a,b,c have the same prime factors so\r\na=p_1 a_1...p_n a_n \r\nb=p_1 b_1...p_n b_n \r\nc=p_1 c_1...p_n c_n \r\nwhere a_i,b_i,c_i>0\r\nWe have from the conditions b|a 3 c|b 3 a|c 3 that\r\nb_i \\leq 3a_i\r\nc_i \\leq 3b_i\r\na_i \\leq 3c_i\r\n\r\n(a+b+c) 13 is a sum of repeating terms of the form a xb y c z where x+y+z=13.\r\nThere is no use considering the case xyz>0 because the term in the sum is divisible by abc. \r\nI want to prove that abc divides each term in the sum.\r\nThis goes to showing that a|b yc z where y+z=11, and the others. it is easy to see that z \\geq 3 or y \\geq 9 or else z+y \\leq 10 (false). Prove the others similarly.", "Solution_3": "[quote=\"orl\"]If positive integers a,b,c are such that b divides a^3, c divides b^3, and a divides c^3, prove that abc divides (a+b+c)^13.[/quote]\nIf a prime $\\forall p\\mid a$ then $p\\mid a \\mid b^3 \\mid c^9$, and it implies $\\text{rad}(a)=\\text{rad}(b)=\\text{rad}(c)$.\nThen $\\upsilon_p\\left((a+b+c)^{13}\\right)\\ge 13\\min\\{\\upsilon_p(a),\\upsilon_p(b),\\upsilon_p(c)\\}$ $=(3^2+3+1)\\min\\{\\upsilon_p(a),\\upsilon_p(b),\\upsilon_p(c)\\}\\ge \\upsilon_p(a)+\\upsilon_p(b)+\\upsilon_p(c)$. []", "Solution_4": "[quote=\"orl\"]If positive integers a,b,c are such that b divides a^3, c divides b^3, and a divides c^3, prove that abc divides (a+b+c)^13.[/quote]\nin fact always this is true: if $ b|a^n $,$ c|b^n $,$ a|c^n $ then we have $ abc|(a+b+c)^{n^2+n+1} $", "Solution_5": "Consider the expansion of $(a+b+c)^{13}$.\nIt consists of 3 types of terms - i) terms consisting of one variable only like $a^{31}$\n ii) terms consisting of 2 variables like $Ka^{m}b^{n}$ with $(m+n) = 31$\n iii) terms consisting of 3 variables like $La^{p}b^{q}c^{r}$ with $(p+q+r+) = 31$\nIts is easy to show that abc divides each of these.\nSo abc divides $(a+b+c)^{13}$", "Solution_6": "What means $rad(a)$ ?", "Solution_7": "[quote=\"TheBernuli\"]What means $rad(a)$ ?[/quote]\nIt's the [url=http://en.wikipedia.org/wiki/Radical_of_an_integer]radical[/url] ;)" } { "Tag": [ "calculus", "geometry" ], "Problem": "Hey,\r\n\r\nThe AP coordinator at a high school near me has blocked some of my friends from taking AP tests because they haven't taken the AP class for that subject and other reasons.\r\n\r\nIs there a way to bypass the AP coordinator at the high school somehow? I have a friend who has been studying calculus on his own for a year and he was denied being able to take the test.", "Solution_1": "Go to a different school? Unfortunately, at a given school, it's more or less up to the school.", "Solution_2": "Is there anyway to bypass the high school AP coordinators?", "Solution_3": "According to the AP site: \r\n\r\n\"If you are a homeschooled student or attend a school that doesn't offer AP, you can still participate. Each year hundreds of students participate through independent study. Some states even sponsor online AP courses.\"\r\n\r\nhttp://www.collegeboard.com/student/testing/ap/reg.html\r\n\r\nI guess it depends on the coordinator.", "Solution_4": "I know for math contests, I've competed at different schools because mine wouldn't offer it. \r\n\r\nCall around AP schools in your area, maybe one of them would be willing to allow you to write with their students.", "Solution_5": "What I've heard is that it depends on whether your school offers the test.\r\nIf they don't offer the test, then it's sorta their fault that they didn't open that test to you guys. On the other hand, they also get some money for proctoring but they would probably rather not proctor 2 people or something.\r\nIf your school doesn't offer the test, you probably could take it at some other school, but you have to ask that other school and turn in their form or something to make sure that Collegeboard knows that you are taking it there.", "Solution_6": "[quote=\"aufha\"]Is there anyway to bypass the high school AP coordinators?[/quote]If you are willing to rock the boat a little, go over their head. Talk to your school's principal. If that doesn't work, talk to somebody in your school's superintendent's office or your representative on the school board. (I'm not saying this is going to work, but it's something to try if you feel strongly enough.)\r\n\r\nUltimately, an AP exam has to be administered through a school.", "Solution_7": "My son (homeschooled) has had a very good experience taking AP tests in ninth grade and tenth grade at public schools in our town that he has never attended. Both schools are willing to offer ANY test in the AP program to students who sign up in advance, even if the school doesn't offer that course. The College Board website \r\n\r\nhttp://www.collegeboard.com/student/testing/ap/reg.html \r\n\r\nmakes reasonably clear how to sign up for AP tests at a different high school. To go take the tests, you would have to be excused from attendance at your regular school, but that is a reasonable schedule conflict for missing class, if your own school won't help you register.", "Solution_8": "When I took AP tests without taking the class, I just signed up for them. I didn't tell anyone anything, and no one checked. Of course, they found out afterwards, but they couldn't do anything about it then. I've heard of schools doing this before, and it seems to usually not work out well. But you should just keep pushing it... go to the counselor, principal, higher people, w/e. It's your risk, not theirs.", "Solution_9": "oh yeah. forgot to mention - like what do you do if they give you trouble because you are in middle school? A middle school kid I know wants to pass out of calc...", "Solution_10": "so what do you do if they reject you because you're a middle school student?", "Solution_11": "Young people of any age can take AP tests. Each year there a few eighth graders who take various AP tests, and even a very few who win AP scholar awards, which means they took multiple AP tests." } { "Tag": [], "Problem": "\u0394\u03b5\u03af\u03c4\u03b5 \u03c3\u03c4\u03bf \u03c3\u03c5\u03bd\u03b7\u03bc\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03c7\u03b5\u03af\u03bf .", "Solution_1": "\u039fk,thanks.\u03a4\u03bf \u03ba\u03b1\u03c4\u03b5\u03b2\u03b1\u03c3\u03b1.....\r\n\r\n\u03a1\u0391\u039b\u039b\u0397\u03a3 :D" } { "Tag": [ "MATHCOUNTS", "modular arithmetic", "geometry", "area of a triangle", "Heron\\u0027s formula" ], "Problem": "2004 MC Target: number 8\r\nA collection [i]S[/i] of integers is defined by the following 3 rules: (I) 2 is in [i]s[/i]; (II) for every [i]x[/i] in [i]S[/i], 3[i]x[/i] and [i]x[/i]+7 are also in [i]S[/i]; (III) No integers except those defined by rules (I) and (II) are in S. What is the smallest integer greater than 2004 which is NOT in [i]S[/i]?\r\n\r\n1992 MC Team: Number 5\r\nOne side of a triangle is divided into segments of lengths 6cm and 8cm by the point of tangency of the inscribed circle. If the radius of the circle is 4 cm, what is the length, in centimeters, of the shortest side of the triangle? [/i]", "Solution_1": "[hide=\"First Problem\"]You take $2\\cdot 3^a \\pmod 7$, with increasing integer values of $a$. Eventually you get (in no particular order) $1,2,3,4,5,6 \\pmod 7$, but no $0\\pmod7$ Therefore the answer is the first multiple of $7$ greater than $2004$ : $\\bold{2009}$. Hope that explains it.[/hide]", "Solution_2": "[hide]The first one:\n\nClearly, 3x will never be divisible by 7 unless x is a multiple of 7, and any 7x + 2 Where x is a whole number will ever be divisible by 7 either, and the first multiple of 7 greater than 2004 is 2009, and a brief check will show that 2005-2008 will work. so 2009[/hide]", "Solution_3": "8\r\n[hide]It was already posted earlier. The answer is 2009 I believe because you can never get a multiple of 7.[/hide]", "Solution_4": "Heron's formula works nicely for #2.", "Solution_5": "Doesn't number two have something to do with the fact that the opposite lines of the triangle of 6 and 8 are also 6 and 8?", "Solution_6": "5\r\n[hide]Two tangents of a circle that meet outside of the circle are equal. We can draw the triangle and label the lengths of the tangents $x, x, 6, 6, 8$, and $8$. Drawing the radii to the tangents we make a bunch of right triangles. We add up the areas of the triangles to get:\n\n$4x + 24 + 32 = 56 + 4x$\n\nWe set this equal to Heron's formula and get:\n\n$56 + 4x = \\sqrt {(14 + x)(8)(6)(x)} \\Rightarrow x = 7$\n\nComes out friendly. The shortest side is then $6 + 7 = 13$.[/hide]", "Solution_7": "[quote=\"NoSoupForYou\"][hide=\"First Problem\"]You take $2\\cdot 3^a \\pmod 7$, with increasing integer values of $a$. Eventually you get (in no particular order) $1,2,3,4,5,6 \\pmod 7$, but no $0\\pmod7$ Therefore the answer is the first multiple of $7$ greater than $2004$ : $\\bold{2009}$. Hope that explains it.[/hide][/quote]\r\n\r\nok, but why did you choose seven to mod?", "Solution_8": "In the problem, 2*3^n + 7x are terms of the sequence, where n and x are non-negative.", "Solution_9": "[quote=\"sirorange\"][hide]The first one:\n\nClearly, 3x will never be divisible by 7 unless x is a multiple of 7, and any 7x + 2 Where x is a whole number will ever be divisible by 7 either, and the first multiple of 7 greater than 2004 is 2009, and a brief check will show that 2005-2008 will work. so 2009[/hide][/quote]\n\na couple of things: first, where did you get 7x+2? \nSecond, you don't give a clear proof that x will never be a multiple of 7 (you say 3x is not unless x is a multiple of 7, but you don't elaborate)\n\n[quote=\"seungsoo\"]In the problem, 2*3^n + 7x are terms of the sequence, where n and x are non-negative.[/quote]\r\n\r\nWhere did you get this?", "Solution_10": "I will attempt to explain it better.\r\n\r\nWe note that by multiplying 2 by 3 a sufficient number of times and then adding enough 7's, we can show that many numbers are in S. Specifically, if we can construct any value mod 7 using $2\\cdot 3^n$ while keeping the numbers reasonably small, we can add some number of 7's to reach any large desired number. Now there are several ways of showing that $2\\cdot 3^n$ can be any value mod 7 other than 0. During the actual competition, guess and check should work out fine. Then you can show that any relatively large number that is not a multiple of 7 is in S (make sure that $2\\cdot 3^n$ does not need to be too large; in this case it doesn't). Furthermore, it is obvious that $2\\cdot 3^n$ is not a multiple of 7, so $2\\cdot 3^n + 7x$ is also not a multiple of 7.", "Solution_11": "I understand why any number divisible by 7 is not in the set, and also why anything except 0 mod 7 is also in the set, but there obviously there are numbers not divisible by 7 that are in the set. \r\n\r\n[quote=\"probablility1.01\"] Then you can show that any relatively large number that is not a multiple of 7 is in S (make sure that does not need to be too large; in this case it doesn't). [/quote]\r\n\r\nHow do you judge if 2004 is a relatively large number? Can you prove that no number less than 2009 is in S in a practical amount of time?", "Solution_12": "Oh, I remember the first one... I got it wrong. :( I think I ran out of time, because I wasn't solving it the optimal way...", "Solution_13": "You test $2\\cdot 3^n$ for $1 \\le n \\le 6$ to check their mods. A good mathcountser should know that $2\\cdot 3^6 < 2004$.", "Solution_14": "[quote=\"Secret Asian\"]2004 MC Target: number 8\nA collection [i]S[/i] of integers is defined by the following 3 rules: (I) 2 is in [i]s[/i]; (II) for every [i]x[/i] in [i]S[/i], 3[i]x[/i] and [i]x[/i]+7 are also in [i]S[/i]; (III) No integers except those defined by rules (I) and (II) are in S. What is the smallest integer greater than 2004 which is NOT in [i]S[/i]?\n [/quote]\r\n\r\n[hide=\"Solution\"]\n1) Mod 7, 3^n.\n\nTo get any number with a remainder of 2, use 2 + 7n.\nTo get any number with a remainder of 6, use 2(3) + 7n.\nTo get any number with a remainder of 4, use 2(3)(3) + 7n.\nTo get any number with a remainder of 5, use 2(3)(3)(3) + 7n.\nTo get any number with a remainder of 1, use 2(3)(3)(3)(3) + 7n.\nTo get any number with a remainder of 3, use 2(3)(3)(3)(3)(3) + 7n.\n\nNotice how you can't get a multiple of 7. [b]2009[/b].\n\n2)\n[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c,d$ be non-negative real numbers. Prove that\r\n\\[a(a-b)(a-c)(a-d)+b(b-a)(b-c)(b-d)+c(c-a)(c-b)(c-d)+d(d-a)(d-b)(d-c)+abcd \\ge 0.\\]", "Solution_1": "\\[=\\sum_{c}x^{4}-\\sum_{s}x^{3}y+\\sum_{s}x^{2}yz-3xyzw \\]\r\nfollows from applying schur on vars taken three at a time", "Solution_2": "No, me@home, you are wrong.\r\nIt follows from the following reasoning: \r\nlet $a=\\min\\{a,b,c,d\\}$ and $b=a+x,$ $c=a+y,$ $d=a+z.$\r\nHence, the hungkhtn's inequality equivalent to\r\n$a^{4}+(x+y+z)a^{3}+(xy+xz+yz)a^{2}+$\r\n$+\\left(\\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)\\right)a+\\sum_{cyc}x^{2}(x-y)(x-z)\\geq0.$ :wink:", "Solution_3": "Yes, the solution of Arqady is true and nice (after expanding, we get two Schur inequality). My solution is to use EMV method, quite the same. Anyway, I really wait for another solution without expanding and also EMV. :blush:" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let triangle ABC. Let X,Y,Z in segment BC,CA,AB respectively such that:\r\nAX=BY=CZ and angle BAX = angle CBY = angle ACZ.\r\nProve that triangle ABC is equilateral", "Solution_1": "I tried but I fail in finding the solution,some one can help me?" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "let $f$ $\\in C^{2}\\left( \\mathbb R,\\mathbb R \\right)$ and suppose that :\t\r\n$i)$ $\\underset{x\\rightarrow +\\infty }{\\lim }f\\left( x\\right) =0$\r\n$ii)$ $f^{\\prime \\prime }=O\\left( \\frac{1}{x^{2}}\\right)$\r\nShow that : $f^{\\prime }=o\\left( \\frac{1}{x}\\right)$", "Solution_1": "Can you please confirm that first one is a big-O and second one is a small-o ?\r\nFor now with the given hypotheses all I can prove is that $f' = O(\\frac{1}{x})$ (big-o).", "Solution_2": "Yes I confirm that the first one is a big $O$ and the second is a small $o$. That's what is given on the document I found on the net.", "Solution_3": "[quote=\"erdos\"]let $f$ $\\in C^{2}\\left( \\mathbb R,\\mathbb R \\right)$ and suppose that :\t\n$i)$ $\\underset{x\\rightarrow +\\infty }{\\lim }f\\left( x\\right) =0$\n$ii)$ $f^{\\prime \\prime }=O\\left( \\frac{1}{x^{2}}\\right)$\nShow that : $f^{\\prime }=o\\left( \\frac{1}{x}\\right)$[/quote]\r\nConsider the Taylor expansion of $f$ at a fixed $x$, in the form $f(x+hx) = f(x) + hxf'(x) + \\frac{h^2x^2}{2}f''(\\xi)$ with $\\xi$ between $x$ and $x+hx$ and $h > 0$. \r\n\r\nAfter rearranging, one obtains $xf'(x) = \\frac{f(x+hx) - f(x)}{h} - \\frac{hx^2}{2}f''(\\xi)$. Given $\\epsilon$, choose $M$ sufficiently large such that $|f(x)| < \\epsilon$ for $x > M$ and then choose $h = \\sqrt{\\epsilon}$. Then $|xf'(x)| \\le 2 \\sqrt{\\epsilon} + \\frac{\\sqrt{\\epsilon}}{2}C$, where $C$ is a bound for $x^2f''(x)$. This can be made as small as we please, so $xf'(x) = o(1)$ and $f'(x) = o\\left(\\frac{1}{x}\\right)$." } { "Tag": [ "geometry", "complex numbers", "projective geometry" ], "Problem": "For the longest time I thought 1/:inf:=0, is that true? (and I apologize if this was discussed before)\r\n\r\nAlso how is operations on :inf: defined? (For example for all real number A which does not contain \":inf:\" in it can you say :inf:+A=:inf:-A=A:inf:=:inf:/A=:inf:?)", "Solution_1": "well for 1/:inf:=0 i would think it would be true because think about it, the graph 1/x has limit 0 for y so 1/:inf: is 0. Also, you can look at patterns... 1/2>1/3>1/4>1/5>...1/:inf:... it keeps getting smaller but still greater than 0 so if you went on forever you would get there.", "Solution_2": "Mmm, but there's also the opposing argument that y=0 is an assymptote in the graph y=1/x so y can never be 0...\r\n\r\nAlso here's this weird proof I came up with which say A/:inf:=1: Let's say A is all real number. So if A/:inf:=0 then this should work for A=:inf:-1. So by substitution you get (:inf:-1)/:inf:=:inf:/:inf:-1/:inf:=1-1/:inf:. Since 1/:inf: is defined as 0 that equals 1-0=0. By transitivity A/:inf:=1...???", "Solution_3": "1/ :inf: isn't anything. It doesn't make sense. limx--> :inf: 1/x=0, but you can't say any more than that.", "Solution_4": "You can't really say any of those, can you? The most you can say is that limx->:inf: 1/x = 0. You can't evaluate infinity.\r\n\r\nI think one needs to remember that infinity is a concept - and :inf: is a symbol for this concept. It's not a number. So none of those operations are defined. You can say that a number A plus a very large number is still a very large number, but not much else.", "Solution_5": "ya, like :inf: = \"a very big number beyond representation by hand\" so A + \"a very big number beyond representation by hand\" yields \"another very big number beyond representation by hand\" which is also :inf:", "Solution_6": "Uhm, not really in my opinion, since \"a very big number beyond representation by hand\" is a rather vague statement. I could say that 1050 fits your description, but 1050 is nowhere near :inf:; and 1050 + A doesn't come close to :inf: either. :)\r\n\r\nOn second thought - what JBL said.", "Solution_7": "The key point: infinity is not a number. It's a concept. It describes what happens when things get bigger than any real number. Now, it makes [i]intuitive[/i] sense to say that 1/ :inf: = 0, 1/0 = :inf:, \r\n:inf: + A = :inf: for any real number A, and many others. However, it does not make [i]mathematical[/i] sense. \r\n\r\nNote also that there are two entirely different sets of ideas about infinity: one is this concept of \"what happens as you take a series of numbers which is increacing and has no real limit.\" The other is an idea about \"how big sets are.\" For example, the natural numbers and the real numbers are both clearly infinite sets -- that is, they have more than any finite number of elements. However, there are also far more real numbers than natural numbers. (In fact, if for every natural number, you took a set the size of the natural numbers, and then counted how many total numbers you had, it would still be smaller than the number of real numbers.) The infinite cardinalities are very important, but they still aren't \"numbers.\"", "Solution_8": "Well the whole infinity operation thing comes into play when I'm quickly trying to figure out the range of a graph and I usually just say infinity plus/minus/to the blank power/multiplied by/divided by/etc. is still infinity and when infinity is on the denominator the whole fraction's 0 just for the simplicity of it and I wanted to see if what I'm doing is correct. Is it? (both the method and the answer I get to as the range)", "Solution_9": "ok, how bout \"a number not displayable by any means of technology to this date,\" that'd be much bigger yet would include a wide range", "Solution_10": "or, rather, an :inf: range, huh? sorry for the double post :)", "Solution_11": "[quote=\"Tare\"]For the longest time I thought 1/:inf:=0, is that true? (and I apologize if this was discussed before)[/quote]\r\n\r\nThere is some discussion of this issue on the Math-Teach forum \r\n\r\nhttp://mathforum.org/epigone/math-teach/choupryshax \r\n\r\nwhere the short answer is no. The slightly longer answer is that infinity is not a number, as other participants here have said.", "Solution_12": "[quote=\"paladin8\"]ok, how bout \"a number not displayable by any means of technology to this date,\" that'd be much bigger yet would include a wide range[/quote]\r\n\r\nNo. It's not a number, not even a big one. It isn't \"displayable\" by [i]any[/i] technology, ever. And Tare, what you are doing is probably not mathematically rigorous -- however, there is likely no reasonable situation in which it would give you the wrong answer. That is, it's not rigorous, but it works.\r\n\r\nNote of course that there is a problem when you deal with something like x :^2:/(x - 1) in that scenario, however.", "Solution_13": "You can't do \"standard\" arithmatic with infinity in that way because it doesn't fit the Ordered Field Axioms that you want regular numbers to satisfy. The ordered field axioms are:\r\n\r\n[u]Addition:[/u]\r\n1) closure under addition\r\n2) commutivity\r\n3) associativity\r\n4) existence of a zero element\r\n5) additive inverses for every element\r\n\r\n[u]Multiplication:[/u]\r\n1) closure under multiplication\r\n2) commutivity\r\n3) associativity\r\n4) existence of a unit element\r\n5) multiplicative inverses for every element\r\n6) distributive law\r\n\r\n[u]Order:[/u]\r\n1) x + y < x + z if y < z\r\n2) xy > 0 if x > 0 and y > 0\r\n\r\nThe rational numbers and real numbers are both ordered fields. Can you see why?\r\n\r\nIn addition, the real numbers have a very special property that you need in order to take limits and do things like calculus. This is called the least upper bound property. I won't go into detail about what that is, but it more or less says that the real numbers don't have any \"gaps\" like the rational numbers do.\r\n\r\nIt's customary to sometimes add the two symbols -:inf: and +:inf: to the real numbers with the conventions: x + :inf: = +:inf: , x - :inf: = -:inf: , and x/ +:inf: = x/ -:inf: = 0, where x is any regular real number. And to leave the expressions :inf: - :inf: , :inf: / :inf: , etc. as undefined. This system is called the extended real numbers, but it is not a field. Can you see why?\r\n\r\nYou can also do a similar thing with complex numbers (which are a field, but not an [i]ordered[/i] field) by adding one element called :inf: that makes the complex \"plane\" into a \"sphere.\" You can make this idea more precise by using projective geometry - but again, this change makes the extended complex numbers no longer a field!!! [b]And the properties of a field are the key property that you need in order to do arithmatic[/b].\r\n\r\nOn a more philosophical level, the reason for the confusion about infinity comes mainly because we confuse the ordinality of a number with the actual number itself. What that means is, there are two different \"meanings\" to a number. One meaning of a number is as an element of a field (usually the ordered real numbers). This looks at the element as an abstract object that obeys the rules of arithmatic. The other meaning is the use of numbers to signify the SIZE of something. These are two different things! For example, the number \"four\" is not the same thing as the set {2, 3, 4, 5} which has a SIZE of four.\r\n\r\nOften times when we think of infinity, we think of it as refering to the relative \"size\" of something. The technical definition of infinity is \"something that is not finite.\" Then finite is definied as the size of something that can be up into a one to one correspondence with a set with a whole number of elements. So you see that even this technical definition of infinity refers to the relative \"size\" of something, not to it as a number in relation to the other real numbers.\r\n\r\nThere are, however, some theories in math that deal with arithmatic of infinite (and infinitessimal) ordinals. These are called non-standard analysis. The most famous such theory is the theory of Surreal Numbers - invented by John Conway while he was trying to understand the game of Go. Some good books you can use to learn about Surreal Numbers are:\r\n\r\nBerlekamp, E. R.; Conway, J. H.; and Guy, R. K. [u]Winning Ways for Your Mathematical Plays, Vol. 1: Games in General.[/u]\r\n\r\nConway, J. H. [u]On Numbers and Games.[/u]\r\n\r\nConway, J. H. and Guy, R. K. [u]The Book of Numbers.[/u]\r\n\r\nConway, J. H. and Jackson, A. [u]\"Budding Mathematician Wins Westinghouse Competition.\"[/u]\r\n\r\nGonshor, H. [u]An Introduction to Surreal Numbers.[/u]\r\n\r\nEbbinghause, et. al. [u]Numbers[/u]\r\n\r\nKnuth, D. [u]Surreal Numbers: How Two Ex-Students Turned on to Pure Mathematics and Found Total Happiness.[/u]\r\n\r\nThe fourth book in that list is about a high school student, Jacob Lourie, who won the Westinghouse Competition with a project on Surreal Numbers. The last book in the list, I believe (but I haven't read it), is a fiction book that uses Surreal Numbers as a theme - sort of like [u]Flatlands[/u]. The book [u]Numbers[/u] is a great book that talks about all kinds of different number systems. Though it's somewhat advanced (probably appropriate for undergraduate level students or above).", "Solution_14": "Hi well just to through in something new to think about well we all know 0 is not negative and not positive and it goes on forever so wouldnt :inf: be 0 since like :inf: 0 could also be seen as a theory since anything divided by 0 is not answerable one of my friends said that the answer to problems divided by 0 is :inf: so technically zero is :inf: right?", "Solution_15": "No, unfortunately that idea doesn't work because zero is defined as a number that when you add it to something the result is that something (i.e. 0 + a = a). But you can see that infinity + something wouldn't be the same something that you added.\r\n\r\nTo see in general why infinitity can't be treated at a number though, take a look at the field axioms that I listed above. Which ones would \"infinity\" not satisfy?", "Solution_16": "I hope nobody said this already, but theoretically :inf: + X= :inf:", "Solution_17": "That's what this entire post is about -- several people have explained why the equation you wrote is problematic." } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AMC 10", "AMC 10 B" ], "Problem": "I know that last year the floor value for the AIME was 7. How about previous years? \r\n\r\nIn the past, did everyone in 10th (or below) who earned the floor value go onto the USAMO? It looks like this year, it is limited to 80 students, and if there are more than 80 then they will factor in the AMC 10 scores, as they do with the AMC 12 participants. I am getting this information from the teachers manual on the AMC site. \r\n\r\nAlso, when they talk about the rules for 10th and below, are they talking about people who took the AMC 10, or are they including 10th (and below) who took the AMC 12?\r\n\r\nThanks!", "Solution_1": "10th and below includes 10th and below who took AMC12. I don't know about the other questions though.", "Solution_2": "2004 - 7\r\n2003 - 8\r\n2002 - 6", "Solution_3": "By the way, here is last year's distribution of USAMO qualifiers by grade:\r\n[code] 7th grade 1\n 8th grade 15 \n 9th grade 25\n10th grade 73\n11th grade 61\n12th grade 87\n13th grade 1\n Total 263[/code]I think all the 10th graders and below who made the AIME floor value were invited to the USAMO. I think the limit of 80 students was in the instructions from last year too, but I don't think it was enforced.", "Solution_4": "Hmm, a question, then.\r\n\r\nIf the 80-person limit is reached, then the USAMO index of each 10th grader is taken, and the 80 highest of these will reach the USAMO. This part I understand. In the calculation of the USAMO index, however, do they only factor in the AMC 10 score, be it A, or B? Or do they factor in whichever test they did better on, be it 10A, 10B, 12A or 12B?\r\n\r\nThis news comes somewhat as a surprise to me, for if I had known that the USAMO index would be a factor for underclassmen, I would have taken the AMC10B rather than the 12, in order to maximize my chances. I really hope this \"new\" (is it new?) rule is enforced.", "Solution_5": "Does anyone know who the 7th grader that qualified is?", "Solution_6": "David Benjamin, Grade 7, William Henry Harrison High School, in West Lafayette, IN.\r\n\r\nTaken from the list in official AMC site", "Solution_7": "He's also dnas on this website, but he doesn't post that often." } { "Tag": [ "trigonometry", "inequalities", "calculus" ], "Problem": "i think this is very strong,\r\ni am not able to prove it :( \r\nProve that for all $ x,y,z$ - positive reals -\r\n$ \\frac {x(x \\plus{} y)}{z(y \\plus{} z)} \\plus{} \\frac {y(y \\plus{} z)}{x(x \\plus{} z)} \\plus{} \\frac {z(z \\plus{} x)}{y(x \\plus{} y)} \\plus{} \\frac {8xyz}{(x \\plus{} y)(y \\plus{} z)(z \\plus{} x)} \\ge 4$\r\nor this one-\r\n$ \\sum_{cyclic} \\frac {a^2}{b^2} \\plus{} 8abc \\ge 4$ (the two equivalent and the first is due to a nice transformation by nayel :wink: )\r\nthe second one has the following conditions\r\n1. $ 0 \\le a,b,c \\le 1$\r\n2. $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 2abc \\equal{} 1$", "Solution_1": "Nayel's transformation is not very much helpful here atleast for me.....\r\n\r\nSubstitute $ a = cosA$, $ b = cosB$, $ c = cos C$ for an acute trinagle $ ABC$.\r\n\r\nLemma 1: $ 4 - 8\\cos A\\cos B\\cos C = 4(\\sum \\cos^2 A )$.\r\nProof: This immediately follows from $ \\cos^2 A + \\cos^2 B + \\cos^2 C = 1 - 2\\cos A\\cos B\\cos C$ which leads us to:\r\n $ 4(\\sum \\cos^2 A ) = 4(1 - 2\\cos A\\cos B\\cos C)$ which in turn implies \r\n $ 4 - 8\\cos A\\cos B\\cos C = 4(\\sum \\cos^2 A )$.\r\n -------------End Lemma -1-------------------\r\n\r\nLemma 2: $ \\cos A\\cos B\\cos C \\le \\frac {1}{8}$ .\r\nProof: This inequality is equivalent to proving that $ \\sin (\\frac {A}{2})\\sin (\\frac {B}{2})\\sin (\\frac {c}{2}) \\le \\frac {1}{8}$\r\n. Now the LHS of the above inequality can be written as $ \\frac {1}{2} [ \\cos({\\frac {A - B}{2}) - cos(\\frac {A + B}{2}) ] sin(\\frac {C}{2}) = \\frac {1}{2} [ \\cos({\\frac {A - B}{2}) - \\sin(\\frac {C}{2}) ] \\sin(\\frac {C}{2}) \\le \\frac {1}{2} [ 1 - \\sin(\\frac {C}{2})] \\sin (\\frac {C}{2}) = \\frac {1}{2} [ \\frac {1}{4} - (\\frac {1}{2} - \\sin\\frac {C}{2})^2 ] \\le \\frac {1}{8}}}$. Equality holds good if $ ABC$ is equilateral.\r\n\r\nProof for Lemma 2 can also be given using Jensen's inequality which I am avoiding since any olympiad math problem can be solved without calculus{Pre calculus tools}. A proof using Cosine rule formulation and applying AM GM in the end is also possible.\r\n\r\n----------------------ENd LEMMA 2-------------------\r\n\r\n\r\nNow we are ready to solve the inequality.\r\n\r\nBy Lemma 1, it is enough to prove the inequality $ \\sum (\\frac {cosA}{cosB})^2 \\ge 4(\\sum cos^2A)$.\r\n\r\nBy the $ AM - GM$ inequality we have \r\n\r\n$ 2 (\\frac {\\cos A}{\\cos B})^2 + (\\frac {\\cos B}{\\cos C})^2 \\ge \\frac {3\\cos ^2 A}{ (\\cos ^2 A\\cos ^2 B\\cos ^2 C)^\\frac {1}{3} } \\ge 12 cos^2 A$. ( Using lemma 2 here )\r\n\r\nAnalogously writing the other two inequalties, we get \r\n\r\n$ 3 \\sum (\\frac {cos A}{cos B})^2\\ge 12(\\sum cos^ 2 A)$and thus we get that$ \\sum (\\frac {cosA}{cosB})^2 \\ge 4(\\sum cos^2 A)$\r\n\r\nThus we end our proof for this inequality." } { "Tag": [ "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Consider the sequence $f_n\\in C^2([0,1])$ such that $f_n(0)=f'_n(0)=0$ and $\\left|f''_n(0)\\right|\\leq 1$. Prove that there is a subsequence of the previos sequence which converges uniformly.", "Solution_1": "Are you sure all you have are conditions on the behavior of $f_n$ in $0$? Because I don't think that suffices. Take, for instance, $f_n$ to be any smooth function which vanishes on $\\left[0,\\frac 13\\right]$ and takes the value $n$ on $\\left[\\frac 23,1\\right]$. Clearly, no subsequence of such a sequence can converge uniformly.", "Solution_2": "I think that in the counterexample given by you $f_n\\notin C^2([0,1])$ ...", "Solution_3": "I did say the functions were supposed to be smooth, as in infinitely differentiable, so it's even better than $C^2$. You can construct smooth functions with those properties.", "Solution_4": "Now I understand. I'll check the source of the problem.", "Solution_5": "I've made the correction. :oops: \r\n\r\nConsider the sequence $f_n\\in C^2([0,1])$ such that $f_n(0)=f'_n(0)=0$ and $\\left|f''_n(x)\\right|\\leq 1$. Prove that there is a subsequence of the previos sequence which converges uniformly.", "Solution_6": "[quote=\"gnosis\"]I've made the correction. :oops: \n\nConsider the sequence $f_n\\in C^2([0,1])$ such that $f_n(0)=f'_n(0)=0$ and $\\left|f''_n(x)\\right|\\leq 1$. Prove that there is a subsequence of the previos sequence which converges uniformly.[/quote]\r\n\r\nI think we can use there Arzela-Ascoli theorem.\r\nFirst,since $f_n(x)=f_n(0)+f_n'(0)x+\\frac{1}{2}f_n''(\\psi)x^2=\\frac{1}{2}f_n''(\\psi)x^2$, so for all $n$ we have that $|f_n(x)| \\leq \\frac{1}{2}$ (*).\r\n\r\nnow, lets prove that the sequence $f_n$ is equicontinious. Since $f_n'(x)=\\int_0^x f_n''(t)dt$, so $|f_n'(x)| \\leq x$.\r\nIn the same manner $f_n(x)=\\int_0^x f_n'(t)dt$, $f_n(y)=\\int_0^y f_n'(t)dt$ which means that $|f_n(x)-f_n(y)|=|\\int_y^x f'(t)dt| \\leq \\frac{1}{2}|x-y|^2$, which proves that the sequence $(f_n(x))_n$ is equicontinous and together with (*) it is bounded by $\\frac{1}{2}$, now Arzela-Ascoli Theorem prove the problem assertion" } { "Tag": [ "inequalities" ], "Problem": "The positive integers a_1, b_1, c_1, a_2, b_2, c_2 are such that \r\na_1+ a_2=31\r\nb_1+b_2=32\r\nc_1+c_2=1994\r\n\r\nProve that the products a_1* b_1* c_1 and a_2* b_2* c_2 are not equal.", "Solution_1": "*bump* can anyone offer some help/a solution/hints? I've spent a good amount of time thinking on the problem and Id like to know whether any of my ideas were leading in the right direction or whether all my failed approaches were nonsense.", "Solution_2": "Really a nice problem! You consider two cases: c_1=c_2=997 and c_1>c_2. The first case is trivial. In the second case gcd(c_1,c_2)<=2 because it must be a divisor of 1994. a_2b_2c_2 must contain all the prime divisors of c_1, so a_2b_2=>c_1/2 and with the same argument: a_1b_1=>c_2/2. Summing up these two inequalities you receive:\r\na_1b_1+a_2b_2=>(c_1+c_2)/2=997. Then you use a_2=31-a_1 and b_2=32-b_1 and get: a_1b_1+(31-a_1)(32-b_1)=>997. If I didn't make a calculation mistake it's the same as: (2a_1-31)(b_1-16)=>501. The left side is atmost (2*30-31)(31-16)=29*15=435<501 which is a contradiction.\r\n\r\nDo you have an other solution(perhaps a shorter one), Iris?" } { "Tag": [ "inequalities", "calculus", "inequalities proposed" ], "Problem": "Given $ a, b, c \\geq\\ 0$. Prove that: $ \\frac{(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)}{a^2\\plus{}b^2\\plus{}c^2} \\geq\\ \\frac{ab(a\\plus{}b)}{a^2\\plus{}b^2} \\plus{} \\frac{bc(b\\plus{}c)}{b^2\\plus{}c^2} \\plus{} \\frac{ca(c\\plus{}a)}{c^2\\plus{}a^2}$\r\n :)", "Solution_1": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. Prove that: $ \\frac {(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2} \\geq\\ \\frac {ab(a \\plus{} b)}{a^2 \\plus{} b^2} \\plus{} \\frac {bc(b \\plus{} c)}{b^2 \\plus{} c^2} \\plus{} \\frac {ca(c \\plus{} a)}{c^2 \\plus{} a^2}$\n :)[/quote]\r\n\\[ \\leftrightarrow 3 \\ge \\frac {ac \\plus{} bc}{a^2 \\plus{} b^2} \\plus{} \\frac {ab \\plus{} bc}{a^2 \\plus{} c^2} \\plus{} \\frac {ac \\plus{} ba}{c^2 \\plus{} b^2}\r\n\\]\r\nI prove it by expanding and Schur inequality :( (It is also reason that I don't post a complete proof\r\nP/s:Sorry all, I also found a mistake in my calculus .Sorry!!!", "Solution_2": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. Prove that: $ \\frac {(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2} \\geq\\ \\frac {ab(a \\plus{} b)}{a^2 \\plus{} b^2} \\plus{} \\frac {bc(b \\plus{} c)}{b^2 \\plus{} c^2} \\plus{} \\frac {ca(c \\plus{} a)}{c^2 \\plus{} a^2}$\n :)[/quote]\r\nTry $ a\\equal{}b\\equal{}1, c\\equal{}2$.\r\n\r\nIf you meant $ \\frac {(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2} \\le\\ \\frac {ab(a \\plus{} b)}{a^2 \\plus{} b^2} \\plus{} \\frac {bc(b \\plus{} c)}{b^2 \\plus{} c^2} \\plus{} \\frac {ca(c \\plus{} a)}{c^2 \\plus{} a^2}$\r\nTry $ a\\equal{}b\\equal{}1,c\\equal{}0.5$ :wink:", "Solution_3": "The following Inequalities are true for all $ a,b,c\\ge0$:\r\n\r\n[b]1/[/b] $ \\frac{(a\\plus{}b\\plus{}c)^3}{3(a^2\\plus{}b^2\\plus{}c^2)}\\ge \\frac {ab(a \\plus{} b)}{a^2 \\plus{} b^2} \\plus{} \\frac {bc(b \\plus{} c)}{b^2 \\plus{} c^2} \\plus{} \\frac {ca(c \\plus{} a)}{c^2 \\plus{} a^2}$\r\n \r\n[b]2/[/b] $ \\frac {ab(a \\plus{} b)}{c^2 \\plus{} ab} \\plus{} \\frac {bc(b \\plus{} c)}{a^2 \\plus{} bc} \\plus{} \\frac {ca(c \\plus{} a)}{b^2 \\plus{} ca}\\ge a\\plus{}b\\plus{}c$", "Solution_4": "[quote=\"dduclam\"]The following Inequalities are true for all $ a,b,c\\ge0$:\n\n[b]2/[/b] $ \\frac {ab(a \\plus{} b)}{c^2 \\plus{} ab} \\plus{} \\frac {bc(b \\plus{} c)}{a^2 \\plus{} bc} \\plus{} \\frac {ca(c \\plus{} a)}{b^2 \\plus{} ca}\\ge a \\plus{} b \\plus{} c$[/quote]\r\nYes, it's equivalent to $ \\sum_{cyc}abc(a^2 \\minus{} b^2)^2\\geq0.$" } { "Tag": [ "algorithm", "floor function" ], "Problem": "1. What does mystery(35) return? What does mystery(101) return?\r\n2. What's the complexity class of mystery()? In other words, what is $ f(x)$ if mystery(value) is $ O(f(x))$ where $ x \\sim \\ln(value)$?\r\n3. Explain what the method mystery() does.\r\n4. Explain why mystery() does what it appears to do. \r\n\r\n\r\n[code]\nlong x, y, n, m, i, q, z;\n\npublic long mystery(long original)\n{\n\t\tx = (int)Math.sqrt(original);\n\t\ty = (int)original / x;\n\n\t\tn = original - x * y;\n\t\tm = n;\n\t\ti = y - x + 1;\n\n\t\tif (n == 0)\n\t\t{\n\t\t\treturn m;\n\t\t}\n\t\tfor(q = 0; q <= x; q++)\n\t\t{\n\t\t\tif((x - q) % m == 0)\n\t\t\t{\n\t\t\t\treturn m;\n\t\t\t}\n\n\t\t\tm += i;\n\t\t\tz = (int)((m - 1) / (x - q - 1));\n\t\t\tm = m - z * (x - q - 1);\n\t\t\ti += z + 2;\n\t\t}\n}[/code]\r\n\r\n[hide]I've figured out that mystery(num) returns an integer factor of num, but I have no idea why it does that, and I have no idea what the complexity class of mystery() is. If anyone can provide some mathematical analysis, I would greatly appreciate it. Thanks! :)[/hide]\r\n\r\nEDIT> This should have gone into Computer Science and informatics... Oops. Sorry!", "Solution_1": "For 1., why don't you just run the program ?", "Solution_2": "Well, probably it makes sense to begin by writing down what the program is actually doing in a way that's not incomprehensible. So, for example, we begin with input $ N$, then set $ x \\equal{} \\lfloor \\sqrt{N}\\rfloor$, $ n \\equal{} \\left\\lfloor \\frac{N}{x}\\right\\rfloor \\equal{} \\left\\lfloor \\frac{N}{\\lfloor \\sqrt{N}\\rfloor}\\right\\rfloor$, and then we compare $ xy$ with $ N$. It seems to not be the case that this returns a factor of $ N$: for example, when you start with $ N \\equal{} k^2$ a perfect square we end up with $ x \\equal{} y \\equal{} k$, $ m \\equal{} n \\equal{} 0$, and then it outputs $ 0$." } { "Tag": [ "factorial", "inequalities", "number theory unsolved", "number theory" ], "Problem": "Given a positive integer u, prove that the equation\r\n\r\nn!=ua-ub have only finitely solutions (n,a,b) in positive integers.", "Solution_1": "[b]Edit: This solution is wrong. Please see the counterexample given by Myth below.[/b]\r\n\r\nconsider u>=4\r\nlet d=a-b\r\nlet p be a prime, p|(u-1)\r\nlet p^m || n!\r\nsince n!=u^b(u^d-1) and gcd(u^b,u^d-1)=1\r\ntherefore p^m || (u^d-1)\r\n=> p^m | (u-1)d\r\n=> d>p^m/u>=p^[n/p]/u (note:m=[n/p]+[n/p^2]+...>=[n/p])\r\n=> n!=u^b(u^d-1)>u^d>u^{p^[n/p]/u}\r\nThe inequality above is true for finitely many values of integer n.\r\n\r\nThe proof for u=2 and u=3 is similar to above with slight modifications.", "Solution_2": "I think statement \"pm | ud-1 ==> pm | (u-1)d\" is false.", "Solution_3": "u>=4,p|(u-1),p^m || (u^d-1) => p^m|(u-1)d\r\n\r\nYou forget about p|(u-1).", "Solution_4": "Let see:\r\nu=7 \\geq 4; p=2 and p|u-1; d=2; m=3; pm|ud-1 ==> pm|(u-1)d, i.e. 8|6*2 ???", "Solution_5": "Myth pointed out an error in my proof. This is the corrected solution.\r\n\r\nconsider u>=3\r\nlet d=a-b\r\nlet p be a prime, p|(u-1)\r\nlet p^m || n!\r\nsince n!=u^b(u^d-1) and gcd(u^b,u^d-1)=1\r\ntherefore p^m || (u^d-1)\r\n=> p^m | (u-1)(u+1)d\r\n=> d>p^m/u^2>=p^[n/p]/u^2 (note:m=[n/p]+[n/p^2]+...>=[n/p])\r\n=> n!=u^b(u^d-1)>u^d>u^{p^[n/p]/u^2}\r\n\r\nThe inequality above is true for finitely many values of integer n because the RHS is of the form x^{y^n},x>1,y>1, it grow very fast compared to LHS n!.\r\n\r\nThe proof for u=2 is similar to above with slight modifications and taking p=3.", "Solution_6": "Will we ever see proof of your new statement \r\n\"p^m || (u^d-1) ==> p^m | (u-1)(u+1)d\" ?", "Solution_7": "[quote=\"Myth\"]Will we ever see proof of your new statement \n\"p^m || (u^d-1) ==> p^m | (u-1)(u+1)d\" ?[/quote]\r\n\r\nThis is the proof when p is odd prime. The proof for p=2 is similar with some modifications.\r\n\r\n[b]lemma: p is odd prime, p^x||(u-1) ==> p^{x+y}||(u^{p^y}-1)\n\nproof:[/b]\r\nlet u=kp^x+1, gcd(k,p)=1\r\nu^{p^y}-1\r\n=(kp^x+1)^{p^y}-1\r\n=kp^yp^x+binom(p^y,2)(kp^x)^2+...\r\nevery term other than kp^xp^y is divisible by p^{x+y+1}\r\ntherefore p^{x+y}||(u^{p^y}-1)\r\n\r\n[b]p^m||(u^d-1) ==> p^m||(u-1)d => p^m|(u-1)(u+1)d\n\nproof:[/b]\r\nlet p^x||(u-1)\r\nlet p^y||d\r\nlet d=p^y.t, gcd(p,t)=1\r\nlet v=u^{p^y}\r\n\r\nby lemma, p^{x+y}||(v-1)\r\nv^{t-1}+...+v+1 is not divisible by p because v^{t-1}+...+v+1==1+...+1+1==t(mod p) and gcd(t,p)=1\r\ntherefore p^{x+y}||v^t-1=u^d-1\r\ncompare with p^m||(u^d-1), we have m=x+y\r\n\r\ntherefore p^x||(u-1),p^y||d ==> p^{x+y}||(u-1)d ==> p^m||(u-1)d ==> p^m|(u-1)(u+1)d", "Solution_8": "Are the Chinese TST 2004 published anywhere? If not, can anyone publish them?", "Solution_9": "recalling mihai manea's favorite fact: \r\n\r\nIf p is an odd prime and a and b are two different integers a=b(mod p) and n is a positive integer then the exponent of p in a^n-b^n is equal to the sum of the exponent of p in a-b and the exponent of p in n. \r\n\r\nSo the only case left is p=2 and i would like to see a proof of the lemma in this case, because for odd primes it is inmediately from mihai's favorite math fact!", "Solution_10": "[quote=\"pascual2004\"]recalling mihai manea's favorite fact [...]\n[/quote]\r\nhey, how did you know that ?! :) do you know him? :P", "Solution_11": "There are two possible cases, n is odd and n is even..\r\n\r\n[b]If a is odd integer and n>0 is even integer and 2^m||(a^n-1), then 2^m||(a-1)(a+1)(n/2).[/b]\r\n\r\nThere are more than one ways to prove this.\r\n\r\nOne way is to let a^2=k.2^x+1, the rest of the proof is like the proof I offered above (for odd prime p).\r\n\r\nAnother way is to factorize. For example, take n=16.\r\na^16-1=(a-1)(a+1)(a^2+1)(a^4+1)(a^8+1)\r\nsince 2||(a^2+1), 2||(a^4+1), 2||(a^8+1)\r\ntherefore 2^m||(a^16-1) ==> 2^m||(a-1)(a+1)(16/2)\r\n\r\n[b]If a is odd integer and n>0 is odd integer and 2^m||(a^n-1), then 2^m||(a-1).[/b]\r\n\r\nThis case is easy.\r\n\r\nCombine all three cases (two cases for p=2 and one for odd prime p), the following result is obtained. I use this result to prove the original problem.\r\np|(a-1),p^m||(a^n-1) ==> p^m|(a-1)(a+1)n\r\n\r\nI don't know if the generalization is also true for p=2, you have to check it.", "Solution_12": "JEJE, No, i dont, but i would like to! at least i Have read he is a great problem solver....I found it in the imo 2001 site, were the contestants had a space to publish their info including favorite math fact....... :D :D :D \r\n\r\nPD: Valentin, yours is empty, so you should check it out, i think it would be pretty cool to have that kind of stuff on every imo :cool:", "Solution_13": "[quote=\"pascual2004\"]PD: Valentin, yours is empty, so you should check it out, i think it would be pretty cool to have that kind of stuff on every imo :cool:[/quote]\r\nyeah, well at that time I was pretty busy buying nike basketball shoes :D ;)", "Solution_14": "I have found generalization also works for 2, that is:\r\nif a and b are odd, and repalce u by a and 1 by b, so i think it is very interesting" } { "Tag": [ "invariant", "modular arithmetic" ], "Problem": "There are a white, b black, and c red chips on a table. In one step, you may choose two chips of different colors and replace each one by a chip of the third color. Find conditions for all chips to become the same color. \r\n\r\nThe solution offered in the book says that a+b+c = 0 (mod 3) and a-b, b-c, a-c = 0 (mod 3) are the requirements, but I've come up with counterexamples to this so it doesn't appear to be correct.", "Solution_1": "What are your counterexamples?\r\n\r\nClearly $a+b+c \\equiv 0 \\mod 3$ is a necessary requirement, since the sum of the chips does not change under the transformation given, and if all chips are eventually of the same colour then $a=b=c$ and $a+b+c=3a \\equiv 0 \\mod 3$.\r\n\r\nIt can be shown that $a-b$ is invariant mod 3 under the transformation (and similarly $b-c$ and $a-c$) since either both $a$ and $b$ decrease by 1 or one decreases by 1 and the other increases by 2. Either way, the difference mod 3 does not change. If, in the end, $a=b=c$, $(a-b) \\equiv 0 \\mod 3$ then it must be to begin with as well.\r\n\r\nedit:... oh, yes- I misread the question", "Solution_2": "I don't think that it is necessary for $a=b=c$. Take $a=2, b=1, c=4$. We trade a and c for 2 b's, so now we have $(a,b,c)=(1,3,3)$, and it's clear from here that we can just trade b's and c's for a's. An even more basic example: just take $b=c$, and we can let a be whatever we want.", "Solution_3": "The answer it gives is the answer to a different problem then the problem you wrote. Go back and check the question: are you sure it says \"for all chips to be the same color\" and not \"for each color to have the same number of chips\"?", "Solution_4": "Literally, it says, \"...Find conditions for all chips to become of the same color.\"", "Solution_5": "So, it is true that [i]one[/i] of the differences must be $\\equiv 0 \\pmod 3$, because of the second half of tim's argument (or else you'll never get two piles both congruent mod 3, which is certainly required to get two piles equal to 0). (Note that he's solving the problem I suggested: getting each pile to the same size, rather than emptying two piles. And he correctly shows that the condition you gave is the condition for that question.)\r\n\r\nIn fact, I believe this condition is good enough: the answer is $(a-b)(b-c)(c-a) \\equiv 0 \\pmod 3$. All you have to do is show that condition is sufficient, which takes a teensy bit of casework: you just have to show how to shift 3 chips from one pile to another pile. Then you can reduce the difference between the two piles which differ by a multiple of 3 to 0, and after that you're golden.", "Solution_6": "Yeah, [b]Phelpedo's [/b]statement of the problem is indeed correct.", "Solution_7": "[quote=\"JBL\"]So, it is true that [i]one[/i] of the differences must be $\\equiv 0 \\pmod 3$, because of the second half of tim's argument (or else you'll never get two piles both congruent mod 3, which is certainly required to get two piles equal to 0). (Note that he's solving the problem I suggested: getting each pile to the same size, rather than emptying two piles. And he correctly shows that the condition you gave is the condition for that question.)\n\nIn fact, I believe this condition is good enough: the answer is $(a-b)(b-c)(c-a) \\equiv 0 \\pmod 3$. All you have to do is show that condition is sufficient, which takes a teensy bit of casework: you just have to show how to shift 3 chips from one pile to another pile. Then you can reduce the difference between the two piles which differ by a multiple of 3 to 0, and after that you're golden.[/quote]\r\n\r\nFor the problem I posted, the condition that one of the differences is 0 mod 3 is indeed both sufficient and necessary." } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "why doesnt LaTex appear in transcripts for me?", "Solution_1": "I don't know. What do you see? What browser are you using?", "Solution_2": "well LaTex appears in half and doesnt appear in the other half. I use internet explorer", "Solution_3": "What version? It works on IE 6 for Windows and IE 5.2 for Mac." } { "Tag": [ "geometry", "geometric transformation", "reflection", "trigonometry", "geometry unsolved" ], "Problem": "Let $\\ A_{1},B_{1},C_{1}$ be poins respectively on the sides $\\ BC,CA,AB$ of a triangle $\\ ABC$. Let $\\ A_{2},B_{2},C_{2}$ be poins respectively reflection of $\\ A,B,C$ in the midpoint of $\\ B_{1}C_{1},C_{1}A_{1},A_{1}B_{1}$. Prove that:\r\n1) $\\ AA_{1},BB_{1},CC_{1}$ are cocurrent if only if $\\ AA_{1},BB_{2},CC_{2}$ are cocurrent.\r\n2) $\\ S(A_{1}B_{1}C_{1})=S(A_{2}B_{2}C_{2})$ with $\\ S(XYZ)$ is the area of the triangle $\\ XYZ$.", "Solution_1": "Nobody help me? :(", "Solution_2": "The first part is easy! Just note that if we let $C_{3}=CC_{2}\\cap A_{1}B_{1}$ then from the sine law applied to the triangles $\\triangle A_{1}CC_{3}$ and $\\triangle B_{1}CC_{3}$ we obtain $\\frac{\\sin\\angle C_{2}CB_{1}}{\\sin\\angle C_{2}CA_{1}}=\\frac{\\sin\\angle C_{3}CB_{1}}{\\sin\\angle C_{3}CA_{1}}=\\frac{CB_{1}}{CA_{1}}$. The product of this equality with two its analogues for another vertices of $ABC$ gives the desired equivalence since the left hand side would be a trig-Ceva product for the lines $AA_{2}, BB_{2}, CC_{2}$ and the right hand side would be a usual Ceva product for the lines $AA_{1}, BB_{1}, CC_{1}$." } { "Tag": [ "algebra", "polynomial", "number theory proposed", "number theory" ], "Problem": "Let $ n$ be an integer greater than 2, and let $ p$ be a prime . Prove that polynomial \r\n$ P(x)\\equal{} x^n \\plus{} p^2 x^{n\\minus{}1}\\plus{}...\\plus{} p^2 x\\plus{}p^2$ cannot be written as the product of two polynomials with integer coefficients.", "Solution_1": "Eisenshtein theorem.", "Solution_2": "[quote=\"Rust\"]Eisenshtein theorem.[/quote]\r\nDo you mean Eisenstein criterion?If so,then you are wrong, Eisenstein criterion works only if $ p^2$ does not divide $ a_0$.", "Solution_3": "For Eisenshtein theorem $ p|a_1,a_2,...,a_{n\\minus{}1}, \\ p^2\\not a_0, p\\not |a_n$.\r\nBut prove for these case $ p^2|a_1,...,a_{n\\minus{}1}, p^3\\not|a_0, p\\not |a_n$ is same.", "Solution_4": "[quote=\"Rust\"]For Eisenshtein theorem $ p|a_1,a_2,...,a_{n \\minus{} 1}, \\ p^2\\not a_0, p\\not |a_n$.\nBut prove for these case $ p^2|a_1,...,a_{n \\minus{} 1}, p^3\\not|a_0, p\\not |a_n$ is same.[/quote]\r\nEisenstein criterion:\r\nIf $ p\\in\\mathbb{P}$ and $ p$ divides $ a_0,a_1,a_2,\\dots,a_n$,but $ p\\not |a_n$ and $ p^2\\not | a_0$ then polynomial \r\n$ P(x) \\equal{} a_nx^n \\plus{} a_{n \\minus{} 1}x^{n \\minus{} 1} \\plus{} \\dots \\plus{} a_1x \\plus{} a_0$ is irreducible over $ \\mathbb{Z}[x]$.\r\nBut in your case $ p^2$ is not a prime." } { "Tag": [ "function", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: \\mathbb{R} \\to \\mathbb{R}$ a function with the property of Darboux. We know also that $ f$ is increasing on $ \\mathbb{R} \\setminus \\mathbb{Q}$. Prove that $ f$ is continous.", "Solution_1": "This is rather inelegant but I'm rusty :blush: \r\n\r\nSuppose by contradiction (ie. that f is not monotonic) there exists x < y for which f(x) > f(y), where x is rational (and y is therefore irrational).\r\n\r\nBoth hypotheses on f along with the inequality f(x) > f(y) implies that f maps the rationals in ]x, y[ onto the open interval ]f(x) < f(y)[. Wrong cardinality for both sets.", "Solution_2": "Suppose that there are $ a \\in \\mathbb{Q}$ and $ b \\in \\mathbb{R} \\setminus \\mathbb{Q}$ such that $ af(b)$, since $ f$ has the Darboux property, if follows that there is $ \\epsilon >0$ such that $ f(x)>f(b)$ , $ \\forall x \\in (a\\minus{}\\epsilon, a\\plus{}\\epsilon)$.But $ \\mathbb{R} \\setminus \\mathbb{Q}$ is dense in $ \\mathbb{R}$, therefore we can always find $ c \\in \\mathbb{R} \\setminus \\mathbb{Q}$ such that $ c \\in (a\\minus{}\\epsilon, a)$ (which implies $ c f(b)$ contradiction. \r\n\r\nWhen you deal with the case $ a,b$ are both in $ \\mathbb{Q}$, you can always find $ m \\in \\mathbb{R} \\setminus \\mathbb{Q}$ such that $ a n because we must choose centers for these disks: either inside D_i or between D_i and a concentric circle with twice D_i's radius. We can divide that area into 6 parts, each a section of the circle with angle 60, and D_i is a seventh part. If we split up the 7n disks into these 7 sections, one section must have at least n disks, and the midpoint of the 60 degree arc of the larger circle is covered by all ther circle centered in that section. However, it is impossible for each section to have exactly n disks since the disks in the outer sections would be tangent to the disk in D_i. Therefore, there must be a point with n+1 of these larger disks covering it." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Given a,b,c>0 and abc=1.Prove that:\r\n$ \\frac {a^2}{b} \\plus{} \\frac {b^2}{c} \\plus{} \\frac {c^2}{a} \\ge a \\plus{} b \\plus{} c \\plus{} \\frac {(a^2c \\plus{} b^2a \\plus{} c^2b \\minus{} 3)^2}{ab \\plus{} bc \\plus{} ca}$", "Solution_1": "[quote=\"babyloverain\"]Given a,b,c>0 and abc=1.Prove that:\n$ \\frac {a^2}{b} \\plus{} \\frac {b^2}{c} \\plus{} \\frac {c^2}{a} \\ge a \\plus{} b \\plus{} c \\plus{} \\frac {(a^2c \\plus{} b^2a \\plus{} c^2b \\minus{} 3)^2}{ab \\plus{} bc \\plus{} ca}$[/quote]\r\nIt is equivalent to $ \\sum_{cyc}(a^3\\minus{}3a^2b\\plus{}6a^2c\\minus{}4abc)\\geq0,$ which is obvious.", "Solution_2": "It is also equivalent to the following:\r\n\r\n$ \\left(\\frac {1}{b} \\plus{} \\frac {1}{c} \\plus{} \\frac {1}{a}\\right)\\left(\\frac {(a \\minus{} b)^2}{b} \\plus{} \\frac {(b \\minus{} c)^2}{c} \\plus{} \\frac {(c \\minus{} a)^2}{a}\\right)$ $ \\ge \\left(\\frac {a \\minus{} b}{b} \\plus{} \\frac {b \\minus{} c}{c} \\plus{} \\frac {c \\minus{} a}{a}\\right)^2$\r\n\r\nwhich is Andrescu lemma (it's no matter whether $ \\frac{a\\minus{}b}{b}$, etc. are negative or positive.) :lol:" } { "Tag": [], "Problem": "One hundred people were surveyed. Of these, $ 87$ indicated they liked Mozart and $ 70$ indicated they liked Bach. What is the minimum number of people surveyed who could have said they liked both Mozart and Bach?", "Solution_1": "The minimum occurs when everyone likes at least one of them, so we have $ 87\\plus{}70\\minus{}x\\equal{}100 \\implies x\\equal{}\\boxed{57}$." } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "Prove or disprove:\r\n\r\n$a,b,c,t>0$\r\n\r\n$\\frac{a^2+tb^2}{ta+b}+\\frac{b^2+tc^2}{tb+c}+\\frac{c^2+ta^2}{tc+a}\\ge a+b+c$", "Solution_1": "Use Cauchy's inequality, since $a,b,c,t>0$\r\n\r\n$\\frac{a^2}{ta+b}+\\frac{b^2}{tb+c}+\\frac{c^2}{tc+a}\\geq \\frac{(a+b+c)^2}{(ta+b)+(tb+c)+(tc+a)}=\\frac{(a+b+c)^2}{(t+1)(a+b+c)}$\r\n\r\n$=\\frac{a+b+c}{t+1}\\ \\cdots [1].$\r\n\r\nSimilarly we have $\\frac{tb^2}{ta+b}+\\frac{tc^2}{tb+c}+\\frac{ta^2}{tc+a}\\geq \\frac{t}{t+1}\\cdot (a+b+c)\\ \\cdots [2].$ Adding the equations $[1],[2]$\r\n\r\nyielding $\\frac{a^2+tb^2}{ta+b}+\\frac{b^2+tc^2}{tb+c}+\\frac{c^2+ta^2}{tc+a}\\geq a+b+c.\\ Q.E.D.$", "Solution_2": "That's amazing!! Good work kunny. Anyway we usually call it Cauchy-Schwarz ;) not Caushy-Schwalz's .. :|", "Solution_3": "[quote=\"conceive\"]That's amazing!! Good work kunny. Anyway we usually call it Cauchy-Schwarz ;) not Caushy-Schwalz's .. :|[/quote]\r\n\r\nI made a typo. Thank you!\r\n\r\nkunny", "Solution_4": "$\\frac{a^2}{ta+b}+\\frac{b^2}{tb+c}+\\frac{c^2}{tc+a}\\geq \\frac{(a+b+c)^2}{(ta+b)+(tb+c)+(tc+a)}$\r\nsorry, can you explain this part? cause im not familiar with cauchy schwarz and i've only seen the application of two variables.. i mean can you explain what you used in your $a_k$, $b_k$ values for the inequality:\r\n$\\left(\\sum_{k=1}^{n}{a_kb_k}\\right)^2\\le \\left(\\sum_{k=1}^{n}{a_k^2}\\right)\\left(\\sum_{k=1}^{n}{b_k^2}\\right)$", "Solution_5": "All right, minsoens. :) \r\n\r\nFor[b] positive numbers [/b]$a,b,c,x,y,z$ use Cauchy-Schwalz's inequality for three variables $x,y,z,$\r\nwe have $\\{(\\sqrt{x})^2+{(\\sqrt{y})^2+\\{(\\sqrt{z})^2\\}\\left\\{\\left(\\frac{a}{\\sqrt{x}}\\right)^2+\\left(\\frac{b}{\\sqrt{y}}\\right)^2+\\left(\\frac{c}{\\sqrt{z}}\\right)^2\\right\\}\\geq \\left (\\sqrt{x}\\cdot \\frac{a}{\\sqrt{x}}+\\sqrt{y}\\cdot \\frac{b}{\\sqrt{y}}+\\sqrt{z}\\cdot \\frac{c}{\\sqrt{z}}\\right)^2,}$ yielding\r\n\r\n\\[ \\boxed{\\frac{a^2}{x}+\\frac{b^2}{y}+\\frac{c^2}{z}\\geq \\frac{(a+b+c)^2}{x+y+z}} \\]", "Solution_6": "oh.. :blush: thanks\r\ni guess it would be possible to get the similar result for any number of variables?", "Solution_7": "Yes, it is :)", "Solution_8": "[quote=\"kunny\"]Use Cauchy-Schwalz's inequality, since $a,b,c,t>0$\n\n$\\frac{a^2}{ta+b}+\\frac{b^2}{tb+c}+\\frac{c^2}{tc+a}\\geq \\frac{(a+b+c)^2}{(ta+b)+(tb+c)+(tc+a)}=\\frac{(a+b+c)^2}{(t+1)(a+b+c)}$\n\n$=\\frac{a+b+c}{t+1}\\ \\cdots [1].$\n\nSimilarly we have $\\frac{tb^2}{ta+b}+\\frac{tc^2}{tb+c}+\\frac{ta^2}{tc+a}\\geq \\frac{t}{t+1}\\cdot (a+b+c)\\ \\cdots [2].$ Adding the equations $[1],[2]$\n\nyielding $\\frac{a^2+tb^2}{ta+b}+\\frac{b^2+tc^2}{tb+c}+\\frac{c^2+ta^2}{tc+a}\\geq a+b+c.\\ Q.E.D.$[/quote]\r\nIn fact, you've used the Inequality of Bergstrom, but it's the same." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "[This is a very interesting problem\uff0cI hope you can enjoy it.] In triangle ABC (a=BC\uff1bb=AC\uff1bc=AB); M is the midpoint of side AC. Prove that inscribed circle of triangle divides segment BM into there segments that are the same length if and only if : a/5=b/10=c/13 :roll:", "Solution_1": "INMO 2005 Problem 1", "Solution_2": "$ CT < CM \\Longleftrightarrow \\frac{_{a \\plus{} b \\minus{} c}}{^2} < \\frac{_b}{^2} \\Longleftrightarrow a < c.$ Then $ BP \\equal{} MQ \\Longleftrightarrow a \\equal{} BC \\equal{} CM \\equal{}\\frac{_b}{^2}.$\r\n$ MQ \\equal{} QP \\equal{} \\frac{_1}{^3}MB \\Longleftrightarrow$ $ MT^2 \\equal{} MQ \\cdot MP \\equal{} \\frac{_2}{^9} MB^2 \\Longleftrightarrow$ $ \\frac{_1}{^4} (c \\minus{} a)^2 \\equal{} \\frac{_2}{^9} \\cdot \\frac{_1}{^4}(2c^2 \\plus{} 2a^2 \\minus{} b^2) \\equal{} \\frac{_1}{^{9}}(c^2 \\minus{} a^2) \\Longleftrightarrow$\r\n$ 9(c\\minus{}a) \\equal{} 4(c\\plus{} a) \\Longleftrightarrow \\frac{_c}{^{13}} \\equal{} \\frac{_a}{^5} \\equal{} \\frac{_b}{^{10}}$", "Solution_3": "it seems the indians have generated a problem from another olympiad to pose it in there national olympiad. is this valid?", "Solution_4": "I am sorry :blush: . I saw this problem on a book , maybe the author of the book has made a mistake." } { "Tag": [ "geometry", "circumcircle", "symmetry", "real analysis", "geometry solved" ], "Problem": "Let A'B'C'D'E' be a convex pentagon and suppose B'C', D'E' meet at A; C'D', E'A' meet at B; etc, so we have a convex pentagon ABCDE obtained from extending the sides of A'B'C'D'E'.\r\n\r\nProve that the second intersections of the neighboring pairs of circumcircles of AC'D' , BD'E', CE'A', DA'B', EB'C' intersect at 5 concyclic points.", "Solution_1": "This is probably one of my favorites :)\r\n\r\nSolution:\r\n\r\nLet the circumcircles of $ DA'B'$ and $ CA'E'$ meet at $ P$. Similarly, let the pairs of circumcircles which meet at $ B',C',D',E'$ have second intersections $ Q,R,S,T$ respectively.\r\n\r\nFirst, let's establish the fact that the second intersections of the circumcircles are outside $ A'B'C'D'E'$. Let $ X$ and $ Y$ be the circumcenters of $ \\triangle DA'B'$ and $ \\triangle CA'E'$. It will suffice to show that\r\n\r\n$ \\angle XA'B' \\plus{} \\angle B'A'E' \\plus{} \\angle YA'E' > \\pi$\r\n$ \\iff \\frac {1}{2}(\\pi \\minus{} \\angle B'XA') \\plus{} \\angle B'A'E' \\plus{} \\frac {1}{2}(\\pi \\minus{} \\angle A'YE') > \\pi$ \r\n$ \\iff \\angle B'A'E' > \\frac {1}{2}(\\angle B'XA' \\plus{} \\angle A'YE') \\equal{} \\angle B'DA' \\plus{} \\angle A'CE'$\r\n\r\nwhich follows from (using exterior angles)\r\n\r\n$ \\angle B'A'E' \\equal{} \\angle B'DA' \\plus{} \\angle DB'C$ \r\n$ \\equal{} \\angle B'DA' \\plus{} (\\angle B'AC \\plus{} \\angle A'CE') > \\angle B'DA' \\plus{} \\angle A'CE'$\r\n\r\n\r\nNow, we show that the points $ A',Q,T,B,E$ are concyclic. Note that\r\n\r\n$ \\angle EQA' \\equal{} \\angle EQB' \\plus{} \\angle B'QA'$\r\n$ \\equal{} (\\pi \\minus{} \\angle EC'B') \\plus{} \\angle B'DA'$\r\n$ \\equal{} \\pi \\minus{} \\angle EBA'$\r\n\r\nbecause $ \\angle EC'B' \\equal{} \\angle B'DA' \\plus{} \\angle EBA'$ again by exterior angles. Thus, $ Q$ lies on the circumcircle of $ A'BE$ and by a similar argument, $ T$ lies on the same circle.\r\n\r\nAgain, by analogy, we have that the points $ A,B',C,R,P$ are also concyclic.\r\n\r\n\r\nFinally, we will show that $ \\angle QRP \\equal{} \\angle QTP$, which will show that $ P,Q,R,T$ are concyclic, and by symmetry, $ Q,P,T,S$ will also be concyclic and the result will follow. We have\r\n\r\n$ \\angle QRP \\equal{} \\angle QRB' \\plus{} \\angle B'RP$\r\n$ \\equal{} \\angle QEB' \\plus{} \\angle PCA'$\r\n$ \\equal{} \\angle QTA' \\plus{} \\angle PTA' \\equal{} \\angle QTP$\r\n\r\nand we're DONE!\r\n\r\n(Wow that write-up was longer than I expeced... on second thought, maybe I should've posted this in the Geometry forum :blush: )", "Solution_2": "[quote=\"ThAzN1\"]\n(Wow that write-up was longer than I expeced... on second thought, maybe I should've posted this in the Geometry forum :blush: )[/quote]\r\n\r\nI'll move it for you ;)", "Solution_3": "Here is my solution (which is similar to ThAnz1 's, but as I use some interesting theorems, so I decide to post it)\r\n [b]Theorem 1[/b] (Miquel 's) \r\n Suppose $a, b, c, d$ are four lines on a plane. Denote by $abc$ the triangle formed by $a, b, c$. Analogously, we define $bcd, cda, abd$. Then the circumcircles of $abc, bcd, cda, abd$ are concurrent.\r\n [b]Theorem 2[/b] (Lebesgue 's)\r\n Let $ABCD$ be a convex cyclic quadrilateral. Let $(a), (b), (c), (d)$ be four circles such that $A$ and $B$, $B$ and $C$, $C$ and $D$, $D$ and $A$ respectively lie on $(a), (b), (c), (d)$. If $A', B', C', D'$ are the second intersections of $(a)$ and $(b)$, $(a)$ and $(b)$, $(c)$ and $(d)$, $(d)$ and $(a)$, then $A'B'C'D'$ is cyclic.\r\n Back to the problem, let $A_1, B_1, C_1, D_1, E_1$ be the second intersections of $(A'E'C)$ and $(A'B'D)$, $(A'B'D)$ and $(B'C'E)$, $(B'C'E)$ and $(C'D'B)$, $(C'D'B)$ and $(D'E'B)$, $(D'E'B)$ and $(E'A'C)$. By applying theorem 1 to four lines $B'A', A'E', E'D', D'C'$, we obtain that $C, E, A', E_1$ lie on a circle. Similarly, we obtain $C, E, A', C_1$ lie on a circle. Hence $CE_1C_1E$ is cyclic.\r\n Consider 4 points $C, A', B', E$ which are collinear in that order (notice that here we can consider $CA'B'E$ is a cyclic quadrilateral). By applying theorem 2 with circumcircles $(CA'E_1A_1), (A'B'A_1B_1), (B'EB_1C_1), (ECC_1E_1)$, we obtain $E_1A_1B_1C_1$ is cyclic. In the same manner, we have $A_1B_1C_1D_1$ is cyclic. Hence 5 points $A_1, B_1, C_1, D_1, E_1$ are on a circle.", "Solution_4": "[url=http://www.math.ust.hk/excalibur/v6_n1.pdf]Here (Mathematical Exaclibur 2001 (vol. 6), no. 1, p. 1-2)[/url] you will find a discussion of this problem and a nice example of political misuse of mathematics...\r\n\r\n Darij", "Solution_5": "In \"Excalibur 2001 (vol 6)\" say this problem was posed by the President Jiang Zemin to the student in a meeting. It is a charming action !" } { "Tag": [ "ceiling function" ], "Problem": "What is the least positive three-digit multiple of 7?", "Solution_1": "The smallest 3-digit integer is 100. When we divide this by 7, we get about 14. Thus our answer is $ 15 \\times 7 \\equal{} \\boxed{105}$.", "Solution_2": "you don't need to divide it, the guess of the solution is possible.\r\n\r\n$ 100\\minus{}70\\equal{}30$, $ 28<30<35$\r\nSo $ 70\\plus{}35\\equal{}\\fbox{105}$", "Solution_3": "[quote=\"isabella2296\"]The smallest 3-digit integer is 100. When we divide this by 7, we get about 14. Thus our answer is $ 15 \\times 7 \\equal{} \\boxed{105}$.[/quote]\r\n\r\nWhat is did (in equation) is this:\r\n\r\n$ \\lceil(\\frac {100}{7})\\rceil \\times 7$\r\n\r\nThis works for everything, so the equation is...\r\n\r\n$ \\lceil(\\frac {\\text{smallest possible number}}{\\text{divisor}})\\rceil \\times \\text{ divisor}$\r\n\r\n\r\n[size=75][color=red]Fixed LaTeX. -Isabella[/color][/size]", "Solution_4": "Note that $ \\lceil$ and $ \\rceil$ means the smallest integer less than or equal to the number inside those symbols.", "Solution_5": "It's smallest integer [b]greater[/b] than or equal to the number in the equation, izzy." } { "Tag": [ "probability" ], "Problem": "Two people play a game in which they take turns rolling a single six sided die. If someone rolls the number, their opponent just rolled or any multiple of that number, then they win; otherwise the game continues and their opponent takes a turn. What is the probability that the first player wins the game on the second turn?", "Solution_1": "Uh from what I can tell of the problem, the second player goes on the second turn, so 0?", "Solution_2": "Perhaps what he meant is that the first player wins the game on the second turn of the first player?\r\nHmm...I'm not sure either.", "Solution_3": "The question, I believe, is asking what is the probability that the first player wins on the second round of rolling the dice. Thats how the problem was worded. I know its pretty confusing :(", "Solution_4": "[hide=\"Solution\"]There are 6 multiples of 1, 3 multiples of 2, 2 multiples of 3, 1 multiple of 4, 1 multiple of 5, and 1 multiple of 6 among the numbers 1 to 6.\n\nHence, the probability of a player winning on any round is $ \\frac{1}{6}\\left(1\\plus{}\\frac{1}{2}\\plus{}\\frac{1}{3}\\plus{}\\frac{1}{6}\\plus{}\\frac{1}{6}\\plus{}\\frac{1}{6}\\right)\\equal{}\\frac{7}{18}.$\n\nIn order for the first player to win on the second round, the second player must lose on the first round and the first player must win on the second round.\n\nThe probability of such event occurring is $ \\frac{11}{18}\\cdot\\frac{7}{18}\\equal{}\\frac{77}{324}.$\n\nAnswer: $ \\frac{77}{324}$[/hide]", "Solution_5": "Thats the answer that I got. The answer key claims that the answer is 59/216 however.", "Solution_6": "The reason that doesn't work is because the first roll of the second player isn't equally likely to be any number. If player 2 rolls a 1, then they only win if the first player rolled was a 1, but if they roll a 6, they win if the first player rolled a 1,2,3, or 6.\r\n\r\nCases after player one and player two roll once\r\n$ 7/18$ chance player two won.\r\n\r\n$ 11/18$ chance player two hasn't won; subcases:\r\n\r\n$ 5/36$ chance player two rolled a 1\r\n$ 1/9$ chance player two rolled a 2\r\n$ 1/9$ chance player two rolled a 3\r\n$ 1/12$ chance player two rolled a 4\r\n$ 1/9$ chance player two rolled a 5\r\n$ 1/18$ chance player two rolled a 6\r\n\r\nCalculating the probability that player 1 wins yields:\r\n\r\n$ (\\frac {5}{36}) (1) \\plus{} (\\frac {1}{9}) (\\frac {1}{2}) \\plus{} (\\frac {1}{9}) (\\frac {1}{3}) \\plus{} (\\frac {1}{12}) (\\frac {1}{6}) \\plus{} (\\frac {1}{9}) (\\frac {1}{6}) \\plus{} (\\frac {1}{18}) (\\frac {1}{6})$\r\n\r\n$ \\frac {5}{36} \\plus{} \\frac {1}{18} \\plus{} \\frac {1}{27} \\plus{} \\frac {1}{72} \\plus{} \\frac {1}{54} \\plus{} \\frac {1}{108}$\r\n\r\n$ \\frac {30}{216} \\plus{} \\frac {12}{216} \\plus{} \\frac {8}{216} \\plus{} \\frac {3}{216} \\plus{} \\frac {4}{216} \\plus{} \\frac {2}{216}$\r\n\r\n$ \\frac {59}{216}$" } { "Tag": [ "arithmetic series", "arithmetic sequence" ], "Problem": "$a_{1}, a_{2}, a_{3}$ are the first three terms in an arithmetic series of reals. Find all (a_1, a_2, a_3) such that\r\n\r\n$1) a_{2}^{2}=a_{1}a_{3}+9$\r\n$2) a_{1}^{4}+a_{2}^{4}+a_{3}^{4}=4737$\r\n\r\nhint:\r\n\r\n[hide]let the middle term be x and denote the other terms using the difference of the sequence[/hide]", "Solution_1": "(2,5,8) and (-8,-5,-2) with the first and third terms switched accounts for all possible comboos.\r\n\r\noh yea, the sign on the middle term doesn't matter", "Solution_2": "Since it is an arithmetic sequence, $a_{2}= \\frac{a_{1}+a_{3}}{2}$\r\n\r\n${a_{2}}^{2}=a_{1}a_{3}+9$\r\n$\\frac{{a_{1}}^{2}+2a_{1}a_{3}+{a_{3}}^{2}}{4}=a_{1}a_{3}+9$\r\n$\\frac{{a_{1}}^{2}-2a_{1}a_{3}+{a_{3}}^{2}}{4}=9$\r\n$\\frac{{a_{1}}^{2}-2a_{1}a_{3}+{a_{3}}^{2}}=36$\r\n$(a_{1}-a_{3})=\\pm6$\r\n$a_{1}=a_{3}\\pm6$\r\n$a_{2}=a_{3}\\pm3$\r\nI'm not sure what to do next... Anybody have any ideas?", "Solution_3": "a better substitution:\r\n\r\n$a_{1}=x-d$\r\n$a_{2}=x$\r\n$a_{3}=x+d$\r\n\r\nworking in equation one first, finding d, then continuing is the idea you should use.", "Solution_4": "How would you prove that (2,5,8), (8,5,2), (-8,-5,-2), (-2,-5,-8) are the only solutions?", "Solution_5": "uh.. after you do the actual work i suggested, you see that:\r\n$a_{2}=\\pm5, \\pm\\sqrt{-61}$\r\n\r\nso by the initial conditions, those four soultuions are clearly the only possiblities.", "Solution_6": "[quote=\"vishalarul\"]Since it is an arithmetic sequence, $a_{2}= \\frac{a_{1}+a_{3}}{2}$\n\n${a_{2}}^{2}=a_{1}a_{3}+9$\n$\\frac{{a_{1}}^{2}+2a_{1}a_{3}+{a_{3}}^{2}}{4}=a_{1}a_{3}+9$\n$\\frac{{a_{1}}^{2}-2a_{1}a_{3}+{a_{3}}^{2}}{4}=9$\n$\\frac{{a_{1}}^{2}-2a_{1}a_{3}+{a_{3}}^{2}}=36$\n$(a_{1}-a_{3})=\\pm6$\n$a_{1}=a_{3}\\pm6$\n$a_{2}=a_{3}\\pm3$\nI'm not sure what to do next... Anybody have any ideas?[/quote]\r\n\r\nAfter doing that you have basically found out that the terms of the sequence have a common difference of 3. Now you can let the first three terms be $x-3,x,x+3$ (or $x,x+3,x+6$, etc.) and solve in the second equation.", "Solution_7": "I guess you are right... Wouldn't it be easier to use $x-3$, $x$, and $x+3$? However, I was worried about taking everything to the 4th power and adding. Isn't it a messy quartic?", "Solution_8": "um, it's obviously a biquartic? and an easily factorable one at that\r\n\r\nand considering you didn't even USE the second equation, I'd think it'd be worth trying to deal with this 'messy quartic' before asking us for help", "Solution_9": "[quote=\"vishalarul\"]I guess you are right... Wouldn't it be easier to use $x-3$, $x$, and $x+3$? However, I was worried about taking everything to the 4th power and adding. Isn't it a messy quartic?[/quote]\r\n\r\nabsolutely not. the odd powers would cancel and the expression could be easily expanded w/ binomial theorem. in this case, it factors nicely into the form (x^2+n)(x^2-m)=0", "Solution_10": "What's a biquadratic?", "Solution_11": "[url]http://mathworld.wolfram.com/QuarticEquation.html[/url]" } { "Tag": [], "Problem": "prove that one is not straight by using contradiction", "Solution_1": "At least this is in the right forum for a change. Seeing as this is topic is probably spam, might as well make it worth a laugh.\r\n\r\n\r\n[youtube]N5UkFfcGQrs[/youtube]", "Solution_2": "show him one of those inapporopriate magazines" } { "Tag": [ "algebra", "polynomial", "LaTeX" ], "Problem": "if ax^2+bx+c has no real roots and a+b+c<0 then prove that c<0", "Solution_1": "It is more beautiful to write in Latex :wink: \r\n\r\nIf $ ax^2\\plus{}bx\\plus{}c$ has no real roots and $ a\\plus{}b\\plus{}c<0$ then prove that $ c<0$", "Solution_2": "how can i do that????\r\nplz help i m new here", "Solution_3": "[hide]Let $ f(x) \\equal{} ax^2\\plus{}bx\\plus{}c$, $ f(1)\\equal{}a\\plus{}b\\plus{}c<0$ so $ f(x) <0$, so $ f(0) \\equal{} c < 0$ as required[/hide]", "Solution_4": "See here. [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url] :lol:" } { "Tag": [ "geometry", "circumcircle", "area of a triangle", "Heron\\u0027s formula" ], "Problem": "Quadrilateral ABCD is inscribed in a circle. AB=6, BC=7, CD=4, and AC=8. What is the area of this circle?", "Solution_1": "Do you mean CD=4?", "Solution_2": "[hide]Well, anyways, it is clear that $ \\bigtriangleup ABC$ is also inscribed in the circle. By Heron's formula, the area of the triangle is $ \\sqrt {s(s - a)(s - b)(s - c}$, where $ s = \\frac {8 + 6 + 7}{2} = \\frac {21}{2}$. The area is then $ \\sqrt {\\frac {21}{2}\\cdot \\frac {5}{2} \\cdot \\frac {9}{2} \\cdot \\frac {7}{2}} = \\sqrt {\\frac {3^3 \\cdot 5\\cdot 7^2}{2^4}} = \\frac {21\\sqrt {15}}{4} = K$.\nThen, the circumradius is given by $ \\frac {abc}{4K} = \\frac {8\\cdot 6\\cdot 7}{21\\sqrt {15}} = \\frac {16}{\\sqrt {15}}$.\nThe area of this circle is therefore $ \\pi {\\left( \\frac {16}{\\sqrt {15} \\right)}^2 = \\boxed{\\frac {256 \\pi}{15}}}$[/hide]" } { "Tag": [ "search" ], "Problem": "Anyone here interested in astronomy :D ? I guess there should be lots of you.\r\n\r\nI propose making a kind of marathon for astronomy, if there are people that enjoy watching the sky and so on :).\r\nAnyhow, anybody here going at the astronomy olympiad? Do you know sites with training problems or anything like?\r\n\r\n\r\nCheers", "Solution_1": "yeah\r\nI'm interested in it\r\nI study astronomy by myself \r\nand I have expriences in astronomy olympiad \r\nit's great to have astronomy marathon :rotfl:", "Solution_2": "It would be great to have an astronomy olympiad.\r\nI have bben to the international astronomy olympiad twice(2005,2006 both golds).\r\nSearch for The Paradoxical Universe on google.It is a problem book with 250 questions on astronomy but it's in Russian>use a good translator and u can figure out the questions pretty easily.", "Solution_3": "if any one can guide about the type of questions for 2nd round of astro olympiad? \r\nwhere can we have sample questions?" } { "Tag": [ "geometry", "number theory unsolved", "number theory" ], "Problem": "1) if 1 + 2^n + 4^n = prime number so prove that n=3^k\r\n2)x^2=a so find x,y,z if a-b=8c-1\r\n y^3=b\r\n z^3=c\r\n3)1000!=x(mod 2003) x=?\r\nAnd one geometry \r\n4)we have a circumscribed quadrilateral.And the tangent points are connected.Prove that the diagonals and this two lines are intersecting at same point", "Solution_1": "Please post different problems in different topics\r\nFor number 1 check\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=150412\r\n\r\nDaniel" } { "Tag": [], "Problem": "Let $a,b>0$ and $E(a,b)=\\sqrt{a\\sqrt{b\\sqrt{a\\sqrt{b...}}}}$. Compute $E^3(a,b)$.", "Solution_1": "By $E^3(a,b)$, do you mean $[E(a,b)]^3$ or $E(E(E(a,b)))$?", "Solution_2": "Well, in the book says $E^3(a,b)$. I think the they meant $[E(a,b)]^3$", "Solution_3": "I think you're right, because it makes the problem easier.\r\n\r\n[hide]$E(a,b)=\\sqrt{a\\sqrt{b\\sqrt{a\\sqrt{b...}}}}$\nLet $x = E(a,b)$:\n$x=\\sqrt{a\\sqrt{bx}}$\n$x^2=a\\sqrt{bx}$\n$x^4=a^2bx$\n$x^3=a^2b$\n$[E(a,b)]^3=a^2b$[/hide]" } { "Tag": [ "group theory", "abstract algebra", "linear algebra", "matrix", "quadratics", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $p,q$ be odd primes such that $q$ is a divisor of $|SL_{2}(Z/pZ)|$. Then any $q$-Sylow of $SL_{2}(Z/pZ)$ is cyclic.", "Solution_1": "The order of $SL_{2}(\\mathbb{Z}/p\\mathbb{Z})$ is $(p-1)p(p+1)$. Since each odd prime divides only one of these factors, it suffices to construct cyclic subgroups of order $p-1,p,p+1$ respectively.\r\n\r\nThe matrices $\\begin{pmatrix}1&a\\\\0&1\\end{pmatrix}$ are a subgroup naturally isomorphic to $\\mathbb{Z}/p\\mathbb{Z}$, which is cyclic of order $p$.\r\nThe matrices $\\begin{pmatrix}a&0\\\\0&\\frac1a\\end{pmatrix}$ for $a\\neq0$ are a subgroup naturally isomorphic to $\\left(\\mathbb{Z}/p\\mathbb{Z}\\right)^{*}$, which is cyclic of order $p-1$.\r\n\r\nFor order $p+1$, we step back slightly. Let $r$ be any quadratic nonresidue mod $p$. The field with $p^{2}$ elements can be written as $\\{a+b\\sqrt{r}\\}$, with $a,b\\in \\mathbb{Z}/p\\mathbb{Z}$. This field is naturally isomorphic to the field of matrices of the form $\\begin{pmatrix}a&rb\\\\b&a\\end{pmatrix}$. The nonzero matrices of this form form a cyclic group (the multiplicative group of the field) with order $(p-1)(p+1)$. The determinant is a surjective homomorphism from this group to $\\left(\\mathbb{Z}/p\\mathbb{Z}\\right)^{*}$. Its kernel (the intersection with $SL_{2}$) therefore has order $p+1$. As a quotient of a cyclic group, it is cyclic.\r\nPutting the pieces together, our cyclic subgroup of order $p+1$ is $\\left\\{\\begin{pmatrix}a&rb\\\\b&a\\end{pmatrix}: a^{2}-rb^{2}=1\\right\\}.$" } { "Tag": [ "geometry" ], "Problem": "[color=red]Here's a problem that we have for homework, I figured out a way to do them, but I was wondering if there was easeir ways, so please reply.\n\nAugustus De Morgan said in 1864, \"At some time in my life the square of my age was the same as the year in which I was that age.\" In what year was De Morgan born? Show how you arrived at your answer.\n\nPlease help.\n[/color]\r\n[color=darkred]\n[i]This thread has been hijacked by Treething to make it more legible =o[/i]\n[/color]", "Solution_1": "Well, he could be 100 or -100, right ;) ?", "Solution_2": "Thanx :lol:", "Solution_3": "Ok, here's another problem:\r\n \r\nPeople used toothpicks to make squares, and on the first design, there was 4 tooth picks and the area was one, and it goes on like this\r\n Toothpick Area\r\n 4 1\r\n 10 3\r\n 18 6\r\n 28 10\r\nand it goes on like that.. and the question is what is the number of tooth picks and the area of the 15th term\r\n\r\nThanx :)", "Solution_4": "LuCy4EvA wrote:Ok, here's another problem:\n \nPeople used toothpicks to make squares, and on the first design, there was 4 tooth picks and the area was one, and it goes on like this\n Toothpick Area\n 4 1\n 10 3\n 18 6\n 28 10\nand it goes on like that.. and the question is what is the number of tooth picks and the area of the 15th term\n\nThanx \n\n\n\nHere's a hint. Look for a pattern (duh). One of the most useful patterns are [hide]triangle numbers[/hide] and they're useful in this case.", "Solution_5": "After 18 years of nonstop work...\n\nSolution with [b] Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee, Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b], [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].\n\n[size=150][b]Problem from post 1[/b][/size]:\n\nWe proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \\in \\mathbb{Z}$. Aqu\u00ed es el idea clave:\n-----\n[b][color=#f00]Lemma:[/color][/b] $0 \\leq \\text{any age} \\leq 130$\n[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love: \n-----\n\nDefine $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:\n\n$a \\leq 42$: His age in $1864$ is $$f(a) \\geq f(42) = 142 > 130,$$ which is absurd.\n$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\\boxed{1806}.$$\n$a \\geq 44$: His age in $1864$ is $$f(a) \\leq f(44) = -28 < 0,$$ which is absurd.\n\nThus, we have exhausted all cases, and are done. $\\blacksquare$\n\n[size=150][b]Problem from post 3[/b][/size]:\n\nWe proceed with engineering. See the attached diagram.\n\nDenote by $\\kappa(n)$ the number of toothpicks in the $n^{\\text{th}}$ term, and denote by $\\tau(n)$ the area of the $n^{\\text{th}}$ term. Note that upon taking the discrete derivative $\\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:\n\n\\begin{tabular}{|c|c|c|c|c|}\\hline\n$n$ & $1$ & $2$ & $3$ \\\\\\hline\n$\\Delta_\\tau(n)$ & $2$ & $3$ & $4$\\\\\\hline\n$\\Delta_\\kappa(n)$ & $6$ & $8$ & $10$\\\\\\hline\n\\end{tabular}\n\u0627\u0644\u0622\u0646 \u0647\u0646\u0627 \u0647\u0648 \u0627\u0644\u0645\u0637\u0627\u0644\u0628\u0629 \u0627\u0644\u0631\u0626\u064a\u0633\u064a\u0629:\n-----\n[color=#f00][b]Lemma[/b][/color]: $\\Delta_\\tau(n)$ and $\\Delta_\\kappa(n)$ are linear functions.\n[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\\boxed{}$\n-----\nChanneling SPIRIT of Ramanujan implies $\\Delta_{\\tau}(n) \\equiv n+1$ and $\\Delta_{\\kappa}(n) \\equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine\n$$\\tau(n) = \\tau(1) +\\sum_{k=1}^{n-1} \\Delta_{\\tau}(k)= 1 + \\sum_{k=1}^{n-1} \\left(k+1\\right) = 1+\\left( -1 + \\frac{n(n+1)}{2} \\right)= \\frac{n(n+1)}{2}$$ \nand $$\\kappa(n) = \\kappa(1) + \\sum_{k=1}^{n-1} \\Delta_{\\kappa}(n) = 4 + \\sum_{k=1}^{n-1}\\left( 2k+4\\right) = 4 + 2 \\cdot \\frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$\nand evaluating at $n=15$ gives a final answer of $\\tau(15)=120$ area and $\\kappa(15)=270$ toothpicks, so we are done. $\\blacksquare$", "Solution_6": "[quote=mira74]\n[b][color=#f00]Lemma:[/color][/b] $0 \\leq \\text{any age} \\leq 130$\n[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love: \n-----[/quote]\n\n[url]https://en.wikipedia.org/wiki/Li_Ching-Yuen[/url] This dude claims to be 256...\n", "Solution_7": "[quote=ghu2024][quote=mira74]\n[b][color=#f00]Lemma:[/color][/b] $0 \\leq \\text{any age} \\leq 130$\n[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love: \n-----[/quote]\n\n[url]https://en.wikipedia.org/wiki/Li_Ching-Yuen[/url] This dude claims to be 256...[/quote]\n\nsorry ghu2024, but Li Ching Yuen looks to be at most 116, unlike my homie braidon, whose mom looks 127 :love:", "Solution_8": "Sar, I have an issue with your steamy first lemma. You said \n[b][color=#f00]Lemma:[/color][/b] $0 \\leq \\text{any age} \\leq 130$\nbut I myself do not believe this. The lemma should be this:\n[b][color=#f00]Lemma:[/color][/b] $-1 \\leq \\text{any age} \\leq 130$\n[i]Proof:[/i] none of us know anyone over 130. However, it is important to note that fetuses are human lives, a well established fact by thousands of mathematicians. This can be supported by work done by the mathematician National Right to Life in their paper [url]http://www.nrlc.org/abortion/wdlb/[/url], and topologist Donald J. Trump [url]https://twitter.com/realdonaldtrump/status/1129954110747422720?lang=en[/url], who builds on work done by Ronald Reagan, who was actually not in fact a mathematician but is well known for his role of being a president. Therefore, it is possible for an age to be negative. \nHowever, I have not met any fetuses under the age of -1. Therefore, the bounds $-1 \\leq \\text{any age} \\leq 130$ are sufficient.\nI am assuming this problem was tackled by someone with a notable lack of olympiad experience, as this is an easy mistake to miss if you have not practice USAMO geometry questions, which typically have similar style proofs. I invite Skrublord420 to proofread my work due to his notable achievements concerning Alphatrion's prime number conjectures.", "Solution_9": "Lmao the girl who originally posted this hasn't visited AOPS in 11 years. Shes probably in her 30s right now", "Solution_10": "What roles does $f'(a)=-2a+1$ play @5above", "Solution_11": "[quote=Treething]Well, he could be 100 or -100, right ;) ?[/quote]\n\nHow can someone be -100?", "Solution_12": "[quote=m1377]Lmao the girl who originally posted this hasn't visited AOPS in 11 years. Shes probably in her 30s right now[/quote]\n\nNah, if she was in middle school, then she was at most 13 yrs old. So, 13+11= 24.", "Solution_13": "but it was posted in 2004, not 2009.", "Solution_14": "So she\u2019s 29. K.", "Solution_15": "Why is her age relevant? ", "Solution_16": "What roles does $f'(a)=-2a+1$ play in mira74's post", "Solution_17": "[quote=Euleramanujan]What roles does $f'(a)=-2a+1$ play in mira74's post[/quote]\n\nHello,\n\nThrough evaluating the derivative of $f$ with respect to $a$, we are able to conclude that the function $f$ is decreasing, allowing us to utilize the inequalities $f(a) \\geq f(42)$ for all $a \\leq 42$ (in the valid age range) and $f(a) \\leq f(44)$ for all $a \\geq 44$.\n\nBest,\n\nmira74", "Solution_18": "the real question is : why did we bump a 16 year old post??", "Solution_19": "[quote=BrightMonkey]the real question is : why did we bump a 16 year old post??[/quote]\n\nto solve the problem!", "Solution_20": "Thank you for clearing up what role $f'(a)=-2a+1$ plays.", "Solution_21": "Also, how did you get $f'(a)=-2a+1$", "Solution_22": "its the derivative of $f(a)$.", "Solution_23": "Oh yeah thanks.", "Solution_24": "[quote=mira74]After 18 years of nonstop work...\n\nSolution with [b] Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee, Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b], [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].\n\n[size=150][b]Problem from post 1[/b][/size]:\n\nWe proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \\in \\mathbb{Z}$. Aqu\u00ed es el idea clave:\n-----\n[b][color=#f00]Lemma:[/color][/b] $0 \\leq \\text{any age} \\leq 130$\n[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love: \n-----\n\nDefine $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:\n\n$a \\leq 42$: His age in $1864$ is $$f(a) \\geq f(42) = 142 > 130,$$ which is absurd.\n$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\\boxed{1806}.$$\n$a \\geq 44$: His age in $1864$ is $$f(a) \\leq f(44) = -28 < 0,$$ which is absurd.\n\nThus, we have exhausted all cases, and are done. $\\blacksquare$\n\n[size=150][b]Problem from post 3[/b][/size]:\n\nWe proceed with engineering. See the attached diagram.\n\nDenote by $\\kappa(n)$ the number of toothpicks in the $n^{\\text{th}}$ term, and denote by $\\tau(n)$ the area of the $n^{\\text{th}}$ term. Note that upon taking the discrete derivative $\\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:\n\n\\begin{tabular}{|c|c|c|c|c|}\\hline\n$n$ & $1$ & $2$ & $3$ \\\\\\hline\n$\\Delta_\\tau(n)$ & $2$ & $3$ & $4$\\\\\\hline\n$\\Delta_\\kappa(n)$ & $6$ & $8$ & $10$\\\\\\hline\n\\end{tabular}\n\u0627\u0644\u0622\u0646 \u0647\u0646\u0627 \u0647\u0648 \u0627\u0644\u0645\u0637\u0627\u0644\u0628\u0629 \u0627\u0644\u0631\u0626\u064a\u0633\u064a\u0629:\n-----\n[color=#f00][b]Lemma[/b][/color]: $\\Delta_\\tau(n)$ and $\\Delta_\\kappa(n)$ are linear functions.\n[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\\boxed{}$\n-----\nChanneling SPIRIT of Ramanujan implies $\\Delta_{\\tau}(n) \\equiv n+1$ and $\\Delta_{\\kappa}(n) \\equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine\n$$\\tau(n) = \\tau(1) +\\sum_{k=1}^{n-1} \\Delta_{\\tau}(k)= 1 + \\sum_{k=1}^{n-1} \\left(k+1\\right) = 1+\\left( -1 + \\frac{n(n+1)}{2} \\right)= \\frac{n(n+1)}{2}$$ \nand $$\\kappa(n) = \\kappa(1) + \\sum_{k=1}^{n-1} \\Delta_{\\kappa}(n) = 4 + \\sum_{k=1}^{n-1}\\left( 2k+4\\right) = 4 + 2 \\cdot \\frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$\nand evaluating at $n=15$ gives a final answer of $\\tau(15)=120$ area and $\\kappa(15)=270$ toothpicks, so we are done. $\\blacksquare$[/quote]\n\nmira74, did you really work for 18 years just to find the answer?", "Solution_25": "@above\n\nNo, I only found part of the solution. The rest was done by [b] Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee, Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b], [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].\n", "Solution_26": "Was this problem seriously this complicated? wow", "Solution_27": "[quote=mira74]After 18 years of nonstop work...\n\nSolution with [b] Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee, Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b], [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].\n\n[size=150][b]Problem from post 1[/b][/size]:\n\nWe proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \\in \\mathbb{Z}$. Aqu\u00ed es el idea clave:\n-----\n[b][color=#f00]Lemma:[/color][/b] $0 \\leq \\text{any age} \\leq 130$\n[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love: \n-----\n\nDefine $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:\n\n$a \\leq 42$: His age in $1864$ is $$f(a) \\geq f(42) = 142 > 130,$$ which is absurd.\n$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\\boxed{1806}.$$\n$a \\geq 44$: His age in $1864$ is $$f(a) \\leq f(44) = -28 < 0,$$ which is absurd.\n\nThus, we have exhausted all cases, and are done. $\\blacksquare$\n\n[size=150][b]Problem from post 3[/b][/size]:\n\nWe proceed with engineering. See the attached diagram.\n\nDenote by $\\kappa(n)$ the number of toothpicks in the $n^{\\text{th}}$ term, and denote by $\\tau(n)$ the area of the $n^{\\text{th}}$ term. Note that upon taking the discrete derivative $\\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:\n\n\\begin{tabular}{|c|c|c|c|c|}\\hline\n$n$ & $1$ & $2$ & $3$ \\\\\\hline\n$\\Delta_\\tau(n)$ & $2$ & $3$ & $4$\\\\\\hline\n$\\Delta_\\kappa(n)$ & $6$ & $8$ & $10$\\\\\\hline\n\\end{tabular}\n\u0627\u0644\u0622\u0646 \u0647\u0646\u0627 \u0647\u0648 \u0627\u0644\u0645\u0637\u0627\u0644\u0628\u0629 \u0627\u0644\u0631\u0626\u064a\u0633\u064a\u0629:\n-----\n[color=#f00][b]Lemma[/b][/color]: $\\Delta_\\tau(n)$ and $\\Delta_\\kappa(n)$ are linear functions.\n[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\\boxed{}$\n-----\nChanneling SPIRIT of Ramanujan implies $\\Delta_{\\tau}(n) \\equiv n+1$ and $\\Delta_{\\kappa}(n) \\equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine\n$$\\tau(n) = \\tau(1) +\\sum_{k=1}^{n-1} \\Delta_{\\tau}(k)= 1 + \\sum_{k=1}^{n-1} \\left(k+1\\right) = 1+\\left( -1 + \\frac{n(n+1)}{2} \\right)= \\frac{n(n+1)}{2}$$ \nand $$\\kappa(n) = \\kappa(1) + \\sum_{k=1}^{n-1} \\Delta_{\\kappa}(n) = 4 + \\sum_{k=1}^{n-1}\\left( 2k+4\\right) = 4 + 2 \\cdot \\frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$\nand evaluating at $n=15$ gives a final answer of $\\tau(15)=120$ area and $\\kappa(15)=270$ toothpicks, so we are done. $\\blacksquare$[/quote]\n\n18 years. :sleeping: ", "Solution_28": "When was AoPS made?..", "Solution_29": "1974@above", "Solution_30": "[quote=BrightMonkey]1974@above[/quote]\n\nreally? How did the computers even function?", "Solution_31": "yeah I searched it up on google.The actual AOPS was actyllu the books", "Solution_32": "[quote=BrightMonkey]the real question is : why did we bump a 16 year old post??[/quote]\n\nSo that (s)he could get 44 upvotes! :P", "Solution_33": "[quote=mira74]After 18 years of nonstop work...\n\nSolution with [b] Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee, Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b], [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].\n\n[size=150][b]Problem from post 1[/b][/size]:\n\nWe proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \\in \\mathbb{Z}$. Aqu\u00ed es el idea clave:\n-----\n[b][color=#f00]Lemma:[/color][/b] $0 \\leq \\text{any age} \\leq 130$\n[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love: \n-----\n\nDefine $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:\n\n$a \\leq 42$: His age in $1864$ is $$f(a) \\geq f(42) = 142 > 130,$$ which is absurd.\n$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\\boxed{1806}.$$\n$a \\geq 44$: His age in $1864$ is $$f(a) \\leq f(44) = -28 < 0,$$ which is absurd.\n\nThus, we have exhausted all cases, and are done. $\\blacksquare$\n\n[size=150][b]Problem from post 3[/b][/size]:\n\nWe proceed with engineering. See the attached diagram.\n\nDenote by $\\kappa(n)$ the number of toothpicks in the $n^{\\text{th}}$ term, and denote by $\\tau(n)$ the area of the $n^{\\text{th}}$ term. Note that upon taking the discrete derivative $\\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:\n\n\\begin{tabular}{|c|c|c|c|c|}\\hline\n$n$ & $1$ & $2$ & $3$ \\\\\\hline\n$\\Delta_\\tau(n)$ & $2$ & $3$ & $4$\\\\\\hline\n$\\Delta_\\kappa(n)$ & $6$ & $8$ & $10$\\\\\\hline\n\\end{tabular}\n\u0627\u0644\u0622\u0646 \u0647\u0646\u0627 \u0647\u0648 \u0627\u0644\u0645\u0637\u0627\u0644\u0628\u0629 \u0627\u0644\u0631\u0626\u064a\u0633\u064a\u0629:\n-----\n[color=#f00][b]Lemma[/b][/color]: $\\Delta_\\tau(n)$ and $\\Delta_\\kappa(n)$ are linear functions.\n[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\\boxed{}$\n-----\nChanneling SPIRIT of Ramanujan implies $\\Delta_{\\tau}(n) \\equiv n+1$ and $\\Delta_{\\kappa}(n) \\equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine\n$$\\tau(n) = \\tau(1) +\\sum_{k=1}^{n-1} \\Delta_{\\tau}(k)= 1 + \\sum_{k=1}^{n-1} \\left(k+1\\right) = 1+\\left( -1 + \\frac{n(n+1)}{2} \\right)= \\frac{n(n+1)}{2}$$ \nand $$\\kappa(n) = \\kappa(1) + \\sum_{k=1}^{n-1} \\Delta_{\\kappa}(n) = 4 + \\sum_{k=1}^{n-1}\\left( 2k+4\\right) = 4 + 2 \\cdot \\frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$\nand evaluating at $n=15$ gives a final answer of $\\tau(15)=120$ area and $\\kappa(15)=270$ toothpicks, so we are done. $\\blacksquare$[/quote]\n\nWouldn't it have been 16 years...", "Solution_34": "https://artofproblemsolving.com/community/q1h43826p277736 How AoPS looked in the early years... o.O", "Solution_35": "[quote=franzliszt][quote=BrightMonkey]the real question is : why did we bump a 16 year old post??[/quote]\n\nSo that (s)he could get 44 upvotes! :P[/quote]\n\n???????", "Solution_36": "franzliszt, congratulations on 2020 posts :).", "Solution_37": "[quote=GoogleNebula]franzliszt, congratulations on 2020 posts :).[/quote]\n\nuh. Congratulations on 58 posts?", "Solution_38": "[quote=m1377]Lmao the girl who originally posted this hasn't visited AOPS in 11 years. Shes probably in her 30s right now[/quote]\n\nlmao I am not even 11...", "Solution_39": "[quote=mira74]After 18 years of nonstop work...\n\nSolution with [b] Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee, Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b], [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].\n\n[size=150][b]Problem from post 1[/b][/size]:\n\nWe proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \\in \\mathbb{Z}$. Aqu\u00ed es el idea clave:\n-----\n[b][color=#f00]Lemma:[/color][/b] $0 \\leq \\text{any age} \\leq 130$\n[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love: \n-----\n\nDefine $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:\n\n$a \\leq 42$: His age in $1864$ is $$f(a) \\geq f(42) = 142 > 130,$$ which is absurd.\n$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\\boxed{1806}.$$\n$a \\geq 44$: His age in $1864$ is $$f(a) \\leq f(44) = -28 < 0,$$ which is absurd.\n\nThus, we have exhausted all cases, and are done. $\\blacksquare$\n\n[size=150][b]Problem from post 3[/b][/size]:\n\nWe proceed with engineering. See the attached diagram.\n\nDenote by $\\kappa(n)$ the number of toothpicks in the $n^{\\text{th}}$ term, and denote by $\\tau(n)$ the area of the $n^{\\text{th}}$ term. Note that upon taking the discrete derivative $\\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:\n\n\\begin{tabular}{|c|c|c|c|c|}\\hline\n$n$ & $1$ & $2$ & $3$ \\\\\\hline\n$\\Delta_\\tau(n)$ & $2$ & $3$ & $4$\\\\\\hline\n$\\Delta_\\kappa(n)$ & $6$ & $8$ & $10$\\\\\\hline\n\\end{tabular}\n\u0627\u0644\u0622\u0646 \u0647\u0646\u0627 \u0647\u0648 \u0627\u0644\u0645\u0637\u0627\u0644\u0628\u0629 \u0627\u0644\u0631\u0626\u064a\u0633\u064a\u0629:\n-----\n[color=#f00][b]Lemma[/b][/color]: $\\Delta_\\tau(n)$ and $\\Delta_\\kappa(n)$ are linear functions.\n[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\\boxed{}$\n-----\nChanneling SPIRIT of Ramanujan implies $\\Delta_{\\tau}(n) \\equiv n+1$ and $\\Delta_{\\kappa}(n) \\equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine\n$$\\tau(n) = \\tau(1) +\\sum_{k=1}^{n-1} \\Delta_{\\tau}(k)= 1 + \\sum_{k=1}^{n-1} \\left(k+1\\right) = 1+\\left( -1 + \\frac{n(n+1)}{2} \\right)= \\frac{n(n+1)}{2}$$ \nand $$\\kappa(n) = \\kappa(1) + \\sum_{k=1}^{n-1} \\Delta_{\\kappa}(n) = 4 + \\sum_{k=1}^{n-1}\\left( 2k+4\\right) = 4 + 2 \\cdot \\frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$\nand evaluating at $n=15$ gives a final answer of $\\tau(15)=120$ area and $\\kappa(15)=270$ toothpicks, so we are done. $\\blacksquare$[/quote]\n\n18 years indeed...", "Solution_40": "[quote=huayd695]Why is her age relevant?[/quote]\n\nidk lol", "Solution_41": "[quote=BrightMonkey]1974@above[/quote]\n\n :o ", "Solution_42": "I wish these people were still on AoPS :/" } { "Tag": [], "Problem": "Find all solutions to the system:\r\n\r\n\\begin{eqnarray*}\r\nx^2+y^2+x+y &=& 8\\\\\r\nxy + x + y &=& 5\r\n\\end{eqnarray*}\r\n\r\nIf you find Intermediate problems easy, please leave this challenge problem for less experienced students to solve.\r\n\r\nSource: Alabama State Contest", "Solution_1": "We have:\r\n$\\left\\{ \\begin{array}{l}\r\n x^2 + y^2 + x + y = 8 \\\\ \r\n xy + x + y = 5 \\\\ \r\n \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l}\r\n \\left( {x + y} \\right)^2 + \\left( {x + y} \\right) - 2xy = 8 \\\\ \r\n xy + \\left( {x + y} \\right) = 5 \\\\ \r\n \\end{array} \\right.$\r\n\r\nWe denote: $s = x + y,p = xy$ and obtain:\r\n$\\left\\{ \\begin{array}{l}\r\n s^2 + s - 2p = 8 \\\\ \r\n s + p = 5 \\\\ \r\n \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l}\r\n p = 5 - s \\\\ \r\n s^2 + 3s - 18 = 0 \\\\ \r\n \\end{array} \\right. \\Leftrightarrow $\r\n\r\n$ \\Leftrightarrow \\left\\{ \\begin{array}{l}\r\n p = 5 - s \\\\ \r\n s = 3 \\\\ \r\n \\end{array} \\right. \\vee \\left\\{ \\begin{array}{l}\r\n p = 5 - s \\\\ \r\n s = - 6 \\\\ \r\n \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l}\r\n s = 3 \\\\ \r\n p = 2 \\\\ \r\n \\end{array} \\right. \\vee \\left\\{ \\begin{array}{l}\r\n s = - 6 \\\\ \r\n p = 11 \\\\ \r\n \\end{array} \\right.$\r\n\r\nFor this reason\r\n$\\left\\{ \\begin{array}{l}\r\n x^2 + y^2 + x + y = 8 \\\\ \r\n xy + x + y = 5 \\\\ \r\n \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l}\r\n x + y = 3 \\\\ \r\n xy = 2 \\\\ \r\n \\end{array} \\right. \\vee \\left\\{ \\begin{array}{l}\r\n x + y = - 6 \\\\ \r\n xy = 11 \\\\ \r\n \\end{array} \\right. \\Leftrightarrow $\r\n\r\n$ \\Leftrightarrow \\left\\{ \\begin{array}{l}\r\n x = 1 \\\\ \r\n y = 2 \\\\ \r\n \\end{array} \\right. \\vee \\left\\{ \\begin{array}{l}\r\n x = 2 \\\\ \r\n y = 1 \\\\ \r\n \\end{array} \\right. \\vee \\left\\{ \\begin{array}{l}\r\n x = - 3 - i\\sqrt 2 \\\\ \r\n y = - 3 + i\\sqrt 2 \\\\ \r\n \\end{array} \\right. \\vee \\left\\{ \\begin{array}{l}\r\n x = - 3 + i\\sqrt 2 \\\\ \r\n y = - 3 - i\\sqrt 2 \\\\ \r\n \\end{array} \\right.$" } { "Tag": [ "trigonometry", "inequalities" ], "Problem": "i saw the following lemma in a soln:\r\n\r\nLet $A,B,C,x,y,z$ be six positive reals s.t. $A+B+C=\\pi$. Prove that \\[ \\sum xy \\sin^2 C \\leq \\dfrac{(x+y+z)^2}{4} .\\]\r\n\r\nin that soln there was no proof and i can't do it.\r\n\r\nps: sorry if a proof was posted before, but i can't find it.", "Solution_1": "Hello Perfect radio!\r\nThe inequality in your post is essentially this inequality:\r\n\r\nX^2 +Y^2 +Z^2 >= 2YZCosP +2ZXCosQ +2XYCosR (* ) where P, Q, R are the angles of a triangle \r\nAdding 2YZ +2ZX +2XY to both sides gives:\r\n\r\n(X+Y+Z)^2 >= 2YZ(1+CosP) +2ZX(1+CosQ) +2XY(1+cosR) =4YZ(CosP/2)^2 +4ZX(CosQ/2)^2 +4XY(CosR/2)^2 (**)\r\n(Using the identity CosP =2(CosP/2)^2 -1 etc)\r\n\r\nNow note that if P, Q and R are the angles of a triangle then A=pi/2 -P/2, B=pi/2-Q/2 and C=pi/2-R/2 are the angles of another triangle and sinA=cosP/2, sinB=CosQ/2 and sinC =CosR/2\r\nSubstituting for CosP/2... in the expression on the right of (**) leads to the inequality we want to prove.\r\n\r\nTo prove (*),This reduces to (X -ZCosQ -YcosR)^2 +(YSinR-ZSinQ)^2 which is positive." } { "Tag": [ "percent" ], "Problem": "I took the January SAT IIs.. and got 800 on both Math IIC and Physics.\r\nWhich is great:), BUT my percentiles aren't fabulous ... 90 and 92.\r\nShould I retake?", "Solution_1": "No, you have received the best possible score. The 90th percentile just means that 9 or 10 percent of the takers received an 800 too. Taking it again won't help.\r\n\r\nDon't worry about percentiles.", "Solution_2": "The only way to raise your percentile when you have already reached the top standard score (800) is to convince all of the tens of thousands of other people who want to take the next test to do badly. In other words, don't worry about your percentile score. \r\n\r\nBut now you see why 800 scores on SAT II tests alone are not impressive enough to ensure getting into a top college. Lots of other people have those scores too. \r\n\r\nGood luck in your application process.", "Solution_3": "Thanks to both of you for the advice. I'm not going to retake. :)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "a man has infinite amount of money. he distribute his money in n people in a multiple of 4,such that each person does not get more than 4n-1 rupees..in how many ways he can do that.", "Solution_1": "[quote=\"ayush_2008\"]a man has infinite amount of money. he distribute his money in n people in a multiple of 4,such that each person does not get more than 4n-1 rupees..in how many ways he can do that.[/quote]\r\n[color=blue]\nLet us consider $ \\ 4$ as an unit or a share $ \\ \\implies$ each person can get at the most $ \\ n \\minus{} 1$ units.\nEvery person can be allotted $ \\ 1$ or $ \\ 2$ or $ \\ 3\\ \\cdots$ or $ \\ n \\minus{} 1$ units $ \\ \\implies$ total ways for a person is $ \\ n \\minus{} 1\\ .$\n$ \\implies$ Total number of ways is $ \\ (n \\minus{} 1)^n$ \n\nAlternatively: \n \nNumber of ways = coefficient of $ \\ x^{n(n \\minus{} 1)}$ in $ \\ \\left(x \\plus{} x^2 \\plus{} x^3 \\plus{} \\cdots \\plus{} x^{n \\minus{} 1}\\right)^n\\left(1 \\plus{} x \\plus{} x^2 \\plus{} x^3 \\plus{} \\cdots\\right)$\n[/color]", "Solution_2": "sir you said that Let us consider 4 as an unit or a share implies each person can get at the most n - 1 units. n-1 units implies that a person will get 4(n-1)>4n-1 Rs.also how can we handle that case in which one person can get 1,2 ,3 4, or any Rs .", "Solution_3": "[quote=\"ayush_2008\"]sir you said that Let us consider 4 as an unit or a share implies each person can get at the most n - 1 units. n-1 units implies that a person will get 4(n-1)>4n-1 Rs.also how can we handle that case in which one person can get 1,2 ,3 4, or any Rs .[/quote]\r\n\r\nSee\r\n$ 4(n\\minus{}1)\\equal{}4n\\minus{}4<4n\\minus{}1$\r\n\r\nFor your second query, I understood that each person is receiving money in a multiple of $ \\ 4$\r\n\r\nIs it your problem or not?", "Solution_4": "[quote=\"makar\"][quote=\"ayush_2008\"]sir you said that Let us consider 4 as an unit or a share implies each person can get at the most n - 1 units. n-1 units implies that a person will get 4(n-1)>4n-1 Rs.also how can we handle that case in which one person can get 1,2 ,3 4, or any Rs .[/quote][/quote]\r\n\r\nSee\r\n$ 4(n \\minus{} 1) \\equal{} 4n \\minus{} 4 < 4n \\minus{} 1$\r\n\r\nFor your second query, I understood that each person is receiving money in a multiple of $ \\ 4$\r\n\r\nIs it your problem or not?[/quote\r\nyes sir it is my problem", "Solution_5": "[quote=\"ayush_2008\"][quote=\"makar\"][quote=\"ayush_2008\"]sir you said that Let us consider 4 as an unit or a share implies each person can get at the most n - 1 units. n-1 units implies that a person will get 4(n-1)>4n-1 Rs.also how can we handle that case in which one person can get 1,2 ,3 4, or any Rs .[/quote]\n\nSee\n$ 4(n \\minus{} 1) \\equal{} 4n \\minus{} 4 < 4n \\minus{} 1$\n\nFor your second query, I understood that each person is receiving money in a multiple of $ \\ 4$\n\nIs it your problem or not?[/quote\nyes sir it is my problem[/quote][/quote]\r\n\r\nThen how can a person will get $ \\ 1,\\ 2,\\ 3,$ or any amount Rs . as every time he is getting in a multiple of $ \\ 4$ . \r\n\r\nIn my calculation suppose person (1) got $ \\ 3$ units or share, then he got Rs. $ \\ 12$.", "Solution_6": "Hello sir \\\r\n sir in this question the total amount of money to be distributed is multiple of 4.not necessary the money given to each person is multiple of 4. for ex if there are three relatives then one can get any amount of money(<4n-1)but total sum should be multiple of four." } { "Tag": [ "combinatorics", "IMO Shortlist", "double counting" ], "Problem": "There are $10001$ students at an university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold:\n\n[i]i.)[/i] Each pair of students are in exactly one club.\n\n[i]ii.)[/i] For each student and each society, the student is in exactly one club of the society.\n\n[i]iii.)[/i] Each club has an odd number of students. In addition, a club with ${2m+1}$ students ($m$ is a positive integer) is \nin exactly $m$ societies.\n\nFind all possible values of $k$.\n\n[i]Proposed by Guihua Gong, Puerto Rico[/i]", "Solution_1": "Easy:\r\n\r\nSuppose $A$ belongs to clubs $C_1,C_2...C_a$, then in these clubs A apears in $\\sum|C_i|-1$ pairs, so we get this sum is equal to $10000$. and since every of these clubs is in $\\frac{|C_i|-1}{2}$ societies, $A$ appears in $5000$ societies.\r\n\r\nNow every Society contains $10001$ students, so we easily get, counting edges in the bypartite graph formed by students and societies, and an edge if a student goes in that society that $k=5000$", "Solution_2": "I think that you could probably eliminate one of conditions (1) and (2) and it would probably still be sufficient to solve the problem (Pascual's solution doesn't look like it really uses condition (2), for example).", "Solution_3": "Is there a proof that those conditions are met :?:", "Solution_4": "[hide=\"Bound\"]Suppose there are $n$ clubs, and let club $C_i$ contain $a_i$ students and be contained in $b_i$ societies. Condition (i) gives \n\\[\n\\sum_{i=1}^n \\binom{a_i}{2} = \\binom{10001}{2}.\n\\]\nCondition (ii) gives \n\\[\n10001 \\cdot k = \\sum_{i=1}^n a_ib_i.\n\\]\nCondition (iii) gives \n\\[\na_ib_i = a\\left( \\frac{a_i-1}{2} \\right) = \\binom{a_i}{2}.\n\\]\nThus the conditions together imply $1001\\cdot k = \\binom{10001}{2}$, or $k=5000$. [/hide]\n\n[hide=\"Construction\"]There is only one club containing all $10001$ students. There are $5000$ societies each containing this one club. All conditions are met. [/hide]\n\n[hide=\"Comments\"]In searching for a construction I tried to use condition (i) as follows: Consider $10001$ lines in a plane in general position; two lines intersect in one point. This suggests letting the intersection points be clubs and lines be students, where a club contains a student exactly when the point lies on the line. Condition (iii) says every intersection of two lines is in fact an intersection of $\\ge 3$ lines (specifically, it is the intersection of an odd number of lines). I remembered a problem which stated that this implies all lines intersect in exactly one point; this is exactly the construction given above. \n\nMy question is: is there a more interesting construction? The wording of the problem suggests there should be \"several\" clubs, not just one. My interpretation suggests the above is the only construction, but it is by no means the only way of finding constructions... [/hide]\n\n[hide=\"@Monkey\"]Pascual's solution uses condition (ii) to conclude each society includes $10001$ students. If no club in some society included a particular student, this student and society would contradict condition (ii). [/hide]", "Solution_5": "Pretty straightforward problem. \n\nLabel the students $s_1, s_2, \\cdots , s_{10001},$ the clubs $C_1, C_2, \\cdots , C_n$ and the societies $S_1, S_2, \\cdots , S_k.$ Let the number of students in club $C_i$ be $2f(C_i)+1,$ and let $g(S_i)$ denote the number of clubs in society $S_i.$\n\nThe last condition implies that club $C_i$ is in exactly $f(C_i)$ societies. Now, for any club $C_i,$ there are $\\dbinom{2f(C_i)+1}{2} = f(C_i) (2f(C_i)+1)$ pairs of students. Since each pair of students is in a unique club, $\\displaystyle \\sum_{i=1}^{n} f(C_i) (2f(C_i)+1)$ must denote the total number of student pairs, so $\\displaystyle \\sum_{i=1}^{n} f(C_i) (2f(C_i)+1) = \\dbinom{10001}{2}.$ Now, each student is contained in exactly one club of a society, so a society contains exactly 10001 students. The total number of students (counted with multiplicity) in all societies is hence $10001 \\times k.$ However, the sum of number of students of all societies is also equal to the sum of the number of students in each club times the number of times that club appears in all the societies, so $10001 \\times k = \\displaystyle \\sum_{i=1}^{n} f(C_i) (2f(C_i)+1).$ Combining them, we easily get that $k=5000.$\n\nFor equality, set $f(C_1)=10001, f(C_2) = f(C_3) = \\cdots = 0$ and $g(S_1) = g(S_2) = \\cdots = g(S_{5000}) = 1$ where $C_1 \\in S_i \\quad \\forall \\ i.$", "Solution_6": "Standard problem. Simply double count the number of ordered triples (P,C,S) where student P is in a club C of society S. The second condition gives $T=10001k$ triples. The third condition gives $T$ is the sum over all clubs of $m(2m+1)$. To make use of the first condition, simply double count the number of triples (P1,P2,C) where students P1, P2 are in the same club C. You get: the sum over all clubs of (2m+1) choose 2 is equal to 10001 choose 2. Multiply both sides by 2 and equate with T to solve for", "Solution_7": "Easy:Count (student,club,society) such that student is in club and club is in society.", "Solution_8": "Nothing different, but I was solving this recently, so here goes :)\n\nLet there be $t$ clubs $C_1 , C_2, \\cdots C_t$.\nConsider the set\n$$X = \\{ (s , C , S) | s \\in C , C \\in S, s \\text{ is a student } , C \\text{ is a club, and } S \\text{ is a society.} \\} $$\n\nWe calculate $|X|$ in two ways. First fix $s$ and $S$. From (ii), there is exactly one club $C_i$ for which $(s,C_i,S) \\in X$. Thus \n$$|X| = 10001k.$$\n\nNow fix the club, say $C_i$. It has $|C_i|$ students, and it belongs to $\\frac{|C_i| - 1}{2} $ societies from (iii). Thus\n$$|X| = \\sum_{i=1}^t \\frac{|C_i|(|C_i|-1)}{2}.$$\n\nNow consider the set\n$$Y = \\{( \\{s_1 , s_2\\} , C) | s_1 , s_2 \\text{ are different students belonging to the club } C.\\}$$\n\nAgain, we calculate $|Y|$ in two ways. First fix $C = C_i$. There are ${|C_i| \\choose 2}$ such pairs of students in the club. Thus\n$$|Y| = \\sum_{i=1}^t {|C_i| \\choose 2} $$\n\nNow fix the pair $\\{s_1 , s_2\\}$. From (i), there is exactly one club $C$ for which $(\\{s_1,s_2\\} , C) \\in Y$. Therefore\n$$|Y| = {10001 \\choose 2}$$\n\nFrom all these relations, we have\n\n$$10001k = |X| = \\sum_{i=1}^t \\frac{|C_i|(|C_i|-1)}{2} = \\sum_{i=1}^t {|C_i| \\choose 2} = |Y| = {10001 \\choose 2}$$\nso that\n$$k = 5000.$$\n\nTo show that $k=5000$ indeed works, consider just $1$ club $C$ (that is, $t=1$) which belongs to $5000$ societies and such that all students are a member of this club. Clearly (i),(ii) and (iii) are satisfied.\n", "Solution_9": "Fix a student $T$ and let $C_1, C_2, \\dots, C_r$ be all the clubs he is in. By the first condition, $$\\sum _{i=1}^r |C_i| = 10000+r$$\nFrom the second condition, each society has exactly all the students in the university. So $T$ appears $k$ times in all $k$ societies exactly once. Moreover, the clubs within each society is disjoint. So $C_1, C_2, \\dots , C_r$ are in different societies.\nFrom the third condition, observe that each $C_i$ appears in $\\frac{|C_i|-1}{2}$ societies. So combining this with the previous equation,$$k = \\sum _{i=1} ^r \\frac{|C_i|-1}{2} = \\frac{10000+r-r}{2} =5000$$\n\nNow a construction for $k=5000$. Put all students into just one club and put this club in each of $5000$ societies. We are done.", "Solution_10": " Let us write n= 10001.Denote by T the set of ordered triples (a,C,S ), where a is a student, C a club, and S a society such that a\u2208C and C\u2208S . We shall count|T |in two different ways. Fix a student a and a society S. By (ii), there is a unique club C such that (a,C,S )\u2208T .Since the ordered pair(a,S ) can be chosen in nk ways,we have that|T |= nk. Now \ufb01x a club C. By (iii), C is in exactly (|C|\u22121)/2 societies, so there are |C|(|C|\u22121)/2 triples from T with second coordinate C. If C is the set of all clubs, we obtain|T |= \u2211C\u2208C |C|(|C|\u22121) 2 . But we also conclude from (i) that \u2211 C\u2208C |C|(|C|\u22121) 2 = n(n\u22121) 2 . Therefore n(n\u22121)/2= nk, i.e., k = (n\u22121)/2= 5000. On the other hand, for k = (n\u22121)/2 there is a desired con\ufb01guration with only one club C that contains all students and k identical societies with only one element (the club C). It is easy to verify that (i)\u2013(iii) hold.", "Solution_11": "[hide=Solution][quote=ISL 2004 C1]There are $10001$ students at an university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold:\n\n[i]i.)[/i] Each pair of students are in exactly one club.\n\n[i]ii.)[/i] For each student and each society, the student is in exactly one club of the society.\n\n[i]iii.)[/i] Each club has an odd number of students. In addition, a club with ${2m+1}$ students ($m$ is a positive integer) is \nin exactly $m$ societies.\n\nFind all possible values of $k$.\n[/quote]\n[b][color=#000]Solution:[/color][/b] Let $x_i$ denote the no. of students in Club $C_i$ and $y_i$ denote the no. of societies in which $C_i$ is contained. Thus Condition (iii) $\\implies$ $x_i=2y_i +1$. Let no. of clubs be $n$. Double Counting the sets\n\\begin{align*} \\mathcal{S} = \\{ \\left( C_i , \\{ S_j, S_k \\} \\right) \\} \\implies \\sum_{i=1}^n \\binom{x_i}{2} = \\binom{10001}{2} \\end{align*}\n\\begin{align*} \\mathcal{T} = \\{ \\left( C_i, S_j, s_k \\right) \\} \\implies 10001 k = \\sum_{i=1}^n x_i y_i = \\sum_{i=1}^n \\frac{x_i (x_i -1)}{2} = \\sum _{i=1}^n \\binom{x_i}{2} \\end{align*}\n\\begin{align*} \\implies \\binom{10001}{2} = 10001 k \\implies \\boxed{k= 5000} \\qquad \\blacksquare \\end{align*}[/hide]", "Solution_12": "The value $k=5000$ is achieved if there is exactly one club, and $5000$ societies consisting of only that club. We show it's the only value.\n\nCondition (i) implies \\[ \\binom{10001}{2} = \\sum_{C \\text{ club}} \\binom{|C|}{2} \\]\nOn the other hand, condition (ii) means that each individual society is a set of clubs whose disjoint union is all $10001$ students, so we have \\[ k \\cdot 10001 = \\sum_{C \\text{ club}} |C| \\cdot \\frac{|C|-1}{2} \\] where $\\frac{|C|-1}{2}$ comes from (iii).\n\nFrom these two equations we see $k = 5000$ is the only possible value.", "Solution_13": "Let $C$ be the number of clubs and let $k_1 , k_2 , \\dots , k_{C}$ be the number of students that each club has.\nFirst, we count the number of triplets $(s_1 , s_2 , c)$ where $s_1 , s_2$ are students whose common club is $c$.\nIf we fix the students, there is only one way to choose $c$ so there are $\\binom{10001}{2}$ triplets.\nIf we fix the $c$ first, then there are exactly $\\binom{k_1}{2} + \\binom{k_2}{2} + \\dots + \\binom{k_C}{2}$ triplets.\nThus, we have that $\\sum_{i=1}^{C} \\binom{k_i}{2} = 10001\\cdot 5000$.\nNow, we count the triplets $(s,c,t)$ where $s$ is a student, $c$ is student's one of the clubs, $t$ is this club's one of the societies.\nIf we fix the student and society, there is only one way to choose $c$ so there are $10001k$ such triplets.\nIf we fix the club, then there are exactly $\\frac{k_1(k_1 - 1)}{2} + \\dots + \\frac{k_C(k_C - 1)}{2} = \\sum_{i=1}^{C} \\binom{k_i}{2} = 10001\\cdot 5000$ such triplets.\nThus, we have $10001k = 10001\\cdot 5000 \\implies k=5000$.\nIt is easy to construct an example for $k=5000$. Let there be only one club and all the students are joined to this club, and this club is in every society. It is clear that this works and we are done. $\\square$", "Solution_14": "We claim that the only possible value of $k$ is $\\boxed{5000}$. This can be achieved if there is one club consisting of everybody, and that club is in $5000$ societies.\n\nConsider a person $A$. Condition (i) implies that the set of clubs containing $A$ contain every other person exactly once. Since every club has an odd number of people, for every club containing $A$ the other students must come in pairs. However, notice that if a club containing $A$ contains $k$ other pairs of students, it will have $2k+1$ people overall and thus must be in $k$ societies. \n\nThere are $10000$ other people, and therefore $5000$ pairs of other people. Thus there are a total of $5000$ societies that the clubs containing $A$ are in. Since no society can contain two clubs with $A$ in them, all $5000$ of these societies are distinct. Furthermore, the presence of another society would imply that there is a society with no club containing $A$, a clear contradiction. The proof is complete.", "Solution_15": "ISL marabot solve\n\nConsider a random student $P$ and let $S$ be the set of clubs $P$ is in. By the first condition, every other student is in exactly one of the clubs in $S$. Now consider some club in $S$. Suppose it has $p$ disjoint pairs of students excluding $P$. Then it must be in $p$ societies. Since there are totally $5000$ pairs, $P$ must be in $5000$ societies. The existence of another society implies that $P$ is not in it, a contradiction. So the only possible value of $k$ is $\\boxed{5000}$. This is achievable if there is exactly one club that consists of everyone and it is in $5000$ societies.", "Solution_16": "success is 99% understanding the problem, 0.99% perspiration and 0.01% inspiration.\n\nLet $S$ be the number of ways to pick a society, a club in that society, and a student in that club. Picking the society and the student, we know that there exists exactly one club that satisfies the conditions. Thus, $S=10001k.$\n\nNow let the clubs have $c_1,c_2,\\dots,c_n$ people, respectively. These clubs belong to $\\frac{c_1-1}{2},\\frac{c_2-1}{2},\\dots,\\frac{c_n-1}{2}$ societies, respectively. Thus, $S=\\sum_{n}^{i=1}{\\binom{c_i}{2}}.$ This is summing the number of pairs of people in any given club over all clubs. However, each pair of people is in exactly one club so this is just $\\binom{10001}{2}.$ Thus, $k=5000.$ This is reachable with just one club with all the people, and this club is in all the societies.", "Solution_17": "The answer is $k=5000$. This is achievable with one club of all $10001$ people in $5000$ societies. This can easily be seen to work.\nLet the clubs be $C_1,C_2, \\cdots, C_k$ and the people be $P_1,P_2, \\cdots , P_{10001}$. We count the number of triples $(P,C,S)$ where $S$ is a society, $C$ is a club in $S$, and $P$ is a person in $C$. Note that if we pick the club first we have that this sum is $$\\sum_{C \\text{ club}} |C_i| \\cdot \\frac{|C_i|-1}{2} = \\sum_{C \\text{ club}} \\binom{|C_i|}{2}$$ using condition (iii). Next note that $$\\sum_{C \\text{ club}} \\binom{|C_i|}{2} = \\sum_{\\text{unordered pair } (P_i, P_j) \\text{ in club } C} 1 = \\binom{10001}{2},$$ where the last equality follows from condition (ii). Next, if we pick the person and society first, the club is automatically chosen so the sum is $10001k$. This means $10001k=\\binom{10001}{2}$, so $k=5000$.", "Solution_18": "Nice problem", "Solution_19": "Let $C_i,$ where $1\\le i\\le \\ell,$ be the size of the clubs. By (i), we see $$\\binom{10001}{2}=\\sum_{i=1}^{\\ell}\\binom{C_i}{2}.$$ Notice by (ii), all students must be part of any society. Hence, if we want to choose a student, a club, and a society (such that the student is in the club is in the society), we can just choose the society and then choose any student, which we can do in $10001k$ ways (this would also determine the club by (ii)). Alternatively, we can proceed by casework on which club the student is in. For each club $C_i,$ we can choose $C_i$ students and $\\frac{C_i-1}{2}$ societies by (iii). Thus, we see $$10001k=\\sum_{i=1}^{\\ell}C_i\\cdot\\frac{C_i-1}{2}.$$ This implies $k=5000.$ This works by letting all the students be in one club which is part of $5000$ societies. $\\square$", "Solution_20": "Let there be $c$ clubs. For each club, let the number of members in it be $a_1,a_2,\\dots,a_c$.\nFrom the first condition, clubs contribute distinct pairs to all student pairs: $$\\binom{10001}{2}=\\binom{a_1}{2}+\\binom{a_2}{2}+\\dots+\\binom{a_c}{2}.$$\nFrom the second and third conditions, each club contributes $(2m+1)m$ student-society pairs: $$10001k=\\binom{a_1}{2}+\\binom{a_2}{2}+\\dots+\\binom{a_c}{2}.$$\nWe see that the two equations have equivalent right hand sides, so $k=\\boxed{5000}$. This can be achieved by just having every student in $1$ club, which is in all $5000$ societies. QED.", "Solution_21": "We claim that the answer is $k=5000$. This can be achieved by having 1 club that contains all of the students and is in all 5000 societies.\\\\\n\nSuppose that there are $n$ clubs, and let club $i$ have $2m_i+1$ students.\\\\\n\nFrom condition (i), we have $${2m_1+1 \\choose 2}+{2m_2+1 \\choose 2}+\\cdots +{2m_n+1 \\choose 2}={10001 \\choose 2},$$ since both sides are equal to the number of pairs of students.\\\\\n\nConsider the number of triples $(a,b,c)$ such that club $b$ contains student $a$ and is in society $c$. By summing over all clubs, this is equal to $$m_1(2m_1+1)+m_2(2m_2+1)\\cdots +m_n(2m_n+1).$$ However, condition (iii) tells us that this is also equal to $10001k.$ Therefore, we have $$m_1(2m_1+1)+m_2(2m_2+1)\\cdots +m_n(2m_n+1)=10001k.$$ However, note that $m_i(2m_i+1)={2m_i+1\\choose 2},$ so combined with our equation from condition (i), we have $${10001\\choose 2}=10001k\\rightarrow k=5000,$$ so we are done.", "Solution_22": "Double count pairs $(s, C, S)$ where student $s$ belongs in club $C$, which belongs in society $S$. For every pair $(s, S)$, there exists a unique $C$, so there are $10001k$ such tripltets. On the other hand, for every $C$, there exsit $m(2m+1)$ pairs for $2m+1$ students in club $C$. But then $$\\sum_C {2m+1 \\choose 2} = \\sum_C m(2m+1) = 5000 \\cdot 10001,$$ so thus we must have $k=5000$. For a construction, put all students in one club, which is part of all $5000$ societies.", "Solution_23": "This solution uses the deep fact that $\\binom{2m+1}{2}=m(2m+1)$.\n\nFor a club $c$ let $s(c)$ denote the number of societies it belongs to. Then $\\sum s(c)(2s(c)+1)=\\sum \\binom{2s(c)+1}{2}$ over all clubs $c$ will count both the number of pairs of students and the number of student-society pairs. There are $10001\\cdot 5000$ of the former so we obtain $k=5000$ only. A construction is to create a single club which belongs for some reason to $5000$ societies and put every student into it.", "Solution_24": "combo is hard\n\nThe answer is $k = 5000$. Let the clubs be $C_1, C_2, C_3 ... C_i$. Then condition i) tells us that \\[ \\dbinom{10001}2 = \\sum \\dbinom{|C_j|}2. \\] We count the number of pairs $(p, q, r)$ such that student $i$ is in club $j$ with society $k$. Condition ii) and iii) together tell us that each club $C_m$ contributes $|C_m|(2|C_m|+1)$ to this number. By summing this across all clubs, we obtain \\[\\sum \\dbinom{|C_j|}2 = 10001k \\]. This implies that $k = 5000$, and a construction is to let all students be in $1$ club with $5000$ societies.", "Solution_25": "This problem made me so mad.\n\nNumber the clubs $1,2, \\ldots, \\ell$. Let $s_i$ and $t_i$ denote the number of students in the $i$'th club. Now, we will do the following genius calculations \\[ 10001k = \\sum s_it_i = \\sum s_i \\left ( \\frac{s_i-1}{2} \\right ) = \\sum \\binom{s_i}{2} = \\binom{10001}{2} \\implies k = 5000\\] Construction for $k=5000$ is just one club with every student and part of $5000$ societies.", "Solution_26": "Let there be $n$ clubs such that the $i$th club has $2s_i+1$ members. We see that $\\sum \\dbinom{2s_i+1}2 = \\dbinom{10001}2$ by the first condition. Consider the sum of the number of people across all societies. By the second condition, this is $10001k$ but by the third condition, this is also $\\sum s_i(2s_i+1) = \\sum \\dbinom{2s_i+1}2 = \\dbinom{10001}2$. Then it follows that $k=5000$.", "Solution_27": "bruh this one is so weird to wrap your mind around even though its standard double counting:\n\nNotice that the number of pairs of students =10001 choose 2 are equal to choosing any club and taking two students, since those two students are only in this single club together, so we sum over all clubs=sum over 2m+1 choose 2=sum over m(2m+1). On the other hand, this is also equivalent to choosing a student and a society because any student-society makes up a distinct club in m(2m+1) ways. Hence 10001C2=2m+1C2=10001k, which readily implies k=5000! $\\blacksquare$", "Solution_28": "Consider the set $\\mathcal{A}=\\{ (s , C , S_{0}) | s \\in C , C \\in S_{0}, s \\text{ is a student } , C \\text{ is a club, and } S_{0} \\text{ is a society.} \\}$\\\\\n\nSo since for each student and each society, the student is exactly in one club of the society we have $|\\mathcal{A}|=10001k$. Also consider a particular club $C_{i}$ , denote $|C_{i}|$ as number of students in it.\\\\\n\nSo we have $|C_{i}|$ many students in a particular club $C_{i}$ , now there are $\\frac{|C_{i}|-1}{2}$ societies. consider total such clubs to be be $n$ we have $\\sum_{i=1}^{i=n} |C_{i}|\\cdot \\left(\\frac{|C_{i}|-1}{2}\\right)=|\\mathcal{A}|$\\\\\n\nAlso, Notice for all clubs we have total pair of distinct students to be $\\sum_{i=1}^{i=n} \\binom{|C_{i}|}{2}$\\\\\n\nand also clearly from $(i)$ there is exactly one club for which two distinct pair of students are in a club so we have $\\binom{10001}{2}=\\sum_{i=1}^{i=n} \\binom{|C_{i}|}{2}$\\\\\n\nso we have $10001k=\\binom{10001}{2} \\implies \\boxed{k=5000}$. $\\blacksquare$\n\nedit: 850th post to this cool problem! :coolspeak:", "Solution_29": "Let there be $n$ clubs, with the $i$th club having $a_i$ members. Then from the first condition, it must be that \n\\[ \\binom{a_1}{2} + \\binom{a_2}{2} + \\dots + \\binom{a_n}{2} = \\binom{10001}{2} \\] \nBut the second and third condition imply that\n\\[ \\binom{a_1}{2} + \\binom{a_2}{2} + \\dots + \\binom{a_n}{2} = 10001k \\] \nIt immediately follows that $k = \\boxed{5000}$. $\\blacksquare$", "Solution_30": "Label the clubs $C_i$ and the societies $S_i$, with corresponding member counts $c_i$ and $s_i$. The first condition tells us the total number of pairings is\n\\[\\binom{10001}{2} = \\sum \\binom{c_i}{2}.\\]\nThe second and third conditions give us the total sum of member counts across all societies as\n\\[10001k = \\sum s_i = \\sum \\frac{c_i-1}{2} \\cdot c_i = \\sum \\binom{c_i}{2}.\\]\n\nEquating these, we find $k=\\boxed{5000}$. $\\blacksquare$", "Solution_31": "Let the clubs be $C_1, C_2, \\ldots, C_j$, with club $C_i$ having $x_i$ members for some odd $x_i$. Then by the first condition, we have \n$$\\binom{10001}{2} = \\sum_{i = 1}^j \\binom{x_i}{2}.$$\nBy the second condition, we have \n$$10001k = \\sum_{i = 1}^j x_i \\cdot \\frac{x_i - 1}{2} = \\sum_{i = 1}^j \\binom{x_i}{2},$$\nby considering the sum of the number of members across all $k$ clubs. Therefore $k = \\tfrac{10000}{2} = 5000$. For a construction, take one and only one club which has all $10001$ students and is in $5000$ societies." } { "Tag": [ "calculus", "integration", "trigonometry", "function", "logarithms", "calculus computations" ], "Problem": "$ \\int \\frac {sin^{2}x}{1 \\minus{} e^{x}}dx$OR$ \\int \\frac {sin^{2}x}{1 \\plus{} e^{x}}dx$", "Solution_1": "I think expand the denominator by using maclaurin series and then by parts each term should work.", "Solution_2": "Mathematica says this involves hypergeometric functions.", "Solution_3": "[quote=\"Timestopper_STG\"]I think expand the denominator by using maclaurin series and then by parts each term should work.[/quote]\r\n\r\n\r\nI don't think it's a good way. If we sub $ sin^{2}x\\equal{}\\left ( \\frac{e^{ix}\\minus{}e^{\\minus{}ix}}{2i} \\right )^{2}$,then do the intergral ,is that right ?", "Solution_4": "centry57... looking at your previous posts, my guess is that your original problems are\r\n$ \\int_{ \\minus{} \\alpha }^{\\alpha }\\frac {\\sin ^{2}x}{1 \\plus{} e^{x}}dx$ and $ \\int_{ \\minus{} \\alpha }^{\\alpha }\\frac {\\sin ^{2}x}{1 \\minus{} e^{x}}dx$\r\nRight?", "Solution_5": "[quote=\"IndoChina\"]centry57... looking at your previous posts, my guess is that your original problems are\n$ \\int_{ \\minus{} \\alpha }^{\\alpha }\\frac {\\sin ^{2}x}{1 \\plus{} e^{x}}dx$ and $ \\int_{ \\minus{} \\alpha }^{\\alpha }\\frac {\\sin ^{2}x}{1 \\minus{} e^{x}}dx$\nRight?[/quote]\r\n\r\nI'm sorry ,I want to know $ \\int\\frac{sin^{2}x}{1\\plus{}e^{x}}dx$", "Solution_6": "Sigh. All the people who think somehow that antiderivatives are more interesting than integrals. Look, this antiderivative exists, except at the places where the denominator is zero. We could always write that antiderivative as a variable-endpoint integral, such as $ \\int_0^x\\frac{\\sin^2t}{1\\plus{}e^t}\\,dt.$\r\n\r\nNow, as to whether that can be written in closed form as an elementary function ... I don't care all that much, but let me ask you this: Do you have some [i]a priori[/i] reason to believe that this antiderivative is an elementary function? Or did you just make it up as something you didn't immediately see a path for? Because if you just make up problems like that, then 99% of the time the answer is going to be \"not an elementary function.\"\r\n\r\nI say that not knowing whether that's the case here or not - but if you don't see a method, the odds are long against you. (Again, I don't even think it's all that interesting a question.)\r\n\r\nIntegrals - meaning definite integrals - make for much more interesting questions.", "Solution_7": "hello, Mathematica says \r\n$ \\int \\frac {\\sin(x)^2}{1 + e^x}\\,dx =$\r\n$ \\frac {1}{8}(4x - ie^{ - 2ix}\\text{Hypergeometric2F1}[2i,1,1 - 2i, - e^x] +$$ ie^{2ix}\\text{Hypergeometric2F1}[2i,1,1 + 2i, - e^x] - 4\\log[1 + e^x])$.\r\nSonnhard.", "Solution_8": "Which is a specific way of saying, \"not an elementary function.\" Which is what I suspected all along." } { "Tag": [ "inequalities", "geometric inequality", "inequalities proposed" ], "Problem": "Given are $\\triangle ABC$ prove that\r\n$sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2}+\\frac{1}{2}(cos^{2}\\frac{A}{2}+cos^{2}\\frac{B}{2}+cos^{2}\\frac{B}{2}-cos\\frac{B}{2}cos\\frac{C}{2}-cos\\frac{C}{2}cos\\frac{A}{2}-cos\\frac{A}{2}cos\\frac{B}{2})\\le \\frac{3}{2}$ :)", "Solution_1": "[quote=\"gemath\"]Given are $\\triangle ABC$ prove that\n$sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2}+\\frac{1}{2}(cos^{2}\\frac{A}{2}+cos^{2}\\frac{B}{2}+cos^{2}\\frac{B}{2}-cos\\frac{B}{2}cos\\frac{C}{2}-cos\\frac{C}{2}cos\\frac{A}{2}-cos\\frac{A}{2}cos\\frac{B}{2})\\le \\frac{3}{2}$ :)[/quote]\r\nIs anybody interested in? :| :P", "Solution_2": "an idea ?:w00tb: or :stretcher: :thumbup:", "Solution_3": "Really is it bored ? :blush:", "Solution_4": "I am bad at inequalities, so the stretcher applies to me :( \r\nat least for now I will make it easier to read (at least for myself) \\[\\frac{3}{2}+\\left(\\mathbf{C}\\frac{A}{2}\\right)\\left(\\mathbf{C}\\frac{B}{2}\\right)+\\left(\\mathbf{C}\\frac{B}{2}\\right)\\left(\\mathbf{C}\\frac{C}{2}\\right)+\\left(\\mathbf{C}\\frac{C}{2}\\right)\\left(\\mathbf{C}\\frac{A}{2}\\right)\\geq \\mathbf{S}\\frac{A}{2}+\\mathbf{S}\\frac{B}{2}+\\mathbf{S}\\frac{C}{2}+\\frac{\\mathbf{C}^{2}\\frac{A}{2}+\\mathbf{C}^{2}\\frac{B}{2}+\\mathbf{C}^{2}\\frac{C}{2}}{2}\\]", "Solution_5": "Thank me@home for your interest\r\n[quote=\"gemath\"]Given are $\\triangle ABC$ prove that\n$sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2}+\\frac{1}{2}(cos^{2}\\frac{A}{2}+cos^{2}\\frac{B}{2}+cos^{2}\\frac{B}{2}-cos\\frac{B}{2}cos\\frac{C}{2}-cos\\frac{C}{2}cos\\frac{A}{2}-cos\\frac{A}{2}cos\\frac{B}{2})\\le \\frac{3}{2}$ :)[/quote]\r\nCan Vasc or anyboy do it ? :D \r\nVasc, are you interested in geometric inequality, I have some hard problems in it, but maybe nobody is interested in !! :(" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Rita drove from her home to work at an average speed of 60 miles per hour and returned home alone the same route at an average speed of 40 miles per hour. If her total driving time for the trip was 2 hours, how many minutes did she take to drive from her job back home?", "Solution_1": "[hide]\nI used an equation. X is the distance between home and work:\n$\\frac{x}{60}+\\frac{x}{40}=2$\n$\\frac{2x}{120}+\\frac{3x}{120}=2$\n$\\frac{5x}{120}=2$\n$x=48$\nSo it took her $\\frac{48}{40}=1.2$ hours, or $72$ minutes, to get back.[/hide]\r\n\r\nHope I did that right.", "Solution_2": "[quote=\"sharkman\"]Rita drove from her home to work at an average speed of 60 miles per hour and returned home alone the same route at an average speed of 40 miles per hour. If her total driving time for the trip was 2 hours, how many minutes did she take to drive from her job back home?[/quote]\r\n\r\n[hide=\"Fast solution\"]She spent $\\frac{60}{40+60}$ of her time going home, so the answer is $2\\cdot \\frac{3}{5}= \\frac{6}{5}$ hours, or $72$ minutes.[/hide]\r\nThis is more like mathcounts level...", "Solution_3": "[quote=\"sharkman\"]Rita drove from her home to work at an average speed of 60 miles per hour and returned home alone the same route at an average speed of 40 miles per hour. If her total driving time for the trip was 2 hours, how many minutes did she take to drive from her job back home?[/quote]\r\n\r\n[color=white]I've seen like, 4 problems which are almost the exact same? Including one in Classroom math or something?[/color]\r\n\r\n[hide=\"My Half-Solution\"]\n\n$\\frac{x}{40}+\\frac{x}{60}=2$\n$\\frac{2x}{120}+\\frac{3x}{120}=2$\n$2x+3x=240$\n$5x=240$\n$x=48$\n$\\frac{48}{40}=\\frac{12}{10}=1.2$\n$72\\mbox{ minutes}$[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "[code]\n\na_{jk} = \n\\begin{cases} \nM & \\text{if} job r_j < k - p\\\\\nw_j(k - d_j) & \\text{if} r_j \\geq k - p\\\\\n\\end{cases}\n[/code]\r\n\r\nI am using the above code (with dollar signs before and after, but I couldn't put those in here since it messed up the text), but it is giving me the errors of \"Undefined Control Sequence\" and \"Misplaced Alignment tab character &.\" What am I doing wrong?\r\nI am using Texnic Center.", "Solution_1": "It works fine here (not sure where 'job' came from) especially if you put a space after 'if':\r\n$a_{jk} = \r\n\\begin{cases} \r\nM & \\text{if } r_j < k - p\\\\ \r\nw_j(k - d_j) & \\text{if } r_j \\geq k - p\\\\ \r\n\\end{cases}$\r\n\r\nIn TeXnicCenter you need the amsmath package loaded to use the cases environment" } { "Tag": [ "geometry", "number theory" ], "Problem": "Please post here ranking of Graduate school in Math of Netherlands. Thank you very much! :)", "Solution_1": "Only Netherland or also Flanders (Belgium, next to it, speaking same language)?\r\n\r\nAnd in what area? I mean the ranking is different for pure maths/applied maths/mathematical physics/... and even within each branch there's difference in which is the best (Algebra/Geometry/Combinatorics/Analysis/NumberTheory/...)\r\n\r\nTell me your field(s) of interest and I'll ask around which university fits you best. :)", "Solution_2": "[quote]\nOnly Netherland or also Flanders (Belgium, next to it, speaking same language)? [/quote]\nAll, if you can help me. :)\n\n[quote=\"Peter\"]\nTell me your field(s) of interest and I'll ask around which university fits you best. :)[/quote]\r\nMy fields: Number Theory, Algebraic Geometry, Non-commutative Geometry. Please post here websites of those schools, if you can. \r\n\r\nThank you very much. :wink:", "Solution_3": "[quote=\"N.T.TUAN\"]Number Theory[/quote]Leiden and Leuven are the two names popping to my mind here (asssuming you're leaning towards algebraic number theory).\n\n[quote=\"N.T.TUAN\"]Non-commutative Geometry[/quote]If you mean moufang quadrangles, generalized hexagons and alike geometry that relies heavily on non-commutative algebra, Ghent is your name.\n(if that's not what you meant, I'll need more explanation)\n\n[quote=\"N.T.TUAN\"]Algebraic Geometry[/quote]I'm not sure about this one, I guess it is common and most universities are working on this? \r\n(at least I know that here in Ghent we have strong people working on it, but no idea about other universities)" } { "Tag": [ "trigonometry", "AMC" ], "Problem": "If sin4x=a express cos8x in terms of a.", "Solution_1": "in spoila:\n\n\n\n[hide] sin4x=a. cos8x=1-2sin:^2:4x, so cos8x=1-2a :^2: . [/hide]", "Solution_2": "right." } { "Tag": [ "AMC", "AIME", "MATHCOUNTS", "USA(J)MO", "USAMO", "Euler", "\\/closed" ], "Problem": "under most of your names, you have some sort of conjecture, or hypothesis, or theory, and a number of question marks, do you guys just put that there? or do you have to do something to get that?", "Solution_1": "the number of posts you have you get a conjecture that is one of the seven famous not proven mathematical conjectures", "Solution_2": "ah, okay, thanks", "Solution_3": "I think you need about 20 or 50 before the first level (P vs. NP) kicks in.\r\n\r\nhttp://www.claymath.org/millennium/ is the website which can tell you about the specific 7 problems (and why they are important).", "Solution_4": "instructors don't have them they have something that says instructors so i am churchill almighty i have both poincare conjecture and instrucotr under my belt.", "Solution_5": "[quote=\"churchilljrhigh\"]instructors don't have them they have something that says instructors so i am churchill almighty i have both poincare conjecture and instrucotr under my belt.[/quote]\r\n\r\nYou really need to learn some netiquette and learn to stop trying to impersonate the AoPS instructors.", "Solution_6": "I have been online for only a short time. I don't know any netiquette. Can you teach me some? BTW I like the mandelbrot set. I did a science fair project on fractals.\r\n\r\nIt is actually for my own good. I want to start having a habit of having good grammar and only posting in the My classes section.", "Solution_7": "I think, as he made it quite clear, he has a problem with your purposefully taking as your avatar an image that makes you appear to be at first glance another user of the forum. There are hundreds of pictures of the mandelbrot fractal available, and you picked the only one that MCrawford uses.\r\n\r\n\r\nIf you want to work on your grammar and understandability, re-read everything you post here to make sure that it actually makes sense.", "Solution_8": "Ok thanyou tokenadult, and jbl. You guys just reminded me that avatars must be for a cause. Kerry must win the BIG DAY ON TUESDAY. So I'm putting a Bush poster and Kerry line.", "Solution_9": "I've heard a little joke going around that the Bush campaign should choose as its slogan \"Bush-Cheney 2004: Don't change horsemen mid-apocalypse.\" ;)", "Solution_10": "What is AOPS?\r\n\r\n\r\n[quote=\"tokenadult\"][quote=\"churchilljrhigh\"]instructors don't have them they have something that says instructors so i am churchill almighty i have both poincare conjecture and instrucotr under my belt.[/quote]\n\nYou really need to learn some netiquette and learn to stop trying to impersonate the AoPS instructors.[/quote]", "Solution_11": "AoPS = \"Art of Problem Solving\"", "Solution_12": "[quote=\"JBL\"]AoPS = \"Art of Problem Solving\"[/quote\r\n\r\nHow do people qualify to teach the online courses?", "Solution_13": "OK you ace the AIME you win national mathcounts you get a p.h.d in mathematics you get an exceptional ranking in USAMO and you write a nyt bestseller math book.", "Solution_14": "Haha. You work for AoPS. You can read about Richard and Mathew in the \"about us\" section of this website.", "Solution_15": "Once you start high school, pre high school math competitions (mathcounts) still count for things?\r\n\r\n\r\n[quote=\"churchilljrhigh\"]OK you ace the AIME you win national mathcounts you get a p.h.d in mathematics you get an exceptional ranking in USAMO and you write a nyt bestseller math book.[/quote]", "Solution_16": "What is a conjecture? How could they exist if they're not proven?\r\n\r\n\r\n[quote=\"churchilljrhigh\"]the number of posts you have you get a conjecture that is one of the seven famous not proven mathematical conjectures[/quote]", "Solution_17": "Not really; they may look good on your resume as with the aformenioned example.\r\n\r\nA conjecture (in English) is something seemingly true, but hasn't been proven yet. They can exist because it could be a trivial fact to some people but can't/haven't been proven yet. (like 1+1=2 is trivial but I'm guessing most people can't \"prove\" that 1+1=2)", "Solution_18": "[quote=\"Mathhelp\"]What is a conjecture? How could they exist if they're not proven?[/quote]\r\n\r\nThe word \"conjecture\" basically means \"educated guess.\" And in mathematics, a conjecture is a statement about what MAY BE a provable mathematical fact, for which no proof has been found. One of the most famous mathematical conjectures is Goldbach's conjecture, from a personal letter from Christian Goldbach (1690-1764) to Leonhard Euler (1707-1783), that every even number greater than or equal to 6 can be written as the sum of two odd prime numbers. BILLIONS of numerical examples have been examined by computer, witbout anyone finding any counterexample, but no one has come up with a proof to cover ALL even numbers greater than or equal to six, which is what is needed to turn this statement from being a conjecture to being a theorem. \r\n\r\nThe theorem proved by Andrew Wiles that is popularly known as Fermat's Last Theorem was really a conjecture until Wiles proved it, but because Fermat claimed to have a proof (which math historians are quite sure he didn't have), it was always informally referred to as a theorem, not as a conjecture. I would like to nudge popular usage in the direction of referring to this statement as the Fermat-Wiles Theorem, giving credit to both the person who posed the problem and the person who solved it.", "Solution_19": "In general, the name of a theorem is the name of the poser rather than the name of the solver.", "Solution_20": "[quote=\"ComplexZeta\"]In general, the name of a theorem is the name of the poser rather than the name of the solver.[/quote]\r\n\r\nThat idea has all sorts of interesting implications.", "Solution_21": "Are you going to start posing tons of hard questions for future generations to solve? Fermat did lots of that. Euler (dis)proved most of them." } { "Tag": [ "geometry", "perimeter" ], "Problem": "This diagram represents a thin belt stretched taut around three pulleys each 2 feet in diameter. The distances between the centers of these pulleys are 6,9, and 13 feet. Using 3.14 for $ \\pi$, express the number of feet in the length of the belt to the nearest hundredth.\n\n[asy]pair rotateleft(pair p){\n return (-p.y, p.x);\n}\nvoid drawCircle(pair center, real radius){\n draw(center+(radius,0)..center+(0,radius)..center-(radius,0)..center-(0,radius)..cycle);\n}\n\npair normalize(pair p){\n return p/sqrt(p.x*p.x + p.y*p.y);\n}\n\nvoid drawTangent(pair left, pair right, real rad){\n pair delta = normalize(rotateleft(right-left))*rad;\n draw(left+delta--right+delta);\n}\n\nreal rad = 1;\npair A=(0,0), B=(9,0), C=(12,-5);\ndrawCircle(A,rad);\ndrawCircle(B,rad);\ndrawCircle(C,rad);\ndraw(A--B--C--cycle,dashed);\ndrawTangent(A,B,rad);\ndrawTangent(B,C,rad);\ndrawTangent(C,A,rad);[/asy]", "Solution_1": "Let the vertices of the triangle be $ A$, $ B$ and $ C$. The length of the belt is the perimeter of the triangle plus the arcs. The total length of the arcs is $ \\frac {3(360) \\minus{} 6(90) \\minus{} 180}{360} \\equal{} 1$ times the circumference of one circle, so the arcs have length $ 2\\pi$ and the perimeter of $ ABC$ is $ 6 \\plus{} 9 \\plus{} 13 \\equal{} 28$, so the belt's length is\r\n\\[ 2\\pi \\plus{} 6 \\plus{} 9 \\plus{} 13\\approx \\boxed{34.28}\r\n\\]", "Solution_2": "how do you know it is one rotation " } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $ x(t)$ be the solution of $ \\frac {dx}{dt} \\equal{} x^2 \\plus{} t^3$ with $ x(0) \\equal{} 0$. Show that for each $ 0 < t < 1$ we have $ x(t) < t^4 \\plus{} t^9.$", "Solution_1": "It suffices to show is that $ y(t)\\equal{}t^4$ is a \"supersolution\", i.e., $ y'\\ge y^2\\plus{}t^3$. This reduces to $ 4t^3\\ge t^8\\plus{}t^3$, which is trivial on $ [0,1]$. I'm pretty sure you meant something else. Perhaps, the equation was $ x'\\equal{}x^2\\plus{}4t^3$ (then you'll really need the $ t^9$ term though the solution remains exactly the same)." } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "I was tutoring today and couldn't show my student how to integrate this :blush: \r\n\r\nmaple gives 1/4*exp(2*x)/(2*x+1) as an answer\r\n\r\nsetting exp(2*x)dx=dv will give increased powers of 1/(2*x+1) in another integral and going the other way gives ln(2*x+1) which doesn't seem to help me.", "Solution_1": "first change variable $u=2x$, and it becomes $\\frac{1}{4}\\int \\frac{ue^u}{(u+1)^2}du$\r\n\r\n\\begin{eqnarray*}\r\n\\int \\frac{ue^u}{(u+1)^2}du &=& \\int \\frac{ue^u}{(u+1)^2}d(u+1)\\\\\r\n&=&-\\int ue^ud\\left(\\frac{1}{u+1}\\right)\\\\\r\n\\mbox{integration by parts}&=&-\\frac{ue^u}{u+1} + \\int \\frac{e^u+ue^u}{u+1}du\\\\\r\n&=& e^u(-\\frac{u}{u+1}) + \\int e^u du = e^u (1-\\frac{u}{u+1})\\\\\r\n&=& \\frac{e^u}{u+1}\r\n\\end{eqnarray*}", "Solution_2": "well... wow\r\n\r\n\r\ni don't understand the second step where 1/(1+u)^2 dissappears.\r\n\r\nalso this it a trick that my highschool calculus student won't understand...\r\ndo you have any simpler suggestions? or a good way of explaining you solution.", "Solution_3": "okay..it is, in fact, changing of variable.\r\n\r\n$d(\\frac{1}{u+1}) = -\\frac{1}{(u+1)^2}du$", "Solution_4": "It's just a notational thing. The obvious thing is to integrate by parts to break off the $\\frac {1}{(1+x)^2}$; do that and the indefinite integral becomes easy to find." } { "Tag": [ "geometry", "calculus", "number theory", "complex analysis" ], "Problem": "Hi, several of the threads here have me wondering what the word on the street is about scoring Olympiad-level competitions. I have read in a lot of places that the IMO explicitly is based on PREcalculus mathematics (although it presupposes a curriculum with much stronger geometry and number theory and combinatorics preparation than is typical in the United States curriculum). But if someone competing at the IMO had an extraordinary background in math, and knew how to use complex analysis (to give a fanciful example) to solve an Olympiad problem, would that be deemed a \"correct\" solution? Or is it only Olympiadly correct to use secondary school mathematics methods to solve the problems? \r\n\r\nJust wondering, that's all.", "Solution_1": "You can use whatever you want to solve the problems. If you want to prove the Riemann Hypothesis on a contest and then use the Riemann Hypothesis in your solution, you'll still get full credit if it's correct.", "Solution_2": "Yeah, you can use calculus or whatever you want. There is one point bad in doing it though, in the fact that if you use a calculus solution and make a small slip, then they're quite likely to give you no marks. You don't often get part marks for something like that.", "Solution_3": "At MOP 2002, someone used Fourier Analysis to solve a problem.", "Solution_4": "Hmm, what are the qualifications of the IMO judges? What if some kid used math that was beyond the knowledge of any of the judges? :D", "Solution_5": "I have wondered that myself on several occasions. I don't know the answer.", "Solution_6": "That is highly improbable. The judges are mostly professional mathematicians.", "Solution_7": "[quote=\"Fierytycoon\"]Hmm, what are the qualifications of the IMO judges? What if some kid used math that was beyond the knowledge of any of the judges? :D[/quote]\n\n[quote=\"JBL\"]That is highly improbable. The judges are mostly professional mathematicians.[/quote]\r\n\r\nInterestingly, this question came up in a [url=http://groups.google.com/groups?&threadm=bmhlgi%24qfa%241%40oravannahka.helsinki.fi&rnum=1]thread on the Usenet sci.math newsgroup[/url], so it's at least a plausible concern to some mathematicians who read about Olympiad competitions in the popular press. I wouldn't lose sleep over it, though. \r\n\r\nhttp://groups.google.com/groups?&threadm=bmhlgi%24qfa%241%40oravannahka.helsinki.fi&rnum=1", "Solution_8": "It is my understanding that at the IMO, scores are determined in the presence of the team leader, who can argue for points. If a judge is not aware of a certain theorem, the leader just has to present the theorem in a credible resource and they have to give credit.\r\n\r\nApparently this came up in 1994; one USA competitor wrote a one line proof citing a theorem identical to the problem statement which the judges were not aware of, but the US was able to find a book in which the theorem was published at a Hong Kong library, thus the competitor recieved full points. The details may be wrong, but something like that occurred.\r\n\r\nOlympiads are designed so that all problems CAN be solved with precalculus mathematics, but they do not NEED to be; all mathematics is fair game. I think that they try to avoid problems which would be easy with higher math but difficult without it." } { "Tag": [ "conics", "ellipse", "integration", "vector", "calculus", "parameterization", "geometry" ], "Problem": "Compute [tex]\\oint_c(y+z)dx + (z-x)dy + (x-y)dz [/tex] using Stoke's theorem, where c is the ellipse [tex]x(t) = asin^2t, \\ y(t) = 2asintcost, z(t) = acos^2t, 0\\leq t \\leq \\pi [/tex]\r\n\r\n\r\nthe version of stoke's theorem I learned is:\r\n[tex]\r\n\\int_c \\overrightarrow{F} \\cdot d\\overrightarrow{r} \r\n= \\int_s curl \\overrightarrow{F} \\cdot d\\overrightarrow{S}\r\n=\\iint_s curl \\overrightarrow{F}\\cdot \\overrightarrow{n} \\cdot dS\r\n[/tex]\r\n\r\nwhere S is the elliptical surface bounded by the curve c, F is a vector field and n is the unit vector pointing out at that point.\r\n\r\nIn this case, [tex]F = [/tex], and I calculated curl F to be [tex]<-2, 0, -2>[/tex].\r\n\r\nSo we have to find\r\n\r\n[tex]\\iint_s <-2, 0, -2> \\cdot \\overrightarrow{n} \\cdot dS [/tex]\r\n\r\nHow would I find [tex]\\overrightarrow {n} [/tex] and dS, and also the bounds of integration for the double integral?", "Solution_1": "Note that the curve lies on the plane $x+z=a.$ It would be useful to rewrite the parameterization as:\r\n\r\n$x=a\\left(\\frac12-\\frac12\\cos2t\\right)$\r\n$y=a\\sin2t$\r\n$z=a\\left(\\frac12+\\frac12\\cos2t\\right)$\r\n\r\nYou have many choices for the surface, but if you have a chance to use a plane, you should, for the sake of simplicity.\r\n\r\nThis plane will be relatively easy to project to a two-dimensional integral.\r\n\r\nIf $z=g(x,y)$ we can write $\\overrightarrow{dS}=\\left(-\\frac{\\partial g}{\\partial x}\\overrightarrow{i}-\\frac{\\partial g}{\\partial y}\\overrightarrow{j}+\\overrightarrow{k} \\right)\\,dx\\,dy.$\r\n\r\n(Or the negative of that, depending on the orientation of the curve.)\r\n\r\nIn our specific case, we have $\\overrightarrow{dS}=(\\overrightarrow{i} +\\overrightarrow{k} )\\,dx\\,dy$ \r\n\r\nand, (using your calculation of the curl, which is correct)\r\n\r\n$\\text{curl}\\overrightarrow{F}\\cdot\\overrightarrow{dS}=-4\\,dx\\,dy.$\r\n\r\nProjected onto the $x$-$y$ plane, we have the curve $x=a\\left(\\frac12- \\frac12\\cos2t\\right),y=a\\sin2t.$ This is an ellipse centered at $\\left(\\frac a2,0\\right)$ with semi-major axis $a$ and semi-minor axis $\\frac a2,$ traced out clockwise in the $x$-$y$ plane. Since clockwise is \"backwards,\" we must reverse the sign in $\\overrightarrow{dS}.$ We now have\r\n\r\n$\\text{curl}\\overrightarrow{F}\\cdot\\overrightarrow{dS}=4\\,dx\\,dy.$\r\n\r\nIntegrate this over the ellipse whose area in the $x$-$y$ plane is $\\frac{\\pi a^2}2$ and our final answer for the integral is simply $2\\pi a^2.$" } { "Tag": [], "Problem": "\u0398\u03b1 \u03b3\u03c1\u03ac\u03c8\u03c9 \u03ac\u03bb\u03bb\u03b7 \u03bc\u03af\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03ac\u03c0\u03c9\u03c2 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ae \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03b7. \r\n\r\n\u0388\u03c3\u03c4\u03c9 \u03c4\u03c5\u03c7\u03b1\u03af\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u0391\u0392\u0393 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03cd\u03c8\u03bf\u03c2 \u0391\u0394. \u0391\u03c0\u03cc \u03c4\u03bf \u0394 \u03c6\u03ad\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b5\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03b51 \u03ba\u03b1\u03b9 \u03b52 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u0391\u0392 \u03ba\u03b1\u03b9 \u0391\u0393 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u0388\u03c0\u03b5\u03b9\u03c4\u03b1 \u03b1\u03c0\u03cc \u03c4\u03bf \u0392 \u03ba\u03b1\u03b9 \u03c4\u03bf \u0393 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03b53 \u03ba\u03b1\u03b9 \u03b54 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03bf\u03b9 \u03bf\u03c0\u03bf\u03af\u03b5\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b1\u03c1\u03ac\u03bb\u03bb\u03b7\u03bb\u03b5\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7\u03bd \u0391\u0394. \u039f\u03b9 \u03b51 \u03ba\u03b1\u03b9 \u03b53 \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u039d \u03b5\u03bd\u03ce \u03bf\u03b9 \u03b52 \u03ba\u03b1\u03b9 \u03b54 \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u039c. \u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b7 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 \u039c\u039d \u03b4\u03b9\u03ad\u03c1\u03c7\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03bf\u03c1\u03b8\u03cc\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u0397 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u0391\u0392\u0393.", "Solution_1": "\u03a4\u03b7\u03bd \u03ad\u03bb\u03c5\u03c3\u03b1 \u03bc\u03b5 \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03ae \u03b1\u03bb\u03bb\u03ac \u03c3\u03c5\u03bd\u03b8\u03b5\u03c4\u03b9\u03ba\u03ac \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03b2\u03c1\u03b5\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b1\u03c1\u03b3\u03ac \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03c3\u03c9 \u03c4\u03ce\u03c1\u03b1.\r\n\r\n\u03a3\u03b7\u03bc\u03b5\u03b9\u03c9\u03c4\u03b5\u03cc\u03bd \u03cc\u03c4\u03b9 \u03b4\u03b5 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce \u03cc\u03c4\u03b9 \u03c4\u03bf \u0394 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c7\u03bd\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03cd\u03c8\u03bf\u03c5\u03c2 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bc\u03b1\u03b9 100% \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 (\u03ba\u03b1\u03b9 \u03bc\u03b5 \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03ae \u03c4\u03bf \u03b1\u03c0\u03ad\u03b4\u03b5\u03b9\u03be\u03b1 \u03ba\u03b9\u03cc\u03bb\u03b1\u03c2) \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7 \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc\u03c4\u03b5\u03c1\u03b7 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03cc\u03c0\u03bf\u03c5 \u03c4\u03bf \u0394 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03c0\u03bf\u03b9\u03bf\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03b7\u03c2 \u0392\u0393 (\u03bf\u03b9 \u03b53 \u03ba\u03b1\u03b9 \u03b54 \u03cc\u03c7\u03b9 \u03c0\u03b1\u03c1\u03ac\u03bb\u03bb\u03b7\u03bb\u03b5\u03c2 \u03c3\u03c4\u03b7\u03bd \u0391\u0394 \u03b1\u03bb\u03bb\u03ac \u03ba\u03ac\u03b8\u03b5\u03c4\u03b5\u03c2 \u03c3\u03c4\u03b7 \u0392\u0393, \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c6\u03c5\u03c3\u03b9\u03ba\u03ac \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7\u03bd \u03b5\u03b9\u03b4\u03b9\u03ba\u03ae \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7)\r\n\r\n\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03b4\u03ce\u03c3\u03b5\u03b9 \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03ac \u03bc\u03b9\u03b1 \u03c3\u03c5\u03bd\u03b8\u03b5\u03c4\u03b9\u03ba\u03ae \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b3\u03b5\u03bd\u03b9\u03ba\u03ae \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7? :?:", "Solution_2": "\u0398\u03b1 \u03b4\u03c9\u03c3\u03c9 \u03bc\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7 (\u03b4\u03c5\u03c3\u03c4\u03c5\u03c7\u03c9\u03c2 \u03bc\u03bf\u03bd\u03bf \u03c3\u03c4\u03b7\u03bd \u03b5\u03b9\u03b4\u03b9\u03ba\u03b7 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b7) \u03bc\u03b5 \u03bf\u03bc\u03bf\u03b9\u03bf\u03c4\u03b7\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u0398\u03b1\u03bb\u03b7.\r\n\r\n\u03a6\u03b5\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c5\u03c8\u03bf\u03c2 BD' \u03ba\u03b1\u03b9 \u03b4\u03b9\u03b1 \u03c4\u03bf\u03c5 N \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03b7 \u03c0\u03c1\u03bf\u03c2 \u03b1\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03c4\u03b5\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd BC (\u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03b5\u03ba\u03c4\u03b1\u03c3\u03b7 \u03c4\u03b7\u03c2) \u03c3\u03c4\u03bf \u039d'\r\n\u03a4\u03c9\u03c1\u03b1 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9:\r\n$ \\angle{CMD} \\equal{} \\angle{C} \\equal{} \\angle{BNN'}, \\angle{BND} \\equal{} \\angle{B}$\r\n\u0391\u03c1\u03b1 \u03c4\u03b1 \u03bf\u03c1\u03b8\u03bf\u03b3\u03c9\u03bd\u03b9\u03b1 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03b1 $ BNN', DAC$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03b9\u03b1, \u03bf\u03c0\u03c9\u03c2 \u03b5\u03c0\u03b9\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b1 $ BDN, DAB$\r\n\u0391\u03c1\u03b1:\r\n$ \\frac{N'B}{AD} \\equal{} \\frac{NB}{DC} \\Rightarrow N'B*DC \\equal{} AD*NB$ (1)\r\n\u03ba\u03b1\u03b9\r\n$ \\frac{NB}{BD} \\equal{} \\frac{DB}{AD} \\Rightarrow AD*NB \\equal{} BD^2$ (2)\r\n\u0391\u03c0\u03bf \u03c4\u03b9\u03c2 2 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03b9\u03b5\u03c2 \u03c3\u03c7\u03b5\u03c3\u03b5\u03b9\u03c2 \u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c0\u03c4\u03b5\u03b9:\r\n$ N'B*DC \\equal{} BD^2 \\Rightarrow \\frac{N'B}{BD} \\equal{} \\frac{BD}{DC}$, \u03b1\u03c0\u03bf \u03bf\u03c0\u03bf\u03c5 \u03bb\u03b1\u03bc\u03b2\u03b1\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c5\u03c0 \u03bf\u03c8\u03b7\u03bd \u03bf\u03c4\u03b9: \r\n$ NN' \\parallel BD' \\parallel DM, NB \\parallel AD \\parallel MC$\r\n\u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c0\u03c4\u03b5\u03b9 \u03bf\u03c4\u03b9 \u03c4\u03b1 \u03b4\u03c5\u03bf \u03c5\u03c8\u03b7 \u03b4\u03b9\u03b1\u03b9\u03c1\u03bf\u03c5\u03bd \u03c4\u03bf \u03b5\u03c5\u03b8\u03b9\u03b3\u03c1\u03b1\u03bc\u03bc\u03bf \u03c4\u03bc\u03b7\u03bc\u03b1 $ MN$ \u03c3\u03c4\u03bf\u03bd \u03b9\u03b4\u03b9\u03bf \u03bb\u03bf\u03b3\u03bf, \u03b1\u03c1\u03b1 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03b9\u03bf \u03c4\u03bf\u03bc\u03b7\u03c2 \u03c4\u03bf\u03c5\u03c2 (\u03b4\u03b7\u03bb\u03b1\u03b4\u03b7 \u03c4\u03bf \u03bf\u03c1\u03b8\u03bf\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf) \u03b1\u03bd\u03b7\u03ba\u03b5\u03b9 \u03c3\u03c4\u03bf \u03c4\u03bc\u03b7\u03bc\u03b1 \u03b1\u03c5\u03c4\u03bf.", "Solution_3": "\u039d\u03bf\u03bc\u03b9\u03b6\u03c9 \u03bf\u03c4\u03b9 \u03b5\u03c7\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03b3\u03b5\u03bd\u03b9\u03ba\u03b7 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b7\r\n\r\n\u0391\u03bd $ K$ \u03c4\u03c5\u03c7\u03b1\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03b9\u03bf \u03c4\u03b7\u03c2 $ BC$ \u03ba\u03b1\u03b9 $ S, S'$ \u03c4\u03b1 \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03c3\u03b7\u03bc\u03b5\u03b9\u03b1 \u03c0\u03bf\u03c5 \u03bf\u03c1\u03b9\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c0\u03b1\u03bd\u03c9 \u03c3\u03c4\u03b9\u03c2 $ {\\epsilon}_3, {\\epsilon}_4$, \u03c4\u03bf\u03c4\u03b5 \u03b1\u03c0\u03bf \u03c4\u03bf $ S$ \u03c6\u03b5\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03b7 \u03c3\u03c4\u03bf \u03c5\u03c8\u03bf\u03c2 $ BD'$ \u03c0\u03bf\u03c5 \u03c4\u03b5\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd $ BC$ \u03c3\u03c4\u03bf $ S''$. \u03a4\u03bf\u03c4\u03b5 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03bf\u03bc\u03bf\u03b9\u03bf\u03c4\u03b7\u03c4\u03b1 \u03c4\u03c9\u03bd $ S''S'B, N'NB$ \u03ba\u03b1\u03b9 \u03c4\u03c9\u03bd $ BND, BS'K$ \u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c0\u03c4\u03b5\u03b9 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 $ \\frac {S''B}{BK} \\equal{} \\frac {N'B}{BD}$ \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03c3\u03c5\u03bd\u03b4\u03b9\u03b1\u03c3\u03bc\u03bf \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b1\u03bd\u03b1\u03bb\u03bf\u03b3\u03b9\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03b2\u03c1\u03b7\u03ba\u03b1\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03c5\u03b5\u03bd\u03b7 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b5\u03bb\u03b9\u03ba\u03b1: $ \\frac {S''B}{BK} \\equal{} \\frac {BD}{DC}$, \u03b1\u03c0\u03bf \u03bf\u03c0\u03bf\u03c5 \u03bf\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03b5\u03b9\u03b4\u03b9\u03ba\u03b7 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c0\u03c4\u03b5\u03b9 \u03bf\u03c4\u03b9 \u03c4\u03b1 \u03c5\u03c8\u03b7 $ AD, BD'$ \u03c4\u03b5\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c0\u03b1\u03bd\u03c9 \u03c3\u03c4\u03bf \u03b5\u03c5\u03b8\u03b9\u03b3\u03c1\u03b1\u03bc\u03bc\u03bf \u03c4\u03bc\u03b7\u03bc\u03b1 $ SS'$ :)\r\n\r\n\u03a0\u03b5\u03b9\u03c4\u03b5 \u03bc\u03bf\u03c5 \u03b1\u03bd \u03c7\u03b1\u03bd\u03b5\u03b9 \u03ba\u03b1\u03c0\u03bf\u03c5 \u03b7 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7...", "Solution_4": "\u03a4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b1\u03bb\u03b7\u03b8\u03b5\u03cd\u03b5\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03c5\u03c7\u03b1\u03af\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ H,$ \u03c3\u03c4\u03bf \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ \\bigtriangleup ABC,$ \u03cc\u03c0\u03bf\u03c5 $ (\\epsilon_{1})\\parallel CH,\\ (\\epsilon_{2}\\parallel BH$ \u03ba\u03b1\u03b9 $ (\\epsilon_{3})\\parallel AD\\parallel (\\epsilon_{4}).$\r\n\r\n\u0391\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03bc\u03b5 \u03c4\u03bf [b][size=100]\u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u0398\u03b1\u03bb\u03ae,[/size][/b] \u03c9\u03c2 \u03ac\u03bc\u03b5\u03c3\u03b7 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03c4\u03bf\u03c5.\r\n\r\n$ \\frac {MC}{ML} \\equal{} \\frac {DC}{DB} \\equal{} \\frac {NK}{NB}$ $ \\Longrightarrow$ $ M,\\ H,\\ N,$ \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03ba\u03b1\u03b9 \u03b7 \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03ad\u03c7\u03b5\u03b9 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af.\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_5": "\u0391\u03c5\u03c4\u03ac \u03b5\u03af\u03bd\u03b1\u03b9... \u039a\u03bf\u03bd\u03c4\u03ac \u03ae\u03bc\u03bf\u03c5\u03bd \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2... \u03a9\u03c1\u03b1\u03af\u03bf\u03c2 Nick. \u03a0\u03bf\u03bb\u03cd \u03cc\u03bc\u03bf\u03c1\u03c6\u03b7 \u03b7 \u03bb\u03cd\u03c3\u03b7. :)" } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Excuse me for being stupid, but I can't prove that the composition of two analytical functions is again analytical. Does the result remain true for entire functions? Could anyone post a proof, in case the assertions are correct? Thanks.", "Solution_1": "Doesn't it simply mean that it's complex differentiable? So then shouldn't it be as easy to prove as the fact that the composition of two differentiable functions is differentiable? :? Or maybe you don't wnt to use the fact that it's analytic iff it's holomorphic?", "Solution_2": "By analytic, do you mean it can be represented by a power series and by entire, do you mean it is complex differentiable? The reason I am asking is that most complex analysis books seem to consider them synonymous...", "Solution_3": "Grobber, do you have a simple proof for the fact that complex differentiable and analytic functions are the same? Sorry, it doesn't seem simple to me. Here by analytic I mean that $f$ is developable in entire series around each point from its domain.", "Solution_4": "No, I do not have a simple proof, that's why I eventually realized that by analyic you actually meant the standard definition of analyticity.", "Solution_5": "[quote=\"harazi\"]Grobber, do you have a simple proof for the fact that complex differentiable and analytic functions are the same? Sorry, it doesn't seem simple to me. Here by analytic I mean that $f$ is developable in entire series around each point from its domain.[/quote]\r\n\r\nthis is a simple consequence of Cauchy's theorem. You can find a simple proof in the book by H. Cartan: th\u00e9orie \u00e9l\u00e9mentaire des fonctions analytiques.\r\n\r\nMichael" } { "Tag": [ "Euler", "limit", "function", "integration", "logarithms", "calculus", "complex analysis" ], "Problem": "Prove: ${\\gamma = \\lim_{n \\to+\\infty}(n-\\Gamma(\\frac1n}))$.", "Solution_1": "${\\lim_{n \\to\\infty}(n-\\Gamma(\\frac1n}))= \\lim_{n \\to\\infty}\\frac{1-\\Gamma(1+\\frac1n)}{\\frac1n}=-\\Gamma'(1).$\r\n\r\nThat doesn't solve the problem but it's an interesting way of restating it.", "Solution_2": "Here is a solution where I use complex analysis (after all, this is where $\\Gamma$ really belongs).\r\n\r\nFirst of, let $H(z) = \\prod_{k=1}^\\infty\\left(1+\\frac{z}{k}\\right)e^{-z/k}$ be the canonical product associated to $-1,-2,-3,\\ldots$. We know this is an entire function and that the product converges absolutely locally uniformly in $\\mathbb{C}$. We have $H(0)=1$ and $H'(0) = 0$. So we get that $H(z) = 1+O(z^{2})$ and thus also $1/H(z) = 1+O(z^{2})$.\r\n\r\nI don't know if this is well known, but in my complex analysis course we showed that $\\Gamma\\left(z\\right) = \\frac{e^{-\\gamma z}}{zH(z)}$. This follows quite easily from the Stirling formula, but of course the Stirling formula is difficult to prove, so this solution might be a bit overkill.\r\n\r\nNow $n-\\Gamma\\left(\\frac{1}{n}\\right) = n\\left(1-\\frac{e^{-\\gamma/n}}{H\\left(\\frac{1}{n}\\right)}\\right) = n\\left(1-\\left(1-\\frac{\\gamma}{n}+O\\left(\\frac{1}{n^{2}}\\right)\\right)\\right) \\rightarrow \\gamma$", "Solution_3": "[quote=\"Kalle\"]Here is a solution where I use complex analysis (after all, this is where $\\Gamma$ really belongs). [/quote]\r\nSure, and it's a nice solution. But I'll leave the topic right here rather than moving it, becuase I think there's still room for an entirely real solution.", "Solution_4": "[quote=\"Kent Merryfield\"]becuase I think there's still room for an entirely real solution.[/quote]\r\n\r\nI think I found such a solution. \r\n\r\nNotice that\r\n\\[\\begin{aligned}n-\\Gamma\\left(\\frac{1}{n}\\right) & = n \\int_{0}^{+\\infty}e^{-x}\\,dx-\\int_{0}^{+\\infty}e^{-x}x^{\\frac{1}{n}-1}\\,dx \\\\ & = n \\int_{0}^{+\\infty}e^{-x}\\,dx-\\int_{0}^{+\\infty}n x^{\\frac{1}{n}}e^{-x}\\,dx \\\\ & =-\\int_{0}^{+\\infty}\\frac{x^{\\frac{1}{n}}-1}{\\frac{1}{n}}e^{-x}\\,dx\\end{aligned}. \\]\r\nHence\r\n\\[\\lim_{n \\to+\\infty}\\left(n-\\Gamma\\left(\\frac{1}{n}\\right)\\right) =-\\lim_{n \\to+\\infty}\\int_{0}^{+\\infty}\\frac{x^{\\frac{1}{n}}-1}{\\frac{1}{n}}e^{-x}\\,dx. \\]\r\nIt's obvious that\r\n\\[\\lim_{n \\to+\\infty}\\frac{x^{\\frac{1}{n}}-1}{\\frac{1}{n}}=\\ln x. \\]\r\nUsing the dominated convergence theorem, we get\r\n\\[\\lim_{n \\to+\\infty}\\int_{0}^{+\\infty}\\frac{x^{\\frac{1}{n}}-1}{\\frac{1}{n}}e^{-x}\\,dx =\\int_{0}^{+\\infty}\\lim_{n \\to+\\infty}\\left(\\frac{x^{\\frac{1}{n}}-1}{\\frac{1}{n}}e^{-x}\\right)dx =\\int_{0}^{+\\infty}\\left(\\ln x\\right) e^{-x}\\,dx \\] and it is well-known that this integral is equal to $-\\gamma$.\r\n\r\n(We can use $g$ as the \"dominating function\", where $g(x) = \\sqrt{x}$ for $0 < x < 1$ and $g(x) = x^{2}$ for $x \\geq 1$.)\r\n\r\nSo, we're done?", "Solution_5": "Arne I think it's not necessery to do all that in order to get $\\underset{n\\rightarrow \\infty }{\\lim }\\left( n-\\Gamma \\left( \\frac{1}{n}\\right) \\right) =-\\int_{0}^{+\\infty }\\left( \\ln x\\right) e^{-x}dx$\r\nKent has proved that : $\\underset{n\\rightarrow \\infty }{\\lim }\\left( n-\\Gamma \\left( \\frac{1}{n}\\right) \\right) =-\\Gamma^{\\prime }\\left( 1\\right) =-\\int_{0}^{+\\infty }\\left( \\ln x\\right) e^{-x}dx$\r\nsince $\\left( \\forall x>0\\right) : \\Gamma^{\\prime }\\left( x\\right) =\\int_{0}^{+\\infty }\\left( \\ln t\\right) t^{x-1}e^{-t}dt$", "Solution_6": "[quote=\"erdos\"]$\\left( \\forall x>0\\right) : \\Gamma^{\\prime }\\left( x\\right) =\\int_{0}^{+\\infty }\\left( \\ln t\\right) t^{x-1}e^{-t}dt$[/quote]\r\n\r\nWell, I didn't know that. Is it hard to prove?", "Solution_7": "[quote=\"Arne\"][quote=\"erdos\"]$\\left( \\forall x>0\\right) : \\Gamma^{\\prime }\\left( x\\right) =\\int_{0}^{+\\infty }\\left( \\ln t\\right) t^{x-1}e^{-t}dt$[/quote]\n\nWell, I didn't know that. Is it hard to prove?[/quote]\r\nNo, you just apply a theorem , I think in english it's called \"DERIVATION UNDER THE INTEGRAL SIGN\" \r\nTry some google with it if you don't know it", "Solution_8": "The difference is really very minor. Kent uses the knowledge of a functional equation to skip a step, and if you want to apply derivation under the integral sign you still must motivate it with dominated convergence.", "Solution_9": "One can use the Weierstrass' formula :\r\n\\[\\forall x > 0, \\quad \\frac 1{\\Gamma(x)}= xe^{\\gamma x}\\prod_{n=1}^{+\\infty}\\left[\\left(1+\\frac{x}{n}\\right)\\right]\\]\r\n\r\nto prove that\r\n\\[\\forall x > 0, \\quad \\frac{\\Gamma'(x)}{\\Gamma(x)}=-\\gamma-\\frac{1}{x}+\\sum_{n=1}^{+\\infty}\\frac x{n(x+n)}.\\]\r\n\r\nHence, for $x = 1$, since $\\Gamma(1) = 1$,\r\n\\[\\Gamma'(1) =-1-\\gamma+\\sum_{n=1}^{+\\infty}\\frac 1{n(n+1)}=-\\gamma.\\]" } { "Tag": [ "MATHCOUNTS", "summer program", "Mathcamp" ], "Problem": "I dont do any programs outside of school and mathcounts so i only have an academic recomendation and just got a second recomendation from my 7th grade teacher is that ok? I mean i think i got 5 test questions and i dont want to not go to camp just because i dont have an outside recomendation.", "Solution_1": "[quote=\"peregrinefalcon88\"]I dont do any programs outside of school and mathcounts so i only have an academic recomendation and just got a second recomendation from my 7th grade teacher is that ok? I mean i think i got 5 test questions and i dont want to not go to camp just because i dont have an outside recomendation.[/quote]\r\n\r\nCan you get a second recommendation from someone who knows you in a non-academic context? That doesn't mean Mathcounts. It could be a friend of the family, or any adult who knows you outside of school. If you absolutely don't have anyone outside of school that's OK, but try to think more broadly before you resort to another teacher!", "Solution_2": "I have already gotten a recomendation from another teacher but I really dont know any of my moms friends. She has so many of them and all i really have to do when they come over is say hello and goodbye. I do have some friends from school but the only friends whose parents i know dont speak english so i dont think i can ask their parents to write me a recomendation either.", "Solution_3": "Do you guys think parents can be recommenders?", "Solution_4": "it can also be a sports coach .... if you play any sort of sports... \r\ni think [private] music teacher might work?", "Solution_5": "Let me make this clear before you all give more suggestions\r\n\r\nI wake up then go to school, go to mathcounts on mondays, come home, do homework and anything else my mom needs then do whatever i want if i have time. Thats my average day.", "Solution_6": "I got my non-academic recommendation from my next-door neighbor. :| \r\n\r\nOut of curiosity, is your Mathcounts coach the one writing the academic recommendation? If you must use a teacher for the non-academic letter, they would probably be the best choice.\r\n\r\nIf that fails . . . the Mathcamp staff is a pretty polyglot group :) . What language do your friend's parents speak?", "Solution_7": "[quote=\"E^(pi*i)=-1\"]I got my non-academic recommendation from my next-door neighbor. :| \n\nOut of curiosity, is your Mathcounts coach the one writing the academic recommendation? If you must use a teacher for the non-academic letter, they would probably be the best choice.\n\nIf that fails . . . the Mathcamp staff is a pretty polyglot group :) . What language do your friend's parents speak?[/quote]\r\n\r\nThese are all great suggestions. If all else fails, use the MathCounts coach as a nonacademic recommendation and make sure to give them the instructions for the nonacademic recommender!\r\n\r\nGood luck!", "Solution_8": "My moms friend speaks chinese but i gave my nonacademic recommendation to my 7th grade math teacher and my academic recomendation to my mathcounts teacher just because my mathcounts teacher probably knows me better academicly and i already asked my 8th grade teacher for a recomendation so i dont want to bother her again but it shouldnt make a difference because there are so few people in almost every class in my school who dont talk during the lesson :P", "Solution_9": "What if my badminton coach happens to be my social sciences teacher? does that count? coz he wrote a bunch of stuff about my academic too", "Solution_10": "my \"outside\" recommendation is from my history teacher, since her period with me is probably the time of the day where i am least concerned with math^^\r\nBut i assume that since she has probably the best shot at assessing my non-math related maturity it shouldn't be that big of a deal.\r\n\r\nIf im wrong, tell me, I can get more people to send in recs.", "Solution_11": "[quote=\"JasmineMoMo\"]What if my badminton coach happens to be my social sciences teacher? does that count? coz he wrote a bunch of stuff about my academic too[/quote]\r\n\r\nThat's fine, so long as he talks about out-of-school stuff as well.", "Solution_12": "[quote=\"e=mc2.kroll\"]my \"outside\" recommendation is from my history teacher, since her period with me is probably the time of the day where i am least concerned with math^^\nBut i assume that since she has probably the best shot at assessing my non-math related maturity it shouldn't be that big of a deal.\n\nIf im wrong, tell me, I can get more people to send in recs.[/quote]\r\n\r\nOur goal is more \"out of school\" than \"non-math;\" it would be nice if you could provide one from someone who's interacted with you outside of school." } { "Tag": [ "linear algebra", "matrix", "vector", "superior algebra", "superior algebra solved" ], "Problem": "assume that $A,B$ be orthognals matrices and we know that $det(A)+det(B)=0$ prove or disprove that $det(A+B)=0$", "Solution_1": "The determinant of a real orthogonal matrix is either 1 or -1, so the condition that $\\det A + \\det B = 0$ is the same as $\\det A \\det B = -1$, or $\\det A^TB = -1$. The eigenvalues of an orthogonal matrix are either 1, -1, or non-real numbers of magnitude 1. The last come in conjugate pairs with product 1, so any orthogonal matrix with determinant -1 must have some eigenvalue be -1.\r\nSince $A^TB$ is orthogonal with determinant -1, there is some vector $x$ such that $A^TBx = -x$. This gives $Bx=-Ax$, so $(A+B)x=0$ and $\\det(A+B)=0$." } { "Tag": [ "calculus", "integration", "inequalities", "function", "Support", "analytic geometry", "graphing lines" ], "Problem": "f:R--->R a function $ C^{2}$ \r\nwith \r\n\r\n$ \\int_{R}f^{2}, \\int_{R}f''^{2}$ are finite \r\n\r\nprove that $ \\int_{R}f'^{2}\\leq \\sqrt{\\int_{R}f^{2} \\int_{R}f''^{2}}$", "Solution_1": "If $ f$ also has compact support, then we can use integration by parts $ \\int_{\\mathbb R} f'(x)f'(x)dx\\equal{}\\minus{}\\int_{\\mathbb R} f(x)f''(x)dx$ followed by Cauchy-Schwarz. In general we get a boundary term $ ff'\\bigg|_{a}^b$, which does not necessarily tend to zero as $ a\\to\\minus{}\\infty$ and $ b\\to\\infty$. However, all we need is $ \\liminf_{x\\to\\infty} |f(x)f'(x)|\\equal{}0$, which can be established as follows. If $ f(x)f'(x)\\ge \\epsilon>0$ for all $ x>b$, then the slope of $ f^2$ in this interval is at least $ 2\\epsilon$, which contradicts the finiteness that $ \\int_b^{\\infty}f(x)^2\\,dx$. If $ f(x)f'(x)<\\minus{}\\epsilon$ for all $ x>b$, then $ f^2$ must eventually become negative, a contradiction. Thus $ \\liminf_{x\\to\\infty} |f(x)f'(x)|\\equal{}0$." } { "Tag": [ "geometry", "circumcircle", "angle bisector", "geometry unsolved" ], "Problem": "Let $ D$ be the point on the base $ BC$ of an isosceles triangle $ ABC$ such that $ BD : DC \\equal{} 2 : 1$, and let $ P$ be the point on the segment $ AD$ such that $ m(BAC) \\equal{} m(BPD)$. Prove that $ m(DPC) \\equal{} m(BAC)/2$.", "Solution_1": "[hide]Let the circumcircle of triangle $ BPD$ intersects side $ AB$ at $ E$ ($ E\\ne B$) and let $ F$ be the point on side $ AC$ such that $ AE\\equal{}AF$. Also let $ \\angle ABC\\equal{}\\angle ACB\\equal{}x$. Note that $ \\triangle AEF$ is also an isosceles triangle, so $ \\angle AEF\\equal{}\\angle AFE\\equal{}x$. We have $ \\angle APE\\equal{}180^{\\circ}\\minus{}\\angle EPD\\equal{}\\angle ABC\\equal{}\\alpha$. Therefore $ A,F,P,E$ lie on a circle. By Miquel's Theorem, $ F,C,P,D$ also lie on a circle. Thus $ \\angle DPC\\equal{}\\angle DFC$. Let $ M$ be the midpoint of $ BC$. Note that $ \\angle DEB\\equal{}\\angle DPB\\equal{}\\angle CAB$, so $ DE\\parallel AC$, and hence $ AE: AB\\equal{}CD: CB\\equal{}1: 3$. Hence we have $ CD: CM\\equal{}2: 3\\equal{}CF: CA$, so $ FD\\parallel AM$, which gives $ FD\\perp DC$. So $ \\angle DFC\\equal{}90^{\\circ}\\minus{}x$. Hence $ \\angle DPC\\equal{}90^{\\circ}\\minus{}x\\equal{}\\frac{180^{\\circ}\\minus{}2x}2\\equal{}\\frac{\\angle BAC}2$. Q.E.D.[/hide]", "Solution_2": "If $ AD$ intersects the circle $ (ABC)$ at $ \\{ A, M \\}$ and $ E$ is the midpoint of $ BM$, the triangle $ \\triangle BPM$ is isosceles, $ MD$ is the angle bisector of $ \\angle BMC$, hence $ MC \\equal{} ME$, and $ PEMC$ is a kite, but $ PE$ is the angle bisector of $ \\angle BPD$, done.\r\n\r\nBest regards,\r\nsunken rock\r\n\r\nP.S. Very nice solution, [b][i]Johan[/i][/b]!", "Solution_3": "We can see that $ P$ is the intersection of $ AD$ with the circle $ (ABQ)$, $ Q$ being the intersection of the tangents to the circle $ (ABC)$. From the equality $ \\angle AQP \\equal{} \\angle ABP \\equal{} \\angle PAC$ we conclude that the quadrilaterals $ AQBP$ and $ CABE$ ( $ E$ being the intersection of $ AD$ with the circle $ (ABC)$ ) are similar, hence $ \\frac{AP}{PB}\\equal{}\\frac{CE}{BE}$, but $ \\frac{CE}{BE}\\equal{}\\frac{CD}{BD}\\equal{}\\frac{1}{2}$, consequently $ BP\\equal{}2\\cdot AP$ ( 1 ).\r\nIf $ M$ is the midpoint of $ BP$, then $ \\triangle ABM \\equal{} \\triangle CAP$, $ \\measuredangle AMP\\equal{}\\measuredangle CPD$ ( 2 ).\r\nBut from $ AP \\equal{} PM$, $ \\measuredangle BAC\\equal{}\\measuredangle DPB \\equal{} 2 \\cdot \\measuredangle AMP\\equal{}\\measuredangle CPD$, hence $ \\measuredangle CPD \\equal{}\\frac{\\measuredangle BAC}{2}$.\r\n\r\nBest regards,\r\nsunken rock", "Solution_4": "The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=39978]link[/url] contains the inverse of the problem asked here.", "Solution_5": "My solution:\n\nLet $ Q \\in BP $ satisfy $ BQ=AP $ .\n\nSince $ \\angle QBA=\\angle BPD-\\angle BAD=\\angle BAC-\\angle BAD=\\angle PAC $ ,\nso we get $ \\triangle ABQ \\cong \\triangle CAP $ . ... $ ( \\& ) $\nSince $ \\text{S}[\\triangle ABP]: \\text{S}[\\triangle ACP]=BD:DC=2:1 $ ,\nso combine with $ ( \\& ) $ we get $ Q $ is the midpoint of $ BP $ and $ \\angle QAP=\\angle PQA $ ,\nhence $ \\angle BAC=\\angle BPD=2\\angle PQA=2\\angle DPC $ .\n\nQ.E.D", "Solution_6": "This is an easy question", "Solution_7": "Here is my solution: (I am not yet acquainted with LateX so I posted a pic of my solution, I am learning LateX)" } { "Tag": [ "trigonometry", "geometry", "angle bisector", "similar triangles", "projective geometry", "geometry proposed" ], "Problem": "Let $ ABC$ be a right triangle with $ \\angle BAC\\equal{}90^{\\circ}$. Let $ AD$ be the angle bisector of $ \\angle BAC$, $ K$ point on $ AD$ such that $ \\angle BKA\\equal{}90^{\\circ}$ and $ H$ the foot of $ A$-altitude. Let $ X$ be the intersection point of $ AB$ and $ HK$. Prove that the intersection point of $ DX$ and $ AC$ is the midpoint of $ AC$.", "Solution_1": "1)suppose that $ ACAB$ by the same way we find $ DVI at the top to LaTeX->PDF. Do Ctrl-F7 then F5. Did that work?", "Solution_2": "Deleting them is fine - they'll be recreated every time you compile the file.\r\n\r\nYou probably won't ever be creating DVI files, but if you do, do this in TeXnicCenter:\r\n\r\n1) Click Build. \r\n2) Click Define Output Profiles\r\n3) Select LaTeX => DVI on the left\r\n4) Click the Viewer tab\r\n5) Put C:\\TEXMF\\MIKTEX\\BIN\\YAP.EXE in the 'Path of executable' field.", "Solution_3": "If you're going to be submitting paper (as opposed to electronic) copies of your documents, DVI is fine. DVI generation is loads faster than PDF generation, and you don't have to shut down YAP every time you want rebuild the document.", "Solution_4": "[quote=\"Osiris\"]If you're going to be submitting paper (as opposed to electronic) copies of your documents, DVI is fine. DVI generation is loads faster than PDF generation, and you don't have to shut down YAP every time you want rebuild the document.[/quote]\r\nNor do you have to shut down adobe reader every time, at least if you're using 6.0.", "Solution_5": "That's interesting. My Adobe Reader seems to lock-read the document, and TeXNicCenter then reports a write failure...", "Solution_6": "[quote=\"confuted\"][quote=\"Osiris\"]If you're going to be submitting paper (as opposed to electronic) copies of your documents, DVI is fine. DVI generation is loads faster than PDF generation, and you don't have to shut down YAP every time you want rebuild the document.[/quote]\nNor do you have to shut down adobe reader every time, at least if you're using 6.0.[/quote]\r\n\r\nThis is one of the things I really like about TeXnicCenter over my old emacs and DOS solution.", "Solution_7": "I downloaded TeXnicCenter today... what a difference from DOS!" } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "f(1),f(2),...........f(n) is a sequence of real numbers.\r\n\r\n f(1) = 2.\r\n\r\n [ f(n+1) ] [ f(1) + f(2) +..................+ f(n) ] = n for all n (natural number)\r\n\r\n Prove that f(2009) > 0.9995\r\n\r\n Sir,please SEND the solution to me [ nuthetishivdeep@gmail.com ] .", "Solution_1": "You already posted this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=302261]here[/url] and it is a current problem . Mods , lock this topic ." } { "Tag": [ "geometry", "algebra", "polynomial" ], "Problem": "A monster buys a camera and a case worth 1/2 of the camera's cost, an ice cream worth 1/2 of the case's value, and a cookie for 1/2 of the ice cream's value, all for $ \\$$105. How much did the cookie cost?\r\n\r\n$ \\textbf{(A)} \\$7\\qquad\\qquad\\textbf{(B)}\r\n\\$14\\qquad\\qquad\\textbf{(C)} \\$56\\qquad\\qquad\\qquad\\textbf{(D)} \r\n\\$57.25\\qquad\\qquad\\qquad\\textbf{(E)} \\$105$\r\n\r\n\r\nHow many distinct solutions exist for the equation $ 2x^3 \\minus{} 5x^2 \\plus{} 4x \\minus{} 1 \\equal{} 0$?\r\n\r\n$ \\textbf{(A)} 0\\qquad\\qquad\\qquad\\textbf{(B)} 1\\qquad\\qquad\\qquad\\textbf{(C)} 2\\qquad\\qquad\\qquad\\textbf{(D)} 3\\qquad\\qquad\\qquad\\textbf{(E)} 4$\r\n\r\n\r\nWhat is the area of a polygon with vertices at: (0,0), (3,4), (6,0), (3,-4), and (0,3)?\r\n\r\n$ \\textbf{(A)} 4.5\\qquad\\qquad\\qquad\\textbf{(B)} 19.5\\qquad\\qquad\\textbf{(C)} 21.5\\qquad\\qquad\\qquad\\textbf{(D)} 24.0\\qquad\\qquad\\qquad\\textbf{(E)} 28.5$\r\n\r\n\r\nWhat is the average of all two-digit integers that satisfy the condition that the number is equal to the sum of the ten's digit and the unit's digit and the product of the ten's digit and the unit's digit?\r\n\r\n$ \\textbf{(A)} 56\\qquad\\qquad\\qquad\\textbf{(B)} 57\\qquad\\qquad\\qquad\\textbf{(C)} 58\\qquad\\qquad\\qquad\\textbf{(D)} 59\\qquad\\qquad\\qquad\\textbf{(E)} 60$\r\n\r\n\r\nWhat is the units digit of the following sum:\r\n\\[ (1!)^2 \\plus{} (2!)^2 \\plus{} \\dots \\plus{} (1337!)^2?\r\n\\]\r\n$ \\textbf{(A)} 5\\qquad\\qquad\\qquad\\textbf{(B)} 6\\qquad\\qquad\\qquad\\textbf{(C)} 7\\qquad\\qquad\\qquad\\textbf{(D)} 9\\qquad\\qquad\\qquad\\textbf{(E)} 0$", "Solution_1": "[hide=\"Problem 11\"]This is just a lot of smaller and smaller fractions. Doing the division results in $ \\boxed{\\textbf{A}}$.[/hide]\n\n[hide=\"Problem 12\"]The quick reaction is to notice that the polynomial is to the third power so it has 3 roots. [b]However[/b], this polynomial has a double root (try root locating and see). Hence, this polynomial only has 2 distinct solutions, so the answer is $ \\boxed{\\textbf{C}}$.[/hide]\n\n[hide=\"Problem 13\"]We divide this into triangles, or we can use Pick's theorem. Either method results in $ 28.5$, or $ \\boxed{\\textbf{E}}$.[/hide]\n\n[hide=\"Problem 14\"]We add least and greatest or whatever and divide by two blah answer is $ \\boxed{\\textbf{D}}$.[/hide]\n\n[hide=\"Problem 15\"]We note that above $ 4!$, the units digit is 0. We compute the last digits of $ (1!)^2\\plus{}(2!)^2\\plus{}(3!)^2\\plus{}(4!)^2$, and our result is $ 7\\implies\\boxed{\\textbf{C}}$.[/hide]" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "MATHCOUNTS", "complementary counting" ], "Problem": "Poll above.\r\n\r\nAs for me, No for AIME, Yes for Nationals", "Solution_1": "You still have 4 years, no worries.\r\nAnyways, yeah, no Nats for me.", "Solution_2": "qualified for both :lol: \r\n\r\nhmm, andrew, jonathan, allison, peijin, jason, me, didn't they all go to chapter?", "Solution_3": "[quote=\"ZhangPeijin\"]Anyways, you still have 4 years to qualify, no worries.[/quote]\r\n\r\nTwo of which will give me a higher chance of qualifying for USAMO :|", "Solution_4": "I qualified for... All three of them... 1, 2, ... oh wait. That just shows how bad I am at counting :( . Peijin is real nice. I should start making fun of him.", "Solution_5": "Yeah, you should complement about how awesome I am, and such.", "Solution_6": "oh yeah, you couldn't count to 5, we should definitely bow down to your awesomeness now", "Solution_7": "Yup, lets see you count 5 dots on a piece of messy paper when you're stressed and you only have 1 minute left on the AMC 10B.", "Solution_8": "I meant so say complement because of the math things and such.\r\nLike the complement of an angle?\r\n-laughter-\r\nAha...", "Solution_9": "[quote=\"ZhangPeijin\"]Yup, lets see you count 5 dots on a piece of messy paper when you're stressed and you only have 1 minute left on the AMC 10B.[/quote]\r\n\r\nWhen I count five dots on a piece of messy paper, I say, \"1, 2, 3, 4, 5\", not ... erm... how else do you count to 5?\r\n\r\nHow many ways can you count from 1 to 5 using 5 numbers from 1 to 10 at most once? (ex. 1, 9, 8, 7, 5)?\r\n\r\nI can always conjure MATHCOUNTS style questions from thin air! :D", "Solution_10": "[quote=\"Yongyi781\"][quote=\"ZhangPeijin\"]Yup, lets see you count 5 dots on a piece of messy paper when you're stressed and you only have 1 minute left on the AMC 10B.[/quote]\n\nWhen I count five dots on a piece of messy paper, I say, \"1, 2, 3, 4, 5\", not ... erm... how else do you count to 5?\n\nHow many ways can you count from 1 to 5 using 5 numbers from 1 to 10 at most once? (ex. 1, 9, 8, 7, 5)?\n\nI can always conjure MATHCOUNTS style questions from thin air! :D[/quote]\r\nNo I mean as I miss one dot, and say 4, and get the question wrong.\r\nThen I look back at my scrap and find 5 dots.\r\n...", "Solution_11": "[quote=\"Yongyi781\"]How many ways can you count from 1 to 5 using 5 numbers from 1 to 10 at most once? (ex. 1, 9, 8, 7, 5)?[/quote]\r\n\r\n1 and 5 are set, so rest is 7C3 so 35. :lol: \r\n\r\nWell anyways, I've never done the AMC's I'm planning to next year, but I'm going to nats.", "Solution_12": "Hmm... The first term would be 1, the last term would be 5. Thus, we can choose from 7 terms to fill in the three. Thus, it would be 7x6x5=$ \\boxed {210}$ ways. :P Or at least I think that's right...\r\n\r\nAaaaaaanyways, I qualified for both, which for my state was pretty easy to accomplish. \r\n\r\nHere's an MC-style problem to think about:\r\nHow many ways can you count from 1 to 5 using 5 numbers from 1 to 10 at most once, where integers next to each other in the sequence cannot have a positive difference of 1? Muhahahaha! An even harder question based on Peijin's inability to count! (no offense)\r\n\r\nEDIT: Beaten. Wait, never mind. Ihatepie, order matters in this case, so it's permutations rather than combinations. Ha!", "Solution_13": "Oops :blush: \r\n\r\nabout yours, 1 and 5 are set.\r\n\r\n1,3,6,8,5\r\n1,3,6,9,5\r\n1,3,6,10,5\r\n1,3,6,2,5\r\n1,3,7,2,5\r\n1,3,7,9,5\r\n1,3,7,10,5\r\n1,3,8,2,5\r\n1,3,8,10,5\r\n1,3,9,2,5\r\n1,3,9,7,5\r\n1,3,10,2,5\r\n1,3,10,3,5\r\n1,3,10,7,5\r\n1,3,10,8,5\r\n1,4,2,7,5\r\n1,4,2,8,5\r\n1,4,2,9,5\r\n1,4,2,10,5\r\n1,4,6,2,5\r\n1,4,6,3,5\r\n1,4,6,8,5\r\n1,4,6,9,5\r\n1,4,6,10,5\r\n1,4,7,2,5\r\n1,4,7,3,5\r\n1,4,7,9,5\r\n1,4,7,10,5\r\n1,4,8,2,5\r\n1,4,8,3,5\r\n1,4,8,10,5\r\n1,4,9,2,5\r\n1,4,9,3,5\r\n1,4,9,7,5\r\n1,4,10,2,5\r\n1,4,10,3,5\r\n1,4,10,7,5\r\n1,4,10,8,5\r\n1,6,2,7,5\r\n1,6,2,8,5\r\n1,6,2,9,5\r\n1,6,2,10,5\r\n1,6,3,8,5\r\n1,6,3,9,5\r\n1,6,3,10,5\r\n1,6,4,2,5\r\n1,6,4,7,5\r\n1,6,4,8,5\r\n1,6,4,9,5\r\n1,6,4,10,5\r\n1,6,8,2,5\r\n1,6,8,3,5\r\n1,6,8,10,5\r\n1,6,9,2,5\r\n1,6,9,3,5\r\n1,6,9,7,5\r\n1,6,10,2,5\r\n1,6,10,3,5\r\n1,6,10,7,5\r\n1,6,10,8,5\r\n1,7,\r\n\r\noh, I give up.", "Solution_14": "Muhahahahahaha, my problem totally pwnages!!!! Mwahahahahaha! Sorry... Hmm, anyone else wanna try? :lol:", "Solution_15": "1 and 5 are set.\r\nThe middle 3 can be any numbers right?\r\nSo its just $ 9\\cdot8\\cdot7\\equal{}504$", "Solution_16": "I would say taking 210 and subtracting all the cases where at least one pair consecutive integers have a difference of 1... but I'm too lazy to do that now :P\r\n\r\nOh, and by the way,\r\n\r\n[quote=\"ZhangPeijin\"]1 and 5 are set.\nThe middle 3 can be any numbers right?\nSo its just $ 9\\cdot8\\cdot7 \\equal{} 504$[/quote]\r\n\r\nBut 1 and 5 are already taken, so there are 8 options left (7 if you don't count 0). So it's $ 8\\cdot7\\cdot6 \\equal{} 336$ or $ 7\\cdot6\\cdot5 \\equal{} 210$. I assumed that 0 was not allowed, but you could've answered both ways.", "Solution_17": "[quote=\"ZhangPeijin\"]1 and 5 are set.\nThe middle 3 can be any numbers right?\nSo its just $ 9\\cdot8\\cdot7 \\equal{} 504$[/quote]\r\nOh, sorry, forgot to mention numbers must be distinct, or did I mention that? Too lazy to scroll up to check... Hmm, Yongyi, I think that's the best way to tackle it, but instead of listing so many, the number of opposite cases is less cuz you know that either the digits are all going up 1 or all going down 1. It's impossible to have, say 1, 2, 3, 2, 5 as one cuz all digits are distinct. That makes things easier.", "Solution_18": "Meh, if you cant use 1, 5, or 0, then it is $ 7\\cdot6\\cdot5$\r\nAre you guys saying there's some other limitation?", "Solution_19": "The numbers next to each other cannot differ by 1. Read my post, not Yongyi's for the question.", "Solution_20": "[quote=\"Math Geek\"]where integers next to each other in the sequence cannot have a positive difference of 1? [/quote]\r\n\r\nwell....", "Solution_21": "I made AIME, but state's on April 5th since it got rescheduled.", "Solution_22": "Oh...\r\nAnyways, gotta do my S&D project thats due tomorrow, solve it later.", "Solution_23": "Hah, if I don't make USAMO this year that will be a major fail. I almost made it last year...", "Solution_24": "The whole of california sends only four to the nationals. Our state is very hard to qualify for nationals.", "Solution_25": "And so is Texas.", "Solution_26": "I take it Cali and Texas are two of the hardest states?\r\n..Aren't they?", "Solution_27": "Wait... for the counting problem it said numbers from 1-10. If you can't use 1 and 5 that leaves 8, not 7.", "Solution_28": "I don't even take AMCs or AIMEs, and I never made Nats. Unfortunately, so sad :mad: :(", "Solution_29": "Start from the back... The number before $ 5$ cannot be $ 4,6$. There are thus $ 6$ choices for that number.\r\n\r\nIf it is $ 2$, then the number before it can be $ 4,6,7,8,9,10$\r\n\r\nIf it is $ 10$, it can be $ 2,3,4,6,7,8$\r\n\r\nFor the other $ 4$ possibilities, there are only $ 5$ other choices since they each get rid of the two numbers next to it. This averages to $ 5.4$ =P\r\n\r\n$ 6*5.4*?$\r\n\r\nDepending on what the third digit was, we can find the second digit, which cannot be 2. Err... ya xD. I would just assume that the ? ends up being 5 since it should be an integer. =). Then the answer is: $ 162$. lol. Probably wrong though since i'm really tired right now\r\n\r\nOr...\r\nComplementary counting, but it still takes too long =P" } { "Tag": [ "number theory", "greatest common divisor", "relatively prime" ], "Problem": "Hey,\r\nI have been a lurker here for a while, but just finally registered because I cannot, for the life of me, figure out how to do this problem. It probably is pretty easy, but I just don't see it. Thanks for your help.\r\n\r\nLet m and n be positive integers. Show that the smallest positive integer a such that m divides kn is k = m/gcd(m,n).\r\n\r\nThanks.", "Solution_1": "Do you mean $k$ instead of $a$? If so, then [hide=\"here is my take on the problem\"]\nNote that this choice of $k$ will always work, it remains only to show that there are no smaller $k$.\n\nLet the product of all the primes that divide $m$ but do not divide $n$ be $q$, and the product of all the primes that divide both $m$ and $n$ be $r$. (repeats included on both occasions). \nNow $\\gcd(m,n)=r$ and $qr=m$. We know that $qr|kn$ and since $\\gcd(q,n)=1$ we have $q|k$. Hence $k \\geq q$. But if $k=\\frac{m}{\\gcd(m,n)}$ then $k=q$, so this is clearly the minimum.[/hide]", "Solution_2": "Thanks, I see how it works now. One more problem that I'm working on is this one:\r\n\r\nSuppose b, m and n are positive integers. If m == n (mod b), prove that gcd(b,m) = gcd(b,n). \r\n\r\nAny input?", "Solution_3": "$m=db+c$\r\n$n=fb+c$\r\n\r\nSuppose a prime $p$ divides $b$. \r\n\r\nThen it divides $db$ and so divides $m$ is and only if it divides $c$. Also, it divides $fb$ and so divides $n$ if and only if it divides $c$. Hence the set of primes that divide both $\\gcd(m,b)$ and $\\gcd(n,b)$ are the same, and by considering whether or not $p^{2}$ divides $c$ etc. we get that $\\gcd(m,b)=\\gcd(n,b)$.", "Solution_4": "$m=n(\\mod{b}) \\rightarrow \\exists{x}\\in \\mathbb{Z}: bx+m=n$\r\n\r\nSo $\\gcd{(b,n)}=\\gcd{(b,bx+m)}=\\gcd{(b,m)}$", "Solution_5": "Ok, I understand all the steps in that second response except the last one. How do you get gcd(b,bx+m) = gcd (b,m)?", "Solution_6": "We know that $\\gcd(a,b)=\\gcd (a,b-a)$\r\n\r\nTo see why, say $d=\\gcd(a,b)$ That means there is an $n$ and $m$ such that $a=dn$ and $b=dm$, and that $\\gcd(n,m)=1$\r\n\r\nSo $\\gcd(a,b-a)=\\gcd(dn,d(m-n))=d\\gcd(n,m-n)$\r\n\r\nBut can you see why $\\gcd(n,m-n)=1$? Because they're relatively prime!\r\nSo we get $\\gcd(a,b)=\\gcd(a,b-a)$\r\n\r\nReturning to our problem: We have \r\n$\\gcd(b,bx+m)=$\r\n$\\gcd(b,bx+m-b)=$\r\n$\\gcd(b,b(x-1)+m)=$\r\n$\\gcd(b,b(x-1)+m-b)=$\r\n$\\gcd(b,b(x-2)+m)=$\r\n$\\cdots=$\r\n$\\gcd(b,b(x-x)+m)=$\r\n$\\gcd(b,m)$\r\n\r\nSorry if this is confusing, its late here in Ireland and im flying to Switzerland in the morning and need to hunt for my passport. Seamus out!", "Solution_7": "Hmmm, well that still seems strange to me. The hint in the back of the book (it's an old textbook I got for hardly anything, can't beat that) say to use the definition of GCD to show that each of them are less than or equal to the other...how would I do that?", "Solution_8": "Anyone? This problem has been driving me crazy all night.", "Solution_9": "[quote=\"redfalcon\"]Thanks, I see how it works now. One more problem that I'm working on is this one:\n\nSuppose b, m and n are positive integers. If m == n (mod b), prove that gcd(b,m) = gcd(b,n). \n\nAny input?[/quote]\r\n\r\n$n=b\\cdot k+m$ for some integer $k$ using the mod condition\r\n\r\n$\\gcd(b,n)=\\gcd(b,b\\cdot k+m)=\\gcd(b,m)$\r\n\r\n(since $\\gcd(a,b)=\\gcd(a,b-a)$)", "Solution_10": "Ok, but then how do I prove that gcd(a,b)=gcd(a,b-a)? Also, how is it done with the definition thing because that has me baffled?\r\n\r\nThanks to everyone for all your help too.", "Solution_11": "To prove that $\\gcd(a,b)=\\gcd(a,b-a)$ just consider all the factors of $\\gcd(a,b)$.\r\n \r\nThere is an elementary rule in number theory that states that if $d|m$ and $d|n$ then $d|mx+ny$ for all integers $x,y$.\r\n\r\nNow since $d|a$ and $d|b$ ($d=\\gcd(a,b)$) $d|a-b$ (here $m=1$,$n=-1$). \r\n\r\nConversely, if $d|a$ and $d|b-a$ then $d|a+(b-a)=b$\r\n\r\nHence $\\gcd(a,b)=\\gcd(a,b-a)$" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "A solitaire game is played with an even number of discs,each coloured red on one side and green on the other side.Each disc is also numbered,and there are two of each number;i.e $\\{1,1,2,2,3,3,...,N,N\\}$ are the labels.The discs sre laid out in rows with each row having at least three discs.A move in this game consists of flipping over simultaneously two discs with the same label.Prove that for every initial deal or lay out there is a sequence of moves that ends with a position in which no row has only red or only green sides showing.", "Solution_1": "First we exclude inductively the case when, for some $k$, both $k$ discs are in the same row.\r\n\r\nIf they're in a 3-disc row (two $k$ discs and an $m$ disc), remove the row and place an $m$ disc in the row containing the other $m$ disc, so that they're of the same color. Solve this case inductively. Then isolate one of the $m$ discs and put it together with two $k$ discs to recreate our row. Clearly the row containing the other $m$ doesn't fail, and we can fix the new row.\r\nIf they're in a 4-disc row of the form $k, k, m, m$ we just remove the row and proceed inductively in the obvious way. If they're in a 4-disc row of the form $k, k, m, l$, place the $l$ disc in the row containing the other $l$, solve this case by the above (we'd done it for $k, k, m$) and put back the $l$. If they're in a 5-disc row, just remove both $k$ discs and proceed inductively.\r\n\r\nSo these cases are done. Now suppose no two discs with the same number are in the same row. We will prove that, if there's a non-zero number of monochromatic rows, we can eventually decrease this number; and this will be enough.\r\n\r\nTake a monochromatic row (we'll call it an F-row, for forbidden) and switch one of its discs. If the other disc with that number is in a row that was either F or doesn't become F because of this move, we've decreased the number of F-rows. Suppose that the row we just fixed is $r_1$, and we made some other row $r_2$ F. Fix $r_2$ (not switching the same number!). If we don't mess up any new rows, we win. Otherwise, fix the new messup $r_3$, and so on. Eventually (after a finite number of moves), either we will win or we will get some $i2, (a,m)=1 and a is a quad residue mod m. prove that the congruence x^2\u2261a (mod m) has exactly 2 solutions if and only if m has a primitive root.\r\n\r\nunknown source", "Solution_1": "Let $m=p_{1}^{k_{1}}...p_{s}^{k_{s}}$, then the ring \\[Z/mZ=\\sum_{i=1}^{s}Z/p_{i}^{k_{i}}.\\]\r\nBecause $(Z/p^{k}Z)^{*}$ had even elements if p is odd or p=2,k>1, number of solutions $x^{2}=a(mod \\ m)$ is $2^{s}$ if m is odd or 4|m, else (when m=2(mod 4)) number of solutions is $2^{s-1}$.", "Solution_2": "[quote=\"Keno20\"]Assume m>2, (a,m)=1 and a is a quad residue mod m. prove that the congruence x^2\u2261a (mod m) has exactly 2 solutions if and only if m has a primitive root.\n\nunknown source[/quote]\r\n\r\nThat isn't true.\r\n\r\nSay, $m=35$.\r\nNow by Gauss' theorem this doesn't have primitive roots.\r\nBut it sure does have solutions to,\r\n$x^{2}\\equiv a \\mod 35$ for some $\\gcd(a,35)=1$\r\nNow if $y$ is solution then $m-y\\not \\equiv y$ is a solution.", "Solution_3": "[quote=\"sylow_theory\"]But it sure does have solutions to,\n$x^{2}\\equiv a \\mod 35$ for some $\\gcd(a,35)=1$[/quote]\r\n\r\nThe problem specifies [b]exactly two[/b] solutions. There are four, I think, in $\\mathbb{Z}_{35}$.", "Solution_4": "[quote=\"t0rajir0u\"][quote=\"sylow_theory\"]But it sure does have solutions to,\n$x^{2}\\equiv a \\mod 35$ for some $\\gcd(a,35)=1$[/quote]\n\nThe problem specifies [b]exactly two[/b] solutions. There are four, I think, in $\\mathbb{Z}_{35}$.[/quote]\r\nAh! I see now.\r\n\r\nI think you are right about Z_35 having 4 solutions.\r\nBecause the polynomial ring Z_35[x] does not obey the rule that there are at most n=2 solutions. For Z_35 is not a field.", "Solution_5": "Prime factorization, CRT kills it, considering $\\mathbb{Z}_{m}$ has primitive roots iff $m$ is a power of a prime or double a power of a prime.", "Solution_6": "I don't know much about rings or fields, so my idea is a bit less nice.\r\n\r\nit's actually quite simple to show that no primitive roots mod m implies the residue equation has 3 or more solutions. \r\n\r\nfor the other way, just use the primitive root to express two solutions to the equation and show that they are the same which follows simply from some very basic properties of primitive roots." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that for all real positive numbers $ x,y,z$, the following inequality is true : \r\n\r\n$ \\frac {x}{13x \\plus{} y \\plus{} z} \\plus{} \\frac {y}{x \\plus{} 13y \\plus{} z} \\plus{} \\frac {z}{x \\plus{} y \\plus{} 13z} \\leq \\frac {1}{5}$", "Solution_1": "write the ineq as:\r\n\r\n$ \\sum{\\frac{y\\plus{}z}{13x\\plus{}y\\plus{}z}} \\geq \\frac{2}{5}$ and then cauchy shwarz :wink:", "Solution_2": "anas, your solution is wrong since a,b,c are[b] reals[/b], not positive reals.", "Solution_3": "I was wrong again , i edited :blush: Sorry great math .", "Solution_4": "That's alright, alex\r\n\r\nHere is my solution\r\n\r\n$ \\frac{a}{12a\\plus{}a\\plus{}b\\plus{}c} \\le \\frac{a}{25}\\left(\\frac{4}{3a}\\plus{} \\frac{1}{a\\plus{}b\\plus{}c}\\right)$" } { "Tag": [ "videos" ], "Problem": "Accordingly,Human there are 46 chromosomes. In\r\nman,x,y.And in woman double x.Each contain a complete \"[b]Genetic[/b] Material[b]\"[/b] that determine the humanly inheritance.Is there any further \"[b]Divisional Components[/b]\"[b](Ex.Genetic Sequencing Arrangement[/b].)taht explain the Physical Make-up of each human being?\r\n\r\n Respectfully,\r\n A.Eddington\r\n\r\n Postscript:\r\n Example Of Physical Attributes:\r\n 5 Feet,3 inches tall,Semi-Slender Body Built,Semi-Oblong Face,Thin Eye Brows,Expressive Eye Ball,Not So Thin Lip,\r\nWhite Skin,Cutely Curve Eye Laces,Black Short Hair,Proportional\r\nNose Figure.\r\n\r\n Same", "Solution_1": "Alright...\r\n\r\n-First off, each chromosome doesn't contain a complete copy of the genetic material. The entire set of chromosomes however, does.\r\n\r\n-The short answer to your question is \"yes\"; there are other things than just the nucleotide sequences that determine the \"physical make-up\" of a human being; things like differential gene expression, environmental interactions, alternative splicing, mitochondrial DNA, and (as always) random chance.\r\n\r\n-Re: your postscript. Many of those things are partially environmentally determined, like height, semi-slender body, etc.", "Solution_2": "Just a small bit to add, as this seems to relate to the question at hand, you might wish to read more about epigenetics and quantum biology.\r\n\r\nOffhand, I know of one video in which epigenetics is at least partially discussed (in reference to the feasibility of a cloned embryo forming 'normal' offspring): \r\n\r\nhttp://mitworld.mit.edu/video/138/\r\n\r\nIt requires RealPlayer to view, but I certainly thought it was an interesting lecture. It addresses some of the ethical and practical issues of 'cloning' and the like." } { "Tag": [ "inequalities", "LaTeX", "inequalities proposed" ], "Problem": "Prove that for positive reals $ a,b,c$ such that $ abc\\equal{}1$,\r\nthe following inequality holds-\r\n$ \\frac{a^3\\plus{}b^3\\plus{}c^3\\plus{}3}{4} \\ge \\frac{b}{a\\plus{}c}\\plus{}\\frac{c}{a\\plus{}b}\\plus{}\\frac{a}{b\\plus{}c}$", "Solution_1": "[quote=\"geniusbliss\"]Prove that for positive reals $ a,b,c$ such that $ abc \\equal{} 1$,\nthe following inequality holds-\n$ \\frac {a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3}{4} \\ge \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {a}{b \\plus{} c}$[/quote]\r\n$ \\frac {a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3}{4} \\ge \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {a}{b \\plus{} c}$\r\n$ \\frac {a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc}{4abc} \\ge \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {a}{b \\plus{} c}$\r\n$ \\frac {\\sum_{cyc}a^3\\plus{}3abc}{4abc} \\ge \\frac{\\sum_{cyc}a^3\\plus{}\\sum_{cyc}a^2(b\\plus{}c)\\plus{}3abc}{\\sum_{cyc}(a^2(b\\plus{}c))\\plus{}2abc}$\r\n$ \\frac {2(\\sum_{cyc}a^3\\minus{}abc)}{8abc} \\ge \\frac{\\sum_{cyc}a^3\\plus{}abc}{\\sum_{cyc}(a^2(b\\plus{}c))\\plus{}2abc}$\r\nbecause of the two inequalities\r\n$ 2(\\sum_{cyc}a^3\\minus{}abc) \\geq \\sum_{cyc}a^3\\plus{}abc$\r\n$ \\sum_{cyc}a^2(b\\plus{}c)\\plus{}2abc \\geq 8abc$\r\nso it has been solved.", "Solution_2": "[quote=\"geniusbliss\"]Prove that for positive reals $ a,b,c$ such that $ abc \\equal{} 1$,\nthe following inequality holds-\n$ \\frac {a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3}{4} \\ge \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {a}{b \\plus{} c}$[/quote]\r\n\r\nyour inequality is equivalent to\r\n$ \\frac {a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3}{4}\\minus{}\\frac{3}{2} \\ge \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {a}{b \\plus{} c}\\minus{}\\frac{3}{2}$\r\n\r\nand using the two flowing identites \r\n\r\n$ \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {a}{b \\plus{} c}\\minus{}\\frac{3}{2}\\equal{}\\sum_{cyc}\\frac{(a\\minus{}b)^2}{(b\\plus{}c)(c\\plus{}a)}$\r\n$ \\frac {a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3}{4}\\minus{}\\frac{3}{2}\\equal{}\\frac{a^3 \\plus{} b^3 \\plus{} c^3\\minus{}3abc}{4}\\equal{}\\frac{(a\\plus{}b\\plus{}c)(\\sum(a\\minus{}b)^2}{8}$\r\n\r\nand since $ \\frac{(a\\minus{}b)^2(a\\plus{}b\\plus{}c)}{8} \\ge \\frac{(a\\minus{}b)^2(a\\plus{}b)}{8} \\ge \\frac{(a\\minus{}b)^2}{(b\\plus{}c)(c\\plus{}a)}$\r\n\r\nbecause $ (a\\plus{}b)(b\\plus{}c)(c\\plus{}a) \\ge 8abc$\r\n\r\nwe are done !", "Solution_3": "[i]another soultion[/i]\r\n\r\nwe have by CS\r\n\r\n$ \\frac {a}{b} \\plus{} \\frac {a}{c} \\ge \\frac {4a}{b \\plus{} c}$\r\n\r\nand hence it sufficent to show that :\r\n\r\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc \\ge \\sum_{cyc}\\left(\\frac {a}{b} \\plus{} \\frac {a}{c}\\right) \\equal{} \\sum_{cyc}ab(a \\plus{} b)$\r\n\r\nwhich is schur :)", "Solution_4": "By AM-GM:\r\n$ \\frac{a^3\\plus{}b^3\\plus{}c^3\\plus{}3}{4}\\ge\\frac{a\\sqrt{a}\\plus{}b\\sqrt{b}\\plus{}c\\sqrt{c}}{2}\\equal{}\\frac{a}{2\\sqrt{bc}}\\plus{}\\frac{b}{2\\sqrt{ac}}\\plus{}\\frac{c}{2\\sqrt{ab}} \\ge\\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{a\\plus{}c}\\plus{}\\frac{c}{a\\plus{}b}$\r\nEasy! :)", "Solution_5": "Just an application of c.s inequality", "Solution_6": "[quote=\"akashram\"]Just an application of c.s inequality[/quote]\r\nwell humour me with the solution then..", "Solution_7": "I am sorry I am not able to use latex properly :D", "Solution_8": "And MRgeniusbliss I have not posted any solution withot working", "Solution_9": "geniusbliss, you could use the $ PM$ to ask Akash Ram to type humorous things for you to laugh(if it really is humorous) and AkashRam, you may also use the $ PM$ to inform geniusbliss about whether your post is genuine and whether you have worked out the solution or not, and it is advisable for you to learn a but of latex. Refer to the AoPs for latex and you can click on the latex formulae of other people's posts and look at the latex commands. I think as far as inequalities are concerned, you need to know how to use \r\n\\sigma, \\frac, \\sqrt frequently. Rest is all formulae using dollar symbols.\r\n\r\nBTW, Abdek, your proof is really nice, superb!!!" } { "Tag": [ "algebra", "polynomial" ], "Problem": "A polynomial P has four roots, 1/4, 1/2, 2, 4. The product of the roots is 1, and P(1) = 1.\r\nFind P(0).", "Solution_1": "$ p(x)\\equal{}a(x\\minus{}\\frac{1}{4})(x\\minus{}\\frac{1}{2})(x\\minus{}2)(x\\minus{}4)$, so $ p(1)\\equal{}1\\equal{}a\\cdot\\frac{3}{4}\\cdot\\frac{1}{2}\\cdot\\minus{}1\\cdot\\minus{}3 \\Rightarrow a\\equal{}\\frac{8}{9}$.\r\n$ p(0)\\equal{}\\frac{8}{9}\\cdot\\minus{}\\frac{1}{4}\\cdot\\minus{}\\frac{1}{2}\\cdot\\minus{}2\\cdot\\minus{}4\\equal{}\\frac{8}{9}$.", "Solution_2": "I wonder why the original problem gives you product of the roots, when you already know the four roots. :huh:" } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "question no.1\r\n$2x-y = 5$\r\nhow to find that x = -3y ?\r\nand how to draw the slope for this ?\r\n\r\ni hope i get fast reply.\r\n\r\nQuestion No.2\r\nyeah the second eqution is \r\nI. $2x-y = 5$ \r\nII. $3y-x = 15$", "Solution_1": "[quote=\"most-wanted\"]question no.1\n$2x-y = 5$\nhow to find that x = -3y ?\nand how to draw the slope for this ?\n\ni hope i get fast reply.\n\nQuestion No.2\nyeah the second eqution is \nI. $2x-y = 5$ \nII. $3y-x = 15$[/quote]\r\nI don't understand what you are saying for the first. For the second I am assuming you are looking for x and y, in which case\r\n[hide]Add y-5 to both sides to get\ny=2x-5. Punching this into the second equation, we have, 6x-15-x=15, which solves to x=6. Plugging back in, we get y=7.[/hide]", "Solution_2": "[quote=\"most-wanted\"]question no.1\n$2x-y = 5$\nhow to find that x = -3y ?\nand how to draw the slope for this ?\n\ni hope i get fast reply.\n\nQuestion No.2\nyeah the second eqution is \nI. $2x-y = 5$ \nII. $3y-x = 15$[/quote]\r\n[hide=\"1\"]$2x-y=5$ is in standard form. To find the slope, you put it into slope-intercept form. $y=2x-5$\nThe slope is the coefficient of $x,$ which is $2$ in this case. Make sure the coefficient of $y$ is $1$ in slope-intercept form.[/hide]\n[hide=\"2\"]$2x-y = 5$\n$y=2x-5$\n\nSubstitute that into the second equation.\n\n$3(2x-5)-x=15$\n$5x=30$\n$x=6$\n\nNow we can find $y=7.$ The answer is $(6,\\ 7).$[/hide]" } { "Tag": [ "trigonometry", "inequalities", "inequalities unsolved" ], "Problem": "$ A,B,C$ are angles of $ \\Delta ABC$,$ u,v,w > 0$are given const,find the maximum and minimum value of the following expression.\r\n\\[ u\\sin{A} \\plus{} v\\sin{B} \\plus{} w\\sin{C}\\]", "Solution_1": "??? \r\n\r\nMin: $ u,v,w\\rightarrow 0$ max: $ u,v,w\\rightarrow \\infty$\r\n\r\nSo it remains to show min and max of sum of sines... That's easy (Jensen for example for one of them)", "Solution_2": ":blush: Sorry,I missed something.I mean $ u,v,w$ are given const.", "Solution_3": ":( Why no one help me....I don't think Jensen work here.", "Solution_4": "im not so sure about this.\r\n[b]my solution:[/b]\r\nWLOG let $ u < = v < = w$ and $ sin A < = sin B < = sin C.$\r\nthen by Chebychev's inequality,\r\n$ u sin A + v sin B + w sin C < = (u + v + w)(sin A + sin B + sin C)/3 (*)$\r\nthen by Jensen's inequality,\r\n$ (*) \\< = (1/2)(u + v + w)\\sqrt3.$\r\nhence we have a maximum.\r\nminimum is obviously when $ A = B = 0,$ and $ C = 180$.", "Solution_5": "@zool007: i think you can not assume $ u \\le v \\le w$ since it's given constant and while $ A,B,C$ runs all along positive real numbers satisfying $ A \\plus{} B \\plus{} C \\equal{} 180^{\\circ}$. or, may be you can work in cases?", "Solution_6": "[quote=\"Raja Oktovin\"]@zool007: i think you can not assume $ u \\le v \\le w$ since it's given constant and while $ A,B,C$ runs all along positive real numbers satisfying $ A \\plus{} B \\plus{} C \\equal{} 180^{\\circ}$. or, may be you can work in cases?[/quote]\r\nyeah, that's why i wasnt sure :D", "Solution_7": "Since $ A,B,C \\in [0,\\pi]$, $ \\sin{A},\\sin{B},\\sin{C} > 0$, so the minimum is $ 0$, attained at $ (A,B,C) = (0,0,\\pi)$ or any of its permutations.\r\n\r\nThen, we note that on $ [0,\\pi], \\sin{x} > 0$ so $ \\frac {d^2x}{dy^2} = - \\sin{x} < 0$, so $ \\sin{x}$ is concave downward on this interval. Thus, by Jensen's Inequality,\r\n\r\n$ u\\sin{A} + v\\sin{B} + w\\sin{C}\\ge(u + v + w)(\\sin{\\frac {A + B + C}{u + v + w}) = (u + v + w)(\\sin{\\frac {\\pi}{u + v + w})}}$", "Solution_8": "[quote=\"Elixir10\"]Since $ A,B,C \\in [0,\\pi]$, $ \\sin{A},\\sin{B},\\sin{C} > 0$, so the minimum is $ 0$, attained at $ (A,B,C) = (0,0,\\pi)$ or any of its permutations.\n\nThen, we note that on $ [0,\\pi], \\sin{x} > 0$ so $ \\frac {d^2x}{dy^2} = - \\sin{x} < 0$, so $ \\sin{x}$ is concave downward on this interval. Thus, by Jensen's Inequality,\n\n$ u\\sin{A} + v\\sin{B} + w\\sin{C}\\ge(u + v + w)(\\sin{\\frac {A + B + C}{u + v + w}) = (u + v + w)(\\sin{\\frac {\\pi}{u + v + w})}}$[/quote]\r\n\r\nI believe that by Jensen's, we get $ LHS\\le (u+v+w)\\sin \\frac{uA+vB+wC}{u+v+w}.$", "Solution_9": "[quote=\"earldbest\"][quote=\"Elixir10\"]Since $ A,B,C \\in [0,\\pi]$, $ \\sin{A},\\sin{B},\\sin{C} > 0$, so the minimum is $ 0$, attained at $ (A,B,C) = (0,0,\\pi)$ or any of its permutations.\n\nThen, we note that on $ [0,\\pi], \\sin{x} > 0$ so $ \\frac {d^2x}{dy^2} = - \\sin{x} < 0$, so $ \\sin{x}$ is concave downward on this interval. Thus, by Jensen's Inequality,\n\n$ u\\sin{A} + v\\sin{B} + w\\sin{C}\\ge(u + v + w)(\\sin{\\frac {A + B + C}{u + v + w}) = (u + v + w)(\\sin{\\frac {\\pi}{u + v + w})}}$[/quote]\n\nI believe that by Jensen's, we get $ LHS\\le (u + v + w)\\sin \\frac {uA + vB + wC}{u + v + w}.$[/quote]\r\n\r\nThen,what's the maximum of $ (u + v + w)\\sin \\frac {uA + vB + wC}{u + v + w}$???", "Solution_10": "My apologies; it appears that earldbest is correct and I am wrong. Unfortunately, this brings us no nearer to solving the problem. :|\r\n\r\n\r\nI think zool007's argument might work, considering that $ A,B,C$ are pretty much flexible. The thing is, will equality always be possible for any values of $ u,v,w$?", "Solution_11": "By Lagrange multipliers, it is easy to see that for the maximum, $ u\\cos A \\equal{} v\\cos B \\equal{} w \\cos C$. So we are looking for a triangle such that $ cosA \\equal{} \\frac ku$ etc. with a certain $ k > 0$.\r\n\r\nBy the identity for any triangle $ \\sum cos^2A \\plus{} 2cosAcosBcosC \\equal{} 1$, we obtain the equation $ k^3\\frac {2}{uvw} \\plus{} k^2\\sum\\frac {1}{u^2} \\minus{} 1 \\equal{} 0$. By Cardano's formula, this may yield a somewhat lengthy formula for $ k$ and then an even longer one for the maximum. I wonder if some ugly terms would eventually cancel out. :( :blush:" } { "Tag": [ "LaTeX" ], "Problem": "When I use \\usepackage[spanish]{babel} the following warning appears:\r\n\r\n[img]http://img245.imageshack.us/img245/4766/sfkjgfjl3.png[/img]\r\n\r\nHow do I fix this?", "Solution_1": "It's telling you that it doesn't know how to hyphenate Spanish words and instead will use English words. You can accept that if the output is OK, otherwise, assuming you are using MiKTeX then\r\n[quote]To select the new language, do:\nStart-> Programs-> MiKTeX 2-> Settings, and select the Languages tab. Select spanish from the list, press the Apply button. Then select the General tab and press the Update Formats button. [/quote]\r\nMore at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=newlang]Using a new language with Babel[/url]" } { "Tag": [], "Problem": "http://v.mercola.com/blogs/public_blog/Shift-Happens----Some-of-the-Most-Amazing-Statistics-You-Never-Knew-11829.aspx\r\n\r\nSome of the most Shocking statistics.", "Solution_1": "What's so shocking about it? :maybe:" } { "Tag": [ "calculus", "limit", "function" ], "Problem": "My teacher told me to use calculator for this question, however i want to know how to do this algebraically, help me.\r\n\r\nQ. Solve 2x=e^(-0.5x).\r\n\r\nThanks~", "Solution_1": "Not possible afaik.", "Solution_2": "By calculus ($2x$ is strictly increasing, $\\mathrm{e}^{-\\frac12x}$ is strictly decreasing) we immediately get that there exists only one root and there must exist at least one root, since $\\lim_{x\\to\\pm\\infty}\\pm2x=\\pm\\infty$, $\\lim_{x\\to-\\infty}\\mathrm{e}^{-\\frac12x}= \\infty$ and $\\lim_{x\\to\\infty}\\mathrm{e}^{-\\frac12x}= 0$. \r\nThe numerical approximation of the only root is $0.40777670940448035$. But I don't think it is an algebraic number or can be expressed by values of elementar functions, because it equals $2u$, where $u\\mathrm{e}^{u}=\\frac14$ and solutions int $w$ of the equation $w\\mathrm{e}^{w}=z$ don't tend to be algebraic or elementar for rational $z$.", "Solution_3": "okai, so i have to use calculator for this!\r\n\r\nanyway, thank you heaps!" } { "Tag": [ "algebra", "polynomial", "Vieta" ], "Problem": "Suppose that $f(x) = x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e$ and that $f(1)=f(2)=f(3)=f(4)=f(5)$. Then $a =$\r\n\r\nA) $-8$\r\nB) $10$\r\nC) $-15$\r\nD) $22$\r\nE) $-35$", "Solution_1": "[hide=\"Hint\"] The polynomial $f(x)-f(1)$ has five roots. [/hide]", "Solution_2": "judging by there being 5 equal f's,\r\n[hide]\nam I wrong to say that f(1)=f(2)=f(3)=f(4)=f(5)=0?\nIf that is the case,\n(x-1)(x-2)(x-3)(x-4)(x-5)=0\nso a= -1+-2+-3+-4+-5=[b]-15[/b][/hide]", "Solution_3": "The official solution uses what t0rajir0u said, and then simply applies Vieta's sum. I don't get why you can do that, do.\r\n\r\nSapphyre571, you got the correct answer, but I think your logic is a bit flawed.", "Solution_4": "It's not necessary that $f(1) = ... = 0$. \r\n\r\n[hide=\"However...\"] $f(1) = 1+a+b+c+d+e = k$. Then\n\n$f(x)-k = (x-1)(x-2)(x-3)(x-4)(x-5) = x^{5}-15x^{4}+...$\n$f(x) = x^{5}-15x^{4}+...+k$\n\nSo the value of $k$ is irrelevant. [/hide]" } { "Tag": [ "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Is there a function,not necessarily continous, which graph is not symmetrical to the y-axis, that suffices following equation:\r\n\r\n$ \\int \\limits_{\\minus{}a}^a f(x) dx \\equal{} 2 \\int \\limits_{\\minus{}a}^0 f(x) dx$", "Solution_1": "Consider:\r\n\r\n$ \\int f(x)\\,dx\\equal{}F(x)$\r\n\r\nthen:\r\n\r\n$ \\int^{a}_{\\minus{}a} f(x)\\,dx\\equal{}F(a)\\minus{}F(\\minus{}a)$\r\n\r\n$ 2\\int^{0}_{\\minus{}a} f(x)\\,dx\\equal{}2F(0)\\minus{}2F(\\minus{}a)$\r\n\r\nnow:\r\n\r\n$ \\frac{d}{da}\\left(F(a)\\minus{}F(\\minus{}a)\\right)^{'}\\equal{}f(a)\\plus{}f(\\minus{}a)$\r\n\r\n$ \\frac{d}{da}\\left(2F(0)\\minus{}2F(\\minus{}a)\\right)^{'}\\equal{}2f(\\minus{}a)$\r\n\r\nso we have:\r\n\r\n$ f(a)\\plus{}f(\\minus{}a)\\equal{}2f(\\minus{}a)$\r\n\r\n$ f(a)\\equal{}f(\\minus{}a)$\r\n\r\nTherefore, no..", "Solution_2": "milin did tacitly assume that $ f$ was continuous when he invoked the Fundamental Theorem of Calculus.\r\n\r\nWe could, of course, start with an even function $ f$ and then change its values in an asymmetric way on a set of points too small to change the integral. In that way, yes there is such a (discontinuous) function - not that that seems to be much of an answer.\r\n\r\nI'm talking about something like this:\r\n\r\n$ f(x)\\equal{}\\begin{cases}0,&x\\equal{}\\frac1n,n\\in\\mathbb{Z}\\\\1&\\text{otherwise}\\end{cases}.$\r\n\r\nAs I said, something of a cheap example." } { "Tag": [ "function" ], "Problem": "Let $ f$ be a function defined on $ \\text{N}_0 \\equal{} \\{ 0,1,2,3,...\\}$ and with values in $ \\text{N}_0$, such that for $ n,m \\in \\text{N}_0$ and $ m \\leq 9, f(10n \\plus{} m) \\equal{} f(n) \\plus{} 11m$ and $ f(0) \\equal{} 0.$ How many solutions are there to the equation $ f(x) \\equal{} 1995$?\r\n\r\nA. None\r\nB. 1\r\nC. 2\r\nD. 11\r\nE. Infinitely many", "Solution_1": "[hide=\"Er...\"] Unless I've misread the problem, $ 11 | f(n)$ for all $ n \\in \\mathbb{N}_0$ (in fact, $ f( \\overline{d_n ... d_2 d_1 d_0}) \\equal{} 11(d_n \\plus{} ... \\plus{} d_2 \\plus{} d_1 \\plus{} d_0)$), so there are no solutions. Did you mean $ f(10n \\plus{} m) \\equal{} 11 f(n) \\plus{} m$? [/hide]", "Solution_2": "[hide=\"Solution\"]\n\nJust like previous poster said, all answers will be in the form of 11 x something so clearly, since 11 doesn't divide 1995, there is no such value or A. [/hide]" } { "Tag": [], "Problem": "When I try to log in to AoPS Wiki, it says that \r\n\r\nLogin error:\r\nYou have not specified a valid user name.\r\n\r\nWhat is the problem here?", "Solution_1": "I imagine it's because you have a slash in your name...", "Solution_2": "Probably; AoPS Wiki does a lot of strange stuff.\r\n\r\nMy contributions don't show (probably because of the underscores).", "Solution_3": "i_like_pie: I fixed the contribution thing for you..." } { "Tag": [], "Problem": "If $x+y=4$ and $xy=2$ then find $x^6+y^6$.\r\n\r\nFind two four digit numbers whose product is $4^8+6^8+9^8$.", "Solution_1": "$(x+y)^2 = 4^2$\r\n$\\Rightarrow x^2 + y^2 = 12$\r\n$(x^2+y^2)^3 = 12^3$\r\n$\\Rightarrow x^6 + y^6 = 1728 - 144 = 1584$", "Solution_2": "Noting that [(3^8)^3-(2^8)^3]/(3^8-2^8)=3^8+6^8+2^8 and 8113.5522=3^8+6^8+2^8", "Solution_3": "Ok I got the first problem, and I kind of get the 2nd one. But can someone explain more on it?", "Solution_4": "\\begin{eqnarray*}\r\n4^8 + 6^8 + 9^8 &=& (2^8)^2 + 2\\cdot 6^8 + (3^8)^2 - 6^8\\\\\r\n&=& (2^8 + 3^8)^2 - 6^8\\\\\r\n&=& (2^8 + 3^8 - 6^4)(2^8 + 3^8 + 6^4)\\\\\r\n&=& 5521 \\cdot 8113\r\n\\end{eqnarray*}" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ n\\in N^*$. Prove that\r\n$ \\sum^n_{j\\equal{}1}\\frac{1}{C^j_n}\\equal{}\\frac{n\\plus{}1}{2^n}\\sum^{n\\minus{}1}_{j\\equal{}0}\\frac{2^j}{j\\plus{}1}$", "Solution_1": "Double post:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225204[/url]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that for all positive integers $ n$: $ \\frac{2n}{3n\\plus{}1} \\le \\displaystyle\\sum_{k\\equal{}n\\plus{}1}^{2n}\\frac{1}{k} \\le \\frac{3n\\plus{}1}{4(n\\plus{}1)}$.", "Solution_1": "[quote=\"moldovan\"]Prove that for all positive integers $ n$: $ \\frac {2n}{3n \\plus{} 1} \\le \\displaystyle\\sum_{k \\equal{} n \\plus{} 1}^{2n}\\frac {1}{k} \\le \\frac {3n \\plus{} 1}{4(n \\plus{} 1)}$.[/quote]\r\n\r\nWe have, for $ 0 < a < x \\le y$, $ \\frac {1}{x} \\plus{} \\frac {1}{y} < \\frac {1}{x \\minus{} a} \\plus{} \\frac {1}{y \\plus{} a}$\r\n$ \\Leftrightarrow (x \\minus{} a)(y \\plus{} a) < xy \\Leftrightarrow 0 < a(y \\minus{} x \\plus{} a)$, which is obviously true.\r\n\r\n\r\nSo, $ \\frac {2n}{3n \\plus{} 1} \\equal{} \\frac {n}{2}(\\dfrac{1}{\\dfrac{2n \\plus{} (n \\plus{} 1)}{2}} \\plus{} \\dfrac{1}{\\dfrac{2n \\plus{} (n \\plus{} 1)}{2}})$\r\n$ \\le \\sum_{k \\equal{} n \\plus{} 1}^{2n}\\frac {1}{k} \\le \\frac {n}{2}(\\frac {1}{n \\plus{} 1} \\plus{} \\frac {1}{2n})$\r\n\r\nNow $ \\frac {n}{2}(\\frac {1}{n \\plus{} 1} \\plus{} \\frac {1}{2n}) \\equal{} \\frac {3n \\plus{} 1}{4(n \\plus{} 1)}$\r\n$ \\Leftrightarrow n(\\frac {2(n \\plus{} 1)}{n \\plus{} 1} \\plus{} \\frac {2(n \\plus{} 1)}{2n}) \\equal{} 3n \\plus{} 1$, which is obviously true, done.\r\n\r\nEquality for left equality holds $ \\Leftrightarrow n \\equal{} 1$\r\nEquality for right equality holds $ \\Leftrightarrow n \\equal{} 1,2$" } { "Tag": [ "Putnam", "linear algebra", "matrix", "induction", "college contests" ], "Problem": "Define a sequence $\\{u_n\\}_{n=0}^{\\infty}$ by $u_0=u_1=u_2=1,$ and thereafter by the condition that\r\n$\\det\\begin{vmatrix} u_n & u_{n+1} \\\\ u_{n+2} & u_{n+3} \\end{vmatrix}=n!$\r\nfor all $n\\ge 0.$ Show that $u_n$ is an integer for all $n.$ (By convention, $0!=1$.)", "Solution_1": "[u][b]The author of this posting is : blahblahblah[/b][/u]\r\n____________________________________________________________________\r\n\r\nWe can show by induction that $u_{n}u_{n+1} = n!$.\r\n\r\nAfter that, we just need to show that $u_n \\vert n! + \\left(n+1\\right)!$, which isn't that difficult, using the above.\r\n\r\nI calculated $u_0$ to $u_{11}$ and still didn't see the above pattern.", "Solution_2": "[quote=\"blahblahblah\"]... and still didn't see the above pattern.[/quote]\r\nIn fairness to blahblahblah, \"above\" refers to another now-vanished post.\r\n\r\nThe pattern is this: $u_{n+2}=(n+1)u_n$, so that $u_{2k}=1\\cdot3\\cdot5\\cdots(2k-1)$ and $u_{2k+1}=2\\cdot4\\cdot6\\cdots(2k).$\r\n\r\nSince the recursion uniquely defines the sequence (proof by induction), once you have shown that the above formulas satisfy the recursion, that suffices as a proof.", "Solution_3": "[quote=\"Kent Merryfield\"][quote=\"blahblahblah\"]... and still didn't see the above pattern.[/quote]\nIn fairness to blahblahblah, \"above\" refers to another now-vanished post.[/quote]\r\n\r\nStrange things happen - in my file, blahblahblah's post was the very first reply to the original message!\r\n\r\n Darij", "Solution_4": "You're correct, darij, the pattern I was referring to was $u_nu_{n+1}=n!$", "Solution_5": "But the problem is that you does not prove that the above solution is the only one to the defined sequence. Probably there are other possible solutions that contain non-integer. Therefore, how could we prove this is the unique solution?", "Solution_6": "The recursion $ u_nu_{n\\plus{}3}\\minus{}u_{n\\plus{}1}u_{n\\plus{}2}\\equal{}n!$ can be solved uniquely: $ u_{n\\plus{}3}\\equal{}\\frac{u_{n\\plus{}1}u_{n\\plus{}2}\\plus{}n!}{u_n}$.", "Solution_7": "A nice problem... We can either guess(?) or with some manipulation conclude that for odd $n$ $u_n$ is the product of all even positive numbers less than $n$ and for even $n$ it is the product of all odd positive numbers less or equal to $n$ . Proof then goes by induction." } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "homothety", "absolute value", "geometry solved" ], "Problem": "In a triangle $ABC$ of the area $S$ , point $H$ is the orthocenter , $D,E,F$ are the feet of the altitudes from $A,B,C$ and $P,Q,R$ are the reflections of $A,B,C$ in $BC,CA,AB$, respectively. The triangles $DEF$ and $PQR$ have the same area $T$ . Given that $T>\\frac 35 S$ , prove that $T=S$.", "Solution_1": "Are you sure the problem is correct? Geometer's Sketchpad does not show that those two triangles can have the same areas...", "Solution_2": "Shobber, it's well possible that the triangles DEF and PQR have the same area. Just check out some obtuse-angled triangles ABC. So the problem is indeed correct.\r\n\r\nI will actually show a stronger assertion:\r\n\r\n[color=blue][b]Problem.[/b] Let ABC be a triangle with area S. Let H be the orthocenter of triangle ABC, and D, E, F the feet of its altitudes from the vertices A, B, C, respectively. Furthermore, let P, Q, R be the reflections of the points A, B, C in the lines BC, CA, AB. Assume that the triangles DEF and PQR have the same (non-directed) area T. Prove that either $T=\\frac35S$ or T = S.[/color]\r\n\r\n[i]Solution.[/i] First some preliminaries:\r\n\r\nLet O be the center and R the radius of the circumcircle of triangle ABC.\r\n\r\nFor an arbitrary point M in the plane of triangle ABC, the triangle formed by the orthogonal projections of the point M on the sides BC, CA, AB of triangle ABC is called the [i]pedal triangle[/i] of the point M with respect to triangle ABC.\r\n\r\nAccording to the Gergonne theorem on pedal triangles (see http://www.mathlinks.ro/Forum/viewtopic.php?t=5658 posts #2, #12 and #13), the area of the pedal triangle of the point M with respect to triangle ABC is $\\frac{S\\cdot\\left|OM^2-R^2\\right|}{4R^2}$; hereby, the expression $\\left|OM^2-R^2\\right|$ is actually the absolute value of the power of the point M with respect to the circumcircle of triangle ABC.\r\n\r\nThe triangle DEF, formed by the feet of the altitudes of triangle ABC, is the pedal triangle of the orthocenter H with respect to triangle ABC; thus, its area equals $\\frac{S\\cdot\\left|OH^2-R^2\\right|}{4R^2}$.\r\n\r\nLet N be the nine-point center of triangle ABC; then, the area of the pedal triangle of this point N with respect to triangle ABC equals $\\frac{S\\cdot\\left|ON^2-R^2\\right|}{4R^2}$.\r\n\r\nThe triangle PQR is the reflection triangle of triangle ABC; according to [url=http://forumgeom.fau.edu/FG2003volume3/FG200311index.html]Darij Grinberg, [i]On the Kosnita point and the reflection triangle[/i], Forum Geometricorum 3 (2003) pages 105-111[/url], Theorem 4, this triangle PQR is the image of the pedal triangle of the nine-point center N of triangle ABC in the homothety with center at the centroid of triangle ABC and factor 4. Hence, the area of triangle PQR is $4^2=16$ times as large as the area of the pedal triangle of the nine-point center N. Since the area of the pedal triangle of the nine-point center N is $\\frac{S\\cdot\\left|ON^2-R^2\\right|}{4R^2}$, we thus obtain that the area of triangle PQR is $16\\cdot\\frac{S\\cdot\\left|ON^2-R^2\\right|}{4R^2}$.\r\n\r\nNow, as we know that both the area of triangle DEF and the area of triangle PQR are equal to T, we get $T=\\frac{S\\cdot\\left|OH^2-R^2\\right|}{4R^2}=16\\cdot\\frac{S\\cdot\\left|ON^2-R^2\\right|}{4R^2}$.\r\n\r\nHence, $\\left|OH^2-R^2\\right|=16\\cdot\\left|ON^2-R^2\\right|$. But it is well-known that the nine-point center N of triangle ABC is the midpoint between the circumcenter O and the orthocenter H. Thus, $OH=2\\cdot ON$, and we get\r\n\r\n$\\left|OH^2-R^2\\right|=16\\cdot\\left|ON^2-R^2\\right|=\\left|16\\cdot ON^2-16R^2\\right|$\r\n$=\\left|4\\cdot\\left(2\\cdot ON\\right)^2-16R^2\\right|=\\left|4\\cdot OH^2-16R^2\\right|$.\r\n\r\nNow, two cases are possible: Either $OH^2-R^2=4\\cdot OH^2-16R^2$ or $OH^2-R^2=-\\left(4\\cdot OH^2-16R^2\\right)$.\r\n\r\nFirst we consider the case when $OH^2-R^2=4\\cdot OH^2-16R^2$. Then, we quickly obtain $OH^2=5\\cdot R^2$. Thus, $T=\\frac{S\\cdot\\left|OH^2-R^2\\right|}{4R^2}=\\frac{S\\cdot\\left|5\\cdot R^2-R^2\\right|}{4R^2}=\\frac{S\\cdot 4R^2}{4R^2}=S$.\r\n\r\nNow, consider the case when $OH^2-R^2=-\\left(4\\cdot OH^2-16R^2\\right)$. Then, we get $OH^2=\\frac{17}{5}\\cdot R^2$. Thus, $T=\\frac{S\\cdot\\left|OH^2-R^2\\right|}{4R^2}=\\frac{S\\cdot\\left|\\frac{17}{5}\\cdot R^2-R^2\\right|}{4R^2}=\\frac{S\\cdot \\frac{12}{5}R^2}{4R^2}=\\frac35S$.\r\n\r\nHence, we get either T = S or $T=\\frac35S$. And the problem is solved.\r\n\r\n darij" } { "Tag": [ "probability" ], "Problem": "Of all the copy machines produced by CopyCo Inc., $ 80\\%$ work correctly, and $ 20\\%$ are broken. A simple copying test is done on all machines before they are sent out. A special paper is put in and copied. When the copied sheet comes out, it either says ``Broken'' or ``Ok'' on it. If the machine is really broken, the copied sheet will always say ``Broken''. However, if the machine is not broken, $ 10\\%$ of the time the test sheet will come out saying ``Broken'' anyway. If a random machine is tested, and the sheet comes out saying, ``Broken'', what is the probability that the machine is really broken?", "Solution_1": "Relative to all possibilities and given that it comes out broken, there is a $ 0.8\\cdot0.1\\equal{}0.08$ probability that it is working, and a $ 0.2$ probability that it is broken. Of this $ 0.2\\plus{}0.08\\equal{}0.28$ probability that this happens, there is a $ \\frac{0.2}{0.28}\\equal{}\\boxed{\\frac57}$ probability that it is really broken." } { "Tag": [ "percent", "symmetry", "probability" ], "Problem": "If a five question True/False test is given and $ 60\\%$ is a passing grade, what is your percent chance of passing if you guess on each question?", "Solution_1": "By symmetry (i.e. it is just as likely to get $ 3$ or more questions right as it is to get $ 3$ or more questions wrong), the answer is $ \\boxed{\\frac12}$.", "Solution_2": "You must get at least 3 of the questions correct to get $ 60\\%$ of the questions correct. Clearly the probability of this happening is the same as getting at least 3 of the questions wrong. These three together encompass all possibilities, so the probability of getting at least 3 right is $ \\boxed{50\\% }$\r\n\r\nmath154, they're slightly different problems :wink:", "Solution_3": "[quote=\"5849206328x\"]math154, they're slightly different problems :wink:[/quote]\r\n\r\nDang, somewhat late? :P But yes, I do apologize for being dyslexic in both of my previous posts." } { "Tag": [ "geometry", "trigonometry", "geometry proposed" ], "Problem": "[color=darkred]Prove that for any point $M$ which belongs to the incircle of the triangle $ABC$ there is the following identity,\n\nwhere $2p=a+b+c\\ :$ $\\boxed{\\ a\\cdot MA^{2}+b\\cdot MB^{2}+c\\cdot MC^{2}=abc+2(p-a)(p-b)(p-c)\\ }\\ \\ (*)\\ .$\n\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=113516 [/color]\r\n\r\n[color=darkblue][b]Proof.[/b] I wanted to verify the [b]Gemath's identity[/b] through this well-known identity. Denote the points $D\\in BC\\ ,\\ E\\in CA\\ ,\\ F\\in AB$ which belong to the incircle $w=C(I,r)$ of the triangle $ABC\\ .$ Thus, $EF=2(p-a)\\sin \\frac{A}{2}$ and $m(\\widehat{EDF})=\\frac{B+C}{2}\\ .$ The Gemath's identity from the topic mentioned above becomes $\\sum 2(p-a)\\sin^{2}\\frac{A}{2}\\cdot MD^{2}=8\\prod \\left[(p-a)\\sin\\frac{A}{2}\\right]\\ ,$ i.e. $\\sum a\\cdot MD^{2}=4(p-a)(p-b)(p-c)\\ .$ \n\nApply the Stewart's theorem to the cevian $[MD]$ in the triangle $BMC\\ : \\ a\\cdot MD^{2}+a(p-b)(p-c)=(p-c)\\cdot MB^{2}+(p-b)\\cdot MC^{2}$ a.s.o.\n\nAdding these three identities obtain $\\sum a\\cdot MD^{2}=\\sum a\\cdot MA^{2}-\\sum [a(p-b)(p-c)]$ $\\implies$ $\\sum a\\cdot MA^{2}=\\sum a(p-b)(p-c)+4(p-a)(p-b)(p-c)\\ .$\n\nProve easily that $(p-a)(p-b)(p-c)=pr^{2}$ and $\\sum a(p-b)(p-c)=2pr(2R-r)\\ .$ In conclusion, $\\sum a\\cdot MA^{2}=2pr(2R+r)=abc+2\\prod (p-a)\\ .$\n\n[b]Remark.[/b] Can obtain directly this identity from the remarkable relation $a\\cdot XA^{2}+b\\cdot XB^{2}+c\\cdot XC^{2}=2p\\cdot XI^{2}+abc$ for $X\\in C(I,r)\\ .$[/color]", "Solution_1": "That is a nice problem! but we can prove simplely in your solution:\r\n$\\sum aMA^{2}=(a+b+c)MI^{2}+\\sum aIA^{2}$\r\n$\\sum aIA^{2}=abc$ and when $M\\in$ incircle $\\Rightarrow MI^{2}=r^{2}$", "Solution_2": "You didn't read the remark from the my last reply.", "Solution_3": "oh sorry Virgil :blush: , I have a nice problem as a present for you:\r\n(maybe it is old but the solution quite new)\r\nGiven are equaliteral triangle $ABC, (O)$ is incircle tagent $BC,CA,AB$ at $A',B',C'$ with any $M$ lies on small arc $B'C'$ and $A'',B'',C''$ are projection of $M$ to the sides prove $\\sqrt{MA''}=\\sqrt{MB''}+\\sqrt{MC''}$\r\nA proof:\r\nThe equal is equaliteral \r\n$MA''=MB''+MC''+2\\sqrt{MB''}\\sqrt{MC''}\\Leftrightarrow (MA''-MB''-MC'')^{2}=4MB''MC''\\Leftrightarrow MA''^{2}+MB''^{2}+MC''^{2}=2(MA''MB''+MB''MC''+MC''MA'')(*)$\r\nwe can prove $\\overrightarrow{MA''}+\\overrightarrow{MB''}+\\overrightarrow{MC''}=\\frac{3}{2}\\overrightarrow{MO}\\Rightarrow MA''^{2}+MB''^{2}+MC''^{2}+2\\sum\\overrightarrow{MB''}{\\overrightarrow{MC''}=\\frac{9}{4}r^{2}\\Rightarrow}$ $MA''^{2}+MB''^{2}+MC''^{2}-\\sum MB''MC''=\\frac{9}{4}r^{2}(1)$(here we notice $\\angle BMC=\\angle CMA=\\angle AMB=120^{0}$)\r\nfrom $MA''+MB''+MC''=3r \\Rightarrow MA''^{2}+MB''^{2}+MC''^{2}+2\\sum MB''MC''=9r^{2}(2)$\r\nChoose fit coefficiens and multiply $(1),(2)$ we will have $(*)$\r\nThe solution maybe quite long but the equal $(1),(2)$ have nice meaning, with it we can creat more nice other equal." } { "Tag": [], "Problem": "\u03a0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce \u03bd\u03b1 \u03bb\u03cd\u03c3\u03c9 \u03b1\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b5\u03b4\u03ce \u03ba\u03b1\u03b9 \u03bc\u03ad\u03c1\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03c4\u03b1 \u03ba\u03b1\u03c4\u03b1\u03c6\u03ad\u03c1\u03bd\u03c9. \u039c\u03ae\u03c0\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03b5 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03c4\u03b5;\r\n\u0394\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $ f(x)$ \u03bc\u03b5 $ x{\\in}R$ \u03ba\u03b1\u03b9 $ f(nx \\plus{} (1 \\minus{} n)y) \\equal{} nf(x) \\plus{} (1 \\minus{} n)f(y)$ \u03ba\u03b1\u03b9 $ n{\\in}[0,1]$. \u039d\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 $ f(0) \\equal{} f(1)$", "Solution_1": "\u0388\u03c3\u03c4\u03c9 \u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $ f(x)\\equal{}ax$ \u03bc\u03b5 $ a\\neq 0$\r\n\r\n\u0388\u03c7\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ f(nx \\plus{} (1 \\minus{} n)y) \\equal{} nf(x) \\plus{} (1 \\minus{} n)f(y) \\iff$\r\n\r\n$ a[nx \\plus{} (1 \\minus{} n)y] \\equal{} nax \\plus{} (1 \\minus{} n)ay \\iff$\r\n\r\n$ anx \\plus{} (1 \\minus{} n)ay \\equal{} nax \\plus{} (1 \\minus{} n)ay$, \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9\r\n\r\n\u0395\u03bd\u03ce \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ f(0)\\equal{}0\\neq a\\equal{}f(1)$", "Solution_2": "\u0394\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 \u03b3\u03b9\u03b1\u03c4\u03af \u03c3\u03b5 \u03b5\u03c0\u03cc\u03bc\u03b5\u03bd\u03bf \u03c5\u03c0\u03bf\u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03bc\u03bf\u03c5 \u03b6\u03b7\u03c4\u03ac\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c9 \u03cc\u03c4\u03b9 $ f(x) \\equal{} f(0)$, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ae", "Solution_3": "gregg o pontios \u03b8\u03ad\u03bb\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b5\u03bd\u03ce \u03b7 $ f(x)\\equal{}ax, \\ a\\neq 0$ \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03af \u03c4\u03b7\u03bd \u03c3\u03c7\u03ad\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03b4\u03c9\u03c3\u03b5\u03c2 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03ae, \u03b4\u03b5\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03af \u03c4\u03bf \u03c3\u03c5\u03bc\u03c0\u03ad\u03c1\u03b1\u03c3\u03bc\u03b1 $ f(0)\\equal{}f(1)$. \u039c\u03ae\u03c0\u03c9\u03c2 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b4\u03af\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03ac\u03bb\u03bb\u03bf \u03c9\u03c2 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03bf \u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03ae? \u0393\u03b9\u03b1\u03c4\u03af \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03c6\u03bf\u03b2\u03ac\u03bc\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03bf pontios \u03ad\u03c7\u03b5\u03b9 \u03b4\u03af\u03ba\u03b9\u03bf. \u03a0\u03bf\u03bb\u03bb\u03ad\u03c2 \u03c6\u03bf\u03c1\u03ad\u03c2 \u03c4\u03c5\u03c7\u03b1\u03af\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2. \u0386\u03bd\u03b8\u03c1\u03c9\u03c0\u03bf\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03bf\u03af \u03c0\u03bf\u03c5 \u03c4\u03b9\u03c2 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ac\u03b6\u03bf\u03c5\u03bd.\r\n\r\n\u03a6\u03b9\u03bb\u03b9\u03ba\u03ac,\r\n\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2", "Solution_4": "\u0394\u03b5\u03bd \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03b5\u03af\u03bc\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2. \u03a4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c4\u03b7\u03bd \u03b2\u03c1\u03ae\u03ba\u03b1 \u03c3\u03c4\u03bf \u03b4\u03b9\u03b1\u03b4\u03af\u03ba\u03c4\u03c5\u03bf \u03b1\u03c0\u03cc \u03c4\u03bf mathsforyou. \r\n\u0391\u03c0\u03bb\u03ac \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bb\u03ad\u03c9 (\u03ba\u03b1\u03b9 \u03af\u03c3\u03c9\u03c2 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c3\u03c4\u03ad\u03ba\u03b5\u03b9 \u03b1\u03bb\u03bb\u03ac \u03ad\u03c4\u03c3\u03b9 \u03c4\u03bf \u03b2\u03bb\u03ad\u03c0\u03c9 :blush: ) \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03b7 f \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ae \u03c4\u03cc\u03c4\u03b5 \u03b5\u03c0\u03b1\u03bb\u03b7\u03b8\u03b5\u03cd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7.", "Solution_5": "gregg prospa8hsa na s apanthsw me pm alla den ta katafera. gia kapoio logo oti pm prospa8w na steilw paramenei sto outbox kai oxi sto sentbox.\r\nMporeis na m dwseis to link ap opou vrhkes thn askhsh?\r\nOso gia to allo me tous migadikous, |1/z_i|*|z_i|=|1/z_i*z_i|=|1|=1 \r\nara 1>3b/a*|1/z_i|pou sunepagetai pollaplasiazontas kai ta 2 melh me a/3b oti a/3b>|1/z_i| \r\nKalutera twra? \r\n\r\nPaidia xilia suggnwmh pou eprepe na to steilw edw pera alla opws proeipa den mporw na steilw pm!an kapoios gnwrizei pws mporw na to dior8wsw auto as m steilei.\r\n\r\nHlias" } { "Tag": [ "inequalities" ], "Problem": "Let $ABC$ be a triangle and $h_a$ be the altitude through $A$. Prove that \\[ (b+c)^2 \\geq a^2 + 4h_a ^2 . \\]", "Solution_1": "Obviously, $ah_a = 2K$ where K is the area. \r\n\r\nThen the ineq transforms into $a^2(b+c-a)(b+c+a) \\ge 4K^2$.\r\n\r\nwhich becomes after Heron and cancelling out, $4a^2 \\ge (b-c+a)(a-b+c)$\r\n\r\nwhich is equivalent to $3a^2 \\ge 0$.", "Solution_2": "Well, isn't that quite brutal? Just try the following:\r\n\r\nLet $H$ be the foot of the altitude $h_a$. Then $(b+c)^2 = b^2+2bc+c^2\\geq BH^2 + h_a^2 + 2BH\\cdot CH + CH^2 + h_a^2\\geq$ $(BH+CH)^2 + h_a^2 = a^2 + h_a^2$.\r\n\r\nSo in fact, the inequality is very unsharp - we only have equality if $h_a=0$.\r\n\r\nPeter", "Solution_3": "The actual official question of RMO-1996 was :\r\n\r\nLet $ ABC$ be a triangle and $ h_a$ the altitude through $ A$. Prove that\r\n \\[ (b\\plus{}c)^{2}\\geq a^{2}\\plus{}4h_{a}^{2}.\\]", "Solution_4": "Let $D$ be the projection of $A$ onto $BC$. Let $DB=m$ and $DC=n$.\n\n$1 \\ge cos(B-C) \\Rightarrow 1 \\ge cosBcosC + sinBsinC \\Rightarrow 1 \\ge \\frac{mn}{bc} + \\frac{h^2_a}{bc} \\Rightarrow bc \\ge mn + h^2_a.$ \n\n$(b+c)^2 = b^2 + 2bc + c^2 \\ge h^2_a + n^2 + 2(mn + h^2_a) + h^2_a + m^2 = n^2+2mn+m^2 + 4h^2_a = a^2+4h^2_a$\n\nEquality holds when $cos(B-C) = 1$ or $\\angle B = \\angle C$", "Solution_5": "I have an easier version of duohead's proof.", "Solution_6": "Let $X$ be the foot of the altitude from $A$ to $BC$\n$BX= c \\cos B$\n$CX= b \\cos C$\n$h_a = b \\sin C = c\\sin B$\n \n$(b+c)^2 \\ge BX^2 + CX^2 + 2 BX \\cdot CX$\n$\\iff b^2 + c^2 + 2bc \\ge b^2 + c^2 + 2 BX \\cdot CX + 2 h_a^ 2$\n$\\iff bc \\ge bc \\cos B \\cos C + bc \\sin B \\sin C$\nNow the inequation can be simplified as\n$bc \\ge bc(\\cos B \\cos C + \\sin B \\sin C) = bc \\cdot cos (B - C)$\nSo we get $1 \\ge \\cos (B - C)$\nSince this is true we get the desired conclusion", "Solution_7": "Any solution with synthetic geometry?\n", "Solution_8": "Take the image of the A about BC and call it A'. Then, draw AC' parallel to BC and AC' = BC, that implies AC'CB is a parallelogram. Notice, angle(A'AC') = 90 also A'C = b and CC' = c. Now, apply triangle inequality in triangle(CA'C') and we are done!!\n" } { "Tag": [ "geometry" ], "Problem": "A circle has an area of 100pi, what is twice the circumference minus 8pi?", "Solution_1": "[url=http://lmgtfy.com/?q=circumference]Circumference[/url]\r\n\r\nThe radius is $ 10$, so the circumference is $ 20\\pi$.\r\n\r\n$ 2(20\\pi) \\minus{} 8\\pi \\equal{} \\boxed{32\\pi}$" } { "Tag": [ "trigonometry", "geometry", "trig identities", "Law of Sines" ], "Problem": "In triangle DEF, an exterior angle at D measures 170\u00ba, and m E > D.\nF is the largest angle, so the side opposite F is ED.\n[/hide]" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let be given a point $M$ inside a quadrilateral $ABCD$.Prove that:\\[ min\\{\\widehat{MAB},\\widehat{MBC},\\widehat{MCD},\\widehat{MDA}\\}\\le\\frac{\\pi}{4} \\]", "Solution_1": "It's a generalization of Brocard Point (?)\r\nIt seems that it is really old enough. You can check this @ India MO 1998. (If I weren't false)" } { "Tag": [], "Problem": "If anyone could help me with the following problem,I would appreciate it :)\r\n\r\n [b]Problem[/b]:\r\n If there is a switch from deposits into currency,what happens to the federal funds rate?\r\n Could you show to me a supply and demand analysis of the market for reserves?\r\n \r\n Thanks in advance.", "Solution_1": "The federal funds rate is set by the Federal Reserve to influence bank overnight rates. That rate wouldn't change just because consumers withdraw their funds. To reign in possible inflation that might arise from increased liquidity outside the banking sector, the FF rate may be raised. However if there is a massive bank run destabilizing the banking system the rate may sharply go down to meet the increased liquidity demands from banks.\nIf you meant if the [i]Federal Reserve[/i] switches its deposits into currency, I would think that they would raise the FF rate instead (if by deposits you mean deposits by banks being held by the Federal Reserve).\n\nYes, I'm necroposting but I thought this question deserved an answer, however late it may be." } { "Tag": [ "induction", "inequalities proposed", "inequalities" ], "Problem": "Let $ a_i, b_i, c_i$ sequences of positive numbers for $ i \\equal{} 1,2,\\dots,n$.\r\n\r\n(1) Prove that\r\n\\[ \\sum \\frac {a_i^2 \\plus{} b_i^2 \\plus{} c_i^2}{a_i \\plus{} b_i \\plus{} c_i} \\geq \\frac {(\\sum a_i)^2 \\plus{} ( \\sum b_i )^2 \\plus{} (\\sum c_i)^2}{\\sum (a_i \\plus{} b_i \\plus{} c_i)}\\]\r\n(2) Prove that\r\n\\[ \\sum \\frac {a_i^2 \\plus{} b_i^2 \\plus{} c_i^2 \\plus{} k(a_ib_i \\plus{} b_ic_i \\plus{} c_ia_i)}{a_i \\plus{} b_i \\plus{} c_i} \\leq \\frac {(\\sum a_i)^2 \\plus{} ( \\sum b_i )^2 \\plus{} (\\sum c_i)^2 \\plus{} k(\\sum a_{i}\\sum b_{i} \\plus{} \\sum b_{i}\\sum c_{i} \\plus{} \\sum c_{i}\\sum a_{i} )}{\\sum (a_i \\plus{} b_i \\plus{} c_i)}\\]\r\nholds for $ k \\equal{} 8$.\r\n\r\n(3)* Is $ k \\equal{} 8$ smallest possible :?:\r\n\r\n(4)* Is $ k \\equal{} 2$ biggest possible such that the [i]converse[/i] of (2) still holds (as in (1) for $ k\\equal{}0$) :?:", "Solution_1": "The first one (for $ k\\equal{}0$) is easy using Cauchy-Schwartz.\r\n\r\nConcerning the general one: It is equivalent to\r\n\\[ (k\\minus{}2)\\cdot\\sum\\frac{a_{i}b_{i}\\plus{}b_{i}c_{i}\\plus{}c_{i}a_{i}}{a_{i}\\plus{}b_{i}\\plus{}c_{i}}\\leq(k\\minus{}2)\\cdot\\frac{\\sum a_{i}\\sum b_{i}\\plus{}\\sum b_{i}\\sum c_{i}\\plus{}\\sum c_{i}\\sum a_{i}}{\\sum (a_{i}\\plus{}b_{i}\\plus{}c_{i})}\\]\r\n\r\nAs you claim in (4) that $ k\\equal{}2$ is the biggest possible value that the converse holds, $ k\\equal{}2$ should be the smallest possible value such that the original one holds...\r\n\r\nSorry I didn't have the time to have a closer look at it...", "Solution_2": "[quote=\"Martin N.\"]The first one (for $ k \\equal{} 0$) \nConcerning the general one: It is equivalent to\n\\[ (k \\minus{} 2)\\cdot\\sum\\frac {a_{i}b_{i} \\plus{} b_{i}c_{i} \\plus{} c_{i}a_{i}}{a_{i} \\plus{} b_{i} \\plus{} c_{i}}\\leq(k \\minus{} 2)\\cdot\\frac {\\sum a_{i}\\sum b_{i} \\plus{} \\sum b_{i}\\sum c_{i} \\plus{} \\sum c_{i}\\sum a_{i}}{\\sum (a_{i} \\plus{} b_{i} \\plus{} c_{i})}\\]\n[/quote]\r\n :blush: I saw of course that for $ k \\equal{} 2$ we have an identity in (2), but it didn't occur to me to just subtract that identity from the general ineq :wallbash:", "Solution_3": "I'll prove it by induction, for $ i\\equal{}1,2$, we have to prove (denoting $ m\\equal{}a\\plus{}b\\plus{}c$, $ n\\equal{}x\\plus{}y\\plus{}z$)\r\n$ \\frac{(a^2\\plus{}b^2\\plus{}c^2)m^2\\plus{}(x^2\\plus{}y^2\\plus{}z^2)n^2}{2}\\geq(ax\\plus{}by\\plus{}cz)mn\\mbox{,}$\r\nwhich is obvious by AM-GM and Cauchy-Schwartz.\r\n\r\nAssuming the statement holds for some $ n$, we get\r\n$ \\sum_{i\\equal{}1}^{n\\plus{}1}\\frac{a_{i}b_{i}\\plus{}b_{i}c_{i}\\plus{}c_{i}a_{i}}{a_{i}\\plus{}b_{i}\\plus{}c_{i}}\\leq\r\n\\frac{\\sum_{i\\equal{}1}^{n} a_{i}\\sum_{i\\equal{}1}^{n} b_{i}\\plus{}\\sum_{i\\equal{}1}^{n} b_{i}\\sum_{i\\equal{}1}^{n} c_{i}\\plus{}\\sum_{i\\equal{}1}^{n} c_{i}\\sum_{i\\equal{}1}^{n} a_{i}}{\\sum_{i\\equal{}1}^{n} a_{i}\\plus{}\\sum_{i\\equal{}1}^{n}b_{i}\\plus{}\\sum_{i\\equal{}1}^{n}c_{i}} \\plus{} \\frac{a_{n\\plus{}1}b_{n\\plus{}1}\\plus{}b_{n\\plus{}1}c_{n\\plus{}1}\\plus{}c_{n\\plus{}1}a_{n\\plus{}1}}{a_{n\\plus{}1}\\plus{}b_{n\\plus{}1}\\plus{}c_{n\\plus{}1}} \\leq \\frac{\\sum_{i\\equal{}1}^{n\\plus{}1} a_{i}\\sum_{i\\equal{}1}^{n\\plus{}1} b_{i}\\plus{}\\sum_{i\\equal{}1}^{n\\plus{}1} b_{i}\\sum_{i\\equal{}1}^{n\\plus{}1} c_{i}\\plus{}\\sum_{i\\equal{}1}^{n\\plus{}1} c_{i}\\sum_{i\\equal{}1}^{n\\plus{}1} a_{i}}{\\sum_{i\\equal{}1}^{n\\plus{}1} a_{i}\\plus{}\\sum_{i\\equal{}1}^{n\\plus{}1}b_{i}\\plus{}\\sum_{i\\equal{}1}^{n\\plus{}1}c_{i}}\\mbox{,}$\r\nand the induction is complete. :) \r\n\r\nit doesn't look nice at all...sorry :blush: \r\n\r\n@spanferkel: how did you prove this?", "Solution_4": "My proof of \\[ \\sum\\frac{a_{i}b_{i}\\plus{}b_{i}c_{i}\\plus{}c_{i}a_{i}}{a_{i}\\plus{}b_{i}\\plus{}c_{i}}\\le \\frac{\\sum a_{i}\\sum b_{i}\\plus{}\\sum b_{i}\\sum c_{i}\\plus{}\\sum c_{i}\\sum a_{i}}{\\sum (a_{i}\\plus{}b_{i}\\plus{}c_{i})}\\]\r\nis based on the same lemma $ \\sum\\frac{x_{i}y_{i}}{x_{i}\\plus{}y_{i}}\\leq\\frac{\\sum x_{i}\\sum y_{i}}{\\sum x_{i}\\plus{}\\sum y_{i}}$ that is used [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1688547#1688547]here[/url] (which is a special case of Minkowski). Just put $ x_i: \\equal{}a_i$ and $ y_i: \\equal{}b_i\\plus{}c_i$, then sum up cyclically. By the way, I had gotten the $ k\\equal{}8$ case the same way via $ x_i: \\equal{}2a_i\\plus{}c_i$ and $ y_i: \\equal{}2b_i\\plus{}c_i$. :)" } { "Tag": [ "function" ], "Problem": "A uniform spring whose unstressed length is $l$ has a force constant $k$ . The spring is cut into two pieces of unstressed length $l_1$ and $l_2$ where $l_1=nl_2$ and $n$ is an integer . What are the corresponding force constant $k_1$ and $k_2$ in terms of $n$ and $k$ ? Does your result seem reasonable for $n=1$ ?", "Solution_1": "I'd say $\\frac{k}{n+1}$ and $\\frac{nk}{n+1}$, don't you think so?", "Solution_2": "What concept do you use ? Is it Hooke's Law ?Cause I cant seem to figure out the right formula to do :(", "Solution_3": "I think you got the answers backward. The shorter spring would have the greater force for a given displacement withing the elastic zone of both springs.", "Solution_4": "Actually, now that I think about it they should be both $k$ no matter how long they are. Should have known this right from the beginning... :)", "Solution_5": "bus -- You're right. However, the shorter spring piece will have the greater force at equal displacement inside the elastic zone of both spring pieces.", "Solution_6": "Actually, I derived the general case of the spring constant when a spring is cut into many pieces by just considering that the extension(tension etc.) was a function of the no. of coils....", "Solution_7": "[quote=\"wizard2378\"]However, the shorter spring piece will have the greater force at equal displacement inside the elastic zone of both spring pieces.[/quote]And why is that? Shouldn't it be the same if they both have the same $k$?" } { "Tag": [ "vector", "inequalities unsolved", "inequalities" ], "Problem": "$ X, Y ,Z$ are three vectors on $ R^3$ prove that \r\n\r\n$ llX \\minus{} Zll \\plus{} llZ \\minus{} Yll \\plus{} llX \\minus{} Zll \\plus{} llX \\plus{} Y \\plus{} Zll > \\equal{} llXll \\plus{} llYll \\plus{} llZll$", "Solution_1": "[quote=\"Axelshmmy\"]$ X, Y ,Z$ are three vectors on $ R^3$ prove that \n\n$ llX \\minus{} Zll \\plus{} llZ \\minus{} Yll \\plus{} llX \\minus{} Zll \\plus{} llX \\plus{} Y \\plus{} Zll > \\equal{} llXll \\plus{} llYll \\plus{} llZll$[/quote]\r\nIt's true in $ \\mathbb R^n$ too.\r\nA squaring kills it. :wink:" } { "Tag": [], "Problem": "Mr. Harter drives from Chicago to Milwaukee on Sunday. On Monday when he makes the return trip, he figures that if he increases his speed by one-fifth, he can make the trip in 80 minutes. How many minutes did the first trip take?", "Solution_1": "Since the distances are equal, we can write \\[ \\left(\\frac{6}{5}r\\right)(80)\\equal{}rt \\Rightarrow t\\equal{}\\boxed{96}\\]" } { "Tag": [ "videos", "Alcumus", "Support" ], "Problem": "It's been several weeks and I haven't seen a new Alcumulus video come out. When will a new one come?", "Solution_1": "Firstly, it's Alcumus. For some reason, a lot people think it's Alcumulus. I don't know why. Just a small thing, unimportant. \r\n\r\nVideos are given to you based on your progress. As you do more problems, the rate at which videos are given reduces dramatically. Thus, cherish the great videos while you are \"Alcumus-young.\" :) \r\n\r\nOr, if you have nothing to do, you could hone your skills rewatching, for the cost of a little Karma. Most people get more the second time. :)", "Solution_2": "[quote=\"r15s11z55y89w21\"]Or, if you have nothing to do, you could hone your skills rewatching, for the cost of a little Karma. [/quote]\nYou don't lose (or gain) Karma as long as you comment on the video every time. You can rate it only the first time, and gain 5 Karma.\n\n[quote=\"maybach\"]When will a new one come?[/quote]\r\nThey made most of them late last year (October-November 2008). Intro to Algebra is probably coming out next year.", "Solution_3": "[quote=\"r15s11z55y89w21\"]\n\nVideos are given to you based on your progress. As you do more problems, the rate at which videos are given reduces dramatically. Thus, cherish the great videos while you are \"Alcumus-young.\" :) \n [/quote]\r\nIs it possible to watch videos on the official Video List? (sometimes it works for me...)" } { "Tag": [ "function", "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "Let $a,b,c$ be non zero constant numbers.\r\nEvaluate \r\n\r\n\\[\\lim_{r\\to\\infty} \\frac{\\cos \\frac{a}{r}-\\cos \\frac{b}{r}\\cos \\frac{c}{r}}{\\sin \\frac{b}{r}\\sin \\frac{c}{r}}\\]", "Solution_1": "[hide]$\\frac{b^2+c^2-a^2}{2bc}$[/hide]", "Solution_2": "$\\sin x \\sim x (x\\to 0)$ and $\\cos x\\sim 1-\\frac{1}{2}x^2 (x\\to 0)$.\r\nHence the limit is\r\n$\\lim_{r\\to\\infty} \\frac{(1-\\frac{a^2}{2r^2}) - (1-\\frac{b^2}{2r^2})(1-\\frac{c^2}{2r^2})}{\\frac{bc}{r^2}} = \r\n\\lim_{r\\to\\infty} \\frac{-2a^2 + 2c^2 + b^2(2-\\frac{c^2}{r^2})}{4bc} = \\frac{b^2+c^2-a^2}{2bc}\r\n$" } { "Tag": [ "calculus", "integration", "algebra", "function", "domain", "Ring Theory", "superior algebra" ], "Problem": "Let R be a commutative ring and I is an ideal of R .Prove that :\r\n(i) The quotient ring R/I is an integral domain if and only if I is a prime ideal.\r\n(ii) The quotient ring R/I is a field if and only if I is a maximal ideal.\r\nCan you prove this clause for me . Thanks a lot .", "Solution_1": "Review the definitions of an integral domain, a prime ideal, a field, and a maximal ideal.\r\n\r\n(i) What does $ ab \\equal{} 0$ mean in $ R/I$?\r\n(ii) What does not having a multiplicative inverse mean in $ R/I$?" } { "Tag": [ "blogs", "trigonometry", "MATHCOUNTS", "algorithm", "AMC", "AIME", "function" ], "Problem": "This is where you can talk about solutions. Request them or answer them, it's up to you. I'll eventually post all the solutions but I'll answer the ones request in the order they are first.\r\n\r\nHere are the answer keys.\r\n\r\n[hide=\"Answer\"]\n1. A\n2. A\n3. B\n4. B\n5. E\n6. E\n7. D\n8. A\n9. B\n10. D\n11. B\n12. B\n13. IGNORE\n14. E\n15. D\n16. A\n17. A\n18. C\n19. D\n20. C\n21. B \n22. C\n23. C\n24. E\n25. B [/hide]\r\n\r\nI put IGNORE for #13 because the correct answer is not one of them (I should be 24 but I put 42).", "Solution_1": "How did you guys get number 14?", "Solution_2": "When we post our answer choices on the blog site, will everyone else see them?\r\nSorry, still kinda confused to what to do.", "Solution_3": "how do you solve the two trig ones? 11 and 13?", "Solution_4": "Yes.\r\n\r\nPeople will be able to see your answers but it doesn't matter as long as you post your own (thus, more of AoPS Honor Code).", "Solution_5": "i remember a few of the problems from a mathcounts practice i had and from the mathcounts problem series. lol.\r\nthe one about the pattern in the square root could be solved in two ways. multiply the two outer numbers and add 1 or multiply the two inner numbers and subtract 1.", "Solution_6": "The trig ones were annyoing.I don't think I got them. Near the end (about 8 or so) my computer crashed, and it took me a while to get it on, so I didn't really answer the last few. Doesn't matter, I couldn't have solved them.", "Solution_7": "Could someone post a solution to #9? I still can't figure it out :blush:", "Solution_8": "I'm going to grade the solutions that I received and send the score back to the owner first. Then, I'll post the solutions so it might be a while.", "Solution_9": "[hide=\"9\"]$ gcd(n^2 \\plus{} 7,n\\plus{}4) \\equal{} gcd(7\\minus{}4n,n\\plus{}4) \\equal{} gcd(23,n\\plus{}4)$ (by Euclid algorithm), and there are $ 86$ numbers divisible by $ 23$ in the range given.[/hide]", "Solution_10": "hm no time to take the test for me but after glancing over the square root pattern one, you could just note that the inside number has a units digit of 1, so the answer must be the answer with 1 as the units digit :P", "Solution_11": "[quote=\"azjps\"][hide=\"9\"]$ gcd(n^2 \\plus{} 7,n \\plus{} 4) \\equal{} gcd(7 \\minus{} 4n,n \\plus{} 4) \\equal{} gcd(23,n \\plus{} 4)$ (by Euclid algorithm), and there are $ 86$ numbers divisible by $ 23$ in the range given.[/hide][/quote]\r\nUgh. I somehow used the Euclidean Algorithm as well but wrote something down wrong so I got gcd(7,16)...so stupid of me.", "Solution_12": "I would have done this Mock AMC if I didn't have a Spanish project to work on...\r\n\r\nAnyway, I just looked at number 14, and if you know something about linear recurrences...\r\n[hide=\"Then...\"]Maybe you can show that $ R_n\\equal{}6R_{n\\minus{}1}\\minus{}R_{n\\minus{}2}$[/hide]\r\nI did this in my head, so I might be wrong, but if anyone can verify that what I am saying is true...", "Solution_13": "That's what I did. Did you get 9 as the answer?\r\n\r\nEdit: Wait, I just read your post and realized that you didn't take it.\r\n\r\n[quote=\"xpmath\"]\nUgh. I somehow used the Euclidean Algorithm as well but wrote something down wrong so I got gcd(7,16)...so stupid of me.[/quote]\r\n\r\nAt first I had $ 7 \\plus{} 16 \\equal{} 25$, and it took me quite a bit of time to figure out what I did wrong. :lol:", "Solution_14": "OMG I thought that exact same thing for about the whole test. I jus went with dividing the num by the denom, but it's the same thing really.", "Solution_15": "[quote=\"Silverfalcon\"]I'm updating solutions to problems that were asked now (and soon too). So, keep asking for solutions for ones that you didn't get! Difficult problems don't benefit you unless you learn how to do them!\n\n[hide=\"Solution for Number 6\"]\nThis is a Mathcounts problem from the book [i]The All-Time Greatest Mathcounts Problems[/i]. You can use log but there is much easier way. It is from 1987 State Individual #9, which is current Target Round (name actually changed in 1988 so this is problem from last Individual Round!).\n\nNote the pattern when the units digit was written:\n\n$ 2^0 \\equal{} 1$ (This is the ODD one that does not belong in pattern)\n$ 2^1 \\equal{} 2$\n$ 2^2 \\equal{} 4$\n$ 2^3 \\equal{} 8$\n$ 2^4 \\equal{} 6$\n$ 2^5 \\equal{} 2$\n$ 2^6 \\equal{} 4$\n\nSo, the pattern is 2,4,8,6.... Note that 11213 = 11212 x 4 + 1. So, the units digit is same as the first term, which is 2. Now, we're going to subtract 1 to get the final answer $ \\fbox{1}$. [/hide]\n\nThere is no need to log. Plus, you can't use it (unless you have $ \\log 2$ and $ \\log 3$ memorized) this year anyway.[/quote]\r\nYes I got the answer this morning using modular arithmetic(not all parts of my solution). Anyway, I don't see why you have to give the information about the number of digits: 3376! It makes me more curious to solve it by hand without using modular and the only method I know relates to number of digit is logarithm function! It's a nice problem :) .", "Solution_16": "i used mod 3 for the one about the jugs, and got two possible cases that i just had to test out. \r\nim glad i saw the simple solution for the one about the units digit of that crazy exponent of 2. that's the only solution that came to my head . :D\r\n\r\n*btw, can anyone give a solution to #17. the first thing that came to my mind when i saw the equations was symmetry, and naturally i added them up. unfortunately i didn't know how to proceed from there as i couldn't find a slick way to factor or manipulate the resulting expression. :blush:", "Solution_17": "[hide=\"17\"]\nSumming all three equations gives $ x^2 \\minus{} 2xy \\plus{} 2y^2 \\plus{} 6yz \\plus{} 9yz^2 \\equal{} (x \\minus{} y)^2 \\plus{} (y \\plus{} 3z)^2 \\equal{} 175$. $ x \\minus{} y, y \\plus{} 3z$ are both integers, and since for all integers $ a$, $ a^2 \\equiv 0, 1 \\pmod{4}$, it follows that the sum of two squares can only be $ \\equiv 0, 1, 2 \\pmod{4}$. However, $ 175 \\equiv 3 \\pmod{4}$, from which it follows that there are no solutions for $ x,y,z$.\n[/hide]\r\n\r\nI skipped 17 when I took it because I tend to have bad experiences with systems of equations on timed tests.", "Solution_18": "thanks azjps! \r\nslick solution!\r\n\r\ni wish i had seen that on the test as well. :P", "Solution_19": "Apparently, I've reached maximum quota for uploading images. So, I'm going to put up the images in my website and put the link here.\r\n\r\nSince you guys are doing well on your own for easy to medium problems, I'm going to put up solutions to some of the later ones. Let's start with the last one, which is absolutely beautiful problem!\r\n\r\nImage:\r\n\r\nhttp://hiddendiscoveries.com/resources/Mock+AMC+A+Number+25.JPG\r\n\r\n[hide=\"Solution for Number 25\"]\nThis is 1993 AIME Number 15. However, when you understand the solution at the end, you'll say, \"Wow, that was pretty difficult but not bad at all!\" Why? It doesn't require any knowledge beyond one year of school geometry. Trust me on that one.\n\nFirst of all, description of the image. \n\n$ r_1$ is the radius of the circle left of $ CH$ and $ r_2$ is the radius of the circle right of $ CH$.\n$ P$ is the point that left circle is tangent to $ AC$ and $ Q$ is the point that left circle is tangent to $ AB$. \n$ x$ is length of $ AH$ and $ y$ is the length of $ BH$. Thus, $ x \\plus{} y \\equal{} c$.\n$ a,b,c$ are standard notation of triangle's sides.\n$ h$ is the length of height.\n\nSecond, some key points.\n\nThis problem is difficult because it's hard to decide where to start. I'm going to start from $ HQ$ and focus on the left circle. Note that $ HQ \\equal{} r_1$ (if you don't get it, see that the segment connecting the center of left circle to $ R$ is a straight line parallel to $ HQ$). Then, $ AQ \\equal{} x \\minus{} r_1$. Since $ P$ and $ Q$ are both points of tangency, $ AQ \\equal{} AP \\equal{} x \\minus{} r_1$. So, $ CP \\equal{} b \\minus{} (x \\minus{} r_1) \\equal{} b \\minus{} x \\plus{} r_1$. Again, using same idea, we know that $ CR \\equal{} h \\minus{} (b \\minus{} x \\plus{} r_1) \\equal{} h \\minus{} b \\plus{} x \\minus{} r_1$. But, note that $ RH \\equal{} r_1$ as well! (This is same way by comparing this segment to the radius of left circle going down and perpendicular to $ AB$).\n\nSo:\n\n$ h \\minus{} b \\plus{} x \\minus{} r_1 \\equal{} r_1 \\rightarrow r_1 \\equal{} \\frac {h \\minus{} b \\plus{} x}{2}$\n\nSimiarly, $ r_2 \\equal{} \\frac {h \\minus{} a \\plus{} y}{2}$. Do NOT just pass by saying, \"Oh okay.\" Work this out on your own so you completely get this!\n\nNow, going back to our expression from the problem:\n\n$ RS \\equal{} |r_1 \\minus{} r_2|$ (since we don't know $ r_1 < r_2$ or opposite; in the diagram, $ r_1 > r_2$ though)\n\n$ \\equal{} |\\frac {1}{2} (h \\minus{} b \\plus{} x \\minus{} h \\plus{} a \\minus{} y)| \\equal{} \\frac {1}{2} \\cdot |(a \\minus{} b) \\plus{} (x \\minus{} y)|$\n\nNow, using Pythagorean Theorem:\n\n$ h^2 \\equal{} b^2 \\minus{} x^2$\n$ h^2 \\equal{} a^2 \\minus{} y^2$\n\nThus, $ b^2 \\minus{} x^2 \\equal{} a^2 \\minus{} y^2$. Using difference of squares and rearranging, we get:\n\n$ x \\minus{} y \\equal{} \\frac {(a \\plus{} b)(a \\minus{} b)}{y \\plus{} x}$\n\nNow, substituting this back:\n\n$ \\equal{} \\frac {1}{2} \\cdot |(a \\minus{} b) \\plus{} \\frac {(a \\plus{} b)(a \\minus{} b)}{x \\plus{} y}|$\n\nRecall that $ x \\plus{} y \\equal{} c$ so:\n\n$ \\equal{} \\frac {1}{2} \\cdot | \\minus{} 1 \\plus{} \\frac {3987}{1995}| \\equal{} \\frac {996}{1995} \\equal{} \\frac {332}{665}$\n\nThus, $ m \\plus{} n \\equal{} 332 \\plus{} 665 \\equal{} \\fbox{997}$. [/hide]", "Solution_20": "For number 25 you can just find angle A in terms of a cosine.\r\n\r\nSo just do the law of cosines backwords. Youll get like 1993^2=1995^2+1994^2-2(1995)(1994)(cosA).\r\n\r\nYoull get cosA=333/665 (make the calculation easier by using cool factoring tricks-also you dont have to solve it all the way because a factor of 1994 from the denominator will cancel out once you do the x/1994=cosA)\r\n\r\nYou can now find the lengths of the two segments broken up by the height\r\n\r\nThen use that ratio to do HA/1994=cosA, so HA=1994(333/665)=998+332/665.\r\n\r\nH is where the height touches AB (look at silverfalcons picture)\r\n\r\nThen you will find like R+x where x (in silverfalcons picture its HA-R) is the segment that doesnt include the big radius but is still on one of the segments broken up by the height on the base (need a picture to explain this). You find that R+x=998+332/665. R is the bigger radius.\r\n\r\nThen you will also find like r+x using the fact that tangents to a circle have equal lengths....just try to use some variables and youll find that r+x where r is the smaller radius. r+x frmo some equation manipulation is 998 (try to set up some equations and subtract them)\r\n\r\nThen do R+x-(r-x)=first answer-second answer which is 998+(332/665)-998= 332/665.\r\nCalculation is a lot harder but since this was number 15 on AIME, 10-15 mintues is actually pretty quick (you can use alot fo (a-b)(a+b) and all that to find cosB which is the hard part).\r\n\r\nHow do you do that one with the basketball and free throws?", "Solution_21": "Use stewart's theorem", "Solution_22": "Please give your full solution, not just hint. Moreover, none of the problems in this mock AMC requires knowledge of Ceva or Stewart's or anything like that. It is just elementary geometry.\r\n\r\nI need a break from difficult problems so I'll put answers for easier problems first. Request any that you want to see the solutions! :) \r\n\r\n[hide=\"Solution for Number 1\"]\nThis is 1993 National Sprint Round #23. It is pretty easy problem to person who's been to math competition for a while. But for someone who's really new, this term is obscure --- telescoping.\n\nBasically, if sequence telescopes, then the terms begin to cancel out.\n\nFirst problem can be written as:\n\n$ \\frac{1}{1 \\cdot 2} \\plus{} \\frac{1}{2 \\cdot 3} \\plus{} \\frac{1}{3 \\cdot 4} \\plus{} \\cdots \\plus{} \\frac{1}{99 \\cdot 100}$\n\nNow, here comes the trick for telescoping.\n\n$ \\frac{1}{n} \\minus{} \\frac{1}{n\\plus{}1} \\equal{} \\frac{1}{n(n\\plus{}1)}$\n\nUsing this idea, we can now note that:\n\n$ \\frac{1}{1} \\plus{} \\left(\\minus{} \\frac{1}{2} \\plus{} \\frac{1}{2} \\minus{} \\frac{1}{3} \\plus{} \\frac{1}{3} \\cdots \\minus{} \\frac{1}{99} \\plus{} \\frac{1}{99}\\right) \\minus{} \\frac{1}{100} \\equal{} \\frac{1}{1} \\minus{} \\frac{1}{100} \\equal{} \\frac{99}{100}$ [/hide]", "Solution_23": "[hide=\"Solution for Number 2\"]\nThis is 1992 State Sprint #10. Honestly, to an average person, this problem is easier than the first one but since this is math competition and telescoping was pretty well-known, I put this one (even though it's state round) as number 2.\n\nIt is simple algebra..\n\n$ 1000 \\minus{} 2x \\equal{} 100 \\plus{} 3x \\rightarrow 900 \\equal{} 5x$\n\nNow, we're trying to find what $ 1000\\minus{}2x$ or $ 100\\plus{}3x$ is. So, instead of finding $ x$, I can multiply both sides by $ \\frac{3}{5}$ to get this:\n\n$ 540 \\equal{} 3x$\n\nAdd 100 to that to get 640.\n\nOf course, there is nothing wrong with finding $ x$. But this is one of those tricks that can save time (especially on the test like this). Because if you find $ x$, you'll have to find $ 3x$ again and well, why not just math in one step then?\n[/hide]\n\n[hide=\"Solution for Number 3\"]\nThis is 1984 State Team #3. I've seen several problems similar to this and as someone already mentioned, there are two ways to approach this problem.\n\nOne way is pattern, and other one is algebra. I'll go to the algebra first.\n\nWe can write the inside as:\n\n$ \\sqrt{1 \\plus{} n(n\\plus{}1)(n\\plus{}2)(n\\plus{}3)}$\n\nGroup the product as $ n(n\\plus{}3)$ and $ (n\\plus{}1)(n\\plus{}2)$ to get the $ n^2 \\plus{} 3n$ and then just set $ y \\equal{} n^2\\plus{}3n$ and solve it. Although this seems like an obvious way in this problem, there are AIME problems that can be solved using this manner. So, if you didn't learn this method, learn it now! It's very effective.\n\nNow, with pattern, note the final result.\n\n5\n11 = 5+6\n19 = 5+6+8\n29 = 5+6+8+10\n\nThis can be also written as $ 5 \\plus{} 2(3 \\plus{} 4 \\plus{} 5 \\plus{} \\cdots)$. Now, since the inside product in the problem is $ 50 \\cdot 51 \\cdot 52 \\cdot 53,$ we can write the result as:\n\n$ 5 \\plus{} 2(3 \\plus{} 4 \\plus{} 5 \\cdots \\plus{} 51) \\equal{} 5 \\plus{} \\left(\\frac{51 \\cdot 52}{2} \\minus{} 2 \\minus{} 1\\right) \\cdot 2 \\equal{} 2651$\n\nNow, here is another important formula found by Gauss.\n\n$ 1 \\plus{} 2 \\plus{} 3 \\plus{} \\cdots \\plus{} n \\equal{} \\frac{n(n\\plus{}1)}{2}$. The proof of this is as follows (using induction; now you don't need proof for AIME or AMC but anyway)\n\nFor $ n \\equal{} 1$, this holds true (1*2/2 = 1).\nNow, assume $ n \\equal{} k$ is true.\nThen, let's prove $ n \\equal{} k\\plus{}1$ is true.\n\n$ 1 \\plus{} 2 \\plus{} 3 \\plus{} \\cdots \\plus{} k \\plus{} k \\plus{} 1 \\equal{} \\frac{k(k\\plus{}1)}{2} \\plus{} k \\plus{} 1 \\equal{} \\frac{k(k\\plus{}1) \\plus{} 2(k\\plus{}1)}{2} \\equal{} \\frac{k^2\\plus{}3k\\plus{}2}{2} \\equal{} \\frac{(k\\plus{}1)(k\\plus{}2)}{2}$\n\nwhich is same as we get on RHS if we set $ n \\equal{} k\\plus{}1$ so thus, this method holds true. \n[/hide]", "Solution_24": "I apologize if I repeat any solutions from earlier in the thread. I'd like to help put answers to all of these problems up so that people may understand them.\r\n\r\n[hide=\"Comment for #3\"]\nYou can also notice that if the sequence is defined as $ \\sqrt{1\\plus{}(n)(n\\plus{}1)(n\\plus{}2)(n\\plus{}3)}$, the final result is $ 1\\plus{}(n\\plus{}3)(n)$. Simple and quick, but useful.\n[/hide]\n\n[hide=\"Solution to 4\"]\nFairly obvious. The maximum possible input into $ f(x)$ to use the $ f(x)\\equal{}x\\plus{}2$ definition is 9. Therefore the answer is $ 11$, $ B$.\n[/hide]\n\n[hide=\"Solution to 5\"]\nI found this somewhat confusingly worded, but it is straightforwards once you realize what exactly the problem is asking.\nLet the arithmetic mean of the $ 12$ numbers be $ x$. Therefore, the sum of the $ 12$ numbers is $ 12x$. $ \\frac{x}{12x} \\cdot 100\\% \\equal{} \\frac{25}{3}\\%$, $ E$.\n[/hide]\n\n[hide=\"Solution to 6\"]\nThis one relies on the pattern that occurs in the units digit of $ 2^n, n>0$. Observe:\n$ 2^{4n\\plus{}1} \\equiv 2 \\bmod{10}$\n$ 2^{4n\\plus{}2} \\equiv 4 \\bmod{10}$\n$ 2^{4n\\plus{}3} \\equiv 8 \\bmod{10}$\n$ 2^{4n} \\equiv 6 \\bmod{10}$\n\nSince $ 11213 \\equiv 1 \\bmod{4}$, the units digit of $ 2^11213\\minus{}1$ is $ 1$.\n[/hide]\n\n[hide=\"Solution to 7\"]\nSince the vendor must split the jugs up in a $ 1: 2$ ratio, the sum of the jugs without the blue punch must be congruent to $ 0 \\bmod{3}$. Since the sum of all the jugs fulfills this condition, only jugs congruent to $ 0 \\bmod{3}$ may be kept for the vendor. This narrows the choices down to a 15 or 21 gallon jug. Seeing that it is impossible to choose jugs so that their capacities sum to $ \\frac{(13\\plus{}15\\plus{}16\\plus{}21\\plus{}21\\plus{}22)}{3}\\minus{}15 \\equal{} 31$, we can conclude that the vendor keeps a 21-gallon jug for himself, meaning the answer is $ D$.[/hide]\n\n[hide=\"Solution to 8\"]\nThis one is easily solved using proportionality. From the given capacity of the tank, the ratio of the radius of the base to the height of the cone is $ \\frac{1}{6}$. If the tank is filled to 1/4 capacity, $ V_1\\equal{}\\frac{1}{4}V_0\\equal{}\\frac{\\pi}{3}r^2h$. Substituting using the fact $ 6r\\equal{}h$, we have $ V_1\\equal{}\\frac{1}{4}V_0\\equal{}\\frac{\\pi}{3}\\frac{1}{6^2}h^3$. Simplifying,\n$ V_0\\equal{}\\frac{\\pi}{3}\\frac{1}{3^2}h^3$\n$ \\frac{\\pi}{3}r_0^2h_0 \\equal{}\\frac{\\pi}{3}\\frac{1}{3^2}h^3$\n$ r_0^2 h_0\\equal{}\\frac{1}{3^2}h^3$\nNow, plug in the initial radius and height.\n$ 16^2 \\cdot 96 \\cdot 3^2 \\equal{} h^3$\nNow, group the cubes on the LHS.\n$ (16^3)(3^3)(2) \\equal{} 2(48^3) \\equal{} h^3$\nand it is remarkably easy to see that h will take the form $ 48\\sqrt[3]{2}$, and thus the answer is $ A$, $ 50$.\n[/hide]", "Solution_25": "[quote=\"Silverfalcon\"][hide=\"Correction for Number 11\"]\nI missed the error while I read the solution but there is an error in this problem. The correct problem should read:\n\n$ \\sin(x \\plus{} y) \\equal{} \\cos(x \\minus{} y)$\n\nNow, in the ARML book, the question has what I have. But, in the solution, it started off with saying $ \\sin(x \\plus{} y) \\minus{} \\cos(x \\minus{} y) \\equal{} 0$... [/hide]\n\n[hide=\"Solution for Number 11\"]\nIn this solution, I'm going to solve the problem assuming that what I have (and the book) is wrong and go with what the book's solution says. There is an answer to $ \\cos(x \\plus{} y)$ but because of answer choice, we'll go with $ \\cos(x \\minus{} y)$.\n\nI won't go into detail but essentially:\n\n$ \\sin(x \\plus{} y) \\minus{} \\cos(x \\minus{} y) \\equal{} (\\sin x \\minus{} \\cos x)(\\cos y \\minus{} \\sin y) \\equal{} 0$\nafter expanding using sine and cosine angle addition formula and factoring by terms.\n\nTherefore:\n\n$ \\sin x \\equal{} \\cos x$ and $ \\cos y \\equal{} \\sin y$\n\nI think that this is the best part of this problem. Because $ \\sin \\frac {\\pi}{4} \\equal{} \\cos \\frac {\\pi}{4}$, we can now say the following:\n\n$ x \\equal{} \\frac {\\pi}{4} \\plus{} n \\pi$\n$ y \\equal{} \\frac {\\pi}{4} \\plus{} n \\pi$\n\nFrom here, we'll consider four cases. \n\nCase 1: $ x \\geq 0$\nSolving $ \\frac {\\pi}{4} \\plus{} n \\pi \\leq 1995$ yields 498 as the quotient with remainder. The presence of remainder is important because this indicates that when $ n \\equal{} 498,$ the resulted line is not a tangent to the circle. So, counting $ n \\equal{} 0$, there are 499 lines resulting in 998 intersections (above the x-axis and below the x-axis).\n\nCase 2: $ y \\geq 0$\nSame as Case 1 so here are another 998 intersections.\n\nCase 3: $ x \\leq 0$\nDo not assume that this is 998 as well! Note that the first line in $ x \\leq 0$ starts off with $ \\frac { \\minus{} 3\\pi}{4}$ and ends with $ \\frac { \\minus{} 1995 \\pi}{4}$. So, this means that although there are 499 lines, the last line is TANGENT, thus only making one intersection. \n\nCase 4: $ y \\leq 0$\nSame as Case 3 so total of $ 499 \\times 2 \\minus{} 1 \\equal{} 997$.\n\nThus, the total number of intersections is $ 998 \\plus{} 998 \\plus{} 997 \\plus{} 997 \\equal{} 3990$ or $ \\fbox{B}$. [/hide][/quote]\r\nI don't get your solution in case 1? Why you can get 498 or 499 when solving that inequality? I got 0<=n<= 634.444 and I just can infer that n has 634 values :( . Someone please explains it to me!", "Solution_26": "There's a couple of typos:\r\n\r\n[quote=\"Silverfalcon\"]\nIn this solution, I'm going to solve the problem assuming that what I have (and the book) is wrong and go with what the book's solution says. There is an answer to $ \\cos(x \\plus{} y)$ but because of answer choice, we'll go with $ \\cos(x \\minus{} y)$.\n\nI won't go into detail but essentially:\n\n$ \\sin(x \\plus{} y) \\minus{} \\cos(x \\minus{} y) \\equal{} (\\sin x \\minus{} \\cos x)(\\cos y \\minus{} \\sin y) \\equal{} 0$\nafter expanding using sine and cosine angle addition formula and factoring by terms.\n\nTherefore:\n\n$ \\sin x \\equal{} \\cos x$ [i][u]or[/u][/i] $ \\cos y \\equal{} \\sin y$\n\nI think that this is the best part of this problem. Because $ \\sin \\frac {\\pi}{4} \\equal{} \\cos \\frac {\\pi}{4}$, we can now say the following:\n\n$ x \\equal{} \\frac {\\pi}{4} \\plus{} n \\pi$\n$ y \\equal{} \\frac {\\pi}{4} \\plus{} n \\pi$\n\nFrom here, we'll consider four cases. \n\nCase 1: $ x \\geq 0$\nSolving $ \\frac {\\pi}{4} \\plus{} n \\pi \\leq \\boxed{\\frac{1995 \\pi}{4}}$ yields 498 as the quotient with remainder. The presence of remainder is important because this indicates that when $ n \\equal{} 498,$ the resulted line is not a tangent to the circle. So, counting $ n \\equal{} 0$, there are 499 lines resulting in 998 intersections (above the x-axis and below the x-axis).\n\nCase 2: $ y \\geq 0$\nSame as Case 1 so here are another 998 intersections.\n\nCase 3: $ x < 0$\nDo not assume that this is 998 as well! Note that the first line in $ x \\leq 0$ starts off with $ \\frac { \\minus{} 3\\pi}{4}$ and ends with $ \\frac { \\minus{} 1995 \\pi}{4}$. So, this means that although there are 499 lines, the last line is TANGENT, thus only making one intersection. \n\nCase 4: $ y < 0$\nSame as Case 3 so total of $ 499 \\times 2 \\minus{} 1 \\equal{} 997$.\n\nThus, the total number of intersections is $ 998 \\plus{} 998 \\plus{} 997 \\plus{} 997 \\equal{} 3990$ or $ \\fbox{B}$. [/quote]\r\n\r\nThe casework at the end isn't really necessary if you see that $ |x|\\equal{}\\left|\\frac {\\pi}{4} \\plus{} n \\pi\\right| \\le \\frac{1995\\pi}{4}$ gives us $ 998$ values of $ n$ and hence $ 998 \\times 2 \\minus{} 1 \\equal{} 1995$ tangent lines, and then double for the $ y$ values.", "Solution_27": "Again, azips, how can you show that n has 998 values?? I just have the same result as I mentioned above :( .Can you show your solution to this inequality in detailed?\r\nPS: Anybody has solution of problem 19 and 21? I really need to see the solution of these problems, especially to 19 when I can't find the number of ways to choose 3 segments form a triangle out of 45C3 sets of 4 segments.", "Solution_28": "I have been busy and forgot to keep my promise of providing solutions to ALL problems in this mock AMC. I'll do so as much as I get chance.\r\n\r\n[hide=\"Solution to Number 24\"]\n\nI'll just type up the official solution from USAMTS website. This was USAMTS problem on Round 4 #5 in Year 10.\n\nNote that this problem's answer holds true regardless of the shape of $ \\triangle ABC$. The solution is nicely done with coordinates (somewhat brute forcing but pretty elegant).\n\n$ \\triangle ABC$ can be placed on a coordinate system with $ A$ on the origin, $ B$ at $ (4x,0)$, and $ C$ at $ (4y,4z)$. By the midpoint theorem, $ D \\equal{} (2x\\plus{}2y, 2z), E \\equal{} (2y,2z), F \\equal{} (2x,0), P \\equal{} (x\\plus{}y,z), Q \\equal{} (2x\\plus{}y,z),$ and $ R \\equal{} (x\\plus{}2y,2z)$.\n\nUsing the distance formula,\n$ AQ^2 \\equal{} (2x\\plus{}y)^2 \\plus{} z^2 \\equal{} 4x^2 \\plus{} 4xy \\plus{} y^2 \\plus{} z^2$\n$ AR^2 \\equal{} (x\\plus{}2y)^2 \\plus{} 4z^2 \\equal{} x^2 \\plus{} 4xy \\plus{} 4y^2 \\plus{} 4z^2$\n$ BP^2 \\equal{} (3x\\minus{}y)^2 \\plus{} z^2 \\equal{} 9x^2 \\minus{} 6xy \\plus{} y^2 \\plus{} z^2$\n$ BR^2 \\equal{} (3x\\minus{}2y)^2 \\plus{} 4z^2 \\equal{} 9x^2\\minus{}12xy\\plus{}4y^2 \\plus{} 4z^2$\n$ CP^2 \\equal{} (x\\minus{}3y)^2 \\plus{} 9z^2 \\equal{} x^2 \\minus{} 6xy \\plus{} 9y^2 \\plus{} 9z^2$\n$ CQ^2 \\equal{} (2x\\minus{}3y)^2 \\plus{} 9z^2 \\equal{} 4x^2 \\minus{} 12xy\\plus{}9y^2 \\plus{} 9z^2$\nAdding these up gives $ 28(x^2 \\minus{} xy \\plus{} y^2 \\plus{} z^2)$. Furthermore,\n\n$ AB^2 \\equal{} 16x^2$\n$ BC^2 \\equal{} 16x^2 \\minus{} 32xy \\plus{} 16y^2 \\plus{} 16z^2$\n$ CA^2 \\equal{} 16y^2 \\plus{} 16z^2$\n\nAdding these up gives $ 32(x^2 \\minus{} xy \\plus{} y^2 \\plus{} z^2)$.\n\nThat makes $ \\displaystyle \\frac{AQ^2 \\plus{} AR^2 \\plus{} BP^2 \\plus{} BR^2 \\plus{} CP^2 \\plus{} CQ^2}{AB^2 \\plus{} BC^2 \\plus{} CA^2} \\equal{} \\frac{28(x^2 \\minus{} xy \\plus{} y^2 \\plus{} z^2)}{32(x^2 \\minus{} xy \\plus{} y^2 \\plus{} z^2)} \\equal{} \\frac{7}{8}$\n\nSince this construction used an arbitrary triangle, the value is independent of the shape of $ \\triangle ABC$. So, $ m\\plus{}n \\equal{} 15$ or E. [/hide]", "Solution_29": "[hide=\"Number 22 OFFICIAL SOLUTION\"]\nThis is 2002 Putnam Problem so don't worry to see this problem because it won't happen. It's pretty nice problem though.\n\nI'll first put the official solution first which is by Putnam committee itself.\n\nQUOTED:\n\nThe probability is 1/99. In fact, we show by induction on $ n$ that after $ n$ shots, the probability of having made any number of shots from 1 to $ n\\minus{}1$ is equal to $ \\frac{1}{n\\minus{}1}$. This is evident for $ n \\equal{} 2$. Given the result for $ n$, we see that the probability of making $ i$ shots after $ n\\plus{}1$ attempts is\n\n$ \\frac{i\\minus{}1}{n} \\cdot \\frac{1}{n\\minus{}1} \\plus{} \\left(1 \\minus{} \\frac{i}{n}\\right) \\cdot \\frac{1}{n\\minus{}1} \\equal{} \\frac{(i\\minus{}1) \\plus{} (n\\minus{}i)}{n(n\\minus{}1)} \\equal{} \\frac{1}{n}$ \n\nas claimed. [/hide]" } { "Tag": [ "geometry", "3D geometry", "function" ], "Problem": "Slove that:$\\sqrt[3]{2x-1}=x \\sqrt[3]{16}-\\sqrt[3]{2x+1}$[list][/list]", "Solution_1": "[hide]\n$\\sqrt[3]{2x-1}+\\sqrt[3]{2x+1}=x\\sqrt[3]{16}$\n$2x-1+3\\sqrt[3]{(2x-1)^{2}(2x+1)}+3\\sqrt[3]{(2x-1)(2x+1)^{2}}+2x+1=16x^{3}$\n$4x+3\\sqrt[3]{(2x-1)^{2}(2x+1)}+3\\sqrt[3]{(2x-1)(2x+1)^{2}}=16x^{3}$\n$4x+3\\sqrt[3]{(2x-1)(2x+1)}(\\sqrt[3]{2x-1}+\\sqrt[3]{2x+1})=16x^{3}$\n$4x+3\\sqrt[3]{(2x-1)(2x+1)}(x\\sqrt[3]{16})=16x^{3}$\n$3x\\sqrt[3]{16(2x-1)(2x+1)}=16x^{3}-4x$\nOne solution is $x=0$.\nSubstitute $k=4x^{2}-1$:\n$3\\sqrt[3]{16k}=4k$\n$27k=4k^{3}$\n$k=0,\\frac{\\pm 3\\sqrt{3}}{2}$\nso i think the solutions are $\\{0,\\pm\\frac{1}{2}},\\pm\\frac{1}{2}\\sqrt{\\frac{2+3\\sqrt{3}}{2}},\\pm\\frac{1}{2}\\sqrt{\\frac{2-3\\sqrt{3}}{2}\\}$ :| \n[/hide]", "Solution_2": "${\\sqrt[3]{2 x-1}=x \\sqrt[3]{16}-\\sqrt[3]{2 x+1}}$\r\n\r\n${x=\\frac{1}{2}\\lor x=\\sqrt{\\frac{1}{4}+\\frac{3 \\sqrt{3}}{8}}}$\r\n\r\nminsoens, Your solution is wrong. Try ${ x = \\frac{-1}{2}}$ and the others.\r\nBefore post a solution, try it on the problem, because of multiple valuable functions \r\n\r\nMy solution is the same, i just rewrite the fraction in other way", "Solution_3": "$\\sqrt[3]{2x-1}=x \\sqrt[3]{16}-\\sqrt[3]{2x+1}$\r\nwait.. it doesn't work? :huh: \r\n\\begin{eqnarray*}\\sqrt[3]{2\\left(-\\frac{1}{2}\\right)-1}&=&\\left(-\\frac{1}{2}\\right)\\sqrt[3]{16}-\\sqrt[3]{2\\left(-\\frac{1}{2}\\right)+1}\\\\ \\sqrt[3]{-1-1}&=&-\\sqrt[3]{2}-\\sqrt[3]{-1+1}\\\\ \\sqrt[3]{-2}&=&-\\sqrt[3]{2}\\end{eqnarray*}\r\nalso i tried $ x=\\frac{1}{2}$ and $\\pm\\frac{1}{2}\\sqrt{\\frac{2+3\\sqrt{3}}{2}}$ on my calculater and i got $-7 \\times 10^{-14}$ and $1 \\times 10^{-13}$", "Solution_4": "[quote=\"minsoens\"][hide]\n$\\sqrt[3]{2x-1}+\\sqrt[3]{2x+1}=x\\sqrt[3]{16}$\n$2x-1+3\\sqrt[3]{(2x-1)^{2}(2x+1)}+3\\sqrt[3]{(2x-1)(2x+1)^{2}}+2x+1=16x^{3}$\n$4x+3\\sqrt[3]{(2x-1)^{2}(2x+1)}+3\\sqrt[3]{(2x-1)(2x+1)^{2}}=16x^{3}$\n$4x+3\\sqrt[3]{(2x-1)(2x+1)}(\\sqrt[3]{2x-1}+\\sqrt[3]{2x+1})=16x^{3}$\n$4x+3\\sqrt[3]{(2x-1)(2x+1)}(x\\sqrt[3]{16})=16x^{3}$\n$3x\\sqrt[3]{16(2x-1)(2x+1)}=16x^{3}-4x$\nOne solution is $x=0$.\nSubstitute $k=4x^{2}-1$:\n$3\\sqrt[3]{16k}=4k$\n$27k=4k^{3}$\n$k=0,\\frac{\\pm 3\\sqrt{3}}{2}$\nso i think the solutions are $\\{0,\\pm\\frac{1}{2}},\\pm\\frac{1}{2}\\sqrt{\\frac{2+3\\sqrt{3}}{2}},\\pm\\frac{1}{2}\\sqrt{\\frac{2-3\\sqrt{3}}{2}\\}$ :| \n[/hide][/quote]\r\nThank your solution", "Solution_5": "I think let $a=\\sqrt[3]{2x-1}$and $b=\\sqrt[3]{2x+1}$", "Solution_6": "We can rewrite the equation as $\\sqrt[3]{2x-1}+\\sqrt[3]{2x+1}+\\sqrt[3]{-16x^{3}}= 0$. Recall the factorization\r\n\\[a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca), \\]\r\nso\r\n\\begin{align*}(2x-1)+(2x+1)+(-16x^{3}) &= 3\\sqrt[3]{(4x^{2}-1)(-16x^{3})}\\\\-16x^{3}+4x &= 3\\sqrt[3]{(4x^{2}-1)(-16x^{3})}\\\\ 4(4x^{2}-1) &= 3\\sqrt[3]{16(4x^{2}-1)}\\\\ \\end{align*}\r\nIf $x\\ne 0$, then let $y =4x^{2}-1$. We have \\[4y = 3\\sqrt[3]{16y}\\implies y=0\\text{ or }y = \\pm\\frac{3}{2}\\sqrt{3}\\] so either $4x^{2}-1=0\\iff x=\\pm\\frac{1}{2}$ or $4x^{2}-1=\\pm\\frac{3}{2}\\sqrt{3}$. I'm too lazy to solve that", "Solution_7": "That doesn't work chess. $a^{2}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)$", "Solution_8": ":oops:" } { "Tag": [ "search" ], "Problem": "George is making spaghetti for dinner. He placed 4.01 kg of water in a pan and brings it to a boil. Before adding the pasta, he adds 58 g of table salt to the water and brings it to a boil. What is the temperature of the salty boiling water? (Its a nice day at sea level so that the pressure is 1.00 atm. Assume negligible evaporation of water. Kb for water is 0.52 degress Celsius/molality.\r\n\r\nFirst you have to find the molality of NaCl:\r\n(58 g NaCl)(1 mole NaCl/58.4425 g NaCl) = .99243 mole NaCl\r\n(.99243 mole NaCl)/(4.01 kg H2O) = .247488 m NaCl\r\nSo, the molality is .247488 m NaCl.\r\n\r\nThen, I think that you have to multiply the molality by the Kb of water (0.52 degrees Celsius).\r\nThat would equal .12869 degrees Celsius.\r\n\r\nFinally, you would add that number to 100 degrees, to get 100.13 degrees Celsius.\r\n\r\nI did a search for this problem and found that the answer was 100.26 degrees Celius. Did I do the problem right, or is the answer I found right?\r\n\r\nI think I found where I messed up. Since there are two ions in NaCl, I multiply (.247488 m NaCl) by (0.52 degress Celsius) by (2.0) to get .25739 degrees Celsius.\r\n\r\nThen I add that to 100 and get the answer that I found, 100.26. Is that answer right?\r\n\r\nPlease help.", "Solution_1": "i'd say that should be correct....the one with the vant hoff factor...100.26!!!", "Solution_2": "You're right it is the particle molarity. If one had molecules with three ions, you would have to multiply it by 3.", "Solution_3": "it's molality that she used... colligative properties..." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all pairs functions $f,g: \\mathbb{N}_{0}\\to\\mathbb{N}_{0}$ such that \\[f(m)-f(n)=(m-n)(g(m)+g(n))\\quad \\forall m,n\\in\\mathbb{N}_{0},\\]\nwhere $\\mathbb{N}_{0}=\\{0,1,2,...\\}$.", "Solution_1": "Hello N.T.TUAN\r\n\r\n [quote=\"N.T.TUAN\"]Find all pairs functions $f,g: \\mathbb{N}_{0}\\to\\mathbb{N}_{0}$ such that\n\\[f(m)-f(n)=(m-n)(g(m)+g(n))\\; \\forall m,n\\in\\mathbb{N}_{0}, \\]\nwhere $\\mathbb{N}_{0}=\\{0,1,2,...\\}$.[/quote]\r\n\r\n$f(m)-f(n)=(m-n)(g(m)+g(n))$ $\\Rightarrow $ $f(m)=m(g(m)+g(0))+f(0)$ $\\Rightarrow $ $f(m)-f(n)=m(g(m)+g(0))-n(g(n)+g(0))$\r\nAnd so : $m(g(m)+g(0))-n(g(n)+g(0))=(m-n)(g(m)+g(n))$, which implies $n(g(m)-g(0))=m(g(n)-g(0))$ and $g(n)=a*n+b$\r\n\r\nThen $f(n)=n(g(n)+g(0))+f(0)=an^{2}+2bn+c$\r\n\r\nAnd it is easy to verify that these necessary conditions work.\r\n\r\nSo the solution is :\r\n$g(n)=a*n+b$\r\n$f(n)=an^{2}+2bn+c$\r\n$a,b,c\\in\\mathbb{N}_{0}$\r\n\r\n-- \r\nPatrick" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "degf(X)= 2007\r\nf(X) is an integer polynomial that have a2007=1\r\nProve that (f(X))^2 - 9 has no more than 2010 roots", "Solution_1": "no more than 2010 roots of what type?", "Solution_2": "And does $ a_{2007}$ refer to the coefficient of $ x^{2007}$ or $ x^0$? (Conventions differ.)" } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "Let be given a triangle $ABC.$ Prove that: \\[a^{2}\\sin^{2}\\frac{A}{2}+b^{2}\\sin^{2}\\frac{B}{2}+c^{2}\\sin^{2}\\frac{C}{2}\\ge\\frac{1}{12}\\cdot (a+b+c)^{2}\\]\r\nIt's really easy problem but do you have a solution without using [b]Chebyshev Inequality[/b]?", "Solution_1": "$a^{2}(1-cosA)+b^{2}(1-cosB)+c^{2}(1-cosC)\\geq \\frac{1}{6}(a+b+c)^{2}$\r\n$\\sum \\frac{2a^{3}bc+a^{5}-a^{3}b^{2}-a^{3}c^{2}}{2abc}\\geq \\frac{1}{6}(a+b+c)^{2}$\r\n$\\sum 2a^{3}bc+a^{5}\\geq \\frac{abc}{3}(a+b+c)^{2}+\\sum (a^{3}b^{2}+a^{3}c^{2})$\r\n\r\nBut $\\sum a^{3}bc \\geq \\frac{abc}{3}(a+b+c)^{2}$ by AM-GM.\r\nAnd I'm too lazy to do the other one, but hopefully it works by some kind of Schur-like thing." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Show that if dimV>1, then the set of normal operators on V is not a subspace of L(V).\r\n\r\nP.S The problem doesn't specific whether V is on C or R. So I think it wants to show both.", "Solution_1": "Seems like all you have to do is find two $ 2 \\times 2$ normal matrices whose sum is not normal. \"Most\" non-trivial choices work, for instance\r\n\r\n$ A \\equal{} \\begin{pmatrix} 1 && 2 \\\\ 2 && 3 \\end{pmatrix}$\r\n\r\n$ B \\equal{} \\begin{pmatrix} 0 && 2 \\\\ \\minus{}2 && 0 \\end{pmatrix}$\r\n\r\nwhich are obviously normal (symmetric and skew-symmetric, respectively) over either $ \\mathbb{R}$ or $ \\mathbb{C}$, and whose sum\r\n\r\n$ A \\plus{} B \\equal{} \\begin{pmatrix} 1 && 4 \\\\ 0 && 3 \\end{pmatrix}$\r\n\r\nis not normal, being triangular and not diagonal. For higher dimensions, just embed these matrices in larger ones in the obvious way.", "Solution_2": "Is A really normal? I think any normal operator has form :blush: \r\n\r\na -b\r\nb a", "Solution_3": "[quote=\"Saradush\"]Is A really normal? I think any normal operator has form :blush: \n\na -b\nb a[/quote]\r\n\r\nNope. A [url=http://en.wikipedia.org/wiki/Normal_operator]normal operator[/url] $ A$ is one which commutes with its adjoint $ A^*$. In matrix terms, this means that $ A$ commutes with its transpose-conjugate, which is just transpose when $ A$ is real. A symmetric matrix $ A$ satisfies $ A\\equal{}A^T$, so it obviously satisfies $ AA^T\\equal{}A^TA$." } { "Tag": [ "number theory", "least common multiple", "modular arithmetic", "quadratics", "induction", "number theory proposed" ], "Problem": "Prove that this expression $ a^2 \\plus{} b^2 \\plus{}c^2$ $ \\equal{}$ $ 7$ does not have any answer in rational numbers.\r\n(a,b,c are rational numbers).", "Solution_1": "First, reduce it to a problem in integers: assume that there's a solution, and let $ d$ be the LCM of the denominators of $ a,b,c$. After multiplying both sides by $ d^2$, the equation takes the form $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 7d^2$, $ a,b,c,d \\in \\mathbb{Z}, d \\neq 0$.\r\nIt is enough to disprove the existence of solution to this equation. The squares mod $ 8$ are $ 0,1,4$.\r\nIf $ d$ is odd, then $ 7d^2 \\equiv 7 \\pmod 8$, but adding 3 (not necessarily distinct) numbers from the set $ \\{ 0,1,4 \\}$ never yields $ 7 \\pmod 8$.\r\nThus, $ d$ is even. $ 7d^2 \\equiv 0,4 \\pmod 8$ and thus $ a,b,c$ are even (adding 3 (not necessarily distinct) numbers from the set $ \\{ 0,1,4 \\}$ yields 0 or 4 iff we take 0 and 4, which are quadratic residues that correspond to even numbers).\r\n\r\nSo for any existing solution $ (a,b,c)$ we can find another solution, $ (\\frac {a}{2},\\frac {b}{2},\\frac {c}{2})$. By induction, $ 2^k |a$ for all $ k$'s, which implies $ a \\equal{} 0$. Likewise, $ b \\equal{} c \\equal{} 0$, which implies $ d \\equal{} 0$, a contradiction.\r\nHence, there are no rational solutions to $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 7$." } { "Tag": [ "function", "logarithms", "algebra", "floor function" ], "Problem": "Hi to all...\r\n\r\nAgain, if any of you have seen this topic posted in another \"topic\", please notify me and I'll go there (thanks).\r\n\r\nThe question is this: \"[i]How many terminal zeros (consecutive zeros appearing at the end of a number) are there, in terms of n, for n![/i]\"?\r\n\r\nIt's a nice easy problem that proved to be interesting. If you have a solution, please explain it as well.\r\n\r\n ;) Try not to make your posts too long, no one wants to read a post that spans for pages and pages...", "Solution_1": "[hide]in base ten, a zero is made by containing a factor of ten, or 5*2 as a product of primes. as every n! n>1 has more twos in its prime factors than fives, the number of fives determine the number of tens which are factors of n!, and therefore the number of zeroes on the end. if you factorise n! into a product of primes:\n\n(2^a)(3^b)(5^c)(7^d)...(p^x) where p is the largest prime factor, then the number of zeroes after n! is c. c in terms of n is\n\nc = [n/5] + [n/5^2] + [n/5^3] + ... to infinity, where [a] is the largest integer such that [a]1$)", "Solution_1": "Hehe, two problems from turkish MO of same \"type\".", "Solution_2": "i cant understand what is the question of the same type on the second round of 1993", "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?t=112708\r\n\r\nnot same year, still funny :)", "Solution_4": "The problem is not quite clear. Can some one from Turkey clarify this problem, especially condition C).", "Solution_5": "The condition $c)$ is problematic above. For the sake of convenience, I'll rewrite the problem.\n\nGiven positive integers $n_1,n_2,\\dots,n_k,a$, satisfying:\n$\\bullet$ $(n_i,n_j)=1$ if $i\\neq j$,\n$\\bullet$ $a^{n_i} \\equiv 1 \\pmod{n_i}$, for every $i$,\n$\\bullet$ $n_i \\nmid a-1$,\n\nshow that the total number of integers $x\\geq 2$, enjoying $a^x \\equiv 1 \\pmod{x}$ is at least $2^{k+1}-2$. \n\nI will post my solution later, unless someone else already posts till then. ", "Solution_6": "Enough waiting time ;) ... Here is the proof, that I've promised to post. \n\nWe begin by a few observations. First, notice that, $a$ is coprime with any $n_i$. Next,\n[b]Claim[/b] None of the $n_i$'s are prime numbers.\n[b]Proof[/b]\nAssume that $n_i$ is prime for some $i$. By Fermat's little theorem, we have $a^{n_i-1}\\equiv 1 \\pmod{n_i}$. Combining this with the second item above, gives us $n_i \\mid a-1$, contradicting with the preamble.$\\square$\n\nThe next claim gives us a way to generate such $x$'s. \n[b]Claim[/b] Let $I$ be a non-empty subset of $\\{1,2,\\dots,k\\}$. Define,\n$$\nx_I \\triangleq \\prod_{i\\in I}n_i.\n$$\nThen, $a^{x_I}\\equiv 1 \\pmod{x_I}$. Hence, we have shown the existence of at least $2^{k}-1$ such numbers.\n[b]Proof[/b]\nFirst, notice that $a^{n_i}\\equiv 1 \\pmod{n_i}$ gives us, $a^{x_I}\\equiv 1 \\pmod{n_i}$, for every $i\\in I$. Since $n_i$'s are coprime, we also have,\n$$\na^{x_I}\\equiv 1 \\pmod{x_I}.\n$$\n\nNext, we will show the existence of a second collection, in a way, analogous to $\\{n_1,n_2,\\dots,n_k\\}$, and therefore will conclude the proof.\n[b]Claim[/b] For every $i$, let $p_i$ be the smallest prime divisor of $n_i$. Then, the following holds:\n$\\bullet$ $n_i>p_i$ for every $i$.\n$\\bullet$ $\\{p_i:i=1,2,\\dots,k\\}$ consists of $k$-distinct primes.\n$\\bullet$ $a \\equiv 1 \\pmod{p_i}$. Hence, $a^{p_i} \\equiv 1\\pmod{p_i}$.\nNoticing that this collection satisfies $a^{p_i}\\equiv 1 \\pmod{p_i}$ for every $i$, we can generate $2^{k}-1$ other such numbers via first taking a non-empty subset $I$ of $\\{1,2,\\dots,k\\}$, and then, letting $y_I = \\prod_{i\\in I}p_i$. Since $n_i>p_i$ for every $i$, and $n_i$'s are coprime, this will generate $2^k-1$ other numbers, enjoying the desired property, and therefore establishing the proof.\n[b]Proof[/b]\nFor each $n_i$, we have already shown that $n_i$ cannot be a prime number, and therefore, $n_i>p_i$. Also, since $n_i$'s are coprime, for any $i\\neq j$, the set of prime divisors of $n_i$ and the set of prime divisors of $n_j$ are disjoint, hence, $p_i$'s are all distinct.\n\nIt remains to prove the last item. For this, note that by Fermat's theorem, we have $a^{p_i-1}\\equiv 1 \\pmod{p_i}$. Now, since $a^{n_i}\\equiv 1 \\pmod{p_i}$, we have,\n$$\na^{(n_i,p_i-1)}\\equiv 1 \\pmod{p_i} \\implies a\\equiv 1\\pmod{p_i},\n$$\nwhere the last assertion holds due to the fact that since $p_i$ is the smallest prime divisor of $n_i$, none of the prime divisors of $p_i-1$ can divide $n_i$, hence they are coprime.\n" } { "Tag": [ "trigonometry", "inequalities unsolved", "inequalities" ], "Problem": "For $ a;b \\in R$.Find min and max of\r\n$ F(a;b)=\\frac{\\sqrt{3}}{4} sin(a-b)+sin(\\frac{\\pi}{6}-a)+sin(b-\\frac{\\pi}{6})$", "Solution_1": "hello, to obtain all critical points you must solve the following equation system with respect to $ a$ and $ b$:\r\n$ \\frac{\\sqrt{3}}{4}\\cos(a\\minus{}b)\\minus{}\\cos\\left(\\frac{\\pi}{6}\\minus{}a\\right)\\equal{}0 \\newline$\r\nand\r\n$ \\minus{}\\frac{\\sqrt{3}}{4}\\cos(a\\minus{}b)\\plus{}\\cos\\left(b\\minus{}\\frac{\\pi}{6}\\right)\\equal{}0$\r\nSonnhard." } { "Tag": [ "integration", "trigonometry", "calculus", "algebra", "polynomial", "function", "quadratics" ], "Problem": "$ \\int\\frac{dx}{4\\minus{}5sinx}$\r\n\r\ni guess i'm supposed to use the weierstrauss method? but i have no idea what that is\r\n\r\nso, help?", "Solution_1": "Make, as usual, $ \\tan \\frac {x}{2} \\equal{} t$.\r\n\r\n[hide=\"Then,\"] $ \\sin x \\equal{} \\frac{2t}{1\\plus{}t^2}$, $ x' \\equal{} \\frac{2}{1\\plus{}t^2}$, and so\n\n$ P \\equal{} \\int \\frac{1}{4\\minus{}5 \\sin x}\\,dx \\equal{} \\int \\frac{dt}{2t^2\\minus{}5t\\plus{}2}$.[/hide]", "Solution_2": "Expanding out the details of Carcul's post:\r\n\r\nSince $ x\\equal{}2\\arctan t,$ we have $ dx\\equal{}\\frac{2}{1\\plus{}t^2}\\,dt.$\r\n\r\n$ \\cos\\left(\\frac{x}{2}\\right)\\equal{}\\frac{1}{\\sqrt{1\\plus{}t^2}}$ and $ \\sin\\left(\\frac{x}{2}\\right)\\equal{}\\frac{t}{\\sqrt{1\\plus{}t^2}}.$\r\n\r\nHence, $ \\cos x\\equal{}\\cos^2\\left(\\frac{x}{2}\\right)\\minus{}\r\n\\sin^2\\left(\\frac{x}{2}\\right)\\equal{}\\frac{1\\minus{}t^2}{1\\plus{}t^2}$ and\r\n\r\n$ \\sin x\\equal{}2\\sin\\left(\\frac{x}{2}\\right)\\cos\\left(\\frac{x}{2}\\right)\\equal{}\r\n\\frac{2t}{1\\plus{}t^2}.$\r\n\r\nMake these substitutions in the original integral, and you will have the integral of a rational function. In this particular case:\r\n\r\n$ \\int\\frac{dx}{4\\minus{}5\\sin x}\\equal{}\\int\\frac{\\frac{2}{1\\plus{}t^2}\\,dt}\r\n{4\\minus{}\\frac{10t}{1\\plus{}t^2}}$\r\n\r\n$ \\equal{}\\int\\frac{2\\,dt}{4\\plus{}4t^2\\minus{}10t}\\equal{}\\int\\frac{dt}{2t^2\\minus{}5t\\plus{}2}.$\r\n\r\nI'll let you finish from there - the remaining steps include factoring the denominator and using partial fractions.\r\n\r\nOne note: that polynomial in the denominator does have real roots, which means that in any definite integral we must be careful about singularities. These singularities were not introduced by the substitution - they were present in the original integral. The original integral has singular points whenever $ \\sin x\\equal{}\\frac45.$\r\n\r\nThis method is guaranteed to turn the integral of any rational function of the usual trigonometric function into a the integral of a rational function. If (and this is a very, very big if) you can factor the denominator of the resulting rational function into linear and quadratic factors (and do so in a closed form), then you can express the integral in elementary terms.\r\n\r\nBut as I said, that is a very big if. The Fundamental Theorem of Algebra guarantees that such a factorization exists, but not that you can find it in closed form.", "Solution_3": "wait, i am so confused\r\n\r\nhow do you make $ tan\\frac{x}{2} \\equal{} t$ ?\r\n\r\nor even....\r\n\r\nwhere did this come from?\r\n\r\n$ x \\equal{} 2arctan^{\\minus{}1}t$ ?\r\n\r\ni honestly have no idea what weierstrauss is so, please explain?", "Solution_4": "Karl Weierstrass (note spelling) was one of the mathematical giants of the 19th century. Of the many things he did, this little manipulation is surely one of the least important. But it does commonly have his name attached to it.\r\n\r\nThis - the Weierstrass method - is simply a certain specific [i]substitution[/i] in an integral. The substitution can either be written \"forwards\" as $ t \\equal{} \\tan\\left(\\frac {x}2\\right)$ or \"backwards\" by solving that equation for $ x$ in terms of $ t.$ The result of that solution is $ x \\equal{} 2\\arctan x$ (or $ x \\equal{} 2\\tan^{ \\minus{} 1}x$ if you prefer. Just don't say $ \\arctan^{ \\minus{} 1}x,$ which isn't what we mean.)\r\n\r\nAs for the explanation - the algebraic manipulations in my previous post are the explanation. There's really nothing deeper here - just those calculations, and the fact that they always \"work\" in a case like this. (If by \"work\" I mean they give you a rational function of $ t$ to integrate; finishing from there has its own complications.)", "Solution_5": "[quote=\"charlestreykang\"]wait, i am so confused\n\nhow do you make $ tan\\frac {x}{2} \\equal{} t$ ?\n\nor even....\n\nwhere did this come from?\n\n$ x \\equal{} 2arctan^{ \\minus{} 1}t$ ?\n\ni honestly have no idea what weierstrauss is so, please explain?[/quote]\r\n\r\nI've seen some \"creative substitutions\" in the forum as well. I found this link that may provide some insight if you're really interested.\r\n\r\n[url=http://pear.math.pitt.edu/Calculus2/week3/3_2.html#3_2li1.html]Good Short Article[/url].\r\n[url=http://pear.math.pitt.edu/Calculus2/week3/3_2li4.html]and Here[/url]." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Given a triangle $ABC$, let $K$ be the midpoint of $AB$ and $L$ a be point on\r\n$AC$ such hat $AL = LC+CB$. Prove that $f(0)$ for all sufficiently large $n$, which means, by our assumption, that $f(n)>f(2n)$ for all sufficiently large $n$. This means that we can find $n$ s.t. $f(n)>f(2n)>f(4n)>\\ldots$, so we get an infinite descending sequence of naturals, which is impossible.\r\n\r\nJust an observation: I think orl posted, some time ago, a problem stating that for any $k$ the numbers $1,2,\\ldots,2^k$ can be permuted by a permutation $\\sigma$ so that there are no $a 0$, we know $ g(0) \\equal{} 2$.\r\n\r\nSwitching the roles of $ a$ and $ b$, it is evident $ g( \\minus{} a) \\equal{} g(a)$. \r\n\r\nHowever, here's the big issue: suppose $ \\sigma : \\mathbb{R} \\to \\mathbb{R}$ is an additive function,\r\ni.e. one satisfying the Cauchy equation $ \\sigma(x \\plus{} y) \\equal{} \\sigma(x) \\plus{} \\sigma(y)$; then\r\n$ g(\\sigma a)g(\\sigma b) \\equal{} g(\\sigma a \\plus{} \\sigma b) \\plus{} g (\\sigma a \\minus{} \\sigma b) \\equal{} g(\\sigma(a \\plus{} b)) \\plus{} g(\\sigma(a \\minus{} b))$.\r\n\r\nIn other words, if $ g$ is a solution, so is $ g \\circ \\sigma$; so we're already going to have that amazingly large\r\ndegree of freedom.\r\n\r\nNot to say it's not solvable (it probably is), but there are definitely an infinitude of ugly solution functions $ g$ if\r\nthere are any nontrivial ones.\r\n\r\n[quote=\"pco\"][quote=\"Raja Oktovin\"]Let $ \\mathbb{R}^ \\plus{}$ be the set of all positive real numbers. Find all functions $ f: \\mathbb{R}^ \\plus{} \\rightarrow \\mathbb{R}^ \\plus{}$ satisfying\n\\[ f(x)f(y) \\equal{} f(xy) \\plus{} f\\left(\\frac {x}{y}\\right)\\]\nfor all positive real numbers $ x$ and $ y$.[/quote]\n\nAt least we have the solutions $ f(x) \\equal{} 2\\cosh(g(\\ln(x)))$ where $ g(x)$ is any solution of the Cauchy equation $ g(x \\plus{} y) \\equal{} g(x) \\plus{} g(y)$\n\nWith continuous solutions $ g(x) \\equal{} ax$, we get $ x^a \\plus{} x^{ \\minus{} a}$ for any real $ a$\n\nThe difficulty (for me) is to see if these are the only solutions or not ...\nTo be continued.[/quote]\r\n\r\nActually, can you show that $ f(x) \\ge 2$ all the time? And, if so, does your reduction $ f(x) \\equal{} 2 \\cosh(g \\ln(x))$ imply $ g(x\\plus{}y) \\equal{} g(x) \\plus{} g(y)$? If so, that's\r\na full solution right there.", "Solution_5": "Actually, might have something from my previous work there... \r\n\r\nSo, subbing in $ b \\equal{} 1$,\r\n\r\n$ g(b\\plus{}1) \\minus{} g(1)g(b) \\plus{} g(b\\minus{}1) \\equal{} 0$,\r\n\r\nor $ g(x) \\minus{} \\alpha g(x\\minus{}1) \\plus{} g(x\\minus{}2) \\equal{} 0$, where $ \\alpha \\equal{} g(1)$. This gives a linear recurrence\r\nwith $ g(0) \\equal{} 2$ and $ g(1) \\equal{} \\alpha$, which has solution $ g(n) \\equal{} \\kappa^n \\plus{} \\kappa^{\\minus{}n}$, where\r\n$ \\kappa \\plus{} \\kappa^{\\minus{}1} \\equal{} \\alpha$. If $ \\alpha < 2$, then $ \\kappa$ is a unit complex number, and $ \\kappa^n \\plus{} \\kappa^{\\minus{}n}$ is eventually negative; this cannot be allowed to happen, so $ \\alpha \\ge 2$ always.\r\n\r\n Let $ \\kappa$ float as a function of $ \\alpha$.\r\n\r\n This is necessarily true for any solution $ g$. \r\n\r\nNow let $ h(x) : \\equal{} g(kx)$; this is also a solution, and so $ h(n) \\equal{} \\kappa^n \\plus{} \\kappa^{\\minus{}n}$ (where\r\n$ \\kappa \\plus{} \\kappa^{\\minus{}1} \\equal{} h(1) \\equal{} g(k)$) This incidentally shows that $ g(x) \\ge 2$ for every $ x$.\r\n\r\nTherefore $ g(nk) \\equal{} \\kappa(g(k))^n \\plus{} \\kappa(g(k))^{\\minus{}n}$ (Here $ \\kappa$ is a function -- excuse\r\nthe abuse of notation).\r\n\r\nSo, if we know $ g$ on any value $ k$, we also know it on any of its integral multiples. In fact, it is the case that $ g(k)$ is recoverable from $ g(nk)$ from the above -- the function taking $ \\kappa$ to $ \\kappa^n \\plus{} \\kappa^{\\minus{}n}$ is invertible modulo the action $ \\kappa \\mapsto 1/\\kappa$ -- from which $ g(qk)$ can be determined for all rational $ q$. \r\n\r\nSo, we're done... and $ g$ is determined from an arbitrary positive function on basis elements of the space of reals over the rationals. \r\n\r\n(If you dig thru the above and actually write out the construction, it's probably something along the lines of what pco wrote above... but what this does do is show we've gotten all the solutions through that line of reasoning.)", "Solution_6": "@Raja Oktovin :\r\n[quote=\"Raja Oktovin\"]I can't do this problem so I posted this here...\n[/quote]\nOk, thanks. Better to post in \"Algebra unsolved Problems\", next time, IMHO.\n\n@Nukular :\n[quote=\"Nukular\"]So, we're done... and is determined from an arbitrary positive function on basis elements of the space of reals over the rationals. \n[/quote]\r\n\r\nThanks a lot for your kind remarks, though I'm not sure to have understood all your conclusions :\r\n\r\nFrom the basis $ g(a)g(b)\\equal{}g(a\\plus{}b)\\plus{}g(a\\minus{}b)$, we get that this is the classical D'Alembert equation (without the continuity requirement) and the solutions $ 2\\cos(ax)$ and $ 2\\cosh(ax)$ for $ x\\in\\mathbb Q$ are classical results. The first can be eliminated since we are looking for positive solutions.\r\n\r\nSo $ f(x)\\equal{}2\\cosh(a\\ln(x))$ $ \\forall x\\in\\mathbb Q$ is a normal result deduced from D'Alembert equation.\r\nAnd $ f(x)\\equal{}2\\cosh(\\sigma(\\ln(x)))$ is a family of solution, as I already stated.\r\n\r\nBut I dont see where (or if) you prove that this is is the unique family, which was my question.\r\n\r\nSorry if I misunderstood or did not see all your demo.", "Solution_7": "The argument has a couple of key lemmas (and, yes, is missing a last step):\r\n\r\nSolutions for $ f(x)f(y) \\equal{} f(xy) \\plus{} f(x/y)$ are directly in bijective correspondence to positive functions $ g$ for which $ g(x \\plus{} y) \\plus{} g(x \\minus{} y) \\equal{} g(x)g(y)$; so I'm\r\ngoing to stick with that latter equation (*), and solve there. (Eliminates the $ \\ln$ term for simplicity).\r\n\r\nLet $ \\kappa : [2,\\infty) \\to [1,\\infty)$ be defined such that $ \\kappa(x) \\plus{} \\kappa(x)^{ \\minus{} 1} \\equal{} x$.\r\n\r\nLemma 1) For any solution $ g$, $ g(n) \\equal{} \\kappa(g(1))^n \\plus{} \\kappa(g(1))^{ \\minus{} n}$ (**) for all integers $ n$. Furthermore, $ g(1) \\ge 2$.\r\n\r\nI proved this in my earlier post, by considering the recurrence $ g(n \\plus{} 2) \\minus{} g(1)g(n \\plus{} 1) \\plus{} g(n) \\equal{} 0$.\r\n \r\nLemma 2) If $ \\sigma$ is a solution to the Cauchy equation, then $ g \\circ \\sigma$ is a solution to (*) whenever $ g$ is a solution.\r\n\r\nLemma 3) If $ g$ is a solution to (*), then $ g(kn) \\equal{} \\kappa(g(k))^n \\plus{} \\kappa(g(k))^{ \\minus{} n}$ (***) for all reals $ k$; furthermore, $ g(k) \\ge 2$ for all real $ k$.\r\nProof: $ \\sigma(x) \\equal{} kx$ is a solution to the Cauchy equation; letting $ h \\equal{} g \\circ \\sigma$ in equation (**) gives equation (***).\r\n\r\nLemma 4) $ g(p/q)$ is determined completely by $ g(1)$, for all integers $ p,q$.\r\nProof: We have $ \\kappa(g(p/q))^q \\plus{} \\kappa(g(p/q))^{ \\minus{} q} \\equal{} g(p) \\equal{} \\kappa(g(1))^p \\plus{} \\kappa(g(1))^{ \\minus{} p}$. There is only one solution for $ g(p/q)$ from this \r\nequation.\r\n\r\nLemma 5) $ g(kx)$ is determined uniquely, given $ g(x)$, for any rational $ k$ and any real $ x$.\r\nProof: Same trick as before, apply Lemma 4 to $ g \\circ (y \\mapsto xy)$.\r\n\r\n------------\r\n\r\nYeah, and I have to still add a lemma which gives $ g(x \\plus{} y)$ from $ g(x)$ and $ g(y)$, provided $ x$ and $ y$ are independent over $ \\mathbb{Q}$. Will \r\nwork on that next.\r\n\r\nEDIT: Was not aware of this being a classical D'Alembert equation; so most of this is probably just reproducing what you already know. But, at least\r\nit's true that we know $ g$ is at least determined by it's footprint on $ \\mathbb{Q} \\minus{}$projective equivalence classes in $ \\mathbb{R}$.\r\n\r\nEDIT 2: Yes, sadly, I believe you get a single additional degree of freedom when you add up two reals from differing rational lines... might be a \r\nton more solutions; though hopefully not :).\r\n\r\nEDIT 3: This can probably be solved if you consider $ g$ to be a positive function from $ \\mathbb{Q}^2 \\to \\mathbb{R}$; if you get more than two degrees of freedom of solutions, then we're going to have a ton here.", "Solution_8": "Heh: or, you could notice\r\n\r\n$ g(a\\plus{}b)g(a\\minus{}b) \\equal{} g(2a) \\plus{} g(2b)$\r\n\r\nand \r\n\r\n$ g(a)g(b) \\equal{} g(a\\plus{}b)\\plus{}g(a\\minus{}b)$.\r\n\r\nWhich should let you get values for both $ g(a\\plus{}b)$ and $ g(a\\minus{}b)$. So you probably can build it up once you know $ g$ on a basis.\r\n\r\nBack to this when I've had some sleep.", "Solution_9": "Or not: \r\n\r\nSketch: \r\n\r\nUse the equations $ g(a)g(b) \\equal{} g(a\\plus{}b)g(a\\minus{}b)$ and $ g(a\\plus{}b)g(a\\minus{}b) \\equal{} g(2a) \\plus{} g(2b)$ to lift $ g$ from disjoint\r\na vector subspace $ V \\subset \\mathbb{R}$ to $ V \\oplus \\mathbb{Q}b$. However, this requires a choice, since\r\nunless one of $ a,b$ is zero, then $ g(a\\plus{}b)$ and $ g(a\\minus{}b)$ are different, and switching the \r\nvalues of $ g(a\\plus{}b)$ and $ g(a\\minus{}b)$ can be accomplished by a reflection across $ b^\\perp$, without interfering\r\nwith other values of $ g$. Once $ g(a\\plus{}b)$ is determined, so is $ g(a \\plus{} qb)$ for rational $ q$, and therefore so is $ g(ra \\plus{} qb)$ for rational $ r$.\r\n\r\nSo, given the value of $ g$ on each basis element of $ \\mathbb{R}$ over $ \\mathbb{Q}$, \r\nwe must also choose, for each basis element (other than the first), an orientation for $ g$ to grow in --\r\ni.e. whether $ g(a\\plus{}b)$ or $ g(a\\minus{}b)$ is the larger. This is a complete family of solutions for $ g$.\r\n\r\nYeah, lots of axiom of choice, but I think it works this time around.", "Solution_10": "[quote=\"Nukular\"] This is a complete family of solutions for $ g$.[/quote]\r\nHemmm, thanks a lot ...\r\nI'll try to read all your demo when it will be finished ...\r\n\r\nIs your family THE complete family of solutions ?\r\n\r\nAnd, if yes, is your family equal to $ \\{2\\cosh(\\sigma(\\ln(x)))\\}$ where $ \\sigma(x)$ is any solution of $ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y)$ ?\r\nThis was my question and I dont understand if you answered it or not.\r\nSorry again if I misunderstood.", "Solution_11": "What I have shown: \r\n\r\n$ g$ is completely determined from its values on a basis for $ \\mathbb{R}$ over $ \\mathbb{Q}$, along with \r\na direction for all but one basis element; this \"direction\" tells us whether $ g(a \\plus{} b)$ or $ g(a \\minus{} b)$ is the larger.\r\nFor any values $ \\ge 2$ chosen on basis elements, along with choices of direction, a unique $ g$ can be \r\nconstructed. So, this is the entire set of such functions $ g$.\r\n\r\nHowever, this \"choice of direction\" is not substantial, since $ g$ is an even function; so it rather chooses\r\nan automorphism of $ \\mathbb{R}$.\r\n\r\nSo... yes, I your $ 2\\cosh(\\sigma(\\ln x))$ is a complete family of functions, provided that my earlier\r\nwork computes $ 2 \\cosh (\\ln x)$ as the only rational solution (I get a [i] unique [/i] solution, so\r\nif you are correct these would coincide).", "Solution_12": "Yes, assuming that nice family $ g(x) \\equal{} 2\\cosh (x)$ for rationals;\r\n\r\n$ g(x\\plus{}y) \\plus{} g(x\\minus{}y) \\equal{} g(x)g(y) : \\equal{} ab$\r\n$ g(x\\plus{}y)g(x\\minus{}y) \\equal{} g(2x)g(2y) : \\equal{} a^2 \\plus{} b^2 \\minus{}4$,\r\n\r\nsolving gives that \r\n\r\n$ g(x\\plus{}y) \\equal{} \\frac{ab \\pm \\sqrt{(a^2\\minus{}4)(b^2\\minus{}4)}}{2} \\equal{} 2\\cosh(x)\\cosh(y) \\pm 2\\sinh(x)\\sinh(y) \\equal{} 2 \\cosh(x \\pm y)$,\r\n\r\nbut that $ \\pm$ depends on our choice of $ \\sigma$, since $ \\cosh$ is an even function.\r\n\r\nSo, definitively, the solution set is $ f(x) \\equal{} 2\\cosh(\\sigma(\\ln x))$.\r\n\r\nI appreciate the constant adversarial criticism to make me provide a simpler more constructive solution :lol:.", "Solution_13": "We see that $ f(x) \\equal{} f(\\frac {1}{x})$; $ f(1) \\equal{} 1$;$ f(x^2) \\equal{} f^2(x) \\minus{} 2$. Thus $ f(x) > \\sqrt {2 \\plus{} \\sqrt {2 \\plus{} ...}} \\equal{} 2$ when the number of root tends to the infinite . It means $ f(x) > 2$ for all positive numbers $ x$.\r\nNotice that $ f(x^2) \\plus{} f(y^2) \\equal{} f(xy)f(\\frac {x}{y})$. It suffices to write $ x^2 \\equal{} xy\\frac {x}{y}$ and $ y^2 \\equal{} \\frac {xy}{\\frac {x}{y}}$ thus $ f^2(x) \\plus{} f^2(y) \\minus{} 4 \\equal{} f(x^2) \\plus{} f(y^2) \\equal{} f(xy)(f(x)(f(y) \\minus{} f(xy))$\r\nNow it implies $ f^2(x) \\plus{} f^2(y) \\plus{} f^2(xy) \\equal{} f(x)f(y)f(xy) \\plus{} 4$\r\nWe know some techniques when proving inequality , now we deduce the existence of continuous function $ g$:$ (0;\\infty )\\to [1;\\infty)$ such that $ f(x) \\equal{} g(x) \\plus{} \\frac {1}{g(x)}$ (since $ f(x) > 2$ above).\r\nThe rest is easy enough; $ g(x) \\equal{} x^k$ is immediately solution. Thus $ f(x) \\equal{} x^k \\plus{} \\frac {1}{x^k}$", "Solution_14": "[quote=\"Allnames\"]We see that $ f(x) \\equal{} f(\\frac {1}{x})$; $ f(1) \\equal{} 1$;$ f(x^2) \\equal{} f^2(x) \\minus{} 2$. Thus $ f(x) > \\sqrt {2 \\plus{} \\sqrt {2 \\plus{} ...}} \\equal{} 2$ when the number of root tends to the infinite . It means $ f(x) > 2$ for all positive numbers $ x$.\nNotice that $ f(x^2) \\plus{} f(y^2) \\equal{} f(xy)f(\\frac {x}{y})$. It suffices to write $ x^2 \\equal{} xy\\frac {x}{y}$ and $ y^2 \\equal{} \\frac {xy}{\\frac {x}{y}}$ thus $ f^2(x) \\plus{} f^2(y) \\minus{} 4 \\equal{} f(x^2) \\plus{} f(y^2) \\equal{} f(xy)(f(x)(f(y) \\minus{} f(xy))$\nNow it implies $ f^2(x) \\plus{} f^2(y) \\plus{} f^2(xy) \\equal{} f(x)f(y)f(xy) \\plus{} 4$\nWe know some techniques when proving inequality , now we deduce the existence of continuous function $ g$:$ (0;\\infty )\\to [1;\\infty)$ such that $ f(x) \\equal{} g(x) \\plus{} \\frac {1}{g(x)}$ (since $ f(x) > 2$ above).\nThe rest is easy enough; $ g(x) \\equal{} x^k$ is immediately solution. Thus $ f(x) \\equal{} x^k \\plus{} \\frac {1}{x^k}$[/quote]\r\n\r\nSadly, $ f$ is not required to be continuous -- see the work above for a full solution (modulo\r\nsome standard vector space knowledge of $ \\mathbb{R}$ over $ \\mathbb{Q}$) .", "Solution_15": "a quick side note: with a condition of continuity the problem appeared in one of the Russian Olympiads. It was either National Olympiad or St. Petersburg olympiad.", "Solution_16": "[quote=\"Erken\"]a quick side note: with a condition of continuity the problem appeared in one of the Russian Olympiads. It was either National Olympiad or St. Petersburg olympiad.[/quote]\r\nIn my notebook; It appeared as St. Petersburg Olympiad problem. And I posted my solution for it ( without reading carefully the condition). I got a badly mistake!" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "prove that there is a finit number of $ (x,y,z)$which they satisfied :\r\n$ \\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z}\\equal{}\\frac{1}{1990}$ and $ x$ ,$ y$ and $ z$ $ \\in N$*", "Solution_1": "Let $ x\\le y\\le z$.\r\nThen $ 1991\\le x\\le 3*1990$. Therefore ${ \\frac{1}{y}+\\frac{1}{z}=\\frac{1}{1990}-\\frac{1}x}\\ge \\frac{1}{1990*1991}$.\r\nIt give $ y\\le 2*1990*1991$ and $ z\\le (1990*1991)(1990*1991+1)$." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "If x0=1 x1=0 x2=2 and x(n+2) = 2x(n) + x(n-1) for n>=1\r\nprove that for any natural m there are two following numbers\r\nthat divides by m.", "Solution_1": "Note that, modulo $ m$, the sequence is periodical, at least from some $ n$, because there are a finite number of possible triples so that at least one will eventually reappear and each term is determined by the three preceeding ones from the recursive relation. In another hand, the recurrence relation shows that we may define the sequence even for negative subscripts. Thus, the complete sequence (infinite in both directions) is periodical modulo $ m$. Let $ T$ be its period.\r\n\r\nBut, direct computations show that $ x_{\\minus{}1} \\equal{} x_{\\minus{}2} \\equal{} 0$. Thus $ x_{T\\minus{}2} \\equal{} x_{T\\minus{}1} \\equal{} 0 \\mod [m]$, as desired.\r\n\r\nPierre." } { "Tag": [ "HMMT" ], "Problem": "Pick a number from between [b]0[/b] and 20. Whoever picks the second highest number wins! All numbers must be integers.\r\n\r\nJust to see what happens, could people please do this and pm me a number? After I get 20 numbers, I will stop. If this has gone for 2 weeks without me getting 20 numbers, I will stop it, unless I am really close. I am going to participate, and to be fair, I have already chosen my number.", "Solution_1": "Just a note: in the original question, 0 was a legal (though illogical) number to guess.", "Solution_2": "Oh. Yeah, if you want, you can pick 0.....", "Solution_3": "6 people so far", "Solution_4": "I still need 13 more numbers! Please do this, it will be quick and easy for you to do, not a waste of time at all :D", "Solution_5": "Today = Two weeks after the contest started. What were the results?", "Solution_6": "[quote=\"E^(pi*i)=-1\"]Today = Two weeks after the contest started. What were the results?[/quote]\r\nyeah, mustafa, were are the results? :wink: :mad:", "Solution_7": "Well, there weren't as many participants as I would like, but here they are....\r\n\r\nMath92:18\r\nDrunner2007:15\r\nNutz4e:16\r\nMustafa:15\r\nE^(pi*i):18\r\nJoml88:18\r\nFireemblem13:15\r\nDeimos:17\r\nIgnite168:18\r\n\r\nCongratulations Deimos, winner of.....\r\n\r\n[hide]The pride and joy of winning :rotfl: [/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "With $ a,b,c > 0$ prove that \r\n $ \\frac {a^2 \\plus{} bc}{(a \\plus{} b)^3} \\plus{} \\frac {b^2 \\plus{} ca}{(b \\plus{} c)^3} \\plus{} \\frac {c^2 \\plus{} ab}{(c\\plus{} a)^3} \\geq \\frac {9}{4(a \\plus{} b \\plus{} c)}$\r\nI have 2 solutions for this ineq!", "Solution_1": "[quote=\"2424\"]With $ a,b,c > 0$ prove that \n $ \\frac {a^2 \\plus{} bc}{(a \\plus{} b)^3} \\plus{} \\frac {b^2 \\plus{} ca}{(b \\plus{} c)^3} \\plus{} \\frac {c^2 \\plus{} ab}{(a \\plus{} b)^3} \\geq \\frac {9}{4(a \\plus{} b \\plus{} c)}$\nI have 2 solutions for this ineq![/quote] \r\nBut it's wrong. Try $ b\\rightarrow \\plus{} \\infty.$ :wink:", "Solution_2": "[quote=\"arqady\"][quote=\"2424\"]With $ a,b,c > 0$ prove that \n $ \\frac {a^2 \\plus{} bc}{(a \\plus{} b)^3} \\plus{} \\frac {b^2 \\plus{} ca}{(b \\plus{} c)^3} \\plus{} \\frac {c^2 \\plus{} ab}{(a \\plus{} b)^3} \\geq \\frac {9}{4(a \\plus{} b \\plus{} c)}$\nI have 2 solutions for this ineq![/quote] \nBut it's wrong. Try $ b\\rightarrow \\plus{} \\infty.$ :wink:[/quote]\r\nWhy? :P \r\nIf no one can prove this ineq,I will write my solution!", "Solution_3": "[quote=\"2424\"][quote=\"arqady\"][quote=\"2424\"]With $ a,b,c > 0$ prove that \n $ \\frac {a^2 \\plus{} bc}{(a \\plus{} b)^3} \\plus{} \\frac {b^2 \\plus{} ca}{(b \\plus{} c)^3} \\plus{} \\frac {c^2 \\plus{} ab}{(a \\plus{} b)^3} \\geq \\frac {9}{4(a \\plus{} b \\plus{} c)}$\nI have 2 solutions for this ineq![/quote] \nBut it's wrong. Try $ b\\rightarrow \\plus{} \\infty.$ :wink:[/quote]\nWhy? :P \nIf no one can prove this ineq,I will write my solution![/quote]\r\nYour solution is obviously wrong. :P Try also $ a \\equal{} c\\rightarrow0^ \\plus{} .$ :wink:", "Solution_4": "[quote=\"arqady\"][/quote][quote=\"2424\"][quote=\"arqady\"][quote=\"2424\"]With $ a,b,c > 0$ prove that \n $ \\frac {a^2 + bc}{(a + b)^3} + \\frac {b^2 + ca}{(b + c)^3} + \\frac {c^2 + ab}{(a + b)^3} \\geq \\frac {9}{4(a + b + c)}$\nI have 2 solutions for this ineq![/quote] \nBut it's wrong. Try $ b\\rightarrow + \\infty.$ :wink:[/quote]\nWhy? :P \nIf no one can prove this ineq,I will write my solution![/quote][quote=\"arqady\"]\nYour solution is obviously wrong. :P Try also $ a = c\\rightarrow0^ + .$ :wink:[/quote]\r\nSorry,I edited my ineq.\r\nyou can see above! :P \r\nand now,It's true! :lol:", "Solution_5": "[quote=\"2424\"]With $ a,b,c > 0$ prove that \n $ \\frac {a^2 \\plus{} bc}{(a \\plus{} b)^3} \\plus{} \\frac {b^2 \\plus{} ca}{(b \\plus{} c)^3} \\plus{} \\frac {c^2 \\plus{} ab}{(c \\plus{} a)^3} \\geq \\frac {9}{4(a \\plus{} b \\plus{} c)}$\nI have 2 solutions for this ineq![/quote]\r\nWe have $ \\sum_{cyc}\\frac{a}{(a\\plus{}b)^2} \\geq \\frac{9}{4(a\\plus{}b\\plus{}c)}$ :P", "Solution_6": "[quote=\"2424\"]\nWe have $ \\sum_{cyc}\\frac {a}{(a \\plus{} b)^2} \\geq \\frac {9}{4(a \\plus{} b \\plus{} c)}$ :P[/quote]\r\nIt's not trues, 2424. Try $ (a, b, c) \\equal{} (1, 2, 0)$ :wink:", "Solution_7": "I think that one of $ a,b,c$ can't be equal to $ 0$ since $ a,b,c>0$ :)" } { "Tag": [ "pigeonhole principle" ], "Problem": "There are $ 16$ persons in a company, each of which likes exactly $ 8$ other persons. Show that there exist two persons who like each other.", "Solution_1": "[hide]\nIsn't this just simple pigeonhole? If the people are nodes and we add a directed edge from $ A$ to $ B$ if $ A$ likes $ B$, there are $ \\binom{16}{2}\\equal{}120$ pairs of people and $ 16*8\\equal{}128$ directed edges, so two edges must come from the same pair . . . [/hide]" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "How can I find all possible Jordan forms for a 6x6 matrix A having as minimal polynomial the $ x^{2}(x \\minus{} 1)^{2}$\r\n\r\nThanks anyone in advance.", "Solution_1": "There are five options:\r\n\r\n[list]$ J\\equal{}J_2(0)\\oplus J_2(1) \\oplus J_2(0)$[/list]\n\n[list]$ J\\equal{}J_2(0)\\oplus J_2(1) \\oplus J_2(1)$[/list]\n\n[list]$ J\\equal{}J_2(0)\\oplus J_2(1) \\oplus J_1(0) \\oplus J_1(0)$[/list]\n\n[list]$ J\\equal{}J_2(0)\\oplus J_2(1) \\oplus J_1(1) \\oplus J_1(1)$[/list]\n\n[list]$ J\\equal{}J_2(0)\\oplus J_2(1) \\oplus J_1(0) \\oplus J_1(1)$[/list]", "Solution_2": "Could you explain me how do you come to this result?\r\n\r\n\r\nThanks in advance.", "Solution_3": "Since $ m_A(\\lambda)\\equal{}x^2(x\\minus{}1)^2$, there are two eigenvalues, $ \\lambda_1\\equal{}0$ and $ \\lambda_2\\equal{}1$. Further on, because $ A^2$ is minimal, which 'destroys' all jordan blocks of $ \\lambda_1$, there is at least one $ 2\\times 2$ jordan block for $ \\lambda_1$ and also there are no greater blocks for $ \\lambda_1$ (otherwise $ A^2$ would not destroy them). The same reasoning for $ \\lambda_2$. Now you just list all the possibilities (using the fact, that the sum of dimensions of all blocks is $ 6$).", "Solution_4": "[quote=\"tmrfea\"]Since $ m_A(\\lambda) \\equal{} x^2(x \\minus{} 1)^2$, there are two eigenvalues, $ \\lambda_1 \\equal{} 0$ and $ \\lambda_2 \\equal{} 1$. Further on, because $ A^2$ is minimal, which 'destroys' all jordan blocks of $ \\lambda_1$, there is at least one $ 2\\times 2$ jordan block for $ \\lambda_1$ and also there are no greater blocks for $ \\lambda_1$ (otherwise $ A^2$ would not destroy them). The same reasoning for $ \\lambda_2$. Now you just list all the possibilities (using the fact, that the sum of dimensions of all blocks is $ 6$).[/quote]\r\n\r\n\r\nThank you very much." } { "Tag": [ "abstract algebra" ], "Problem": "Hi! I have a question (propably stupid...):\r\n\r\nWhy, when each element of group is in order 2 ($ \\forall a\\in G: a^2 \\equal{} 1$), does exist an homomorphism (diffrent from identity)?", "Solution_1": "This is not High School Basics. Mods, please move this topic.", "Solution_2": "You're right, sorry:).", "Solution_3": "[quote=\"Kibu\"]Hi! I have a question (propably stupid...):\n\nWhy, when each element of group is in order 2 ($ \\forall a\\in G: a^2 \\equal{} 1$), does exist an homomorphism (diffrent from identity)?[/quote]No, if you let $ \\varphi$ be a homomorphism, $ 1\\equal{}\\varphi(1)\\equal{}\\varphi(a^2)\\equal{}\\varphi(a\\cdot a)\\equal{}\\varphi(a)\\varphi(a)$, which implies that $ \\varphi(a)$ is its own inverse for all $ a\\in G$." } { "Tag": [ "algebra", "polynomial", "binomial theorem" ], "Problem": "I've been thinking about this problem with no luck:\r\n\r\nGiven a polynomial p(x), prove that you can also write an equivalent polynomial in terms of u where u = x + k for any k, where k is an arbitrary real constant.\r\n\r\nThis proof is trivial if you know how to use Taylor Series from calculus. Can you come up with another proof?", "Solution_1": "..Maybe I'm missing something, but wouldn't p(x) = p(u-k), and so our proof is complete?", "Solution_2": "You should say a little bit more. Perhaps write ``Thus we can replace x by u-k in each term of p(x) and then expand by the binomial theorem to get q(u), a polynomial in u.''", "Solution_3": "[quote=\"ComplexZeta\"]You should say a little bit more. Perhaps write ``Thus we can replace x by u-k in each term of p(x) and then expand by the binomial theorem to get q(u), a polynomial in u.''[/quote]\r\n\r\nYeah, same deal. But in any case, my point was: the problem is sorta trivial. :D", "Solution_4": "*smacks forehead*\r\n\r\nIt's always the simple ones...:(" } { "Tag": [], "Problem": "Two pencils and one pen cost $ 80$ cents, and one pencil and\ntwo pens cost $ \\$1.15$. How many cents would three pencils\ncost?", "Solution_1": "Let the cost of a pencil be $ n$, and the cost of a pen by $ m$. Then:\r\n\r\n\\begin{align*}2x+y&=80\\\\x+2y&=115\\end{align*}\r\n\r\nMultiply the first equation by $ 2$, and subtract the second:\r\n\r\n\\begin{align*}4x+2y&=160\\\\x+2y&=115\\\\3x&=45\\end{align*}\r\n\r\nThus, three pencils cost $ 45$ cents.", "Solution_2": "[quote=\"AIME15\"]Let the cost of a pencil be $ n$, and the cost of a pen by $ m$. Then:\n\n\\begin{align*}2x + y & = 80 \\\\\nx + 2y & = 115\\end{align*}\n\nMultiply the first equation by $ 2$, and subtract the second:\n\n\\begin{align*}4x + 2y & = 160 \\\\\nx + 2y & = 115 \\\\\n3x & = 45\\end{align*}\n\nThus, three pencils cost $ 45$ cents.[/quote]\r\nummm do you mean x for n and y for m?", "Solution_3": "We can also add the 2 equations to get that the cost of 3 pencils and 3 pens is 195 cents. Dividing by 3, one pencil and 1 pen cost 65 cents. Sicne 2 pencils and 1 pen (80) costs 1 pencil(15) more than 1 pencil and 1 pen(65), 1 pencil costs 15, meaning 3 cost 45." } { "Tag": [ "induction", "algebra proposed", "algebra" ], "Problem": "Viet Nam MO (group B)\r\n let a be a real number # 0 and the sequence of real numbers (x_n) is defined by : \r\n x1 = 0 , x_(n+1)(x_n+a)=a+1 , n=1,2,3...\r\n find the x_n ? \r\n \r\nThe solution from THTT (Mathematics and Youth of Vienamese) is very\r\n horrible !\r\nI solve it very easily and better than that way.\r\nDo you have nice solution ?", "Solution_1": "At least it works by induction !\r\n\r\nx(2*n+1) = (a+1)a y(n) / z(n)\r\nand\r\nx(2*n) = (a+1)/a z(n) / y(n-1)\r\n\r\nwith z(n) = (a+1)^(2n) - ((a+1)^(2n)-1) / (a+2)\r\nand y(n) = ((a+1)^(2n)-1) / ((a+1)^2-1)\r\n\r\nit can be still be simplified (a bit) but it is \"convenient\" on this form to work by induction.\r\n\r\n\r\nBut probably it's smarter to set b=a+1, then it is immediate to see that\r\n\r\nx(n) = b (b^(n-2) - b^(n-3) + ... + (-1)^n) / (b^(n-1) - b^(n-2) + ... - (-1)^n) works :) \r\n\r\nIt can again be simplified a bit but it look nice this way.\r\n\r\n\r\nWhat's it your method A1lqdSchool ?", "Solution_2": "Here is my solution :\r\nif a=-2 ,we have x_n=(n-1)/n;n=1,2,3...\r\nif a #-2\r\nConsider f(x)=(x+a+1)/(x-1)\r\nIt is easily to see that : f(x_(n+1))=-(a+1)*f(x_n)\r\nso f(x_n)=(-1)^n*(a+1)^n*f(0)=(-a-1)^(n+1)\r\nhence x_n=[(-1)^(n-2)+(a+1)^(n-1)](a+1)/[(-1)^(n-1)+(a+1)^n] , n=1,2,3.." } { "Tag": [ "geometry" ], "Problem": "Bughouse is i think the best strategy game (but it goes in the categories of chess)...anyone else play bughouse?", "Solution_1": "[quote=\"7h3.D3m0n.117\"]Bughouse is i think the best strategy game (but it goes in the categories of chess)...anyone else play bughouse?[/quote]\r\n\r\nYeah I have. It's really fun especially when you want to waste time.(usually when you wait for your next match up post-up)", "Solution_2": "For everyone's information:\r\n\r\nBughouse is a varient of chess played by partnerships of 2 people on two boards. In each partnership, one partner gets white and the other partner gets black. When one partner captures a piece, he passes that piece to the other partner. If you have a piece passed to you, you may drop that piece onto the chessboard instead of making a move.\r\n\r\nThe only thing you may not do while dropping a piece is dropping a pawn on the first or last ranks.\r\n\r\nWhen one person loses, his entire team loses.\r\n\r\nCommunication between partners is generally allowed, for example, when one partner tells the other partner to get a knight so that he can go for checkmate in 1 or something. In inexperienced settings, a partner can also give play advice to the other partner.\r\n\r\nTypically bughouse games are played 5 min/side (two clocks are needed).\r\nAlso, because of the fast time controls, kings can be captured (there would be too much confusion if illegal moves had to be retracted), causing loss of game.\r\n\r\n----\r\n\r\nBughouse is similar and different to regular chess. Bughouse pawns and knights are worth a bit more than they are in normal chess. Also, positional weaknesses in bughouse are often exploited much more ruthlessly, as one can just drop pieces in the holes in his opponent's positions rather than needing to maneuver his own pieces into the holes.", "Solution_3": "[quote=\"randomdragoon\"]For everyone's information:\n\nBughouse is a varient of chess played by partnerships of 2 people on two boards. In each partnership, one partner gets white and the other partner gets black. When one partner captures a piece, he passes that piece to the other partner. If you have a piece passed to you, you may drop that piece onto the chessboard instead of making a move.\n\nThe only thing you may not do while dropping a piece is dropping a pawn on the first or last ranks.\n\nWhen one person loses, his entire team loses.\n\nCommunication between partners is generally allowed, for example, when one partner tells the other partner to get a knight so that he can go for checkmate in 1 or something. In inexperienced settings, a partner can also give play advice to the other partner.\n\nTypically bughouse games are played 5 min/side (two clocks are needed).\nAlso, because of the fast time controls, kings can be captured (there would be too much confusion if illegal moves had to be retracted), causing loss of game.\n\n----\n\nBughouse is similar and different to regular chess. Bughouse pawns and knights are worth a bit more than they are in normal chess. Also, positional weaknesses in bughouse are often exploited much more ruthlessly, as one can just drop pieces in the holes in his opponent's positions rather than needing to maneuver his own pieces into the holes.[/quote]\r\n\r\nthat was the shortest set of bughouse rules i've ever heard :lol: \r\n\r\nnice job\r\n\r\nand besides, the best opening is (assuming black plays e5)\r\n\r\n1. e4 e5\r\n2. Bc4 (black does something)\r\n3. Bxf7. \r\n\r\nThen you can drop knights, develop your queen, etc. Now white has the tempo and he needs to keep it...so he can sack sometimes rook for checks to keep the pressure on", "Solution_4": "i always thought bughouse was more elementary. Most people in my area don't play it. They grow out of it after elementary school. \r\nSerious chess is more popular here.", "Solution_5": "[quote=\"now a ranger\"]\nSerious chess is more popular here.[/quote]\r\n\r\nchess64 and my chess instructor never played bhouse...he said it was a bad way for rookies to develop strong chess skills (but we always play it after a couple rounds of serious chess)", "Solution_6": "haha that is true. \r\n\r\nI quit mostly because my parents wouldn't let me have a coach because it was expensive. :(", "Solution_7": "[quote=\"now a ranger\"]I quit mostly because my parents wouldn't let me have a coach because it was expensive. :([/quote]\r\n\r\ntoo expensive? wow...chess and i had a science teacher in middle school that had a club that taught anyone who was interested about chess for free...", "Solution_8": "well our schools don't have clubs. And i looked , and the cheapest coaches were like bad, and they cost 50 + dollars/hour. Anyway, I quit already and at that time I had too much time and didn't do math much. this year i like math a lot, so i literally forgot about chess and haven't played at all since last year." } { "Tag": [ "percent" ], "Problem": "Just in time for the World Series... \r\n\r\nA baseball player has been officially at bat 322 times and has a batting average of .289. He figures that he will have 53 times at bat for the rest of the season. (Don't count walks.) What batting average does he have to maintain for the rest of the season in order to end with a season average of .300?", "Solution_1": "[hide]He's already batted 322*0.289 or 93 times.\n\n\n\nHe need to bat 3/10*375 or 113 times to get an average of 0.300\n\n\n\n113-93 = 20\n\n\n\nSo the baseball player need to having a batting average of 0.377 for the rest of the season.[/hide]", "Solution_2": "yay red sox won!!!!!!!!!!!!!!!\n\n\n\n[hide].0377[/hide]", "Solution_3": "Hey baseball fan, batting average is expressed in percent :P ." } { "Tag": [ "MATHCOUNTS", "AMC", "AIME", "geometry", "ARML" ], "Problem": "What scores did you guys get?\r\n\r\nMy team got 53.25 (which in my bad chapter was 1st)\r\n\r\nMy team-\r\nMe: 1st-28/16=44\r\nOther guy: 11th-16/14=30\r\nOther guy: 12th-14/16=30\r\nOther guy: 16th-15/14=29\r\nTeam:20", "Solution_1": "aaaaaaaaaaaaaa", "Solution_2": "64 I think...\r\n\r\nTop: 46\r\nMe: 45 BLEGH.\r\nTop's brother: 43\r\nOther person : 42", "Solution_3": "I think I already posted in the other topic but,\r\n45, 44, 43, 41. Perfect team round, 63.25.\r\nI think the chapter is too easy to tell which team/individual is the best, so I guess we have to wait for states.", "Solution_4": "well my team got a 62.25 I think we lost to the first place team by about 0.25\r\nsomeone in our team got one wrong on target :mad:", "Solution_5": "It was, but recently, MATHCOUNTS posted the tests on their site: http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=47\r\n\r\nso we can talk about it.", "Solution_6": "NEWAYS...\r\n46-ME YAY\r\n44-3rd\r\n44-6th\r\n41-10th\r\nteam-9 *DONT ASK*\r\nu do the math 4 total *lol*", "Solution_7": "I wasn't on the team, but I got a 44 (which was like 20th place?)\r\n\r\nOur team was in a tie for 5th place, and we had TWO people that got 37's (HORRIBLE!), so we totally failed.\r\n\r\nSo nobody from our school is going to States.\r\n\r\nOy Miller4Math, are you Jonny (or Johnny), Marcus, that girl, or that tall Asian one?", "Solution_8": "[quote=\"happyme\"]I wasn't on the team, but I got a 44 (which was like 20th place?)\n\nOur team was in a tie for 5th place, and we had TWO people that got 37's (HORRIBLE!), so we totally failed.\n\nSo nobody from our school is going to States.\n\nOy Miller4Math, are you Jonny (or Johnny), Marcus, that girl, or that tall Asian one?[/quote]\r\n\r\nomg, a 44 won't get you into state?!?!?!?!?!?!?!, which state are you in?", "Solution_9": "California, I think.", "Solution_10": "Ugh, I got beat my stevenmeow (not a good sign for state!)\r\n\r\nMe-45 (1st)\r\nOther ppl: 45 (2nd), 41 (7th), 40 (10th)\r\n\r\nTeam 9/10=18\r\n\r\nTeam Total 60.75 ( :huh: poll needs decimals)", "Solution_11": "Its ok, i've been getting an EXTREMELY long streak of perfects *even @ practices* and im sure to crash someday and get a 42... hopefully ill get a 42 on an AIME-level state test", "Solution_12": "oh crap my team is going to get pwnt xD\r\nme - 46\r\nother ppl - 39, 38, 32", "Solution_13": "[quote=\"stevenmeow\"]Its ok, i've been getting an EXTREMELY long streak of perfects *even @ practices* and im sure to crash someday and get a 42... hopefully ill get a 42 on an AIME-level state test[/quote]\r\n\r\n\r\ncrap....and i havn't made a single perfect since i don't know when....\r\n :( :( \r\n\r\n45\r\n43\r\n36\r\n34\r\n\r\n\r\nteam round: 8\r\n\r\ni think...", "Solution_14": "[quote=\"ra5249\"]oh crap my team is going to get pwnt xD\nme - 46\nother ppl - 39, 38, 32[/quote]\r\nDon't worry, my team:\r\nMe - 44 = 3rd place\r\nOther teammates - around 30 : didn't place.\r\nHeh, this is off topic, but for old states, i've gotten:\r\n42\r\n43\r\n45\r\n2008: 46 jk\r\nBut I'm improving :lol:", "Solution_15": "My team was really, really, really bad! \r\nMe-45, 1st\r\n3rd place, 36\r\n6th place, 30\r\n?th place, 28\r\nTeam- 5 (obviously only one person was working *cough*)\r\nTeam score=44.75\r\nYup, we're really bad! :P But somehow we still got first place.", "Solution_16": "My team was decent, but we screwed up on the team round. People were doing the problems they were weak in, and we missed 2. Careless mistakes killed us too.", "Solution_17": "Hmm, our team's scores were\r\n46\r\n46\r\n45\r\n34.\r\nAnd we got 8 on the team round...\r\nSo we got 58.75\r\nAnd we lost to Diamond.\r\nBut if I were on the team it would've been 61.25...\r\nMeh, I guess you couldn't really tell, since something may have happened if I were on the team, like me getting a ridiculously low score and causing the team to get a score lower than 8 or something, but you never know...", "Solution_18": "[quote=\"Stargirl240\"]My team was really, really, really bad! \nMe-45, 1st\n3rd place, 36\n6th place, 30\n?th place, 28\nTeam- 5 (obviously only one person was working *cough*)\nTeam score=44.75\nYup, we're really bad! :P But somehow we still got first place.[/quote]\r\nSo are we. Lasst year we got 7th place team and this year we were way worse.\r\n\r\nThis is what we got last year:\r\nMe: 29 (6th)\r\n13th place: 26\r\nVERY BAD PLACE: <20\r\nANOTHER BAD***: <20\r\nHORRIBLE AT MATH: 11 (he only got 5 sprint and 3 target!)\r\n\r\nSUCKERS!!!!! :mad: :mad: :mad:\r\n\r\n\r\nAAAAAAAAAAAAAAARRRRRRRRRRRRRRRRRGGGGGGGGGGGGGGGGGGGGGGGHHHHHHHHHHHHHHHHHHHH!!!!!!!!!", "Solution_19": "hehe my team didnt do so well...but our chapter phails so we made states! :D \r\n\r\nMe: 43 (1st)\r\n34 (10th)\r\n27\r\n22\r\n\r\n4 on the team round... :ninja: dont ask\r\n39.5 total", "Solution_20": "wow, \r\ncaptain:44\r\nme:41\r\nzolojetto:35\r\nother member:31\r\n6 on team...49.75...how skilled are we? :rotfl:", "Solution_21": "I'm on the same team as math154. Our school pwned chapter but this is Missouri so the competition was very weak.\r\n\r\n1st-45 (me)\r\n2nd-44(another kid at our school)\r\n3rd-42(math154)\r\n4th-42(kid from another school)\r\n5th-41?(kid from another school)\r\n6th-40?(2nd place's sister)\r\n\r\nTeam Round: 8 (missed number 10 and a stupid mistake somewhere...)\r\nWe beat second place team by like 15. :rotfl:", "Solution_22": "thats not very weak...im in pennsylvannia\r\n44\r\n44\r\n42\r\n41\r\n40\r\n40\r\n38\r\n38\r\n38\r\n38", "Solution_23": "[quote]I'm on the same team as math154. Our school pwned chapter but this is Missouri so the competition was very weak.\n\n1st-45 (me)\n2nd-44(another kid at our school)\n3rd-42(math154)\n4th-42(kid from another school)\n5th-41?(kid from another school)\n6th-40?(2nd place's sister)\n\nTeam Round: 8 (missed number 10 and a stupid mistake somewhere...)\nWe beat second place team by like 15. :rotfl: [/quote]\r\n\r\nwiseidiot, i think 5th was also 42. i won the tiebreaker. 6th-8th was 41 i think. i made so many careless mistakes...our team screwed up cause this guy who sucked at geometry was doing 10...and this is STL so the competition is a little harder than in the rest of the state\r\n\r\nabacadaea, the reason the scores were so close this year was because of how easy the test was. last year, for example, the state cutoff was about ~30.\r\n\r\nand when was the last time Missouri was even in the top 23 at nats???", "Solution_24": "[quote=\"math154\"]\nand when was the last time Missouri was even in the top 23 at nats???[/quote]\r\n\r\nHere is some Missouri history :\r\n\r\n2006 - 20th at Nats\r\n2005 - 7th at Nats\r\n2005 - 16th at Nats\r\n2004 - 2nd at Nats & 3rd best individual\r\n\r\nLooks pretty good to me!!", "Solution_25": "I don't see how that's bad. It's much better than my state, Wyoming!", "Solution_26": "At chapter, I got a 44, and our team had 44, 43, 40, 37, with two others not on the official team getting 44 and 40. But first place got a perfect score. You might've heard of him...name's Darryl Wu?", "Solution_27": "I guess so.", "Solution_28": "[quote=\"grant.ghs\"]At chapter, I got a 44, and our team had 44, 43, 40, 37, with two others not on the official team getting 44 and 40. But first place got a perfect score. You might've heard of him...name's Darryl Wu?[/quote]\r\n\r\nlol. \r\n\r\nat the museum, bobby and the rest of texas were chasing darryl wanting his autograph.... ra tried to sell me bobbies b4 he lost ..... lol", "Solution_29": "wait, lost what?\r\nand the entire texas team got darryl's autograph yehhhhh", "Solution_30": "cd, I would guess?\r\nBut wow it seems that Bobby won written, team, AND masters... which... more than makes up for that.[/i]", "Solution_31": "It makes up for MORE than Darryl! :D \r\n\r\nSomeday if I made Nats next year I'd ask for Bobby's autograph :P", "Solution_32": "Yep me too...\r\nOh wait I got 5th in states in 8th grade.\r\nI can try getting held back tho. I'll still be normally aged, lol.\r\nand btw, reread my other post: [quote=\"academicsrule\"]which... MORE THAN makes up[/quote]", "Solution_33": "darn, i lost darryl's autograph, i blame it on alanchou.\r\n\r\nThis has been a public shaming of alan chou, thank you.", "Solution_34": "[quote=\"ra5249\"]wait, lost what?\nand the entire texas team got darryl's autograph yehhhhh[/quote]\r\noh ya. that was pathetic lol... You guys were like \" SCORE!!! \"... -_-...\r\nkemps was fun... ELEVEN.", "Solution_35": "[quote=\"dingzhou\"]darn, i lost darryl's autograph, i blame it on alanchou.\n\nThis has been a public shaming of alan chou, thank you.[/quote]\r\n\r\nkinda sad that 8th graders (with exception of bobby) are getting an autograph from a 6th grader. Kinda like the time last year, when someone on one of the ARML teams from indiana saw Kevin chen during arml and asked him for his autograph.", "Solution_36": "you're just jealous that the tx team didn't ask you for autograph. :P", "Solution_37": "deep down, i know you, bobby, kl, ra, darryl, etc. all want my autograph. :roll:", "Solution_38": "eh. Ding doesn't want yours. Because you're associated with me. lol", "Solution_39": "we got a 62.75 team, i think, becuz we had a sucky guy who got a 37\r\n\r\nOh, yeah, on another unrelated thing at nats, Texas [i]half-rhymes[/i] with breakfast" } { "Tag": [ "search", "geometry", "number theory" ], "Problem": "Did anyone every make a mock canadian competition? It might be fun to do something like a mock Euclid. Could someone write it? Also, there shouldn't be any sign-ups, people just PM their answers to the writer (not me). :D\r\n\r\nEDIT: all I found in the search was a 'math battle' a while back.", "Solution_1": "If someone else makes one, I'll join. If I get VERY board, I'll try and make a contest. But I'm never board. ever. Except today.", "Solution_2": "[quote=\"rzsolt\"]If someone else makes one, I'll join. If I get VERY board, I'll try and make a contest. But I'm never board. ever. Except today.[/quote]\r\ndoes that mean you're making the contest? :w00t:", "Solution_3": "I'll make one. I've been digging though some old problems and I have tons of gems that I'd love to share with you guys.\r\n\r\nProbably I would write something harder than Euclid (at least without as many easy, boring problems as Euclid). And you know me... I'd put lots of ad hoc puzzles on there rather than hardcore geometry, number theory, or inequalities.", "Solution_4": "[quote=\"Elyot\"]I'll make one. I've been digging though some old problems and I have tons of gems that I'd love to share with you guys.\nProbably I would write something harder than Euclid (at least without as many easy, boring problems as Euclid). And you know me... I'd put lots of ad hoc puzzles on there rather than hardcore geometry, number theory, or inequalities.[/quote]\r\nAlright! :lol: What are you going to call it? You should probably start a new thread.", "Solution_5": "why don't you make a canadian open math challenge instead?" } { "Tag": [ "logarithms", "function", "algebra proposed", "algebra" ], "Problem": "a) Prove that there are no one-to-one (injective) functions $f: \\mathbb{N} \\to \\mathbb{N}\\cup \\{0\\}$ such that \r\n\\[ f(mn) = f(m)+f(n) , \\ \\forall \\ m,n \\in \\mathbb{N}. \\] \r\n\r\nb) Prove that for all positive integers $k$ there exist one-to-one functions $f: \\{1,2,\\ldots,k\\}\\to\\mathbb{N}\\cup \\{0\\}$ such that $f(mn) = f(m)+f(n)$ for all $m,n\\in \\{1,2,\\ldots,k\\}$ with $mn\\leq k$. \r\n\r\n[i]Mihai Baluna[/i]", "Solution_1": "a) It is clear that if such a function does exist then :\r\n1) $f(1)=0$ \r\n2) For each prime $p$ and non-negative integer $a$, we have $f(p^a) = af(p).$\r\n3) For pairwise distinct primes $p_1, \\cdots, p_k$, and non-negative integers $a_1, \\cdots ,a_k$, we have $f(p_1^{a_1} \\cdots p_k^{a_k} ) = a_1f(p_1) + \\cdots a_kf(p_k).$\r\n\r\nNow if $f$ is injective, $f(2)f(3) \\neq 0$ (from 1) ), so that $f(2^{f(3)} ) = f(3)f(2) = f(3^{f(2)})$ (from 2) ), which is impossible since $2^{f(3)} \\neq 3^{f(2)}$.\r\n\r\nb) Let $k$ be given, and let $p$ be a prime greater than all the exponents appearing in the prime decompositions of the numbers $1,2, \\cdots, k.$\r\nLet $f(1)=0.$\r\nLet $p_1, \\cdots, p_m$ be the primes not greater than $k$. Thus each of the numbers $2,3, \\cdots, k$ only decomposes with the $p_i$'s.\r\nFor each $i$, let $f(p_i) = p^i.$\r\nThen, for all noin-primes $t \\leq k$, define $f(t)$ with 3) above.\r\nIt is clear that $f(mn) = f(m)+f(n)$ for all $m,n$ such that $mn \\leq k$.\r\nMoreover, using the unicity of the decomposition in base $p$, the equality $f(t) = f(t')$ is equivalent to the fact that $t$ and $t'$ have the same prime decompositions, which is $t=t'$ and $f$ is injective.\r\n\r\nWow! I am now a 2000-man. :) :alien:\r\n\r\nPierre." } { "Tag": [ "\\/closed" ], "Problem": "Is there a limit on the number of people for WOOT?", "Solution_1": "[quote=\"kimby_102\"]Is there a limit on the number of people for WOOT?[/quote]\r\n\r\nNot officially. However, if it looked like it were to exceed the very high side of our expectations, we might set a maximum." } { "Tag": [ "articles", "calculus" ], "Problem": "Could someone give a brief history of Russia and Europe? I read the article at Wikipedia but it is too long and involved. I just need a brief summary of the history. Thank you.", "Solution_1": "The Mongols conquered the place, terrorized the people, and took everything of value. The prince of Moscow threw them off, terrorized the people, and took everything of value. A few hundred years later, the Communists threw off the tsar, terrorized the people, and took everything of value.\r\n(Based on what my favorite high school history teacher said)\r\n\r\nThat's probably not what you really wanted, but what you really wanted probably doesn't exist. If you do want to build an outline of Russian history, start with the Mongolian invasion; the fact that the name \"Russia\" comes from river vikings a few centuries earlier is a lot less important.", "Solution_2": "I have a very simply question : when did the territory of what we call Russia today get a Russian - or let's make it simpler - Slavic (relative or absolute) majority? :maybe:", "Solution_3": "Could you also summarize the history of Eastern Europe (countries like Croatia, Kosovo, Yugoslavia, etc)? Thank you.", "Solution_4": "[b][i][/b][/i]No.", "Solution_5": "[quote=\"mathwizarddude\"]Could you also summarize the history of Eastern Europe (countries like Croatia, Kosovo, Yugoslavia, etc)? Thank you.[/quote]\r\n\r\nwah wah reading wikipedia is just too much work!", "Solution_6": "Well, here is the world history in one sentence: [i]people were born, suffered, and died[/i]. (I leave it to you to dig up the source of this quote :P). The history of Russia was once summarized in one word by Karamsin: [i]\u0432\u043e\u0440\u0443\u044e\u0442[/i].", "Solution_7": "[quote=\"mathwizarddude\"]Could you also summarize the history of Eastern Europe (countries like Croatia, Kosovo, Yugoslavia, etc)? Thank you.[/quote]\r\n\r\nOh, this could be a good one... Just replace the Mongols, Tsarist Russia and Communists in jmerry's with\r\n\r\n-Illyrians\r\n-Romans\r\n-Slavs\r\n-Hungarians\r\n-Turks\r\n-Austro-Hungarians\r\n-Yugoslav royalists\r\n-Yugoslav communists\r\n\r\nrespectively.\r\n\r\nHonestly, what's the use of such condensed history?", "Solution_8": "Could someone give me a brief summary of calculus? The wikipedia article was too long and involved." } { "Tag": [ "function", "quadratics", "algebra", "functional equation", "algebra solved" ], "Problem": "Find all functions $f$ from positive integers to positive integers such that, for all positive integers $m$ and $n$, \\[(2^m+1)\\cdot f(n)\\cdot f(2^m n)=2^m\\cdot f^2(n)+f^2(2^m n)+(2^m-1)^2 n\\]", "Solution_1": "I can't send solution but answer is\r\nFor every odd natural $n$ suppose $s$ is a divisor of $n$.Now for every nonnegative $m$:\r\n$\\displaystyle f(2^mn)= 2^ms + \\frac ns$", "Solution_2": "how to prove ?", "Solution_3": "Well notice that:\r\n\\[ (2^mf(n)-f(2^mn))(f(2^mn)-f(n))=n(2^m-1)^2\\]", "Solution_4": "the result of Omid Hatami is right\r\n\r\nalternatively, the functional equation can be interpreted as a quadratic equation with variable $f(2^{m}n)$, and the first step is to apply the well-known forula for quadratic equations.\r\n\r\n(a very bad functional equation, in my opinion.. i tried to show that $f$ was increasing during the test ;) )" } { "Tag": [ "geometry", "perpendicular bisector", "angle bisector" ], "Problem": "In the figure to the right, $ ABC$ is an equilateral triangle. The circles with centers $ B$ and $ C$ both pass through points $ A$ and $ D$. What is the measure of angle $ BAD$?\n\n[asy]import graph;\ndraw((0,0)--(30,0)--(15,26)--cycle);\ndraw(Circle((30,0),30));\ndraw(Circle((15,26),30));\ndot((0,0));\ndot((15,26));\ndot((45,26));\ndot((30,0));\nlabel(\"$A$\",(0,0),SW);\nlabel(\"$B$\",(30,0),SE);\nlabel(\"$C$\",(15,26),N);\nlabel(\"$D$\",(45,26),NE);[/asy]", "Solution_1": "triangle ABC is equilateral because thats how your supposed to construct an equilateral using compass + straightedge. AD is an angle biscetor, so 1 and x_(n+1)=x_n+(x_n-1)^1/2. What's the limit of x_n/2^n as n->infinity? Does ne1 have any idea? I thought of trigonometric functions at the beginning, but if we put 1/x_n=a_n, getting a_n<1 for any n, and a_n=sinx we get, after the calculations, a_(n+1)=tan(x/2), so we don't get sth with sin, but with tan. Any other ideas?", "Solution_1": "Are you sure of the terms of the problem :? \r\nEither I'm completly wrong (becoming an habit :( ) either it's not too difficult.\r\n\r\n\r\nIf x_(n+1)=x_(n) + (x_(n) - 1)^1/2\r\n\r\nThen x(n) is strictly increasing, for n >= n0, x(n) > 4, so sqrt(x(n)-1) <= x(n)/2 and x(n+1) <= 3/2 x(n)\r\n\r\nNow this proves x(n)/2^n -> 0 when n ->+oo.\r\n\r\n\r\nI even think that x(n)/a^n -> 0 when n ->+oo whenever a > 1.", "Solution_2": "Isn't this the correct version :\r\n\r\na1 > 1, an+1 = an + (an2 - 1)1/2 ?\r\n\r\nI saw the following similar problem somewhere but it's much easier :\r\n\r\na1 = 1, an+1 = an + (an2 + 1)1/2.\r\nCompute limn->:inf:an2-n.\r\n\r\nIn the latter case trig substitution works just fine.\r\nBut this is not the case for the problem Grobber posted.", "Solution_3": "Arne is right, I forgot to write \"^2\". Indeed Arne, that's the problem I was refering to." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "find all functions $ f(x)$ such that it is continuos on $ [0;1]$,differentiable on the open interval $ (0;1)$ and the following two conditions are satisfied:\r\na)$ 20f(x)\\plus{}11f(x)\\plus{}2009 \\leq 0$ with all $ x \\in (0;1)$\r\nb)$ f(0)\\equal{}f(1)\\equal{}\\frac{\\minus{}2009}{11}$", "Solution_1": "I think you have to write the correct statement ... :cool:", "Solution_2": "My problem is true...", "Solution_3": "[quote=\"lhplhp\"]find all functions $ f(x)$ such that it is continuos on $ [0;1]$,differentiable on the open interval $ (0;1)$ and the following two conditions are satisfied:\na)$ 20f(x) \\plus{} 11f(x) \\plus{} 2009 \\leq 0$ with all $ x \\in (0;1)$\nb)$ f(0) \\equal{} f(1) \\equal{} \\frac { \\minus{} 2009}{11}$[/quote]\r\nThe a) condition is equivalent to $ f(x) \\leq \\minus{}\\frac{2009}{31}$.\r\nWhat's the interest ? Too many functions saisfay the conditions :!: \r\n :cool:", "Solution_4": "No.No Here is a problem in MYM in my country(T 12/379) so [b]Dont help him [/b]\r\nSee here ,please http://www.mathlinks.ro/viewtopic.php?t=168836&start=20" } { "Tag": [ "calculus", "algebra", "polynomial", "derivative", "function", "calculus computations" ], "Problem": "diffretiate,y=ax^2", "Solution_1": "what does that mean??", "Solution_2": "Calculus problems go in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]Calculus Computations and Tutorials[/url] section of the Forum.\r\n\r\nThe answer to your question can be found in:\r\n1. [url=http://en.wikipedia.org/wiki/Differentiation_rules#The_polynomial_or_elementary_power_rule]Wikipedia: \"Differentiation Rules\"[/url]\r\n2. [url=http://en.wikipedia.org/wiki/Table_of_derivatives#Derivatives_of_simple_functions]Any table of derivatives[/url]\r\n3. [url=http://en.wikipedia.org/wiki/Calculus_with_polynomials#The_power_rule]Wikipedia: \"Calculus with polynomials\"[/url]\r\n4. Your calculus book\r\n5. A billion other places" } { "Tag": [ "probability", "AMC" ], "Problem": "Someday.... somewhere.... the qualifier list is, but here in the real world, everyone is sad; the qualifier list isnae cognizable yet...... (bad adaption of poetry.)\r\n\r\nBilly", "Solution_1": "It'll never be out. Sorry.", "Solution_2": "See they made the final decision at latest at 2.30p.m.(Nebraska zone) \r\nHow late or How long will the AMC directors work today? Till the Nebraska time zone 4p.m.? 5p.m.? or 6p.m.? What's the probability they can work out the list today (to corellate the CEEB)?", "Solution_3": "Hehe, I am exactly right!", "Solution_4": "hm.... 30 minutes ago!", "Solution_5": "Today, Cuz it is already out!!! yaya" } { "Tag": [ "geometry", "trapezoid", "geometry unsolved" ], "Problem": "Let $ ABCD$ be an inscribable and circumscribable quadrilateral. Through the center $ I$ of the incircle of the quadrilateral $ ABCD$ there is constructsd a line parallel to the line $ BC$, which intersects the sides $ AB$ and $ CD$ at the points $ K$ and $ L$, respectively.\r\nProve that $ KL \\equal{} \\frac {AB \\plus{} BC \\plus{} CD \\plus{} DA}{4}$.", "Solution_1": "That sounds interesting!\r\nBut, what's so specific about line BC?\r\n[i]As the KL-expression ist symmetric w. r. to all four side-lenghths, we have indeed an even more striking result for the four parallels to the sides going through I.[/i] :maybe:", "Solution_2": "$ P$ is tangency point of $ (I)$ with $ BC.$ $ IP$ cuts $ (I)$ again at $ Q.$ Tangent to $ (I)$ at $ Q$ cuts $ AB, CD$ at $ M, N$ and $ MN \\parallel BC.$ $ KL$ is midline of the tangential trapezoid $ BCNM,$\r\n\r\n$ KL \\equal{} \\frac {MN \\plus{} BC}{2} \\equal{} \\frac {MN \\plus{} BC \\plus{} \\overline{BM} \\plus{} \\overline{CN}}{4}.$\r\n\r\n$ AB, CD$ intersect at $ X.$ $ ABCD$ is cyclic, $ DA$ is antiparallel to $ BC$ WRT the angle $ \\angle BXC.$ Therefore, the $ \\triangle AXD \\sim \\triangle NXM$ are oppositely similar. Since they have common X-excircle $ (I),$ they are oppositely congruent, $ DA \\equal{} MN$ and $ \\overline{MA} \\equal{} \\minus{} \\overline{ND},$ $ \\overline{MA} \\plus{} \\overline{ND} \\equal{} 0.$ Substituting this to the above expression for $ KL,$\r\n\r\n$ KL \\equal{} \\frac {MN \\plus{} BC \\plus{} \\overline{BM} \\plus{} \\overline{MA} \\plus{} \\overline{CN} \\plus{} \\overline{ND}}{4} \\equal{} \\frac {DA \\plus{} BC \\plus{} BA \\plus{} CD}{4}$" } { "Tag": [ "geometry", "incenter", "analytic geometry", "trigonometry", "geometry proposed" ], "Problem": "[color=darkblue]Let $ABC$ be a triangle for which the projection $D$ of the vertex $A$ to the its opposite side-line $BC$ belongs to the line $IG$,\nwhere $I$ is the incenter and $G$ is the centroid for the given triangle. Ascertain the maximum value of the angle $\\widehat{BAC}$.[/color][hide=\"Here is a direction.\"]$D\\in IG\\Longleftrightarrow b+c=p\\sqrt 2$, where $2p=a+b+c\\ .$[/hide]", "Solution_1": "Let $A'$ be the orthogonal proyection of $A$ in $\\overline{BC}$, so the barycentric coordinates of $A'$ wrt $\\triangle ABC$ are $([A'CB]: [A'CA]: [A'AB])$ that is $(0: h_{a}\\frac{b.cosC}{2}: h_{a}\\frac{c.cosB}{2})=(0: senBcosC: senCcosB)$. As for the condition of $A'\\in \\overrightarrow{IG}$ then $0=\\begin{displaymath}\\left| \\begin{array}{ccc}0&senBcosC&senCcosB \\\\ a & b & c \\\\ 1 & 1 & 1 \\end{array}\\right|=\\left| \\begin{array}{ccc}0&senBcosC&senCcosB \\\\ 0 & b-a & c-a \\\\ 1 & 1 & 1 \\end{array}\\right|=\\left| \\begin{array}{ccc}senBcosC&SenCcosB\\\\ b-a & c-a \\end{array}\\right|=(senBcosC)(c-a)-(senCcosB)(b-a)$. And because of $cosB=\\frac{a^{2}+c^{2}-b^{2}}{2ac}$ and $cosC=\\frac{a^{2}+b^{2}-c^{2}}{2ab}$. That is $\\frac{b(c-a)(a^{2}+b^{2}-c^{2})}{2ab}=\\frac{c(b-a)(a^{2}+c^{2}-b^{2})}{2ac}$ so $a^{2}c+b^{2}c-c^{3}-ab^{2}+ac^{2}=a^{2}b+bc^{2}-ac^{2}-b^{3}+ab^{2}$ and then $bc(b-c)+(b^{3}-c^{3})=2a(b^{2}-c^{2})+a^{2}(b-c)$ so, a solution is $b=c$. If $b \\neq c$ it becomes $b^{2}+c^{2}+2bc=2a(b+c)+a^{2}$ an then $2b^{2}+2c^{2}+4bc=2a(b+c)+a^{2}+b^{2}+c^{2}+2bc$ and then $2(b+c)^{2}=(a+b+c)^{2}=4p^{2}$ and then $(b+c)^{2}=2p^{2}$ thus $b+c=\\sqrt[]{2}p$.\r\n\r\nIt means that $(2-\\sqrt[]{2})(b+c)=\\sqrt[]{2}a$, that is $(2-\\sqrt[]{2})(senB+senC)=\\sqrt[]{2}senA$ thus $(2-\\sqrt[]{2})(2sen\\frac{B+C}{2}cos\\frac{B-C}{2})=\\sqrt[]{2}senA$ so, making the sustitution $B+C=\\pi-A$ it becomes $(2-\\sqrt[]{2})sen(B+\\frac{A}{2})=\\sqrt[]{2}sen\\frac{A}{2}$ that is $sen\\frac{A}{2}\\leq \\frac{2-\\sqrt[]{2}}{\\sqrt[]{2}}$ thus $\\frac{A}{2}\\leq arcsen\\frac{2-\\sqrt[]{2}}{\\sqrt[]{2}}$ so $A \\leq 2arcsen\\frac{2-\\sqrt[]{2}}{\\sqrt[]{2}}$. $\\blacksquare$", "Solution_2": "[quote=\"Virgil Nicula\"][color=darkblue]Let $ABC$ be a triangle for which the projection $D$ of the vertex $A$ in the its opposite side-line $BC$ belongs to the line $IG$,\nwhere $I$ is the incenter and $G$ is the centroid for the given triangle. Ascertain the maximum value of the angle $\\widehat{BAC}$.[/color][/quote]\r\n[color=darkred][b]Huchs[/b], you demonstrated that you know very well the barycentrical coordinates.\nIn my opinion, is better to work with the barycentrical coordinates than with the triangle complex coordinates (a monstrous tautology !).[/color]\r\n\r\n[color=darkblue][b]The my proof.[/b] Suppose $b\\ne c$ and denote the projections $T$, $S$ of the points $I$, $G$ respectively in the side $[BC]$.\nProve easily that $GS=\\frac{1}{3}\\cdot h_{a}$, $\\boxed{\\ DT=\\frac{p-a}{a}\\cdot |b-c|\\ }$, $MD=\\frac{b+c}{2a}\\cdot |b-c|$ and $DS=\\frac{2}{3}\\cdot DM\\Longrightarrow \\boxed{\\ DS=\\frac{b+c}{3a}\\cdot |b-c|\\ }\\ .$\n\n$1.\\blacktriangleright$ Therefore, $D\\in IG$ $\\Longleftrightarrow$ $\\frac{DT}{DS}=\\frac{IT}{GS}$ $\\Longleftrightarrow$ $h_{a}(p-a)=r(b+c)$ $\\Longleftrightarrow$ $2p(p-a)=a(b+c)$ $\\Longleftrightarrow$ $(b+c)^{2}-a^{2}=2a(b+c)$ $\\Longleftrightarrow$ $a^{2}+2(b+c)a-(b+c)^{2}=0$ $\\Longleftrightarrow$ $a=(\\sqrt 2-1)(b+c)$ $\\Longleftrightarrow$ $\\boxed{\\ b+c=p\\sqrt 2\\ }\\ .$\n\n$2.\\blacktriangleright$ Denote $\\frac{b}{c}=t\\ .$ Thus, $a=(\\sqrt 2-1)(b+c)$ $\\Longleftrightarrow$ $a^{2}=(3-2\\sqrt 2)(b+c)^{2}$ $\\Longleftrightarrow$ $b^{2}+c^{2}-2bc\\cos A=(3-2\\sqrt 2)(b^{2}+c^{2}+2bc)$ $\\Longleftrightarrow$ $t^{2}-2t\\cos A+1=(3-2\\sqrt 2)\\left(t^{2}+2t+1\\right)$ $\\Longleftrightarrow$ $(\\sqrt 2-1)t^{2}-(\\cos A+3-2\\sqrt 2)t+(\\sqrt 2-1)=0\\ \\ (1)\\ .$ But $b^{2}+c^{2}-a^{2}=$ $b^{2}+c^{2}-(3-2\\sqrt 2)(b^{2}+c^{2}+2bc)=$ $2(\\sqrt 2-1)\\left[b^{2}-(\\sqrt 2-1)bc+c^{2}\\right]>0$ $\\Longleftrightarrow$ $\\cos A>0$ $\\Longleftrightarrow$ $A<90^{\\circ}\\ .$ Thus, for the equation $(1)$, $\\Delta =(\\cos A+3-2\\sqrt 2)^{2}-4(\\sqrt 2-1)^{2}\\ge 0$ $\\Longleftrightarrow$ $\\cos A+3-2\\sqrt 2\\ge 2(\\sqrt 2-1)$ $\\Longleftrightarrow$ $\\cos A>4\\sqrt 2-5$ $\\Longleftrightarrow$ $\\boxed{\\ A\\le \\arccos (4\\sqrt 2-5)\\ }\\ .$\n\n[b]Remark.[/b] The Hucht's answer is $A\\le 2\\arcsin (\\sqrt 2-1)$ what is identically with the my answer $A\\le\\arccos (4\\sqrt 2-5)$ because $1-2(\\sqrt 2-1)^{2}=4\\sqrt 2-5\\ .$[/color]", "Solution_3": "[quote=\"Virgil Nicula\"]\n[color=darkred][b]Huchs[/b], you demonstrated that you know very well the barycentrical coordinates.\nIn my opinion, is better to work with the barycentrical coordinates than with the triangle complex coordinates (a monstrous tautology !).[/color]\n[/quote]\n\nLOL thanks!, but I as looking for the use of triangle complex coordinates because I'd like to prove some theorems (like Lester theorem), because when I looked at the Lester's page I got surprised. That's why I'd like to leran it!\n\nCapitan Mandarina!\n\nPS. \n[quote=\"Virgil Nicula\"]\n[color=darkred][b]Huchs[/b], ...[/color]\n[/quote]\r\nIt's huch[b]t[/b], with [b]t[/b]." } { "Tag": [ "trigonometry", "algebra", "function", "domain" ], "Problem": "Solve for $x$:\r\n\r\n1. $4\\cos^3 x+2\\cos^2x-3\\cos x-1=0$ ($0\\le x<\\pi$)\r\n\r\n2. $\\csc x+\\sec x=2\\sqrt{2}$ ($0\\le x<\\pi$)\r\n\r\n3. $\\sin 3x(\\sin 3x-\\cos x)=\\sin x(\\sin x-\\cos 3x)$ ($0\\le x<\\pi$)\r\n\r\n4. $\\cos x+\\sqrt{3}\\sin x+1=0$ ($0\\le x<2\\pi$)\r\n\r\n5. $\\tan^2 x+2\\tan x-1=0$ ($-\\pi\\le x<\\pi$)", "Solution_1": "5.\r\n\r\n$tan^2x+2tanx-1=0$\r\n$2tanx=1-tan^2x$\r\n$\\dfrac{2tanx}{1-tan^2x}=1$\r\n\r\nwe know that \r\n$\\dfrac{2tanx}{1-tan^2x}=tan2x$\r\n$.^..$\r\n$tan2x=1$\r\n\r\n$-\\pi\\leq x<\\pi$\r\n\r\n$x=\\dfrac{\\pi}{8}$\r\n$x=\\dfrac{5\\pi}{8}$\r\n$x=-\\dfrac{7\\pi}{8}$\r\n$x=-\\dfrac{3\\pi}{8}$", "Solution_2": "4.\r\n$cosx+\\sqrt{3}sinx+1=0$\r\n$cosx+\\sqrt{3}sinx=-1$\r\n\r\n\r\n$(cosx+\\sqrt{3}sinx)^2=(-1)^2$\r\n$cos^2x+2\\sqrt{3}sinxcosx+3sin^2x=1$\r\nwe know that\r\n$sin^2x+cos^2x=1$\r\n$.^..$\r\n$cos^2x+2\\sqrt{3}sinxcosx+3sin^2x=sin^2x+cos^2x$\r\n$2\\sqrt{3}sinxcosx=-2sin^2x$\r\n$tanx=-\\sqrt{3}$\r\n\r\n$x=\\dfrac{2\\pi}{3}+k\\pi$$k=1$", "Solution_3": "Your answer for #5 is correct except that $k=0,-1$, not $k=0,1$.\r\n(Watch out for the domain :roll: )\r\n\r\n[b]EDIT[/b]- Also, it should be $\\dfrac{\\pi}{8}+\\dfrac{k\\pi}{2}$ ($k=0,-1$)\r\n\r\nFor #4, I think you're on the right track, but your answer is not correct.\r\n\r\nI don't get this part:\r\n[quote=\"felipesa\"]\n$cos^2x+2\\sqrt{3}sinxcosx+3sin^2x=sin^2x+cos^2x$\n$2\\sqrt{3}sinxcosx=-2sin^2x$\n$tanx=-\\sqrt{3}$[/quote]\r\nHow did you get from the 2nd line to the 3rd line?\r\nDid you divide both sides by $\\cos^2 x$? :?", "Solution_4": "[quote=\"frt\"]Your answer for #5 is correct except that $k=0,-1$, not $k=0,1$.\n(Watch out for the domain :roll: )\n\n[b]EDIT[/b]- Also, it should be $\\dfrac{\\pi}{8}+\\dfrac{k\\pi}{2}$ ($k=0,-1$)\n\nFor #4, I think you're on the right track, but your answer is not correct.\n\nI don't get this part:\n[quote=\"felipesa\"]\n$cos^2x+2\\sqrt{3}sinxcosx+3sin^2x=sin^2x+cos^2x$\n$2\\sqrt{3}sinxcosx=-2sin^2x$\n$tanx=-\\sqrt{3}$[/quote]\nHow did you get from the 2nd line to the 3rd line?\nDid you divide both sides by $\\cos^2 x$? :?[/quote]\r\n\r\nNO \r\nI did\r\n$2\\sqrt{3}sinxcosx=-2sin^2x$\r\n$\\sqrt{3}=\\dfrac{-2sin^2x}{2sinxcosx}$\r\n$\\sqrt{3}=-\\dfrac{sinx}{cosx}$\r\n$tanx=-\\sqrt{3}$", "Solution_5": "wow nobody solved 1. yet, but i ll do it:\r\n\r\nFirst, one should notice that\r\n$4\\cos^3 x -3\\cos x = \\cos 3x$, and\r\n$2\\cos^2 x -1 = \\cos 2x$\r\n\r\ntherefore \r\n$\\cos 3x+\\cos 2x =0\\implies \\cos 3x = \\cos(2x+\\pi)$ \r\nso actually $3x=2x+\\pi +2\\pi k ;\\ k\\in\\mathbb Z$ or $2\\pi-3x=2x+\\pi +2\\pi k ;\\ k\\in\\mathbb Z$\r\nbut since $x\\in[0;\\pi)$ we only have\r\n$3x=2x+\\pi \\implies x=\\pi$ (not really a solution because of $x\\in[0;\\pi)$\r\n$2\\pi-3x=2x+\\pi \\implies x=\\dfrac{\\pi}5$, or\r\n$2\\pi - 3x = 2x-\\pi \\implies x= \\dfrac{3\\pi}{5}$", "Solution_6": "2.\r\n\r\n$\\dfrac{1}{sinx}+\\dfrac{1}{cosx}=2\\sqrt2$\r\n\r\n$\\dfrac{sinx+cosx}{sinxcosx}=2\\sqrt2$\r\n\r\n$(\\dfrac{sinx+cosx}{sinxcosx})^2=(2\\sqrt2)^2$\r\n\r\n$\\dfrac{sin^2x+cos^2x+2sinxcosx}{sin^2xcos^2x}=8$\r\n\r\n$\\dfrac{1+sin2x}{sen^2xcos^2x}=8$\r\n\r\n$1+sin2x=2sinxcosx2sinxcos2$\r\n\r\n$1+sin2x=2sin^22x$\r\n\r\n$2sin^22x-sin2x-1=0$\r\n\r\n$sin2x=1$ or $sin2x=-1/2$\r\n\r\nif $x\\in[0,\\pi)$\r\n\r\nso \r\n$x=\\dfrac{\\pi}{4}$and$x=\\dfrac{11 \\pi}{12}$ :)", "Solution_7": "Can anyone tell me what is the relevance of $\\pi$? in angles, what does it stand for?", "Solution_8": "[quote=\"riddler\"]Can anyone tell me what is the relevance of $\\pi$? in angles, what does it stand for?[/quote]\r\n\r\nused to measure angles in radians.\r\n\r\nHere is a formula $\\dfrac{M}{180}=\\dfrac{a}{\\pi}$\r\n\r\nwhere $M$ is the measure of the angle in degrees and $a$ its measure in radians!\r\nok? :)", "Solution_9": "well angles can be expressed in terms of the so called arc length.\r\n\r\nthe arc length is defined as the length of the arc over a circle with the radius 1 where the angle goes from the center.\r\n\r\nso $2\\pi = 360^{\\circ}$ because that would be a full circle and the circumfence $c=2\\pi r$ and $r=1$. then the arc length is propotional to the angle, hence $\\dfrac{\\alpha}{360^{\\circ}} = \\dfrac{x}{2\\pi}$ where $\\alpha$ is in degree and $x$ in radiants ( a diffrent name for arc length, the more common one actually)", "Solution_10": "Oh yeah I think we learnt that in physics (angular motion) just forgot", "Solution_11": "peeta- Yes, that's correct :clap2: There is also a way to solve it without knowing the formula for triple angle.\r\n\r\nfelipesa- Your answer for #2 is now correct :D \r\n\r\nAlso, for #4, your work looks fine to me, but somehow it's not quite correct. $\\dfrac{2\\pi}{3}$ doesn't work, but $\\dfrac{5\\pi}{3}$ does work. (Does anyone know why?) In addition, there's one more solution...", "Solution_12": "#3\r\n\r\n$sin3x(sin3x-cosx)=sinx(sinx-cos3x)$\r\n\r\n$\\dfrac{sin3x}{sinx}=\\dfrac{sinx-cos3x}{sin3x-cosx}$\r\n\r\n$\\dfrac{sinx(-1+4cos^2x)}{sinx}=\\dfrac{sinx-cosx(1-4sen^2x}{sinx(-1+4cos^2x)-cosx}$\r\n\r\n$sin^2x(-1+4cos^2x)^2-sinxcosx(-1+cos^2x)=sin^2x-sinxcosx(1-4sin^2x)$\r\n\r\n$sin^2x(-1+4cos^2x)^2-sin^2x=senxcosx(-1+4cos^2x-(1-4sin^2x))$\r\n\r\n$-8sin^2xcos^2x+16sin^2xcos^4x=2sinxcosx$\r\n\r\n$8sin^2xcos^2x(2cos^2x-1)=2sinxcosx$\r\n\r\n$4sen2xcos2x=2$\r\n\r\n$8sinxcosx(1-2cos^2x)=2$\r\n\r\n$-4sinxcosx+8sinxcos^3x=1$\r\n\r\n$sinx(-4cosx+8cos^3x)=1$\r\n\r\n$sin4x=1$\r\n\r\n$x \\in [0,\\pi)$\r\n\r\n\r\n$x=\\dfrac{\\pi}{8}$\r\n$x=\\dfrac{5\\pi}{8}$\r\n\r\nwe can have more two but in this domain $x \\in [0,2\\pi)$\r\n$x=\\dfrac{9\\pi}{8}$\r\n$x=\\dfrac{13\\pi}{8}$\r\n\r\nFor this domain $x \\in [0,\\pi)$ only two solutions are possible.\r\n$x=\\dfrac{\\pi}{8}$\r\n$x=\\dfrac{5\\pi}{8}$", "Solution_13": "Well... you are missing 3 more solutions :( \r\n$\\sin 4x=1$ even gives you one of them (because $0\\le x<\\pi\\Longleftrightarrow 0\\le 4x<4\\pi$)\r\n\r\n\r\nEDIT (since felipesa edited the answer a bit)-\r\nClose, but it's $x=\\dfrac{\\pi}{8}+\\dfrac{k\\pi}{2}$, where $k=0,1$ ;)\r\n\r\nFind two more!\r\n\r\nEDIT 2- Think about the domain one more time!", "Solution_14": "Ok, since the posts are getting complicated, I'll clarify which ones we got so far.\r\n\r\npeeta solved #1.\r\nfelipesa solved #2 and #5.\r\n\r\nfelipsa tried #3 and #4 as well, but he's still missing some solutions.\r\n\r\n[hide=\"Tips for 3\"]\nfelipesa got $\\dfrac{\\pi}{8}$ and $\\dfrac{5\\pi}{8}$, and these are correct. However, there are two more solutions that are NOT in the form of $\\dfrac{\\pi}{8}+\\dfrac{k\\pi}{2}$, where$k$ is an integer. Try something different. For example, what if you expand the original equation and simplify it? :) [/hide]\n\n[hide=\"Tips for 4\"]\nfelipesa gave $\\dfrac{2\\pi}{3}$ and $\\dfrac{5\\pi}{3}$ as solutions, but only $\\dfrac{5\\pi}{3}$ works. In addition, there's one more solution. Think about the following: $\\cos x+\\sqrt{3}\\sin x=a\\sin (x-b)$ ;) [/hide]\r\n\r\n\r\nAnyone else got #3 or #4?", "Solution_15": "here is 4:\r\n\r\n$\\cos x + \\sqrt 3 \\sin x +1 =0 \\implies \\cos x=- ( \\sqrt 3 \\sin x +1)$\r\n\r\n$\\implies \\cos ^2 x = 1- \\sin ^2x = 3\\sin ^2 x + 2\\sqrt 3 \\sin x +1$\r\n$\\implies 4\\sin x ( \\sin x + \\dfrac{\\sqrt 3}{2}) = 0$\r\n$\\implies \\sin x= 0$ or $\\sin x = -\\dfrac{\\sqrt 3}{2}$\r\n\r\nbecause $x\\in[0;2\\pi)$:\r\n$x_1 = 0$\r\n$x_2 = \\pi$\r\n$x_3 = \\dfrac{4\\pi}{3}$\r\n$x_4 = \\dfrac{ 5\\pi}{3}$\r\n\r\nbecause of the squaring we have to select the correct solutions, which are after plugging in $x_2$ and $x_4$", "Solution_16": "#3 :) \r\n\r\n$\\sin^23x-\\sin 3x\\cos x=\\sin^2x-\\sin x\\cos 3x$\r\n\r\n$\\implies (1-\\cos 6x)-(\\sin 4x+\\sin 2x)=(1-\\cos 2x)-(\\sin 4x-\\sin 2x)$\r\n\r\n$\\implies 2\\sin 2x=\\cos 2x-\\cos 6x$\r\n\r\n$\\implies 2\\sin 2x=2\\sin 4x\\sin2x$\r\n\r\n$\\implies \\sin 2x(\\sin 4x -1)=0$\r\n\r\nFor $\\sin 2x=0$ , $2x=n\\pi +(-1)^n\\times 0=n\\pi \\implies x=\\dfrac{n\\pi}{2}$\r\n\r\nSo we obtain $\\boxed{x=0}$ , $\\boxed{\\dfrac{\\pi}{2}}$\r\n\r\nFor $\\sin 4x=1$ , $4x=n\\pi +(-1)^n\\times \\dfrac{\\pi}{2}$\r\n\r\nSo $\\boxed{x=\\dfrac{\\pi}{8}}$ , $\\boxed{\\dfrac{5\\pi}{8}}$", "Solution_17": "#4 :) \r\n\r\n$\\cos x+\\sqrt 3\\sin x+1=0$\r\n\r\n$\\implies 2\\sin(x+\\dfrac{\\pi}{6})=-1$\r\n\r\n$\\implies \\sin(x+\\dfrac{\\pi}{6})=\\dfrac{-1}{2}$\r\n\r\n$\\implies x+\\dfrac{\\pi}{6}=n\\pi+(-1)^n\\times \\dfrac{-\\pi}{6}$\r\n\r\nHence $x=n\\pi+(-1)^n\\times \\dfrac{-\\pi}{6}-\\dfrac{\\pi}{6}$\r\n\r\nputting $n=1,2$ and obtain $\\boxed{x=\\pi}$ , $\\boxed{\\dfrac{5\\pi}{3}}$", "Solution_18": "[quote=\"shyong\"]\nFor $\\sin 2x=0$ , $2x=n\\pi +(-1)^n\\times 0=n\\pi \\implies x=\\dfrac{n\\pi}{2}$\n\nSo we obtain $\\boxed{x=0}$ , $\\boxed{\\dfrac{\\pi}{2}}$\n\nFor $\\sin 4x=1$ , $4x=n\\pi +(-1)^n\\times \\dfrac{\\pi}{2}$\n\nSo $\\boxed{x=\\dfrac{\\pi}{8}}$ , $\\boxed{\\dfrac{5\\pi}{8}}$[/quote]\r\n\r\nDo you know why I dont find $sin2x=0$ in my solution?\r\n\r\nthanks", "Solution_19": "It's because you divided both sides by $\\sin{x}\\cos{x}(=\\dfrac{1}{2}\\sin{2x})$ in your solution. To avoid it, you have to factor it out ;) \r\n\r\nAnyway, the answers peeta and shyong got were correct :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a, b, c$ three non-negative real numbers. Prove the following inequality\r\n\r\n \\[ \\sum_{cyc}\\frac{a^{2}(b+c)}{b^{2}+bc+c^{2}}\\geq\\frac{2}{3}(a+b+c). \\]", "Solution_1": "[quote=\"Cezar Lupu\"]Let $a, b, c$ three non-negative real numbers. Prove the following inequality\n\n \\[ \\sum_{cyc}\\frac{a^{2}(b+c)}{b^{2}+bc+c^{2}}\\geq\\frac{2}{3}(a+b+c). \\][/quote]\r\n\r\ni think that this was posted by darij some time ago and harazi gave a very nice proof!!\r\n :)", "Solution_2": "A stronger inequality is the following\r\n $\\sum_{cyc}\\frac{a^{2}(b+c)}{b^{2}+c^{2}}\\geq a+b+c.$", "Solution_3": "Thank you for reminding me...\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=14577\r\n\r\n darij", "Solution_4": "[quote=\"Vasc\"]A stronger inequality is the following\n $\\sum_{cyc}\\frac{a^{2}(b+c)}{b^{2}+c^{2}}\\geq a+b+c.$[/quote]\r\nNICE INEQUALITY.\r\nAnd i Sloved by caughy.\r\n${\\sum_{cyc}\\frac{a^{2}(b+c)}{b^{2}+c^{2}} (\\sum a^2(b+c)(b^{2}+c^{2}}) ) \\ge (\\sum a^2(b+c))^2$", "Solution_5": "To zhaobin: :) I solved it in the same way as yours.\r\nthe Inquality come to be Shur's Ineq. :)", "Solution_6": "Harazi's solution in the link is also very nice ;)", "Solution_7": "I agrre with you :)", "Solution_8": "hey $\\sum_{cyc} \\frac{a^2(b+c)}{b^2+c^2}\\geq \\sum_{cyc} a \\leftrightarrow \\sum_{sym} a^6 b \\geq \\sum_{sym} a^5 b^2$", "Solution_9": "You are right! Check demands no more, than 5 minutes! :D" } { "Tag": [ "LaTeX", "calculus" ], "Problem": "Grandma bought 2 candles. The red is 1 cm longer than the blue candle. In the afternoon of the Day of Christmas at 17:30 she lit the red candle, at 19:00 she lit the blue candle, also, and let them burn until they were finished. The two candles had the same length at 21:30. The red was finished at 23:30, and the blue was finished at 23:00. How long was the red candle originally?\r\n\r\nABACUS International Math Challenge for 7th and 8th graders", "Solution_1": "4 hours to burn the red to the length of the blue one which burned 2.5 hours. So the red one burned one more inch.\r\n\r\n1.5 hours to burn an inch\r\n\r\nsince the red one burned for 6 hours it had 4 inches", "Solution_2": "[hide]hm... I got the blue candle burned for 4 hours while the red candle burned for 6 hours\n\nso each cm burns for 2 hours, so the red candle would be 3cm long.... not sure if it's right....[/hide]", "Solution_3": "[hide]R=B+1\nRed burns for 6 hrs\nBlue burns for 4 hrs\nRed=3 in\nBlue=2 in[/hide]\r\nAt least, that's what I think. I might be wrong though", "Solution_4": "I got that too but it doesn't satisfy the condition for when the candles are the same height.", "Solution_5": "[quote=\"236factorial\"]Grandma bought 2 candles. The red is 1 cm longer than the blue candle. In the afternoon of the Day of Christmas at 17:30 she lit the red candle, at 19:00 she lit the blue candle, also, and let them burn until they were finished. The two candles had the same length at 21:30. The red was finished at 23:30, and the blue was finished at 23:00. How long was the red candle originally?\n\nABACUS International Math Challenge for 7th and 8th graders[/quote]\r\n\r\n[hide]\n\n$r(t)=r_0+\\dot{r}t$\n\n$b(\\tau)=b_0+\\dot{b}\\tau$\n\n$r_0 = b_0+1$\n\n$r(4)=b(2.5)$\n\n$r(6)=r_0+6\\dot{r} =0 \\Rightarrow \\dot{r} =\\frac{-r_0}{6} =\\frac{-(b_0+1)}{6}$\n\n$b(4)=b_0+4\\dot{b} =0 \\Rightarrow \\dot{b} =\\frac{-b_0}{4}$\n\n$r(4)=b(2.5) \\Rightarrow b_0+1-4\\cdot\\frac{(b_0+1)}{6} =b_0-2.5\\cdot\\frac{b_0}{4}$\n\n$\\Rightarrow \\frac{1}{3}(b_0+1)=\\frac{1.5}{4}b_0$\n\n$\\frac{1}{3}=b_0(\\frac{1.5}{4}-\\frac{1}{3})$\n\n$12\\times\\frac{1}{3}=12\\times b_0(\\frac{1.5}{4}-\\frac{1}{3})$\n\n$4=b_0(4.5-4)$\n\n$4=\\frac{b_0}{2} \\Rightarrow b_0 = 8$\n\n$r_0=b_0 + 1 = 8 + 1 = 9\\text{ cm }\\blacksquare$\n\n[/hide]", "Solution_6": "Well done, but a little long. Could there possibly be a shorter way to do this?", "Solution_7": "[quote=\"236factorial\"]Well done, but a little long. Could there possibly be a shorter way to do this?[/quote]\r\n\r\nThanks, everyone for your kind words. I don't know whether there is a shorter method or not. I shall put no restrictions on anyone's cleverness -- if there is a shorter way someone ought to find it.\r\n\r\nFor myself, I only solved the problem as an example of how to systematically attack such problems and force them to give a correct answer :D It's the teacher in me I suppose.", "Solution_8": "So you a teacher?", "Solution_9": "[quote=\"nat mc\"]So you a teacher?[/quote]\r\n\r\nYes, though I am not currently employed as a teacher. Teaching is the one thing I would do for free though, I love it.", "Solution_10": "[hide=\"How I did it\"]\nBlue candle: 2.5 hours same height 1.5 hours\nRed candle: 4 hours same height 2 hours\nThey burn at different rates, since after they were the same height, it burned differently. So:\n1.5/2=3/4\n2.5/3=.8333333333\nAns x 4 = 3.33333333\n4 - Ans = 2/3\n6/(2/3)=9\n[color=darkblue][u][b][size=200]9[/size][/b][/u][/color][/hide]", "Solution_11": "didn't like the numbers, try $\\LaTeX$", "Solution_12": "I didn't quite get your answer either.", "Solution_13": "[quote=\"nat mc\"]didn't like the numbers, try $\\text{\\LaTeX}$[/quote]\r\nBlue candle: 2.5 hours....same height.....1.5 hours \r\nRed candle: 4 hours.......same height.....2 hours \r\nThey burn at different rates, since after they were the same height, it burned differently. So: \r\n$\n1.5 \\div 2 \\equiv \\frac{3}{4}\n2.5 \\div 3 \\equiv .8333333333\n Ans \\times 4 \\equiv 3.33333333 \n4 - Ans \\equiv \\frac{2}{3} \n6/(2/3) \\equiv 9 \nAnswer: 9$\r\n\r\nsorry, but right now I can't remember where those numbers came from....I can't explain it right now.", "Solution_14": "i think the part that confuses me is that I can't see where the 1 cm longer part went.", "Solution_15": "[quote=\"Dr. No\"][quote=\"nat mc\"]So you a teacher?[/quote]\n\nYes, though I am not currently employed as a teacher. Teaching is the one thing I would do for free though, I love it.[/quote]\r\nOh that's why you know calculas and all this complicateed math", "Solution_16": "[quote=\"math92\"][quote=\"Dr. No\"][quote=\"nat mc\"]So you a teacher?[/quote]\n\nYes, though I am not currently employed as a teacher. Teaching is the one thing I would do for free though, I love it.[/quote]\nOh that's why you know calculas and all this complicateed math[/quote]\r\n\r\nGuilty as charged :) though being a calculus show-off is not at all why I do this.\r\n\r\nThe one thing I hope students will get from my solutions, is that if you solve the problem in general first (using variables instead of specific numerical values), and then \"plug-in the numbers\" after you're done, you'll very seldom go wrong.", "Solution_17": "[quote=\"Dr. No\"][/quote][quote=\"math92\"][quote=\"Dr. No\"][quote=\"nat mc\"]So you a teacher?[/quote]\n\nYes, though I am not currently employed as a teacher. Teaching is the one thing I would do for free though, I love it.[/quote]\nOh that's why you know calculas and all this complicateed math[/quote][quote=\"Dr. No\"]\n\nGuilty as charged :) though being a calculus show-off is not at all why I do this.\n\nThe one thing I hope students will get from my solutions, is that if you solve the problem in general first (using variables instead of specific numerical values), and then \"plug-in the numbers\" after you're done, you'll very seldom go wrong.[/quote]\r\n\r\nSo, your solution was somewhat calculus?!!\r\nI'm a little too stupid for calculus :blush:", "Solution_18": "I think that I understood his solution. However, it's less beneficial for me to know it thatn to solve it by simpler means.", "Solution_19": "[quote=\"4everwise\"]I think that I understood his solution. However, it's less beneficial for me to know it thatn to solve it by simpler means.[/quote]\r\n\r\nMainly the symbols threw me off. Sometimes I didn't know what they stood for.", "Solution_20": "[quote=\"236factorial\"][quote=\"4everwise\"]I think that I understood his solution. However, it's less beneficial for me to know it thatn to solve it by simpler means.[/quote]\n\nMainly the symbols threw me off. Sometimes I didn't know what they stood for.[/quote]\r\n\r\nThe only thing in my solution that required knowledge of anything beyond Algebra I was where I used the symbols $\\dot{r}$ or $\\dot{b}$. These meant the rates at which the candles, red and blue respectively, were burning. Also $r_0$ and $b_0$ meant the lengths of the candles at time, t = 0. This is all shorthand and shouldn't bother anybody too much." } { "Tag": [], "Problem": "Who made it to the next round? I didn't, and I got 19/25 on the test...\r\n\r\n[size=150][color=blue]\nMOD EDIT:I added the Country[/color][/size]", "Solution_1": "are you sure about that? cuz i got a 14.25 and i made it...you can check your official score on webassign if you took it online...", "Solution_2": "We dont do webassign, is there a way I can still find out my score?", "Solution_3": "do specify the name of ur country :D", "Solution_4": "I'm pretty sure we're talking about the US.\r\n\r\nAnyway, yeah, I think we're talking on CC too...(OVB = piccolojunior?)\r\n\r\nI think you should check to make sure you got a 19 - they have the answers up. If you want a copy of the test, I can e-mail that to you. (Or, well, once I get your e-mail address, anyway) Otherwise, your teacher got an e-mail with scores. (Or, well, our teacher did).", "Solution_5": "My teacher only knows who qualify for next round. How can I know my score?", "Solution_6": "How many perfect scores around the whole country?", "Solution_7": "Where do you get all this information? I can't find an updated (or detailed) website about the US physics team tests.", "Solution_8": "I also got through to the next round, however my teacher wasn't able to tell me what my score was. Anyone kows how I can find out my score?", "Solution_9": "does anyone have an electronic copy of the 2008 F=ma contest (multiple choice as well as free response parts)? could someone possibly PM it to me?", "Solution_10": "Can anyone forward it to me if they have it too (:\r\nThanks", "Solution_11": "I'm curious to see the new format of the preliminary round, as I only have access to previous years when it was not exclusively mechanics. If any one has a digital copy of this year's F=MA exam (2008), could you please send me a copy at Tommy2030@gmail.com . It would be greatly appreciated.\r\n\r\n-Tommy" } { "Tag": [ "function", "geometry", "3D geometry", "logarithms" ], "Problem": "Well this is something like Kebian's 24 but it is different \r\n\r\nHere you have to use all the numbers 0,1,2,3,4,5,6,7,8,9 [u]in the given order only[/u] to form a number n. \r\n\r\nUse of all mathematical functions lik log , in , ! etc. is allowed .\r\n\r\nYou can use square root without using the '2' but for cube roots fourth roots etc. you have to use a '3' or '4' etc.\r\n\r\nThe first person to answer will get 5 points and the next 4 will get 4,3,2,1 in that order.\r\n\r\nIn case you post more than one answer you will get a '*' which\r\nwill decide who is first in case of a tie.\r\n\r\nToday's number is 44. :huh:", "Solution_1": "[hide](1*2)+3+4+5+6+7+8+9[/hide]", "Solution_2": "$\\log 1 +2+3+4+5+6+7+8+9$", "Solution_3": "nice problem....i keep getting 42 instead of 41...argggggggggh....", "Solution_4": "$((1+ 2) \\cdot 3 + 4!) -5+6-7+8+9$", "Solution_5": "[quote=\"ch1n353ch3s54a1l\"]$1!\\cdot 2! \\cdot 3! \\cdot 4! -5+6-7+8+9$[/quote]\r\n\r\nThats... um... 299...", "Solution_6": "[quote=\"Klebian\"][quote=\"ch1n353ch3s54a1l\"]$1!\\cdot 2! \\cdot 3! \\cdot 4! -5+6-7+8+9$[/quote]\n\nThats... um... 299...[/quote]\r\n\r\nSorry. *EDITED*", "Solution_7": "0!-1+2+3+4+5+6+7+8+9\r\n\r\nto chinesechess4all, you have to use 0", "Solution_8": "[quote=\"biffanddoc\"]0!-1+2+3+4+5+6+7+8+9\n\nto chinesechess4all, you have to use 0[/quote]\r\nOk then. \r\n$0+((1+ 2) \\cdot 3 + 4!) -5+6-7+8+9$\r\n :)", "Solution_9": "$0!+1-2+3!+\\sqrt{4}+5+6+7+8+9$", "Solution_10": "$(\\log(0+1+2+3+4)+5) \\times 6 +7 -8 +9$\r\nto vihang, don't I get 5 points because my solution was the first to use 0?", "Solution_11": "Oh, come on, don't be cheap, all of them could just do \" 0 +\" and be done too....", "Solution_12": "New number is 248.\r\n\r\nWill post the scores later.", "Solution_13": "$0+(((1\\cdot 2)^3)\\cdot 4 -5+6)\\cdot 7 +8+9$", "Solution_14": "[quote=\"Klebian\"]Oh, come on, don't be cheap, all of them could just do \" 0 +\" and be done too....[/quote]\r\nWell maybe if they had read the rules better :roll:", "Solution_15": "on the honor code:\r\n$0+(-1+2+34)(5)-6+7+8(9)$\r\n$0+1 \\cdot 2^3 \\cdot (-4(5)+6(7))+8(9)$\r\n$0+(-1+23+4)(5+6)+7+8+9$\r\n$0+(1+23)(4)+56+7+89$\r\n$0+1 \\cdot 2+3 \\cdot 4+5!+6 \\cdot 7+8 \\cdot 9$\r\n\r\n[hide=\"EDIT\"]\nsorry, but i got in through a link and i didn't realize this was a middle school forum.\ni don't think my answers should count because i'm in highshool\n :oops: \n[/hide]", "Solution_16": "[quote=\"minsoens\"]\nsorry, but i got in through a link and i didn't realize this was a middle school forum.\ni don't think my answers should count because i'm in highshool\n :oops: \n[/quote]\r\nWell, try this one then :) : http://www.mathlinks.ro/Forum/viewtopic.php?p=396980#p396980", "Solution_17": "same here, i'm high school...meh\r\n[hide]$0+1*2^3(4+5+6+7+8-9+10)$[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "Suppose $ a_1 \\plus{} a_2 \\plus{} \\ldots \\plus{} a_n \\equal{} 2n^2(3 \\plus{} a_{n\\plus{}1} \\plus{} a_{n\\plus{}2} \\plus{} \\ldots \\plus{} a_{50})$ holds for $ n \\in \\{1, 2, \\ldots , 50\\}.$ Find $ a_1.$\r\n\r\n[hide=\"answer\"]\n10002[/hide]\r\n\r\nHow do you do this problem?", "Solution_1": "[hide]\n$ a_1\\plus{}a_2...a_50 \\equal{} 2.(50)^2.(3)$\n$ \\Rightarrow a_1 \\equal{} 2(3 \\plus{} 6.(50)^2 \\minus{} a_1)$\n$ \\Rightarrow a_1 \\equal{} 2 \\plus{} 4.(50)^2 \\equal{} 10002$ :lol: [/hide]", "Solution_2": "a1 + a2 + ..... + an = (2n^2)(3 + a_(n+1) + .... + a50)\r\n\r\n=> a1 = (2) (3 + a2 + .... + a50)...............(1)\r\n\r\nalso,\r\na1 + a2 + ....... + a50 = 3*2(50)^2\r\n\r\n=> a2 + .... + a50 = 6 * (50)^2 - a1 ........................(2).\r\n\r\nsubstituting (2) in (1),\r\n\r\na1 = 2 (3 + 6*(50)^2 - a1)\r\n\r\n3a1 = 6 + 12*(50)^2\r\n\r\na1 = 2 + 4*(50)^2\r\n\r\n = 2 + 10000\r\n\r\n = 10002. :)", "Solution_3": "[quote=\"hurdler\"]Suppose $ a_1 \\plus{} a_2 \\plus{} \\ldots \\plus{} a_n \\equal{} 2n^2(3 \\plus{} a_{n \\plus{} 1} \\plus{} a_{n \\plus{} 2} \\plus{} \\ldots \\plus{} a_{50})$ holds for $ n \\in \\{1, 2, \\ldots , 50\\}.$ Find $ a_1.$\n\n[hide=\"answer\"]\n10002[/hide]\n\nHow do you do this problem?[/quote]\r\nHow can the last term on the RHS be a_50? It doesn't satisfy the pattern of that sequence(when n=49 or n=50),isn't it?", "Solution_4": "[quote=\"Attila the Hun\"]$ a_1 \\plus{} a_2 \\plus{} ..... \\plus{} a_n \\equal{} (2n^2)(3 \\plus{} a_{n\\plus{}1} \\plus{} .... \\plus{} a_{50})$\n\n=> $ a_1 \\equal{} (2) (3 \\plus{} a_2 \\plus{} \\hdots \\plus{} a_{50})$ (1)\n\nalso,\n$ a_1 \\plus{} a_2 \\plus{} \\hdots \\plus{} a_{50} \\equal{} 3 \\cdot 2(50)^2$\n\n=> $ a_2 \\plus{} .... \\plus{} a_{50} \\equal{} 6 * (50)^2 \\minus{} a_1$ (2).\n\nsubstituting (2) in (1),\n\n$ a_1 \\equal{} 2 (3 \\plus{} 6*(50)^2 \\minus{} a_1)$\n\n$ 3a_1 \\equal{} 6 \\plus{} 12*(50)^2$\n\n$ a_1 \\equal{} 2 \\plus{} 4*(50)^2$\n\n = $ 2 \\plus{} 10000$\n\n = $ 10002$. :)[/quote]\r\n\r\nJust LaTeXifying (Coining a new word, if I may), for your convieniece.", "Solution_5": "[quote=\"BOGTRO\"][quote=\"Attila the Hun\"]$ a_1 \\plus{} a_2 \\plus{} ..... \\plus{} a_n \\equal{} (2n^2)(3 \\plus{} a_{n \\plus{} 1} \\plus{} .... \\plus{} a_{50})$\n\n=> $ a_1 \\equal{} (2) (3 \\plus{} a_2 \\plus{} \\hdots \\plus{} a_{50})$ (1)\n\nalso,\n$ a_1 \\plus{} a_2 \\plus{} \\hdots \\plus{} a_{50} \\equal{} 3 \\cdot 2(50)^2$\n\n=> $ a_2 \\plus{} .... \\plus{} a_{50} \\equal{} 6 * (50)^2 \\minus{} a_1$ (2).\n\nsubstituting (2) in (1),\n\n$ a_1 \\equal{} 2 (3 \\plus{} 6*(50)^2 \\minus{} a_1)$\n\n$ 3a_1 \\equal{} 6 \\plus{} 12*(50)^2$\n\n$ a_1 \\equal{} 2 \\plus{} 4*(50)^2$\n\n = $ 2 \\plus{} 10000$\n\n = $ 10002$. :)[/quote]\n\nJust LaTeXifying (Coining a new word, if I may), for your convieniece.[/quote]\r\n\r\n\r\nthanks BOGTRO. I don't have LaTeX. :oops:", "Solution_6": "You don't need to have anything installed to use $ \\text{\\LaTeX}$. All you need to do is put dollar signs around your math, with [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]commands and other stuff, which are not hard to learn[/url].", "Solution_7": "[quote=\"ghjk\"][quote=\"hurdler\"]Suppose $ a_1 \\plus{} a_2 \\plus{} \\ldots \\plus{} a_n \\equal{} 2n^2(3 \\plus{} a_{n \\plus{} 1} \\plus{} a_{n \\plus{} 2} \\plus{} \\ldots \\plus{} a_{50})$ holds for $ n \\in \\{1, 2, \\ldots , 50\\}.$ Find $ a_1.$\n\n[hide=\"answer\"]\n10002[/hide]\n\nHow do you do this problem?[/quote]\nHow can the last term on the RHS be a_50? It doesn't satisfy the pattern of that sequence(when n=49 or n=50),isn't it?[/quote]\r\n\r\nfor n=49 or n=50, this problem probably means that no terms exist on the RHS that are of the form a_i, so the right hand side would become $ 2n^2(3)$ when n=49 or n=50." } { "Tag": [ "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "1. sinx/1-cosx. \r\n(1-cosx)(sinx)'-(sinx)(1-cosx)'\r\n-----------------------------------\r\n (1-cosx)^2\r\n\r\n(1-cosx)(cosx)-(sinx)(sinx)\r\n--------------------------------\r\n (1-cosx)^2\r\n\r\n(cosx-cosx^2)-sinx^2\r\n--------------------------------\r\n (1-cosx)^2\r\n\r\nthe answer comes out to be 1/cos(x)-1.\r\n\r\n2. 1+sinx/2-cosx\r\n(2-cosx)(1+sinx)'-(1+sinx)(2-cosx)'\r\n---------------------------------------\r\n(2-cosx)^2\r\n\r\n(2-cosx)(cosx)-(1+sinx)(sinx)\r\n---------------------------------------\r\n(2-cosx)^2\r\n\r\n2cosx-cosx^2-sinx-sinx^2\r\n---------------------------------------\r\n(2-cosx)^2\r\n\r\nthe answer comes out to be (2cosx -sinx -1)/(2-cosx)^2.\r\n\r\n3.sinx+cosx/sinx-cosx\r\n \r\n(sinx-cosx)(sinx+cosx)'-(sinx+cosx)(sinx-cosx)'\r\n-----------------------------------------------\r\n(sinx-cosx)^2\r\n\r\n(sinx-cosx)(cos-sinx)-(sinx+cosx)(cosx+sinx)\r\n-----------------------------------------------\r\n(sinx-cosx)^2\r\n\r\nand after a whole of crossing out, I got (sinxcosx+sinxcosx-sinxcosx+sinxcosx)/(sinx-cosx)^2\r\n\r\nbut the answer says -2/(sinx-cosx)^2", "Solution_1": "[quote=\"red218\"]\n3.sinx+cosx/sinx-cosx\n[/quote]\r\n$ f(x) = \\frac{\\sin x+\\cos x}{\\sin x-\\cos x}\\cdot \\frac{\\sin x+\\cos x}{\\sin x+\\cos x}$\r\nThus,\r\n$ f(x) = \\frac{\\sin^{2}x+\\cos^{2}x+2\\sin x \\cos x}{\\sin^{2}x-\\cos^{2}x}=-\\frac{1+\\sin 2x}{\\cos 2x}=-\\sec 2x-\\tan 2x$\r\n\r\nNow that is easier to differentiate." } { "Tag": [ "function", "LaTeX", "real analysis", "real analysis unsolved" ], "Problem": "[code] y y^{'} +y =f \ny and f are functions in code H^s\nthere is any solution of this ode?\n[/code]", "Solution_1": "hello, can you use $ \\text{\\LaTeX}$ please?\r\nSonnhard." } { "Tag": [ "group theory", "abstract algebra", "search", "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "prove that in the ring $ (\\mathbb Q, \\plus{} ,.)$ hasn't a maximal ideal .\r\n\r\nwhere $ \\plus{} : \\text{\\ usual addition } ,\\cdot: a.b \\equal{} 0,\\forall a,b\\in R$\r\n\r\nAlso, does anyone know another example ?\r\n\r\n :)", "Solution_1": "You should say that your \"rings\" are in fact without unity :wink: \r\n\r\nWell, the only nontrivial ideal condition is that it's closed under addition (the multiplicative part just requires it to contain $ 0$). So you want to show that $ \\mathbb Q$ has no maximal proper subgroup.", "Solution_2": "very nice :D , using the correspondence theorem and $ M$ is maximal subgroup iff $ G/M$ is simple . we get $ \\mathbb Q$ hasn't maximal subgroup.\r\n\r\nI'm thinking about $ \\mathbb{Z}_{p^{\\infty}}$ group , with the same multiplication above .(i.e. $ ({\\mathbb{Z}_{p^{\\infty}}, + ,.)}$ ring \r\n\r\n\r\nIs there any maximal ideal in it ?\r\n\r\nThank you ZetaX :)", "Solution_3": "A quick google search turns up [url=http://sierra.nmsu.edu/morandi/notes/NoMaxIdeals.pdf]this note[/url]." } { "Tag": [ "geometry" ], "Problem": "Find the minimum area of the triangle whose vertices are $A(-1,1,2)$, $B(1,2,3)$, and $C(t,1,1)$. ($t$ is a real number.)", "Solution_1": "$\\overrightarrow {AB}=(2,1,1), \\overrightarrow {AC}=(t+1,0,-1),$ thus\r\n\r\n$\\triangle {ABC}=\\frac{1}{2}\\sqrt{|\\overrightarrow {AB}|^2|\\overrightarrow {AC}|^2-(\\overrightarrow{AB}\\cdot \\overrightarrow {AC})^2}=\\frac{1}{2}\\sqrt{6\\{(t+1)^2+1\\}-\\{2(t+1)-1\\}^2}$\r\n\r\n${=\\frac{1}{2}\\sqrt{2(t+2)^2+3}\\geq \\frac{\\sqrt{3}}{2}},$ the equality holds when $t=-2.$ Therefore desired answer is $\\frac{\\sqrt{3}}{2}.$" } { "Tag": [ "geometry" ], "Problem": "Well, there really haven't been many major math competitions, so until there are, let's talk about some sports teams in Wisconsin. (mustafa I did this for you!)", "Solution_1": "Bucks Suck\r\n\r\nGo Pistons\r\n\r\nBrewers Suck\r\n\r\nGo Tigers\r\n\r\nPackers Suck\r\n\r\nSo do the Lions", "Solution_2": "More like\r\n\r\nBrewers Suck\r\nTigers Suck\r\nGo Red Sox (Daisuke Matsuzaka is the shiz)\r\n\r\nPackers Suck\r\nLions Suck\r\nGo Pats\r\n\r\nBucks Suck\r\nPistons Suck\r\nOh wait so do the Celtics", "Solution_3": "Well, the Red Wings are good and there aren't really any hockey teams in Wisconsin. And the Bruins suck.", "Solution_4": "Well we've owned football for the last six years so i don't care.", "Solution_5": "Brewers-used to be good, may be good in near future\r\nBucks-the epitomy of mediocrity\r\nPackers-used to be good, not out of playoff hunt yet\r\nUW Badger Basketball-\"the shiz\"\r\nMarquette Basketball-almost \"the shiz\"\r\nBadger Men's Hockey-were \"the shiz\"\r\nBadger Football- \"the super deluxe ultra shiz\"\r\n\r\nThanks for letting me use your phrase LEETSPEAK :rotfl:", "Solution_6": "Dang. When I saw this topic, I thought you misspelled Wii sports. Boo.", "Solution_7": "[quote=\"Klebian\"]Dang. When I saw this topic, I thought you misspelled Wii sports. Boo.[/quote]Well, WI sports are just as shizzy as Wii sports! :D", "Solution_8": "[quote=\"13375P34K43V312\"]Well we've owned football for the last six years so i don't care.[/quote]\r\n\r\nare you talking about the patriots...its kinda funny what i think\r\n\r\n[code]\n Boston (Ma area) St. Louis \n\nBaseball Love them Hate them\n\n\nFootball Hate them Love them\n\n[/code]\r\n\r\n-jorian", "Solution_9": "Indeed. I was once a Ram bandwagon fan in the days of Kurt Warner", "Solution_10": "lol, that was an annoying game against the patriots :mad: [url]http://en.wikipedia.org/wiki/Super_Bowl_XXXVI[/url]\r\n-jorian", "Solution_11": "\uff22\uff32\uff25\uff34\uff34\u3000\uff26\uff21\uff36\uff32\uff25\uff1a\u3000\uff32\uff25\uff34\uff29\uff32\uff25\uff0e", "Solution_12": "The Packers are going to win the Super Bowl this year after a magical elf brings Brett Favre a special lotion that makes him younger. :roll:", "Solution_13": "Well, the Packers are close to the playoffs, the Bucks are going for 6 straight tonight, the Badgers play #12 Arkansas in 5 days, Badger Basketball is #4, and Marquette is #19. Wisconsin sports are looking pretty darn good.", "Solution_14": "mavericks=awesome\r\ncowboys=almost awesome\r\nstars=close to awesome\r\nrangers=awesome for the first half of the season, then they pretty much die", "Solution_15": "badgersw win again why am i not surprised FOWIST GUMP", "Solution_16": "That was far too close for comfort, almost blowing a 12 point lead, but the Badgers prevailed over Ohio State, no matter how good their cafeteria food may be.", "Solution_17": "[quote=\"Klebian\"]Dang. When I saw this topic, I thought you misspelled Wii sports. Boo.[/quote]\r\n\r\nSame with me." } { "Tag": [ "MATHCOUNTS", "Support" ], "Problem": "uhh.\r\n\r\nI totally want to go to Mathcounts, but my school doesn't have any math team, so I can't go to Mathcounts. (I took amc at some high school :( )\r\n\r\nWhat can I do to go to MC?\r\n\r\nShould I register with the private school?", "Solution_1": "i'm fairly sure you can do mathcounts without a team. The people who do mathcounts without a team are called Individuals. As an individual, the competition is exactly the same, except you dont have a team(which didn't bother me at states; i was an individual) and thus the team round isnt graded. At some places you have an option of taking the team round with other individuals, but it wont really matter. However, many people(i would think) benefit from having a team, since you can interact with your teammates and talk math. But that what i do on AoPS :lol: . So basically, you can still do mathcounts without a team, if that's what you are asking.", "Solution_2": "You could request a team be made for your school. Ask a math teacher to be the coach, perhaps. You've got a whole year...", "Solution_3": "I'm lucky to be on a team, and even have someone else on my team to go to nats with. How else would you spend the like 2 HOURS in between team and countdown other than messing around with your team???? It definitely helps me to relax in new surroundings when I have someone else there that is new too. So here's to teams!!! :lol:\r\n\r\np.s.: fantasy lover, what grade and state r u?", "Solution_4": "yeah i already requested to principal to make a math team but he said that there's too few students who is good at math...\r\n\r\nand we don't have any math coach (my school is not very good)\r\n\r\nP.S. im in 7th grade, nj.", "Solution_5": "maybe even you could be the coach! usually the person who wants something has to put more effort than everybody else. And the other people in your school dont have to be good at math, they just have to want to learn and they have to like it. One of the main purposes of mathcounts is to \"promote interest in middle school problem solving skill\" so if they aren't good at math already, then it's easier to teach them something :lol:", "Solution_6": "[quote=\"FantasyLover\"]yeah i already requested to principal to make a math team but he said that there's too few students who is good at math...[/quote]\r\n\r\nThat's too bad. I always geared my teams for those who wanted to participate, not those who were best, or even really good at math. The point is to learn something from it. \r\n\r\nThe last team I coached really wasn't an official school team. (OK, I've honestly never had an official school team, I've always just made the teams on my own with no support from the school.) It was just several kids who wanted to do it. They weren't even the best kids. My top participant probably was about 8th best in math at the school. This out of a school of only 80 7th and 8th graders combined. Some of the participants I would put near the midline 50th percentile in math ability at the school. However, they wanted to do it, so that was good enough for me! Our practices were occasional and very informal, but the students would learn some good math every now and then. Perhaps your school could do something like that.\r\n\r\nNow if I could only have my dream of being able to have a fulltime load of daily mathcounts teaching...oh man, that would be great. I could compete against Boyd and Frost! Oh well, dream on.", "Solution_7": "W000000000000T!!!! I MIGHT GO TO MATHCOUNTS!!!!" } { "Tag": [ "algebra", "function", "domain" ], "Problem": "What is the domain of the function $ f(x) \\equal{} (2x\\minus{}3)^{x^2 \\minus{} 9x \\plus{} 20}$ where $ f(x)$ is an exponential function?", "Solution_1": "The function is defined if $ 2x \\minus{} 3 \\geq 0 \\equal{}> x \\geq \\frac{3}{2}$, and also $ \\forall x \\in N$.", "Solution_2": "[quote=\"Carcul\"]The function is defined if $ 2x \\minus{} 3 \\geq 0 \\equal{} > x \\geq \\frac {3}{2}$, and also $ \\forall x \\in N$.[/quote]\r\n\r\nDid you mean $ 2x\\minus{}3>0$ and $ x>3/2$? because it needs to be an exponential function...\r\n\r\nI am not familiar with notations: $ \\forall x \\in N$, what does it mean?", "Solution_3": "$ \\forall x\\in\\mathbb{N}$ means \"for all $ x$ in $ \\mathbb{N}$\".", "Solution_4": "I don't see why $ 2x\\minus{}3$ needs to be positive.\r\nMy guess is the domain is all reals, where $ x\\not\\equal{}\\frac{3}{2}$ and $ x\\not\\equal{}2$, because in those values the function becomes linear.", "Solution_5": "By the definition of exponential function, the base must be positive and not equal to 1, so the domain will be $ {x|x>\\frac {3}{2} and x\\not=2}$\r\n\r\nbtw, if a negative number is raised by irrational powers, is it always a complex number? For rational number, this occurs only when it has a even factor in the denominator.", "Solution_6": "[hide=\"ANSWER SHEET\"] $ x>3/2$ or $ x\\in(3,\\plus{}\\infty)$[/hide]" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": "1. Let M, N be mxn and nxm matrices, respectively. If c is an eigenvalue for MN then it's also an eigenvalue for NM. That's easy to see....but show by example that this isn't necessairly true if c=0.\r\n\r\n2. If A is the inverse of I-NM (identity minus NM) and B is the inverse of I-MN then find the expression for B in terms of A, M, N...\r\n\r\ni got stuck here...\r\ncheers and thanks for help", "Solution_1": "[quote=\"omnyx\"]If c is an eigenvalue for MN then it's also an eigenvalue for NM. That's easy to see....but show by example that this isn't necessairly true if c=0.[/quote]\r\n\r\nI don't have an example handy, but it's possible to construct $ M, N$ so that $ NM$ is an invertible matrix while $ MN$ is not.", "Solution_2": "Let $ M\\equal{}\\begin{bmatrix}1\\\\1\\end{bmatrix}$ and $ N\\equal{}\\begin{bmatrix}1&1\\end{bmatrix}.$\r\n\r\nThen $ MN\\equal{}\\begin{bmatrix}1&1\\\\1&1\\end{bmatrix}$ is not invertible, while $ NM\\equal{}[2]$ is invertible.\r\n\r\n$ MN$ has characteristic polynomial $ x(x\\minus{}2)$ while $ NM$ has characteristic polynomial $ x\\minus{}2.$\r\n\r\nI don't feel like searching right now, but the proof has appeared on this forum that $ MN$ and $ NM$ have the same characteristic polynomial, except that the one with larger dimension has the polynomial \"padded\" with enough extra factors of $ x$ to make up the difference in degree." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "The following inequalities are true or false?\r\n$ (1)$\r\n\\[ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {8abc(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)(a^2\\plus{} b^2\\plus{} c^2\\plus{}ab\\plus{}bc\\plus{}ca)} \\ge 2\\]\r\nfor all $ a,b,c\\in[\\sqrt2\\minus{}1,1]$, with equality when $ (a,b,c)\\sim (1,1,1);(1,1,\\sqrt2\\minus{}1).$\r\n\r\n$ (2)$\r\n\\[ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {4abc(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)(a^2\\plus{} b^2\\plus{} c^2)} \\le 2\\]\r\nfor all $ a,b,c$ be sidelengths of a triangle, with equality when $ (a,b,c)\\sim (1,1,1) or (1,1,0) or (1,1,\\sqrt2).$", "Solution_1": "[quote=\"dduclam\"]The following inequalities are true or false?\n$ (1)$\n\\[ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {8abc(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca)} \\ge 2\\]\nfor all $ a,b,c\\in[\\sqrt2 \\minus{} 1,1]$, with equality when $ (a,b,c)\\sim (1,1,1);(1,1,\\sqrt2 \\minus{} 1).$\n\n$ (2)$\n\\[ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{a \\plus{} c} \\plus{} \\frac {c}{a \\plus{} b} \\plus{} \\frac {4abc(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)} \\le 2\\]\nfor all $ a,b,c$ be sidelengths of [b]an acute[/b] triangle, with equality when $ (a,b,c)\\sim (1,1,1) or (1,1,0) or (1,1,\\sqrt2).$[/quote]\r\nThese inequalities are true. But I think they are not very nice. Here is my proof:\r\n\r\n(1) Write the inequality as\r\n$ \\sum a(a \\plus{} b)(a \\plus{} c) \\plus{} \\frac {8abc(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca} \\ge 2(a \\plus{} b)(b \\plus{} c)(c \\plus{} a),$\r\nor\r\n$ \\sum a(a \\plus{} b)a \\plus{} c) \\plus{} 4abc \\minus{} 2(a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\ge 4abc\\left[ 1 \\minus{} \\frac {2(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca}\\right].$\r\nThis is equvalent to\r\n$ \\sum a(a \\minus{} b)(a \\minus{} c) \\ge \\frac {4abc}{a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca} \\sum (a \\minus{} b)(a \\minus{} c),$\r\nor\r\n$ \\sum a(a \\minus{} b)(a \\minus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} ac \\minus{} 3bc) \\ge 0.$\r\nLet $ X \\equal{} a(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} ac \\minus{} 3bc)$ and $ Y,Z$ are similar. Since $ a,b,c \\in [\\sqrt {2} \\minus{} 1,1],$ we have $ X,Y,Z \\ge 0.$ Indeed,\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} ac \\minus{} 3bc \\ge a^2 \\plus{} ab \\plus{} ac \\minus{} bc \\ge \\left(\\sqrt {2} \\minus{} 1\\right)^2 \\plus{} \\left(\\sqrt {2} \\minus{} 1\\right) (b \\plus{} c) \\minus{} bc \\ge \\left(\\sqrt {2} \\minus{} 1\\right)^2bc \\plus{} \\left(\\sqrt {2} \\minus{} 1\\right) (b \\plus{} c)\\sqrt {bc} \\minus{} bc \\ge \\left(\\sqrt {2} \\minus{} 1\\right)^2bc \\plus{} 2\\left(\\sqrt {2} \\minus{} 1\\right) bc \\minus{} bc \\equal{} 0.$\r\nNow, let us assume that $ a \\ge b \\ge c.$ We have\r\n$ Z(c \\minus{} a)(c \\minus{} b) \\ge 0,$ $ a(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} ac \\minus{} 3bc) \\ge b(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} bc \\plus{} ba \\minus{} 3ac).$\r\nTherefore\r\n$ \\sum X(a \\minus{} b)(a \\minus{} c) \\ge X(a \\minus{} b)(a \\minus{} c) \\plus{} Y(b \\minus{} c)(b \\minus{} a) \\equal{} Y(a \\minus{} b)(a \\minus{} c) \\plus{} Y(b \\minus{} c)(b \\minus{} a) \\equal{} Y(a \\minus{} b)^2 \\ge 0.$\r\n\r\n(2) Write the inequality as\r\n$ \\sum a(a \\plus{} b)(a \\plus{} c) \\plus{} \\frac {4abc(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2} \\le 2(a \\plus{} b)(b \\plus{} c)(c \\plus{} a),$\r\nor\r\n$ 4abc \\left( 1 \\minus{} \\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}\\right) \\ge \\sum a(a \\plus{} b)(a \\plus{} c) \\plus{} 4abc \\minus{} 2(a \\plus{} b)(b \\plus{} c)(c \\plus{} a).$\r\nThis is equivalent to\r\n$ \\frac {4abc}{a^2 \\plus{} b^2 \\plus{} c^2} \\sum (a \\minus{} b)(a \\minus{} c) \\ge \\sum a(a \\minus{} b)(a \\minus{} c),$\r\nor\r\n$ \\sum a(a \\minus{} b)(a \\minus{} c)(4bc \\minus{} a^2 \\minus{} b^2 \\minus{} c^2) \\ge 0.$\r\nBecause $ 4bc \\minus{} a^2 \\minus{} b^2 \\minus{} c^2 \\equal{} (b^2 \\plus{} c^2 \\minus{} a^2) \\minus{} 2(b \\minus{} c)^2,$ it can be written as\r\n$ \\sum a(b^2 \\plus{} c^2 \\minus{} a^2)(a \\minus{} b)(a \\minus{} c) \\minus{} 2\\sum a(a \\minus{} b)(a \\minus{} c)(b \\minus{} c)^2 \\ge 0,$\r\nor\r\n$ \\sum a(b^2 \\plus{} c^2 \\minus{} a^2)(a \\minus{} b)(a \\minus{} c) \\ge 0.$\r\nWithout loss of generality, we may assume that $ a \\ge b \\ge c.$ Then, we have $ a \\minus{} c \\ge \\frac {a}{b}(b \\minus{} c)$ and $ a \\minus{} c \\ge \\frac {b}{c}(a \\minus{} b)$ (becaue $ b \\plus{} c \\ge a$). According to these two inequalities, we have\r\n$ a(b^2 \\plus{} c^2 \\minus{} a^2)(a \\minus{} b)(a \\minus{} c) \\ge \\frac {a^2(b^2 \\plus{} c^2 \\minus{} a^2)(a \\minus{} b)(b \\minus{} c)}{b},$\r\nand\r\n$ c(a^2 \\plus{} b^2 \\minus{} c^2)(c \\minus{} a)(c \\minus{} b) \\ge \\frac {c^2(a^2 \\plus{} b^2 \\minus{} c^2)(a \\minus{} b)(b \\minus{} c)}{b}.$\r\nTherefore\r\n$ \\sum a(b^2 \\plus{} c^2 \\minus{} a^2)(a \\minus{} b)(a \\minus{} c) \\ge \\frac {(a \\minus{} b)(b \\minus{} c)}{b} [a^2(b^2 \\plus{} c^2 \\minus{} a^2) \\plus{} c^2(a^2 \\plus{} b^2 \\minus{} c^2) \\minus{} b^2(c^2 \\plus{} a^2 \\minus{} b^2)] \\equal{} \\frac {(a \\minus{} b)(b \\minus{} c)}{b}(a^2 \\plus{} b^2 \\minus{} c^2)(b^2 \\plus{} c^2 \\minus{} a^2) \\ge 0.$\r\nThe proof is completed.\r\n\r\nRemark. (2) does not hold for any triangle, it can only hold when $ a,b,c$ are the side-lengths of an acute triangle. That's the reason why I changed your statement a little." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Prove that are not matrix $A$ and $B$ $(n$x$n)$ such $AB-BA=I$", "Solution_1": "if $A$ and $B$ exist such as $AB-BA=I_{n}$, then if we apply Trace in this equality, we get that $Tr(AB-BA)=Tr(I_{n})$, so $0=n$, which isn't possible. so there is no $A$ and $B$ ... i used the propriety that $Tr(AB)=Tr(BA)$", "Solution_2": "This does depend on field; the above proof works in any field of characteristic zero (such as the rational, real, or complex numbers). In a field of finite characteristic, such as the integers mod $p$ (with $p|n$), the result can fail.\r\n\r\nAn example for $n=2$, in the field of integers mod $2$:\r\n$A=\\begin{bmatrix}1&1\\\\0&1\\end{bmatrix}, B=\\begin{bmatrix}1&0\\\\1&1\\end{bmatrix}$.\r\n$AB=\\begin{bmatrix}0&1\\\\1&1\\end{bmatrix}, BA=\\begin{bmatrix}1&1\\\\1&0\\end{bmatrix}, AB-BA=AB+BA=I$", "Solution_3": "[quote]$AB=\\begin{bmatrix}0&1\\\\1&1\\end{bmatrix}, BA=\\begin{bmatrix}1&1\\\\1&0\\end{bmatrix}, AB-BA=AB+BA=I$[/quote]\r\nyou sure this is right?? Forgive me if i'm wrong but $AB-BA \\neq AB+BA\\neq I$\r\n$\\begin{bmatrix}0&1\\\\1&1\\end{bmatrix}+\\begin{bmatrix}1&1\\\\1&0\\end{bmatrix}=\\begin{bmatrix}1&2\\\\2&1\\end{bmatrix}\\neq\\begin{bmatrix}1&0\\\\0&1\\end{bmatrix}$\r\n$\\begin{bmatrix}0&1\\\\1&1\\end{bmatrix}-\\begin{bmatrix}1&1\\\\1&0\\end{bmatrix}=\\begin{bmatrix}-1&0\\\\0&1\\end{bmatrix}\\neq\\begin{bmatrix}1&0\\\\0&1\\end{bmatrix}$", "Solution_4": "yes but he did not say what field we were working on, so i took the good old one", "Solution_5": "In the field I was working with, $2=0$. That's the key to my example. Over a general field, my $AB-BA$ is $\\begin{bmatrix}1&0\\\\0&-1\\end{bmatrix}$, but $1=-1$ in the field with two elements." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "The goodfood catering company finds that competitors cater lunch for a group of 100 people for $\\$5$ each. The manager of Goodfood calculates that for each 25 cent discount per lunch, it is possible to sell an additional 10 lunches. If each lunch costs Goodfood $\\$2$ to prepare, how many lunches should be prepared to maximize the profit?", "Solution_1": "Where are you having difficulties? It seems like you havne't even thought about the problem and expect someone to do your homework ." } { "Tag": [ "function", "geometric series", "algebra unsolved", "algebra" ], "Problem": "Prove that the sum $0,1+0,01+0,002+0,0003+0,00005+0,000008+0,0000013+...$\r\nconverges and calculate it.", "Solution_1": "I assume the $n$-th term is $\\frac{F_n}{10^n}$.\r\nThen just use Binet's forumla, split the two summands in it to two geometrical sums, done.", "Solution_2": "[quote=\"ZetaX\"]I assume the $n$-th term is $\\frac{F_n}{10^n}$.\nThen just use Binet's forumla, split the two summands in it to two geometrical sums, done.[/quote]\r\n\r\nCan you explain the solution with detail please??? And another way without using that formula????", "Solution_3": "Here $F_n$ denotes the $n$th Fibonacci number. Binet's formula for Fibonacci numbers says that $n$th Fibonacci number equals\r\n\\[ F_n=\\frac{(1+\\sqrt{5})^n-(1-\\sqrt{5})^n}{2^n\\sqrt{5}}. \\]", "Solution_4": "[quote=\"puuhikki\"]Here $F_n$ denotes the $n$th Fibonacci number. Binet's formula for Fibonacci numbers says that $n$th Fibonacci number equals\n\\[ F_n=\\frac{(1+\\sqrt{5})^n-(1-\\sqrt{5})^n}{2^n\\sqrt{5}}. \\][/quote]\r\n\r\nAll right, and how can I use this to solve the original problem???? I have to prove 1\u00ba that the sum converges.....", "Solution_5": "Just set it into the sum and look what happens! It gets the sum of two converging geometric series (or majorize it by $\\sum_{n=1}^\\infty \\frac{1}{5^n}$).\r\n\r\nAnother way:\r\nThe generating function $f(x)=\\sum_{n=0}^\\infty F_n x^n$ fulfills $f(x)- x = \\sum_{n=2}^\\infty F_n x^n= \\sum_{n=2}^\\infty F_{n-1} x^n + \\sum_{n=2}^\\infty F_{n-2} x^n = x f(x) + x^2 f(x)$, giving $f(x) = \\frac{x}{1-x-x^2}$.\r\nNow set $x= \\frac{1}{10}$ and you get $\\sum_{n=0}^\\infty \\frac{F_n}{10^n} = f\\left( \\frac{1}{10} \\right) = \\frac{10}{89}$", "Solution_6": "Thanks ZetaX" } { "Tag": [ "logarithms", "function", "algebra unsolved", "algebra" ], "Problem": "Slove this equation: $ x\\equal{}2^{\\frac{x^2\\minus{}1}{3}}$", "Solution_1": "It is clear that this answer $ x\\equal{}0$.\r\nIf $ x<0 \\ \\Rightarrow \\left\\{\\begin{array}{ll} x<0 \\\\ 2^{\\frac{x^2\\minus{}1}{3}>0} \\end{array}\\right.$.\r\nSo, $ x>0$\r\n\r\n$ x \\equal{} 2^{\\frac{x^2\\minus{}1}{3}} \\iff \\log_2 x \\equal{} \\frac{x^2\\minus{}1}{3}$.\r\n\r\nLet $ f(x) \\equal{} \\log_2 x \\minus{} \\frac{x^2\\minus{}1}{3}$.\r\n$ f '(x) \\equal{} \\frac{1}{x \\ln 2} \\minus{} \\frac{2}{3}x$\r\n$ f ''(x) \\equal{} \\minus{}\\frac{1}{x^2 \\ln 2} \\minus{} \\frac{2}{3}$\r\nSo one extremum. $ \\left(x \\equal{} \\sqrt{\\frac{3}{2 \\ln 2}}\\right)$\r\n$ \\left(\\frac{9}{5}<\\sqrt{\\frac{3}{2 \\ln 2}}<\\frac{18}{7} \\ \\ \\left( \\because 1\\minus{}\\frac{1}{2}\\plus{}\\frac{1}{3}\\minus{}\\frac{1}{4}<\\ln 2<1\\minus{}\\frac{1}{2}\\plus{}\\frac{1}{3} \\right)\\right)$\r\nTherefore, a solution is two at most.\r\n\r\nIf $ x\\equal{}1 \\Rightarrow f(x)\\equal{}0 \\iff 1\\equal{}2^{\\frac{1^2\\minus{}1}{3}}\\equal{}2^0\\equal{}1$.\r\nIf $ x\\equal{}2 \\Rightarrow f(x)\\equal{}0 \\iff 2\\equal{}2^{\\frac{2^2\\minus{}1}{3}}\\equal{}2^1\\equal{}2$.\r\n\r\nThus, $ x\\equal{}1, 2$", "Solution_2": "Just observe that RHS is convex, and 1, 2 are solutions.\r\n\r\nThe convexity follows by composition of convex functions $ f \\circ g$ where $ f$ is increasing." } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Let (an) and (bn) be two sequences with values in R* so that:\r\nsin(a_n+b_n)=2a_n+3b_n for all n natural.Prove that if lim a_n=0 then:\r\na)lim b_n=0;\r\nb)lim a_n/b_n=-2", "Solution_1": "An other try :\r\n\r\n. since a(n) -> 0, b(n) is bounded.\r\n\r\n. Let b(f(n)) a convergent subsequence of (exists by Bolzano-Weierstrass) -> b\r\nThen sinb = 3b and b=0\r\n\r\n. So b(n) has only one convergent sub-sequence and is bounded so is convergent and b(n) -> 0.\r\n\r\n. sin (a(n)+b(n)) = a(n) + b(n) + O((a(n)+b(n))^3) = 2a(n) + 3b(n) (O = big 'O' notation)\r\nor a(n)/b(n) + 2 = O((a(n)+b(n))^3 / b(n))\r\n\r\n. (a(n)+b(n))^3 / b(n) = a(n)^3/b(n) + O(b(n)) so it's enough to prove a(n)^3/b(n) -> 0 and it's enough to prove a(n)/b(n) is bounded.\r\n\r\n. |3 + 2a(n)/b(n)| <= |1 + a(n)/b(n)| (|sinx|<=|x|) which is possible only if a(n)/b(n) [-2..-1]" } { "Tag": [], "Problem": "If the mr and mrs smith plans to organise sports lesson for they children. The number of soccer lesson is 3, the number of baseball lesson is 4 and the number of baskeball lesson is 5. If each child can enter a maximum of 2 lessons. Find the minimum number of children in the smith family.\r\n\r\n\r\n[hide]answer=6[/hide]", "Solution_1": "[hide]If it is the minimum number of children, they all need to go to 2 lessons. Since it really doesn't matter which lesson they go to, we add up all the lessons and divide by 2 to get 6.[/hide]", "Solution_2": "[hide]There are 12 lessons total, and if everyone takes two lessons, that gives the minimum, 6.[/hide]", "Solution_3": "[hide=\"Really easy\"]There are a total of $12$ lessons. It really doesn't matter which class each child goes to, so the answer is $\\frac{12}{2}= 6$. [/hide]\r\n\r\nYou complicate the question by saying: Child A can't go to basketball lessons, Child B can't go to soccer lessons, etc..." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ x$, $ y$ and $ z$ are positive real numbers prove the inequality:\r\n\r\n$ \\frac {xy}{xy \\plus{} x \\plus{} y} \\plus{} \\frac {yz}{yz \\plus{} y \\plus{} z} \\plus{} \\frac {zx}{zx \\plus{} x \\plus{} y}\\leq\\frac {x}{2x \\plus{} z} \\plus{} \\frac {y}{2y \\plus{} x} \\plus{} \\frac {z}{2z \\plus{} y}$", "Solution_1": "[quote=\"delegat\"]If $ x$, $ y$ and $ z$ are positive real numbers prove the inequality:\n\n$ \\frac {xy}{xy \\plus{} x \\plus{} y} \\plus{} \\frac {yz}{yz \\plus{} y \\plus{} z} \\plus{} \\frac {zx}{zx \\plus{} x \\plus{} y}\\leq\\frac {x}{2x \\plus{} z} \\plus{} \\frac {y}{2y \\plus{} x} \\plus{} \\frac {z}{2z \\plus{} y}$[/quote]\r\n\r\nThe above is wrong . Just try $ x\\equal{}y\\equal{}z\\equal{}2$ . The following is nice and true \r\n$ \\frac {xy}{xy \\plus{} x^2 \\plus{} y^2} \\plus{} \\frac {yz}{yz \\plus{} y^2 \\plus{} z^2} \\plus{} \\frac {zx}{zx \\plus{} x^2 \\plus{} y^2}\\leq\\frac {x}{2x \\plus{} z} \\plus{} \\frac {y}{2y \\plus{} x} \\plus{} \\frac {z}{2z \\plus{} y}$\r\n\r\nbut please do not discuss it !! (it is from a running contest)", "Solution_2": "[quote=\"silouan\"][quote=\"delegat\"]If $ x$, $ y$ and $ z$ are positive real numbers prove the inequality:\n\n$ \\frac {xy}{xy \\plus{} x \\plus{} y} \\plus{} \\frac {yz}{yz \\plus{} y \\plus{} z} \\plus{} \\frac {zx}{zx \\plus{} x \\plus{} y}\\leq\\frac {x}{2x \\plus{} z} \\plus{} \\frac {y}{2y \\plus{} x} \\plus{} \\frac {z}{2z \\plus{} y}$[/quote]\n\nThe above is wrong . Just try $ x \\equal{} y \\equal{} z \\equal{} 2$ . The following is nice and true \n$ \\frac {xy}{xy \\plus{} x^2 \\plus{} y^2} \\plus{} \\frac {yz}{yz \\plus{} y^2 \\plus{} z^2} \\plus{} \\frac {zx}{zx \\plus{} x^2 \\plus{} y^2}\\leq\\frac {x}{2x \\plus{} z} \\plus{} \\frac {y}{2y \\plus{} x} \\plus{} \\frac {z}{2z \\plus{} y}$\n\nbut please do not discuss it !! (it is from a running contest)[/quote]\r\n\r\nI think you maybe made a typo....\r\nMaybe it should be like the following:\r\n$ \\frac {xy}{xy \\plus{} x^2 \\plus{} y^2} \\plus{} \\frac {yz}{yz \\plus{} y^2 \\plus{} z^2} \\plus{} \\frac {zx}{zx \\plus{} z^2 \\plus{} x^2}\\leq\\frac {x}{2x \\plus{} z} \\plus{} \\frac {y}{2y \\plus{} x} \\plus{} \\frac {z}{2z \\plus{} y}$ :maybe:", "Solution_3": "I've found the following similar result :\r\n\r\n$ \\frac {xy}{x^2 \\plus{} y(z \\plus{} x)} \\plus{} \\frac {yz}{y^2 \\plus{} z(x \\plus{} y)} \\plus{} \\frac {zx}{z^2 \\plus{} x(y \\plus{} z)}\\le \\frac {x}{2x \\plus{} z} \\plus{} \\frac {y}{2y \\plus{} z} \\plus{} \\frac {z}{2z \\plus{} x}$\r\n\r\nMay be we can discuss this one ." }